🧠:[<think>Okay, so I have this problem here. Let me read it again and try to understand what's going on. There are 2^m cards, each initially with the number 1 on them. The operation we perform is: each time, choose two different cards. If the numbers on these two cards are a and b, erase those numbers and write a + b on both cards. We need to prove that after performing 2^{m-1} * m operations, the sum of all the numbers on the cards is at least 4^m.Hmm. Let's start by unpacking the problem. So initially, we have 2^m cards, each with 1. The sum starts at 2^m * 1 = 2^m. Each operation involves picking two cards, adding their numbers, and writing the sum back on both. So each operation replaces two numbers a and b with two numbers a + b. Therefore, each operation increases the total sum by (a + b) - (a + b) = 0? Wait, that can't be. Wait, hold on. If you have two numbers, a and b, and you replace both with a + b, then the total sum becomes (a + b) + (a + b) = 2a + 2b. But originally, it was a + b. So the sum increases by 2a + 2b - (a + b) = a + b. So each operation increases the sum by a + b. Therefore, each operation's contribution to the sum is equal to the sum of the two numbers chosen. Therefore, the key is that each operation increases the total sum by the sum of the two numbers selected. So the strategy would be to maximize the total sum as quickly as possible. However, since we have a fixed number of operations (2^{m-1} * m), we need to figure out how the sum can grow over these operations. But the problem says to prove that after 2^{m-1} * m operations, the sum is at least 4^m. Let me note that 4^m is (2^m)^2, which is the square of the initial number of cards. The initial sum is 2^m, so we need to show that through these operations, the sum can be squared. That seems like a significant increase. Wait, but maybe there's a pattern here. Let me try with small m and see what happens. Let's take m = 1. Then we have 2^1 = 2 cards, each with 1. The number of operations is 2^{0} * 1 = 1 operation. Let's perform the operation: pick both cards, replace each with 1 + 1 = 2. The sum becomes 2 + 2 = 4, which is 4^1 = 4. So that works.For m = 2. We have 4 cards, each with 1. The number of operations is 2^{1} * 2 = 4 operations. Let's see:Start: sum = 4.After each operation, the sum increases by a + b. Let's try to maximize the sum. So perhaps, in each operation, pick the two largest numbers. But initially, all numbers are 1.First operation: pick any two 1s. Sum becomes 2 + 2 + 1 + 1 = 6. So sum increases by 2.Second operation: Now we have two 2s and two 1s. Let's pick a 2 and a 1. Then replacing them with 3 and 3. Sum becomes 3 + 3 + 2 + 1 = 9. Wait, but sum was 6, now increases by 3 (2 + 1). So sum is 9.Third operation: Now we have two 3s, a 2, and a 1. Let's pick two 3s. Replace them with 6 and 6. Then the sum becomes 6 + 6 + 2 + 1 = 15. Sum increased by 6 (3 + 3).Fourth operation: Now we have two 6s, a 2, and a 1. Let's pick the two 6s. Replace with 12 and 12. Sum becomes 12 + 12 + 2 + 1 = 27. So after 4 operations, sum is 27. 4^2 = 16, and 27 is indeed greater than 16. So that works.But maybe there's a way to get even a higher sum? Let me check.Alternatively, in the second operation, maybe pair the two remaining 1s. Let's see.First operation: two 1s become 2s. Sum 6.Second operation: the other two 1s become 2s. Now all four cards are 2s. Sum 8.Third operation: pair two 2s to make 4s. Sum becomes 4 + 4 + 2 + 2 = 12.Fourth operation: pair the other two 2s to make 4s. Now all cards are 4s. Sum 16. Which is exactly 4^2. So in this case, depending on the operations, you can get either 16 or 27. But the problem states "at least 4^m", so even if you do the minimal operations, you still get 16. But in reality, depending on how you pair, you can get more. So the problem is just asking to prove that after 2^{m-1}*m operations, regardless of the choices (since the problem doesn't specify that we need to maximize the sum, just that after that many operations, the sum is at least 4^m). Wait, but does the problem specify that you can choose the operations optimally, or is it regardless of the operations? Let me check.The problem says: "choose two different cards" each time, but it doesn't specify that we get to choose them in a particular way. Wait, but in problems like this, usually, you have to prove that no matter how you perform the operations (i.e., regardless of the choices), the result holds. But the wording here is a bit ambiguous. Let me check again."Prove: After 2^{m-1} m operations, the sum of the numbers on all cards is at least 4^{m}."Hmm. The problem doesn't specify any particular strategy, so perhaps we need to show that regardless of the sequence of operations (as long as each operation is valid, i.e., choosing two different cards each time), the sum will be at least 4^m after 2^{m-1} m operations. So even in the worst case, where the sum increases as slowly as possible, after that many operations, it's still at least 4^m. Alternatively, maybe the problem is intended that you can choose the operations in such a way to reach at least 4^m, but given that in the m=2 case, if you pair 1s first and then pair the resulting 2s, you can get 16, which is exactly 4^2. So maybe the minimal possible sum after that many operations is 4^m, and any other way gives a higher sum. So the problem might be that regardless of the operations, you can't have a sum lower than 4^m. But in the m=2 case, if you pair the 2s first, you can get higher sums, but if you pair 1s and 2s, you can get a lower sum? Wait, in m=2, with 4 operations, if you pair the two 1s first, then pair a 2 and a 1, then pair two 3s, then pair two 6s, you end up with sum 27. But if you pair 1s first, then pair 2s, then pair 2s again, you get sum 16. Wait, so depending on the pairing, you can get different sums. Wait, but how is that possible? Because the operations are adding a + b to the total sum each time. So if you have a sequence of operations, the total sum is the initial sum plus the sum of a_i + b_i over each operation, where a_i and b_i are the numbers on the two cards chosen in the i-th operation.Therefore, the total sum after k operations is 2^m + sum_{i=1}^k (a_i + b_i). So the problem is to show that after k = 2^{m-1} m operations, the total sum is at least 4^m, which is (2^m)^2. So perhaps there is an invariant or a lower bound on the sum that can be proven regardless of the choices of pairs. Maybe using induction on m? Let me try that.Base case m=1: As we saw, after 1 operation, sum is 4, which is 4^1. So holds.Assume it holds for m. Let's see for m+1.But wait, m is the exponent in 2^m. Hmm. Maybe another approach.Let’s note that each operation increases the sum by a + b, where a and b are the numbers on the two selected cards. So the total sum after k operations is S = 2^m + sum_{i=1}^k (a_i + b_i).We need to show that after k = 2^{m-1} * m operations, S >= 4^m.So 4^m is (2^m)^2. The initial sum is 2^m. So we need to have 2^m + sum_{i=1}^k (a_i + b_i) >= (2^m)^2.Thus, sum_{i=1}^k (a_i + b_i) >= (2^m)^2 - 2^m = 2^m (2^m - 1).Therefore, the problem reduces to showing that the total increments over k operations is at least 2^m (2^m - 1).But how to bound this sum? Perhaps by considering the average value of a_i + b_i over the operations?Alternatively, maybe we can model this as a process where each operation merges two numbers into their sum, but since both cards are replaced by the sum, it's like combining two elements. Wait, but in the standard "combining" process, you might replace two numbers with their sum, but here you replace both with the sum. So it's different. So each operation replaces two numbers a, b with two numbers a + b. Therefore, each operation increases the sum by a + b, but also the multiplicity of numbers. Wait, maybe not. Let me think.Alternatively, think of each card as a node, and when you perform an operation on two cards, you are essentially creating a connection between them, and both now carry the sum. But I'm not sure.Alternatively, consider the sum of squares or some other invariant. Maybe the sum of squares? Let me think. If we consider the sum of the squares of the numbers on the cards, how does it change with each operation?Suppose we have two numbers a and b. Originally, their squares are a² + b². After the operation, both numbers become a + b, so the sum of squares becomes 2(a + b)². The difference is 2(a + b)² - (a² + b²) = 2(a² + 2ab + b²) - a² - b² = 2a² + 4ab + 2b² - a² - b² = a² + 4ab + b². So the sum of squares increases by a² + 4ab + b². Alternatively, maybe this is useful? But the problem is about the linear sum, not the sum of squares. Hmm. Maybe another invariant.Alternatively, think of the process as building a binary tree. Each operation corresponds to combining two nodes into a parent node with value a + b. But since we are writing a + b on both cards, it's a bit different. Maybe not a binary tree.Alternatively, note that each time we perform an operation, we can imagine that the two cards are being linked in some way, and their combined contribution is being doubled. Wait, if we have two cards with a and b, and we replace them with two cards with a + b, then effectively, the "potential" of these two cards is now 2(a + b). Maybe we can model this as a process where each operation doubles some quantity.Alternatively, think in terms of expectations. Maybe not. Hmm.Wait, let's try to model the sum. Let S_t be the sum after t operations. Then S_0 = 2^m. Then S_{t+1} = S_t + a + b, where a and b are the numbers on the two selected cards. So each operation adds a + b to the sum. Therefore, the problem is to show that after k = 2^{m-1} m operations, S_k >= 4^m.Alternatively, perhaps we can model this as a recurrence relation. Let me see for small m.For m=1: S_0=2, after 1 operation, S_1=4. Which is 4^1.For m=2: As earlier, depending on operations, sum can be 16 or higher. But the minimal sum seems to be 16.Wait, maybe there's a way to pair cards such that each operation's a + b is as small as possible. If we try to minimize the sum, then we would pair the two smallest numbers each time. But even so, after a certain number of operations, the minimal possible sum would be 4^m. Let's see.Alternatively, let's consider that each operation is combining two numbers. But since you replace two numbers with two copies of their sum, the number of cards remains the same: 2^m. Wait, but the number of cards never changes. Wait, is that right? Each operation replaces two cards with two cards. So yes, the number of cards is always 2^m. So that's an important point. The number of cards remains constant throughout the process. Each operation just modifies two of them.So perhaps we can model the numbers on the cards as variables x_1, x_2, ..., x_{2^m}, each starting at 1. Each operation picks two indices i and j, sets x_i = x_j = x_i + x_j. The sum S = sum x_i, and we need to show that after 2^{m-1} m operations, S >= 4^m.Hmm. Let's consider the product of all the numbers on the cards. Let’s denote P = product x_i. How does P change when we perform an operation? If we replace x_i and x_j (which were a and b) with (a + b) and (a + b), then the new product is P * (a + b)^2 / (a * b). So the product becomes P * (a + b)^2 / (a b). Therefore, the ratio of the new product to the old product is (a + b)^2 / (a b). Therefore, each operation multiplies the product by (a + b)^2 / (a b). To maximize the product, we need to maximize this ratio each time. Alternatively, if we want to find a lower bound on the product, we can consider the minimal possible ratio over all possible operations. But the problem is about the sum, not the product. However, perhaps there's a relationship between the sum and the product via the AM ≥ GM inequality.Indeed, the arithmetic mean of the numbers is S / 2^m, and the geometric mean is P^{1 / 2^m}. By AM ≥ GM, S / 2^m >= P^{1 / 2^m}, so S >= 2^m * P^{1 / 2^m}. Therefore, if we can find a lower bound on P, we can get a lower bound on S.But perhaps we can track how P evolves through operations. Since each operation replaces two numbers a and b with two copies of a + b, then the product becomes P * (a + b)^2 / (a b). Therefore, each operation multiplies the product by (a + b)^2 / (a b). Let's denote this factor as (a + b)^2 / (a b) = (a/b + b/a + 2). Since a and b are positive numbers, this factor is at least 4 by AM ≥ GM: (a + b)^2 / (ab) >= 4. Equality when a = b.Therefore, each operation multiplies the product by at least 4. Therefore, after k operations, the product P_k >= P_0 * 4^k. Since initially, P_0 = 1^{2^m} = 1. Therefore, P_k >= 4^k. Then, by AM ≥ GM, S_k >= 2^m * (P_k)^{1 / 2^m} >= 2^m * (4^k)^{1 / 2^m} = 2^m * 4^{k / 2^m} = 2^m * (2^2)^{k / 2^m} = 2^m * 2^{2k / 2^m} = 2^{m + 2k / 2^m}.We need to show that S_k >= 4^m = 2^{2m}. Therefore, we require:2^{m + 2k / 2^m} >= 2^{2m}Which simplifies to:m + 2k / 2^m >= 2mSubtract m from both sides:2k / 2^m >= mMultiply both sides by 2^{m} / 2:k >= m * 2^{m - 1}Which is exactly the number of operations given: k = 2^{m - 1} * m. Therefore, equality holds. Therefore, the lower bound from AM ≥ GM gives exactly S_k >= 4^m.Therefore, this seems to be the way to go. Each operation multiplies the product by at least 4, leading to P_k >= 4^k. Then, using AM ≥ GM, the sum S_k is at least 2^m * (4^k)^{1 / 2^m} = 2^{m + 2k / 2^m}. Plugging in k = m 2^{m - 1}, we get exponent m + 2*(m 2^{m - 1}) / 2^m = m + (m 2^{m}) / 2^m = m + m = 2m. Therefore, S_k >= 2^{2m} = 4^m. Hence proved.But wait, we need to ensure that the multiplicative factor for the product is indeed at least 4 each time. As I thought earlier, for any two positive numbers a and b, (a + b)^2 / (ab) >= 4. Since (a + b)^2 = a^2 + 2ab + b^2, so (a + b)^2 / (ab) = a/b + b/a + 2 >= 2*sqrt(a/b * b/a) + 2 = 2 + 2 = 4, by AM >= GM. Equality holds when a = b. Therefore, each operation multiplies the product by at least 4. Therefore, after k operations, P_k >= 4^k.Therefore, using AM >= GM on the final product gives us S_k >= 2^m * (4^k)^{1/2^m} = 2^{m + 2k / 2^m}. Plugging in the given k = m 2^{m - 1}, this exponent becomes m + 2*(m 2^{m - 1}) / 2^m = m + (m 2^{m}) / 2^m = m + m = 2m. Therefore, S_k >= 2^{2m} = 4^m. Therefore, the required result.So, the key idea is that each operation, regardless of which two cards you pick, causes the product of all card numbers to increase by a factor of at least 4. Then, by using the inequality between the arithmetic and geometric means, we can relate the sum to the product and show that the sum must be at least 4^m after the required number of operations.But let me verify with m=2 again. For m=2, k=4. Then 4^k=4^4=256. The product P_4 >= 256. Then by AM >= GM, S_4 >= 4*(256)^{1/4}=4*(4)=16, which matches 4^2=16. However, in practice, if you pair 1s first, then pair 2s, you end up with product 4*4*4*4=256, so P=256, which gives S=16. But if you pair differently, you can get higher products. For example, pairing 1 and 1 first, then 2 and 1, then 3 and 3, then 6 and 6, the product would be 6*6*2*1=72, which is less than 256. Wait, but that contradicts the earlier assertion that P_k >=4^k=256.Wait, this inconsistency suggests an error in my reasoning. Wait, no. Wait, in the m=2 case, if you pair two 1s first, turning them into 2s, then the product becomes 2*2*1*1=4. Then, if you pair a 2 and a 1, replacing them with 3 and 3, the product becomes 3*3*2*1=18. Then, pairing two 3s into 6 and 6, product becomes 6*6*2*1=72. Then, pairing two 6s into 12 and 12, product becomes 12*12*2*1=288. So after 4 operations, product is 288, which is greater than 4^4=256. Wait, but in this case, the product after 4 operations is 288, which is greater than 256. But in the other strategy, where you pair the 1s first, then pair the 2s, the product after each operation would be:First operation: 2*2*1*1=4.Second operation: 2*2*2*2=16.Third operation: 4*4*2*2=64.Fourth operation: 4*4*4*4=256.So here, after four operations, the product is 256, which is exactly 4^4. Therefore, depending on the pairing, the product can be as low as 256 or higher. But according to the earlier logic, each operation multiplies the product by at least 4, so after 4 operations, the product should be at least 4^4=256, which matches. The case where you pair different numbers leads to a higher product, but the minimal product is 256. Therefore, regardless of the operations, the product is at least 4^k, hence the sum is at least 4^m. Therefore, the lower bound holds.Therefore, my initial reasoning is correct. Each operation multiplies the product by at least 4, so after k operations, the product is at least 4^k. Then, using AM >= GM, the sum is at least 2^m * (4^k)^{1/2^m} = 4^m. Therefore, the proof holds.Final Answerboxed{4^m}</think>We start with (2^m) cards, each initially labeled with the number 1. The operation involves choosing two cards with numbers (a) and (b), and replacing both with the number (a + b). This operation increases the sum of all numbers by (a + b). We need to prove that after (2^{m-1} m) operations, the sum of the numbers on all cards is at least (4^m). Key Steps:1. Initial Sum: The initial sum of the numbers on the cards is (2^m).2. Sum Increase: Each operation increases the sum by (a + b), where (a) and (b) are the numbers on the chosen cards. 3. Product Analysis: Consider the product (P) of all numbers on the cards. Each operation replaces (a) and (b) with (a + b), multiplying the product by (frac{(a + b)^2}{ab}). By the AM-GM inequality, (frac{(a + b)^2}{ab} geq 4), so each operation multiplies the product by at least 4.4. Product Lower Bound: After (k) operations, the product (P_k) is at least (4^k). Initially, (P_0 = 1), so (P_k geq 4^k).5. AM-GM Inequality: Using the AM-GM inequality on the final product, the sum (S_k) satisfies: [ S_k geq 2^m left( P_k right)^{1/2^m} ] Substituting (P_k geq 4^k), we get: [ S_k geq 2^m left( 4^k right)^{1/2^m} = 2^m cdot 2^{2k/2^m} = 2^{m + 2k/2^m} ]6. Substitute (k): Given (k = 2^{m-1} m), we substitute: [ m + frac{2 cdot 2^{m-1} m}{2^m} = m + m = 2m ] Thus, (S_k geq 2^{2m} = 4^m). Conclusion:After (2^{m-1} m) operations, the sum of the numbers on all cards is at least (4^m).[boxed{4^m}],<think>Okay, so I need to prove that after performing 2^{m-1} * m operations on 2^m cards, each initially labeled with the number 1, the sum of all the numbers on the cards is at least 4^m. Hmm, let's try to unpack this step by step.First, let me make sure I understand the problem correctly. We start with 2^m cards, each with a 1. Each operation involves picking two different cards, say with numbers a and b, then erasing those numbers and writing a + b on both cards. So after each operation, the two selected cards now have the same number, which is the sum of their original numbers. The total number of operations performed is 2^{m-1} * m, and we need to show that the sum of all numbers on the cards after these operations is at least 4^m. Wait a second, the sum after each operation: when you replace a and b with a + b on both cards, the total sum increases by (a + b) * 2 - (a + b) = (a + b). Because before, the sum had a + b, and after, it's 2*(a + b), so the difference is 2*(a + b) - (a + b) = (a + b). So each operation increases the total sum by a + b. Therefore, each operation adds the sum of the two chosen cards to the total. Interesting.So the total sum after k operations would be the initial sum plus the sum of all the a + b's from each operation. Since the initial sum is 2^m * 1 = 2^m. Then each operation adds some amount to this. The question is, after 2^{m-1} * m operations, what is the minimal possible total sum?Wait, but the problem says "the sum of the numbers on all cards is at least 4^m". 4^m is (2^2)^m = 2^{2m}, which is (2^m)^2. So starting from 2^m, after 2^{m-1} * m operations, we need the sum to be at least (2^m)^2. Hmm. So the initial sum is S_0 = 2^m. Each operation increases the sum by a + b, where a and b are the numbers on the two selected cards. The key is to figure out the minimal possible total sum after 2^{m-1} * m operations. To minimize the total sum, we need to choose a and b as small as possible in each operation. Because the lower the a and b, the less the sum increases each time.Therefore, to minimize the total sum, we would want to always pick the two smallest numbers available on the cards. That way, the increase in the sum each time is minimized. If we can model this process, then perhaps we can find a lower bound on the total sum, which would be at least 4^m.Alternatively, maybe there's an invariant or some kind of potential function that can be used here. Since the problem involves combining numbers and tracking the sum, perhaps there's a way to relate the number of operations to the growth of the sum.Let me consider small cases first to get some intuition.Let's take m = 1. Then we have 2^1 = 2 cards, each with 1. The number of operations is 2^{0} * 1 = 1 operation. After one operation: pick the two cards, both 1. Replace both with 1 + 1 = 2. The total sum is 2 + 2 = 4. 4^1 = 4. So the sum is exactly 4. So for m=1, it works.m = 2: 2^2 = 4 cards, each 1. Number of operations: 2^{1} * 2 = 4 operations. Let's see. We need to perform 4 operations. Let's try to minimize the sum. Let's see:First operation: pick two 1s. Replace them with 2 each. Sum becomes 2 + 2 + 1 + 1 = 6.Second operation: pick the next two 1s. Replace them with 2 each. Sum becomes 2 + 2 + 2 + 2 = 8.But we need to do 4 operations. Wait, but after two operations, all cards are 2. Then, for the third operation, pick two 2s. Replace them with 4 each. Sum becomes 4 + 4 + 2 + 2 = 12.Fourth operation: pick the remaining two 2s. Replace them with 4 each. Sum becomes 4 + 4 + 4 + 4 = 16. Which is 4^2 = 16. So again, it's exactly 16. So the minimal sum is achieved by pairing up the smallest numbers each time. But in this case, the minimal sum is exactly 4^m. So maybe in general, the minimal sum after 2^{m-1} * m operations is exactly 4^m. So perhaps the problem is to show that regardless of the strategy, the sum is at least 4^m, but if you do the minimal strategy, you get exactly 4^m.Wait, but the problem says "at least 4^m", so maybe if you do other operations, the sum could be higher. For example, if instead of always combining the smallest numbers, you sometimes combine larger numbers, then the sum increases by more each time. So the minimal sum is achieved by the greedy strategy of always combining the two smallest numbers. Therefore, the total sum is at least 4^m, with equality when you follow the minimal strategy.Therefore, the key is to show that even if you do the worst possible (i.e., minimal) strategy, you still end up with 4^m. So perhaps the problem is equivalent to showing that after 2^{m-1} * m operations, even if you always combine the smallest two numbers each time, the total sum is 4^m.Alternatively, maybe there's a more mathematical way to model the sum. Let's think of each operation as replacing two numbers a and b with two copies of a + b. Therefore, the sum increases by (a + b). So, each operation adds a + b to the total sum. Therefore, the total sum after k operations is S = 2^m + sum_{i=1}^k (a_i + b_i), where a_i and b_i are the numbers on the two cards chosen in the i-th operation.To minimize S, we need to minimize the sum of a_i + b_i over all operations. Therefore, the minimal S is 2^m + sum_{i=1}^{2^{m-1} m} (a_i + b_i), minimized over all possible choices of pairs.Hence, the problem reduces to showing that this minimal sum is at least 4^m. So if we can show that sum_{i=1}^{2^{m-1} m} (a_i + b_i) >= 4^m - 2^m, then the total sum S >= 4^m.Alternatively, perhaps there's a way to track the product of all numbers on the cards or some other invariant. Let me think.Wait, another approach: think about the sum and the number of cards. Wait, each operation doesn't change the number of cards. Wait, wait, the problem says: "choose two different cards. If the numbers on these two cards are a and b, then erase these two numbers and write the number a + b on both cards." Wait, so you erase the two numbers a and b, and replace both cards with a + b. So you still have 2^m cards after each operation. The number of cards never decreases. So each operation replaces two numbers a and b with two numbers (a + b). So the count of cards remains 2^m throughout. Therefore, the number of cards is always 2^m. So the number of operations can be as many as desired? Wait, but in the problem statement, we perform exactly 2^{m - 1} * m operations. So after 2^{m - 1} * m operations, we still have 2^m cards, each with some number, and the sum of these numbers is at least 4^m.So perhaps the key is to model this process as a kind of weighted tree or to track the sum in terms of some other parameter.Alternatively, maybe consider that each operation is similar to a transformation on the cards. Let's see: when you replace a and b with two copies of a + b, the total sum increases by a + b, as we established. So each operation increases the sum. The more operations you do, the higher the sum. But here, we have a fixed number of operations: 2^{m -1 } * m. So we need to show that regardless of the choices made in the operations (i.e., even if we try to minimize the sum), the total sum is at least 4^m.Alternatively, maybe using induction. Let's try induction on m.Base case: m = 1. As we saw, 2 cards, 1 operation. The sum becomes 4, which is 4^1. So it holds.Assume that for some m = k, after 2^{k -1} * k operations on 2^k cards, the sum is at least 4^k. Now, consider m = k + 1. So we have 2^{k + 1} cards. Need to perform 2^{(k +1) - 1} * (k + 1) = 2^{k} * (k +1) operations.But how does the induction step work here? Maybe divide the 2^{k + 1} cards into two groups of 2^k each. Apply the induction hypothesis to each group? But the operations involve choosing two cards, which could be from the same group or different groups. Hmm, not sure.Alternatively, perhaps consider that each operation can be thought of as "merging" two numbers. If we track the sum and the number of operations, maybe there's a relation between the number of operations and the growth of the sum.Alternatively, think of the process as building a binary tree. Each operation merges two nodes into a parent node with value a + b, but since we replace both original nodes with a + b, it's like having two copies of the parent node. Wait, but maybe this is getting too abstract.Alternatively, note that 4^m is (2^m)^2. The initial sum is 2^m. If we can show that the sum grows by a factor of 2^m over the operations, which is 2^m / 2^m = 1. Wait, no. Wait, 4^m is (2^m)^2, which is 2^{2m}. The initial sum is 2^m, so we need to increase it by a factor of 2^m. But over 2^{m -1} * m operations. Hmm, perhaps not directly.Wait, let me think about the sum S after k operations. Each operation increases S by the sum of the two numbers chosen. Therefore, if we can find a lower bound on each a + b in the operations, then we can find a lower bound on S.But if we want to minimize S, we need to minimize each a + b. Therefore, the minimal S is obtained by always choosing the two smallest numbers in each operation. Therefore, the minimal sum S_min is equal to the initial sum plus the sum of the a + b's when always combining the two smallest numbers.Therefore, if we can model the minimal S_min as 4^m, then the problem is solved. Since in the case m=1 and m=2, this gives exactly 4^m. So perhaps in general, after 2^{m -1} * m operations of always combining the two smallest numbers, we end up with all cards being 2^m, so the sum is 2^m * 2^m = 4^m.Wait, but how?Wait, for m=1: 2 cards. After 1 operation: both become 2. Sum is 4.For m=2: 4 cards. After 4 operations, as we saw, all cards become 4. Sum is 16.For m=3: 8 cards. Let's see. Start with 8 ones. Number of operations needed: 2^{2} * 3 = 12 operations.If we follow the minimal strategy:First 4 operations: pair up all the 1s into 2s. Each operation combines two 1s into 2s. 4 operations: 8 / 2 = 4 pairs. After that, all cards are 2. Sum is 8 * 2 = 16.Next, pair up the 2s into 4s. Each operation combines two 2s into 4s. For 8 cards, each operation affects 2 cards, so to convert all 8 cards from 2s to 4s, we need 8 / 2 = 4 operations. Wait, but 4 operations would only process 8 cards. But each operation replaces two cards. Wait, no, each operation takes two cards and turns them into two 4s. So starting with 8 twos:Operation 5: combine two 2s into two 4s. Now we have 6 twos and 2 fours. Sum is 6*2 + 2*4 = 12 + 8 = 20.Operation 6: combine two 2s into two 4s. Now 4 twos and 4 fours. Sum is 4*2 + 4*4 = 8 + 16 = 24.Operation 7: combine two 2s into two 4s. Now 2 twos and 6 fours. Sum is 2*2 + 6*4 = 4 + 24 = 28.Operation 8: combine the last two 2s into two 4s. Now all 8 cards are 4s. Sum is 8*4 = 32.But we need 12 operations. So after 8 operations, all cards are 4s. Then we have 12 - 8 = 4 operations remaining.Operations 9-12: Now we have all 4s. Each operation combines two 4s into two 8s. Each such operation increases the sum by 4 + 4 = 8. So starting from sum 32:Operation 9: combine two 4s into two 8s. Now 6 fours and 2 eights. Sum is 6*4 + 2*8 = 24 + 16 = 40.Operation 10: combine two 4s into two 8s. Now 4 fours and 4 eights. Sum = 16 + 32 = 48.Operation 11: combine two 4s into two 8s. Now 2 fours and 6 eights. Sum = 8 + 48 = 56.Operation 12: combine the last two 4s into two 8s. Now all 8 cards are 8s. Sum is 8*8 = 64.Which is 4^3 = 64. So again, after 12 operations, the sum is exactly 64. So it seems that in general, if you pair up the numbers at each stage, doubling them each time, after m doublings, you get each number as 2^m, and the sum is (2^m)^2 = 4^m.But how does the number of operations relate to m? For m=1: 1 operation (2^{0}*1). For m=2: 4 operations (2^{1}*2). For m=3: 12 operations (2^{2}*3). So the pattern seems to be 2^{m -1} * m. Let's verify:For m=1: 2^{0}*1=1, correct.For m=2: 2^{1}*2=4, correct.For m=3: 2^{2}*3=12, correct.Therefore, the number of operations required to perform m doublings is 2^{m -1} * m. Each doubling phase requires 2^{m -1} operations? Wait, in the m=3 case, to go from 1s to 2s: 4 operations (2^{2}). Then from 2s to 4s: 4 operations (2^{2}), and from 4s to 8s: 4 operations (2^{2}), total 12 = 3*4 = 3*2^{2} = 2^{3 -1}*3. Ah, so each doubling step requires 2^{m -1} operations? Wait, for m=3, each doubling step (from 1 to 2, 2 to 4, 4 to 8) requires 4 operations, which is 2^{3 -1} = 4. And there are m=3 such steps. Therefore, total operations 3*4 = 12. Similarly, for m=2: each doubling step (1 to 2, 2 to 4) requires 2^{2 -1}=2 operations per step, and m=2 steps, total 2*2=4 operations. Yes. So in general, for m, each of the m doubling steps requires 2^{m -1} operations, so total operations m * 2^{m -1}.Therefore, if you structure the operations into m phases, each phase doubling the numbers from x to 2x, and each phase requiring 2^{m -1} operations, then after m phases, you end up with all numbers being 2^m, sum 4^m. Therefore, this is the minimal sum, and any other strategy would result in a higher sum. Therefore, the sum after m * 2^{m -1} operations is at least 4^m.Therefore, the key idea is that the minimal total sum is achieved by systematically doubling the numbers in each card through pairing them up in each phase, requiring m phases each with 2^{m -1} operations. Thus, regardless of the strategy, the sum cannot be lower than this minimal case, which gives exactly 4^m.To formalize this, perhaps we can use induction. Suppose that after k phases (each phase corresponding to doubling the numbers), the sum is (2^k)^2, and we need m phases. Each phase requires 2^{m -1} operations, so total operations m * 2^{m -1}.Alternatively, another approach is to track the product of the numbers on the cards. Wait, but the problem is about the sum, not the product. However, maybe using the AM-GM inequality? Since the arithmetic mean is greater than or equal to the geometric mean. But we have 2^m numbers. The sum is S, so the arithmetic mean is S / 2^m. The geometric mean is (product of all numbers)^(1 / 2^m). Therefore, S / 2^m >= (product)^(1 / 2^m). Therefore, S >= 2^m * (product)^(1 / 2^m). But how does the product evolve?Wait, let's consider the product of all numbers. When we perform an operation, replacing a and b with two copies of a + b. So the new product is (a + b)^2 / (a * b) times the old product. Because we remove a * b and multiply by (a + b)^2. Therefore, the product becomes product * (a + b)^2 / (a * b). Therefore, the product is multiplied by (a + b)^2 / (a * b) each operation.To minimize the sum, we need to choose a and b such that (a + b) is minimized, but this also affects the product. However, it's not clear how to relate this to the sum. Perhaps another angle.Alternatively, note that each operation is similar to a step in a merging process where the total "energy" or something doubles. Wait, if we track the sum S, each operation increases S by a + b. If we can relate the total sum to a function that grows exponentially with the number of operations, but since the number of operations is 2^{m -1} * m, maybe we can find a lower bound.Alternatively, model the process as follows: every time you perform an operation on two cards with numbers a and b, you are effectively redistributing the values but increasing the total sum. The minimal increase happens when a and b are as small as possible. So, if we can show that even with the minimal increases, after 2^{m-1}*m operations, the sum reaches 4^m, then we are done.As observed in the small cases, if you always pair the smallest numbers, you end up with all numbers being 2^m after m phases. Each phase corresponds to doubling the numbers. Let's see:Phase 1: Combine all 1s into 2s. This requires 2^{m -1} operations. For example, with 2^m cards, pairing them up into 2^{m -1} pairs, each operation creating two 2s. So after 2^{m -1} operations, all cards are 2s. Wait, no. Wait, each operation processes two cards, replacing them with two 2s. Therefore, for 2^m cards, each operation affects two cards, so to convert all 2^m cards from 1s to 2s, you need 2^{m -1} operations. Because each operation converts two 1s into two 2s. So 2^{m -1} operations * 2 cards per operation = 2^m cards processed. Therefore, Phase 1: 2^{m -1} operations, resulting in all 2s.Phase 2: Similarly, combine all 2s into 4s. Again, 2^{m -1} operations. Wait, but in the case of m=3, we needed 4 operations (2^{3 -1}) to convert all 8 cards from 2s to 4s. Wait, but 8 cards: each operation converts two 2s into two 4s. So to process all 8 cards, you need 4 operations. Similarly, for 2^m cards, converting from 2s to 4s would take 2^{m -1} operations.But in the m=3 example, we needed 3 phases (1→2, 2→4, 4→8), each with 4 operations (2^{3 -1}), total 12 operations. So in general, m phases, each with 2^{m -1} operations, totaling m * 2^{m -1} operations. Therefore, after m phases, each doubling the numbers, the numbers go from 1 to 2, to 4, ..., up to 2^m, and the sum is 2^m * 2^m = 4^m.Therefore, if you follow this systematic pairing strategy, you reach the sum 4^m after exactly m * 2^{m -1} operations. Therefore, since this is the minimal sum, any other strategy would result in a higher sum. Therefore, the sum after m * 2^{m -1} operations is at least 4^m.Therefore, the conclusion is that regardless of the strategy used, the sum cannot be lower than 4^m after that many operations, hence proving the required result.Final Answerboxed{4^m}</think>We start with (2^m) cards, each labeled with the number 1. Each operation involves choosing two different cards with numbers (a) and (b), erasing these numbers, and writing (a + b) on both cards. We need to prove that after (2^{m-1} m) operations, the sum of the numbers on all cards is at least (4^m).1. Initial Sum: The initial sum of the numbers on the cards is (2^m) since there are (2^m) cards each labeled 1.2. Effect of Each Operation: Each operation increases the total sum by (a + b), where (a) and (b) are the numbers on the chosen cards. This is because replacing (a) and (b) with two copies of (a + b) increases the sum by (2(a + b) - (a + b) = a + b).3. Minimizing the Sum: To minimize the total sum after (2^{m-1} m) operations, we should always choose the two smallest numbers available. This strategy minimizes the increase in the sum each time.4. Phases of Operations: The process can be divided into (m) phases, each doubling the numbers on the cards. Each phase requires (2^{m-1}) operations: - Phase 1: Convert all 1s to 2s, requiring (2^{m-1}) operations. - Phase 2: Convert all 2s to 4s, requiring another (2^{m-1}) operations. - Continue this process until after (m) phases, all numbers are (2^m).5. Total Operations: Each phase requires (2^{m-1}) operations, and there are (m) phases, totaling (2^{m-1} cdot m) operations.6. Resulting Sum: After (m) phases, each card will be labeled (2^m). The total sum is (2^m cdot 2^m = 4^m).7. Conclusion: Since the minimal strategy results in a sum of (4^m), any other strategy will result in a sum that is at least (4^m).Thus, after (2^{m-1} m) operations, the sum of the numbers on all cards is at least (4^m).[boxed{4^m}]]

🔑:2. Let the product of all numbers on the cards after the k-th operation be P_{k}. Suppose in the (k+1)-th operation, the numbers a and b are replaced by two (a+b). Then in P_{k+1}, ab is replaced by (a+b)^{2}, while other factors remain unchanged.Since (a+b)^{2} geqslant 4ab, we have P_{k+1} geqslant 4P_{k}.Given that P_{0}=1, by mathematical induction we getP_{k} geqslant 4^{k}(k in mathbf{N}) text {. }In particular, P_{2^{m-1} m} geqslant 4^{2^{m-1} m}=left(2^{m}right)^{2^{m}}.By the AM-GM inequality, after 2^{m-1} m operations, the sum of all numbers on the cards is at least2^{m} cdot sqrt[2^{m}]{P_{2^{m-1}}} geqslant 2^{m} times 2^{m}=4^{m} .

🧠:[<think>Okay, so there's this problem about four brothers who inherited a convex quadrilateral plot of land. They connected the midpoints of the opposite sides, which divided the land into four quadrilaterals. Three of the brothers got 360, 720, and 900 square meters, and we need to find out what the fourth brother got. Hmm, interesting. Let me try to visualize this.First, I need to recall what a convex quadrilateral is. It's a four-sided figure where all interior angles are less than 180 degrees, and the sides don't cross each other. Now, when they connect the midpoints of opposite sides, that should form some internal lines inside the quadrilateral. I imagine that connecting midpoints would divide the original quadrilateral into smaller sections. But how exactly?Wait, maybe I should draw a rough sketch. Let me think: if you have a convex quadrilateral ABCD, with AB, BC, CD, DA as sides. The midpoints of opposite sides would be, say, the midpoint of AB and CD, and the midpoint of BC and DA. Connecting these midpoints would create two lines inside the quadrilateral. These lines intersect each other, right? So that intersection point would be a central point, and the four quadrilaterals would be around that center.Alternatively, connecting midpoints of sides in a quadrilateral often relates to the Varignon theorem, which states that the midpoints of the sides of any quadrilateral form a parallelogram. But in this case, they're connecting midpoints of opposite sides, not adjacent ones. Wait, connecting midpoints of opposite sides – so that would form two lines instead of a parallelogram. Let me confirm.Suppose we have quadrilateral ABCD. Let M be the midpoint of AB, N the midpoint of BC, P the midpoint of CD, and Q the midpoint of DA. The Varignon parallelogram is MNPQ. But the problem says connecting midpoints of opposite sides. So perhaps connecting M (midpoint of AB) to P (midpoint of CD), and N (midpoint of BC) to Q (midpoint of DA). Then those two lines MP and NQ intersect at some point, dividing the quadrilateral into four smaller quadrilaterals.Alternatively, maybe it's connecting midpoints of both pairs of opposite sides, so two lines connecting midpoints of AB and CD, and midpoints of AD and BC. Depending on the quadrilateral, these lines might intersect at the centroid or some other central point.Wait, maybe it's like this: connecting the midpoints of opposite sides creates two segments that bisect each other. In a parallelogram, the midpoints would create diagonals that bisect each other, but in a general convex quadrilateral, connecting midpoints of opposite sides would still result in two segments that intersect each other at the midpoint of both segments. So that intersection point would be the midpoint of both connecting lines.Therefore, connecting midpoints of opposite sides divides the original quadrilateral into four smaller quadrilaterals. Each of these smaller quadrilaterals would have an area related to the original. But how?Alternatively, maybe each of the four quadrilaterals has an area that's a quarter of the original? But the given areas are 360, 720, 900, and an unknown. If they added up to the total area, but 360 + 720 + 900 + x = total area. However, the problem is probably trickier than that, because if you just connect midpoints, the areas might not be equal.Wait, let's think about this. If we connect the midpoints of opposite sides, the lines formed are called the bimedians. The intersection of the bimedians divides them into halves, and this point is the centroid of the quadrilateral. The centroid divides each bimedian in the ratio 1:1. So each of the four quadrilaterals formed would have areas that are related in some proportion.Alternatively, perhaps each of the four quadrilaterals has an area equal to a quarter of the original? But if the three given areas are 360, 720, 900, which sum to 1980, and if the fourth is x, then total area would be 1980 + x. If each quarter is equal, then 1980 + x = 4x, which would mean 1980 = 3x, so x=660. But 360, 720, 900, 660 – these aren't equal, so that's not possible. Therefore, the areas are not equal, so that approach is wrong.Hmm. Maybe the areas are related in another way. Let me recall if there's a theorem or formula related to connecting midpoints of opposite sides in a convex quadrilateral.Wait, when you connect the midpoints of opposite sides in a convex quadrilateral, the two bimedians (the midlines) intersect at the centroid, and the four resulting quadrilaterals have equal areas. Wait, is that true?Wait, actually, in a convex quadrilateral, the two bimedians divide the quadrilateral into four regions of equal area. So if that's the case, then each of the four quadrilaterals should have the same area. But in the problem, the areas are given as 360, 720, 900, which are not equal. So that contradicts.Therefore, my assumption must be wrong. Maybe connecting the midpoints of opposite sides doesn't result in four equal areas. Wait, maybe the problem is that they connect midpoints of opposite sides and form four quadrilaterals, but they are not the ones formed by the bimedians. Wait, maybe there's another way to connect midpoints.Alternatively, maybe they connect midpoints of adjacent sides. Wait, but the problem says connecting midpoints of opposite sides. Let me check again.Original problem: "By connecting the midpoints of the opposite sides of the plot, they divided the inheritance into four quadrilaterals." So, connect midpoints of opposite sides. So for quadrilateral ABCD, opposite sides are AB and CD, BC and DA. So connect midpoint of AB to midpoint of CD, and midpoint of BC to midpoint of DA. So two lines, which are the bimedians.These two lines intersect at the centroid of the quadrilateral. Then, these two lines divide the quadrilateral into four smaller quadrilaterals. So each of these four quadrilaterals is adjacent to a side of the original quadrilateral and to the centroid.If I can recall, in a convex quadrilateral, the areas of these four smaller quadrilaterals are equal. Wait, but in that case, the problem's given areas would not make sense. So perhaps my memory is incorrect.Wait, let's do an example. Take a square. Connect midpoints of opposite sides. The two lines would be the vertical and horizontal midlines, intersecting at the center. This divides the square into four smaller squares, each of equal area. So in this case, yes, all four areas are equal.But what if the quadrilateral is not a square? Let's take a rectangle that's not a square. Connecting midpoints of opposite sides (the longer sides and the shorter sides) would still divide it into four smaller rectangles, each with equal area, right? Because in a rectangle, the midlines divide it into four equal smaller rectangles. So same as the square.Wait, but what if it's a different convex quadrilateral, like a trapezoid? Let's say an isosceles trapezoid. Connecting midpoints of the two bases (the two parallel sides) and the two legs. The midline connecting the midpoints of the bases is the midline of the trapezoid, which is equal to the average of the two bases. The midline connecting the midpoints of the legs – since it's isosceles, those midpoints connected would be another line. Where they intersect would be the centroid. But would the areas of the four quadrilaterals be equal?In a trapezoid, let's say the two bases are of length 4 and 8, and the height is 6. The area is (4+8)/2 * 6 = 36. The midline connecting the midpoints of the bases would be a horizontal line at height 3, length (4+8)/2 = 6. The midline connecting the midpoints of the legs – since it's isosceles, the legs are equal in length. The midpoints of the legs, when connected, would form a vertical line through the center. The intersection at the center. Then, the four quadrilaterals would each be a rectangle and two trapezoids? Wait, maybe not. Let me visualize.Wait, in the trapezoid, connecting midpoints of the two bases (the top and bottom) gives a horizontal line in the middle. Connecting midpoints of the legs (left and right sides) gives a vertical line through the center. So this divides the trapezoid into four smaller trapezoids. Each of the four would have the same area? Let's compute.Original area is 36. Each of the four parts: if the horizontal midline divides the trapezoid into two trapezoids each of area 18. Then, the vertical midline divides each of those into two, so each would be 9. Wait, but the original trapezoid has height 6. The horizontal midline is at height 3. So the upper trapezoid has bases 4 and 6 (since midline is average of the two bases), and height 3. So area (4+6)/2 * 3 = 15. The lower trapezoid would have bases 6 and 8, height 3, area (6+8)/2 * 3 = 21. Then, dividing each of these with the vertical midline. But how?Wait, the vertical midline would split the trapezoid into left and right parts. For the upper trapezoid (area 15), splitting vertically would divide it into two equal areas? If the midline is vertical, then each part would have half the area, so 7.5 each. Similarly, the lower trapezoid (21) would split into 10.5 each. So then the four areas would be 7.5, 7.5, 10.5, 10.5. Which are not equal. Therefore, in this case, connecting midpoints of opposite sides does not result in four regions of equal area. Therefore, my initial thought was wrong.Therefore, the areas of the four quadrilaterals formed by connecting midpoints of opposite sides in a convex quadrilateral are not necessarily equal. So the problem must involve some relationship between the areas of these four regions.So how do we relate the areas? Let's think. Let me try to model the quadrilateral.Let the original convex quadrilateral be ABCD. Let E be the midpoint of AB, F the midpoint of BC, G the midpoint of CD, H the midpoint of DA. Now, connecting the midpoints of opposite sides: connect E to G and F to H. These two lines EG and FH intersect at point O, which is the centroid of the quadrilateral.Now, the four quadrilaterals formed are:1. AEOH (adjacent to vertex A)2. BFOE (adjacent to vertex B)3. CFOG (adjacent to vertex C)4. DGOH (adjacent to vertex D)Wait, maybe not exactly, but some quadrilaterals adjacent to each original side.Alternatively, perhaps the four quadrilaterals are each bounded by two midlines and two sides of the original quadrilateral. But I need to figure out the relationship between their areas.Alternatively, maybe each of these four quadrilaterals has an area that is 1/4 of the original quadrilateral. But as we saw in the trapezoid example, that's not the case. So that approach is invalid.Wait, maybe there is a relationship where the sum of the areas of opposite quadrilaterals equals half the area of the original quadrilateral. So if the original quadrilateral has area S, then, say, the area of the first and third quadrilaterals sum to S/2, and the second and fourth also sum to S/2. But I need to verify this.Alternatively, maybe each of the four quadrilaterals has an area equal to half the area of the original quadrilateral minus something. Wait, perhaps we need to use vectors or coordinate geometry to model this.Let me try coordinate geometry. Let's assign coordinates to the quadrilateral. Let me place the quadrilateral in a coordinate system. Let’s suppose the original quadrilateral has vertices at points A, B, C, D. Let’s assign coordinates to these points. Let me choose coordinates such that the centroid O is at the origin. Wait, but maybe that complicates things.Alternatively, let me pick coordinates such that point A is at (0, 0), B is at (2a, 0), C is at (2b, 2c), and D is at (2d, 2e). Choosing even coordinates so that midpoints have integer coordinates. Then, the midpoints would be:E: midpoint of AB: (a, 0)F: midpoint of BC: (a + b, c)G: midpoint of CD: (b + d, c + e)H: midpoint of DA: (d, e)Now, connecting E to G: line from (a, 0) to (b + d, c + e)And connecting F to H: line from (a + b, c) to (d, e)The intersection point O of EG and FH can be found by solving the equations of these two lines.But this might be a bit complicated. Alternatively, maybe there's a formula or theorem that relates the areas of these four quadrilaterals.Wait, here's a thought. The lines connecting the midpoints of opposite sides (the bimedians) intersect at the centroid of the quadrilateral. The centroid divides each bimedian into two equal parts. Therefore, each bimedian is bisected by the other. So the point O is the midpoint of both EG and FH.Therefore, the four quadrilaterals formed are all parallelograms? Wait, no. Wait, in the Varignon theorem, connecting midpoints of sides gives a parallelogram, but here we are connecting midpoints of opposite sides. So the lines EG and FH are the bimedians, and they bisect each other. So each of the four quadrilaterals is a quadrilateral with two sides along the original quadrilateral and two sides along the bimedians.Alternatively, maybe each of the four quadrilaterals is a kite? Not necessarily.Wait, perhaps the areas of these four quadrilaterals can be related to the areas of triangles formed by the centroid.Alternatively, maybe the four quadrilaterals each have areas equal to half of the original quadrilateral's area divided by two, but that seems vague.Wait, let's think in terms of vectors. Let’s denote the position vectors of A, B, C, D as vectors a, b, c, d. The midpoints E, F, G, H have position vectors (a + b)/2, (b + c)/2, (c + d)/2, (d + a)/2. The bimedians EG and FH have direction vectors (c + d - a - b)/2 and (d + a - b - c)/2. The intersection point O is the midpoint of both EG and FH, so its position vector is (a + b + c + d)/4.Therefore, the centroid O is the average of the four vertices.Now, the four quadrilaterals are:1. Between A, E, O, H2. Between B, F, O, E3. Between C, G, O, F4. Between D, H, O, GEach of these quadrilaterals is a quadrilateral with vertices at a vertex of the original quadrilateral, a midpoint of a side, the centroid, and another midpoint.To compute the area of each quadrilateral, perhaps we can use coordinate geometry or vector cross products.Suppose we compute the area of quadrilateral AEOH. This quadrilateral has vertices at A, E, O, H.Expressed in vectors, these points are:A: aE: (a + b)/2O: (a + b + c + d)/4H: (d + a)/2So the quadrilateral AEOH is a polygon with vertices a, (a + b)/2, (a + b + c + d)/4, (d + a)/2.To find the area of this quadrilateral, perhaps we can divide it into two triangles: A-E-O and O-H-A.Wait, but maybe it's easier to use the shoelace formula if we assign coordinates.Let me assign coordinates to the quadrilateral. Let me set A at (0, 0), B at (2, 0), D at (0, 2), and C at (2, 2). Wait, but that's a square. Let me choose a different quadrilateral to test.Let me pick a trapezoid with A(0,0), B(4,0), C(6,4), D(2,4). So this is a trapezoid with bases AB (4 units) and CD (4 units as well, from (2,4) to (6,4)), and legs AD and BC.Wait, but in this case, the midpoints:E is midpoint of AB: (2,0)F is midpoint of BC: midpoint of (4,0) and (6,4): (5,2)G is midpoint of CD: midpoint of (6,4) and (2,4): (4,4)H is midpoint of DA: midpoint of (2,4) and (0,0): (1,2)Connecting E(2,0) to G(4,4): this line goes from (2,0) to (4,4). The equation is y = 2x - 4.Connecting F(5,2) to H(1,2): this line is horizontal at y=2.These two lines intersect at y=2. Substitute y=2 into EG's equation: 2 = 2x - 4 => x=3. So intersection point O is (3,2).Now, compute the areas of the four quadrilaterals.1. Quadrilateral AEOH: A(0,0), E(2,0), O(3,2), H(1,2)Using shoelace formula:Coordinates: (0,0), (2,0), (3,2), (1,2), (0,0)Area = 1/2 | (0*0 + 2*2 + 3*2 + 1*0) - (0*2 + 0*3 + 2*1 + 2*0) | = 1/2 | (0 + 4 + 6 + 0) - (0 + 0 + 2 + 0)| = 1/2 |10 - 2| = 4.2. Quadrilateral BFOE: B(4,0), F(5,2), O(3,2), E(2,0)Coordinates: (4,0), (5,2), (3,2), (2,0), (4,0)Area = 1/2 | (4*2 + 5*2 + 3*0 + 2*0) - (0*5 + 2*3 + 2*2 + 0*4) | = 1/2 | (8 + 10 + 0 + 0) - (0 + 6 + 4 + 0)| = 1/2 |18 - 10| = 4.3. Quadrilateral CFOG: C(6,4), F(5,2), O(3,2), G(4,4)Coordinates: (6,4), (5,2), (3,2), (4,4), (6,4)Area = 1/2 | (6*2 + 5*2 + 3*4 + 4*4) - (4*5 + 2*3 + 2*4 + 4*6) | = 1/2 | (12 + 10 + 12 + 16) - (20 + 6 + 8 + 24)| = 1/2 |50 - 58| = 4.Wait, that's 1/2 | -8 | = 4. Hmm.4. Quadrilateral DGOH: D(2,4), G(4,4), O(3,2), H(1,2)Coordinates: (2,4), (4,4), (3,2), (1,2), (2,4)Area = 1/2 | (2*4 + 4*2 + 3*2 + 1*4) - (4*4 + 4*3 + 2*1 + 2*2) | = 1/2 | (8 + 8 + 6 + 4) - (16 + 12 + 2 + 4)| = 1/2 |26 - 34| = 4.Wait a minute, all four areas are equal to 4 in this trapezoid. But the original area of the trapezoid is ((4 + 4)/2)*4 = 16. Each of the four quadrilaterals has area 4, which is 1/4 of the total area. But in the previous example with the trapezoid, when I took a different trapezoid, the areas weren't equal. But here, in this specific trapezoid, they are equal. Why is that?Wait, in this case, the trapezoid is actually a parallelogram. Because AB is from (0,0) to (4,0), and CD is from (2,4) to (6,4). Wait, but CD is from (2,4) to (6,4), which is also length 4, same as AB. The sides AD and BC: AD is from (0,0) to (2,4), which is a line with slope 2, and BC is from (4,0) to (6,4), which also has slope 2. So AD and BC are parallel. Therefore, this trapezoid is actually a parallelogram. So in a parallelogram, connecting midpoints of opposite sides divides it into four regions of equal area.But in a non-parallelogram trapezoid, like the earlier example with bases 4 and 8, the areas weren't equal. So in a parallelogram, the four regions are equal, but in a general quadrilateral, they are not. Therefore, the problem must involve a general convex quadrilateral, so the areas can be different.But in the problem, the three given areas are 360, 720, and 900. So how can we find the fourth?Wait, maybe there is a relationship between the areas. For instance, in any convex quadrilateral, when you connect the midpoints of opposite sides, the four resulting quadrilaterals satisfy that the sum of the areas of two opposite quadrilaterals equals the sum of the areas of the other two. Or some other relation.Alternatively, maybe the areas are related proportionally based on the lengths of the bimedians or something else.Wait, perhaps we can use barycentric coordinates or affine geometry. But maybe a simpler approach.Wait, let's suppose that the original quadrilateral is divided into four parts by the two bimedians. Let the areas of these four parts be S1, S2, S3, S4. Then, perhaps there's a relation among S1, S2, S3, S4.In the example above with the parallelogram, all areas were equal. In another example, maybe with a kite, the areas might have some relationship.Wait, let's think of another example. Let’s take a convex quadrilateral that's not a parallelogram. Let's choose A(0,0), B(2,0), C(3,2), D(1,3). Then, the midpoints:E is midpoint of AB: (1,0)F is midpoint of BC: (2.5,1)G is midpoint of CD: (2,2.5)H is midpoint of DA: (0.5,1.5)Connecting E(1,0) to G(2,2.5): line from (1,0) to (2,2.5). The equation can be parametrized as x = 1 + t, y = 0 + 2.5t, for t from 0 to 1.Connecting F(2.5,1) to H(0.5,1.5): line from (2.5,1) to (0.5,1.5). The slope is (1.5 - 1)/(0.5 - 2.5) = 0.5/(-2) = -0.25. Equation: y - 1 = -0.25(x - 2.5).Find intersection point O of EG and FH.From EG: x = 1 + t, y = 2.5t.From FH: y = -0.25x + 0.625 + 1 = -0.25x + 1.625.Substitute x and y from EG into FH equation:2.5t = -0.25(1 + t) + 1.6252.5t = -0.25 - 0.25t + 1.6252.5t + 0.25t = 1.625 - 0.252.75t = 1.375t = 0.5Therefore, O is at x = 1 + 0.5 = 1.5, y = 2.5 * 0.5 = 1.25.Now, compute the areas of the four quadrilaterals.1. AEOH: A(0,0), E(1,0), O(1.5,1.25), H(0.5,1.5)Using shoelace formula:Coordinates: (0,0), (1,0), (1.5,1.25), (0.5,1.5), (0,0)Area = 1/2 | (0*0 + 1*1.25 + 1.5*1.5 + 0.5*0) - (0*1 + 0*1.5 + 1.25*0.5 + 1.5*0) | = 1/2 | (0 + 1.25 + 2.25 + 0) - (0 + 0 + 0.625 + 0) |= 1/2 |3.5 - 0.625| = 1/2 * 2.875 = 1.43752. BFOE: B(2,0), F(2.5,1), O(1.5,1.25), E(1,0)Coordinates: (2,0), (2.5,1), (1.5,1.25), (1,0), (2,0)Area = 1/2 | (2*1 + 2.5*1.25 + 1.5*0 + 1*0) - (0*2.5 + 1*1.5 + 1.25*1 + 0*2) |= 1/2 | (2 + 3.125 + 0 + 0) - (0 + 1.5 + 1.25 + 0) |= 1/2 |5.125 - 2.75| = 1/2 * 2.375 = 1.18753. CFOG: C(3,2), F(2.5,1), O(1.5,1.25), G(2,2.5)Coordinates: (3,2), (2.5,1), (1.5,1.25), (2,2.5), (3,2)Area = 1/2 | (3*1 + 2.5*1.25 + 1.5*2.5 + 2*2) - (2*2.5 + 1*1.5 + 1.25*2 + 2.5*3) |= 1/2 | (3 + 3.125 + 3.75 + 4) - (5 + 1.5 + 2.5 + 7.5) |= 1/2 |13.875 - 16.5| = 1/2 * 2.625 = 1.31254. DGOH: D(1,3), G(2,2.5), O(1.5,1.25), H(0.5,1.5)Coordinates: (1,3), (2,2.5), (1.5,1.25), (0.5,1.5), (1,3)Area = 1/2 | (1*2.5 + 2*1.25 + 1.5*1.5 + 0.5*3) - (3*2 + 2.5*1.5 + 1.25*0.5 + 1.5*1) |= 1/2 | (2.5 + 2.5 + 2.25 + 1.5) - (6 + 3.75 + 0.625 + 1.5) |= 1/2 |8.75 - 11.875| = 1/2 * 3.125 = 1.5625Total area of the four quadrilaterals: 1.4375 + 1.1875 + 1.3125 + 1.5625 = 5.5Original quadrilateral area: Let's compute it using shoelace formula.Coordinates: A(0,0), B(2,0), C(3,2), D(1,3), A(0,0)Area = 1/2 | (0*0 + 2*2 + 3*3 + 1*0) - (0*2 + 0*3 + 2*1 + 3*0) | = 1/2 | (0 + 4 + 9 + 0) - (0 + 0 + 2 + 0)| = 1/2 |13 - 2| = 5.5. So that checks out.So the four quadrilaterals have areas: approx 1.4375, 1.1875, 1.3125, 1.5625.Looking for a pattern here. Let me check if opposite quadrilaterals sum to the same.AEOH (1.4375) and CFOG (1.3125): sum to 2.75BFOE (1.1875) and DGOH (1.5625): sum to 2.75So total area 5.5, each pair of opposite quadrilaterals sums to 2.75, which is half of 5.5.So perhaps in a convex quadrilateral, the areas of the two pairs of opposite quadrilaterals formed by connecting midpoints of opposite sides each sum to half the area of the original quadrilateral.In the problem statement, three areas are given: 360, 720, and 900. Let's denote the four quadrilaterals as S1, S2, S3, S4.If opposite quadrilaterals sum to half the total area, then S1 + S3 = S2 + S4 = total area / 2.Given three areas, we can find the fourth by seeing which pair they belong to.Suppose S1 = 360, S2 = 720, S3 = 900. Then, if S1 and S3 are opposite, their sum is 360 + 900 = 1260. Then S2 + S4 = 720 + S4 = 1260, so S4 = 1260 - 720 = 540.Alternatively, if S1 and S2 are opposites: 360 + 720 = 1080, then S3 + S4 = 900 + S4 = 1080, so S4 = 180. But 180 seems too small compared to the others.Alternatively, if S2 and S3 are opposites: 720 + 900 = 1620, then S1 + S4 = 360 + S4 = 1620, so S4 = 1260. But then the total area would be 1620 * 2 = 3240. But 360 + 720 + 900 + 1260 = 3240, which is possible.But how do we know which ones are opposite? The problem states that the four brothers received plots of land measuring 360, 720, 900, and the fourth. It doesn't specify the order. But since the problem refers to connecting midpoints of opposite sides, the four quadrilaterals would be adjacent to each original vertex. So the first brother's plot (360) is adjacent to vertex A, the second (720) adjacent to B, the third (900) adjacent to C, and the fourth adjacent to D. Then, the opposite quadrilaterals would be A and C, B and D. Therefore, the areas of A and C sum to half the total, and B and D sum to the other half.Therefore, if the areas adjacent to A, B, C are 360, 720, 900, then the area adjacent to D is x. Then:360 (A) + 900 (C) = 720 (B) + x (D) = total area / 2Therefore, 360 + 900 = 1260 = total area / 2 ⇒ total area = 2520. Then, 720 + x = 1260 ⇒ x = 540. So the fourth brother received 540 m².Alternatively, if the three given areas are not in order, but since the problem states the first three brothers, it might be in order around the quadrilateral. However, without loss of generality, as long as opposite quadrilaterals sum to half the area, we can pair them accordingly.Alternatively, if we don't know the order, but given that three areas are 360, 720, 900, possible pairings:Case 1: 360 and 720 are opposites. Then their sum is 1080 = total / 2 ⇒ total 2160. The remaining pair would be 900 and x, which should sum to 1080 ⇒ x=180. But 180 is much smaller than the others. Possible? Maybe.Case 2: 360 and 900 are opposites. Sum 1260 ⇒ total 2520. The other pair 720 and x must sum to 1260 ⇒ x=540.Case 3: 720 and 900 are opposites. Sum 1620 ⇒ total 3240. The other pair 360 and x sum to 1620 ⇒ x=1260.So depending on how the areas are arranged, x could be 180, 540, or 1260. But which one is correct?The problem states that the brothers divided the land by connecting midpoints of opposite sides. The four resulting quadrilaterals are adjacent to the original vertices. Therefore, if we traverse the original quadrilateral, the areas would be in the order of the vertices. So if we have quadrilateral ABCD, the four plots would be adjacent to A, B, C, D respectively. Therefore, if the first three brothers got the plots adjacent to A, B, C, then the fourth is adjacent to D. Then, the opposite plots are A-C and B-D. Hence, S1 + S3 = S2 + S4. Therefore, given S1=360, S2=720, S3=900, then S4= 360 + 900 - 720 = 540. Thus, the fourth brother received 540 m².Therefore, the answer should be 540.Final Answerboxed{540}</think>The problem involves four brothers inheriting a convex quadrilateral plot of land, which is divided into four smaller quadrilaterals by connecting the midpoints of opposite sides. Three of the brothers received plots of 360, 720, and 900 square meters, and we need to determine the area of the fourth plot.Key steps:1. Understanding the Division: Connecting the midpoints of opposite sides of a convex quadrilateral forms two bimedians, which intersect at the centroid of the quadrilateral. This division creates four smaller quadrilaterals.2. Area Relationship: The areas of the four smaller quadrilaterals formed by the bimedians have a specific relationship. The sum of the areas of opposite quadrilaterals is equal. This is because the centroid divides each bimedian into two equal parts, leading to the sum of the areas of opposite quadrilaterals being half of the total area of the original quadrilateral.3. Applying the Relationship: Given three areas (360, 720, and 900), we need to determine the fourth area. By checking the possible pairings of opposite quadrilaterals, we find that the correct pairing (considering the order of the brothers' plots adjacent to the vertices of the original quadrilateral) leads to the following: - The sum of the areas of the first and third plots (360 and 900) is 1260. - The sum of the areas of the second and fourth plots (720 and the unknown) must also be 1260. - Therefore, the fourth area is (1260 - 720 = 540).Thus, the fourth brother received (boxed{540}) square meters of land.,<think>Okay, so there's this problem about four brothers who inherited a convex quadrilateral plot of land. By connecting the midpoints of the opposite sides, they divided the land into four quadrilaterals. Three of the brothers got plots measuring 360, 720, and 900 square meters, and we need to find out what the fourth brother got. Hmm, interesting.First, let me visualize this. A convex quadrilateral, so four sides, all interior angles less than 180 degrees. Connecting the midpoints of opposite sides... Wait, how exactly are they connecting them? If it's a quadrilateral, there are two pairs of opposite sides. So if you connect the midpoints of each pair of opposite sides, you would get two lines. But connecting both pairs would intersect somewhere inside the quadrilateral, dividing it into four smaller quadrilaterals. Is that how it works?Let me think. If you take a convex quadrilateral ABCD, with midpoints of sides AB, BC, CD, DA being, say, E, F, G, H. Then connecting midpoints of opposite sides would mean connecting E to G and F to H? But in that case, connecting midpoints of opposite sides would create two lines intersecting at the centroid of the quadrilateral? Maybe. Wait, but connecting midpoints of sides in a quadrilateral usually forms a parallelogram, according to the midpoint theorem. Hmm, maybe not in this case. Let me recall.The Varignon theorem states that the midpoints of the sides of any quadrilateral form a parallelogram, called the Varignon parallelogram. So if you connect the midpoints of each side in order, you get a parallelogram. But in this problem, they are connecting midpoints of opposite sides, not adjacent ones. So perhaps they connect the midpoint of AB to the midpoint of CD, and midpoint of BC to midpoint of DA. Then those two lines would intersect at some point inside the quadrilateral, dividing it into four smaller quadrilaterals.Okay, so each of these four quadrilaterals is bounded by half of each of the two connecting lines. Now, each brother gets one of these four quadrilaterals. The areas of three are given, and we need to find the fourth.I need to figure out a relationship between the areas of these four quadrilaterals. Maybe there's a symmetry or proportion here. Since we're dealing with midpoints, the lines connecting them might be related to midlines in triangles, which are parallel to the base and half its length. But in a quadrilateral, connecting midpoints of opposite sides... Hmm.Alternatively, perhaps the area ratios can be determined by some properties of the original quadrilateral. Let me try to think of a coordinate system. Maybe assign coordinates to the original quadrilateral and compute the areas of the four smaller quadrilaterals.Let’s denote the original quadrilateral as ABCD, with coordinates A, B, C, D. Let's assign coordinates for simplicity. Let me set point A at (0, 0). Let’s say point B is at (2a, 0), so the midpoint of AB is at (a, 0). Then point D can be at (0, 2d), so the midpoint of AD is at (0, d). Then point C would be somewhere in the plane to make the quadrilateral convex. Let's say point C is at (2c, 2e). Then the midpoint of BC is at ( (2a + 2c)/2, (0 + 2e)/2 ) = (a + c, e). The midpoint of CD is at ( (2c + 0)/2, (2e + 2d)/2 ) = (c, e + d). Now, connecting midpoints of opposite sides: midpoint of AB (a, 0) to midpoint of CD (c, e + d), and midpoint of BC (a + c, e) to midpoint of DA (0, d). The intersection of these two lines would divide the original quadrilateral into four smaller quadrilaterals. Each of these smaller quadrilaterals should have an area that relates to the original quadrilateral's area.But this might get complicated with coordinates. Maybe there's a better approach. Let's think in terms of vectors or areas ratios.Another approach: since we are connecting midpoints, maybe each of these lines divides the original quadrilateral into two parts with equal areas? Wait, connecting midpoints of opposite sides... If that's the case, then each line would split the quadrilateral into two regions of equal area. Then the intersection of the two lines would create four regions, each of which would have a quarter of the total area? But that seems too simplistic because the problem states different areas: 360, 720, 900, and the fourth one. If each was a quarter, they would all be equal. So that can't be right.Wait, maybe it's not that each line divides the area equally, but instead the four quadrilaterals formed have areas that relate in a specific way.Alternatively, perhaps the areas of the four smaller quadrilaterals add up to the area of the original quadrilateral. So if we can find a relation between them, maybe the sum of all four is the total area, and then we can find the missing one.But the problem is, we need to find the fourth area given the other three, but we don't know the total area. Unless there's a relation between the areas of the four parts.Alternatively, maybe the areas of opposite quadrilaterals are equal. Let me think. If the original quadrilateral is divided by two lines connecting midpoints of opposite sides, then maybe opposite quadrilaterals have the same area? If that's the case, then the missing area would be equal to one of the given areas. But in the problem, the three given areas are 360, 720, 900. None of them are equal, so perhaps this is not the case.Alternatively, maybe the sum of the areas of two opposite quadrilaterals equals the sum of the areas of the other two. But again, without knowing the total area, it's hard to see.Wait, maybe using vectors. Let me try that.Let’s denote the original quadrilateral as ABCD. Let’s assign vectors to the points. Let’s let vectors A, B, C, D be points in the plane.The midpoints:Midpoint of AB: M1 = (A + B)/2Midpoint of BC: M2 = (B + C)/2Midpoint of CD: M3 = (C + D)/2Midpoint of DA: M4 = (D + A)/2Now, connecting midpoints of opposite sides: so connecting M1 to M3 and M2 to M4. The intersection point of these two lines would divide the original quadrilateral into four smaller quadrilaterals.Let’s denote the intersection point as O.So each of the four quadrilaterals is: AM1OM4, BM1OM2, CM2OM3, DM3OM4 (assuming the connections). Wait, maybe not exactly. Let me think.If you connect M1 to M3 and M2 to M4, then the intersection point O divides each connecting line into two segments. The four quadrilaterals would be:1. Between A, M1, O, M4.2. Between M1, B, M2, O.3. Between O, M2, C, M3.4. Between M4, O, M3, D.So quadrilaterals: AM1OM4, M1BM2O, OM2CM3, M4OM3D.Now, to find the areas of these four quadrilaterals. Maybe we can find relations between them.Alternatively, since all midpoints are involved, perhaps the areas are related to the area of the original quadrilateral divided in some proportion.Alternatively, perhaps each of the four quadrilaterals has an area equal to half of the area of one of the four triangles formed by the original quadrilateral's diagonals? Wait, but we aren't given the diagonals here.Wait, maybe the key is that connecting midpoints of opposite sides creates a parallelogram, as per Varignon's theorem, but in this case, connecting midpoints of opposite sides instead of adjacent ones. Wait, Varignon's parallelogram is formed by connecting midpoints of adjacent sides. If we connect midpoints of opposite sides, maybe we get another parallelogram?Wait, let me check. Suppose we have a convex quadrilateral, connect the midpoints of opposite sides. Then the figure formed by these two connecting lines is two lines intersecting each other. But Varignon's theorem is about connecting midpoints of consecutive sides, which forms a parallelogram. Here, connecting midpoints of opposite sides would form two segments that intersect each other, but not necessarily a parallelogram. So maybe this is a different configuration.Alternatively, perhaps the key is that the lines connecting midpoints of opposite sides bisect each other. Let me think. If we have a quadrilateral, and connect midpoints of AB and CD, and midpoints of BC and DA, then those two lines will bisect each other. That is, their intersection point is the midpoint of both connecting lines. Is that always true?Hmm, perhaps so. Let me test with a simple quadrilateral, like a square. If you connect midpoints of opposite sides in a square, those lines would be the same as the diagonals of the square, which intersect at the center and bisect each other. So in that case, the intersection point is the center. Similarly, in a rectangle, connecting midpoints of opposite sides would result in lines that cross at the center. So maybe in any convex quadrilateral, connecting midpoints of opposite sides results in two lines that bisect each other, meaning their intersection is the midpoint of both lines.Therefore, the four smaller quadrilaterals would each have areas that are related in some way.Wait, in the case of a square, each of the four quadrilaterals would be congruent, so each would have a quarter of the total area. But in the problem, the areas are different, so it's not a square. So maybe the key is that in any convex quadrilateral, the areas of the four quadrilaterals formed by connecting midpoints of opposite sides satisfy a certain relationship.Alternatively, maybe the sum of the areas of opposite quadrilaterals is equal. For example, AM1OM4 + OM2CM3 = M1BM2O + M4OM3D. If that's the case, then given three areas, we can find the fourth. Let's see.But how do we know if that's true? Let me consider a simple case. Take a trapezoid. Suppose it's a trapezoid with bases of length 4 and 8, and height 6. The area would be (4+8)/2 * 6 = 36. Now, connecting midpoints of opposite sides. Let's assign coordinates.Let’s say the trapezoid has vertices at A(0,0), B(4,0), C(6,6), D(2,6). Then midpoints:Midpoint of AB: (2,0)Midpoint of BC: (5,3)Midpoint of CD: (4,6)Midpoint of DA: (1,3)Connecting midpoints of AB and CD: line from (2,0) to (4,6).Connecting midpoints of BC and DA: line from (5,3) to (1,3).These two lines intersect at (3,3), which is the midpoint of both connecting lines. Now, the four quadrilaterals:1. Between A(0,0), midpoint AB (2,0), intersection (3,3), midpoint DA (1,3). Let's compute its area.Coordinates: (0,0), (2,0), (3,3), (1,3). This is a quadrilateral. To find its area, we can divide it into two triangles or use the shoelace formula.Shoelace formula: (0*0 + 2*3 + 3*3 + 1*0) - (0*2 + 0*3 + 3*1 + 3*0) = (0 + 6 + 9 + 0) - (0 + 0 + 3 + 0) = 15 - 3 = 12. So area is 12/2 = 6.2. Between midpoint AB (2,0), B(4,0), midpoint BC (5,3), intersection (3,3).Coordinates: (2,0), (4,0), (5,3), (3,3). Shoelace: (2*0 + 4*3 + 5*3 + 3*0) - (0*4 + 0*5 + 3*3 + 3*2) = (0 + 12 + 15 + 0) - (0 + 0 + 9 + 6) = 27 - 15 = 12. Area is 6.3. Between intersection (3,3), midpoint BC (5,3), C(6,6), midpoint CD (4,6).Coordinates: (3,3), (5,3), (6,6), (4,6). Shoelace: (3*3 + 5*6 + 6*6 + 4*3) - (3*5 + 3*6 + 6*4 + 6*3) = (9 + 30 + 36 + 12) - (15 + 18 + 24 + 18) = 87 - 75 = 12. Area 6.4. Between midpoint DA (1,3), intersection (3,3), midpoint CD (4,6), D(2,6).Coordinates: (1,3), (3,3), (4,6), (2,6). Shoelace: (1*3 + 3*6 + 4*6 + 2*3) - (3*3 + 3*4 + 6*2 + 6*1) = (3 + 18 + 24 + 6) - (9 + 12 + 12 + 6) = 51 - 39 = 12. Area 6.Wait, in this trapezoid example, all four quadrilaterals have the same area of 6, but the total area of the trapezoid was 36. But 4*6 = 24, which is less than 36. That doesn't add up. Wait, what's wrong here?Wait, no. The original trapezoid area is 36, but the four quadrilaterals formed by connecting the midpoints of opposite sides only cover part of the trapezoid? Or maybe the shoelace calculation is wrong.Wait, let's check again. First quadrilateral: (0,0), (2,0), (3,3), (1,3). Shoelace formula:List the coordinates in order:(0,0), (2,0), (3,3), (1,3), back to (0,0).Sum of x_i y_{i+1}:0*0 + 2*3 + 3*3 + 1*0 = 0 + 6 + 9 + 0 = 15Sum of y_i x_{i+1}:0*2 + 0*3 + 3*1 + 3*0 = 0 + 0 + 3 + 0 = 3Area: |15 - 3| / 2 = 12 / 2 = 6. That's correct.But then the total area covered by the four quadrilaterals is 4*6 = 24. But the original trapezoid area is 36. So where is the remaining 12?Ah, because connecting the midpoints of opposite sides divides the original quadrilateral into four smaller quadrilaterals and a central parallelogram? Wait, in the example, the lines intersect at (3,3), but in the trapezoid, maybe there's a central region as well? Wait, in my coordinates, the lines connecting midpoints of AB to CD and midpoints of BC to DA intersect at (3,3), and the four quadrilaterals each have area 6, but maybe there are also four triangles in the middle?Wait, no. Let me check again. If we have a trapezoid with vertices A(0,0), B(4,0), C(6,6), D(2,6). Connecting midpoints of AB (2,0) to CD (4,6), which is the line from (2,0) to (4,6), and midpoints of BC (5,3) to DA (1,3), which is the horizontal line from (5,3) to (1,3). Their intersection is at (3,3). So the figure is divided into four quadrilaterals:1. A(0,0), (2,0), (3,3), (1,3)2. (2,0), B(4,0), (5,3), (3,3)3. (3,3), (5,3), C(6,6), (4,6)4. (1,3), (3,3), (4,6), D(2,6)But when I computed each of their areas, they each gave 6, but the original trapezoid's area is 36, so clearly something is wrong here. Wait, but maybe the original trapezoid is not 36. Let me recalculate the area of the trapezoid.Coordinates: A(0,0), B(4,0), C(6,6), D(2,6). Using shoelace formula:List the coordinates: A(0,0), B(4,0), C(6,6), D(2,6), back to A(0,0).Sum of x_i y_{i+1}:0*0 + 4*6 + 6*6 + 2*0 = 0 + 24 + 36 + 0 = 60Sum of y_i x_{i+1}:0*4 + 0*6 + 6*2 + 6*0 = 0 + 0 + 12 + 0 = 12Area: |60 - 12| / 2 = 48 / 2 = 24. Oh! So the trapezoid's area is actually 24. Then the four quadrilaterals each have area 6, which adds up to 24. So that makes sense. I must have miscalculated the area earlier. So in this case, each quadrilateral has equal area. So in a trapezoid, connecting midpoints of opposite sides divides it into four regions of equal area. Interesting. So in that case, each brother would get an equal area. But in the problem given, the areas are different: 360, 720, 900. So that suggests that the original quadrilateral is not a trapezoid or any symmetric shape. Therefore, the areas of the four regions can vary.But in the trapezoid example, they were equal. So maybe in general, for any quadrilateral, connecting midpoints of opposite sides divides it into four regions with equal areas? But no, the problem states different areas, so that must not be the case. Wait, but in the trapezoid example, the areas were equal. Maybe in some cases they can be different. Let me try with another quadrilateral.Take a convex quadrilateral that's not symmetric. Let's say A(0,0), B(4,0), C(6,2), D(1,3). Let's compute midpoints.Midpoint of AB: (2,0)Midpoint of BC: (5,1)Midpoint of CD: (3.5, 2.5)Midpoint of DA: (0.5, 1.5)Now, connecting midpoints of AB (2,0) to CD (3.5, 2.5): the line from (2,0) to (3.5,2.5)Connecting midpoints of BC (5,1) to DA (0.5,1.5): the line from (5,1) to (0.5,1.5)Find the intersection point O of these two lines.First, find equations of the two lines.Line 1: from (2,0) to (3.5,2.5). The slope is (2.5 - 0)/(3.5 - 2) = 2.5 / 1.5 = 5/3. Equation: y - 0 = (5/3)(x - 2), so y = (5/3)x - 10/3.Line 2: from (5,1) to (0.5,1.5). The slope is (1.5 - 1)/(0.5 - 5) = 0.5 / (-4.5) = -1/9. Equation: y - 1 = (-1/9)(x - 5). So y = (-1/9)x + 5/9 + 1 = (-1/9)x + 14/9.Find intersection O by solving:(5/3)x - 10/3 = (-1/9)x + 14/9Multiply both sides by 9 to eliminate denominators:15x - 30 = -x + 1416x = 44x = 44/16 = 11/4 = 2.75Then y = (5/3)(11/4) - 10/3 = (55/12) - (40/12) = 15/12 = 5/4 = 1.25So intersection O is at (2.75, 1.25).Now, compute the areas of the four quadrilaterals.First quadrilateral: A(0,0), midpoint AB (2,0), O(2.75,1.25), midpoint DA (0.5,1.5). Let's use shoelace formula.Coordinates: (0,0), (2,0), (2.75,1.25), (0.5,1.5), back to (0,0).Sum of x_i y_{i+1}:0*0 + 2*1.25 + 2.75*1.5 + 0.5*0 = 0 + 2.5 + 4.125 + 0 = 6.625Sum of y_i x_{i+1}:0*2 + 0*2.75 + 1.25*0.5 + 1.5*0 = 0 + 0 + 0.625 + 0 = 0.625Area: |6.625 - 0.625| / 2 = 6 / 2 = 3Second quadrilateral: midpoint AB (2,0), B(4,0), midpoint BC (5,1), O(2.75,1.25).Coordinates: (2,0), (4,0), (5,1), (2.75,1.25), back to (2,0).Shoelace:Sum x_i y_{i+1}:2*0 + 4*1 + 5*1.25 + 2.75*0 = 0 + 4 + 6.25 + 0 = 10.25Sum y_i x_{i+1}:0*4 + 0*5 + 1*2.75 + 1.25*2 = 0 + 0 + 2.75 + 2.5 = 5.25Area: |10.25 - 5.25| /2 = 5 /2 = 2.5Third quadrilateral: O(2.75,1.25), midpoint BC (5,1), C(6,2), midpoint CD (3.5,2.5).Coordinates: (2.75,1.25), (5,1), (6,2), (3.5,2.5), back to (2.75,1.25).Shoelace:Sum x_i y_{i+1}:2.75*1 + 5*2 + 6*2.5 + 3.5*1.25 = 2.75 + 10 + 15 + 4.375 = 32.125Sum y_i x_{i+1}:1.25*5 + 1*6 + 2*3.5 + 2.5*2.75 = 6.25 + 6 + 7 + 6.875 = 26.125Area: |32.125 - 26.125| /2 = 6 /2 = 3Fourth quadrilateral: midpoint DA (0.5,1.5), O(2.75,1.25), midpoint CD (3.5,2.5), D(1,3).Coordinates: (0.5,1.5), (2.75,1.25), (3.5,2.5), (1,3), back to (0.5,1.5).Shoelace:Sum x_i y_{i+1}:0.5*1.25 + 2.75*2.5 + 3.5*3 + 1*1.5 = 0.625 + 6.875 + 10.5 + 1.5 = 19.5Sum y_i x_{i+1}:1.5*2.75 + 1.25*3.5 + 2.5*1 + 3*0.5 = 4.125 + 4.375 + 2.5 + 1.5 = 12.5Area: |19.5 - 12.5| /2 = 7 /2 = 3.5So the four areas are 3, 2.5, 3, 3.5. Total area is 3 + 2.5 + 3 + 3.5 = 12. Let's check the original quadrilateral's area.Original quadrilateral coordinates: A(0,0), B(4,0), C(6,2), D(1,3). Shoelace formula:Sum x_i y_{i+1}:0*0 + 4*2 + 6*3 + 1*0 = 0 + 8 + 18 + 0 = 26Sum y_i x_{i+1}:0*4 + 0*6 + 2*1 + 3*0 = 0 + 0 + 2 + 0 = 2Area: |26 - 2| /2 = 24 /2 = 12. Correct.So in this case, the four quadrilaterals have areas 3, 2.5, 3, 3.5, which are not equal. So in this asymmetric quadrilateral, the areas are different. So the problem's scenario is possible.Now, the question is: given three of the areas, 360, 720, 900, how do we find the fourth?Looking at the example I just did, the areas were 3, 2.5, 3, 3.5. Hmm. Is there a pattern or relation between them?Let me see. Let's denote the four areas as S1, S2, S3, S4. In my example, S1=3, S2=2.5, S3=3, S4=3.5. Is there any relation like S1 + S3 = S2 + S4? 3 + 3 = 6, 2.5 + 3.5 = 6. So yes, their sums are equal. Similarly, in the trapezoid example, all four areas were equal, so their sums would also be equal.So perhaps in general, S1 + S3 = S2 + S4. If that's the case, then given three areas, we can compute the fourth.In the problem, the three given areas are 360, 720, 900. Assuming that the missing area is x. If the rule is S1 + S3 = S2 + S4, then depending on which areas correspond to which positions.But in the problem statement, they mention the first three brothers received plots of 360, 720, and 900. The fourth is to be found. The problem doesn't specify the order of the plots, so we need to figure out if there's a way to order them such that the relation S1 + S3 = S2 + S4 holds.Alternatively, maybe all four areas are in proportion. But in my asymmetric example, the areas were 3, 2.5, 3, 3.5. The products of opposite areas: 3*3 = 9 and 2.5*3.5 = 8.75. Not equal. The sums of adjacent areas: 3 + 2.5 = 5.5, 2.5 + 3 = 5.5, 3 + 3.5 = 6.5, 3.5 +3 = 6.5. Hmm, no, not a pattern.But in that example, S1 + S3 = S2 + S4, so 3 + 3 = 2.5 + 3.5 = 6.So maybe that is a general rule. Let me check another example.Take another quadrilateral. Let’s say A(0,0), B(2,0), C(3,2), D(1,3). Midpoints:AB midpoint: (1,0)BC midpoint: (2.5,1)CD midpoint: (2,2.5)DA midpoint: (0.5,1.5)Connecting (1,0) to (2,2.5) and (2.5,1) to (0.5,1.5). Find intersection O.First line: from (1,0) to (2,2.5). Slope: (2.5 - 0)/(2 - 1) = 2.5/1 = 2.5. Equation: y = 2.5x - 2.5.Second line: from (2.5,1) to (0.5,1.5). Slope: (1.5 -1)/(0.5 -2.5) = 0.5/(-2) = -0.25. Equation: y -1 = -0.25(x -2.5). So y = -0.25x + 0.625 +1 = -0.25x +1.625.Find intersection:2.5x -2.5 = -0.25x +1.6252.75x = 4.125x = 4.125 /2.75 = 1.5Then y = 2.5*1.5 -2.5 = 3.75 -2.5 =1.25. So O(1.5,1.25).Now compute areas of four quadrilaterals.First quadrilateral: A(0,0), (1,0), O(1.5,1.25), (0.5,1.5). Shoelace:Coordinates: (0,0), (1,0), (1.5,1.25), (0.5,1.5), back to (0,0).Sum x_i y_{i+1}:0*0 +1*1.25 +1.5*1.5 +0.5*0 =0 +1.25 +2.25 +0=3.5Sum y_i x_{i+1}:0*1 +0*1.5 +1.25*0.5 +1.5*0 =0 +0 +0.625 +0=0.625Area: |3.5 -0.625|/2=2.875/2=1.4375Second quadrilateral: (1,0), B(2,0), (2.5,1), O(1.5,1.25). Shoelace:(1,0), (2,0), (2.5,1), (1.5,1.25), back to (1,0).Sum x_i y_{i+1}:1*0 +2*1 +2.5*1.25 +1.5*0 =0 +2 +3.125 +0=5.125Sum y_i x_{i+1}:0*2 +0*2.5 +1*1.5 +1.25*1 =0 +0 +1.5 +1.25=2.75Area: |5.125 -2.75|/2=2.375/2=1.1875Third quadrilateral: O(1.5,1.25), (2.5,1), C(3,2), (2,2.5). Shoelace:(1.5,1.25), (2.5,1), (3,2), (2,2.5), back to (1.5,1.25).Sum x_i y_{i+1}:1.5*1 +2.5*2 +3*2.5 +2*1.25=1.5 +5 +7.5 +2.5=16Sum y_i x_{i+1}:1.25*2.5 +1*3 +2*2 +2.5*1.5=3.125 +3 +4 +3.75=13.875Area: |16 -13.875|/2=2.125/2=1.0625Fourth quadrilateral: (0.5,1.5), O(1.5,1.25), (2,2.5), D(1,3). Shoelace:(0.5,1.5), (1.5,1.25), (2,2.5), (1,3), back to (0.5,1.5).Sum x_i y_{i+1}:0.5*1.25 +1.5*2.5 +2*3 +1*1.5=0.625 +3.75 +6 +1.5=11.875Sum y_i x_{i+1}:1.5*1.5 +1.25*2 +2.5*1 +3*0.5=2.25 +2.5 +2.5 +1.5=8.75Area: |11.875 -8.75|/2=3.125/2=1.5625Total area: 1.4375 +1.1875 +1.0625 +1.5625=5.25Check original quadrilateral's area:A(0,0), B(2,0), C(3,2), D(1,3). Shoelace:Sum x_i y_{i+1}:0*0 +2*2 +3*3 +1*0=0 +4 +9 +0=13Sum y_i x_{i+1}:0*2 +0*3 +2*1 +3*0=0 +0 +2 +0=2Area: |13 -2|/2=11/2=5.5. Wait, but our total from the four quadrilaterals was 5.25. That's a discrepancy. Hmm, this suggests an error in calculations.Wait, let me recalculate the original quadrilateral's area.Coordinates: A(0,0), B(2,0), C(3,2), D(1,3), back to A(0,0).Shoelace sum:x_i y_{i+1}: 0*0 + 2*2 +3*3 +1*0 =0 +4 +9 +0=13y_i x_{i+1}:0*2 +0*3 +2*1 +3*0=0 +0 +2 +0=2Area: |13 -2| /2=11/2=5.5. But according to the four quadrilaterals, total area is 5.25. So where is the missing 0.25?Hmm, this suggests a miscalculation in the areas of the quadrilaterals. Let me check the first quadrilateral again.First quadrilateral: (0,0), (1,0), (1.5,1.25), (0.5,1.5).Shoelace:x: 0,1,1.5,0.5,0y:0,0,1.25,1.5,0Sum x_i y_{i+1}:0*0 +1*1.25 +1.5*1.5 +0.5*0 =0 +1.25 +2.25 +0=3.5Sum y_i x_{i+1}:0*1 +0*1.5 +1.25*0.5 +1.5*0=0 +0 +0.625 +0=0.625Area: |3.5 -0.625|/2=2.875/2=1.4375. Correct.Second quadrilateral: (1,0), (2,0), (2.5,1), (1.5,1.25)Shoelace:x:1,2,2.5,1.5,1y:0,0,1,1.25,0Sum x_i y_{i+1}:1*0 +2*1 +2.5*1.25 +1.5*0=0 +2 +3.125 +0=5.125Sum y_i x_{i+1}:0*2 +0*2.5 +1*1.5 +1.25*1=0 +0 +1.5 +1.25=2.75Area: |5.125 -2.75|/2=2.375/2=1.1875. Correct.Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5)Shoelace:x:1.5,2.5,3,2,1.5y:1.25,1,2,2.5,1.25Sum x_i y_{i+1}:1.5*1 +2.5*2 +3*2.5 +2*1.25=1.5 +5 +7.5 +2.5=16Sum y_i x_{i+1}:1.25*2.5 +1*3 +2*2 +2.5*1.5=3.125 +3 +4 +3.75=13.875Area: |16 -13.875|/2=2.125/2=1.0625. Correct.Fourth quadrilateral: (0.5,1.5), (1.5,1.25), (2,2.5), (1,3)Shoelace:x:0.5,1.5,2,1,0.5y:1.5,1.25,2.5,3,1.5Sum x_i y_{i+1}:0.5*1.25 +1.5*2.5 +2*3 +1*1.5=0.625 +3.75 +6 +1.5=11.875Sum y_i x_{i+1}:1.5*1.5 +1.25*2 +2.5*1 +3*0.5=2.25 +2.5 +2.5 +1.5=8.75Area: |11.875 -8.75|/2=3.125/2=1.5625. Correct.Total area: 1.4375 +1.1875 +1.0625 +1.5625=5.25. But original area is 5.5. So discrepancy of 0.25. Why?Ah, perhaps due to overlapping or calculation errors. Wait, maybe the way I divided the original quadrilateral is incorrect. Wait, when connecting midpoints of opposite sides, perhaps there's a central region as well? But in my previous examples, connecting the two midlines divided the quadrilateral into four regions with total area equal to the original. Wait, but in this case, it's not. There's a missing area of 0.25.Wait, perhaps the problem is that the quadrilaterals as defined don't cover the entire original quadrilateral? Or perhaps my coordinates were entered incorrectly.Wait, original quadrilateral A(0,0), B(2,0), C(3,2), D(1,3). When connecting midpoints of AB (1,0) to CD (2,2.5), and midpoints of BC (2.5,1) to DA (0.5,1.5), their intersection at (1.5,1.25). So the four quadrilaterals should cover the entire original quadrilateral. But according to the calculations, they don't. Where is the mistake?Wait, maybe one of the quadrilaterals is not a simple quadrilateral but overlaps or something. Alternatively, perhaps I made a mistake in the shoelace formula for one of the quadrilaterals.Let me check the fourth quadrilateral again: (0.5,1.5), (1.5,1.25), (2,2.5), (1,3)Coordinates in order: (0.5,1.5), (1.5,1.25), (2,2.5), (1,3), back to (0.5,1.5).Sum x_i y_{i+1}:0.5*1.25 +1.5*2.5 +2*3 +1*1.5 =0.625 +3.75 +6 +1.5=11.875Sum y_i x_{i+1}:1.5*1.5 +1.25*2 +2.5*1 +3*0.5=2.25 +2.5 +2.5 +1.5=8.75Difference:11.875 -8.75=3.125. Absolute value divided by 2:1.5625. Correct.Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5). Shoelace again:Sum x_i y_{i+1}:1.5*1 +2.5*2 +3*2.5 +2*1.25=1.5 +5 +7.5 +2.5=16Sum y_i x_{i+1}:1.25*2.5 +1*3 +2*2 +2.5*1.5=3.125 +3 +4 +3.75=13.875Difference:16 -13.875=2.125. Area:1.0625. Correct.Second quadrilateral: (1,0), (2,0), (2.5,1), (1.5,1.25). Shoelace:Sum x_i y_{i+1}:1*0 +2*1 +2.5*1.25 +1.5*0=0 +2 +3.125 +0=5.125Sum y_i x_{i+1}:0*2 +0*2.5 +1*1.5 +1.25*1=0 +0 +1.5 +1.25=2.75Difference:5.125 -2.75=2.375. Area:1.1875. Correct.First quadrilateral: (0,0), (1,0), (1.5,1.25), (0.5,1.5). Shoelace:Sum x_i y_{i+1}:0*0 +1*1.25 +1.5*1.5 +0.5*0=0 +1.25 +2.25 +0=3.5Sum y_i x_{i+1}:0*1 +0*1.5 +1.25*0.5 +1.5*0=0 +0 +0.625 +0=0.625Difference:3.5 -0.625=2.875. Area:1.4375. Correct.Total area:1.4375 +1.1875 +1.0625 +1.5625=5.25. But original area is5.5. So where is the missing 0.25?Ah, perhaps the intersection point O is not part of the original quadrilateral? No, because O is inside the original quadrilateral. Alternatively, maybe my coordinates had a mistake. Let me check the original quadrilateral again.Original quadrilateral: A(0,0), B(2,0), C(3,2), D(1,3). Let me plot these points. A is at origin, B is at (2,0), C is at (3,2), D is at (1,3). Connecting these points in order: A to B to C to D to A. The figure is a convex quadrilateral. Computing its area with shoelace formula gave 5.5, which should be correct. However, the sum of the four quadrilaterals is 5.25. There's a discrepancy of 0.25. This is confusing.Wait, maybe I missed a region? When connecting midpoints of opposite sides, perhaps there's an overlapping area or another region? But in the previous trapezoid example, it added up correctly. What's different here?Alternatively, perhaps the lines connecting midpoints of opposite sides do not divide the quadrilateral into four regions but into four regions and a central parallelogram? But in the trapezoid example, the lines intersected at the center and divided the trapezoid into four quadrilaterals with no central region. But in this asymmetric case, maybe there is a central region?Wait, no. If you connect two lines inside a quadrilateral, their intersection divides it into four regions. So in this case, the two lines connecting midpoints of opposite sides intersect at O, so the four regions are as computed. But why does the area not add up?Alternatively, maybe I made a mistake in the coordinates of the original quadrilateral. Let me recalculate the original quadrilateral's area again.Coordinates: A(0,0), B(2,0), C(3,2), D(1,3). Shoelace formula:List the coordinates in order:A(0,0), B(2,0), C(3,2), D(1,3), back to A(0,0).Compute sum of x_i y_{i+1}:0*0 (A to B) + 2*2 (B to C) + 3*3 (C to D) + 1*0 (D to A) = 0 + 4 + 9 + 0 = 13Sum of y_i x_{i+1}:0*2 (A to B) + 0*3 (B to C) + 2*1 (C to D) + 3*0 (D to A) = 0 + 0 + 2 + 0 = 2Area: |13 - 2| /2 =11/2=5.5. Correct.So where is the mistake? The four quadrilaterals' areas sum to 5.25, but the original area is 5.5. So missing 0.25.Wait, perhaps rounding errors? In the calculation of the intersection point O?Wait, O was at (1.5,1.25). Let me verify the equations.First line: from (1,0) to (2,2.5). The slope is 2.5/1=2.5. Equation: y=2.5x -2.5.Second line: from (2.5,1) to (0.5,1.5). Slope: (1.5-1)/(0.5-2.5)=0.5/-2=-0.25. Equation: y -1 = -0.25(x -2.5). So y= -0.25x +0.625 +1= -0.25x +1.625.Solving 2.5x -2.5= -0.25x +1.625.2.5x +0.25x=1.625 +2.5.2.75x=4.125.x=4.125/2.75=1.5. Then y=2.5*1.5 -2.5=3.75 -2.5=1.25. Correct.So O is correctly calculated. Therefore, coordinates are accurate.Hmm, this is perplexing. There must be a mistake in the assumption that the four quadrilaterals cover the entire original area. But according to the way we divided the original quadrilateral with two lines, it should be divided into four regions. Unless some areas were double-counted or missed.Wait, let me check visually. Original quadrilateral A(0,0), B(2,0), C(3,2), D(1,3). Connecting midpoints (1,0) to (2,2.5) and (2.5,1) to (0.5,1.5), intersecting at (1.5,1.25). So the four regions are:1. A(0,0), (1,0), (1.5,1.25), (0.5,1.5)2. (1,0), B(2,0), (2.5,1), (1.5,1.25)3. (1.5,1.25), (2.5,1), C(3,2), (2,2.5)4. (0.5,1.5), (1.5,1.25), (2,2.5), D(1,3)Plotting these, it seems like they should cover the entire original quadrilateral. Maybe the shoelace formula missed something due to the order of the coordinates?Let me try calculating the area of the first quadrilateral again, ensuring that the coordinates are ordered correctly (clockwise or counter-clockwise).First quadrilateral: A(0,0), (1,0), (1.5,1.25), (0.5,1.5). Let's order them clockwise.Coordinates:A(0,0) -> (1,0) -> (1.5,1.25) -> (0.5,1.5) -> A(0,0)This is counter-clockwise. Shoelace formula requires coordinates to be ordered either all clockwise or all counter-clockwise. Let's check if that's the case.Plotting the points:A(0,0) is bottom-left.(1,0) is to the right on the x-axis.(1.5,1.25) is up and to the right.(0.5,1.5) is up and to the left.Connecting these in order makes a counter-clockwise quadrilateral, so shoelace formula should work.Similarly for the others.Wait, maybe one of the quadrilaterals is actually a complex polygon that intersects itself, leading to negative areas? But in the shoelace formula, taking the absolute value should handle that. Hmm.Alternatively, maybe the intersection point O is not inside all of the original quadrilateral? No, in a convex quadrilateral, connecting midpoints of opposite sides should intersect inside.This is confusing. Maybe there's a bug in my calculations. Let me use another method to calculate the area of the fourth quadrilateral.Fourth quadrilateral: (0.5,1.5), (1.5,1.25), (2,2.5), (1,3)We can divide this quadrilateral into two triangles: (0.5,1.5), (1.5,1.25), (2,2.5) and (0.5,1.5), (2,2.5), (1,3)First triangle area:Using determinant formula: 0.5*| (0.5*(1.25 -2.5) +1.5*(2.5 -1.5) +2*(1.5 -1.25) ) |Wait, maybe better to use vectors.Vector from (0.5,1.5) to (1.5,1.25): (1, -0.25)Vector from (0.5,1.5) to (2,2.5): (1.5,1)Area of the parallelogram: |(1)(1) - (-0.25)(1.5)| = |1 +0.375|=1.375. So triangle area is 0.6875Second triangle: (0.5,1.5), (2,2.5), (1,3)Vectors from (0.5,1.5) to (2,2.5): (1.5,1)From (0.5,1.5) to (1,3): (0.5,1.5)Area of parallelogram: |1.5*1.5 -1*0.5| = |2.25 -0.5|=1.75. Triangle area:0.875Total area:0.6875 +0.875=1.5625. Which matches the shoelace result. So that's correct.Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5)Divide into two triangles:(1.5,1.25), (2.5,1), (3,2) and (1.5,1.25), (3,2), (2,2.5)First triangle area:Vectors from (1.5,1.25) to (2.5,1): (1, -0.25)From (1.5,1.25) to (3,2): (1.5,0.75)Parallelogram area: |1*0.75 - (-0.25)*1.5| = |0.75 +0.375|=1.125. Triangle area:0.5625Second triangle:Vectors from (1.5,1.25) to (3,2): (1.5,0.75)From (1.5,1.25) to (2,2.5): (0.5,1.25)Parallelogram area: |1.5*1.25 -0.75*0.5|= |1.875 -0.375|=1.5. Triangle area:0.75Total area:0.5625 +0.75=1.3125. Wait, but earlier shoelace gave 1.0625. Contradiction. This suggests an error in my method.Wait, no, perhaps the division into triangles is not correct. If the quadrilateral is not convex, the shoelace formula might not work. Let me check if the quadrilateral is convex.Coordinates: (1.5,1.25), (2.5,1), (3,2), (2,2.5). Plotting these points:(1.5,1.25) -> (2.5,1) -> (3,2) -> (2,2.5) -> (1.5,1.25). It seems convex. So why discrepancy?Alternatively, maybe the two triangles overlap. Wait, no. If we divide the quadrilateral into two triangles along a diagonal, there should be no overlap.First triangle: (1.5,1.25), (2.5,1), (3,2)Second triangle: (1.5,1.25), (3,2), (2,2.5)Computing first triangle area via determinant:Area = 0.5*| (1.5*(1 -2) +2.5*(2 -1.25) +3*(1.25 -1) ) |=0.5*|1.5*(-1) +2.5*(0.75) +3*(0.25)|=0.5*| -1.5 +1.875 +0.75 | =0.5*|1.125|=0.5625Second triangle:Area =0.5*|1.5*(2 -2.5) +3*(2.5 -1.25) +2*(1.25 -2)|=0.5*|1.5*(-0.5) +3*(1.25) +2*(-0.75)|=0.5*| -0.75 +3.75 -1.5 | =0.5*|1.5|=0.75Total area:0.5625 +0.75=1.3125. But shoelace gave 1.0625. What's wrong here?Wait, perhaps I made a mistake in the determinant formula. Let me use the shoelace formula for the third quadrilateral again.Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5), back to (1.5,1.25).Shoelace sum:x_i y_{i+1}:1.5*1 +2.5*2 +3*2.5 +2*1.25=1.5 +5 +7.5 +2.5=16y_i x_{i+1}:1.25*2.5 +1*3 +2*2 +2.5*1.5=3.125 +3 +4 +3.75=13.875Area: |16 -13.875| /2=2.125 /2=1.0625. Correct.But according to the triangle areas, it should be 1.3125. This contradiction suggests that my division into triangles is flawed.Ah, because if the quadrilateral is not convex, the diagonal might not lie entirely inside the quadrilateral, leading to overlapping areas. But we assumed it's convex. Wait, the original quadrilateral is convex, but the smaller quadrilaterals might not be.Wait, looking at the coordinates of the third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5). Plotting these points, does this form a convex quadrilateral?(1.5,1.25) to (2.5,1): downward slope.(2.5,1) to (3,2): upward slope.(3,2) to (2,2.5): leftward and upward.(2,2.5) to (1.5,1.25): downward slope.This seems convex. So why the discrepancy?Wait, perhaps I made a mistake in the determinant calculation. Let me use vectors for the first triangle.First triangle: (1.5,1.25), (2.5,1), (3,2)Vectors:From (1.5,1.25) to (2.5,1): (1, -0.25)From (1.5,1.25) to (3,2): (1.5,0.75)The area is half the magnitude of the cross product:|(1)(0.75) - (-0.25)(1.5)| = |0.75 +0.375| =1.125. Half is 0.5625. Correct.Second triangle: (1.5,1.25), (3,2), (2,2.5)Vectors:From (1.5,1.25) to (3,2): (1.5,0.75)From (1.5,1.25) to (2,2.5): (0.5,1.25)Cross product: (1.5)(1.25) - (0.75)(0.5) =1.875 -0.375=1.5. Half is 0.75. Correct.Total area:0.5625 +0.75=1.3125. But shoelace says 1.0625. This is a contradiction.Ah! I realize now. The shoelace formula calculates the area based on the order of the points, assuming non-intersecting polygon. However, if the points are ordered incorrectly, the shoelace formula can give the wrong area. Let me check the order of the points in the third quadrilateral.Third quadrilateral's coordinates: (1.5,1.25), (2.5,1), (3,2), (2,2.5)If we plot these points:1. (1.5,1.25)2. (2.5,1)3. (3,2)4. (2,2.5)Connecting them in this order: 1 ->2 ->3 ->4 ->1. Is this a non-intersecting polygon? Let's see.From (1.5,1.25) to (2.5,1): downward.From (2.5,1) to (3,2): upward to the right.From (3,2) to (2,2.5): leftward and upward.From (2,2.5) to (1.5,1.25): downward to the left.This should form a convex quadrilateral. Therefore, the shoelace formula should work. But why the discrepancy?Wait, perhaps I made a mistake in transcribing the coordinates. Let me re-enter them:Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5)Shoelace formula step-by-step:List the coordinates:1. (1.5,1.25)2. (2.5,1)3. (3,2)4. (2,2.5)5. (1.5,1.25)Sum of x_i y_{i+1}:1.5*1 +2.5*2 +3*2.5 +2*1.25 =1.5 +5 +7.5 +2.5=16Sum of y_i x_{i+1}:1.25*2.5 +1*3 +2*2 +2.5*1.5=3.125 +3 +4 +3.75=13.875Difference:16 -13.875=2.125. Area=1.0625. Correct.But when split into two triangles, the total is 1.3125. This is impossible. What's going on here?Wait, perhaps the two triangles overlap? How? If the quadrilateral is convex, splitting it into two triangles along a diagonal should not cause overlap.Wait, maybe I split it along the wrong diagonal. Let me try splitting along another diagonal.Third quadrilateral: (1.5,1.25), (2.5,1), (3,2), (2,2.5)Split into triangles (1.5,1.25), (2.5,1), (2,2.5) and (1.5,1.25), (2,2.5), (3,2)First triangle: (1.5,1.25), (2.5,1), (2,2.5)Area via determinant:0.5*|1.5*(1 -2.5) +2.5*(2.5 -1.25) +2*(1.25 -1)|=0.5*|1.5*(-1.5) +2.5*(1.25) +2*(0.25)|=0.5*| -2.25 +3.125 +0.5 | =0.5*|1.375|=0.6875Second triangle: (1.5,1.25), (2,2.5), (3,2)Area:0.5*|1.5*(2.5 -2) +2*(2 -1.25) +3*(1.25 -2.5)|=0.5*|1.5*0.5 +2*0.75 +3*(-1.25)|=0.5*|0.75 +1.5 -3.75| =0.5*| -1.5 | =0.75Total area:0.6875 +0.75=1.4375. Still not matching shoelace's 1.0625.This is perplexing. There must be something fundamentally wrong in my approach. Perhaps the original assumption that connecting midpoints of opposite sides divides the quadrilateral into four regions whose areas sum to the original area is incorrect? But in the trapezoid example, it worked.Alternatively, maybe my coordinate system is flawed. I think the problem here is that in asymmetric quadrilaterals, the lines connecting midpoints of opposite sides can create overlapping regions or some other complexity, leading to discrepancies. However, in the trapezoid example, it worked perfectly. So maybe the key is that in certain quadrilaterals, the areas add up, and in others, they don't? That seems unlikely.Wait, perhaps the correct approach here is to accept that there is a relationship between the areas, such as S1 + S3 = S2 + S4, as observed in the trapezoid and the first asymmetric example. In the first asymmetric example, S1=3, S2=2.5, S3=3, S4=3.5. So S1 + S3=6, S2 + S4=6. Similarly, in the problem with the brothers, if the areas are 360, 720, 900, and x, then maybe 360 + 900 = 720 + x. So 1260 = 720 + x → x = 540. Alternatively, if the pairing is different, like 360 + 720 = 900 + x → 1080 = 900 + x → x=180. Or 360 + x = 720 + 900 → x=1260. But which pairing is correct?In the example where S1 + S3 = S2 + S4, the opposite areas summed to equal totals. So perhaps, in the original problem, we need to determine which of the given areas are opposite each other.But the problem statement doesn't specify the order of the plots. So the brothers could be arranged in any order around the original quadrilateral. Therefore, we can't be sure which areas are opposite unless there's a different hint.However, in the examples I tried, the opposite areas added to the same sum. So perhaps in the problem, regardless of the order, the missing area plus one of the given areas equals the sum of the other two.But how?Wait, in the first example: areas 3, 2.5, 3, 3.5. Sum of all areas:12. So pairs of opposite areas summed to 6 each. Therefore, S1 + S3 = S2 + S4.In the problem, given three areas:360,720,900, missing x. If the rule is S1 + S3 = S2 + S4, then depending on which ones are paired, different x's result. But without knowing the positions, we need another approach.Alternatively, maybe all four areas satisfy S1 + S2 + S3 + S4 = Total area, but we don't know the total. But if there's a relation between them like S1 + S3 = S2 + S4 = Total/2, then we can compute Total = (S1 + S3)*2, and then subtract the three given areas to find the fourth.But in the first example, Total =12, which is indeed 6*2. But in the second example, the areas summed to 5.25 but the Total was5.5. So this approach might not work.Alternatively, maybe there's a different relation. Wait, in the first example (trapezoid), all four areas were equal. In the asymmetric example, two areas were equal (3 and 3), and the other two were 2.5 and 3.5. The sum of the two pairs of opposite areas were equal. So S1 + S3 = S2 + S4. If that's a general rule, then regardless of the shape, connecting midpoints of opposite sides creates four regions where the sums of opposite areas are equal.Therefore, in the problem, the fourth area can be found by ensuring that the sum of two opposite areas equals the sum of the other two.But since we don't know which ones are opposite, we need to find which pairing of the given areas allows this.Assuming the three given areas are adjacent, but it's also possible that two of them are opposite. But since we don't have information about the arrangement, the problem might be designed such that the fourth area is uniquely determined regardless of the pairing.Let’s test this. Suppose the three given areas are S1=360, S2=720, S3=900. Then S4=x.If S1 + S3 = S2 + S4 → 360 +900 =720 +x → 1260=720 +x →x=540.Alternatively, if S1 + S2 = S3 +S4 →360+720=900 +x →1080=900 +x →x=180.Alternatively, S1 + S4 = S2 + S3 →360 +x=720+900 →x=1260.But which one is correct?In the examples, the opposite areas were those across the intersection point. So in the trapezoid, each area was equal. In the asymmetric example, S1=3 opposite S3=3, and S2=2.5 opposite S4=3.5. So the rule was S1 + S3 = S2 + S4.Similarly, in the problem, if the areas are arranged such that 360 and 900 are opposite, and 720 and x are opposite, then 360 +900=720 +x →x=540.Alternatively, if 360 and 720 are opposite, then 360+720=900 +x →x=180.Or if 360 and x are opposite, then 360 +x=720 +900 →x=1260.Therefore, the answer depends on which areas are considered opposite. But the problem doesn't specify the arrangement.However, given that the problem is from a competition or textbook, it's likely that the intended answer is 540, assuming that the given areas are placed such that 360 and 900 are opposite. Because in the examples, the areas that were across from each other had the same sum. Since 360 +900=1260 and 720 +540=1260, which balances.Alternatively, if the problem is designed such that the four areas form a proportion where the product of opposite areas are equal:360*900=720*x →x=(360*900)/720=450. But this is different.Alternatively, maybe the areas are related to the midlines, and each area is the average of adjacent areas. But this is speculative.Alternatively, recall that in the trapezoid example, the four areas were all equal, each being 1/4 of the total area. But in the asymmetric example, the total area was 12, and the areas were 3,2.5,3,3.5. Their sum is 12, and indeed 3+3=6, 2.5+3.5=6.So in general, regardless of the shape, the sum of opposite areas is equal, each equal to half of the total area. Therefore, in the problem, the fourth area can be found by pairing the given areas such that two of them sum to the same as the other two.Given three areas:360,720,900. The fourth area x must satisfy that the sum of two areas equals the sum of the other two. So possible pairings:Case 1: 360 + 900 = 720 +x →x=540Case 2:360 +720 =900 +x →x=180Case 3:360 +x =720 +900 →x=1260But the problem states that the plot is divided into four quadrilaterals. If the fourth area is 540, then total area would be 360+720+900+540=2520. If it's 180, total is 2160. If it's 1260, total is 3240.But unless there's a reason to choose one over the others, the problem is ambiguous. However, in the examples I worked through, the rule was that opposite areas sum to equal amounts. Therefore, if we assume that the three given areas include two that are opposite and one adjacent, the missing area must pair with the remaining one to balance the sum.But how to determine which given areas are opposite?In the problem statement, they divided the inheritance into four quadrilaterals by connecting midpoints of opposite sides. The first three brothers received 360, 720, and 900. Assuming the areas are labeled in order around the original quadrilateral, the fourth area would be opposite the first one. But without knowing the order, it's impossible to be sure.Alternatively, perhaps the areas are related to the original quadrilateral's properties. For example, the area of the entire plot is the sum of the four areas. If the lines connecting midpoints of opposite sides divide the quadrilateral into four regions whose areas are related to the original's area.But wait, in the trapezoid example, the total area was 24, and each of the four regions was 6. In the first asymmetric example, total area 12, regions summed to 12. In the problematic example with total area 5.5, but computed areas summed to5.25, there was a discrepancy, but that might have been due to calculation error.But assuming that in general, the four regions sum to the total area, then the fourth area is total area minus the sum of the three given. But since we don't know the total area, this approach doesn't help. However, if the total area can be inferred from the given areas.But in the examples where opposite areas summed to half the total area, the total area would be twice the sum of any pair of opposite areas. So if we can figure out which pair of given areas are opposite, we can compute the total area and then find the fourth.Given three areas:360,720,900. The possible pairs:360+720=1080360+900=1260720+900=1620Assuming that one of these is half the total area, then the total would be 2160, 2520, or 3240.Then the fourth area would be:If total is 2160: fourth area=2160 - (360+720+900)=180If total is2520: fourth area=2520 - (360+720+900)=540If total is3240: fourth area=3240 - (360+720+900)=1260But which total is correct?In the examples, the sum of opposite areas was equal to half the total. So if we assume that two of the given areas are opposite, their sum would be half the total area, hence the total is twice their sum. Then the fourth area is the other half minus the third given area.But since we have three given areas, we don't know which two are opposite. Hence, three possibilities.But in competition problems, often such problems are designed so that the fourth area is uniquely determined. Hence, there must be a different approach.Alternatively, recall that connecting]

🔑:Solution. Let the vertices of the quadrilateral be A, B, C, and D, and the midpoints of the sides be E, F, G, and H as shown in Figure 1. The intersection point of the two midlines M is a median in the four triangles formed by the segments connecting M to the vertices, which divide the quadrilateral. These segments thus bisect the area of each of these triangles. From this, it follows that[E A F M]+[H M G C]=[D E M H]+[M F B G]=frac{t}{2}where t is the area of the quadrilateral A B C D. ([EAFM] denotes the area of the quadrilateral with vertices E, A, F, and M, and the same applies to the other terms.) This follows from the fact that each pair of the four triangles has one member on the left side and the other on the right side of the above sum.Figure 1For simplicity, let's choose the unit of area to be 180 mathrm{~m}^{2}. Then the areas of the known quadrilaterals are 2, 4, and 5 units, respectively. According to equation (1), there are three possible cases depending on which of the three known quadrilaterals is diagonally opposite the unknown quadrilateral. These cases are shown in Figures 2/a, 2/b, and 2/c. We will show that two of these cases are not possible.t=12, frac{t}{8}=frac{3}{2}>1t=18, frac{t}{8}=frac{9}{4}>2Figure 2/aFigure 2/bFigure 2/cIt is known that the midpoints of a quadrilateral's sides form a parallelogram whose area is half that of the quadrilateral. The midlines of the quadrilateral A B C D are diagonals in this parallelogram E F G H, which divides it into four triangles of area frac{t}{8}. Each of these triangles contains one of the four equal shares, so each of their areas is greater than frac{t}{8} (Figure 3). This is only true in the second arrangement, shown in Figure 2/b, where the fourth sibling's share is only 3 units, or 540 mathrm{~m}^{2}.Figure 3We must still show that this result is possible and that there are no additional conditions that would rule it out. To do this, we provide a convex quadrilateral in which the areas of the individual sub-quadrilaterals are 2, 4, 5, and 3 units, respectively.Start with a quadrilateral A B C D whose diagonals intersect in the ratios 5: 9 and 1: 13 as shown in Figure 4. Such a convex quadrilateral can clearly be constructed. Then frac{[A B D]}{[C D B]}=frac{1}{13}, since the heights corresponding to the common base B D of the two triangles are in the ratio 1: 13. Similarly, we get frac{[A C D]}{[C A B]}=frac{5}{9}. If the area of the quadrilateral A B C D is 14 units, then its diagonals divide it into triangles of areas 1 and 13, and 5 and 9 units, respectively, as shown in Figure 5. The segments connecting the midpoints of the sides of the quadrilateral are midlines in these triangles, so the areas of the triangles formed at the vertices of the quadrilateral A B C D are one-fourth of these values. Finally, the frac{t}{2}=7 unit area parallelogram E F G H is divided by its diagonals into triangles of area frac{7}{4} (Figure 6). The areas of the individual parts are thusbegin{aligned}& {[M E A F]=[E A F]+[E F M]=frac{1}{4}+frac{7}{4}=2} & {[M F B G]=[F B G]+[F G M]=frac{9}{4}+frac{7}{4}=4} & {[M G C H]=[G C H]+[G H M]=frac{13}{4}+frac{7}{4}=5, quad text { and finally }} & {[M H D E]=[H D E]+[H E M]=frac{5}{4}+frac{7}{4}=3}end{aligned}Thus, the fourth sibling received a share of 540 mathrm{~m}^{2}.Figure 5Figure 6

🧠:[<think>Okay, so I have this probability problem here. Let me read it again and make sure I understand what's being asked. The problem says: When the voltage increases, a break in the electrical circuit may occur due to the failure of one of three elements connected in series. The probabilities of failure of the elements are 0.2, 0.3, and 0.4, respectively. We need to determine the probability that the circuit will not break.Hmm, right. So, first off, elements connected in series. From what I remember, in electrical circuits, components connected in series mean that if any one of them fails, the entire circuit breaks. So, the circuit breaks if at least one of the elements fails. Therefore, the circuit will not break only if none of the elements fail. Is that right?Let me verify. Suppose we have three elements A, B, C connected in series. If any one of them fails, the circuit is broken. So, the circuit works only if all three elements are working. So, the probability that the circuit does not break is the probability that all three elements do not fail. That makes sense.Therefore, the problem reduces to finding the probability that all three elements work, i.e., none of them fail. To find this, we need to compute the probability that element 1 does not fail AND element 2 does not fail AND element 3 does not fail. Since they are in series, their failures might be independent events? Wait, the problem doesn't specify whether the failures are independent or dependent. Hmm, this is a crucial point. In probability problems, unless stated otherwise, we often assume independence. So, perhaps I can assume that the failures of the three elements are independent events. If that's the case, then the probability that all three do not fail is the product of their individual probabilities of not failing.Let me check the given probabilities. The probabilities of failure are 0.2, 0.3, and 0.4. Therefore, the probabilities of not failing would be 1 - 0.2 = 0.8 for the first element, 1 - 0.3 = 0.7 for the second, and 1 - 0.4 = 0.6 for the third. So, if they are independent, multiplying these together: 0.8 * 0.7 * 0.6.Calculating that: 0.8 times 0.7 is 0.56. Then, 0.56 times 0.6. Let me compute that. 0.56 * 0.6. Hmm, 0.5 * 0.6 is 0.3, and 0.06 * 0.6 is 0.036. So, adding those together, 0.3 + 0.036 = 0.336. So, 0.336 is the probability that none of the elements fail, meaning the circuit doesn't break. So, 0.336 is the answer?Wait, but let me just make sure that I didn't make any miscalculations. Let's do the multiplication step by step:First element: 0.8 (probability of not failing)Second element: 0.7Third element: 0.6Multiply 0.8 and 0.7: 0.8 * 0.7 = 0.56Then multiply 0.56 by 0.6: 0.56 * 0.6Breaking it down: 0.5 * 0.6 = 0.3, 0.06 * 0.6 = 0.036. So, 0.3 + 0.036 = 0.336. Yes, that's correct. So, 0.336 is the probability.But hold on, is there another way to approach this problem? Maybe by considering the probability that at least one element fails and subtracting that from 1? Let's see. The probability that the circuit breaks is the probability that at least one element fails. Then, 1 minus that probability would give the probability that the circuit does not break. However, calculating P(at least one failure) is more complicated because you have to account for overlaps. That is, P(A ∨ B ∨ C) = P(A) + P(B) + P(C) - P(A ∧ B) - P(A ∧ C) - P(B ∧ C) + P(A ∧ B ∧ C). But if the failures are independent, then P(A ∧ B) = P(A) * P(B), and similarly for the others. So, maybe we can calculate it both ways to check.Let's try that. Let me define the events:Let F1 be the event that element 1 fails (probability 0.2)F2: element 2 fails (0.3)F3: element 3 fails (0.4)Then, the probability that the circuit breaks is P(F1 ∨ F2 ∨ F3). Which is equal to P(F1) + P(F2) + P(F3) - P(F1 ∧ F2) - P(F1 ∧ F3) - P(F2 ∧ F3) + P(F1 ∧ F2 ∧ F3).Assuming independence, since we can multiply the probabilities for the intersections.So, let's compute each term:P(F1) = 0.2P(F2) = 0.3P(F3) = 0.4P(F1 ∧ F2) = 0.2 * 0.3 = 0.06P(F1 ∧ F3) = 0.2 * 0.4 = 0.08P(F2 ∧ F3) = 0.3 * 0.4 = 0.12P(F1 ∧ F2 ∧ F3) = 0.2 * 0.3 * 0.4 = 0.024Now, plug these into the formula:P(F1 ∨ F2 ∨ F3) = 0.2 + 0.3 + 0.4 - 0.06 - 0.08 - 0.12 + 0.024Let's compute step by step:0.2 + 0.3 = 0.50.5 + 0.4 = 0.9Now subtract 0.06: 0.9 - 0.06 = 0.84Subtract 0.08: 0.84 - 0.08 = 0.76Subtract 0.12: 0.76 - 0.12 = 0.64Add 0.024: 0.64 + 0.024 = 0.664Therefore, the probability that the circuit breaks is 0.664. Therefore, the probability that the circuit does not break is 1 - 0.664 = 0.336. Which matches the previous result. Good, so that's a check.So, both methods lead to the same answer, 0.336. Therefore, the probability that the circuit will not break is 0.336.But just to be thorough, let's think again about the assumption of independence. The problem states that the elements are connected in series, but does that necessarily imply independence? In real life, maybe the failure of one element could affect the others, especially if the voltage increase is the common cause. For example, if the voltage increases, maybe the stress on the remaining elements increases when one fails, but in a series circuit, if one element fails, the circuit is broken and current stops, so perhaps the other elements aren't stressed anymore. So, maybe in this case, the failures are independent because the elements are in series, so the failure of one doesn't cause the others to fail because the circuit is already broken. Therefore, the initial assumption of independence is valid here.Alternatively, if the elements were in parallel, the situation would be different. But here, series connection. So, each element's failure is an independent event. Therefore, multiplying the probabilities is correct.Therefore, I can be confident that the answer is 0.336. To present it as a fraction or decimal? The problem doesn't specify, but since the given probabilities are decimals, likely expecting a decimal answer. Alternatively, 0.336 can be written as 336/1000, which simplifies to 42/125. Let me check:336 ÷ 8 = 421000 ÷ 8 = 125So, 42/125. Let me verify that 42 divided by 125 is 0.336.125 × 0.336: 125 × 0.3 = 37.5, 125 × 0.036 = 4.5. 37.5 + 4.5 = 42. So yes, 42/125 is 0.336. So, 42/125 is the fractional form. The problem doesn't specify, but both are correct. However, in the example number 6.8, maybe they expect a decimal. Let me check if 0.336 is the exact value. Since each of the individual probabilities is exact (0.2, 0.3, 0.4), which are exact decimals, then 0.8 * 0.7 * 0.6 = 0.336 is exact. So, 0.336 is exact here.Alternatively, if we were to write it as a percentage, it would be 33.6%, but the question just asks for the probability, so 0.336 is acceptable. So, the answer is 0.336. Therefore, the probability that the circuit will not break is 0.336.Final AnswerThe probability that the circuit will not break is boxed{0.336}.</think>To determine the probability that the circuit will not break when the voltage increases, we need to consider the probabilities of each element not failing. The elements are connected in series, meaning the circuit will not break only if all three elements do not fail. Given the probabilities of failure for each element are 0.2, 0.3, and 0.4, the probabilities of each element not failing are:- Probability element 1 does not fail: (1 - 0.2 = 0.8)- Probability element 2 does not fail: (1 - 0.3 = 0.7)- Probability element 3 does not fail: (1 - 0.4 = 0.6)Assuming the failures of the elements are independent events, the probability that all three elements do not fail is the product of their individual probabilities of not failing:[0.8 times 0.7 times 0.6]Calculating this step-by-step:[0.8 times 0.7 = 0.56][0.56 times 0.6 = 0.336]Thus, the probability that the circuit will not break is (boxed{0.336}).,<think>Okay, so I need to figure out the probability that the electrical circuit doesn't break when the voltage increases. The problem says there are three elements connected in series, and each can fail with probabilities 0.2, 0.3, and 0.4 respectively. If any one of them fails, the circuit breaks. So, the question is asking for the probability that none of them fail, right?First, I should recall how probabilities work for independent events in series. Since the elements are connected in series, the circuit won't break only if all three elements don't fail. That means I need to find the probability that all three elements work properly.For each element, the probability that it doesn't fail should be 1 minus the probability that it fails. So for the first element with a failure probability of 0.2, the probability of not failing is 1 - 0.2 = 0.8. Similarly, the second element has a 0.3 chance of failing, so it has a 0.7 chance of working. The third element with a 0.4 failure probability has a 0.6 chance of not failing.Since the elements are independent, the combined probability that all three don't fail is the product of their individual probabilities. That would be 0.8 * 0.7 * 0.6. Let me compute that step by step. First, 0.8 multiplied by 0.7. 0.8 times 0.7 is 0.56. Then, 0.56 multiplied by 0.6. Let's do that: 0.56 * 0.6. Breaking it down, 0.5 * 0.6 is 0.3, and 0.06 * 0.6 is 0.036. Adding those together gives 0.3 + 0.036 = 0.336. So, 0.56 * 0.6 = 0.336.Therefore, the probability that all three elements work and the circuit doesn't break is 0.336. Let me double-check my calculations to make sure I didn't make an arithmetic error. First element: 1 - 0.2 = 0.8 (correct).Second element: 1 - 0.3 = 0.7 (correct).Third element: 1 - 0.4 = 0.6 (correct).Multiplying them: 0.8 * 0.7 = 0.56 (correct). Then 0.56 * 0.6. Let me calculate this again. 0.5 * 0.6 = 0.3, 0.06 * 0.6 = 0.036. Sum is 0.336 (correct). So yes, 0.336 seems right.Alternatively, converting everything to fractions might help avoid decimal errors. Let's see:0.8 is 4/5, 0.7 is 7/10, and 0.6 is 3/5.Multiplying these fractions: (4/5) * (7/10) * (3/5) = (4*7*3)/(5*10*5) = (84)/(250). Simplify that, divide numerator and denominator by 2: 42/125. Let me compute 42 divided by 125. 125 goes into 42 zero times. Add a decimal point, 125 goes into 420 three times (3*125=375), subtract 375 from 420 gives 45. Bring down a zero to make 450. 125 goes into 450 three times again (3*125=375), subtract to get 75. Bring down a zero to make 750. 125 goes into 750 exactly 6 times (6*125=750). So 42/125 is 0.336. That matches the decimal calculation. So both methods give the same result. That gives me confidence that the answer is correct.Alternatively, another way to think about it is to calculate the probability that at least one element fails and subtract that from 1. But that might be more complicated because the events are not mutually exclusive. For three elements, calculating the probability that at least one fails would involve inclusion-exclusion principle: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A∩B) - P(A∩C) - P(B∩C) + P(A∩B∩C). Then subtract that from 1. Let me try that method to verify.First, P(A) = 0.2, P(B) = 0.3, P(C) = 0.4. Assuming independence, the probability that both A and B fail is P(A∩B) = 0.2 * 0.3 = 0.06. Similarly, P(A∩C) = 0.2 * 0.4 = 0.08, and P(B∩C) = 0.3 * 0.4 = 0.12. The probability all three fail is P(A∩B∩C) = 0.2 * 0.3 * 0.4 = 0.024.So applying inclusion-exclusion:P(A ∪ B ∪ C) = 0.2 + 0.3 + 0.4 - 0.06 - 0.08 - 0.12 + 0.024Calculating step by step:0.2 + 0.3 = 0.50.5 + 0.4 = 0.9Now subtract:0.9 - 0.06 = 0.840.84 - 0.08 = 0.760.76 - 0.12 = 0.64Then add 0.024:0.64 + 0.024 = 0.664So the probability that at least one element fails is 0.664. Therefore, the probability that the circuit does not break is 1 - 0.664 = 0.336. Which matches the previous result. So both methods confirm that the answer is 0.336. Therefore, the probability that the circuit will not break is 0.336. To represent that as a fraction, as we saw earlier, it's 42/125. But the question doesn't specify the format, and since the original probabilities were given as decimals, it's probably appropriate to present the answer as a decimal. 0.336 is the decimal form. Just to ensure that all steps are logical: since the circuit is in series, any single failure breaks the circuit. Therefore, the circuit remains intact only if all elements do not fail. Since the elements are independent, their non-failure probabilities multiply. Correct. So the method is sound. Alternatively, if the elements were in parallel, the approach would be different, but the problem specifies they are connected in series. So series implies that all must work for the circuit to work. Therefore, the calculation is correct.So after verifying through two different methods and checking the logic, I believe the answer is 0.336.Final AnswerThe probability that the circuit will not break is boxed{0.336}.</think>To determine the probability that the circuit will not break, we need to find the probability that all three elements do not fail. The probabilities of each element not failing are calculated as follows:- The first element has a failure probability of 0.2, so the probability of not failing is (1 - 0.2 = 0.8).- The second element has a failure probability of 0.3, so the probability of not failing is (1 - 0.3 = 0.7).- The third element has a failure probability of 0.4, so the probability of not failing is (1 - 0.4 = 0.6).Since the elements are independent, the combined probability that all three do not fail is the product of their individual probabilities:[0.8 times 0.7 times 0.6]Calculating step by step:1. Multiply the probabilities of the first two elements: [ 0.8 times 0.7 = 0.56 ]2. Multiply the result by the probability of the third element: [ 0.56 times 0.6 = 0.336 ]Thus, the probability that the circuit will not break is (boxed{0.336}).]

🔑:Solution. Let events A_{1}, A_{2}, A_{3} denote the failure of the first, second, and third elements, respectively. Their probabilities are given as: Pleft(A_{1}right)=0.2; Pleft(A_{2}right)=0.3; Pleft(A_{3}right)=0.4. Then the probabilities of the opposite events bar{A}_{1}, bar{A}_{2}, bar{A}_{3} (the first, second, and third elements did not fail) are:Pleft(overline{A_{1}}right)=1-Pleft(A_{1}right)=0.8; Pleft(A_{2}right)=0.7; Pleft(A_{3}right)=0.6.The event A, which consists of the circuit not being broken, is the conjunction of independent events bar{A}_{1}, bar{A}_{2}, bar{A}_{3}: A=bar{A}_{1} cdot bar{A}_{2} cdot bar{A}_{3}. Therefore, by formula (5) we get:P(A)=Pleft(bar{A}_{1}right) cdot Pleft(bar{A}_{2}right) cdot Pleft(bar{A}_{3}right)=0.8 cdot 0.7 cdot 0.6=0.336

🧠:[<think>Okay, so I have this geometry problem here. Let me read it again to make sure I understand. The problem says: Circle C is externally tangent to the circle defined by the equation x² + y² - 2x = 0 and is tangent to the line x + √3 y = 0 at the point Q(3, -√3). We need to find the equation of circle C.Alright, let's start by recalling some concepts. The equation of a circle in standard form is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. The given circle is x² + y² - 2x = 0. I should rewrite that in standard form to find its center and radius. To do that, I can complete the square for the x terms.Starting with x² - 2x + y² = 0. For the x terms: x² - 2x, take half of the coefficient of x, which is -1, square it to get 1, so add and subtract 1. That gives (x² - 2x + 1) - 1 + y² = 0 → (x - 1)² + y² = 1. So the given circle has center at (1, 0) and radius 1.Okay, so circle C is externally tangent to this circle. External tangency means the distance between their centers is equal to the sum of their radii. Let me note that. Let's call the center of circle C as (h, k) and its radius as r. Then the distance between (h, k) and (1, 0) should be equal to r + 1 (since the given circle has radius 1). So, √[(h - 1)² + (k - 0)²] = r + 1. That's one equation.Next, circle C is tangent to the line x + √3 y = 0 at the point Q(3, -√3). Since Q is the point of tangency, two things must be true: first, Q lies on circle C, so substituting x=3 and y=-√3 into the equation of circle C should satisfy it. Second, the line x + √3 y = 0 is tangent to the circle at Q, which means the radius of circle C at point Q is perpendicular to the tangent line. Therefore, the center of circle C (h, k) lies along the line perpendicular to x + √3 y = 0 at point Q.Let me elaborate on that. The tangent line at Q is x + √3 y = 0. The slope of this line is -1/√3. Therefore, the slope of the perpendicular line is the negative reciprocal, which is √3. So the line perpendicular to the tangent at Q has slope √3 and passes through Q(3, -√3). Therefore, the center (h, k) of circle C must lie on this perpendicular line.So, let's write the equation of that perpendicular line. Using point-slope form: y - (-√3) = √3(x - 3) → y + √3 = √3(x - 3) → y = √3 x - 3√3 - √3 → y = √3 x - 4√3. Wait, let me check the arithmetic. Starting from y + √3 = √3(x - 3). So y = √3(x - 3) - √3. Distribute the √3: y = √3 x - 3√3 - √3 = √3 x - 4√3. Yes, that's correct. So the center (h, k) lies on the line y = √3 x - 4√3.So we can parameterize the center as (h, √3 h - 4√3). So h is the x-coordinate, and the y-coordinate is expressed in terms of h.Now, since the point Q(3, -√3) is on the circle C, substituting into the circle's equation gives:(3 - h)² + (-√3 - k)² = r². But since k = √3 h - 4√3, substitute that into the equation:(3 - h)² + (-√3 - (√3 h - 4√3))² = r².Let's simplify this expression step by step.First, compute the second term: -√3 - (√3 h - 4√3) = -√3 - √3 h + 4√3 = (-√3 + 4√3) - √3 h = 3√3 - √3 h = √3(3 - h). So the second term squared becomes [√3(3 - h)]² = 3(3 - h)².So substituting back into the equation:(3 - h)² + 3(3 - h)² = r² → 4(3 - h)² = r².So that's another equation: r² = 4(3 - h)². Therefore, r = 2|3 - h|. Since radius is positive, r = 2|3 - h|.Now, we have another equation from the external tangency condition. The distance between the centers is equal to the sum of the radii. The centers are (h, k) and (1, 0). So:√[(h - 1)² + (k - 0)²] = r + 1.But since k = √3 h - 4√3, substitute that in:√[(h - 1)² + (√3 h - 4√3)^2] = r + 1.We already have an expression for r in terms of h: r = 2|3 - h|. So substitute that in:√[(h - 1)^2 + (√3 h - 4√3)^2] = 2|3 - h| + 1.Now, we need to solve this equation for h. Let's first compute the left-hand side (LHS) and the right-hand side (RHS) expressions.Starting with LHS: √[(h - 1)^2 + (√3 h - 4√3)^2]. Let's expand the terms inside the square root.First term: (h - 1)^2 = h² - 2h + 1.Second term: (√3 h - 4√3)^2. Factor out √3: [√3(h - 4)]² = (√3)^2 (h - 4)^2 = 3(h - 4)^2 = 3(h² - 8h + 16).So adding both terms inside the square root:h² - 2h + 1 + 3h² - 24h + 48 = (h² + 3h²) + (-2h -24h) + (1 + 48) = 4h² - 26h + 49.Therefore, LHS = √(4h² - 26h + 49).RHS: 2|3 - h| + 1. Let's note that |3 - h| = |h - 3|. Depending on whether h is less than 3 or greater than or equal to 3, the absolute value will resolve differently. However, since the circle is externally tangent to the given circle, we might need to consider the position. Let me think.Given that the given circle is centered at (1,0) with radius 1, and our circle C is externally tangent to it. If Q is at (3, -√3), which is to the right and below the given circle. So maybe circle C is located near Q, so h is likely to be greater than 3, so |3 - h| = h - 3. But let's check that.Wait, if h is greater than 3, then 3 - h is negative, so |3 - h| = h - 3. If h is less than 3, it's 3 - h. But given that the tangent point Q is at (3, -√3), and the center is along the line y = √3 x - 4√3. Let's plug x=3 into that line: y = √3*3 -4√3 = 3√3 -4√3 = -√3. So the point Q(3, -√3) is actually on the line itself. Wait, that makes sense, because the center lies along the line perpendicular to the tangent at Q, so the line from Q to the center is the perpendicular line. So if we follow the line y = √3 x - 4√3, starting at Q(3, -√3), moving along that line would take us towards the center. So depending on the direction, the center can be either on one side or the other. Wait, but the line perpendicular to the tangent at Q is the radial line, so the center must lie along that line. Since Q is the point of tangency, the center is along that line. Therefore, the center (h, k) is somewhere along y = √3 x - 4√3, either moving from Q in the direction of the normal vector. But given that the circle is tangent to the line at Q, the center must be located at a distance r along the normal line from Q. Wait, perhaps another approach. The distance from the center (h, k) to the line x + √3 y = 0 is equal to the radius r. But since Q is the point of tangency, the distance from (h, k) to the line is equal to the radius, and Q is the closest point on the line to the center. Therefore, the distance from (h, k) to the line is equal to the radius, and Q is exactly that closest point. Therefore, Q is the foot of the perpendicular from (h, k) to the line. Which is another way of saying that (h, k) lies along the line perpendicular to x + √3 y = 0 through Q, which is what we already have. So the center is along that line y = √3 x - 4√3.So going back to the problem. Let's suppose h can be either greater than 3 or less than 3. Let's proceed without assuming. But let's see. The given circle is at (1,0). The center of circle C is along y = √3 x - 4√3. Let's see where that line is. At x=1, y=√3*1 -4√3= -3√3. So when x=1, y=-3√3. The line is going upwards with slope √3. So at x=3, y= -√3 as given. If h increases beyond 3, y increases further. For example, at h=4, y=√3*4 -4√3=0. At h=5, y=√3*5 -4√3=√3. So the line crosses the x-axis at h=4. So if the center is at (4,0), which is interesting because the original circle is at (1,0). But (4,0) is 3 units away from (1,0). If the radius of circle C is, say, 2|3 -4|=2*1=2, then the distance between centers would be 3, which should equal r +1, so 2 +1=3. That works. Wait, maybe (4,0) is a possible center? Let me check.If center is (4,0), then the radius is 2|3 -4|=2. Then check if the distance from (4,0) to (1,0) is 3, which is 4 -1 =3, and 3 = 2 +1, which satisfies the external tangency. Also, check if (3, -√3) is on the circle: (3-4)^2 + (-√3 -0)^2 =1 +3=4=2². Yes, that works. Also, check if the line x + √3 y =0 is tangent to the circle. The distance from center (4,0) to the line x +√3 y=0 is |4 +0|/√(1 +3)=4/2=2, which is equal to the radius. So that works. Therefore, (4,0) with radius 2 is a solution. But wait, the problem says "circle C is externally tangent to the circle x² + y² -2x=0". So if the center is at (4,0), radius 2, then the distance between centers is 3, which is equal to 1 +2, so that's external tangent. That seems to satisfy all conditions. But wait, is there another circle?Wait, let's check if there's another possible solution. Suppose h <3. Let's take h=2. Then k=√3*2 -4√3= -2√3. Then the radius r=2|3 -2|=2*1=2. Then the center is (2, -2√3). Check the distance to (1,0): √[(2-1)^2 + (-2√3 -0)^2] =√[1 + 12]=√13. Then the sum of radii would be 2 +1=3. But √13≈3.605, which is not equal to 3. So that doesn't satisfy the external tangency condition. So h=2 is invalid. How about h=3? Then k=√3*3 -4√3= -√3. Then the center is (3, -√3), which is the point Q. Then radius r=2|3 -3|=0. A circle with radius 0 doesn't make sense. So h cannot be 3. So if h is less than 3, then |3 - h|=3 -h, so r=2(3 -h). Then, the distance between centers is √[(h -1)^2 + (√3 h -4√3)^2] which we found to be √(4h² -26h +49). Then set equal to r +1=2(3 -h)+1=7 -2h. So we have √(4h² -26h +49)=7 -2h. Let's square both sides:4h² -26h +49 = (7 -2h)^2 =49 -28h +4h².Subtract 4h² from both sides: -26h +49 =49 -28h.Add 28h to both sides: 2h +49=49.Subtract 49: 2h=0→h=0. Then k=√3*0 -4√3= -4√3. Then radius r=2|3 -0|=6. Check distance between (0, -4√3) and (1,0): √[(0 -1)^2 + (-4√3 -0)^2]=√[1 + 48]=√49=7. Sum of radii is 6 +1=7, which matches. Then check if (3, -√3) is on the circle: (3 -0)^2 + (-√3 -(-4√3))^2=9 + (3√3)^2=9 +27=36=6². Correct. Also, check the distance from center (0, -4√3) to the line x +√3 y=0: |0 + √3*(-4√3)| /√(1 +3)= | -4*3 | /2=12/2=6, which is equal to the radius. So this also satisfies all conditions. So we have two circles: one at (4,0) with radius 2, and another at (0, -4√3) with radius 6.Wait, but the problem states "circle C is externally tangent". So both solutions are externally tangent. Let me confirm. For the first circle at (4,0) with radius 2: distance between centers is 3, sum of radii is 2+1=3, so yes, externally tangent. For the second circle at (0, -4√3) with radius 6: distance between centers is 7, sum of radii is 6 +1=7, also externally tangent. So both solutions are valid.But the problem says "the circle C". Hmm, maybe there are two circles satisfying the conditions. But let me check if both circles are tangent to the line at Q(3, -√3). Wait, both centers (4,0) and (0, -4√3) lie on the line y=√3 x -4√3. Let's check for (4,0): y=√3*4 -4√3=4√3 -4√3=0. Correct. For (0, -4√3): y=√3*0 -4√3=-4√3. Correct. So both centers are on the required line.But the problem states "is tangent to the line x +√3 y=0 at point Q(3, -√3)". So both circles pass through Q and are tangent to the line there. Let me confirm. For the first circle: (4,0) with radius 2. The point Q is (3, -√3). The distance from (4,0) to Q is √[(3-4)^2 + (-√3 -0)^2] = √[1 +3]=2, which is equal to the radius. For the second circle: (0, -4√3) with radius 6. Distance from center to Q is √[(3-0)^2 + (-√3 -(-4√3))^2] = √[9 + (3√3)^2] = √[9 +27]=√36=6, which is the radius. So both circles pass through Q, and as verified earlier, the distance from the centers to the line is equal to their radii, so they are tangent at Q. Therefore, both circles satisfy the conditions.But the problem says "find the equation of circle C". It might be expecting both answers, but perhaps only one is valid based on the diagram. Wait, let me visualize.The given circle is at (1,0) with radius 1. The line x +√3 y=0 has a negative slope and passes through the origin. The point Q(3, -√3) is on this line. Let me calculate coordinates:For the line x +√3 y=0: when x=3, y= -√3, which matches Q. So Q is in the fourth quadrant. The given circle is centered at (1,0), radius 1, so it touches the origin? Wait, no: (1,0) with radius 1, so it goes from x=0 to x=2 on the x-axis, and y from -1 to 1. So it's a circle sitting on the x-axis, centered at (1,0), touching the origin.The line x +√3 y=0 passes through the origin and goes through Q(3, -√3). So it's a line going from the origin down into the fourth quadrant. The point Q is far out in the fourth quadrant. The two possible circles are:1. Center (4,0), radius 2: this circle is to the right of the given circle, sitting on the x-axis, touching the given circle externally. It also touches the line at Q(3, -√3). Since the center is on the x-axis, the circle extends from x=2 to x=6, and y from -2 to 2. The point Q(3, -√3) is on its edge.2. Center (0, -4√3), radius 6: this circle is far below the origin, centered at (0, -4√3). It would enclose the given circle, but since it's externally tangent, we need to check. Wait, no: external tangency means they touch at one point, but if the other circle is much larger, perhaps they can still be externally tangent. Wait, the distance between centers is 7, and the sum of radii is 6 +1=7, so they just touch externally. The circle at (0, -4√3) with radius 6 would extend from x=-6 to x=6, y from -4√3 -6 to -4√3 +6. The point Q(3, -√3) is on its edge. However, the line x +√3 y=0 is tangent to this circle at Q. So both circles satisfy the conditions.But the problem might require both answers, but in the original problem statement, it's stated as "find the equation of circle C". If it's expecting both, we should present both. However, sometimes such problems might have two solutions, but depending on the relative positions. Wait, let me check if both circles are externally tangent. The first one at (4,0) is clearly externally tangent. The second one at (0, -4√3) is 7 units away from (1,0), which equals 6 +1, so that's also external. So both are valid.Therefore, the answer should include both circles. Let me write their equations.First circle: center (4,0), radius 2. Equation: (x -4)^2 + y^2 =4.Second circle: center (0, -4√3), radius 6. Equation: x² + (y +4√3)^2 =36.Therefore, these are the two possible equations.But let me check if the problem specifies any particular orientation or if there's a unique solution. Wait, the problem says "circle C is externally tangent to the circle x² + y² -2x=0". Both circles are externally tangent. It doesn't specify anything else, so both are valid. Hence, the answer should include both equations.But let me check once more for calculation errors.For the first circle (h=4):- Center (4,0). Radius 2.- Distance to given circle's center (1,0): 3, which is 2 +1.- Distance from center to the line: |4 +0| /2=2, which is the radius.- Passes through Q(3, -√3): (3-4)^2 + (-√3)^2=1 +3=4. Correct.For the second circle (h=0):- Center (0, -4√3). Radius 6.- Distance to given circle's center: √[(0 -1)^2 + (-4√3 -0)^2]=√[1 + 48]=7=6 +1.- Distance from center to the line: |0 +√3*(-4√3)| /2= |-12| /2=6. Correct.- Passes through Q(3, -√3): (3-0)^2 + (-√3 +4√3)^2=9 + (3√3)^2=9 +27=36. Correct.Yes, both are correct. So the problem has two solutions.But let me check again why when solving the equation √(4h² -26h +49)=2|3 -h| +1, we got two solutions. When we considered h >3, |3 -h|=h -3, so the equation becomes √(4h² -26h +49)=2(h -3) +1=2h -6 +1=2h -5. Then square both sides:4h² -26h +49 = (2h -5)^2=4h² -20h +25.Subtract 4h²: -26h +49 = -20h +25.Move terms: -26h +49 +20h -25=0 → -6h +24=0 → -6h= -24 → h=4. So h=4.Then when h <3, |3 -h|=3 -h, so equation becomes √(4h² -26h +49)=2(3 -h) +1=6 -2h +1=7 -2h. Then square both sides:4h² -26h +49 = (7 -2h)^2=49 -28h +4h².Subtract 4h²: -26h +49 =49 -28h.Add 28h: 2h +49=49 → 2h=0 →h=0. Then k= -4√3. So that's where the two solutions came from. Therefore, both are valid.Thus, the answer should have both equations. But let me check the problem statement again. It says "circle C is externally tangent..." but maybe in the original problem, there's a diagram or some additional constraint? Since the user hasn't provided a diagram, we have to assume both are possible. Therefore, the correct answer is both circles. So I need to present both equations.Final AnswerThe equations of circle C are boxed{(x - 4)^2 + y^2 = 4} and boxed{x^2 + left(y + 4sqrt{3}right)^2 = 36}.</think>Given that circle ( C ) is externally tangent to the circle ( x^2 + y^2 - 2x = 0 ) and is tangent to the line ( x + sqrt{3}y = 0 ) at point ( Q(3, -sqrt{3}) ), we need to find the equation of circle ( C ).First, we rewrite the given circle's equation in standard form:[ x^2 + y^2 - 2x = 0 ]Completing the square for ( x ):[ (x - 1)^2 + y^2 = 1 ]This circle has center ((1, 0)) and radius (1).Circle ( C ) with center ((h, k)) and radius ( r ) must satisfy the following conditions:1. The distance between the centers ((h, k)) and ((1, 0)) is ( r + 1 ).2. The point ( Q(3, -sqrt{3}) ) lies on circle ( C ).3. The line ( x + sqrt{3}y = 0 ) is tangent to circle ( C ) at ( Q ), so the center ((h, k)) lies on the line perpendicular to ( x + sqrt{3}y = 0 ) at ( Q ).The slope of the line ( x + sqrt{3}y = 0 ) is ( -frac{1}{sqrt{3}} ). The slope of the perpendicular line is ( sqrt{3} ). Using point-slope form, the equation of the perpendicular line through ( Q ) is:[ y + sqrt{3} = sqrt{3}(x - 3) ][ y = sqrt{3}x - 4sqrt{3} ]Thus, the center ((h, k)) of circle ( C ) lies on this line, so ( k = sqrt{3}h - 4sqrt{3} ).Substituting ( Q(3, -sqrt{3}) ) into the circle's equation:[ (3 - h)^2 + (-sqrt{3} - k)^2 = r^2 ]Substituting ( k = sqrt{3}h - 4sqrt{3} ):[ (3 - h)^2 + (sqrt{3}(3 - h))^2 = r^2 ][ (3 - h)^2 + 3(3 - h)^2 = r^2 ][ 4(3 - h)^2 = r^2 ][ r = 2|3 - h| ]The distance between centers ((h, k)) and ((1, 0)) is:[ sqrt{(h - 1)^2 + (sqrt{3}h - 4sqrt{3})^2} = r + 1 ]Substituting ( r = 2|3 - h| ):[ sqrt{4h^2 - 26h + 49} = 2|3 - h| + 1 ]Solving this equation, we consider two cases:1. ( h geq 3 ): ( |3 - h| = h - 3 )[ sqrt{4h^2 - 26h + 49} = 2h - 5 ]Solving gives ( h = 4 ), leading to center ((4, 0)) and radius (2).2. ( h < 3 ): ( |3 - h| = 3 - h )[ sqrt{4h^2 - 26h + 49} = 7 - 2h ]Solving gives ( h = 0 ), leading to center ((0, -4sqrt{3})) and radius (6).Both circles satisfy all conditions. Their equations are:[ boxed{(x - 4)^2 + y^2 = 4} ][ boxed{x^2 + left(y + 4sqrt{3}right)^2 = 36} ],<think>Okay, so I need to find the equation of circle C. Let me start by understanding the problem. Circle C is externally tangent to another circle defined by the equation x² + y² - 2x = 0, and it's also tangent to the line x + √3 y = 0 at the point Q(3, -√3). Hmm, okay. Let me break this down step by step.First, let me recall the general equation of a circle: (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. So, I need to find h, k, and r for circle C.The given circle is x² + y² - 2x = 0. Let me rewrite this in standard form by completing the square. Group the x terms: x² - 2x + y² = 0. To complete the square for the x terms, take half of -2, which is -1, square it to get 1, and add it to both sides: (x² - 2x + 1) + y² = 1. So, this becomes (x - 1)² + y² = 1. Therefore, the center of the given circle is (1, 0) and its radius is 1.Okay, so circle C is externally tangent to this circle. External tangency means that the distance between their centers is equal to the sum of their radii. Let's denote the center of circle C as (h, k) and its radius as r. Then, the distance between (h, k) and (1, 0) should be equal to r + 1 (since the radius of the given circle is 1). So, the first equation is:√[(h - 1)² + (k - 0)²] = r + 1.Squaring both sides to eliminate the square root:(h - 1)² + k² = (r + 1)². ... (1)Second condition: circle C is tangent to the line x + √3 y = 0 at point Q(3, -√3). Since the circle is tangent to the line at point Q, this point lies on both the circle and the line. Also, the tangent at Q implies that the line from the center of the circle to Q is perpendicular to the given line.First, let's verify that Q is on the line x + √3 y = 0. Plugging in x = 3 and y = -√3:3 + √3*(-√3) = 3 - 3 = 0. Yes, that's correct. So, point Q is indeed on the line.Since Q is the point of tangency, the center (h, k) of circle C must lie along the line perpendicular to x + √3 y = 0 at point Q. Let me find the slope of the given line. The line is x + √3 y = 0. Let's rearrange it to slope-intercept form:√3 y = -x → y = (-1/√3)x.So, the slope of the line is -1/√3. The slope of the perpendicular line is the negative reciprocal, which would be √3. Therefore, the center (h, k) lies along the line passing through Q(3, -√3) with slope √3.Let me write the equation of this perpendicular line. Using point-slope form:y - (-√3) = √3(x - 3)Simplify:y + √3 = √3(x - 3)→ y = √3 x - 3√3 - √3→ y = √3 x - 4√3.Therefore, the center (h, k) of circle C must lie on the line y = √3 x - 4√3. So, we can express k in terms of h: k = √3 h - 4√3. ... (2)Also, since point Q(3, -√3) lies on circle C, plugging into the circle's equation:(3 - h)² + (-√3 - k)² = r². ... (3)Now, we have three equations: (1), (2), and (3). Let's substitute equation (2) into equations (1) and (3) to reduce variables.First, substitute k = √3 h - 4√3 into equation (1):(h - 1)² + (√3 h - 4√3)^2 = (r + 1)^2.Let me compute each term.First term: (h - 1)^2 = h² - 2h + 1.Second term: (√3 h - 4√3)^2. Factor out √3: [√3 (h - 4)]² = (√3)^2 (h - 4)^2 = 3(h² - 8h + 16).So, the left-hand side of equation (1) becomes:h² - 2h + 1 + 3(h² - 8h + 16) = h² -2h +1 +3h² -24h +48 = 4h² -26h +49.So, equation (1) becomes:4h² -26h +49 = (r +1)^2. ... (1a)Now, substitute k = √3 h -4√3 into equation (3):(3 - h)^2 + (-√3 - (√3 h -4√3))^2 = r².Let me simplify the terms inside.First term: (3 - h)^2 = ( -h + 3)^2 = h² -6h +9.Second term: (-√3 - √3 h +4√3) = ( (-√3 +4√3) - √3 h ) = (3√3 - √3 h ) = √3 (3 - h).So, the second term is [√3 (3 - h)]^2 = (3)(3 - h)^2.Therefore, equation (3) becomes:h² -6h +9 + 3(3 - h)^2 = r².Compute 3(3 - h)^2: 3*(9 -6h + h²) =27 -18h +3h².So, equation (3) becomes:h² -6h +9 +27 -18h +3h² = r²Combine like terms:(1h² +3h²) + (-6h -18h) + (9 +27) = r²→ 4h² -24h +36 = r². ... (3a)Now, we have equations (1a) and (3a):From (1a): 4h² -26h +49 = (r +1)^2.From (3a): 4h² -24h +36 = r².Let me write (r +1)^2 as r² + 2r +1. So equation (1a) becomes:4h² -26h +49 = r² + 2r +1.But from equation (3a), we have r² =4h² -24h +36. Substitute that into the above equation:4h² -26h +49 = (4h² -24h +36) + 2r +1.Simplify the right-hand side:4h² -24h +36 +2r +1 =4h² -24h +37 +2r.Now, set equal to left-hand side:4h² -26h +49 =4h² -24h +37 +2r.Subtract 4h² from both sides:-26h +49 = -24h +37 +2r.Bring all terms to left-hand side:-26h +49 +24h -37 -2r =0.Simplify:(-26h +24h) + (49 -37) -2r =0→ -2h +12 -2r=0Divide both sides by -2:h -6 +r=0→ h + r =6.So, equation (4): h + r =6. Therefore, r =6 - h.Now, substitute r =6 - h into equation (3a): 4h² -24h +36 = (6 - h)^2.Compute (6 - h)^2 =36 -12h +h².Therefore, equation (3a) becomes:4h² -24h +36 =36 -12h +h².Subtract right-hand side from both sides:4h² -24h +36 -36 +12h -h² =0.Simplify:3h² -12h =0.Factor:3h(h -4)=0.Therefore, h=0 or h=4.So, two possible solutions for h: 0 and 4. Let's check each case.Case 1: h=0.From equation (4): r=6 -0=6.From equation (2): k=√3*0 -4√3= -4√3.Therefore, center is (0, -4√3) and radius 6. Let me check if this circle satisfies the external tangency condition with the given circle (1,0) radius 1.Compute the distance between centers: sqrt[(0 -1)^2 + (-4√3 -0)^2] = sqrt[1 + 16*3] = sqrt[1 +48] = sqrt[49]=7.Sum of radii: 6 +1=7. So, 7=7, which satisfies the external tangency condition.Also, check that the circle passes through Q(3, -√3):(3 -0)^2 + (-√3 - (-4√3))^2 =9 + (3√3)^2=9 +27=36, which is 6². Correct.Case 2: h=4.From equation (4): r=6 -4=2.From equation (2): k=√3*4 -4√3=4√3 -4√3=0.So, center is (4,0) and radius 2. Check the distance between (4,0) and (1,0): sqrt[(4-1)^2 +0]=sqrt[9]=3. Sum of radii:2 +1=3. So, that's good for external tangency.Check if Q(3, -√3) lies on this circle:(3 -4)^2 + (-√3 -0)^2=(-1)^2 + ( -√3)^2=1 +3=4=2². Correct.Wait, so both circles satisfy the conditions? But the problem says "circle C is externally tangent", so both are possible? But maybe there's another condition. Wait, the tangent line: is there a case where the circle is on the other side of the line? Let me check.First, for center (0, -4√3), radius 6. The tangent line at Q(3, -√3). The line from center (0, -4√3) to Q(3, -√3) should be perpendicular to the given line x + √3 y =0.Compute the slope of the line connecting (0, -4√3) to Q(3, -√3):Slope = [ -√3 - (-4√3) ] / (3 -0) = (3√3)/3 = √3. Which is the slope of the perpendicular line as calculated before, so that's correct.For the other circle (4,0), radius 2. The line from (4,0) to Q(3, -√3) has slope [ -√3 -0 ] / (3 -4) = (-√3)/(-1)=√3. Also correct. So both centers lie along the perpendicular line at Q, as required.Therefore, both circles satisfy the given conditions. Wait, but the problem states "circle C is externally tangent to the circle x² + y² -2x=0". Since both circles are externally tangent, but perhaps depending on their positions, maybe both are valid solutions?Wait, the original circle is at (1,0) radius 1. The first solution circle is at (0, -4√3) radius 6. The other is at (4,0) radius 2.Visually, the first circle is much larger, located below the original circle, and the second is to the right. Both are externally tangent. Hmm. But the tangent line is x + √3 y =0, which passes through the origin with slope -1/√3. The point Q(3, -√3) is on this line.If we draw both circles, the first one (0, -4√3) with radius 6 would extend from x= -6 to x=6, y from -4√3 -6 to -4√3 +6. The point Q(3, -√3) is in the upper part of this circle. The tangent at Q is towards the line x + √3 y=0, which is correct.The second circle (4,0) with radius 2 is centered at (4,0), so it goes from x=2 to x=6, y from -2 to 2. The point Q(3, -√3) is on the lower part of this circle. The tangent line at Q is the same line x + √3 y=0, which also is a tangent here.But the problem statement says "circle C is externally tangent". Both are externally tangent. So perhaps both are valid solutions? The problem might have two solutions, but the question says "find the equation of circle C". Maybe I missed a constraint?Wait, let me check if both circles are tangent to the line at Q. The first circle: center (0, -4√3), radius 6. The distance from center to the line x + √3 y =0 should be equal to radius. Let me verify.The formula for distance from point (h,k) to line ax + by + c =0 is |ah + bk + c| / sqrt(a² + b²). The given line is x + √3 y =0, so a=1, b=√3, c=0.Distance from (0, -4√3) to the line:|1*0 + √3*(-4√3) +0| / sqrt(1 + 3) = |0 -4*3| /2 = | -12 | /2 =6. Which is equal to the radius. So that's correct.For the second circle (4,0), distance to the line:|1*4 + √3*0 +0| /2 = |4|/2=2, which is equal to radius 2. Correct.So both circles are valid. But the problem states "circle C is externally tangent to the circle x² + y² -2x=0 and is tangent to the line x + √3 y=0 at point Q(3, -√3)". It doesn't specify which one. So maybe both are acceptable. But the problem asks for "the equation of circle C". Hmm. Maybe there's a unique solution? Let me check again.Wait, perhaps when they say "externally tangent", in some contexts, it might mean that the circles are not overlapping and not one inside the other. But both solutions here are externally tangent. However, maybe the position relative to the tangent line?Looking at the original circle (1,0) radius 1. The line x + √3 y =0 passes through the origin, going downwards to the right. The point Q(3, -√3) is on the fourth quadrant. The first circle is centered at (0, -4√3), which is far below the origin, so the tangent point Q is above the center. The second circle is centered at (4,0), so the tangent point Q is to the left and below the center. Depending on the orientation, maybe both are possible.But the problem didn't specify any other conditions, so maybe both are valid. However, the original problem might expect only one answer. Let me check the problem again: "Given that circle C is externally tangent to the circle x²+y²−2x=0 and is tangent to the line x+√3y=0 at point Q(3, −√3), find the equation of circle C."Wait, perhaps there is a miscalculation here. Let me double-check the equations.Starting again:After substituting equations (1a) and (3a):We found h=0 or h=4, leading to two circles.But maybe one of them is actually internally tangent? Wait, external tangent requires the distance between centers equals sum of radii. For h=0, center (0, -4√3) radius 6. Distance to (1,0) is sqrt(1 + (4√3)^2)=sqrt(1 + 48)=sqrt(49)=7, which is 6 +1=7. Correct.For h=4, center (4,0) radius 2. Distance to (1,0) is 3, which is 2 +1=3. Correct. So both are externally tangent.Therefore, both solutions are valid. But the problem says "find the equation of circle C", implying possibly only one. But perhaps the answer requires both. Let me check if the problem was maybe translated from Chinese and "externally tangent" is explicitly only one. Wait, the original problem is in English as given. Maybe the user expects both equations. Wait, but in the original problem statement, the circle is tangent to the line at Q(3, -√3). Let me visualize.First circle: (0, -4√3) with radius 6. The tangent at Q(3, -√3) would have the line x + √3 y =0 as tangent. The center is below Q, so the circle is large, encompassing Q and going below. The tangent at Q is on the upper part of the circle.Second circle: (4,0) with radius 2. The center is to the right of Q, and the circle is small, with Q on its lower left. The tangent line at Q is on the lower part of the circle.But the problem doesn't specify any other conditions. So perhaps both are acceptable. However, in some textbooks, sometimes problems like this assume that the circle is located on a particular side of the tangent line, but since Q is given, maybe both are valid.But let me check the original problem again: "externally tangent to the circle x²+y²−2x=0". The original circle is at (1,0). For the first solution (0, -4√3), the other circle is below and left, so external tangent. The second solution (4,0) is to the right, also external. So both are external.Therefore, the answer should have two circles. But since the user's instruction says "put your final answer within boxed{}", which usually indicates a single answer. Maybe I made a mistake? Wait, let's check if maybe one of the circles is actually not tangent at Q. Wait, we checked that both circles pass through Q and the distance from center to the line is equal to radius, so they are tangent at Q. So both are correct. Therefore, the problem might have two solutions. But perhaps in the original problem's context, considering the position, only one is valid. Wait, maybe the circle (4,0) is actually intersecting the original circle?Wait, the original circle is at (1,0) radius 1. The circle at (4,0) radius 2. The distance between centers is 3, which is equal to 1 +2, so they are externally tangent. So no intersection. Similarly, the other circle is 7 units away, sum of radii is 7, so also externally tangent.Therefore, both are valid. But the problem might expect both answers. However, in the initial problem statement, it's mentioned "circle C", singular. So maybe both circles satisfy the condition, and the answer should present both equations. But I need to check if the user expects multiple answers.Wait, the user instruction says: "find the equation of circle C". If there are two possible circles, the answer should include both. Let me check my work again.Wait, but when solving the equations, we had h=0 and h=4. Let me check if h=4, the center (4,0) and radius 2. Then, plugging into the original equations.First, distance to the given circle's center (1,0): 3 units, which is 2 +1. Correct.The tangent at Q(3, -√3). The distance from (4,0) to the line x + √3 y=0 is |4 +0| /2=2, which is the radius. Correct.Therefore, it's valid.Similarly, the other circle is also valid.Therefore, the answer should have two equations. However, perhaps the problem is designed so that one of them is invalid for some reason. Wait, let me check the tangent line. For the circle (4,0), tangent at Q(3, -√3). The line from center (4,0) to Q(3, -√3) has slope √3, as we saw, which is perpendicular to the tangent line's slope -1/√3. Correct.Similarly for the other circle. Therefore, both circles are valid. Hence, the problem has two solutions.But the original problem says "find the equation of circle C", so maybe both are required. However, since the user hasn't specified, and given that often problems like this expect both solutions, perhaps I should present both.Wait, but maybe I made an error in calculation when solving for h. Let me check again.From equation (4): h + r =6.From equation (3a):4h² -24h +36 = r².Substitute r=6 -h into equation (3a):4h² -24h +36 = (6 -h)^2 =36 -12h +h².Subtracting RHS from LHS:4h² -24h +36 -36 +12h -h²=3h² -12h=0.Factor:3h(h -4)=0. So h=0 or h=4. Correct.So, no mistake here. Therefore, two valid solutions.But perhaps the problem expects both equations. However, given the initial problem statement, maybe only one is intended? Let me think. The point Q is (3, -√3). Let's see, for circle (4,0), the point Q is (3, -√3), which is left and below the center. The tangent line at Q is sloping downward to the right. For the other circle, center (0, -4√3), point Q is above and to the right. Depending on the direction of the tangent line, maybe both are possible.But given that the problem gives a specific tangent point, maybe both circles are valid. Therefore, the answer should include both equations.But the user instruction says "put your final answer within boxed{}". Usually, boxed{} is used for a single answer. Maybe the problem assumes only one solution, but according to my calculations, there are two.Wait, maybe I missed something. Let me check the coordinates again.For the circle at (4,0), radius 2. The tangent point Q(3, -√3). Let me check the slope of the radius. The slope from (4,0) to (3, -√3) is (-√3 -0)/(3 -4)= (-√3)/(-1)=√3. The tangent line's slope is -1/√3, which is the negative reciprocal. Correct.For the other circle, slope from (0, -4√3) to Q(3, -√3) is [ -√3 - (-4√3) ]/(3 -0)= (3√3)/3=√3. Same as before, perpendicular slope is -1/√3. Correct.So, both are okay.Therefore, the answer should be two equations:(x - 0)^2 + (y +4√3)^2 =6^2 → x² + (y +4√3)^2 =36.and(x -4)^2 + (y -0)^2 =2^2 → (x -4)^2 + y² =4.But let me check if these are the correct forms.Yes, the first circle: center (0, -4√3), radius 6:(x -0)^2 + (y - (-4√3))^2 =6² →x² + (y +4√3)^2=36.Second circle: center (4,0), radius 2:(x -4)^2 + y²=4.But the problem statement says "circle C is externally tangent". Maybe in some contexts, only the circle that doesn't enclose the original circle is considered. Wait, the original circle has radius 1. The first solution has radius 6, which is larger, and the center is further away. The second solution has radius 2, which is larger than 1, but the center is only 3 units away. But both are externally tangent.Alternatively, maybe the problem is from an exam where only one answer is expected, and perhaps based on the position of Q(3, -√3), which is in the fourth quadrant. The original circle is in the first and fourth quadrants (center at (1,0)). The circle centered at (4,0) is in the first and fourth quadrants as well, while the other circle is centered in the third/fourth quadrant.But the tangent line x + √3 y=0 passes through the origin and goes to the fourth and second quadrants. The point Q is in the fourth quadrant. The circle centered at (4,0) is in the fourth/first, so the tangent at Q is on the side towards the original circle. The other circle is below, so the tangent is away from the original circle.But the problem doesn't specify direction. Therefore, both are correct.But since the user asked to put the final answer within a single box, maybe I need to check if there is an error in my process or if the problem indeed has two solutions. But according to the calculations, two circles satisfy all the given conditions.However, let me verify once again with different approaches.Alternative approach: Since the circle is tangent to the line at Q(3, -√3), its center lies along the line perpendicular to x + √3 y =0 at Q, which we found as y = √3 x -4√3. So, parametrize the center as (t, √3 t -4√3). The radius is the distance from the center to Q, which is sqrt[(t -3)^2 + (√3 t -4√3 +√3)^2].Compute that distance:sqrt[(t -3)^2 + (√3 t -3√3)^2].= sqrt[(t -3)^2 + [√3(t -3)]^2]= sqrt[(t -3)^2 +3(t -3)^2]= sqrt[4(t -3)^2]= 2|t -3|.Therefore, the radius r =2|t -3|.Also, the distance from center (t, √3 t -4√3) to the original circle's center (1,0) must be equal to r +1.Compute the distance squared between (t, √3 t -4√3) and (1,0):(t -1)^2 + (√3 t -4√3 -0)^2= (t -1)^2 + (√3(t -4))^2= (t² -2t +1) +3(t² -8t +16)= t² -2t +1 +3t² -24t +48=4t² -26t +49.This must equal (r +1)^2. But r=2|t -3|. Therefore:4t² -26t +49 = (2|t -3| +1)^2.Case 1: t -3 ≥0 → t ≥3. Then |t -3|=t -3, so:4t² -26t +49 = (2(t -3) +1)^2 = (2t -6 +1)^2=(2t -5)^2=4t² -20t +25.Set equal:4t² -26t +49 =4t² -20t +25.Subtract 4t² from both sides:-26t +49 = -20t +25.→ -6t = -24 → t=4. Which is ≥3. Valid.Case 2: t -3 <0 → t <3. Then |t -3|=3 -t, so:4t² -26t +49 = (2(3 -t) +1)^2 = (6 -2t +1)^2=(7 -2t)^2=49 -28t +4t².Set equal:4t² -26t +49 =4t² -28t +49.Subtract 4t² and 49:-26t =-28t →2t=0→t=0. Which is <3. Valid.Therefore, t=0 and t=4. So centers at (0, -4√3) and (4,0), radii 2|0 -3|=6 and 2|4 -3|=2. So same results.Therefore, there are indeed two solutions. Therefore, the problem has two possible circles. The user might need to present both.But the problem says "the equation of circle C". Maybe in the problem's original context, only one is intended. Perhaps the user needs to specify both. Since the problem didn't mention anything else, but given the calculations, two circles are possible.Therefore, I should present both equations as answers.Final AnswerThe equations of circle C are boxed{x^2 + (y + 4sqrt{3})^2 = 36} and boxed{(x - 4)^2 + y^2 = 4}.</think>Given that circle ( C ) is externally tangent to the circle ( x^2 + y^2 - 2x = 0 ) and is tangent to the line ( x + sqrt{3}y = 0 ) at point ( Q(3, -sqrt{3}) ), we need to find the equation of circle ( C ).1. Rewrite the given circle equation: The circle ( x^2 + y^2 - 2x = 0 ) can be rewritten in standard form by completing the square: [ (x - 1)^2 + y^2 = 1 ] This gives the center ((1, 0)) and radius (1).2. Determine the line perpendicular to the tangent line: The line ( x + sqrt{3}y = 0 ) has a slope of (-frac{1}{sqrt{3}}). The perpendicular slope is (sqrt{3}). The line perpendicular to ( x + sqrt{3}y = 0 ) passing through ( Q(3, -sqrt{3}) ) is: [ y = sqrt{3}x - 4sqrt{3} ] The center of circle ( C ) must lie on this line.3. Set up equations for the center and radius: Let the center of circle ( C ) be ((h, k)). Since it lies on the line ( y = sqrt{3}x - 4sqrt{3} ), we have ( k = sqrt{3}h - 4sqrt{3} ).4. Use the external tangency condition: The distance between the centers of the circles must be equal to the sum of their radii: [ sqrt{(h - 1)^2 + k^2} = r + 1 ] Squaring both sides: [ (h - 1)^2 + k^2 = (r + 1)^2 ]5. Use the tangency condition at point ( Q ): The point ( Q(3, -sqrt{3}) ) lies on circle ( C ): [ (3 - h)^2 + (-sqrt{3} - k)^2 = r^2 ]6. Substitute ( k = sqrt{3}h - 4sqrt{3} ) into the equations: - For the external tangency condition: [ (h - 1)^2 + (sqrt{3}h - 4sqrt{3})^2 = (r + 1)^2 ] - For the tangency condition at ( Q ): [ (3 - h)^2 + (-sqrt{3} - (sqrt{3}h - 4sqrt{3}))^2 = r^2 ]7. Solve the equations: After solving the system of equations, we find two possible solutions for ( h ): - ( h = 0 ) leading to center ((0, -4sqrt{3})) and radius (6) - ( h = 4 ) leading to center ((4, 0)) and radius (2)8. Verify both solutions: Both circles satisfy the conditions of being externally tangent to the given circle and tangent to the line at ( Q ).The equations of circle ( C ) are:[boxed{x^2 + (y + 4sqrt{3})^2 = 36} quad text{and} quad boxed{(x - 4)^2 + y^2 = 4}]]

🔑:Since circle C is externally tangent to the circle x^2+y^2-2x=0, the distance between the centers of the two circles equals the sum of their radii. Also, since circle C is tangent to the line x+sqrt{3}y=0 at point Q(3, -sqrt{3}), the line connecting the center of circle C to point Q(3, -sqrt{3}) is perpendicular to the line x+sqrt{3}y=0, which means its slope is sqrt{3}. Let the center of circle C be (a, b), then we have the system of equations:begin{cases} frac{b+sqrt{3}}{a-3}=sqrt{3} sqrt{(a-1)^2+b^2}=1+frac{|a+sqrt{3}b|}{2} end{cases}Solving this system, we get a=4, b=0, r=2 or a=0, b=-4sqrt{3}, r=6.Therefore, the equation of circle C is either (x-4)^2+y^2=4 or x^2+(y+4sqrt{3})^2=36.boxed{(x-4)^2+y^2=4 text{ or } x^2+(y+4sqrt{3})^2=36}

🧠:[<think>Okay, so I have this problem here where two quadratic equations, x² + a x + b = 0 and x² + c x + d = 0, have the same root α, and a is not equal to c. I need to find what α is. Hmm, let's see. Let me start by recalling some algebra.Since α is a common root of both equations, plugging α into both equations should satisfy them. That means:First equation: α² + a α + b = 0Second equation: α² + c α + d = 0So both of these equations equal zero. If I subtract the second equation from the first one, maybe that will help. Let me try that.Subtracting the second equation from the first:(α² + a α + b) - (α² + c α + d) = 0 - 0Simplifying the left side:α² - α² + a α - c α + b - d = 0Which simplifies to:(a - c)α + (b - d) = 0So, (a - c)α + (b - d) = 0. Now, since a ≠ c (given in the problem), that means (a - c) is not zero. So we can solve for α.Let me write that step out:(a - c)α = d - bTherefore,α = (d - b)/(a - c)Wait, but another way to write that is α = (b - d)/(c - a). Let me check the algebra again to make sure.Starting from the subtraction:First equation minus second equation:(α² + a α + b) - (α² + c α + d) = 0So, α² cancels out. Then:a α - c α + b - d = 0Factor α:α(a - c) + (b - d) = 0So, α(a - c) = d - bTherefore,α = (d - b)/(a - c)Which is the same as (b - d)/(c - a) because multiplying numerator and denominator by -1. Both are correct, but depending on how you want to present it. The problem doesn't specify any particular form, so either is acceptable. Let me verify with an example to make sure.Suppose I have two quadratics:First: x² + 2x + 1 = 0. Roots are x = -1 (double root).Second: x² + 3x + 2 = 0. Roots are x = -1 and x = -2.So here, a = 2, b = 1, c = 3, d = 2. The common root is α = -1.Using the formula I derived:α = (d - b)/(a - c) = (2 - 1)/(2 - 3) = 1/(-1) = -1. Correct.Alternatively, (b - d)/(c - a) = (1 - 2)/(3 - 2) = (-1)/1 = -1. Also correct. So both forms work.Therefore, the answer should be α = (d - b)/(a - c) or α = (b - d)/(c - a). Since the problem is in terms of a, b, c, d, either is acceptable, but maybe we should present it as (d - b)/(a - c) since that's the direct result from the equation. Let me check the problem statement again to see if they specify anything about the form. The problem just says "Then α =", so either form is okay. However, to present it neatly, maybe factor out the negative sign. Let's see:α = (d - b)/(a - c) is same as α = (b - d)/(c - a). Which one is more standard? Maybe (b - d)/(c - a) is similar to slope formula, but I don't think it matters. Both are algebraically equivalent.Alternatively, since the problem states that a ≠ c, so denominators are non-zero, so either way is fine. The answer is uniquely determined. So the answer is α equals (d - b) divided by (a - c). Therefore, boxed form would be boxed{dfrac{d - b}{a - c}}.But let me think again. When I subtracted the equations, was there another way to approach this? Maybe using Vieta's formula? Let me see.If both quadratics have a common root α, then let the other roots be β and γ for the first and second equations respectively.So for the first equation: x² + a x + b = 0, the roots are α and β. Then by Vieta's formula:α + β = -aαβ = bSimilarly, for the second equation: x² + c x + d = 0, roots are α and γ. Vieta's gives:α + γ = -cαγ = dSo, we have:From first equation: β = -a - αFrom second equation: γ = -c - αAlso, from the products:From first equation: αβ = b ⇒ α(-a - α) = b ⇒ -a α - α² = b ⇒ α² + a α + b = 0 (which is consistent with original equation)Similarly, from second equation: αγ = d ⇒ α(-c - α) = d ⇒ -c α - α² = d ⇒ α² + c α + d = 0 (also consistent)But how does this help? Maybe subtract the two equations again?Alternatively, since both equations equal to zero when x = α, so set them equal to each other?Wait, if both equations equal zero at x = α, then:α² + a α + b = 0andα² + c α + d = 0Therefore, set them equal to each other:α² + a α + b = α² + c α + dSubtract α² from both sides:a α + b = c α + dBring terms with α to one side:a α - c α = d - bFactor α:α(a - c) = d - bThen:α = (d - b)/(a - c)Same result as before. So whether using Vieta's or directly subtracting equations, we get the same answer. Therefore, that must be the correct solution. The key was recognizing that since α is a common root, substituting into both equations and then equating the two results allows us to solve for α. Since a ≠ c, the coefficient of α is non-zero, so we can safely divide.Therefore, the answer is α = (d - b)/(a - c). Box that.Final Answerboxed{dfrac{d - b}{a - c}}</think>Given the quadratic equations (x^2 + ax + b = 0) and (x^2 + cx + d = 0) (with (a neq c)) that have the same root (alpha), we need to find (alpha).Since (alpha) is a common root of both equations, substituting (alpha) into both equations gives:[alpha^2 + aalpha + b = 0]and[alpha^2 + calpha + d = 0]Subtracting the second equation from the first:[(alpha^2 + aalpha + b) - (alpha^2 + calpha + d) = 0]Simplifying the left side:[alpha^2 - alpha^2 + aalpha - calpha + b - d = 0][(a - c)alpha + (b - d) = 0]Solving for (alpha):[(a - c)alpha = d - b][alpha = frac{d - b}{a - c}]Thus, the value of (alpha) is:[boxed{dfrac{d - b}{a - c}}],<think>Okay, so I need to find the value of α that is a common root of the two quadratic equations: x² + a x + b = 0 and x² + c x + d = 0, where a is not equal to c. Hmm, let me think about how to approach this. First, since α is a root of both equations, plugging α into both equations should satisfy them. That means:For the first equation: α² + aα + b = 0For the second equation: α² + cα + d = 0So, if I subtract the second equation from the first, maybe I can eliminate α². Let's try that:(α² + aα + b) - (α² + cα + d) = 0 - 0Simplifying the left side:α² - α² + aα - cα + b - d = 0Which simplifies to:(a - c)α + (b - d) = 0Okay, so this gives me a linear equation in terms of α. Since a ≠ c, which is given, the coefficient of α is not zero. So I can solve for α by dividing both sides by (a - c):α = (d - b)/(a - c)Wait, let me check the signs again. The equation was (a - c)α + (b - d) = 0, so moving (b - d) to the other side:(a - c)α = d - bTherefore, α = (d - b)/(a - c)Alternatively, if I factor out a negative sign from numerator and denominator, it could also be written as (b - d)/(c - a), but both are equivalent. Let me verify with an example to ensure I didn't make a sign mistake.Suppose a = 1, c = 2, b = 3, d = 4. Then according to this formula, α = (4 - 3)/(1 - 2) = 1/(-1) = -1. Let's check if α = -1 is a root of both equations:First equation: (-1)² + 1*(-1) + 3 = 1 -1 +3 = 3 ≠ 0. Hmm, that's not zero. Did I do something wrong here?Wait, maybe I need to choose different values where α is actually a common root. Let me think. Let's try to create a scenario where α is known. Let's pick α = 2. Let’s set up equations such that 2 is a root.First equation: (x - 2)(x - m) = x² - (2 + m)x + 2m. So here, a = -(2 + m), b = 2m.Second equation: (x - 2)(x - n) = x² - (2 + n)x + 2n. So c = -(2 + n), d = 2n.So according to the formula, α = (d - b)/(a - c) = (2n - 2m)/[-(2 + m) - (-(2 + n))] = 2(n - m)/[-2 - m + 2 + n] = 2(n - m)/(n - m) = 2. Which is correct. So in this case, the formula works. But in my previous example, why did it not work? Because I randomly chose a, b, c, d without ensuring that α exists. If I choose arbitrary a, b, c, d, there might not be a common root. So the formula gives the value of α in terms of the coefficients under the condition that such a common root exists. So if a common root exists, then α must be (d - b)/(a - c). So, going back, the answer should be α = (d - b)/(a - c). Let me check another example. Let’s say first equation is x² + 3x + 2 =0, roots are -1 and -2. Second equation x² + 4x + 3=0, roots are -1 and -3. The common root is -1. Let's compute (d - b)/(a - c) = (3 - 2)/(3 - 4)=1/(-1)=-1. Correct. So that works. Another example: First equation x² -5x +6=0, roots 2 and 3. Second equation x² -6x +8=0, roots 2 and 4. Common root is 2. Compute (d - b)/(a - c) = (8 -6)/(-5 - (-6))=2/(1)=2. Correct again. Hence, the formula seems to hold. Therefore, the value of α is (d - b)/(a - c). But just to make sure, let me derive it again. If α is a root of both quadratics:α² + aα + b = 0 ...(1)α² + cα + d = 0 ...(2)Subtract equation (2) from equation (1):(α² + aα + b) - (α² + cα + d) = 0 - 0Which simplifies to (a - c)α + (b - d) = 0Then, solving for α:α = (d - b)/(a - c)Yes, that's consistent. So the answer is α = (d - b)/(a - c). Alternatively, in some textbooks, they might present it as (b - d)/(c - a), but since (d - b) = -(b - d) and (a - c) = -(c - a), it's the same value. Therefore, the final answer is α = (d - b)/(a - c). So boxed, that would be boxed{dfrac{d - b}{a - c}}.Final AnswerThe common root α is boxed{dfrac{d - b}{a - c}}.</think>Given the quadratic equations (x^2 + ax + b = 0) and (x^2 + cx + d = 0) with (a neq c) and a common root (alpha), we need to find the value of (alpha).Since (alpha) is a root of both equations, substituting (alpha) into both equations gives:1. (alpha^2 + aalpha + b = 0)2. (alpha^2 + calpha + d = 0)Subtracting the second equation from the first to eliminate (alpha^2):[(alpha^2 + aalpha + b) - (alpha^2 + calpha + d) = 0]Simplifying this, we get:[(a - c)alpha + (b - d) = 0]Solving for (alpha) (since (a neq c)):[alpha = frac{d - b}{a - c}]Thus, the common root (alpha) is boxed{dfrac{d - b}{a - c}}.]

🔑:3. frac{d-b}{a-c}

🧠:[<think>Alright, so I need to solve this problem where the function y equals a squared to the power of 2x plus 2 times a to the power of x minus 9, and they say that a is greater than 0 and not equal to 1. The function has a maximum value of 6 on the interval [-1,1]. I need to find the value of a. Hmm, okay. Let me try to break this down.First, let's write down the function properly. The function is y = a^{2x} + 2a^x - 9. They mention that a > 0 and a ≠ 1. So, since a is positive and not equal to 1, this is an exponential function. The interval given is [-1,1], which is for x. So, the maximum value of y on this interval is 6, and we have to find a.Since the function is in terms of a^x, maybe substituting t = a^x would simplify things. Let's try that. Let t = a^x. Then, since a^{2x} is (a^x)^2, which is t^2. So, substituting, the function becomes y = t^2 + 2t - 9. That's a quadratic in terms of t. Nice, quadratics are easier to handle.But we have to consider the interval for x. Since x is in [-1,1], we need to find the corresponding range of t. Because a^x is an exponential function, depending on whether a is greater than 1 or between 0 and 1, the function will be increasing or decreasing. Let's note that.If a > 1, then a^x is increasing, so when x is in [-1,1], t will range from a^{-1} to a^1, which is [1/a, a]. If 0 < a < 1, then a^x is decreasing, so the range of t will be [a, 1/a]. So, depending on the value of a, the interval for t changes. But since a ≠ 1, it's either increasing or decreasing.But since the problem doesn't specify whether a is greater than 1 or less than 1, I might have to consider both cases. Hmm. Let's keep that in mind.Now, the function y is a quadratic in t: y = t² + 2t - 9. The quadratic opens upwards because the coefficient of t² is positive. So, the minimum of the quadratic is at the vertex, and it tends to infinity as t increases. But since we're looking for the maximum value on a closed interval, the maximum must occur at one of the endpoints of the interval for t. Wait, but quadratics that open upwards have a minimum, so the maximum on a closed interval would indeed be at one of the endpoints. That's a key point.Therefore, the maximum value of y on the interval for t (which is either [1/a, a] or [a, 1/a], depending on a) is achieved at one of the endpoints. Therefore, substituting the endpoints into the quadratic will give the maximum value. The problem states that this maximum value is 6. So, we need to set y equal to 6 at one of the endpoints and solve for a.So, let's formalize that. Let's denote the endpoints as t1 and t2. If a > 1, then t1 = 1/a and t2 = a. If 0 < a < 1, then t1 = a and t2 = 1/a. In either case, the endpoints are 1/a and a. Since a ≠ 1, these two endpoints are different.So, regardless of whether a is greater than 1 or between 0 and 1, the endpoints of t are 1/a and a, just in different order. Therefore, we can consider both cases by just taking the two endpoints and setting y equal to 6 for each, then solving for a. However, we need to check which of the two endpoints gives the maximum value. Since the quadratic is increasing for t > -1 (vertex at t = -1), so if both endpoints are greater than -1, then the maximum is at the larger endpoint.Wait, let me find the vertex of the quadratic. The quadratic is y = t² + 2t - 9. The vertex is at t = -b/(2a) = -2/(2*1) = -1. The vertex is at t = -1, and since the quadratic opens upwards, the minimum is at t = -1. So, the function decreases from negative infinity up to t = -1 and then increases from t = -1 to infinity. But in our case, the domain of t is [1/a, a] or [a, 1/a], depending on a. So, if the interval of t includes -1, then the minimum would be at t = -1, but since a > 0, and t = a^x, which is always positive, because a is positive. So, t is always positive. Therefore, the interval [1/a, a] or [a, 1/a] is entirely in the positive real numbers. Therefore, the vertex at t = -1 is outside our interval. Therefore, on the interval of t being positive, the quadratic is increasing because the vertex is at t = -1, which is to the left of our interval. Therefore, the quadratic is increasing on the entire interval of t. Therefore, the maximum value of y on the interval is at the larger endpoint, and the minimum at the smaller endpoint.Therefore, regardless of whether a is greater than 1 or less than 1, the larger endpoint is a if a > 1, and 1/a if 0 < a < 1. Wait, if a > 1, then the interval is [1/a, a], so the larger endpoint is a. If 0 < a < 1, then the interval is [a, 1/a], so the larger endpoint is 1/a. Therefore, depending on the value of a, the maximum of y is at x = 1 (if a > 1) or x = -1 (if 0 < a < 1). Wait, x is in [-1,1], so when a > 1, a^x is increasing, so maximum at x=1 (t = a), and minimum at x=-1 (t = 1/a). If 0 < a < 1, then a^x is decreasing, so maximum at x=-1 (t = 1/a) and minimum at x=1 (t = a). Therefore, the maximum value of y is either at x=1 (if a > 1) or at x=-1 (if 0 < a < 1).Therefore, in either case, the maximum value is at the larger of the two endpoints (either a or 1/a). So, if a > 1, maximum at a, if 0 < a < 1, maximum at 1/a. But since the problem says the maximum value is 6, we can set up equations for both cases and see which one gives a valid solution.So, let's consider the two cases:Case 1: a > 1. Then the maximum is at x=1, which is t = a. So, substituting t = a into the quadratic: y = a² + 2a - 9 = 6. So, the equation is a² + 2a - 9 = 6 => a² + 2a - 15 = 0. Solve this quadratic equation: a = [-2 ± sqrt(4 + 60)] / 2 = [-2 ± sqrt(64)] / 2 = [-2 ± 8]/2. Since a > 1, we take the positive root: (-2 + 8)/2 = 6/2 = 3. The other root is (-2 -8)/2 = -10/2 = -5, which is negative, so discard. So, a = 3 is a candidate.Case 2: 0 < a < 1. Then the maximum is at x=-1, which is t = 1/a. Substituting t = 1/a into the quadratic: y = (1/a)² + 2*(1/a) - 9 = 6. So, 1/a² + 2/a - 9 = 6 => 1/a² + 2/a - 15 = 0. Let's let b = 1/a, so since 0 < a < 1, then b > 1. Then the equation becomes b² + 2b -15 = 0. Which is the same equation as in case 1: b² + 2b -15 =0. Solving, b = [-2 ± sqrt(4 +60)]/2 = [-2 ± 8]/2. Since b > 1, take the positive root: (-2 +8)/2 = 6/2 = 3. The other root is -5, which we discard. Therefore, b = 3, so a = 1/b = 1/3. But 1/3 is between 0 and 1, so this is valid.Therefore, there are two possible solutions: a = 3 or a = 1/3. Wait, but the problem states "has a maximum value of 6 on the interval [-1,1]". So, both a=3 and a=1/3 would give the maximum value of 6 at different endpoints. But we need to check if these values actually yield the maximum as 6, and also ensure that there are no higher values within the interval. Wait, but we already considered that the quadratic is increasing on the interval of t, so if the maximum is at the larger endpoint, which is either a or 1/a, then substituting those gives 6. Therefore, both a=3 and a=1/3 satisfy this condition. However, the problem says "the function has a maximum value of 6". So, maybe both are valid? Wait, but let's check.Wait, when a=3, the function is y = 3^{2x} + 2*3^x -9. On the interval x ∈ [-1,1], the maximum is at x=1: y(1) = 3² + 2*3 -9 = 9 +6 -9=6. Correct. Similarly, when a=1/3, the function is y = (1/3)^{2x} + 2*(1/3)^x -9. The maximum is at x=-1: y(-1) = (1/3)^{-2} + 2*(1/3)^{-1} -9 = 9 + 6 -9 =6. Correct. So both a=3 and a=1/3 seem to satisfy the condition. But the problem asks for "then a=", implying a unique answer. Hmm, this is confusing. Maybe I made a mistake here?Wait, let's check if both a=3 and a=1/3 are valid solutions. The problem says "a >0, a ≠1". Both 3 and 1/3 satisfy this. So why does the problem expect a unique answer? Wait, maybe I need to check if there are other extrema within the interval. Wait, we considered the quadratic in t, which is increasing on the interval of t, which is positive. But maybe there are critical points within the interval for x? Wait, but the original function is in terms of x. Let's compute the derivative of y with respect to x and see if there are any critical points within [-1,1]. If there are, then perhaps the maximum could be at those points instead of the endpoints.Wait, good point. Let me compute the derivative. So, y = a^{2x} + 2a^x -9. The derivative dy/dx = 2a^{2x} ln a + 2a^x ln a = ln a (2a^{2x} + 2a^x). To find critical points, set dy/dx =0. So, ln a (2a^{2x} + 2a^x) =0. Since a ≠1, ln a ≠0. If a >1, ln a >0; if 0 <a <1, ln a <0. The term 2a^{2x} + 2a^x is always positive because a^x is always positive. Therefore, dy/dx =0 only if 2a^{2x} + 2a^x =0, but since a^x is positive, this sum is always positive. Therefore, there are no critical points where the derivative is zero. Therefore, the function is either always increasing or always decreasing in x. Wait, but wait, when a >1, a^x is increasing, so the function y = a^{2x} +2a^x -9 is also increasing. Similarly, if 0 <a <1, a^x is decreasing, so the function y is decreasing in x. Therefore, in both cases, the function is monotonic on the interval [-1,1], so the maximum is at one endpoint and the minimum at the other. Therefore, the maximum is at x=1 if a >1 and at x=-1 if 0 <a <1. So, in that case, both solutions a=3 and a=1/3 are valid. But the problem says "then a=". So maybe both answers are acceptable? But the problem is probably expecting a single answer, so perhaps I need to check which one of them is correct?Wait, maybe there is a mistake here. Let me verify again. When a=3, the maximum is at x=1, which is 6. When a=1/3, the maximum is at x=-1, which is also 6. So both are valid. But perhaps there's something else? Wait, maybe the problem is translated from Chinese, and the original might have a unique answer. Wait, maybe in the problem statement, there's a typo, or maybe I need to check my reasoning again.Wait, let's verify the function for both values of a over the interval.First, take a=3. Then the function is y=3^{2x} + 2*3^x -9. At x=1, y=9 +6 -9=6. At x=-1, y=3^{-2} + 2*3^{-1} -9=1/9 + 2/3 -9= (1 +6 -81)/9= (-74)/9 ≈ -8.222. So, indeed, the maximum is 6 at x=1. There are no higher values in between because the function is increasing.Now take a=1/3. The function is y=(1/3)^{2x} + 2*(1/3)^x -9. At x=-1, y=(1/3)^{-2} + 2*(1/3)^{-1} -9=9 +6 -9=6. At x=1, y=(1/3)^2 + 2*(1/3) -9=1/9 + 2/3 -9=(1 +6 -81)/9= (-74)/9 ≈ -8.222. Similarly, the function is decreasing here, so maximum at x=-1. So both a=3 and a=1/3 work. Therefore, the answer should be both 3 and 1/3? But the problem asks for "a=", implying a single answer. So maybe I made a mistake in the problem statement?Wait, let me check the problem again. It says: "the function y=a^{2x} +2a^x -9(a>0, a≠1), has a maximum value of 6 on the interval [-1,1], then a= "So, according to the problem statement, both a=3 and a=1/3 satisfy the condition. Therefore, unless there is a restriction that I missed, both are valid. However, perhaps in the original problem, there is a unique answer, so maybe I need to check my steps again.Wait, but wait, perhaps the problem is translated incorrectly, or maybe there is a misunderstanding. Let me check again.The function is given as y=a^{2x} +2a^x -9. The interval is [-1,1]. The maximum value on this interval is 6. We found that when a>1, maximum at x=1, and when 0<a<1, maximum at x=-1. Both cases give solutions a=3 and a=1/3. So, perhaps both are acceptable. But the problem might expect both answers. However, since the problem is presented as a single answer with a box, maybe in the original problem, the answer is both 3 and 1/3, but in the translation, it's presented as a single answer. Alternatively, maybe I made a miscalculation somewhere.Wait, let me check the equations again. For case 1 (a >1):y = a² + 2a -9 =6 => a² + 2a -15=0. Solution a=3 and a=-5. Discard a=-5. So a=3.For case 2 (0 <a <1):y = (1/a)² + 2*(1/a) -9 =6 => 1/a² +2/a -15=0. Let b=1/a, then equation is b² +2b -15=0. Solutions b=3 and b=-5. Since b=1/a>1, then b=3, so a=1/3.So, both a=3 and a=1/3 are solutions. Therefore, the answer should be a=3 or a=1/3.But maybe the problem expects both answers? In Chinese textbooks, sometimes they expect multiple answers, but in the format here, with boxed{}, it's usually for a single answer. Maybe the problem has a unique solution? Wait, perhaps the function can have both maxima at x=1 and x=-1, but in reality, the function's behavior might differ. Wait, no, we already checked that for a=3, the function is increasing, so maximum at x=1; for a=1/3, the function is decreasing, so maximum at x=-1. Both yield the maximum value of 6. Therefore, both are correct.But since the problem states "has a maximum value of 6 on the interval [-1,1]", and since both a=3 and a=1/3 satisfy that, the answer should be both. However, in the given format, maybe we need to write both answers. But the user instruction says "put your final answer within boxed{}", which usually indicates a single answer. Hmm.Wait, maybe I made a mistake in assuming both solutions are valid. Let me check for a=3 and a=1/3, is the maximum indeed 6 and there are no points in the interval where y exceeds 6?For a=3: y=3^{2x} +2*3^x -9. Let's check the value at x=1: 9 +6 -9=6. At x=0: 1 +2 -9= -6. At x=-1: 1/9 +2/3 -9= (1 +6 -81)/9= -74/9≈-8.22. So the maximum is indeed 6 at x=1.For a=1/3: y=(1/3)^{2x} +2*(1/3)^x -9. At x=-1:9 +6 -9=6. At x=0:1 +2 -9=-6. At x=1:1/9 +2/3 -9≈-8.22. So maximum at x=-1 is 6. So both are correct.Therefore, the answer should be a=3 or a=1/3. But in the problem statement, maybe the answer is expecting both, but since it's presented as a single answer, maybe I need to write both in the box. However, in standard practice, if multiple answers are possible, they should be listed. But maybe the original problem had a different condition, or perhaps there is an error in translation. Alternatively, maybe the problem is from a source where only one answer is expected, but according to the mathematical reasoning, both are correct.Wait, maybe there's a mistake in substituting t. Let me confirm. When a>1, t ranges from 1/a to a, so the quadratic is increasing on that interval, so the maximum is at t=a. For a<1, t ranges from a to 1/a, and since quadratic is increasing, the maximum is at t=1/a. Therefore, substituting those gives us the equations leading to a=3 and a=1/3. Therefore, both solutions are valid. Therefore, unless the problem has additional constraints, both answers are correct.But since the user might expect a single answer, maybe there's something wrong here. Wait, perhaps the problem is from an exam where only one answer is correct, so maybe I need to check again.Alternatively, perhaps I made a mistake in assuming the maximum is at the endpoints. Let's check the function for a=3 on the entire interval. The function y=9^x +2*3^x -9. At x=1, 9 +6 -9=6. What about x=0.5? y=sqrt(9) + 2*sqrt(3) -9=3 + 3.464 -9≈-2.536. At x=0:1 +2 -9=-6. At x=-0.5:1/3 +2/(sqrt(3)) -9≈0.333 +1.154 -9≈-7.513. So the function is increasing, so maximum at x=1. Similarly for a=1/3, decreasing function, so maximum at x=-1. Therefore, both are correct.Therefore, since the problem doesn't specify any other constraints, both a=3 and a=1/3 are solutions. Therefore, the answer should be a=3 or a=1/3. However, in the problem statement, maybe due to translation or formatting, it's presented as a single answer, but mathematically both are correct. Therefore, perhaps the answer is boxed{dfrac{1}{3}} and boxed{3}. But since the user might expect one answer, but according to the problem, both are correct. Wait, in the original problem, maybe it's multiple choice, but since we don't have that context, as a solver, I need to state both solutions.But since the user instruction says "put your final answer within boxed{}", which usually is for a single answer, but perhaps here we need to box both. But standard practice in mathematics is to present all solutions. However, in Chinese math problems, sometimes even if there are two solutions, they might expect both written. But how to represent that in a box? Maybe separate boxes? But the instruction says "put your final answer within boxed{}". Maybe the problem actually has a unique solution, and my reasoning is flawed.Wait, let me check the equations again. For case 1: a² + 2a -15=0 gives a=3 or a=-5, but a>0 so a=3. For case 2:1/a² + 2/a -15=0. Let me multiply both sides by a²:1 + 2a -15a²=0. That is, -15a² +2a +1=0. Multiply by -1:15a² -2a -1=0. Then solving for a: a = [2 ± sqrt(4 +60)]/(30)= [2 ±8]/30. So a=(10)/30=1/3 or a=(-6)/30=-1/5. Since a>0, a=1/3. So that's correct. Therefore, both solutions are valid.Therefore, the answer is a=3 or a=1/3. But given the user instruction, maybe we need to write both inside the box, separated by a comma. However, standard Latex boxing is for a single answer. Maybe the problem expects only one solution, so perhaps I made a mistake here. Alternatively, the problem might have a unique solution if we consider the function's minimum or another condition.Wait, the problem says "has a maximum value of 6". If the function is monotonic, then yes, the maximum is at the endpoint. But if the function had a critical point inside the interval, maybe the maximum could be there. But we already checked that the derivative doesn't have any zeros because dy/dx = ln a (2a^{2x} +2a^x) which is never zero because 2a^{2x} +2a^x is always positive and ln a is not zero (a≠1). Therefore, no critical points inside the interval, so function is strictly monotonic, so maximum at the endpoints. Therefore, both solutions are valid.Therefore, the answer is both 3 and 1/3. But since the problem is presented as a single answer question, maybe the answer is boxed{3} or boxed{dfrac{1}{3}}? However, in many cases, unless specified, both solutions are correct. But since the problem says "then a=", maybe there is a unique answer, which makes me think maybe I made a mistake.Wait, maybe there's a miscalculation in substituting t. Let me check again.When a >1, t ranges from 1/a to a. The quadratic y = t² +2t -9 is increasing on that interval, so maximum at t=a, leading to equation a² +2a -15=0.For 0 <a <1, t ranges from a to 1/a. The quadratic is increasing, so maximum at t=1/a, equation (1/a)^2 +2*(1/a) -15=0, which simplifies to same as above, leading to a=1/3.So both a=3 and a=1/3 are valid. Therefore, the answer is both. Therefore, in the box, we can write both answers. Maybe comma-separated inside the box.But in standard math notation, if multiple answers, we use multiple boxes. But according to the user instruction, "put your final answer within boxed{}", so perhaps the answer is two boxed answers. But the user might not accept that. Alternatively, maybe the problem is expecting only one answer, and there's a mistake in the problem statement.Alternatively, maybe there is a misunderstanding. Let me read the problem again."If the function y=a^{2x} +2a^x -9(a>0, a ≠1), has a maximum value of 6 on the interval [-1,1], then a="So, the function has a maximum value of 6 on [-1,1]. If the function is monotonic, then the maximum is at one endpoint. If a>1, then it's increasing, maximum at x=1. If a<1, decreasing, maximum at x=-1. Both give valid solutions. Therefore, both a=3 and a=1/3 are correct. So, unless the problem specifies further constraints, both are correct.But since the problem is probably expecting a single numerical answer, maybe due to the source's answer being one of them, but according to our analysis, both are correct. However, given that the problem is presented in Chinese, and sometimes Chinese math problems might consider only the positive solution greater than 1, but in this case, both 3 and 1/3 are positive and satisfy the conditions. Therefore, I think the correct answer is both.But given the user's instruction, perhaps I need to write both answers inside the box separated by a comma. However, the standard format is one box per answer, but since the user said "put your final answer within boxed{}", maybe they expect multiple boxed answers. Alternatively, if the original problem had a unique answer, there might be an error in my reasoning.Alternatively, perhaps the problem is in the translation, and actually, the function is y = a^{2x} + 2a^{x} - 9, and the maximum value is 6, but maybe in the original problem, it's a minimum value? Because if it's the minimum value, then the solution would be unique. Let me check.If the problem had a minimum value of 6, then for a>1, the minimum is at x=-1 (t=1/a): y=1/a² +2/a -9=6. Which leads to 1/a² +2/a -15=0, same as before, leading to a=1/3. For a<1, the minimum is at x=1 (t=a): y=a² +2a -9=6, leading to a=3. So, if it were a minimum, then a=1/3 (for a>1) and a=3 (for a<1). But since a>1 and a<1, you would have a=3 (a>1) and a=1/3 (a<1). But since the problem says maximum value, then both are valid.Alternatively, maybe the problem is mistranslated, and it's supposed to be a minimum. If that's the case, then unique answers. But given the current problem statement, it's maximum.Given that, I think the correct answer is both 3 and 1/3. Therefore, the answer is boxed{3} and boxed{dfrac{1}{3}}.But since the user might expect one answer, I need to check again.Wait, maybe the original problem has a unique answer because when a=1/3, although the maximum at x=-1 is 6, when we check other points, maybe the function exceeds 6? But we already checked that for a=1/3, the function at x=-1 is 6, and since it's decreasing, all other points are less than 6. Similarly for a=3, the function at x=1 is 6, and increasing, so all other points are less than 6. So both are valid.Therefore, the conclusion is that a=3 or a=1/3. Therefore, the answer is both.But given the instruction to "put your final answer within boxed{}", maybe the problem expects both answers in separate boxes. So, I will present both.Final AnswerThe values of ( a ) are boxed{3} and boxed{dfrac{1}{3}}.</think>Given the function ( y = a^{2x} + 2a^x - 9 ) where ( a > 0 ) and ( a neq 1 ), we need to find the value of ( a ) such that the function has a maximum value of 6 on the interval ([-1, 1]).1. Substitution: - Let ( t = a^x ). Then ( a^{2x} = t^2 ), so the function becomes ( y = t^2 + 2t - 9 ).2. Range of ( t ): - If ( a > 1 ), ( t ) ranges from ( frac{1}{a} ) to ( a ). - If ( 0 < a < 1 ), ( t ) ranges from ( a ) to ( frac{1}{a} ).3. Quadratic Analysis: - The quadratic function ( y = t^2 + 2t - 9 ) opens upwards (vertex at ( t = -1 )), which is outside the interval since ( t ) is always positive. - Therefore, the maximum value occurs at the endpoint of the interval for ( t ).4. Case Analysis: - Case 1 ( ( a > 1 )): - Maximum at ( t = a ). - Solving ( a^2 + 2a - 9 = 6 ) gives ( a^2 + 2a - 15 = 0 ). - Solutions: ( a = 3 ) (since ( a = -5 ) is discarded). - Case 2 ( ( 0 < a < 1 )): - Maximum at ( t = frac{1}{a} ). - Solving ( left(frac{1}{a}right)^2 + 2left(frac{1}{a}right) - 9 = 6 ) gives ( frac{1}{a^2} + frac{2}{a} - 15 = 0 ). - Let ( b = frac{1}{a} ), then ( b^2 + 2b - 15 = 0 ). - Solutions: ( b = 3 ) (since ( b = -5 ) is discarded), so ( a = frac{1}{3} ).5. Verification: - For ( a = 3 ), the function has a maximum value of 6 at ( x = 1 ). - For ( a = frac{1}{3} ), the function has a maximum value of 6 at ( x = -1 ).Both values satisfy the given condition. Therefore, the values of ( a ) are (boxed{3}) and (boxed{dfrac{1}{3}}).,<think>Okay, let's try to solve this problem. The function given is ( y = a^{2x} + 2a^x - 9 ), where ( a > 0 ) and ( a neq 1 ). We need to find the value of ( a ) such that the function has a maximum value of 6 on the interval ([-1, 1]).First, let me understand the problem. The function is defined in terms of ( a^x ). Since ( a > 0 ) and ( a neq 1 ), this is an exponential function. The interval given is from -1 to 1 for ( x ). The function is supposed to have a maximum value of 6 on this interval, so somewhere between ( x = -1 ) and ( x = 1 ), the highest point of the function is 6. We need to find the value of ( a ) that makes this true.Hmm. Let me think about how to approach this. Since the function involves ( a^{2x} ) and ( a^x ), maybe it's helpful to make a substitution to simplify the expression. Let me let ( t = a^x ). Then, ( a^{2x} = (a^x)^2 = t^2 ). So substituting, the function becomes:( y = t^2 + 2t - 9 ).But we need to remember that ( t = a^x ). Since ( a > 0 ), ( t ) will always be positive. Also, the interval for ( x ) is from -1 to 1. So when ( x ) is in that interval, ( t ) will take values from ( a^{-1} ) to ( a^1 ), which is ( frac{1}{a} ) to ( a ).So the problem reduces to finding ( a ) such that the quadratic function ( y = t^2 + 2t - 9 ) has a maximum value of 6 when ( t ) is in the interval ([1/a, a]).Wait a second, quadratic functions have either a maximum or a minimum depending on the leading coefficient. Here, the coefficient of ( t^2 ) is positive (1), so the parabola opens upwards, meaning it has a minimum point, not a maximum. That seems contradictory. If the quadratic opens upwards, the vertex is the minimum point, so the maximum of the quadratic on the interval ([1/a, a]) would occur at one of the endpoints, right?Therefore, the maximum value of ( y ) on the interval ([1/a, a]) would be either at ( t = 1/a ) or at ( t = a ). So, we can set up equations for both endpoints, compute ( y ) at each, set the maximum of those two equal to 6, and solve for ( a ).But before jumping into that, let me confirm. Since the quadratic has a minimum, the maximum on the interval would indeed be at one of the endpoints. Therefore, we need to evaluate ( y ) at both ( t = a ) and ( t = 1/a ), compare them, and set the larger one equal to 6.So, let's compute ( y(a) ) and ( y(1/a) ):First, at ( t = a ):( y(a) = a^2 + 2a - 9 ).At ( t = 1/a ):( y(1/a) = (1/a)^2 + 2(1/a) - 9 = frac{1}{a^2} + frac{2}{a} - 9 ).Now, depending on the value of ( a ), either ( y(a) ) or ( y(1/a) ) will be larger. Since we need the maximum of these two to be 6, we can set both possibilities equal to 6 and see which one gives a valid solution.So, we have two cases:Case 1: ( a^2 + 2a - 9 = 6 ).Case 2: ( frac{1}{a^2} + frac{2}{a} - 9 = 6 ).We need to solve both equations for ( a ) and check which solutions satisfy ( a > 0 ), ( a neq 1 ), and also ensure that the maximum is indeed 6.Let's start with Case 1:Case 1: ( a^2 + 2a - 9 = 6 ).Simplify:( a^2 + 2a - 15 = 0 ).This is a quadratic equation. Let's solve for ( a ):Using quadratic formula:( a = frac{ -2 pm sqrt{(2)^2 - 4(1)(-15)} }{2(1)} = frac{ -2 pm sqrt{4 + 60} }{2} = frac{ -2 pm sqrt{64} }{2} = frac{ -2 pm 8 }{2} ).So, two solutions:1. ( a = frac{ -2 + 8 }{2 } = frac{6}{2} = 3 ).2. ( a = frac{ -2 - 8 }{2 } = frac{ -10 }{2 } = -5 ).But ( a > 0 ), so we discard -5. So, potential solution ( a = 3 ).Now, check if this is valid. We need to make sure that when ( a = 3 ), the maximum value of ( y ) on the interval is indeed 6. But since we set ( y(a) = 6 ), we need to check if ( y(1/a) ) is less than or equal to 6.Compute ( y(1/a) ) when ( a = 3 ):( y(1/3) = (1/3)^2 + 2*(1/3) - 9 = 1/9 + 2/3 - 9 = (1 + 6)/9 - 9 = 7/9 - 9 = (7 - 81)/9 = -74/9 ≈ -8.222 ).Which is definitely less than 6. So, the maximum would indeed be at ( t = a = 3 ), which gives y = 6. So, a = 3 is a valid solution.Now, check Case 2:Case 2: ( frac{1}{a^2} + frac{2}{a} - 9 = 6 ).Simplify:( frac{1}{a^2} + frac{2}{a} - 15 = 0 ).Multiply both sides by ( a^2 ) to eliminate denominators:( 1 + 2a - 15a^2 = 0 ).Rearranged:( -15a^2 + 2a + 1 = 0 ).Multiply both sides by -1 to make it easier:( 15a^2 - 2a - 1 = 0 ).Now, solve this quadratic equation for ( a ):Using quadratic formula:( a = frac{2 pm sqrt{(-2)^2 - 4(15)(-1)}}{2*15} = frac{2 pm sqrt{4 + 60}}{30} = frac{2 pm sqrt{64}}{30} = frac{2 pm 8}{30} ).So, two solutions:1. ( a = frac{2 + 8}{30} = frac{10}{30} = frac{1}{3} ).2. ( a = frac{2 - 8}{30} = frac{-6}{30} = -frac{1}{5} ).Again, ( a > 0 ), so discard the negative solution. Potential solution ( a = 1/3 ).Check if this is valid. When ( a = 1/3 ), compute ( y(a) ):( y(1/3) = (1/3)^2 + 2*(1/3) - 9 = 1/9 + 2/3 - 9 = same as before, which is -74/9 ≈ -8.222 ).Wait, but in this case, we set ( y(1/a) = 6 ). Let me check:Wait, when ( a = 1/3 ), ( 1/a = 3 ). So, compute ( y(1/a) = y(3) = 3^2 + 2*3 -9 = 9 +6 -9 = 6. So, that's correct. But then, also compute ( y(a) = y(1/3) = -74/9 ≈ -8.222 ). So, the maximum is indeed 6 at ( t = 3 ), which corresponds to ( x = log_a(3) ).Wait, but since ( a = 1/3 ), ( x = log_{1/3}(3) ). Since ( (1/3)^x = 3 ), so ( x = -1 ). Therefore, ( t = 3 ) when ( x = -1 ). So, the maximum is at ( x = -1 ), which is within the interval [-1, 1]. So, that's acceptable. So, ( a = 1/3 ) is another potential solution.But now, we need to check if in this case, the maximum is indeed 6. Since when ( a = 1/3 ), the function's maximum value on the interval is 6, achieved at ( x = -1 ). So, both ( a = 3 ) and ( a = 1/3 ) seem to satisfy the equation. But we need to check if both are valid.Wait, but let's think again. When we set ( t = a^x ), with ( x in [-1,1] ), and ( a ) can be either greater than 1 or between 0 and 1. If ( a > 1 ), then ( a^x ) is increasing, so ( t ) ranges from ( a^{-1} ) to ( a^1 ). If ( 0 < a < 1 ), then ( a^x ) is decreasing, so ( t ) ranges from ( a^{1} ) to ( a^{-1} ). So in either case, the interval for ( t ) is [min(a, 1/a), max(a, 1/a)].Therefore, when ( a > 1 ), the interval for ( t ) is [1/a, a], and when ( 0 < a <1 ), it's [a, 1/a].But since we made the substitution ( t = a^x ), regardless of ( a ), the interval for ( t ) is between 1/a and a when a >1, or between a and 1/a when a <1.But in both cases, the endpoints are 1/a and a. So, for a=3, the t-interval is [1/3, 3], and for a=1/3, it's also [1/3, 3], since 1/a = 3. So in both cases, the t-interval is [1/3, 3]. Therefore, for both a=3 and a=1/3, the interval for t is [1/3, 3]. Then, the function y(t) = t^2 + 2t -9 is a quadratic with minimum at t = -1 (vertex), but since our interval is [1/3, 3], which is to the right of the vertex. Since the parabola opens upwards, the minimum is at t = -1, but our interval is all positive. Therefore, on [1/3, 3], the function is increasing, because the vertex is at t = -1, and to the right of that, the function is increasing. Wait, is that true? Let me check the derivative.The derivative of y with respect to t is y’ = 2t + 2. Setting this equal to zero gives t = -1, which is the critical point. So, for t > -1, the function is increasing. Since our interval is t in [1/3, 3], which is all greater than -1, the function is increasing on this interval. Therefore, the minimum is at t = 1/3 and the maximum at t = 3. Therefore, if the maximum is 6, then when t=3, y=6. So, solving y(3) = 6.Wait, but in the problem statement, the maximum is 6. So according to this, for both a=3 and a=1/3, the maximum of y on the interval [1/3, 3] is at t=3, which is y=6. Therefore, both a=3 and a=1/3 would lead to the maximum value 6. But wait, when a=1/3, how does t=3 relate to x?Wait, if a=1/3, then t = a^x = (1/3)^x. To get t=3, we have (1/3)^x = 3 => 3^{-x} = 3^1 => -x = 1 => x = -1. So, t=3 corresponds to x=-1, which is in the interval [-1,1]. Similarly, for a=3, t=3 corresponds to x=1, since 3^1 =3. So in both cases, the maximum occurs at the endpoints of the interval x=-1 or x=1.But the problem states that the maximum is 6 on the interval [-1,1]. Therefore, both a=3 and a=1/3 satisfy this condition.Wait, but the problem asks for "the value of a". So does that mean there are two solutions? However, the problem is stated as "has a maximum value of 6", implying there might be a unique solution. Let me check the problem again.Wait, the problem says "the function [...] has a maximum value of 6 on the interval [-1,1], then a= "So maybe both 3 and 1/3 are solutions? Let me check.Wait, but let's think again. When a=3, the function y = 3^{2x} + 2*3^x -9. Let's check the maximum at x=1:y(1) = 3^2 + 2*3 -9 = 9 +6 -9=6. Correct. Then at x=-1:y(-1) = 3^{-2} + 2*3^{-1} -9= 1/9 + 2/3 -9≈0.111+0.666-9≈-8.222. So, maximum at x=1 is 6.Similarly, when a=1/3, the function is y = (1/3)^{2x} + 2*(1/3)^x -9. At x=-1:y(-1) = (1/3)^{-2} +2*(1/3)^{-1} -9= 9 + 6 -9=6. At x=1:y(1)= (1/3)^2 +2*(1/3) -9≈0.111 +0.666 -9≈-8.222. So maximum at x=-1 is 6.Therefore, both a=3 and a=1/3 make the function have a maximum of 6 on [-1,1]. So why does the problem imply a single answer? Maybe I need to check if both are valid. Let me check the original substitution.Wait, when a=3, the t-interval is [1/3, 3], and when a=1/3, the t-interval is [1/3, 3] as well. So in both cases, t ranges from 1/3 to 3, and since the function y(t) is increasing on that interval, maximum at t=3, which is 6. Therefore, both a=3 and a=1/3 lead to the maximum value of 6. Therefore, there are two solutions. But the problem says "then a=", which might suggest a single answer. Perhaps there's a mistake in my reasoning.Wait, perhaps when a=1/3, even though t=3 corresponds to x=-1, which is in the interval, but maybe the function could have a critical point inside the interval. Wait, the original function is in terms of x. Maybe I need to check if there are any critical points in x in [-1,1].So, let's compute the derivative of y with respect to x and set it to zero to find critical points.Given ( y = a^{2x} + 2a^x -9 ).Derivative dy/dx = 2a^{2x} ln(a) + 2a^x ln(a) = ln(a) (2a^{2x} + 2a^x ).Set derivative equal to zero:ln(a)(2a^{2x} + 2a^x ) = 0.Since a ≠1 and a>0, ln(a) ≠0 (because a ≠1). Therefore, 2a^{2x} + 2a^x = 0.But 2a^{2x} + 2a^x = 2a^x(a^x +1 ). Since a^x is always positive, and a^x +1 is also positive. Therefore, 2a^x(a^x +1 ) is always positive. Therefore, there are no critical points. Therefore, the function has no critical points, so the maximum must occur at one of the endpoints, x=-1 or x=1.Therefore, both a=3 and a=1/3 are valid solutions. But the problem says "then a=", which is singular. Maybe the problem expects both answers? But in the Chinese curriculum, sometimes problems might expect both solutions. However, in the original problem statement, it might depend on the context.Wait, but let me check again. If a=3, then the maximum occurs at x=1. If a=1/3, the maximum occurs at x=-1. Both are valid. Therefore, there are two solutions. But the problem might have a typo, or perhaps there is a constraint that I missed.Looking back: the problem states "a >0, a ≠1". So both 3 and 1/3 satisfy this. Therefore, both are valid. So the answer should be a=3 or a=1/3. However, the way the problem is phrased "then a= " suggests a single answer. Maybe in the problem's source, there is only one answer, but according to my calculations, there are two. Let me check my calculations again.Wait, when solving Case 1: ( a^2 + 2a -15 =0 ), we got a=3 and a=-5. Discarded a=-5. Case 2: ( 15a^2 -2a -1=0 ), we got a=1/3 and a=-1/5. Discarded a=-1/5. So, both a=3 and a=1/3. Therefore, two solutions.But maybe in the problem statement, the interval is [-1,1], and when a=1/3, the maximum is at x=-1, and when a=3, it's at x=1. Both are within the interval, so both are acceptable. So the answer should be both 3 and 1/3. But the problem says "then a=", maybe expecting both? But how to write two answers in boxed notation. Alternatively, maybe there's a mistake in my analysis.Wait, let me check the functions again. For a=3:Function: y=3^{2x} +2*3^x -9. At x=1, y=9+6-9=6. At x=0, y=1+2-9=-6. At x=-1, y≈0.111 +0.666 -9≈-8.222. So the function is increasing from x=-1 to x=1, because the base a=3>1, so the exponentials increase. Therefore, the maximum is at x=1.For a=1/3:Function: y=(1/3)^{2x} +2*(1/3)^x -9. At x=-1, y=9 +6 -9=6. At x=0, y=1 +2 -9=-6. At x=1, y≈0.111 +0.666 -9≈-8.222. Since a=1/3<1, the exponentials decrease as x increases, so the function is decreasing from x=-1 to x=1, so maximum at x=-1.Therefore, both a=3 and a=1/3 make the function achieve maximum 6 at x=1 and x=-1 respectively, which are within the interval. So both are valid.But the problem might expect both answers. However, if I recall correctly, sometimes substitution can lead to extraneous solutions, but in this case, both seem valid. Therefore, the answer should be a=3 or a=1/3. But how to present this?Wait, maybe in the problem's original source, there are two answers, but here the user might expect me to write both. However, since the user instruction says "put your final answer within boxed{}", which usually indicates a single answer. Maybe the problem is designed such that a=1/3 and a=3 are reciprocal, so maybe they are the same in some sense. But in reality, they are distinct.Alternatively, perhaps I made an error in assuming both solutions are valid. Wait, let's check the quadratic function y = t^2 + 2t -9 on the interval [1/3, 3]. Since it's increasing on that interval, as we saw earlier, then if we have the maximum at t=3, which is y=6, that's correct. But in this substitution, when a=3, t=3 corresponds to x=1, and when a=1/3, t=3 corresponds to x=-1. So both of these are valid, but the substitution didn't consider the relationship between a and the interval. Wait, but since the substitution is just a way to rewrite the function, and the original function's maximum is determined by x in [-1,1], which maps to t in [1/a, a], both cases are covered.Alternatively, perhaps the problem considers the function's behavior over the interval. For example, if a>1, the function y increases from x=-1 to x=1, so maximum at x=1. If 0 <a <1, the function decreases from x=-1 to x=1, so maximum at x=-1. So both cases give the maximum at different endpoints. Since the problem states that the maximum is 6, both a=3 and a=1/3 are solutions.Therefore, the answer is a=3 or a=1/3. But the problem says "then a=". In Chinese math problems, sometimes reciprocal answers are considered the same in some contexts, but here they are different. Maybe the answer is both 3 and 1/3. But the user might expect two answers. However, given the way the problem is presented, I need to check if I made a miscalculation.Wait, let's see. If we go back to the equations:Case 1: a^2 + 2a -15 =0 --> a=3Case 2: 15a^2 -2a -1=0 --> a=1/3Therefore, two solutions. Hence, the answer is a=3 or a=1/3.But maybe the problem expects both answers. Therefore, the answer is boxed{frac{1}{3}} and boxed{3}. But since the user might expect one answer, but according to my analysis, there are two. Wait, maybe in the original problem, there is a constraint that a>1 or 0 <a <1, but the problem says a>0, a≠1. So both are valid.Alternatively, maybe when substituting t = a^x, and we set y(t) =6, we get two solutions for a. Therefore, the answer should include both 3 and 1/3. But since the problem is presented as "then a=", perhaps in the original problem, there is only one answer, but according to my calculations, there are two.Wait, let me check with a=2. Suppose a=2, what happens?Then, y = 2^{2x} + 2*2^x -9. At x=1: 4 +4 -9= -1. At x=-1: 1/4 +1 -9= -7.75. So maximum at x=1 is -1, which is less than 6. So not a solution.If a=4, y(1)=16 +8 -9=15. Which is larger than 6. So not a solution. So a=3 is correct.If a=1/2, then y(-1)= (1/2)^{-2} +2*(1/2)^{-1} -9=4 +4 -9= -1. At x=1, y(1)=1/4 +1 -9= -7.75. So maximum at x=-1 is -1, which is less than 6. Not a solution.So only a=3 and a=1/3 give the maximum of 6. Hence, both are correct.But the problem might have a mistake. Alternatively, maybe when the problem says "the function has a maximum value of 6", perhaps the function must attain 6 as the maximum, but not necessarily only at one point. Since both a=3 and a=1/3 give the maximum of 6, they are both acceptable. Therefore, the answer is a=3 or a=1/3.But in the Chinese context, maybe only one answer is expected, but mathematically, there are two. Therefore, maybe the answer is boxed{dfrac{1}{3}} or boxed{3}. However, considering the problem statement didn't specify further constraints, both are correct. Since the user hasn't specified multiple answers, but typically in such problems, unless specified, both solutions should be considered. However, in many textbooks, sometimes only the positive solution greater than 1 is considered, but that's not necessarily the case.Wait, to double-check, if we consider the original function: when a=3, the function is increasing on [-1,1], so the maximum is at x=1. When a=1/3, the function is decreasing on [-1,1], so the maximum is at x=-1. Both give y=6. Therefore, both are correct.Therefore, the answer should be both 3 and 1/3.But how to present this? If the system allows multiple boxed answers, then both. If not, maybe the problem expects both. But the user instruction says "put your final answer within boxed{}", which is singular. Maybe the problem in the original source had both answers, but here the user might have mistyped or missed something. Alternatively, maybe I made a mistake in substitution.Alternatively, perhaps there's a restriction in the problem that I didn't consider. Wait, let's check the substitution again.We set t = a^x, so x = log_a(t). The interval x ∈ [-1,1] translates to t ∈ [a^{-1}, a] if a>1, and t ∈ [a, a^{-1}] if 0 <a <1. But since the quadratic in t is y = t^2 + 2t -9, which has its vertex at t = -1, which is not in the interval [a^{-1}, a] or [a, a^{-1}]. Since a>0, and a≠1, so the interval for t is either increasing or decreasing. But as we saw earlier, in both cases, the quadratic is increasing on the interval, because the derivative is 2t + 2, which is positive for t > -1. Since t is always positive (as a>0 and a≠1), the derivative is always positive, so the function is increasing on t >0. Therefore, regardless of a>1 or 0 <a <1, the function y(t) is increasing on the interval [min(a,1/a), max(a,1/a)]. Therefore, the maximum occurs at t = max(a,1/a). Therefore, if a>1, then max(a,1/a)=a, so maximum at t=a. If 0 <a <1, then max(a,1/a)=1/a, so maximum at t=1/a. But when a=3>1, maximum at t=a=3. When a=1/3 <1, maximum at t=1/a=3. Therefore, in both cases, the maximum is at t=3, which gives y=6. Therefore, both a=3 and a=1/3 lead to t=3 as the maximum point, hence y=6. So both are solutions.Therefore, the answer is a=3 and a=1/3. However, since the problem asks for "a=", and in the format boxed{}, maybe we need to write both answers. But in the Chinese math problems, sometimes when reciprocal solutions exist, they are both considered, but maybe here, depending on the problem's expectation. But strictly mathematically, both are correct.Therefore, I think the correct answers are a=3 and a=1/3.Final AnswerThe value of ( a ) is boxed{frac{1}{3}} or boxed{3}.</think>Given the function ( y = a^{2x} + 2a^x - 9 ) where ( a > 0 ) and ( a neq 1 ), we need to find the value of ( a ) such that the function has a maximum value of 6 on the interval ([-1, 1]).First, we make a substitution ( t = a^x ). This transforms the function into ( y = t^2 + 2t - 9 ). The interval for ( x ) from (-1) to (1) translates to ( t ) ranging from ( frac{1}{a} ) to ( a ) if ( a > 1 ), or from ( a ) to ( frac{1}{a} ) if ( 0 < a < 1 ).Since the quadratic function ( y = t^2 + 2t - 9 ) opens upwards (coefficient of ( t^2 ) is positive), it has a minimum at ( t = -1 ). However, since ( t ) must be positive, the function is increasing on the interval ([ min(a, frac{1}{a}), max(a, frac{1}{a}) ]). Therefore, the maximum value occurs at the right endpoint of the interval.We consider two cases for the maximum value at the endpoints:1. Case 1: Maximum at ( t = a ) [ a^2 + 2a - 9 = 6 implies a^2 + 2a - 15 = 0 ] Solving this quadratic equation: [ a = frac{-2 pm sqrt{64}}{2} = frac{-2 pm 8}{2} ] This gives solutions ( a = 3 ) (since ( a > 0 )).2. Case 2: Maximum at ( t = frac{1}{a} ) [ frac{1}{a^2} + frac{2}{a} - 9 = 6 implies frac{1}{a^2} + frac{2}{a} - 15 = 0 ] Multiplying by ( a^2 ): [ 1 + 2a - 15a^2 = 0 implies 15a^2 - 2a - 1 = 0 ] Solving this quadratic equation: [ a = frac{2 pm sqrt{64}}{30} = frac{2 pm 8}{30} ] This gives solutions ( a = frac{1}{3} ) (since ( a > 0 )).Both solutions ( a = 3 ) and ( a = frac{1}{3} ) are valid. When ( a = 3 ), the maximum occurs at ( x = 1 ), and when ( a = frac{1}{3} ), the maximum occurs at ( x = -1 ). Both satisfy the condition of having a maximum value of 6 on the interval ([-1, 1]).Thus, the values of ( a ) are boxed{frac{1}{3}} and boxed{3}.]

🔑:7. 3 or frac{1}{3} Let t=a^{x}, y=t^{2}+2 t-9(t>0), when a>1, t=a, y_{text {max }}=a^{2}+2 a-9=6, a=3; when 0<a<1, left(frac{1}{a}right)^{2}+frac{2}{a}-9=0, a=frac{1}{3}.