🧠:Alright, let me try to figure out this problem. So, we need to find the sum A, which is the sum of the floor of 7^k divided by 8 from k=1 to k=2020. Then, we have to find the remainder when A is divided by 50. Hmm, okay. Let me start by recalling what the floor function does. The floor of a number is the greatest integer less than or equal to that number. So, for example, if 7^k divided by 8 is not an integer, the floor function would subtract the decimal part, right?First, maybe I should compute some small terms to see if there's a pattern. Let's compute [7^1/8], [7^2/8], [7^3/8], etc., and see if there's a cycle or something that repeats, which could help in summing up the terms for such a large exponent (2020 terms!).Starting with k=1: 7^1 = 7. 7 divided by 8 is 0.875. The floor of that is 0. So, [7/8] = 0.k=2: 7^2 = 49. 49 divided by 8 is 6.125. Floor of that is 6. So, [49/8] = 6.k=3: 7^3 = 343. 343 divided by 8 is 42.875. Floor is 42. So, [343/8] = 42.k=4: 7^4 = 2401. Divided by 8: 2401 /8 = 300.125. Floor is 300.k=5: 7^5 = 16807. Divided by 8: 16807/8 = 2100.875. Floor is 2100.Wait, maybe I can see a pattern here. Let's note these results:k | 7^k | 7^k /8 | [7^k /8]---|-----|-------|--------1 | 7 | 0.875 | 02 | 49 | 6.125 | 63 | 343 | 42.875 | 424 | 2401 | 300.125 | 3005 | 16807 | 2100.875 | 21006 | 117649 | 14706.125 | 14706...Looking at the decimal parts: 0.875, 0.125, 0.875, 0.125, 0.875, 0.125... Wait, alternating between 0.875 and 0.125? Let me check:7^1 /8 = 0.875, remainder 7 (since 7 = 8*0 +7)7^2 =49. 49 divided by 8 is 6*8=48, remainder 1. So 49/8 =6.125. So decimal part 0.125.7^3 =343. 343 divided by 8: 8*42=336, remainder 7. So 343/8=42.875. Decimal 0.875.7^4=2401. Divided by 8: 8*300=2400, remainder 1. So 2401/8=300.125. Decimal 0.125.Similarly, 7^5=16807. Divided by 8: 8*2100=16800, remainder 7. So decimal 0.875.Yes, so the remainder when 7^k is divided by 8 alternates between 7 and 1, starting with 7 when k=1, then 1 when k=2, 7 when k=3, etc. So the pattern is 7,1,7,1,... for the remainders. Therefore, the decimal part of 7^k/8 is 7/8=0.875 when k is odd and 1/8=0.125 when k is even. So the floor of 7^k/8 is (7^k - remainder)/8. Therefore, [7^k/8] = (7^k - remainder)/8. Since the remainder alternates between 7 and 1, then:For odd k: [7^k/8] = (7^k -7)/8For even k: [7^k/8] = (7^k -1)/8Therefore, the sum A can be written as the sum over k=1 to 2020 of [7^k/8] = sum_{k=1}^{2020} (7^k - r_k)/8, where r_k is 7 if k is odd, 1 if k is even.Therefore, A = (1/8) * [sum_{k=1}^{2020} 7^k - sum_{k=1}^{2020} r_k]Compute sum_{k=1}^{2020} 7^k: this is a geometric series.Sum = 7*(7^{2020} -1)/(7-1) = (7^{2021} -7)/6Then sum_{k=1}^{2020} r_k: since r_k alternates between 7 and 1. Let's note that from k=1 to 2020, there are 1010 odd k and 1010 even k (since 2020 is even). So for each odd k, r_k=7, so sum over odd k: 7*1010. For even k, r_k=1, sum over even k:1*1010. Therefore, total sum of r_k =7*1010 +1*1010=8*1010=8080.Therefore, A=(1/8)*[(7^{2021}-7)/6 -8080] = (7^{2021}-7 -6*8080)/48But maybe we can write this as [ (7^{2021} -7)/6 -8080 ] /8But perhaps there is a better way. Let me check.Alternatively, since A = sum_{k=1}^{2020} [7^k/8] = sum_{k=1}^{2020} (7^k - r_k)/8 = (1/8)(sum_{k=1}^{2020}7^k - sum_{k=1}^{2020}r_k)So we have:A = [ (7^{2021} -7)/6 - 8080 ] /8Simplify numerator:(7^{2021} -7 -6*8080)/6*8 = (7^{2021} -7 -48480)/48But 6*8080=48480. Therefore, A=(7^{2021} -7 -48480)/48But this seems complicated. However, we need A mod 50. So perhaps instead of computing A directly, we can compute the numerator modulo (48*50)=2400, but maybe even that is not necessary. Alternatively, since we need A mod 50, and 48 and 50 are coprime? Wait, 48 and 50 have gcd 2. Hmm. So maybe it's better to compute A modulo 50 by handling the expression.Alternatively, perhaps compute A modulo 50. Let's note that 50 and 48 have gcd 2. Therefore, to compute A=(7^{2021} -7 -48480)/48 mod 50.But let's see. Let me compute numerator modulo (48*50)=2400. Because if I can compute the numerator modulo 2400, then divide by 48 to get A modulo 50.Yes, since 48*50=2400, and if I compute numerator modulo 2400, then (numerator)/48 mod 50 is equal to (numerator mod 2400)/48 mod 50.So let me proceed step by step.First, compute the numerator: 7^{2021} -7 -48480. Let's compute this modulo 2400.Compute 7^{2021} mod 2400, then subtract 7 and 48480, and then take modulo 2400.But first, compute 7^{2021} mod 2400. To compute this, perhaps use Euler's theorem, but 7 and 2400 are coprime. Euler totient function φ(2400)=φ(2^5 *3*5^2)=2400*(1-1/2)*(1-1/3)*(1-1/5)=2400*(1/2)*(2/3)*(4/5)=2400*(1/2)=1200; 1200*(2/3)=800; 800*(4/5)=640. So φ(2400)=640. Therefore, 7^640 ≡1 mod 2400. Therefore, 7^{2021}=7^{640*3 + 101}= (7^{640})^3 *7^{101}≡1^3 *7^{101}≡7^{101} mod2400.So need to compute 7^101 mod2400.Compute 7^101 mod2400. This might still be tedious. Maybe use exponentiation by squaring.Alternatively, compute 7^1 mod2400=77^2=497^4=49^2=2401≡1 mod2400Wait, 7^4=2401≡1 mod2400. That's a key point!Therefore, 7^4 ≡1 mod2400. Therefore, 7^(4k)≡1^k=1 mod2400. Then 7^101=7^(4*25 +1)= (7^4)^25 *7^1 ≡1^25 *7=7 mod2400. Therefore, 7^101 ≡7 mod2400.Therefore, 7^2021 ≡7 mod2400.Therefore, numerator is 7 -7 -48480 ≡0 -48480 mod2400.Compute -48480 mod2400. Since 48480 divided by2400: 2400*20=48000. 48480-48000=480. So 48480=2400*20 +480. Therefore, -48480≡-480 mod2400. Then -480 mod2400=2400-480=1920.Therefore, numerator≡1920 mod2400.Therefore, A=(1920)/48 mod50. 1920/48=40. Therefore, A≡40 mod50.Wait, that seems straightforward. Let me check again.So, 7^2021 ≡7 mod2400. Then numerator=7 -7 -48480= -48480. Then -48480 mod2400= -48480 +2400*21= -48480 +50400=1920. Then 1920/48=40. Therefore, A≡40 mod50.But let me verify this step by step, as it's critical.First, φ(2400)=640, so 7^640≡1 mod2400. Then 2021 divided by 640: 640*3=1920. 2021-1920=101. So 7^2021≡7^101 mod2400. Then, 7^4=2401≡1 mod2400. Therefore, 101 divided by4: 4*25=100. 101-100=1. Therefore, 7^101=(7^4)^25 *7^1≡1^25 *7=7 mod2400. So yes, 7^101≡7 mod2400. Therefore, 7^2021≡7 mod2400. So numerator=7 -7 -48480= -48480.Now, -48480 mod2400. Since 2400*20=48000. So 48480-48000=480. Therefore, 48480=2400*20 +480. Therefore, -48480= -2400*20 -480≡-480 mod2400. Then -480 +2400=1920. So -48480≡1920 mod2400. Then, numerator≡1920 mod2400. Then A=(1920)/48=40. 40 mod50=40.Therefore, the remainder when A is divided by50 is40. Hmm, that seems correct. But let me check with another approach to verify.Alternatively, perhaps compute A mod50 directly. Since A=(sum [7^k/8]) from k=1 to2020. Let's note that [7^k/8]=(7^k -r_k)/8, where r_k=7 if k odd,1 if even. Therefore, A=(sum (7^k -r_k)/8)= (sum7^k)/8 - sum(r_k)/8.Sum7^k from1 to2020 is(7^{2021}-7)/6. Sumr_k=8080 as before. So A=( (7^{2021}-7)/6 -8080 )/8.We need to compute this modulo50. Let's compute numerator modulo (6*8*50)=2400, but we already did that. Alternatively, compute each part modulo50*8=400 or something else. But since the previous method worked, perhaps it's correct.Alternatively, let me compute A mod50 by splitting into two parts: sum [7^k/8] mod50.Each term [7^k/8] = (7^k -r_k)/8. Therefore, [7^k/8] ≡(7^k -r_k)/8 mod50.But division by8 modulo50 requires that 8 and50 are coprime? No, gcd(8,50)=2. So division by8 is only possible if the numerator is even. Let's see. Since 7^k is odd, and r_k is 7 or1, both odd. So 7^k -r_k is even. Therefore, (7^k -r_k)/8 is an integer, and when taking modulo50, we can compute (7^k -r_k)/8 mod50.Therefore, perhaps we can compute each term modulo50*8=400, sum them up, and divide by8? Wait, but sum of terms each (7^k -r_k)/8 mod50. Alternatively, compute the total sum modulo50*8=400, then divide by8.Alternatively, compute A= sum_{k=1}^{2020} [7^k/8] = sum (7^k -r_k)/8. Then A mod50= [sum (7^k -r_k)/8 ] mod50. Let's compute (sum7^k - sumr_k)/8 mod50.Sum7^k mod(8*50)=400. Let's compute sum7^k from1 to2020 mod400. Similarly, sumr_k=8080. Compute 8080 mod(8*50)=400. 8080 divided by400: 400*20=8000, 8080-8000=80. So sumr_k≡80 mod400.Then, (sum7^k -80)/8 mod50.Compute sum7^k mod400. Let's compute sum7^k from1 to2020 mod400. Note that sum7^k=7*(7^{2020}-1)/6. So we need to compute (7*(7^{2020}-1)/6 ) mod400.First, compute 7^{2020} mod(400*6/ gcd(6,400)). Wait, maybe better to compute 7^{2020} mod2400, but that seems complex. Alternatively, compute 7^{2020} mod400 and mod6, then use Chinese Remainder Theorem?Wait, perhaps first compute 7^{2020} mod400. Let's compute 7^φ(400)=7^160≡1 mod400, since φ(400)=φ(2^4*5^2)=400*(1-1/2)*(1-1/5)=400*(1/2)*(4/5)=160. So 7^160≡1 mod400. Then 2020=160*12 + 100. So 7^{2020}= (7^{160})^12 *7^{100}≡1^12 *7^{100}≡7^{100} mod400.Compute 7^{100} mod400. Let's use exponentiation by squaring:7^1=7 mod4007^2=497^4=49^2=2401≡1 mod4007^8≡1^2=1 mod400Similarly, 7^16≡1, etc. So 7^4≡1 mod400. Therefore, 7^100=7^(4*25)= (7^4)^25≡1^25=1 mod400. Therefore, 7^{2020}≡1 mod400. Therefore, sum7^k from1 to2020=7*(7^{2020}-1)/6≡7*(1 -1)/6≡0 mod400.Therefore, sum7^k≡0 mod400.Then sum7^k -80≡-80 mod400.Thus, (sum7^k -80)/8≡(-80)/8≡-10 mod50. Since 400/8=50. Therefore, -10 mod50=40.Thus, A≡40 mod50.Same result as before. Therefore, the remainder is40. Therefore, the answer is 40.Final Answerboxed{40}

✅:To solve the problem, we need to find the remainder when ( A = left[frac{7}{8}right] + left[frac{7^2}{8}right] + cdots + left[frac{7^{2019}}{8}right] + left[frac{7^{2020}}{8}right] ) is divided by 50. First, we observe that the floor function (left[frac{7^k}{8}right]) can be expressed as (frac{7^k - r_k}{8}), where ( r_k ) is the remainder when ( 7^k ) is divided by 8. The remainders alternate between 7 and 1 for odd and even ( k ) respectively. Thus, we can write:[ A = frac{1}{8} left( sum_{k=1}^{2020} 7^k - sum_{k=1}^{2020} r_k right) ]The sum of the geometric series (sum_{k=1}^{2020} 7^k) is:[ sum_{k=1}^{2020} 7^k = frac{7(7^{2020} - 1)}{6} ]The sum of the remainders (sum_{k=1}^{2020} r_k) is calculated as follows: there are 1010 odd ( k ) and 1010 even ( k ), so:[ sum_{k=1}^{2020} r_k = 7 times 1010 + 1 times 1010 = 8 times 1010 = 8080 ]Thus, we have:[ A = frac{1}{8} left( frac{7^{2021} - 7}{6} - 8080 right) ]To find ( A mod 50 ), we compute the numerator modulo ( 2400 ) (since ( 48 times 50 = 2400 )) and then divide by 48. First, we compute ( 7^{2021} mod 2400 ). Using Euler's theorem, we find ( 7^{2021} equiv 7 mod 2400 ). Therefore, the numerator is:[ 7 - 7 - 48480 equiv -48480 mod 2400 ][ -48480 equiv 1920 mod 2400 ]Dividing 1920 by 48 gives:[ frac{1920}{48} = 40 ]Thus, the remainder when ( A ) is divided by 50 is:[ boxed{40} ]

🔑:1. To find (A), we need to compute the sum: [ A = leftlfloor frac{7}{8} rightrfloor + leftlfloor frac{7^2}{8} rightrfloor + cdots + leftlfloor frac{7^{2019}}{8} rightrfloor + leftlfloor frac{7^{2020}}{8} rightrfloor ]2. Note that (frac{7^{2k-1}}{8}) and (frac{7^{2k}}{8}) are not integers for any positive integer (k).3. According to the floor function definition, we have: [ leftlfloor frac{7^{2k-1}}{8} rightrfloor + leftlfloor frac{7^{2k}}{8} rightrfloor ] We need to bound this expression by: [ 7^{2k-1} - 2 < leftlfloor frac{7^{2k-1}}{8} rightrfloor + leftlfloor frac{7^{2k}}{8} rightrfloor < 7^{2k-1} ]4. For any positive integer (k), we have: [ leftlfloor frac{7^{2k-1}}{8} rightrfloor + leftlfloor frac{7^{2k}}{8} rightrfloor = 7^{2k-1} - 1 ] This results from: [ 7 times 7^{2k-2} - 1 = 7 times 49^{k-1} - 1 = 7 times (50 - 1)^{k-1} - 1 ]5. Considering modulo 50, we get: [ 7 times (50 - 1)^{k-1} - 1 equiv 7 times (-1)^{k-1} - 1 pmod{50} ]6. Summing these over (k) from 1 to 1010, since: [ A = sum_{k=1}^{1010}left[ frac{7^{2k-1}}{8} right] + left[ frac{7^{2k}}{8} right] ] we get: [ A equiv sum_{k=1}^{1010} left( 7 times (-1)^{k-1} - 1 right) pmod{50} ]7. Evaluating the sum, split the terms: [ sum_{k=1}^{1010} 7 times (-1)^{k-1} - sum_{k=1}^{1010} 1 ] The first sum, (sum_{k=1}^{1010} 7 times (-1)^{k-1}), is: [ 7(1 - 1 + 1 - 1 + cdots + 1 - 1) ] There are 1010 terms, so: [ = 7(505) + 7(-505) = 0 ]8. Hence: [ 7 times 505 - 1010 = -510 ] and [ -510 equiv -10 equiv 40 pmod{50} ]9. Conclusion: [ boxed{40} ]

🧠:Okay, so I need to prove that among all the pairs of five real numbers whose squares sum to 1, the smallest squared difference between any two numbers is at most 1/10. Hmm. Let me try to break this down step by step.First, the problem states that we have five real numbers a₁, a₂, a₃, a₄, a₅ such that the sum of their squares is 1. We need to show that the minimum of (a_i - a_j)² for i ≠ j is less than or equal to 1/10. In other words, no matter how we choose these five numbers, there must be at least two numbers whose squared difference is at most 1/10. Let me think about how to approach this. Maybe using the pigeonhole principle? If the numbers are spread out, then some differences have to be small. Since there are five numbers, maybe arranging them in some intervals?Alternatively, perhaps considering the numbers as points on the real line and trying to maximize the minimal distance between any two points, given that their squares sum to 1. If we can find the maximum possible minimal distance, then if that maximum is less than or equal to 1/10, the original statement is proved.Wait, but the problem is about the minimal squared difference, not the minimal distance. But since squaring is a monotonic function for non-negative numbers, the minimal squared difference would correspond to the minimal distance squared. So perhaps I can think in terms of distances.So, maybe to rephrase: given five points on the real line with the sum of squares equal to 1, prove that the smallest distance between any two points is at most sqrt(1/10), which is approximately 0.316. So, the square of that distance is 1/10. So, if I can show that the minimal distance between two points is at most sqrt(1/10), then the square will be at most 1/10.Alternatively, suppose all the differences (a_i - a_j)^2 are greater than 1/10. Then, can we get a contradiction with the sum of squares being 1?Alternatively, perhaps using an averaging argument. If the numbers are spread out, then their squares would have to be large. But since the sum is fixed at 1, maybe there's a balance between how spread out they are and how large their squares can be.Wait, maybe arranging the numbers in some optimal way. For example, if we want to maximize the minimal difference between any two numbers, given the sum of squares constraint, then how would we place them? Maybe equally spaced? But equally spaced around zero?Alternatively, maybe place them symmetrically around zero. Let me try to think of five numbers arranged symmetrically to maximize the minimal distance. For example, suppose we have numbers at -x, -y, 0, y, x. Then the differences would be x - y, y - 0, etc. Maybe this symmetry could help.But maybe a better approach is to use the pigeonhole principle. Suppose we divide the real line into intervals such that if two numbers are in the same interval, their difference is small. Then, with five numbers, by pigeonhole, two must be in the same interval, hence their difference is small.But how to choose the intervals? The problem is the real line is unbounded, but the sum of squares is 1, so all numbers must lie within a certain range. For example, by Cauchy-Schwarz, the maximum absolute value of any number is 1, since if all other numbers are zero, then one number can be 1. Wait, but actually, the maximum possible value for any a_i is 1, since if you have one number as 1 and the rest zero, the sum is 1. But if multiple numbers are non-zero, each of their absolute values is less than or equal to 1.But perhaps even tighter. If the sum of squares is 1, then each a_i is between -1 and 1. So all numbers lie in the interval [-1,1]. If we can partition [-1,1] into four intervals such that the length of each interval is sqrt(1/10), then by pigeonhole principle, two numbers must lie in the same interval, so their difference is at most sqrt(1/10), hence squared difference at most 1/10. But let's check if four intervals each of length sqrt(1/10) can cover [-1,1].The total length of four intervals would be 4*sqrt(1/10) ≈ 4*0.316 ≈ 1.264. But the interval [-1,1] has length 2, which is approximately 1.264 < 2, so that approach doesn't work. So maybe partitioning into more intervals? Wait, but with five numbers, the pigeonhole principle would require four intervals to ensure two numbers in the same interval. But if the intervals aren't sufficient to cover the entire range, then maybe the numbers could lie in the gaps between intervals. Hmm, not sure.Alternatively, maybe use the concept of packing. If we have five points in [-1,1], what's the maximum minimal distance between them? If the minimal distance is greater than sqrt(1/10), then the sum of squares would exceed 1. Let's try to model this.Suppose all pairwise squared differences are greater than 1/10. Then, the minimal distance between any two points is greater than sqrt(1/10). Let's try to arrange five points in [-1,1] with minimal distance d. What's the maximum possible d such that the sum of squares is 1?Alternatively, perhaps use an energy minimization approach. If the numbers are spread out as much as possible, the sum of squares would be large, so to minimize the sum, the numbers should be close together. But we need to maximize the minimal distance while keeping the sum of squares equal to 1.Wait, this seems a bit conflicting. To maximize the minimal distance, the points should be as spread out as possible, but that would require some points to be further from zero, hence increasing the sum of squares. So, the problem is to find the maximal minimal distance such that the sum of squares does not exceed 1. Then, if we can show that this maximal minimal distance is at most sqrt(1/10), then the original statement holds.Alternatively, perhaps consider the case where all the numbers are equally spaced. Let me try to model this. Suppose the five numbers are equally spaced points. Let's assume they are ordered: a₁ ≤ a₂ ≤ a₃ ≤ a₄ ≤ a₅. If they are equally spaced, then the differences between consecutive terms are the same. Let's denote the common difference as δ. So, a₂ = a₁ + δ, a₃ = a₂ + δ = a₁ + 2δ, etc., up to a₅ = a₁ + 4δ.But since the numbers can be placed anywhere on the real line, maybe symmetrically around zero to minimize the sum of squares. Let's suppose the numbers are symmetric around zero. So, if we have five numbers, maybe two pairs symmetric around zero and the middle one at zero. For example: -x, -y, 0, y, x. Then, the differences would be y - (-y) = 2y, x - y, etc. But perhaps if we have equally spaced points around zero. Wait, with five points, equally spaced would have positions like -2d, -d, 0, d, 2d. Then, the differences between consecutive terms are d, and the total spread is 4d. The sum of squares would be 2*( (2d)^2 + d^2 ) + 0 = 2*(4d² + d²) = 10d². If the sum of squares is 1, then 10d² = 1 => d² = 1/10 => d = 1/√10 ≈ 0.316. So in this case, the minimal difference between consecutive terms is d = 1/√10, so the minimal squared difference is (1/√10)^2 = 1/10. Therefore, in this arrangement, the minimal squared difference is exactly 1/10. Hence, this shows that the minimal squared difference can be as large as 1/10. Therefore, the problem statement is to show that you cannot have all squared differences larger than 1/10; in other words, the minimal squared difference cannot exceed 1/10.But wait, in this example, the minimal squared difference is exactly 1/10. Therefore, this example achieves the bound. Therefore, the maximum possible minimal squared difference is 1/10, hence proving that the least of the (a_i - a_j)^2 cannot exceed 1/10. Therefore, this example serves as an extremal case.But to make this rigorous, perhaps assume that all pairwise squared differences are greater than 1/10, then derive a contradiction with the sum of squares being 1. Let me try that.Assume that for all i ≠ j, (a_i - a_j)^2 > 1/10. Then, the minimal distance between any two numbers is greater than sqrt(1/10). Let's order the numbers in increasing order: a₁ < a₂ < a₃ < a₄ < a₅. Then, the consecutive differences are a₂ - a₁ > sqrt(1/10), a₃ - a₂ > sqrt(1/10), a₄ - a₃ > sqrt(1/10), a₅ - a₄ > sqrt(1/10). Therefore, the total span from a₁ to a₅ is greater than 4*sqrt(1/10). Let's compute 4*sqrt(1/10) ≈ 4*0.316 ≈ 1.264. So the entire span is greater than 1.264. But since the numbers are real numbers with sum of squares 1, each number must lie within [-1,1], because if any |a_i| > 1, then its square would exceed 1, making the total sum exceed 1. But wait, actually, that's not necessarily true. For example, if one number is 1.5, but others are negative, the sum could still be 1. Wait, no, because squares are non-negative. If one number is 1.5, then its square is 2.25, which already exceeds 1, so that's impossible. Therefore, all numbers must lie within [-1,1]. Therefore, the total span from a₁ to a₅ cannot exceed 2, since the maximum distance from -1 to 1 is 2. But according to the assumption, the span is greater than 1.264, which is possible. So that alone doesn't give a contradiction.Alternatively, maybe consider the positions of the numbers. If the numbers are spread out with minimal differences greater than sqrt(1/10), their positions must be spaced out, so they can't be too close to each other, which might force some of them to be too large in absolute value, making the sum of squares exceed 1.Let me try to model this. Suppose we have five numbers ordered as a₁ ≤ a₂ ≤ a₃ ≤ a₄ ≤ a₅. Assume that each consecutive difference is at least d (where d > sqrt(1/10)). Then, we can write:a₂ ≥ a₁ + da₃ ≥ a₂ + d ≥ a₁ + 2da₄ ≥ a₃ + d ≥ a₁ + 3da₅ ≥ a₄ + d ≥ a₁ + 4dTherefore, the total span is a₅ - a₁ ≥ 4d. Now, since all numbers are in [-1,1], the maximum possible span is 2 (from -1 to 1). Therefore, 4d ≤ 2 => d ≤ 0.5. But sqrt(1/10) ≈ 0.316, which is less than 0.5, so this doesn't contradict. Therefore, this line of reasoning isn't sufficient.Alternatively, maybe use the concept of variance. The sum of squares is related to the variance. Let's recall that for numbers a₁, ..., a₅, the variance is (1/5)Σ(a_i - μ)^2, where μ is the mean. But since we have the sum of squares fixed, perhaps we can relate the pairwise differences to the variance.Alternatively, consider that the sum of squares can be written in terms of the pairwise differences. There's a formula that relates the sum of squares to the pairwise squared differences. Let me recall that formula.We know that Σ_{i=1}^n (a_i - μ)^2 = (1/n)Σ_{i<j} (a_i - a_j)^2. Wait, is that correct?Wait, actually, the sum of squared deviations from the mean is equal to (1/n) times the sum of all pairwise squared differences. Let me check:Σ_{i=1}^n (a_i - μ)^2 = (1/2n) Σ_{i=1}^n Σ_{j=1}^n (a_i - a_j)^2.But maybe I need to verify this.Alternatively, expand Σ_{i<j} (a_i - a_j)^2. Let's compute that:Σ_{i<j} (a_i - a_j)^2 = (1/2) [Σ_{i=1}^n Σ_{j=1}^n (a_i - a_j)^2 - Σ_{i=1}^n (a_i - a_i)^2]But since (a_i - a_j)^2 is symmetric in i and j, and the diagonal terms are zero, so:= (1/2) Σ_{i≠j} (a_i - a_j)^2.But expanding (a_i - a_j)^2 gives a_i² - 2a_i a_j + a_j². Therefore, summing over all i ≠ j:Σ_{i≠j} (a_i - a_j)^2 = Σ_{i≠j} (a_i² - 2a_i a_j + a_j²) = Σ_{i≠j} a_i² + Σ_{i≠j} a_j² - 2 Σ_{i≠j} a_i a_j.But Σ_{i≠j} a_i² = (n-1) Σ_{i=1}^n a_i², since each a_i² is summed (n-1) times. Similarly for Σ_{i≠j} a_j². Therefore:= 2(n-1) Σ a_i² - 2 Σ_{i≠j} a_i a_j.But Σ_{i≠j} a_i a_j = (Σ a_i)^2 - Σ a_i².Therefore, substituting:= 2(n-1) Σ a_i² - 2[(Σ a_i)^2 - Σ a_i²] = 2(n-1)Σ a_i² - 2(Σ a_i)^2 + 2 Σ a_i² = 2n Σ a_i² - 2(Σ a_i)^2.Therefore, Σ_{i<j} (a_i - a_j)^2 = (1/2)[2n Σ a_i² - 2(Σ a_i)^2] = n Σ a_i² - (Σ a_i)^2.So, in our case, n = 5, so Σ_{i<j} (a_i - a_j)^2 = 5 Σ a_i² - (Σ a_i)^2.But we know that Σ a_i² = 1. So, Σ_{i<j} (a_i - a_j)^2 = 5*1 - (Σ a_i)^2 = 5 - (Σ a_i)^2.Therefore, the sum of all pairwise squared differences is 5 minus the square of the sum of the numbers.But how does this help us? We need to relate the minimal squared difference to this total sum. Since there are C(5,2) = 10 pairs, if all squared differences are greater than 1/10, then the total sum would be greater than 10*(1/10) = 1. But according to the formula, the total sum is 5 - (Σ a_i)^2. So, 5 - (Σ a_i)^2 > 1 => (Σ a_i)^2 < 4 => |Σ a_i| < 2. But this doesn't directly lead to a contradiction, because the sum of the numbers can indeed be less than 2. For example, if all numbers are zero except one which is 1, the sum is 1. If numbers are symmetric around zero, the sum is zero. So this approach might not work.Wait, but if all pairwise squared differences are greater than 1/10, then the total sum Σ_{i<j} (a_i - a_j)^2 > 10*(1/10) = 1. But we also have Σ_{i<j} (a_i - a_j)^2 = 5 - (Σ a_i)^2. Therefore, 5 - (Σ a_i)^2 > 1 => (Σ a_i)^2 < 4 => |Σ a_i| < 2. But this is possible. For example, if the numbers sum to 1, then (Σ a_i)^2 = 1 < 4. So this does not give a contradiction. Therefore, this approach is insufficient.Alternatively, perhaps use an energy maximization approach. If we fix the sum of squares, what is the maximum possible minimal squared difference? We saw earlier that arranging the numbers symmetrically as -2d, -d, 0, d, 2d gives a minimal difference of d, and the sum of squares is 2*(4d² + d²) = 10d². Setting this equal to 1 gives d² = 1/10, so d = 1/√10. Therefore, in this case, the minimal squared difference is d² = 1/10, so this configuration achieves exactly the bound. Hence, this suggests that 1/10 is the maximal minimal squared difference, so in any other configuration, the minimal squared difference must be less than or equal to 1/10.But to make this rigorous, we need to show that any other configuration either has a smaller minimal difference or violates the sum of squares constraint. Perhaps using the concept of majorization or some inequality.Alternatively, consider that if we have five numbers with sum of squares 1, then by the pigeonhole principle, at least two of them must lie within a certain interval. To formalize this, let's consider the interval [-1,1] where all numbers lie. If we divide this interval into four subintervals of equal length, then by pigeonhole, two numbers must lie in the same subinterval. The length of each subinterval would be 2/4 = 0.5. The maximum squared difference within a subinterval is (0.5)^2 = 0.25, which is larger than 1/10, so this doesn't help.Alternatively, divide the interval into more subintervals. Let's say we divide [-1,1] into m subintervals. Then, if m is chosen such that the minimal number of intervals needed to ensure that two numbers are in the same interval would give a maximal length per interval such that the squared difference is at most 1/10.Let me calculate how many intervals we need. Suppose each interval has length sqrt(1/10). Then, the number of intervals needed to cover [-1,1] would be 2 / sqrt(1/10) = 2 * sqrt(10) ≈ 6.324. So, approximately 7 intervals. But since we have five numbers, by pigeonhole, if we have 4 intervals, two numbers must lie in the same interval. Wait, no: with five numbers and m intervals, the pigeonhole principle says that at least one interval contains at least ceiling(5/m) numbers. So, if m=4, then ceiling(5/4)=2, so at least one interval has two numbers. Therefore, if we can divide [-1,1] into four intervals each of length sqrt(1/10), then two numbers must lie within sqrt(1/10) of each other. However, the total length of four intervals would be 4*sqrt(1/10) ≈ 1.264, which is less than 2, so we can't cover [-1,1] with four intervals of length sqrt(1/10). Therefore, this approach fails.Alternatively, maybe use overlapping intervals. If we use intervals that overlap, but even then, ensuring coverage is tricky. Alternatively, perhaps use a probabilistic method or consider the positions relative to each other.Wait, another idea: consider the numbers as points on the real line. To maximize the minimal distance between any two points, given the sum of squares constraint, we should place the points as far apart as possible without exceeding the sum of squares. The configuration I mentioned earlier (-2d, -d, 0, d, 2d) seems optimal for this. Let's verify:Sum of squares is 2*(4d² + d²) = 10d². Set to 1: d² = 1/10. So in this case, the minimal distance is d = 1/√10, which gives the squared difference of 1/10. If we try any other configuration, maybe with uneven spacing, would that allow for a larger minimal distance?Suppose we have five numbers with minimal distance greater than 1/√10. Let's try to arrange them such that all consecutive differences are greater than 1/√10. For example, place them at positions x₁, x₂, x₃, x₄, x₅ where each x_{i+1} - x_i > 1/√10. Then, the total spread would be more than 4*(1/√10) ≈ 1.264. However, as previously noted, the numbers must lie within [-1,1], so the total spread can't exceed 2. But 1.264 is less than 2, so this is possible. Therefore, this line of reasoning is insufficient.Alternatively, consider that the sum of squares would be larger if the numbers are spread out. For example, in the symmetric configuration with minimal distance 1/√10, the sum of squares is exactly 1. If we spread the numbers further apart, the sum of squares would increase beyond 1, which contradicts the constraint. Therefore, the symmetric configuration is the one that maximizes the minimal distance while keeping the sum of squares equal to 1. Hence, any other configuration with a larger minimal distance would require a larger sum of squares, which is impossible. Therefore, the minimal squared difference cannot exceed 1/10.This seems promising, but I need to formalize it. Suppose there exists a configuration where all pairwise squared differences are greater than 1/10. Then, the minimal distance is greater than 1/√10. If we can show that the sum of squares in such a configuration must be greater than 1, then this would contradict the given condition, thereby proving the result.So, assume that all pairs of numbers have squared differences greater than 1/10. Then, the minimal distance between any two numbers is greater than sqrt(1/10). Let's order the numbers as a₁ < a₂ < a₃ < a₄ < a₅. Then, a_{i+1} - a_i > sqrt(1/10) for i=1,2,3,4.To model this, let's try to construct such a configuration and compute the sum of squares. Let's set a₁ = x, a₂ = x + d, a₃ = x + 2d, a₄ = x + 3d, a₅ = x + 4d, where d > sqrt(1/10). Then, the numbers are equally spaced with spacing d. The sum of squares is:x² + (x + d)^2 + (x + 2d)^2 + (x + 3d)^2 + (x + 4d)^2.Expand this:= x² + (x² + 2xd + d²) + (x² + 4xd + 4d²) + (x² + 6xd + 9d²) + (x² + 8xd + 16d²)= 5x² + (2xd + 4xd + 6xd + 8xd) + (d² + 4d² + 9d² + 16d²)= 5x² + 20xd + 30d².To minimize the sum of squares, we can choose x such that the derivative with respect to x is zero. Taking derivative:d/dx (5x² + 20xd + 30d²) = 10x + 20d = 0 => x = -2d.So, the minimal sum occurs when x = -2d. Substituting back:Sum = 5*(-2d)^2 + 20*(-2d)*d + 30d² = 5*4d² - 40d² + 30d² = 20d² - 40d² + 30d² = 10d².So, the minimal sum of squares is 10d². Wait, this is the same as the symmetric case! Therefore, if we set x = -2d, the numbers become -2d, -d, 0, d, 2d, which is exactly the symmetric configuration I considered earlier. Therefore, the sum of squares is 10d². For the sum to be 1, d² = 1/10, hence d = 1/√10. Therefore, this configuration achieves the sum of squares 1 with minimal distance d = 1/√10. Therefore, any attempt to increase d beyond 1/√10 would result in the sum of squares exceeding 1. Hence, it's impossible to have all pairwise squared differences greater than 1/10, because that would require d > 1/√10, leading to sum of squares > 1. Therefore, this proves that the minimal squared difference must be at most 1/10.To ensure this is rigorous, we need to confirm that the symmetric configuration is indeed the one that minimizes the sum of squares for a given minimal distance. Since we derived that by fixing the minimal distance d and equally spacing the numbers around zero, the sum of squares is minimized when the numbers are symmetrically placed. Therefore, any other configuration with the same minimal distance d would have a larger sum of squares, hence making it impossible to have d > 1/√10 without exceeding the sum of squares constraint.Therefore, the conclusion is that the least of the squared differences cannot exceed 1/10, as required.

✅:To prove that the least of the numbers (left(a_{i} - a_{j}right)^{2}) for five real numbers (a_1, a_2, a_3, a_4, a_5) with (sum_{i=1}^5 a_i^2 = 1) does not exceed (frac{1}{10}), we proceed as follows:Step 1: Symmetric Configuration AnalysisConsider five numbers arranged symmetrically around 0: (-2d, -d, 0, d, 2d). The sum of squares is:[(-2d)^2 + (-d)^2 + 0^2 + d^2 + (2d)^2 = 4d^2 + d^2 + 0 + d^2 + 4d^2 = 10d^2.]Given (sum_{i=1}^5 a_i^2 = 1), we solve (10d^2 = 1 implies d = frac{1}{sqrt{10}}). Here, the minimal squared difference between consecutive terms is (left(d - 0right)^2 = d^2 = frac{1}{10}).Step 2: Optimality of the Symmetric ConfigurationAssume there exists a configuration where all pairwise squared differences exceed (frac{1}{10}). Order the numbers (a_1 < a_2 < a_3 < a_4 < a_5) with consecutive differences (a_{i+1} - a_i > frac{1}{sqrt{10}}). The minimal sum of squares occurs when the numbers are symmetrically placed (derived via optimization), yielding a sum of (10d^2). If (d > frac{1}{sqrt{10}}), the sum exceeds 1, contradicting the given condition. Thus, no such configuration exists.ConclusionThe symmetric configuration achieves the maximal minimal squared difference of (frac{1}{10}). Any attempt to increase this minimal difference forces the sum of squares to exceed 1. Hence, the least of the (left(a_i - a_jright)^2) must satisfy:[boxed{frac{1}{10}}]

🔑:1. Assume an ordered sequence of numbers: Without loss of generality, let us assume that the sequence of the five real numbers is ordered, such that: [ a_{1} leq a_{2} leq a_{3} leq a_{4} leq a_{5} ]2. Define the least difference ( m ): Let ( m ) be the least value of ( left|a_{i} - a_{j}right| ) for ( i neq j ). Since ( m ) is the minimum difference between any two distinct terms of the sequence, it follows that: [ a_{i+1} - a_{i} geq m quad text{for} quad i = 1, 2, ldots, 4 ]3. Generalize the difference: For any pair ( i ) and ( j ) where ( i > j ), we can generalize the difference as follows: [ a_{i} - a_{j} geq (i - j) m ]4. Sum of squares of differences: Now we compute the sum of the squares of all differences ( left(a_{i} - a_{j}right)^2 ) for ( i > j ). Given that there are ( 10 ) such pairs in total for ( 5 ) numbers: [ sum_{i > j} left(a_{i} - a_{j}right)^{2} geq m^{2} sum_{i > j} (i - j)^{2} ] The sum of ( (i - j)^2 ) over all distinct pairs is computed as: [ sum_{i > j} (i - j)^2 = (1^{2} + 1^{2} + 1^{2} + 1^{2}) + (2^{2} + 2^{2} + 2^{2}) + (3^{2} + 3^{2}) + (4^{2}) = 1 cdot 4 + 4 cdot 3 + 9 cdot 2 + 16 cdot 1 = 50 ] Therefore: [ sum_{i > j} left(a_{i} - a_{j}right)^{2} geq 50m^{2} ]5. Use the problem's condition: Given the problem condition that the sum of the squares of the five numbers equals 1: [ sum_{i=1}^{5} a_{i}^{2} = 1 ] Using the identity for the sum of squares of differences: [ sum_{i > j} left(a_{i} - a_{j}right)^{2} = 5 sum_{i=1}^{5} a_{i}^{2} - left(a_{1} + a_{2} + a_{3} + a_{4} + a_{5}right)^{2} ] Since the sum of the squares is ( 1 ) and the quantity ( left(a_{1} + a_{2} + a_{3} + a_{4} + a_{5}right)^{2} geq 0 ), we have: [ sum_{i > j} left(a_{i} - a_{j}right)^{2} leq 5 ]6. Combining the inequalities: From the inequalities derived, we have: [ 50m^{2} leq 5 ]7. Solving for ( m^{2} ): Solving the inequality: [ m^{2} leq frac{5}{50} = frac{1}{10} ]8. Conclusion: Therefore, the least value of ( left|a_{i} - a_{j}right|^{2} ) where ( i neq j ) does not exceed ( frac{1}{10} ): [ m^{2} leq frac{1}{10} ] (boxed{frac{1}{10}})

🧠:Okay, so I need to find the minimum value of AD in an isosceles trapezoid ABCD where AD is parallel to BC, angle B is 45 degrees. There's a point P on side BC such that the area of triangle PAD is 1/2 and angle APD is 90 degrees. Hmm, let me break this down step by step.First, let me visualize the trapezoid. Since it's isosceles, the legs AB and CD are equal in length, and the base angles are equal. Given angle B is 45 degrees, that means angle C is also 45 degrees because it's an isosceles trapezoid. The bases are AD and BC, with AD being the one we need to minimize.Let me sketch this mentally. AD is the top base, BC is the bottom base. Points A and D are on the top, B and C on the bottom. The legs AB and CD make 45-degree angles with the base BC. So if I can figure out the heights and lengths of the bases, maybe I can relate them to the area of triangle PAD and the right angle at APD.Since it's an isosceles trapezoid, the height can be found if I know the lengths of the bases. Let me denote the height as h. Then, the legs AB and CD can be calculated using the sine of 45 degrees. Since angle B is 45, and the height is h, then AB = h / sin(45°) = h * √2. Similarly, the difference between the bases BC and AD can be related to the horizontal component of the legs. Each leg contributes h * cos(45°) = h * √2/2 to the horizontal extension on each side. Therefore, BC = AD + 2*(h * √2/2) = AD + h√2. So BC = AD + h√2.Alternatively, maybe I can express h in terms of BC and AD. Let me denote AD as x, which is what we need to find the minimum of. Then BC = x + h√2. But I need another relation to connect these variables.The area of triangle PAD is given as 1/2. The area of a triangle is ½ * base * height. Here, if I take AD as the base, then the height would be the perpendicular distance from point P to AD. But since AD is parallel to BC, and the height of the trapezoid is h, the maximum height from BC to AD is h. However, point P is on BC, so the height from P to AD would be h. Wait, but if P is on BC, then the distance from P to AD is h, regardless of where P is on BC? Wait, no. Because the distance from any point on BC to AD is the height of the trapezoid, h. So regardless of where P is on BC, the vertical distance from P to AD is h. Therefore, the area of triangle PAD would be ½ * AD * h = 1/2. So ½ * x * h = 1/2, which simplifies to x * h = 1. So x * h = 1. So h = 1/x.That's a useful relation. So the height of the trapezoid is inversely proportional to AD. So if AD is larger, h is smaller, and vice versa. But we need to find the minimum AD, so that would correspond to maximizing h. But h can't be arbitrary because the trapezoid has to have BC as the lower base. Let me check if this reasoning is correct.Wait, the area of triangle PAD is ½ * base (AD) * height (distance from P to AD). Since AD is parallel to BC, all points on BC are at a distance h from AD. Therefore, regardless of where P is on BC, the height is h, so the area is ½ * AD * h. Therefore, this must equal 1/2, so AD * h = 1. Therefore, x * h = 1. So h = 1/x. That's correct.So h = 1/x. Now, we need to relate this to the other condition: angle APD is 90 degrees. So angle at P is 90 degrees, meaning triangle APD is right-angled at P.Wait, angle APD is 90 degrees. So point P is such that when you connect A to P and D to P, the angle at P is 90 degrees. So triangle APD is right-angled at P. Therefore, by the converse of Thales' theorem, P lies on the circle with diameter AD. But since P is on BC, the intersection of BC with the circle with diameter AD will give possible positions of P. However, since we need the minimal AD, perhaps the circle must be tangent to BC? Because the minimal AD would correspond to the case where the circle with diameter AD just touches BC, so there is exactly one such point P. If the circle intersects BC at two points, then AD could be smaller, but if it's tangent, that's the minimal AD. Wait, but since we are given that such a point P exists, the circle must intersect BC. But to find the minimal AD, perhaps we need the case where the circle is tangent to BC, so only one point P exists. Therefore, maybe the minimal AD is when the circle with diameter AD is tangent to BC. Let me explore this.But first, let's set up coordinates to model this problem. Let's place the trapezoid in coordinate system to make calculations easier.Let me place point A at the origin (0,0). Since AD is parallel to BC, and the trapezoid is isosceles, we can set up coordinates as follows:- Let AD be along the x-axis from (0,0) to (x,0), so D is at (x,0).- The height of the trapezoid is h, so points B and C will be located at (a, h) and (x + a, h) respectively, since it's an isosceles trapezoid. Wait, but the legs AB and CD should be equal in length and symmetric.Wait, perhaps it's better to center BC relative to AD. Since it's an isosceles trapezoid, the legs AB and CD are symmetric with respect to the vertical axis through the midpoint of AD and BC. Therefore, if AD has length x, then BC has length x + 2*(h * cot(45°)) because each side extends beyond AD by h * cot(45°) = h*1 = h. Wait, angle B is 45 degrees, so the horizontal component of the leg AB is h * cot(45°) = h. Therefore, the length of BC is AD + 2h. Therefore, BC = x + 2h. Wait, but earlier I thought BC = x + h√2, but that might be incorrect.Wait, let's re-examine. In an isosceles trapezoid with bases AD and BC, legs AB and CD. The angles at B and C are 45 degrees. The height is h. Then, the horizontal projection from B to the base AD is h * cot(45°) = h * 1 = h. Therefore, each of the legs extends h units horizontally beyond AD. Therefore, the length of BC is AD + 2h. So BC = x + 2h. So that's the correct relation. Therefore, BC = x + 2h.But earlier, I thought BC = x + h√2, which must be wrong. Let's check. If the legs have length AB = CD = L, then the height h = L * sin(45°) = L * √2/2. The horizontal component is L * cos(45°) = L * √2/2. Therefore, the extension on each side is L * √2/2, so total BC = AD + 2*(L * √2/2) = AD + L√2. But since h = L * √2/2, then L = h * 2/√2 = h√2. Therefore, BC = x + (h√2)*√2 = x + 2h. Ah, so both approaches agree. Therefore, BC = x + 2h.Therefore, BC = x + 2h. So we can write BC = x + 2h. But we also have h = 1/x from the area condition. So substituting h = 1/x into BC, we get BC = x + 2*(1/x) = x + 2/x.But we might need another relation from the right angle at P. Let's proceed with coordinates.Let me set coordinate system with point A at (0,0), D at (x,0). The height of the trapezoid is h = 1/x. Then, points B and C are located at (h, h) and (x - h, h), since the horizontal extensions are h on each side. Wait, no. Wait, the horizontal extension on each side is h, so from point A (0,0), moving h units to the right along the base BC would reach point B. Wait, actually, in an isosceles trapezoid, the legs are AB and CD. So if AD is from (0,0) to (x,0), then the legs go up to BC. The horizontal distance from A to B is h, since angle at B is 45 degrees. The height is h, so the horizontal component is h. Therefore, point B is at (h, h), and point C is at (x - h, h). Therefore, BC is from (h, h) to (x - h, h), so the length of BC is (x - h) - h = x - 2h. Wait, that contradicts earlier. Wait, this seems conflicting.Wait, maybe I need to think carefully. Let me use coordinates again.Set point A at (0,0), D at (x,0). The height of the trapezoid is h. The legs AB and CD are inclined at 45 degrees. The horizontal component of each leg is h * cot(45°) = h. Therefore, point B is located at (h, h), since from A (0,0), moving h units to the right and h units up. Similarly, point C is located at (x - h, h), since from D (x,0), moving h units to the left and h units up. Therefore, BC spans from (h, h) to (x - h, h), so the length of BC is (x - h) - h = x - 2h. Wait, so BC = x - 2h. But earlier we had BC = x + 2h. That's contradictory. So which is correct?Wait, if the horizontal component from each side is h, then BC is shorter than AD by 2h. Because from A, moving h to the right to B, and from D, moving h to the left to C. Therefore, BC is AD minus 2h. So BC = x - 2h. But that seems counterintuitive. Wait, in a trapezoid, typically the lower base is longer. If AD is the upper base and BC is the lower base, then BC should be longer. Wait, but in the problem statement, it just says AD || BC, not specifying which is upper or lower. Maybe I assumed AD is the upper base, but if AD is the lower base, then BC would be the upper base. Wait, the problem says "In the isosceles trapezoid ABCD, AD || BC, angle B = 45°". Usually, trapezoid notation is such that consecutive letters are adjacent vertices. So ABCD, so sides AB, BC, CD, DA. So AD and BC are the two bases. Since angle B is 45°, which is at vertex B, which is adjacent to base BC. So if the trapezoid is oriented with BC as the lower base, then angle at B is 45°, so the legs are AB and CD. Then, BC is the lower base, AD is the upper base. Then, the height is h, and the horizontal projections from the legs would make BC longer than AD. Wait, so if the legs slope outwards, then BC would be longer.But according to the coordinate system, if we place AD as the upper base at height h, then BC would be the lower base. Wait, perhaps I need to adjust the coordinate system.Wait, let's start over. Let me define the trapezoid with BC as the lower base, and AD as the upper base. So points B and C are on the lower base BC, and points A and D are on the upper base AD. The legs are AB and CD. Angle at B is 45°, so angle between base BC and leg AB is 45°. So if we place point B at the origin (0,0), then BC is along the x-axis to (c,0). Then, point A is somewhere above B, forming a 45° angle. The height of the trapezoid is h, so the y-coordinate of A and D is h. Then, AB has length h / sin(45°) = h√2. The horizontal component from B to A is h / tan(45°) = h. So point A would be at (h, h). Then, since the trapezoid is isosceles, the leg CD is also inclined at 45°, so point D would be at (c - h, h). Therefore, the upper base AD is from (h, h) to (c - h, h), so AD has length (c - h) - h = c - 2h. The lower base BC is from (0,0) to (c,0), so length c. Therefore, AD = BC - 2h. So AD = c - 2h. Therefore, BC = AD + 2h. So if we denote AD as x, then BC = x + 2h. So in this coordinate system, BC is longer than AD by 2h. Therefore, BC = x + 2h.But earlier, from the area of triangle PAD, we have x * h = 1. So h = 1/x. Therefore, BC = x + 2*(1/x) = x + 2/x.Now, point P is on BC. Since BC is the lower base from (0,0) to (c,0), which in terms of x is from (0,0) to (x + 2h,0) = (x + 2/x, 0). Wait, but in the coordinate system where AD is upper base at (h, h) to (c - h, h), which is (h, h) to (x + 2h - h, h) = (x + h, h). Wait, this is getting a bit confusing. Let me clarify the coordinate system.Let me fix the coordinate system as follows:- Let the lower base BC be along the x-axis from (0,0) to (b,0), where b = BC.- The upper base AD is parallel to BC and is at height h. Since the trapezoid is isosceles, AD is centered above BC. Therefore, the coordinates of A and D are ((b - a)/2, h) and ((b + a)/2, h), where a = AD. But given angle B is 45°, which is at point B (0,0). The leg AB connects B (0,0) to A ((b - a)/2, h). The angle at B is 45°, so the slope of AB is (h - 0)/[( (b - a)/2 - 0)] = 2h / (b - a). Since angle B is 45°, this slope is tan(45°) = 1. Therefore, 2h / (b - a) = 1 => 2h = b - a => b = a + 2h.Which aligns with the earlier conclusion that BC = AD + 2h. So BC = a + 2h. Therefore, the coordinates of A are ((b - a)/2, h) = (( (a + 2h) - a )/2, h ) = (h, h). Similarly, coordinates of D are ((b + a)/2, h) = ( (a + 2h + a)/2, h ) = (a + h, h).So to recap:- B is at (0,0).- C is at (a + 2h, 0).- A is at (h, h).- D is at (a + h, h).Now, point P is on BC, so its coordinates can be parameterized. Let's let P be at (t, 0) where t ranges from 0 to a + 2h.Given that angle APD is 90°, the vectors PA and PD should be perpendicular. Let's write that condition.Coordinates:- Point A: (h, h)- Point D: (a + h, h)- Point P: (t, 0)Vector PA = A - P = (h - t, h - 0) = (h - t, h)Vector PD = D - P = (a + h - t, h - 0) = (a + h - t, h)Their dot product should be zero:(h - t)(a + h - t) + h * h = 0Let me compute that:= (h - t)(a + h - t) + h²First expand (h - t)(a + h - t):= h(a + h - t) - t(a + h - t)= h(a + h) - h t - t(a + h) + t²= h(a + h) - h t - a t - h t + t²= h(a + h) - (2h + a)t + t²Therefore, the dot product becomes:h(a + h) - (2h + a)t + t² + h² = 0Simplify:= h(a + h) + h² - (2h + a)t + t²= h a + h² + h² - (2h + a)t + t²= h a + 2h² - (2h + a)t + t²So the equation is:t² - (2h + a)t + h a + 2h² = 0This is a quadratic equation in t. For real solutions to exist, the discriminant must be non-negative.Discriminant D:= [-(2h + a)]² - 4 * 1 * (h a + 2h²)= (2h + a)^2 - 4(h a + 2h²)Expand (2h + a)^2:= 4h² + 4h a + a²Subtract 4(h a + 2h²):= 4h² + 4h a + a² - 4h a - 8h²= (4h² - 8h²) + (4h a - 4h a) + a²= -4h² + 0 + a²= a² - 4h²Therefore, discriminant D = a² - 4h²For real solutions, D ≥ 0 => a² - 4h² ≥ 0 => a² ≥ 4h² => a ≥ 2h.But we also have from the area condition that a * h = 1 => h = 1/a. Therefore, substituting h = 1/a into a ≥ 2h gives:a ≥ 2*(1/a) => a² ≥ 2 => a ≥ √2So this gives a lower bound of √2 for a. But is this the minimal value? Wait, but we need to check whether there exists a point P on BC such that angle APD is 90°, and the discriminant must be non-negative. So if a ≥ √2, then D is non-negative, so real solutions exist. Therefore, the minimal value of a is √2? But wait, let's check.Wait, if a = √2, then h = 1/√2. Then, BC = a + 2h = √2 + 2*(1/√2) = √2 + √2 = 2√2. So BC = 2√2, and the coordinates of points:- B: (0,0)- C: (2√2, 0)- A: (h, h) = (1/√2, 1/√2)- D: (a + h, h) = (√2 + 1/√2, 1/√2)Now, point P is on BC, which is from (0,0) to (2√2,0). The quadratic equation for t becomes:t² - (2h + a)t + h a + 2h² = 0Substituting a = √2, h = 1/√2:2h + a = 2*(1/√2) + √2 = √2 + √2 = 2√2h a + 2h² = (1/√2)*√2 + 2*(1/√2)^2 = 1 + 2*(1/2) = 1 + 1 = 2Thus, the equation becomes:t² - 2√2 t + 2 = 0Compute discriminant D = (2√2)^2 - 4*1*2 = 8 - 8 = 0Therefore, t = [2√2]/2 = √2. So point P is at (√2, 0), which is the midpoint of BC since BC is from 0 to 2√2. So in this case, P is the midpoint. Therefore, when a = √2, there is exactly one point P (midpoint of BC) such that angle APD is 90°, and the area of triangle PAD is 1/2. Therefore, this suggests that the minimal AD is √2. However, I need to confirm that there isn't a smaller a satisfying the conditions. For example, if a < √2, then discriminant D = a² - 4h² = a² - 4*(1/a)^2 = a² - 4/a². For D ≥ 0, a² - 4/a² ≥ 0 => a^4 - 4 ≥ 0 => a^4 ≥ 4 => a ≥ √2. Therefore, the minimal a is indeed √2. Hence, the answer is √2. But let me verify with another approach.Alternatively, since we have vectors PA and PD perpendicular, maybe we can use coordinate geometry to find the relation.Given points A(h, h), D(a + h, h), P(t, 0). Then PA ⋅ PD = 0.As earlier:PA = (h - t, h)PD = (a + h - t, h)Dot product: (h - t)(a + h - t) + h² = 0Let me rearrange:(h - t)(a + h - t) = -h²Expand left side:= (h - t)(a + h - t)Let me set s = t - h, then:= (-s)(a - s)= -s(a - s) = -a s + s²Set equal to -h²:-a s + s² = -h² => s² - a s + h² = 0But s = t - h, so:(t - h)^2 - a(t - h) + h² = 0Expand:t² - 2h t + h² - a t + a h + h² = 0Combine like terms:t² - (2h + a) t + 2h² + a h = 0Which is the same quadratic as before. Therefore, discriminant D = (2h + a)^2 - 4*(2h² + a h) = 4h² + 4a h + a² - 8h² -4a h = a² -4h². Therefore, same result.Thus, for real solutions, a² -4h² ≥ 0. Since h =1/a, then a² -4/a² ≥0 => a^4 -4 ≥0 => a^4 ≥4 => a≥√2. Therefore, minimal a is √2. Hence, the minimal AD is √2. Therefore, the answer is √2. But let me check with an example.Suppose a = √2, h =1/√2. Then BC = √2 + 2*(1/√2) = √2 + √2 = 2√2. Then P is at t = [2h + a ±√(a² -4h²)]/2. Since discriminant is zero when a=√2, t = [2*(1/√2) + √2]/2 = [√2 + √2]/2 = (2√2)/2 = √2. So P is at (√2,0), which is the midpoint of BC (since BC is from 0 to 2√2). Therefore, APD is right-angled at P, and the area is ½ * AD * h = ½ * √2 * (1/√2) = ½ *1=1/2, which matches the given area. Therefore, all conditions are satisfied. Therefore, the minimal AD is indeed √2. So the answer should be √2, boxed as boxed{sqrt{2}}.Final AnswerThe minimum value of (AD) is (boxed{sqrt{2}}).

✅:Given an isosceles trapezoid (ABCD) with (AD parallel BC) and (angle B = 45^circ), we need to find the minimum value of (AD) given that the area of (triangle PAD) is (frac{1}{2}) and (angle APD = 90^circ) for some point (P) on (BC).1. Area Condition: - The area of (triangle PAD) is (frac{1}{2}). - The height of the trapezoid (h) is the perpendicular distance from (BC) to (AD). - The area of (triangle PAD) is given by (frac{1}{2} times AD times h = frac{1}{2}), leading to (AD times h = 1). Thus, (h = frac{1}{AD}).2. Coordinate System and Geometry: - Place (A) at ((0, 0)), (D) at ((x, 0)), (B) at ((h, h)), and (C) at ((x - h, h)). - The length of (BC) is (x - 2h), but since (BC = AD + 2h), we have (BC = x + 2h).3. Right Angle Condition: - Point (P) on (BC) must satisfy (angle APD = 90^circ). - Using vectors, the coordinates of (P) are parameterized, and the dot product condition for perpendicular vectors is applied. - This leads to the quadratic equation in (t) (the x-coordinate of (P)): (t^2 - (2h + x)t + hx + 2h^2 = 0).4. Discriminant Analysis: - The discriminant of the quadratic equation must be non-negative for real solutions. - Substituting (h = frac{1}{x}), the discriminant condition simplifies to (x^2 - 4left(frac{1}{x}right)^2 geq 0), leading to (x^4 geq 4) or (x geq sqrt{2}).5. Verification: - When (x = sqrt{2}), the height (h = frac{1}{sqrt{2}}), and the quadratic equation yields a single solution, confirming the existence of point (P) such that (angle APD = 90^circ).Thus, the minimum value of (AD) is (boxed{sqrt{2}}).

🔑:1. Given that (ABCD) is an isosceles trapezoid with (AD parallel BC) and (angle B = 45^circ).2. Let the lengths of the bases be (AD = x) and (BC = y). Since (angle APD = 90^circ), triangle (APD) is right-angled at (P) and the area of (triangle PAD) is given to be ( frac{1}{2} ).3. Observing that (angle PAD = 90^circ), we can use the Pythagorean theorem. In this right triangle, if we denote the height from (D) to (P) as (h), we have: [ text{Area of } triangle PAD = frac{1}{2} times AD times h = frac{1}{2} times x times h = frac{1}{2} ] Solving for (h): [ x times h = 1 implies h = frac{1}{x} ]4. Since (angle B = 45^circ), the height of the trapezoid from (AD) perpendicularly to (BC) can be found due to the properties of 45-degree right triangles where the legs are equal: [ h = frac{y - x}{2} ] We already found (h = frac{1}{x}) from the area of (triangle PAD), so: [ frac{1}{x} = frac{y - x}{2} ]5. Solving for (y): [ 2 times frac{1}{x} = y - x implies frac{2}{x} = y - x implies y = x + frac{2}{x} ]6. Since we need to find the minimum value of (AD = x), we use calculus. Let: [ f(x) = x + frac{2}{x} ]7. To find the minimum, take the derivative of (f(x)) and set it to zero: [ f'(x) = 1 - frac{2}{x^2} ] Setting the derivative to zero: [ 1 - frac{2}{x^2} = 0 implies frac{2}{x^2} = 1 implies x^2 = 2 implies x = sqrt{2} ]8. We confirm this is a minimum by checking the second derivative: [ f''(x) = frac{4}{x^3} ] Since (f''(x) > 0) for (x = sqrt{2}), this is indeed a minimum. Conclusion:The minimum value of (AD) is (boxed{sqrt{2}}).

🧠:Okay, let me try to figure out this problem step by step. The problem is about finding the maximum area of the inscribed circle (incircle) of triangle F1PQ, where F1 is the left focus of the given ellipse, and points P and Q are where a line l passes through the right focus F2 and intersects the ellipse. First, I need to recall some properties of ellipses. The standard equation of an ellipse is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. In this case, the given ellipse is x²/4 + y²/3 = 1, so a² = 4 and b² = 3. Therefore, a = 2 and b = √3. Next, I need to find the coordinates of the foci F1 and F2. For an ellipse, the distance from the center to each focus is c, where c² = a² - b². Plugging in the values, c² = 4 - 3 = 1, so c = 1. Since the ellipse is centered at the origin (0,0), the foci are located at (-c, 0) and (c, 0), which means F1 is (-1, 0) and F2 is (1, 0).Now, the line l passes through the right focus F2 (1, 0) and intersects the ellipse at points P and Q. The problem asks for the maximum area of the inscribed circle of triangle F1PQ. To approach this, I need to consider the following steps:1. Parametrize the line l passing through F2 (1,0): Since the line passes through (1,0), I can write its equation in the form y = m(x - 1), where m is the slope. Alternatively, if the line is vertical, its equation is x = 1, but in that case, there would only be one intersection point with the ellipse (since vertical lines through the center would intersect at two points symmetric about the x-axis). However, since the problem mentions two points P and Q, maybe the line isn't vertical. Wait, but actually, even a vertical line through F2 (1,0) would intersect the ellipse at two points: substituting x=1 into the ellipse equation gives (1)/4 + y²/3 = 1, so y² = 3*(1 - 1/4) = 3*(3/4) = 9/4, so y = ±3/2. Therefore, the points would be (1, 3/2) and (1, -3/2). So even vertical lines would intersect at two points. So maybe parametrizing the line with a slope m is okay, and then considering vertical as a separate case.2. Find the points P and Q where line l intersects the ellipse: Substitute the equation of the line into the ellipse equation and solve for x and y. This will give the coordinates of P and Q in terms of m (or whatever parameter we use).3. Determine the coordinates of P and Q in terms of m: Once we have the parametric equations, we can find expressions for P and Q.4. Calculate the area of triangle F1PQ: To find the area of the triangle, we can use the formula for the area given three vertices. Alternatively, using vectors or determinants.5. Find the inradius of triangle F1PQ: The area of the inscribed circle (incircle) is πr², where r is the inradius. The inradius of a triangle is given by r = A/s, where A is the area of the triangle and s is the semi-perimeter. Therefore, to maximize the area of the incircle, we need to maximize r, which depends on A/s. So we need to maximize A/s.6. Express A and s in terms of m (or another parameter): Once we have expressions for A and s in terms of the slope m, we can then find the maximum value of A/s.7. Use calculus to find the maximum: Take the derivative of A/s with respect to m, set it equal to zero, and solve for m. Then verify it's a maximum.Alternatively, there might be a geometric interpretation or property that could simplify the problem without calculus. But since it's about maximizing a quantity that depends on the triangle's configuration, calculus might be necessary.Let me proceed step by step.Step 1: Equation of the line lAssuming the line is not vertical, let's use slope m. The line passes through (1, 0), so equation is y = m(x - 1).If the line is vertical, x = 1, which we already saw intersects the ellipse at (1, ±3/2). We can consider this as a separate case.Step 2: Find intersection points P and QSubstitute y = m(x - 1) into the ellipse equation:x²/4 + [m²(x - 1)²]/3 = 1Multiply through by 12 to eliminate denominators:3x² + 4m²(x - 1)² = 12Expand the terms:3x² + 4m²(x² - 2x + 1) = 12Combine like terms:3x² + 4m²x² - 8m²x + 4m² = 12Group x² terms:(3 + 4m²)x² - 8m²x + (4m² - 12) = 0This is a quadratic in x: Ax² + Bx + C = 0, whereA = 3 + 4m²B = -8m²C = 4m² - 12To find x-coordinates of P and Q, solve quadratic equation:x = [8m² ± sqrt(64m^4 - 4*(3 + 4m²)*(4m² - 12))]/[2*(3 + 4m²)]Let me compute discriminant D:D = 64m^4 - 4*(3 + 4m²)(4m² - 12)First compute the product term:4*(3 + 4m²)(4m² - 12) = 4*[12m² - 36 + 16m^4 - 48m²] = 4*[16m^4 - 36m² - 36] = 64m^4 - 144m² - 144Therefore, D = 64m^4 - (64m^4 - 144m² - 144) = 64m^4 -64m^4 +144m² +144 = 144m² +144 = 144(m² +1)So discriminant D = 144(m² +1), which is positive for all m, which makes sense because the line passes through the ellipse.Thus, x = [8m² ± 12√(m² +1)]/[2*(3 + 4m²)] = [4m² ± 6√(m² +1)]/(3 + 4m²)Therefore, the x-coordinates of P and Q are:x1 = [4m² + 6√(m² +1)]/(3 + 4m²)x2 = [4m² - 6√(m² +1)]/(3 + 4m²)Corresponding y-coordinates are y = m(x -1), so:For x1: y1 = m([4m² + 6√(m² +1)]/(3 + 4m²) -1) = m([4m² +6√(m² +1) -3 -4m²]/(3 +4m²)) = m([6√(m² +1) -3]/(3 +4m²))Similarly for x2: y2 = m([4m² -6√(m² +1) -3 -4m²]/(3 +4m²)) = m([-6√(m² +1) -3]/(3 +4m²)) = -m([6√(m² +1) +3]/(3 +4m²))So points P and Q are:P = ( [4m² +6√(m² +1)]/(3 +4m²), m[6√(m² +1) -3]/(3 +4m²) )Q = ( [4m² -6√(m² +1)]/(3 +4m²), -m[6√(m² +1) +3]/(3 +4m²) )These expressions look complicated. Maybe there's a smarter way to parametrize the line. Alternatively, use parametric equations for the line with a parameter t, but I don't know if that would simplify things.Alternatively, perhaps parametrize the line in terms of the angle θ instead of slope m. Let’s consider θ as the angle the line makes with the x-axis. Then the slope m = tanθ. Maybe trigonometric substitution could simplify the expressions.Let me try that. Let m = tanθ, so √(m² +1) = secθ. Then, expressions become:x1 = [4tan²θ + 6secθ]/(3 +4tan²θ)Similarly, x2 = [4tan²θ -6secθ]/(3 +4tan²θ)But I'm not sure if this helps yet. Alternatively, let's note that 3 +4m² = 3 +4tan²θ = 3(1) +4tan²θ. Maybe express in terms of sin and cos.Alternatively, perhaps parametrize the line using a parameter t. Let’s say the line passes through F2(1,0) and has direction vector (cosθ, sinθ). Then parametric equations are x =1 + t cosθ, y = 0 + t sinθ. Substitute into ellipse equation:( (1 + t cosθ)^2 )/4 + ( (t sinθ)^2 ) /3 =1Expand:[1 + 2t cosθ + t² cos²θ]/4 + [t² sin²θ]/3 =1Multiply through by 12:3[1 + 2t cosθ + t² cos²θ] +4[t² sin²θ] =12Expand:3 +6t cosθ +3t² cos²θ +4t² sin²θ =12Combine like terms:3 +6t cosθ + t²(3cos²θ +4 sin²θ) -12 =0Simplify:-9 +6t cosθ + t²(3cos²θ +4 sin²θ) =0This is a quadratic in t:[3cos²θ +4 sin²θ] t² +6cosθ t -9 =0Let me compute discriminant D:D = (6cosθ)^2 -4*(3cos²θ +4 sin²θ)*(-9) =36cos²θ +36*(3cos²θ +4 sin²θ)=36cos²θ +108cos²θ +144 sin²θ =144cos²θ +144 sin²θ =144(cos²θ + sin²θ)=144Therefore, solutions for t:t = [ -6cosθ ±12 ]/[2*(3cos²θ +4 sin²θ) ] = [ -6cosθ ±12 ]/(6cos²θ +8 sin²θ )Simplify numerator:First solution: -6cosθ +12 =6(-cosθ +2)Second solution: -6cosθ -12 = -6(cosθ +2)Denominator: 6cos²θ +8 sin²θ =6cos²θ +8(1 -cos²θ)=8 -2cos²θTherefore, t = [6(-cosθ +2)]/(8 -2cos²θ) and t = [-6(cosθ +2)]/(8 -2cos²θ)Wait, but 8 -2cos²θ is denominator, so:t1 = [6(2 -cosθ)]/(8 -2cos²θ)t2 = [-6(cosθ +2)]/(8 -2cos²θ)Therefore, the parameter t for points P and Q are t1 and t2.Thus, coordinates of P and Q:For t1: x =1 + t1 cosθ, y = t1 sinθFor t2: x =1 + t2 cosθ, y = t2 sinθBut perhaps this is more manageable? Maybe. Let's see.Alternatively, note that the product of t1 and t2 can be found from quadratic equation: t1 * t2 = c/a = (-9)/(3cos²θ +4 sin²θ). But maybe not helpful.Alternatively, perhaps compute the distance between P and Q, and the distances from F1 to P and Q, then compute the semi-perimeter and area.But this seems complicated.Alternatively, note that the area of triangle F1PQ can be computed using the determinant formula.Given three points F1(-1,0), P(x1,y1), Q(x2,y2), the area A is (1/2)| (x1 +1)(y2 -0) - (x2 +1)(y1 -0) | = (1/2)| (x1 +1)y2 - (x2 +1)y1 |.Alternatively, using vectors: F1P × F1Q /2.But maybe using coordinates is better.But since expressions for x1, x2, y1, y2 are complicated, this might get messy.Alternatively, recall that the area of triangle can also be expressed as (1/2)*base*height. If we can find a suitable base and height.Alternatively, use parametric equations with θ and try to express the area in terms of θ.Let me try this approach.Using the parametric equations with angle θ. Let’s define the line passing through F2(1,0) at angle θ with the x-axis. Then, points P and Q are obtained by moving along this line in both directions. The parameter t represents the distance from F2 along the line. So the points P and Q correspond to t1 and t2 as found earlier.So, the coordinates of P and Q are:For t1:x_P = 1 + t1 cosθy_P = t1 sinθFor t2:x_Q = 1 + t2 cosθy_Q = t2 sinθBut since t2 is negative, this might represent the point in the opposite direction from F2.Given that t1 and t2 are the roots, and from earlier:t1 = [6(2 -cosθ)] / [8 -2cos²θ]t2 = [ -6(2 +cosθ) ] / [8 -2cos²θ]Wait, actually, let me recheck:Earlier, we had t1 = [6(2 -cosθ)] / [8 -2cos²θ]But let's re-express:Wait, in denominator, 8 -2cos²θ = 2(4 - cos²θ). Hmm.But perhaps we can compute coordinates of P and Q in terms of θ.So x_P =1 + t1 cosθ =1 + [6(2 -cosθ) cosθ]/[8 -2cos²θ]Similarly, y_P = t1 sinθ = [6(2 -cosθ) sinθ]/[8 -2cos²θ]Similarly, x_Q =1 + t2 cosθ =1 + [ -6(2 +cosθ) cosθ ]/[8 -2cos²θ ]y_Q = t2 sinθ = [ -6(2 +cosθ) sinθ ]/[8 -2cos²θ ]Simplify x_P:x_P = [ (8 -2cos²θ) +6(2 -cosθ)cosθ ] / [8 -2cos²θ ]Compute numerator:8 -2cos²θ +6(2cosθ -cos²θ) =8 -2cos²θ +12cosθ -6cos²θ =8 +12cosθ -8cos²θSimilarly, x_P = (8 +12cosθ -8cos²θ)/(8 -2cos²θ)Factor numerator and denominator:Numerator: 8(1 -cos²θ) +12cosθ =8sin²θ +12cosθ. Wait, but 8 -8cos²θ is 8sin²θ. Wait:Wait, original numerator:8 +12cosθ -8cos²θ = -8cos²θ +12cosθ +8.Not sure if factoring helps. Similarly for x_Q:x_Q =1 + [ -6(2 +cosθ)cosθ ]/[8 -2cos²θ ] = [8 -2cos²θ -6(2cosθ +cos²θ) ]/[8 -2cos²θ ]Compute numerator:8 -2cos²θ -12cosθ -6cos²θ =8 -12cosθ -8cos²θSo x_Q = (8 -12cosθ -8cos²θ)/(8 -2cos²θ) = [ -8cos²θ -12cosθ +8 ]/[8 -2cos²θ ]Similarly, the expressions for x_P and x_Q are similar but with cosθ terms having opposite signs.Similarly, y_P and y_Q will have sinθ terms with different signs.But maybe there is a better way. Let's compute the vectors F1P and F1Q.F1 is (-1,0). So coordinates of P relative to F1:F1P = (x_P +1, y_P -0) = (x_P +1, y_P)Similarly, F1Q = (x_Q +1, y_Q )The area A of triangle F1PQ is (1/2)| F1P × F1Q |, where × denotes the cross product.So A = (1/2)| (x_P +1)y_Q - (x_Q +1)y_P |.Let me compute this expression.First, compute (x_P +1) and (x_Q +1):From earlier, x_P = [8 +12cosθ -8cos²θ]/[8 -2cos²θ], so x_P +1 = [8 +12cosθ -8cos²θ +8 -2cos²θ]/[8 -2cos²θ] = [16 +12cosθ -10cos²θ]/[8 -2cos²θ]Wait, wait, no. Wait, x_P is already computed as [8 +12cosθ -8cos²θ]/[8 -2cos²θ], so x_P +1 = [8 +12cosθ -8cos²θ +8 -2cos²θ]/[8 -2cos²θ] = [16 +12cosθ -10cos²θ]/[8 -2cos²θ]Similarly, x_Q +1 = [8 -12cosθ -8cos²θ]/[8 -2cos²θ] +1 = [8 -12cosθ -8cos²θ +8 -2cos²θ]/[8 -2cos²θ] = [16 -12cosθ -10cos²θ]/[8 -2cos²θ]But this seems complicated. Let me instead express x_P +1 and x_Q +1 in terms of t1 and t2:x_P =1 + t1 cosθ => x_P +1 =2 + t1 cosθSimilarly, x_Q +1 =2 + t2 cosθBut y_P = t1 sinθ and y_Q = t2 sinθ.Therefore, the area A = (1/2)| (2 + t1 cosθ) t2 sinθ - (2 + t2 cosθ) t1 sinθ |.Factor sinθ:A = (1/2)| sinθ [ (2 + t1 cosθ)t2 - (2 + t2 cosθ)t1 ] |.Compute the expression inside the brackets:(2 + t1 cosθ)t2 - (2 + t2 cosθ)t1 =2t2 + t1 t2 cosθ -2t1 - t1 t2 cosθ =2(t2 - t1) + t1 t2 cosθ - t1 t2 cosθ =2(t2 - t1).Therefore, A = (1/2)| sinθ * 2(t2 - t1)| = | sinθ (t2 - t1) |.But since t2 is negative and t1 is positive (since t1 corresponds to moving in one direction from F2 and t2 in the opposite), t2 - t1 is negative, so the absolute value becomes |sinθ| (t1 - t2).Therefore, A = |sinθ| (t1 - t2).Now, from earlier, t1 and t2 are roots of the quadratic equation:[3cos²θ +4 sin²θ] t² +6cosθ t -9 =0Thus, the difference t1 - t2 can be computed using the roots formula.For a quadratic equation at² + bt +c =0, the roots are t1 and t2, so t1 - t2 = sqrt(D)/a, where D is discriminant.But since t1 > t2 (as t1 is positive and t2 is negative), t1 - t2 = sqrt(D)/a.From earlier, discriminant D =144, so sqrt(D) =12. Wait, earlier when we parametrized with m, the discriminant was 144(m² +1), but when we parametrized with θ, the discriminant was 144.Wait, in the parametrization with θ, discriminant D=144, so sqrt(D)=12, and the coefficient a =3cos²θ +4 sin²θ.Thus, t1 - t2 = sqrt(D)/a =12 / (3cos²θ +4 sin²θ).Therefore, area A = |sinθ| * 12 / (3cos²θ +4 sin²θ).Simplify:A = 12 |sinθ| / (3cos²θ +4 sin²θ)This is a much simpler expression for the area! So A = 12 |sinθ| / (3cos²θ +4 sin²θ)Since θ can range from 0 to π (as beyond that, it's symmetric), and sinθ is non-negative in [0, π], so |sinθ| = sinθ.Therefore, A = 12 sinθ / (3cos²θ +4 sin²θ)Now, this is a function A(θ) that we need to maximize. However, the problem is not just to maximize the area of the triangle, but the area of its incircle. Therefore, I need to relate the inradius r to the area A and semi-perimeter s.Recall that for a triangle, the inradius r = A / s, where s = (a + b + c)/2.Therefore, the area of the incircle is πr² = π(A / s)². So to maximize the area of the incircle, we need to maximize (A / s)², which is equivalent to maximizing A / s.Therefore, the problem reduces to maximizing A / s, where A is the area of triangle F1PQ, and s is its semi-perimeter.So, first, we need expressions for A and s in terms of θ.We already have A(θ) =12 sinθ / (3cos²θ +4 sin²θ)Now, need to find s(θ). To find s, we need the lengths of the three sides: F1P, F1Q, PQ.Let’s compute these lengths.First, compute F1P and F1Q.F1 is at (-1,0). Points P and Q are located at (x_P, y_P) and (x_Q, y_Q).From earlier parametrization using θ:Coordinates of P: x_P =1 + t1 cosθ, y_P = t1 sinθCoordinates of Q: x_Q =1 + t2 cosθ, y_Q = t2 sinθTherefore, vector F1P = (x_P +1, y_P) = (2 + t1 cosθ, t1 sinθ)So length F1P = sqrt[ (2 + t1 cosθ)^2 + (t1 sinθ)^2 ]Similarly, length F1Q = sqrt[ (2 + t2 cosθ)^2 + (t2 sinθ)^2 ]Length PQ = sqrt[ (x_P - x_Q)^2 + (y_P - y_Q)^2 ] = sqrt[ (t1 cosθ - t2 cosθ)^2 + (t1 sinθ - t2 sinθ)^2 ] = |t1 - t2| * sqrt[ cos²θ + sin²θ ] = |t1 - t2| *1 = |t1 - t2|But since t1 > t2 (as t1 is positive, t2 is negative), PQ = t1 - t2 = same as before, which was 12 / (3cos²θ +4 sin²θ). Wait, earlier we had t1 - t2 = sqrt(D)/a =12 / a, where a =3cos²θ +4 sin²θ. So yes, PQ=12/(3cos²θ +4 sin²θ).Therefore, PQ=12/(3cos²θ +4 sin²θ).Now, compute F1P and F1Q:Let’s compute F1P first.F1P = sqrt[ (2 + t1 cosθ)^2 + (t1 sinθ)^2 ]Expand:= sqrt[4 +4 t1 cosθ + t1² cos²θ + t1² sin²θ ]= sqrt[4 +4 t1 cosθ + t1² (cos²θ + sin²θ) ]= sqrt[4 +4 t1 cosθ + t1² ]Similarly for F1Q:F1Q= sqrt[ (2 + t2 cosθ)^2 + (t2 sinθ)^2 ] = sqrt[4 +4 t2 cosθ + t2² ]Now, recall that t1 and t2 are roots of the quadratic equation:(3cos²θ +4 sin²θ)t² +6cosθ t -9 =0So, t1 + t2 = -6cosθ / (3cos²θ +4 sin²θ)t1 t2 = -9 / (3cos²θ +4 sin²θ)Also, we know t1 - t2 =12 / (3cos²θ +4 sin²θ) =12 / a, where a=3cos²θ +4 sin²θLet me denote a=3cos²θ +4 sin²θ to simplify expressions.Thus, t1 + t2 = -6cosθ /at1 t2 = -9/at1 - t2 =12/aAlso, t1² + t2² = (t1 + t2)^2 -2t1 t2 = (36cos²θ /a²) -2*(-9/a)=36cos²θ /a² +18/aSimilarly, t1² + t2² = (36cos²θ +18a)/a²But maybe instead of that, let's compute F1P and F1Q.Compute F1P = sqrt(4 +4 t1 cosθ + t1² )Similarly, F1Q = sqrt(4 +4 t2 cosθ + t2² )Let’s compute F1P + F1Q + PQ, which is the perimeter, then s = (F1P + F1Q + PQ)/2.But this seems complicated. Maybe there's a smarter way.Alternatively, note that since we have expressions for t1 and t2 in terms of θ, maybe we can express F1P and F1Q in terms of θ.Alternatively, let me compute F1P squared:F1P² =4 +4 t1 cosθ + t1²But from quadratic equation:(3cos²θ +4 sin²θ)t1² +6cosθ t1 -9=0 => t1² = (-6cosθ t1 +9)/(3cos²θ +4 sin²θ )So F1P² =4 +4 t1 cosθ + [ (-6cosθ t1 +9 ) /a ]Where a =3cos²θ +4 sin²θTherefore,F1P² =4 +4 t1 cosθ - (6cosθ t1)/a +9/a=4 + [4 t1 cosθ - (6cosθ t1)/a ] +9/aFactor t1 cosθ:=4 + t1 cosθ [4 -6/a ] +9/aBut this seems messy. Alternatively, maybe use specific values.Alternatively, let's compute F1P and F1Q using the expressions for t1 and t2.Recall that t1 = [6(2 -cosθ)] /at2= [ -6(2 +cosθ) ] /aWhere a=3cos²θ +4 sin²θ.Therefore, F1P = sqrt(4 +4 t1 cosθ + t1² )Plug t1:First compute 4 +4 t1 cosθ + t1²:4 +4*(6(2 -cosθ)/a)*cosθ + [6(2 -cosθ)/a]^2=4 + (24(2 -cosθ)cosθ)/a +36(2 -cosθ)^2 /a²Similarly for F1Q:F1Q = sqrt(4 +4 t2 cosθ + t2² )= sqrt(4 +4*(-6(2 +cosθ)/a)*cosθ + [ -6(2 +cosθ)/a ]² )= sqrt(4 -24(2 +cosθ)cosθ /a +36(2 +cosθ)^2 /a² )This is getting very complicated. Maybe instead of trying to compute s directly, we can find a relationship between A and s.Alternatively, use the formula that relates inradius r = A/s, so if we can express s in terms of A, maybe we can find a relationship.But I need expressions for F1P, F1Q, PQ.We already have PQ=12/aAnd A=12 sinθ /aBut we still need F1P and F1Q.Alternatively, maybe there's a formula that connects A, r, and s. However, without knowing s, it's difficult. Alternatively, perhaps use Heron's formula: A = sqrt[s(s -a)(s -b)(s -c)], but that might not help here.Alternatively, think about the triangle F1PQ. Maybe there's a property related to ellipses or foci that can be used.Recall that in an ellipse, the sum of distances from any point on the ellipse to the two foci is constant, equal to 2a. Here, a=2, so for any point on the ellipse, say P, F1P + F2P =4.But F2 is (1,0), so F2P is the distance from (1,0) to P, which is the same as the parameter t1, because when we parametrized the line through F2 with parameter t, t is the distance from F2 along the line. Wait, no. The parameter t is not necessarily the Euclidean distance unless the direction vector is a unit vector. In our parametrization earlier, we used direction vector (cosθ, sinθ), which is a unit vector. Therefore, t is the actual Euclidean distance from F2. Therefore, F2P = |t1|, since t1 is positive, so F2P = t1. Similarly, F2Q = |t2|, but t2 is negative, so F2Q = |t2| = -t2.Therefore, since P is on the ellipse, F1P + F2P =2a=4, so F1P + t1=4 => F1P=4 - t1.Similarly, for Q, since Q is on the ellipse, F1Q + F2Q=4. But F2Q = |t2| = -t2 (since t2 is negative). Therefore, F1Q + (-t2)=4 => F1Q=4 + t2.Therefore, we have:F1P =4 - t1F1Q=4 + t2PQ= t1 - t2Therefore, the perimeter of triangle F1PQ is:F1P + F1Q + PQ = (4 - t1) + (4 + t2) + (t1 - t2) =8.Wait, that's interesting! The perimeter is always 8, regardless of θ. Therefore, the semi-perimeter s=8/2=4.But if the perimeter is constant, then the semi-perimeter s is constant (equal to 4). Therefore, the inradius r = A/s = A/4. Therefore, the area of the incircle is πr² = π(A/4)². Therefore, to maximize the area of the incircle, we need to maximize A, since s is constant.Wait, this is a crucial realization! If the perimeter is always 8, then s=4, so r = A/4, and maximizing r is equivalent to maximizing A. Therefore, the maximum area of the incircle is π*(A_max/4)^2, where A_max is the maximum area of triangle F1PQ.Therefore, the problem reduces to finding the maximum area of triangle F1PQ, which we already have as A=12 sinθ / (3cos²θ +4 sin²θ). So now we just need to maximize A(θ)=12 sinθ / (3cos²θ +4 sin²θ).Therefore, the maximum area of the incircle is π*(A_max /4)^2, so we just need to find A_max.Therefore, the problem simplifies to maximizing A(θ)=12 sinθ / (3cos²θ +4 sin²θ)Let me focus on maximizing this function.Let’s denote f(θ)= sinθ / (3cos²θ +4 sin²θ)We need to maximize f(θ), then multiply by 12 to get A_max.To maximize f(θ), let's write it in terms of a single trigonometric function.Note that denominator:3cos²θ +4 sin²θ=3(1 - sin²θ)+4 sin²θ=3 + sin²θ.Therefore, denominator=3 + sin²θ.Therefore, f(θ)= sinθ / (3 + sin²θ)So, we need to maximize sinθ / (3 + sin²θ)Let me set x=sinθ, where x ∈ [-1,1], but since θ ∈ [0, π], x ∈ [0,1]. Because in the parametrization earlier, θ is the angle from the x-axis upwards or downwards, but since we're taking sinθ as positive (since θ ∈ [0, π]), x ∈ [0,1].Therefore, we can consider the function g(x)=x/(3 +x²), where x ∈ [0,1], and find its maximum.Take derivative: g’(x)= [1*(3 +x²) -x*(2x)]/(3 +x²)^2= [3 +x² -2x²]/(3 +x²)^2= (3 -x²)/(3 +x²)^2Set derivative equal to zero: 3 -x²=0 => x²=3 => x=√3. But √3 ≈1.732, which is greater than 1. Therefore, in the interval [0,1], the maximum occurs at the critical points or endpoints. Since there are no critical points in [0,1] (since x=√3 is outside), the maximum occurs at the endpoint x=1.Therefore, maximum of g(x)=1/(3 +1)=1/4.But wait, let me verify:At x=0, g(0)=0At x=1, g(1)=1/4Since derivative g’(x) is positive in [0,1] (since 3 -x² >0 for x² <3, which is always true in [0,1]), the function g(x) is increasing on [0,1]. Therefore, maximum at x=1, which is 1/4.Therefore, maximum value of f(θ)=1/4, achieved when sinθ=1, i.e., θ=π/2.Therefore, maximum A=12*(1/4)=3.Therefore, the maximum area of the triangle is 3, so the inradius r= A/s=3/4.Therefore, the area of the incircle is π*(3/4)^2=9π/16.Wait, but let me check this. If A_max=3, then r=3/4, and area of incircle is π*(3/4)^2=9π/16.But let's verify this result by checking when θ=π/2.When θ=π/2, the line l is vertical, x=1, intersecting the ellipse at (1, ±3/2). Therefore, points P=(1, 3/2) and Q=(1, -3/2). Then, triangle F1PQ has vertices at (-1,0), (1,3/2), (1,-3/2).Compute the area of this triangle. The base can be PQ, which is the distance between (1,3/2) and (1,-3/2), which is 3. The height is the distance from F1(-1,0) to the line PQ, which is the horizontal distance from x=-1 to x=1, which is 2. Therefore, area A=(1/2)*base*height=(1/2)*3*2=3. Which matches our earlier result.The semi-perimeter s=4, so inradius r=3/4. Therefore, the area of the incircle is π*(3/4)^2=9π/16.Therefore, the maximum area of the inscribed circle is 9π/16.But wait, let me make sure there isn't a case where A could be larger. For example, when the line is not vertical, could the area be larger?But according to our analysis, since the function g(x)=x/(3 +x²) is increasing on [0,1], the maximum occurs at x=1, which corresponds to θ=π/2, vertical line. Therefore, the maximum area of the triangle is indeed 3, leading to the incircle area of 9π/16.But wait, another way to compute the inradius for the specific case when the triangle is isoceles.In the case where θ=π/2, the triangle F1PQ is isoceles with base PQ=3 and two equal sides F1P and F1Q.Compute the lengths:F1P= distance from (-1,0) to (1, 3/2)=sqrt[(1 - (-1))² + (3/2 -0)^2]=sqrt[4 +9/4]=sqrt[25/4]=5/2.Similarly, F1Q=5/2. So sides are 5/2, 5/2, 3.Semi-perimeter s=(5/2 +5/2 +3)/2=(5 +3)/2=8/2=4, which matches earlier result.Area A=3. Inradius r=A/s=3/4. Therefore, the incircle area is π*(3/4)^2=9π/16.Therefore, the maximum area is 9π/16.But let me check another case. For example, take θ=0, which is the horizontal line through F2(1,0). But θ=0 corresponds to the line y=0, which is the x-axis. This line intersects the ellipse at x=±2, y=0. But since it's passing through F2(1,0), the points P and Q would be (1,0) and (2,0). But (1,0) is F2, so the line intersects the ellipse at F2 and (2,0). But then the triangle F1PQ would degenerate into a line segment from F1(-1,0) to (2,0) via F2(1,0), so area zero. Which makes sense, as sinθ=0, so A=0.Similarly, if we take θ approaching 0, the area approaches zero. Therefore, the maximum at θ=π/2 seems correct.Another test case: take θ=π/4. Let's compute A.θ=π/4, sinθ=√2/2≈0.707, cosθ=√2/2≈0.707Denominator a=3cos²θ +4 sin²θ=3*(0.5)+4*(0.5)=1.5+2=3.5Therefore, A=12*(√2/2)/3.5=12*(√2/2)/3.5=6√2/3.5≈6*1.414/3.5≈8.485/3.5≈2.424, which is less than 3. So indeed, the maximum at θ=π/2 is larger.Therefore, conclusion: The maximum area of the inscribed circle is 9π/16.But wait, I need to confirm whether the maximum of A(θ)=3 indeed occurs only at θ=π/2, or if there could be another maximum.Given that we transformed the problem to maximizing x/(3 +x²) over x ∈ [0,1], and found that the maximum is at x=1, which corresponds to sinθ=1, θ=π/2, so there are no other critical points. Therefore, the maximum is indeed at θ=π/2.Therefore, the final answer should be 9π/16.Final Answerboxed{dfrac{9}{16}pi}

✅:Given the ellipse (frac{x^2}{4} + frac{y^2}{3} = 1) with foci (F_1) and (F_2) at ((-1, 0)) and ((1, 0)) respectively, we need to find the maximum area of the inscribed circle of (triangle F_1PQ) where a line (l) passes through (F_2) and intersects the ellipse at points (P) and (Q).1. Equation of the Line and Intersection Points: - The line passing through (F_2(1, 0)) can be parametrized using an angle (theta) with the x-axis, giving the parametric equations (x = 1 + t costheta) and (y = t sintheta). - Substituting these into the ellipse equation and solving for (t), we find the roots (t_1) and (t_2) which correspond to the points (P) and (Q).2. Area of (triangle F_1PQ): - The area (A) of (triangle F_1PQ) is derived using the cross product formula, leading to (A = frac{12 sintheta}{3cos^2theta + 4sin^2theta}).3. Semi-perimeter and Inradius: - The perimeter of (triangle F_1PQ) is found to be constant (8) due to the properties of the ellipse, leading to a semi-perimeter (s = 4). - The inradius (r) is given by (r = frac{A}{s}).4. Maximizing the Area: - The function (A = frac{12 sintheta}{3cos^2theta + 4sin^2theta}) is maximized by analyzing it in terms of (sintheta) and (costheta). The maximum occurs at (theta = frac{pi}{2}), where the line is vertical, yielding the maximum area (A = 3).5. Maximum Inradius and Area of Incircle: - The maximum inradius (r) is (frac{3}{4}), leading to the maximum area of the inscribed circle as (pi left(frac{3}{4}right)^2 = frac{9}{16}pi).Thus, the maximum area of the inscribed circle of (triangle F_1PQ) is (boxed{dfrac{9}{16}pi}).

🔑:Given the ellipse equation:[frac{x^{2}}{4} + frac{y^{2}}{3} = 1]The foci of the ellipse, denoted as (F_1) and (F_2), can be found using the formula for the foci of an ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1):1. Identify (a) and (b): [ a^2 = 4 quad text{(semi-major axis squared)} ] [ b^2 = 3 quad text{(semi-minor axis squared)} ]2. Calculate (c) using (c^2 = a^2 - b^2): [ c^2 = a^2 - b^2 = 4 - 3 = 1 implies c = 1 ] The foci are at: [ F_1(-1, 0), quad F_2(1, 0) ] Step-by-Step Solution to Find the Largest Area:Consider a line (l) passing through the right focus (F_2). The general form of this line:[l: x = my + 1] Finding Intercepts with the EllipseSubstitute (x = my + 1) into the ellipse equation:[frac{(my + 1)^2}{4} + frac{y^2}{3} = 1]This leads to:[frac{m^2 y^2 + 2my + 1}{4} + frac{y^2}{3} = 1]Clearing the fraction, multiply through by 12:[3(m^2 y^2 + 2my + 1) + 4y^2 = 12]Simplify:[3m^2 y^2 + 6my + 3 + 4y^2 = 12][(3m^2 + 4) y^2 + 6my - 9 = 0]Let the points of intersection be (P(x_1, y_1)) and (Q(x_2, y_2)). By Vieta's formulas:[y_1 + y_2 = -frac{6m}{3m^2 + 4}][y_1 y_2 = -frac{9}{3m^2 + 4}] Calculating the Area of (triangle F_1 PQ):The area of (triangle F_1 PQ) can be given by:[S_{triangle F_1 PQ} = frac{1}{2} |F_1 F_2| |y_1 - y_2|]Calculate ( |y_1 - y_2| ):[ |y_1 - y_2| = sqrt{(y_1 + y_2)^2 - 4y_1 y_2}][= sqrt{left(-frac{6m}{3m^2 + 4}right)^2 - 4left(-frac{9}{3m^2 + 4}right)}][= sqrt{frac{36m^2}{(3m^2 + 4)^2} + frac{36}{3m^2 + 4}}][= sqrt{frac{36(m^2 + 1)}{(3m^2 + 4)^2}}]Thus:[S_{triangle F_1 PQ} = frac{1}{2} cdot 2 cdot sqrt{frac{36(m^2 + 1)}{(3m^2 + 4)^2}}][= 12 cdot sqrt{frac{m^2 + 1}{(3m^2 + 4)^2}}]Further, simplify:[sqrt{frac{m^2 + 1}{(3m^2 + 4)^2}} = frac{sqrt{m^2 + 1}}{3m^2 + 4}][S_{triangle F_1 PQ} = 12 frac{sqrt{m^2 + 1}}{3m^2 + 4}] Maximizing the Area:To maximize (frac{m^2 + 1}{(3m^2 + 4)^2}):[frac{m^2 + 1}{(3m^2 + 4)^2} = frac{1}{(3m^2+4)^2} leq frac{1}{16}]Finally, the inradius ( r ) of the incircle is given by:[r = frac{2S_{triangle F_1 PQ}}{8} leq frac{3}{4}]Therefore, the maximum area of the incircle:[A = pi r^2 = pi left(frac{3}{4}right)^2 = frac{9pi}{16}] Conclusion:(boxed{frac{9pi}{16}})

🧠:Okay, so I need to calculate the sum of this series: 1/1 + 2/2 + 3/4 + 4/8 + 5/16 + ... + n/(2^{n-1}). And then find what it approaches as n goes to infinity. Hmm, let's start by understanding the pattern here.First, let me write out the terms more clearly. The first term is 1 divided by 1, which is 1. The second term is 2 divided by 2, which is also 1. The third term is 3 divided by 4, so 0.75. The fourth term is 4 divided by 8, which is 0.5, and the fifth term is 5 divided by 16, which is 0.3125, and so on. So each term is (k)/(2^{k-1}) where k starts from 1 and goes up to n. But the question also asks what happens as n increases without bound, so we need to find the sum to infinity.I remember that for infinite series, if the terms decrease sufficiently fast, the series might converge. Let me check if this series converges. The terms here are of the form k/(2^{k-1}). Let's see, 2^{k-1} grows exponentially, while the numerator grows linearly. Since exponential growth outpaces linear growth, the terms go to zero as k approaches infinity. So the necessary condition for convergence (that the terms approach zero) is satisfied. But we still need to check if the series actually converges and find its sum.To compute the sum, maybe I can use the formula for the sum of a geometric series or a related series. I recall that there's a standard result for the sum of k x^{k} from k=1 to infinity, which is x/(1 - x)^2, for |x| < 1. Wait, that might be helpful here. Let me think.But in our case, the general term is k / 2^{k - 1}. Let's rewrite that as k * (1/2)^{k - 1}. So, that's k * (1/2)^{k - 1} = k * (1/2)^{k} * 2 = 2k*(1/2)^{k}. Therefore, the sum from k=1 to n of k/(2^{k-1}) is equal to 2 times the sum from k=1 to n of k/(2^{k}).So, if I can compute the sum from k=1 to infinity of k/(2^k), then multiply by 2 to get the original sum. That makes sense. Let me verify this step again. Let's take the original term: k/(2^{k-1}) = k/(2^{k} / 2) = 2k/(2^{k}) = 2*(k/(2^k)). Yes, so the original sum is 2 times the sum of k/(2^k) from k=1 to n. Therefore, if we can find the sum of k/(2^k), multiplying by 2 gives the desired result.So, let's focus on computing S = sum_{k=1}^∞ k/(2^k). I remember there is a formula for this. Let me recall how to derive it. Let me write S = sum_{k=1}^∞ k x^k, evaluated at x = 1/2. The formula for S is x/(1 - x)^2 when |x| < 1. So substituting x = 1/2, we get S = (1/2)/(1 - 1/2)^2 = (1/2)/(1/2)^2 = (1/2)/(1/4) = 2. Wait, so sum_{k=1}^∞ k/(2^k) = 2. Then the original sum is 2*2 = 4? Wait, but that seems too high. Let me check with partial sums.Wait, let's compute the sum up to n=5. The original series up to n=5 is 1 + 1 + 0.75 + 0.5 + 0.3125 = 1 + 1 = 2, plus 0.75 is 2.75, plus 0.5 is 3.25, plus 0.3125 is 3.5625. If the infinite sum is 4, then as n increases, the sum approaches 4. Let me check with n=6. The next term is 6/(2^5) = 6/32 = 0.1875. So adding that gives 3.5625 + 0.1875 = 3.75. Next term: 7/64 ≈ 0.109375, total ≈ 3.859375. Then 8/128 = 0.0625, total ≈ 3.921875. Then 9/256 ≈ 0.03515625, total ≈ 3.95703125. So it's approaching 4. So maybe the sum does converge to 4. That seems plausible.Wait, but let me recheck the steps. The sum S = sum_{k=1}^∞ k/(2^k) = 2. Then the original sum is 2*S = 4. So that seems correct.But let me derive the formula for S from scratch to be sure. Let me start with the infinite geometric series: sum_{k=0}^∞ x^k = 1/(1 - x) for |x| < 1. Then, differentiate both sides with respect to x. The derivative of the left side is sum_{k=0}^∞ k x^{k - 1} = 1/(1 - x)^2. Therefore, multiplying both sides by x gives sum_{k=0}^∞ k x^k = x/(1 - x)^2. But the sum from k=0 to infinity of k x^k is 0 + x + 2x^2 + 3x^3 + ... So starting from k=1, it's sum_{k=1}^∞ k x^k = x/(1 - x)^2. Therefore, substituting x = 1/2, we get sum_{k=1}^∞ k (1/2)^k = (1/2)/(1 - 1/2)^2 = (1/2)/(1/4) = 2. So yes, that is correct. Therefore, the original sum is 2*2 = 4.Therefore, the sum as n approaches infinity is 4.Wait, but let me verify this again with another approach. Let's let S be the sum from k=1 to infinity of k/(2^{k - 1}). Then, S = sum_{k=1}^∞ k/(2^{k - 1}) = sum_{k=1}^∞ k * 2^{ - (k - 1) } = sum_{k=1}^∞ k * 2^{1 - k} = 2 sum_{k=1}^∞ k / 2^{k}.Which is the same as 2S', where S' is sum_{k=1}^∞ k/2^k, which we found to be 2, so 2*2=4. So that's consistent.Alternatively, maybe we can use recurrence relations. Let me try writing the sum S_n = sum_{k=1}^n k/(2^{k - 1}) and try to find a recursive formula.But that might be more complicated. Alternatively, let's think of S = 1 + 2/2 + 3/4 + 4/8 + 5/16 + ... Let's write S as:S = 1 + (2)/2 + (3)/4 + (4)/8 + (5)/16 + ... Multiply both sides by 2:2S = 2 + 2 + 3/2 + 4/4 + 5/8 + ... Subtract the original S from this:2S - S = S = (2 + 2 + 3/2 + 4/4 + 5/8 + ...) - (1 + 2/2 + 3/4 + 4/8 + 5/16 + ...)Let's compute term by term:First term: 2 - 1 = 1Second term: 2 - 1 (since 2 - 2/2 = 2 -1 =1)Third term: 3/2 - 3/4 = 3/4Fourth term: 4/4 - 4/8 = 1 - 0.5 = 0.5Fifth term: 5/8 - 5/16 = 5/16And so on. So we have S = 1 + 1 + 3/4 + 1/2 + 5/16 + ... Wait, but maybe this approach is getting messy. Let me try a different way. Let's see:If S = sum_{k=1}^∞ k/(2^{k-1}) = sum_{k=1}^∞ k*(1/2)^{k-1}Let me let x = 1/2. Then S = sum_{k=1}^∞ k x^{k - 1} = sum_{k=1}^∞ k x^{k - 1}But sum_{k=1}^∞ k x^{k - 1} is the derivative of sum_{k=0}^∞ x^k, which is 1/(1 - x). The derivative is 1/(1 - x)^2. Therefore, sum_{k=1}^∞ k x^{k - 1} = 1/(1 - x)^2. Therefore, sum_{k=1}^∞ k x^{k} = x/(1 - x)^2. Wait, but in our case, S = sum_{k=1}^∞ k x^{k - 1} = 1/(1 - x)^2. Since x = 1/2, then S = 1/(1 - 1/2)^2 = 1/(1/2)^2 = 4. So that's the same result as before. Therefore, the sum is 4. So regardless of the method, whether manipulating the series directly or using calculus, we arrive at the sum being 4. Therefore, as n approaches infinity, the sum approaches 4.But just to make sure there's no mistake in the algebra, let me verify once more. Let me write S = sum_{k=1}^∞ k/(2^{k - 1}) = sum_{k=1}^∞ k * 2^{1 - k} = 2 sum_{k=1}^∞ k / 2^{k} = 2 * S', where S' = sum_{k=1}^∞ k / 2^{k}. As established before, S' = 2. Therefore, 2 * 2 = 4. Correct.Alternatively, let me compute S' again. Let S' = sum_{k=1}^∞ k / 2^k. Let's write S' = 1/2 + 2/4 + 3/8 + 4/16 + ... Multiply both sides by 2: 2S' = 1 + 2/2 + 3/4 + 4/8 + 5/16 + ... Now subtract S' from 2S':2S' - S' = S' = 1 + (2/2 - 1/2) + (3/4 - 2/4) + (4/8 - 3/8) + ... Simplifying each term:= 1 + (1/2) + (1/4) + (1/8) + (1/16) + ... This is a geometric series with first term 1 and ratio 1/2. The sum is 1 / (1 - 1/2) = 2. Therefore, S' = 2, so original S = 2*2 = 4. Yep, that's another way to look at it. So this confirms the result again.Therefore, the sum approaches 4 as n increases without bound.Final AnswerThe sum of the series approaches boxed{4} as the number of terms increases without bound.

✅:To calculate the sum of the series frac{1}{1}+frac{2}{2}+frac{3}{4}+frac{4}{8}+frac{5}{16}+ldots+frac{n}{2^{n-1}}and determine what it approaches as the number of terms increases without bound, we start by recognizing the general term of the series as (frac{k}{2^{k-1}}) for (k) from 1 to (n).We rewrite the general term as (2 cdot frac{k}{2^k}). This allows us to express the sum of the series as twice the sum of the series (sum_{k=1}^{infty} frac{k}{2^k}).To find (sum_{k=1}^{infty} frac{k}{2^k}), we use the known result for the sum of the series (sum_{k=1}^{infty} kx^k), which is (frac{x}{(1 - x)^2}) for (|x| < 1). Substituting (x = frac{1}{2}), we get:sum_{k=1}^{infty} k left(frac{1}{2}right)^k = frac{frac{1}{2}}{(1 - frac{1}{2})^2} = frac{frac{1}{2}}{left(frac{1}{2}right)^2} = frac{frac{1}{2}}{frac{1}{4}} = 2Thus, the sum (sum_{k=1}^{infty} frac{k}{2^k} = 2). Multiplying this result by 2 gives the sum of the original series:2 times 2 = 4Therefore, the sum of the series approaches (boxed{4}) as the number of terms increases without bound.

🔑:1. Consider the given series and write it out explicitly: [ sum_{k=1}^{n} frac{k}{2^{k-1}} = frac{1}{1} + frac{2}{2} + frac{3}{4} + frac{4}{8} + frac{5}{16} + ldots + frac{n}{2^{n-1}} ]2. To simplify the analysis, rewrite the summation with a common denominator: [ sum_{k=1}^{n} frac{k}{2^{k-1}} = frac{1 cdot 2^{n-1} + 2 cdot 2^{n-2} + 3 cdot 2^{n-3} + ldots + (n-1) cdot 2 + n}{2^{n-1}} ]3. Focus on simplifying the numerator separately. Break down individual terms of the form ( k cdot 2^{n-k} ): [ begin{aligned} 2 cdot 2^{n-2} &= 2^{n-2} + 2^{n-2}, 3 cdot 2^{n-3} &= 2^{n-3} + 2^{n-3} + 2^{n-3}, &vdots (n-1) cdot 2 &= 2 + 2 + ldots + 2 quad text{(n-1 terms)}, n cdot 1 &= 1 + 1 + ldots + 1 quad text{(n terms)}. end{aligned} ]4. Rearrange the expression to group similar powers of (2): [ begin{aligned} & (2^{n-1} + 2^{n-2} + ldots + 1) + (2^{n-2} + 2^{n-3} + ldots + 1) + ldots & + (2^{n-3} + ldots + 1) + ldots + (2 + 1) + 1 end{aligned} ]5. Sum each geometric series within the brackets: [ sum_{k=0}^{n-1} 2^k = frac{2^n - 1}{2 - 1} = 2^n - 1 ]6. Simplify the overall series: [ sum_{k=0}^{n-2} 2^k + sum_{k=0}^{n-3} 2^k + ldots + 1 = (2^n - 1) + (2^{n-1} - 1) + (2^{n-2} - 1) + ldots + 1 ]7. Combine terms: [ sum_{m=0}^{n-1} (2^{n-m} - 1) = 2^n + 2^{n-1} + 2^{n-2} + ldots + 2 + 1 - (n+1) ]8. This further simplifies to: [ 2^{n+1} - 1 - (n+1) ]9. Divide by (2^{n-1}) to get back to the original fraction's form: [ frac{2^{n+1} - (n + 2)}{2^{n-1}} = 4 - frac{n + 2}{2^{n-1}} ]10. Analyze the second term as ( n to infty ): [ lim_{n to infty} left(4 - frac{n+2}{2^{n-1}}right) = 4 ]Therefore, the sum of the series approaches:[boxed{4}]

🧠:Okay, let me try to figure out this geometry problem. So, we have an isosceles triangle ABC where angles at C and B are both 40 degrees. That means the triangle has two equal angles, so the sides opposite those angles should be equal. Wait, in a triangle, the sides opposite equal angles are equal. So if angles at B and C are both 40 degrees, then sides opposite them, which would be AC and AB, right? Wait, angle at B is 40°, so the side opposite is AC. Angle at C is 40°, so the side opposite is AB. Therefore, AC = AB. So triangle ABC is isosceles with AB = AC. Wait, but the problem says it's an isosceles triangle ABC. Usually, the order matters, so ABC is isosceles with base BC or something? Hmm, maybe I should clarify.Wait, angles at B and C are both 40°, so angle at A must be 100°, since the sum of angles in a triangle is 180°. So angle BAC is 100°, and angles at ABC and ACB are 40° each. Therefore, sides AB and AC are equal because their opposite angles are equal (angles at C and B). Wait, no, hold on. The side opposite angle B is AC, and the side opposite angle C is AB. Since angles B and C are equal, sides AC and AB are equal. Therefore, AB = AC. So triangle ABC is isosceles with AB = AC. Therefore, vertex at A, with legs AB and AC, and base BC. So BC is the base.Now, the problem says: On the ray AC, mark off the segment AD such that AD = BC. We need to find the measure of angle BDC.Let me try to draw this mentally. So triangle ABC with AB = AC, angles at B and C are 40°, angle at A is 100°. Then, on the ray AC, we mark point D such that AD = BC. Since AC is a ray, D is beyond point C on the extension of AC. Because AD is longer than AC if D is beyond C. Because AD = BC. Since ABC is isosceles with AB = AC, BC can be calculated using the Law of Sines or Cosines. Maybe we can find the lengths in terms of AB or AC.But maybe instead of actual lengths, we can use some geometric constructions or properties. The key is to find angle BDC. So points B, D, and C. So triangle BDC. Need to find angle at D.Hmm. Let me try to sketch this.First, triangle ABC: AB = AC = let's say length 'x'. Then, BC can be found using the Law of Cosines in triangle ABC. Since angle at A is 100°, sides AB and AC are 'x', then BC² = AB² + AC² - 2*AB*AC*cos(angle BAC). So BC² = x² + x² - 2x² cos(100°) = 2x²(1 - cos(100°)). So BC = x*sqrt(2(1 - cos100°)). Alternatively, using Law of Sines: BC / sin(100°) = AB / sin(40°). Since AB = x, BC = x * sin(100°) / sin(40°). Since sin(100°) = sin(80°), and sin(80°)/sin(40°) = 2 cos(40°). So BC = x * 2 cos(40°). Therefore, BC = 2x cos(40°). So BC is twice AB times cosine of 40 degrees.Now, AD is equal to BC. So AD = 2x cos(40°). Since AD is on the ray AC, and AC = x, then D is beyond point C such that AD = AC + CD = x + CD = 2x cos40°. Therefore, CD = 2x cos40° - x = x(2 cos40° - 1). So the length of CD is x(2 cos40° - 1). Hmm. But maybe instead of working with lengths, we can find some angles or use triangle congruence or similarity.Alternatively, perhaps constructing triangle BDC and using some properties. Let's think about triangle BDC. We need angle at D. Points B, D, and C. If we can find some relationships between the sides or angles in this triangle, maybe we can find angle BDC.Alternatively, maybe we can construct some auxiliary lines or points. For example, since AD = BC, and AD is on the ray AC, maybe there's a triangle where these sides correspond. Let me think. If we can create a triangle congruent to triangle ABC or some other triangle, that might help.Alternatively, consider point D such that AD = BC. Since BC is known in terms of AB, and AD is along AC extended. Maybe triangle ABD or something else is congruent or similar.Wait, in triangle ABC, AB = AC = x, BC = 2x cos40°, as we saw earlier. Then AD = BC = 2x cos40°, and since AC = x, then AD is AC + CD = x + CD = 2x cos40°, so CD = 2x cos40° - x. So CD = x(2 cos40° - 1). Let's compute 2 cos40° - 1 numerically. cos40° is approximately 0.7660, so 2*0.7660 = 1.532; 1.532 - 1 = 0.532. So CD ≈ 0.532x. So D is a little beyond C on the extension of AC.But maybe instead of numbers, we can find an angle. Let's think about triangle BDC. To find angle BDC, we might need information about sides BD, DC, BC. Alternatively, maybe some other triangles involving D.Alternatively, maybe using the Law of Sines or Cosines in triangle BDC.But to use Law of Sines or Cosines, we need some side lengths or angles. Let's see.First, BC is known in terms of x: BC = 2x cos40°.CD is x(2 cos40° - 1).What about BD? To find BD, we might need to look at triangle ABD or CBD.Wait, point D is on the extension of AC beyond C. So coordinates might help here. Let me try coordinate geometry.Let me place point A at the origin (0,0). Since AB = AC = x, and angle at A is 100°, maybe we can set up coordinates.But maybe a better coordinate system: Let me place point A at (0,0). Let me set AC along the positive x-axis. So point C is at (x, 0). Then point B is somewhere in the plane. Since AB = AC = x, and angle at A is 100°, point B can be located at (x cos100°, x sin100°). Wait, no. If we place A at (0,0), and AC along the x-axis to (x,0), then AB is length x, making an angle of 100° at A. So point B would be at (x cos100°, x sin100°). Then, point C is at (x,0). Then, BC can be computed using the distance formula between (x cos100°, x sin100°) and (x,0).But maybe this is getting too algebraic. Let's proceed.Coordinates:A: (0,0)C: (c,0) where c is the length AC, which we can take as 1 for simplicity.Wait, let's set AC = 1 unit for simplicity. Then AB = 1 as well. Then BC can be calculated as 2 cos40°, since BC = 2*AB*cos40°, as before. Wait, if AB = AC = 1, then using Law of Cosines:BC² = 1² + 1² - 2*1*1*cos(100°) = 2 - 2 cos100°. So BC = sqrt(2 - 2 cos100°). Alternatively, using Law of Sines: BC / sin100° = 1 / sin40°, so BC = sin100° / sin40° = sin80° / sin40° = 2 cos40°, as we had before. So BC = 2 cos40°, approximately 2*0.7660 ≈ 1.532.So AD = BC = 2 cos40°, and since AC = 1, then AD = AC + CD = 1 + CD = 2 cos40°, so CD = 2 cos40° - 1 ≈ 1.532 - 1 = 0.532. So D is at (1 + 0.532, 0) = (1.532, 0) if we set AC along the x-axis from (0,0) to (1,0).Wait, but in coordinate system, if AC is from (0,0) to (1,0), then point C is at (1,0). Then point B is at (cos100°, sin100°), since AB = 1 and angle at A is 100°. Let's confirm:If AB = 1, angle at A is 100°, so coordinates of B are (cos100°, sin100°). Then coordinates of D would be on the ray AC extended beyond C. Since AC is along the x-axis from (0,0) to (1,0), the ray AC is the positive x-axis. So point D is at (1 + (2 cos40° - 1), 0) = (2 cos40°, 0). Because AD = 2 cos40°, and AC = 1, so CD = 2 cos40° - 1. So D is at (2 cos40°, 0).So coordinates:A: (0,0)B: (cos100°, sin100°)C: (1,0)D: (2 cos40°, 0)Now, we need to find angle BDC. That is, angle at point D between points B, D, and C. So angle at D between points B, D, C. So in terms of coordinates, we can compute vectors DB and DC and find the angle between them.First, let's compute coordinates:Point D: (2 cos40°, 0)Point C: (1, 0)Point B: (cos100°, sin100°)Compute vectors:Vector DB = B - D = (cos100° - 2 cos40°, sin100° - 0) = (cos100° - 2 cos40°, sin100°)Vector DC = C - D = (1 - 2 cos40°, 0 - 0) = (1 - 2 cos40°, 0)Now, angle BDC is the angle between vectors DB and DC. The angle can be found using the dot product formula:cos(theta) = (DB • DC) / (|DB| |DC|)First, compute DB • DC = (cos100° - 2 cos40°)(1 - 2 cos40°) + (sin100°)(0) = (cos100° - 2 cos40°)(1 - 2 cos40°)Compute |DB| = sqrt( (cos100° - 2 cos40°)^2 + (sin100°)^2 )Compute |DC| = sqrt( (1 - 2 cos40°)^2 + 0 ) = |1 - 2 cos40°|So let's compute these terms step by step.First, compute cos100° and cos40°.Note that cos100° = cos(180° - 80°) = -cos80° ≈ -0.1736cos40° ≈ 0.7660So approximate values:cos100° ≈ -0.1736cos40° ≈ 0.7660Compute 2 cos40° ≈ 1.532So:DB • DC ≈ (-0.1736 - 1.532)(1 - 1.532) = (-1.7056)(-0.532) ≈ (-1.7056)*(-0.532) ≈ 0.907Compute |DB|:sqrt( (-1.7056)^2 + (sin100°)^2 ) ≈ sqrt( (2.909) + (0.9848)^2 ) ≈ sqrt(2.909 + 0.9698) ≈ sqrt(3.8788) ≈ 1.9695Compute |DC| = |1 - 1.532| = |-0.532| = 0.532Then cos(theta) ≈ 0.907 / (1.9695 * 0.532) ≈ 0.907 / (1.047) ≈ 0.866So theta ≈ arccos(0.866) ≈ 30°So angle BDC is approximately 30 degrees. Hmm, that seems familiar. But let's check if this is exact.Wait, 0.866 is approximately sqrt(3)/2 ≈ 0.8660, which is cosine of 30°, so angle BDC is 30°. But is this exact?Alternatively, maybe there's a geometric construction that leads to 30° exactly.Let me try to see if this can be derived without approximate calculations.Alternative approach:Given triangle ABC with AB = AC, angles at B and C are 40°, angle at A is 100°. We need to construct AD = BC on ray AC, find angle BDC.Perhaps triangle BDC is a 30-60-90 triangle? Or maybe there's an equilateral triangle involved.Alternatively, construct point E such that triangle ADE is congruent or similar to some other triangle.Alternatively, consider triangle BCD. Since AD = BC, and AD is along AC, perhaps there's a reflection or rotation that maps BC to AD.Wait, since AD = BC and in triangle ABC, AB = AC, maybe triangle ABD is congruent to triangle BDC or something.Alternatively, let's consider triangle ABD and triangle BDC.Wait, AB = AC = 1 (for simplicity), AD = BC = 2 cos40°, so BD is a common side. Hmm, not sure.Alternatively, since AD = BC, and AB = AC, maybe triangle ABD is congruent to triangle BCD? Let's check.In triangle ABD: AB = 1, AD = BC = 2 cos40°, angle at A is angle between AB and AD. AB is from A to B, and AD is along AC extended. The angle at A between AB and AD is angle BAC + angle CAD. But angle BAC is 100°, and angle CAD is 180° since AD is on the ray AC beyond C. Wait, no. Wait, angle at A between AB and AD would actually be angle BAD. Since AD is along AC extended, the angle between AB and AD is angle BAC', where BAC' is 100° plus the angle between AC and AD. Wait, but AD is along the same line as AC, so angle between AB and AD is still 100°, but in the other direction? Wait, maybe not. Wait, point D is along ray AC, which is the same line as AC beyond C. So from point A, the direction to C is along AC, and D is further along that direction. So angle BAD is the same as angle BAC, which is 100°, because AD is just an extension of AC. Wait, no. From point A, AB forms an angle of 100° with AC, but when you go to AD, which is beyond C on AC, the angle between AB and AD would still be 100°, right? Because AD is colinear with AC. So angle BAD = 100°.In triangle ABD, sides AB = 1, AD = 2 cos40°, angle BAD = 100°. Then BD can be found by the Law of Cosines:BD² = AB² + AD² - 2*AB*AD*cos(angle BAD)BD² = 1² + (2 cos40°)^2 - 2*1*(2 cos40°)*cos(100°)Compute this:BD² = 1 + 4 cos²40° - 4 cos40° cos100°Hmm, let's try to compute this expression.Note that cos100° = -cos80°, so:BD² = 1 + 4 cos²40° - 4 cos40°*(-cos80°) = 1 + 4 cos²40° + 4 cos40° cos80°Also, cos80° = 2 cos²40° - 1 (using double angle formula: cos2θ = 2cos²θ -1, so θ=40°, cos80°=2cos²40° -1).Therefore, substituting:BD² = 1 + 4 cos²40° + 4 cos40° (2 cos²40° -1 )= 1 + 4 cos²40° + 8 cos³40° - 4 cos40°Hmm, this is getting complicated. Maybe there's a trigonometric identity that can simplify this.Alternatively, compute numerically:cos40° ≈ 0.7660, cos80° ≈ 0.1736Then:BD² ≈ 1 + 4*(0.7660)^2 - 4*(0.7660)*(-0.1736)≈ 1 + 4*(0.5868) + 4*(0.7660)*(0.1736)≈ 1 + 2.3472 + 4*(0.1332)≈ 1 + 2.3472 + 0.5328 ≈ 1 + 2.88 ≈ 3.88Therefore, BD ≈ sqrt(3.88) ≈ 1.9695, which matches the earlier coordinate calculation.Then, in triangle BDC, sides:BD ≈ 1.9695BC = 2 cos40° ≈ 1.532CD = 2 cos40° -1 ≈ 0.532So sides of triangle BDC: BD ≈1.9695, DC≈0.532, BC≈1.532Using the Law of Sines on triangle BDC:sin(angle BDC)/BC = sin(angle B)/DCWait, no. Law of Sines states that:sin(angle BDC)/BC = sin(angle BCD)/BD = sin(angle CBD)/DCBut perhaps we can use Law of Sines here.Wait, angle at D is what we need: angle BDC.Using Law of Sines:sin(angle BDC)/BC = sin(angle DBC)/DCBut I don't know angle DBC. Alternatively, use Law of Cosines.Law of Cosines on triangle BDC:BC² = BD² + CD² - 2*BD*CD*cos(angle BDC)We can solve for cos(angle BDC):cos(angle BDC) = (BD² + CD² - BC²) / (2*BD*CD)Plugging in the approximate values:BD² ≈ 3.88CD² ≈ (0.532)^2 ≈ 0.283BC² ≈ (1.532)^2 ≈ 2.347So:cos(angle BDC) ≈ (3.88 + 0.283 - 2.347) / (2 * 1.9695 * 0.532)≈ (1.816) / (2 * 1.9695 * 0.532)Calculate denominator: 2 *1.9695 *0.532 ≈ 2 *1.047 ≈ 2.094So cos(angle BDC) ≈1.816 /2.094≈ 0.867Which is approximately 30°, as before.But does this angle come out exactly 30°? Let's see.If angle BDC is 30°, then cos(30°) = sqrt(3)/2 ≈0.866, which is close to our approximate 0.867. So it's likely that the exact angle is 30 degrees.But to confirm this without approximation, we need to find an exact geometric construction.Let me think differently. Suppose we construct point E on AB such that AE = BC. Then, maybe triangle AED is congruent to triangle BDC or something. Not sure.Alternatively, since AD = BC and AB = AC, perhaps triangle ABD is congruent to triangle CBA or something. Let's check.In triangle ABC: AB = AC =1, BC = 2 cos40°, angle at A =100°, angles at B and C =40°.In triangle ABD: AB=1, AD=BC=2 cos40°, angle at A=100°, as we saw. If we can prove that triangle ABD is similar or congruent to some other triangle, maybe we can find some relations.Alternatively, maybe consider triangle CBD. Wait, CD = AD - AC = 2 cos40° -1. In triangle CBD, sides BC=2 cos40°, CD=2 cos40° -1, BD is some length.Alternatively, construct an equilateral triangle somewhere. Since 30° is part of an equilateral triangle split.Alternatively, let's consider the circumcircle of triangle BDC. Maybe some cyclic quadrilateral properties.Alternatively, let's recall that in triangle ABC, AB=AC=1, BC=2 cos40°. So BC is twice the projection of AB onto the base. Hmm.Wait, if we consider triangle ABC with AB=AC=1, angle at A=100°, then the length BC can be calculated as 2 sin50°, since in an isosceles triangle with apex angle 100°, the base BC is 2*AB*sin(angle/2). Wait, the formula for the base of an isosceles triangle is 2*leg*sin(theta/2), where theta is the apex angle. So here, theta=100°, so BC=2*1*sin50°≈2*0.7660≈1.532, which matches earlier results. So BC=2 sin50°.But 50° is complementary to 40°, so sin50°=cos40°, which is why BC=2 cos40°, as we had before.So BC=2 cos40°, and AD=BC=2 cos40°. So AD=2 cos40°, AC=1, so CD=2 cos40° -1.Now, let's try to find angle BDC.Wait, since AD=BC and AB=AC, perhaps triangle BCD is isosceles or has some other properties.Alternatively, let's consider triangle BDC. If we can show that BD=CD, then angle BDC would be equal to angle BCD. But BD was approximately 1.9695, CD≈0.532, so not equal. So that's not the case.Alternatively, maybe triangle BDC has angles 30°, some angle, and so on.Alternatively, since we're getting angle BDC≈30°, maybe there's a way to show that angle BDC=30° exactly.Let me try to use exact trigonometric identities.Let me denote angle BDC as θ. We need to find θ.Using Law of Sines on triangle BDC:BC / sinθ = BD / sin(angle BCD) = CD / sin(angle CBD)But we need to relate these angles.Alternatively, note that in triangle ABC, angle at C is 40°, and point D is on the extension of AC beyond C. So angle BCD is supplementary to angle ACB if D were on the other side, but here D is on the extension beyond C, so angle BCD is actually equal to 180° - angle ACB? Wait, no.Wait, angle at C in triangle BDC is angle BCD. Since point D is on the extension of AC beyond C, angle BCD is the angle at C between BC and CD. Since AC is a straight line, angle ACB in triangle ABC is 40°, but angle BCD is a different angle. Wait, in triangle ABC, angle at C is 40°, which is angle ACB. When we go to point D, which is on the extension of AC beyond C, the angle at C between B, C, D is angle BCD. To find angle BCD, we can consider that line AC is straight, so angle ACB + angle BCD = 180°, but that's not necessarily true because point B is not on the line AC.Wait, no. Point B is a vertex of the triangle, so it's not on line AC. So angle BCD is just the angle at point C between points B, C, D.To find angle BCD, maybe we can use some other triangles or laws.Alternatively, in triangle ABC, we can find coordinates or use vectors.Wait, let's try to compute angle BCD.In coordinate system:Point C is at (1,0)Point B is at (cos100°, sin100°)Point D is at (2 cos40°, 0)So vector CB = B - C = (cos100° -1, sin100° -0) = (cos100° -1, sin100°)Vector CD = D - C = (2 cos40° -1, 0 -0) = (2 cos40° -1, 0)Angle BCD is the angle at point C between vectors CB and CD. So we can compute it using the dot product.cos(angle BCD) = (CB • CD) / (|CB| |CD|)Compute CB • CD = (cos100° -1)(2 cos40° -1) + (sin100°)(0) = (cos100° -1)(2 cos40° -1)Compute |CB| = sqrt( (cos100° -1)^2 + (sin100°)^2 )= sqrt( cos²100° - 2 cos100° +1 + sin²100° )= sqrt( (cos²100° + sin²100°) - 2 cos100° +1 )= sqrt(1 - 2 cos100° +1 )= sqrt(2 - 2 cos100° )Which is equal to BC, which is 2 cos40°, as before. Wait, sqrt(2 - 2 cos100° ) = BC.But 2 - 2 cos100° = 4 sin²50°, since 1 - cosθ = 2 sin²(θ/2). Therefore, sqrt(4 sin²50° ) = 2 sin50° = 2 cos40°, since sin50°=cos40°.So |CB| = 2 cos40°, which matches BC.|CD| = 2 cos40° -1Therefore, angle BCD:cos(angle BCD) = [ (cos100° -1)(2 cos40° -1) ] / [2 cos40°*(2 cos40° -1) ]= [ (cos100° -1)(2 cos40° -1) ] / [2 cos40°*(2 cos40° -1) ]Cancel (2 cos40° -1):cos(angle BCD) = (cos100° -1) / (2 cos40° )Now, compute cos100° -1:cos100° = -cos80°, so cos100° -1 = -cos80° -1 = -(cos80° +1)Therefore:cos(angle BCD) = [ - (cos80° +1) ] / (2 cos40° )= - (cos80° +1 ) / (2 cos40° )Let me compute numerator:cos80° +1 = 1 + cos80°Using the identity 1 + cosθ = 2 cos²(θ/2), so 1 + cos80° = 2 cos²40°Therefore:cos(angle BCD) = - (2 cos²40° ) / (2 cos40° ) = - cos40°So cos(angle BCD) = - cos40°, which means angle BCD = 180° -40°=140°, since cosine is negative and angle in triangle can't be more than 180°. Wait, but angle in triangle can't be 140° if other angles are already... Wait, in triangle BDC, if angle BCD is 140°, then the sum of angles would be θ (angle BDC) + angle BCD + angle CBD = 180°. So θ +140° + angle CBD =180°, so θ + angle CBD=40°. But we need to find θ.Alternatively, maybe I made a mistake in calculation.Wait, let's recast the calculation:We had cos(angle BCD) = -cos40°, which implies angle BCD=180°-40°=140°, since cosine is -cos40°, which is equal to cos(180°-40°)=cos140°. Therefore, angle BCD=140°.So in triangle BDC, angles are:angle BCD=140°, angle BDC=θ, angle CBD=φ.Sum of angles: 140° + θ + φ=180°, so θ + φ=40°.If we can find φ, then θ=40° - φ.Alternatively, use Law of Sines:In triangle BDC,BD / sin(angle BCD) = BC / sinθ = CD / sinφFrom earlier, BD≈1.9695, BC≈1.532, CD≈0.532.So:BD / sin(140°) ≈ 1.9695 / sin140° ≈ 1.9695 / 0.6428≈3.065BC / sinθ ≈1.532 / sinθCD / sinφ ≈0.532 / sinφSo:3.065 ≈1.532 / sinθ → sinθ ≈1.532 /3.065≈0.5So sinθ≈0.5 → θ≈30° or 150°, but in the triangle, angle BDC can't be 150° since angle BCD is already 140°, so θ=30°.Therefore, angle BDC=30°, exactly.So the measure of angle BDC is 30 degrees.This matches the approximate calculation we did earlier. Therefore, the answer is 30°.Final AnswerThe measure of (angle BDC) is (boxed{30}) degrees.

✅:In the isosceles triangle (ABC) with (angle ACB = angle ABC = 40^circ), we know that (AB = AC) and (angle BAC = 100^circ). We need to find the measure of (angle BDC) where (AD = BC) and (D) is on the ray (AC).1. Identify Key Lengths and Angles: - Since (AB = AC), triangle (ABC) is isosceles with (AB = AC). - Using the Law of Sines in (triangle ABC), we find (BC = 2 cdot AB cdot cos(40^circ)). - Given (AD = BC), we extend (AC) to point (D) such that (AD = BC).2. Coordinate System and Vectors: - Place (A) at the origin ((0,0)), (C) at ((1,0)), and (B) at ((cos(100^circ), sin(100^circ))). - Point (D) is on the extension of (AC) at ((2 cos(40^circ), 0)).3. Vectors and Dot Product: - Calculate vectors (overrightarrow{DB}) and (overrightarrow{DC}). - Use the dot product to find the angle between vectors (overrightarrow{DB}) and (overrightarrow{DC}).4. Law of Sines and Cosines: - Using the Law of Cosines in (triangle BDC), we find the cosine of (angle BDC). - Simplify using trigonometric identities and exact values, leading to (cos(angle BDC) = frac{sqrt{3}}{2}), which corresponds to (30^circ).5. Conclusion: - The measure of (angle BDC) is found to be (30^circ) through exact trigonometric calculations and vector analysis.The measure of (angle BDC) is (boxed{30}) degrees.

🔑:1. Given that ( triangle ABC ) is an isosceles triangle with ( angle ACB = angle ABC = 40^circ ).2. Measure a line segment ( AD ) on the ray ( AC ) such that ( AD = BC ).3. Let point ( E ) on the plane be such that ( triangle ABC ) is congruent to ( triangle EDA ), and ( E ) is on the same side of line ( AD ) as ( B ). The point ( E ) is unique and well-defined. Since ( triangle ABC cong triangle EDA ): [ AE = ED = AB = AC ]4. We know that: [ angle ADE = angle EAD = angle ABC = angle ACB = 40^circ ] Because the sum of the angles in a triangle is (180^circ), it follows that: [ angle BAC = 180^circ - 2 times 40^circ = 100^circ ] Therefore: [ angle BAE = angle BAC - angle EAD = 100^circ - 40^circ = 60^circ ]5. Since ( AE = AB ) and ( angle BAE = 60^circ ), this forms an equilateral triangle ( triangle EAB ). Thus, all sides are equal: [ EA = AB = AE = ED ]6. This implies ( E ) is the circumcenter of ( triangle ABD ) because the distances from ( E ) to ( A ), ( B ), and ( D ) are all equal.7. ( angle BDC ) is the inscribed angle subtended by chord ( AB ) in the circumcircle of ( triangle ABD ). The central angle ( angle AEB ) subtended by the same chord ( AB ) is ( 60^circ ) (because ( triangle EAB ) is equilateral).8. Since the inscribed angle ( angle BDC ) is half of the central angle: [ angle BDC = frac{angle AEB}{2} = frac{60^circ}{2} = 30^circ ] Conclusion:[boxed{30^circ}]