🧠:[<think>Okay, let's see. I need to find the coordinates of vector p in the basis {i, j, k} given that its coordinates in the basis {a, b, c} are (8, 6, 4). The vectors a, b, and c are defined as a = i + j, b = j + k, and c = k + i. Hmm, so the problem is about changing the basis of a vector from one basis to another. I think this involves linear combinations. Let me recall.When a vector is expressed in terms of a basis, it's a linear combination of the basis vectors. So, in the basis {a, b, c}, vector p is 8a + 6b + 4c. But since a, b, and c are themselves expressed in terms of i, j, k, I can substitute those expressions into the equation and then combine like terms to get p in terms of i, j, k. That should give me the coordinates in the standard basis.Let me write that down step by step. First, express p in terms of a, b, c:p = 8a + 6b + 4cNow substitute the definitions of a, b, c:a = i + jb = j + kc = k + iSo substituting these into the equation for p:p = 8(i + j) + 6(j + k) + 4(k + i)Now, I need to distribute the coefficients:p = 8i + 8j + 6j + 6k + 4k + 4iNow combine like terms. Let's group the i terms, j terms, and k terms together.For i: 8i + 4i = 12iFor j: 8j + 6j = 14jFor k: 6k + 4k = 10kSo putting it all together:p = 12i + 14j + 10kTherefore, the coordinates of p in the {i, j, k} basis should be (12, 14, 10). Let me check the answer options. Option A is (12, 14, 10), which matches my result. Let me double-check my steps to make sure I didn't make a mistake.First, substituting the basis vectors: correct. Then distributing the coefficients: 8a becomes 8i +8j, 6b becomes 6j +6k, 4c becomes 4k +4i. Then combining like terms: 8i +4i is 12i, 8j +6j is 14j, 6k +4k is 10k. Yes, that seems right. So the answer should be A.Wait, let me just think again. Is there a possibility that the problem is using a different convention, like the order of the basis vectors? The original basis is {a, b, c}, which are given as a = i + j, b = j + k, c = k + i. So when they give coordinates (8,6,4), that's 8a +6b +4c. So the process is correct. Another way to think about it is to represent this as a matrix multiplication. Let me try that approach to verify.If we consider the basis vectors a, b, c as columns in a matrix, then the coordinates in the new basis can be converted to the standard basis by multiplying this matrix with the coordinate vector. So, the matrix M whose columns are a, b, c is:M = [a | b | c] = [ [1, 0, 1], [1, 1, 0], [0, 1, 1] ]Wait, let's see. Each basis vector is a column in the matrix. So:a = i + j = (1, 1, 0)b = j + k = (0, 1, 1)c = k + i = (1, 0, 1)Therefore, matrix M is:[1 0 11 1 00 1 1]Then, the coordinates in the standard basis would be M * [8, 6, 4]^T.Calculating that:First component: 1*8 + 0*6 + 1*4 = 8 + 0 +4 =12Second component:1*8 +1*6 +0*4=8 +6 +0=14Third component:0*8 +1*6 +1*4=0 +6 +4=10So, yes, the result is (12,14,10), which is option A. Therefore, both methods confirm that the answer is A.I don't see any errors here. The key was to express the given vector in terms of the original basis vectors and then substitute the expressions of a, b, c in terms of i, j, k. Alternatively, using the change of basis matrix and multiplying by the coordinate vector. Both approaches lead to the same result. So, I'm confident the correct answer is A.Final Answerboxed{A}</think>Given that the coordinates of vector ( p ) in the basis ({a, b, c}) are ((8, 6, 4)), where ( a = i + j ), ( b = j + k ), and ( c = k + i ), we need to find the coordinates of vector ( p ) in the basis ({i, j, k}).First, express ( p ) in terms of the basis vectors ( a ), ( b ), and ( c ):[p = 8a + 6b + 4c]Substitute the definitions of ( a ), ( b ), and ( c ):[a = i + j, quad b = j + k, quad c = k + i]Substituting these into the expression for ( p ):[p = 8(i + j) + 6(j + k) + 4(k + i)]Distribute the coefficients:[p = 8i + 8j + 6j + 6k + 4k + 4i]Combine like terms for ( i ), ( j ), and ( k ):[p = (8i + 4i) + (8j + 6j) + (6k + 4k) = 12i + 14j + 10k]Thus, the coordinates of vector ( p ) in the basis ({i, j, k}) are ((12, 14, 10)).To verify, we can use the change of basis matrix. The matrix ( M ) with columns ( a ), ( b ), and ( c ) is:[M = begin{bmatrix}1 & 0 & 1 1 & 1 & 0 0 & 1 & 1end{bmatrix}]Multiplying this matrix by the coordinate vector ((8, 6, 4)):[begin{bmatrix}1 & 0 & 1 1 & 1 & 0 0 & 1 & 1end{bmatrix}begin{bmatrix}8 6 4end{bmatrix}=begin{bmatrix}1 cdot 8 + 0 cdot 6 + 1 cdot 4 1 cdot 8 + 1 cdot 6 + 0 cdot 4 0 cdot 8 + 1 cdot 6 + 1 cdot 4end{bmatrix}=begin{bmatrix}12 14 10end{bmatrix}]Both methods confirm that the coordinates of ( p ) in the basis ({i, j, k}) are ((12, 14, 10)).[boxed{A}],<think>Okay, so I need to find the coordinates of vector p in the basis {i, j, k} given that in the basis {a, b, c} it's (8, 6, 4). The vectors a, b, c are defined as a = i + j, b = j + k, and c = k + i. Hmm, right. So, this is a change of basis problem. Let me recall how that works.First, if I have a vector expressed in one basis, say {a, b, c}, and I want to express it in another basis, like the standard basis {i, j, k}, I need to express each of the original basis vectors in terms of the standard basis and then combine them according to the coordinates given.So, vector p in the basis {a, b, c} is 8a + 6b + 4c. Since a, b, c are given in terms of i, j, k, I can substitute those in and then combine like terms.Let's write that out step by step.First, substitute each basis vector:a = i + jb = j + kc = k + iSo, p = 8a + 6b + 4cSubstituting:p = 8(i + j) + 6(j + k) + 4(k + i)Now, distribute the coefficients:p = 8i + 8j + 6j + 6k + 4k + 4iCombine like terms for each basis vector:For i: 8i + 4i = 12iFor j: 8j + 6j = 14jFor k: 6k + 4k = 10kSo, putting it all together:p = 12i + 14j + 10kTherefore, the coordinates of p in the standard basis {i, j, k} are (12, 14, 10). Let me check the options given. Option A is (12, 14, 10). That seems right. Wait, but let me make sure I didn't mix up any coefficients.Wait, 8a is 8i +8j, correct. 6b is 6j +6k, yes. 4c is 4k +4i. Then adding up i: 8+4=12, j:8+6=14, k:6+4=10. Yep, that's correct. So the answer should be A. Let me check if the options are in order. The problem says "coordinates of vector p in the basis {i, j, k}" so (12,14,10) is A. So the answer is A.Wait, but let me think again. Sometimes, change of basis can involve inverse matrices or something, but in this case, since we're going from the given basis to the standard basis, we just need to express the linear combination. So since p is 8a +6b +4c, and each a, b, c are expressed in terms of i, j, k, substitution is straightforward. So the computation I did seems correct. So I think A is the right answer. Let me just verify once more.Compute each component:i component: 8 (from a) + 4 (from c) = 12j component: 8 (from a) + 6 (from b) = 14k component: 6 (from b) + 4 (from c) = 10Yes, that's exactly what I did. So (12,14,10) is correct, which is option A. So I think the answer is A.Final Answerboxed{A}</think>Given that the coordinates of vector ( p ) in the basis ({a, b, c}) are ((8, 6, 4)), where ( a = i + j ), ( b = j + k ), and ( c = k + i ), we need to find the coordinates of vector ( p ) in the basis ({i, j, k}).First, we express vector ( p ) in terms of the basis ({a, b, c}):[p = 8a + 6b + 4c]Substituting the expressions for ( a ), ( b ), and ( c ):[p = 8(i + j) + 6(j + k) + 4(k + i)]Distributing the coefficients:[p = 8i + 8j + 6j + 6k + 4k + 4i]Combining like terms for each basis vector ( i ), ( j ), and ( k ):- For ( i ): ( 8i + 4i = 12i )- For ( j ): ( 8j + 6j = 14j )- For ( k ): ( 6k + 4k = 10k )Thus, the coordinates of vector ( p ) in the basis ({i, j, k}) are ((12, 14, 10)).[boxed{A}]]

🔑:AnalysisThis problem involves the addition and subtraction operations of spatial vectors and their coordinate representations. From the given information, we can deduce that p = 12i + 14j + 10k, which leads to the solution of the problem.Step-by-step Solution1. According to the problem, we have p = 8a + 6b + 4c.2. Substitute the given values of a, b, and c into the equation: p = 8(i + j) + 6(j + k) + 4(k + i).3. Distribute the scalars: p = 8i + 8j + 6j + 6k + 4k + 4i.4. Combine like terms: p = 12i + 14j + 10k.Therefore, the coordinates of vector p in the basis {i, j, k} is (12, 14, 10).Hence, the correct answer is boxed{A}.

🧠:[<think>Alright, so I need to figure out the maximum difference between two numbers in the list after applying the "Make-My-Day" procedure 2018 times starting with {20, 1, 8}. Let me try to understand what this procedure does step by step.First, the example given: starting with {1, 3, 8}, applying the procedure once gives {11, 9, 4}. Let me verify that. The first number is the sum of the other two: 3 + 8 = 11. The second number is 1 + 8 = 9. The third is 1 + 3 = 4. Yep, that checks out. Then applying it again: 9 + 4 = 13, 11 + 4 = 15, 11 + 9 = 20. So the new list is {13, 15, 20}. The maximum difference there is 20 - 13 = 7. Interesting.So each time, we replace each number with the sum of the other two. Let me try to model this. Let's denote the list as a triplet (a, b, c). After one "Make-My-Day" procedure, the triplet becomes (b + c, a + c, a + b). Let me write that down:First iteration: (a1, b1, c1) = (b0 + c0, a0 + c0, a0 + b0)Second iteration: (a2, b2, c2) = (b1 + c1, a1 + c1, a1 + b1)And so on.So, given the starting triplet (20, 1, 8), which we'll call (a0, b0, c0), we need to find (a2018, b2018, c2018) and then compute the maximum difference between any two of them.But 2018 is a large number. Manually computing each step isn't feasible. So I need to find a pattern or a formula that can let me compute the triplet after n steps.Let me see if I can find a pattern or some invariant here. Let's look at the sum of the numbers in each iteration. Maybe that's useful.In the first example, the initial sum was 1 + 3 + 8 = 12. After one iteration, the sum was 11 + 9 + 4 = 24. Then the next iteration sum was 13 + 15 + 20 = 48. So each time, the sum is doubling. Let me check if that's a general rule.Suppose we have a triplet (a, b, c). The sum S = a + b + c. After applying the procedure, the new triplet is (b + c, a + c, a + b). The new sum S' = (b + c) + (a + c) + (a + b) = 2a + 2b + 2c = 2S. So indeed, the sum doubles each time. Therefore, the sum after n iterations will be S_n = 2^n * S_0, where S_0 is the initial sum.So for our problem, the initial triplet is (20, 1, 8). The initial sum S0 = 20 + 1 + 8 = 29. Therefore, after 2018 iterations, the sum will be 2^2018 * 29. That's a huge number, but maybe we don't need the exact values, just the differences.However, the problem asks for the maximum difference between two numbers on the list after 2018 iterations. Hmm. So even if the numbers are getting huge, perhaps their differences follow a certain pattern.Let me check the example again. First iteration: {1, 3, 8} becomes {11, 9, 4}. The differences here are 11 - 9 = 2, 11 - 4 = 7, 9 - 4 = 5. So max difference is 7. Next iteration: {13, 15, 20}, differences are 2, 7, 5 again. Wait, no: 15 - 13 = 2, 20 - 13 = 7, 20 - 15 = 5. Same differences as before? Interesting.Wait, so in the first step, the differences were 2, 5, 7. Then in the next step, same differences but in different order? Let me verify. From {1,3,8}, the first step is {11,9,4}. The differences here are 11-9=2, 11-4=7, 9-4=5. Then next step {13,15,20}, differences: 15-13=2, 20-13=7, 20-15=5. So same set of differences, just attached to different numbers. So the differences themselves are cycling? Or maybe after two steps, the differences cycle?Wait, maybe the differences follow a cycle of their own. Let me check the next iteration. If we apply Make-My-Day to {13,15,20}, the next triplet would be 15+20=35, 13+20=33, 13+15=28. So {35,33,28}. The differences are 35-33=2, 35-28=7, 33-28=5. Again, same differences. So after each iteration, the differences remain the same? But the numbers themselves are growing.Wait, but in the first step, the differences changed. Wait, original triplet {1,3,8} had differences 2,5,7 (between 3-1, 8-3, 8-1). Then after one iteration, the differences were 2,5,7 but in different positions. So perhaps the set of differences is preserved, but their positions change. Then in the next iteration, same thing.Wait, but let's check the differences more carefully.Original triplet: 1, 3, 8. Differences: 3-1=2, 8-3=5, 8-1=7.After first Make-My-Day: 11,9,4. Differences: 9-4=5, 11-9=2, 11-4=7.So the differences are the same, but the order is different.Then next iteration: 13,15,20. Differences: 15-13=2, 20-15=5, 20-13=7. Again, same differences, different order.So the differences cycle through permutations of the original differences. So if the differences stay the same set, just ordered differently each time, then the maximum difference would always be the same, which is 7 in this example. Wait, but in the example, the maximum difference was 7, then 7 again. So maybe, in general, the maximum difference remains constant after each iteration? But that seems counterintuitive because the numbers are getting bigger.Wait, but in the example, the differences are preserved. Let me see:First triplet: 1, 3, 8. Differences: 2,5,7.After one iteration: 11,9,4. Differences: 5,2,7.After second iteration: 13,15,20. Differences: 2,5,7.Third iteration: 35,33,28. Differences: 2,5,7.Wait, actually, the differences cycle between {2,5,7} and {5,2,7}? Wait, no. Wait, first iteration, differences are 2,5,7. Then next iteration, differences are 2,5,7 again? Wait, in the second iteration, {13,15,20}, the differences are 2 (15-13), 5 (20-15), 7 (20-13). So same as original. Then third iteration, triplet is 15+20=35, 13+20=33, 13+15=28. So {35,33,28}, differences: 2 (35-33), 5 (33-28), 7 (35-28). So same differences again. So regardless of the iteration, the differences are always 2,5,7. So in the example, the maximum difference is always 7, regardless of how many times you apply the procedure. Therefore, the maximum difference remains constant? That's fascinating.But wait, let's check with another example. Suppose we start with a different triplet. Let's say {2, 3, 5}. Then applying Make-My-Day once would give {8, 7, 5}. The differences here are 1, 3, 3. Then applying again: {12, 13, 15}. Differences here are 1, 2, 3. Then again: {28, 27, 25}. Differences: 1, 3, 3. Wait, now the differences are cycling between different sets. So in this case, the differences aren't preserved. So maybe in the original problem, the example had differences that were preserved, but in other cases, it's not.So why in the first example, the differences remain the same? Let me analyze this.Suppose we have a triplet (a, b, c). After one iteration, it becomes (b + c, a + c, a + b). Let's compute the differences between the new numbers.First difference: (b + c) - (a + c) = b - a.Second difference: (a + c) - (a + b) = c - b.Third difference: (b + c) - (a + b) = c - a.Wait, so the differences after one iteration are (b - a, c - b, c - a). But the original differences were (b - a, c - b, c - a) as well. Wait, that can't be.Wait, original differences: between a, b, c. Let's order them first. Let's say a < b < c. Then the differences would be b - a, c - b, and c - a. But after the iteration, the new triplet is (b + c, a + c, a + b). Let's order the new triplet. Let's see:If a < b < c, then a + b < a + c < b + c. So the new triplet is ordered as (a + b, a + c, b + c). The differences would be (a + c) - (a + b) = c - b, (b + c) - (a + c) = b - a, and (b + c) - (a + b) = c - a. So the differences are (c - b, b - a, c - a). Which are the same as the original differences (b - a, c - b, c - a), but reordered. So the set of differences is preserved, but their order changes.Therefore, the differences themselves are cyclically permuted with each iteration. Therefore, the maximum difference remains the same, because the differences are the same in magnitude, just attached to different pairs.Wait, so in the example, the differences {2, 5, 7} are preserved as a set, just their positions change. Therefore, the maximum difference is always 7, no matter how many iterations. Therefore, in the original problem, if this holds, then the maximum difference after 2018 iterations would be the same as the maximum difference after the first iteration.But let's verify with the initial problem's starting triplet: {20, 1, 8}.Let's compute the first iteration. The new triplet will be (1 + 8, 20 + 8, 20 + 1) = (9, 28, 21). Wait, hold on: the procedure replaces each number with the sum of the other two, so first number is 1 + 8 = 9, second is 20 + 8 = 28, third is 20 + 1 = 21. So new triplet is {9, 28, 21}. Let's compute the differences here. Let's order them: 9, 21, 28. Differences: 21 - 9 = 12, 28 - 21 = 7, 28 - 9 = 19. So maximum difference is 19.Original triplet was {20, 1, 8}. Ordered: 1, 8, 20. Differences: 7, 12, 19. So same differences as after one iteration, just reordered. Wait, original differences were 7 (8 - 1), 12 (20 - 8), 19 (20 - 1). After one iteration, differences are 12 (21 - 9), 7 (28 - 21), 19 (28 - 9). So same set of differences, {7, 12, 19}, just reordered. Therefore, the maximum difference is still 19.So if we do another iteration, the differences should still be {7, 12, 19}, but reordered again.Let's check. Take the triplet {9, 28, 21}. Applying Make-My-Day: each number is replaced by the sum of the other two.First number: 28 + 21 = 49.Second number: 9 + 21 = 30.Third number: 9 + 28 = 37.So new triplet: {49, 30, 37}. Ordering them: 30, 37, 49. Differences: 7 (37 - 30), 12 (49 - 37), 19 (49 - 30). Again, same differences. So maximum difference is 19.So regardless of the number of iterations, the set of differences {7, 12, 19} is preserved. Therefore, the maximum difference between any two numbers will always be 19, no matter how many times we apply the Make-My-Day procedure.But wait, the original problem starts with {20, 1, 8}, which has differences 7, 12, 19. After each iteration, the differences remain the same set, just reordered. Therefore, even after 2018 iterations, the maximum difference will still be 19.But let me confirm with another iteration. Starting from {49, 30, 37}:Next iteration: 30 + 37 = 67, 49 + 37 = 86, 49 + 30 = 79. So triplet {67, 86, 79}. Ordering: 67, 79, 86. Differences: 12 (79 - 67), 7 (86 - 79), 19 (86 - 67). Yep, same set. Max difference 19.Therefore, no matter how many iterations, the maximum difference stays 19. Therefore, the answer should be 19. But wait, let me check the initial steps again carefully.Wait, starting with {20, 1, 8}:First iteration: {9, 28, 21}. Max difference 19.Second iteration: {49, 30, 37}. Max difference 19.Third iteration: {67, 86, 79}. Max difference 19.Fourth iteration: 30 + 37 = 67, 49 + 37 = 86, 49 + 30 = 79. Wait, that's the same as the third iteration. Wait, no. Wait, third iteration was {67, 86, 79}. Applying Make-My-Day again:First number: 86 + 79 = 165Second number: 67 + 79 = 146Third number: 67 + 86 = 153So triplet {165, 146, 153}. Ordering: 146, 153, 165. Differences: 7 (153 - 146), 12 (165 - 153), 19 (165 - 146). Again, same differences. So max difference 19.Therefore, regardless of the number of iterations, the maximum difference remains 19. Therefore, even after 2018 iterations, the maximum difference is 19. Hence, the answer is 19.But let me check why in the initial example, the maximum difference was preserved. It seems that when we apply the Make-My-Day operation, the differences between the numbers are preserved as a set. Therefore, the maximum difference, which is the largest element of that set, remains the same.This is because each difference in the new triplet corresponds to a difference in the original triplet. For example, in the first step, when you replace each number with the sum of the other two, the new numbers are combinations of the original numbers, and their differences relate to the original differences.Let me formalize this. Let’s suppose we have three numbers a, b, c. Without loss of generality, let's assume a < b < c. Then the differences are (b - a, c - b, c - a). After one iteration, the new triplet is (b + c, a + c, a + b). Let's order these new numbers:a + b < a + c < b + c (since b < c)So ordered, the new triplet is (a + b, a + c, b + c). Now compute the differences:(a + c) - (a + b) = c - b(b + c) - (a + c) = b - a(b + c) - (a + b) = c - aSo the new differences are (c - b, b - a, c - a), which is a permutation of the original differences (b - a, c - b, c - a). Therefore, the set of differences is preserved, and hence the maximum difference remains the same.Therefore, no matter how many times we apply the Make-My-Day procedure, the set of differences between the numbers remains the same, just reordered. Thus, the maximum difference between any two numbers in the list remains constant through all iterations. Therefore, even after 2018 iterations, the maximum difference will be the same as the maximum difference in the original list.But wait, in the original problem, the starting list is {20, 1, 8}. Let's order that: 1, 8, 20. The differences are 7 (8 - 1), 12 (20 - 8), and 19 (20 - 1). Therefore, the maximum difference is 19. Therefore, after any number of iterations, including 2018, the maximum difference remains 19.Therefore, the answer is 19.Final Answerboxed{19}</think>Starting with the list ({20, 1, 8}), the "Make-My-Day" procedure replaces each number with the sum of the other two. We need to determine the maximum difference between two numbers after 2018 consecutive applications of this procedure.1. Initial Analysis: - The initial list is ({20, 1, 8}). - The sum of the numbers in the list doubles with each application of the procedure. - The differences between the numbers are key to solving the problem.2. First Iteration: - Applying the procedure once: (20 + 1 rightarrow 21), (20 + 8 rightarrow 28), (1 + 8 rightarrow 9). - New list: ({9, 28, 21}). - Differences: (28 - 9 = 19), (21 - 9 = 12), (28 - 21 = 7). - Maximum difference: (19).3. Second Iteration: - Applying the procedure again: (28 + 21 rightarrow 49), (9 + 21 rightarrow 30), (9 + 28 rightarrow 37). - New list: ({49, 30, 37}). - Differences: (49 - 30 = 19), (37 - 30 = 7), (49 - 37 = 12). - Maximum difference: (19).4. General Pattern: - Each iteration preserves the set of differences ({7, 12, 19}) but reorders them. - The maximum difference remains (19) regardless of the number of iterations.5. Conclusion: - After any number of iterations, the maximum difference between two numbers in the list remains (19).Thus, the maximum difference between two numbers on the list after 2018 consecutive "Make-My-Day" procedures is (boxed{19}).,<think>Okay, so I need to figure out the maximum difference between two numbers in the list after applying the "Make-My-Day" procedure 2018 times starting with {20, 1, 8}. Let's break down the problem step by step.First, let me make sure I understand the procedure correctly. The "Make-My-Day" procedure replaces each number in the list with the sum of the other two. So, if we have a list {a, b, c}, the new list becomes {b + c, a + c, a + b}. Let's verify this with the example given: starting with {1, 3, 8}, applying the procedure once gives {3 + 8, 1 + 8, 1 + 3} which is {11, 9, 4}. Applying it again, we take those new numbers: {9 + 4, 11 + 4, 11 + 9} which simplifies to {13, 15, 20}. Yep, that matches the example. So I got that part right.Now, starting with {20, 1, 8}, we need to apply this procedure 2018 times and find the maximum difference between any two numbers in the resulting list. That's a lot of iterations, so doing it manually isn't feasible. I need to look for a pattern or perhaps find a formula that can help predict the state of the list after many applications.Let me start by applying the procedure a few times manually and see if I can spot a pattern.First iteration:Original list: {20, 1, 8}Apply Make-My-Day: {1 + 8, 20 + 8, 20 + 1} = {9, 28, 21}So after 1st iteration: {9, 28, 21}Maximum difference here is 28 - 9 = 19Second iteration:Current list: {9, 28, 21}Apply Make-My-Day: {28 + 21, 9 + 21, 9 + 28} = {49, 30, 37}After 2nd iteration: {49, 30, 37}Maximum difference: 49 - 30 = 19Third iteration:Current list: {49, 30, 37}Apply Make-My-Day: {30 + 37, 49 + 37, 49 + 30} = {67, 86, 79}After 3rd iteration: {67, 86, 79}Maximum difference: 86 - 67 = 19Fourth iteration:Current list: {67, 86, 79}Apply Make-My-Day: {86 + 79, 67 + 79, 67 + 86} = {165, 146, 153}After 4th iteration: {165, 146, 153}Maximum difference: 165 - 146 = 19Wait a minute, the maximum difference is staying at 19 each time? Let me check again.First iteration difference: 28 - 9 = 19Second iteration: 49 - 30 = 19Third iteration: 86 - 67 = 19Fourth iteration: 165 - 146 = 19So, the maximum difference between any two numbers is remaining 19 after each iteration. That's interesting. But how can that be? Each time, the numbers are getting larger, but the difference between the largest and smallest remains 19? Let's check the fifth iteration to confirm.Fifth iteration:Current list: {165, 146, 153}Apply Make-My-Day: {146 + 153, 165 + 153, 165 + 146} = {299, 318, 311}After 5th iteration: {299, 318, 311}Maximum difference: 318 - 299 = 19Yes, still 19. So the difference remains constant? That's unexpected. Let me try with another example to see if this is a general property or specific to the starting numbers.Take the example given in the problem: starting with {1, 3, 8}. After first iteration, {11, 9, 4}, difference is 11 - 4 = 7. Then next iteration gives {13, 15, 20}, difference is 20 - 13 = 7. Wait, so here the difference remains 7? Let me check the next iteration.Third iteration: {15 + 20, 13 + 20, 13 + 15} = {35, 33, 28}, difference 35 - 28 = 7. Yep, same difference. So in that case as well, the maximum difference remains the same.So this seems to be a general property of the "Make-My-Day" procedure. The maximum difference between any two numbers in the list remains constant after each iteration. If that's the case, then no matter how many times we apply the procedure, the maximum difference will always be the same as the original maximum difference.Wait, but in the original problem, starting with {20, 1, 8}, the initial maximum difference is 20 - 1 = 19. Then after each iteration, even though the numbers are changing, the maximum difference remains 19. Therefore, after 2018 iterations, the maximum difference would still be 19. But let me verify this with another example to be sure.Suppose we start with {5, 2, 3}. The maximum difference is 5 - 2 = 3.First iteration: {2 + 3, 5 + 3, 5 + 2} = {5, 8, 7}. Maximum difference is 8 - 5 = 3.Second iteration: {8 + 7, 5 + 7, 5 + 8} = {15, 12, 13}. Max difference 15 - 12 = 3.Third iteration: {12 + 13, 15 + 13, 15 + 12} = {25, 28, 27}. Max difference 28 - 25 = 3.So again, the difference remains the same. Therefore, it's safe to conclude that the "Make-My-Day" procedure preserves the maximum difference between any two numbers in the list. Therefore, no matter how many times we apply the procedure, the maximum difference remains equal to the original maximum difference.But wait, in the starting example, {1, 3, 8} had an initial maximum difference of 7 (8 - 1). After first iteration, {11, 9, 4}, which has 11 - 4 = 7. Then next iteration {13, 15, 20} with 20 - 13 = 7. So again, same. So seems like the difference is preserved.But in the original problem, starting with {20, 1, 8}, the initial maximum difference is 20 - 1 = 19. Then after each iteration, the max difference is 19. Therefore, after 2018 iterations, it's still 19.But let me check once again with the first steps for the original problem to confirm.Original list: {20, 1, 8}, max difference 19.First iteration: {1 + 8, 20 + 8, 20 + 1} = {9, 28, 21}, sorted: 9, 21, 28. Max difference 28 - 9 = 19.Second iteration: {28 + 21, 9 + 21, 9 + 28} = {49, 30, 37}, sorted: 30, 37, 49. Max difference 49 - 30 = 19.Third iteration: {30 + 37, 49 + 37, 49 + 30} = {67, 86, 79}, sorted: 67, 79, 86. Max difference 86 - 67 = 19.Fourth iteration: {79 + 86, 67 + 86, 67 + 79} = {165, 153, 146}, sorted: 146, 153, 165. Max difference 165 - 146 = 19.So same as before, so it's consistent.But why is this happening? Maybe there's a mathematical reason behind this. Let me think.Suppose we have three numbers a, b, c. Let's sort them so that a ≤ b ≤ c. Then the maximum difference is c - a.After applying Make-My-Day, we get the new numbers: b + c, a + c, a + b. Let's see what the new maximum and minimum would be.The new numbers are (b + c), (a + c), (a + b). Let's sort them. The smallest would be a + b, since a ≤ b ≤ c. The next would be a + c, and the largest would be b + c. So sorted: (a + b), (a + c), (b + c). So the maximum difference now is (b + c) - (a + b) = c - a. Which is the same as before. So the maximum difference remains c - a.Therefore, regardless of the numbers, each iteration preserves the maximum difference between the original largest and smallest numbers.Therefore, no matter how many times we apply the Make-My-Day procedure, the maximum difference between any two numbers in the list remains equal to the original maximum difference in the starting list.In the problem, the starting list is {20, 1, 8}. The original maximum difference is 20 - 1 = 19. Therefore, after any number of iterations, including 2018, the maximum difference will still be 19. Therefore, the answer should be 19.But just to make sure there isn't a case where this doesn't hold, let's consider a different starting list where numbers are not in order.Suppose starting list is {5, 10, 2}. Original max difference is 10 - 2 = 8.First iteration: {10 + 2, 5 + 2, 5 + 10} = {12, 7, 15}. Max difference 15 - 7 = 8.Second iteration: {7 + 15, 12 + 15, 12 + 7} = {22, 27, 19}. Max difference 27 - 19 = 8.Third iteration: {19 + 27, 22 + 27, 22 + 19} = {46, 49, 41}. Max difference 49 - 41 = 8.Yes, so again, same result. Therefore, the maximum difference is preserved regardless of the order of the numbers in the list. Therefore, the key insight is that the difference between the largest and smallest numbers remains invariant under the Make-My-Day transformation.Therefore, the answer must be 19.Final Answerboxed{19}</think>Starting with the list ({20, 1, 8}), we need to determine the maximum difference between two numbers after applying the "Make-My-Day" procedure 2018 times. The procedure replaces each number with the sum of the other two. First, we observe the effect of the procedure on the list through a few iterations:- Initial list: ({20, 1, 8}), maximum difference is (20 - 1 = 19).- After 1st iteration: ({9, 28, 21}), maximum difference is (28 - 9 = 19).- After 2nd iteration: ({49, 30, 37}), maximum difference is (49 - 30 = 19).- After 3rd iteration: ({67, 86, 79}), maximum difference is (86 - 67 = 19).- After 4th iteration: ({165, 146, 153}), maximum difference is (165 - 146 = 19).- After 5th iteration: ({299, 318, 311}), maximum difference is (318 - 299 = 19).We notice that the maximum difference remains constant at 19 after each iteration. To generalize, for any list ({a, b, c}) sorted as (a leq b leq c), the maximum difference is (c - a). After applying the procedure, the new list is ({b + c, a + c, a + b}). Sorting these new values, the smallest is (a + b) and the largest is (b + c). The new maximum difference is ((b + c) - (a + b) = c - a), which is the same as the original maximum difference.Thus, the maximum difference between two numbers remains invariant under the "Make-My-Day" procedure. Therefore, after 2018 iterations, the maximum difference will still be the original maximum difference of 19.[boxed{19}]]

🔑:1. Let's denote the initial list of numbers as (a, b, c). The "Make-My-Day" procedure transforms this list into (b+c, a+c, a+b).2. We need to find the maximum difference between any two numbers in the list after applying the "Make-My-Day" procedure multiple times.3. First, let's analyze the transformation: [ (a, b, c) longrightarrow (b+c, a+c, a+b) ]4. Calculate the differences between the new numbers: [ |(b+c) - (a+c)| = |b - a| ] [ |(a+c) - (a+b)| = |c - b| ] [ |(a+b) - (b+c)| = |a - c| ]5. The maximum difference between any two numbers in the new list is: [ text{Max}(|b - a|, |c - b|, |a - c|) ]6. Notice that the maximum difference between any two numbers in the list remains the same after each "Make-My-Day" transformation. This is because the differences |b - a|, |c - b|, and |a - c| are preserved.7. Given the initial list (20, 1, 8), we calculate the initial differences: [ |20 - 1| = 19 ] [ |1 - 8| = 7 ] [ |8 - 20| = 12 ]8. The maximum difference in the initial list is: [ text{Max}(19, 7, 12) = 19 ]9. Since the maximum difference remains the same after each "Make-My-Day" transformation, the maximum difference after 2018 transformations is still 19.The final answer is boxed{19}

🧠:[<think>Okay, let's see. I need to prove that |AT| = |MC| in this geometry problem. There are a lot of midpoints and perpendiculars here. Let me start by drawing a diagram to visualize the problem. First, triangle ABC with altitude AD. So D is the foot of the perpendicular from A to BC. Then, M is the midpoint of AB, N is the midpoint of AD, and P is the midpoint of BC. Next, K is the foot of the perpendicular from D to AC. Then, T is a point on the extension of KD beyond D such that DT = MN + DK. Also, given that MP = 2*KN, I need to show AT = MC.Hmm. Let me note down all the given information step by step.1. AD is an altitude, so AD ⊥ BC.2. M is the midpoint of AB, so AM = MB.3. N is the midpoint of AD, so AN = ND.4. P is the midpoint of BC, so BP = PC.5. K is the foot of the perpendicular from D to AC, so DK ⊥ AC.6. T is on the extension of KD beyond D, so T is such that DT = MN + DK.7. Given that MP = 2*KN.8. Need to prove AT = MC.Since there are midpoints, maybe using midline theorem or properties of midsegments in triangles could help. Coordinates might also be useful here. Let me consider coordinate geometry.Let me place the triangle in coordinate system to make it easier. Let me set point A at (0, 0). Since AD is an altitude, let's let AD be vertical. Let me assign coordinates such that D is at (0, d) for some d. Wait, but altitude is from A to BC, so if A is at (0,0), then AD is the altitude, so BC is horizontal? Wait, no. If AD is an altitude, then BC is the base, and AD is perpendicular to BC. So maybe setting coordinate system such that BC is on the x-axis and D is the foot of the altitude from A to BC. Then A is at (0, h) for some h, D is at (0,0), and BC is on the x-axis. Wait, but then AD would be vertical, BC horizontal. Maybe that's a good setup.Let me try this:Let me set point D at the origin (0, 0). Then, since AD is an altitude, A is somewhere along the y-axis. Let's say A is at (0, a) for some a > 0. Then BC is on the x-axis, since AD is perpendicular to BC. Let me denote coordinates:- D: (0, 0)- A: (0, a)- B and C are on the x-axis. Let's denote B as (b, 0) and C as (c, 0). Since D is the foot of the altitude from A to BC, D must lie on BC. Wait, BC is the base, so D is the foot from A to BC. But if BC is on the x-axis from (b, 0) to (c, 0), then the foot of the perpendicular from A (0, a) to BC would be (0,0) only if BC is from (-k, 0) to (k, 0) or something? Wait, maybe I need to adjust coordinates.Wait, if AD is the altitude, then D is on BC. So BC is a horizontal line (if we take AD vertical), and D is the foot. So if I set D at (0,0), then BC is the x-axis, and A is (0, a). Then B and C are points on the x-axis, so B can be (p, 0) and C can be (q, 0). Then AD is from (0,a) to (0,0). Then midpoints:M is the midpoint of AB. So coordinates of M would be ((p/2), (a/2)).N is the midpoint of AD. So coordinates of N would be (0, a/2).P is the midpoint of BC. Since B is (p,0) and C is (q,0), midpoint P is ((p+q)/2, 0).K is the foot of the perpendicular from D to AC. So AC is from A(0,a) to C(q,0). The equation of AC can be found. Then K is the foot of perpendicular from D(0,0) to AC. Let's compute coordinates of K.First, equation of AC. Points A(0,a) and C(q,0). Slope of AC is (0 - a)/(q - 0) = -a/q. Therefore, equation of AC is y = (-a/q)x + a.The foot of perpendicular from D(0,0) to AC. The slope of AC is -a/q, so the slope of the perpendicular is q/a. Therefore, equation of DK (perpendicular from D to AC) is y = (q/a)x.Intersection point K is where the two lines meet: (-a/q)x + a = (q/a)x.Multiply both sides by aq to eliminate denominators:-a^2 x + a^2 q = q^2 xBring terms with x to one side:-a^2 x - q^2 x = -a^2 qx(-a^2 - q^2) = -a^2 qx = (-a^2 q)/(-a^2 - q^2) = (a^2 q)/(a^2 + q^2)Then y = (q/a)x = (q/a)*(a^2 q)/(a^2 + q^2) = (a q^2)/(a^2 + q^2)Therefore, coordinates of K are ( (a^2 q)/(a^2 + q^2), (a q^2)/(a^2 + q^2) )Hmm, that seems complicated. Maybe there's a better coordinate system. Wait, perhaps if I set specific coordinates for simplicity. Let me assign coordinates such that BC is symmetric with respect to the y-axis? Let me try that.Suppose I set point B as (-b, 0) and C as (b, 0). Then BC is from (-b, 0) to (b, 0), so midpoint P is (0,0). Wait, but D is the foot of the altitude from A to BC. If BC is from (-b,0) to (b,0), then the altitude from A would be along the y-axis, so D is (0,0). But then P, the midpoint of BC, is also (0,0). So in this case, D and P coincide? But the problem states that P is the midpoint of BC, so if BC is from (-b, 0) to (b, 0), then P is indeed (0,0), which is the same as D. But in the problem statement, D is the foot of altitude, and P is the midpoint of BC, which may or may not coincide. If they do, then perhaps some relations hold. However, in the problem, there is a point K, which is the foot of perpendicular from D to AC, which in this case would be from (0,0) to AC. If A is (0,a), then AC is from (0,a) to (b,0). The foot of the perpendicular from (0,0) to AC would be some point K.But maybe this symmetric case is too restrictive? Because in the problem, MP = 2*KN. If D and P coincide, then MP is the distance from M to P (which is D here). Let me see.Wait, in the symmetric case where B is (-b,0) and C is (b,0), then coordinates:A(0,a), D(0,0), M is the midpoint of AB, which is (-b/2, a/2). N is the midpoint of AD, which is (0, a/2). P is the midpoint of BC, which is (0,0). So MP is the distance from M(-b/2, a/2) to P(0,0). The distance MP is sqrt( (b/2)^2 + (a/2)^2 ) = (1/2)sqrt(a^2 + b^2).KN is the distance from K to N. Let's compute K. In this symmetric case, AC is from (0,a) to (b,0). The equation of AC is y = (-a/b)x + a. The foot of perpendicular from D(0,0) to AC is point K.Using the formula from before, coordinates of K would be ( (a^2 b)/(a^2 + b^2), (a b^2)/(a^2 + b^2) )So K is at ( (a^2 b)/(a^2 + b^2), (a b^2)/(a^2 + b^2) )N is at (0, a/2). So distance KN is sqrt( [ (a^2 b/(a^2 + b^2) - 0 )^2 + ( a b^2/(a^2 + b^2) - a/2 )^2 ] )Let me compute that.First, x-coordinate difference: a²b / (a² + b²)y-coordinate difference: (ab²/(a² + b²)) - a/2 = a[ b²/(a² + b²) - 1/2 ] = a[ (2b² - a² - b²) / 2(a² + b²) ] = a[ (b² - a²) / 2(a² + b²) ) ] = a(b² - a²)/(2(a² + b²))Therefore, distance KN:sqrt[ (a^4 b²)/(a² + b²)^2 + (a² (b² - a²)^2 )/(4(a² + b²)^2) ) ]Factor out a²/(4(a² + b²)^2):sqrt[ (4a² b² + (b² - a²)^2 ) * a²/(4(a² + b²)^2) ) ]Compute numerator inside sqrt:4a² b² + (b² - a²)^2 = 4a² b² + b^4 - 2a² b² + a^4 = 2a² b² + b^4 + a^4 = (a^2 + b^2)^2Therefore, sqrt[ ( (a^2 + b^2)^2 ) * a²/(4(a² + b²)^2) ) ] = sqrt( a²/4 ) = a/2So distance KN = a/2Then MP = (1/2)sqrt(a² + b²), and given that MP = 2*KN, so:(1/2)sqrt(a² + b²) = 2*(a/2) => (1/2)sqrt(a² + b²) = aMultiply both sides by 2:sqrt(a² + b²) = 2aSquare both sides:a² + b² = 4a² => b² = 3a² => b = a*sqrt(3)So in this symmetric case, the condition MP = 2*KN implies that BC has length 2b = 2a*sqrt(3), and coordinates:A(0,a), B(-a√3, 0), C(a√3, 0), D(0,0), M(-a√3/2, a/2), N(0, a/2), P(0,0), K is as computed above.Wait, with b = a√3, coordinates of K become:( (a² * a√3)/(a² + (a√3)^2 ), (a * (a√3)^2 )/(a² + (a√3)^2 ) )Compute denominator: a² + 3a² = 4a²Therefore, x-coordinate of K: (a^3√3)/(4a²) = (a√3)/4y-coordinate of K: (a * 3a²)/4a² = 3a/4So K is ( (a√3)/4, 3a/4 )Then, MN is the distance between M(-a√3/2, a/2) and N(0, a/2). That's sqrt( (a√3/2)^2 + 0 ) = (a√3)/2DK is the distance from D(0,0) to K(a√3/4, 3a/4). Which is sqrt( (a√3/4)^2 + (3a/4)^2 ) = sqrt( 3a²/16 + 9a²/16 ) = sqrt(12a²/16 ) = sqrt(3a²/4 ) = (a√3)/2So MN = (a√3)/2 and DK = (a√3)/2, so MN + DK = a√3Thus, DT = MN + DK = a√3. Since T is on the extension of KD beyond D, and KD is from K(a√3/4, 3a/4) to D(0,0). The direction from K to D is towards the origin. Extending beyond D would be in the direction opposite to K from D. So direction vector from D to K is (a√3/4, 3a/4). Therefore, beyond D, direction vector is (-a√3/4, -3a/4). So to get T, starting at D(0,0), move in the direction opposite to K by length DT = a√3.But the direction vector from K to D is (-a√3/4, -3a/4). The length of DK is (a√3)/2, as computed earlier. So to move a√3 beyond D in that direction, we need to scale the direction vector.Let me parametrize the line KD. Starting at K(a√3/4, 3a/4), going towards D(0,0). The parametric equations can be written as:x = a√3/4 - (a√3/4)ty = 3a/4 - (3a/4)twhere t=0 is at K and t=1 is at D. To go beyond D, we need t > 1.The direction vector is (-a√3/4, -3a/4). The length of the direction vector is sqrt( (a√3/4)^2 + (3a/4)^2 ) = sqrt( 3a²/16 + 9a²/16 ) = sqrt(12a²/16 ) = (a√3)/2, which matches DK.We need to move a distance of DT = a√3 beyond D. Since the direction vector has length (a√3)/2, moving a distance a√3 would require scaling by 2. So T is D plus 2*(direction vector from K to D). Wait, direction vector from K to D is (-a√3/4, -3a/4). So scaling by 2 gives (-a√3/2, -3a/2). Therefore, T is D(0,0) plus (-a√3/2, -3a/2), so T is (-a√3/2, -3a/2).Wait, but let me confirm. If direction from K to D is vector (-a√3/4, -3a/4), then beyond D, we go in the same direction. So starting at D(0,0), moving in direction (-a√3/4, -3a/4) for some multiple. Since DT = a√3, and the length of the direction vector is (a√3)/2, the number of steps needed is (a√3) / ( (a√3)/2 ) = 2. So we go 2 steps in the direction beyond D. Therefore, T is D + 2*(-a√3/4, -3a/4) = (-a√3/2, -3a/2). So coordinates of T are (-a√3/2, -3a/2).Now, need to compute |AT| and |MC|.First, coordinates of A(0,a) and T(-a√3/2, -3a/2). The distance AT is sqrt( ( -a√3/2 - 0 )^2 + ( -3a/2 - a )^2 )Simplify:x-component: (-a√3/2)^2 = (3a²)/4y-component: (-5a/2)^2 = 25a²/4Total: (3a² + 25a²)/4 = 28a²/4 = 7a²Thus, AT = sqrt(7a²) = a√7Now, |MC|. Coordinates of M(-a√3/2, a/2) and C(a√3, 0). Distance MC is sqrt( ( a√3 + a√3/2 )^2 + ( 0 - a/2 )^2 )Compute x-component: ( (3a√3)/2 )^2 = (9*3a²)/4 = 27a²/4y-component: (-a/2)^2 = a²/4Total: (27a² + a²)/4 = 28a²/4 = 7a²Thus, MC = sqrt(7a²) = a√7Therefore, |AT| = |MC|, which is what needed to be proved.Wait, so in this symmetric coordinate system, with the given conditions, it works out. Therefore, the general case should hold as well. But since I used a specific coordinate system where BC is symmetric about the y-axis, and the given conditions led to |AT| = |MC|, does this hold in general?But in the problem statement, there is no specification about the triangle being isoceles or any particular type. However, by choosing coordinates where BC is symmetric, we were able to satisfy the given condition MP = 2*KN, which forced BC to have length related to a. So maybe in general, regardless of the triangle's shape, the relation MP = 2*KN imposes a similar condition, leading to |AT| = |MC|.Alternatively, since the problem is purely based on lengths and midpoints, coordinate geometry might be the right approach here. Given that the problem is likely designed to be solved with coordinate geometry, and the relations hold in this symmetric case (which satisfies the given condition MP = 2*KN), it's sufficient to show the result in this case. Because if the problem is true in general, then in any coordinate system it should hold, so choosing a convenient one is allowed.Alternatively, maybe there's a synthetic proof, but given the complexity of the relations, coordinates might be the way to go.Alternatively, using vectors.Let me try a vector approach.Let me denote vectors from point D as the origin. So set D at the origin. Let’s assign coordinates:Let’s let D be (0,0). Since AD is an altitude, AD is vertical. Let’s let A be (0, a), and B be (b, 0), C be (c, 0). Then, midpoints:M is midpoint of AB: ( (b + 0)/2, (0 + a)/2 ) = (b/2, a/2 )N is midpoint of AD: (0/2, a/2 ) = (0, a/2 )P is midpoint of BC: ( (b + c)/2, 0 )K is the foot of perpendicular from D(0,0) to AC. AC is from A(0,a) to C(c,0). The parametric equation of AC is (0 + tc, a - ta) for t in [0,1]. The foot K can be found by projecting D onto AC.The vector form: Let vector AC = (c, -a). The projection of vector DK (which is vector OK, since D is origin) onto AC is:K = proj_{AC}(OD) where OD is the vector from O to D, but D is origin. Wait, no. Wait, we need to project point D (which is the origin) onto line AC.Alternatively, the formula for the foot of the perpendicular from a point (x0,y0) to the line ax + by + c = 0 is given by:But maybe using vectors is better.Coordinates of A(0,a), C(c,0). Vector AC is (c, -a). The line AC can be parametrized as A + t*AC = (0, a) + t(c, -a).We need to find t such that the vector from D(0,0) to K is perpendicular to AC.So K = (tc, a - ta). Then vector DK = (tc, a - ta). This must be perpendicular to AC, which is (c, -a). So their dot product is zero:(tc)*c + (a - ta)*(-a) = 0t c² - a(a - ta) = 0t c² - a² + t a² = 0t(c² + a²) = a²t = a² / (a² + c²)Thus, coordinates of K are ( (a² c)/(a² + c² ), a - (a³)/(a² + c² ) ) = ( (a² c)/(a² + c² ), (a(a² + c² ) - a³ )/(a² + c² ) ) = ( (a² c)/(a² + c² ), (a c² )/(a² + c² ) )Therefore, K( (a² c)/(a² + c² ), (a c² )/(a² + c² ) )Now, compute MN and DK.Coordinates of M( b/2, a/2 ), N(0, a/2 ). So vector MN is ( -b/2, 0 ). The length |MN| = b/2.Wait, coordinates of M are (b/2, a/2 ), N is (0, a/2 ). So the distance MN is sqrt( (b/2 - 0)^2 + (a/2 - a/2 )^2 ) = sqrt( b²/4 ) = |b| / 2. Since lengths are positive, |MN| = b/2.But wait, in our coordinate system, B is at (b,0). If the triangle is not necessarily isoceles, then b can be positive or negative, but distance is absolute. So |MN| = |b|/2.Wait, but in our previous symmetric case, B was at (-a√3,0), so |MN| was | -a√3 / 2 | = a√3 /2. But in the general case here, if we take B at (b,0), then |MN| = |b|/2.Similarly, DK is the distance from D(0,0) to K( (a² c)/(a² + c² ), (a c² )/(a² + c² ) ). So |DK| = sqrt( [ (a² c / (a² + c² ))^2 + (a c² / (a² + c² ))^2 ] )Factor out (a² c²)/(a² + c² )²:sqrt( (a² c² (a² + c² )) / (a² + c² )² ) ) = sqrt( (a² c² ) / (a² + c² ) ) = (a c)/sqrt(a² + c² )Therefore, |DK| = (a c)/sqrt(a² + c² )Then, |DT| = |MN| + |DK| = |b|/2 + (a c)/sqrt(a² + c² )But T is on the extension of KD beyond D, so we need to find coordinates of T such that DT = MN + DK, in the direction opposite to K from D.In vector terms, the direction from K to D is D - K = ( - (a² c)/(a² + c² ), - (a c² )/(a² + c² ) )The unit vector in this direction is ( - (a² c)/(a² + c² ), - (a c² )/(a² + c² ) ) divided by |DK|.Since |DK| = (a c)/sqrt(a² + c² ), the unit vector is:( - (a² c)/(a² + c² ) , - (a c² )/(a² + c² ) ) / ( (a c)/sqrt(a² + c² ) ) )= ( - a / sqrt(a² + c² ), - c / sqrt(a² + c² ) )Therefore, the vector from D to T is this unit vector multiplied by |DT| = |MN| + |DK|.So vector DT = ( - a / sqrt(a² + c² ), - c / sqrt(a² + c² ) ) * ( |b|/2 + (a c)/sqrt(a² + c² ) )Therefore, coordinates of T are D + vector DT = (0,0) + DT = ( - a (|b|/2 + (a c)/sqrt(a² + c² )) / sqrt(a² + c² ), - c ( |b|/2 + (a c)/sqrt(a² + c² ) ) / sqrt(a² + c² ) )Simplify:First component:- a ( |b|/2 + (a c)/sqrt(a² + c² ) ) / sqrt(a² + c² )= - ( a |b| / 2 + a² c / sqrt(a² + c² ) ) / sqrt(a² + c² )= - ( a |b| / (2 sqrt(a² + c² )) + a² c / (a² + c² ) )Similarly, second component:- ( c |b| / (2 sqrt(a² + c² )) + a c² / (a² + c² ) )This is getting really complicated. Maybe I need to use the given condition MP = 2*KN.Compute MP and KN.Coordinates of M( b/2, a/2 ), P( (b + c)/2, 0 )So vector MP is ( (b + c)/2 - b/2, 0 - a/2 ) = ( c/2, -a/2 )Thus, |MP| = sqrt( (c/2)^2 + (a/2)^2 ) = (1/2)sqrt(a² + c² )Coordinates of K( (a² c)/(a² + c² ), (a c² )/(a² + c² ) ), N(0, a/2 )Distance KN:sqrt( ( (a² c)/(a² + c² ) - 0 )^2 + ( (a c² )/(a² + c² ) - a/2 )^2 )Compute x-coordinate: (a² c)/(a² + c² )y-coordinate difference: (a c² / (a² + c² ) - a/2 ) = a( c² / (a² + c² ) - 1/2 ) = a( (2 c² - a² - c² ) / (2(a² + c² )) ) = a( (c² - a² ) / (2(a² + c² )) )Thus, distance KN:sqrt( (a^4 c²)/(a² + c² )² + (a² (c² - a² )² )/(4(a² + c² )² ) )Factor out (a²)/(4(a² + c² )² ):sqrt[ (4 a² c² + (c² - a² )² ) * a² / (4(a² + c² )² ) ]Compute numerator inside sqrt:4 a² c² + c⁴ - 2 a² c² + a⁴ = 2 a² c² + c⁴ + a⁴ = (a² + c² )²Therefore, distance KN = sqrt( (a² + c² )² * a² / (4(a² + c² )² ) ) = sqrt( a² / 4 ) = a/2Thus, |KN| = a/2Given that |MP| = 2 |KN|:(1/2)sqrt(a² + c² ) = 2*(a/2 ) => sqrt(a² + c² ) = 2aSquare both sides: a² + c² = 4a² => c² = 3a² => c = a√3 or c = -a√3Assuming c is positive (since it's a coordinate on the x-axis, but depending on the triangle's orientation). Let's take c = a√3. Then, coordinates of C are (a√3, 0 )Therefore, in this case, c = a√3. Then, coordinates of K:( (a² * a√3 ) / (a² + 3a² ), (a * (a√3 )² ) / (a² + 3a² ) )Simplify denominator: 4a²First component: (a³√3 ) / (4a² ) = (a√3 ) / 4Second component: (a * 3a² ) / (4a² ) = (3a ) / 4So K is ( a√3 /4, 3a/4 )Coordinates of T, given that |MN| = |b|/2. Since in our general case, coordinates of B are (b,0). But from the earlier condition MP = 2*KN led to c = a√3. However, in this coordinate system, what is b?Wait, MP is (1/2)sqrt(a² + c² ). But in our earlier work, the condition MP = 2*KN led to c² = 3a². But in this general case, with B at (b,0) and C at (c,0), the midpoint P is at ( (b + c)/2, 0 ). Then, MP is distance from M( b/2, a/2 ) to P( (b + c)/2, 0 ), which is sqrt( ( (b + c)/2 - b/2 )² + ( 0 - a/2 )² ) = sqrt( (c/2 )² + ( -a/2 )² ) = (1/2)sqrt( c² + a² )Given that MP = 2*KN, and KN is a/2, so (1/2)sqrt(c² + a² ) = 2*(a/2 ) => sqrt(c² + a² ) = 2a => c² + a² = 4a² => c² = 3a² => c = a√3.Therefore, regardless of the position of B (coordinate b), the condition MP = 2*KN forces c = a√3. Therefore, point C must be at (a√3, 0 ). The coordinate of B is still (b, 0 ), which hasn't been constrained yet.Wait, but in this case, how do we determine the position of B? The problem didn't specify anything else, so perhaps the position of B can be arbitrary, but with C fixed at (a√3, 0 ) due to the condition MP = 2*KN. However, this seems contradictory because MP depends on both B and C.Wait, no. Wait, in the coordinate system where D is at (0,0), A is at (0,a), B is at (b,0), and C is at (c,0). The midpoint P is at ( (b + c)/2, 0 ). The distance MP is between M( b/2, a/2 ) and P( (b + c)/2, 0 ), which is sqrt( ( (b + c)/2 - b/2 )² + (0 - a/2 )² ) = sqrt( (c/2 )² + ( a/2 )² ) = (1/2)sqrt( c² + a² )But KN is a/2, as we computed. So MP = (1/2)sqrt( c² + a² ) = 2*(a/2 ) => sqrt( c² + a² ) = 2a => c² = 3a².Therefore, regardless of where B is, the coordinate of C must satisfy c² = 3a². Therefore, in this coordinate system, C is fixed at (a√3, 0 ), and B is at (b,0), but with no constraint on b. However, point D is the foot of the altitude from A to BC. Wait, but in this coordinate system, D is at (0,0), which is the foot of the altitude from A(0,a) to BC (the x-axis). Therefore, BC must lie on the x-axis, and D is (0,0), so the altitude from A to BC is AD, which is vertical. Therefore, BC is the x-axis from B(b,0) to C(c,0), with D(0,0) lying on BC. But for D to be the foot of the altitude from A, BC must contain D. However, in this coordinate system, BC is from B(b,0) to C(c,0), and D is at (0,0), so D lies on BC only if 0 is between b and c. Wait, but in the general case, if we have BC as the x-axis from B(b,0) to C(c,0), then the foot of the altitude from A(0,a) is the projection on BC, which is (0,0) only if BC is the x-axis and the altitude is vertical, which would require that the line BC is horizontal, and the altitude is vertical. For that, the projection of A onto BC must be (0,0). But in order for the projection of A(0,a) onto BC (the x-axis) to be D(0,0), BC must be the x-axis, and the altitude from A is the y-axis. Therefore, this coordinate system is valid, with D(0,0) being the foot of the altitude from A(0,a) to BC on the x-axis, with B(b,0) and C(c,0), and D(0,0) lying between B and C.But given that D is the foot of the altitude, BC must be such that AD is perpendicular to BC. In this coordinate system, that's already satisfied since AD is vertical and BC is horizontal.But in this setup, the position of B is arbitrary except that D(0,0) must lie on BC. Wait, but if BC is from B(b,0) to C(c,0), and D is at (0,0), then 0 must be between b and c (assuming b and c have opposite signs), unless D coincides with B or C. However, in the problem statement, D is the foot of the altitude, so it should lie on BC between B and C, assuming ABC is a non-degenerate triangle.But given that in our coordinate system, we've set D at (0,0), and BC is the x-axis from B(b,0) to C(c,0) with 0 between b and c, i.e., b < 0 < c. Then, with this, we can have a proper triangle.However, from the condition MP = 2*KN, we derived c = a√3. Therefore, in this coordinate system, C is fixed at (a√3, 0 ). Then, B is at (b,0), with b < 0. But does this affect the generality of the proof? Because the position of B hasn't been constrained by the condition MP = 2*KN, except that c = a√3.Wait, but wait. If we have BC with midpoint P at ( (b + c)/2, 0 ). But with c = a√3, P is at ( (b + a√3 )/2, 0 ). Then, MP is the distance from M(b/2, a/2 ) to P( (b + a√3 )/2, 0 ). So:MP = sqrt( ( ( (b + a√3 )/2 - b/2 )^2 + ( 0 - a/2 )^2 ) = sqrt( ( a√3 / 2 )^2 + ( -a/2 )^2 ) = sqrt( (3a² /4 ) + (a² /4 ) ) = sqrt(a²) = aBut from the given condition, MP = 2*KN = 2*(a/2 ) = a. So this holds true regardless of b. Wait, this is interesting. So if we fix c = a√3, then MP = a regardless of the value of b. But earlier, when we computed MP in terms of c, we had (1/2)sqrt(a² + c² ). If c = a√3, then sqrt(a² + 3a² ) = sqrt(4a² ) = 2a, so (1/2)*2a = a. Thus, MP = a, which matches 2*KN = a.Therefore, in this coordinate system, once we set c = a√3, MP automatically becomes a, satisfying MP = 2*KN. Therefore, the position of B can be anywhere on the x-axis left of D (i.e., b < 0), and the condition MP = 2*KN will still hold because it's enforced by c = a√3. However, in the problem statement, the positions of B and C are such that D is the foot of the altitude from A to BC, so BC must be positioned such that D lies between B and C.Therefore, in this coordinate system, with A(0,a), D(0,0), C(a√3,0), and B(b,0) where b < 0, we can proceed to compute |AT| and |MC|.First, coordinates of K: as computed earlier, K is ( a√3 /4, 3a/4 )MN is the distance from M(b/2, a/2 ) to N(0, a/2 ). This is |b/2 - 0| = |b/2 |. Since b is negative, |MN| = -b/2.DK is the distance from D(0,0) to K(a√3 /4, 3a/4 ). Which is sqrt( (a√3 /4 )² + (3a/4 )² ) = sqrt( 3a² /16 + 9a² /16 ) = sqrt(12a² /16 ) = (2a√3)/4 = (a√3)/2.Therefore, DT = |MN| + |DK| = (-b/2 ) + (a√3)/2.But T is on the extension of KD beyond D. Direction from K to D is D - K = (-a√3 /4, -3a/4 ). To move beyond D by DT = (-b/2 + a√3 /2 ), which is a length. Wait, but DT is a length, so it must be positive. Since |MN| = |b/2 | and |DK| = (a√3)/2, DT is |MN| + |DK| = |b/2 | + (a√3)/2. However, in our coordinate system, b < 0, so |MN| = -b/2. Therefore, DT = (-b/2 ) + (a√3)/2. Since b is negative, -b/2 is positive, so DT is positive.To find coordinates of T, starting from D(0,0), move in the direction opposite to K. The direction vector from K to D is (-a√3 /4, -3a/4 ), which has length |DK| = (a√3)/2. We need to move a distance of DT = (-b/2 + a√3 /2 ) in this direction. The unit vector in the direction from K to D is (-a√3 /4, -3a/4 ) divided by (a√3 /2 ), which is:( (-a√3 /4 ) / (a√3 /2 ), (-3a/4 ) / (a√3 /2 ) ) = ( (-√3 /4 ) / (√3 /2 ), (-3/4 ) / (√3 /2 ) ) = ( (-1/2 ), (-3/(2√3 )) ) = ( -1/2, -√3 / 2 )Therefore, the unit vector is ( -1/2, -√3 / 2 )Then, vector DT is this unit vector multiplied by DT = (-b/2 + a√3 /2 )So coordinates of T are D + vector DT = (0,0) + ( -1/2 * (-b/2 + a√3 /2 ), -√3 /2 * (-b/2 + a√3 /2 ) )Wait, no. Wait, the unit vector is ( -1/2, -√3 / 2 ), and we need to scale it by DT = (-b/2 + a√3 /2 ). So:vector DT = ( -1/2 * ( (-b/2 + a√3 /2 ) ), -√3 /2 * ( (-b/2 + a√3 /2 ) ) )Wait, no. Wait, the unit vector multiplied by the scalar DT gives the displacement from D to T. Therefore:T_x = 0 + (-1/2) * ( (-b/2 + a√3 /2 ) ) = (1/2)( b/2 - a√3 /2 ) = b/4 - a√3 /4T_y = 0 + (-√3 /2 ) * ( (-b/2 + a√3 /2 ) ) = (√3 /2 )( b/2 - a√3 /2 ) = (√3 b )/4 - ( a * 3 )/4Therefore, coordinates of T are ( b/4 - a√3 /4, (√3 b)/4 - (3a)/4 )Now, compute |AT| and |MC|.Coordinates of A(0,a) and T( b/4 - a√3 /4, (√3 b)/4 - (3a)/4 )Distance AT:sqrt( ( b/4 - a√3 /4 - 0 )² + ( (√3 b)/4 - 3a/4 - a )² )Simplify:x-component: ( b - a√3 ) /4y-component: ( √3 b - 3a - 4a ) /4 = ( √3 b - 7a ) /4Thus, distance AT squared:[ (b - a√3 )² + ( √3 b - 7a )² ] / 16Expand:Numerator:(b² - 2a√3 b + 3a² ) + ( 3b² - 14a√3 b + 49a² )= b² - 2a√3 b + 3a² + 3b² - 14a√3 b + 49a²= 4b² - 16a√3 b + 52a²Factor:4b² - 16a√3 b + 52a² = 4(b² - 4a√3 b + 13a² )Now, compute |MC|.Coordinates of M( b/2, a/2 ) and C( a√3, 0 )Distance MC:sqrt( ( a√3 - b/2 )² + ( 0 - a/2 )² )= sqrt( ( a√3 - b/2 )² + ( -a/2 )² )Square:( a√3 - b/2 )² + ( a/2 )²Expand:3a² - a√3 b + b²/4 + a²/4= 3a² + a²/4 - a√3 b + b²/4= (13a²)/4 - a√3 b + b²/4Multiply numerator and denominator by 4 to compare with AT squared:AT squared: [4b² - 16a√3 b + 52a² ] /16 = [b² - 4a√3 b +13a² ] /4MC squared: (13a²)/4 - a√3 b + b²/4 = ( b² -4a√3 b +13a² ) /4Thus, AT squared = MC squared. Therefore, |AT| = |MC|Hence proved.So regardless of the value of b (as long as the coordinate system satisfies the given conditions), the distance AT equals MC. Therefore, the original statement is proved.Final Answerboxed{|AT| = |MC|}</think>Given triangle (ABC) with altitude (AD), midpoints (M), (N), and (P) of (AB), (AD), and (BC) respectively, foot of perpendicular (K) from (D) to (AC), and point (T) on the extension of (KD) such that (DT = MN + DK). We need to prove ( |AT| = |MC| ) given ( |MP| = 2 cdot |KN| ).1. Coordinate Setup: - Place (D) at the origin ((0, 0)), (A) at ((0, a)), (B) at ((b, 0)), and (C) at ((c, 0)). - Midpoints: (Mleft(frac{b}{2}, frac{a}{2}right)), (Nleft(0, frac{a}{2}right)), (Pleft(frac{b+c}{2}, 0right)).2. Foot of Perpendicular (K): - Equation of (AC): (y = -frac{a}{c}x + a). - Foot of perpendicular from (D(0,0)) to (AC) gives (Kleft(frac{a^2 c}{a^2 + c^2}, frac{a c^2}{a^2 + c^2}right)).3. Distances: - ( |MN| = frac{|b|}{2} ). - ( |DK| = frac{ac}{sqrt{a^2 + c^2}} ).4. Given Condition: - ( |MP| = frac{1}{2} sqrt{a^2 + c^2} ). - ( |KN| = frac{a}{2} ). - Given ( |MP| = 2 cdot |KN| implies sqrt{a^2 + c^2} = 2a implies c = asqrt{3} ).5. Coordinates of (K): - With (c = asqrt{3}), (Kleft(frac{asqrt{3}}{4}, frac{3a}{4}right)).6. Point (T): - (DT = |MN| + |DK| = frac{|b|}{2} + frac{asqrt{3}}{2}). - Coordinates of (T) found by extending (KD) beyond (D).7. Distances (AT) and (MC): - Coordinates of (T) derived as (left(frac{b - asqrt{3}}{4}, frac{sqrt{3}b - 3a}{4}right)). - Calculate ( |AT| ) and ( |MC| ): - ( |AT| = sqrt{left(frac{b - asqrt{3}}{4}right)^2 + left(frac{sqrt{3}b - 7a}{4}right)^2} ) - ( |MC| = sqrt{left(asqrt{3} - frac{b}{2}right)^2 + left(frac{a}{2}right)^2} )8. Verification: - Both distances simplify to (sqrt{frac{4b^2 - 16asqrt{3}b + 52a^2}{16}}), confirming ( |AT| = |MC| ).Thus, the final answer is (boxed{|AT| = |MC|}).,<think>Okay, let's see. I need to prove that |AT| = |MC| in this geometric setup. Hmm. Let me start by drawing a diagram to visualize the problem. So, we have triangle ABC with altitude AD. That means AD is perpendicular to BC. Then M, N, and P are midpoints of AB, AD, and BC respectively. K is the foot of the perpendicular from D to AC. Then T is a point on the extension of KD beyond D such that DT = MN + DK. Also, we're given that MP = 2*KN. I need to show AT = MC. First, let's note all the midpoints. Since M is the midpoint of AB, and N is the midpoint of AD, that connects M and N as midpoints of two sides. Similarly, P is the midpoint of BC, which is the third side. Hmm. Maybe coordinate geometry can help here? Assign coordinates to the triangle and compute all these lengths.Let me place triangle ABC in a coordinate system. Let's set point A at (0, 0) for simplicity. Since AD is an altitude, BC must be horizontal if AD is vertical. Wait, but if AD is the altitude, then AD is perpendicular to BC. Let me assign coordinates such that BC is on the x-axis. Wait, but if I put A at (0, 0), then AD would be the vertical altitude. Let's see. Let's assign coordinates:Let’s let point A be at (0, 0). Let’s let D be at (0, h) since it's the altitude from A to BC. Wait, no. If AD is the altitude, then D is on BC, and AD is perpendicular to BC. So BC is the base, and AD is the height. So, perhaps setting point D at the origin? Wait, maybe another approach.Alternatively, let me set BC horizontal. Let’s set point B at (b, 0), point C at (c, 0), so BC is along the x-axis. Then the altitude AD from A to BC will meet BC at D. Since AD is an altitude, AD is vertical if BC is horizontal. Wait, but unless A is above BC. Let me suppose A is at (d, h), and D is the foot of the altitude from A to BC. Then coordinates of D would be (d, 0). Because the altitude from A to BC is vertical? No, unless BC is horizontal. Wait, maybe coordinate geometry is complicating things. Alternatively, maybe using vectors or properties of midpoints and parallelograms?Alternatively, since M, N, P are midpoints, maybe using midline theorems. Also, since MP = 2*KN, perhaps there's a relation between segments MP and KN. Hmm. Let's try coordinate geometry step by step.Let me assign coordinates:Let’s place point D at the origin (0,0) because it's the foot of the altitude from A. Then since AD is the altitude, A is at (0, h) for some h. Let’s say BC is on the x-axis, so D is (0,0), B is (-b, 0), C is (c, 0). Wait, but if D is the foot of the altitude from A, then AD is perpendicular to BC, which is horizontal. So BC is horizontal, and AD is vertical. So yes, A is (0, h), D is (0,0), and BC is on the x-axis from (-b, 0) to (c, 0). Then AB is from (0, h) to (-b, 0), and AC is from (0, h) to (c, 0). Let's keep it general. Then midpoints:M is the midpoint of AB. So coordinates of M would be [(0 + (-b))/2, (h + 0)/2] = (-b/2, h/2).N is the midpoint of AD. AD is from (0, h) to (0,0), so midpoint N is (0, h/2).P is the midpoint of BC. BC is from (-b, 0) to (c, 0), so midpoint P is [( -b + c ) / 2, 0].K is the foot of the perpendicular from D to AC. D is (0,0), AC is from (0, h) to (c, 0). Let's find the equation of line AC. The slope of AC is (0 - h)/(c - 0) = -h/c. Therefore, the equation of AC is y = (-h/c)x + h.The foot of the perpendicular from D(0,0) to AC is point K. To find K, we can use the formula for the projection of a point onto a line. The formula for foot of perpendicular from (x0, y0) to line ax + by + c = 0 is:K(x) = (b(bx0 - ay0) - ac) / (a² + b²)K(y) = (a(-bx0 + ay0) - bc) / (a² + b²)But maybe it's easier to parametrize. Let me write line AC as y = (-h/c)x + h. The direction vector of AC is (c, -h). A perpendicular direction would be (h, c). So the line perpendicular to AC through D(0,0) has direction (h, c). Therefore, parametric equations for KD (the perpendicular) is x = ht, y = ct. We need to find the intersection point K between this line and AC.So substitute x = ht, y = ct into AC's equation:ct = (-h/c)(ht) + hct = -h² t / c + hMultiply both sides by c:c² t = -h² t + hct(c² + h²) = hct = hc / (c² + h²)Therefore, coordinates of K:x = h * t = h * (hc / (c² + h²)) = h² c / (c² + h²)y = c * t = c * (hc / (c² + h²)) = h c² / (c² + h²)So K is at (h² c / (c² + h²), h c² / (c² + h²))That seems complicated, but maybe manageable.Now, T is a point on the extension of KD beyond D such that DT = MN + DK.First, let's compute MN and DK.MN: M is (-b/2, h/2), N is (0, h/2). So vector MN is (0 - (-b/2), h/2 - h/2) = (b/2, 0). So |MN| = b/2.Wait, |MN| is the length of MN, which is sqrt( (b/2)^2 + 0^2 ) = b/2. So MN = b/2.DK: D is (0,0), K is (h² c / (c² + h²), h c² / (c² + h²)). So vector DK is (h² c / (c² + h²), h c² / (c² + h²)), and |DK| is sqrt( (h² c / (c² + h²))² + (h c² / (c² + h²))² )= sqrt( h^4 c² + h² c^4 ) / (c² + h² )= h c sqrt( h² + c² ) / (c² + h² )= h c / sqrt(c² + h² )Wait, because sqrt(h^4 c² + h² c^4) = h c sqrt( h² + c² )Then denominator is (c² + h²). So |DK| = h c / sqrt(c² + h² )Therefore, |DT| = |MN| + |DK| = b/2 + h c / sqrt(c² + h² )But T is on the extension of KD beyond D. So from K, going through D, then extending past D by |DT|.Wait, no. Wait, KD is the segment from K to D. The extension beyond D would be beyond D from K. Wait, KD is from K to D. So if we extend KD beyond D, then T is such that DT = MN + DK, but direction?Wait, the problem says: "T is a point on extension of line KD (over point D) such that |DT| = |MN| + |DK|". So starting from D, going in the direction opposite to K for a distance |MN| + |DK|. Wait, KD is from K to D. The extension over D would be beyond D, away from K. So vector from K to D is DK. So to go beyond D from D in the same line, direction is from K to D. So T is beyond D along the line KD, distance from D is |DT| = |MN| + |DK|. So DT is a segment from D to T, with length |MN| + |DK|, in the direction opposite to K (since extension is over D). Wait, maybe parametrize.Let me parametrize line KD. Since K is at (h² c / (c² + h²), h c² / (c² + h²)), and D is (0,0), the direction vector from K to D is (-h² c / (c² + h²), -h c² / (c² + h²)). So parametric equations for line KD beyond D (in the direction from D opposite to K) would be:x = 0 - t*(h² c / (c² + h²))y = 0 - t*(h c² / (c² + h²))for t ≥ 0.So point T is at:x = - t*(h² c / (c² + h²))y = - t*(h c² / (c² + h²))and the distance from D to T is t * |DK|, since |DK| is h c / sqrt(c² + h²). Wait, |DK| is h c / sqrt(c² + h²). So if we set |DT| = |MN| + |DK| = b/2 + h c / sqrt(c² + h²), then t must satisfy t * (h c / sqrt(c² + h²)) ) = b/2 + h c / sqrt(c² + h² )Therefore,t = (b/2 + h c / sqrt(c² + h² )) / (h c / sqrt(c² + h² )) )= (b/2) / (h c / sqrt(c² + h² )) + 1= (b/2) * sqrt(c² + h² ) / (h c ) + 1Therefore, coordinates of T:x = - [ (b/2) * sqrt(c² + h² ) / (h c ) + 1 ] * (h² c / (c² + h² ))Similarly for y.This is getting really messy. Maybe there's a better approach.Alternatively, perhaps using vectors. Let me assign vectors to the points.Let’s denote vectors with bold letters.Let’s set coordinate system with D at the origin (0,0). Then vector AD is vertical. Let’s let vector AD be (0, h), so point A is at (0, h). Let’s let point B be at (b, 0) and point C be at (c, 0). So BC is from (b, 0) to (c, 0), midpoint P is at ((b + c)/2, 0).Midpoint M of AB: coordinates ((b)/2, h/2).Midpoint N of AD: coordinates (0, h/2).K is the foot of the perpendicular from D(0,0) to AC. Vector AC is from A(0, h) to C(c, 0), so parametric equation of AC is (ct, h - h t) for t from 0 to 1.Wait, the line AC can be parametrized as (c t, h (1 - t)) where t ∈ [0,1]. The foot of the perpendicular from D(0,0) to AC can be found by projecting D onto AC.In vector terms, the projection of vector OD (which is (0,0)) onto AC is given by:OK = ( (OD - OA) · AC / |AC|² ) * AC + OAWait, OD is (0,0), OA is (0, h). Vector AC is (c, -h). So vector DA is (0, h) - (0,0) = (0, h). Hmm, maybe better to use standard projection formula.The formula for the foot of the perpendicular from point P to line through points Q and R is:K = Q + ((P - Q) · (R - Q)) / |R - Q|² * (R - Q)In this case, P is D(0,0), Q is A(0, h), R is C(c, 0). So vector QR is (c, -h). Then:K = A + [(D - A) · (C - A)] / |C - A|² * (C - A)Compute (D - A) = (0 - 0, 0 - h) = (0, -h). (C - A) = (c, -h). Then:(D - A) · (C - A) = 0 * c + (-h) * (-h) = h²|C - A|² = c² + h²Therefore, K = A + (h² / (c² + h²)) * (c, -h)Which gives coordinates:K_x = 0 + (h² / (c² + h²)) * c = c h² / (c² + h²)K_y = h + (h² / (c² + h²)) * (-h) = h - h³ / (c² + h²) = (h(c² + h²) - h³) / (c² + h²) = h c² / (c² + h²)So K is ( c h² / (c² + h² ), h c² / (c² + h² ) )Same as before.Then vector DK is K - D = ( c h² / (c² + h² ), h c² / (c² + h² ) )Length DK is sqrt( (c h² / (c² + h² ))^2 + (h c² / (c² + h² ))^2 )= (h c / (c² + h² )) sqrt( h² + c² )= (h c / (c² + h² )) * sqrt(c² + h² )= h c / sqrt(c² + h² )So |DK| = h c / sqrt(c² + h² )Then |MN| is the distance between M and N. M is midpoint of AB: (b/2, h/2). N is midpoint of AD: (0, h/2). So distance MN is sqrt( (b/2 - 0)^2 + (h/2 - h/2)^2 ) = b/2Therefore, |DT| = |MN| + |DK| = b/2 + h c / sqrt(c² + h² )Now, T is along the line KD extended beyond D by this distance. Since direction from K to D is given by vector DK, which is ( c h² / (c² + h² ), h c² / (c² + h² ) ). The direction from D to T is opposite to DK, so direction vector is -DK. So parametric equation for DT: starting at D(0,0), moving in direction -DK, scaled appropriately.But the length DT is |DT|. Let me find the unit vector in the direction from K to D. The vector DK is ( c h², h c² ) / (c² + h² ). Its magnitude is h c / sqrt(c² + h² ). Therefore, the unit vector in the direction from K to D is ( c h², h c² ) / (c² + h² ) divided by (h c / sqrt(c² + h² )) = ( c h², h c² ) / (c² + h² ) * sqrt(c² + h² ) / (h c ) = ( c h sqrt(c² + h² ), c² sqrt(c² + h² ) ) / (c² + h² ) * 1 / h c ?Wait, maybe better to note that direction from D to T is the same as direction from D to K reversed. Wait, no. KD is from K to D. So extending beyond D, the direction is from D away from K. Wait, KD is from K to D. So beyond D, the direction is opposite to the vector DK. So the direction vector is -DK. Therefore, to get from D to T, we need to move in direction -DK for length |DT|.But the vector -DK has magnitude |DK|, so to move a distance of |DT| in that direction, we need to scale the unit vector in that direction by |DT|.Thus, point T is D + (-DK unit vector) * |DT|But DK unit vector is DK / |DK| = ( c h² / (c² + h² ), h c² / (c² + h² )) / ( h c / sqrt(c² + h² ) )= ( c h² , h c² ) / (c² + h² ) * sqrt(c² + h² ) / (h c )= ( c h² * sqrt(c² + h² ), h c² * sqrt(c² + h² ) ) / [ (c² + h² ) h c ]= ( h sqrt(c² + h² ), c sqrt(c² + h² ) ) / (c² + h² )= ( h, c ) / sqrt(c² + h² )Therefore, the unit vector in the direction DK is ( h, c ) / sqrt(c² + h² )Therefore, the unit vector in the direction opposite (from D to T) is - ( h, c ) / sqrt(c² + h² )Thus, point T is D + |DT| * ( - ( h, c ) / sqrt(c² + h² ) )Since D is at (0,0), coordinates of T are:T_x = 0 - |DT| * h / sqrt(c² + h² )T_y = 0 - |DT| * c / sqrt(c² + h² )But |DT| = b/2 + h c / sqrt(c² + h² )Thus,T_x = - ( b/2 + h c / sqrt(c² + h² ) ) * h / sqrt(c² + h² )T_y = - ( b/2 + h c / sqrt(c² + h² ) ) * c / sqrt(c² + h² )Simplifying T_x:= - [ (b/2) h / sqrt(c² + h² ) + h² c / (c² + h² ) ]Similarly, T_y:= - [ (b/2) c / sqrt(c² + h² ) + h c² / (c² + h² ) ]This is very complicated. Maybe there's a smarter coordinate system. Let me consider specific coordinates to make the problem symmetric or choose values for b, c, h to simplify.Alternatively, perhaps set some variables to specific values. For instance, let me set h = 1 for simplicity, c = 1, and adjust b accordingly. Wait, but then we have to satisfy MP = 2*KN. Maybe not. Alternatively, set coordinate system such that AC is along some axis. Hmm.Alternatively, since the problem is to show |AT| = |MC|, maybe we can compute |AT| and |MC| in terms of coordinates and show they are equal under the given condition MP = 2*KN.First, let's compute |MP| and |KN|.Point M is midpoint of AB: ( (-b)/2, h/2 )Wait, wait. Wait in our coordinate system, point B is at (b, 0), point C is at (c, 0). Wait no, earlier I set D at (0,0), so BC is from (b, 0) to (c, 0)? Wait, hold on. Wait, in the coordinate system where D is (0,0), which is the foot of the altitude from A(0, h) to BC. Then BC is the base, but if we set D at (0,0), then BC can be located anywhere along the x-axis, but AD is vertical from (0, h) to (0,0). Wait, but in that case, BC can be anywhere, but in order for AD to be the altitude, BC must be horizontal, and D is the foot. However, unless AD is the altitude, BC can be placed such that D is the foot. Wait, in reality, in triangle ABC, AD is the altitude, so BC is the base, and D is the foot on BC. Therefore, BC is a horizontal line, and AD is vertical. Then coordinates make sense as I set before.Wait, perhaps to simplify, set D at (0,0), A at (0, h), B at (-b, 0), and C at (c, 0). Then BC is from (-b, 0) to (c, 0), midpoint P is ((c - b)/2, 0). Then midpoint M of AB is at ( (-b)/2, h/2 ). Midpoint N of AD is (0, h/2). Then compute MP and KN.Compute MP: distance from M( (-b)/2, h/2 ) to P( (c - b)/2, 0 )Coordinates difference: x: (c - b)/2 - (-b)/2 = (c - b + b)/2 = c/2y: 0 - h/2 = -h/2Therefore, |MP| = sqrt( (c/2)^2 + ( - h/2 )^2 ) = (1/2) sqrt( c² + h² )Compute KN: distance from K to N. Coordinates of K: ( c h² / (c² + h² ), h c² / (c² + h² ) )Coordinates of N: (0, h/2 )Difference in coordinates:x: 0 - c h² / (c² + h² ) = - c h² / (c² + h² )y: h/2 - h c² / (c² + h² ) = ( h(c² + h² ) / 2(c² + h² ) ) - ( 2 h c² / 2(c² + h² ) ) = [ h(c² + h² ) - 2 h c² ] / 2(c² + h² )= h( - c² + h² ) / 2(c² + h² )Therefore, |KN| = sqrt( [ - c h² / (c² + h² ) ]^2 + [ h( - c² + h² ) / 2(c² + h² ) ]^2 )= sqrt( c² h^4 / (c² + h² )² + h² ( - c² + h² )² / 4(c² + h² )² )= [ h / (c² + h² ) ] * sqrt( c² h² + ( - c² + h² )² / 4 )Let’s compute the term inside the sqrt:c² h² + ( (-c² + h² )² ) / 4First, expand (-c² + h² )² = c^4 - 2 c² h² + h^4So term becomes:c² h² + (c^4 - 2 c² h² + h^4 ) / 4= (4 c² h² + c^4 - 2 c² h² + h^4 ) / 4= (c^4 + 2 c² h² + h^4 ) / 4= (c² + h² )² / 4Therefore, |KN| = [ h / (c² + h² ) ] * sqrt( (c² + h² )² / 4 )= [ h / (c² + h² ) ] * (c² + h² ) / 2= h / 2So |KN| = h / 2Wait, that's a surprise. So regardless of c, |KN| = h/2.But we have the given condition that |MP| = 2 * |KN|But we found |MP| = (1/2) sqrt(c² + h² ) and |KN| = h/2. Therefore,(1/2) sqrt(c² + h² ) = 2*(h/2 ) = hMultiply both sides by 2:sqrt(c² + h² ) = 2 hSquare both sides:c² + h² = 4 h²Therefore, c² = 3 h² => c = h sqrt(3 )So c = sqrt(3 ) h. Therefore, coordinates of C are ( sqrt(3 ) h, 0 )Hmm, interesting. So under the given condition, c = sqrt(3 ) h. So we can now substitute c = sqrt(3 ) h into the coordinates.Let me do that. Let’s set h as a parameter, say h = 1 for simplicity. Then c = sqrt(3 ). So coordinates:A(0, 1), D(0,0), B(-b, 0), C(sqrt(3 ), 0). Wait, but earlier we set D at (0,0), which is the foot of altitude from A. Wait, but in that case, BC is from (-b, 0) to (sqrt(3 ), 0). However, in this case, we need to relate b. Wait, but in our calculation, we didn't use the value of b except in the coordinates of B and M. However, from the condition, we found c = sqrt(3 ) h, but b is still arbitrary? Wait, but in the problem statement, there was no given relation between AB and other sides. So perhaps b is determined by the triangle? Wait, but we might have multiple triangles satisfying the conditions, but in this case, with c = sqrt(3 ) h, but b is still arbitrary. Wait, but let's check.Wait, but since the altitude is AD, which is from A(0, h) to D(0,0). Then BC is the base. So BD and DC are segments of BC. Wait, but in our coordinate system, we have BC from (-b, 0) to (c, 0). But the foot of the altitude from A is D(0,0). So BD = | -b - 0 | = b, DC = |c - 0| = c. So in this case, BD = b, DC = c. But in the problem statement, there was no given condition on BD and DC, so they can vary. However, given MP = 2*KN led us to c = sqrt(3 ) h. But what about b?Wait, but in our calculation of |KN|, the value of b didn't factor in. Because KN is the distance from K to N, and K is determined by the foot from D to AC, which is independent of B. So perhaps b is arbitrary? But in the coordinates of point T, which depends on b through |DT| = b/2 + h c / sqrt(c² + h² )But given that c = sqrt(3 ) h, let's substitute that.If c = sqrt(3 ) h, then sqrt(c² + h² ) = sqrt(3 h² + h² ) = 2 hTherefore, |DK| = h c / sqrt(c² + h² ) = h * sqrt(3 ) h / (2 h ) = ( sqrt(3 ) / 2 ) hAlso, |MN| = b / 2Therefore, |DT| = b/2 + sqrt(3 ) h / 2But in our coordinate system, h is the height from A to BC, which we can set as 1 for simplicity. Let’s set h = 1, so c = sqrt(3 )Therefore, coordinates:A(0, 1), D(0,0), B(-b, 0), C(sqrt(3 ), 0), M( -b/2, 1/2 ), N(0, 1/2 ), P( (sqrt(3 ) - b ) / 2, 0 )K is the foot from D(0,0) to AC. Since AC is from (0,1) to (sqrt(3 ),0), which we already calculated. For h=1, c=sqrt(3 ), K is:K_x = c h² / (c² + h² ) = sqrt(3 ) * 1 / ( 3 + 1 ) = sqrt(3 ) / 4K_y = h c² / (c² + h² ) = 1 * 3 / 4 = 3/4So K is ( sqrt(3 ) / 4, 3/4 )Then T is the point on extension of KD beyond D such that |DT| = |MN| + |DK| = b/2 + sqrt(3 ) / 2Coordinates of T: as calculated before,T_x = - ( b/2 + sqrt(3 ) / 2 ) * h / sqrt(c² + h² )But h=1, c=sqrt(3 ), sqrt(c² + h² )=2. So,T_x = - ( b/2 + sqrt(3 ) / 2 ) * 1 / 2 * 1 ?Wait, no. Wait earlier formula:T_x = - ( |DT| ) * h / sqrt(c² + h² )Since h=1, sqrt(c² + h² )=2, soT_x = - ( (b/2 + sqrt(3 ) / 2 ) ) * 1 / 2 = - (b + sqrt(3 )) / 4Similarly,T_y = - ( (b/2 + sqrt(3 ) / 2 ) ) * c / sqrt(c² + h² )c=sqrt(3 ), sqrt(c² + h² )=2,So T_y = - ( (b + sqrt(3 )) / 2 ) * sqrt(3 ) / 2 = - ( (b + sqrt(3 )) sqrt(3 ) ) / 4Therefore, coordinates of T are ( - (b + sqrt(3 )) / 4, - ( (b + sqrt(3 )) sqrt(3 ) ) / 4 )Now, need to compute |AT| and |MC|.Point A is (0,1). T is ( - (b + sqrt(3 )) / 4, - ( (b + sqrt(3 )) sqrt(3 ) ) / 4 )Distance AT:sqrt[ ( - (b + sqrt(3 )) / 4 - 0 )² + ( - ( (b + sqrt(3 )) sqrt(3 ) ) / 4 - 1 )² ]= sqrt[ ( (b + sqrt(3 ))² / 16 ) + ( ( - (b + sqrt(3 )) sqrt(3 ) / 4 - 1 ) )² ]Let me compute each term:First term: ( (b + sqrt(3 ))² ) / 16Second term:( - (b + sqrt(3 )) sqrt(3 ) / 4 - 1 )²= ( - sqrt(3 )(b + sqrt(3 )) / 4 - 1 )²= ( - ( sqrt(3 ) b + 3 ) / 4 - 1 )²= ( - ( sqrt(3 ) b + 3 ) / 4 - 4/4 )²= ( - ( sqrt(3 ) b + 3 + 4 ) / 4 )²= ( - ( sqrt(3 ) b + 7 ) / 4 )²= ( ( sqrt(3 ) b + 7 ) / 4 )²Therefore, |AT| = sqrt[ (b + sqrt(3 ))² / 16 + ( sqrt(3 ) b + 7 )² / 16 ]= (1/4) sqrt[ (b + sqrt(3 ))² + ( sqrt(3 ) b + 7 )² ]Expand inside the sqrt:First term: (b + sqrt(3 ))² = b² + 2 b sqrt(3 ) + 3Second term: ( sqrt(3 ) b + 7 )² = 3 b² + 14 sqrt(3 ) b + 49Sum:b² + 2 b sqrt(3 ) + 3 + 3 b² + 14 sqrt(3 ) b + 49= 4 b² + 16 sqrt(3 ) b + 52Factor:4 b² + 16 sqrt(3 ) b + 52= 4( b² + 4 sqrt(3 ) b + 13 )Hmm, maybe this factors further? Let me check discriminant:b² + 4 sqrt(3 ) b + 13Discriminant: (4 sqrt(3 ))^2 - 4*1*13 = 48 - 52 = -4 < 0, so it doesn't factor over reals. So |AT| = (1/4) sqrt(4(b² + 4 sqrt(3 ) b +13 )) = (1/4)*2*sqrt(b² +4 sqrt(3 ) b +13 ) = (1/2) sqrt(b² +4 sqrt(3 ) b +13 )Now compute |MC|. Point M is midpoint of AB: (-b/2, 1/2 ). Point C is (sqrt(3 ), 0 )Distance MC:sqrt[ ( sqrt(3 ) + b/2 )² + ( 0 - 1/2 )² ]= sqrt[ ( b/2 + sqrt(3 ) )² + ( -1/2 )² ]= sqrt[ (b² /4 + b sqrt(3 ) + 3 ) + 1/4 ]= sqrt[ b² /4 + b sqrt(3 ) + 3 + 1/4 ]= sqrt[ (b² + 4 b sqrt(3 ) + 13 ) /4 ]= (1/2) sqrt( b² +4 b sqrt(3 ) +13 )Which is the same as |AT|. Therefore, |AT| = |MC|. Hence proved.Wait, but this is under the condition that c = sqrt(3 ) h and h=1. But in general, since we set h=1 for simplicity, but the relation c = sqrt(3 ) h comes from the condition MP = 2*KN. So in general, even if h is arbitrary, the same algebra would show that |AT| and |MC| both equal (1/2) sqrt( b² +4 b sqrt(3 ) h +13 h² ). Wait, but in our case, with c = sqrt(3 ) h, but we also have that c is determined by the condition MP = 2*KN. Wait, in our coordinate system, c was chosen to satisfy the condition. Therefore, in the problem statement, it's given that MP = 2*KN, which led us to c = sqrt(3 ) h. Then under this condition, we found that |AT| = |MC|. So the proof holds.Therefore, the answer is proved.Final Answerboxed{ |AT| = |MC| }</think>Given triangle (ABC) with (AD) as an altitude, midpoints (M), (N), and (P) of (AB), (AD), and (BC) respectively, and (K) as the foot of the perpendicular from (D) to (AC). Point (T) is on the extension of (KD) beyond (D) such that ( |DT| = |MN| + |DK| ). We need to prove ( |AT| = |MC| ) given ( |MP| = 2 cdot |KN| ).1. Coordinate Setup: - Place (D) at the origin ((0, 0)), (A) at ((0, h)), (B) at ((-b, 0)), and (C) at ((c, 0)). - Midpoints: (Mleft(-frac{b}{2}, frac{h}{2}right)), (N(0, frac{h}{2})), (Pleft(frac{c - b}{2}, 0right)).2. Foot of Perpendicular (K): - Line (AC) has equation (y = -frac{h}{c}x + h). - Parametrize line (KD) with direction vector ((h, c)). - Solve for intersection (K) on (AC): (Kleft(frac{ch^2}{c^2 + h^2}, frac{hc^2}{c^2 + h^2}right)).3. Length Calculations: - (|MN| = frac{b}{2}). - (|DK| = frac{hc}{sqrt{c^2 + h^2}}). - Given (|MP| = 2 cdot |KN|), calculate (|MP| = frac{1}{2}sqrt{c^2 + h^2}) and (|KN| = frac{h}{2}). - Solve for (c): (c = sqrt{3}h).4. Coordinates of (T): - (|DT| = frac{b}{2} + frac{sqrt{3}}{2}h). - Parametrize (T) beyond (D): (Tleft(-frac{(b + sqrt{3}h)}{4}, -frac{(b + sqrt{3}h)sqrt{3}}{4}right)).5. Distance Calculations: - Compute (|AT|) and (|MC|): - (|AT| = frac{1}{2}sqrt{b^2 + 4sqrt{3}bh + 13h^2}). - (|MC| = frac{1}{2}sqrt{b^2 + 4sqrt{3}bh + 13h^2}).Since both distances are equal, we conclude ( |AT| = |MC| ).[boxed{|AT| = |MC|}]]

🔑:1. Identify Key Elements and Relationships: - Let AD be the altitude of triangle ABC. - M, N, and P are midpoints of AB, AD, and BC, respectively. - K is the foot of the perpendicular from D to AC. - T is a point on the extension of KD such that |DT| = |MN| + |DK|. - Given: |MP| = 2 cdot |KN|.2. Establish Initial Relationships: - Since N is the midpoint of AD, |AN| = |ND| = frac{1}{2}|AD|. - Since M is the midpoint of AB, |AM| = |MB| = frac{1}{2}|AB|. - Since P is the midpoint of BC, |BP| = |PC| = frac{1}{2}|BC|.3. Use Given Condition |MP| = 2 cdot |KN|: - MP is the midsegment of triangle ABC parallel to AC and half its length, so |MP| = frac{1}{2}|AC|. - KN is the altitude from D to AC, so |KN| = frac{1}{2}|AD|. - Given |MP| = 2 cdot |KN|, we have: [ frac{1}{2}|AC| = 2 cdot frac{1}{2}|AD| implies |AC| = 2|AD|. ]4. Analyze triangle ADC: - Since |AC| = 2|AD|, triangle ADC is a 30^circ-60^circ-90^circ triangle. - In a 30^circ-60^circ-90^circ triangle, the side opposite the 30^circ angle is half the hypotenuse, and the side opposite the 60^circ angle is frac{sqrt{3}}{2} times the hypotenuse.5. Relate DK and DC: - Since triangle ADC is a 30^circ-60^circ-90^circ triangle, |DC| = sqrt{3}|AD|. - K is the foot of the perpendicular from D to AC, so |DK| = frac{sqrt{3}}{2}|AD|.6. Use Given Condition |DT| = |MN| + |DK|: - MN is the midsegment of triangle ABD parallel to BD and half its length, so |MN| = frac{1}{2}|BD|. - Since |BD| = |DC| (as D is the midpoint of BC), |MN| = frac{1}{2}|DC| = frac{1}{2}sqrt{3}|AD|. - Therefore, |DT| = |MN| + |DK| = frac{1}{2}sqrt{3}|AD| + frac{sqrt{3}}{2}|AD| = sqrt{3}|AD|.7. Relate DT and PC: - Since |PC| = frac{1}{2}|BC| = frac{1}{2}|DC| = frac{1}{2}sqrt{3}|AD|. - Given |DT| = sqrt{3}|AD|, we have |DT| = 2|PC|.8. Analyze triangle MPC and triangle ADT: - angle ACM = alpha implies angle MCP = 30^circ - alpha. - Since MP parallel AC, angle PMC = alpha. - angle MPC = 150^circ. - angle KDC = 60^circ = angle BDT and angle ADB = 90^circ. - triangle MPC cong triangle ADT by AA similarity (angles are equal).9. Conclude |AT| = |MC|: - Since triangle MPC cong triangle ADT, corresponding sides are equal. - Therefore, |AT| = |MC|.The final answer is boxed{|AT| = |MC|}.

🧠:[<think>Okay, so I have this problem here: Given n ≥ 3 points in a plane, not all collinear, find the least possible number of triangles formed by these points. The vertices of the triangles must not be collinear, of course. Hmm, interesting. Let me try to figure this out step by step.First off, triangles are formed by three non-collinear points. So, if all the points were in general position (meaning no three are collinear), the number of triangles would just be the combination of n points taken 3 at a time, which is C(n,3). But the problem says the points are not all collinear, which means there could be some collinear points, but not all of them. So, we need to arrange the points in such a way that we have as many collinear triples as possible to minimize the number of triangles. Because each collinear triple doesn't form a triangle, so subtracting those from the total would give the number of triangles. So, our goal is to maximize the number of collinear triples to minimize the total triangles.So, the formula for the minimum number of triangles would be Total possible triangles (C(n,3)) minus the maximum number of collinear triples. Therefore, the problem reduces to finding the maximum number of collinear triples possible given n points, not all collinear. Now, how do we maximize the number of collinear triples? Intuitively, if we have as many points as possible on a single line, that would create the most collinear triples. However, since the problem states that the points are not all collinear, we can't have all n points on one line. So, the maximum number of points we can have on a single line is n-1, with one point off the line. But wait, if we have n-1 points on a line, then the number of collinear triples would be C(n-1,3). Then the remaining point can't form any collinear triples with the others because it's not on that line, so it can only form triangles. Therefore, the number of triangles would be C(n,3) - C(n-1,3). Let me check this.For example, let's take n=4. If three points are on a line and the fourth is off, then the total number of triangles is C(4,3) - C(3,3) = 4 - 1 = 3. Which is correct: each of the three triples that include the off-line point form a triangle, and the three points on the line don't. So yes, that works.Similarly, for n=5. If four points are on a line and the fifth is off, then collinear triples are C(4,3)=4. Total triangles would be C(5,3)-4=10-4=6. Let's verify: the point off the line can form triangles with any two points on the line. There are C(4,2)=6 such pairs, each forming a triangle with the off-line point. Additionally, the four points on the line have no other triangles. So total triangles are 6, which matches. So that seems correct.But wait, what if we have more than one line with multiple points? For example, two lines each with two points. But with n=4, if you have two lines each with two points, that's two pairs. But then each line only has two points, so there's no collinear triples. Therefore, all C(4,3)=4 triples would form triangles, which is worse (i.e., more triangles) than the case where we have three collinear points. So indeed, putting as many points as possible on a single line seems better for minimizing the number of triangles.But maybe there's a case where having two lines with more points can create more collinear triples? Let's think for n=5. Suppose we have two lines: one with three points and another with two points. Then, the collinear triples would be C(3,3)=1 on the first line, and no collinear triples on the second line. So total collinear triples=1, so triangles would be C(5,3)-1=10-1=9. But if we have four points on a line and one off, we have 4 collinear triples, so triangles=6. That's better. So even with two lines, you don't get as many collinear triples as putting n-1 points on a line.Therefore, the strategy of putting as many points as possible on a single line (n-1 points) and one point off seems optimal. Then the number of triangles is C(n,3) - C(n-1,3). Let me compute that.C(n,3) is n(n-1)(n-2)/6. C(n-1,3) is (n-1)(n-2)(n-3)/6. So subtracting them:n(n-1)(n-2)/6 - (n-1)(n-2)(n-3)/6 = (n-1)(n-2)/6 [n - (n-3)] = (n-1)(n-2)/6 [3] = (n-1)(n-2)/2.Wait, that simplifies to (n-1)(n-2)/2. Let me check with n=4: (4-1)(4-2)/2=3*2/2=3, which matches. For n=5: (5-1)(5-2)/2=4*3/2=6, which also matches. So the formula gives us (n-1)(n-2)/2 triangles. But is this the minimal number?Wait, let me think again. Suppose for n=6. If we put 5 points on a line and 1 off, then the number of collinear triples is C(5,3)=10. Total triangles would be C(6,3)-10=20-10=10. Alternatively, if we arrange the points into two lines with three points each. Then each line would have C(3,3)=1 collinear triple, so total collinear triples=2. Therefore, total triangles=20-2=18, which is more. So still worse. Alternatively, maybe three lines with two points each, but then there are no collinear triples, so all 20 triangles. So definitely, putting as many points as possible on a single line gives the minimal number of triangles.But wait, maybe another configuration? For example, three lines with two points each and one line with three points. Wait, but with n=6, if we have three points on one line and the other three points arranged such that each is on a different line, but then that might not be possible. Wait, actually, arranging points on multiple lines can sometimes create more collinear triples? Let's see.Suppose for n=6, we have two lines each with three points. Then each line has C(3,3)=1 collinear triple. So total collinear triples=2. Therefore, triangles=20-2=18. But if we have five points on one line and one off, we have 10 collinear triples, so triangles=10. So definitely, the single line with n-1 points is better. Alternatively, what if we have three points on one line and three points on another line, but the two lines intersect at a point, so one point is shared between the two lines. Then, total points would be 3 + 3 -1 =5, which is less than 6. So to have six points, maybe three lines each with three points, intersecting at a common point? But three lines each with three points would have the common intersection point plus two other points on each line. So total points would be 1 + 2*3=7 points. So that's more than 6. So that's not possible. Maybe two lines intersecting, each with three points. The intersection point is shared, so total points would be 3 + 3 -1=5. Still not enough. So perhaps, not possible for n=6. So arranging in multiple lines doesn't help in increasing collinear triples beyond what's possible with a single line.Therefore, the initial approach seems valid: putting n-1 points on a line and one point off the line. Then the number of triangles is (n-1)(n-2)/2. But wait, let me confirm for another n. Let's take n=3. The problem states n≥3. If n=3, the points are not all collinear. Wait, but with three points, if they are not all collinear, they form a triangle. So the number of triangles is 1. Using our formula: (3-1)(3-2)/2=2*1/2=1. That works. So the formula holds.But wait, is there a case where arranging points on multiple lines could produce more collinear triples? For example, n=6. Suppose we have three lines, each with three points, but overlapping? Wait, but arranging three lines each with three points would require more than six points unless there's overlap. For instance, if three lines all pass through a common point, and each line has two other points. Then total points would be 1 + 2*3=7. Still too many. Alternatively, if two lines have three points each, intersecting at a common point. Then total points are 3 + 3 -1=5. So not enough for n=6. Hmm. So seems like it's not possible to arrange six points on two lines with three points each without overlapping. Unless the lines are parallel. If two parallel lines each have three points. Then total points are six, and each line has three points. Then, each line has one collinear triple, so total collinear triples=2. So triangles=20-2=18. But if you have five points on one line and one off, you get 10 triangles. So 10 is still less. So even with two lines, you can't beat the single line configuration.Alternatively, maybe three lines with two points each. Then, no collinear triples, so all C(6,3)=20 triangles. Worse. So the conclusion is that the maximum number of collinear triples is achieved when we have n-1 points on a line and one point off. Therefore, the minimal number of triangles is (n-1)(n-2)/2.Wait, but let me check if there's another configuration that allows more collinear triples. For example, maybe two lines, each with k points, such that C(k,3) + C(m,3) is more than C(n-1,3), where k + m = n + l, where l is the number of intersection points. Hmm, this seems complicated.Suppose for n=6, can we have two lines with three points each and one line with two points? Wait, but how would that work? If two lines have three points each, intersecting at a point, so total points are 3 + 3 -1 =5. Then adding another line with two points, but those points could be on existing lines or new. If we add a new point off both lines, then total points become 6. Then, how many collinear triples do we have? The two original lines each have three points, so each contributes one collinear triple. The new line with two points doesn't contribute. So total collinear triples=2. So triangles=20-2=18. But if we have five points on a line and one off, collinear triples=10, so triangles=10. So 10 is still better.Alternatively, suppose we have three lines with two points each. Then no collinear triples. All 20 triangles. Worse. Alternatively, three lines each with three points. But as before, that would require more than six points. So not possible.Therefore, for n=6, the maximum number of collinear triples is C(5,3)=10, so triangles=10. So the formula seems to hold.Wait, but let's think for n=7. If we put six points on a line and one off, collinear triples=C(6,3)=20. Total triangles=C(7,3)-20=35-20=15. Alternatively, if we arrange into two lines, say four points on one line and three on another, intersecting at a point. Then total points=4+3-1=6. Not enough. To get seven points, maybe two lines with four points each, intersecting at a point. Total points=4+4-1=7. Then collinear triples on each line: C(4,3)=4, so total collinear triples=8. Therefore, total triangles=C(7,3)-8=35-8=27, which is worse than 15. Alternatively, three lines with three points each. But three lines with three points each would require, if all intersect at a common point, 1 + 2*3=7 points. Then each line has three points, contributing one collinear triple each. So total collinear triples=3. Therefore, triangles=35-3=32. Still worse. So again, the single line configuration gives fewer triangles.Therefore, it seems that arranging the points with as many as possible on a single line (n-1 points) gives the maximum number of collinear triples, hence minimal number of triangles. So the minimal number of triangles is (n-1)(n-2)/2.But wait, wait. Let me verify this with another example. Let's take n=5. If we have four points on a line and one off, collinear triples=C(4,3)=4. Total triangles=10-4=6. Alternatively, if we have three points on one line and two on another, intersecting at a point. Total points=3+2-1=4. Not enough. To get five points, maybe three points on one line and three on another, intersecting at a point. Then total points=3+3-1=5. Each line has three points, so collinear triples=1 each. Total collinear triples=2. Therefore, triangles=10-2=8, which is more than 6. So again, single line is better.So, in all these cases, arranging n-1 points on a line and one off gives the minimal number of triangles. Therefore, the minimal number is (n-1)(n-2)/2. But wait, the problem states "n points which are not all collinear." So, when n=3, we have three points not all collinear, so they form a triangle. The minimal number is 1. Which matches (3-1)(3-2)/2=1. For n=4, it's 3. For n=5, it's 6. For n=6, 10. So this seems to be forming a sequence of triangular numbers: 1, 3, 6, 10,... which are C(n-1,2). Wait, indeed, (n-1)(n-2)/2 is equal to C(n-1,2). So the formula can also be written as C(n-1,2).But wait, why is that? Let me think. If you have n-1 points on a line and one off, the number of triangles is the number of ways to choose two points on the line and one off. Which is C(n-1,2)*1= C(n-1,2). That's exactly the number of triangles. Because each triangle must include the off-line point and two points from the line. Since the off-line point can't form a collinear triple with any two points on the line. So yes, that makes sense.Therefore, the minimal number of triangles is indeed C(n-1,2) = (n-1)(n-2)/2. But wait, let me confirm with another configuration. Suppose we have n=5, with two lines: one line with three points and another line with three points intersecting at a point. So total points=3+3-1=5. Then, how many triangles do we have? The collinear triples: each line has C(3,3)=1, so total collinear triples=2. Total triangles=C(5,3)-2=10-2=8. But the minimal is supposed to be 6. So this configuration gives more triangles, so not minimal. Therefore, the minimal is indeed when you have n-1 points on a line and one off.Therefore, the answer should be (n-1)(n-2)/2. Which is also equivalent to C(n-1,2). Let me check for n=3: (3-1)(3-2)/2=1. Correct. n=4: 3*2/2=3. Correct. So yes, seems consistent.But wait, hold on. What if we have more than one off-line point? For example, n=5: four points on a line and one off. Triangles=6. Alternatively, three points on a line and two off. How many triangles would that be? Let's calculate.Total triangles: C(5,3) - C(3,3) - C(2,3) - ... Wait, no. If there are three points on a line, collinear triples=1. The two off-line points. Each pair of off-line points with any two on the line. Wait, but if two points are off the line, they can form triangles with the three on the line. Let's compute.Total collinear triples: 1 (from the three on the line). Then, the two off-line points: any three points including two off-line and one on would not be collinear, since the two off-line points are not on the line. Similarly, three points consisting of one off-line and two on the line: those are not collinear. So the only collinear triple is the three on the line. Therefore, total triangles=C(5,3)-1=10-1=9. But if we have four on a line and one off, triangles=6. So 6 is better. Therefore, even with two off-line points, you get more triangles. Therefore, the minimal is achieved when you have as few off-line points as possible, i.e., one.Therefore, the minimal number of triangles is achieved when we have n-1 points on a line and one point off, leading to (n-1)(n-2)/2 triangles.Wait, but another thought. Suppose we have two lines with multiple points. For example, in the case of n=6, maybe three lines each with three points, arranged such that each pair of lines intersects at a unique point, forming a triangle configuration. But I think this would require more points. Let's see: three lines each intersecting the other two at distinct points. Then each line has two intersection points and one other point. But that would lead to three lines with three points each, and total points being 3*2=6? Wait, no. Let me visualize. If three lines form a triangle, each side of the triangle is a line. Then, each line (side) has two points (the vertices). But if we need three points on each line, we need to add a point on each side. So, each side would have three points: two vertices and one midpoint. Then total points: each side has three points, but the vertices are shared between two sides. So total points: 3 vertices + 3 midpoints =6 points. So, three lines each with three points, forming a triangle with midpoints. Then, how many collinear triples do we have? Each side has three points, so three collinear triples. Therefore, total collinear triples=3. So total triangles=C(6,3)-3=20-3=17. But if we have five points on a line and one off, collinear triples=10, triangles=10. So 10 is better. So even in this configuration, the number of triangles is higher. Therefore, still, the single line with n-1 points and one off is better.Therefore, after considering various configurations, it seems that the minimal number of triangles is indeed (n-1)(n-2)/2. So the answer should be boxed{dfrac{(n-1)(n-2)}{2}}.Final AnswerThe least possible number of triangles is boxed{dfrac{(n-1)(n-2)}{2}}.</think>Given ( n ge 3 ) points in a plane, not all collinear, we need to find the least possible number of triangles formed by these points. To minimize the number of triangles, we need to maximize the number of collinear triples. The optimal configuration is to place ( n-1 ) points on a single line and the remaining point off this line. This configuration maximizes the number of collinear triples, which are formed by choosing any three points from the ( n-1 ) collinear points.The total number of triangles formed by ( n ) points is given by the combination formula ( binom{n}{3} ). The number of collinear triples is ( binom{n-1}{3} ). Subtracting the collinear triples from the total gives the number of triangles.The formula for the minimum number of triangles is:[binom{n}{3} - binom{n-1}{3}]Calculating this:[binom{n}{3} - binom{n-1}{3} = frac{n(n-1)(n-2)}{6} - frac{(n-1)(n-2)(n-3)}{6}][= frac{(n-1)(n-2)}{6} left[ n - (n-3) right]][= frac{(n-1)(n-2)}{6} cdot 3][= frac{(n-1)(n-2)}{2}]Thus, the least possible number of triangles is (boxed{dfrac{(n-1)(n-2)}{2}}).,<think>Alright, let's tackle this problem. So, we have n points in a plane, not all collinear, and we need to find the least possible number of triangles that can be formed by these points. The question is asking for the minimum number of triangles, right? So, basically, we want to arrange these n points in such a way that as few triangles as possible are formed. Hmm, interesting.First, let's recall that a triangle is formed by three non-collinear points. Since the points are not all collinear, there must be at least one triangle. But we need to find the minimum number. So, maybe if we arrange as many points as possible on a single line, then the remaining points off that line would form triangles with pairs of points from the line. But how does that work exactly?Let me think. Suppose we have n points, and we put k points on a straight line, and the remaining (n - k) points off that line. Then, any triangle would require either three points not on the line or two points from the line and one point off the line. If all the off-line points are themselves collinear, then they might not form triangles among themselves. Wait, but if there are at least three off-line points and they're collinear, then they still form a line, so triangles would have to involve points from both lines. But maybe we can have the off-line points arranged in a way that they don't form another line.But hold on, the problem states that the points are not all collinear, so there's at least one point not on the main line. But if we have multiple points off the main line, but they are arranged such that they don't form another line, then triangles can be minimized. Wait, actually, if we have multiple points off the main line but not collinear, then those points can form triangles with each other. So perhaps to minimize triangles, we need to minimize the number of points off the main line. Because each point off the main line can form a triangle with each pair of points on the main line. So, if we have m points off the main line, then each of those m points combined with any two points on the main line will form a triangle. So, the number of triangles would be m multiplied by the number of pairs on the main line, which is C(k, 2) = k(k - 1)/2. Plus, if there are points off the main line, if they are more than two, then they might form triangles among themselves. Wait, so if there are m points off the main line, then the number of triangles among themselves is C(m, 3), but only if those m points are not all collinear. Since the problem states that not all points are collinear, but we can have some collinear.Wait, but if we arrange all the m points off the main line on another straight line, then those m points would be collinear, so they wouldn't form triangles among themselves. So, if we have two lines: one with k points and another with m points, where k + m = n, then the total number of triangles would be m * C(k, 2) + k * C(m, 2). Because each point on the first line can pair with two points on the second line to form a triangle, and vice versa. Wait, no. If the m points are on another line, then triangles can be formed by taking two points from the first line and one from the second, or two from the second line and one from the first. So, the total number would be C(k, 2)*m + C(m, 2)*k. But if the m points are not on a line, then there might be additional triangles among the m points. So, to minimize the number of triangles, we need to arrange the m points on another line. Therefore, the minimal number of triangles would be when we split the points into two collinear sets, one of size k and the other of size m = n - k. Then, the number of triangles is C(k, 2)*m + C(m, 2)*k.So, the problem reduces to choosing k such that this expression is minimized. Let's formalize this:Total triangles T = (k choose 2)*(n - k) + (n - k choose 2)*kSimplify this expression:First term: [k(k - 1)/2]*(n - k)Second term: [(n - k)(n - k - 1)/2]*kSo, T = [k(k - 1)(n - k) + (n - k)(n - k - 1)k]/2Factor out k(n - k):T = [k(n - k){(k - 1) + (n - k - 1)}]/2Simplify inside the braces:(k - 1) + (n - k - 1) = n - 2Therefore, T = [k(n - k)(n - 2)]/2So, T = (n - 2)/2 * k(n - k)Therefore, the expression simplifies to T = (n - 2)/2 * k(n - k)Now, we need to minimize T with respect to k, where k is an integer between 1 and n - 1 (since we need at least one point on each line to have two lines).Wait, but actually, the original problem states that the points are not all collinear. So, we need at least one point not on the main line, so m = n - k ≥ 1, hence k ≤ n - 1, and k ≥ 2? Wait, no. If we have k points on the main line, then the other m = n - k points must be on another line. But if m = 1, then that one point is not on the main line, so it's okay. So, k can be from 1 to n - 1, but actually, if k = 1, then the main line has only one point, which doesn't make sense because a line needs at least two points. Wait, actually, in our setup, the two lines each need to have at least two points? No, maybe not. If one line has k points (k ≥ 2), and the other line has m = n - k points (m ≥ 1), but even if m = 1, then we can still form triangles by taking two points from the main line and the one point from the other line. However, if m = 1, then the second line is just a single point, so it's not a line, but we can arrange that single point anywhere not on the main line.Wait, perhaps the initial assumption that arranging the off-line points on another line is not necessary. If we have m points not on the main line, we can place them anywhere else, not necessarily on another line. But if we place them on another line, then we prevent them from forming triangles among themselves. So, to minimize the number of triangles, it's better to place the off-line points on another line, so that they don't form triangles among themselves.But if we have m points on a second line, then in order for those m points to not form triangles among themselves, that second line must be straight. So, yes, arranging them on a line would prevent them from forming triangles. Therefore, the minimal number of triangles is achieved when we split the points into two lines, with k and m points, and then the total number of triangles is C(k,2)*m + C(m,2)*k, which simplifies to (n - 2)/2 *k(n -k) as we derived earlier.Therefore, we need to find integer k between 1 and n - 1 that minimizes the expression k(n -k). Wait, but since T = (n - 2)/2 * k(n -k), and (n - 2)/2 is a constant for given n, we can just minimize k(n -k).So, k(n -k) is a quadratic function in k, which is maximized when k = n/2, but since we want to minimize it, it would be minimized at the endpoints. Wait, no. The quadratic k(n -k) is a downward-opening parabola, so it has maximum at k = n/2 and minima at the endpoints k = 1 and k = n -1. Therefore, the minimal value is 1*(n -1) = n -1. But wait, if k(n -k) is minimized at k=1 or k =n -1, then the minimal value is n -1. But that seems to suggest T = (n -2)/2*(n -1). But let's check with k=1:If k=1, then m =n -1. Then T = (1*0/2)*(n -1) + [(n -1)(n -2)/2]*1 = 0 + (n -1)(n -2)/2 = (n -1)(n -2)/2.But if k=2, m =n -2, then T = (2*1/2)*(n -2) + [(n -2)(n -3)/2]*2 = (1)*(n -2) + (n -2)(n -3) = (n -2)[1 + (n -3)] = (n -2)(n -2) = (n -2)^2.Wait, so when k=1, T=(n -1)(n -2)/2. When k=2, T=(n -2)^2. Let's compare these two:(n -1)(n -2)/2 vs. (n -2)^2Which is smaller? Let's subtract them:(n -2)^2 - (n -1)(n -2)/2 = (n -2)[ (n -2) - (n -1)/2 ] = (n -2)[ (2(n -2) - (n -1)) /2 ] = (n -2)[ (2n -4 -n +1)/2 ] = (n -2)(n -3)/2Which is positive for n >=4. So, (n -2)^2 is larger than (n -1)(n -2)/2 when n >=4. Therefore, for n >=4, T is smaller when k=1 than when k=2. Hmm, that's interesting.Wait, but when k=1, that would mean one point on the main line and n -1 points on another line. But a line with one point? That's not a line. Wait, actually, in our initial model, we had two lines: one with k points and another with m =n -k points. But if k=1, then the main line has only one point, which is not a line. So, maybe our initial assumption is wrong.Wait, actually, if we split the points into two lines, each line must have at least two points. Because otherwise, if one line has only one point, then that point is just a single point, not a line. So, perhaps the minimal k is 2. Therefore, k can range from 2 to n -2. So, in that case, the minimal value of k(n -k) would be when k=2 or k=n -2. Let's check that.If k=2, then m =n -2. Then T = (2*1/2)*(n -2) + [(n -2)(n -3)/2]*2 = (1)*(n -2) + (n -2)(n -3) = (n -2)[1 + (n -3)] = (n -2)(n -2) = (n -2)^2.Similarly, if k =n -2, then m=2, and T = [ (n -2)(n -3)/2 ]*2 + [2*1/2]*(n -2) = (n -2)(n -3) + 1*(n -2) = (n -2)(n -3 +1) = (n -2)(n -2) = same as above.If we take k=3, then m =n -3. Then T = (3*2/2)*(n -3) + [ (n -3)(n -4)/2 ]*3 = 3(n -3) + 3(n -3)(n -4)/2 = 3(n -3)[1 + (n -4)/2] = 3(n -3)(n -2)/2.Compare that to T when k=2: (n -2)^2.So, 3(n -3)(n -2)/2 vs. (n -2)^2.Which is smaller?Let's compute for n >=5:Take n=5:k=2: (5 -2)^2 = 9k=3: 3*(5 -3)*(5 -2)/2 = 3*2*3/2 = 9So, equal.n=6:k=2: (6 -2)^2 =16k=3: 3*(6 -3)*(6 -2)/2 =3*3*4/2=18Which is larger. So, T=18 vs. 16. So, T is larger for k=3.Similarly, for k=4:But n=6, k=4 would be same as k=2 (since m=2). So, symmetric. So, it's the same as k=2. So, for even n, the minimal is achieved at k=2 and k=n-2, but for odd n, maybe the middle k gives higher T.Wait, but in n=5, k=2 and k=3 give same T. So, maybe for odd n, when k=(n-1)/2, it gives same as k=(n+1)/2. But in terms of minimal, maybe k=2 is still better for even and odd.Wait, let's check n=4:For n=4, k=2, m=2.T= (2*1/2)*2 + (2*1/2)*2= 2 + 2=4Alternatively, k=1, but as before, k=1 would lead to m=3, but k=1 is invalid as a line.Wait, actually for n=4, the minimal T. If we split into two lines of 2 points each:Triangles formed by two points on one line and one on the other: 2*C(2,2)*2 = 2*1*2=4. So total triangles=4.Alternatively, if we put 3 points on a line and 1 off. Then, triangles would be C(3,2)*1 + C(1,2)*3= 3*1 + 0=3. But wait, but in this case, the single point off the line is not on a line, so the other line is just a single point? Wait, but if we have three points on a line and one point off, then we can't form another line with the single point. So, in that case, the triangles are C(3,2)*1 =3. So, that's less than 4. But then, why did our earlier formula give T=(n -2)/2 *k(n -k). For k=3, n=4:T=(4 -2)/2 *3*(4 -3)=2/2 *3*1=3. So, that's correct. But when we split into two lines of 2 each, the formula gives T=(4 -2)/2 *2*(4 -2)=2/2 *2*2=4. So, which one is better? The case with three points on a line and one off gives fewer triangles (3 triangles vs. 4). But according to the problem statement, the points are not all collinear. So, in the case of three on a line and one off, that's acceptable. So, why does the two lines of two each give more triangles? Because in that case, each line has two points, so the triangles formed are two from one line and one from the other. For two lines of two each: each line contributes C(2,2)*2=1*2=2 triangles, so total 4. But the three-on-a-line and one-off gives only 3 triangles. So, actually, splitting into a line of three and a single point gives fewer triangles. Therefore, our initial assumption that splitting into two lines each with at least two points was necessary might be wrong.Wait, maybe the formula T = (n -2)/2 *k(n -k) holds regardless of whether the m points are on a line or not. Wait, no. The formula was derived under the assumption that the m points are on a line. If the m points are not on a line, then they might form triangles among themselves, which would add C(m,3) triangles. Therefore, to minimize the number of triangles, it's better to have the m points on a line. So, if we have k points on a line and m =n -k points on another line, then the number of triangles is C(k,2)*m + C(m,2)*k. However, if the m points are not on a line, then the number of triangles would be higher. Therefore, to minimize, we need to have the m points on a line.But when m=1, you can't have a line with one point, so in that case, the single point is not on a line. Therefore, when m=1, the number of triangles is C(k,2)*1 + C(1,2)*k = C(k,2)*1 + 0 = C(k,2). Which is k(k -1)/2. So, for example, if n=4, k=3, m=1: triangles=3*2/2=3. Which is the case we had earlier.Similarly, if n=5, k=4, m=1: triangles=4*3/2=6. But if we split into two lines, say k=3 and m=2: triangles= C(3,2)*2 + C(2,2)*3= 3*2 +1*3=6 +3=9. So, which is worse. So, having a single point off the line gives fewer triangles. Hence, the minimal number is achieved when we have as many points as possible on a single line, and the remaining points off the line. But the remaining points can't form a line themselves (unless they are only one point). So, the minimal number of triangles would be when we have k points on a line and m =n -k points off the line, with m as small as possible. But m must be at least 1. So, the minimal m is 1, leading to k =n -1. Wait, but if we have k =n -1 points on a line and 1 point off, then the number of triangles is C(n -1, 2)*1 + C(1, 2)*(n -1)= [ (n -1)(n -2)/2 ]*1 + 0 = (n -1)(n -2)/2.Alternatively, if we split the points into two lines with k and m points, where m >=2, then the number of triangles is C(k,2)*m + C(m,2)*k. For example, if k=2 and m=2, then triangles=1*2 +1*2=4. For n=4, this is higher than 3. For n=5, k=3, m=2: triangles=3*2 +1*3=6 +3=9, which is higher than if we have k=4, m=1: triangles=6. So, indeed, arranging the points as a single line plus one point off gives fewer triangles.Therefore, perhaps the minimal number of triangles is achieved when we have n -1 points on a line and 1 point off the line. Then, the number of triangles is C(n -1, 2)*1 = (n -1)(n -2)/2.But wait, let's check for n=4: (4 -1)(4 -2)/2=3*2/2=3, which matches. For n=5: 4*3/2=6. If instead, we have two lines with k=2 and m=3. Then, triangles= C(2,2)*3 + C(3,2)*2=1*3 +3*2=3 +6=9, which is higher. So, yes, the minimal is achieved when we have n -1 points on a line and 1 point off.But wait, what if n=6. Then, according to this, the minimal number of triangles would be (6 -1)(6 -2)/2=5*4/2=10. But if we arrange 4 points on a line and 2 points on another line. Then, the number of triangles would be C(4,2)*2 + C(2,2)*4=6*2 +1*4=12 +4=16, which is more than 10. Alternatively, if we arrange 5 points on a line and 1 off, triangles= C(5,2)*1=10, same as above. Alternatively, if we have 3 points on a line and 3 points on another line, triangles= C(3,2)*3 + C(3,2)*3=3*3 +3*3=9 +9=18, which is even higher. So, yes, the minimal is 10 in that case.But wait, is there a way to get fewer than (n -1)(n -2)/2 triangles?Wait, let's think. Suppose we have n -2 points on a line, and 2 points off the line. Then, the number of triangles would be C(n -2, 2)*2 + C(2, 2)*(n -2) = [ (n -2)(n -3)/2 ]*2 +1*(n -2) = (n -2)(n -3) + (n -2) = (n -2)(n -3 +1) = (n -2)(n -2) = (n -2)^2.Compare this to the case with n -1 points on a line and 1 off: triangles=(n -1)(n -2)/2. For n >=5, (n -1)(n -2)/2 vs. (n -2)^2.Which is smaller?Take n=5: (4)(3)/2=6 vs. (3)^2=9. So, 6 <9.n=6: 5*4/2=10 vs. 16: 10 <16.So, indeed, (n -1)(n -2)/2 is smaller than (n -2)^2. Therefore, having one point off the line is better.Alternatively, what if we have n -3 points on a line and 3 points off. Then, triangles would be C(n -3,2)*3 + C(3,2)*(n -3). So, [ (n -3)(n -4)/2 ]*3 + 3*(n -3)/2*2. Wait, compute:C(n -3,2)*3 = [ (n -3)(n -4)/2 ]*3C(3,2)*(n -3) = 3*(n -3)Total triangles = 3(n -3)(n -4)/2 + 3(n -3) = 3(n -3)[ (n -4)/2 +1 ] = 3(n -3)(n -2)/2Compare to (n -1)(n -2)/2.Which is smaller? 3(n -3)(n -2)/2 vs. (n -1)(n -2)/2.Divide both by (n -2)/2: 3(n -3) vs. (n -1)So, 3(n -3) vs. (n -1)For n=6: 3(3)=9 vs.5: 9>5n=7: 3*4=12 vs.6:12>6So, 3(n -3) is larger than (n -1) for n >=6. Therefore, triangles would be larger in this case. Hence, the minimal is still achieved when we have n -1 points on a line and 1 off.Wait, so in general, the minimal number of triangles is (n -1)(n -2)/2. But let's check n=3. For n=3, since they are not all collinear, they must form a triangle. So, number of triangles is 1. According to the formula: (3 -1)(3 -2)/2=2*1/2=1. Correct.n=4: (4 -1)(4 -2)/2=3*2/2=3. Which is correct, as we saw.Therefore, perhaps the minimal number is (n -1)(n -2)/2.But wait, let's check another configuration. Suppose for n=4, instead of 3 points on a line and 1 off, we have two lines with two points each. Then, we get 4 triangles. But 3 is less than 4, so the previous configuration is better.Alternatively, suppose we have 2 points on a line, and 2 points off forming another line. Then, triangles formed by two from each line. But as we saw, that gives 4 triangles. Which is more than the 3 from the 3-and-1 split. Therefore, the 3-and-1 split is better.Therefore, the minimal number is indeed (n -1)(n -2)/2.Wait, but let's think again. Is there a configuration where we can have even fewer triangles? For example, if we have more than two lines. Let's say, three lines. For example, in n=4: three lines: two lines with two points each and the fourth point alone. Wait, but arranging four points with three lines? Hmm, not sure. Wait, but if we have three lines, each with two points, but that would require six points. For four points, maybe two lines with two points each. But then, the triangles formed would be as before.Alternatively, arrange three points on one line, and the fourth point not on that line. Then, the triangles formed are C(3,2)*1=3. So, same as before. So, regardless of the number of lines, if we have one point off the main line, we get the same number of triangles.Alternatively, maybe arranging the points in some non-linear way to minimize triangles. For example, place multiple points on different lines such that no three points are collinear. But in that case, the number of triangles would be C(n,3), which is maximum, so that's not helpful. Therefore, to minimize triangles, we need to maximize collinearity. But since not all points are collinear, we need at least one point off the main line.Therefore, the minimal number of triangles is achieved when we have n -1 collinear points and 1 point off the line. Therefore, the number of triangles is C(n -1, 2)*1 = (n -1)(n -2)/2.But wait, the problem states "which are not all collinear". So, if we have n -1 points on a line and 1 point off, that satisfies the condition. Then, the number of triangles is indeed (n -1 choose 2)*1 = (n -1)(n -2)/2.But wait, is there a case where arranging the points in two lines with more than one point off the main line could lead to fewer triangles? Let's check for n=5.If we have 4 points on a line and 1 off: triangles= C(4,2)*1=6.If we have 3 points on a line and 2 points on another line: triangles= C(3,2)*2 + C(2,2)*3=3*2 +1*3=6 +3=9.Which is higher. Alternatively, arrange 3 points on a line, 1 point on another line, and 1 point on a third line. Then, triangles would be C(3,2)*2 + C(1,2)*3 + C(1,2)*3= 3*2 +0 +0=6. Same as the first case. Wait, but in this case, the two off-line points are on separate lines. Then, triangles formed by those two points and one from the main line: for each off-line point, C(3,2)*1 triangles. So, two off-line points would give 3*2=6 triangles. So, same total. But actually, each pair from the main line with each off-line point. So, 3 points on the main line, each pair can form a triangle with each off-line point. So, two off-line points: each contributes C(3,2)=3 triangles, so total 6 triangles. So, same as if the two off-line points were on a separate line. But in this case, the two off-line points are not on a line, so they don't form triangles among themselves. If they were on a line, they still wouldn't form a triangle because two points don't form a line. Wait, if two points are on a line, they can't form a triangle by themselves. So, whether the two off-line points are on a line or not, the number of triangles formed among themselves is zero (since you need three points for a triangle). So, arranging two off-line points on a line doesn't prevent any triangles, since there are no triangles to prevent. Therefore, in that case, the number of triangles would still be 6, same as if they were not on a line. However, if we have three off-line points, arranging them on a line would prevent triangles among themselves. But in the case of two off-line points, it's irrelevant.Therefore, for n=5, putting two off-line points on a separate line or not doesn't affect the number of triangles. The total is still 6. Wait, but wait, if two off-line points are on a line, then can we form a triangle with one from the main line and two from the second line? But the second line has only two points. So, to form a triangle, you need three non-collinear points. If two points are on a line, then any three points that include those two and another not on that line would form a triangle. So, if we have two points on a second line, then triangles formed by one from the main line and two from the second line would be C(2,2)*3=3 triangles. But also, triangles formed by two from the main line and one from the second line: C(3,2)*2=3*2=6. So total triangles=3+6=9. Wait, this is different. If the two off-line points are on a separate line, then the number of triangles would be C(3,2)*2 + C(2,2)*3=6 +3=9. But if the two off-line points are not on a line, then the number of triangles is C(3,2)*2=6. Wait, why is that? Because if the two off-line points are not on a line, then you can't form a triangle with two from the second line and one from the main line, since the two off-line points are not collinear. Wait, no. If the two off-line points are not collinear, then any three points consisting of two from the main line and one off-line, or one from the main line and two off-line would form a triangle. Wait, but if two off-line points are not collinear, then taking those two with a third from the main line would form a triangle. Wait, but if the two off-line points are not on a line, then they are just two separate points. So, to form a triangle, you need three points. So, any three points that are not all collinear. If you take two off-line points and one from the main line, since the two off-line points are not collinear with the main line (assuming they are placed off the main line), then those three points form a triangle. So, the number of triangles would be C(3,2)*2 (two from main line and one off-line) + C(2,2)*3 (one from main line and two off-line) = same as if the two off-line points were on a line. Wait, but if the two off-line points are not on a line, then C(2,2)*3 is still 3 triangles, but those triangles are formed by the two off-line points and one from the main line. However, if the two off-line points are not collinear, then each pair of off-line points with a main line point forms a triangle, but since there's only two off-line points, each triangle is formed by those two and a main line point. But there are three main line points, so each of the three main line points can pair with the two off-line points to form a triangle? Wait, no, a triangle requires three points. So, if you have two off-line points, say A and B, and three main line points C, D, E. Then, the triangles would be:1. A, C, D2. A, C, E3. A, D, E4. B, C, D5. B, C, E6. B, D, EPlus:7. A, B, C8. A, B, D9. A, B, ESo, total 9 triangles. Wait, but if the two off-line points are not on a line, then triangles 7-9 are still triangles because three points (A, B, C) are not collinear (since A and B are off the main line, and C is on the main line). So, even if A and B are not on a line, they form triangles with each main line point. Therefore, regardless of whether the two off-line points are collinear or not, the number of triangles is the same: 9. But earlier, we thought it might be 6. So, this contradicts our previous reasoning. Therefore, perhaps our initial formula was wrong.Wait, let's clarify. If we have k points on a line and m points off the line, whether the m points are on another line or not, the number of triangles is:Triangles with two points on the main line and one off: C(k,2)*m.Triangles with one point on the main line and two off: k*C(m,2).But if the m points are not on a line, then C(m,2) is the number of pairs of off-line points, each pair forming a triangle with each main line point. So, regardless of whether the off-line points are collinear or not, C(m,2)*k triangles exist. However, if the m points are on a line, then those pairs are collinear, so they don't form triangles. Wait, no. If two points are on a line, but the third is on another line, then the three points are not collinear. For example, if we have two lines intersecting at a point, and two points on each line. Then, taking two points from one line and one from the other forms a triangle. So, even if the two off-line points are on another line, the triangles formed by two from one line and one from the other are still non-collinear. Therefore, the number of triangles is still C(k,2)*m + C(m,2)*k, regardless of whether the off-line points are on a line or not. Wait, that seems to contradict the previous example where two off-line points not on a line would form triangles among themselves with a main line point.Wait, in the case where the two off-line points are on another line, say line L2, and the main line is L1. Then, any two points from L1 and one from L2 form a triangle, because L1 and L2 are different lines. Similarly, any two points from L2 and one from L1 also form a triangle. So, even if the off-line points are on a line, the triangles are still formed. Therefore, the total number of triangles is indeed C(k,2)*m + C(m,2)*k, regardless of the configuration.But earlier, when we considered two off-line points not on a line, we thought that triangles formed by those two and a main line point would exist, but that would actually still be the case even if they are on a line. Because three points, two on one line and one on another, are not collinear. Therefore, the number of triangles is the same whether the off-line points are on a line or not. Therefore, the formula holds in both cases. Therefore, the minimal number of triangles is achieved when we minimize the value of C(k,2)*m + C(m,2)*k, where k + m =n, and k >=2, m >=1.But earlier, we saw that when m=1, the number of triangles is C(k,2)*1 + C(1,2)*k = C(k,2) =k(k -1)/2. So, if we have k =n -1 and m=1, triangles=(n -1)(n -2)/2. If we have k=2 and m=n -2, triangles= C(2,2)*(n -2) + C(n -2,2)*2=1*(n -2) + [ (n -2)(n -3)/2 ]*2= (n -2) + (n -2)(n -3)= (n -2)(n -2) = (n -2)^2. So, compare these two: (n -1)(n -2)/2 vs. (n -2)^2.Which is smaller? For n >=4:(n -1)(n -2)/2 vs. (n -2)^2.Let's divide both by (n -2):(n -1)/2 vs. (n -2)Multiply both sides by 2:(n -1) vs. 2(n -2)n -1 vs. 2n -4Subtract n -1 from both sides:0 vs. n -3So, for n >3, 2(n -2) >n -1. Therefore, (n -1)(n -2)/2 < (n -2)^2.Therefore, the minimal number of triangles is achieved when m=1 and k=n -1, leading to (n -1)(n -2)/2 triangles.But wait, earlier when n=5, we saw that arranging two lines of two and three points gave more triangles than arranging four and one. But according to the formula, even if the three points are on a line, the number of triangles would still be C(3,2)*2 + C(2,2)*3=3*2 +1*3=9. Whereas arranging four and one gives 6 triangles. Therefore, the minimal is indeed when m=1.Therefore, the minimal number of triangles is (n -1)(n -2)/2.But let's check for n=3: (3 -1)(3 -2)/2=2*1/2=1. Correct.For n=4:3*2/2=3. Correct.For n=5:4*3/2=6. Correct.But wait, is this the absolute minimal? Is there no other configuration where we can get fewer triangles?Suppose we have three lines. For example, for n=5: three lines with two points each and one point left. Wait, but arranging five points on three lines. Let's see:- Two points on line L1.- Two points on line L2.- One point on line L3.Then, the number of triangles would be:- Two points from L1 and one from L2: C(2,2)*2=2.- Two points from L2 and one from L1: C(2,2)*2=2.- Two points from L1 and one from L3: C(2,2)*1=1.- Two points from L2 and one from L3: C(2,2)*1=1.- One point from L1 and two from L3: C(1,2)*2=0.Similarly for others.Total triangles=2 +2 +1 +1=6.Same as the previous minimal. Wait, but the single point on L3 can form triangles with pairs from L1 and L2. So, this configuration also gives six triangles. But maybe there's a different configuration with fewer triangles?Alternatively, three points on a line, and two other points forming another line with one of the three points. So, three points on line L1, and two points on line L2 intersecting L1 at one point. Then, how many triangles would that form?The three points on L1: A, B, C.The two points on L2: A, D, E.Now, triangles can be formed by:- Two from L1 and one from L2 (excluding A): pairs (B,C) with D or E. So, 1*2=2.- Two from L2 and one from L1 (excluding A): pairs (D,E) with B or C. So,1*2=2.- Triangles involving A: for example, A, B, D; A, B, E; A, C, D; A, C, E; and A, D, E. But A, D, E are collinear, so not a triangle. So, triangles involving A are 4.Total triangles=2 +2 +4=8. Which is more than 6.So, in this configuration, we get more triangles. Therefore, the minimal remains 6.Therefore, it seems that the minimal number of triangles is indeed (n -1)(n -2)/2, achieved by placing n -1 points on a line and one point off.Therefore, the answer should be boxed{dfrac{(n-1)(n-2)}{2}}.Final AnswerThe least possible number of triangles is boxed{dfrac{(n-1)(n-2)}{2}}.</think>To find the least possible number of triangles formed by ( n ) points in a plane (not all collinear), we need to consider configurations that minimize the number of triangles. Key steps:1. Maximizing Collinearity: To minimize the number of triangles, we should maximize the number of collinear points. This reduces the number of triangles formed since collinear points do not form triangles.2. Single Line with ( n-1 ) Points: The optimal configuration is to place ( n-1 ) points on a single line and the remaining 1 point off this line. This setup ensures that the only triangles formed are those involving the single off-line point and pairs of points from the main line.3. Calculating Triangles: The number of triangles formed by this configuration is given by the number of ways to choose 2 points from the ( n-1 ) collinear points and combining them with the single off-line point. This is calculated as: [ binom{n-1}{2} = frac{(n-1)(n-2)}{2} ]Thus, the least possible number of triangles is (boxed{dfrac{(n-1)(n-2)}{2}}).]

🔑:1. Understanding the Problem: We need to find the least possible number of triangles that can be formed by ( n ) points in the plane, where not all points are collinear. Recall that a triangle is formed by three non-collinear points.2. Initial Consideration: If all ( n ) points were collinear, no triangles could be formed. However, since not all points are collinear, at least one point must be off the line formed by any subset of collinear points.3. Optimal Configuration: To minimize the number of triangles, we should maximize the number of collinear points. Suppose ( n-1 ) points are collinear, and one point is not on this line. This configuration will help us achieve the minimum number of triangles.4. Counting Triangles: - If ( n-1 ) points are collinear, any triangle must include the one point that is not collinear with the others. - The number of ways to choose 2 points from the ( n-1 ) collinear points is ( binom{n-1}{2} ). - Each pair of collinear points, combined with the one non-collinear point, forms a unique triangle.5. Verification: - The number of triangles formed in this configuration is exactly ( binom{n-1}{2} ). - This is because each triangle must include the non-collinear point, and the other two vertices must be chosen from the ( n-1 ) collinear points.6. General Case: - If we add a new point that is not on any existing line, the number of triangles increases by ( binom{n}{2} ). - If the new point is on a line with ( k ) points, the number of new triangles formed is ( binom{n+1-k}{2} + (k-1)(n+1-k) ).7. Minimizing the Number of Triangles: - The expression ( binom{n+1-k}{2} + (k-1)(n+1-k) ) simplifies to ( frac{(n+1-k)(n+k-2)}{2} ). - The term ( 3k - k^2 ) attains its minimum value when ( k ) is as large as possible, i.e., ( k = n ). - When ( k = n ), the increase in the number of triangles is ( n-1 ), which is less than ( binom{n}{2} ).8. Conclusion: - The least number of triangles is achieved when ( n-1 ) points are collinear, and one point is not collinear with them. - The number of triangles in this configuration is ( binom{n-1}{2} ).[boxed{binom{n-1}{2}}]

🧠:[<think>Okay, so I need to figure out the positional relationship between these two circles. The problem gives me two equations:First circle: ( x^2 + y^2 = -4y )Second circle: ( (x - 1)^2 + y^2 = 1 )The options are intersecting, disjoint, externally tangent, or internally tangent. Let me recall how to analyze circles from their equations.First, I remember that the general form of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where ((h, k)) is the center and ( r ) is the radius. So I need to rewrite both given equations into this standard form.Starting with the first circle: ( x^2 + y^2 = -4y ). Hmm, let me rearrange this. Let me move the -4y to the left side:( x^2 + y^2 + 4y = 0 )Now, to complete the square for the y-terms. The x terms are already squared, so just focus on y. The coefficient of y is 4. Take half of 4, which is 2, and square it, getting 4. So add and subtract 4 on the left side:( x^2 + (y^2 + 4y + 4 - 4) = 0 )Simplify:( x^2 + (y + 2)^2 - 4 = 0 )Move the -4 to the right side:( x^2 + (y + 2)^2 = 4 )So the first circle has center at (0, -2) and radius 2. Okay, that makes sense. Let me verify:Original equation: ( x^2 + y^2 = -4y )If I plug in the center (0, -2), left side is (0 + (-2)^2 = 4), right side is -4*(-2) = 8. Wait, that doesn't match. Wait, that seems wrong. Wait, but in the standard form, the equation is ( x^2 + (y + 2)^2 = 4 ), so center (0, -2) and radius 2. Let me check if that's correct.Let me expand the standard form equation:( x^2 + (y + 2)^2 = 4 )Expanding the (y + 2)^2 gives ( y^2 + 4y + 4 ), so:( x^2 + y^2 + 4y + 4 = 4 )Subtract 4 from both sides:( x^2 + y^2 + 4y = 0 )Which is the original equation. So that is correct. So the center is (0, -2) and radius 2. Wait, but when I plug in the center (0, -2) into the original equation:Left side: (0^2 + (-2)^2 = 4)Right side: -4*(-2) = 8But 4 ≠ 8. Hmm, but that's okay because the center isn't a point on the circle unless the radius is 0, which it's not. The center is just the center, so plugging the center into the original equation wouldn't satisfy it unless the radius is zero. So that's fine.Okay, moving on to the second circle: ( (x - 1)^2 + y^2 = 1 ). This is already in standard form. So the center is at (1, 0) and the radius is 1, since the right side is 1 squared.So now, I have two circles:1. Circle C1: Center (0, -2), radius 22. Circle C2: Center (1, 0), radius 1Now, to find their positional relationship, I need to calculate the distance between their centers and compare it with the sum and difference of their radii.First, calculate the distance between the centers. The centers are (0, -2) and (1, 0). Using the distance formula:Distance ( d = sqrt{(1 - 0)^2 + (0 - (-2))^2} = sqrt{1^2 + 2^2} = sqrt{1 + 4} = sqrt{5} approx 2.236 )Sum of the radii: ( r1 + r2 = 2 + 1 = 3 )Difference of the radii: ( |r1 - r2| = |2 - 1| = 1 )Now, depending on the relationship between d and these two values:- If d > r1 + r2: The circles are disjoint (no intersection, outside each other)- If d = r1 + r2: Externally tangent- If |r1 - r2| < d < r1 + r2: Intersecting- If d = |r1 - r2|: Internally tangent- If d < |r1 - r2|: One circle is inside the other (disjoint, but one inside the other)So let's compute:Our d is sqrt(5) ≈ 2.236Sum of radii: 3, which is greater than d ≈2.236Difference: 1, which is less than d ≈2.236So since 1 < 2.236 < 3, then the circles are intersecting.Wait, but let me confirm. Let's see:sqrt(5) is approximately 2.236, which is less than 3 (sum of radii) and greater than 1 (difference). So according to the criteria, that would mean the circles intersect at two points.Therefore, the answer should be A: Intersecting.But let me double-check my calculations to be sure.First, centers:C1: (0, -2), radius 2C2: (1, 0), radius 1Distance between centers:Δx = 1 - 0 = 1Δy = 0 - (-2) = 2Distance: sqrt(1^2 + 2^2) = sqrt(1 + 4) = sqrt(5) ≈2.236. Correct.Sum of radii: 2 + 1 = 3. So sqrt(5) ≈2.236 < 3. Therefore, the circles are close enough to intersect. Since the distance between centers is less than the sum of radii, they must intersect.Additionally, the difference of radii is 1, which is less than sqrt(5). So yes, since the distance is between the difference and the sum, they intersect at two points.But just to visualize, maybe draw a rough sketch:C1 is centered at (0, -2) with radius 2. So it goes from (0, -2 -2)= (0, -4) to (0, -2 +2)= (0, 0). So it touches the origin (0,0) on the top and goes down to (0, -4).C2 is centered at (1,0) with radius 1. So it goes from (1 -1, 0) = (0,0) to (1 +1, 0) = (2, 0). So it touches the origin (0,0) on the left and goes to (2,0) on the right.So C1 touches (0,0) and C2 also touches (0,0). Wait, so do they both pass through the origin? Let me check:For C1: Plug (0,0) into the equation (x^2 + y^2 = -4y). Left side: 0 + 0 = 0. Right side: -4*0 = 0. So yes, (0,0) is on C1.For C2: Plug (0,0) into ((x - 1)^2 + y^2 = 1). Left side: ( -1)^2 + 0 = 1. Right side: 1. So (0,0) is on C2 as well. Therefore, both circles pass through the origin. Therefore, they intersect at least at (0,0). But if they intersect at another point, then it's two points.Wait, but if they are tangent, they would intersect at one point. But here, they both pass through (0,0). Let me check if there's another intersection point.Alternatively, perhaps solving the equations simultaneously would help.Let me solve the two equations:First equation: ( x^2 + y^2 = -4y )Second equation: ( (x - 1)^2 + y^2 = 1 )Let me subtract the second equation from the first:Left side: ( x^2 + y^2 - [(x - 1)^2 + y^2] = x^2 - (x^2 - 2x + 1) = x^2 - x^2 + 2x -1 = 2x -1 )Right side: ( -4y - 1 )So we have:2x - 1 = -4y - 1Simplify:2x = -4yDivide both sides by 2:x = -2ySo we can substitute x = -2y into one of the equations. Let's substitute into the second equation:( (x - 1)^2 + y^2 = 1 )Substitute x = -2y:( (-2y - 1)^2 + y^2 = 1 )Expand (-2y -1)^2:= ( -2y -1 )^2 = (2y + 1)^2 = 4y^2 +4y +1So equation becomes:4y^2 +4y +1 + y^2 = 1Combine like terms:5y^2 +4y +1 =1Subtract 1 from both sides:5y^2 +4y =0Factor:y(5y +4)=0So solutions:y=0 or 5y +4=0 => y= -4/5So when y=0, x= -2*0=0. So one point is (0,0), as we saw.When y= -4/5, x= -2*(-4/5)=8/5.So the other intersection point is (8/5, -4/5). Let's verify these points in both equations.First, (0,0):First equation: 0 +0 = -4*0 => 0=0. Correct.Second equation: (0-1)^2 +0=1. 1=1. Correct.Second point (8/5, -4/5):First equation: (8/5)^2 + (-4/5)^2 = 64/25 +16/25=80/25=16/5.Right side: -4*(-4/5)=16/5. So 16/5=16/5. Correct.Second equation: (8/5 -1)^2 + (-4/5)^2 = (3/5)^2 + (16/25)=9/25 +16/25=25/25=1. Correct.Therefore, the circles intersect at two points: (0,0) and (8/5, -4/5). Therefore, they are intersecting circles, so the answer is A.But wait, the options given are A: Intersecting, B: Disjoint, C: Externally tangent, D: Internally tangent.Therefore, the correct answer is A.But let me check once more in case I made a mistake in calculation.We found two points of intersection, so definitely intersecting. So answer is A.But just to ensure that when I calculated the distance between centers, which is sqrt(5) ≈2.236, and compared to sum of radii 3, which is greater, so yes, circles can intersect.Alternatively, if the distance were equal to the sum or difference, they would be tangent. Since sqrt(5) is approximately 2.236, which is less than 3, so intersecting. So answer is A.Final Answerboxed{A}</think>To determine the positional relationship between the circles (x^2 + y^2 = -4y) and ((x - 1)^2 + y^2 = 1), we first convert them to standard form.For the first circle (x^2 + y^2 = -4y):- Rearrange and complete the square: (x^2 + (y + 2)^2 = 4)- Center: ((0, -2))- Radius: 2For the second circle ((x - 1)^2 + y^2 = 1):- Already in standard form- Center: ((1, 0))- Radius: 1Next, calculate the distance between the centers:- Distance (d = sqrt{(1 - 0)^2 + (0 - (-2))^2} = sqrt{1 + 4} = sqrt{5} approx 2.236)Compare this distance with the sum and difference of the radii:- Sum of radii: (2 + 1 = 3)- Difference of radii: (|2 - 1| = 1)Since (1 < sqrt{5} < 3), the circles intersect at two points. Solving the equations simultaneously confirms two intersection points: ((0, 0)) and ((frac{8}{5}, -frac{4}{5})).Thus, the positional relationship is intersecting.[boxed{A}],<think>Okay, so I need to figure out the positional relationship between two circles given by their equations. The options are intersecting, disjoint, externally tangent, or internally tangent. Let me start by recalling what each of these terms means.First, intersecting circles mean they cross each other at two distinct points. Disjoint circles don't intersect at all and are separate from each other. Externally tangent means they touch at exactly one point but don't intersect otherwise, and they're separate. Internally tangent is when one circle is inside the other and they touch at exactly one point.Alright, so the given circles are:1. ( x^2 + y^2 = -4y )2. ( (x - 1)^2 + y^2 = 1 )Hmm, the first equation looks a bit strange. Let me rewrite it to standard circle form. The standard form of a circle is ( (x - h)^2 + (y - k)^2 = r^2 ), where (h, k) is the center and r is the radius.Starting with the first circle: ( x^2 + y^2 = -4y ). Let me complete the square for the y-terms. So, moving the -4y to the left side:( x^2 + y^2 + 4y = 0 )Now, to complete the square for the y terms: ( y^2 + 4y ). Take half of 4, which is 2, square it to get 4. So add and subtract 4:( x^2 + (y^2 + 4y + 4 - 4) = 0 )This simplifies to:( x^2 + (y + 2)^2 - 4 = 0 )Then:( x^2 + (y + 2)^2 = 4 )So the first circle has center at (0, -2) and radius 2. Okay, that makes sense.The second circle is given as ( (x - 1)^2 + y^2 = 1 ). That's already in standard form. So its center is at (1, 0) and radius is 1.Now, to determine their positional relationship, I need to find the distance between the centers and compare it to the sum and difference of the radii.First, let's find the distance between the centers. The centers are at (0, -2) and (1, 0). The distance formula is:( d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )Plugging in:( d = sqrt{(1 - 0)^2 + (0 - (-2))^2} = sqrt{1 + 4} = sqrt{5} approx 2.236 )Now, the radii of the circles are 2 and 1. Let's compute the sum and the difference of the radii.Sum: 2 + 1 = 3Difference: |2 - 1| = 1So, if the distance between centers is equal to the sum, they are externally tangent. If it's equal to the difference, they are internally tangent. If the distance is greater than the sum, they are disjoint; if it's less than the sum but greater than the difference, they intersect; if it's less than the difference, then one is entirely inside the other (also disjoint).Here, the distance is sqrt(5) ≈ 2.236. Let's compare this to sum (3) and difference (1):1 < 2.236 < 3So, the distance is between the difference and the sum. Wait, but hold on, actually, the difference is 1, sum is 3. If the distance is less than the difference, that would mean one is inside the other. If the distance is greater than the sum, they are separate. If it's between, they intersect.Wait, actually, no. Let me recall the exact criteria.Let me double-check:If d > r1 + r2: circles are separate (disjoint).If d = r1 + r2: externally tangent.If |r1 - r2| < d < r1 + r2: intersecting.If d = |r1 - r2|: internally tangent.If d < |r1 - r2|: one circle is inside the other (disjoint).So in our case, d ≈ 2.236, r1 + r2 = 3, |r1 - r2| = 1.Since 1 < 2.236 < 3, the circles intersect at two distinct points. Therefore, the answer should be A: Intersecting.But let me verify once more to be sure. Maybe I made a mistake in calculating something.First, centers:First circle: (0, -2), radius 2.Second circle: (1, 0), radius 1.Distance between centers: sqrt((1-0)^2 + (0 - (-2))^2) = sqrt(1 + 4) = sqrt(5) ≈ 2.236.Sum of radii: 3, which is greater than sqrt(5) ≈ 2.236, so sum of radii is larger, so circles must intersect.Wait, but hold on, the radius of the first circle is 2, the second is 1. So the larger circle is centered at (0, -2), radius 2. The smaller circle is centered at (1, 0), radius 1.So the distance between centers is sqrt(5) ≈ 2.236. The sum is 3, which is more than sqrt(5), and the difference is 1, which is less than sqrt(5). Therefore, since the distance is between the difference and the sum, they intersect. So answer is A.Wait, but let me visualize this. The first circle is centered at (0, -2) with radius 2, so it goes from y = -2 - 2 = -4 to y = -2 + 2 = 0. So it touches the origin (0, 0). The second circle is centered at (1, 0) with radius 1, so it goes from x=0 to x=2 on the x-axis, and y from -1 to 1. The first circle touches the origin, and the second circle is centered at (1,0) with radius 1, so the origin (0,0) is on the edge of the second circle. Therefore, the two circles meet at the origin. But is there another point of intersection?Wait, if they both pass through the origin, then that's one point. But maybe they have another point.Wait, let's check. Let me solve the equations simultaneously.First circle equation: ( x^2 + y^2 = -4y )Second circle equation: ( (x - 1)^2 + y^2 = 1 )Let me expand the second equation:( x^2 - 2x + 1 + y^2 = 1 )Simplify:( x^2 + y^2 - 2x = 0 )From the first equation, ( x^2 + y^2 = -4y ). Substitute this into the second equation:( (-4y) - 2x = 0 )Which gives:( -4y - 2x = 0 )Divide both sides by -2:( 2y + x = 0 )So, x = -2y.Now, substitute x = -2y into one of the original equations. Let's use the first one:( (-2y)^2 + y^2 = -4y )Simplify:( 4y^2 + y^2 = -4y )Which is:( 5y^2 = -4y )Bring all terms to one side:( 5y^2 + 4y = 0 )Factor:( y(5y + 4) = 0 )So, y = 0 or y = -4/5.When y = 0, x = -2(0) = 0. So point (0, 0).When y = -4/5, x = -2(-4/5) = 8/5.Therefore, the two circles intersect at (0, 0) and (8/5, -4/5). So two distinct points. Therefore, they intersect, so answer is A.But wait, let me confirm the second point. Let's compute when y = -4/5, x = 8/5. Let's check if this point lies on both circles.First circle: ( x^2 + y^2 = (8/5)^2 + (-4/5)^2 = 64/25 + 16/25 = 80/25 = 16/5 ). On the right side of the first equation: -4y = -4*(-4/5) = 16/5. So yes, equal. So that point is on the first circle.Second circle: ( (x - 1)^2 + y^2 = (8/5 - 5/5)^2 + (-4/5)^2 = (3/5)^2 + (16/25) = 9/25 + 16/25 = 25/25 = 1 ). Which matches the second equation. So yes, that point is also on the second circle. Therefore, two intersection points. So definitely intersecting.But wait, the options are A: Intersecting, B: Disjoint, C: Externally tangent, D: Internally tangent. So answer is A.But just to double-check, maybe I made a mistake in the calculation somewhere. Let me go through again.Distance between centers: sqrt(1^2 + 2^2) = sqrt(5) ≈ 2.236. Sum of radii is 3, which is bigger than sqrt(5). So since the distance is less than the sum, and the difference is 1, and the distance is greater than the difference (since 2.236 > 1), they must intersect. Therefore, the answer is A.Alternatively, maybe the first circle is actually a real circle? Wait, the original equation was x² + y² = -4y. If we rearrange, x² + y² +4y =0. Completing the square gives x² + (y+2)^2 =4. So yes, it's a real circle with center (0,-2) and radius 2. The second circle is (x-1)^2 + y² =1, center (1,0), radius 1. So they are two real circles.Another way to check if they intersect is to calculate the distance between centers and see if it's less than the sum of the radii. Since sqrt(5) ≈2.236 < 3 (sum), so yes, they must intersect. But also, the distance is greater than the difference of radii (1), so they are not tangent, but intersecting at two points.Therefore, the correct answer is A: Intersecting.Final Answerboxed{A}</think>To determine the positional relationship between the circles (x^2 + y^2 = -4y) and ((x - 1)^2 + y^2 = 1), we first convert them to standard form.1. For the first circle (x^2 + y^2 = -4y): - Rewrite as (x^2 + y^2 + 4y = 0). - Complete the square for (y): (x^2 + (y + 2)^2 - 4 = 0). - This gives (x^2 + (y + 2)^2 = 4). The center is ((0, -2)) and the radius is 2.2. The second circle ((x - 1)^2 + y^2 = 1) is already in standard form. The center is ((1, 0)) and the radius is 1.Next, we calculate the distance between the centers:- The centers are ((0, -2)) and ((1, 0)).- Using the distance formula: (d = sqrt{(1 - 0)^2 + (0 - (-2))^2} = sqrt{1 + 4} = sqrt{5} approx 2.236).We compare this distance with the sum and difference of the radii:- Sum of the radii: (2 + 1 = 3)- Difference of the radii: (|2 - 1| = 1)Since (1 < sqrt{5} < 3), the distance between the centers is between the sum and difference of the radii, indicating that the circles intersect at two distinct points.To confirm, solving the equations simultaneously:- Substitute (x = -2y) from (2y + x = 0) into the first circle's equation: - (5y^2 + 4y = 0) - Solutions: (y = 0) and (y = -frac{4}{5}) - Corresponding points: ((0, 0)) and (left(frac{8}{5}, -frac{4}{5}right))Both points satisfy both circle equations, confirming two intersection points. Thus, the circles intersect.[boxed{A}]]

🔑:Solution: The center of the circle x^2+y^2=-4y is C_1(0, -2), and its radius r_1= frac{1}{2} sqrt{16}=2, The center of the circle (x-1)^2+y^2=1 is C_2(1, 0), and its radius r_2=1, Since |C_1C_2|= sqrt{(1-0)^2+(0+2)^2}= sqrt{5}, and 2-1 < sqrt{5} < 2+1, Therefore, the positional relationship between the circle x^2+y^2=-4y and the circle (x-1)^2+y^2=1 is intersecting. Hence, the correct option is: boxed{A}. By calculating the centers and radii of both circles, and observing that the distance between the centers is greater than the absolute value of the difference of the radii and less than the sum of the radii, we can determine the positional relationship between the two circles. This question examines the method of determining the positional relationship between two circles, which is a basic problem. When solving, it is important to carefully read the problem and pay attention to the properties of circles and the reasonable application of the formula for the distance between two points.

🧠:[<think>Okay, so I need to find the sum of 1/(1 + x_k) for k from 1 to 2016, where each x_k is a root of the equation x^2017 = 1 except for x = 1. That is, the 2017th roots of unity excluding 1. Hmm, roots of unity, right? They are complex numbers equally spaced around the unit circle in the complex plane. So, these x_k are the primitive 2017th roots of unity. First, I remember that for roots of unity, there are some nice properties and formulas that might help here. Maybe something related to symmetric sums or using the properties of polynomials. Let me think.The equation x^2017 - 1 = 0 has roots 1, x_1, x_2, ..., x_2016. So, we can write the polynomial as (x - 1)(x - x_1)(x - x_2)...(x - x_2016). But perhaps factoring out (x - 1), we get (x - 1) times the product over (x - x_k) for k from 1 to 2016. So, (x^2017 - 1)/(x - 1) = product_{k=1}^{2016} (x - x_k). That polynomial is the cyclotomic polynomial for 2017, which is a prime, so all roots except 1 are primitive.But how does that help with the sum 1/(1 + x_k)? Maybe I can relate this sum to the coefficients of some polynomial. Alternatively, use complex analysis or residues, but that might be too advanced. Wait, perhaps a substitution?Let me consider the sum S = sum_{k=1}^{2016} 1/(1 + x_k). If I let y_k = x_k, then S = sum_{k=1}^{2016} 1/(1 + y_k). Maybe I can find a clever way to express this sum using the properties of the roots.Alternatively, consider that 1/(1 + x_k) can be rewritten in terms of something related to the roots. Let me manipulate the expression. Note that 1/(1 + x_k) = (1 - x_k)/(1 - x_k^2) if that helps? Wait, not sure.Alternatively, let's consider that for each root x_k, x_k^2017 = 1. So, x_k ≠ -1, since (-1)^2017 = -1 ≠ 1, so none of the x_k are equal to -1. Therefore, 1 + x_k ≠ 0, so each term is defined.Maybe consider the sum S = sum_{k=1}^{2016} 1/(1 + x_k). Let me consider substituting z = x, so the equation is z^2017 = 1. Let’s denote ω = e^(2πi/2017), so that the roots are ω^m for m from 0 to 2016. Then, x_k = ω^k for k from 1 to 2016.Therefore, the sum becomes sum_{m=1}^{2016} 1/(1 + ω^m). Hmm, maybe I can relate this to some known sum. Alternatively, use symmetry. Since the roots are symmetrically distributed, maybe pair terms where ω^m and ω^(2017 - m). Let's see.Note that ω^(2017 - m) = ω^(-m) because ω^2017 = 1. So, for each m, there's a term 1/(1 + ω^m) and 1/(1 + ω^(-m)). Let's add these two terms together: 1/(1 + ω^m) + 1/(1 + ω^(-m)).Let’s compute that:1/(1 + ω^m) + 1/(1 + ω^{-m}) = [1 + ω^{-m} + 1 + ω^m] / [(1 + ω^m)(1 + ω^{-m})]Simplify numerator: 2 + ω^m + ω^{-m}Denominator: (1 + ω^m)(1 + ω^{-m}) = 1 + ω^m + ω^{-m} + 1 = 2 + ω^m + ω^{-m}Therefore, each pair sums to (2 + ω^m + ω^{-m}) / (2 + ω^m + ω^{-m}) ) = 1.So each pair adds up to 1. But wait, how many such pairs are there?Since 2017 is prime and odd, 2016 is even. So, the number of roots excluding 1 is 2016, which is even. Therefore, the roots can be paired as (ω^m, ω^{-m}) for m = 1 to 1008. Each pair contributes 1 to the sum, so total sum S would be 1008. But wait, that seems too straightforward. Let me check.Wait, hold on. If we pair each m from 1 to 2016 with its inverse, but since 2017 is prime, each ω^m is distinct and ω^{-m} is distinct from ω^m except when m = 2017 - m, but since 2017 is odd, there's no m such that m ≡ -m mod 2017, except m=0, which is excluded. So, all roots come in pairs. Therefore, the total number of pairs is 2016/2 = 1008, and each pair sums to 1. Therefore, the total sum S is 1008. Wait, is that correct?But wait, let me check with a smaller case. Let's take n = 3, so equation x^3 = 1, roots 1, ω, ω^2 where ω = e^(2πi/3). Then, the sum would be 1/(1 + ω) + 1/(1 + ω^2). Let's compute that.First, ω = -1/2 + i√3/2, ω^2 = -1/2 - i√3/2. So, 1 + ω = 1 + (-1/2 + i√3/2) = 1/2 + i√3/2. Similarly, 1 + ω^2 = 1/2 - i√3/2. So, 1/(1 + ω) + 1/(1 + ω^2) = [1/(1/2 + i√3/2)] + [1/(1/2 - i√3/2)].Multiply numerator and denominator by the conjugate:First term: (1/2 - i√3/2)/[(1/2)^2 + (√3/2)^2] = (1/2 - i√3/2)/1 = 1/2 - i√3/2Second term: (1/2 + i√3/2)/1 = 1/2 + i√3/2Adding them together: (1/2 - i√3/2) + (1/2 + i√3/2) = 1. So, the sum is 1. But according to our previous logic, n = 3, so number of non-1 roots is 2, which is even, so pairs would be 1, sum is 1. Which matches. So for n=3, the answer is 1. Similarly, for n=5, sum would be 2? Let's test with n=5.But n=5 is more complex. Let me think. If n=5, the sum S would be sum_{k=1}^4 1/(1 + x_k), where x_k are the 5th roots of unity except 1. Let's compute this.Using the same pairing idea, each pair would add 1, so since there are 4 terms, 2 pairs, sum is 2. Alternatively, let's compute directly. Let ω = e^(2πi/5). Then, the sum is 1/(1 + ω) + 1/(1 + ω^2) + 1/(1 + ω^3) + 1/(1 + ω^4). Note that ω^3 = ω^{-2}, ω^4 = ω^{-1}. So, pairing terms:1/(1 + ω) + 1/(1 + ω^{-1}) and 1/(1 + ω^2) + 1/(1 + ω^{-2}).Each pair should sum to 1. Let's check the first pair: 1/(1 + ω) + 1/(1 + ω^{-1}) = [1/(1 + ω) + 1/(1 + ω^{-1})] = [ (1 + ω^{-1} + 1 + ω) / ( (1 + ω)(1 + ω^{-1}) ) ] = [2 + ω + ω^{-1} ] / [2 + ω + ω^{-1}] = 1. Similarly for the second pair. Therefore, total sum is 2. So, yes, the formula seems to hold: for n roots (prime), sum is (n - 1)/2. Wait, n=3: (3 -1)/2=1. n=5: (5 -1)/2=2. So for 2017, which is prime, the sum would be (2017 -1)/2=2016/2=1008. So, answer is 1008. Therefore, boxed{1008}. But wait, let me confirm this with another approach because sometimes these symmetry arguments can be misleading. Let's try using algebraic manipulation.Consider the polynomial f(x) = x^2017 - 1. Its roots are 1, x_1, ..., x_2016. Then, f(x) = (x - 1)(x - x_1)...(x - x_2016). Let’s consider the sum S = sum_{k=1}^{2016} 1/(1 + x_k). To compute this sum, perhaps consider the derivative of the polynomial or logarithmic derivatives.Recall that sum_{k=1}^{n} 1/(a - x_k) = f'(a)/f(a), where f(x) = product_{k=1}^n (x - x_k). Wait, but in our case, we have sum_{k=1}^{2016} 1/(1 + x_k) = sum_{k=1}^{2016} 1/(1 - (-x_k)). So, if we let y_k = -x_k, then the sum becomes sum_{k=1}^{2016} 1/(1 - y_k). If we have a polynomial whose roots are y_k = -x_k, then that polynomial would be product_{k=1}^{2016} (x - y_k) = product_{k=1}^{2016} (x + x_k). Let's find such a polynomial.Given that x_k are roots of x^2017 - 1 = 0, except x=1. So, x^2017 -1 = (x - 1)product_{k=1}^{2016} (x - x_k). Then, if we substitute x with -x, we get (-x)^2017 -1 = (-1)^2017 x^2017 -1 = -x^2017 -1. Therefore, -x^2017 -1 = (-1)(x^2017 +1) = ( -x -1 ) product_{k=1}^{2016} (-x - x_k). Hmm, not sure.Alternatively, the polynomial with roots y_k = -x_k would be product_{k=1}^{2016} (x + x_k). Let me see. If x_k satisfies x^2017 = 1, then y_k = -x_k satisfies y^2017 = (-1)^2017. Since 2017 is odd, this is (-1)^2017 = -1. Therefore, the polynomial satisfied by y_k is y^2017 + 1 = 0. However, note that y = -1 would be a root here, but x_k are roots of x^2017 =1, excluding x=1, so y_k = -x_k would exclude y=-1 only if x_k=1, but x=1 is excluded. Wait, x=1 is excluded from the roots x_k, so y_k = -x_k are roots of y^2017 +1=0, except possibly y=-1. Wait, y=-1 would be a root if x=1 is plugged into y=-x, but x=1 is excluded, so y=-1 is not among the y_k. Wait, but y^2017 +1=0 has roots y = e^(πi(2m +1)/2017) for m from 0 to 2016. However, x_k are roots of x^2017=1 except x=1, so x_k = e^(2πik/2017) for k from 1 to 2016. Therefore, y_k = -x_k = e^(2πik/2017 + πi) = e^(πi(2k/2017 +1)) = e^(πi( (2k + 2017)/2017 )) = e^(πi(2k + 2017)/2017). But 2k + 2017 = 2017 + 2k, so exponents are πi(1 + 2k/2017). These correspond to roots of y^2017 +1=0. Therefore, the polynomial with roots y_k is indeed y^2017 +1, but note that y^2017 +1 factors as product_{k=0}^{2016} (y - e^(πi(2k +1)/2017)). But y_k = -x_k = e^(πi(2k +1)/2017 + πi(2k)/2017 )? Wait, maybe not. Let me check.Wait, x_k = e^(2πik/2017), so y_k = -x_k = e^(2πik/2017 + πi) = e^(πi(2k/2017 +1)) = e^(πi( (2k +2017)/2017 )) = e^(πi( (2k +2017)/2017 ). Since 2017 is a prime, 2k +2017 modulo 4034 (since denominator is 2017, numerator can be considered modulo 4034). But maybe this is overcomplicating.The key point is that the roots y_k are the roots of y^2017 +1=0. Therefore, the polynomial is y^2017 +1, which has roots y_k = e^(πi(2m +1)/2017) for m from 0 to 2016. However, our y_k are specifically y_k = -x_k where x_k are the roots of x^2017=1 except x=1, so y_k are the roots of y^2017 +1=0 except y=-1 (since x=1 would correspond to y=-1, but x=1 is excluded). Therefore, the polynomial with roots y_k is (y^2017 +1)/(y +1). Let me verify that. If we remove y=-1 from y^2017 +1, which is (y +1)(y^2016 - y^2015 + y^2014 - ... - y +1). Therefore, the polynomial is y^2016 - y^2015 + ... - y +1. Therefore, product_{k=1}^{2016} (y - y_k) = y^2016 - y^2015 + y^2014 - ... - y +1.Therefore, if we let Q(y) = y^2016 - y^2015 + ... - y +1. Then, the sum we need is sum_{k=1}^{2016} 1/(1 - y_k), since 1/(1 + x_k) = 1/(1 - y_k) where y_k = -x_k. Therefore, the sum S = sum_{k=1}^{2016} 1/(1 - y_k). Now, there is a formula in complex analysis that says that for a polynomial Q(y) with roots y_1, y_2, ..., y_n, the sum over 1/(a - y_k) is Q'(a)/Q(a). Therefore, in this case, S = Q'(1)/Q(1). Let me compute Q(1) and Q'(1).First, Q(y) = y^2016 - y^2015 + y^2014 - ... - y +1. Let's evaluate Q(1):Q(1) = 1^2016 -1^2015 +1^2014 - ... -1 +1. Since each term alternates sign, starting with +1 (since the first term is y^2016 which is coefficient +1), then -1, +1, ..., -1, +1. Since there are 2017 terms (from y^2016 down to y^0). Wait, 2016 degree polynomial would have 2017 coefficients. Let's check the number of terms. Q(y) is y^2016 - y^2015 + y^2014 - ... - y +1. Let's count the number of terms. The exponents go from 2016 down to 0, so 2017 terms. The coefficients alternate starting with +1, then -1, +1, etc. Therefore, when evaluated at y=1, the terms alternate: +1 -1 +1 -1 ... +1. Since 2017 terms, the last term is +1. Let's see:Number of terms: 2017. Starting with +1, then alternating. For even index terms: if we pair them, each pair is +1 -1 = 0. Since 2017 is odd, there are (2017 -1)/2 = 1008 pairs and one last term. Wait, actually, starting from the first term: term 1: +1, term2: -1, term3:+1,..., term2017:+1. So, the total sum is (number of +1 terms) - (number of -1 terms). How many +1 terms are there? Since the first term is +1, and alternates every term. Since there are 2017 terms, the number of +1 terms is ceiling(2017/2) = 1009, and number of -1 terms is floor(2017/2) = 1008. Therefore, Q(1) = 1009 -1008 =1. So Q(1)=1.Now compute Q'(y). Q'(y) = 2016y^2015 -2015y^2014 +2014y^2013 - ... -1. Then, Q'(1) = 2016*1^2015 -2015*1^2014 +2014*1^2013 - ... -1. So, Q'(1) = 2016 -2015 +2014 -2013 + ... +2 -1. Let's compute this sum.This is an alternating sum starting from 2016 -2015 +2014 -2013 + ... +2 -1. Let's pair the terms:(2016 -2015) + (2014 -2013) + ... + (2 -1). Each pair is 1, and how many pairs are there? Since the total number of terms is 2016 (from 2016 down to 1), so 2016 terms, which is 1008 pairs. Therefore, Q'(1) = 1008*1 = 1008.Therefore, S = Q'(1)/Q(1) = 1008 /1 = 1008. Therefore, the sum is indeed 1008. Therefore, the answer is boxed{1008}. But let me check once again with another method. Suppose I use the formula for sum_{k=1}^{n-1} 1/(1 + x_k) where x_k are the n-th roots of unity except 1. If this formula holds as (n -1)/2, which in the case of prime n, which 2017 is prime, but actually n here is 2017. Wait, in our case, n=2017, and the answer is (2017 -1)/2=1008. So, seems like the general formula is (n -1)/2. Let me check with n=3, which gave us 1, which is (3-1)/2=1. n=5 gives (5-1)/2=2, which also matches. So perhaps in general, for the equation x^n =1, the sum over 1/(1 + x_k) for k=1 to n-1 is (n -1)/2. Is that a general formula?Wait, let's test for n=4. The roots of x^4=1 are 1, i, -1, -i. Excluding 1, the roots are i, -1, -i. Then, sum 1/(1 + i) + 1/(1 + (-1)) +1/(1 + (-i)). Compute each term:1/(1 + i) = (1 - i)/ (1 +1) = (1 -i)/21/(1 -1) is undefined, but wait, x= -1 is a root. Wait, for n=4, x^4=1, roots are 1, -1, i, -i. So excluding 1, the roots are -1, i, -i. So sum is 1/(1 + (-1)) + 1/(1 + i) + 1/(1 + (-i)).But 1/(1 + (-1)) is 1/0, which is undefined. So, in this case, the formula would fail because x=-1 is a root, making denominator zero. Therefore, the formula (n -1)/2 only works when none of the roots x_k is equal to -1, which is when n is odd. Because if n is even, then -1 is an n-th root of unity. So for n even, the sum would include a term 1/(1 + (-1)) which is undefined. Therefore, in our problem, n=2017 is odd, so -1 is not a root, since 2017 is odd, so x^2017 =1 implies x=1, but other roots are not -1. Therefore, for n odd, the sum is (n -1)/2, and for n even, the sum is undefined (or requires a different treatment). Therefore, in our case, n=2017 is odd, so answer is (2017 -1)/2=1008. Therefore, the answer is boxed{1008}.Final Answerboxed{1008}</think>Given the equation (x^{2017} = 1), the roots are the 2017th roots of unity, which are (1, x_1, x_2, ldots, x_{2016}). We need to find the sum (sum_{k=1}^{2016} frac{1}{1 + x_k}).1. Roots of Unity: The roots (x_k) are the primitive 2017th roots of unity, excluding 1. These roots are symmetrically distributed around the unit circle in the complex plane.2. Pairing Roots: For each root (x_k), there exists a corresponding root (x_{2017 - k} = x_k^{-1}). Pairing these roots, we consider the sum (frac{1}{1 + x_k} + frac{1}{1 + x_k^{-1}}).3. Simplifying the Pair: [ frac{1}{1 + x_k} + frac{1}{1 + x_k^{-1}} = frac{1}{1 + x_k} + frac{x_k}{x_k + 1} = frac{1 + x_k}{1 + x_k} = 1 ] Each pair sums to 1.4. Number of Pairs: Since there are 2016 roots (excluding 1), they form 1008 pairs. Each pair contributes 1 to the sum, leading to a total sum of 1008.5. Verification Using Polynomials: Considering the polynomial (Q(y) = y^{2016} - y^{2015} + y^{2014} - cdots - y + 1) with roots (y_k = -x_k), the sum (sum_{k=1}^{2016} frac{1}{1 - y_k}) is calculated using the derivative of (Q(y)) evaluated at 1: [ Q(1) = 1 quad text{and} quad Q'(1) = 2016 - 2015 + 2014 - cdots - 1 = 1008 ] Thus, the sum is (frac{Q'(1)}{Q(1)} = 1008).Therefore, the final answer is (boxed{1008}).,<think>Okay, so I have this problem here: Given that the 2017 roots of the equation ( x^{2017} = 1 ) are 1, ( x_1, x_2, ldots, x_{2016} ). Then I need to find the sum ( sum_{k=1}^{2016} frac{1}{1 + x_k} ). Hmm, okay. Let me try to figure this out step by step.First, I know that the equation ( x^{2017} = 1 ) has 2017 roots, which are the 2017th roots of unity. These are evenly spaced around the unit circle in the complex plane. One of them is 1, and the others are complex numbers. The problem is asking for the sum of ( 1/(1 + x_k) ) where each ( x_k ) is a root except 1.So, I need to compute ( sum_{k=1}^{2016} frac{1}{1 + x_k} ). Let me recall that sometimes when dealing with roots of unity, we can use properties of polynomials or symmetric sums to evaluate such expressions. Maybe I can relate this sum to the roots of a certain polynomial?Let's start by considering the polynomial ( x^{2017} - 1 = 0 ). The roots are 1, ( x_1, x_2, ldots, x_{2016} ). If I factor out ( (x - 1) ), I get ( x^{2017} - 1 = (x - 1)(x^{2016} + x^{2015} + cdots + x + 1) ). So the other roots ( x_1, x_2, ldots, x_{2016} ) are roots of the polynomial ( P(x) = x^{2016} + x^{2015} + cdots + x + 1 ).Now, I need to compute ( sum_{k=1}^{2016} frac{1}{1 + x_k} ). Let me think about how to relate this sum to the roots of ( P(x) ). A common technique is to use the formula for the sum of reciprocals. If I have a polynomial ( Q(x) = prod_{k=1}^n (x - a_k) ), then the sum ( sum_{k=1}^n frac{1}{c - a_k} ) can be related to the derivative of ( Q(x) ) evaluated at ( c ). Specifically, ( sum_{k=1}^n frac{1}{c - a_k} = frac{Q'(c)}{Q(c)} ). But in our case, we have ( frac{1}{1 + x_k} ), which is similar to ( frac{1}{c - a_k} ) where ( c = -1 ) and ( a_k = -x_k ). Wait, maybe I need to adjust the polynomial accordingly.Let me consider substituting ( y = x ), so that ( 1 + x = 1 + y ). But maybe a substitution would help here. Let me try to consider the polynomial whose roots are ( -x_k ). If ( x_k ) are roots of ( P(x) ), then ( -x_k ) would be roots of ( P(-y) = (-y)^{2016} + (-y)^{2015} + cdots + (-y) + 1 ). Let's compute that:( P(-y) = (-y)^{2016} + (-y)^{2015} + cdots + (-y) + 1 )Since the exponents go from 2016 down to 0, the coefficients alternate in sign. Specifically, the coefficient of ( y^{2016} ) is positive, ( y^{2015} ) is negative, and so on, until the constant term is 1. However, this might complicate things. Alternatively, perhaps I can relate the sum ( sum frac{1}{1 + x_k} ) to the derivative of the polynomial ( P(x) ).Recall that for a polynomial ( Q(x) = prod_{k=1}^n (x - a_k) ), the sum ( sum frac{1}{c - a_k} = frac{Q'(c)}{Q(c)} ). So, if I can write a polynomial whose roots are ( a_k = -x_k ), then ( 1/(1 + x_k) = 1/(1 - (-x_k)) = 1/(c - a_k) ) where ( c = 1 ). Therefore, if I let ( Q(y) = prod_{k=1}^{2016} (y - (-x_k)) = prod_{k=1}^{2016} (y + x_k) ), then the sum ( sum_{k=1}^{2016} frac{1}{1 + x_k} = sum_{k=1}^{2016} frac{1}{1 - (-x_k)} = frac{Q'(1)}{Q(1)} ).Therefore, if I can compute ( Q(1) ) and ( Q'(1) ), then the desired sum is ( Q'(1)/Q(1) ). Now, what is ( Q(y) )?Since ( Q(y) = prod_{k=1}^{2016} (y + x_k) ). But note that the original polynomial ( P(x) = prod_{k=1}^{2016} (x - x_k) ). So, substituting ( x = -y ), we get ( P(-y) = prod_{k=1}^{2016} (-y - x_k) = (-1)^{2016} prod_{k=1}^{2016} (y + x_k) = Q(y) ). Since 2016 is even, ( (-1)^{2016} = 1 ). Therefore, ( Q(y) = P(-y) ).Therefore, ( Q(y) = P(-y) = (-y)^{2016} + (-y)^{2015} + cdots + (-y) + 1 ). Let's compute this. Let's write out ( P(-y) ):( P(-y) = (-y)^{2016} + (-y)^{2015} + cdots + (-y) + 1 ).But note that ( (-y)^k = (-1)^k y^k ). So, each term alternates in sign:( P(-y) = y^{2016} - y^{2015} + y^{2014} - cdots - y + 1 ).So, ( Q(y) = y^{2016} - y^{2015} + y^{2014} - cdots - y + 1 ).Therefore, ( Q(1) = 1^{2016} - 1^{2015} + 1^{2014} - cdots - 1 + 1 ). Let's compute this.The polynomial ( Q(1) ) would be:1 - 1 + 1 - 1 + ... -1 +1.Since there are 2017 terms (from degree 0 to 2016). Wait, let's check:Wait, the original polynomial ( P(x) = x^{2016} + x^{2015} + cdots + x + 1 ). Therefore, ( P(-y) ) is of degree 2016, with coefficients alternating signs starting from positive. So, the coefficients are:Term 2016: +1Term 2015: -1Term 2014: +1...Term 1: -1 (since 2015 is odd, so term 1 is -1)Term 0: +1Therefore, when evaluating at y=1, each term becomes +1 -1 +1 -1 + ... +1. How many terms are there?From degree 0 to 2016, that's 2017 terms. Since 2017 is odd, the number of +1 and -1 terms: starting with +1 (for the highest degree term 2016), then -1, +1, etc. So, for 2017 terms:Number of +1 terms: (2017 + 1)/2 = 1009Number of -1 terms: 2017 - 1009 = 1008Therefore, Q(1) = (1009)(1) + (1008)(-1) = 1009 - 1008 = 1.So, Q(1) = 1.Now, we need to compute Q'(y). Let's find the derivative of Q(y):Q(y) = y^{2016} - y^{2015} + y^{2014} - ... - y + 1Then Q'(y) = 2016y^{2015} - 2015y^{2014} + 2014y^{2013} - ... -1 + 0Evaluating Q'(1):Q'(1) = 2016(1)^{2015} - 2015(1)^{2014} + 2014(1)^{2013} - ... -1So, Q'(1) = 2016 - 2015 + 2014 - 2013 + ... -1This is an alternating sum starting from 2016, subtracting 2015, adding 2014, subtracting 2013, and so on, down to subtracting 1.Let me see how this sequence works. Let's pair the terms:(2016 - 2015) + (2014 - 2013) + ... + (2 - 1)Each pair is 1, and how many such pairs are there?From 2016 down to 1, stepping by 2 each time. Since 2016 is even, the total number of terms is 2016, which is 2016 terms. Wait, no. Wait, the sum is 2016 - 2015 + 2014 - 2013 + ... + 2 - 1.Each pair (2016 - 2015), (2014 - 2013), ..., (2 - 1) is 1. The number of such pairs is 2016 / 2 = 1008 pairs. Each pair contributes 1, so the total sum is 1008 * 1 = 1008.Wait, so Q'(1) = 1008.Therefore, the sum ( sum_{k=1}^{2016} frac{1}{1 + x_k} = frac{Q'(1)}{Q(1)} = frac{1008}{1} = 1008 ).But wait, hold on. The problem says "Then ( sum_{k=1}^{2016} frac{1}{1 + x_k} ) = ...". But according to this calculation, the answer is 1008. However, let me verify this because sometimes when dealing with roots of unity, there might be another approach that gives a different result, so I need to confirm.Alternatively, maybe there's another way to compute this sum. Let's consider another approach.Let me recall that for roots of unity, except 1, we can use properties of complex numbers. Let me denote ( omega = e^{2pi i /2017} ), so that all roots can be written as ( omega^k ) for ( k = 0, 1, ..., 2016 ), where ( omega^0 = 1 ), and the other roots are ( omega^1, omega^2, ..., omega^{2016} ).Therefore, we can write the sum as ( sum_{k=1}^{2016} frac{1}{1 + omega^k} ).To compute this sum, perhaps we can use some identity or symmetry.Another idea: For a root ( omega^k neq 1 ), we have ( omega^k ) is a primitive 2017th root of unity. Let me consider that ( sum_{k=0}^{2016} frac{1}{1 + omega^k} = frac{1}{1 + 1} + sum_{k=1}^{2016} frac{1}{1 + omega^k} ). Therefore, the desired sum is ( sum_{k=1}^{2016} frac{1}{1 + omega^k} = sum_{k=0}^{2016} frac{1}{1 + omega^k} - frac{1}{2} ).If I can compute the sum over all roots, then subtract 1/2 to get the desired sum.So, let's compute ( S = sum_{k=0}^{2016} frac{1}{1 + omega^k} ).To compute S, we can use the formula for sums over roots of unity. Let's recall that for a root of unity ( omega ), sums of the form ( sum_{k=0}^{n-1} frac{1}{a + omega^k} ) can sometimes be evaluated using geometric series or other techniques.Alternatively, consider that ( frac{1}{1 + omega^k} = frac{1}{1 + omega^k} cdot frac{1 - omega^k}{1 - omega^k} = frac{1 - omega^k}{1 - omega^{2k}} ), but this might not help directly.Alternatively, note that ( frac{1}{1 + omega^k} = frac{1}{2} left( 1 + frac{1 - omega^k}{1 + omega^k} right) ), but I'm not sure.Wait, here's another approach. Let me consider that:Let me recall the identity that for roots of unity ( omega neq 1 ), ( sum_{k=0}^{n-1} frac{1}{1 - omega^k} = frac{n}{2} ). But our case is ( frac{1}{1 + omega^k} ), which is different. Hmm.Alternatively, note that ( 1 + omega^k = frac{omega^{k} - (-1)}{omega^k - (-1)} cdot omega^k - (-1) ). Wait, perhaps that's not helpful.Wait, maybe use the identity that ( sum_{k=0}^{n-1} frac{1}{1 - omega^k} = frac{n}{2} ) when n is odd? Wait, maybe not.Alternatively, consider the following. Let me set n = 2017, which is prime. Let's consider the sum ( S = sum_{k=0}^{n-1} frac{1}{1 + omega^k} ).Let me write this as ( S = sum_{k=0}^{n-1} frac{1}{1 + omega^k} ).Note that ( omega^k ) for k = 0 to n-1 are all distinct roots of ( x^n - 1 = 0 ).Now, consider multiplying numerator and denominator by ( 1 - omega^k ):( frac{1}{1 + omega^k} = frac{1 - omega^k}{(1 + omega^k)(1 - omega^k)} = frac{1 - omega^k}{1 - omega^{2k}} ).But then, ( 1 - omega^{2k} = 0 ) when ( 2k equiv 0 mod n ). Since n = 2017 is prime, 2k ≡ 0 mod n implies k ≡ 0 mod n, which is k=0. But for k=0, the term is ( frac{1}{1 + 1} = 1/2 ), so perhaps we need to handle that separately.But this seems complicated. Let me think differently.Consider that the sum S = sum_{k=0}^{n-1} 1/(1 + ω^k). Let me substitute m = -k mod n. Since ω^{-k} = conjugate(ω^k). But not sure if that helps.Alternatively, note that for each k ≠ 0, ω^k is a primitive nth root, so they come in pairs? Wait, n is prime, so all roots except 1 are primitive.Alternatively, consider that S = 1/2 + sum_{k=1}^{n-1} 1/(1 + ω^k). So we need to compute the sum from k=1 to n-1.Wait, but maybe we can use symmetry. Let me pair each term k with term n - k.Note that ω^{n - k} = ω^{-k} = conjugate(ω^k). Therefore, 1 + ω^{n - k} = 1 + conjugate(ω^k) = conjugate(1 + ω^k). Therefore, 1/(1 + ω^{n - k}) = conjugate(1/(1 + ω^k)). Therefore, the sum of these two terms is 2 * Re(1/(1 + ω^k)).But since the sum is over all k from 1 to n-1, which is symmetric, maybe the imaginary parts cancel out, and the total sum is real. However, this might not necessarily help unless we can compute the real parts.Alternatively, consider that sum_{k=1}^{n-1} 1/(1 + ω^k) is equal to (n - 1)/2. But this is a guess. Wait, let me check for a smaller n.Let me test with n=3. Let n=3, then roots of x^3=1 are 1, ω, ω^2, where ω = e^{2πi/3}. Compute sum_{k=1}^2 1/(1 + ω^k).Compute 1/(1 + ω) + 1/(1 + ω^2). Let's compute this.First, ω = -1/2 + i√3/2, so 1 + ω = 1/2 + i√3/2. The reciprocal is (1/2 - i√3/2)/( (1/2)^2 + (√3/2)^2 ) = (1/2 - i√3/2)/1 = 1/2 - i√3/2.Similarly, 1 + ω^2 = 1 + (-1/2 - i√3/2) = 1/2 - i√3/2. The reciprocal is (1/2 + i√3/2)/1 = 1/2 + i√3/2.Adding these two: (1/2 - i√3/2) + (1/2 + i√3/2) = 1. So the sum is 1. For n=3, the sum is 1. But (n - 1)/2 = 1, which matches.Similarly, test n=5. Let's see. Wait, n=5, sum from k=1 to 4 of 1/(1 + ω^k). If the formula holds, the sum should be (5 -1)/2 = 2.But let's compute it. However, this might take longer, but let's try.Alternatively, if for n=3, the sum is 1, which is (3 - 1)/2 = 1. For n=5, maybe it's 2. Let's check:Wait, let's compute the sum for n=5. The roots are 1, ω, ω^2, ω^3, ω^4. Compute S = 1/(1 + ω) + 1/(1 + ω^2) + 1/(1 + ω^3) + 1/(1 + ω^4).Note that ω^k and ω^{n -k} are conjugates. So 1/(1 + ω) + 1/(1 + ω^4) = 2 Re(1/(1 + ω)). Similarly for the other pair. Let me compute Re(1/(1 + ω)).Let ω = e^{2πi/5} = cos(72°) + i sin(72°). Then 1 + ω = 1 + cos(72°) + i sin(72°). The modulus squared is (1 + cos72°)^2 + (sin72°)^2 = 1 + 2cos72° + cos²72° + sin²72° = 2 + 2cos72°.Therefore, 1/(1 + ω) = (1 + cos72° - i sin72°)/ (2 + 2cos72°). The real part is (1 + cos72°)/(2 + 2cos72°) = [1 + cos72°]/[2(1 + cos72°)] = 1/2. Therefore, each pair contributes 2*(1/2) = 1. So there are two pairs, so total sum is 2. Therefore, S=2, which is (5 -1)/2=2. So the formula holds here as well.So this suggests that for prime n, the sum ( sum_{k=1}^{n-1} frac{1}{1 + omega^k} = frac{n - 1}{2} ). Wait, but in our problem, n=2017, which is prime, so the sum would be (2017 -1)/2 = 2016/2 = 1008. Which matches the previous result. Therefore, the answer is 1008.Wait, but in the first approach, using the polynomial, we got Q'(1)/Q(1) = 1008/1 = 1008. So both methods confirm that the sum is 1008. Therefore, the answer is 1008.But let me just verify once again with n=3. For n=3, the sum is 1, and 3-1/2=1, which is correct. For n=5, the sum is 2, and 5-1/2=2, correct. So seems the formula holds. Therefore, for general odd prime n, the sum is (n -1)/2. Therefore, for n=2017, the answer is (2017 -1)/2 = 2016/2 = 1008.Hence, the answer is 1008.Alternatively, another approach. Let's recall that the sum over all roots (excluding 1) of ( 1/(1 + x_k) ) can be related to the sum over all roots of unity.Note that ( sum_{k=1}^{2016} frac{1}{1 + x_k} ).Let me consider the following identity. Let me recall that:For a polynomial ( f(x) = prod_{k=1}^m (x - a_k) ), then ( sum_{k=1}^m frac{1}{c - a_k} = frac{f'(c)}{f(c)} ).In our case, the polynomial whose roots are ( x_k ) is ( P(x) = x^{2016} + x^{2015} + cdots + x + 1 ). So, if we want to compute ( sum_{k=1}^{2016} frac{1}{1 + x_k} ), which is equivalent to ( sum_{k=1}^{2016} frac{1}{1 - (-x_k)} ). So if we let ( Q(x) = prod_{k=1}^{2016} (x - (-x_k)) ), then the sum is ( frac{Q'(1)}{Q(1)} ).But as established earlier, ( Q(x) = P(-x) ), so ( Q(1) = P(-1) ).Wait, let's compute ( P(-1) ):( P(-1) = (-1)^{2016} + (-1)^{2015} + cdots + (-1) + 1 ).Again, this is an alternating sum of 1's and -1's. Since 2016 is even, the sequence starts with 1 (from ( (-1)^{2016} = 1 )), then -1, +1, ..., ending with +1 (the constant term). The number of terms is 2017. As before, this alternates starting with 1, so the sum is 1 -1 +1 -1 + ... +1. Since 2017 is odd, there are 1009 terms of 1 and 1008 terms of -1. Therefore, ( P(-1) = 1009*1 + 1008*(-1) = 1009 - 1008 = 1 ). Therefore, ( Q(1) = P(-1) = 1 ).Now, ( Q(x) = P(-x) ), so ( Q'(x) = -P'(-x) ). Therefore, ( Q'(1) = -P'(-1) ).Compute ( P'(-1) ). First, compute ( P'(x) = 2016x^{2015} + 2015x^{2014} + cdots + 2x + 1 ).Therefore, ( P'(-1) = 2016(-1)^{2015} + 2015(-1)^{2014} + cdots + 2(-1) + 1 ).Simplify each term:- ( 2016(-1)^{2015} = 2016(-1)^{2015} = 2016(-1) )- ( 2015(-1)^{2014} = 2015(1) )- ( 2014(-1)^{2013} = 2014(-1) )- ...- ( 2(-1)^1 = 2(-1) )- ( 1(-1)^0 = 1(1) )So, the sequence is:-2016 + 2015 -2014 + 2013 - ... -2 +1.Let's reorganize the terms:(-2016 + 2015) + (-2014 + 2013) + ... + (-2 +1).Each pair in parentheses is (-1). How many such pairs are there?Starting from 2016 down to 2, stepping by 2 each time. The number of terms from 2016 to 2 is 2016 -1 = 2015 terms, but since we are pairing two at a time, the number of pairs is (2016)/2 = 1008 pairs. Wait, 2016 is the coefficient of the first term, which is paired with 2015, then 2014 with 2013, etc., down to 2 paired with 1. Wait, but 2 is the coefficient of x, so exponent 1. Wait, the polynomial derivative P'(x) has terms from x^{2015} down to x^0 (the constant term 1). So the total number of terms is 2016 (from 2016x^{2015} down to 1). Therefore, when we pair them as above:First pair: -2016 +2015Second pair: -2014 +2013...Last pair: -2 +1Each pair is (-n + (n-1)) = -1.Number of pairs: Since starting from 2016 and going down by 2 each time, but each pair uses two consecutive coefficients. The total number of terms is 2016 (from 2016 down to 1). Therefore, the number of pairs is 2016 / 2 = 1008 pairs. Each contributes -1, so total sum is 1008*(-1) = -1008.Therefore, ( P'(-1) = -1008 ).Therefore, ( Q'(1) = -P'(-1) = -(-1008) = 1008 ).Therefore, the sum is ( Q'(1)/Q(1) = 1008/1 = 1008 ).Hence, this method again leads us to 1008. So regardless of the approach—whether using complex roots and pairing terms, or using polynomial derivatives—we arrive at the same conclusion.Therefore, the answer is ( boxed{1008} ).Final Answerboxed{1008}</think>Given the equation ( x^{2017} = 1 ), the roots are the 2017th roots of unity, which are ( 1, x_1, x_2, ldots, x_{2016} ). We need to find the sum ( sum_{k=1}^{2016} frac{1}{1 + x_k} ).First, we note that the roots ( x_1, x_2, ldots, x_{2016} ) are the roots of the polynomial ( P(x) = x^{2016} + x^{2015} + cdots + x + 1 ). We consider the polynomial ( Q(y) = prod_{k=1}^{2016} (y + x_k) ), which is equivalent to ( P(-y) ).Evaluating ( Q(1) ):[Q(1) = P(-1) = (-1)^{2016} + (-1)^{2015} + cdots + (-1) + 1]This results in an alternating sum of 1's and -1's. Since there are 2017 terms (an odd number), the sum is ( 1 ).Next, we compute the derivative ( Q'(y) ):[Q'(y) = frac{d}{dy} P(-y) = -P'(-y)]Evaluating ( Q'(1) ):[Q'(1) = -P'(-1)]The derivative ( P'(x) ) is:[P'(x) = 2016x^{2015} + 2015x^{2014} + cdots + 2x + 1]Evaluating ( P'(-1) ):[P'(-1) = 2016(-1)^{2015} + 2015(-1)^{2014} + cdots + 2(-1) + 1]This results in an alternating sum of terms, which simplifies to (-1008). Therefore:[Q'(1) = -P'(-1) = 1008]Thus, the sum ( sum_{k=1}^{2016} frac{1}{1 + x_k} ) is:[frac{Q'(1)}{Q(1)} = frac{1008}{1} = 1008]The final answer is:[boxed{1008}]]

🔑:5.1008 .Given x_{k}=mathrm{e}^{frac{2 pi m}{2017} mathrm{i}}(k=1,2, cdots, 2016), we knowbegin{array}{l}overline{x_{k}}=mathrm{e}^{frac{-2 k pi}{2017} mathrm{i}}=mathrm{e}^{frac{2(2017-k) pi mathrm{i}}{2017} mathrm{i}}=x_{2017-k} . text { Then } frac{1}{1+x_{k}}+frac{1}{1+x_{2017-k}} =frac{1}{1+x_{k}}+frac{1}{1+overline{x_{k}}} =frac{1}{1+x_{k}}+frac{x_{k} overline{x_{k}}}{x_{k} overline{x_{k}}+overline{x_{k}}}=1 . text { Therefore, } sum_{k=1}^{2016} frac{1}{1+x_{k}}=1008 .end{array}