🧠:[<think>Okay, so I have this problem here about a box of fruit. Let me read it again: there are 20 apples, 10 oranges, and no other fruit. When a fruit is randomly chosen from the box, what is the probability that it's an orange? The options are A through E, with fractions like 1/10 up to 2/3. Alright, probability question. Let me think how to approach this.First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, the favorable outcome is picking an orange, and the total outcomes are picking any fruit. So, the formula should be Probability = Number of Oranges / Total Number of Fruits. Let me check if that's right. Yeah, that seems basic. So, if there are 10 oranges and 20 apples, the total number of fruits is 20 + 10. Let me compute that: 20 apples plus 10 oranges equals 30 fruits total. So, total fruits are 30.Therefore, the probability of picking an orange would be 10 divided by 30. Let me write that down: 10/30. Hmm, can that be simplified? Yeah, both numerator and denominator are divisible by 10. Wait, 10 divided by 10 is 1, and 30 divided by 10 is 3. So, 10/30 simplifies to 1/3. So, the probability is 1/3. Let me check the answer options. Option D is 1/3. So, that should be the answer.Wait, let me make sure I didn't make a mistake. So, apples are 20, oranges 10. Total fruits 30. So oranges are 10 out of 30, which is indeed 1/3. Hmm. Is there another way to think about this? Maybe using ratios? The ratio of apples to oranges is 20:10, which simplifies to 2:1. So, for every 3 fruits, 2 are apples and 1 is an orange. So, the chance of picking an orange is 1 out of 3. Yep, that matches. So, 1/3 is correct. So, D is the answer. Let me just verify once again. 10 oranges over 30 total is 1/3. Yes, that seems right. I don't think I made any mistakes here. The answer must be D.Final Answerboxed{D}</think>To determine the probability that a randomly chosen fruit from the box is an orange, we start by noting the number of each type of fruit. There are 20 apples and 10 oranges, making a total of 30 fruits.The probability of choosing an orange is the number of oranges divided by the total number of fruits. This can be calculated as follows:[text{Probability} = frac{text{Number of Oranges}}{text{Total Number of Fruits}} = frac{10}{30} = frac{1}{3}]Thus, the probability that the chosen fruit is an orange is (frac{1}{3}), which corresponds to option D.[boxed{D}],<think>Okay, so I need to find the probability that a randomly chosen fruit from the box is an orange. The box has 20 apples, 10 oranges, and no other fruit. Let me think about how to approach this.First, probability is generally calculated as the number of favorable outcomes divided by the total number of possible outcomes. In this case, the favorable outcomes are picking an orange, and the total outcomes are picking any fruit from the box.So, the number of oranges is 10, and the total number of fruits should be the sum of apples and oranges. Let me verify that. The problem says there are 20 apples and 10 oranges, so total fruits would be 20 + 10. Let me add that: 20 + 10 equals 30. So, there are 30 fruits in total.Therefore, the probability of picking an orange should be the number of oranges divided by the total number of fruits. That would be 10 divided by 30. Let me write that out as a fraction: 10/30. Now, simplifying that fraction, both numerator and denominator can be divided by 10. So, 10 divided by 10 is 1, and 30 divided by 10 is 3. So, simplifying gives 1/3.Wait, but let me check the answer choices. The options are (A) 1/10, (B) 1/20, (C) 1/30, (D) 1/3, (E) 2/3. So, 1/3 is option D. That seems straightforward, but maybe I made a mistake somewhere.Let me go through it again. Number of oranges: 10. Total fruits: 20 apples + 10 oranges = 30. Probability is 10/30 = 1/3. Hmm, that seems correct. Alternatively, maybe I need to consider something else? Are there any other factors here? The problem states that the fruit is chosen randomly, so each fruit has an equal chance of being selected. Since all fruits are either apples or oranges, and they are the only ones in the box, there's no trick here with other fruits. So, 10 out of 30 is indeed the probability, which simplifies to 1/3. So the answer must be D. Let me check if there's a common mistake people make here. For example, sometimes problems might ask for the probability of not getting an orange, but that's not the case here. Also, if someone misreads the number of oranges as 10 and the total as 20, they might think it's 10/20, but the total is 30. So, 10/30 is correct. Yeah, so D seems right.Final Answerboxed{D}</think>To find the probability that a randomly chosen fruit from the box is an orange, we start by noting the number of each type of fruit: there are 20 apples and 10 oranges. The total number of fruits in the box is the sum of apples and oranges:[20 + 10 = 30]The probability of choosing an orange is the number of oranges divided by the total number of fruits:[frac{10}{30} = frac{1}{3}]Thus, the probability that the fruit chosen is an orange is (boxed{D}).]

🔑:Each of the 30 pieces of fruit in the box is equally likely to be chosen. Since there are 10 oranges in the box, then the probability that the chosen fruit is an orange is frac{10}{30} or frac{1}{3}.#

🧠:[<think>Okay, so I need to find the value of b for the quadratic function y = 2x² - bx -1. The problem says that the graph passes through the points (3, y₀) and (9, y₀). Hmm, that means when x is 3 and when x is 9, the y-value is the same, y₀. Since both points are on the graph of the quadratic, plugging x=3 and x=9 into the equation should give the same y-value. So, let me write that out. For x = 3:y₀ = 2*(3)² - b*(3) - 1= 2*9 - 3b - 1= 18 - 3b - 1= 17 - 3bFor x = 9:y₀ = 2*(9)² - b*(9) - 1= 2*81 - 9b - 1= 162 - 9b - 1= 161 - 9bSince both expressions equal y₀, I can set them equal to each other:17 - 3b = 161 - 9bNow, solve for b. Let's bring all terms to one side. Let's add 9b to both sides:17 - 3b + 9b = 161That simplifies to:17 + 6b = 161Now subtract 17 from both sides:6b = 161 - 176b = 144Divide both sides by 6:b = 144 / 6b = 24Wait, so b is 24? Let me check the options. The options are A: -24, B: -12, C: 12, D: 24. So D is 24, which matches my calculation. But let me verify again to make sure I didn't make a mistake. Let's substitute b=24 into the original equations for x=3 and x=9 and see if they give the same y₀.First, for x=3:y₀ = 2*(3)^2 -24*3 -1= 2*9 -72 -1= 18 -72 -1= -55For x=9:y₀ = 2*(9)^2 -24*9 -1= 2*81 -216 -1= 162 -216 -1= -55Yep, both give y₀ = -55. So that checks out. Therefore, the value of b is 24, which is option D.Alternatively, another way to approach this problem is to recognize that in a quadratic function, if two different x-values give the same y-value, then those x-values are symmetric with respect to the vertex. The axis of symmetry of a quadratic given by y = ax² + bx + c is at x = -b/(2a). In this case, the quadratic is y = 2x² -bx -1, so a = 2, and the axis of symmetry is at x = -(-b)/(2*2) = b/4. Since the points (3, y₀) and (9, y₀) are symmetric about the axis of symmetry, the midpoint of x=3 and x=9 should be equal to the axis of symmetry. Midpoint between 3 and 9 is (3 + 9)/2 = 12/2 = 6. Therefore, the axis of symmetry is x = 6. So setting b/4 = 6, solving for b gives b = 6*4 = 24. That's another way to get the answer without solving the equations, using the properties of the quadratic function. This also leads to b=24, which is option D. So both methods confirm that the answer is D:24.Final Answerboxed{D}</think>The quadratic function ( y = 2x^2 - bx - 1 ) passes through the points ((3, y_0)) and ((9, y_0)). This means the y-values for ( x = 3 ) and ( x = 9 ) are the same. First, we substitute ( x = 3 ) into the function:[y_0 = 2(3)^2 - b(3) - 1 = 18 - 3b - 1 = 17 - 3b]Next, we substitute ( x = 9 ) into the function:[y_0 = 2(9)^2 - b(9) - 1 = 162 - 9b - 1 = 161 - 9b]Setting the two expressions for ( y_0 ) equal to each other:[17 - 3b = 161 - 9b]Solving for ( b ):[17 - 3b + 9b = 161 implies 17 + 6b = 161 implies 6b = 144 implies b = 24]We can also verify this by recognizing the axis of symmetry of the quadratic function. The midpoint of ( x = 3 ) and ( x = 9 ) is ( 6 ). The axis of symmetry for ( y = 2x^2 - bx - 1 ) is given by ( x = frac{b}{4} ). Setting this equal to 6 gives:[frac{b}{4} = 6 implies b = 24]Thus, the value of ( b ) is (boxed{D}).,<think>Okay, let's see here. I need to find the value of b in the quadratic function y = 2x² - bx - 1, given that the graph passes through both (3, y₀) and (9, y₀). Hmm. So, both these points have the same y-value, y₀, but different x-values, 3 and 9. Since the graph is a quadratic, which is a parabola, and parabolas are symmetric around their vertex. The fact that two points have the same y-value means they are symmetric with respect to the vertex's axis of symmetry. So, the axis of symmetry for a parabola given by y = ax² + bx + c is at x = -b/(2a). Wait, in this case, the quadratic is y = 2x² - bx - 1. Comparing to the standard form y = ax² + bx + c, here a is 2, the coefficient of x², and the linear coefficient is -b, so the standard form's b is actually -b here. Therefore, the axis of symmetry should be at x = -(-b)/(2*2) = b/4. Is that right? Let me check again. Standard form: y = ax² + bx + c. So, the axis of symmetry is x = -B/(2A), where B is the coefficient of x. In this problem, the quadratic is y = 2x² - bx - 1. So, A is 2, and B is -b. Therefore, the axis of symmetry is x = -(-b)/(2*2) = b/(4). Yes, that's correct. Now, since the points (3, y₀) and (9, y₀) are on the parabola and have the same y-value, they must be equidistant from the axis of symmetry. Therefore, the axis of symmetry is exactly halfway between x = 3 and x = 9. Let's compute the midpoint between 3 and 9. The midpoint is (3 + 9)/2 = 12/2 = 6. So, the axis of symmetry is x = 6. But we also found earlier that the axis of symmetry is x = b/4. Therefore, setting them equal: b/4 = 6. Solving for b, we get b = 6*4 = 24. So, b is 24, which is option D. Wait, but let me verify this to make sure. Let me substitute x = 3 and x = 9 into the equation with b = 24 and check if the y-values are the same. For x = 3:y = 2*(3)^2 - 24*(3) -1 = 2*9 -72 -1 = 18 -72 -1 = -55.For x = 9:y = 2*(9)^2 -24*(9) -1 = 2*81 -216 -1 = 162 -216 -1 = -55.Yes, both give y = -55, so they do have the same y-value. Therefore, b = 24 is correct. So the answer should be D: 24. Alternatively, another way to approach this is by setting the two expressions equal since they both equal y₀. So, substituting x = 3 and x = 9 into the equation gives:For (3, y₀):y₀ = 2*(3)^2 - b*(3) -1 = 18 - 3b -1 = 17 - 3b.For (9, y₀):y₀ = 2*(9)^2 - b*(9) -1 = 162 - 9b -1 = 161 - 9b.Since both equal y₀, we can set them equal to each other:17 - 3b = 161 - 9b.Solving for b:17 - 3b = 161 - 9b -3b + 9b = 161 -17 6b = 144 b = 144 /6 b = 24.Same result. So that confirms it. So, answer D:24 is correct.Final Answerboxed{D}</think>Given the quadratic function ( y = 2x^2 - bx - 1 ) which passes through the points ( (3, y_0) ) and ( (9, y_0) ), we need to find the value of ( b ).Since the points ( (3, y_0) ) and ( (9, y_0) ) have the same ( y )-value, they are symmetric with respect to the axis of symmetry of the parabola. The axis of symmetry for a quadratic function ( y = ax^2 + bx + c ) is given by ( x = -frac{B}{2A} ), where ( A ) is the coefficient of ( x^2 ) and ( B ) is the coefficient of ( x ). For the given function ( y = 2x^2 - bx - 1 ), we have ( A = 2 ) and ( B = -b ). Therefore, the axis of symmetry is:[x = frac{-(-b)}{2 cdot 2} = frac{b}{4}]The midpoint of ( x = 3 ) and ( x = 9 ) is:[frac{3 + 9}{2} = 6]Setting the axis of symmetry equal to this midpoint:[frac{b}{4} = 6]Solving for ( b ):[b = 6 times 4 = 24]To verify, we substitute ( x = 3 ) and ( x = 9 ) into the quadratic function with ( b = 24 ):- For ( x = 3 ): [ y = 2(3)^2 - 24(3) - 1 = 18 - 72 - 1 = -55 ]- For ( x = 9 ): [ y = 2(9)^2 - 24(9) - 1 = 162 - 216 - 1 = -55 ]Both points yield the same ( y )-value, confirming ( b = 24 ).Thus, the value of ( b ) is (boxed{D}).]

🔑:Given the quadratic function y=2x^{2}-bx-1, and it passes through the points (3, y_{0}) and (9, y_{0}), we can determine the value of b by understanding the properties of a quadratic function's graph.1. The axis of symmetry formula for a quadratic function ax^2 + bx + c is x = -frac{b}{2a}. However, in this context, we should correct the axis of symmetry formula to x = frac{b}{2a}, considering the standard form provided in the solution.2. Given a = 2, we substitute into the axis of symmetry formula:[x = frac{b}{2a} = frac{b}{2 cdot 2} = frac{b}{4}]3. Since the graph passes through (3, y_{0}) and (9, y_{0}), and these points are equidistant from the axis of symmetry, the axis of symmetry is exactly in the middle of x = 3 and x = 9. Thus, the axis of symmetry is at x = 6.4. Setting the axis of symmetry equal to 6, we solve for b:[6 = frac{b}{4}]5. Solving for b gives us:[b = 6 cdot 4 = 24]Therefore, the value of b is boxed{24}, which corresponds to choice boxed{text{D}}.

🧠:[<think>Okay, so I need to figure out the minimum number of table tennis balls required to place in 10 boxes, each following these rules: each box has a different number of balls, no box has fewer than 11 balls, none can have exactly 17 balls, and none can have a number that's a multiple of 6. The goal is to find the smallest total number of balls needed while satisfying all these conditions.First, let's list out the constraints to make sure I understand them all:1. 10 boxes, each with a different number of balls.2. Each box must have at least 11 balls.3. No box can have exactly 17 balls.4. No box can have a number of balls that is a multiple of 6.5. We need the minimum total number of balls.So, starting with the first few numbers above 11, but avoiding 17 and multiples of 6. Let's list numbers starting from 11 upwards, crossing out those that are forbidden.Starting numbers:11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, ...Now, let's identify which numbers are excluded:- 12: multiple of 6 (12 ÷ 6 = 2) → excluded- 17: exactly 17 → excluded- 18: multiple of 6 (18 ÷ 6 = 3) → excluded- 24: multiple of 6 → excluded- Etc. So every number that's 6n (where n is an integer) is excluded.So allowed numbers are 11, 13, 14, 15, 16, 19, 20, 21, 22, 23, 25, 26, etc., as long as they are not multiples of 6 and not 17.We need 10 different numbers, each at least 11, avoiding 17 and multiples of 6. So the strategy is to pick the 10 smallest numbers that meet the criteria, and sum them up. However, we have to skip 17 and any multiples of 6.Let's list them step by step:Starting from 11:1. 11 – allowed (not 17, not multiple of 6)2. 12 – excluded (multiple of 6)3. 13 – allowed4. 14 – allowed5. 15 – allowed (15 ÷ 6 = 2.5, not integer)6. 16 – allowed7. 17 – excluded8. 18 – excluded (multiple of 6)9. 19 – allowed10. 20 – allowed (20 ÷ 6 ≈ 3.333)11. 21 – allowed (21 ÷ 6 = 3.5)12. 22 – allowed13. 23 – allowed14. 24 – excluded (multiple of 6)15. 25 – allowed16. 26 – allowed17. 27 – excluded (27 ÷ 6 = 4.5, but wait, 27 is 6*4 + 3, so it's not a multiple of 6. Wait, 6*4=24, 6*5=30. So 27 is not a multiple of 6. Hmm, maybe I made a mistake here. 27 is 3*9, but not 6*n. So 27 is allowed. Wait, 27 ÷ 6 is 4.5, so it's not a multiple of 6. So 27 is allowed. So maybe my earlier thought was wrong. Let's check.Wait, multiples of 6 are numbers like 6, 12, 18, 24, 30, etc. So 27 is not a multiple of 6, so it's allowed. So in that case, numbers like 21, 22, 23, 25, 26, 27, etc., are all allowed.So starting from 11, let's list the allowed numbers:11, 13, 14, 15, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30 (30 is excluded), 31, etc.So the first 10 allowed numbers after 11 would be:1. 112. 13 (skipping 12)3. 144. 155. 166. 19 (skipping 17 and 18)7. 208. 219. 2210. 23Wait, that's 10 numbers: 11,13,14,15,16,19,20,21,22,23. Let me check if all of these are allowed.- 11: allowed- 13: allowed- 14: allowed (14 ÷6 ≈ 2.333)- 15: allowed- 16: allowed- 19: allowed- 20: allowed- 21: allowed (21 ÷6 =3.5)- 22: allowed- 23: allowedYes, none of these are 17 or multiples of 6. So these 10 numbers are acceptable. Let's sum them up to find the total number of balls.Calculating the sum:11 + 13 = 2424 +14 = 3838 +15 = 5353 +16 = 6969 +19 = 8888 +20 = 108108 +21 = 129129 +22 = 151151 +23 = 174Wait, so the total sum is 174? Let me verify each addition step by step:1. 112. 11 +13 =243. 24 +14 =384. 38 +15 =535. 53 +16 =696. 69 +19 =887. 88 +20 =1088. 108 +21 =1299. 129 +22 =15110. 151 +23 =174Yes, that's correct. So the total would be 174 balls.But wait, maybe there is a way to have a lower total by using numbers that are allowed but arranged in a way that the numbers are smaller. For example, maybe if instead of using 23, we can use a lower number that we might have skipped because of the constraints. Let me check if there is a number between 11 and 23 that we excluded but could have been included if arranged differently.Wait, starting from 11:11, then 12 is excluded. Then 13,14,15,16. Then 17 excluded, 18 excluded. Then 19,20,21,22,23. So between 11 and 23, the excluded numbers are 12,17,18,24. So the numbers we used are indeed the smallest possible 10 numbers that meet the criteria. So 11,13,14,15,16,19,20,21,22,23.But let me check if there's a possibility that using higher numbers earlier allows for smaller numbers later. Wait, no, because all numbers have to be distinct and we have to use the smallest possible to minimize the total.Wait, maybe not. Let me think again. Suppose we use the 10 smallest numbers that are allowed. That should give the minimal total. So 11 is the smallest, then 13 is next, then 14,15,16, then after 16, the next is 19 (since 17 and 18 are excluded). Then 20,21,22,23. So that's correct.But let me check if there's a number between 16 and 19 that is allowed. For example, 17 is excluded, 18 is excluded. So next after 16 is 19. So that's correct.Alternatively, could we arrange the numbers differently, like skipping some higher numbers to use lower ones? Wait, no, since we have to use 10 different numbers, each as small as possible. So using the 10 smallest possible numbers would give the minimal total.Therefore, the total would be 174. However, let me check another angle. Suppose if we need 10 distinct numbers starting from 11, avoiding 17 and multiples of 6. Let's list the allowed numbers in order:1. 112. 133. 144. 155. 166. 197. 208. 219. 2210. 23Yes, these are the first 10 allowed numbers. Therefore, their sum is indeed 174. Let me confirm once more.Alternatively, maybe there's a mistake in the exclusion. Let's check each number:11: allowed13: allowed14: 14 divided by 6 is 2.333, not a multiple. Allowed.15: 15 ÷6=2.5, not a multiple. Allowed.16: same, 16 ÷6≈2.666. Allowed.19: allowed.20: 20 ÷6≈3.333, allowed.21: 21 ÷6=3.5, allowed.22: 22 ÷6≈3.666, allowed.23: allowed.All these numbers are not 17 and not multiples of 6. So yes, correct.Therefore, the minimal total is 174. But wait, let me check if maybe we can use a higher number but skip a larger number to get a smaller sum. For example, perhaps replacing 23 with a smaller number that we missed. Wait, after 16, we have 17 and 18 excluded. Next is 19, 20,21,22,23. So all numbers from 19 to 23 except multiples of 6. But 24 is excluded. So yes, 19,20,21,22,23 are allowed.Alternatively, could there be a number between, say, 20 and 23 that's excluded? No, because 24 is the next multiple of 6 after 18, which is 24, so between 19 and 23, all numbers are allowed except 17 and 18. So 19,20,21,22,23 are allowed.So that sequence is correct. Therefore, the total is 174. So I think that's the answer. But let me check once again with another approach.Alternative approach: The problem is similar to finding the sum of the first 10 numbers in the sequence of allowed numbers starting from 11, excluding 17 and multiples of 6.The allowed numbers starting from 11 are:11, (12 excluded), 13,14,15,16, (17 excluded), (18 excluded),19,20,21,22,23, (24 excluded),25,26,27,28,29, (30 excluded),31,...So the 10 numbers are 11,13,14,15,16,19,20,21,22,23. Sum them up:11 +13 =2424+14=3838+15=5353+16=6969+19=8888+20=108108+21=129129+22=151151+23=174Same result. Therefore, I think 174 is the minimal total.But let me check if there's any alternative set of numbers where maybe substituting a higher number with a lower number that's allowed but wasn't in the initial sequence. For instance, is 25 allowed? Yes, but it's higher than 23, so using 25 would only increase the total. Similarly, 26 is higher. So there's no benefit in using higher numbers.Alternatively, maybe starting from a higher number than 11? But the problem states each box must have no fewer than 11, so starting at 11 is necessary. Therefore, the minimal starting point is 11, and then the next allowed numbers in order.Therefore, the minimal total number of table tennis balls needed is 174.Final Answerboxed{174}</think>To determine the minimum number of table tennis balls needed to place in 10 boxes, each following the constraints of having no fewer than 11 balls, not exactly 17, not a multiple of 6, and each containing a different number, we start by listing the allowed numbers:1. Allowed numbers starting from 11: - 11, 13, 14, 15, 16 (excluding 12, 17, 18) - 19, 20, 21, 22, 23 (excluding 24)2. Verification: - Each number is checked to ensure it is not 17 or a multiple of 6. - The numbers 11, 13, 14, 15, 16, 19, 20, 21, 22, and 23 are all valid.3. Summing the numbers: - Calculate the sum step-by-step: [ 11 + 13 + 14 + 15 + 16 + 19 + 20 + 21 + 22 + 23 ] - Detailed addition: [ 11 + 13 = 24 24 + 14 = 38 38 + 15 = 53 53 + 16 = 69 69 + 19 = 88 88 + 20 = 108 108 + 21 = 129 129 + 22 = 151 151 + 23 = 174 ]Thus, the minimum total number of table tennis balls needed is (boxed{174}).,<think>Okay, so I have this problem where I need to place table tennis balls into 10 boxes. Each box has some specific requirements: the number of balls in each box can't be fewer than 11, can't be 17, and can't be a multiple of 6. Also, all the boxes must have different numbers of balls. The question is asking for the minimum total number of table tennis balls needed. Hmm, let me think about how to approach this.First, let's list out the requirements again to make sure I have them all:1. 10 boxes.2. Each box has a different number of balls.3. Each box must have at least 11 balls.4. No box can have exactly 17 balls.5. The number of balls in each box cannot be a multiple of 6.6. We need the minimal total number of balls.Alright, so starting with point 3, each box has at least 11 balls. Since all numbers must be different, if we ignore the other restrictions, the minimal total would be the sum of numbers from 11 to 20. Let me check that: 11+12+13+...+20. The formula for the sum of consecutive numbers is (n/2)*(first + last). Here, n=10, first=11, last=20. So sum = (10/2)*(11+20) = 5*31 = 155. So 155 would be the minimal if there were no other restrictions. But we have to exclude numbers that are 17 or multiples of 6. So let's see which numbers between 11 and 20 are forbidden.First, numbers that are multiples of 6 between 11 and 20: 12, 18. Wait, 6*2=12, 6*3=18. So 12 and 18 are multiples of 6. Then also, we can't have 17. So the forbidden numbers in the 11-20 range are 12, 17, 18.So if we were to start from 11 and go up, the numbers 11,12,13,14,15,16,17,18,19,20. Out of these, 12,17,18 are forbidden. So we need to replace those three numbers with the next available numbers that aren't forbidden and also not already used.So instead of 12, 17, 18, we need to use higher numbers. Let's list the original 10 numbers and then see which ones need to be replaced.Original numbers: 11,12,13,14,15,16,17,18,19,20.Forbidden in these: 12,17,18. So we need to replace these three with the next numbers after 20 that are not forbidden. Let's see what those would be.After 20, the next numbers are 21,22,23, etc. Now, check if these are allowed. The rules are: not 17 (but we're past that), not a multiple of 6. So 21: 21 is not a multiple of 6 (6*3=18, 6*4=24), so 21 is okay. 22: 22 is not a multiple of 6. 23: same. 24: multiple of 6 (6*4=24), so forbidden. 25: okay. 26: okay. 27: 6*4.5=27, not a multiple, so okay. 28: okay. 29: okay. 30: multiple of 6. So the next numbers after 20 that are allowed are 21,22,23,25,26,27, etc.So replacing the three forbidden numbers (12,17,18) with the next three allowed numbers: 21,22,23. Wait, but 20 is already included. So after 20, next numbers are 21,22,23. So replacing each forbidden number with the next available number. Let's see:Original set: 11,12,13,14,15,16,17,18,19,20.Remove 12,17,18, so remaining numbers are 11,13,14,15,16,19,20. That's 7 numbers. We need 10, so we need to add 3 more numbers. The next numbers after 20 are 21,22,23. Check if they are allowed:21: not 17, not multiple of 6. Allowed.22: same.23: same.So the new set would be 11,13,14,15,16,19,20,21,22,23.Wait, but let's check if all numbers are distinct and meet the requirements. Let's list them:11: okay.13: okay.14: okay.15: okay.16: okay.19: okay.20: okay.21: okay.22: okay.23: okay.So all numbers are distinct, none are 17, none are multiples of 6. So that works. Now, what's the total here?Let me calculate the sum:11 +13=2424+14=3838+15=5353+16=6969+19=8888+20=108108+21=129129+22=151151+23=174So total is 174. But wait, originally, the minimal sum without restrictions was 155. So by replacing 12,17,18 with 21,22,23, we added 21-12=9, 22-17=5, 23-18=5. So total added: 9+5+5=19. So 155 +19=174. Yep.But is this the minimal possible? Let me check. Maybe there's a way to replace the forbidden numbers with smaller numbers instead of going all the way up to 21,22,23. Wait, but numbers must be different, so if we can't use 12,17,18, perhaps there are other numbers below 20 that are allowed. Let me check the numbers from 11 to 20 again, excluding 12,17,18. So allowed numbers are 11,13,14,15,16,19,20. So 7 numbers. We need 10 boxes, so need 3 more numbers. The next numbers after 20 are 21,22,23, but maybe there's a way to use numbers between 11 and 20 that we haven't considered.Wait, maybe instead of replacing 12,17,18 with 21,22,23, perhaps we can use other numbers in the 11-20 range that are allowed but were not in the original sequence. But in the original sequence, we started from 11 and went up by 1 each time. If we skip the forbidden numbers, maybe there are other numbers in that range that aren't consecutive. Wait, no. If we have to have different numbers, starting at 11, the next numbers are 12,13, etc., but if 12 is forbidden, we need to skip to 13. But in the problem, it just requires each box to have a different number, not necessarily consecutive. So maybe instead of taking 11,12,13,...,20, we can pick 10 different numbers starting from 11, avoiding 12,17,18. But in that case, the numbers would be 11,13,14,15,16,19,20, and then we need three more numbers above 20. Because from 11 to 20, excluding 12,17,18, we have 7 numbers. So the other three numbers have to be 21,22,23. So that seems unavoidable. Therefore, the minimal sum would indeed be 174. But let me check again.Wait, is there a way to use numbers below 21 that are allowed but not in the original 11-20 sequence? For example, if instead of starting at 11, maybe use numbers higher than 11 but still within 11-20, but that are not forbidden. Wait, no. The numbers from 11 to 20 are fixed. The forbidden ones are 12,17,18. So we have to exclude those. So even if we rearrange, we can't get more numbers in the 11-20 range. Therefore, 7 numbers allowed, 3 needed from above 20. Hence, 21,22,23.But maybe there's a way to use smaller numbers by overlapping or something? Wait, but all numbers must be different, and starting from 11. If we can't use 12,17,18, then the next available numbers are indeed 21,22,23.Wait, but another thought: perhaps instead of replacing 12,17,18 with 21,22,23, which are the next three numbers, maybe replacing them with higher numbers but with some smaller increments? Wait, no. Because we have to have distinct numbers, so once we take 21,22,23, they are the next ones. There's no smaller numbers available since 11-20 is already used except for the forbidden ones. So I think 21,22,23 are indeed the minimal replacements.But let's check: for example, if instead of replacing 12 with 21, can we replace it with another number between 11 and 20 that is allowed? But between 11 and 20, allowed numbers are 11,13,14,15,16,19,20. So 12 is forbidden, 17 and 18 forbidden. So if we have 11,13,14,15,16,19,20, that's 7 numbers. So we need three more. The next numbers are 21,22,23. So I don't see any way to get numbers lower than 21. Unless we can repeat numbers, but the problem says numbers must be different. So no.Therefore, the total sum is 174. But let me verify the calculation again step by step to make sure I didn't make a mistake.Original numbers (excluding forbidden):11,13,14,15,16,19,20,21,22,23.Sum:11 +13 =2424 +14=3838 +15=5353 +16=6969 +19=8888 +20=108108 +21=129129 +22=151151 +23=174Yes, that's 174.But wait, maybe there's a different combination where the numbers added are smaller? Let's see.Suppose instead of replacing 12,17,18 with 21,22,23, could we use some numbers that are not consecutive but have a lower total?For example, replacing 12,17,18 with 21,22,23 adds 9,5,5 respectively. But maybe if we can find numbers that are allowed and closer to the original numbers, but still unique. Wait, but the next allowed numbers after 20 are 21,22,23, since 24 is a multiple of 6 (24=6*4), so 24 is forbidden. Then 25 is allowed, but 25 is higher than 21, so using 21,22,23 is better for minimizing the total.Alternatively, if we skip some numbers above 20 but that might lead to higher total. For example, using 21,22,25 would be 21+22+25=68 instead of 21+22+23=66. So worse.Alternatively, 21,23,25? That's even higher. So no, 21,22,23 are the minimal.Alternatively, check if 24 is allowed. 24 is a multiple of 6, so no. So the next numbers after 20 are 21,22,23,25,26,27, etc., with 24 forbidden.Wait, 25 is allowed. But 25 is higher than 21,22,23, so replacing 12,17,18 with 21,22,23 is better.Therefore, I think 174 is indeed the minimal total.But let me check another approach. Maybe if instead of starting at 11, can we start higher but arrange the numbers in a way that avoids the forbidden numbers without having to go too high?Wait, but starting higher would mean all numbers are higher, leading to a larger total. For example, if instead of starting at 11, start at 13, but the problem says each box must have no fewer than 11, so starting at 11 is necessary for minimal total.Alternatively, perhaps using some numbers below 21 that are not in the original sequence but allowed? Wait, but between 11 and 20, the allowed numbers are 11,13,14,15,16,19,20. There's no others. So we can't get more numbers there. So we have to go above 20 for the remaining three.Alternatively, is there a way to use numbers above 20 that are smaller than 21,22,23? But no, numbers above 20 start at 21. So 21 is the next.Wait, another thought: maybe overlapping with the existing numbers. For example, if we can rearrange the numbers such that some numbers are higher but others are lower? But since all numbers have to be at least 11 and distinct, and we already used the lowest possible (11,13,14,15,16,19,20), there's no way to have a lower number than 11. So we can't get lower than that.Alternatively, if we could replace a higher number in the original set with a lower number, but the original set already starts at 11. So replacing, say, 20 with a lower number isn't possible because all lower numbers are already used or forbidden.Wait, for example, if we have the original numbers: 11,12,13,14,15,16,17,18,19,20. If we remove 12,17,18, we need to add three numbers. If instead of adding 21,22,23, can we replace 12,17,18 with numbers like 24,25,26? But that would make the total even higher. So that's worse. Therefore, the minimal is indeed 21,22,23.Alternatively, maybe 19 is already in the set, but if we remove 19 and replace it with something else? But that complicates things. Wait, no. The original set after removal has 11,13,14,15,16,19,20. So 19 is allowed. If we remove 19 and add a higher number, that would increase the total. Not helpful.Alternatively, if we can have duplicate numbers, but the problem says numbers must be different, so duplicates aren't allowed.Therefore, after considering all possibilities, it seems that replacing 12,17,18 with 21,22,23 is the minimal way to get 10 distinct numbers meeting all the requirements, leading to a total of 174.But wait, let me just check if there are any other numbers between 11 and 20 that are forbidden that I might have missed.Forbidden numbers are:- Less than 11: but the problem says no fewer than 11, so 11 is the minimum.- Exactly 17.- Multiples of 6: 12, 18 (since 6*2=12 and 6*3=18; 6*4=24 is above 20).So only 12,17,18 are forbidden between 11 and 20. So that's correct.Therefore, yes, 174 seems correct. Wait, but let me check once more if there is a smarter combination.Suppose instead of replacing 12,17,18 with 21,22,23, can we replace them with numbers that result in a smaller total? For example, if instead of replacing 12 with 21 (difference +9), maybe replace 17 with a lower number? But 17 is in the middle; we can't replace it with a lower number because all numbers below 17 are already used (11,13,14,15,16). They are already in the set. So we can't replace 17 with a lower number, since those are already taken. Similarly, 18 is next, but numbers above 18 are 19,20, which are already in the set. So we have to replace 12,17,18 with numbers above 20.Alternatively, perhaps if we shift some numbers higher to allow lower replacements? Wait, I don't see how. For example, if we take a higher number for one box, maybe that allows a lower number for another? But all lower numbers are already used or forbidden. For example, 11 is the lowest. If we don't use 11, but start at 12, but 12 is forbidden. So 11 is necessary. Then next is 13, since 12 is forbidden. So 13 is next. Then 14,15,16,17 is forbidden, so 19,20. Then we need three more numbers. So even if we tried to rearrange, we still end up needing 21,22,23.Wait, another angle: the problem says "each box containing no fewer than 11", so 11 is allowed. But maybe using numbers above 11 but skipping some forbidden ones could allow a lower total? But since the numbers have to be distinct, and 11 is the lowest possible, not using 11 would require starting at 13, which would make the total higher. For example, starting at 13,14,15,16,19,20,21,22,23,24. But 24 is forbidden. So 13,14,15,16,19,20,21,22,23,25. That sum would be way higher than 174. So definitely worse.Therefore, using 11 is necessary for minimal total. Hence, the original approach is correct.Therefore, the minimal total number of table tennis balls needed is 174. So the answer should be boxed{174}.Final Answerboxed{174}</think>To solve the problem of placing table tennis balls into 10 boxes with specific requirements, we need to ensure each box has a different number of balls, no fewer than 11, not 17, and not a multiple of 6. The goal is to find the minimum total number of balls required.1. Identify the forbidden numbers between 11 and 20: - Multiples of 6: 12, 18 - The number 172. List the allowed numbers between 11 and 20: - 11, 13, 14, 15, 16, 19, 20 (7 numbers)3. Replace the forbidden numbers (12, 17, 18) with the next available numbers above 20: - The next numbers after 20 that are not multiples of 6 are 21, 22, 234. Calculate the sum of the allowed numbers: - Original allowed numbers: 11, 13, 14, 15, 16, 19, 20 - Replacement numbers: 21, 22, 235. Sum the numbers: [ 11 + 13 + 14 + 15 + 16 + 19 + 20 + 21 + 22 + 23 ] - Breaking down the sum step-by-step: [ 11 + 13 = 24 24 + 14 = 38 38 + 15 = 53 53 + 16 = 69 69 + 19 = 88 88 + 20 = 108 108 + 21 = 129 129 + 22 = 151 151 + 23 = 174 ]Thus, the minimum total number of table tennis balls needed is (boxed{174}).]

🔑:【Analysis】Starting from 11, find 10 numbers that are neither multiples of 17 nor multiples of 6, then add them up.【Solution】Solution: The smallest 10 numbers that meet the conditions are:11,13,14,15,16,19,20,21,22,23;Therefore, at least 11+13+14+15+16+19+20+21+22+23=174 (units) are needed.Answer: At least 174 table tennis balls are needed.The answer is: 174.

🧠:[<think>Alright, so I have this geometry problem here. Let me read it again to make sure I get all the details.There's a circle with center O and radius R. Inside this circle, there's a fixed point A, and the distance from O to A is a. The task is to fold the paper so that a point A' on the circumference of the circle coincides exactly with point A. Each fold leaves a straight line crease. We need to find the set of all such crease lines as A' moves around the circumference of the circle.Hmm, okay. So, folding the paper such that a point A' on the circle overlaps with point A inside the circle. Each time we do this, there's a crease line. We need to find the collection of all these crease lines.First, I need to recall how folding a paper maps one point to another. When you fold a paper so that two points coincide, the crease line is the perpendicular bisector of the segment joining those two points. Is that right? Let me confirm. Yes, because folding over the perpendicular bisector would reflect one point to the other. So, the crease line is indeed the perpendicular bisector of AA' when we fold A' onto A.Therefore, for each point A' on the circumference, the crease line is the perpendicular bisector of the segment AA'. So, if I can find the locus of all these perpendicular bisectors as A' moves around the circle, that should give me the set of all crease lines.But wait, the set of points on all the crease lines... Hmm. The problem says "find the set of points on all the crease lines." So, not the lines themselves, but all the points that lie on any of these crease lines. So, the union of all the crease lines. That's a bit different. So, I need to find the envelope or the union of all these perpendicular bisectors as A' moves around the circle.Alternatively, maybe it's a specific curve that all these crease lines touch or form. Let me think.Alternatively, perhaps it's another circle or an ellipse or something like that. Let me try to visualize.Given that A is inside the circle, and A' is on the circumference, each crease is the perpendicular bisector of AA'. So, each crease line is the set of points equidistant from A and A'. So, all points on the crease line are equidistant to A and A'. Therefore, the crease line is the locus of points equidistant to A and A', which is the perpendicular bisector.So, if we take all such perpendicular bisectors for every A' on the circle, what is the union of all these lines? Hmm. Alternatively, perhaps the set of all points that are equidistant to A and some point on the circle. So, the union would be the set of all points P such that there exists a point A' on the circle where PA = PA'. So, the set of points P for which PA = PA' for some A' on the circle.Alternatively, maybe the set of points P such that the difference of distances from P to O and P to A is constant? Wait, not sure.Alternatively, if I consider the set of all points equidistant from A and a moving point on the circle. So, the locus of points P such that PA = PA' where A' is on the circle. So, this is like the set of points P where PA' - PA = 0 for some A' on the circle. But A' is moving on the circle. Hmm. Maybe this forms another conic section.Alternatively, maybe I can parametrize point A' on the circle. Let's suppose coordinates. Let me place the circle with center O at the origin for simplicity. Let’s set up a coordinate system where O is at (0,0), and point A is at (a,0) since OA = a. Then the circle has equation x² + y² = R². Point A is (a,0). Then, any point A' on the circumference can be represented as (R cos θ, R sin θ) where θ is the angle parameter.Now, the crease line is the perpendicular bisector of segment AA'. Let's find the equation of this perpendicular bisector.First, midpoint M of AA' is ((a + R cos θ)/2, (0 + R sin θ)/2).The slope of AA' is (R sin θ - 0)/(R cos θ - a) = (R sin θ)/(R cos θ - a). Therefore, the slope of the perpendicular bisector is the negative reciprocal, which is -(R cos θ - a)/(R sin θ).So, the equation of the perpendicular bisector (the crease line) is:(y - (R sin θ)/2) = [-(R cos θ - a)/(R sin θ)] (x - (a + R cos θ)/2)This seems complicated, but maybe we can simplify it or find an envelope.Alternatively, perhaps it's easier to consider the set of all such crease lines. The problem is to find the set of all points lying on any of these crease lines.Alternatively, perhaps each crease line is tangent to a particular conic. For instance, the envelope of these crease lines might form an ellipse or hyperbola.Alternatively, let's consider that for each crease line, which is the set of points equidistant from A and A', so all points P such that PA = PA'. So, the crease line is the perpendicular bisector between A and A'.The set of all points P such that PA = PA' for some A' on the circle.Alternatively, for a given P, if there exists an A' on the circle such that PA = PA', then P lies on the perpendicular bisector of AA' for some A' on the circle.Alternatively, this is equivalent to saying that P lies on the radical axis of the circle centered at A with radius PA and the original circle. But since PA = PA', the radical axis would be the set of points P where power with respect to both circles is equal.Wait, perhaps I need to think in terms of the locus of P such that there exists A' on the original circle where PA = PA'. Let me write this equation:PA = PA'PA' is the distance from P to A', which is on the original circle. So, PA'^2 = (x - R cos θ)^2 + (y - R sin θ)^2, and PA^2 = (x - a)^2 + y^2.Set them equal:(x - R cos θ)^2 + (y - R sin θ)^2 = (x - a)^2 + y^2Expand both sides:Left side: x² - 2xR cos θ + R² cos² θ + y² - 2yR sin θ + R² sin² θRight side: x² - 2ax + a² + y²Subtract right side from left side:(-2xR cos θ + R² cos² θ - 2yR sin θ + R² sin² θ) - (-2ax + a²) = 0Simplify:-2xR cos θ - 2yR sin θ + R² (cos² θ + sin² θ) + 2ax - a² = 0Since cos² θ + sin² θ = 1, this becomes:-2xR cos θ - 2yR sin θ + R² + 2ax - a² = 0Rearranged:2ax - 2xR cos θ - 2yR sin θ + (R² - a²) = 0Divide both sides by 2:ax - xR cos θ - yR sin θ + (R² - a²)/2 = 0Let me write this as:x(a - R cos θ) + y(-R sin θ) + (R² - a²)/2 = 0This is the equation of the crease line for a particular θ. Now, the set of all such lines as θ varies is what we need, and the union of all these lines is the set of points we are to find.But how do we find the union of all such lines? Maybe by finding an equation that is satisfied by all points lying on any of these lines. Let's see.Suppose (x, y) is a point that lies on at least one crease line. Then there exists some θ such that:x(a - R cos θ) + y(-R sin θ) + (R² - a²)/2 = 0Let me rearrange this equation:x(a) - x R cos θ - y R sin θ + (R² - a²)/2 = 0Bring terms with θ to one side:- x R cos θ - y R sin θ = -x a - (R² - a²)/2Multiply both sides by -1:x R cos θ + y R sin θ = x a + (R² - a²)/2Let me denote the right-hand side as a constant for given x and y:x R cos θ + y R sin θ = C, where C = x a + (R² - a²)/2Now, the equation x R cos θ + y R sin θ = C must have a solution for θ. The maximum value of the left-hand side is sqrt( (x R)^2 + (y R)^2 ) by the Cauchy-Schwarz inequality, since cos θ and sin θ are coefficients. So, for this equation to have a solution θ, we need that |C| ≤ sqrt( (x R)^2 + (y R)^2 )Therefore:|x a + (R² - a²)/2| ≤ sqrt( (x R)^2 + (y R)^2 )Square both sides to eliminate the absolute value and the square root:(x a + (R² - a²)/2)^2 ≤ (x R)^2 + (y R)^2Expand the left-hand side:x² a² + x a (R² - a²) + ( (R² - a²)/2 )² ≤ x² R² + y² R²Bring all terms to the right-hand side:0 ≤ x² R² + y² R² - x² a² - x a (R² - a²) - ( (R² - a²)^2 ) /4Factor x² terms:x² (R² - a²) + y² R² - x a (R² - a²) - ( (R² - a²)^2 ) /4 ≥ 0Factor out (R² - a²) from the x terms:(R² - a²)(x² - a x) + y² R² - ( (R² - a²)^2 ) /4 ≥ 0Let me complete the square for the x terms. Let's look at x² - a x:x² - a x = x² - a x + (a²)/4 - (a²)/4 = (x - a/2)^2 - (a²)/4Therefore:(R² - a²)[ (x - a/2)^2 - (a²)/4 ] + y² R² - ( (R² - a²)^2 ) /4 ≥ 0Expand this:(R² - a²)(x - a/2)^2 - (R² - a²)(a²)/4 + y² R² - (R² - a²)^2 /4 ≥ 0Let me collect like terms:(R² - a²)(x - a/2)^2 + y² R² - [ (R² - a²)a² + (R² - a²)^2 ] /4 ≥ 0Factor out (R² - a²) from the terms in the brackets:(R² - a²)(x - a/2)^2 + y² R² - (R² - a²)(a² + R² - a²)/4 ≥ 0Simplify the term inside the brackets:a² + R² - a² = R², so:(R² - a²)(x - a/2)^2 + y² R² - (R² - a²)(R²)/4 ≥ 0Factor out (R² - a²) from the first and third terms:(R² - a²)[ (x - a/2)^2 - R²/4 ] + y² R² ≥ 0Hmm, this is getting a bit messy. Maybe there's a better approach. Let me think.Alternatively, let's consider that the crease lines are the perpendicular bisectors of AA', so the set of points equidistant from A and A'. Therefore, the set of all such points P where PA = PA' for some A' on the circle. So, this is equivalent to the locus of points P such that there exists a point A' on the circle with PA = PA'.Alternatively, the set of points P such that the circle centered at P with radius PA intersects the original circle ⊙O. Because if there's a point A' on both circles, then PA = PA' and OA' = R. So, the intersection of two circles: ⊙P with radius PA and ⊙O with radius R must be non-empty. Therefore, the condition is that the distance between P and O is less than or equal to R + PA, and greater than or equal to |R - PA|.But PA is variable here because P is arbitrary. Wait, perhaps not. Wait, in this problem, for each P, PA is fixed, but we are looking for P such that there exists A' on ⊙O where PA = PA'. So, the circles ⊙P (radius PA) and ⊙O (radius R) must intersect. The set of all such P is the locus where for each P, the distance from P to O satisfies |R - PA| ≤ PO ≤ R + PA.But I need to relate this to find the equation. Let me consider coordinates again. Let’s place O at (0,0), A at (a,0). Let P be (x,y). Then PA = sqrt( (x - a)^2 + y^2 ), and PO = sqrt( x^2 + y^2 ). The condition is that there exists A' on ⊙O such that PA = PA', which is equivalent to the intersection of ⊙O and ⊙P (radius PA). So, the distance between centers PO must satisfy |R - PA| ≤ PO ≤ R + PA.So,|R - PA| ≤ PO ≤ R + PABut PA = sqrt( (x - a)^2 + y^2 ) and PO = sqrt( x^2 + y^2 )So,| R - sqrt( (x - a)^2 + y^2 ) | ≤ sqrt( x^2 + y^2 ) ≤ R + sqrt( (x - a)^2 + y^2 )This seems complicated, but maybe squaring both sides of the inequalities could help. Let's first consider the right inequality:sqrt( x^2 + y^2 ) ≤ R + sqrt( (x - a)^2 + y^2 )Square both sides:x² + y² ≤ R² + 2 R sqrt( (x - a)^2 + y^2 ) + (x - a)^2 + y²Simplify:x² + y² ≤ R² + 2 R sqrt( (x - a)^2 + y^2 ) + x² - 2 a x + a² + y²Cancel x² and y²:0 ≤ R² + 2 R sqrt( (x - a)^2 + y^2 ) - 2 a x + a²Rearranged:2 R sqrt( (x - a)^2 + y^2 ) ≥ 2 a x - a² - R²Divide both sides by 2 R:sqrt( (x - a)^2 + y^2 ) ≥ (2 a x - a² - R²)/(2 R )But I'm not sure if this leads me anywhere. Let's check the left inequality:| R - sqrt( (x - a)^2 + y^2 ) | ≤ sqrt( x^2 + y^2 )Which gives two cases:Case 1: R - sqrt( (x - a)^2 + y^2 ) ≤ sqrt( x^2 + y^2 )Case 2: sqrt( (x - a)^2 + y^2 ) - R ≤ sqrt( x^2 + y^2 )Case 1: R ≤ sqrt( (x - a)^2 + y^2 ) + sqrt( x^2 + y^2 )Which is always true because by triangle inequality, sqrt( (x - a)^2 + y^2 ) + sqrt( x^2 + y^2 ) ≥ |a|, but since R is a positive constant and a < R (since A is inside the circle), this inequality would always hold?Wait, maybe not necessarily. Hmm.Case 2: sqrt( (x - a)^2 + y^2 ) - R ≤ sqrt( x^2 + y^2 )Rearranged:sqrt( (x - a)^2 + y^2 ) ≤ sqrt( x^2 + y^2 ) + RWhich is also always true by triangle inequality. So, the left inequality is always satisfied. Therefore, the real condition comes from the right inequality.Wait, but maybe I made a mistake here. The original condition is that there exists A' on the circle such that PA = PA'. Therefore, the two circles ⊙O and ⊙P must intersect. The necessary and sufficient condition for two circles to intersect is that the distance between their centers is less than or equal to the sum of their radii and greater than or equal to the absolute difference of their radii.In this case, the circles are ⊙O with radius R and ⊙P with radius PA. Therefore, the condition is |R - PA| ≤ PO ≤ R + PA.But since we need that for some A', i.e., that such a point exists, then this condition must hold. Therefore, the set of points P such that |R - PA| ≤ PO ≤ R + PA.But how do we translate this into an equation?Wait, maybe the boundary of this region is when PO = R + PA or PO = |R - PA|. But since PA is variable, but in our problem, PA is fixed for each P. Wait, no. Wait, for each P, PA is fixed, and we are looking for P such that there exists A' on ⊙O with PA' = PA. So, the intersection condition is |R - PA| ≤ PO ≤ R + PA. So, this is a region in the plane bounded by the loci PO = R + PA and PO = |R - PA|.So, the set of all points P such that PO ≤ R + PA and PO ≥ |R - PA|.But we need to find the union of all crease lines, which are the perpendicular bisectors of AA' where A' is on ⊙O. So, the crease lines are precisely the radical axes of the circles ⊙O and ⊙P (with radius PA) when they intersect. The radical axis of two intersecting circles is the set of points with equal power with respect to both circles, which is the line we found earlier.But the union of all radical axes as P varies such that ⊙O and ⊙P intersect... Wait, but in our problem, P is varying such that there exists A' with PA' = PA, so P is such that ⊙P (radius PA) intersects ⊙O. So, the radical axes (crease lines) are the lines along which the points have equal power with respect to ⊙O and ⊙P. But the union of all such radical axes would be the set of all points that lie on any radical axis of ⊙O and some ⊙P where ⊙P intersects ⊙O.Alternatively, perhaps the union of all such lines forms another conic section. For instance, an ellipse with foci at O and A.Wait, if we consider that for each crease line (radical axis), the set of points equidistant to O and A might relate to an ellipse or hyperbola. Wait, the radical axis is the set of points with equal power with respect to both circles. The power with respect to ⊙O is PO² - R², and the power with respect to ⊙P is PA² - PA² = 0. Wait, no: the power with respect to ⊙P would be PP² - PA², but PP is zero, so it's -PA². Wait, maybe I confused the circles. Let's clarify:The radical axis between ⊙O (center O, radius R) and ⊙P (center P, radius PA) is the set of points X such that XO² - R² = XP² - PA².So, XO² - XP² = R² - PA².But this is the equation of the radical axis. For each P, this is a line. The union of all such lines as P moves such that ⊙O and ⊙P intersect (i.e., |PO - PA| ≤ R ≤ PO + PA) would form a surface. However, we need to find the union of all these lines.Alternatively, perhaps we can express this equation in terms of coordinates. Let me try again with coordinates.Let O be at (0,0) and A be at (a,0). Let P be a point (h,k). Then, the radical axis between ⊙O and ⊙P is given by:XO² - R² = XP² - PA²For any point X on the radical axis:XO² - XP² = R² - PA²Expressed in coordinates:(x² + y²) - [ (x - h)^2 + (y - k)^2 ] = R² - [ (h - a)^2 + k² ]Expand the left side:x² + y² - (x² - 2 h x + h² + y² - 2 k y + k²) = R² - (h² - 2 a h + a² + k²)Simplify left side:2 h x + 2 k y - h² - k² = R² - h² + 2 a h - a² - k²Cancel terms:2 h x + 2 k y = R² + 2 a h - a²Divide both sides by 2:h x + k y = (R² + 2 a h - a²)/2But h and k are coordinates of P, which must satisfy |PO - PA| ≤ R ≤ PO + PA. Wait, this might not be helpful. Alternatively, since for each crease line (radical axis), there exists a point P (the center of ⊙P) such that this equation holds. However, the problem is to find all points X that lie on any such radical axis. So, for each X, there exists a P such that h x + k y = (R² + 2 a h - a²)/2 and |PO - PA| ≤ R ≤ PO + PA.This seems too abstract. Maybe there is another way.Wait, let's recall that the set of points equidistant from A and A' (for A' on ⊙O) is the same as the set of points P such that PA = PA'. This is equivalent to P lying on the perpendicular bisector of AA', as we thought earlier. Therefore, the set of all such points P is the union of all perpendicular bisectors of segments joining A to points A' on ⊙O.Alternatively, consider that for each A', the perpendicular bisector of AA' is the set of points equidistant to A and A'. Therefore, the union of all these bisectors is the set of points P such that there exists a point A' on ⊙O with PA = PA'.Alternatively, this is equivalent to the set of points P for which the circle centered at P with radius PA intersects the original circle ⊙O. So, the set of points P where ⊙P (radius PA) and ⊙O intersect.Therefore, the locus of points P such that the distance between P and O is less than or equal to R + PA and greater than or equal to |R - PA|.But how to convert this condition into an equation?Alternatively, maybe we can think of P such that there exists A' on ⊙O where PA = PA', which implies that P lies on the perpendicular bisector of OA'.Wait, OA' is a radius of the circle. The perpendicular bisector of OA' would pass through the midpoint of OA' and be perpendicular to OA'. But OA' is a radius, so the midpoint is (R/2 cos θ, R/2 sin θ), and the slope is - (R cos θ)/(R sin θ) = - cot θ. Wait, but OA' is from O to A', which is (R cos θ, R sin θ). The perpendicular bisector of OA' is the line perpendicular to OA' at its midpoint. However, in our problem, the crease line is the perpendicular bisector of AA', not OA'.Wait, maybe another approach. Since P lies on the perpendicular bisector of AA', then PA = PA'. Let me consider vectors. Let O be at the origin, A at (a,0), A' at (R cos θ, R sin θ). Then, the midpoint M of AA' is [(a + R cos θ)/2, (0 + R sin θ)/2]. The direction of AA' is (R cos θ - a, R sin θ), so the direction of the perpendicular bisector is (-R sin θ, R cos θ - a). So, the crease line has direction perpendicular to AA'.But I need the equation of the crease line. Wait, earlier we had:(y - (R sin θ)/2) = [-(R cos θ - a)/(R sin θ)] (x - (a + R cos θ)/2)But maybe instead of trying to find the envelope, we can parametrize θ and find the equation.Alternatively, notice that the set of crease lines are all the tangents to a certain conic. For instance, the envelope of these lines might be an ellipse. To check this, we can compute the envelope.The equation of the crease line was:ax - x R cos θ - y R sin θ + (R² - a²)/2 = 0Let me write this as:x(a - R cos θ) - y R sin θ + (R² - a²)/2 = 0Let me denote this as L(θ) = 0. To find the envelope of the family of lines L(θ), we can use the method of solving for L(θ) = 0 and dL/dθ = 0.So, compute the derivative of L with respect to θ:dL/dθ = x R sin θ - y R cos θ = 0So, the envelope is the set of points (x, y) that satisfy both:x(a - R cos θ) - y R sin θ + (R² - a²)/2 = 0andx R sin θ - y R cos θ = 0Let me solve these equations.From the second equation: x R sin θ = y R cos θ => y = x tan θLet me substitute y = x tan θ into the first equation.First equation:x(a - R cos θ) - (x tan θ) R sin θ + (R² - a²)/2 = 0Simplify:x a - x R cos θ - x R sin θ tan θ + (R² - a²)/2 = 0Note that sin θ tan θ = sin θ (sin θ / cos θ) = sin² θ / cos θSo,x a - x R cos θ - x R (sin² θ / cos θ) + (R² - a²)/2 = 0Combine the terms with x R cos θ:x a - x R [ cos θ + (sin² θ)/cos θ ] + (R² - a²)/2 = 0Simplify inside the brackets:cos θ + (sin² θ)/cos θ = (cos² θ + sin² θ)/cos θ = 1/cos θTherefore:x a - x R / cos θ + (R² - a²)/2 = 0Rearranged:x (a - R / cos θ) = - (R² - a²)/2Therefore,x = [ - (R² - a²)/2 ] / (a - R / cos θ )Hmm, this seems a bit complicated. Let me see if I can express cos θ in terms of y. From the second equation, y = x tan θ, so tan θ = y / x. Therefore, cos θ = x / sqrt(x² + y²). Wait, but tan θ = y / x implies θ is the angle made by the point (x, y). Wait, but θ here is the parameter for the point A' on the circle. Hmm, maybe not directly related.Alternatively, from y = x tan θ, we can express sin θ = y / sqrt(x² + y²) and cos θ = x / sqrt(x² + y²). But θ here is the angle parameter for A', so perhaps different from the angle of the point (x, y). Hmm, maybe this substitution is not valid.Wait, but in the envelope equations, for each θ, the point (x, y) lies on both L(θ) and dL/dθ = 0. So, (x, y) depends on θ, but we need to eliminate θ to find the envelope.Let me denote t = θ. Then, from the second equation:x R sin t - y R cos t = 0 => x sin t - y cos t = 0 => y = x tan tSo, we can write y = x tan t. Then, substitute into the first equation:x(a - R cos t) - y R sin t + (R² - a²)/2 = 0Replace y with x tan t:x(a - R cos t) - x tan t * R sin t + (R² - a²)/2 = 0Simplify the second term:x tan t * R sin t = x R sin t * (sin t / cos t) = x R sin² t / cos tSo, the equation becomes:x a - x R cos t - x R sin² t / cos t + (R² - a²)/2 = 0Factor x:x [ a - R cos t - R sin² t / cos t ] + (R² - a²)/2 = 0Simplify the expression in the brackets:a - R cos t - R sin² t / cos t = a - R [ cos t + sin² t / cos t ]As before, cos t + sin² t / cos t = (cos² t + sin² t)/cos t = 1 / cos tTherefore,a - R / cos tSo, the equation is:x ( a - R / cos t ) + (R² - a²)/2 = 0Therefore,x ( a - R / cos t ) = - (R² - a²)/2Solve for cos t:x a - x R / cos t = - (R² - a²)/2Bring x a to the right:- x R / cos t = - (R² - a²)/2 - x aMultiply both sides by -1:x R / cos t = (R² - a²)/2 + x aTherefore,cos t = x R / [ (R² - a²)/2 + x a ]Similarly, from y = x tan t, we can express tan t = y / x, so sin t = y / sqrt(x² + y²) and cos t = x / sqrt(x² + y²), but here we have cos t expressed in terms of x. Let me see.Let me denote:cos t = x R / [ (R² - a²)/2 + x a ]But we also know that sin² t + cos² t = 1.From y = x tan t, tan t = y / x => sin t = (y / x) cos tTherefore,sin t = (y / x) * [ x R / ( (R² - a²)/2 + x a ) ] = y R / ( (R² - a²)/2 + x a )Now, using sin² t + cos² t = 1:[ y R / ( (R² - a²)/2 + x a ) ]² + [ x R / ( (R² - a²)/2 + x a ) ]² = 1Factor out [ R / ( (R² - a²)/2 + x a ) ]²:[ R² / ( (R² - a²)/2 + x a )² ] ( y² + x² ) = 1Therefore,R² (x² + y² ) = [ (R² - a²)/2 + x a ]²Expand the right-hand side:(R² - a²)^2 /4 + x a (R² - a²) + x² a²Therefore,R² (x² + y² ) = (R² - a²)^2 /4 + x a (R² - a²) + x² a²Bring all terms to the left-hand side:R² x² + R² y² - (R² - a²)^2 /4 - x a (R² - a²) - x² a² = 0Factor x² terms:x² ( R² - a² ) + R² y² - x a ( R² - a² ) - ( R² - a² )² /4 = 0Factor out ( R² - a² ) from the first, third, and fourth terms:( R² - a² ) [ x² - x a - ( R² - a² ) /4 ] + R² y² = 0Now, let's complete the square for the x terms inside the brackets:x² - x a - ( R² - a² ) /4Complete the square for x² - x a:x² - x a = x² - x a + (a²)/4 - (a²)/4 = (x - a/2)^2 - (a²)/4So,( R² - a² ) [ (x - a/2)^2 - (a²)/4 - ( R² - a² ) /4 ] + R² y² = 0Simplify inside the brackets:( x - a/2 )^2 - [ a² + R² - a² ] /4 = ( x - a/2 )^2 - R² /4Therefore, the equation becomes:( R² - a² ) ( (x - a/2)^2 - R² /4 ) + R² y² = 0Expand this:( R² - a² )(x - a/2)^2 - ( R² - a² ) R² /4 + R² y² = 0Multiply through by 4 to eliminate denominators:4( R² - a² )(x - a/2)^2 - ( R² - a² ) R² + 4 R² y² = 0Rearrange terms:4( R² - a² )(x - a/2)^2 + 4 R² y² = ( R² - a² ) R²Divide both sides by ( R² - a² ) (assuming R ≠ a, which is true since A is inside the circle, so a < R ):4 (x - a/2)^2 + (4 R² / ( R² - a² )) y² = R²Divide both sides by 4:(x - a/2)^2 + ( R² / ( R² - a² )) y² = R² /4Multiply numerator and denominator of the y-term by 1:(x - a/2)^2 / ( R² /4 ) + y² / ( ( R² - a² ) /4 ) = 1This is the equation of an ellipse centered at (a/2, 0) with semi-major axis R/2 and semi-minor axis sqrt( ( R² - a² ) ) /2.Therefore, the envelope of the crease lines is an ellipse with center at (a/2, 0), major axis length R, minor axis length sqrt( R² - a² ), and equation:(x - a/2)^2 / ( R² /4 ) + y² / ( ( R² - a² ) /4 ) = 1Simplify denominators:4(x - a/2)^2 / R² + 4 y² / ( R² - a² ) = 1But in standard form, it's:( (x - a/2)^2 ) / ( (R/2)^2 ) + ( y^2 ) / ( ( sqrt(R² - a²)/2 )^2 ) = 1Yes, so this is an ellipse with center at (a/2, 0), semi-major axis R/2 along the x-axis, and semi-minor axis sqrt(R² - a²)/2.Therefore, the set of all crease lines forms an ellipse as their envelope. But the problem says "find the set of points on all the crease lines." The envelope is the set of points where each crease line is tangent to the ellipse. However, the union of all crease lines would cover the area inside the ellipse? Wait, but in reality, each crease line is a tangent to this ellipse. Therefore, the set of all points on the crease lines is the entire ellipse itself, because each point on the ellipse is touched by exactly one crease line (tangent). But wait, no—the envelope is the ellipse, which is the set of points where each crease line touches the ellipse, but the crease lines themselves extend infinitely. However, the problem might be considering the crease lines only within the original circle or the paper. Wait, the problem doesn't specify, but since it's a fold of the paper, the crease line would be the entire line, but the physical paper is a plane, so all crease lines are straight lines in the plane. The set of all points on all crease lines is the union of all these tangent lines to the ellipse. But the union of all tangent lines to an ellipse is the entire plane except the interior of the ellipse. Because each tangent line touches the ellipse at one point, and lines extend to infinity. But in our case, the crease lines are produced by folding the paper so that A' maps to A. However, physically, the crease line would be the intersection of two planes when folding, but in the 2D plane, it's just the line. However, since all crease lines are tangents to the ellipse, their union would cover the plane except the inside of the ellipse. But that can't be, since for example, points near O might not be on any crease line. Wait, but according to the previous analysis, the envelope is the ellipse, but the union of all tangents to an ellipse is the entire plane except the interior of the ellipse. However, this contradicts with the physical intuition. Maybe the correct answer is the ellipse itself.Wait, no. Let me think again. The crease lines are the perpendicular bisectors of AA'. Each crease line is the set of points equidistant to A and A'. So, if a point lies on any crease line, then it is equidistant to A and some A' on the circle. Therefore, the set of all such points is the union of all perpendicular bisectors of segments AA', which are the crease lines. This is the same as the set of points P such that there exists a point A' on the circle with PA = PA'.Now, according to our previous derivation, this set of points is the ellipse given by the equation above. However, the envelope of the crease lines is the ellipse, but the union of the crease lines (all points on any crease line) is not the ellipse, but all points outside or on the ellipse. Wait, but when we derived the envelope, we found that the crease lines are tangent to the ellipse. So, each crease line touches the ellipse at exactly one point, but extends infinitely in both directions. Therefore, the union of all crease lines would cover the region outside the ellipse, but not sure. Wait, but if you have all tangent lines to an ellipse, then every point outside the ellipse lies on two tangent lines, and every point on the ellipse lies on one tangent line, and every point inside the ellipse lies on no tangent line. Therefore, the union of all crease lines (which are the tangent lines to the ellipse) would be all points outside and on the ellipse. But this contradicts with the previous idea that the set of points P such that there exists A' on the circle with PA = PA' is the ellipse. But how?Wait, there's a confusion here. The set of points equidistant to A and some A' on the circle (i.e., lying on the perpendicular bisector of AA') is not the same as the set of points P such that PA = PA' for some A' on the circle. Wait, actually, they are the same. Because for a point P to be equidistant to A and A', it must lie on the perpendicular bisector of AA'. Therefore, the union of all crease lines is precisely the set of points P such that PA = PA' for some A' on the circle. Which we found is the ellipse. Wait, but no. If P is on the perpendicular bisector of AA', then PA = PA', but different A' give different P? Wait, no. Each crease line is a set of points equidistant to A and a particular A'. So, each crease line is a line of points equidistant to A and A'. The union of all such lines is all points that are equidistant to A and some A' on the circle. Which is equivalent to saying all points P such that there exists A' on the circle with PA = PA'. But according to the previous analysis, this is the ellipse. Wait, but how is that possible? Because for each A', the crease line (perpendicular bisector) is a line, and the union of all such lines would be a 2D region, but according to the envelope, it's an ellipse. But there must be a mistake here.Wait, no. Wait, in the problem, when you fold the paper so that A' maps to A, the crease line is the perpendicular bisector of AA'. So, each crease line is a line in the plane. The set of all crease lines is the collection of all such perpendicular bisectors as A' moves around the circle. The question is asking for the set of all points that lie on any of these crease lines. So, the union of all these lines.However, according to the previous calculation, this union is the set of points P such that PA = PA' for some A' on the circle. Which, in turn, corresponds to an ellipse. Therefore, the set of all such points P forms an ellipse. Therefore, the answer is an ellipse.Wait, but an ellipse is a 1-dimensional curve, but the union of lines is a 2-dimensional region. There's a contradiction here. Therefore, my mistake must be in assuming that PA = PA' implies P is on the ellipse, but actually, PA = PA' for some A' on the circle is a different condition.Wait, let me clarify:If P is equidistant to A and A', then PA = PA', so P is on the perpendicular bisector of AA'. So, the union of all such P for all A' is the set of all P such that there exists A' on the circle with PA = PA'.But this is equivalent to the set of points P where the circle centered at P with radius PA intersects the original circle ⊙O. The set of such points P is called the set of centers of circles intersecting ⊙O with radius PA, which depends on P.Wait, this is getting too abstract. Let me consider specific points. For example, take A at (a, 0). What is the set of points P such that PA = PA' for some A' on ⊙O.If A is at (a,0), then when A' is at (R,0), the perpendicular bisector of AA' is the vertical line x = (a + R)/2. So, points on this line are equidistant to A and A'=(R,0).Similarly, when A' is at (-R,0), the perpendicular bisector is x = (a - R)/2.When A' is at (0,R), the perpendicular bisector is the line that's the perpendicular bisector of the segment from (a,0) to (0,R). The midpoint is (a/2, R/2), and the slope of AA' is (R - 0)/(0 - a) = -R/a, so the slope of the perpendicular bisector is a/R. Therefore, the equation is y - R/2 = (a/R)(x - a/2).As A' moves around the circle, these perpendicular bisectors form various lines. The envelope of these lines is an ellipse, as we derived earlier. However, the union of all these lines would cover the entire plane except for the interior of this ellipse. Because for any point outside the ellipse, there are two crease lines passing through it (corresponding to two points A' on the circle), and for points on the ellipse, there is exactly one crease line tangent at that point, and for points inside the ellipse, there are no crease lines.But according to the problem statement, we are to find the set of points on all the crease lines. If crease lines are only the tangent lines to the ellipse, then the set of points on these lines would be all points on the tangents, which is the entire plane except the interior of the ellipse. But the problem says "When A' covers all points on the circumference, find the set of points on all the crease lines." So, likely, the answer is the ellipse itself. But according to the previous analysis, the envelope is the ellipse. However, the envelope is the set of points where each crease line is tangent, which is a subset of the union of all crease lines.But maybe the problem is actually asking for the envelope, i.e., the set of all points where the crease lines touch, which is the ellipse. But the problem says "the set of points on all the crease lines," which would be all points lying on any crease line, which is the union of all crease lines. If each crease line is a tangent to the ellipse, then the union of all tangents to an ellipse is the entire plane except the interior of the ellipse. But this contradicts with the mathematical result that the set of points P such that PA = PA' for some A' is the ellipse.Wait, no. Let me re-examine the equations. When we derived the condition, we found that the set of points P such that there exists A' with PA = PA' is the ellipse given by:4(x - a/2)^2 / R² + 4 y² / ( R² - a² ) = 1But how does this reconcile with the idea that the union of crease lines (perpendicular bisectors) is the set of points on those lines?There must be a misunderstanding. Let me think again. If P is a point on any crease line, then P is equidistant to A and some A' on the circle. Therefore, such P must satisfy PA = PA', which is exactly the condition we analyzed. However, when we solved that, we ended up with the ellipse equation. Therefore, the set of all such points P forms an ellipse. However, this is contradictory because a crease line is a line, which contains infinitely many points, each of which is equidistant to A and some A' (but different points on the crease line would correspond to different A's? Wait, no. Wait, for a given crease line (perpendicular bisector of AA'), every point on that line is equidistant to A and A'. So, all points on that line satisfy PA = PA'. Therefore, the union of all crease lines is the set of all points P such that PA = PA' for some A' on the circle. Which is exactly the same as the ellipse we derived. Therefore, this suggests that the union of all crease lines is the ellipse. But how? A line is a one-dimensional set, and the union of lines would typically be a two-dimensional set, unless all lines are tangent to a curve. But according to the math, the set of all points P with PA = PA' for some A' is an ellipse. Therefore, the answer must be an ellipse.Wait, perhaps I was confused earlier. If every point on the ellipse is equidistant to A and some A' on the circle, then the ellipse is precisely the set of points P such that PA = PA' for some A'. But the crease lines are the lines that consist of all points equidistant to A and a fixed A'. Therefore, each crease line corresponds to a particular A', and is the set of points equidistant to A and A'. Therefore, the union of all crease lines is the set of points P such that PA = PA' for some A', which is the ellipse. Therefore, the answer is the ellipse.Therefore, the set of points on all the crease lines is the ellipse with center at (a/2, 0), major axis R/2, and minor axis sqrt(R² - a²)/2.But let me confirm with a specific example. Let's take R = 2, a = 1. Then the center of the ellipse is at (0.5, 0), major axis 1 (since R/2 = 1), and minor axis sqrt(4 - 1)/2 = sqrt(3)/2 ≈ 0.866. So, the ellipse equation is:( (x - 0.5)^2 ) / 1 + ( y^2 ) / ( 3/4 ) = 1Which simplifies to:(x - 0.5)^2 + (4/3) y² = 1Let's pick a point on this ellipse. For example, take x = 0.5, y = sqrt(3)/2. This point satisfies (0 + (4/3)(3/4)) = 1, which is correct. Now, check if there exists an A' such that PA = PA'. P = (0.5, sqrt(3)/2). PA = distance from (0.5, sqrt(3)/2) to A=(1,0):sqrt( (0.5 - 1)^2 + ( sqrt(3)/2 - 0 )^2 ) = sqrt( 0.25 + 0.75 ) = sqrt(1) = 1PA' should be 1. So, A' must be a point on the original circle (radius 2) such that distance from P to A' is 1. Let's check if such a point exists. The circle around P with radius 1 must intersect the original circle (radius 2 centered at O=(0,0)). The distance between O and P is sqrt(0.5² + (sqrt(3)/2)^2 ) = sqrt(0.25 + 0.75) = sqrt(1) = 1. So, the distance between centers is 1, and radii are 1 and 2. Therefore, the circles intersect because 1 + 1 = 2, which is equal to the distance between centers. So, they touch externally at one point. Therefore, there is exactly one point A' where PA' = 1 and OA' = 2. Therefore, the point P lies on the crease line which is the perpendicular bisector of AA', where A' is that touching point. Therefore, this confirms that points on the ellipse are precisely those points that lie on a crease line. Therefore, the set of all points on the crease lines is the ellipse.Therefore, the answer should be an ellipse. The equation of the ellipse, as derived, is:(x - a/2)^2 / (R² /4) + y² / ((R² - a²)/4) = 1Simplify:( (x - a/2)^2 ) / (R/2)^2 + ( y^2 ) / ( (sqrt(R² - a²)/2 )^2 ) = 1Therefore, the set of points is an ellipse with center at the midpoint of OA, which is (a/2, 0), major axis R/2, minor axis sqrt(R² - a²)/2, and major axis along the line OA.Hence, the answer is an ellipse with center at the midpoint of OA, semi-major axis R/2, and semi-minor axis sqrt(R² - a²)/2.Final AnswerThe set of points forms an ellipse, which can be written as boxed{dfrac{(x - frac{a}{2})^2}{(frac{R}{2})^2} + dfrac{y^2}{(frac{sqrt{R^2 - a^2}}{2})^2} = 1}.</think>Given a circle (odot O) with radius (R) and a fixed point (A) inside the circle such that (OA = a), we need to find the set of points on all the crease lines formed by folding the paper so that a point (A') on the circumference coincides with (A).1. Understanding the Crease Line: - When folding the paper such that (A') coincides with (A), the crease line is the perpendicular bisector of the segment (AA'). - The crease line consists of all points equidistant from (A) and (A').2. Equation of the Perpendicular Bisector: - For a point (A') on the circumference of the circle, the midpoint (M) of (AA') is (left(frac{a + R cos theta}{2}, frac{R sin theta}{2}right)). - The slope of (AA') is (frac{R sin theta}{R cos theta - a}), so the slope of the perpendicular bisector is (-frac{R cos theta - a}{R sin theta}).3. Deriving the Envelope of Crease Lines: - The equation of the crease line (perpendicular bisector) is derived and simplified. - Using the method of envelopes, we solve for the set of points that satisfy both the crease line equation and its derivative with respect to (theta).4. Resulting Ellipse: - The envelope of the crease lines forms an ellipse. - The equation of the ellipse is derived using the conditions for the radical axis and the power of a point with respect to the circle.5. Equation of the Ellipse: - The ellipse is centered at (left(frac{a}{2}, 0right)) with a semi-major axis of (frac{R}{2}) and a semi-minor axis of (frac{sqrt{R^2 - a^2}}{2}).The final answer is the ellipse given by:[boxed{dfrac{(x - frac{a}{2})^2}{(frac{R}{2})^2} + dfrac{y^2}{(frac{sqrt{R^2 - a^2}}{2})^2} = 1}],<think>Okay, so there's this geometry problem here. Let me try to visualize it first. There's a circle with center O and radius R, and a fixed point A inside the circle. The distance from O to A is a, which is less than R since A is inside. Now, the task is to fold the paper such that a point A' on the circumference of the circle coincides exactly with point A. Each fold leaves a crease line, and we need to find the set of all such crease lines as A' moves around the circumference. Hmm, interesting.Alright, folding a point onto another point... I remember that when you fold a paper to make two points coincide, the crease line is the perpendicular bisector of the segment joining those two points. So in this case, if we're folding A' onto A, the crease should be the perpendicular bisector of the segment AA'. That makes sense. So for each A' on the circumference, the crease line is the perpendicular bisector of AA'.Now, the question is, as A' moves around the entire circumference of the circle, what is the set of all points that lie on any of these crease lines? So, essentially, the union of all these perpendicular bisectors. Hmm. I need to find the locus of all such crease lines.Let me think. Maybe I can parametrize point A'. Let's set up a coordinate system. Let me place point O at the origin for simplicity. Let’s assume that OA is along the x-axis. So point A has coordinates (a, 0). Then point A' can be any point on the circle of radius R, so its coordinates can be written as (R cos θ, R sin θ) where θ varies from 0 to 2π.The crease line is the perpendicular bisector of AA'. So, first, let's find the coordinates of the midpoint of AA'. The midpoint M would be ((a + R cos θ)/2, (0 + R sin θ)/2) = ((a + R cos θ)/2, (R sin θ)/2).The slope of AA' is (R sin θ - 0)/(R cos θ - a) = (R sin θ)/(R cos θ - a). Therefore, the slope of the perpendicular bisector (the crease line) is the negative reciprocal, which is (a - R cos θ)/(R sin θ). So the crease line has slope (a - R cos θ)/(R sin θ) and passes through the midpoint M.So, using point-slope form, the equation of the crease line is:y - (R sin θ)/2 = [(a - R cos θ)/(R sin θ)] (x - (a + R cos θ)/2)That's a bit complicated. Maybe we can simplify this equation.Multiply both sides by R sin θ to eliminate the denominator:(R sin θ)(y - (R sin θ)/2) = (a - R cos θ)(x - (a + R cos θ)/2)Let me expand both sides.Left side: R sin θ * y - (R sin θ * R sin θ)/2 = R sin θ * y - (R² sin² θ)/2Right side: (a - R cos θ)x - (a - R cos θ)(a + R cos θ)/2First, note that (a - R cos θ)(a + R cos θ) = a² - (R cos θ)² = a² - R² cos² θSo right side becomes: (a - R cos θ)x - (a² - R² cos² θ)/2Putting left and right sides together:R sin θ * y - (R² sin² θ)/2 = (a - R cos θ)x - (a² - R² cos² θ)/2Let me move all terms to the left side:R sin θ * y - (R² sin² θ)/2 - (a - R cos θ)x + (a² - R² cos² θ)/2 = 0Combine the constant terms:(-R² sin² θ + a² - R² cos² θ)/2 = (a² - R² (sin² θ + cos² θ))/2 = (a² - R²)/2So the equation simplifies to:- (a - R cos θ)x + R sin θ * y + (a² - R²)/2 = 0Multiply through by -1 to make it look cleaner:(a - R cos θ)x - R sin θ * y + (R² - a²)/2 = 0So the equation of the crease line is:(a - R cos θ)x - R sin θ * y + (R² - a²)/2 = 0Hmm. This is the equation for each crease line corresponding to a particular θ. To find the set of all such lines, maybe we can eliminate θ from the equation.Let me denote cos θ = c and sin θ = s. But we know that c² + s² = 1. Let's rewrite the equation:(a - R c)x - R s y + (R² - a²)/2 = 0Let me rearrange terms:(a x + (R² - a²)/2) = R c x + R s ySo:R c x + R s y = a x + (R² - a²)/2Let’s write this as:R (c x + s y) = a x + (R² - a²)/2Hmm. Let me see. The left side is R times the dot product of (x, y) with (c, s). Since (c, s) is a unit vector, this is R times the projection of (x, y) onto the direction of (c, s). But not sure if that helps.Alternatively, let me square both sides to eliminate θ. But since c = cos θ and s = sin θ, and they are related by c² + s² = 1, maybe squaring can help.But wait, before that, let's isolate the terms with c and s.Let me bring all terms to one side:R c x + R s y - a x - (R² - a²)/2 = 0Let me group terms with c and s:(R x) c + (R y) s - a x - (R² - a²)/2 = 0Let me denote this as:(R x) c + (R y) s = a x + (R² - a²)/2Let me think of this as an equation in variables c and s with the constraint c² + s² = 1. So this is a linear equation in c and s, which can be thought of as a line in the c-s plane. The solutions (c, s) must lie on both this line and the unit circle. So, there's a solution only if the line intersects the unit circle. Since θ is real, there must be solutions here. But maybe this approach can help.Alternatively, the equation can be written as:(R x) cos θ + (R y) sin θ = a x + (R² - a²)/2This resembles an equation of the form A cos θ + B sin θ = C, which can be solved for θ. The maximum and minimum values of A cos θ + B sin θ are ±√(A² + B²). Therefore, for the equation to have a solution θ, we must have |C| ≤ √(A² + B²). Therefore, in our case,|a x + (R² - a²)/2| ≤ √[(R x)^2 + (R y)^2]Square both sides:(a x + (R² - a²)/2)^2 ≤ R² x² + R² y²Expand the left side:a² x² + (R² - a²) a x + (R² - a²)^2 /4 ≤ R² x² + R² y²Bring all terms to the right:0 ≤ R² x² + R² y² - a² x² - (R² - a²) a x - (R² - a²)^2 /4Simplify the terms:(R² - a²) x² + R² y² - (R² - a²) a x - (R² - a²)^2 /4 ≥ 0Factor out (R² - a²) from the first and third terms:(R² - a²)(x² - a x) + R² y² - (R² - a²)^2 /4 ≥ 0Hmm, let's complete the square for the x terms in x² - a x.x² - a x = x² - a x + (a²/4) - (a²/4) = (x - a/2)^2 - (a²/4)So substitute back:(R² - a²)[(x - a/2)^2 - (a²/4)] + R² y² - (R² - a²)^2 /4 ≥ 0Expanding:(R² - a²)(x - a/2)^2 - (R² - a²)(a²/4) + R² y² - (R² - a²)^2 /4 ≥ 0Combine the constant terms:- (R² - a²)(a²/4) - (R² - a²)^2 /4 = - (R² - a²)/4 [a² + (R² - a²)] = - (R² - a²)/4 (R²) = - R² (R² - a²)/4So the inequality becomes:(R² - a²)(x - a/2)^2 + R² y² - R² (R² - a²)/4 ≥ 0Divide both sides by R² (assuming R ≠ 0, which it is since it's a circle):[(R² - a²)/R²](x - a/2)^2 + y² - (R² - a²)/4 ≥ 0Multiply through by 4 to eliminate denominators:4[(R² - a²)/R²](x - a/2)^2 + 4 y² - (R² - a²) ≥ 0Hmm, maybe this is the equation of an ellipse or some conic section. Let's see.Let me rearrange:4[(R² - a²)/R²](x - a/2)^2 + 4 y² ≥ (R² - a²)Divide both sides by (R² - a²):4/R² (x - a/2)^2 + 4 y² / (R² - a²) ≥ 1Hmm, so this resembles the equation of an ellipse. Wait, but it's an inequality. So the set of points (x, y) satisfying this inequality would lie on or outside the ellipse. But in our original problem, we are considering all crease lines, so the points on all crease lines must satisfy equality, because for each crease line, the points on it satisfy the equation when θ is such that the crease exists. Therefore, the equality case must hold, i.e.,4/R² (x - a/2)^2 + 4 y² / (R² - a²) = 1Wait, but let's check that. The inequality came from squaring both sides, which can sometimes introduce extraneous solutions, but since we derived the condition for θ to exist, the equality case would correspond to the envelope of all possible crease lines.So perhaps the set of all crease lines form the envelope of this ellipse. Wait, but an envelope of lines is typically a curve, but here the question says "the set of points on all the crease lines", which would be all points lying on any of the crease lines. If all crease lines are tangent to an ellipse, then the set of points on the crease lines would be the ellipse itself. Because each crease line touches the ellipse at exactly one point, and as the crease lines vary, their union is the ellipse.Wait, but actually, no. If the crease lines are all tangent to an ellipse, then the ellipse is the envelope of the crease lines, but the set of points on all crease lines would be all the points on all those tangent lines, which is the region covered by those lines. However, if the envelope is an ellipse, the lines are tangent to it, so the union of the lines is the area outside the ellipse? Wait, no. Wait, in the case of an ellipse, each tangent line touches the ellipse at one point, so the union of all tangent lines would be the set of all points that are on any tangent line to the ellipse, which is the entire plane except the interior of the ellipse. But in our case, maybe it's different. Wait, perhaps not. Let me think.Wait, in our problem, the crease lines are the perpendicular bisectors of AA' as A' moves around the circle. The set of all such perpendicular bisectors... I recall that the set of perpendicular bisectors of segments from a fixed point to points on a circle is an ellipse. Specifically, the ellipse with foci at O and A, but maybe not. Wait, actually, the set of midpoints M forms a circle called the midpoint circle. Wait, but the crease lines themselves, being the perpendicular bisectors, would envelope another conic.Alternatively, maybe the set of crease lines envelope an ellipse. Let me recall that the envelope of the perpendicular bisectors of segments joining a fixed point and a moving point on a circle is an ellipse. Yes, that seems familiar.Alternatively, here's another approach. The crease line is the set of points equidistant from A and A'. Therefore, the crease line is the locus of points P such that PA = PA'. So, if we consider all such crease lines as A' moves on the circumference, then the set of all points that lie on any crease line is the set of points P such that there exists a point A' on the circle where PA = PA'.Therefore, the union of all such crease lines is the set of points P for which there exists a point A' on the circle such that P is equidistant from A and A'. That is, PA = PA'.But this set of points P is the set of points where P lies on the perpendicular bisector of AA' for some A' on the circle. So, the question is asking for the union of all such perpendicular bisectors.Alternatively, this is equivalent to the set of centers of circles passing through A and tangent to the given circle ⊙O. Because if you have a circle passing through A and tangent to ⊙O at A', then its center must lie on the perpendicular bisector of AA' and also on the line OA' (since it must be equidistant from O and A' for tangency). Wait, maybe not exactly.Wait, the center of a circle tangent to ⊙O at A' must lie on the line OA', at a distance of R ± r from O, where r is the radius of the smaller circle. But if the circle passes through A, then the distance from the center to A is equal to the radius. Hmm, perhaps that's a more complicated way.Alternatively, going back to the problem: we need all points P such that there exists A' on ⊙O with PA = PA'. So, for each such P, there exists a point A' on ⊙O where PA' = PA.Therefore, the locus of points P such that the circle centered at P with radius PA intersects the original circle ⊙O. So, the set of points P where the distance from P to O is less than or equal to R + PA, but I need to think differently.Alternatively, for each P, the set of A' such that PA' = PA is the intersection of two circles: ⊙O with radius R and ⊙P with radius PA. So, for such an A' to exist, the two circles must intersect. Therefore, the condition is that |PO - PA| ≤ R ≤ PO + PA.But since A is a fixed point inside ⊙O, and P varies, perhaps this can lead us to an equation.Wait, let's try to express the condition that there exists A' on ⊙O such that PA' = PA. So, A' lies on both ⊙O and ⊙P with radius PA. Therefore, the intersection of ⊙O and ⊙P is non-empty. Therefore, the distance between centers O and P must satisfy |R - PA| ≤ OP ≤ R + PA.But since A is a fixed point, PA is variable depending on P. Hmm, not sure if this helps.Alternatively, fix O and A. Let’s denote the coordinates. Let’s place O at the origin and A at (a, 0). Let P be (x, y). Then PA = sqrt( (x - a)^2 + y^2 ). The circle ⊙P with radius PA is (X - x)^2 + (Y - y)^2 = (x - a)^2 + y^2. The intersection of this circle with ⊙O (X² + Y² = R²) must be non-empty.Subtracting the equations of the two circles:X² + Y² - R² - [(X - x)^2 + (Y - y)^2 - ( (x - a)^2 + y^2 )] = 0Simplify:X² + Y² - R² - [X² - 2 X x + x² + Y² - 2 Y y + y² - (x² - 2 a x + a² + y²)] = 0Simplify term by term:X² + Y² - R² - X² + 2 X x - x² - Y² + 2 Y y - y² + x² - 2 a x + a² + y² = 0Simplify:- R² + 2 X x + 2 Y y - 2 a x + a² = 0Rearranged:2 x X + 2 y Y = 2 a x + R² - a²Divide both sides by 2:x X + y Y = a x + (R² - a²)/2This is the equation of a line in terms of X and Y. This line is the radical axis of the two circles ⊙O and ⊙P. For the circles to intersect, this radical axis must intersect the circle ⊙O. Therefore, the distance from the center O (0,0) to the radical axis must be less than or equal to R.The distance from O to the line x X + y Y = a x + (R² - a²)/2 is |0 + 0 - (a x + (R² - a²)/2)| / sqrt(x² + y²). Wait, but X and Y are variables here. Wait, no, actually, the radical axis equation is in X and Y, so the distance from O(0,0) to this line is | - (a x + (R² - a²)/2 ) | / sqrt(x² + y²). For the radical axis to intersect ⊙O, this distance must be ≤ R.Therefore:| - (a x + (R² - a²)/2 ) | / sqrt(x² + y²) ≤ RSquare both sides:(a x + (R² - a²)/2 )² ≤ R² (x² + y²)Which is the same inequality we derived earlier! So this tells us that the set of points P(x, y) for which there exists a point A' on ⊙O such that PA' = PA (i.e., P lies on the perpendicular bisector of AA') is the set of points satisfying (a x + (R² - a²)/2 )² ≤ R² (x² + y²). And the boundary of this region is when equality holds, which is the ellipse equation we found earlier.But the problem asks for the set of points on all the crease lines. Each crease line is a perpendicular bisector of AA', so each crease line is part of the radical axis between ⊙O and ⊙P where P is such that PA = PA'. But the radical axis exists only when the circles intersect, which corresponds to the inequality above.But the set of points on all crease lines would be all points that lie on any of these radical axes. So, this would be the union of all such lines. However, each line itself is a set of points satisfying the equation we derived for the crease line, which depends on θ. However, through our analysis, we found that all such lines are tangent to the ellipse defined by the equality case.Wait, if the crease lines are all tangent to this ellipse, then the envelope of the crease lines is the ellipse, and the set of points on all crease lines would be the union of all these tangent lines, which would form the ellipse's exterior? Wait, but no. Actually, the envelope is the ellipse itself. Each crease line touches the ellipse at one point, so the set of all points on all crease lines would be the union of all lines tangent to the ellipse. However, the union of all tangent lines to an ellipse is the entire plane except for the interior of the ellipse. But in our case, is that true?Wait, let's think. If you have an ellipse, any point outside the ellipse has two tangent lines to the ellipse, and any point on the ellipse has one tangent line, and any point inside has none. Therefore, the union of all tangent lines to an ellipse is the entire plane outside the ellipse. However, in our case, the crease lines are not all the tangent lines to the ellipse, but only those that are perpendicular bisectors of AA' for some A' on ⊙O.So perhaps not all tangent lines to the ellipse are crease lines, only a subset. Therefore, the union of the crease lines might only cover part of the plane. However, according to our previous result, the envelope of the crease lines is the ellipse given by the equation 4/R² (x - a/2)^2 + 4 y² / (R² - a²) = 1. If we can confirm that each crease line is tangent to this ellipse, then the set of points on all crease lines is the union of all these tangent lines, which would form a region. But if the crease lines are moving around as A' moves, their union might actually form the ellipse. Wait, that doesn't make sense. The union of lines tangent to an ellipse is not the ellipse itself; the ellipse is the envelope.Wait, no. Each tangent line touches the ellipse at one point, but the line itself extends infinitely in both directions. So the union of all tangent lines would cover the area outside the ellipse. The ellipse itself is just the envelope, the boundary where these lines are tangent.But the problem states "the set of points on all the crease lines". So, if each crease line is a tangent to the ellipse, then the union of all crease lines is the set of all points lying on any tangent line of the ellipse. But depending on the ellipse, this could be the entire plane except the interior. But in our case, maybe it's different because the crease lines are only those that are perpendicular bisectors of AA', which might limit the possible tangent lines.But according to our previous equation, the envelope of the crease lines is the ellipse given by 4/R² (x - a/2)^2 + 4 y² / (R² - a²) = 1. Let me write that in standard form.First, divide both sides by 4:(R² - a²)/R² (x - a/2)^2 + y² / (R² - a²) = 1/4Wait, no. Wait, let's go back.Earlier, we had:4[(R² - a²)/R²](x - a/2)^2 + 4 y² / (R² - a²) = 1Let me write this as:[(x - a/2)^2]/(R²/(4(R² - a²))) + [y²]/[(R² - a²)/4] = 1So this is an ellipse with center at (a/2, 0), semi-major axis sqrt[(R² - a²)/4] = (sqrt(R² - a²))/2 along the y-axis, and semi-minor axis R/(2 sqrt(R² - a²)) along the x-axis. Wait, hold on.Wait, the standard form of an ellipse is [(x - h)^2]/A² + [(y - k)^2]/B² = 1. Comparing:[(x - a/2)^2]/(R²/(4(R² - a²))) + [y²]/[(R² - a²)/4] = 1So the denominator under (x - a/2)^2 is R²/(4(R² - a²)), so A² = R²/(4(R² - a²)), so A = R/(2 sqrt(R² - a²))Similarly, the denominator under y² is (R² - a²)/4, so B² = (R² - a²)/4, so B = sqrt(R² - a²)/2Therefore, the semi-major axis is B = sqrt(R² - a²)/2 (since R² - a² is positive, and assuming R > a, which it is because A is inside the circle), and the semi-minor axis is A = R/(2 sqrt(R² - a²))Hmm, interesting. So this is the ellipse equation. Therefore, the envelope of the crease lines is this ellipse. So each crease line is tangent to this ellipse. Therefore, the set of points on all crease lines is the set of all points lying on any tangent line of this ellipse. But the problem is likely expecting the answer to be this ellipse itself. Wait, but the set of points on the crease lines would be all points that are on any of the tangent lines, which is not the ellipse, but the union of all those lines. However, in some contexts, the envelope is considered the solution. Wait, maybe the problem is asking for the envelope, which is the ellipse. Let me check the problem statement again:"When A' covers all points on the circumference, find the set of points on all the crease lines."Hmm. If "the set of points on all the crease lines" is interpreted as the union of all crease lines, that would be all points that are on any crease line. However, if each crease line is tangent to the ellipse, then the union of all these tangent lines is the entire plane except the interior of the ellipse, but this seems too broad. But in our case, perhaps the crease lines only cover the ellipse. Wait, no, each crease line is an infinite line, so their union would be a region. But maybe the ellipse is the boundary of this region.Alternatively, perhaps the correct answer is that the set of points is the ellipse itself. But why? Because each crease line touches the ellipse, but the points on the crease lines include all points on those tangent lines. So unless there's a misinterpretation here.Wait, let's think differently. If we take any point on the crease line, it must satisfy PA = PA' for some A' on the circle. So, such a point P is equidistant to A and A', which lies on the circle. Therefore, the set of all such points P is the set of points for which there exists an A' on the circle such that PA = PA'. Which is equivalent to saying that P lies on the perpendicular bisector of OA' for some A' on the circle. Wait, no, perpendicular bisector of AA'.Alternatively, the set of all such points P is the locus of points where the distance from P to A is equal to the distance from P to some point on the circle. This is equivalent to the set of points P such that the circle centered at P with radius PA intersects the original circle ⊙O. So the set of all such P is the region where the distance from P to O is less than or equal to R + PA (which is always true since PA can be any positive number) but also |PA - PO| ≤ R by the reverse triangle inequality. Wait, maybe that's the case.Wait, the circles ⊙P (radius PA) and ⊙O (radius R) must intersect. Therefore, the distance between their centers OP must satisfy |R - PA| ≤ OP ≤ R + PA. But PA is the radius of ⊙P. So we have |R - PA| ≤ OP ≤ R + PA. However, PA is the distance from P to A, which is another point. So this becomes:|R - |P - A|| ≤ |P - O| ≤ R + |P - A|This is a condition on the point P. But how does this define a region?Alternatively, considering that O and A are fixed points, this inequality defines the set of points P such that the difference between their distances to O and A is at most R. Wait, | |P - O| - |P - A| | ≤ R. That resembles the definition of a hyperbola, but with a inequality. However, when you have | |P - O| - |P - A| | = constant, that's a hyperbola. But here we have an inequality, so it would be the region between the two branches of the hyperbola. However, in our case, the constant is R, which is the radius of the original circle.But since A is inside ⊙O, OA = a < R. Therefore, the set of points P where | |P - O| - |P - A| | ≤ R would be... Hmm. Let's think. If we consider that the maximum value of | |P - O| - |P - A| | is OA = a (by the triangle inequality), so if R ≥ a, which it is, then the inequality | |P - O| - |P - A| | ≤ R is always true. Wait, because | |P - O| - |P - A| | ≤ |O - A| = a ≤ R. Therefore, the inequality is always satisfied. Therefore, the condition |R - PA| ≤ OP is automatically satisfied?Wait, perhaps my reasoning is flawed here. Let me step back.The circles ⊙O (radius R) and ⊙P (radius PA) must intersect. So, the distance between their centers OP must satisfy |R - PA| ≤ OP ≤ R + PA. Therefore, for each P, this inequality must hold. Therefore, the set of all P is the set of points for which there exists a PA (which is the distance from P to A) such that |R - PA| ≤ OP ≤ R + PA.But PA is dependent on P. So for each P, PA is fixed, so the question is whether OP satisfies |R - PA| ≤ OP ≤ R + PA. But since OP and PA are distances from P to O and A, respectively, which are two fixed points, this condition relates OP and PA.Let me write this as:OP - PA ≤ R and PA - OP ≤ RBut OP - PA ≤ R is always true because by triangle inequality, OP ≤ PA + OA, so OP - PA ≤ OA = a < R. Similarly, PA - OP ≤ OA = a < R. Therefore, both inequalities are automatically satisfied. Wait, that seems contradictory. If OP - PA ≤ OA and PA - OP ≤ OA, then |OP - PA| ≤ OA. Therefore, |OP - PA| ≤ a. But R is greater than a. So, the condition |R - PA| ≤ OP is not automatically satisfied.Wait, maybe my confusion arises from mixing different inequalities. Let's clarify.The circles ⊙O (radius R) and ⊙P (radius PA) intersect if and only if |R - PA| ≤ OP ≤ R + PA.But we need this to hold for some PA. However, PA is not a variable here; PA is determined by the position of P. Wait, no, PA is the distance from P to A, which is fixed once P is fixed. Therefore, for a given P, PA is fixed, and OP is also fixed. Therefore, the circles intersect if and only if |R - PA| ≤ OP ≤ R + PA.But since OP and PA are specific values for each P, whether this inequality holds depends on the position of P. Therefore, the set of points P for which there exists A' on ⊙O such that PA' = PA (i.e., circles intersect) is the set of points P where |R - PA| ≤ OP.Because the upper bound OP ≤ R + PA is always true by triangle inequality (OP ≤ PA + OA, but OA = a < R, so PA + OA ≤ PA + R, but this might not be directly helpful). Wait, actually, OP ≤ R + PA is automatically true because by triangle inequality, OP ≤ PA + OA = PA + a. But since a < R, PA + a < PA + R, so OP ≤ PA + a < PA + R, so OP ≤ PA + R is always true. Therefore, the lower bound |R - PA| ≤ OP is the critical condition.Therefore, the condition for intersection is OP ≥ |R - PA|. Since R ≥ PA - OP is always true (from triangle inequality), the main condition is OP ≥ R - PA.Therefore, the set of points P is defined by OP ≥ R - PA.But OP is the distance from O to P, and PA is the distance from P to A. Let's write this inequality in coordinates. Let’s place O at (0,0) and A at (a,0). Then P is (x,y). Then OP = sqrt(x² + y²), PA = sqrt( (x - a)^2 + y^2 ). The inequality is sqrt(x² + y²) ≥ R - sqrt( (x - a)^2 + y^2 )Let’s denote PA = d. Then the inequality becomes OP ≥ R - d, where d = PA.Rearranged: OP + d ≥ R. So OP + PA ≥ R.But OP + PA is the sum of distances from P to O and from P to A. So the set of points P where OP + PA ≥ R.But since R is a constant greater than a, and OP + PA ≥ R, this defines the region outside or on the ellipse with foci at O and A, and major axis length R. Wait, an ellipse is defined as the set of points where the sum of distances to two foci is constant. Here, we have an inequality: OP + PA ≥ R. So this would be the region outside or on the ellipse with foci at O and A where the sum of distances is R. However, since R > OA = a, such an ellipse is non-degenerate. Wait, but the sum R must be greater than the distance between the foci, which is a. Since R > a, this is possible.Therefore, the boundary of this region is the ellipse with foci at O and A, with major axis length R. Therefore, the set of points P such that OP + PA ≥ R is the region outside or on this ellipse. But in our problem, we need the set of points P that lie on any crease line, i.e., the union of all crease lines. Each crease line is the radical axis of ⊙O and ⊙P where OP + PA ≥ R. However, the radical axis exists only when the circles intersect, which corresponds to OP ≤ R + PA, which is always true, but also OP ≥ |R - PA|. Since we are considering crease lines where A' exists, so the radical axis must exist, i.e., the inequality OP ≥ R - PA must hold. Therefore, the set of points P on crease lines is the region where OP + PA ≥ R, which is the region outside or on the ellipse with foci O and A, sum of distances R.But the problem says "find the set of points on all the crease lines". If each crease line is a radical axis, which is a line, then the union of all such lines is the region where OP + PA ≥ R. But this is an area, not a curve. However, this contradicts our previous result where we derived an ellipse as the envelope.Alternatively, maybe the correct answer is the ellipse which is the envelope of the crease lines. However, the problem states "the set of points on all the crease lines", which is the union of all the crease lines. Each crease line is a straight line, so their union would be a region, not a curve. But the problem might be expecting the ellipse as the answer, considering that each crease line is tangent to it, so all crease lines combined form the ellipse as their envelope. But in Chinese or Japanese problems, sometimes "the set of crease lines" refers to their envelope, which is the ellipse.But let me verify with a specific case. Let's take R = 2, a = 1, so OA = 1. Then the ellipse equation we derived earlier would be:[(x - 1/2)^2]/( (2)^2 / (4*(2^2 - 1^2)) ) + [y^2]/( (2^2 - 1^2)/4 ) = 1Calculating denominators:For x-term: 4 / (4*(4 - 1)) = 4 / 12 = 1/3. Wait, no. Wait, the standard form equation we had was:[(x - a/2)^2]/(R²/(4(R² - a²))) + [y²]/[(R² - a²)/4] = 1Plugging in R = 2, a = 1:[(x - 0.5)^2]/(4/(4*(4 - 1))) + [y²]/[(4 - 1)/4] = 1Simplify:[(x - 0.5)^2]/(4/(12)) + [y²]/(3/4) = 1Which is:[(x - 0.5)^2]/(1/3) + [y²]/(3/4) = 1Multiply numerator and denominator:3(x - 0.5)^2 + (4/3)y² = 1Hmm, this is an ellipse centered at (0.5, 0) with semi-major axis sqrt(3/4) along y-axis and semi-minor axis sqrt(1/3) along x-axis.Alternatively, maybe there's a mistake here. Let me recheck.Earlier, we had:4[(R² - a²)/R²](x - a/2)^2 + 4 y² / (R² - a²) = 1For R = 2, a = 1:4[(4 - 1)/4](x - 0.5)^2 + 4 y² / (4 - 1) = 1Calculates to:4*(3/4)*(x - 0.5)^2 + 4*(y²)/3 = 1Which is:3(x - 0.5)^2 + (4/3)y² = 1Yes, that's correct. So this is an ellipse. Let's see if this ellipse is the same as the one defined by OP + PA = R.With R = 2, the ellipse defined by OP + PA = 2. Let's compute OP + PA for a point on the derived ellipse.Take a point on the major axis, say along y=0. The ellipse equation becomes 3(x - 0.5)^2 = 1 => (x - 0.5)^2 = 1/3 => x = 0.5 ± sqrt(1/3). So x ≈ 0.5 ± 0.577. Let's take x ≈ 0.5 + 0.577 ≈ 1.077. Then OP = sqrt(1.077² + 0²) ≈ 1.077. PA = sqrt( (1.077 - 1)^2 + 0 ) ≈ 0.077. Then OP + PA ≈ 1.077 + 0.077 ≈ 1.154, which is less than R = 2. That doesn't make sense, because the ellipse defined by OP + PA = R should have points where OP + PA = 2. So clearly, my previous assumption that the region OP + PA ≥ R corresponds to the derived ellipse is incorrect. Therefore, there must be a mistake in my reasoning.Wait, this inconsistency suggests that my earlier conclusion is wrong. So perhaps the envelope of the crease lines is not the same as the ellipse defined by OP + PA = R. Let's dig deeper.Alternatively, let's consider a specific example where point A is at the center O. If a = 0, so A coincides with O. Then folding any point A' on the circle onto O would result in the crease line being the perpendicular bisector of OA', which is the line perpendicular to OA' at its midpoint. The set of all such crease lines would be all lines that are perpendicular to radii at their midpoints. The envelope of these lines is known as the midcircle, which in this case would be a circle of radius R/2. So when A is at the center, the set of crease lines envelope a circle of radius R/2.But according to our previous ellipse equation, if a = 0, then the equation becomes:4[(R² - 0)/R²](x - 0/2)^2 + 4 y² / (R² - 0) = 1Simplifies to:4[R²/R²]x² + 4 y² / R² = 1 => 4x² + (4 y²)/R² = 1If R is the radius, say R = 2, then this becomes 4x² + y² = 1, which is an ellipse, but when a = 0, we expect a circle of radius 1 (if R = 2). Wait, but 4x² + y² = 1 is not a circle. So this contradicts our expectation, meaning that somewhere in the derivation, there's an error.This suggests that my previous result is incorrect. So where did I go wrong?Let me retrace the steps.We started with the crease line as the perpendicular bisector of AA', derived its equation, squared both sides to eliminate θ, and ended up with an ellipse equation. However, in the case where a = 0, this should reduce to a circle, but it doesn't. Therefore, my mistake must be in the squaring step, which might have introduced extraneous solutions or incorrectly derived the envelope.Alternatively, maybe the set of crease lines when a = 0 is indeed a circle, but according to the derived equation, it's an ellipse. This inconsistency suggests an error.Let me re-examine the case when A is at the center O. So a = 0. Then point A is O. The crease line is the perpendicular bisector of OA', which is the line perpendicular to OA' at its midpoint. The midpoint of OA' is at (R/2 cos θ, R/2 sin θ). The crease line is perpendicular to OA', so its slope is -cot θ. Therefore, the equation of the crease line is y - (R/2 sin θ) = -cot θ (x - R/2 cos θ)Simplify:y = -cot θ x + (R/2 cos θ) cot θ + (R/2 sin θ)= -cot θ x + (R/2) (cos θ * cot θ + sin θ)= -cot θ x + (R/2)( (cos² θ)/sin θ + sin θ )= -cot θ x + (R/2)( (cos² θ + sin² θ)/sin θ )= -cot θ x + (R/2)(1/sin θ )Multiply both sides by sin θ:y sin θ = -cos θ x + R/2Rearrange:x cos θ + y sin θ = R/2This is the equation of the crease line when A is at O. The envelope of these lines as θ varies is found by differentiating with respect to θ and solving. Let’s compute the envelope.Given the family of lines x cos θ + y sin θ = R/2.To find the envelope, take the derivative with respect to θ:- x sin θ + y cos θ = 0So we have the system:1. x cos θ + y sin θ = R/22. -x sin θ + y cos θ = 0From equation 2: -x sin θ + y cos θ = 0 => y cos θ = x sin θ => y = x tan θSubstitute into equation 1:x cos θ + (x tan θ) sin θ = R/2 => x cos θ + x sin² θ / cos θ = R/2 => x (cos² θ + sin² θ)/cos θ = R/2 => x / cos θ = R/2 => x = (R/2) cos θSimilarly, from y = x tan θ => y = (R/2) cos θ * tan θ = (R/2) sin θTherefore, the envelope is parametrized by x = (R/2) cos θ, y = (R/2) sin θ, which is a circle of radius R/2 centered at the origin. So when A is at O, the envelope is indeed a circle of radius R/2, as expected.But according to the previous ellipse equation when a = 0, we had:4x² + (4 y²)/R² = 1If R = 2, this becomes 4x² + y² = 1, which is not a circle. But we know the correct envelope is a circle. Therefore, my previous derivation must have an error.Looking back at the earlier steps, when a = 0, the equation of the crease line was:(a - R cos θ)x - R sin θ * y + (R² - a²)/2 = 0With a = 0, this becomes:(-R cos θ)x - R sin θ * y + (R²)/2 = 0Which simplifies to:R cos θ x + R sin θ y = R²/2Divide both sides by R:cos θ x + sin θ y = R/2Which is the same as we derived earlier, and the envelope is a circle. Therefore, the previous ellipse equation must be incorrect when a = 0. Therefore, there's a mistake in the process where I squared both sides. Let me check that part again.Starting from the crease line equation:(R sin θ)(y - (R sin θ)/2) = (a - R cos θ)(x - (a + R cos θ)/2)When a = 0, this becomes:R sin θ y - (R² sin² θ)/2 = (-R cos θ)(x - (R cos θ)/2 )Expand right side:- R cos θ x + (R² cos² θ)/2Bring all terms to the left:R sin θ y - (R² sin² θ)/2 + R cos θ x - (R² cos² θ)/2 = 0Factor R:R (sin θ y + cos θ x) - (R²/2)(sin² θ + cos² θ) = 0Since sin² θ + cos² θ = 1:R (sin θ y + cos θ x) - R²/2 = 0Divide by R:sin θ y + cos θ x - R/2 = 0Which is the same as cos θ x + sin θ y = R/2, confirming the correct equation. Therefore, when I squared both sides in the general case, I must have made a mistake.Wait, when a ≠ 0, the process of squaring both sides of the equation may have introduced extraneous solutions, leading to an incorrect ellipse. Therefore, my previous result is invalid.Therefore, I need a different approach. Let's consider that when folding point A' to A, the crease line is the perpendicular bisector of AA'. The set of all such perpendicular bisectors as A' moves on the circle is the set of lines equidistant from A and some point on the circle. This is analogous to the concept of the envelope of the perpendicular bisectors of segments joining a fixed point and a moving point on a circle.I recall that the envelope of the perpendicular bisectors of segments from a fixed point to points on a circle is an ellipse with foci at the fixed point and the center of the circle. Wait, in our case, the fixed point is A and the center is O. So maybe the envelope is an ellipse with foci at O and A.But when A is at O, the envelope should be a circle, as we saw. An ellipse with both foci at the same point is a circle. So that makes sense. Therefore, in general, the envelope is an ellipse with foci at O and A, where the sum of distances from any point on the ellipse to O and A is constant.Wait, but in our specific case, what would be the constant? When A is at O, the envelope is a circle of radius R/2. For an ellipse with foci coinciding, the constant sum is 2b, where b is the radius. In that case, 2b = R/2 * 2 = R. Wait, no. When the foci coincide, the ellipse becomes a circle with radius equal to the semi-major axis. So if the sum is R, then the semi-major axis would be R/2. Which matches the case when a = 0, the envelope is a circle of radius R/2. Therefore, the constant sum for the ellipse is R. Therefore, the envelope of the crease lines is the ellipse with foci at O and A, where the sum of distances from any point on the ellipse to O and A is equal to R. Therefore, the equation of this ellipse is OP + AP = R.Therefore, the set of points on all the crease lines is the ellipse with foci at O and A, with major axis length R. This makes sense, because when you fold A' onto A, the crease line is the set of points equidistant to A and A', and as A' moves, the envelope of these lines is the ellipse where the sum of distances to O and A is R.This also resolves the earlier contradiction: when a = 0, the ellipse becomes a circle of radius R/2, as expected. For the general case, the envelope is an ellipse with foci at O and A, major axis length R.Therefore, the answer should be that the set of all crease lines forms an ellipse with foci at O and A, where the sum of the distances from any point on the ellipse to O and A is equal to R.But let me confirm this with another example. Suppose R = 3, a = 1. Then the ellipse should have foci at O and A, with sum of distances R = 3. The distance between the foci is OA = 1, so the semi-major axis length is R/2 = 1.5, and the semi-minor axis length is sqrt( (R/2)^2 - (OA/2)^2 ) = sqrt( 2.25 - 0.25 ) = sqrt(2) ≈ 1.414.So the ellipse equation would be:( (x - 0.5)^2 )/( (1.5)^2 ) + ( y^2 )/( (sqrt(2))^2 ) = 1=> ( (x - 0.5)^2 )/2.25 + y² / 2 = 1Which is the same as:( (x - 0.5)^2 )/( (3/2)^2 ) + y² / ( (sqrt(2))^2 ) = 1This makes sense for an ellipse with major axis 3 and distance between foci 1.Comparing this with our previous incorrect equation:When R = 3 and a = 1, our previous result gave:4[(9 - 1)/9](x - 0.5)^2 + 4 y² / (9 - 1) = 1Which is:4*(8/9)*(x - 0.5)^2 + 4 y² /8 = 1Simplifies to:(32/9)(x - 0.5)^2 + 0.5 y² = 1Or:( (x - 0.5)^2 )/(9/32) + y² / 2 = 1This does not match the correct ellipse equation. Therefore, my previous approach was flawed.Given that, the correct envelope is the ellipse with foci at O and A, where the sum of distances is R. Therefore, the set of points on all the crease lines is this ellipse.Therefore, the final answer is an ellipse with foci at O and A, with major axis length R.But how to express this in terms of the given variables?The standard equation of an ellipse with foci at O and A (assuming O is at the origin and A is at (a, 0)) is:sqrt(x² + y²) + sqrt( (x - a)^2 + y² ) = RThis is the equation in terms of distances. Squaring both sides to derive the standard form:sqrt(x² + y²) + sqrt( (x - a)^2 + y² ) = RLet’s denote OP = sqrt(x² + y²) and PA = sqrt( (x - a)^2 + y² )So OP + PA = RSquaring both sides:(OP)^2 + 2 OP PA + (PA)^2 = R²But OP² + PA² = x² + y² + (x - a)^2 + y² = 2 x² - 2 a x + a² + 2 y²Therefore:2 x² - 2 a x + a² + 2 y² + 2 OP PA = R²Isolate 2 OP PA:2 OP PA = R² - (2 x² - 2 a x + a² + 2 y² )Square both sides again:4 OP² PA² = [ R² - 2 x² + 2 a x - a² - 2 y² ]²But OP² = x² + y² and PA² = (x - a)^2 + y². Therefore:4 (x² + y²)( (x - a)^2 + y² ) = [ R² - 2 x² + 2 a x - a² - 2 y² ]²This equation would expand to the standard form of the ellipse. However, this is quite complex. Instead, the answer is best expressed as the ellipse with foci at O and A, with major axis length R.Therefore, the set of points on all the crease lines is the ellipse with foci at O and A, where the sum of the distances to O and A is equal to R. This ellipse is sometimes called the envelope of the crease lines, but since each crease line is tangent to this ellipse, the points on the crease lines collectively form the ellipse as their envelope.However, the problem asks for "the set of points on all the crease lines", which technically are all the points lying on any of the crease lines. But since each crease line is tangent to the ellipse, the union of all these tangent lines would cover the area outside the ellipse. But in reality, the crease lines themselves are the tangent lines to the ellipse, meaning that the envelope is the ellipse itself, and hence the set of points on the crease lines includes all points on these tangent lines, which is not just the ellipse. However, in such folding problems, the locus of the creases (their envelope) is considered the solution, which is the ellipse.Given the problem's context and typical mathematical problem conventions, the expected answer is the ellipse with foci at O and A where the sum of distances is R. Therefore, the set of points on all the crease lines is this ellipse.Final Answerboxed{text{The set of points is an ellipse with foci } O text{ and } A text{ and major axis length } R}</think>To solve the problem of finding the set of points on all the crease lines formed by folding a point ( A' ) on the circumference of a circle ( odot O ) with radius ( R ) onto a fixed point ( A ) inside the circle, we start by noting that each crease line is the perpendicular bisector of the segment ( AA' ). 1. Perpendicular Bisector Analysis: - The crease line for each ( A' ) is the perpendicular bisector of ( AA' ). - This crease line is equidistant from ( A ) and ( A' ).2. Parametrization and Equation Derivation: - Place ( O ) at the origin and ( A ) at coordinates ( (a, 0) ). - Parametrize ( A' ) on the circumference as ( (R cos theta, R sin theta) ). - The midpoint ( M ) of ( AA' ) is ( left( frac{a + R cos theta}{2}, frac{R sin theta}{2} right) ). - The slope of ( AA' ) is ( frac{R sin theta}{R cos theta - a} ), so the slope of the perpendicular bisector is ( frac{a - R cos theta}{R sin theta} ).3. Equation of the Crease Line: - Using the point-slope form, the equation of the crease line is derived and simplified to: [ (a - R cos theta)x - R sin theta y + frac{R^2 - a^2}{2} = 0 ]4. Envelope of Crease Lines: - The set of all crease lines forms an envelope, which is found by eliminating the parameter ( theta ). - The envelope of these crease lines is an ellipse with foci at ( O ) and ( A ), where the sum of the distances from any point on the ellipse to ( O ) and ( A ) is equal to ( R ).5. Verification and Conclusion: - When ( A ) coincides with ( O ), the envelope reduces to a circle of radius ( R/2 ), confirming the correctness. - For general cases, the envelope is an ellipse with foci at ( O ) and ( A ), major axis length ( R ).Thus, the set of points on all the crease lines is an ellipse with foci ( O ) and ( A ) and major axis length ( R ).[boxed{text{The set of points is an ellipse with foci } O text{ and } A text{ and major axis length } R}]]

🔑:Solve: Establish a rectangular coordinate system as shown in Figure 4. Then we have A(a, 0). Suppose when folding, point A^{prime}(R cos alpha, R sin alpha) on odotleft(right. ) coincides with point A. The fold line is the line M N, which is the perpendicular bisector of segment A A^{prime}. Let P(x, y) be any point on M N. Then left|P A^{prime}right|=|P A|. That is,(x-R cos alpha)^{2}+(y-R sin alpha)^{2}=(x-a)^{2}+y^{2} text {. }Therefore,frac{x cos alpha+y sin alpha}{sqrt{x^{2}+y^{2}}}=frac{R^{2}-a^{2}+2 a x}{2 R sqrt{x^{2}+y^{2}}} .We getsin (theta+alpha)=frac{R^{2}-a^{2}+2 a x}{2 R sqrt{x^{2}+y^{2}}} text {, }Given quad sin theta=frac{x}{sqrt{x^{2}+y^{2}}}, cos theta=frac{y}{sqrt{x^{2}+y^{2}}}.Thusleft|frac{R^{2}-a^{2}+2 a x}{2 R-x^{2}+y^{2}}right| leqslant 1Simplifying, we getfrac{(2 x-a)^{2}}{R^{2}}+frac{4 y^{2}}{R^{2}-a^{2}} geqslant 1 text {. }Therefore, the set of points sought is frac{(2 x-a)^{2}}{R^{2}}+frac{4 y^{2}}{R^{2}-a^{2}}=1 and its exterior (including the boundary).

🧠:[<think>Okay, so I need to find p(5) where p(x) is a polynomial that satisfies the equation p(x) - p'(x) = x² + 2x + 1. Hmm, let's start by understanding what this problem is asking. First, p(x) is a polynomial, and when we subtract its derivative p'(x) from it, we get the quadratic polynomial x² + 2x + 1. My goal is to determine the value of this polynomial at x = 5. Since p(x) is a polynomial, and the equation involves its derivative, maybe I can assume a general form for p(x) and then find its coefficients. Let's see. The right-hand side is a quadratic, so maybe p(x) is a quadratic as well? But wait, if I take the derivative of a quadratic, it becomes linear. Then p(x) - p'(x) would be a quadratic minus a linear term, which is still a quadratic. So maybe p(x) is a quadratic polynomial. Let me check that.Suppose p(x) is a quadratic: p(x) = ax² + bx + c. Then p'(x) = 2ax + b. So p(x) - p'(x) = ax² + bx + c - (2ax + b) = ax² + (b - 2a)x + (c - b). According to the problem, this should equal x² + 2x + 1. Therefore, we can set up equations by equating the coefficients:For the x² term: a = 1.For the x term: b - 2a = 2.For the constant term: c - b = 1.Now, substituting a = 1 into the second equation: b - 2(1) = 2 ⇒ b = 4.Substituting b = 4 into the third equation: c - 4 = 1 ⇒ c = 5.Therefore, p(x) would be x² + 4x + 5. Let me verify this. Compute p(x) - p'(x):p(x) = x² + 4x + 5p'(x) = 2x + 4So p(x) - p'(x) = (x² + 4x + 5) - (2x + 4) = x² + 2x + 1, which matches the given equation. Great, so this seems to work. Therefore, p(x) is x² + 4x + 5, and p(5) would be 5² + 4*5 + 5 = 25 + 20 + 5 = 50. Wait, but hold on a second. Let me double-check this calculation.Calculating p(5):5 squared is 25. 4 times 5 is 20. Then adding 5. So 25 + 20 is 45, plus 5 is 50. Yep, that's correct. So p(5) is 50. Hmm, so maybe that's the answer? But wait, the problem didn't specify the degree of p(x). What if p(x) is a higher-degree polynomial? Let's see. Suppose, for example, p(x) is a cubic. Let's test that.Suppose p(x) is a cubic: p(x) = ax³ + bx² + cx + d. Then p'(x) = 3ax² + 2bx + c. Then p(x) - p'(x) = ax³ + bx² + cx + d - (3ax² + 2bx + c) = ax³ + (b - 3a)x² + (c - 2b)x + (d - c). But the right-hand side is x² + 2x + 1, which has no cubic term. Therefore, the coefficient of x³ in p(x) - p'(x) must be zero. That implies a = 0. But if a = 0, then p(x) is a quadratic, which brings us back to the previous case. So even if we assume a higher degree polynomial, the cubic term's coefficient must be zero. Similarly, higher-degree terms would also have to have coefficients zero because their derivatives would produce lower-degree terms, but the right-hand side has only up to x². Therefore, p(x) cannot have degree higher than 2, because higher-degree terms would result in higher-degree terms in p(x) - p'(x), which are not present in the given equation. So p(x) must be a quadratic. Therefore, the solution we found earlier is the only possible one, and p(5) is indeed 50.Wait, but just to be thorough, let's try assuming p(x) is a first-degree polynomial. Suppose p(x) = mx + n. Then p'(x) = m. So p(x) - p'(x) = mx + n - m = mx + (n - m). But the right-hand side is x² + 2x + 1, which is quadratic. So there's no way a linear polynomial can satisfy this equation. Therefore, p(x) must be at least quadratic. Since the right-hand side is quadratic, and we've already considered that higher-degree terms in p(x) would require their coefficients to be zero, then p(x) is indeed a quadratic. Therefore, our original solution holds.Alternatively, another approach could be using the method of undetermined coefficients, which is essentially what I did. But in this case, since the nonhomogeneous term is a quadratic polynomial, and the operator is L[p] = p - p', the particular solution can be found by assuming a quadratic polynomial. Alternatively, maybe using linear algebra or differential equations techniques? Wait, this is a first-order linear differential equation. The equation p(x) - p'(x) = x² + 2x + 1 can be rewritten as p'(x) - p(x) = -x² - 2x - 1. The standard approach for solving such equations is to find an integrating factor. Let me try that.The differential equation is p'(x) - p(x) = -x² - 2x - 1. Rewriting it as p'(x) + (-1)p(x) = -x² - 2x - 1. The integrating factor would be e^(∫-1 dx) = e^(-x). Multiply both sides by the integrating factor:e^(-x) p'(x) - e^(-x) p(x) = -e^(-x)(x² + 2x + 1)The left-hand side is the derivative of (e^(-x) p(x)). So:d/dx [e^(-x) p(x)] = -e^(-x)(x² + 2x + 1)Integrate both sides:e^(-x) p(x) = -∫ e^(-x)(x² + 2x + 1) dx + CSo p(x) = e^x [ -∫ e^(-x)(x² + 2x + 1) dx + C ]Now, we can compute the integral ∫ e^(-x)(x² + 2x + 1) dx. Let's note that x² + 2x + 1 is (x + 1)^2. So integrating e^(-x)(x + 1)^2 dx. Let's use integration by parts. Let me set u = (x + 1)^2, dv = e^(-x) dx. Then du = 2(x + 1) dx, and v = -e^(-x). Then:∫ u dv = uv - ∫ v du = -e^(-x)(x + 1)^2 + ∫ 2(x + 1)e^(-x) dxNow, compute the remaining integral ∫ 2(x + 1)e^(-x) dx. Let's use integration by parts again. Let u = 2(x + 1), dv = e^(-x) dx. Then du = 2 dx, v = -e^(-x):∫ 2(x + 1)e^(-x) dx = -2(x + 1)e^(-x) + ∫ 2 e^(-x) dx = -2(x + 1)e^(-x) - 2 e^(-x) + CPutting it all together:Original integral = -e^(-x)(x + 1)^2 + [ -2(x + 1)e^(-x) - 2 e^(-x) ] + CSimplify:= -e^(-x)(x + 1)^2 - 2e^(-x)(x + 1) - 2 e^(-x) + CFactor out -e^(-x):= -e^(-x)[ (x + 1)^2 + 2(x + 1) + 2 ] + CLet's expand the terms inside the brackets:(x + 1)^2 = x² + 2x + 12(x + 1) = 2x + 2Adding these together with the 2:x² + 2x + 1 + 2x + 2 + 2 = x² + 4x + 5Therefore, the integral becomes:= -e^(-x)(x² + 4x + 5) + CTherefore, going back to the expression for p(x):p(x) = e^x [ - ( -e^(-x)(x² + 4x + 5) + C ) ] + CWait, let me check that again. Wait, the original equation after integrating was:e^(-x) p(x) = - [ -e^(-x)(x² + 4x + 5) + C ] + C?Wait, let's retrace. The integral was:∫ e^(-x)(x + 1)^2 dx = -e^(-x)(x² + 4x + 5) + CWait, but in the previous step, we had:Original integral = -e^(-x)(x² + 4x + 5) + CBut in the differential equation, the integral was multiplied by -1, so:e^(-x) p(x) = - [ integral ] + C = - [ -e^(-x)(x² + 4x + 5) + C ] + C?Wait, no. Wait, the equation after integrating both sides was:e^(-x) p(x) = - ∫ e^(-x)(x² + 2x + 1) dx + CAnd we computed ∫ e^(-x)(x² + 2x + 1) dx = -e^(-x)(x² + 4x + 5) + CTherefore, substituting back:e^(-x) p(x) = - [ -e^(-x)(x² + 4x + 5) + C ] + CWait, but the integral is equal to -e^(-x)(x² + 4x + 5) + C, so:e^(-x) p(x) = - [ -e^(-x)(x² + 4x + 5) + C ] + CBut hold on, the integral on the RHS is computed with a constant of integration, but when we do the indefinite integral in solving the differential equation, the constant from the integral gets absorbed into the overall constant C. So actually, let's write:∫ e^(-x)(x² + 2x + 1) dx = -e^(-x)(x² + 4x + 5) + CTherefore:e^(-x) p(x) = - [ -e^(-x)(x² + 4x + 5) + C ] + DWait, but constants can be combined. Let me clarify.The integral is:∫ e^(-x)(x² + 2x + 1) dx = -e^(-x)(x² + 4x + 5) + CTherefore, substituting into the equation:e^(-x) p(x) = - [ -e^(-x)(x² + 4x + 5) + C ] + DWhich simplifies to:e^(-x) p(x) = e^(-x)(x² + 4x + 5) - C + DBut -C + D is just another constant, let's call it K. So:e^(-x) p(x) = e^(-x)(x² + 4x + 5) + KMultiply both sides by e^x:p(x) = x² + 4x + 5 + K e^xBut here, p(x) is supposed to be a polynomial. However, K e^x is an exponential function, which is not a polynomial unless K = 0. Therefore, to have p(x) as a polynomial, we must set K = 0. Hence, p(x) = x² + 4x + 5. This matches the earlier result. Therefore, even when approaching this as a differential equation, we get the same polynomial solution. Therefore, p(5) = 5² + 4*5 + 5 = 25 + 20 + 5 = 50.Therefore, both methods lead to the same answer. So I can be confident that p(5) is indeed 50. Just to recap: The problem gives a polynomial p(x) such that p(x) minus its derivative equals a quadratic. By assuming p(x) is a quadratic polynomial and equating coefficients, we find that p(x) is x² + 4x + 5. Checking with the differential equation approach confirms this solution. Therefore, evaluating at x = 5 gives 50.Final Answerboxed{50}</think>Given the polynomial ( p(x) ) such that ( p(x) - p'(x) = x^2 + 2x + 1 ), we need to compute ( p(5) ).First, we assume ( p(x) ) is a quadratic polynomial: ( p(x) = ax^2 + bx + c ). Then, its derivative is ( p'(x) = 2ax + b ). Substituting these into the given equation:[p(x) - p'(x) = ax^2 + bx + c - (2ax + b) = ax^2 + (b - 2a)x + (c - b)]We compare coefficients with ( x^2 + 2x + 1 ):1. For the ( x^2 ) term: ( a = 1 )2. For the ( x ) term: ( b - 2a = 2 ) which gives ( b - 2 = 2 ) hence ( b = 4 )3. For the constant term: ( c - b = 1 ) which gives ( c - 4 = 1 ) hence ( c = 5 )Thus, the polynomial is ( p(x) = x^2 + 4x + 5 ).To verify, we compute ( p(x) - p'(x) ):[p(x) = x^2 + 4x + 5 quad text{and} quad p'(x) = 2x + 4][p(x) - p'(x) = (x^2 + 4x + 5) - (2x + 4) = x^2 + 2x + 1]This matches the given equation. Therefore, ( p(x) = x^2 + 4x + 5 ) is correct. Evaluating at ( x = 5 ):[p(5) = 5^2 + 4 cdot 5 + 5 = 25 + 20 + 5 = 50]Thus, the final answer is (boxed{50}).,<think>Okay, let's see. I need to find p(5) where p(x) is a polynomial that satisfies p(x) - p'(x) = x² + 2x +1. Hmm, okay. So, this is a differential equation involving a polynomial. The equation is p(x) minus its derivative equals this quadratic polynomial. My goal is to find p(5). First, let's recall that if p(x) is a polynomial, then its derivative p'(x) is also a polynomial of one degree less. So, if the right-hand side of the equation is a quadratic (degree 2), then p(x) must be a quadratic or maybe higher? Wait, let's think. If p(x) is a quadratic, then p'(x) would be linear. Then p(x) - p'(x) would be a quadratic minus a linear term, so it would still be a quadratic. But the right-hand side is x² + 2x +1, which is quadratic. So, maybe p(x) is a quadratic polynomial. Let me try assuming that p(x) is a quadratic.Let's suppose p(x) = ax² + bx + c. Then p'(x) = 2ax + b. So, according to the equation, p(x) - p'(x) = ax² + bx + c - (2ax + b) = ax² + (b - 2a)x + (c - b). This should be equal to x² + 2x +1. Therefore, we can set up equations by equating the coefficients.For the x² term: a = 1.For the x term: b - 2a = 2. Since a =1, this becomes b - 2 = 2, so b = 4.For the constant term: c - b = 1. Since b =4, then c -4 =1, so c =5.Therefore, p(x) would be x² +4x +5. Let me check this. If p(x) is x² +4x +5, then p'(x) is 2x +4. Then p(x) - p'(x) is (x² +4x +5) - (2x +4) = x² +2x +1. That's exactly the right-hand side. So that works. Therefore, p(x) = x² +4x +5, and p(5) would be 25 +20 +5 =50. Wait, 25 +20 is 45, plus 5 is 50. So, the answer is 50? Hmm, but wait, before I get too confident, let me make sure that there isn't a higher degree polynomial that could satisfy this equation. For example, what if p(x) is a cubic polynomial? Let me check.Suppose p(x) is cubic: p(x) = ax³ + bx² + cx + d. Then p'(x) is 3ax² +2bx +c. Then p(x) - p'(x) = ax³ + bx² + cx + d - (3ax² +2bx +c) = ax³ + (b -3a)x² + (c -2b)x + (d -c). But the right-hand side is x² +2x +1, which has no cubic term. Therefore, the coefficient of x³ must be zero. So, a=0. If a=0, then p(x) is just a quadratic, which reduces to the previous case. So, there's no need for a cubic term. Similarly, higher degree terms would also result in higher degree terms in p(x) - p'(x), which are not present here. Therefore, the only solution is the quadratic polynomial. So, p(x) is indeed x² +4x +5, and p(5) is 5² +4*5 +5 =25 +20 +5=50. Therefore, the answer should be 50.But wait, let me think again. Maybe there's a constant term or something missing. Wait, the derivative of a cubic term would have introduced a quadratic term in p'(x), but since the original equation only has up to x², perhaps the cubic coefficient must be zero. So, yeah, that's exactly right. So, even if we had assumed a cubic, the cubic coefficient would have to be zero, leading us back to the quadratic. Therefore, the only possible polynomial is quadratic, and we found the correct coefficients. Therefore, p(5) is 50. Alternatively, another way to approach this problem is by using the method of undetermined coefficients. Let me try that as a cross-check. Let's suppose that p(x) is a general polynomial. Since the equation p(x) - p'(x) = x² +2x +1 is given, and the right-hand side is a quadratic, we can assume that p(x) is a quadratic polynomial. Let p(x) = Ax² + Bx + C. Then p'(x) = 2Ax + B. So, p(x) - p'(x) = Ax² + Bx + C -2Ax -B = Ax² + (B - 2A)x + (C - B). Set this equal to x² +2x +1, so we have:Ax² + (B - 2A)x + (C - B) = 1x² +2x +1.Therefore, equating coefficients:A =1,B -2A =2,C - B =1.From the first equation, A=1.Substitute into the second equation: B -2*1 =2 => B=4.Substitute B=4 into the third equation: C -4=1 => C=5.Therefore, p(x) =x² +4x +5. Then p(5)=5² +4*5 +5=25+20+5=50. So, same result.Alternatively, maybe use linear algebra or differential equations approach. The equation p - p’ = q, where q is a given polynomial. For linear differential equations like this, the general solution is the particular solution plus the solution to the homogeneous equation p - p’ =0. The homogeneous solution would be p_h(x) = Ce^x. However, since we are looking for polynomial solutions, the homogeneous solution must be zero (since e^x is not a polynomial unless C=0). Therefore, the only solution is the particular solution, which is the quadratic polynomial we found. Therefore, p(x)=x² +4x +5, so p(5)=50. Hence, confident the answer is 50.Final Answerboxed{50}</think>To find ( p(5) ) where ( p(x) ) is a polynomial satisfying ( p(x) - p'(x) = x^2 + 2x + 1 ), we start by assuming ( p(x) ) is a quadratic polynomial. Let ( p(x) = ax^2 + bx + c ). Then, the derivative ( p'(x) = 2ax + b ).Substituting ( p(x) ) and ( p'(x) ) into the given equation, we get:[ax^2 + bx + c - (2ax + b) = x^2 + 2x + 1]Simplifying the left-hand side, we have:[ax^2 + (b - 2a)x + (c - b)]Equating the coefficients of corresponding terms from both sides, we get:[begin{cases}a = 1 b - 2a = 2 c - b = 1end{cases}]Solving these equations step-by-step:1. From ( a = 1 ), we substitute ( a ) into the second equation: [ b - 2(1) = 2 implies b = 4 ]2. Substituting ( b = 4 ) into the third equation: [ c - 4 = 1 implies c = 5 ]Thus, the polynomial is ( p(x) = x^2 + 4x + 5 ). Evaluating this at ( x = 5 ):[p(5) = 5^2 + 4(5) + 5 = 25 + 20 + 5 = 50]Therefore, the final answer is (boxed{50}).]

🔑:Answer: 50 Observe that p(x) must be quadratic. Let p(x)=a x^{2}+b x+c. Comparing coefficients gives a=1, b-2 a=2, and c-b=1. So b=4, c=5, p(x)=x^{2}+4 x+5 and p(5)=25+20+5=50.

🧠:[<think>Okay, so I need to solve this geometry problem. Let me first try to visualize it. We have an acute-angled triangle ABC. That means all three angles are less than 90 degrees. Then, they've drawn the altitudes AA₁ and BB₁. So, altitude from A is AA₁, which meets BC at A₁, and altitude from B is BB₁, which meets AC at B₁. Got that.Now, the problem mentions the bisector of the exterior angle at vertex C. Let me recall that an exterior angle bisector is different from the internal angle bisector. The exterior angle at C would be the angle formed by one side of the triangle and the extension of the other side. Since the triangle is acute-angled, the exterior angle at C would be adjacent to angle C, right? So, if we take angle C, which is acute, the exterior angle would be 180 degrees minus angle C. The bisector of this exterior angle would split that exterior angle into two equal parts.This bisector intersects the lines AB and A₁B₁ at points L and K, respectively. So, first, the exterior angle bisector at C meets AB at L, and the same bisector meets A₁B₁ at K. Then, it's given that CL = 2 CK. We need to find the measure of angle C.Hmm. Let me start by drawing a rough sketch in my mind. Triangle ABC, acute. Altitudes from A and B, so their feet are on the opposite sides. The exterior angle bisector at C... Let me recall that in a triangle, the internal and external angle bisectors at a vertex are perpendicular to each other. Wait, no. The internal angle bisector and the external angle bisector are supplementary. If the internal angle is θ, the external angle is 180 - θ, so the bisectors would each split their respective angles into θ/2 and (180 - θ)/2. So, the angle between the internal and external bisectors would be θ/2 + (180 - θ)/2 = 90 degrees. So, they are perpendicular. That's an important point.So, the internal angle bisector and the external angle bisector at C are perpendicular. That might come into play later.Now, the problem says the external angle bisector at C intersects AB at L and A₁B₁ at K. Then CL = 2 CK. So, points K and L are along the external angle bisector, with K between C and L, since CL is twice CK. So, CK is half of CL. So, the distance from C to K is half the distance from C to L. So, K is the midpoint of CL? Wait, no. If CL = 2 CK, then CK = CL / 2, which would mean that K divides CL in the ratio CK:KL = 1:1. Wait, but CL is the entire length from C to L, and CK is from C to K. So, if CL is twice CK, then K is halfway between C and L. So, yes, K is the midpoint of CL.But maybe there's another way to look at this. Since the external angle bisector meets AB at L and A₁B₁ at K, then CK is a segment from C to K on the bisector, and CL is from C to L on the same bisector, but extended further to meet AB. So, CK is a part of CL, with CL being longer. So, CK is half of CL. Therefore, K divides CL in the ratio 1:1. So, K is the midpoint of CL.Hmm, maybe. Let me confirm. If CL = 2 CK, then CK is half of CL. So, starting from point C, moving along the external bisector, we reach K first, then L is further out, such that the distance from C to L is twice the distance from C to K. So, the distance from K to L is equal to CK. So, KL = CK. So, CK:KL = 1:1, so K is the midpoint of CL. Got it.Now, the challenge is to relate this ratio to angle C. The problem states that in this configuration, angle C has a measure that satisfies CL = 2 CK, and we need to find that angle.Given that this is a geometry problem involving angle bisectors, altitudes, and ratios, perhaps some properties of similar triangles, trigonometric ratios, or coordinate geometry could be useful here.Let me consider possible approaches:1. Coordinate Geometry: Assign coordinates to the triangle ABC, find equations of the altitudes, the external angle bisector, find points L and K, compute distances CL and CK, set up the ratio, and solve for angle C.2. Trigonometry: Use trigonometric identities, the Angle Bisector Theorem, properties of altitudes, and relations in triangle to express CL and CK in terms of angle C, then set up the equation CL = 2 CK.3. Synthetic Geometry: Find similar triangles, use properties of midpoints, angle bisectors, orthocenters, etc., to derive relations leading to angle C.Each approach has its pros and cons. Coordinate geometry might be more straightforward but could involve lengthy computations. Trigonometry might be more elegant but requires knowing the right identities. Synthetic geometry could be tricky but might provide a more insightful solution.Let me try coordinate geometry first, as it can sometimes make the problem more mechanical.Let me place triangle ABC in the coordinate plane. Let me set point C at the origin (0,0) to simplify calculations. Let me denote angle C as γ, which we need to find. Let me let side AB be opposite angle C, so sides AC and BC are adjacent to angle C.But perhaps a better coordinate system would be to set point C at the origin, point B along the x-axis, and point A somewhere in the plane. Let me try that.Let me define:- Point C: (0, 0)- Point B: (b, 0) for some b > 0- Point A: (d, e) for some d, e > 0 (since the triangle is acute-angled, all angles less than 90°, so coordinates should be such that all altitudes are inside the triangle)But maybe even better to use specific coordinates based on angle C. Let me see.Alternatively, since we need to deal with altitudes AA₁ and BB₁, perhaps coordinate geometry can help by setting coordinates so that these altitudes can be easily calculated.Alternatively, let me consider triangle ABC with coordinates:Let me let point C be at (0, 0).Let me let side BC be along the x-axis, so point B is (c, 0), and point C is (0,0).Let me let point A be at (a, b), so that triangle ABC has coordinates:- A: (a, b)- B: (c, 0)- C: (0, 0)Since the triangle is acute-angled, all angles are less than 90°, so the coordinates must satisfy certain conditions.But maybe this is getting too vague. Let me try to set specific coordinates. Let me assume angle C is γ, and sides adjacent to angle C are length a and b. Wait, in standard notation, sides opposite angles A, B, C are a, b, c respectively. But maybe this can cause confusion. Alternatively, let me denote sides as follows:Let me denote:- Length of AC = m- Length of BC = n- Length of AB = pAngle at C is γ.But maybe coordinate system is still better.Let me proceed step by step.Set point C at (0, 0).Let me place point B at (1, 0), so BC is along the x-axis from (0,0) to (1, 0).Let me place point A somewhere in the first quadrant, say (d, e), so that triangle ABC is acute-angled.Then, the altitude from A to BC is AA₁. Since BC is along the x-axis, the altitude from A is vertical? Wait, no. The altitude from A to BC is a perpendicular line from A to BC. Since BC is horizontal, the altitude will be vertical if BC is horizontal. Wait, no. If BC is along the x-axis from (0,0) to (1, 0), then the altitude from A (d, e) to BC is a vertical line only if BC is horizontal. Wait, the slope of BC is 0, so the altitude from A to BC must be vertical? Wait, the altitude from A to BC is perpendicular to BC. Since BC is horizontal, the altitude is vertical. So, the foot of the altitude from A, which is A₁, is the projection of A onto BC, which is (d, 0). Similarly, the altitude from B to AC.Wait, let's confirm:If BC is along the x-axis from (0,0) to (1,0), then the altitude from A (d, e) to BC is the vertical line dropping from (d, e) to (d, 0), so A₁ is (d, 0).Similarly, the altitude from B to AC. The line AC goes from (0,0) to (d, e). The slope of AC is e/d, so the altitude from B (1, 0) to AC must be perpendicular to AC. So, the slope of the altitude is -d/e. The equation of AC is y = (e/d)x. The equation of the altitude from B is y - 0 = (-d/e)(x - 1). The foot of this altitude, which is B₁, is the intersection of this altitude with AC.So, solving for intersection:From AC: y = (e/d)xFrom altitude: y = (-d/e)(x - 1)Set equal:(e/d)x = (-d/e)(x - 1)Multiply both sides by d e to eliminate denominators:e^2 x = -d^2 (x - 1)e^2 x = -d^2 x + d^2Bring all terms to left:e^2 x + d^2 x - d^2 = 0x (e^2 + d^2) = d^2x = d^2 / (d^2 + e^2)Then y = (e/d)x = (e/d)(d^2 / (d^2 + e^2)) = (e d) / (d^2 + e^2)Therefore, B₁ is at (d^2 / (d^2 + e^2), e d / (d^2 + e^2))Okay, so coordinates of B₁ are (d²/(d² + e²), e d/(d² + e²))Similarly, A₁ is (d, 0)Now, the line A₁B₁ is the line connecting A₁ (d, 0) and B₁ (d²/(d² + e²), e d/(d² + e²))Let me compute the equation of line A₁B₁.First, compute the slope:Slope m = [e d/(d² + e²) - 0] / [d²/(d² + e²) - d] = [e d / (d² + e²)] / [ (d² - d(d² + e²)) / (d² + e²) ]Simplify denominator:d² - d(d² + e²) = d² - d³ - d e² = -d³ + d² - d e² = -d(d² - d + e²)Wait, maybe better to compute numerator and denominator step by step.Numerator: e d / (d² + e²)Denominator: d²/(d² + e²) - d = [d² - d(d² + e²)] / (d² + e²) = [d² - d³ - d e²] / (d² + e²) = [ -d³ + d² - d e² ] / (d² + e²) = -d(d² - d + e²) / (d² + e²)So, slope m = [e d / (d² + e²)] / [ -d(d² - d + e²) / (d² + e²) ] = [e d / (d² + e²)] * [ (d² + e²) / (-d(d² - d + e²)) ) ] = e d / (-d(d² - d + e²)) ) = -e / (d² - d + e²)Therefore, the slope of line A₁B₁ is -e / (d² - d + e²)Now, equation of line A₁B₁: Using point A₁ (d, 0):y - 0 = m (x - d)So, y = [ -e / (d² - d + e²) ] (x - d )Now, the external angle bisector at C. Let me recall that the external angle bisector of angle C will bisect the exterior angle formed by extending side BC or AC. Since angle C is at (0,0), the sides are CB along the x-axis to (1,0) and CA to (d, e). The exterior angle at C can be formed by extending one of these sides. Since the triangle is acute, both exterior angles at C are greater than 90°, but the external angle bisector would depend on which side we extend.Wait, actually, in standard terms, the external angle bisector is the bisector of the angle formed by one side and the extension of the other side. So, for angle C, we can extend side BC beyond C or extend side AC beyond C. But in this problem, the external angle bisector at C is mentioned without specification. However, in the context of intersecting AB and A₁B₁, it must be the external angle bisector that is on the "external" side relative to the triangle.Given that the triangle is acute-angled, the external angle bisector at C would be the one that's outside the triangle. Since the triangle is acute, the external angle bisector would go away from the triangle. To determine which side to extend, maybe it's the side opposite to angle C? Wait, angle C is at the origin between sides CB and CA. The external angle bisector would be the bisector of the angle formed by extending either CB or CA.But since the problem mentions that the bisector intersects AB and A₁B₁, which are both inside or related to the triangle, perhaps the external angle bisector is the one that is on the opposite side of AB. Let me think.Alternatively, perhaps the external angle bisector at C is the bisector of the angle formed by extending side AC beyond C, and keeping side BC as is. Wait, but the external angle can be formed by extending either AC or BC. Since the problem doesn't specify, perhaps we need to figure out which one would intersect AB and A₁B₁.Wait, if we extend BC beyond C, the external angle at C would be between BC extended and AC. Alternatively, if we extend AC beyond C, the external angle would be between AC extended and BC. Either way, the external angle bisector would be a line that splits that external angle into two equal parts.But the problem says that this external angle bisector intersects AB at L and A₁B₁ at K. So, the external angle bisector must cross both AB and A₁B₁. Therefore, depending on which external angle bisector we take (extending BC or AC), the direction of the bisector would be different.Given that the problem is in an acute-angled triangle, and considering the usual notation, perhaps the external angle bisector is the one that is on the side opposite to the triangle's interior. Let me think.Alternatively, perhaps coordinate geometry can resolve this.Given that point C is at (0,0), side CB is along the x-axis to (1,0), and point A is at (d, e). Then, the external angle at C can be formed by extending, say, CB beyond C to the left, creating an external angle between the extended CB and the original side CA. The bisector of this external angle would be a line starting at C and going into the plane such that it splits this external angle.Alternatively, extending AC beyond C would create an external angle between extended AC and CB. Depending on which extension we take, the external angle bisector would be different.But since the problem states that the external angle bisector intersects AB and A₁B₁, which are segments in the plane of the triangle, the external angle bisector must be intersecting those segments. So, depending on where AB and A₁B₁ are located, the external angle bisector must pass through them.Given that AB is the side opposite angle C, and A₁B₁ is the segment connecting the feet of the altitudes from A and B. Given that the triangle is acute, A₁ and B₁ lie on the sides BC and AC respectively, so A₁B₁ is a segment inside the triangle.Therefore, the external angle bisector must pass through AB (the side opposite) and A₁B₁ (a segment inside the triangle). Therefore, the external angle bisector in question must be the one that is directed towards the interior of the triangle in such a way that it intersects both AB and A₁B₁. Wait, but external angle bisectors are generally outside the triangle. However, depending on the triangle's shape, maybe it can intersect AB and A₁B₁.Wait, this is getting a bit confusing. Maybe coordinate geometry can help.Let me attempt to compute the equation of the external angle bisector at C.First, we need to determine the direction of the external angle bisector.At point C (0,0), the two sides are CB along the positive x-axis to (1,0) and CA to (d, e). The external angle can be formed by either extending CB beyond C or extending CA beyond C.Case 1: Extend CB beyond C to the left, creating an external angle between the extended CB (which is along the negative x-axis) and side CA. The bisector of this external angle will be a line that splits the angle between the negative x-axis and the line CA.Case 2: Extend CA beyond C, creating an external angle between the extended CA (beyond C) and side CB (along the positive x-axis). The bisector of this external angle will split the angle between the extended CA and CB.Since the problem says the external angle bisector intersects AB and A₁B₁, which are both on the "original" side of the triangle, perhaps the external angle bisector in Case 1 is the correct one, because in Case 2, extending CA beyond C would take the external angle bisector away from the triangle, whereas in Case 1, extending CB beyond C, the external angle bisector might intersect AB and A₁B₁.Alternatively, let's calculate both possibilities.First, let's compute the external angle bisector when we extend CB beyond C.Original sides at C:- CB along positive x-axis: direction vector (1,0)- CA to (d, e): direction vector (d, e)External angle is formed by extending CB beyond C, so the extended CB is along the negative x-axis. The external angle is between the negative x-axis and CA.The bisector of this external angle can be found using the angle bisector formula.The angle bisector between two lines with direction vectors u and v can be found by normalizing the vectors and adding them. For external angle bisector, we take the negative of one vector.So, for the external angle between the negative x-axis (direction vector (-1, 0)) and CA (direction vector (d, e)), the bisector direction vector would be proportional to (-1, 0) normalized plus (d, e) normalized.Wait, the angle bisector lies in the angle formed by the two lines. The external angle bisector direction can be calculated as follows:If we have two lines with direction vectors v1 and v2, the external angle bisector direction is proportional to v1/|v1| - v2/|v2|.Wait, actually, for internal angle bisector, it's v1/|v1| + v2/|v2|, and for external, it's v1/|v1| - v2/|v2|. Hmm, I need to confirm.Alternatively, using the formula for angle bisectors in terms of slopes.Given two lines with slopes m1 and m2, the angle bisectors can be found using the formula:If the lines are given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the angle bisectors are given by (a1x + b1y + c1)/sqrt(a1² + b1²) = ±(a2x + b2y + c2)/sqrt(a2² + b2²)But in our case, the two lines forming the external angle at C are:1. The extended CB beyond C, which is the negative x-axis: line y = 0, x ≤ 0.2. The original side CA, which goes from C(0,0) to A(d, e): line y = (e/d)xWe need the external angle bisector between these two lines.So, the two lines are:Line 1: y = 0 (for x ≤ 0)Line 2: y = (e/d)xThe external angle bisector will be the bisector of the angle between Line 1 (extended CB) and Line 2 (CA), outside the triangle.Using the angle bisector formula:For two lines L1: a1x + b1y + c1 = 0 and L2: a2x + b2y + c2 = 0, the bisectors are given by:(a1x + b1y + c1)/sqrt(a1² + b1²) = ±(a2x + b2y + c2)/sqrt(a2² + b2²)In our case, Line 1 (extended CB) is y = 0 for x ≤ 0. But as a line, it's y = 0, with the condition x ≤ 0. However, as a line without restriction, it's y = 0.Line 2 is y - (e/d)x = 0.So, L1: 0x + 1y + 0 = 0L2: (-e/d)x + 1y + 0 = 0Thus, the bisectors satisfy:(0x + 1y + 0)/sqrt(0 + 1) = ±[ (-e/d)x + 1y + 0 ] / sqrt( (e/d)^2 + 1 )Simplify:y = ±[ (-e/d x + y ) / sqrt( (e²/d²) + 1 ) ]Multiply both sides by sqrt( (e²/d²) + 1 ):y sqrt( (e²/d²) + 1 ) = ±( -e/d x + y )Bring all terms to left:y sqrt( (e²/d²) + 1 ) ± e/d x ∓ y = 0Factor y:y [ sqrt( (e²/d²) + 1 ) ∓ 1 ] ± e/d x = 0This is getting complicated. Let me compute the external angle bisector direction using vectors.The external angle is between the negative x-axis and the line CA.The direction vector of the negative x-axis is (-1, 0). The direction vector of line CA is (d, e). The external angle bisector direction should be a vector that is a combination of the unit vectors in these directions, subtracted because it's external.So, unit vector along negative x-axis: (-1, 0)Unit vector along CA: (d, e)/sqrt(d² + e²)Then, the external bisector direction is (-1, 0) - (d, e)/sqrt(d² + e²)But I think the formula for external angle bisector between two lines with direction vectors u and v is to take the normalized u minus normalized v.So, direction vector of the external bisector: u/|u| - v/|v|In this case, u is along the negative x-axis: (-1, 0), so |u| = 1.v is along CA: (d, e), |v| = sqrt(d² + e²)So, direction vector: (-1, 0) - (d, e)/sqrt(d² + e²)So, the external angle bisector direction is:( -1 - d/sqrt(d² + e²), 0 - e/sqrt(d² + e²) )This seems complex, but perhaps we can parametrize the external angle bisector.Alternatively, since it's an angle bisector, maybe we can use the ratio of distances.In angle bisector theorem, the external angle bisector divides the opposite side externally in the ratio of the adjacent sides.Wait, yes, the Angle Bisector Theorem states that the external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides.So, in triangle ABC, the external angle bisector of angle C will meet the line AB at point L such that AL / LB = AC / CB.But since it's an external division, the ratio is negative.Wait, the Angle Bisector Theorem for external angle bisector says that the external angle bisector divides the opposite side externally in the ratio of the adjacent sides.So, in our case, AL / LB = AC / BC, but as an external division.Therefore, coordinates of point L can be found using external division.Given points A (d, e) and B (1, 0), and ratio AC / CB = m / n, where m = AC = sqrt(d² + e²), n = BC = 1 (since we placed B at (1,0)).Therefore, the coordinates of L can be found by externally dividing AB in the ratio m:n = sqrt(d² + e²):1.The formula for external division: if a point divides AB externally in the ratio λ:1, then the coordinates are [ (B - λ A) / (1 - λ) ]Wait, no. The external division formula is similar to the internal division, but with a negative sign.If a point L divides AB externally in the ratio AL:LB = m:n, then coordinates of L are:( (n * Ax - m * Bx)/(n - m), (n * Ay - m * By)/(n - m) )In our case, ratio AL:LB = AC:CB = sqrt(d² + e²):1Therefore, coordinates of L are:( (1 * d - sqrt(d² + e²) * 1 ) / (1 - sqrt(d² + e²)), (1 * e - sqrt(d² + e²) * 0 ) / (1 - sqrt(d² + e²)) )Simplify:x-coordinate: (d - sqrt(d² + e²)) / (1 - sqrt(d² + e²))y-coordinate: e / (1 - sqrt(d² + e²))Hmm, that seems messy, but maybe simplifying.Alternatively, rationalize the denominator:Multiply numerator and denominator by (1 + sqrt(d² + e²)):x-coordinate: [ (d - sqrt(d² + e²))(1 + sqrt(d² + e²)) ] / [ (1 - sqrt(d² + e²))(1 + sqrt(d² + e²)) ]Denominator becomes 1 - (d² + e²)Numerator:d(1) + d sqrt(d² + e²) - sqrt(d² + e²) - (d² + e²)= d + d sqrt(d² + e²) - sqrt(d² + e²) - d² - e²Similarly, y-coordinate:[ e (1 + sqrt(d² + e²)) ] / [ 1 - (d² + e²) ]This seems too complicated. Maybe there's a better approach.Alternatively, since we need CL = 2 CK, which implies that K is the midpoint of CL. So, if we can express coordinates of L and K, we can set up the relation that K is the midpoint.But we need to find the coordinates of K, which is the intersection of the external angle bisector with A₁B₁.Given that we have equations for the external angle bisector and for A₁B₁, we can solve for their intersection point K.Alternatively, maybe using vectors.Alternatively, perhaps there's a property or theorem that relates the external angle bisector with the orthocenter or the orthic triangle (the triangle formed by the feet of the altitudes).But since A₁B₁ is part of the orthic triangle, maybe there's a relation.Alternatively, maybe using trigonometric Ceva's theorem for concurrent lines.Alternatively, let's consider the ratio CL/CK = 2. Since K is on A₁B₁, perhaps using Menelaus' theorem on triangle A₁B₁C or some other triangle.But this is getting too vague.Alternatively, let's consider specific cases. Maybe assume angle C is 60°, see if the ratio CL/CK is 2, and check. If not, try 90°, but the triangle is acute, so angle C must be less than 90°. Wait, the problem states it's acute-angled, so all angles less than 90°, so angle C is less than 90°. Wait, but the problem is asking for angle C, which might be a special angle like 60°, 45°, 30°, etc.Alternatively, maybe angle C is 60°, and in that case, the ratio CL/CK becomes 2. Let's test this.Assume angle C is 60°. Let me construct triangle ABC with angle C = 60°, sides AC and BC of some length, compute the external angle bisector, find points L and K, compute CL and CK, check if CL = 2 CK.But this might take time, but maybe with specific coordinates.Let me choose coordinates such that angle C is 60°, sides AC and BC are of equal length for simplicity, making triangle ABC an acute triangle.Let me set point C at (0,0), point B at (1,0), and point A at (0.5, (√3)/2). This makes triangle ABC equilateral with all angles 60°, but that might be too specific. Wait, but in an equilateral triangle, all altitudes are also medians and angle bisectors, so maybe the external angle bisector would have specific properties.But in an equilateral triangle, the external angle bisector at C would be the same as the internal angle bisector of the external angle, which would be 120°, so the bisector would split it into two 60° angles. But in that case, the external angle bisector would be a line making 60° with side CB.But in this case, if we compute points L and K, maybe CL = 2 CK?Wait, but in an equilateral triangle, all the altitudes are equal, and A₁ and B₁ would coincide with midpoints, so A₁B₁ would be the midline, but then K would be the midpoint of CL. If CL = 2 CK, then that would hold, but in an equilateral triangle, the external angle bisector might not intersect AB at L and A₁B₁ at K such that CL = 2 CK. Maybe not. Let's check.Wait, in an equilateral triangle with side length 1, coordinates:C: (0,0)B: (1,0)A: (0.5, √3/2)Altitude from A is the vertical line to BC, which is the x-axis. So, A₁ is (0.5, 0).Altitude from B to AC: since AC is from (0,0) to (0.5, √3/2). The slope of AC is (√3/2 - 0)/(0.5 - 0) = √3. So, the altitude from B is perpendicular to AC, so slope -1/√3.Equation of AC: y = √3 xEquation of altitude from B(1,0): y = -1/√3 (x - 1)Find intersection B₁:Set √3 x = -1/√3 (x - 1)Multiply both sides by √3:3x = -(x - 1)3x = -x + 14x = 1x = 1/4y = √3 * 1/4 = √3 /4Thus, B₁ is (1/4, √3/4)Line A₁B₁ connects (0.5, 0) and (1/4, √3/4)Slope of A₁B₁: (√3/4 - 0)/(1/4 - 0.5) = (√3/4)/(-1/4) = -√3Equation: y - 0 = -√3 (x - 0.5)So, y = -√3 x + (√3 / 2)External angle bisector at C: Let's compute it.The external angle at C is 180° - 60° = 120°, so the bisector splits it into two 60° angles.Since in the equilateral triangle, the external angle bisector at C would be the line that is 60° from both the extended side and the original side.Wait, but in the equilateral triangle, all external angles are 120°, and their bisectors would be lines making 60° with the sides.But in this case, since the triangle is equilateral, the external angle bisector at C (which is 120°) would be a line that is at 60° to both the extended CB and the original CA.But in coordinate terms, since CB is along the x-axis from (0,0) to (1,0), extending CB beyond C would go to the left along the negative x-axis. The external angle between the negative x-axis and CA (which is at 60° from CB) would be 120°, so the bisector would be 60° from each.Therefore, the external angle bisector at C would be a line making 60° with the negative x-axis, which is equivalent to 120° from the positive x-axis.So, its slope is tan(120°) = tan(180° - 60°) = -tan(60°) = -√3.Thus, the external angle bisector at C is the line y = -√3 x.But in the equilateral triangle, the external angle bisector at C is y = -√3 x. Now, let's find points L and K.Point L is the intersection of the external angle bisector with AB.Line AB in the equilateral triangle connects (0.5, √3/2) and (1,0). The equation of AB can be found.Slope of AB: (0 - √3/2)/(1 - 0.5) = (-√3/2)/0.5 = -√3Equation of AB: y - √3/2 = -√3(x - 0.5)Simplify:y = -√3 x + (√3 * 0.5) + √3/2 = -√3 x + √3/2 + √3/2 = -√3 x + √3So, equation of AB: y = -√3 x + √3Intersection with external angle bisector y = -√3 x:Set -√3 x = -√3 x + √3This implies 0 = √3, which is impossible. Therefore, in the equilateral triangle, the external angle bisector at C (y = -√3 x) does not intersect AB, which contradicts the problem's statement. Hence, angle C cannot be 60°, at least in the equilateral case.Wait, that's odd. Maybe in the equilateral triangle, the external angle bisector is parallel to AB? Since both have slope -√3, which would mean they are parallel and never meet. But AB in our coordinates has equation y = -√3 x + √3, and the external angle bisector is y = -√3 x. Indeed, parallel lines, so they don't intersect. Hence, in the equilateral triangle, there is no point L, which contradicts the problem's condition. Therefore, angle C cannot be 60°, or maybe my coordinate choice is flawed.Alternatively, perhaps in my coordinate system for the equilateral triangle, the external angle bisector at C is not y = -√3 x. Let me re-examine.Wait, angle at C is 60°, so the external angle is 120°, and the external angle bisector should split it into two 60° angles. If CB is along the positive x-axis, and we extend it beyond C to the negative x-axis, the external angle is between the negative x-axis and CA.The direction of CA is from C(0,0) to A(0.5, √3/2), which is 60° above the x-axis.The external angle is between the negative x-axis and the line CA, which is 60° from the negative x-axis. The external angle is 120°, so the bisector would be 60° from each side.Therefore, from point C(0,0), the external angle bisector would be at 60° from the negative x-axis, which is 180° - 60° = 120° from the positive x-axis. So, the slope is tan(120°) = -√3, which gives the line y = -√3 x. However, this line is parallel to AB, as AB also has slope -√3. Hence, they never meet, which means no point L exists. Therefore, in the equilateral triangle, the problem's conditions are not satisfied. So angle C cannot be 60°.Therefore, maybe angle C is 90°? But the triangle is acute, so angle C must be less than 90°. Hence, angle C must be something else.Perhaps 45°? Let me test with angle C = 45°.Let me construct triangle ABC with angle C = 45°, sides AC and BC such that angle C is 45°, and triangle is acute.Let me place C at (0,0), B at (1,0), and A somewhere in the first quadrant such that angle C is 45°.To make angle C = 45°, the coordinates of A should satisfy tan(45°) = 1 = opposite/adjacent in some right triangle, but since angle C is between sides CB and CA, we need the angle between CB (along x-axis) and CA to be 45°. Therefore, the slope of CA is tan(45°) = 1, so CA is the line y = x. Therefore, point A can be at (1,1), making CA along y = x. But then, the triangle ABC would have points C(0,0), B(1,0), A(1,1). Let's check the angles.In this triangle:- CB is from (0,0) to (1,0), length 1.- CA is from (0,0) to (1,1), length √2.- AB is from (1,1) to (1,0), length 1.So, triangle with sides 1, 1, √2. Wait, this is a right-angled triangle at B, since AB is vertical from (1,0) to (1,1), and BC is horizontal from (0,0) to (1,0). So, angle at B is 90°, but the problem states it's acute-angled. Therefore, this won't work.So, perhaps another configuration. Let me choose point A such that angle C is 45°, but triangle remains acute.Let me set point A at (1,1). Then, CB is (0,0) to (1,0), CA is (0,0) to (1,1), and AB is (1,1) to (1,0). As above, this is right-angled at B, which is not allowed. So, let me choose A at (2,1). Then:- CB is length 1.- CA is from (0,0) to (2,1), length sqrt(4 + 1) = sqrt(5).- AB is from (2,1) to (1,0), length sqrt(1 + 1) = sqrt(2).Now, angle at C: between CB (1,0) and CA (2,1). The angle can be computed using the dot product.Dot product of CB and CA vectors:CB vector is (1,0), CA vector is (2,1).Dot product = 1*2 + 0*1 = 2.|CB| = 1, |CA| = sqrt(5)cos(angle C) = 2 / (1 * sqrt(5)) = 2/sqrt(5) ≈ 0.894, which corresponds to angle C ≈ 26.565°, which is less than 45°, so not 45°. Hmm.Alternatively, to get angle C = 45°, set point A at (1,1), but as we saw, it creates a right angle at B. Alternatively, use the Law of Cosines.Suppose angle C = 45°, sides AC = m, BC = n, and AB = p.By Law of Cosines:p² = m² + n² - 2mn cos(γ)Where γ = 45°.To keep the triangle acute, all angles must be less than 90°, so we need to ensure that for sides m, n, p, the squares of the sides satisfy the conditions for acute angles.Alternatively, let me choose sides AC = 2, BC = 2, angle C = 45°, then compute AB.By Law of Cosines:AB² = 2² + 2² - 2*2*2*cos(45°) = 8 - 8*(√2/2) = 8 - 4√2 ≈ 8 - 5.656 ≈ 2.344, so AB ≈ 1.531. Then, the other angles can be computed using Law of Sines.But this might be getting too involved. Maybe coordinate geometry isn't the best approach here.Let me try a different method, perhaps using trigonometric identities and properties of angle bisectors.Given that CL = 2 CK, and K is the midpoint of CL, as established earlier.Since K is on A₁B₁ and L is on AB, and both K and L lie on the external angle bisector of angle C, perhaps we can use the Angle Bisector Theorem in some way.The external angle bisector of angle C meets AB at L, and by the Angle Bisector Theorem, as mentioned earlier, AL / LB = AC / BC, but externally.So, AL / LB = AC / BC, with external division.Similarly, since K is on A₁B₁, which is the orthic line (the line connecting the feet of the altitudes), perhaps there's a relation between the ratios in the orthic triangle.Alternatively, consider triangles CKA₁ and CLB₁ or something similar.Alternatively, since A₁ and B₁ are feet of the altitudes, maybe we can relate angles or use similarity.Let me recall that in a triangle, the orthocenter H is the intersection of altitudes. Since we have altitudes AA₁ and BB₁, their intersection is the orthocenter H.But not sure if that helps directly.Alternatively, since A₁B₁ is the orthic line, maybe there's a relation between the angles in triangle A₁B₁C and the original triangle.Alternatively, use coordinates with C at the origin, but let me try to assign coordinates more strategically.Let me assume angle C is γ, which we need to find.Let me place point C at (0,0), side CB along the x-axis, point B at (c, 0), and point A somewhere in the plane. Let me use trigonometric coordinates based on angle γ.Let me denote:- CB = a- CA = b- Angle at C is γThen, coordinates of point B can be (a, 0), and coordinates of point A can be (b cos γ, b sin γ )This way, angle at C is γ, with CA of length b, CB of length a, and AB can be computed via the Law of Cosines.Now, altitudes AA₁ and BB₁ are drawn.Altitude from A to BC: since BC is along the x-axis, the foot A₁ is the projection of A onto BC, which has coordinates (b cos γ, 0 )Similarly, altitude from B to AC: the line AC goes from (0,0) to (b cos γ, b sin γ ). The slope of AC is (b sin γ)/(b cos γ) = tan γ. Therefore, the slope of the altitude from B is -cot γ.Equation of AC: y = tan γ xEquation of altitude from B(a, 0): y - 0 = -cot γ (x - a )Intersection point B₁ is the foot of the altitude from B to AC.Solving for intersection:y = tan γ xy = -cot γ (x - a )Set equal:tan γ x = -cot γ (x - a )Multiply both sides by cot γ:tan γ * cot γ x = - (x - a )Since tan γ * cot γ = 1,x = -x + a2x = ax = a / 2Then y = tan γ (a / 2 )So, coordinates of B₁ are ( a / 2, (a / 2 ) tan γ )Therefore, coordinates of A₁: (b cos γ, 0 )Coordinates of B₁: ( a / 2, (a / 2 ) tan γ )Now, line A₁B₁ connects (b cos γ, 0 ) and ( a / 2, (a / 2 ) tan γ )Let me find the equation of line A₁B₁.Slope of A₁B₁:m = [ (a/2 tan γ - 0 ) ] / [ (a/2 - b cos γ ) ] = ( (a/2) tan γ ) / ( a/2 - b cos γ )Equation of A₁B₁ using point A₁(b cos γ, 0 ):y - 0 = m (x - b cos γ )So, y = [ (a tan γ ) / (a - 2b cos γ ) ] (x - b cos γ )Now, the external angle bisector of angle C.As before, the external angle bisector at C divides the opposite side AB externally in the ratio of the adjacent sides: AL / LB = AC / CB = b / a (external division)Using the external division formula, coordinates of L are:L = ( (a * b cos γ - b * a ) / (a - b ), (a * b sin γ - b * 0 ) / (a - b ) )Wait, wait. Let's recall that external division in ratio m:n gives coordinates:( (n * Ax - m * Bx)/(n - m), (n * Ay - m * By)/(n - m) )In our case, ratio AL:LB = AC:CB = b:a, so m = b, n = a.Points A(b cos γ, b sin γ ) and B(a, 0 )Coordinates of L:x = (a * b cos γ - b * a ) / (a - b ) = (ab cos γ - ab ) / (a - b ) = ab (cos γ - 1 ) / (a - b )y = (a * b sin γ - b * 0 ) / (a - b ) = (ab sin γ ) / (a - b )Simplify x-coordinate:ab (cos γ - 1 ) / (a - b ) = ab ( - (1 - cos γ ) ) / (a - b ) = -ab (1 - cos γ ) / (a - b )Similarly, coordinates of L:( -ab (1 - cos γ ) / (a - b ), ab sin γ / (a - b ) )Now, the external angle bisector passes through C(0,0 ) and L. Therefore, its parametric equation is:x = t * ( -ab (1 - cos γ ) / (a - b ) )y = t * ( ab sin γ / (a - b ) )For t >= 0, this parametrizes the external angle bisector from C to L and beyond.Now, point K is the intersection of this external angle bisector with A₁B₁.We need to find the parameter t where the external angle bisector intersects A₁B₁.Equation of external angle bisector:x = t * ( -ab (1 - cos γ ) / (a - b ) )y = t * ( ab sin γ / (a - b ) )Equation of A₁B₁:y = [ (a tan γ ) / (a - 2b cos γ ) ] (x - b cos γ )Substitute x and y from external angle bisector into this equation:t * ( ab sin γ / (a - b ) ) = [ (a tan γ ) / (a - 2b cos γ ) ] ( t * ( -ab (1 - cos γ ) / (a - b ) ) - b cos γ )Simplify:Left side: t * ( ab sin γ / (a - b ) )Right side: (a tan γ / (a - 2b cos γ )) * ( - t ab (1 - cos γ ) / (a - b ) - b cos γ )Let me compute the right side step by step.First, factor out terms:= (a tan γ / (a - 2b cos γ )) * [ - t ab (1 - cos γ ) / (a - b ) - b cos γ ]= (a tan γ / (a - 2b cos γ )) * [ - ( t ab (1 - cos γ ) + b cos γ (a - b ) ) / (a - b ) ]= - (a tan γ / (a - 2b cos γ )) * [ t ab (1 - cos γ ) + b cos γ (a - b ) ] / (a - b )Now, equate left and right sides:t * ( ab sin γ / (a - b ) ) = - (a tan γ / (a - 2b cos γ )) * [ t ab (1 - cos γ ) + b cos γ (a - b ) ] / (a - b )Multiply both sides by (a - b ):t ab sin γ = - (a tan γ / (a - 2b cos γ )) [ t ab (1 - cos γ ) + b cos γ (a - b ) ]Simplify tan γ = sin γ / cos γ:t ab sin γ = - (a sin γ / cos γ / (a - 2b cos γ )) [ t ab (1 - cos γ ) + b cos γ (a - b ) ]Multiply both sides by cos γ:t ab sin γ cos γ = - a sin γ / (a - 2b cos γ ) [ t ab (1 - cos γ ) + b cos γ (a - b ) ]Divide both sides by sin γ (assuming sin γ ≠ 0, which it isn't since γ is acute):t ab cos γ = - a / (a - 2b cos γ ) [ t ab (1 - cos γ ) + b cos γ (a - b ) ]Now, expand the bracket:= - a / (a - 2b cos γ ) [ t ab (1 - cos γ ) + b cos γ (a - b ) ]Factor b from the bracket:= - a b / (a - 2b cos γ ) [ t a (1 - cos γ ) + cos γ (a - b ) ]Now, equation becomes:t ab cos γ = - a b / (a - 2b cos γ ) [ t a (1 - cos γ ) + cos γ (a - b ) ]Cancel ab from both sides:t cos γ = - a / (a - 2b cos γ ) [ t a (1 - cos γ ) + cos γ (a - b ) ]Multiply both sides by (a - 2b cos γ ):t cos γ (a - 2b cos γ ) = - a [ t a (1 - cos γ ) + cos γ (a - b ) ]Expand left side:t a cos γ - 2 t b cos² γ = - a [ t a (1 - cos γ ) + a cos γ - b cos γ ]Expand the right side:= - a t a (1 - cos γ ) - a (a cos γ - b cos γ )= - t a² (1 - cos γ ) - a² cos γ + a b cos γNow, bring all terms to left side:t a cos γ - 2 t b cos² γ + t a² (1 - cos γ ) + a² cos γ - a b cos γ = 0Factor terms with t:t [ a cos γ - 2b cos² γ + a² (1 - cos γ ) ] + a² cos γ - a b cos γ = 0Let me compute the coefficient of t:= a cos γ - 2b cos² γ + a² (1 - cos γ )= a cos γ + a² - a² cos γ - 2b cos² γ= a² + a cos γ - a² cos γ - 2b cos² γ= a² (1 - cos γ ) + a cos γ (1 - a ) - 2b cos² γWait, perhaps better to factor differently.Alternatively, group terms:= a² (1 - cos γ ) + a cos γ - 2b cos² γSimilarly, the constant term:a² cos γ - a b cos γ = a cos γ (a - b )So, the equation is:t [ a² (1 - cos γ ) + a cos γ - 2b cos² γ ] + a cos γ (a - b ) = 0Solve for t:t = - [ a cos γ (a - b ) ] / [ a² (1 - cos γ ) + a cos γ - 2b cos² γ ]Let me factor numerator and denominator.Numerator: - a cos γ (a - b )Denominator: a² (1 - cos γ ) + a cos γ - 2b cos² γ= a² - a² cos γ + a cos γ - 2b cos² γ= a² + (-a² cos γ + a cos γ ) - 2b cos² γ= a² + a cos γ (1 - a ) - 2b cos² γHmm, this seems complicated. Maybe factor terms with cos γ:= a² + a cos γ (1 - a ) - 2b cos² γAlternatively, factor cos² γ:= a² - 2b cos² γ + a cos γ (1 - a )Not helpful. Perhaps substitute specific ratios between a and b. Since the problem doesn't specify sides, maybe there's a relation between a and b that comes from CL = 2 CK.Recall that CL = 2 CK, which implies that the parameter t for point L is twice the parameter t for point K.Wait, in the parametrization of the external angle bisector, points K and L are along the line parametrized by t. Since K is between C and L, and CL = 2 CK, then the parameter t for L is twice that for K.Wait, in the parametrization:For point K, parameter t = t_kFor point L, parameter t = t_l = 2 t_kBut in our earlier calculation, we derived t for point K (intersection with A₁B₁). However, point L is further along the external angle bisector, so t_l is larger. Therefore, if CL = 2 CK, then t_l = 2 t_k.But from the problem statement, point L is the intersection with AB, and point K is the intersection with A₁B₁. Therefore, in our parametrization, the parameter t for L is different from that of K.But in our earlier calculation, we found t for K (intersection with A₁B₁). If we can also find t for L (intersection with AB), then we can set t_l = 2 t_k.But wait, point L is on AB, which we already found using the external division. The coordinates of L were expressed in terms of a, b, and γ. However, in the parametrization of the external angle bisector, we can also express L's coordinates.Wait, in the parametrization:x = t * ( -ab (1 - cos γ ) / (a - b ) )y = t * ( ab sin γ / (a - b ) )But we also have coordinates of L from the external division:L = ( -ab (1 - cos γ ) / (a - b ), ab sin γ / (a - b ) )Wait, comparing this to the parametrization, it's when t = 1. Therefore, in the parametrization, point L corresponds to t = 1.But that can't be right, because in the parametrization, t is a scaling factor from C along the external angle bisector. If t = 1 corresponds to point L, then t_k for point K would be t = 1/2, since CL = 2 CK implies CK = CL / 2, so t_k = t_l / 2 = 1/2.But in our previous calculation, solving for t gave us t_k = [some expression]. Therefore, setting t_l = 1 (from parametrization) and t_k = 1/2, and equate the expression for t_k to 1/2.From earlier, we had:t_k = - [ a cos γ (a - b ) ] / [ a² (1 - cos γ ) + a cos γ - 2b cos² γ ]But since t_k should be 1/2, we set:- [ a cos γ (a - b ) ] / [ a² (1 - cos γ ) + a cos γ - 2b cos² γ ] = 1/2Multiply both sides by denominator:- a cos γ (a - b ) = (1/2) [ a² (1 - cos γ ) + a cos γ - 2b cos² γ ]Multiply both sides by 2:-2 a cos γ (a - b ) = a² (1 - cos γ ) + a cos γ - 2b cos² γExpand left side:-2 a² cos γ + 2 a b cos γ = a² - a² cos γ + a cos γ - 2b cos² γBring all terms to left side:-2 a² cos γ + 2 a b cos γ - a² + a² cos γ - a cos γ + 2b cos² γ = 0Combine like terms:(-2a² cos γ + a² cos γ ) + (2a b cos γ ) + (-a² ) + (-a cos γ ) + (2b cos² γ ) = 0= (-a² cos γ ) + 2a b cos γ - a² - a cos γ + 2b cos² γ = 0Factor terms:= -a² cos γ - a² + (-a cos γ ) + 2a b cos γ + 2b cos² γ= -a² (cos γ + 1 ) - a cos γ + 2a b cos γ + 2b cos² γ= -a² (cos γ + 1 ) + cos γ ( -a + 2a b ) + 2b cos² γHmm, this seems complicated. Maybe factor out common terms.Alternatively, let's rearrange terms:- a² (cos γ + 1 ) - a cos γ + 2a b cos γ + 2b cos² γ = 0Factor -a from the first two terms:= -a [ a (cos γ + 1 ) + cos γ ] + 2a b cos γ + 2b cos² γ = 0= -a [ a cos γ + a + cos γ ] + 2a b cos γ + 2b cos² γ = 0= -a² cos γ - a² - a cos γ + 2a b cos γ + 2b cos² γ = 0This brings us back to the same expression.Perhaps factor terms with cos γ:= cos γ (-a² - a + 2a b ) + (-a² ) + 2b cos² γ = 0Let me write this as:2b cos² γ + cos γ ( -a² - a + 2a b ) - a² = 0This is a quadratic equation in cos γ. Let me denote x = cos γ.Then, the equation becomes:2b x² + ( -a² - a + 2a b ) x - a² = 0Let me write this as:2b x² + (2a b - a² - a ) x - a² = 0This quadratic equation can be solved for x in terms of a and b. However, since the problem doesn't specify a and b, this suggests that the equation must hold for any a and b, which is only possible if the coefficients of corresponding terms are zero. But this seems unlikely. Alternatively, there might be a relation between a and b that allows this equation to hold for a specific x (i.e., cos γ).Alternatively, perhaps the equation can be satisfied for a ratio of a/b.Let me assume k = a/b. Let me set k = a/b, so a = k b.Substitute a = k b into the equation:2b x² + (2 k b * b - (k b )² - k b ) x - (k b )² = 0Simplify:2b x² + (2 k b² - k² b² - k b ) x - k² b² = 0Divide both sides by b (assuming b ≠ 0):2 x² + (2 k b - k² b - k ) x - k² b = 0Wait, this still has b in the coefficients, which complicates things. Maybe this approach isn't fruitful.Alternatively, perhaps there's a special angle γ where this equation holds regardless of a and b, which would imply that the coefficients of a², a, etc., are zero.Looking at the original equation:- a² (cos γ + 1 ) + cos γ ( -a + 2a b ) + 2b cos² γ = 0But this seems impossible unless coefficients of a², a, and constants are zero:For the coefficient of a²:- (cos γ + 1 ) = 0 ⇒ cos γ = -1, which is impossible since γ is acute.For the coefficient of a:cos γ ( -1 + 2b ) = 0 ⇒ Either cos γ = 0 or b = 1/2. But cos γ = 0 implies γ = 90°, which is not acute. If b = 1/2, then for the constant term:2b cos² γ = 2*(1/2)*cos² γ = cos² γ. So, equation becomes:0 + 0 + cos² γ = 0 ⇒ cos γ = 0, which again implies γ = 90°, which is invalid.Therefore, there must be a specific relation between a and b for this to hold, which depends on γ. Therefore, perhaps the only solution is when the equation holds for specific ratios of a/b and γ.This seems complicated. Maybe there's a smarter approach.Let me recall that in the problem statement, the ratio CL = 2 CK is given, and we need to find angle C. Perhaps this ratio implies a specific angle, regardless of the triangle's side lengths.Let me consider using trigonometric identities with the law of sines or cosines.Alternatively, consider triangle CK L.Since K is the midpoint of CL, then CK = KL.Given that K is on A₁B₁ and L is on AB, both on the external angle bisector.Maybe there are similar triangles involved.Alternatively, consider the homothety (similarity transformation) centered at C that maps K to L. Since CK = 1/2 CL, the scale factor is 2.Therefore, the homothety with center C and factor 2 maps K to L. Therefore, A₁B₁ is mapped to AB under this homothety.Therefore, AB is the image of A₁B₁ under a homothety of factor 2 centered at C.Therefore, AB is parallel to A₁B₁ and twice as long.If AB is parallel to A₁B₁, then the slope of AB equals the slope of A₁B₁.Let me verify this.From earlier, in the coordinate system where C is at (0,0), B at (a,0), and A at (b cos γ, b sin γ ), the slope of AB is:(0 - b sin γ ) / (a - b cos γ ) = -b sin γ / (a - b cos γ )Slope of A₁B₁ is:From coordinates of A₁ (b cos γ, 0 ) and B₁ (a/2, (a/2 ) tan γ ), slope is:( (a/2 ) tan γ - 0 ) / ( a/2 - b cos γ ) = ( (a/2 ) tan γ ) / ( a/2 - b cos γ )If AB is parallel to A₁B₁, their slopes must be equal:-b sin γ / (a - b cos γ ) = ( (a/2 ) tan γ ) / ( a/2 - b cos γ )Simplify RHS:( (a/2 ) tan γ ) / ( a/2 - b cos γ ) = ( (a/2 )( sin γ / cos γ ) ) / ( a/2 - b cos γ )Multiply numerator and denominator by 2:= ( a sin γ / cos γ ) / ( a - 2b cos γ )Thus, equating slopes:- b sin γ / (a - b cos γ ) = ( a sin γ / cos γ ) / ( a - 2b cos γ )Multiply both sides by (a - b cos γ )(a - 2b cos γ ):- b sin γ (a - 2b cos γ ) = ( a sin γ / cos γ ) (a - b cos γ )Divide both sides by sin γ (assuming sin γ ≠ 0):- b (a - 2b cos γ ) = ( a / cos γ ) (a - b cos γ )Multiply both sides by cos γ:- b cos γ (a - 2b cos γ ) = a (a - b cos γ )Expand left side:- a b cos γ + 2b² cos² γ = a² - a b cos γBring all terms to left:- a b cos γ + 2b² cos² γ - a² + a b cos γ = 0Simplify:2b² cos² γ - a² = 0Thus:2b² cos² γ = a²Therefore:cos² γ = a² / (2b² )cos γ = a / (b sqrt(2 ) )But since cos γ must be between 0 and 1 (as γ is acute), this implies that a / (b sqrt(2 )) ≤ 1 ⇒ a ≤ b sqrt(2 )But a and b are lengths of sides CB and CA, respectively.Additionally, from the Law of Cosines in triangle ABC:AB² = a² + b² - 2ab cos γBut we also have from the above relation, 2b² cos² γ = a² ⇒ cos γ = a / (b sqrt(2 ))Substitute into Law of Cosines:AB² = a² + b² - 2ab * (a / (b sqrt(2 )) )Simplify:AB² = a² + b² - 2a² / sqrt(2 )= a² + b² - sqrt(2 ) a²But AB must be a real positive length, so this expression must be positive. However, this doesn't immediately help us find γ.But from the previous result, cos γ = a / (b sqrt(2 ) )But we need to relate this to the angle γ. However, unless there's a relation between a and b, we cannot determine γ uniquely. Therefore, this suggests that the problem's condition CL = 2 CK imposes a relation between sides a and b, and angle γ, but to find γ uniquely, additional constraints must come from the triangle being acute and the configuration of the points.Alternatively, perhaps the only possibility is when cos γ = 1/√2, which implies γ = 45°, but we saw earlier that in such a case, the triangle might not satisfy the acute condition. Wait, if cos γ = 1/√2, then γ = 45°, which is acute. Let's check if this satisfies the equation 2b² cos² γ = a².If cos γ = 1/√2, then:2b² (1/2 ) = a² ⇒ b² = a² ⇒ b = aTherefore, sides CB and CA are equal: a = b.Thus, triangle ABC is isoceles with AC = BC = a, angle C = 45°.In this case, let's verify if CL = 2 CK.Let me construct such a triangle.Let me set a = b = 1 for simplicity. So, CB = 1, CA = 1, angle C = 45°, so coordinates:C: (0,0)B: (1,0)A: ( cos 45°, sin 45° ) = (√2/2, √2/2 )Now, compute altitudes AA₁ and BB₁.Altitude from A to BC: since BC is along x-axis from (0,0) to (1,0), the foot A₁ is the projection of A onto BC, which is (√2/2, 0 )Altitude from B to AC: line AC is from (0,0) to (√2/2, √2/2 ), slope 1. The altitude from B(1,0) is perpendicular, slope -1. Equation: y = -x + 1Intersection with AC: line AC is y = xIntersection at solving y = x and y = -x + 1:x = -x + 1 ⇒ 2x = 1 ⇒ x = 0.5 ⇒ y = 0.5Thus, B₁ is (0.5, 0.5 )Line A₁B₁ connects (√2/2, 0 ) and (0.5, 0.5 )Compute slope:m = (0.5 - 0 ) / (0.5 - √2/2 ) = 0.5 / (0.5 - √2/2 )Rationalize denominator:Multiply numerator and denominator by 2:= 1 / (1 - √2 )Multiply numerator and denominator by (1 + √2 ):= (1 + √2 ) / (1 - 2 ) = - (1 + √2 )Thus, slope m = - (1 + √2 )Equation of A₁B₁: y - 0 = - (1 + √2 )(x - √2/2 )Now, the external angle bisector at C.Angle at C is 45°, external angle is 135°, bisected into 67.5°.Direction of the external angle bisector: since internal angle bisector is at 22.5° from CB, but external angle bisector would be 67.5° from CB (extended).But since the external angle bisector needs to intersect AB and A₁B₁, let's compute its equation.Given that the external angle bisector divides AB externally in the ratio AC:CB = 1:1 (since AC = CB = 1), so AL / LB = 1/1 externally.Using external division formula for AB with ratio 1:1:Coordinates of L:x = (1*1 - 1*1 ) / (1 - 1 ) → undefined, but this suggests that the external division is at infinity, which can't be.Wait, this is because when AC = CB, the external angle bisector is parallel to AB.Wait, if AC = CB, then the external angle bisector at C is parallel to AB. But in this case, AB has slope:From A(√2/2, √2/2 ) to B(1,0 ), slope = (0 - √2/2 ) / (1 - √2/2 ) = (-√2/2 ) / (1 - √2/2 )Rationalizing the denominator:Multiply numerator and denominator by (1 + √2/2 ):= (-√2/2 )(1 + √2/2 ) / [ (1)^2 - (√2/2 )^2 ] = (-√2/2 - (√2 * √2 )/4 ) / (1 - 0.5 ) = (-√2/2 - 0.5 ) / 0.5 = (-√2 - 1 ) / 1 = - ( √2 + 1 )Which matches the slope of A₁B₁ computed earlier: - (1 + √2 )Thus, in this isoceles triangle with angle C = 45°, the external angle bisector is parallel to AB and A₁B₁, hence it does not intersect AB, which contradicts the problem's condition. Therefore, angle C cannot be 45°.This suggests that my previous approach has a mistake, or perhaps the assumption that a = b is invalid.Alternatively, maybe the correct angle is 60°, but we saw earlier in the equilateral triangle it doesn't work. Alternatively, maybe 30°.Let me try angle C = 60° with a ≠ b.Assume angle C = 60°, so cos γ = 0.5.From the equation derived earlier: 2b² cos² γ = a² ⇒ 2b² (0.25 ) = a² ⇒ a² = 0.5 b² ⇒ a = b / √2Thus, CB = a = b / √2, CA = b.Let me set b = √2, then a = 1.Coordinates:C: (0,0)B: (1,0 )A: (√2 cos 60°, √2 sin 60° ) = (√2 * 0.5, √2 * (√3 / 2 )) = (√2/2, (√6)/2 )Compute altitudes:Altitude from A to BC: foot at A₁, which is (√2/2, 0 )Altitude from B to AC:Equation of AC: from (0,0 ) to (√2/2, √6/2 ). Slope = (√6/2 ) / (√2/2 ) = √3.Thus, equation of AC: y = √3 xEquation of altitude from B(1,0 ) is perpendicular, slope = -1/√3.Equation: y = -1/√3 (x - 1 )Intersection with AC:√3 x = -1/√3 (x - 1 )Multiply both sides by √3:3x = - (x - 1 )3x = -x + 14x = 1 ⇒ x = 1/4y = √3 * 1/4 = √3/4Thus, B₁ is,<think>Okay, so I have this geometry problem here. Let me try to understand it step by step. It's about an acute-angled triangle ABC. That means all three angles are less than 90 degrees. Then, they mention that the altitudes AA₁ and BB₁ are drawn. So, altitude from A to BC is AA₁, and altitude from B to AC is BB₁. These altitudes intersect BC and AC at points A₁ and B₁ respectively. Got that.Next, the problem says that the bisector of the exterior angle at vertex C intersects the lines AB and A₁B₁ at points L and K respectively. Hmm, let me visualize that. So, vertex C has an exterior angle. The exterior angle at C would be formed by one side adjacent to C (say, BC) and the extension of the other side (AC). Wait, actually, the exterior angle could be formed by extending either BC or AC. Let me clarify: in a triangle, each vertex has two exterior angles, depending on which side is extended. But since they mention "the" bisector of the exterior angle at C, maybe there's a specific one they're referring to. Wait, maybe it's the exterior angle adjacent to angle C. Let me think. In triangle ABC, angle at C is between sides AC and BC. The exterior angle would be if we extend either AC beyond C or BC beyond C. But since they are talking about the bisector of the exterior angle, maybe it's the one that's not adjacent to the triangle's interior. So, for example, if we extend side BC beyond C, forming an exterior angle at C, which is 180 degrees minus angle C. Then, the bisector of that exterior angle would split it into two equal parts. Similarly, if we extend AC beyond C, we get another exterior angle, which is also 180 minus angle C. Wait, but regardless of which side we extend, the exterior angle's measure is 180 - angle C. So the bisector would split that into two angles of (180 - angle C)/2 each. Hmm. But the problem doesn't specify which side is extended. Maybe it's the extension opposite to AB? Wait, the bisector of the exterior angle at C should intersect AB and A₁B₁. Let me try to sketch this mentally.First, triangle ABC is acute-angled. Let me consider the exterior angle at C when we extend BC beyond C. So, the exterior angle would be between BC extended and AC. Then, the bisector of this angle would start at C and go outwards, splitting the exterior angle into two equal parts. Similarly, if we extend AC beyond C, we get another exterior angle between AC extended and BC, and its bisector. However, the problem says that this bisector intersects AB and A₁B₁ at points L and K. So, depending on which direction the bisector goes, it might intersect AB first, then A₁B₁. Wait, if we bisect the exterior angle at C (let's say extending BC beyond C), then the bisector would go towards the exterior, but would it intersect AB? Let me think.Alternatively, maybe the exterior angle is formed by extending side AC beyond C, so the exterior angle is between AC extended and BC. Then, the bisector of that angle would start at C and go into the plane. Let's consider coordinates to make this more concrete. Maybe setting up coordinate axes will help. Let me place point C at the origin (0,0), point B along the x-axis, and point A somewhere in the plane. Then, construct the altitudes and the bisector.But perhaps there's a better approach. Let me recall that in triangle problems, sometimes using properties of angle bisectors, similar triangles, or trigonometric identities can help. The key here is that CL = 2 CK. So, the distance from C to L is twice the distance from C to K. Since L and K are points where the exterior angle bisector intersects AB and A₁B₁ respectively, there might be a ratio involved here that we can exploit.First, let's note that A₁B₁ is the orthocenter-related segment? Wait, in an acute triangle, the orthocenter lies inside the triangle. The altitudes AA₁ and BB₁ intersect at the orthocenter H. Then, A₁B₁ would be the segment connecting the feet of the altitudes from A and B. So, A₁B₁ is part of the orthic triangle. Hmm.Alternatively, maybe triangle A₁B₁C is related to the original triangle. Let me try to recall that in some cases, the orthic triangle (formed by the feet of the altitudes) has properties similar to the original triangle. But not sure if that's directly applicable here.But back to the problem. The bisector of the exterior angle at C meets AB at L and A₁B₁ at K. Then, CL = 2 CK. So, perhaps we can use the Angle Bisector Theorem here. The Angle Bisector Theorem states that the bisector of an angle (interior or exterior) divides the opposite side into segments proportional to the adjacent sides.Wait, in the case of an exterior angle bisector, the theorem might be slightly different. Let me recall. For an exterior angle bisector, the Angle Bisector Theorem would state that the bisector divides the opposite side externally in the ratio of the adjacent sides. So, for example, if we have triangle ABC, and we bisect the exterior angle at C, then the bisector would intersect line AB at some point L such that AL / LB = AC / BC. But since it's an external division, the ratio would be AL / LB = AC / BC, but with a negative sign, indicating external division.But in this problem, the bisector intersects AB at L. So, according to the theorem, AL / LB = AC / BC. Hmm. Let me check.Yes, the exterior angle bisector theorem states that the exterior angle bisector divides the opposite side externally in the ratio of the adjacent sides. So, in this case, point L divides AB externally in the ratio AC : BC. That is, AL / LB = AC / BC, but since it's an external division, the point L is located on the extension of AB beyond B or A, depending on the ratio. But since triangle ABC is acute-angled, and the exterior angle is being bisected, maybe L is on AB itself? Wait, not necessarily. Wait, the external division would place L outside the segment AB. But the problem states that the bisector intersects AB at L. So, perhaps L is on segment AB. Hmm, this might not always be the case. Wait, maybe depending on the angle.Wait, if we have an exterior angle bisector, it should start at C and go outside the triangle. So, if we extend BC beyond C, the bisector of the exterior angle at C (between BC extended and AC) would go in some direction. If the bisector intersects AB at L, then L must be on AB. So, perhaps in this configuration, the exterior angle bisector does intersect AB at some point L between A and B. But that might depend on the triangle's angles.Alternatively, maybe in an acute-angled triangle, the exterior angle bisector at C (which is an obtuse angle since it's 180 - angle C) would intersect AB extended beyond A or B. Wait, the problem states that the bisector intersects AB and A₁B₁ at L and K. So, L is on AB, not its extension, and K is on A₁B₁.Wait, but maybe in an acute triangle, the exterior angle bisector can intersect AB within the segment. Let me take an example. Suppose angle C is 60 degrees. Then, the exterior angle is 120 degrees, and the bisector would split it into two 60-degree angles. So, the bisector would form a 60-degree angle with the extended side. Depending on the triangle's configuration, this bisector might intersect AB somewhere.Alternatively, perhaps using coordinate geometry will help here. Let me set up coordinates. Let me place point C at the origin (0, 0), point B at (c, 0), and point A at (d, e), where d and e are positive because the triangle is acute-angled. Then, the altitude from A to BC (which is the x-axis) would be the vertical line from A to BC, so A₁ would be the projection of A onto BC. Similarly, the altitude from B to AC would be BB₁, which is the line from B perpendicular to AC. The coordinates might get messy, but maybe manageable.Alternatively, use trigonometric relationships. Let me denote angle at C as γ. Since the triangle is acute, γ < 90 degrees. The exterior angle at C would then be 180 - γ, and its bisector would split it into two angles of (180 - γ)/2 each. The direction of the bisector can be determined based on the sides.Alternatively, let me consider vectors. The exterior angle bisector direction can be represented as a vector that is the sum of the unit vectors along the external angle's sides. So, if we extend BC beyond C, the external angle is between BC (from C to B to extension) and AC. Wait, this might be getting too vague. Let me try to think of it another way.Suppose in triangle ABC, we have the exterior angle at C formed by extending BC to a point D, so angle DCA is the exterior angle. The bisector of angle DCA would then be a line that splits angle DCA into two equal angles. The problem states that this bisector intersects AB at L and A₁B₁ at K. Then, CL = 2 CK. So, CK is half of CL. Therefore, point K is the midpoint of CL? Wait, not necessarily, because they are along the same line, but CK is half of CL. So, CK = (1/2) CL, which would mean that K divides CL internally in the ratio 1:1? Wait, if CL = 2 CK, then CK is half of CL. So, if we consider points along the bisector from C, then K is halfway between C and L. So, K is the midpoint of CL. Therefore, if we can show that for some angle C, this midpoint property holds, given the other conditions.But how does this relate to the triangle's angles? Maybe we can use similarity or some ratio properties. Let me think.Since L is on AB and K is on A₁B₁, and CL = 2 CK, perhaps we can relate the lengths using coordinates or vectors.Let me try coordinate geometry. Let me set up a coordinate system with point C at (0, 0). Let me let side AB be in the plane, and set coordinates such that point C is at (0,0), point B is at (1,0), and point A is at (0,1). Wait, but then triangle ABC would be right-angled at C. But the problem states it's acute-angled, so maybe that's not a good idea. Let me instead place point C at (0,0), point B at (b, 0), and point A at (a, c), where a, b, c are positive. Then, we can compute the altitudes and the bisector.But this might get complicated. Alternatively, maybe use barycentric coordinates or something. Wait, maybe another approach.Since AA₁ and BB₁ are altitudes, then A₁ is the foot from A to BC, and B₁ is the foot from B to AC. Then, A₁B₁ is the segment connecting these two feet. The line A₁B₁ is called the orthic line or something similar. Maybe there's a property about this line.Alternatively, consider triangle A₁B₁C. This is the pedal triangle of the orthocenter, but since the triangle is acute, the orthocenter is inside the triangle. Hmm.Alternatively, since CL = 2 CK, and K is on A₁B₁, maybe using homothety. A homothety centered at C that maps K to L would have a ratio of 2. So, if such a homothety exists, then L is the image of K under a homothety with center C and factor 2. But for this to map A₁B₁ to AB, the lines AB and A₁B₁ would have to be parallel, which they aren't necessarily. Hmm, maybe not.Wait, but if CL = 2 CK, then the homothety centered at C with ratio 2 would send K to L. So, if K is on A₁B₁, then L must be on the image of A₁B₁ under this homothety. The image of A₁B₁ under homothety centered at C with ratio 2 would be a line parallel to A₁B₁ but twice as far from C. However, L is on AB. So, unless AB is parallel to A₁B₁, which would require that the homothety sends A₁B₁ to AB. Wait, but if AB is not parallel to A₁B₁, then the intersection point L is on AB, but this might not hold. Therefore, maybe AB is parallel to A₁B₁? If that's the case, then triangle ABC would have AB parallel to A₁B₁. Is that possible?Wait, in general, in a triangle, the orthic line (A₁B₁) is not parallel to AB unless the triangle is equilateral? Wait, maybe not even then. Let me check. In an equilateral triangle, all altitudes are also medians and angle bisectors. So, the feet of the altitudes would be the midpoints, so A₁ and B₁ would be midpoints, so A₁B₁ would be midline parallel to AB, and indeed, in that case, A₁B₁ is parallel to AB and half its length. So, in an equilateral triangle, A₁B₁ is parallel to AB and half its length. Then, if C is the third vertex, and we have a homothety centered at C with ratio 2, mapping A₁B₁ to AB. Then, any point K on A₁B₁ would map to L on AB such that CL = 2 CK. So, if angle C is 60 degrees, as in an equilateral triangle, this would hold. Therefore, angle C is 60 degrees. So, maybe the answer is 60 degrees.But wait, in the problem, it's not necessarily an equilateral triangle. So, is this only true for angle C being 60 degrees? Let me check. Suppose angle C is 60 degrees. Then, in an equilateral triangle, it's true. But suppose angle C is something else. Let me take a triangle with angle C = 60 degrees, but not equilateral. Let's say triangle ABC with angle C = 60 degrees, sides AC and BC not equal. Then, would CL = 2 CK?Alternatively, maybe the condition CL = 2 CK only holds when angle C is 60 degrees. Let me test this with coordinate geometry.Let me set up coordinates. Let me place point C at (0,0). Let me set angle C as 60 degrees. Let me assume sides AC and BC have lengths a and b. Then, coordinates of A can be (a, 0), and coordinates of B can be (b cos 60°, b sin 60°) = (b/2, (b√3)/2). Wait, but angle at C is 60 degrees. Wait, if I place point C at (0,0), point A at (a, 0), and point B at (d, e), then angle at C is 60 degrees, so the cosine of angle C is (AC² + BC² - AB²)/(2*AC*BC) = 0.5. So, (a² + (d² + e²) - ( (a - d)^2 + e^2 )) / (2*a*sqrt(d² + e²)) ) = 0.5. Let me compute this.Numerator: a² + d² + e² - ( (a - d)^2 + e² ) = a² + d² + e² - (a² - 2ad + d² + e² ) = a² + d² + e² - a² + 2ad - d² - e² = 2ad.Therefore, numerator is 2ad. Denominator is 2*a*sqrt(d² + e²).So, 2ad / (2a sqrt(d² + e²)) ) = d / sqrt(d² + e²) = 0.5.Therefore, d / sqrt(d² + e²) = 0.5. Squaring both sides: d² / (d² + e²) = 0.25 => 4d² = d² + e² => 3d² = e² => e = d√3.So, coordinates of B are (d, d√3). So, to simplify, let me set d = 1, so B is (1, √3). Then, AC is length a, from (0,0) to (a,0). AB is from (a,0) to (1, √3). The distance AB squared is (1 - a)^2 + (√3 - 0)^2 = (1 - 2a + a²) + 3 = a² - 2a + 4.But angle at C is 60 degrees. So, in this setup, the triangle has angle C = 60 degrees. Let's choose a specific a. Let me take a = 2. Then, point A is (2, 0), point B is (1, √3), point C is (0,0). Now, let's find the altitudes AA₁ and BB₁.Altitude from A to BC: First, BC is from (0,0) to (1, √3). The equation of BC is y = √3 x. The altitude from A(2,0) to BC is perpendicular to BC. The slope of BC is √3, so the slope of the altitude is -1/√3. The equation of the altitude is y - 0 = -1/√3 (x - 2). So, y = (-1/√3)x + 2/√3.Intersection point A₁ is where this altitude meets BC. So, solve the two equations:y = √3 xy = (-1/√3)x + 2/√3Set equal:√3 x = (-1/√3)x + 2/√3Multiply both sides by √3:3x = -x + 24x = 2 => x = 0.5Then, y = √3 * 0.5 = √3/2So, A₁ is (0.5, √3/2)Similarly, altitude from B to AC. AC is the x-axis from (0,0) to (2,0). The equation of AC is y = 0. The altitude from B(1, √3) to AC is vertical? Wait, no. Since AC is horizontal, the altitude from B to AC is vertical. Wait, but AC is horizontal, so the altitude from B to AC is a vertical line. Wait, AC is the x-axis, so the altitude from B is the perpendicular line to AC, which is vertical. So, since AC is horizontal, the altitude from B is vertical, so it's the line x = 1, intersecting AC at (1,0). Therefore, B₁ is (1,0).Therefore, A₁ is (0.5, √3/2), B₁ is (1,0). Then, line A₁B₁ connects (0.5, √3/2) to (1,0). Let me find the equation of A₁B₁.Slope of A₁B₁: (0 - √3/2)/(1 - 0.5) = (-√3/2)/0.5 = -√3.Equation: y - √3/2 = -√3(x - 0.5)Simplify: y = -√3 x + (√3 * 0.5) + √3/2 = -√3 x + √3/2 + √3/2 = -√3 x + √3So, equation of A₁B₁ is y = -√3 x + √3Now, the exterior angle bisector at C. Let's find the bisector of the exterior angle at C. Since angle at C is 60 degrees, the exterior angle is 120 degrees, so the bisector will split it into two 60-degree angles. The direction of the exterior angle bisector can be found using the Angle Bisector Theorem or by vectors.Since we're at point C(0,0), and the exterior angle is formed by extending BC beyond C. The sides forming the exterior angle are CB (from C to B) and the extension of AC beyond C. Wait, actually, the exterior angle can be considered in two ways. Wait, in our coordinate system, angle at C is between AC (along positive x-axis) and BC (going to (1, √3)). The exterior angle can be formed by extending AC beyond C (along negative x-axis) or extending BC beyond C (which would go towards negative direction). Wait, perhaps we need to clarify.But the problem states the bisector of the exterior angle at C. So, depending on which side we extend, the direction of the bisector changes. Let's suppose we extend AC beyond C to a point D, making the exterior angle at C between AC extended (CD) and BC. Then, the bisector of that exterior angle would be a line from C that splits the angle between CD and BC into two equal parts. Alternatively, if we extend BC beyond C, then the exterior angle is between BC extended and AC, and the bisector would split that angle.But in our coordinate system, if we extend AC beyond C to the left (negative x-axis), the exterior angle between CD (extension) and CB (from C to B(1, √3)) is 120 degrees. The bisector of this angle would be a line making 60 degrees with each side.Alternatively, using vector methods. The direction vector of the exterior angle bisector can be found by normalizing the vectors along the two sides forming the angle and then adding them. Wait, the internal angle bisector direction is given by the sum of the unit vectors of the two sides. For the external angle bisector, it's the difference?Wait, let me recall. For internal angle bisector between two vectors, the direction is the sum of the unit vectors. For external angle bisector, it's the difference. So, if we have two vectors forming the angle, then the internal bisector direction is (u/|u| + v/|v|), and the external bisector is (u/|u| - v/|v|).So, in our case, the exterior angle at C is between the extension of AC and BC. Let's define the two vectors: one is along the extension of AC beyond C, which is the negative x-axis, direction vector (-1, 0). The other is along CB towards B, which is the vector from C to B: (1, √3). The unit vectors are (-1, 0) and (1, √3)/|BC|. The length of BC is sqrt(1² + (√3)^2) = sqrt(1 + 3) = 2. So, unit vectors are (-1, 0) and (1/2, √3/2).Then, the external angle bisector direction is (-1, 0) - (1/2, √3/2) = (-1 - 1/2, 0 - √3/2) = (-3/2, -√3/2). To make this a direction vector, we can multiply by 2 to eliminate fractions: (-3, -√3). So, the direction vector is (-3, -√3). But we can also take the positive direction; since it's a line, direction doesn't matter. So, the bisector line from C(0,0) has direction (-3, -√3), which can be simplified by dividing by -√3: direction (3/√3, 1) = (√3, 1). Wait, maybe not. Wait, actually, direction vectors can be scaled. So, (-3, -√3) is equivalent to (3, √3) in the opposite direction. Wait, but since we're dealing with an external angle bisector, the direction should be such that it's going outward from the triangle. In our coordinate system, the external angle is formed by extending AC beyond C (to the left) and BC. The bisector should be going into the lower-left quadrant.But let's parametrize the bisector. Let me use the direction vector (-3, -√3). So, parametric equations for the bisector are x = -3t, y = -√3 t, where t >= 0.Now, we need to find where this bisector intersects AB and A₁B₁.First, find point L on AB. The line AB in our coordinate system is from A(2,0) to B(1, √3). Let's find the equation of AB.Slope of AB: (√3 - 0)/(1 - 2) = √3 / (-1) = -√3Equation: y - 0 = -√3(x - 2) => y = -√3 x + 2√3Now, parametric equations for the bisector: x = -3t, y = -√3 tFind intersection point L between bisector and AB.Set y = -√3 t equal to y = -√3 x + 2√3So:-√3 t = -√3 (-3t) + 2√3Simplify:-√3 t = 3√3 t + 2√3Bring all terms to left:-√3 t - 3√3 t - 2√3 = 0 => -4√3 t - 2√3 = 0 => -4t - 2 = 0 (divided both sides by √3)So, -4t = 2 => t = -0.5But t was supposed to be >= 0 in the parametrization of the bisector. Negative t would mean going in the opposite direction. So, t = -0.5 gives us point L at x = -3*(-0.5) = 1.5, y = -√3*(-0.5) = (√3)/2. So, L is at (1.5, √3/2). But in our coordinate system, AB is from (2,0) to (1, √3). The point (1.5, √3/2) is the midpoint of AB. Wait, yes, the midpoint between A(2,0) and B(1, √3) is ((2+1)/2, (0 + √3)/2) = (1.5, √3/2). So, L is the midpoint of AB. Interesting.Wait, so in this case, with angle C = 60 degrees, the exterior angle bisector intersects AB at its midpoint. Then, CL is the distance from C(0,0) to L(1.5, √3/2). Let's compute CL:CL = sqrt( (1.5)^2 + (√3/2)^2 ) = sqrt(2.25 + 3/4 ) = sqrt(2.25 + 0.75) = sqrt(3) ≈ 1.732.Now, point K is the intersection of the bisector with A₁B₁. The line A₁B₁ is from A₁(0.5, √3/2) to B₁(1, 0), equation y = -√3 x + √3 (as computed earlier). The bisector is parametrized as x = -3t, y = -√3 t. Wait, but earlier when we found L, we had to use t negative, which might be confusing. Alternatively, perhaps parametrize the bisector differently. Let's use a different parameter.Alternatively, since we found that the bisector intersects AB at L(1.5, √3/2), which corresponds to t = -0.5 in the previous parametrization, but let's instead parametrize the bisector starting at C and going through L. Since L is on the bisector, we can parametrize from C to L. So, vector CL is (1.5, √3/2), so direction vector (3, √3). So, parametric equations: x = 3s, y = √3 s, where s >= 0. Then, point L is at s = 0.5, since 3*0.5 = 1.5 and √3*0.5 = √3/2.Now, find intersection K of bisector with A₁B₁. The bisector is x = 3s, y = √3 s. Line A₁B₁ is y = -√3 x + √3.Set y = √3 s equal to y = -√3 x + √3, with x = 3s:√3 s = -√3*(3s) + √3√3 s = -3√3 s + √3Bring all terms to left:√3 s + 3√3 s - √3 = 0 => 4√3 s - √3 = 0 => 4s - 1 = 0 => s = 1/4Therefore, point K is at x = 3*(1/4) = 3/4, y = √3*(1/4) = √3/4Then, CK is the distance from C(0,0) to K(3/4, √3/4):CK = sqrt( (3/4)^2 + (√3/4)^2 ) = sqrt(9/16 + 3/16 ) = sqrt(12/16 ) = sqrt(3/4 ) = (√3)/2 ≈ 0.866.CL was sqrt(3) ≈ 1.732, which is indeed 2 times CK. So, in this case, CL = 2 CK holds. Therefore, when angle C is 60 degrees, the condition is satisfied. Therefore, the angle C is 60 degrees.But the problem states "In an acute-angled triangle ABC...", and we tested with angle C = 60 degrees, which is acute, and the condition holds. Therefore, angle C must be 60 degrees.But to confirm that this is the only possible angle, we need to ensure that no other angle would satisfy CL = 2 CK. Let's suppose angle C is not 60 degrees. Let's take angle C = 90 degrees, but wait, the triangle has to be acute-angled, so angle C cannot be 90. Let's take angle C = 45 degrees and check.Wait, but that would be tedious. Alternatively, notice that in the above example with angle C = 60 degrees, the ratio CL / CK = 2. If we can show that this ratio uniquely determines angle C as 60 degrees, then that's the answer.Alternatively, in general, for angle C, the ratio CL / CK can be expressed in terms of trigonometric functions involving angle C, leading to an equation that solves angle C = 60 degrees.Alternatively, note that in the example above, the key was that A₁B₁ was the midline of the triangle, hence parallel to AB and half its length. But this happens when the triangle is equilateral, but in our case, the triangle wasn't equilateral (AC = 2, BC = 2, AB was... Wait, in our example, AC was 2 units (from (0,0) to (2,0)), BC was 2 units (from (0,0) to (1, √3)), so AC = 2, BC = 2, AB distance is sqrt( (1 - 2)^2 + (√3 - 0)^2 ) = sqrt(1 + 3) = 2. So, triangle ABC with AC = BC = AB = 2, so equilateral triangle. Wait, but in my coordinate system, I set a = 2, but then in computation, AB turned out to be 2 as well. So, the triangle is equilateral. Hmm. Therefore, in that case, A₁B₁ is the midline, hence parallel to AB and half its length, leading to homothety with ratio 2. But the problem didn't specify the triangle is equilateral, just acute-angled. So, perhaps the only solution is when angle C is 60 degrees, regardless of the triangle being equilateral or not.Wait, but in my example, the triangle was equilateral. Let me take a non-equilateral triangle with angle C = 60 degrees and check.Let me take point C at (0,0), point A at (3,0), point B at (1, √3). Then, angle at C is 60 degrees. Let's verify:Vectors CA = (3,0), CB = (1, √3). The angle between them is given by:cos γ = (CA · CB) / (|CA| |CB|) = (3*1 + 0*√3)/ (3 * 2) = 3 / 6 = 0.5, so γ = 60 degrees. Correct.Now, find altitudes AA₁ and BB₁.Altitude from A(3,0) to BC. First, find equation of BC. Points B(1, √3) and C(0,0). Slope of BC is (√3 - 0)/(1 - 0) = √3. Equation: y = √3 x.Altitude from A is perpendicular to BC, so slope -1/√3. Equation: y - 0 = -1/√3 (x - 3). So, y = (-1/√3)x + 3/√3.Intersection with BC (y = √3 x):Set equal: √3 x = (-1/√3)x + 3/√3Multiply both sides by √3:3x = -x + 34x = 3 => x = 3/4, y = √3 * 3/4 = 3√3/4. So, A₁ is (3/4, 3√3/4).Altitude from B(1, √3) to AC. AC is from (0,0) to (3,0), along x-axis. The altitude is vertical? No. Wait, AC is horizontal, so the altitude from B to AC is vertical line. Wait, no. Wait, the altitude from B to AC is perpendicular to AC. Since AC is horizontal, the altitude is vertical. Therefore, equation is x = 1. Intersection with AC (y=0) is (1,0), so B₁ is (1,0).Now, line A₁B₁: from (3/4, 3√3/4) to (1,0). Let's find its equation.Slope: (0 - 3√3/4)/(1 - 3/4) = (-3√3/4)/(1/4) = -3√3.Equation: y - 3√3/4 = -3√3(x - 3/4)Simplify:y = -3√3 x + (9√3)/4 + 3√3/4= -3√3 x + 12√3/4= -3√3 x + 3√3Now, the exterior angle bisector at C. Let's find its equation.Vectors forming the exterior angle: extend AC beyond C to D(-1,0). The exterior angle at C is between CD and CB. The bisector direction can be found by vector subtraction.Unit vectors: CD is (-1,0), normalized is (-1,0). CB is (1, √3), normalized is (1/2, √3/2). The external bisector direction is (-1,0) - (1/2, √3/2) = (-3/2, -√3/2). Direction vector (-3, -√3).Parametric equations for bisector: x = -3t, y = -√3 t.Find intersection with AB. Equation of AB: from A(3,0) to B(1, √3).Slope of AB: (√3 - 0)/(1 - 3) = √3 / (-2) = -√3/2Equation: y - 0 = -√3/2 (x - 3)=> y = (-√3/2)x + (3√3)/2Intersection with bisector:Set y = -√3 t equal to y = (-√3/2)x + (3√3)/2But x = -3t, so substitute:-√3 t = (-√3/2)(-3t) + (3√3)/2Simplify:-√3 t = (3√3/2) t + (3√3)/2Multiply both sides by 2 to eliminate denominators:-2√3 t = 3√3 t + 3√3Bring terms with t to left:-2√3 t - 3√3 t = 3√3-5√3 t = 3√3Divide both sides by √3:-5t = 3 => t = -3/5Then, coordinates of L are x = -3*(-3/5) = 9/5, y = -√3*(-3/5) = 3√3/5. So, L is at (9/5, 3√3/5). Check if this is on AB.AB is from (3,0) to (1, √3). Let's verify:Parametrize AB as x = 3 - 2s, y = 0 + √3 s, where s in [0,1].Set x = 9/5 = 3 - 2s => 2s = 3 - 9/5 = 6/5 => s = 3/5Then y should be √3*(3/5) = 3√3/5, which matches. So, L is indeed on AB.Now, distance CL: from (0,0) to (9/5, 3√3/5):CL = sqrt( (81/25) + (27/25) ) = sqrt(108/25) = (6√3)/5 ≈ 2.078Now, find K as intersection of bisector with A₁B₁.Equation of A₁B₁ is y = -3√3 x + 3√3.Bisector parametrized as x = -3t, y = -√3 t. Wait, but again, t is negative. Alternatively, parametrize from C through L with direction vector (9/5, 3√3/5). Parametric equations: x = (9/5)s, y = (3√3/5)s, where s >= 0. When s=1, it's point L.Intersection with A₁B₁:Set y = (3√3/5)s equal to y = -3√3 x + 3√3, where x = (9/5)s.So:(3√3/5)s = -3√3*(9/5)s + 3√3Multiply both sides by 5:3√3 s = -27√3 s + 15√3Bring terms with s to left:3√3 s + 27√3 s = 15√330√3 s = 15√3Divide both sides by √3:30s = 15 => s = 0.5Therefore, point K is at x = (9/5)*0.5 = 9/10, y = (3√3/5)*0.5 = 3√3/10.Distance CK: from (0,0) to (9/10, 3√3/10):CK = sqrt( (81/100) + (27/100) ) = sqrt(108/100) = sqrt(27/25) = (3√3)/5 ≈ 1.039Then, CL = (6√3)/5 ≈ 2.078, which is twice CK ≈ 1.039. So, CL = 2 CK holds here as well. This triangle is not equilateral (AC = 3, BC = 2, AB = sqrt( (1-3)^2 + (√3 - 0)^2 ) = sqrt(4 + 3) = sqrt(7) ≈ 2.645), but angle C is 60 degrees, and CL = 2 CK holds. Therefore, angle C = 60 degrees is the answer regardless of the triangle being equilateral or not.Therefore, the angle C must be 60 degrees. So, the answer is 60 degrees, which in boxed form is boxed{60^circ}.Final AnswerThe angle ( C ) is boxed{60^circ}.</think>In an acute-angled triangle (ABC), altitudes (AA_1) and (BB_1) are drawn. The bisector of the exterior angle at vertex (C) intersects the lines (AB) and (A_1B_1) at points (L) and (K) respectively, with (CL = 2CK). We need to find the angle (C).1. Understanding the Problem: - The triangle is acute-angled. - Altitudes (AA_1) and (BB_1) are drawn. - The exterior angle bisector at (C) intersects (AB) at (L) and (A_1B_1) at (K). - Given (CL = 2CK), we need to determine the angle (C).2. Key Observations: - The exterior angle bisector theorem and properties of similar triangles are considered. - The coordinates of points (A), (B), and (C) are set up to analyze the problem geometrically. - The orthocenter and the orthic triangle are considered to understand the positions of (A_1) and (B_1).3. Coordinate Geometry Approach: - Placing (C) at the origin ((0,0)), (B) at ((1, sqrt{3})), and (A) at ((2, 0)) forms a triangle with angle (C = 60^circ). - Calculating the equations of altitudes (AA_1) and (BB_1), and finding their feet (A_1) and (B_1). - Determining the equation of the line (A_1B_1).4. Exterior Angle Bisector: - The exterior angle bisector at (C) (which is (120^circ)) splits into two (60^circ) angles. - Using vector methods and parametric equations to find the intersections (L) and (K).5. Verification: - Calculating the coordinates of (L) and (K) and verifying the distances (CL) and (CK). - Confirming that (CL = 2CK) holds true for angle (C = 60^circ).6. Conclusion: - Through coordinate geometry and vector analysis, it is shown that the condition (CL = 2CK) holds true when angle (C) is (60^circ), regardless of the triangle being equilateral or not.Thus, the angle (C) is (boxed{60^circ}).]

🔑:As is known, triangle C A_{1} B_{1} is similar to triangle C A B with a similarity coefficient of cos angle C.Let's assume that point L lies on the extension of side A B beyond point B. We will prove that point C lies between K and L. Suppose this is not the case. Then, if P is a point on the extension of side A C beyond point C, and Q is a point on the extension of side A B beyond point A, then angle A_{1} C K = angle L C P,angle C A_{1} K = 180^{circ} - angle C A_{1} B_{1} = 180^{circ} - angle C A B = angle C A Q, which means the sum of two angles of triangle C A_{1} K equals the sum of two exterior angles of triangle A C L, i.e., more than 180^{circ}. Contradiction.Notice that the angles of triangles C A L and C A_{1} K at vertex C are equal. Moreover, angle A = angle C A_{1} B_{1}. Therefore, these triangles are similar. Hence,cos angle C = C A_{1 / C A} = C K_{/} / C L = 1 / 2## Answer60^{circ}.