🧠:<think>Alright, so I've got this problem here about a right triangle, triangle PQR. They tell me that the tangent of angle Q is 0.5, and the length of side QP is 16. I need to find the length of side QR. Hmm, okay, let me try to visualize this.First off, since it's a right triangle, one of the angles is 90 degrees. But they don't specify which angle that is. They mention angle Q, so I guess angle Q is one of the acute angles, not the right angle. So, either P or R must be the right angle. I think it's important to figure out which one is the right angle because that will determine which sides are opposite and adjacent to angle Q.Wait, in triangle PQR, the vertices are P, Q, and R. So, the sides are PQ, QR, and RP. If angle Q is one of the angles, then the sides opposite and adjacent to angle Q would be PR and PQ, respectively. So, if I consider angle Q, then side PR is opposite to angle Q, and side PQ is adjacent to angle Q. That makes sense because in a right triangle, the sides are labeled based on their relation to the angle.They gave me that tan Q is 0.5. Remember, tangent is opposite over adjacent. So, tan Q = opposite / adjacent = PR / PQ. They told me that PQ is 16, so I can set up the equation: 0.5 = PR / 16. To find PR, I can rearrange that: PR = 0.5 * 16 = 8. So, PR is 8 units long.Now, I need to find QR. QR is the hypotenuse of the triangle because in a right triangle, the side opposite the right angle is the hypotenuse. But wait, I need to confirm which side is the hypotenuse. If angle Q is not the right angle, then the hypotenuse must be the side opposite the right angle. So, if angle P or angle R is the right angle, then the hypotenuse would be QR or PR, respectively.Wait, let me think again. In triangle PQR, if angle Q is not the right angle, then either angle P or angle R is the right angle. If angle P is the right angle, then the sides would be: PQ and PR as the legs, and QR as the hypotenuse. If angle R is the right angle, then the sides would be QR and PR as the legs, and PQ as the hypotenuse. But they told me that PQ is 16, and we found PR is 8. So, if angle P is the right angle, then QR is the hypotenuse, and we can use the Pythagorean theorem to find its length.Alternatively, if angle R is the right angle, then PQ would be the hypotenuse, but they gave us PQ as 16, and we found PR as 8. That would mean QR is one of the legs, but that doesn't make sense because if angle R is the right angle, then QR and PR would be the legs, and PQ would be the hypotenuse. But they gave us PQ as 16, and we found PR as 8, so QR would be the other leg. Wait, but then we can still use the Pythagorean theorem to find QR.Wait, no, hold on. If angle R is the right angle, then side PQ is the hypotenuse, which is 16, and sides PR and QR are the legs. We found PR as 8, so we can find QR using Pythagorean theorem: PQ² = PR² + QR². So, 16² = 8² + QR². That would be 256 = 64 + QR², so QR² = 256 - 64 = 192, so QR = sqrt(192) = 8*sqrt(3). But wait, that's different from what I thought earlier.But earlier, I assumed angle P was the right angle, making QR the hypotenuse. So, which one is it? The problem doesn't specify where the right angle is. Hmm, that's confusing. Maybe I need to figure it out based on the given information.Wait, in the problem statement, they mention triangle PQR, and they say "in a right triangle." They don't specify which angle is the right angle. So, maybe I need to assume that angle Q is not the right angle, and the right angle is at either P or R. But without more information, it's ambiguous.Wait, but in the initial problem, they gave tan Q = 0.5. If angle Q were the right angle, then tan Q would be undefined because tangent of 90 degrees is undefined. So, angle Q cannot be the right angle. Therefore, the right angle must be at either P or R.So, if angle Q is not the right angle, then the right angle is at P or R. Let me consider both cases.Case 1: Right angle at P. Then, sides PQ and PR are the legs, and QR is the hypotenuse.Case 2: Right angle at R. Then, sides PR and QR are the legs, and PQ is the hypotenuse.Given that tan Q = 0.5, which is opposite over adjacent. If the right angle is at P, then side PR is opposite to angle Q, and side PQ is adjacent to angle Q. So, tan Q = PR / PQ = 0.5, which gives PR = 0.5 * PQ = 0.5 * 16 = 8, as I found earlier.Then, QR would be the hypotenuse, so QR = sqrt(PQ² + PR²) = sqrt(16² + 8²) = sqrt(256 + 64) = sqrt(320) = 8*sqrt(5).Alternatively, if the right angle is at R, then side PR is opposite to angle Q, and side QR is adjacent to angle Q. So, tan Q = PR / QR = 0.5. We know PR is 8, so 0.5 = 8 / QR, which would give QR = 8 / 0.5 = 16. But wait, they already told us that PQ is 16. So, if QR is also 16, that would mean that sides PQ and QR are both 16, making it an isosceles right triangle, but with angle Q having tan 0.5, which is not 1, so that can't be.Wait, that doesn't make sense. If the right angle is at R, then PR is opposite to angle Q, and QR is adjacent to angle Q. So, tan Q = PR / QR = 0.5. We found PR = 8, so QR = PR / 0.5 = 8 / 0.5 = 16. But PQ is given as 16, which would mean that in this case, sides PQ and QR are both 16, making triangle PQR an isosceles right triangle with legs of 16 and hypotenuse PR. But wait, PR was found to be 8, which contradicts because in an isosceles right triangle, the legs are equal, and the hypotenuse is leg * sqrt(2). So, if legs are 16, hypotenuse would be 16*sqrt(2), not 8.Therefore, this case leads to a contradiction because PR cannot be both 8 and 16*sqrt(2). So, this must mean that the right angle is not at R, but rather at P. Therefore, QR is the hypotenuse, and its length is 8*sqrt(5).So, to recap, since angle Q cannot be the right angle (as tan Q is defined), and if we assume the right angle is at P, then QR is the hypotenuse, and using the Pythagorean theorem, we find QR = 8*sqrt(5). If we assume the right angle is at R, it leads to a contradiction, so that case is invalid. Therefore, the correct length of QR is 8*sqrt(5).I think that makes sense. So, the key steps were:1. Determine that angle Q is not the right angle because tan Q is defined.2. Consider both possibilities for the right angle (at P or R).3. Use the given tan Q = 0.5 to find the opposite side PR = 8.4. Apply the Pythagorean theorem in both cases to find QR.5. Discard the case where the right angle is at R because it leads to a contradiction.6. Conclude that QR = 8*sqrt(5).Yeah, that seems solid. I don't think I made any mistakes in the calculations, but let me double-check.Given tan Q = 0.5 = opposite / adjacent = PR / PQ = 8 / 16 = 0.5. That checks out.Then, QR² = PQ² + PR² = 256 + 64 = 320, so QR = sqrt(320) = sqrt(64*5) = 8*sqrt(5). Yep, that's correct.And in the other case, assuming right angle at R, we get QR = 16, which conflicts with PR being 8 and PQ being 16, making it impossible for PR to be both 8 and 16*sqrt(2). So, that case is invalid.Therefore, the length of QR is indeed 8*sqrt(5).</think>

🧠:<think>Okay, so I have this problem here where I need to find the largest real number ( d ) such that the inequality[x_1^2 + x_2^2 + dots + x_{51}^2 geq dleft(frac{x_1 + x_2 + dots + x_{51}}{51}right)^2]holds whenever ( x_1, x_2, ldots, x_{51} ) are real numbers with the condition that their sum is equal to the median ( M ). That is, ( x_1 + x_2 + cdots + x_{51} = M ) and ( M ) is the median of these numbers.First, let me understand the problem step by step. Since there are 51 numbers, which is an odd count, the median ( M ) will be the 26th term when the numbers are arranged in non-decreasing order. So, if I sort the numbers such that ( x_1 leq x_2 leq ldots leq x_{51} ), then ( M = x_{26} ).Given that the sum of all these numbers is equal to the median, which is ( M ), so:[x_1 + x_2 + cdots + x_{51} = M]I need to find the largest ( d ) such that the inequality holds. The inequality relates the sum of squares of the numbers to the square of their average. So, it's comparing the sum of squares to a multiple of the square of the mean.I recall that there's an inequality called the Cauchy-Schwarz inequality which might be useful here. The Cauchy-Schwarz inequality in its basic form states that for any real numbers ( a_i ) and ( b_i ):[left( sum_{i=1}^{n} a_i b_i right)^2 leq left( sum_{i=1}^{n} a_i^2 right) left( sum_{i=1}^{n} b_i^2 right)]If I set all ( b_i = 1 ), then it simplifies to:[left( sum_{i=1}^{n} a_i right)^2 leq n left( sum_{i=1}^{n} a_i^2 right)]Which can be rearranged as:[sum_{i=1}^{n} a_i^2 geq frac{1}{n} left( sum_{i=1}^{n} a_i right)^2]In this problem, ( n = 51 ), so applying Cauchy-Schwarz, we get:[x_1^2 + x_2^2 + dots + x_{51}^2 geq frac{1}{51} left( x_1 + x_2 + dots + x_{51} right)^2]But in our case, the sum ( x_1 + x_2 + dots + x_{51} = M ), so substituting that in:[x_1^2 + x_2^2 + dots + x_{51}^2 geq frac{M^2}{51}]Now, the inequality we need is:[x_1^2 + x_2^2 + dots + x_{51}^2 geq d left( frac{M}{51} right)^2]So, substituting the lower bound from Cauchy-Schwarz into this, we have:[frac{M^2}{51} geq d left( frac{M}{51} right)^2]Let me simplify this inequality. First, note that ( M ) is non-zero because it's the median, and the sum of all numbers is ( M ). So, ( M ) can't be zero unless all numbers are zero, but that's a trivial case. Assuming ( M neq 0 ), we can divide both sides by ( frac{M^2}{51^2} ):[frac{M^2}{51} div left( frac{M^2}{51^2} right) geq d]Simplifying the left side:[frac{M^2}{51} times frac{51^2}{M^2} = 51]So, we get:[51 geq d]This suggests that ( d ) must be less than or equal to 51. Therefore, the largest possible value for ( d ) is 51.But wait, I should verify if this bound is achievable. That is, is there a set of numbers where the inequality becomes an equality? If so, then 51 is indeed the largest possible ( d ).For the Cauchy-Schwarz inequality, equality holds when all the ( x_i ) are equal. So, if all ( x_i = M ), then the sum is ( 51M ), but in our case, the sum is ( M ). So, if all ( x_i = M ), the sum would be ( 51M ), which is not equal to ( M ) unless ( M = 0 ). But if ( M = 0 ), then all ( x_i = 0 ), which is a trivial case.Hmm, so in the non-trivial case where ( M neq 0 ), can we have equality? Maybe not directly, but perhaps we can construct a case where the sum is ( M ) and the sum of squares is exactly ( frac{M^2}{51} ).Wait, but if all ( x_i = frac{M}{51} ), then the sum would be ( 51 times frac{M}{51} = M ), which satisfies the condition. And the sum of squares would be ( 51 times left( frac{M}{51} right)^2 = frac{M^2}{51} ). So, in this case, the sum of squares is exactly ( frac{M^2}{51} ), which would make the inequality an equality when ( d = 51 ).Therefore, ( d = 51 ) is achievable, so it's indeed the largest possible value.But let me think again. The problem states that ( M ) is the median. So, in the case where all ( x_i = frac{M}{51} ), the median would be ( frac{M}{51} ), but the sum is ( M ). So, in this case, ( M = 51 times frac{M}{51} = M ), which is consistent. So, the median is ( frac{M}{51} ), but wait, that would mean ( M = frac{M}{51} ), implying ( M = 0 ). So, this is only possible if ( M = 0 ).Hmm, that's a problem. So, if ( M ) is the median, and all ( x_i = frac{M}{51} ), then the median is ( frac{M}{51} ), which must equal ( M ). Therefore, ( M = frac{M}{51} ), leading to ( M = 0 ). So, in the non-trivial case where ( M neq 0 ), we can't have all ( x_i ) equal because that would force ( M = 0 ).Therefore, maybe the case where all ( x_i ) are equal is only possible when ( M = 0 ). So, in non-trivial cases, we can't achieve equality through that method.So, perhaps I need another approach to find the maximum ( d ). Maybe I should consider specific configurations of the numbers where the sum is ( M ) and the median is ( M ), and then compute the sum of squares to find the minimal possible value, which would give me the maximum ( d ).Let me try to construct such a configuration. Since ( M ) is the median, which is the 26th term, that means at least 26 numbers are greater than or equal to ( M ), and at least 26 numbers are less than or equal to ( M ).Given that the sum of all numbers is ( M ), which is equal to the median, perhaps we can set as many numbers as possible to ( M ) and adjust the others accordingly.Let me suppose that the first 25 numbers are as small as possible, and the last 25 numbers are as large as possible, with the 26th number being ( M ). But since the sum is ( M ), which is the median, maybe we can set the first 25 numbers to be equal to some value ( a ), the 26th number is ( M ), and the last 25 numbers are equal to some value ( b ).So, let me denote:- ( x_1 = x_2 = ldots = x_{25} = a )- ( x_{26} = M )- ( x_{27} = x_{28} = ldots = x_{51} = b )Now, the sum of all numbers is:[25a + M + 25b = M]So,[25a + 25b = 0 implies a + b = 0 implies b = -a]Therefore, ( b = -a ). So, the numbers are symmetric around zero in some sense.Now, the sum of squares is:[25a^2 + M^2 + 25b^2 = 25a^2 + M^2 + 25a^2 = 50a^2 + M^2]We need to express this in terms of ( M ). Since ( b = -a ), and the median is ( M ), which is the 26th term, so ( M ) must be greater than or equal to ( a ) and less than or equal to ( b ). But since ( b = -a ), if ( a ) is negative, ( b ) is positive.Wait, let's think about the ordering. If ( a leq M leq b ), but ( b = -a ). So, if ( a ) is negative, ( b ) is positive, so ( M ) must be between ( a ) and ( b ). But since ( M ) is the median, it's the 26th term, so it's the middle value.But in our setup, ( x_{26} = M ), so ( M ) is the median. So, as long as ( a leq M leq b ), which is ( a leq M leq -a ). So, ( M ) must be between ( a ) and ( -a ).But since ( a ) is the value of the first 25 numbers, which are less than or equal to ( M ), and ( b = -a ) is the value of the last 25 numbers, which are greater than or equal to ( M ).So, ( a leq M leq -a ). Let me solve for ( a ).From ( a leq M leq -a ), we have:1. ( a leq M )2. ( M leq -a )From the first inequality, ( a leq M ).From the second inequality, ( M leq -a implies a leq -M ).So combining these, ( a leq min(M, -M) ).But ( M ) can be positive or negative. Let's consider two cases.Case 1: ( M geq 0 )Then, ( min(M, -M) = -M ). So, ( a leq -M ).But since ( a leq M ) as well, and ( M geq 0 ), ( a leq -M leq M ). So, ( a leq -M ).Case 2: ( M < 0 )Then, ( min(M, -M) = M ). So, ( a leq M ).But since ( M < 0 ), ( a leq M ).In either case, ( a ) is bounded above by ( -|M| ).But perhaps instead of getting bogged down in inequalities, I can express ( a ) in terms of ( M ).From the sum equation:[25a + 25b = 0 implies a = -b]So, ( a = -b ). Therefore, the sum of squares becomes:[50a^2 + M^2]We need to express this in terms of ( M ). But how?Wait, we can relate ( a ) and ( M ) through the median condition. Since ( M ) is the median, it must be greater than or equal to the first 25 numbers and less than or equal to the last 25 numbers.So, ( a leq M leq b ). But since ( b = -a ), this becomes ( a leq M leq -a ).So, ( a leq M leq -a ). Let's solve for ( a ).From ( a leq M ) and ( M leq -a ), we get:1. ( a leq M )2. ( M leq -a implies a leq -M )So, combining these, ( a leq min(M, -M) ).But ( min(M, -M) = -|M| ). So, ( a leq -|M| ).But ( a ) can be any real number, so to minimize the sum of squares, we need to find the minimal possible value of ( 50a^2 + M^2 ).Wait, actually, since ( a leq -|M| ), the minimal value of ( a^2 ) occurs when ( a ) is as close to zero as possible, but given the constraints.Wait, no. Since ( a leq -|M| ), the smallest ( a ) can be is negative infinity, but that would make ( a^2 ) very large, which would increase the sum of squares. So, to minimize the sum of squares, we need to maximize ( a ), i.e., make ( a ) as large as possible, given the constraints.So, the maximum value ( a ) can take is ( a = -|M| ).Wait, let me clarify. If ( a leq -|M| ), then the maximum ( a ) can be is ( -|M| ). So, setting ( a = -|M| ) would give the minimal sum of squares because increasing ( a ) beyond that would violate the constraints.Wait, no, actually, ( a leq -|M| ) implies that ( a ) is less than or equal to a negative number. So, the largest ( a ) can be is ( -|M| ). So, if I set ( a = -|M| ), then ( b = |M| ).So, let's set ( a = -|M| ) and ( b = |M| ). Then, the sum of squares becomes:[50a^2 + M^2 = 50|M|^2 + M^2 = 51M^2]Wait, that's interesting. So, the sum of squares is ( 51M^2 ).But in our inequality, we have:[x_1^2 + x_2^2 + dots + x_{51}^2 geq d left( frac{M}{51} right)^2]Substituting the sum of squares we just found:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq d frac{M^2}{51^2}]Assuming ( M neq 0 ), we can divide both sides by ( M^2 ):[51 geq frac{d}{51^2}]Wait, that can't be right. Let me check my substitution.Wait, the left side is ( 51M^2 ), and the right side is ( d times left( frac{M}{51} right)^2 = d times frac{M^2}{2601} ).So, the inequality becomes:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ) (assuming ( M neq 0 )):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]But 51 times 2601 is a very large number, which contradicts our earlier conclusion that ( d leq 51 ). So, something is wrong here.Wait, perhaps I made a mistake in setting ( a = -|M| ). Let me go back.We have ( a leq -|M| ), so ( a ) is less than or equal to a negative number. Therefore, ( a ) is negative, and ( b = -a ) is positive.So, if I set ( a = -k ) where ( k geq |M| ), then ( b = k ).So, the sum of squares is:[25a^2 + M^2 + 25b^2 = 25k^2 + M^2 + 25k^2 = 50k^2 + M^2]We need to find the minimal value of ( 50k^2 + M^2 ) given that ( k geq |M| ).To minimize this expression, we should choose the smallest possible ( k ), which is ( k = |M| ).So, substituting ( k = |M| ):[50|M|^2 + M^2 = 51M^2]So, the minimal sum of squares is ( 51M^2 ).Therefore, the inequality becomes:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ) (assuming ( M neq 0 )):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]But 51 times 2601 is 132,651, which is way larger than 51. This suggests that ( d ) can be as large as 132,651, which contradicts our earlier result.Wait, this can't be right. There must be a mistake in my reasoning.Let me go back to the problem statement. It says that ( M ) is the median, and the sum of all numbers is ( M ). So, in our setup, we have 25 numbers equal to ( a ), one number equal to ( M ), and 25 numbers equal to ( b ), with ( a leq M leq b ) and ( 25a + M + 25b = M implies 25(a + b) = 0 implies a + b = 0 implies b = -a ).So, ( a leq M leq -a ). Therefore, ( a leq M leq -a implies a leq M leq -a implies a leq -|M| ).Wait, if ( M ) is positive, then ( a leq -M ). If ( M ) is negative, then ( a leq M ).But in either case, ( a leq -|M| ).So, the minimal sum of squares is achieved when ( a = -|M| ), which gives ( b = |M| ), and the sum of squares is ( 51M^2 ).But then, substituting into the inequality:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]Which is ( 132,651 geq d ). But this can't be right because earlier, using Cauchy-Schwarz, we found ( d leq 51 ). So, there's a contradiction here.Wait, perhaps I misapplied the Cauchy-Schwarz inequality. Let me double-check.The Cauchy-Schwarz inequality says:[left( sum x_i right)^2 leq 51 sum x_i^2]Which rearranges to:[sum x_i^2 geq frac{1}{51} left( sum x_i right)^2]Given that ( sum x_i = M ), this becomes:[sum x_i^2 geq frac{M^2}{51}]So, in our problem, we have:[sum x_i^2 geq d left( frac{M}{51} right)^2]So, substituting the lower bound from Cauchy-Schwarz:[frac{M^2}{51} geq d left( frac{M^2}{2601} right)]Simplify:[frac{1}{51} geq frac{d}{2601}]Multiply both sides by 2601:[frac{2601}{51} geq d]Calculate ( 2601 / 51 ):51 times 51 is 2601, so ( 2601 / 51 = 51 ).Therefore, ( 51 geq d ).So, the maximum ( d ) is 51.But in the earlier setup, when I set ( a = -|M| ) and ( b = |M| ), the sum of squares was ( 51M^2 ), which when substituted into the inequality gave ( d leq 132,651 ), which is way larger than 51. So, why is there a discrepancy?Ah, I see. Because in that setup, the sum of squares is ( 51M^2 ), which is much larger than the lower bound given by Cauchy-Schwarz, which is ( frac{M^2}{51} ). So, in that specific case, the sum of squares is much larger, so the inequality holds for a much larger ( d ). But since we need the inequality to hold for all possible configurations, the minimal sum of squares across all configurations will determine the maximum ( d ).Wait, so the minimal sum of squares is ( frac{M^2}{51} ), which occurs when all ( x_i ) are equal to ( frac{M}{51} ). But earlier, I realized that if all ( x_i = frac{M}{51} ), then the median would be ( frac{M}{51} ), which must equal ( M ), implying ( M = 0 ). So, in the non-trivial case where ( M neq 0 ), we can't have all ( x_i ) equal.Therefore, the minimal sum of squares in the non-trivial case is higher than ( frac{M^2}{51} ). So, perhaps the minimal sum of squares is achieved when as many variables as possible are equal to the median, but given the constraints.Wait, let me think differently. Maybe I can use Lagrange multipliers to minimize the sum of squares subject to the constraints that the sum is ( M ) and the median is ( M ).But that might be complicated. Alternatively, perhaps I can consider that since the median is ( M ), at least 26 numbers are greater than or equal to ( M ), and at least 26 numbers are less than or equal to ( M ).So, in the minimal sum of squares case, we can set 25 numbers to be as small as possible, 25 numbers as large as possible, and the 26th number is ( M ). But to minimize the sum of squares, we should set the 25 numbers less than or equal to ( M ) to be as large as possible, and the 25 numbers greater than or equal to ( M ) to be as small as possible.Wait, that might make sense. So, to minimize the sum of squares, we should set the first 25 numbers to be equal to ( M ), and the last 25 numbers to be equal to ( M ), but that would make all numbers equal to ( M ), which would make the sum ( 51M ), not ( M ). So, that's not possible.Alternatively, perhaps set the first 25 numbers to be as large as possible without exceeding ( M ), and the last 25 numbers to be as small as possible without going below ( M ). But since the sum is ( M ), which is much smaller than ( 51M ), we need to have some numbers below ( M ) and some above.Wait, perhaps the minimal sum of squares occurs when 25 numbers are as small as possible, 25 numbers are as large as possible, and the median is ( M ). But given that the sum is ( M ), which is the median, we have to balance the numbers.Wait, let me consider the following configuration:- 25 numbers are equal to ( a )- 1 number is equal to ( M )- 25 numbers are equal to ( b )With ( a leq M leq b ) and ( 25a + M + 25b = M implies 25(a + b) = 0 implies a + b = 0 implies b = -a ).So, in this case, the sum of squares is ( 25a^2 + M^2 + 25b^2 = 50a^2 + M^2 ).To minimize the sum of squares, we need to minimize ( 50a^2 + M^2 ). Since ( a leq M leq -a ), we have ( a leq -|M| ).So, the minimal value occurs when ( a = -|M| ), which gives ( b = |M| ), and the sum of squares becomes ( 50|M|^2 + M^2 = 51M^2 ).But wait, if ( a = -|M| ), then ( M ) must be greater than or equal to ( a ), which is ( -|M| ). So, if ( M ) is positive, ( a = -M ), and ( M geq a = -M implies M geq -M implies M geq 0 ), which is consistent.Similarly, if ( M ) is negative, ( a = M ), and ( M leq -M implies M leq 0 ), which is also consistent.So, in either case, the minimal sum of squares is ( 51M^2 ).Therefore, substituting into the inequality:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ) (assuming ( M neq 0 )):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]But 51 times 2601 is 132,651, which is way larger than 51. This suggests that ( d ) can be as large as 132,651, but that contradicts our earlier result from Cauchy-Schwarz.Wait, I think the confusion arises because in this specific configuration, the sum of squares is much larger than the lower bound given by Cauchy-Schwarz. Therefore, in this case, the inequality holds for a much larger ( d ), but we need the inequality to hold for all possible configurations, not just this one.So, the minimal sum of squares across all configurations is ( frac{M^2}{51} ), which occurs when all ( x_i ) are equal to ( frac{M}{51} ). But as we saw earlier, this configuration only works when ( M = 0 ), which is a trivial case.Therefore, in non-trivial cases where ( M neq 0 ), the minimal sum of squares is higher. So, perhaps the minimal sum of squares in non-trivial cases is ( 51M^2 ), which would imply that ( d ) can be as large as 51.Wait, but in the configuration where all ( x_i = frac{M}{51} ), the sum of squares is ( frac{M^2}{51} ), but this only works when ( M = 0 ). So, in non-trivial cases, the minimal sum of squares is higher, which would mean that the minimal value of ( sum x_i^2 ) is ( frac{M^2}{51} ) only when ( M = 0 ).Therefore, in non-trivial cases, the minimal sum of squares is higher, so the inequality ( sum x_i^2 geq d left( frac{M}{51} right)^2 ) must hold for the minimal sum of squares across all possible configurations.But since in some configurations, the sum of squares can be as low as ( frac{M^2}{51} ) (when ( M = 0 )), but in non-trivial cases, it's higher, the maximum ( d ) that works for all configurations is 51, because in the case where ( M = 0 ), the sum of squares is ( frac{M^2}{51} = 0 ), and the right-hand side is also 0, so the inequality holds.But wait, if ( M = 0 ), then the median is 0, and the sum is 0. So, all numbers sum to 0, and the median is 0. In this case, the sum of squares is non-negative, and the right-hand side is 0, so the inequality holds for any ( d ). But we need the inequality to hold for all ( M ), including non-zero ( M ).Therefore, the minimal sum of squares in non-zero ( M ) cases is higher, so the maximum ( d ) is determined by the minimal sum of squares in non-zero cases.Wait, perhaps I need to consider the minimal sum of squares when ( M neq 0 ).In the configuration where 25 numbers are ( a ), one is ( M ), and 25 are ( b ), with ( a = -|M| ) and ( b = |M| ), the sum of squares is ( 51M^2 ). So, in this case, the sum of squares is ( 51M^2 ), which is much larger than ( frac{M^2}{51} ).But if I can find a configuration where the sum of squares is lower than ( 51M^2 ), then that would give a higher lower bound for ( d ).Wait, perhaps if I set more numbers equal to ( M ). For example, if I set 26 numbers equal to ( M ), and the remaining 25 numbers to some value ( a ), then the sum would be ( 26M + 25a = M implies 25a = -25M implies a = -M ).So, in this case, the sum of squares is ( 26M^2 + 25a^2 = 26M^2 + 25M^2 = 51M^2 ). So, same as before.Alternatively, if I set 25 numbers equal to ( M ), one number equal to ( M ), and 25 numbers equal to ( a ), but that's the same as before.Wait, perhaps if I set more numbers equal to ( M ), but then the sum would require the remaining numbers to be negative, which would increase the sum of squares.Alternatively, what if I set some numbers above ( M ) and some below, but not symmetrically. For example, set some numbers slightly above ( M ) and some slightly below, but in such a way that the sum is ( M ).But this might complicate things. Alternatively, perhaps the minimal sum of squares is indeed ( 51M^2 ), achieved when 25 numbers are ( -|M| ), one is ( M ), and 25 are ( |M| ).Therefore, in that case, the sum of squares is ( 51M^2 ), which when substituted into the inequality gives:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]But this is ( 132,651 geq d ), which is way larger than 51. So, this suggests that ( d ) can be as large as 132,651, but that contradicts our earlier result.Wait, perhaps I'm making a mistake in interpreting the inequality. Let me go back to the original inequality:[x_1^2 + x_2^2 + dots + x_{51}^2 geq d left( frac{x_1 + x_2 + dots + x_{51}}{51} right)^2]Given that ( x_1 + x_2 + dots + x_{51} = M ), this becomes:[x_1^2 + x_2^2 + dots + x_{51}^2 geq d left( frac{M}{51} right)^2]So, the left side is the sum of squares, and the right side is ( d times left( frac{M}{51} right)^2 ).In the configuration where 25 numbers are ( -|M| ), one is ( M ), and 25 are ( |M| ), the sum of squares is ( 51M^2 ). So, substituting:[51M^2 geq d left( frac{M}{51} right)^2]Simplify:[51M^2 geq frac{d M^2}{2601}]Divide both sides by ( M^2 ):[51 geq frac{d}{2601}]Multiply both sides by 2601:[51 times 2601 geq d]But 51 times 2601 is 132,651, so ( d leq 132,651 ). But this can't be right because in the Cauchy-Schwarz case, we have ( d leq 51 ).Wait, perhaps I'm confusing the direction of the inequality. Let me think again.The Cauchy-Schwarz inequality gives a lower bound on the sum of squares, which is ( frac{M^2}{51} ). So, the sum of squares is always at least ( frac{M^2}{51} ). Therefore, the inequality ( sum x_i^2 geq d left( frac{M}{51} right)^2 ) must hold for all configurations, meaning that ( d ) must be such that ( frac{M^2}{51} geq d left( frac{M^2}{2601} right) ).Simplify:[frac{1}{51} geq frac{d}{2601}]Multiply both sides by 2601:[frac{2601}{51} geq d]Calculate ( 2601 / 51 ):51 times 51 is 2601, so ( 2601 / 51 = 51 ).Therefore, ( 51 geq d ).So, the maximum ( d ) is 51.But in the specific configuration where the sum of squares is ( 51M^2 ), substituting into the inequality gives ( 51M^2 geq d times frac{M^2}{2601} implies 51 geq frac{d}{2601} implies d leq 51 times 2601 ), which is a much larger number. But since the inequality must hold for all configurations, including the one where the sum of squares is minimal, which is ( frac{M^2}{51} ), we have ( d leq 51 ).Therefore, the largest possible ( d ) is 51.But wait, in the configuration where all ( x_i = frac{M}{51} ), the sum of squares is ( frac{M^2}{51} ), which is the minimal sum of squares. However, in this case, the median is ( frac{M}{51} ), which must equal ( M ), implying ( M = 0 ). So, in non-trivial cases where ( M neq 0 ), the sum of squares is higher than ( frac{M^2}{51} ), but the inequality must hold for all cases, including the trivial case where ( M = 0 ).Therefore, the minimal sum of squares across all configurations is ( frac{M^2}{51} ), which occurs when ( M = 0 ). But since the inequality must hold for all ( M ), including non-zero ( M ), the maximum ( d ) is determined by the minimal sum of squares in non-zero ( M ) cases.Wait, but if ( M neq 0 ), the minimal sum of squares is higher than ( frac{M^2}{51} ), so the inequality ( sum x_i^2 geq d left( frac{M}{51} right)^2 ) will hold for a larger ( d ) in those cases. However, since we need the inequality to hold for all ( M ), including ( M = 0 ), the maximum ( d ) is determined by the case where the sum of squares is minimal, which is ( frac{M^2}{51} ) when ( M = 0 ). But in that case, both sides are zero, so the inequality holds for any ( d ).Wait, this is getting confusing. Let me try to approach it differently.Let me consider the ratio:[frac{sum x_i^2}{left( frac{sum x_i}{51} right)^2} = frac{sum x_i^2}{left( frac{M}{51} right)^2}]We need this ratio to be at least ( d ) for all configurations. Therefore, the minimal value of this ratio across all configurations will be the maximum ( d ).So, ( d ) is the minimal value of ( frac{sum x_i^2}{left( frac{M}{51} right)^2} ).Therefore, ( d = min left( frac{sum x_i^2}{left( frac{M}{51} right)^2} right) ).So, to find ( d ), we need to find the minimal possible value of ( frac{sum x_i^2}{left( frac{M}{51} right)^2} ).Given that ( M ) is the median, and ( sum x_i = M ).From Cauchy-Schwarz, we know that ( sum x_i^2 geq frac{M^2}{51} ), so the ratio ( frac{sum x_i^2}{left( frac{M}{51} right)^2} geq 51 ).Therefore, the minimal value of the ratio is 51, achieved when ( sum x_i^2 = frac{M^2}{51} ). But as we saw earlier, this is only possible when ( M = 0 ).In non-zero ( M ) cases, the ratio is higher than 51. Therefore, the minimal ratio across all configurations is 51, achieved when ( M = 0 ).Therefore, the largest ( d ) such that the inequality holds for all configurations is 51.So, the answer is ( boxed{51} ).</think>

🧠:<think>Alright, so I've got this problem here: I need to find the minimum value of ( a^2 - ab + b^2 ) given that ( a ) and ( b ) are real numbers and ( 4 leq a^2 + b^2 leq 9 ). Hmm, okay. Let me try to figure this out step by step.First, I remember that expressions involving ( a^2 ) and ( b^2 ) often relate to circles or ellipses in the coordinate plane. Since ( a^2 + b^2 ) is given, that's the equation of a circle with radius ( r ) where ( r^2 ) is between 4 and 9. So, ( r ) is between 2 and 3. That makes sense.Now, the expression I need to minimize is ( a^2 - ab + b^2 ). Hmm, that looks similar to ( a^2 + b^2 ), but with a subtraction of ( ab ). Maybe I can rewrite this expression in a different form to make it easier to handle.I recall that sometimes completing the square or using trigonometric identities can help simplify such expressions. Let me try expressing ( a ) and ( b ) in terms of polar coordinates since ( a^2 + b^2 = r^2 ) is a natural fit for that. So, let me set ( a = r cos theta ) and ( b = r sin theta ), where ( r ) is between 2 and 3, and ( theta ) is the angle.Substituting these into the expression ( a^2 - ab + b^2 ), I get:[a^2 - ab + b^2 = (r cos theta)^2 - (r cos theta)(r sin theta) + (r sin theta)^2]Simplifying each term:[= r^2 cos^2 theta - r^2 cos theta sin theta + r^2 sin^2 theta]Factor out ( r^2 ):[= r^2 (cos^2 theta - cos theta sin theta + sin^2 theta)]I know that ( cos^2 theta + sin^2 theta = 1 ), so substituting that in:[= r^2 (1 - cos theta sin theta)]Hmm, okay. So now the expression is ( r^2 (1 - cos theta sin theta) ). I need to find the minimum value of this expression given that ( 2 leq r leq 3 ).Looking at ( cos theta sin theta ), I remember that ( sin 2theta = 2 sin theta cos theta ), so ( sin theta cos theta = frac{1}{2} sin 2theta ). Let me substitute that in:[= r^2 left(1 - frac{1}{2} sin 2theta right)]So now, the expression becomes ( r^2 left(1 - frac{1}{2} sin 2theta right) ). To find the minimum value, I need to consider both ( r ) and ( theta ).Since ( r ) is between 2 and 3, and ( sin 2theta ) varies between -1 and 1, the term ( 1 - frac{1}{2} sin 2theta ) will vary. Let's analyze this term:The maximum value of ( sin 2theta ) is 1, and the minimum is -1. So:- When ( sin 2theta = 1 ), ( 1 - frac{1}{2} times 1 = frac{1}{2} )- When ( sin 2theta = -1 ), ( 1 - frac{1}{2} times (-1) = frac{3}{2} )Therefore, the term ( 1 - frac{1}{2} sin 2theta ) varies between ( frac{1}{2} ) and ( frac{3}{2} ).So, the entire expression ( r^2 left(1 - frac{1}{2} sin 2theta right) ) will be minimized when ( r^2 ) is as small as possible and the term ( 1 - frac{1}{2} sin 2theta ) is as small as possible.Wait, but hold on. If I want to minimize the entire expression, I need to consider both factors. Since ( r^2 ) is positive, and the term ( 1 - frac{1}{2} sin 2theta ) is also positive, the minimum occurs when both are minimized.But actually, ( r^2 ) is between 4 and 9, so to minimize the product, I should take the smallest ( r^2 ) and the smallest value of ( 1 - frac{1}{2} sin 2theta ).So, the smallest ( r^2 ) is 4, and the smallest value of ( 1 - frac{1}{2} sin 2theta ) is ( frac{1}{2} ). Therefore, the minimum value of the expression would be ( 4 times frac{1}{2} = 2 ).Wait, but let me double-check that. If ( r^2 = 4 ) and ( sin 2theta = 1 ), then:[4 times left(1 - frac{1}{2} times 1right) = 4 times frac{1}{2} = 2]Yes, that seems correct. So, the minimum value is 2.But just to be thorough, let me consider if there's a case where a larger ( r ) could lead to a smaller value. For example, if ( r^2 = 9 ) and ( sin 2theta = 1 ), then:[9 times left(1 - frac{1}{2} times 1right) = 9 times frac{1}{2} = 4.5]Which is larger than 2, so that doesn't help. Similarly, if ( r^2 = 3 ) (wait, ( r ) is between 2 and 3, so ( r^2 ) is between 4 and 9). So, no, the minimum is indeed when ( r ) is smallest and ( sin 2theta ) is largest.Alternatively, maybe I can approach this problem using algebra without trigonometric substitution. Let me try that.I know that ( a^2 - ab + b^2 ) can be rewritten as ( frac{3}{4}(a - b)^2 + frac{1}{4}(a + b)^2 ). Wait, is that correct? Let me check:Let me expand ( frac{3}{4}(a - b)^2 + frac{1}{4}(a + b)^2 ):[frac{3}{4}(a^2 - 2ab + b^2) + frac{1}{4}(a^2 + 2ab + b^2)][= frac{3}{4}a^2 - frac{3}{2}ab + frac{3}{4}b^2 + frac{1}{4}a^2 + frac{1}{2}ab + frac{1}{4}b^2]Combine like terms:[= left( frac{3}{4} + frac{1}{4} right) a^2 + left( -frac{3}{2} + frac{1}{2} right) ab + left( frac{3}{4} + frac{1}{4} right) b^2][= a^2 - ab + b^2]Yes, that works. So, ( a^2 - ab + b^2 = frac{3}{4}(a - b)^2 + frac{1}{4}(a + b)^2 ). Hmm, interesting. So, it's a sum of squares, which is always non-negative.But how does that help me find the minimum? Maybe I can use the constraint ( 4 leq a^2 + b^2 leq 9 ) to find bounds on ( a^2 - ab + b^2 ).Alternatively, I can consider using the method of Lagrange multipliers since this is a constrained optimization problem. Let me try that approach.Let me define the function to minimize as ( f(a, b) = a^2 - ab + b^2 ), and the constraint is ( g(a, b) = a^2 + b^2 - k = 0 ), where ( k ) is between 4 and 9.Wait, but actually, the constraint is ( 4 leq a^2 + b^2 leq 9 ), so it's an inequality. Hmm, so maybe I need to consider both the interior and the boundary.But since we're looking for the minimum, it's likely to occur on the boundary because the function ( f(a, b) ) is convex. Let me check if ( f(a, b) ) is convex.The Hessian matrix of ( f ) is:[H = begin{bmatrix}2 & -1 -1 & 2end{bmatrix}]The eigenvalues of this matrix are ( 2 - 1 = 1 ) and ( 2 + 1 = 3 ), both positive, so the function is convex. Therefore, the minimum occurs at the boundary of the feasible region, which is when ( a^2 + b^2 = 4 ) or ( a^2 + b^2 = 9 ).But since we're looking for the minimum, it's likely to occur at the smallest possible ( a^2 + b^2 ), which is 4. So, let's set ( a^2 + b^2 = 4 ) and find the minimum of ( f(a, b) ).Using Lagrange multipliers, set up the equations:[nabla f = lambda nabla g]Compute the gradients:[nabla f = begin{bmatrix} 2a - b -a + 2b end{bmatrix}, quad nabla g = begin{bmatrix} 2a 2b end{bmatrix}]So, setting ( nabla f = lambda nabla g ):[2a - b = 2lambda a quad (1)][- a + 2b = 2lambda b quad (2)]And the constraint:[a^2 + b^2 = 4 quad (3)]From equation (1):[2a - b = 2lambda a implies 2a - 2lambda a = b implies b = 2(1 - lambda) a]From equation (2):[- a + 2b = 2lambda b implies -a = 2lambda b - 2b implies -a = 2b(lambda - 1)]But from equation (1), ( b = 2(1 - lambda) a ). Let's substitute this into equation (2):[- a = 2 times 2(1 - lambda) a times (lambda - 1)]Simplify:[- a = 4(1 - lambda)(lambda - 1) a]Note that ( (1 - lambda)(lambda - 1) = -(1 - lambda)^2 ), so:[- a = -4(1 - lambda)^2 a]Multiply both sides by -1:[a = 4(1 - lambda)^2 a]Assuming ( a neq 0 ), we can divide both sides by ( a ):[1 = 4(1 - lambda)^2]Take square roots:[1 - lambda = pm frac{1}{2}]So, two cases:1. ( 1 - lambda = frac{1}{2} implies lambda = frac{1}{2} )2. ( 1 - lambda = -frac{1}{2} implies lambda = frac{3}{2} )Let's consider each case.Case 1: ( lambda = frac{1}{2} )From equation (1):[b = 2(1 - frac{1}{2}) a = 2 times frac{1}{2} a = a]So, ( b = a ). Substitute into constraint (3):[a^2 + a^2 = 4 implies 2a^2 = 4 implies a^2 = 2 implies a = pm sqrt{2}]Thus, ( a = sqrt{2} ), ( b = sqrt{2} ) or ( a = -sqrt{2} ), ( b = -sqrt{2} ).Compute ( f(a, b) = a^2 - ab + b^2 ):[= 2 - (sqrt{2})(sqrt{2}) + 2 = 2 - 2 + 2 = 2]Case 2: ( lambda = frac{3}{2} )From equation (1):[b = 2(1 - frac{3}{2}) a = 2 times (-frac{1}{2}) a = -a]So, ( b = -a ). Substitute into constraint (3):[a^2 + (-a)^2 = 4 implies 2a^2 = 4 implies a^2 = 2 implies a = pm sqrt{2}]Thus, ( a = sqrt{2} ), ( b = -sqrt{2} ) or ( a = -sqrt{2} ), ( b = sqrt{2} ).Compute ( f(a, b) = a^2 - ab + b^2 ):[= 2 - (sqrt{2})(-sqrt{2}) + 2 = 2 + 2 + 2 = 6]So, in this case, the value is 6.Comparing both cases, the minimum value is 2, and the maximum is 6 on the boundary ( a^2 + b^2 = 4 ).Wait, but earlier I thought the minimum could be 2, but I also need to check if a larger ( r ) could give a smaller value. Let me see.Suppose ( a^2 + b^2 = 9 ). Let's try to see what the minimum of ( f(a, b) ) is on this boundary.Using the same method, set up Lagrange multipliers with ( g(a, b) = a^2 + b^2 - 9 = 0 ).Compute gradients:[nabla f = begin{bmatrix} 2a - b -a + 2b end{bmatrix}, quad nabla g = begin{bmatrix} 2a 2b end{bmatrix}]Set ( nabla f = lambda nabla g ):[2a - b = 2lambda a quad (1)][- a + 2b = 2lambda b quad (2)]Same as before. So, following the same steps:From equation (1):[b = 2(1 - lambda) a]From equation (2):[- a = 2b(lambda - 1)]Substitute ( b = 2(1 - lambda) a ):[- a = 4(1 - lambda)(lambda - 1) a = -4(1 - lambda)^2 a]So,[- a = -4(1 - lambda)^2 a implies 1 = 4(1 - lambda)^2 implies (1 - lambda)^2 = frac{1}{4} implies 1 - lambda = pm frac{1}{2}]Thus, ( lambda = frac{1}{2} ) or ( lambda = frac{3}{2} ).Case 1: ( lambda = frac{1}{2} )Then, ( b = a ). Substitute into ( a^2 + b^2 = 9 ):[2a^2 = 9 implies a^2 = frac{9}{2} implies a = pm frac{3}{sqrt{2}}]Compute ( f(a, b) = a^2 - ab + b^2 ):[= frac{9}{2} - frac{9}{2} + frac{9}{2} = frac{9}{2}]Case 2: ( lambda = frac{3}{2} )Then, ( b = -a ). Substitute into ( a^2 + b^2 = 9 ):[2a^2 = 9 implies a^2 = frac{9}{2} implies a = pm frac{3}{sqrt{2}}]Compute ( f(a, b) = a^2 - ab + b^2 ):[= frac{9}{2} - (-frac{9}{2}) + frac{9}{2} = frac{9}{2} + frac{9}{2} + frac{9}{2} = frac{27}{2}]So, on the boundary ( a^2 + b^2 = 9 ), the minimum value of ( f(a, b) ) is ( frac{9}{2} = 4.5 ), which is larger than the minimum we found on the boundary ( a^2 + b^2 = 4 ), which was 2.Therefore, the overall minimum value of ( a^2 - ab + b^2 ) is 2.Wait, but just to make sure, is there any point inside the region ( 4 < a^2 + b^2 < 9 ) where ( f(a, b) ) could be smaller than 2? Since ( f(a, b) ) is convex, the minimum should occur on the boundary, so I think 2 is indeed the minimum.Alternatively, another approach is to note that ( a^2 - ab + b^2 ) can be expressed as ( frac{3}{4}(a - b)^2 + frac{1}{4}(a + b)^2 ), which is always non-negative. But since we have a constraint on ( a^2 + b^2 ), the minimum occurs when ( (a - b)^2 ) is minimized.Wait, but actually, ( frac{3}{4}(a - b)^2 + frac{1}{4}(a + b)^2 ) is equal to ( a^2 - ab + b^2 ), so to minimize this, we need to minimize both terms. However, since both terms are squares, they are non-negative, so the minimum occurs when both are as small as possible.But given the constraint ( a^2 + b^2 geq 4 ), we can't have both ( a ) and ( b ) too small. So, the minimum occurs when ( a^2 + b^2 = 4 ) and ( a = b ), as we saw earlier, leading to ( f(a, b) = 2 ).Another way to think about it is using the Cauchy-Schwarz inequality or other inequalities, but I think the methods I've used so far are sufficient.Just to recap:1. Expressed ( a ) and ( b ) in polar coordinates, transformed the expression, and found that the minimum occurs at ( r = 2 ) and ( sin 2theta = 1 ), giving a minimum value of 2.2. Used Lagrange multipliers on the boundary ( a^2 + b^2 = 4 ) and found the minimum value of 2.3. Checked the other boundary ( a^2 + b^2 = 9 ) and found a higher minimum value of 4.5.4. Considered the expression as a sum of squares and reasoned that the minimum occurs at the smallest possible ( a^2 + b^2 ) with ( a = b ).All approaches lead to the conclusion that the minimum value is 2.I think I'm confident now that 2 is indeed the correct minimum value.</think>

🧠:<think>Okay, so I have this problem here: I need to find how many integer values of x will make a triangle with sides 15, 20, and x have all acute angles. The options are A) 9, B) 10, C) 11, D) 12. Hmm, okay, let me think about how to approach this.First, I remember that for a triangle to have all acute angles, none of its angles can be 90 degrees or more. That means each angle must be less than 90 degrees. I also recall that in a triangle, the largest angle is opposite the longest side. So, if all angles are acute, then the square of the longest side must be less than the sum of the squares of the other two sides. That comes from the Law of Cosines, right?So, let me break this down. I need to consider two cases: one where x is the longest side, and another where 20 is the longest side. Because depending on the value of x, either x or 20 could be the longest side.First, let's consider the triangle inequality theorem. The sum of any two sides must be greater than the third side. So, for sides 15, 20, and x, the inequalities are:1. 15 + 20 > x2. 15 + x > 203. 20 + x > 15Simplifying these:1. 35 > x2. x > 53. x > -5 (which is always true since side lengths are positive)So, combining these, x must be greater than 5 and less than 35. So, x is in (5, 35). But we need integer values, so x can be from 6 to 34 inclusive. That gives us 29 possible integer values for x. But not all of these will result in an acute triangle.Now, let's get back to the acute angle condition. As I thought earlier, if x is the longest side, then x^2 < 15^2 + 20^2. Let me calculate that:15^2 = 22520^2 = 400So, 225 + 400 = 625Therefore, if x is the longest side, x^2 < 625, which means x < sqrt(625) = 25. So, x must be less than 25. But x also has to be greater than 20 because if x is the longest side, it must be greater than 20. So, in this case, x must be in (20, 25). Since x is an integer, x can be 21, 22, 23, 24. That's 4 values.Now, the other case is when 20 is the longest side. That happens when x is less than or equal to 20. In this case, for the triangle to be acute, 20^2 < 15^2 + x^2. Let's compute that:20^2 = 40015^2 = 225So, 400 < 225 + x^2Subtracting 225 from both sides: 175 < x^2Therefore, x^2 > 175Taking square roots: x > sqrt(175) ≈ 13.228Since x must be an integer, x must be at least 14. But x also has to be less than or equal to 20 in this case. So, x can be 14, 15, 16, 17, 18, 19, 20. That's 7 values.Now, combining both cases: when x is the longest side (4 values) and when 20 is the longest side (7 values), we have a total of 4 + 7 = 11 integer values of x that satisfy the conditions.Wait, let me double-check to make sure I didn't miss anything. So, x must be greater than 5 and less than 35, but also, for the triangle to be acute, x must be either between 14 and 20 or between 21 and 24. So, integers from 14 to 24 inclusive. Let's count them: 14,15,16,17,18,19,20,21,22,23,24. That's 11 numbers. Okay, that seems right.I think I considered all the necessary conditions. The triangle inequality gives the range for x, and then the acute angle conditions further restrict x to specific ranges. Combining those gives the total number of integer solutions. So, I'm confident the answer is 11.Final Answerboxed{C}</think>