🧠:[<think>Okay, so I need to figure out the minimum number of numbers I must select from the first 100 natural numbers (1 to 100) to ensure that at least two of them add up to a multiple of 10. Hmm, let's see. This seems like a problem that involves the Pigeonhole Principle. The idea is that if you have more pigeons than pigeonholes, at least two pigeons have to share a hole. First, I should understand how pairs of numbers can sum to a multiple of 10. If two numbers add up to 10, 20, 30, ..., 100, then their sum is a multiple of 10. So, for example, 1 and 9, 2 and 8, 3 and 7, etc., each add up to 10. Similarly, numbers like 11 and 19 would add up to 30, which is also a multiple of 10. So, the key here is to look at the remainders when numbers are divided by 10. That is, their residues modulo 10.Each number from 1 to 100 can be categorized by its remainder when divided by 10. The possible remainders are 0, 1, 2, ..., 9. So, there are 10 residue classes. If I can pair numbers from complementary residues such that their sum is a multiple of 10, that should work. For instance, a number with remainder 1 and another with remainder 9 (since 1 + 9 = 10), a number with remainder 2 and another with remainder 8, etc. The remainder 0 is self-complementary because 0 + 0 = 0, which is a multiple of 10. Similarly, remainder 5 pairs with itself since 5 + 5 = 10. So, the pairs would be (0,0), (1,9), (2,8), (3,7), (4,6), and (5,5). Each of these pairs sums to a multiple of 10. Therefore, to avoid having two numbers that sum to a multiple of 10, we must not select both numbers from any of these pairs. So, how many numbers can we select without having such a pair? Let's think about each residue class. For residues 1 and 9, we can choose numbers from one of these classes but not both. Similarly for 2 and 8, 3 and 7, 4 and 6. For residue 0 and 5, we can have at most one number each because selecting two from 0 or two from 5 would give a pair summing to a multiple of 10.Therefore, to maximize the number of numbers selected without any pair summing to 10, we can take:- From each complementary pair (1,9), (2,8), (3,7), (4,6): choose all numbers from one residue in the pair. For example, choose all numbers ≡1 mod 10 and none from ≡9 mod 10, or vice versa. Similarly for the others.- From residue 0: at most one number.- From residue 5: at most one number.So, how many numbers are there in each residue class from 1 to 100?Each residue from 0 to 9 occurs exactly 10 times in the numbers 1 to 100. Wait, let's check. For residues 1 to 9: numbers 1-9, 11-19, ..., 91-99. Each residue occurs 10 times. The number 100 is ≡0 mod 10. So, residue 0 occurs 10 times (10, 20, ..., 100). Similarly, residue 5 occurs 10 times (5, 15, ..., 95). Therefore, if we are to select the maximum number of numbers without any two summing to a multiple of 10, we need to pick the larger size between each complementary pair. For example, between residues 1 and 9, both have 10 numbers each. So, we can pick all 10 from residue 1 or all 10 from residue 9. Similarly for the other pairs. So, for each of the four pairs (1,9), (2,8), (3,7), (4,6), we can take 10 numbers each. Then, from residue 0 and residue 5, we can take at most 1 each. So, total maximum numbers without any pair summing to 10 would be 4*10 + 1 + 1 = 42 numbers.Wait, let me verify that. For each of the four complementary pairs, we can take 10 numbers from one residue in the pair. So, 4 pairs * 10 numbers = 40. Then, from residue 0 and residue 5, we can take at most 1 each. So, 40 + 1 + 1 = 42. So, 42 numbers can be selected without having any two that sum to a multiple of 10. Therefore, if we select one more number, that is 43, we must have at least two numbers whose sum is a multiple of 10. But wait, let's make sure. If we have 42 numbers, which is the maximum possible without any pair summing to 10, then adding one more number would force us to have a pair. Because the next number must belong to one of the residue classes we've already selected from, or to the residue 0 or 5. But let's check the residues. Suppose we have selected all numbers from residues 1,2,3,4 (and none from 9,8,7,6 respectively). Then residue 0 and 5 each have 1 number. So, total is 10+10+10+10 +1+1=42. If we add another number, it can be from residues 0,5,5,6,7,8,9. Wait, residues 5 and 0 already have one each, so adding another to residue 0 or 5 would give two in that residue, which would sum to a multiple of 10. Alternatively, adding a number from residue 6 would pair with residue 4 (since 6+4=10), but we already have all numbers from residue 4. Similarly, adding a number from residue 7 would pair with residue 3, which we have all of. So, yes, any additional number would create a pair that sums to a multiple of 10.Therefore, the minimal number needed is 43. So, the answer should be 43.Wait, but let me check again. The total numbers in the four complementary pairs are 10 each. So, by selecting 10 from each of four pairs (but only one side of each pair), we have 40 numbers. Then, residue 0 and 5 have 10 numbers each, but we can take only one from each. So, 40 +1 +1 =42. Hence, 42 numbers can be selected without any two summing to 10. Therefore, selecting 43 numbers must force at least two numbers to sum to 10.Therefore, the answer is 43. But let me verify with an example. Suppose we have 42 numbers: 10 numbers from each of residues 1,2,3,4, and one number from residue 0 and one from residue 5. Then, none of these numbers sum to a multiple of 10. Because 1+9=10, but there's no 9s; 2+8=10, no 8s; 3+7=10, no 7s; 4+6=10, no 6s; residue 0 has only one number, so no two 0s; residue 5 has only one number, so no two 5s. So, indeed, no pairs. Then, adding one more number. If that number is from residue 0, we have two 0s, which sum to 0, a multiple of 10. If it's from residue 5, same thing. If it's from residue 6, then 6 pairs with 4 (since 6+4=10), but we already have all 4s. Similarly, residue 7 pairs with 3, which we have; residue 8 pairs with 2; residue 9 pairs with 1. So, adding any number from residues 5,0,6,7,8,9 would create a pair. But since in our original selection we have all residues 1,2,3,4, and one each of 0 and 5, the next number has to be from one of the other residues, which would create a pair. Therefore, 43 numbers must contain such a pair.But wait, hold on, the residues 0 and 5 each have 10 numbers in the original 1-100. So, when we take one number from residue 0 and one from residue 5, there are still 9 numbers left in residue 0 and 9 in residue 5. So, if we take another number from residue 0 or 5, that would give two numbers in that residue, which sum to a multiple of 10. Similarly, if we take a number from residue 6, which we haven't taken any, but residue 4 is already all taken. So, 6 +4=10, and since residue 4 is all taken, any number from residue 6 would pair with residue 4. Similarly for 7 and 3, 8 and 2, 9 and 1.Therefore, 43 numbers would indeed create a pair. Therefore, the minimal number is 43. So, the answer is boxed{43}.But hold on, wait. Let me think once more. Let me check how many numbers are in each residue class. From 1 to 100:- Numbers congruent to 0 mod 10: 10, 20, ..., 100. That's 10 numbers.- Similarly, each residue from 1 to 9 occurs exactly 10 times: for example, residue 1: 1,11,21,...,91 (10 numbers), same for others.So, each residue has 10 numbers. So, if we take one from residue 0 and one from residue 5, and 10 each from residues 1,2,3,4, that's 10*4 +1+1=42. Then, any additional number must come from residues 5,0,6,7,8,9. But residues 5 and 0 already have one each. If the next number is from residue 0 or 5, we have two in that residue, summing to a multiple of 10. If it's from 6,7,8,9, then since residues 4,3,2,1 are already fully included, their sums with 6,7,8,9 would be 10,10,10,10 respectively. So, for example, 6 +4=10, but we already have all 4s. So, any number from residue 6 would pair with an existing 4. Similarly for 7 and 3, 8 and 2, 9 and 1. Therefore, yes, the 43rd number would create such a pair.Therefore, the minimal number is 43. So, the answer should be 43.But let me check another way. Suppose we use the Pigeonhole Principle. The number of pairs that sum to 10 is... Wait, but maybe another approach. If we consider the possible residues, and how they can combine.Alternatively, think of the problem as forming pairs of numbers that sum to 10. There are 50 such pairs in 1-100? Wait, no. For each residue pair (1,9), (2,8), etc., there are 10 numbers in each residue. So, the number of pairs that sum to 10 would be combinations between these residues. For example, between residue 1 and residue 9, there are 10*10=100 possible pairs. But perhaps that's complicating.Alternatively, the problem is similar to ensuring that in the selected subset, there are two numbers a and b such that (a + b) mod 10 = 0. So, the question is, what's the minimal number to ensure that such a pair exists.To avoid such a pair, the subset must not contain both a number and its complement modulo 10. So, the maximum size of such a subset is the sum of the maximum sizes of each complementary pair without overlapping. For residues 1 and 9, the maximum is 10 (all from one residue), same for 2 and 8, etc. For residues 0 and 5, the maximum is 1 each. So, total maximum is 10+10+10+10+1+1=42. Therefore, 42 is the maximal size without such a pair. Therefore, 43 is the minimal number needed to guarantee a pair. Therefore, the answer is 43.Yes, that seems consistent. So, I think the answer is boxed{43}.Final Answerboxed{43}</think>To determine the minimum number of numbers that must be selected from the natural numbers 1 to 100 to ensure that there are definitely two numbers whose sum is a multiple of 10, we use the Pigeonhole Principle and consider the residues modulo 10.Each number from 1 to 100 can be categorized by its remainder when divided by 10, resulting in residues 0, 1, 2, ..., 9. We need to avoid selecting numbers from complementary residues that sum to a multiple of 10. The complementary pairs are (0,0), (1,9), (2,8), (3,7), (4,6), and (5,5).To maximize the number of selected numbers without any pair summing to 10:- For each complementary pair (1,9), (2,8), (3,7), (4,6), we can select all 10 numbers from one residue in each pair.- For residues 0 and 5, we can select at most 1 number each.Thus, the maximum number of numbers that can be selected without any pair summing to 10 is:[ 10 + 10 + 10 + 10 + 1 + 1 = 42 ]Selecting one more number (43 in total) will force at least two numbers to be from complementary residues, ensuring their sum is a multiple of 10. Therefore, the answer is:[boxed{43}],<think>Okay, let me try to figure out this problem. So, the question is: From the natural numbers 1 to 100, how many numbers do I need to select at least to ensure that there are definitely two numbers whose sum is a multiple of 10? Hmm, interesting. I need to use the pigeonhole principle here, probably. Let me recall that the pigeonhole principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. So, applying this idea to the problem, I need to figure out the "pigeonholes" and the "pigeons" in this context.First, maybe I should consider the remainders when numbers are divided by 10. Because if two numbers add up to a multiple of 10, their remainders when divided by 10 must add up to 10. For example, if one number leaves a remainder of 3, the other should leave a remainder of 7, since 3 + 7 = 10. Similarly, remainders 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5. So, these are the pairs that sum up to 10. Also, 0 and 0, since 0 + 0 = 0, which is a multiple of 10.So, let's categorize the numbers from 1 to 100 based on their remainder when divided by 10. The possible remainders are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These are the pigeonholes. Each number will fall into one of these categories. Now, if we can ensure that two numbers are in complementary pairs (i.e., their remainders add up to 10) or both have remainder 0, then their sum will be a multiple of 10.Now, let's see how many numbers are in each remainder category from 1 to 100.For remainder 0: The numbers are 10, 20, 30, ..., 100. That's 10 numbers.For each of the other remainders (1 through 9), since 100 divided by 10 is 10, each remainder from 1 to 9 will also have 10 numbers. For example, remainder 1: 1, 11, 21, ..., 91; that's 10 numbers. Similarly for the others.So, the counts are:- Remainder 0: 10 numbers- Remainder 1: 10 numbers- Remainder 2: 10 numbers- ...- Remainder 9: 10 numbersNow, the key is to figure out the maximum number of numbers we can select without having any two that sum to a multiple of 10. Then, the minimum number needed to guarantee such a pair would be one more than that maximum.So, let's try to build the largest possible set without any two numbers summing to a multiple of 10. To do this, we need to avoid selecting numbers from both a remainder pair (like 1 and 9, 2 and 8, etc.), and also avoid selecting more than one number from remainder 0.First, let's look at the pairs:- (1,9)- (2,8)- (3,7)- (4,6)- (5,5)- (0,0)But since we can't have two numbers from the same pair. For example, if we pick a number with remainder 1, we can't pick any number with remainder 9, and vice versa. Similarly, for 2 and 8, etc. For the pair (5,5), since both numbers would need to have remainder 5, so we can have at most one number from remainder 5. Wait, actually, no. If we have two numbers with remainder 5, their sum would be 10, which is a multiple of 10. So, actually, even two numbers with remainder 5 would sum to 10, which is a multiple of 10. So, we can have at most one number from remainder 5 as well.Similarly, for remainder 0, if we have two numbers with remainder 0, their sum is 0 + 0 = 0, which is a multiple of 10. So, we can have at most one number from remainder 0.Therefore, to construct the largest set without any two numbers summing to a multiple of 10, we need to pick numbers such that:- From each complementary pair of remainders (1-9, 2-8, 3-7, 4-6), we pick numbers from only one remainder in each pair.- From remainder 5, we can pick at most one number.- From remainder 0, we can pick at most one number.So, let's see. There are 4 pairs: (1,9), (2,8), (3,7), (4,6). For each pair, we can choose all numbers from one remainder or the other. Since each remainder in a pair has 10 numbers, choosing all from one side gives 10 numbers per pair. For example, if I choose all numbers with remainder 1 (10 numbers) and none from remainder 9, then that's safe. Similarly for the other pairs.For the remainder 5, we can pick at most 1 number. So, instead of 10, we can only take 1.For remainder 0, same thing, at most 1 number.So, calculating the maximum possible numbers without any two summing to a multiple of 10:4 pairs * 10 numbers each = 40 numbersPlus 1 number from remainder 5 = 1Plus 1 number from remainder 0 = 1Total = 40 + 1 + 1 = 42 numbers.Wait, that's 42 numbers. Therefore, if we pick 42 numbers, it's possible that we have 10 from each of 4 pairs, 1 from remainder 5, and 1 from remainder 0. But wait, but each pair (1,9), etc., are 10 numbers each. Let me verify.Each pair (like 1 and 9) has 10 numbers in each remainder. So, if I choose to take all numbers from remainder 1 (10 numbers) and none from 9, then that's 10. Similarly, if I take all from remainder 2 (10 numbers) and none from 8, that's another 10. Then, 3 and 7: take all from 3, none from 7, another 10. 4 and 6: take all from 4, none from 6, another 10. Then, from remainder 5: 1 number. From remainder 0: 1 number. So, 10 + 10 + 10 + 10 + 1 + 1 = 42. Okay, that seems correct.Therefore, 42 numbers can be selected without having any two numbers sum to a multiple of 10. So, if we select 43 numbers, then by the pigeonhole principle, we must have at least two numbers that sum to a multiple of 10. Because there's no way to have 43 numbers without violating one of the maximums. So, 43 is the minimum number needed to ensure that such a pair exists.Wait, but let me double-check. Let's think of the pigeonholes here. The pigeonholes are the different cases where adding two numbers would result in a multiple of 10. But how exactly does the pigeonhole principle apply here?Alternatively, we can think of the pairs as the pigeonholes. Wait, maybe not. The way we constructed the problem is by considering the maximum set without the desired property, so the size of that set plus one is the minimal number needed to guarantee the property.So, since 42 is the maximum size of a set without any two numbers summing to a multiple of 10, then 43 numbers must contain at least two numbers that sum to a multiple of 10.Therefore, the answer should be 43. But let me check again to make sure.Alternatively, maybe there's a different way to categorize the numbers. Let's see.Another approach: consider the residues mod 10. Each number can be categorized by its remainder when divided by 10. As we did earlier, we have residues 0 through 9.To avoid having two numbers that sum to a multiple of 10, we need to ensure that:- No two numbers are in residue pairs that add up to 10. That is, for residues (1,9), (2,8), (3,7), (4,6), (5,5), (0,0).So, if we pick numbers such that we don't pick both residues from any of these pairs.For the pairs (1,9), (2,8), (3,7), (4,6), each pair consists of two different residues, each with 10 numbers. So, for each of these four pairs, we can pick at most 10 numbers from one residue in the pair. For the residue pair (5,5), since it's the same residue, we can pick at most 1 number. Similarly, for residue pair (0,0), we can pick at most 1 number.Therefore, maximum size is:4 pairs * 10 numbers each (from one side of the pair) = 40Plus 1 number from residue 5 = 1Plus 1 number from residue 0 = 1Total = 42. So, same as before.Thus, 42 numbers is the largest set without any two numbers summing to 10. Therefore, 42 + 1 = 43 is the minimal number to ensure such a pair exists.But let me check if there is any possibility of having a larger set. Suppose instead of taking 10 numbers from each pair, maybe some pairs have less, but others have more? Wait, but since each pair has 10 numbers in each residue, taking all 10 from one side gives the maximum. If you take less, then the total would be smaller. For example, taking 9 from residue 1 and 9 from residue 9 would lead to 18, but that would create pairs that sum to 10. Wait, no, if you take numbers from both residues in a pair, then their sum would be 10. So, to avoid that, you must choose all numbers from only one residue in the pair. Therefore, maximum you can take from each pair is 10. So, 4 pairs * 10 = 40. Then 1 from 5, 1 from 0. Total 42. So, 42 is indeed the maximum.Therefore, the answer should be 43. But wait, let me confirm with an example. Suppose I take 10 numbers with remainder 1, 10 with remainder 2, 10 with remainder 3, 10 with remainder 4, 1 with remainder 5, and 1 with remainder 0. That's 10+10+10+10+1+1=42. None of these numbers when added together give a multiple of 10. Because, for example, adding a remainder 1 and remainder 2 gives 3, which is not 0 mod 10. Adding remainder 1 and remainder 1 gives 2, etc. Similarly, remainder 5 added to any of the others: 5+1=6, 5+2=7, ..., 5+4=9, 5+5=10 (but there's only one 5). 0 added to anything: 0+1=1, ..., 0+4=4, 0+5=5, and 0+0=0 (but only one 0). So, there's no pair in this set that adds to a multiple of 10. Therefore, 42 is indeed possible. Therefore, 43 is the minimal number that guarantees such a pair.But wait, another thought: the numbers from 1 to 100 include 100 numbers. If we pick 43 numbers, then according to the pigeonhole principle, we must have some pair. Wait, but maybe there's a different approach where you can have more than 42 numbers without such a pair. But according to our previous reasoning, 42 is the maximum. Let me check with a smaller case.Suppose we have numbers from 1 to 10. Then, how many numbers do we need to pick to ensure that two sum to 10? Let's see. The pairs are (1,9), (2,8), (3,7), (4,6), (5,5), (10,10). So, similar logic. The maximum set without pairs would be taking one from each pair. Wait, in 1 to 10:- 1 and 9: pick one- 2 and 8: pick one- 3 and 7: pick one- 4 and 6: pick one- 5: pick one- 10: pick oneBut in numbers 1 to 10, there's only one number for each remainder. So, to avoid pairs, you can pick at most one from each pair. So, maximum 5 +1 (for 5) +1 (for 10) = 7? Wait, but in numbers 1 to 10:If you pick 1, 2, 3, 4, 5, 10, that's 6 numbers. Any two of these don't sum to 10. For example, 1+2=3, etc. If you pick 1, 2, 3, 4, 5, 6, 10, that's 7. Wait, 4 and 6 sum to 10. So, actually, in this smaller case, maximum is 6. Hmm. So, applying the same logic as before, for the pairs (1,9), (2,8), (3,7), (4,6), each pair in 1-10 has only one number. So, if we pick 1,2,3,4, and avoid their pairs (9,8,7,6). But since 9,8,7,6 are separate numbers. Wait, maybe in 1-10, the maximum set without any pair summing to 10 is 5 numbers: one from each pair. Wait, 1 and 9 are a pair. If you pick 1, you can't pick 9. So, in 1-10, you can pick 1,2,3,4,5,10. Wait, but 5 and 10: 5 + 5 isn't there (only one 5), and 10 + 10 isn't there (only one 10). So, 6 numbers. But 1,2,3,4,5,10: none of these sum to 10. So, 6 numbers. Then, if you pick 7 numbers, you have to have a pair. Indeed, in 1-10, 7 numbers must include two that sum to 10. So, the answer would be 7. But according to the formula, 4 pairs (each pair has 1 element in 1-10), so maximum would be 4 (from pairs) +1 (from 5) +1 (from 0) = 6. Then 6+1=7. So, similar logic applies.Therefore, in the case of 1-100, following the same logic, 42 +1 =43.So, seems like the answer is 43. Let me check another example. Suppose I have numbers from 1 to 20. Then, similar logic. Each remainder 0-9 has two numbers. So, for pairs (1,9), (2,8), etc. To avoid pairs, you can take two numbers from each of four remainders (e.g., 1,2,3,4), one number from remainder 5, and one number from remainder 0. Then total is 2*4 +1 +1= 10. Then, 10+1=11 would force a pair. Let's see. Wait, in 1-20, remainder 0 has two numbers (10 and 20). If I pick one of them, say 10. Then, remainders 1 has two numbers (1,11), 2 has (2,12), etc. If I pick both numbers from remainder 1 (1 and 11), then if I also pick any number from remainder 9 (9 or 19), their sum would be 1+9=10, 1+19=20, 11+9=20, 11+19=30. So, if I pick both from remainder 1, then I can't pick any from remainder 9. Wait, so to maximize, for each pair, pick all from one side. For example, pick both numbers from remainder 1 (2 numbers), none from 9. Similarly, both from 2, none from 8. Both from 3, none from 7. Both from 4, none from 6. Then, one from 5, one from 0. Total: 2*4 +1 +1= 10. So, 10 numbers. Then, the next number (11th) would have to come from either a remainder already chosen (but if you have two from each, you can't have another). Wait, in 1-20, if you take 2 from 1, 2 from 2, 2 from 3, 2 from 4, 1 from 5, 1 from 0, that's 10 numbers. The next number would have to be from remainder 5, 0, or from the other remainders (9,8,7,6). But if you pick from 9,8,7,6, then those pair with 1,2,3,4. So, if you already have two numbers in, say, remainder 1, adding a number from remainder 9 would create a sum with each of the two in remainder 1. For example, 1 +9=10, 11 +9=20. So, both sums are multiples of 10. Therefore, adding any number from the complementary remainders would create a pair. Similarly, adding another number from remainder 0 would create 10 +10=20. So, if you have two zeros, their sum is 20. But in 1-20, there are two zeros: 10 and 20. If you already have one (say 10), adding another (20) would make 10+20=30, which is a multiple of 10. So, yes, same thing.Therefore, in 1-20, 10 numbers can be picked without any pair summing to 10, and 11th would force such a pair. So, applying the same logic, it seems the formula holds.Therefore, returning to the original problem: 100 numbers, grouped into remainders 0-9, each with 10 numbers. The maximum set without any pair summing to 10 is 4*10 +1 +1=42. Therefore, the answer is 42 +1=43.But let me check once again for possible errors. Suppose we have 42 numbers: 10 from each of four pairs (say remainders 1,2,3,4), 1 from 5, and 1 from 0. Then, in this set, there are 10+10+10+10+1+1=42 numbers. None of them sum to a multiple of 10 because:- Numbers from remainder 1: their sums with others would be 1+1=2, 1+2=3, ..., 1+4=5, 1+5=6, 1+0=1. None are multiples of 10.- Similarly, numbers from remainder 2: 2+2=4, 2+3=5, ..., 2+5=7, 2+0=2.- Numbers from remainder 3: 3+3=6, ..., 3+5=8, 3+0=3.- Numbers from remainder 4: 4+4=8, 4+5=9, 4+0=4.- The number from remainder 5: 5 + any other remainder: 5+1=6, ..., 5+4=9, 5+0=5.- The number from remainder 0: 0 + any other remainder: same as the other remainders.Therefore, indeed, no two numbers in this set sum to a multiple of 10. Therefore, 42 is possible. Therefore, 43 is the minimal number needed.Therefore, the answer is 43.Final Answerboxed{43}</think>To determine the minimum number of numbers that must be selected from the natural numbers 1 to 100 to ensure that there are definitely two numbers whose sum is a multiple of 10, we use the pigeonhole principle and consider the remainders when numbers are divided by 10.1. Categorize Numbers by Remainders: - The possible remainders when divided by 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. - Each remainder category (0 through 9) contains 10 numbers.2. Pairs of Remainders: - Two numbers sum to a multiple of 10 if their remainders are complementary (e.g., 1 and 9, 2 and 8, etc.) or both are 0. - The pairs are (0, 0), (1, 9), (2, 8), (3, 7), (4, 6), and (5, 5).3. Maximizing the Set Without Complementary Pairs: - For each complementary pair (1-9, 2-8, 3-7, 4-6), we can select all numbers from one remainder in each pair (10 numbers each). - For the pair (5, 5), we can select at most 1 number. - For the remainder 0, we can select at most 1 number.4. Calculating the Maximum Set Size: - 4 pairs * 10 numbers each = 40 numbers. - 1 number from remainder 5. - 1 number from remainder 0. - Total = 40 + 1 + 1 = 42 numbers.5. Applying the Pigeonhole Principle: - If we select 42 numbers, it is possible to avoid any pair summing to a multiple of 10. - Selecting one more number (43 in total) ensures that at least two numbers will form a pair whose sum is a multiple of 10.Thus, the minimum number of numbers that must be selected to ensure there are definitely two numbers whose sum is a multiple of 10 is (boxed{43}).]

🔑:Answer: 43

🧠:[<think>Okay, let's tackle this math problem. So I need to calculate (56 × 0.57 × 0.85) divided by (2.8 × 19 × 1.7). Hmm, that looks a bit complicated with all those numbers, but maybe there's a way to simplify it before diving into the multiplication and division. Let me think...First, I remember that when you have a fraction like this with multiplication in both the numerator and denominator, you can sometimes cancel out common factors or rearrange the numbers to make the calculation easier. Maybe breaking down each number into its prime factors would help? Or perhaps converting the decimals to fractions? Let me try both approaches and see which one works better.Starting with the numerator: 56 × 0.57 × 0.85.56 is a whole number, so that's straightforward. 0.57 and 0.85 are decimals. Converting them to fractions might be useful. Let's see:0.57 is 57/100, and 0.85 is 85/100. So the numerator becomes:56 × (57/100) × (85/100). Similarly, the denominator is 2.8 × 19 × 1.7. Let's convert those decimals as well:2.8 is 28/10, and 1.7 is 17/10. So the denominator becomes (28/10) × 19 × (17/10).So the entire expression is:[56 × (57/100) × (85/100)] / [(28/10) × 19 × (17/10)]Hmm, that looks a bit messy, but maybe we can rearrange the fractions. Let me rewrite the division as multiplication by the reciprocal:[56 × (57/100) × (85/100)] × [(10/28) × (1/19) × (10/17)]Now, let's see if we can cancel out some terms here. Let's look for common factors between numerators and denominators.First, notice that 56 and 28 are in the numerator and denominator respectively. 56 divided by 28 is 2. So we can simplify 56/28 to 2/1. Let me note that:56 × (10/28) = (56/28) × 10 = 2 × 10 = 20. Wait, but hold on, in the current setup, the 56 is in the first part of the numerator, and the 28 is in the denominator's first part. So when we multiply them, it's 56 × (10/28) = (56/28) × 10 = 2 × 10 = 20. That's correct.So replacing 56 × (10/28) with 20. Then moving on.Next, looking at 57 and 19. 57 divided by 19 is 3. So 57 in the numerator and 19 in the denominator can be simplified. Let's see:(57/100) × (1/19) = (57/19) × (1/100) = 3 × (1/100) = 3/100.So that's another simplification. So far, combining the two simplifications:We have 20 (from 56×10/28) multiplied by 3/100 (from 57×1/19) multiplied by the remaining terms.The remaining terms in the numerator are 85/100 from the original numerator and 10/17 from the denominator. So let's look at 85 and 17. 85 divided by 17 is 5. So:(85/100) × (10/17) = (85/17) × (10/100) = 5 × (1/10) = 5/10 = 1/2.So putting it all together:20 × (3/100) × (1/2)Let me compute that step by step.First, 20 multiplied by 3 is 60. Then 60 divided by 100 is 0.6. Then 0.6 multiplied by 1/2 is 0.3.Wait, so is the final result 0.3? Let me verify that because that seems straightforward, but I need to make sure I didn't skip any steps.Let me retrace:Original expression:(56 × 0.57 × 0.85) / (2.8 × 19 × 1.7)Converted decimals to fractions:Numerator: 56 × 57/100 × 85/100Denominator: 28/10 × 19 × 17/10Then rewriting as multiplication by reciprocal:56 × 57/100 × 85/100 × 10/28 × 1/19 × 10/17Group terms:(56 × 10/28) × (57 × 1/19) × (85 × 10/17) × (1/100 × 1/100)Wait, but in my earlier steps, I had different groupings, but perhaps this way is clearer.Calculating each group:56 × 10/28: 56 divided by 28 is 2, so 2 ×10=20.57 × 1/19: 57 divided by19 is3.85 ×10/17: 85 divided by17 is5, 5×10=50.Then the remaining fractions: 1/100 ×1/100=1/10,000? Wait, hold on. Wait, the original fractions were 1/100 and 1/100 in the numerator (from 0.57 and 0.85) and in the denominator we had 10/10 from the 2.8 and 1.7. Wait, perhaps I need to check again.Wait, let me clarify:Original numerator fractions: 57/100 and 85/100.Original denominator fractions: 28/10 and 17/10.So when we take reciprocal, it's 10/28 and 10/17. Then the remaining terms:In the numerator, we have 56, 57, 85.In the denominator, we have 28, 19, 17.But when written as multiplication by reciprocal, it's:56 × 57/100 × 85/100 × 10/28 × 1/19 × 10/17.So grouping terms:(56 ×10/28) × (57 ×1/19) × (85 ×10/17) × (1/100 ×1/100)Wait, but the 10s in the reciprocal come from converting 28/10 to 10/28 and 17/10 to 10/17. So each of those denominators contributes a 10 in the reciprocal numerator. So those are two separate 10s. So the total 10s from the reciprocal are 10 ×10=100. So in total, the 10s from the reciprocal multiply to 100.But the original fractions from the numerator (57/100 and 85/100) contribute 100 ×100=10,000 in the denominator. So overall, the denominators are 10,000 and the numerators from the reciprocal are 100, so 100/10,000=1/100. Wait, this might not be the right way to think about it.Alternatively, let's track each component:Starting with 56 × (57/100) × (85/100) × (10/28) × (1/19) × (10/17)Let me rearrange the multiplication:(56 × 10/28) × (57 × 1/19) × (85 ×10/17) × (1/100 ×1/100)So:First group: 56 ×10 /28 = (56/28)×10 = 2×10=20Second group:57 ×1 /19 =57/19=3Third group:85 ×10 /17= (85/17)×10=5×10=50Fourth group:1/(100×100)=1/10,000So combining all these:20 ×3 ×50 ×1/10,000Calculating 20×3=6060×50=3,0003,000 ×1/10,000=3,000/10,000=0.3So yes, that's 0.3. So that seems correct.Alternatively, if I do the calculation directly without simplifying:First calculate numerator:56×0.57×0.8556×0.57: Let's compute that.56×0.5=28, 56×0.07=3.92, so 28+3.92=31.92Then 31.92×0.85: Let's compute that.31.92×0.8=25.536, 31.92×0.05=1.596, so total 25.536+1.596=27.132So numerator is 27.132Denominator:2.8×19×1.7First compute 2.8×19: 2×19=38, 0.8×19=15.2, so total 38+15.2=53.2Then 53.2×1.7: 50×1.7=85, 3.2×1.7=5.44, so total 85+5.44=90.44So denominator is 90.44Then the result is 27.132 /90.44Let me compute that division.27.132 ÷90.44. Hmm, since 27.132 is less than 90.44, the result is less than 1. Let's convert this to a fraction:27.132 /90.44. Multiply numerator and denominator by 1000 to eliminate decimals: 27132/90440.Simplify this fraction. Let's divide numerator and denominator by 4: 27132 ÷4=6783, 90440 ÷4=22610.So 6783/22610. Check if they have a common divisor. Let's see:Divide numerator and denominator by 7: 6783 ÷7=969, 22610 ÷7≈3230. Hmm, 22610 ÷7=3230 (7×3230=22610). So 969/3230.Check if 969 and 3230 have a common divisor. Let's see:969 ÷3=323, 3230 ÷3≈1076.66, not integer. So 3 is a divisor of numerator but not denominator.Check GCD(969,3230). Let's apply Euclidean algorithm.3230 ÷969=3 times (969×3=2907), remainder 3230-2907=323.Then GCD(969,323). 969 ÷323=3 times (323×3=969), remainder 0. So GCD is 323.Therefore, divide numerator and denominator by 323.969 ÷323=3, 3230 ÷323=10. So simplified to 3/10=0.3.So the result is 3/10, which is 0.3. So both methods give the same answer. That's reassuring.So whether I simplify first or compute directly and then simplify the fraction, I end up with 0.3. Therefore, the answer should be 0.3, or 3/10. Since the question doesn't specify the form, but given the original numbers have decimals, 0.3 is appropriate. However, to check, maybe the problem expects a fractional answer. 0.3 is equal to 3/10, so both are correct. But the problem includes decimals in the original expression, so 0.3 is fine. However, let me confirm once again.Alternatively, in the first method, after simplifying all the terms, we had 20 × 3 × 50 /10,000. Wait, 20×3=60, 60×50=3,000. 3,000 divided by 10,000 is 0.3. Correct.Another way: Let's write all the numbers as fractions over 1.Numerator:56/1 × 57/100 × 85/100Denominator:28/10 × 19/1 × 17/10So when dividing, multiply by reciprocal:56/1 × 57/100 × 85/100 × 10/28 × 1/19 ×10/17Now, rearrange the fractions:(56/28) × (57/19) × (85/17) × (10/100) × (10/100) × (1/1) × (1/1)Simplify each part:56/28 = 257/19 = 385/17 = 510/100 = 1/10, and another 10/100 =1/10So multiplying these together:2 ×3 ×5 × (1/10) × (1/10) = 30 × (1/100) = 30/100 = 0.3Perfect, same result. So regardless of the method, the answer is 0.3.Therefore, the final answer is 0.3, or in boxed form, boxed{0.3}Final Answerboxed{0.3}</think>To calculate ((56 times 0.57 times 0.85) div (2.8 times 19 times 1.7)), we can simplify the expression by converting decimals to fractions and rearranging terms.First, convert the decimals to fractions:- (0.57 = frac{57}{100})- (0.85 = frac{85}{100})- (2.8 = frac{28}{10})- (1.7 = frac{17}{10})Rewrite the original expression as:[frac{56 times frac{57}{100} times frac{85}{100}}{frac{28}{10} times 19 times frac{17}{10}}]Convert the division to multiplication by the reciprocal:[56 times frac{57}{100} times frac{85}{100} times frac{10}{28} times frac{1}{19} times frac{10}{17}]Simplify by grouping and canceling common factors:1. (56 times frac{10}{28} = 2 times 10 = 20)2. (frac{57}{19} = 3)3. (frac{85}{17} times 10 = 5 times 10 = 50)4. Combine the remaining fractions: (frac{1}{100} times frac{1}{100} = frac{1}{10000})Multiply the simplified terms:[20 times 3 times 50 times frac{1}{10000} = frac{3000}{10000} = 0.3]Thus, the final answer is:[boxed{0.3}],<think>Alright, let me try to figure out this problem. So, the question is to calculate (56 × 0.57 × 0.85) divided by (2.8 × 19 × 1.7). Hmm, okay. Let's break this down step by step. First, maybe I should handle the numerator and the denominator separately before dividing them. That might make it easier. Let's start with the numerator: 56 multiplied by 0.57 multiplied by 0.85. Then the denominator is 2.8 multiplied by 19 multiplied by 1.7. Hmm. But before jumping into multiplication, maybe there's a way to simplify this expression by canceling out common factors or using fractions. That could save some time instead of multiplying all those decimals. Let me think.Looking at the numbers in the numerator and denominator, I see 56 in the numerator and 2.8 in the denominator. 56 divided by 2.8 is something... Let me check. 2.8 times 20 is 56, right? Because 2.8 times 10 is 28, so times 20 is 56. So 56 divided by 2.8 is 20. That's a nice whole number. So maybe I can simplify 56/2.8 first. Let me write that down.So, 56 ÷ 2.8 = 20. That's straightforward. So that simplifies part of the problem. Then, moving on to the next numbers. The numerator has 0.57 and the denominator has 19. 0.57 and 19. Hmm, 19 is a prime number, but 0.57 is 57/100. So, maybe 57 and 19 have a common factor. 57 divided by 19 is 3. Exactly. So 57 ÷ 19 = 3. Therefore, 0.57 ÷ 19 would be (57/100) ÷ 19 = 57/(100×19) = 3/100. Because 57 ÷ 19 is 3. So that simplifies 0.57 ÷ 19 to 0.03. Wait, let me confirm that. If I have 0.57 divided by 19, that's the same as 0.57 times (1/19). Since 57 is 19×3, so 0.57 is 19×3×0.01. Therefore, 0.57 ÷ 19 would be 3×0.01, which is 0.03. Yes, that's correct. So 0.57 ÷ 19 = 0.03. So that simplifies another part. Then, the remaining numbers are 0.85 in the numerator and 1.7 in the denominator. 0.85 divided by 1.7. Let's see. 1.7 goes into 0.85 how many times? Well, 1.7 times 0.5 is 0.85. Because 1.7 times 0.5 is half of 1.7, which is 0.85. So 0.85 ÷ 1.7 = 0.5. So putting all those simplifications together:(56/2.8) × (0.57/19) × (0.85/1.7) = 20 × 0.03 × 0.5.Wait, hold on. Let me make sure. The original expression is (56 × 0.57 × 0.85) divided by (2.8 × 19 × 1.7). So, that can be rewritten as (56/2.8) × (0.57/19) × (0.85/1.7). Yes, that's correct. So each term in the numerator is divided by each term in the denominator. So, breaking them apart like that simplifies the calculation. So, 56 divided by 2.8 is 20. Then 0.57 divided by 19 is 0.03. Then 0.85 divided by 1.7 is 0.5. Then multiplying those together: 20 × 0.03 × 0.5.Let's compute that step by step. First, 20 × 0.03. 20 times 0.03 is 0.6. Because 20 × 0.01 is 0.2, so 0.03 is three times that, which is 0.6. Then, 0.6 × 0.5. That's half of 0.6, which is 0.3. So, the entire expression simplifies to 0.3. Wait, that seems straightforward, but let me verify to make sure I didn't make any mistakes in the steps. Let's check each division again.First division: 56 ÷ 2.8. Let me do this longhand. 2.8 goes into 56 how many times? 2.8 × 10 = 28. 28 × 2 = 56. So 2.8 × 20 = 56. So yes, 56 ÷ 2.8 = 20. Correct.Second division: 0.57 ÷ 19. As mentioned, 0.57 is 57/100. 57 ÷ 19 = 3, so 57/100 ÷ 19 = 3/100 = 0.03. Correct.Third division: 0.85 ÷ 1.7. 1.7 × 0.5 = 0.85. So yes, 0.85 ÷ 1.7 = 0.5. Correct.Then multiplying those results: 20 × 0.03 × 0.5. Let's do 20 × 0.03 first. 20 × 0.03 is 0.6. Then 0.6 × 0.5 is 0.3. So, yes, 0.3. Alternatively, I can rearrange the multiplication. Since multiplication is commutative, I can group them differently. For example, 20 × 0.5 first, which is 10, and then 10 × 0.03 is 0.3. Same result.So, the final answer is 0.3. Let me check once again by calculating numerator and denominator separately and then dividing.First, numerator: 56 × 0.57 × 0.85. Let's compute each step.56 × 0.57. Let's do 56 × 0.5 = 28, 56 × 0.07 = 3.92. So 28 + 3.92 = 31.92. Then 31.92 × 0.85. Hmm. Let's compute 31.92 × 0.8 = 25.536 and 31.92 × 0.05 = 1.596. Adding those gives 25.536 + 1.596 = 27.132. So numerator is 27.132.Denominator: 2.8 × 19 × 1.7. Let's compute 2.8 × 19 first. 2 × 19 = 38, 0.8 × 19 = 15.2. Adding those gives 38 + 15.2 = 53.2. Then 53.2 × 1.7. Compute 53.2 × 1 = 53.2, 53.2 × 0.7 = 37.24. Adding those gives 53.2 + 37.24 = 90.44. So denominator is 90.44.Now divide numerator by denominator: 27.132 ÷ 90.44. Let's see. Let me compute this division. 27.132 divided by 90.44. Hmm, 90.44 goes into 27.132 less than once. Multiply both numerator and denominator by 1000 to eliminate decimals: 27132 ÷ 90440. Simplify by dividing numerator and denominator by 4: 6783 ÷ 22610. Hmm, maybe not helpful. Let me do decimal division.Divide 27.132 by 90.44. Let me write it as 27.132 / 90.44. Let's approximate this. 90.44 × 0.3 = 27.132. Because 90.44 × 0.3: 90 × 0.3 = 27, 0.44 × 0.3 = 0.132. So 27 + 0.132 = 27.132. Exactly! So 27.132 divided by 90.44 is 0.3. So that confirms the answer is indeed 0.3. Therefore, both methods give the same result. The simplified way by breaking down each division step first and then multiplying gives 0.3, and calculating numerator and denominator separately and then dividing also gives 0.3. So that gives me confidence that the answer is correct.Another way to verify is to use fractions throughout, converting decimals to fractions. Let me try that approach to cross-check.First, convert all decimals to fractions:56 is 56/1.0.57 is 57/100.0.85 is 85/100 = 17/20.2.8 is 28/10 = 14/5.19 is 19/1.1.7 is 17/10.So the original expression is:(56/1 × 57/100 × 17/20) ÷ (14/5 × 19/1 × 17/10)Dividing by a fraction is multiplying by its reciprocal, so:(56 × 57 × 17) / (1 × 100 × 20) × (5 × 1 × 10) / (14 × 19 × 17)Now, let's write it as:(56 × 57 × 17 × 5 × 10) / (100 × 20 × 14 × 19 × 17)Notice that 17 appears in both numerator and denominator, so they cancel out:(56 × 57 × 5 × 10) / (100 × 20 × 14 × 19)Now, let's simplify each part:First, let's look for common factors.56 and 14: 56 ÷ 14 = 4. So 56 = 14 × 4. So replace 56 with 14 × 4:(14 × 4 × 57 × 5 × 10) / (100 × 20 × 14 × 19)Cancel out the 14 in numerator and denominator:(4 × 57 × 5 × 10) / (100 × 20 × 19)Now, 4 in numerator and 20 in denominator: 4/20 = 1/5. So replace 4 with 1 and 20 with 5:(1 × 57 × 5 × 10) / (100 × 5 × 19)Wait, let's see:Wait, denominator is 100 × 20 × 19. If we have 4 in numerator and 20 in denominator, 4/20 = 1/5. So 4 × (rest of numerator) divided by (20 × rest of denominator). So:(1 × 57 × 5 × 10) / (100 × 5 × 19)Wait, maybe that's a bit confusing. Let me reorganize:After canceling 14, the numerator is 4 × 57 × 5 × 10, denominator is 100 × 20 × 19. So, numerator: 4 × 57 × 5 × 10.Denominator: 100 × 20 × 19.Now, let's factor each number:Numerator:4 = 2²57 = 3 × 195 = 510 = 2 × 5Denominator:100 = 2² × 5²20 = 2² × 519 = 19So numerator factors: 2² × 3 × 19 × 5 × 2 × 5 = 2³ × 3 × 5² × 19Denominator factors: 2² × 5² × 2² × 5 × 19 = 2⁴ × 5³ × 19Now, divide numerator by denominator:(2³ × 3 × 5² × 19) / (2⁴ × 5³ × 19)Cancel out common factors:2³ / 2⁴ = 1/25² / 5³ = 1/519 /19 = 1So remaining factors: 3 / (2 × 5) = 3/10 = 0.3So that gives the same result, 0.3. So using fractions and prime factors also leads us to the answer 0.3. Therefore, all methods confirm that the result is 0.3. I think that's solid. I don't see any mistakes in the reasoning. So I'm confident that the answer is 0.3.Final Answerboxed{0.3}</think>To calculate (frac{56 times 0.57 times 0.85}{2.8 times 19 times 1.7}), we can simplify the expression by breaking it down into smaller parts and canceling common factors.First, simplify each division separately:1. ( frac{56}{2.8} = 20 )2. ( frac{0.57}{19} = 0.03 )3. ( frac{0.85}{1.7} = 0.5 )Next, multiply the results of these simplified divisions:[ 20 times 0.03 times 0.5 ]Calculating step by step:1. ( 20 times 0.03 = 0.6 )2. ( 0.6 times 0.5 = 0.3 )Thus, the final result is:[boxed{0.3}]]

🔑:Reference answer: 0.3Key points: Simplified calculation - adding and removing parentheses

🧠:[<think>Okay, so I need to figure out the values of m and n such that the monomials 3a^{m+1}b and -b^{n-1}a³ can be combined. The options are given, so maybe I can work through it step by step.First, I remember that monomials can only be combined if they are like terms. Like terms have the same variables raised to the same powers. So, the variables a and b in both terms need to have the same exponents for the terms to be combined. That means the exponent of a in the first term should equal the exponent of a in the second term, and similarly for b.Let me write down the exponents for each variable in both monomials.First monomial: 3a^{m+1}b. Let's break it down:- The coefficient is 3.- The variable a has an exponent of m + 1.- The variable b has an exponent of 1 (since it's written as b, which is b¹).Second monomial: -b^{n-1}a³. Let's break this one down too:- The coefficient is -1.- The variable b has an exponent of n - 1.- The variable a has an exponent of 3.Wait, even though the order of the variables is different (a first in the first term, b first in the second term), that doesn't matter because multiplication is commutative. So the order of variables doesn't affect whether they're like terms. The important part is the exponents for each variable.So, for the two monomials to be combined, the exponents of a must be equal in both terms, and the exponents of b must be equal in both terms.Therefore, setting the exponents equal:For variable a:Exponent in first term: m + 1Exponent in second term: 3So, m + 1 = 3For variable b:Exponent in first term: 1Exponent in second term: n - 1So, 1 = n - 1Now, solving these equations:From the a exponents:m + 1 = 3Subtract 1 from both sides:m = 3 - 1m = 2From the b exponents:1 = n - 1Add 1 to both sides:n = 1 + 1n = 2Wait, so m is 2 and n is 2? Let me check the answer options. The options are:A: 2, 2B: 3, 2C: 2, 0D: 3, 0So, according to this, the answer should be A: 2, 2. But let me double-check my work because sometimes it's easy to mix up exponents.First monomial: a^{m+1}b. So a's exponent is m+1, b's exponent is 1.Second monomial: -b^{n-1}a³. So a's exponent is 3, b's exponent is n-1.To combine them, exponents must match. Therefore:m + 1 = 3 → m = 2n - 1 = 1 → n = 2Yes, that seems correct. So m=2, n=2. So option A. Hmm, but let me check the options again. Wait, maybe I made a mistake in the variable order? Let me check the original problem again.The monomials are "3a^{m+1}b" and "-b^{n-1}a³". So yes, the first term has a^{m+1} times b, and the second term has b^{n-1} times a³. So exponents for a and b are as I considered. So m=2 and n=2. So answer A.But wait, the answer options given include C: 2, 0 and D:3,0, which makes me think maybe I did something wrong. Let me check again.Wait, maybe I misread the problem. Let me confirm:The problem states that the monomials can be combined, so they must be like terms. Therefore, their corresponding exponents must be equal.So, in the first term, 3a^{m+1}b, the exponents are (m+1) for a and 1 for b.In the second term, -b^{n-1}a³, the exponents are (n-1) for b and 3 for a.Therefore, to combine them, the exponents for a must be equal: m + 1 = 3, so m=2.The exponents for b must be equal: n - 1 = 1, so n=2.Therefore, m=2, n=2, which is answer A.But the options are A:2,2; B:3,2; C:2,0; D:3,0. So A is the correct answer. Wait, but maybe there's a trick here? Let me check the original problem again.Wait, the first monomial is 3a^{m+1}b. So that is 3 multiplied by a^(m+1) multiplied by b. The second monomial is -b^{n-1}a³. So that is -1 multiplied by b^(n-1) multiplied by a^3.So variables are a and b in both cases, with exponents:First term: a^(m+1), b^1Second term: a^3, b^(n-1)Therefore, for the terms to be like terms, the exponents of a must be equal, so m + 1 = 3, leading to m = 2. Similarly, exponents of b must be equal, so 1 = n - 1, leading to n = 2. So answer is A:2,2.But maybe I'm missing something? Let me verify with the answer choices. If the answer is A, then that's the case. But let's see why other options might be tempting.Option B: m=3, n=2. If m=3, then in the first term, the exponent of a would be 3 + 1 = 4, but in the second term, it's 3. Not equal. So exponents of a would differ. So that's not possible.Option C: m=2, n=0. If n=0, then the exponent of b in the second term is 0 -1 = -1. But in the first term, exponent of b is 1. So exponents of b would be 1 and -1. Not equal. So that's not possible.Option D: m=3, n=0. Then exponent of a in first term: 3+1=4 vs 3 in second term. Exponent of b: 1 vs 0-1=-1. Neither matches. So definitely not.Therefore, only option A works. Therefore, the answer is A.Wait, but the original problem says "can be combined". So if the exponents are equal, then they can be combined. So A is correct.But wait, hold on, maybe there's a different interpretation? For example, sometimes people might confuse coefficients with exponents? But no, the problem states that the exponents are m+1 and n-1.Alternatively, perhaps the variables were switched? Wait, in the first term it's a^{m+1}b, which is a^(m+1) * b^1. The second term is -b^{n-1}a³, which is b^{n-1} * a^3. So variables are in different order, but that doesn't matter because multiplication is commutative. So the variables are the same, and exponents for each variable must match. So exponents for a must be m+1 and 3, exponents for b must be 1 and n-1. So equations m+1=3 and 1=n-1. Solving gives m=2, n=2. So answer A.But the problem is given as multiple choice, and the answer is supposed to be one of A, B, C, D. So in that case, A is the correct one. Hmm. The initial problem must be correct, so I think that's the case.But wait, the original problem is in Chinese? Maybe translation issues? But the problem is given here in English: "the monomials 3a^{m+1}b and -b^{n-1}a³ can be combined, the values of m and n are respectively ( )." So, no, translation seems accurate.Therefore, I think the correct answer is A:2,2.But wait, looking back at the options, maybe the original problem had a typo? For example, if in the second monomial, it's a^{n-1} instead of b^{n-1}, but no, the user wrote "-b^{n-1}a³". So as per the given problem, it's correct.Alternatively, maybe I misread the exponents. Let me check again.First monomial: 3a^{m+1}b. So the exponent on a is m +1. The exponent on b is 1.Second monomial: -b^{n -1}a³. So exponent on b is n -1. Exponent on a is 3.Therefore, equate exponents for a: m +1 =3, so m=2.Equate exponents for b: 1 = n -1, so n=2.Therefore, answer is A:2,2.Therefore, the correct answer should be A.But looking at the options again, the problem might be different. Wait, the options given are A: 2,2; B:3,2; C:2,0; D:3,0.But according to my calculation, it's A. However, maybe the answer is different. Wait, maybe the problem is in the order of the exponents. Let me check once more.Wait, if the first term is 3a^{m+1}b, which is 3 * a^{m+1} * b^1.Second term is -1 * b^{n -1} * a^3. Which is -a^3 * b^{n -1}.Therefore, for the terms to be like terms, the exponents of a must be equal and exponents of b must be equal.So:For a: m +1 = 3 ⇒ m=2For b: 1 = n -1 ⇒ n=2Thus, m=2 and n=2. Therefore, answer A. So I don't see any mistake here.But perhaps the original problem was different? For example, if the first term was 3a^{m+1}b^{n} and the second term was -b^{n-1}a³, then we would have different equations. But as per the given problem, the first term is 3a^{m+1}b and the second is -b^{n-1}a³.Therefore, unless there's a mistake in my reading, the answer is A. But since the options include A, which is 2,2, which matches my calculation, I think that's correct.But maybe the answer provided is different? Hmm. Wait, the user is asking me to provide the answer, not that they have the answer already. So according to my reasoning, A is correct.But wait, maybe the question is from a source where the answer is different, but given the problem as stated, A is correct.Alternatively, maybe combining the terms requires the coefficients to be combined, so after setting exponents equal, but maybe the problem is implying something else?Wait, no. If two monomials can be combined, that means they are like terms, so they can be added or subtracted. Therefore, their variables and exponents must be identical. So yes, setting exponents equal is the way to go.Therefore, I think the answer is A:2,2.But maybe the original problem had a different exponent? Let me check again.Original problem: "the monomials 3a^{m+1}b and -b^{n-1}a³ can be combined"Yes, so as written, first term has a^{m+1} and b^1, second term has b^{n -1} and a^3. Thus, m+1=3 and 1=n-1. Solving gives m=2, n=2. Answer A.Therefore, I think the answer is A.But wait, looking at the options again: C is 2,0. If n=0, then exponent of b in the second term would be -1, which is different from 1. So that's not possible.Similarly, D:3,0: exponent of a would be 3+1=4 vs 3. Not equal. So no.B:3,2: exponent of a would be 3+1=4 vs 3. Not equal. So no.Therefore, only A works.Therefore, I think the correct answer is A:2,2.Final Answerboxed{A}</think>To determine the values of ( m ) and ( n ) such that the monomials ( 3a^{m+1}b ) and ( -b^{n-1}a^3 ) can be combined, we need to ensure that the exponents of the corresponding variables ( a ) and ( b ) are equal in both terms.1. Exponents of ( a ): - In the first term ( 3a^{m+1}b ), the exponent of ( a ) is ( m + 1 ). - In the second term ( -b^{n-1}a^3 ), the exponent of ( a ) is ( 3 ). - Setting these equal gives: ( m + 1 = 3 ). - Solving for ( m ): ( m = 3 - 1 = 2 ).2. Exponents of ( b ): - In the first term ( 3a^{m+1}b ), the exponent of ( b ) is ( 1 ). - In the second term ( -b^{n-1}a^3 ), the exponent of ( b ) is ( n - 1 ). - Setting these equal gives: ( 1 = n - 1 ). - Solving for ( n ): ( n = 1 + 1 = 2 ).Thus, the values of ( m ) and ( n ) are ( 2 ) and ( 2 ) respectively. The correct answer is (boxed{A}).,<think>Okay, so I need to figure out the values of m and n such that the monomials 3a^{m+1}b and -b^{n-1}a^3 can be combined. Hmm, combining monomials usually means they have to be like terms, right? Like terms have the same variables raised to the same powers. So, for these two monomials to be combined, the exponents of a and b in both expressions must be equal. Let me check that.First, let's write down both monomials clearly:1. 3a^{m+1}b2. -b^{n-1}a^3Wait, the variables in the second monomial are written in a different order, but the order doesn't matter because multiplication is commutative. So, exponents of a and b just need to match in both terms. Let me separate the variables and their exponents.For the first monomial, the exponent of a is (m + 1) and the exponent of b is 1 (since it's just b to the first power). For the second monomial, the exponent of a is 3 and the exponent of b is (n - 1). To combine these terms, the exponents of a must be equal in both, and the exponents of b must also be equal. So we can set up equations:For the exponents of a:m + 1 = 3For the exponents of b:1 = n - 1Now, solving these equations:Starting with the exponent of a:m + 1 = 3Subtract 1 from both sides:m = 3 - 1m = 2Now for the exponent of b:1 = n - 1Add 1 to both sides:1 + 1 = nn = 2Wait, so m is 2 and n is 2? Let me check the answer choices. Option A is 2, 2. Option B is 3, 2. Option C is 2, 0. Option D is 3, 0. So according to this, the answer should be A: 2, 2.But hold on, let me double-check. Let's substitute m = 2 and n = 2 into the original monomials.First monomial: 3a^{2 + 1}b = 3a^3bSecond monomial: -b^{2 - 1}a^3 = -b^1a^3 = -a^3bYes, so both terms have a^3 and b^1, so they are like terms. So combining them would be 3a^3b - a^3b = (3 - 1)a^3b = 2a^3b. So that works. So m = 2 and n = 2, which is option A.But wait, the options given include C: 2, 0. Let me check if that's possible. If n was 0, then the exponent of b in the second monomial would be n - 1 = 0 - 1 = -1. So the second monomial would be -b^{-1}a^3. But that would mean the term has b to the negative power, which would make it not a monomial anymore, or at least not a monomial with non-negative exponents. Since the problem states that these are monomials, typically monomials are considered to have non-negative integer exponents. So having a negative exponent might not be allowed. Therefore, n can't be 0. So option C is invalid because of the negative exponent.Similarly, option D: m = 3, n = 0. If m = 3, then the exponent of a in the first monomial is 3 + 1 = 4. The second monomial's a exponent is 3, so they wouldn't match. So that's not possible. Option B: m = 3, n = 2. Let's check that. If m = 3, the first monomial would have a^{3+1} = a^4. The second monomial has a^3, so exponents of a wouldn't match. So that can't be combined. So option B is invalid. Therefore, the correct answer must be A: 2, 2.Wait, but I need to check the problem again. The problem says the monomials can be combined. So even if n was 0, leading to a negative exponent, but the problem doesn't specify that exponents have to be non-negative. Wait, but in standard definitions, monomials are terms with non-negative integer exponents. So if the problem is following that convention, then n can't be 0 because that would give a negative exponent. However, if the problem allows negative exponents, then n could be 0. But in standard math problems, unless stated otherwise, monomials have non-negative exponents. Therefore, n - 1 must be non-negative, so n - 1 ≥ 0 ⇒ n ≥ 1. But in our earlier solution, n = 2, which is okay. If we considered n = 0, then n - 1 = -1, which would not be allowed. So option C is invalid. So the correct answer is A: 2, 2.Wait, but the answer options include C: 2, 0. Maybe I made a mistake? Let me check again.Wait, the problem says the monomials can be combined. If combining here just requires that the variables are the same, regardless of exponents? But no, that's not the case. To combine monomials, they must be like terms, which requires the same variables with the same exponents. So exponents of a must be equal, and exponents of b must be equal.So, first monomial: exponents of a: m + 1, exponent of b: 1.Second monomial: exponents of a: 3, exponent of b: n - 1.Therefore, equations:m + 1 = 3 ⇒ m = 2.n - 1 = 1 ⇒ n = 2.So indeed, m = 2, n = 2. Therefore, answer A. But why is option C present? Let's check option C: m = 2, n = 0.If n = 0, then the exponent of b in the second monomial is -1. So the second monomial would be -b^{-1}a^3, which is -a^3 / b. But that is not a monomial in the standard sense because of the negative exponent. So even if we considered combining them, the terms wouldn't be like terms because one has b and the other has b^{-1}. Therefore, they can't be combined. Therefore, option C is invalid.Therefore, the answer is A. But wait, looking at the problem again, maybe I misread something. Let me confirm.The problem says: "the monomials 3a^{m+1}b and -b^{n-1}a^{3} can be combined"So combining is possible only if they are like terms, which requires the exponents of a and b to match. So yes, m + 1 = 3 and n - 1 = 1. So m = 2, n = 2. So answer A. Therefore, the correct answer is A.But wait, in the answer options given, the original question is in Chinese perhaps? Wait, the user wrote the problem in English. The options are A: 2, 2; B:3,2; C:2,0; D:3,0. So according to the reasoning, A is correct.But I recall sometimes problems where when they say "can be combined" maybe they mean after simplification? Wait, but these are monomials. There's no simplification unless they are like terms. So unless there's a different interpretation.Alternatively, maybe the exponents are not supposed to be matched? Wait, no. Monomials can be combined only if they are like terms. Therefore, variables and exponents must match. So 3a^{m+1}b and -b^{n-1}a^3 can be combined only if the exponents for a and b are equal in both. Therefore, m + 1 = 3 and n - 1 = 1. Hence, m = 2, n = 2.Therefore, the answer is A. But maybe the answer is C? Let me check again.Wait, let me substitute m = 2 and n = 0. Then first monomial: 3a^{2 + 1}b = 3a^3b. Second monomial: -b^{0 - 1}a^3 = -b^{-1}a^3. So that's -a^3 / b. These are not like terms, so they can't be combined. So C is invalid.Alternatively, if m = 2, n = 0, but the exponents of a in the first term is 3, and in the second term is 3, so exponents of a match, but exponents of b do not. First term has b^1, second term has b^{-1}. So variables don't match. Hence, they can't be combined. So C is invalid.Similarly, option D: m = 3, n = 0. Then first monomial: 3a^{3 +1}b = 3a^4b. Second monomial: -b^{-1}a^3. Different exponents for a and b. So can't be combined.Option B: m = 3, n = 2. Then first monomial: 3a^{4}b. Second monomial: -b^{1}a^3 = -a^3b. Different exponents for a, 4 vs 3, so can't be combined.So only option A works. Therefore, the answer is A.But wait, the original problem in Chinese might have a different answer? Wait, the user presented the problem in English. The options are A to D as given. So unless there's a mistake in my reasoning.Wait, but let's check again the exponents.First monomial: 3a^{m+1}b is equivalent to 3a^{m+1}b^1.Second monomial: -b^{n-1}a^3 is equivalent to -a^3b^{n-1}.To combine, the exponents of a and b must be the same. Therefore, for a: m + 1 = 3; for b: 1 = n - 1. Solving gives m = 2, n = 2. So answer A. That seems correct.Alternatively, maybe I made a mistake in the order of the variables? Wait, no, variables can be written in any order. The exponents are per variable. So a^3b is the same as ba^3. So order doesn't matter.Therefore, my conclusion is that the answer is A. However, looking back at the options, perhaps there's a mistake in the problem or the options. Wait, in the problem statement, is it possible that the first monomial is 3a^{m+1}b and the second is -b^{n-1}a^3. So maybe when they say combined, it's allowed even if the coefficients are different? But no, combining terms requires the same variables and exponents. Coefficients can be different, but variables and exponents must be the same. So the coefficients just add up. Therefore, the answer is A. So I think the answer should be A: 2,2. But the given options include C and D, which suggests maybe there's a different approach. Wait, let me check again.Wait, in the problem statement, the monomials are 3a^{m+1}b and -b^{n-1}a^{3}. Maybe the exponents are being considered differently? For example, if there was a typo and the variables are supposed to be multiplied in a certain way? Let me re-express the monomials.First monomial: 3 * a^{m+1} * b.Second monomial: -1 * b^{n-1} * a^3.So variables are a and b. Exponents must be the same for both variables in both monomials. So the exponent of a in the first is m + 1, in the second is 3. So m + 1 = 3, leading to m = 2. Exponent of b in the first is 1, in the second is n - 1. So 1 = n - 1, leading to n = 2. So m = 2, n = 2. Answer A.Therefore, the answer is A. So why does option C exist? Maybe a common mistake is to set m + 1 = 3 and n - 1 = 1, but miscalculating n? Wait, n - 1 = 1 gives n = 2. So not sure why option C is 2, 0. Maybe someone thought that exponent for b in the second term is 0 - 1 = -1, but then thought that n - 1 = 1 would mean n = 0? That would be incorrect.Alternatively, perhaps there was a different interpretation. Maybe combining the monomials through multiplication? But the problem says "can be combined", which typically refers to addition or subtraction. If they are multiplied, then they can always be combined, but that would be a different operation. But the term "combined" in algebra usually refers to addition or subtraction. So unless specified otherwise, combining means adding or subtracting like terms.Therefore, given that, the answer should be A: 2,2.But let me check the problem again: "can be combined". If combining means something else here, but I think in any case, for monomials to be combined (in the sense of addition or subtraction), they must be like terms. Therefore, exponents must match. So answer A.Thus, the answer is boxed{A}.Final Answerboxed{A}</think>To determine the values of ( m ) and ( n ) such that the monomials ( 3a^{m+1}b ) and ( -b^{n-1}a^3 ) can be combined, we need to ensure that they are like terms. Like terms have the same variables raised to the same powers.1. For the first monomial ( 3a^{m+1}b ): - The exponent of ( a ) is ( m + 1 ). - The exponent of ( b ) is 1.2. For the second monomial ( -b^{n-1}a^3 ): - The exponent of ( a ) is 3. - The exponent of ( b ) is ( n - 1 ).To combine these monomials, the exponents of ( a ) and ( b ) must be equal in both terms. This gives us the following equations:- For the exponents of ( a ): [ m + 1 = 3 ] Solving this, we get: [ m = 3 - 1 = 2 ]- For the exponents of ( b ): [ 1 = n - 1 ] Solving this, we get: [ n = 1 + 1 = 2 ]Substituting ( m = 2 ) and ( n = 2 ) into the original monomials, we get:- ( 3a^{2+1}b = 3a^3b )- ( -b^{2-1}a^3 = -a^3b )Both terms are like terms with exponents ( a^3 ) and ( b^1 ), confirming they can be combined. Thus, the values of ( m ) and ( n ) are respectively ( 2 ) and ( 2 ), corresponding to option A.[boxed{A}]]

🔑:To combine monomials, the variables and their exponents must match. Given the monomials 3a^{m+1}b and -b^{n-1}a^{3}, we can set the exponents of a and b equal to each other to find m and n.For a, we have:[m + 1 = 3]Solving for m gives us:[m = 3 - 1][m = 2]For b, we have:[n - 1 = 1]Solving for n gives us:[n = 1 + 1][n = 2]Therefore, the values of m and n are 2 and 2, respectively.So, the correct answer is boxed{text{A}}.

🧠:[<think>Okay, let's try to figure out how many pairs of skew lines there are between a cube's diagonal and its edges. The options are A: 2, B: 3, C: 6, D: 12. Hmm, skew lines are lines that don't intersect and aren't parallel. So in a cube, which is a 3D shape, there should be several edges that are neither intersecting nor parallel to the diagonal. First, I need to visualize a cube. A cube has 12 edges, right? Each edge is the same length. Now, the problem mentions a cube's diagonal. There are two types of diagonals in a cube: face diagonals and space diagonals. But since the question mentions the diagonal forming skew lines with edges, I think they're talking about a space diagonal. Because a face diagonal lies on a face of the cube, and edges on the same face would either intersect or be parallel. But a space diagonal goes from one corner of the cube to the opposite corner, passing through the interior of the cube.So, let's confirm: a space diagonal is not on any face, right? It connects vertices that aren't on the same face. For example, in a cube with vertices labeled from (0,0,0) to (1,1,1), a space diagonal would be from (0,0,0) to (1,1,1). Now, edges of the cube are along the axes. So, each edge is either along the x, y, or z-axis. Skew lines are lines that are neither parallel nor intersecting. Since the space diagonal isn't parallel to any edge (because the edges are along the axes, and the space diagonal has direction vector (1,1,1)), which isn't parallel to any axis. So, the space diagonal isn't parallel to any edge. Now, do any edges intersect with the space diagonal?Well, the space diagonal starts at a vertex and ends at the opposite vertex. So, it passes through those two vertices. The edges connected to those two vertices would intersect the space diagonal at the start or end points. So, for example, if the space diagonal is from (0,0,0) to (1,1,1), then the edges connected to (0,0,0) are along the x, y, and z-axes from that vertex. Similarly, edges connected to (1,1,1) are along the negative x, y, and z-axes. So, the space diagonal intersects those six edges at the two vertices. Therefore, those six edges intersect the space diagonal, so they can't form skew lines. The other edges of the cube are not connected to either end of the space diagonal. Since the cube has 12 edges in total, and 6 of them intersect the space diagonal at its endpoints, the remaining 6 edges don't intersect the space diagonal. But are those remaining edges parallel to the space diagonal? Let's check. The direction vector of the space diagonal is (1,1,1), and the edges are along the coordinate axes with direction vectors (1,0,0), (0,1,0), (0,0,1). The space diagonal isn't a scalar multiple of any of these, so it's not parallel to any edge. Therefore, all the edges that don't intersect the space diagonal are skew with it. So, there are 12 total edges, minus 6 edges that intersect the diagonal, leaving 6 edges that are skew. Therefore, the number of skew pairs is 6. Wait, but let me double-check. Wait, the question says "the cube's diagonal and the edges of the cube can form n pairs of skew lines". So, each pair is between the diagonal and one edge. So, if the space diagonal is one line, and each edge is another line, then each pair is (diagonal, edge). So, for each edge, check if it's skew with the diagonal. We concluded that 6 edges are skew with the diagonal, so there are 6 such pairs. So the answer should be C: 6. But wait, the options given are A: 2, B: 3, C: 6, D: 12. Hmm, maybe my reasoning is wrong. Let me think again. Wait, perhaps the question is asking for all pairs of skew lines where one is a diagonal and the other is an edge. So, if there is only one space diagonal considered here, then each edge is checked against this diagonal. But in a cube, there are four space diagonals. Wait, hold on. Wait, maybe I misunderstood the problem. Wait, the problem says "a cube's diagonal and the edges of the cube can form n pairs of skew lines". So, does this refer to one specific diagonal and all edges, or all diagonals and all edges? The wording is a bit ambiguous. But since it says "the cube's diagonal", maybe it refers to one diagonal. But in a cube, there are multiple diagonals. Wait, maybe it's referring to a single space diagonal. Let's assume that. So, if we take one space diagonal, how many edges are skew to it? As before, edges connected to the endpoints of the diagonal (there are 3 edges at each endpoint, total 6 edges) intersect the diagonal. The other 6 edges are not connected to either endpoint, so they don't intersect. And since none of the edges are parallel to the diagonal (the diagonal's direction is (1,1,1), edges are axis-aligned), so those 6 edges are skew. Therefore, 6 pairs. So, answer is C: 6. But let me check again. Let me take a specific cube. Let's say we have a cube with vertices at (0,0,0) to (1,1,1). Let's take the space diagonal from (0,0,0) to (1,1,1). The edges connected to (0,0,0) are along the x, y, z axes to (1,0,0), (0,1,0), (0,0,1). Similarly, edges connected to (1,1,1) go to (0,1,1), (1,0,1), (1,1,0). So those 6 edges intersect the diagonal. The other edges are: From (1,0,0) to (1,1,0), (1,0,1), and (0,0,0) but (0,0,0) is already considered. Wait, no. Let's list all edges not connected to (0,0,0) or (1,1,1). Each vertex not on the diagonal: the other six vertices. Wait, no. The cube has 8 vertices. The two endpoints of the diagonal are (0,0,0) and (1,1,1). The other six vertices are (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0). Each of these is connected by edges. The edges that are not connected to (0,0,0) or (1,1,1) would be the edges between these six vertices. But in reality, all edges are connected to either (0,0,0) or (1,1,1) or both? Wait, no. For example, the edge from (0,0,1) to (0,1,1) is connected to neither (0,0,0) nor (1,1,1). Similarly, the edge from (1,0,0) to (1,1,0) is connected to neither. Wait, actually:Wait, (0,0,1) is connected to (0,0,0), so the edge from (0,0,1) to (0,1,1) is connected to (0,0,1), which is connected to (0,0,0). Wait, maybe not. The edge from (0,0,1) to (0,1,1) is along the y-axis at x=0, z=1. So that edge is not connected to (0,0,0) or (1,1,1). Similarly, edge from (1,0,0) to (1,1,0) is along the y-axis at x=1, z=0. Not connected to (0,0,0) or (1,1,1). So, edges that are not connected to either endpoint of the space diagonal: Let's count. Each of the six vertices not on the diagonal has three edges, but some of those edges connect back to the diagonal's endpoints. Wait, maybe a better way: total edges is 12. Edges connected to (0,0,0): 3. Edges connected to (1,1,1): another 3. However, these two sets of edges are distinct because (0,0,0) and (1,1,1) are opposite corners. So, total edges connected to either endpoint is 3 + 3 = 6. The remaining 12 - 6 = 6 edges are not connected to either (0,0,0) or (1,1,1). Therefore, those 6 edges don't intersect the space diagonal. Are those 6 edges parallel to the space diagonal? The space diagonal has direction vector (1,1,1). The edges are along the coordinate axes, so direction vectors (1,0,0), (0,1,0), (0,0,1). Since (1,1,1) isn't a scalar multiple of any of these, none of the edges are parallel to the space diagonal. Therefore, the remaining 6 edges are skew to the space diagonal. Therefore, the number of skew pairs is 6. So the answer is C: 6. But wait, let me verify with an example. Let's take the space diagonal from (0,0,0) to (1,1,1). Take an edge that's not connected to either end, like the edge from (0,1,0) to (0,1,1). This edge is along the z-axis at x=0, y=1. Does this edge intersect the space diagonal? The space diagonal is parametrized as (t, t, t) for t from 0 to 1. The edge is parametrized as (0, 1, s) for s from 0 to 1. To check for intersection, set (t, t, t) = (0, 1, s). So t must be 0 from the x-coordinate, but then the y-coordinate t would be 0, which is not 1. So no intersection. And they're not parallel. So this edge is skew with the diagonal. Another example: the edge from (1,0,0) to (1,1,0). This is along the y-axis at x=1, z=0. Parametrize as (1, s, 0). The diagonal is (t, t, t). Setting equal: 1 = t, s = t, 0 = t. From the z-coordinate, t=0, but then x-coordinate would be 0, which is not 1. So no intersection. Not parallel. So skew. Similarly, the edge from (0,0,1) to (0,1,1) is skew. So, yes, all these 6 edges not connected to the diagonal's endpoints are skew. Therefore, answer is 6. Wait, but the options include D:12. Maybe if they consider all space diagonals? There are 4 space diagonals in a cube. Each space diagonal would have 6 skew edges. So 4 diagonals * 6 edges = 24, but that's not an option. Wait, but the question says "a cube's diagonal", singular. So probably just one diagonal. So 6 edges. So answer is C:6. Therefore, I think the correct answer is C.Final Answerboxed{C}</think>To determine the number of pairs of skew lines formed by a cube's space diagonal and its edges, we start by noting that a cube has 12 edges. A space diagonal connects two opposite vertices and is not parallel to any edge. 1. Identify edges intersecting the space diagonal: - A space diagonal connects two vertices, each connected to three edges. For example, the space diagonal from ((0,0,0)) to ((1,1,1)) intersects the edges connected to ((0,0,0)) and ((1,1,1)). - Each vertex is connected to three edges, so there are (3 + 3 = 6) edges intersecting the space diagonal.2. Remaining edges: - The cube has 12 edges in total. Subtracting the 6 edges that intersect the space diagonal, we have (12 - 6 = 6) edges left.3. Check for parallelism: - The direction vector of the space diagonal is ((1,1,1)), which is not parallel to any edge direction vectors ((1,0,0)), ((0,1,0)), or ((0,0,1)). Thus, the remaining edges are neither parallel nor intersecting with the space diagonal, making them skew lines.4. Verification: - Each of the remaining 6 edges is confirmed to be skew with the space diagonal by checking parametrization and ensuring no intersection and non-parallel direction vectors.Thus, the number of pairs of skew lines formed by a cube's space diagonal and its edges is (boxed{C}).,<think>Okay, so I need to figure out how many pairs of skew lines are formed by a cube's diagonal and the edges of the cube. The options are 2, 3, 6, or 12. Hmm, let me start by recalling what a skew line is. Skew lines are lines that are neither parallel nor intersecting. They exist in three-dimensional space, right? So in a cube, which is a 3D shape, there should be some skew lines. First, let me visualize a cube. A cube has 12 edges, right? Each edge is connected to two vertices. Now, the problem mentions a cube's diagonal. Wait, which diagonal are they talking about? There are face diagonals and the space diagonal. The space diagonal is the one that goes from one corner of the cube to the opposite corner, passing through the interior of the cube. The face diagonals are the ones that lie on the faces of the cube. But the problem just says "a cube's diagonal," so maybe it's referring to the space diagonal? Let me check. If they meant a face diagonal, the answer might be different. But generally, when someone refers to a diagonal of a cube without specifying, they might mean the space diagonal. Let me assume it's a space diagonal for now.So, if we take a space diagonal of the cube, how many edges of the cube are skew to this diagonal? Each edge of the cube can either intersect the diagonal, be parallel to it, or be skew to it. Since the diagonal is a space diagonal, it's not parallel to any edge because all edges are either along the x, y, or z-axis, and the space diagonal has components in all three axes. So, the direction vectors can't be scalar multiples of each other, so they aren't parallel. Now, do any edges intersect the space diagonal? The space diagonal connects two opposite vertices. So, the edges that meet at those two vertices would intersect the diagonal at those vertices. For example, if the space diagonal goes from vertex (0,0,0) to (1,1,1), then the edges connected to (0,0,0) are along the x, y, and z-axes from that vertex. Similarly, the edges connected to (1,1,1) go towards (0,1,1), (1,0,1), and (1,1,0). So, those six edges (three at each end) intersect the space diagonal at its endpoints. Therefore, those six edges are not skew to the diagonal. But the cube has 12 edges total. So, subtracting those six edges that intersect the diagonal, we have 12 - 6 = 6 edges left. Are these remaining edges skew to the diagonal? Let's check. Since the diagonal isn't parallel to any edge, and if an edge doesn't intersect the diagonal, then they must be skew. Wait, but let's make sure. Let's take an example. Suppose the space diagonal is from (0,0,0) to (1,1,1). Take an edge that's not connected to either of these two vertices. For example, the edge from (0,0,1) to (0,1,1). This edge is on the top face of the cube (assuming z-axis is up). The space diagonal goes from (0,0,0) to (1,1,1). Does this edge intersect the diagonal? Let's parametrize both lines. The space diagonal can be parametrized as (t, t, t) for t from 0 to 1. The edge from (0,0,1) to (0,1,1) can be parametrized as (0, s, 1) where s goes from 0 to 1. To see if they intersect, we set the coordinates equal: t = 0, t = s, and t = 1. But t can't be both 0 and 1, so there's no solution. Therefore, they don't intersect. Since they aren't parallel either, they are skew. Similarly, another edge not connected to the endpoints: say the edge from (1,0,0) to (1,0,1). Parametrize this as (1,0,r) where r goes from 0 to 1. Setting equal to the diagonal: t = 1, t = 0, t = r. Again, t can't be both 1 and 0, so no intersection. Therefore, skew. So, the remaining 6 edges are all skew to the space diagonal. Therefore, n equals 6. Wait, but the answer options have C as 6. So is the answer C? But let me double-check. Wait, but maybe the problem is asking for the number of pairs. If the diagonal is one line, and each edge is another line, then each pair is (diagonal, edge). So, since there are 6 edges that are skew to the diagonal, then there are 6 such pairs. Therefore, n equals 6. So the answer is C: 6. But hold on, maybe the problem is considering all space diagonals? Wait, the question says "a cube's diagonal". So maybe it's referring to one specific space diagonal. A cube has four space diagonals. But if the problem is talking about one space diagonal, then the answer is 6. If it's considering all space diagonals, the answer might be different, but the question says "a cube's diagonal", singular. So probably just one. Wait, but maybe they are talking about all pairs of skew lines where one is any space diagonal and the other is any edge? But the wording is "the diagonal and the edges", so maybe only one diagonal is considered. Hmm. Let me read the problem again: "A cube's diagonal and the edges of the cube can form n pairs of skew lines. Then n equals..." So the diagonal is one line, and the edges are the other lines. So each pair is (diagonal, edge). So if we fix one diagonal, how many edges are skew to it? As we computed earlier, 6. Therefore, n=6. But let me confirm once more. Let's count. The cube has 12 edges. The diagonal connects two vertices. Each vertex has three edges. So the two vertices connected by the diagonal have 3 edges each, total 6 edges connected to them. These 6 edges intersect the diagonal at the vertices. The remaining 12 - 6 = 6 edges do not intersect the diagonal. Since none of the edges are parallel to the diagonal (edges are axis-aligned, diagonal is not), so those 6 edges are skew. Therefore, 6 pairs. So the answer is C:6. Wait, but maybe I made a mistake here. Let me think again. Each edge is a line. The diagonal is a line. So for each edge, check if it's skew with the diagonal. So total 12 edges. For each edge, check if it's skew with the diagonal. How many are skew? As before, edges connected to the endpoints of the diagonal (each endpoint has 3 edges, so 6 edges total) intersect the diagonal. So they are not skew. The other 6 edges: do they intersect the diagonal? Let's take an example. Take a space diagonal from (0,0,0) to (1,1,1). Take an edge that's not connected to these points, like from (0,1,0) to (0,1,1). Let's see if this edge intersects the diagonal. The edge is along the z-axis at x=0, y=1. The diagonal is the line x=y=z. To see if they intersect, we need a point where x=y=z and x=0, y=1. But x can't be both 0 and 1. So no intersection. Since the edge is along the z-axis, which is not parallel to the diagonal's direction (1,1,1). Therefore, this edge is skew to the diagonal. Similarly, the edge from (1,0,0) to (1,0,1) is along the z-axis at x=1, y=0. The diagonal would have x=y=z, so if x=1, then y=1 and z=1, but the edge is at y=0, so no intersection. Hence, skew. So all 6 edges not connected to the diagonal's endpoints are skew. Therefore, 6 pairs. Therefore, the answer should be 6, which is option C. But wait, maybe I'm missing something. Let me check another edge. Take the edge from (0,1,0) to (1,1,0). This edge is along the x-axis at y=1, z=0. The diagonal is (t,t,t). So to intersect, we need t = x-coordinate, t = y-coordinate =1, t = z-coordinate =0. So t must be 1 and 0, which is impossible. Therefore, no intersection. Not parallel, so skew. Another edge: (1,0,1) to (1,1,1). This is along the y-axis at x=1, z=1. The diagonal at (t,t,t). To intersect, t=1, t= y-coordinate, t=1. So t=1. But the edge goes from (1,0,1) to (1,1,1). So when t=1, the diagonal is at (1,1,1), which is the endpoint of the edge. So does that count as intersecting? Wait, the edge is from (1,0,1) to (1,1,1), so the point (1,1,1) is on both the edge and the diagonal. But (1,1,1) is an endpoint of the edge. Therefore, the edge and the diagonal intersect at that point. Wait, but earlier I thought edges connected to the diagonal's endpoints were the ones intersecting. But this edge is connected to (1,1,1), which is an endpoint of the diagonal. So yes, that edge intersects the diagonal at (1,1,1). So, that edge is not skew. Wait, but in my previous count, I subtracted 6 edges: the three edges at each endpoint of the diagonal. The diagonal has two endpoints, each connected to three edges. So total 6 edges. So in this case, the edge from (1,0,1) to (1,1,1) is connected to (1,1,1), which is an endpoint of the diagonal, so it's one of those six edges. Therefore, that edge is not skew. So, in our previous calculation, we accounted for that. Therefore, the remaining edges are 12 - 6 = 6, which are not connected to either endpoint of the diagonal. Hence, they don't intersect the diagonal and are not parallel, so skew. Therefore, yes, the answer is 6. So option C. But let me check once again. Maybe there's a different interpretation. Maybe the problem is asking for the number of skew lines pairs where one is any space diagonal and the other is any edge. In that case, since there are four space diagonals, each contributing 6 skew edges, but then we have to consider overlaps. But the problem says "a cube's diagonal", singular. So probably one diagonal. So the answer is 6. Alternatively, if "a cube's diagonal" refers to all space diagonals, but that would be plural. The problem is in Chinese maybe, but the translation is "a cube's diagonal", so singular. So I think the answer is 6. Therefore, option C.Final Answerboxed{C}</think>To determine the number of pairs of skew lines formed by a cube's diagonal and the edges of the cube, we start by understanding that skew lines are neither parallel nor intersecting. A cube has 12 edges and a space diagonal connects two opposite vertices.1. Identify the space diagonal: A space diagonal connects two vertices of the cube that are not on the same face. For example, from vertex ((0,0,0)) to ((1,1,1)).2. Edges intersecting the space diagonal: Each vertex of the cube is connected to three edges. The two vertices connected by the space diagonal each have three edges, totaling (3 + 3 = 6) edges. These edges intersect the space diagonal at its endpoints.3. Remaining edges: The cube has 12 edges in total. Subtracting the 6 edges that intersect the space diagonal, we have (12 - 6 = 6) edges left.4. Check for skew lines: The remaining 6 edges do not intersect the space diagonal and are not parallel to it (since edges are axis-aligned and the space diagonal has a direction vector ((1,1,1))). Therefore, these 6 edges are skew to the space diagonal.Thus, the number of pairs of skew lines formed by a cube's diagonal and the edges of the cube is (boxed{C}).]

🔑:Analysis: Draw a cube and make a diagonal. By understanding the definition of skew lines, we can determine that there are 6 pairs of skew lines.Review: Drawing is more intuitive and helps in solving the problem. This question tests knowledge of skew lines and is considered a basic question.boxed{text{C}}

🧠:[<think>Alright, let's try to tackle this geometry problem step by step. It's about an isosceles triangle with some points and vectors, and we need to find the range of a certain dot product. Hmm, okay. Let me parse the problem again.We have an isosceles triangle ABC where AC = BC = √5. So, that means triangle ABC has two sides equal: AC and BC. Therefore, the base is AB, and the two equal legs are AC and BC. Points D, E, F are on sides AB, BC, CA respectively, with AD = DB = EF = 1. So, D is the midpoint of AB since AD = DB = 1, which also tells me that AB must be 2 units long. Wait, AD and DB are both 1, so AB = 2. But AC and BC are both √5. So triangle ABC is isosceles with sides AB = 2, AC = BC = √5. Let me verify if that makes sense. Let's compute the length of AB to check if it's consistent with the triangle inequality.Wait, in an isosceles triangle with two sides √5 and base AB = 2, the triangle inequality should hold. The sum of AB and AC should be greater than BC, but AC and BC are equal. So 2 + √5 > √5, which is true. Similarly, √5 + √5 > 2? 2√5 ≈ 4.47 > 2, which is true. So that's okay.So, triangle ABC has vertices A and B separated by 2 units, and both C to A and C to B are √5. Points D is the midpoint of AB, since AD = DB = 1. Then E is on BC, F is on CA, such that EF = 1. The problem states that the vectors DE · DF ≤ 25/16, and we need to find the range of the dot product EF · BA.First, maybe I should draw a coordinate system to model this triangle. Let me place the triangle in a coordinate system to make the calculations easier.Since it's an isosceles triangle with base AB = 2, let's place points A and B on the x-axis. Let me set point A at (-1, 0) and point B at (1, 0), so that the midpoint D is at (0, 0). Wait, but AD = DB = 1. If AB is 2 units, then yes, from -1 to 1 on the x-axis, the distance is 2. Then point C is somewhere above the x-axis. We need to find the coordinates of point C such that AC = BC = √5.Let me calculate the coordinates of C. Let's assume point C is at (0, h) because the triangle is isosceles with base AB, so the vertex C should be along the y-axis. Then AC is the distance from (-1, 0) to (0, h), which is √[(0 - (-1))² + (h - 0)²] = √[1 + h²]. Similarly, BC is the distance from (1, 0) to (0, h), which is also √[1 + h²]. Given AC = BC = √5, so √[1 + h²] = √5. Then 1 + h² = 5, so h² = 4, so h = 2. Therefore, point C is at (0, 2).So, coordinates:A: (-1, 0)B: (1, 0)C: (0, 2)D: midpoint of AB, which is (0, 0)Okay, that seems good.Now, points E and F are on BC and CA respectively, such that EF = 1. We need to find the range of EF · BA given that DE · DF ≤ 25/16.First, let's recall that vectors BA and EF. BA is the vector from B to A, which is A - B = (-1 - 1, 0 - 0) = (-2, 0). So BA vector is (-2, 0). So the dot product EF · BA will be the projection of EF onto BA multiplied by the magnitude of BA, but since BA is along the x-axis, maybe this simplifies things.But let's see. Maybe parameterize points E and F.Since E is on BC and F is on CA. Let me parameterize these points.First, parametrize BC. The line from B(1, 0) to C(0, 2). Let's let E divide BC in some ratio. Let's use a parameter t for E, where t is the fraction from B to C. So when t = 0, E is at B(1, 0), and when t = 1, E is at C(0, 2). So E can be represented as (1 - t, 0 + 2t) = (1 - t, 2t).Similarly, parametrize point F on CA. The line from C(0, 2) to A(-1, 0). Let's use a parameter s for F, where s is the fraction from C to A. So F can be represented as (0 + (-1)s, 2 - 2s) = (-s, 2 - 2s).So E is (1 - t, 2t) and F is (-s, 2 - 2s).Given that EF = 1. The distance between E and F should be 1.So let's compute the distance between E(1 - t, 2t) and F(-s, 2 - 2s):EF² = [(-s - (1 - t))² + (2 - 2s - 2t)²] = [(-s -1 + t)² + (2 - 2s - 2t)²] = (t - s -1)² + (2 - 2s - 2t)² = 1² = 1.So we have the equation:(t - s -1)² + (2 - 2s - 2t)² = 1.This is one equation relating parameters t and s. But we also have another condition: DE · DF ≤ 25/16.First, let's find vectors DE and DF.Point D is (0, 0). Point E is (1 - t, 2t). So vector DE is E - D = (1 - t, 2t).Similarly, vector DF is F - D = (-s, 2 - 2s).Therefore, the dot product DE · DF = (1 - t)(-s) + (2t)(2 - 2s) = -s(1 - t) + 4t(1 - s).Simplify:= -s + st + 4t - 4ts= (-s + 4t) + (st - 4ts)= -s + 4t - 3tsSo DE · DF = -s + 4t - 3ts ≤ 25/16.So we have the constraint:- s + 4t - 3ts ≤ 25/16.But we also have the other equation from EF = 1:(t - s -1)^2 + (2 - 2s - 2t)^2 = 1.So now we have two equations involving s and t. The problem is to find the range of EF · BA. Let's compute EF · BA.Vector EF is F - E = (-s - (1 - t), (2 - 2s) - 2t) = (-s -1 + t, 2 - 2s - 2t).Vector BA is A - B = (-1 - 1, 0 - 0) = (-2, 0).So the dot product EF · BA = (-s -1 + t)(-2) + (2 - 2s - 2t)(0) = 2(s +1 - t) + 0 = 2(s +1 - t).Therefore, EF · BA = 2(s +1 - t).So we need to find the range of 2(s +1 - t) given the constraints:1. (t - s -1)^2 + (2 - 2s - 2t)^2 = 1.2. -s + 4t - 3ts ≤ 25/16.So the problem reduces to finding the possible values of 2(s +1 - t) given these two constraints on s and t.Therefore, our goal is to find the maximum and minimum of 2(s +1 - t) subject to the constraints:1. (t - s -1)^2 + (2 - 2s - 2t)^2 = 1,2. -s + 4t - 3ts ≤ 25/16.So perhaps we can first express s in terms of t or vice versa from the first equation, then substitute into the second inequality and the expression to be maximized/minimized. But this might be complicated due to the quadratic equation. Alternatively, we can use Lagrange multipliers or parameterization.Alternatively, notice that the first equation is a quadratic in s and t, representing a circle or ellipse. Let's see:Expand the first equation:(t - s -1)^2 + (2 - 2s - 2t)^2 = 1.Let me expand each term:First term: (t - s -1)^2 = (t - s -1)^2 = ( - s + t -1 )² = s² + (t -1)^2 - 2s(t -1).Wait, expanding:Let me write it as ( (t - s) -1 )² = (t - s)² - 2(t - s) + 1.Second term: (2 - 2s - 2t)^2 = [ -2s -2t + 2 ]² = [ -2(s + t) + 2 ]² = 4(s + t -1)^2.Therefore, the first equation becomes:(t - s)² - 2(t - s) + 1 + 4(s + t -1)^2 = 1.Let me let u = t - s and v = s + t.Then the equation becomes:u² - 2u + 1 + 4(v -1)^2 = 1.But maybe this substitution complicates things. Alternatively, let's expand all terms.First term: (t - s -1)^2 = t² - 2ts + s² - 2t + 2s +1.Second term: (2 - 2s - 2t)^2 = 4s² + 8st + 4t² - 8s - 8t +4.Adding both terms:t² - 2ts + s² - 2t + 2s +1 + 4s² +8st +4t² -8s -8t +4 = 1.Combine like terms:t² +4t² =5t²-2ts +8st=6sts² +4s²=5s²-2t -8t= -10t2s -8s= -6s1 +4=5So total equation:5t² +6st +5s² -10t -6s +5 =1Subtract 1:5t² +6st +5s² -10t -6s +4 =0.Hmm, this is a quadratic equation in s and t, representing a conic section. Maybe an ellipse.This seems a bit complicated. Maybe instead of expanding, we can parameterize s and t using another parameter.Alternatively, let's try to express one variable in terms of the other.From the first equation:(t - s -1)^2 + (2 - 2s - 2t)^2 =1.Let me set x = t - s -1 and y = 2 - 2s - 2t. Then x² + y² =1.So, x and y lie on the unit circle. Therefore, x = cosθ, y = sinθ for some θ.But we can express s and t in terms of x and y.From x = t - s -1 => t - s = x +1.From y = 2 - 2s -2t => 2s + 2t = 2 - y => s + t =1 - y/2.So we have two equations:t - s = x +1s + t =1 - y/2We can solve for s and t:Adding the two equations:(t - s) + (s + t) = x +1 +1 - y/22t = x +2 - y/2=> t = (x +2 - y/2)/2Similarly, subtracting the two equations:(s + t) - (t - s) = (1 - y/2) - (x +1)2s = - y/2 -x=> s = (-x - y/2)/2But since x² + y² =1, perhaps we can express s and t in terms of θ where x=cosθ, y=sinθ.So,t = (cosθ + 2 - (sinθ)/2)/2s = (-cosθ - (sinθ)/2)/2Therefore,t = (2 + cosθ - (1/2) sinθ)/2s = ( - cosθ - (1/2) sinθ ) /2Simplify:t = 1 + (cosθ)/2 - (sinθ)/4s = - (cosθ)/2 - (sinθ)/4Thus, s and t are expressed in terms of θ.Now, our expression to maximize/minimize is 2(s +1 - t).Plugging s and t:2( [ - (cosθ)/2 - (sinθ)/4 ] +1 - [1 + (cosθ)/2 - (sinθ)/4 ] )Simplify inside the brackets:= [ - (cosθ)/2 - (sinθ)/4 +1 -1 - (cosθ)/2 + (sinθ)/4 ]= [ - (cosθ)/2 - (sinθ)/4 - (cosθ)/2 + (sinθ)/4 ]= [ - (cosθ)/2 - (cosθ)/2 + (-sinθ)/4 + sinθ/4 ]= [ -cosθ + 0 ]= -cosθTherefore, 2(s +1 - t ) = 2*(-cosθ) = -2cosθ.So, the expression EF · BA is equal to -2cosθ. Therefore, the range of EF · BA is the range of -2cosθ, given the constraints on θ.But θ must satisfy the second inequality: DE · DF ≤ 25/16.Recall that DE · DF = -s +4t -3ts ≤25/16.So we need to express DE · DF in terms of θ as well.We already have expressions for s and t in terms of θ:s = - (cosθ)/2 - (sinθ)/4t =1 + (cosθ)/2 - (sinθ)/4Let me compute DE · DF = -s +4t -3ts.First, compute each term:-s = (cosθ)/2 + (sinθ)/44t =4[1 + (cosθ)/2 - (sinθ)/4] =4 + 2cosθ - sinθ-3ts = -3[ (- (cosθ)/2 - (sinθ)/4 )(1 + (cosθ)/2 - (sinθ)/4 ) ]First, compute t*s:t*s = [1 + (cosθ)/2 - (sinθ)/4 ] * [ - (cosθ)/2 - (sinθ)/4 ]Multiply term by term:=1*(- (cosθ)/2) +1*(- (sinθ)/4 ) + (cosθ)/2*(- (cosθ)/2) + (cosθ)/2*(- (sinθ)/4 ) + (- (sinθ)/4 )*(- (cosθ)/2) + (- (sinθ)/4 )*(- (sinθ)/4 )= - (cosθ)/2 - (sinθ)/4 - (cos²θ)/4 - (cosθ sinθ)/8 + (sinθ cosθ)/8 + (sin²θ)/16Simplify:= - (cosθ)/2 - (sinθ)/4 - (cos²θ)/4 + 0 (since - (cosθ sinθ)/8 + (cosθ sinθ)/8 cancels) + (sin²θ)/16So t*s = - (cosθ)/2 - (sinθ)/4 - (cos²θ)/4 + (sin²θ)/16Therefore, -3ts = 3[ (cosθ)/2 + (sinθ)/4 + (cos²θ)/4 - (sin²θ)/16 ]Now, putting all together:DE · DF = -s +4t -3ts= [ (cosθ)/2 + (sinθ)/4 ] + [4 + 2cosθ - sinθ ] + 3[ (cosθ)/2 + (sinθ)/4 + (cos²θ)/4 - (sin²θ)/16 ]Let me compute term by term:First part: (cosθ)/2 + (sinθ)/4Second part: 4 + 2cosθ - sinθThird part: 3*( (cosθ)/2 + (sinθ)/4 + (cos²θ)/4 - (sin²θ)/16 )Let's compute each part:First part: (cosθ)/2 + (sinθ)/4Second part: 4 + 2cosθ - sinθThird part:3*( (cosθ)/2 ) = (3cosθ)/23*( (sinθ)/4 ) = (3sinθ)/43*( (cos²θ)/4 ) = (3cos²θ)/43*( - (sin²θ)/16 ) = - (3sin²θ)/16So, third part: (3cosθ)/2 + (3sinθ)/4 + (3cos²θ)/4 - (3sin²θ)/16Now, sum all three parts:First + Second + Third:[ (cosθ)/2 + (sinθ)/4 ] + [4 + 2cosθ - sinθ ] + [ (3cosθ)/2 + (3sinθ)/4 + (3cos²θ)/4 - (3sin²θ)/16 ]Combine like terms:Constants: 4cosθ terms:cosθ/2 + 2cosθ + 3cosθ/2 = (1/2 + 2 + 3/2)cosθ = (1/2 + 4/2 + 3/2)cosθ = (8/2)cosθ =4cosθsinθ terms:sinθ/4 - sinθ + 3sinθ/4 = (1/4 -1 +3/4)sinθ = ( (1 +3)/4 -1 )sinθ = (1 -1 )sinθ = 0cos²θ terms:3cos²θ/4sin²θ terms:-3sin²θ/16So overall:DE · DF =4 +4cosθ + (3/4)cos²θ - (3/16)sin²θSimplify the sin²θ term using identity sin²θ =1 - cos²θ:=4 +4cosθ + (3/4)cos²θ - (3/16)(1 - cos²θ)=4 +4cosθ + (3/4)cos²θ -3/16 + (3/16)cos²θCombine constants:4 - 3/16 = 64/16 -3/16 =61/16Combine cosθ terms: 4cosθCombine cos²θ terms:(3/4 + 3/16)cos²θ = (12/16 +3/16)cos²θ =15/16 cos²θThus,DE · DF =61/16 +4cosθ + (15/16)cos²θWe are given that DE · DF ≤25/16.Therefore,61/16 +4cosθ + (15/16)cos²θ ≤25/16Multiply both sides by16 to eliminate denominators:61 +64cosθ +15cos²θ ≤25Subtract25:36 +64cosθ +15cos²θ ≤0Which is:15cos²θ +64cosθ +36 ≤0Let me write this as:15cos²θ +64cosθ +36 ≤0Let’s let x = cosθ. Then the inequality becomes:15x² +64x +36 ≤0We need to solve this quadratic inequality.First, find roots of 15x² +64x +36 =0.Using quadratic formula:x = [ -64 ±√(64² -4*15*36) ]/(2*15)Compute discriminant:64² =40964*15*36=2160Discriminant=4096 -2160=1936√1936=44Thus,x = [ -64 ±44 ]/30So,x1=( -64 +44 )/30=(-20)/30= -2/3x2=( -64 -44 )/30=(-108)/30= -18/5= -3.6But cosθ can only be in [-1,1], so x2= -3.6 is invalid. Therefore, the quadratic 15x² +64x +36 crosses zero at x=-2/3 and x=-3.6, but since x must be ≥-1, the interval where the inequality holds is between x=-1 and x=-2/3.But wait, the quadratic opens upwards since the coefficient of x² is positive (15). Therefore, the inequality 15x² +64x +36 ≤0 holds between the roots. However, the only valid interval in x ∈ [-1,1] is x ∈ [-1, -2/3], since x=-3.6 is outside.Therefore, the inequality DE · DF ≤25/16 is equivalent to cosθ ∈ [-1, -2/3].Therefore, θ must satisfy cosθ ∈ [-1, -2/3].But we need to relate this to our expression EF · BA = -2cosθ.Since EF · BA = -2cosθ, and cosθ ∈ [-1, -2/3], then:When cosθ = -1, EF · BA = -2*(-1) = 2.When cosθ = -2/3, EF · BA = -2*(-2/3) =4/3 ≈1.333...But since cosθ ranges from -1 to -2/3, then EF · BA ranges from 4/3 to 2.But wait, when cosθ decreases from -2/3 to -1, EF · BA increases from 4/3 to 2. Therefore, the range is [4/3, 2].Wait, but hold on. The original problem says "the range of EF · BA". So the minimum is 4/3 and maximum is 2? But let's verify this.Wait, cosθ is in [-1, -2/3], so -cosθ is in [2/3,1], so EF · BA = -2cosθ is in [4/3, 2].Yes, that seems correct. Therefore, the range is from 4/3 to 2. Therefore, the answer should be [4/3, 2], which in boxed form is boxed{left[ dfrac{4}{3}, 2 right]}.But wait, let me check for possible mistakes.First, we parameterized E and F, expressed EF · BA as -2cosθ, and found that due to the constraint DE · DF ≤25/16, cosθ is restricted to [-1, -2/3], hence EF · BA ranges from 4/3 to 2. Seems okay.But perhaps I should verify with specific points.For example, when cosθ = -1, θ = π. Then, x = cosθ = -1, y = sinθ =0.From earlier expressions:s = - (cosθ)/2 - (sinθ)/4 = - (-1)/2 -0 =1/2t =1 + (cosθ)/2 - (sinθ)/4 =1 + (-1)/2 -0=1 -1/2=1/2So s=1/2, t=1/2.Then point E is (1 - t, 2t) = (1 -1/2, 2*(1/2))=(1/2,1)Point F is (-s, 2 -2s)=(-1/2, 2 -2*(1/2))=(-1/2,1)So EF is from E(1/2,1) to F(-1/2,1). The vector EF is (-1,0), length is 1, which matches EF=1.Then EF · BA = (-1,0)·(-2,0)=2, which matches the upper limit.Another case: cosθ = -2/3, then sinθ=±√(1 - (4/9))=±√(5/9)=±√5/3. But need to check the sign.But wait, θ is such that cosθ = -2/3. Depending on θ, sinθ could be positive or negative. However, in our parametrization, θ was used to parameterize the unit circle x=cosθ, y=sinθ, where θ is the angle for the unit circle. But in our setup, x and y are related to s and t through the equations x = t - s -1 and y =2 -2s -2t. Therefore, θ can be in any range as per the unit circle, but the constraints on s and t (since they are points on the sides BC and CA) may restrict θ to certain values. However, since we already accounted for the constraints DE · DF ≤25/16 leading to cosθ ∈ [-1, -2/3], perhaps sinθ can be positive or negative. Wait, but when cosθ = -2/3, sinθ is sqrt(1 - (4/9))=sqrt(5/9)=sqrt5/3 or -sqrt5/3. But in our case, does sinθ have a sign? Let's check.If cosθ =-2/3, then from our expressions:s = - (cosθ)/2 - (sinθ)/4 = - (-2/3)/2 - (sinθ)/4 =1/3 - (sinθ)/4t=1 + (cosθ)/2 - (sinθ)/4=1 + (-2/3)/2 - (sinθ)/4=1 -1/3 - (sinθ)/4=2/3 - (sinθ)/4But E is on BC: BC goes from (1,0) to (0,2). So E=(1 -t, 2t). Let's plug t=2/3 - (sinθ)/4:So E_x=1 - [2/3 - (sinθ)/4 ]=1 -2/3 + (sinθ)/4=1/3 + (sinθ)/4E_y=2t=2[2/3 - (sinθ)/4 ]=4/3 - (sinθ)/2Similarly, F=(-s, 2 -2s)=-[1/3 - (sinθ)/4 ] , 2 -2[1/3 - (sinθ)/4 ]= (-1/3 + (sinθ)/4, 2 -2/3 + (sinθ)/2 )=( -1/3 + (sinθ)/4, 4/3 + (sinθ)/2 )Now, since E and F must lie on BC and CA respectively, their coordinates must satisfy certain conditions. For example, for E on BC: the x-coordinate must be between 0 and1, and y-coordinate between0 and2. Similarly, F on CA: x between -1 and0, y between0 and2.So for E:E_x=1/3 + (sinθ)/4 must be between0 and1. If sinθ= sqrt5/3, then E_x=1/3 + (sqrt5)/12 ≈1/3 +0.206≈0.539, which is okay. If sinθ=-sqrt5/3, then E_x=1/3 - sqrt5/12≈0.333 -0.206≈0.127, still between0 and1.Similarly, E_y=4/3 - (sinθ)/2. If sinθ=sqrt5/3, E_y≈4/3 -0.745/2≈4/3 -0.372≈1.333 -0.372≈0.961, which is between0 and2. If sinθ=-sqrt5/3, E_y≈4/3 - (-0.745)/2≈1.333 +0.372≈1.705, still okay.For F:F_x=-1/3 + (sinθ)/4. If sinθ=sqrt5/3, F_x≈-0.333 +0.206≈-0.127, which is between -1 and0. If sinθ=-sqrt5/3, F_x≈-0.333 -0.206≈-0.539, also okay.F_y=4/3 + (sinθ)/2. If sinθ=sqrt5/3, F_y≈1.333 +0.372≈1.705; if sinθ=-sqrt5/3,≈1.333 -0.372≈0.961. Both okay.So points E and F are valid for both sinθ positive and negative. Therefore, θ can correspond to angles where sinθ is positive or negative, but since we have a range for cosθ, the sinθ can vary accordingly. However, in our expression for EF · BA, which is -2cosθ, the value depends only on cosθ. So even though θ can have different sinθ values, as long as cosθ is within [-1, -2/3], EF · BA will be within [4/3, 2]. Therefore, the range is indeed [4/3, 2].Therefore, the answer is boxed{left[ dfrac{4}{3}, 2 right]}.Final Answerboxed{left[ dfrac{4}{3}, 2 right]}</think>Given an isosceles triangle ( triangle ABC ) with ( AC = BC = sqrt{5} ) and ( AB = 2 ). Points ( D ), ( E ), and ( F ) are on sides ( AB ), ( BC ), and ( CA ) respectively, such that ( AD = DB = 1 ) and ( EF = 1 ). We need to find the range of ( overrightarrow{EF} cdot overrightarrow{BA} ) given ( overrightarrow{DE} cdot overrightarrow{DF} leq frac{25}{16} ).1. Coordinates Setup: - ( A(-1, 0) ), ( B(1, 0) ), ( C(0, 2) ), and ( D(0, 0) ). - Parameterize points ( E ) and ( F ) on ( BC ) and ( CA ) respectively: - ( E(1 - t, 2t) ) - ( F(-s, 2 - 2s) )2. Distance Constraint: - ( EF = 1 ) gives the equation: [ (t - s - 1)^2 + (2 - 2s - 2t)^2 = 1 ]3. Dot Product Constraint: - Vectors ( overrightarrow{DE} ) and ( overrightarrow{DF} ): - ( overrightarrow{DE} = (1 - t, 2t) ) - ( overrightarrow{DF} = (-s, 2 - 2s) ) - Dot product ( overrightarrow{DE} cdot overrightarrow{DF} ): [ -s + 4t - 3ts leq frac{25}{16} ]4. Expression Simplification: - Express ( s ) and ( t ) in terms of ( theta ): - ( t = 1 + frac{costheta}{2} - frac{sintheta}{4} ) - ( s = -frac{costheta}{2} - frac{sintheta}{4} ) - Dot product ( overrightarrow{EF} cdot overrightarrow{BA} ): [ 2(s + 1 - t) = -2costheta ]5. Range Determination: - Solve the quadratic inequality from the constraint: [ 15cos^2theta + 64costheta + 36 leq 0 ] - Solutions for ( costheta ) are ( [-1, -frac{2}{3}] ).6. Final Range: - Convert ( costheta ) range to ( overrightarrow{EF} cdot overrightarrow{BA} ): [ costheta in [-1, -frac{2}{3}] implies -2costheta in left[ frac{4}{3}, 2 right] ]The range of ( overrightarrow{EF} cdot overrightarrow{BA} ) is (boxed{left[ dfrac{4}{3}, 2 right]}).,<think>Okay, so I have this geometry problem here. Let me try to parse it step by step. The problem is about an isosceles triangle ABC where AC equals BC, both being √5. Points D, E, F are on sides AB, BC, and CA respectively, with AD = DB = EF = 1. Then, given that the dot product of vectors DE and DF is less than or equal to 25/16, we need to find the range of the dot product of vectors EF and BA.First, let me sketch the triangle. Since ABC is isosceles with AC = BC = √5, the base must be AB, right? Because in an isosceles triangle, the two equal sides are the legs, and the unequal side is the base. So, AC and BC are the legs, each of length √5, and AB is the base. Therefore, vertex C is the apex.Now, points D, E, F are on AB, BC, CA respectively. AD = DB = 1. So, since AD and DB are both 1, the entire length of AB is AD + DB = 1 + 1 = 2. So, AB is 2 units long. So, triangle ABC has sides AC = BC = √5 and AB = 2. Let me check if that makes sense. Let me compute the length of AB using the distance formula if I assign coordinates to the triangle. Maybe that would help.Let me place the triangle in a coordinate system. Let's put point A at (-1, 0) and point B at (1, 0), so that AB is along the x-axis with length 2. Then point C, being the apex of the isosceles triangle, should be somewhere along the y-axis. Let me compute the coordinates of C. Since AC = √5, the distance from A(-1,0) to C(0, h) is √[(0 - (-1))² + (h - 0)²] = √[1 + h²] = √5. Therefore, 1 + h² = 5 => h² = 4 => h = 2 or h = -2. Since it's a triangle above the base AB, h is positive. So, point C is at (0, 2).So, coordinates: A(-1, 0), B(1, 0), C(0, 2). Then, point D is on AB with AD = 1. Since AB is from (-1, 0) to (1, 0), length AB is 2. So, D is the midpoint? Wait, AD = 1. Since AB is 2 units, AD is half of AB. So, D is the midpoint of AB. Wait, but AB is from -1 to 1 on the x-axis. The midpoint would be (0,0). But that would make AD = distance from (-1,0) to (0,0) which is 1. Similarly, DB is from (0,0) to (1,0), which is also 1. So, D is at (0,0). Okay, that makes sense.So, D is the origin (0,0). Then points E and F are on BC and CA respectively, such that EF = 1. Also, we have to consider vectors DE and DF, compute their dot product, and given that this dot product is ≤ 25/16, find the range of the dot product EF · BA.Hmm. Let me break this down.First, let's parameterize points E and F. Since E is on BC and F is on CA, let me find their coordinates in terms of parameters.Point B is (1, 0), and point C is (0, 2). So, the line BC can be parameterized. Let me use a parameter t for point E. Let me let E divide BC in some ratio. Similarly, F divides CA in some ratio. Let me choose parameters for E and F.Alternatively, since EF = 1, maybe there's a relation between their positions? Hmm, not sure yet. Let me try to parameterize E and F first.Let me parametrize point E on BC. Since BC goes from (1,0) to (0,2), we can write E as a linear combination:E = B + t*(C - B) = (1, 0) + t*(-1, 2) = (1 - t, 0 + 2t) = (1 - t, 2t), where t ∈ [0,1].Similarly, point F is on CA. CA goes from C(0,2) to A(-1,0). Let me parameterize F similarly:F = C + s*(A - C) = (0,2) + s*(-1, -2) = (0 - s, 2 - 2s) = (-s, 2 - 2s), where s ∈ [0,1].So, E is (1 - t, 2t) and F is (-s, 2 - 2s). Then, vector EF is F - E = (-s - (1 - t), (2 - 2s) - 2t) = (-s -1 + t, 2 - 2s - 2t).Given that EF has length 1, so:|EF|² = (-s -1 + t)² + (2 - 2s - 2t)² = 1.Let me expand this:First component squared: (t - s -1)²Second component squared: (2 - 2s - 2t)² = [2(1 - s - t)]² = 4(1 - s - t)²Therefore, the equation becomes:(t - s -1)² + 4(1 - s - t)² = 1.Let me denote u = s + t. Then, the equation becomes:(t - s -1)^2 + 4(1 - u)^2 = 1.But t - s -1 = (t + s) - 2s -1 = u - 2s -1. Hmm, maybe this substitution complicates things. Alternatively, let me expand both terms:First term: (t - s -1)^2 = (t - s)^2 - 2(t - s) + 1Second term: 4(1 - s - t)^2 = 4[(1 - (s + t))^2] = 4[1 - 2(s + t) + (s + t)^2]So, sum of both terms:(t - s)^2 - 2(t - s) + 1 + 4 - 8(s + t) + 4(s + t)^2 = 1Combine constants: 1 + 4 = 5Then:(t - s)^2 -2(t - s) -8(s + t) + 4(s + t)^2 +5 =1Simplify:(t - s)^2 +4(s + t)^2 -2(t - s) -8(s + t) +5 -1=0So,(t - s)^2 +4(s + t)^2 -2(t - s) -8(s + t) +4=0Hmm, this is getting messy. Maybe I need a different approach.Alternatively, since EF = 1, maybe we can express this constraint in terms of s and t. But I need another equation to relate s and t. Wait, but there's another condition given: the dot product of DE and DF is ≤25/16. So maybe first, I need to express vectors DE and DF in terms of s and t, compute their dot product, set it ≤25/16, and then find the corresponding range for EF · BA.But BA is the vector from B to A, which is A - B = (-1, 0) - (1, 0) = (-2, 0). So, BA is (-2, 0). The vector EF is already given above as (-s -1 + t, 2 - 2s - 2t). Therefore, the dot product EF · BA is:(-s -1 + t)*(-2) + (2 - 2s - 2t)*0 = (-s -1 + t)*(-2) + 0 = 2(s +1 - t).So, EF · BA = 2(s +1 - t). So, our target is to find the range of 2(s +1 - t), given that DE · DF ≤25/16 and EF=1.Therefore, I need to express DE · DF in terms of s and t, set up the inequality, and then find the possible values of 2(s +1 - t).So, first, compute vectors DE and DF.Point D is at (0,0).Point E is (1 - t, 2t), so vector DE is E - D = (1 - t, 2t).Similarly, point F is (-s, 2 - 2s), so vector DF is F - D = (-s, 2 - 2s).Therefore, the dot product DE · DF is:(1 - t)(-s) + (2t)(2 - 2s) = -s(1 - t) + 4t - 4ts= -s + st +4t -4ts= -s + (st -4ts) +4t= -s -3ts +4tFactor terms:= -s(1 +3t) +4tSo, DE · DF = -s(1 +3t) +4t.We are given that this is ≤25/16.So, -s(1 +3t) +4t ≤25/16.Additionally, we have the EF=1 condition, which gives:(t - s -1)^2 +4(1 - s - t)^2 =1.So, we have two equations involving s and t:1. (t - s -1)^2 +4(1 - s - t)^2 =12. -s(1 +3t) +4t ≤25/16Our variables are s and t, which are parameters between 0 and1, since E is on BC and F is on CA.Our goal is to find the range of EF · BA = 2(s +1 - t).So, we need to find all possible values of 2(s +1 - t) given that s and t satisfy equation 1 and inequality 2.Therefore, the problem reduces to optimizing 2(s +1 - t) under the constraints:1. (t - s -1)^2 +4(1 - s - t)^2 =12. -s(1 +3t) +4t ≤25/16So, this is an optimization problem with a quadratic constraint and a linear inequality constraint. To find the range of 2(s +1 - t), we need to find its maximum and minimum values subject to these constraints.Let me denote x = s and y = t. Then, variables x and y are in [0,1]?Wait, but since E is on BC and F is on CA, when parameterizing E as (1 - t, 2t), t ranges from 0 to1, since when t=0, E is at B(1,0), and t=1, E is at C(0,2). Similarly, F is parameterized as (-s, 2 - 2s), so s=0 gives F=C(0,2), and s=1 gives F=A(-1,0). So, yes, s and t are in [0,1].So, x, y ∈ [0,1].So, constraints:1. (y - x -1)^2 +4(1 - x - y)^2 =12. -x(1 +3y) +4y ≤25/16Objective function: 2(x +1 - y)So, first, let's analyze the first constraint. Let me expand it:(y - x -1)^2 +4(1 - x - y)^2 =1Let me compute each term:First term: (y - x -1)^2 = [-(x - y +1)]^2 = (x - y +1)^2Second term: 4(1 - x - y)^2So, total: (x - y +1)^2 +4(1 - x - y)^2 =1Let me set u = x + y. Then, the second term becomes 4(1 - u)^2. The first term is (x - y +1)^2. Let me also set v = x - y. Then, the first term is (v +1)^2. But maybe substitution isn't the way to go here. Alternatively, expand the terms:First term: (x - y +1)^2 = (x - y)^2 + 2(x - y) +1Second term:4(1 - x - y)^2 =4[(1 - (x + y))^2] =4[1 - 2(x + y) + (x + y)^2]So, sum:(x - y)^2 +2(x - y) +1 +4 -8(x + y) +4(x + y)^2 =1Combine constants:1 +4=5Then:(x - y)^2 +2(x - y) -8(x + y) +4(x + y)^2 +5=1Simplify:(x - y)^2 +4(x + y)^2 +2(x - y) -8(x + y) +4=0Let me compute each term:(x - y)^2 =x² -2xy + y²4(x + y)^2 =4x² +8xy +4y²2(x - y)=2x -2y-8(x + y)= -8x -8ySo, combining all terms:(x² -2xy + y²) + (4x² +8xy +4y²) + (2x -2y) + (-8x -8y) +4=0Combine like terms:x² +4x² =5x²-2xy +8xy=6xyy² +4y²=5y²2x -8x= -6x-2y -8y= -10ySo, total:5x² +6xy +5y² -6x -10y +4=0So, the first constraint is:5x² +6xy +5y² -6x -10y +4=0Hmm, that's a quadratic equation in x and y. It might represent an ellipse or some conic section. Let me check.The general quadratic equation Ax² + Bxy + Cy² + Dx + Ey + F =0.Here, A=5, B=6, C=5, D=-6, E=-10, F=4.To determine the type of conic, we can compute the discriminant B² -4AC.B² -4AC=36 -100= -64 <0, so it's an ellipse.Therefore, our first constraint is an ellipse in the xy-plane (with x and y in [0,1]). The second constraint is a linear inequality: -x(1 +3y) +4y ≤25/16.So, the problem is to find the maximum and minimum of 2(x +1 - y) over the ellipse 5x² +6xy +5y² -6x -10y +4=0 and under the inequality -x(1 +3y) +4y ≤25/16, with x,y ∈ [0,1].This seems complicated. Maybe parametrize the ellipse?Alternatively, since it's a quadratic constraint, we can use Lagrange multipliers to find extrema of the objective function 2(x +1 - y) subject to the ellipse constraint. But we also have the inequality constraint. Hmm.Alternatively, since the problem is likely symmetric or can be transformed into a simpler coordinate system, maybe we can rotate the coordinate system to eliminate the cross term (the xy term).The ellipse equation is 5x² +6xy +5y² -6x -10y +4=0.The quadratic terms are 5x² +6xy +5y². The matrix for the quadratic form is:[5 33 5]The eigenvalues can be found by solving det([5 - λ, 3], [3,5 - λ])=0.So, (5 - λ)^2 -9=0 → λ² -10λ +25 -9=λ² -10λ +16=0.Solutions: λ=(10 ±√(100 -64))/2=(10 ±6)/2= (16)/2=8 or (4)/2=2. So eigenvalues 8 and 2. Therefore, the ellipse can be transformed into standard form by rotating the coordinate system.The angle of rotation θ satisfies tan(2θ)= B/(A - C). Here, B=6, A - C=0, since A=C=5. Therefore tan(2θ)= ∞, so 2θ=π/2 → θ=π/4. Therefore, rotate the coordinate system by 45 degrees.Let me perform a rotation by θ=45 degrees. Let:x = (x' - y')/√2y = (x' + y')/√2But this might get messy. Alternatively, let u = x + y and v = x - y. Then:x = (u + v)/2y = (u - v)/2Substitute into the quadratic equation:5x² +6xy +5y² -6x -10y +4=0Compute x²:[(u + v)/2]^2 = (u² +2uv +v²)/4xy = [(u + v)/2][(u - v)/2] = (u² - v²)/4y² = [(u - v)/2]^2 = (u² -2uv +v²)/4So, plugging into the equation:5*(u² +2uv +v²)/4 +6*(u² - v²)/4 +5*(u² -2uv +v²)/4 -6*(u + v)/2 -10*(u - v)/2 +4=0Multiply through by 4 to eliminate denominators:5(u² +2uv +v²) +6(u² - v²) +5(u² -2uv +v²) -12(u + v) -20(u - v) +16=0Expand:5u² +10uv +5v² +6u² -6v² +5u² -10uv +5v² -12u -12v -20u +20v +16=0Combine like terms:u²:5u² +6u² +5u²=16u²uv:10uv -10uv=0v²:5v² -6v² +5v²=4v²u terms:-12u -20u= -32uv terms:-12v +20v=8vConstants:+16So, equation becomes:16u² +4v² -32u +8v +16=0Divide all terms by 4:4u² +v² -8u +2v +4=0Complete the square for u and v.For u:4u² -8u =4(u² -2u) =4[(u -1)^2 -1]For v:v² +2v = (v +1)^2 -1So, substituting back:4[(u -1)^2 -1] +[(v +1)^2 -1] +4=0Expand:4(u -1)^2 -4 + (v +1)^2 -1 +4=0Combine constants: -4 -1 +4= -1So,4(u -1)^2 + (v +1)^2 -1=0 → 4(u -1)^2 + (v +1)^2=1So, this is an ellipse centered at (u=1, v=-1) with semi-axes lengths 1/2 along the u-axis and 1 along the v-axis.Since u =x + y and v =x - y.Therefore, the ellipse equation in terms of u and v is 4(u -1)^2 + (v +1)^2=1.So, this is an ellipse with center at u=1, v=-1, major axis along v (since the coefficient for v is smaller? Wait, no: the semi-major axis is determined by the reciprocal of the square root of the coefficient. So, for the term 4(u -1)^2, the semi-axis length is 1/sqrt(4)=1/2, and for (v +1)^2, the semi-axis length is 1. Therefore, it's an ellipse stretched along the v-axis (the vertical axis in u-v coordinates).But how does this help us? Well, perhaps we can parametrize the ellipse.Let me parametrize the ellipse as:u -1 = (1/2)cosθv +1 = sinθTherefore,u =1 + (1/2)cosθv =-1 + sinθThen, since u =x + y and v =x - y,We have:x + y =1 + (1/2)cosθx - y =-1 + sinθSolving for x and y:Add the two equations:2x = (1 + (1/2)cosθ) + (-1 + sinθ) = (1 -1) + (1/2)cosθ + sinθ = (1/2)cosθ + sinθTherefore,x = (1/2)( (1/2)cosθ + sinθ ) = (1/4)cosθ + (1/2)sinθSubtract the two equations:2y = (1 + (1/2)cosθ) - (-1 + sinθ) =1 + (1/2)cosθ +1 - sinθ =2 + (1/2)cosθ - sinθTherefore,y = (2 + (1/2)cosθ - sinθ)/2 =1 + (1/4)cosθ - (1/2)sinθSo, x and y in terms of θ:x = (1/4)cosθ + (1/2)sinθy =1 + (1/4)cosθ - (1/2)sinθWhere θ ∈ [0, 2π).So, now we can express s = x and t = y. So, s and t are parametrized in terms of θ. Therefore, we can express everything in terms of θ.Our objective function is 2(s +1 - t) =2(x +1 - y) =2[ x +1 - y ].Plugging in x and y:2[ (1/4 cosθ +1/2 sinθ ) +1 - (1 +1/4 cosθ -1/2 sinθ ) ]Simplify inside the brackets:(1/4 cosθ +1/2 sinθ ) +1 -1 -1/4 cosθ +1/2 sinθThe +1 and -1 cancel.1/4 cosθ -1/4 cosθ +1/2 sinθ +1/2 sinθ =0 + sinθSo, the objective function becomes 2*sinθ.Therefore, EF · BA =2 sinθ. So, our target is to find the range of 2 sinθ, given the other constraint.But wait, there's another constraint: -s(1 +3t) +4t ≤25/16.Expressed in terms of θ:-s(1 +3t) +4t = -x(1 +3y) +4yWe have x and y in terms of θ. Let me compute this expression.First, compute 1 +3y:1 +3y =1 +3[1 + (1/4)cosθ - (1/2)sinθ ]=1 +3 + (3/4)cosθ - (3/2)sinθ=4 + (3/4)cosθ - (3/2)sinθThen, -x(1 +3y)= -[ (1/4 cosθ +1/2 sinθ ) ] * [4 + (3/4)cosθ - (3/2)sinθ ]Then, adding 4y:+4y=4[1 + (1/4)cosθ - (1/2)sinθ ]=4 + cosθ -2 sinθSo, total expression:- x(1 +3y) +4y= -[ (1/4 cosθ +1/2 sinθ ) * (4 + (3/4)cosθ - (3/2)sinθ ) ] +4 + cosθ -2 sinθFirst, compute the product term:Let me denote A =1/4 cosθ +1/2 sinθand B=4 + (3/4)cosθ - (3/2)sinθThen, A * B = (1/4 cosθ)(4) + (1/4 cosθ)(3/4 cosθ) + (1/4 cosθ)(-3/2 sinθ) + (1/2 sinθ)(4) + (1/2 sinθ)(3/4 cosθ) + (1/2 sinθ)(-3/2 sinθ)Compute term by term:1. (1/4 cosθ)(4)=cosθ2. (1/4 cosθ)(3/4 cosθ)=3/16 cos²θ3. (1/4 cosθ)(-3/2 sinθ)= -3/8 cosθ sinθ4. (1/2 sinθ)(4)=2 sinθ5. (1/2 sinθ)(3/4 cosθ)=3/8 sinθ cosθ6. (1/2 sinθ)(-3/2 sinθ)= -3/4 sin²θSo, summing all terms:cosθ +3/16 cos²θ -3/8 cosθ sinθ +2 sinθ +3/8 cosθ sinθ -3/4 sin²θSimplify:cosθ +3/16 cos²θ + (-3/8 +3/8)cosθ sinθ +2 sinθ -3/4 sin²θThe cosθ sinθ terms cancel.So, remaining terms:cosθ +3/16 cos²θ +2 sinθ -3/4 sin²θTherefore, A*B=cosθ +3/16 cos²θ +2 sinθ -3/4 sin²θThen, -A*B= -cosθ -3/16 cos²θ -2 sinθ +3/4 sin²θAdding 4 + cosθ -2 sinθ:Total expression:(-cosθ -3/16 cos²θ -2 sinθ +3/4 sin²θ ) +4 + cosθ -2 sinθSimplify:-cosθ + cosθ =0-3/16 cos²θ-2 sinθ -2 sinθ = -4 sinθ3/4 sin²θ+4So, altogether:-3/16 cos²θ -4 sinθ +3/4 sin²θ +4Therefore, the inequality is:-3/16 cos²θ -4 sinθ +3/4 sin²θ +4 ≤25/16Multiply both sides by 16 to eliminate denominators:-3 cos²θ -64 sinθ +12 sin²θ +64 ≤25Bring all terms to the left:-3 cos²θ -64 sinθ +12 sin²θ +64 -25 ≤0Simplify constants:64 -25=39So:-3 cos²θ -64 sinθ +12 sin²θ +39 ≤0Now, recall that cos²θ =1 - sin²θ. Substitute:-3(1 - sin²θ) -64 sinθ +12 sin²θ +39 ≤0Expand:-3 +3 sin²θ -64 sinθ +12 sin²θ +39 ≤0Combine like terms:3 sin²θ +12 sin²θ=15 sin²θ-64 sinθ-3 +39=36So:15 sin²θ -64 sinθ +36 ≤0Therefore, the inequality reduces to:15 sin²θ -64 sinθ +36 ≤0Let me write this quadratic in terms of z = sinθ:15 z² -64 z +36 ≤0Solve quadratic inequality 15 z² -64 z +36 ≤0.First, find roots:z = [64 ±√(64² -4*15*36)]/(2*15)Compute discriminant:64²=40964*15*36=2160Discriminant=4096 -2160=1936√1936=44Thus,z=(64 ±44)/30So, two roots:z=(64 +44)/30=108/30=18/5=3.6But sinθ cannot exceed 1, so this root is invalid.Other root:z=(64 -44)/30=20/30=2/3≈0.6667Therefore, the quadratic 15 z² -64 z +36 is a parabola opening upwards (since coefficient of z² is positive), so it is ≤0 between its roots. But one root is at z=2/3 and the other at z=18/5, which is invalid. Since sinθ ∈ [-1,1], the inequality 15 z² -64 z +36 ≤0 holds only for z between 2/3 and 18/5. But since z=sinθ ≤1, the valid interval is z ∈ [2/3,1]. Therefore, the inequality 15 sin²θ -64 sinθ +36 ≤0 is equivalent to sinθ ≥2/3.Wait, no. Since the parabola opens upwards, it's ≤0 between the roots. But since only z=2/3 is within the valid range, the inequality holds for z ∈ [2/3,1].Wait, but let's confirm.If we plot 15 z² -64 z +36, it's positive outside [2/3, 18/5], but within z ∈ [2/3,1], since 18/5=3.6 >1, the inequality is true for z ∈ [2/3,1]. Therefore, the original inequality -s(1 +3t) +4t ≤25/16 is equivalent to sinθ ≥2/3.Therefore, our problem now is to find the range of 2 sinθ, given that sinθ ≥2/3 and θ parametrizes the ellipse.But wait, EF · BA =2 sinθ, and we need to find its range when sinθ ≥2/3. But θ is such that points E and F are on BC and CA respectively, with EF=1. So, the ellipse parametrization already accounts for the EF=1 constraint. The additional constraint is sinθ ≥2/3. So, essentially, we need to find the maximum and minimum of 2 sinθ over the portion of the ellipse where sinθ ≥2/3.But since the ellipse is parametrized with θ, and the sinθ ≥2/3, then the possible values of θ are those where sinθ ≥2/3. So, the range of 2 sinθ would be from 2*(2/3)=4/3 up to the maximum possible value of 2 sinθ on the ellipse.But wait, we need to check if sinθ can actually reach 1 on the ellipse.Wait, from the parametrization of the ellipse, θ is an angle parameter, but does θ correspond to the actual angle in the ellipse parametrization? Or is it just a parameter?Wait, in our parametrization:u =1 + (1/2)cosθv =-1 + sinθBut u and v are related to x and y through:u =x + yv =x - yBut x and y are parameters in [0,1], so u ranges from 0+0=0 to1+1=2, and v ranges from -1 to1 (since x - y can be between -1 and1 when x,y∈[0,1]).But our ellipse equation in u and v is 4(u -1)^2 + (v +1)^2=1.So, the ellipse is centered at u=1, v=-1, with u ranging from 1 - 1/2 to1 +1/2, so u ∈ [0.5,1.5], and v ranging from -1 -1 to -1 +1, so v ∈ [-2,0].But x and y are both between0 and1. So, u =x + y ∈ [0,2], but the ellipse restricts u to [0.5,1.5]. Similarly, v=x - y ∈ [-1,1], but the ellipse restricts v to [-2,0]. However, since x and y are between0 and1, v=x - y ∈ [-1,1], so intersection with ellipse's v ∈ [-2,0] gives v ∈ [-1,0]. So, the actual feasible region for u and v is the intersection of the ellipse and u ∈ [0,2], v ∈ [-1,0].But maybe this is complicating.Alternatively, since we have a parametrization in terms of θ, we can see what θ corresponds to points on the ellipse where x and y are in [0,1].Given x = (1/4)cosθ + (1/2)sinθ and y=1 + (1/4)cosθ - (1/2)sinθ, we need to ensure x ∈ [0,1] and y ∈ [0,1].So, constraints:For x:0 ≤(1/4)cosθ + (1/2)sinθ ≤1For y:0 ≤1 + (1/4)cosθ - (1/2)sinθ ≤1Let me analyze these inequalities.First, for y:Lower bound:1 + (1/4)cosθ - (1/2)sinθ ≥0Upper bound:1 + (1/4)cosθ - (1/2)sinθ ≤1 → (1/4)cosθ - (1/2)sinθ ≤0 → (1/4)cosθ ≤ (1/2)sinθ → (1/2)cosθ ≤ sinθ → tanθ ≥1/2So, tanθ ≥1/2. Which implies θ ∈ [arctan(1/2), π/2) or θ ∈ [arctan(1/2)+π, 3π/2). But since θ is a parameter for the ellipse, which is 2π periodic, but given the original problem, we might need to consider θ in a certain range.But this is getting too involved. Alternatively, since we have parametrized θ and we need to find the maximum and minimum of 2 sinθ with sinθ ≥2/3, perhaps we can proceed as follows.Given that sinθ ≥2/3, and θ is such that x and y are in [0,1]. We need to find the allowable θ where sinθ ≥2/3, and then find the maximum and minimum of 2 sinθ in this region.But first, we need to verify if sinθ can reach up to 1. Let's check.In our parametrization:sinθ is part of the parametrization, but due to the constraints on x and y, θ is restricted.For example, let's check when sinθ=1. Then θ=π/2.Then x=(1/4)*0 + (1/2)*1=1/2y=1 + (1/4)*0 - (1/2)*1=1 -1/2=1/2So, x=1/2, y=1/2. Both in [0,1]. So, this is allowed.Similarly, check when sinθ=2/3. Then θ=arcsin(2/3)≈0.7297 radians.Compute x and y:x=(1/4)cosθ + (1/2)(2/3)= (1/4)sqrt(1 - (4/9)) +1/3= (1/4)(sqrt(5/9)) +1/3= (1/4)(√5 /3)+1/3≈(1/4)(2.236/3)+1/3≈0.186 +0.333≈0.519Similarly, y=1 + (1/4)cosθ - (1/2)(2/3)=1 + (1/4)(sqrt(5)/3) -1/3≈1 +0.186 -0.333≈0.853Which is still within [0,1]. So, the point is valid.Therefore, sinθ can range from 2/3 to1, and thus 2 sinθ can range from4/3 to2.But wait, is that correct? Let me verify.Wait, the maximum value of sinθ is1, so 2 sinθ=2. The minimum is2/3, so2*(2/3)=4/3. So, the range would be [4/3,2]. But we need to confirm if all values between4/3 and2 are achievable given the constraints on x and y.But to be thorough, we need to check if when sinθ varies from2/3 to1, whether there are any restrictions due to x and y staying within [0,1].Wait, earlier when we checked at sinθ=1, x=1/2 and y=1/2, which is within [0,1]. At sinθ=2/3, we had x≈0.519 and y≈0.853, which is also within [0,1].Is there a possibility that for some θ with sinθ between2/3 and1, x or y might exceed [0,1]?Let's check when sinθ is maximum, which is1. We already saw x=1/2, y=1/2.If we take θ approaching π/2 from below, sinθ approaching1, x approaches1/2, y approaches1/2.If we take θ=arcsin(2/3), which is≈0.7297 radians, then as θ increases from arcsin(2/3) to π/2, sinθ increases from2/3 to1, x decreases from≈0.519 to0.5, and y decreases from≈0.853 to0.5.Wait, that seems counterintuitive. Wait, when θ increases, sinθ increases. Let's see.Wait, in the parametrization:x=(1/4)cosθ + (1/2)sinθy=1 + (1/4)cosθ - (1/2)sinθAs θ increases from arcsin(2/3) to π/2:cosθ decreases from sqrt(1 - (4/9))=sqrt(5)/3≈0.745 to0.sinθ increases from2/3 to1.So, x=(1/4)cosθ + (1/2)sinθAt θ=arcsin(2/3):x=(1/4)(sqrt(5)/3) + (1/2)(2/3)≈0.745/4 +0.333≈0.186 +0.333≈0.519At θ=π/2:x=(1/4)(0) + (1/2)(1)=0.5Similarly, y=1 + (1/4)cosθ - (1/2)sinθAt θ=arcsin(2/3):y≈1 +0.186 -0.333≈0.853At θ=π/2:y=1 +0 -0.5=0.5So, as θ increases from arcsin(2/3) to π/2, x decreases from≈0.519 to0.5, and y decreases from≈0.853 to0.5. Both x and y remain within [0,1]. Similarly, if θ decreases from arcsin(2/3) to where?Wait, θ is a parameter for the ellipse. If we consider θ beyond arcsin(2/3) downwards, but the constraint sinθ ≥2/3 restricts θ to [arcsin(2/3), π/2] in the upper half, but since sinθ is positive in [0, π], but due to the ellipse parametrization, perhaps θ is limited.Wait, but we need to check if there are other intervals where sinθ ≥2/3. For example, in the lower half of the ellipse, but since the parametrization might not cover that due to x and y constraints.But in our previous check, when θ=3π/2, sinθ=-1, but x and y would be:x=(1/4)cos(3π/2)+(1/2)sin(3π/2)=0 + (1/2)(-1)=-0.5, which is invalid since x≥0. So, only a portion of the ellipse is valid.Therefore, the valid θ values are those where x and y stay within [0,1]. From our earlier analysis, when sinθ increases from2/3 to1, x and y stay within [0,1]. What about decreasing θ below arcsin(2/3)?If we take θ less than arcsin(2/3), say θ=0, sinθ=0:x=(1/4)(1) + (1/2)(0)=0.25y=1 + (1/4)(1) -0=1.25, which is invalid since y>1. Therefore, θ cannot go below a certain value without making y>1.Wait, but our earlier upper bound on y was y ≤1. The upper bound on y is from the parametrization. Let's check when y=1:y=1 + (1/4)cosθ - (1/2)sinθ=1So, (1/4)cosθ - (1/2)sinθ=0 → (1/4)cosθ=(1/2)sinθ → tanθ=(1/4)/(1/2)=1/2. So θ=arctan(1/2)≈0.4636 radians≈26.565 degrees.At θ=arctan(1/2):sinθ=1/√(1 +4)=1/√5≈0.447 <2/3≈0.666.So, when θ=arctan(1/2), sinθ≈0.447 <2/3. Therefore, in the valid parametrization where y≤1, θ ranges from arctan(1/2) to some upper limit where x or y becomes0.Wait, let's find when y=0:y=1 + (1/4)cosθ - (1/2)sinθ=0→ (1/4)cosθ - (1/2)sinθ= -1Multiply by4:cosθ -2 sinθ= -4But cosθ -2 sinθ= -4. The maximum value of cosθ -2 sinθ is sqrt(1 +4)=sqrt(5)≈2.236. Therefore, this equation has no solution. Therefore, y cannot reach0.Similarly, when does x=0:x=(1/4)cosθ + (1/2)sinθ=0→ (1/4)cosθ= - (1/2)sinθ→ tanθ= -1/2But θ ∈ [arctan(1/2), π/2] in the valid parametrization where y ≤1. tanθ= -1/2 would be in a different quadrant. Therefore, x cannot be0 in the valid θ range.Similarly, x=1:(1/4)cosθ + (1/2)sinθ=1Multiply by4:cosθ +2 sinθ=4But maximum value of cosθ +2 sinθ is sqrt(1 +4)=sqrt(5)≈2.236 <4. Therefore, no solution. Thus x cannot reach1.Therefore, the valid θ range is from θ=arctan(1/2)≈0.4636 radians up to θ where y=1/2. Wait, no. Wait, since when θ increases beyond π/2, sinθ starts decreasing, but we already saw that at θ=π/2, y=0.5. As θ increases past π/2, sinθ decreases, but cosθ becomes negative.Wait, but in our parametrization, θ is just a parameter for the ellipse, not necessarily the actual angle in the coordinate system. So, maybe θ ranges from0 to2π, but due to constraints on x and y, only a subset is valid.But this is getting too complicated. Given that we have a constraint sinθ ≥2/3 and our earlier analysis showed that when sinθ ≥2/3, x and y are within [0,1], then the valid θ for our problem is θ ∈ [arcsin(2/3), π/2], and in this interval, sinθ increases from2/3 to1, making 2 sinθ increase from4/3 to2.But wait, θ=arcsin(2/3) is≈0.7297 radians, and θ=π/2 is≈1.5708 radians. In this interval, sinθ increases, so 2 sinθ increases from4/3 to2. Therefore, the range of EF · BA=2 sinθ is [4/3,2]. But we need to confirm if this is the case.However, the problem states that the dot product DE · DF ≤25/16. We translated this into sinθ ≥2/3. So, given this constraint, EF · BA=2 sinθ must be ≥4/3. What is the upper limit?When does DE · DF reach25/16? When sinθ=2/3. Because our inequality is DE · DF ≤25/16 corresponds to sinθ≥2/3. Wait, no. Wait, we had:DE · DF= -s(1 +3t) +4t= expression that led to15 sin²θ -64 sinθ +36 ≤0 → sinθ ≥2/3. So, DE · DF ≤25/16 corresponds to sinθ ≥2/3.Therefore, when sinθ=2/3, DE · DF=25/16. When sinθ increases beyond2/3, DE · DF becomes less than25/16. Wait, no, hold on.Wait, originally, the inequality is DE · DF ≤25/16. We transformed this into sinθ ≥2/3. So, when sinθ=2/3, DE · DF=25/16, and when sinθ increases beyond2/3, DE · DF decreases, since the expression DE · DF is a quadratic that opens upwards, so when sinθ increases, the value goes below25/16. Therefore, the permissible values are sinθ ≥2/3, which correspond to DE · DF ≤25/16.Therefore, the range of EF · BA=2 sinθ is [4/3, maximum value].But what is the maximum value of2 sinθ on the ellipse under the constraint x,y∈[0,1].Earlier, we saw that at θ=π/2, sinθ=1, and x=0.5, y=0.5, which are within [0,1]. So, maximum EF · BA=2*1=2.But wait, let me verify with the actual coordinates.When θ=π/2:E is(1 - t, 2t). Wait, no, E is parameterized as(1 - t, 2t). Wait, but we have t=y=0.5.Wait, no. Wait, when θ=π/2, we have:From parametrization:x=(1/4)cosθ + (1/2)sinθ=0 + (1/2)(1)=0.5=sy=1 + (1/4)cosθ - (1/2)sinθ=1 +0 -0.5=0.5=tSo, s=0.5 and t=0.5.Therefore, point E is on BC: E=(1 - t, 2t)= (1 -0.5, 2*0.5)=(0.5,1)Point F is on CA: F=(-s, 2 -2s)=(-0.5, 2 -1)=(-0.5,1)Therefore, vector EF= F - E=(-0.5 -0.5,1 -1)=(-1,0)So, EF=(-1,0), which has length1, which matches the given condition EF=1.Then, BA vector is(-2,0), so EF · BA=(-1,0)·(-2,0)=2 +0=2. Which matches 2 sinθ=2*1=2.Similarly, at θ=arcsin(2/3):sinθ=2/3, cosθ=√(1 - (4/9))=√(5/9)=√5/3≈0.745s=x=(1/4)(√5/3) + (1/2)(2/3)= (√5)/12 +1/3≈0.186 +0.333≈0.519t=y=1 + (1/4)(√5/3) - (1/2)(2/3)=1 + √5/12 -1/3≈1 +0.186 -0.333≈0.853Point E=(1 - t, 2t)= (1 -0.853, 2*0.853)=≈(0.147,1.706)Point F=(-s, 2 -2s)=≈(-0.519,2 -1.038)=≈(-0.519,0.962)Vector EF= F - E≈(-0.519 -0.147,0.962 -1.706)=≈(-0.666,-0.744)Wait, the length of EF should be1. Let me check:(-0.666)^2 + (-0.744)^2≈0.443 +0.553≈0.996≈1. Correct.Vector BA=(-2,0)So, EF · BA=(-0.666)(-2) + (-0.744)(0)=1.332≈4/3≈1.333. Close enough considering rounding.Thus, EF · BA≈1.332≈4/3 when sinθ=2/3.Therefore, the range of EF · BA is from4/3 to2.But the problem asks for the range of EF · BA given that DE · DF ≤25/16. Since DE · DF ≤25/16 corresponds to sinθ ≥2/3, which in turn gives EF · BA=2 sinθ ∈ [4/3,2].Therefore, the answer should be [4/3,2], which in boxed form is boxed{[dfrac{4}{3}, 2]}.But let me cross-verify once more to ensure no mistakes.From the parametrization, we found that EF · BA=2 sinθ, and the constraint DE · DF ≤25/16 imposes sinθ ≥2/3. The maximum value of sinθ on the valid portion of the ellipse is1, leading to EF · BA=2*1=2. The minimum value is sinθ=2/3, leading to EF · BA=4/3.Therefore, the range is [4/3,2].Final AnswerThe range of overrightarrow{E F} cdot overrightarrow{B A} is boxed{left[ dfrac{4}{3}, 2 right]}.</think>Given an isosceles triangle ( triangle ABC ) with ( AC = BC = sqrt{5} ) and ( AB = 2 ), points ( D, E, F ) are on sides ( AB, BC, CA ) respectively such that ( AD = DB = EF = 1 ). We need to find the range of ( overrightarrow{EF} cdot overrightarrow{BA} ) given ( overrightarrow{DE} cdot overrightarrow{DF} leq frac{25}{16} ).1. Coordinates and Parameterization: - Place ( A ) at ( (-1, 0) ), ( B ) at ( (1, 0) ), and ( C ) at ( (0, 2) ). - ( D ) is the midpoint of ( AB ), hence ( D ) is at ( (0, 0) ). - Parameterize points ( E ) on ( BC ) as ( (1 - t, 2t) ) and ( F ) on ( CA ) as ( (-s, 2 - 2s) ).2. Vectors and Dot Product: - Vectors ( overrightarrow{DE} = (1 - t, 2t) ) and ( overrightarrow{DF} = (-s, 2 - 2s) ). - Compute ( overrightarrow{DE} cdot overrightarrow{DF} = -s(1 + 3t) + 4t ).3. Constraint Analysis: - Given ( overrightarrow{DE} cdot overrightarrow{DF} leq frac{25}{16} ), we derive the quadratic inequality ( 15 sin^2 theta - 64 sin theta + 36 leq 0 ). - Solving this inequality gives ( sin theta geq frac{2}{3} ).4. Objective Function: - The objective function ( overrightarrow{EF} cdot overrightarrow{BA} = 2(s + 1 - t) ) translates to ( 2 sin theta ).5. Range Determination: - The valid range of ( sin theta ) is ( left[ frac{2}{3}, 1 right] ). - Therefore, the range of ( 2 sin theta ) is ( left[ frac{4}{3}, 2 right] ).The range of ( overrightarrow{EF} cdot overrightarrow{BA} ) is (boxed{left[ dfrac{4}{3}, 2 right]}).]

🔑:5. left[frac{4}{3}, 2right].Taking D as the origin, the rays D B, D C as the positive directions of the x, y axes respectively to establish a Cartesian coordinate system, then A(-1,0), B(1,0), C(0,2). Let point Eleft(x_{1}, y_{1}right), Fleft(x_{2}, y_{2}right), where, y_{1}=-2 x_{1}+2, y_{2}=2 x_{2}+2. Let the midpoint of line segment E F be Mleft(x_{0}, y_{0}right), thenleft{begin{array}{l}y_{1}-y_{2}=-2left(x_{1}+x_{2}right)=-4 x_{0} x_{1}-x_{2}=2-frac{y_{1}+y_{2}}{2}=2-y_{0} .end{array}right.By E F=1, we getleft(4 x_{0}right)^{2}+left(2-y_{0}right)^{2}=1Thus, left(4 x_{0}right)^{2}=1-left(2-y_{0}right)^{2} geqslant 0, solving gives1 leqslant y_{0} leqslant 3Also,overrightarrow{D E} cdot overrightarrow{D F}=frac{1}{4}left((overrightarrow{D E}+overrightarrow{D F})^{2}-(overrightarrow{D E}-overrightarrow{D F})^{2}right)=overrightarrow{D M}^{2}-frac{1}{4} overrightarrow{E F}^{2}=overrightarrow{D M}^{2}-frac{1}{4} leqslant frac{25}{16},which means overrightarrow{D M}^{2} leqslant frac{29}{16}, orx_{0}^{2}+y_{0}^{2} leqslant frac{29}{16} .Substituting equation (1) into equation (3), eliminating x_{0}, and rearranging gives 15 y_{0}^{2}-4 y_{0}-32 leqslant 0, solving gives-frac{8}{5} leqslant y_{0} leqslant frac{4}{3}Combining equations (2) and (4), we get 1 leqslant y_{0} leqslant frac{4}{3}, then frac{2}{3} leqslant x_{1}-x_{2} leqslant 1, thereforeoverrightarrow{E F} cdot overrightarrow{B A}=left(x_{2}-x_{1}, y_{2}-y_{1}right) cdot(-2,0)=2left(x_{1}-x_{2}right) inleft[frac{4}{3}, 2right]

🧠:[<think>Alright, so I need to find the remainder when 77^{77} minus 15 is divided by 19. Hmm, modular arithmetic, right? Let me remember how to approach this. First, maybe simplify 77 modulo 19, because if I can find 77 mod 19, that would make the exponent easier to handle. Let me compute 19 times 4: 19*4 is 76. So 77 divided by 19 is 4 with a remainder of 1. So 77 ≡ 1 mod 19. That means 77^{77} ≡ 1^{77} ≡ 1 mod 19. Wait, that seems straightforward. So 77 is congruent to 1 modulo 19, so raising it to any power would still be 1. Then subtracting 15 would give me 1 - 15 = -14 mod 19. But I need a positive remainder, so adding 19 to -14 gives 5. So the remainder is 5? Let me check my steps again to make sure I didn't make a mistake.First step: Find 77 mod 19. 19*4=76, so 77-76=1. Correct. So 77 ≡1 mod19. Then 77^{77} is 1^{77}=1 mod19. Then subtract 15: 1 -15= -14. Since we want the remainder when divided by 19, -14 +19=5. That seems right. But let me verify with another method to ensure I didn't overlook anything. Maybe using Fermat's little theorem?Fermat's little theorem states that if p is prime and a is not divisible by p, then a^{p-1} ≡1 mod p. 19 is prime, and 77 is 1 mod19, so 1 is not divisible by 19. So 1^{18} ≡1 mod19. Wait, but exponent is 77. Let me see. 77 divided by 18. 18*4=72, remainder 5. So 1^{77}=1^{18*4 +5}= (1^{18})^4 *1^5 ≡1^4 *1^5=1*1=1 mod19. So that still gives 1. So same result. So 77^{77}≡1 mod19. Then subtract 15 gives -14≡5 mod19.Alternatively, maybe compute 77^{77} mod19 directly without simplifying the base first? Let me try that. Wait, but 77 is congruent to 1, so even if I do it directly, 77^1≡1, 77^2≡1*1=1, and so on. So regardless of the exponent, it's always 1. So yeah, that's consistent. So the key insight was reducing the base modulo 19 first.But let me check another angle. Suppose I didn't simplify 77 mod19 first. Maybe compute 77^{77} mod19 using exponentiation by squaring? Let's see. But 77 mod19=1, so maybe that's complicating it. Alternatively, since 77 is 1 mod19, exponentiation won't change that. So 1^77=1. So subtract 15 is same as before.Alternatively, maybe think about negative numbers? For instance, 77 - 19*4=1, but 77≡1 mod19. Alternatively, maybe 77 is -18 mod19? Wait, 19*4=76, so 77=76+1, which is 19*4 +1, so 77≡1 mod19. So not negative. So that's not helpful here.Alternatively, perhaps compute 77^{77} as (19*4 +1)^{77} and expand using binomial theorem. But since every term except the last one would have a factor of 19, so modulo 19, all those terms would be 0, leaving only the last term, which is 1^{77}=1. So again, same result.Therefore, regardless of the method, 77^{77} mod19 is 1. Then subtract 15 gives 1 -15 = -14. Now, -14 mod19. To make this positive, add 19: -14 +19=5. Therefore, the remainder is 5. Therefore, the answer is 5. Wait, but let me double-check with actual computation. Maybe compute 77 mod19 again. 19*4=76, 77-76=1. Yes. So 77 is 1 mod19, so 77^{77} is 1^{77}=1. 1-15=-14. -14 mod19. Since remainders are positive, add 19: -14+19=5. So yes, 5 is correct. Alternatively, verify with another example. Suppose we have 20^{20} -5 mod19. 20≡1 mod19, so 1^{20}=1, 1-5=-4≡15 mod19. Which is correct. So same principle. So I think my steps are correct here.So, the remainder is 5. Therefore, the answer should be 5. But wait, let me just compute 77 -15 first? Wait, no. The expression is 77^{77} -15. So need to compute 77^{77} mod19 first, then subtract 15 mod19, then take the result modulo19. So steps: 77 ≡1 mod19. So 77^{77}≡1 mod19. Then subtract 15: 1 -15 ≡-14 mod19. Then -14 +19=5 mod19. So yes.Alternatively, write as (77^{77} mod19 -15 mod19) mod19. Which is (1 -15) mod19= (-14) mod19=5. Correct. So all methods point to 5. I think that's correct. But let me confirm once more with a different approach.Suppose I compute 77 mod19=1. Then 77^{77} mod19=1^{77} mod19=1. Then subtract 15: 1-15=-14. Then -14 mod19. To get the positive equivalent, since 19*0=0, -14 is between -19 and 0, so add 19: 19-14=5. Hence, 5. So all right. So answer is 5. Wait, but maybe I made a mistake in the initial step. Let me check again. 19*4=76. 77-76=1. Therefore, 77≡1 mod19. Correct. So 77^{77}≡1^77≡1. 1-15=-14. -14 mod19. To convert to positive remainder: 19-14=5. So yes. So 5 is the remainder. Therefore, the answer is 5.Alternatively, think in terms of congruence classes. Since 77≡1 mod19, so 77^{77}≡1^{77}≡1. Then 1 -15≡-14≡5 mod19. Yep. So that's consistent. Alternatively, maybe compute 77^{77} as follows: since 77≡1 mod19, 77^1≡1, 77^2≡1*1=1, etc., so every exponent is 1. Therefore, 77^{77}≡1. Then subtract 15: -14≡5 mod19. Same result.Therefore, I think the answer is 5. Therefore, I can confidently say the remainder is 5 when 77^{77}-15 is divided by 19.Final AnswerThe remainder is boxed{5}.</think>To find the remainder when (77^{77} - 15) is divided by 19, we start by simplifying (77 mod 19). First, we note that:[ 77 div 19 = 4 text{ remainder } 1 ]Thus, [ 77 equiv 1 mod 19 ]Next, we raise 1 to the power of 77:[ 77^{77} equiv 1^{77} equiv 1 mod 19 ]We then subtract 15 from this result:[ 1 - 15 = -14 ]To convert (-14) to a positive remainder modulo 19, we add 19:[ -14 + 19 = 5 ]Therefore, the remainder when (77^{77} - 15) is divided by 19 is (boxed{5}).,<think>Okay, so I need to find the remainder when 77^77 minus 15 is divided by 19. Hmm, modular arithmetic, right? Let me remember how to approach this. Since we're dealing with modulo 19, maybe I can simplify 77 modulo 19 first. Let me check: 19 times 4 is 76, so 77 divided by 19 is 4 with a remainder of 1. So, 77 ≡ 1 mod 19. That makes things easier. So, 77^77 mod 19 would be the same as 1^77 mod 19, which is just 1. Then subtract 15, so 1 - 15 = -14. But since we want a positive remainder, we need to add 19 to -14, which gives 5. So the remainder is 5? Wait, let me verify step by step.First, confirm 77 mod 19. 19*4=76, 77-76=1. Yes, 77 ≡1 mod19. So 77^77 ≡1^77=1 mod19. Then subtract 15: 1 -15= -14. Now, -14 mod19. To make this positive, add 19: -14+19=5. So the remainder should be 5. Let me check with smaller exponents to make sure. Suppose 77^1 -15=77-15=62. 62 divided by 19: 19*3=57, 62-57=5. So remainder 5. That matches. What about 77^2 -15. 77^2 is 5929. 5929 -15=5914. Now divide by 19. Let me compute 5914 divided by 19. 19*300=5700, 5914-5700=214. 19*10=190, 214-190=24. 19*1=19, 24-19=5. So total is 300+10+1=311, remainder 5. Again remainder 5. Wait, so even 77^2 -15 gives remainder 5. So maybe for any exponent, 77^n ≡1 mod19, so 1 -15 ≡-14≡5 mod19. So regardless of the exponent, as long as it's 77^n, it's 1^n=1 mod19. Then subtract 15, same result. Therefore, the remainder is always 5. So answer should be 5. But let me test with n=3. Let's compute 77^3 -15. 77^3 =77*77*77=77*5929. Wait, that's a big number. Maybe compute modulo 19 step by step. 77 ≡1 mod19, so 77^3 ≡1^3=1 mod19. Then 1 -15= -14 ≡5 mod19. So even 77^3 -15 ≡5 mod19. So regardless of the exponent, it's 5. So yeah, the answer is 5. Wait, but just to be thorough, maybe check with n=0? 77^0=1. 1 -15= -14 ≡5 mod19. So even n=0, which is interesting. So yeah, regardless of the exponent, as long as 77 ≡1 mod19, any power would be 1. Therefore, 77^77 -15 ≡1 -15 ≡-14≡5 mod19. So the remainder is 5. I think that's correct. Let me check another approach. Maybe using Fermat's little theorem. Since 19 is prime, for any integer a not divisible by 19, a^(18)≡1 mod19. So 77 ≡1 mod19, so 77^18≡1^18=1 mod19. Then 77^77=77^(4*18 + 5)= (77^18)^4 *77^5 ≡1^4 *1^5=1 mod19. So again, 77^77 ≡1 mod19. Then subtract 15 gives -14≡5 mod19. So same answer. So this method also confirms it. Therefore, the remainder is 5. Yeah, I think that's solid. So the answer is 5.Final AnswerThe remainder is boxed{5}.</think>To find the remainder when (77^{77} - 15) is divided by 19, we start by simplifying (77 mod 19). Since (19 times 4 = 76), we have:[77 equiv 77 - 76 = 1 mod 19]Thus, (77^{77} equiv 1^{77} mod 19). Since (1^{77} = 1), we get:[77^{77} equiv 1 mod 19]Next, we subtract 15 from this result:[1 - 15 = -14]To convert (-14) to a positive remainder modulo 19, we add 19:[-14 + 19 = 5]We can verify this result using Fermat's Little Theorem. Since 19 is prime, (a^{18} equiv 1 mod 19) for any integer (a) not divisible by 19. Here, (77 equiv 1 mod 19), so:[77^{77} equiv 1^{77} equiv 1 mod 19]Subtracting 15 gives:[1 - 15 equiv -14 equiv 5 mod 19]Thus, the remainder when (77^{77} - 15) is divided by 19 is (boxed{5}).]

🔑:Since 76 = 19 times 4, we can rewrite 77^{77} - 15 as follows:77^{77} - 15 = (76 + 1)^{77} - 15Using the Binomial Theorem, we can expand the expression as:(76 + 1)^{77} = C_{77}^{0}76^{77} + C_{77}^{1}76^{76} + C_{77}^{2}76^{75} + ... + C_{77}^{76}76 + 1Now, subtract 15:(76 + 1)^{77} - 15 = C_{77}^{0}76^{77} + C_{77}^{1}76^{76} + C_{77}^{2}76^{75} + ... + C_{77}^{76}76 + (1 - 15)Notice that all terms in the expansion contain a factor of 76 (and thus, 19) except for the last term (1 - 15) = -14. Therefore, the remainder when 77^{77} - 15 is divided by 19 is the same as the remainder when -14 is divided by 19.To find this remainder, we can add 19 to -14:-14 + 19 = 5So, the remainder is boxed{5}.