🧠:Okay, so I need to prove that in an odd-order Latin square (that's a square table with numbers 1 to n where each number appears exactly once in each row and column) which is symmetric with respect to its main diagonal (the one from top-left to bottom-right), the main diagonal must contain all the numbers from 1 to n when n is odd. Hmm, let's start by recalling what a Latin square is and what it means for it to be symmetric about the main diagonal.First, a Latin square of order n has each number from 1 to n exactly once in every row and every column. Symmetry with respect to the main diagonal means that for any cell (i, j), the number there is the same as in cell (j, i). So, if I pick any cell above the diagonal, its mirror image below the diagonal has the same number. The main diagonal itself consists of cells where i = j, so those cells are their own mirrors.Now, the problem states that if n is odd and the Latin square is symmetric about the main diagonal, then the main diagonal must contain all the numbers from 1 to n. Let me think about how to approach this. Maybe I can assume the contrary and reach a contradiction? Or perhaps use some properties of Latin squares and symmetry.Let me consider small cases first. For example, let n = 3, which is odd. If we have a symmetric Latin square of order 3, does the main diagonal necessarily contain all numbers 1, 2, 3?Let me try to construct such a square. Let's say the diagonal has 1, 2, 3. Then the cells (1,1)=1, (2,2)=2, (3,3)=3. Because of symmetry, the cell (1,2) must equal cell (2,1), and similarly for others. Let's attempt to fill it:First row: 1, a, bSecond row: a, 2, cThird row: b, c, 3Since each row and column must have 1,2,3, let's fill in the first row. The first row already has 1, so a and b must be 2 and 3. Let's set a=2, then b=3. Then the first row is [1,2,3]. Then the second row starts with 2 (from symmetry), then has 2 in (2,2), but wait, that would mean the second row has two 2s, which is invalid. So that's a problem.Wait, maybe choosing a different arrangement. Let's try a different diagonal. Suppose the diagonal is [1,1,1]. But that's impossible because each row must have distinct numbers. So the diagonal must have distinct numbers? Wait, but the problem wants to prove that the diagonal contains all numbers, so in the case of n=3, the diagonal must have 1,2,3 in some order. Wait, but in the example above, when I tried to put 1,2,3 on the diagonal, I ran into a conflict. Maybe my approach is wrong.Wait, let me try constructing a symmetric Latin square of order 3. Let's start with the diagonal entries. Let's set (1,1)=1, (2,2)=2, (3,3)=3. Now, since the square is symmetric, cell (1,2) = cell (2,1), cell (1,3)=cell(3,1), cell(2,3)=cell(3,2).Let me try to fill in the first row. The first row is 1, a, b. The second row starts with a, 2, c. The third row starts with b, c, 3.Now, in the first row, we need 1, 2, 3. Since we already have 1, a and b must be 2 and 3. Let's set a=2, then b=3. Then the first row is [1,2,3]. But then the second row starts with 2, and the second element is 2, which is a conflict. So that doesn't work. Therefore, maybe a=3 and b=2. Then first row is [1,3,2]. Then the second row starts with 3, but then the second row needs to have 1,2,3. The first element is 3, second is 2, so third must be 1. So c=1. Then third row starts with 2, has c=1, and third element 3. So third row is [2,1,3]. Now check columns:Column 1: 1,3,2 – which has 1,3,2: okay.Column 2: 3,2,1 – okay.Column 3: 2,1,3 – okay.So the square is:1 3 23 2 12 1 3This is a symmetric Latin square with the diagonal [1,2,3]. So here the diagonal does contain all numbers. If I try to make the diagonal have a repeated number, it's impossible because each row must have distinct numbers. Therefore, for n=3, the diagonal must contain all numbers.Similarly, let's try n=5. Suppose we have a symmetric Latin square of order 5. The claim is that the main diagonal must contain all numbers 1-5. If we assume that one number is missing from the diagonal, say 1 is missing, then each diagonal cell must be 2,3,4,5. But each row and column must still contain 1. However, since the square is symmetric, if 1 is not on the diagonal, then wherever 1 appears in cell (i,j), it must also appear in (j,i). So 1s come in pairs symmetrically around the diagonal. But since n is odd, there are an odd number of cells not on the diagonal. Wait, each pair (i,j) and (j,i) for i≠j accounts for two cells. The total number of off-diagonal cells is n(n-1), which is even when n is odd (since n-1 is even). So 1s would have to be placed in pairs, but the total number of 1s needed is n, which is odd. However, if 1 is not on the diagonal, then all 1s are in pairs, making an even number, but n is odd. Contradiction. Therefore, 1 must appear on the diagonal. Similarly, all numbers must appear on the diagonal.Wait, that seems like a possible approach. Let's formalize this.Suppose that the diagonal does not contain all numbers. Then there exists at least one number k in 1,2,...,n that does not appear on the diagonal. Since the square is symmetric, every occurrence of k must be in pairs (i,j) and (j,i) where i≠j. Therefore, the number of times k appears in the square is twice the number of such pairs. However, in a Latin square, each number must appear exactly n times, once in each row and column. But if k is not on the diagonal, it must appear an even number of times (since each occurrence off the diagonal is paired), but n is odd. This is a contradiction because n is odd and even ≠ odd. Therefore, every number must appear at least once on the diagonal. Since there are exactly n diagonal cells and n numbers, each number must appear exactly once on the diagonal. Hence, the diagonal contains all numbers 1 to n.Wait, that seems to make sense. Let me check for n=3 again. If a number is missing from the diagonal, say 1 is not there, then 1 has to appear in pairs (i,j) and (j,i). But since there are 3 rows and columns, 1 needs to appear 3 times. If none are on the diagonal, it's impossible because 3 is odd, and pairs would give even numbers. Therefore, at least one 1 must be on the diagonal. Similarly for other numbers. Since each number must appear exactly once on the diagonal, the diagonal is a permutation of 1 to n.Therefore, the general proof would be:Assume n is odd. Suppose for contradiction that there exists a number k in {1,2,...,n} that does not appear on the main diagonal. Since the Latin square is symmetric, every occurrence of k must be in pairs (i,j) and (j,i) with i≠j. Thus, the total number of occurrences of k is even. However, in a Latin square, each number appears exactly n times (once per row), and n is odd. This is a contradiction because even ≠ odd. Therefore, every number must appear at least once on the diagonal. Since there are n numbers and n diagonal cells, each number appears exactly once on the diagonal. Hence, the diagonal contains all numbers 1 to n.Yes, that seems correct. Let me verify with another example. Take n=5. Suppose we try to construct a symmetric Latin square without a particular number on the diagonal. Say, number 1. Then 1 must appear in pairs (i,j) and (j,i). But since there are 5 rows, we need 5 occurrences of 1. However, pairing them off would require an even number, but 5 is odd. Hence, impossible. Therefore, one of the 1s must be on the diagonal. The same applies to all numbers. Therefore, each number must be on the diagonal exactly once. Therefore, the diagonal is a permutation of 1 to n.Therefore, the proof is established by counting the parity of the number of occurrences of each number, leveraging the symmetry and the oddness of n.

✅:To prove that the main diagonal of an odd-order symmetric Latin square contains all numbers from 1 to ( n ), we proceed as follows:1. Assumption of Symmetry and Latin Property: Given an ( n times n ) Latin square (( n ) odd) symmetric about the main diagonal, each number ( 1, 2, ldots, n ) appears exactly ( n ) times (once per row and column). Symmetry implies ( a_{i,j} = a_{j,i} ) for all ( i, j ).2. Contradiction via Parity Argument: Suppose a number ( k ) does not appear on the main diagonal. By symmetry, every occurrence of ( k ) must lie in pairs ((i, j)) and ((j, i)) where ( i neq j ). Thus, the total number of ( k )'s is even. However, since ( k ) must appear exactly ( n ) times (an odd number), this leads to a contradiction. Hence, ( k ) must appear at least once on the diagonal.3. Conclusion for All Numbers: Repeating this argument for every ( k in {1, 2, ldots, n} ), each number must occupy at least one diagonal cell. As there are exactly ( n ) diagonal cells and ( n ) distinct numbers, each number appears exactly once on the diagonal.Final Answer: The main diagonal of an odd-order symmetric Latin square must contain all numbers ( 1, 2, ldots, n ). (boxed{text{The diagonal contains all numbers } 1 text{ to } n})

🔑:1. Assumption and Problem Restatement: Given an ( n times n ) square matrix filled with numbers from 1 to ( n ) such that each number appears exactly once in each row and each column, we need to prove that if ( n ) is odd and the matrix is symmetric with respect to the diagonal running from the top-left to the bottom-right corner, then each number from 1 to ( n ) appears exactly once on this diagonal.2. Matrix Symmetry Consideration: Because the matrix is symmetric with respect to the specified diagonal, for any element ( a_{ij} ) located at ( (i, j) ), there exists an equal element ( a_{ji} ) at the symmetric position across the diagonal. This implies ( a_{ij} = a_{ji} ).3. Counting Elements on and off the Diagonal: Since ( n ) is odd, the total number of cells in the matrix is ( n^2 ). There are ( n ) cells on the main diagonal. We examine the distribution of numbers both on and off this diagonal: - Off the diagonal, the matrix contains ( n(n-1) ) cells, all of which contain ( n-1 ) occurrences of each number from 1 to ( n ). - Each number from 1 to ( n ) must appear ( n ) times in total within the entire matrix.4. Implication of Symmetry: Since the matrix is symmetric and each row and column contains all numbers from 1 to ( n ) exactly once, and given the numbers off-diagonal must sum up correctly, we analyze the diagonal elements: - Due to symmetry, for each element ( a_{ij} = a_{ji} ), the numbers off the diagonal come in pairs: - This pairing implies that each number must balance to appear ( n-1 ) times away from the diagonal among an even number of symmetric positions. - Therefore, the remaining single instance of each number must necessarily be on the diagonal.5. Conclusion: By considering the fill constraint (each number from 1 to ( n ) appearing exactly once per row and column), symmetry constraint (( a_{ij} = a_{ji} )), and count balancing (odd vs even distribution), we infer that each number from 1 to ( n ) appears exactly once on the main diagonal of the matrix. Thus, we have proved that in an ( n times n ) matrix (with ( n ) odd), symmetric about the diagonal from the top-left to the bottom-right, all numbers from 1 to ( n ) appear on the diagonal exactly once.[boxed{text{Each number from 1 to ( n ) appears exactly once on the diagonal.}}]

🧠:Okay, so I have this problem here: A square is cut into rectangles. Prove that the sum of the areas of the circles circumscribed around each rectangle is not less than the area of the circle circumscribed around the square. Hmm, interesting. Let me try to work through this step by step.First, let me visualize the problem. There's a square, and it's been divided into several rectangles. For each of these rectangles, we can draw a circumscribed circle, which would be the circle that passes through all four vertices of the rectangle. The area of each such circle is calculated, and we need to show that when we add up all these areas, the total is at least as large as the area of the circumscribed circle around the original square.Alright, so first, let's recall some basic facts. For a rectangle, the circumscribed circle has a diameter equal to the diagonal of the rectangle. That makes sense because the diagonal is the longest distance between two vertices, and the circle's diameter needs to span that to encompass all four corners. So, the radius of the circumscribed circle around a rectangle would be half the length of the diagonal.Given a rectangle with sides of length a and b, the diagonal d is sqrt(a² + b²) by the Pythagorean theorem. Therefore, the radius r is sqrt(a² + b²)/2, and the area of the circle would be πr² = π(a² + b²)/4.So, for each rectangle, the area of its circumscribed circle is π times (a² + b²) divided by 4. Therefore, if we have multiple rectangles resulting from cutting the square, the sum of the areas of their circumscribed circles would be the sum of π(ai² + bi²)/4 for each rectangle i.Our original square has side length, let's say, s. Then its diagonal is s*sqrt(2), so the radius of its circumscribed circle is s*sqrt(2)/2, and the area is π*(s²*2)/4 = πs²/2. So we need to show that the sum over all rectangles of π(ai² + bi²)/4 is at least πs²/2. Since π/4 is a common factor, if we divide both sides by π/4, the inequality reduces to the sum of (ai² + bi²) over all rectangles being at least 2s².Therefore, the problem simplifies to proving that when a square of side length s is divided into rectangles, the sum of (ai² + bi²) for all rectangles is at least 2s². That seems more manageable. Let me confirm this transformation: original inequality is sum(π(ai² + bi²)/4) ≥ πs²/2. Multiply both sides by 4/π: sum(ai² + bi²) ≥ 2s². Yes, that's correct. So now the problem is algebraic: sum of squares of sides of rectangles in a partition of a square is at least 2s².So, how can we approach this? Let's think about the sum of ai² + bi² over all rectangles. Each rectangle has two sides, ai and bi. Since the square is partitioned into these rectangles, the sum of the areas of the rectangles is s². The area of each rectangle is ai*bi, so sum(ai*bi) = s².But we need to relate sum(ai² + bi²) to this. Let's recall that for any rectangle, the quantity ai² + bi² is related to the diagonal. Since we already used that for the circle's radius. But how does this relate to the area?Perhaps we can use the Cauchy-Schwarz inequality or some other inequality to relate these terms. Let me think. For each rectangle, we have ai and bi. Maybe applying the inequality that for positive numbers x and y, x² + y² ≥ 2xy. That's the AM-QM inequality. So, for each rectangle, ai² + bi² ≥ 2ai*bi. Therefore, sum(ai² + bi²) ≥ 2 sum(ai*bi) = 2s². Wait, that's exactly what we need! Because sum(ai*bi) is the total area of the square, which is s². Therefore, sum(ai² + bi²) ≥ 2s². Hence, the required inequality holds.Wait, is that possible? It seems almost too straightforward. Let me check again. For each rectangle, ai² + bi² ≥ 2ai*bi, by AM ≥ GM. Because (ai² + bi²)/2 ≥ (ai*bi), so multiplying both sides by 2, we get ai² + bi² ≥ 2ai*bi. Therefore, summing over all rectangles, sum(ai² + bi²) ≥ 2 sum(ai*bi). Since sum(ai*bi) is the total area, which is s², then sum(ai² + bi²) ≥ 2s². Hence, the inequality holds. Therefore, the sum of the areas of the circumscribed circles is sum(π(ai² + bi²)/4) ≥ π/4 * 2s² = πs²/2, which is the area of the circumscribed circle around the square. Thus, the proof is complete.But wait a minute, is there a catch here? Because when you divide a square into rectangles, each rectangle's sides are parts of the square's sides. So, for example, if the square is divided into smaller rectangles by horizontal and vertical lines, then each rectangle's sides are some fractions of the square's side. However, in this problem, the square is cut into rectangles in any possible way, not necessarily by axis-aligned cuts. Wait, the problem just says "a square is cut into rectangles," so it might be any partition into rectangles, not necessarily with sides parallel to the square's sides. Hmm, but rectangles by definition have sides at right angles, so unless the square is rotated, but even then, the rectangles must have their sides aligned with the square's sides? Wait, actually, no. If you rotate a rectangle, its sides are not aligned with the square's sides. But if the square is partitioned into rectangles, which can be placed at any orientation, then the problem becomes more complex. Wait, but in the problem statement, is there any restriction on the orientation of the rectangles?Wait, the problem says "a square is cut into rectangles." Typically, when such problems are posed, unless specified otherwise, the cuts are along straight lines, but the rectangles can be of any size and position, but in the standard orientation (sides parallel to the square sides). However, the problem doesn't specify, so we might need to consider the general case where the rectangles can be rotated.Wait, but hold on. If the rectangles are allowed to be rotated, then their sides might not be aligned with the square's sides. But in that case, how is the square partitioned into such rectangles? It's a bit more complicated. However, in most partition problems, especially when dealing with axis-aligned rectangles, the assumption is that the rectangles are placed without rotation. So perhaps here, the rectangles are axis-aligned. Let me check the original problem statement again: it just says "a square is cut into rectangles." There's no mention of orientation, so perhaps we need to consider the general case.But in that case, how do we handle rotated rectangles? For example, if a rectangle is rotated inside the square, its own sides would form some angle with the square's sides, and the circumscribed circle would depend on its diagonal. However, the key point here is that regardless of the rectangle's orientation, the circumscribed circle's area depends only on the rectangle's diagonal length. Because the circle is circumscribed around the rectangle, so the diameter is the rectangle's diagonal, regardless of its orientation. Therefore, even if the rectangle is rotated, the diagonal length remains the same. Therefore, the area of the circumscribed circle is still π*(diagonal)^2/4 = π*(a² + b²)/4, where a and b are the side lengths of the rectangle. So even if the rectangle is rotated, the formula holds. Therefore, regardless of the orientation of the rectangles in the square, the area of their circumscribed circles depends only on their side lengths.Therefore, the problem reduces to showing that for any partition of the square into rectangles (of any orientation), the sum over each rectangle of (a_i² + b_i²) is at least 2s², where s is the side length of the square.But in that case, the earlier approach using the AM ≥ GM inequality still holds. Because for each rectangle, regardless of its orientation, a_i² + b_i² ≥ 2a_i b_i. Therefore, summing over all rectangles, sum(a_i² + b_i²) ≥ 2 sum(a_i b_i) = 2s². Therefore, the inequality holds, regardless of how the square is partitioned into rectangles. Therefore, the conclusion follows.But wait, this seems too general. Let me test it with an example. Suppose the square is divided into four smaller squares, each of side length s/2. Then each small square has a_i = b_i = s/2. Then for each small square, a_i² + b_i² = (s²/4 + s²/4) = s²/2. There are four such squares, so the total sum is 4*(s²/2) = 2s², which matches the inequality. So equality holds here.Another example: divide the square into two equal rectangles by cutting it along the middle. Each rectangle is s by s/2. Then for each rectangle, a_i² + b_i² = s² + (s²/4) = 5s²/4. Sum over two rectangles: 2*(5s²/4) = 5s²/2, which is 2.5s², which is greater than 2s². So the inequality holds.Another example: divide the square into n thin rectangles, each of size s by s/n. Then each rectangle has a_i = s, b_i = s/n. Then a_i² + b_i² = s² + s²/n². Sum over n rectangles: n*(s² + s²/n²) = n s² + s²/n. For n ≥ 1, this is always greater than 2s². Let's check n=1: 1*s² + s²/1 = 2s², which is equality. Wait, but n=1 is the original square itself. So if we have just the square, then sum(a² + b²) = 2s². So equality holds when there's only one rectangle, the square itself. For any other partition, the sum is greater. So in this example, when n=2, sum is 2s² + s²/2 = 2.5s², which is greater. For n=3, 3s² + s²/3 ≈ 3.333s², still greater. So seems like the inequality holds.Wait, but let's test a different partition. Suppose we divide the square into rectangles of varying sizes. For example, divide the square into two rectangles: one very thin and one almost square. Let's say one rectangle is s by ε, and the other is s by (s - ε). Then sum(a_i² + b_i²) would be [s² + ε²] + [s² + (s - ε)^2]. Which is s² + ε² + s² + s² - 2sε + ε² = 3s² - 2sε + 2ε². Comparing this to 2s²: 3s² - 2sε + 2ε² - 2s² = s² - 2sε + 2ε². For very small ε, say ε approaching 0, this becomes s² - 0 + 0 = s² > 0. So the sum is still greater. So even if one rectangle is very thin, the sum is still larger than 2s². That seems to hold.Another edge case: divide the square into a single rectangle, which is the square itself. Then sum(a² + b²) = 2s², which is equality. If we divide it into more than one rectangle, sum exceeds 2s².Therefore, the key idea is that the sum of (a_i² + b_i²) over all rectangles is minimized when there's only one rectangle, the original square. For any partition into more rectangles, this sum increases. Hence, the inequality holds.But why does this happen? Because the act of partitioning a rectangle into smaller rectangles necessarily introduces more edges, and the sum of squares tends to increase due to the convexity of the square function. For example, if you split a rectangle into two smaller ones, you replace one pair of sides with two pairs, which may have a larger total sum.Wait, let's formalize this a bit. Suppose we have a rectangle of sides x and y, and we split it into two smaller rectangles. Depending on how we split it, either along the x or y side, the sum of (a_i² + b_i²) may increase.For example, split a rectangle x by y into two rectangles along the x-axis: resulting in two rectangles of sizes a by y and (x - a) by y. Then the original sum is x² + y². After splitting, the sum becomes [a² + y²] + [(x - a)² + y²] = a² + (x - a)^2 + 2y². Compare this to the original x² + y²: The difference is [a² + (x - a)^2 + 2y²] - [x² + y²] = a² + x² - 2ax + a² + 2y² - x² - y² = 2a² - 2ax + y². Hmm, this doesn't immediately look positive. Wait, let's plug in numbers. Let x = 1, y = 1, a = 0.5. Then original sum is 1 + 1 = 2. After splitting, sum is [0.25 + 1] + [0.25 + 1] = 0.25 + 1 + 0.25 + 1 = 2.5. So 2.5 - 2 = 0.5 > 0. So the sum increases.Alternatively, if we split along the y-axis: two rectangles of size x by b and x by (y - b). Then sum becomes [x² + b²] + [x² + (y - b)^2] = 2x² + b² + (y - b)^2. Original sum is x² + y². Difference is 2x² + b² + y² - 2by + b² - x² - y² = x² + 2b² - 2by. If we take x =1, y=1, b=0.5, then difference is 1 + 0.5 - 1 = 0.5, so sum increases by 0.5 again.So splitting a rectangle into two smaller ones increases the sum of squares of sides by some positive amount. Hence, each time you split, the total sum increases. Therefore, the minimal total sum is achieved when there are no splits, i.e., the original square. Therefore, for any partition into multiple rectangles, the sum must be greater than or equal to 2s², with equality only when there's a single rectangle (the square itself).Therefore, the inequality sum(a_i² + b_i²) ≥ 2s² holds, with equality iff the partition is trivial (only the original square). Therefore, translating back to the original problem, the sum of the areas of the circumscribed circles is at least the area of the circumscribed circle around the square.Therefore, the proof is complete. But let me check again if there's any case where splitting might not increase the sum. Suppose we split a square into four smaller squares. Each smaller square has side s/2. Then sum(a_i² + b_i²) for each small square is (s/2)^2 + (s/2)^2 = s²/2. Four such squares give a total sum of 4*(s²/2) = 2s², which is equality. Wait, but splitting a square into four smaller squares requires making two cuts, so the minimal sum is still achieved here? But according to the previous reasoning, splitting should increase the sum.Wait, this seems contradictory. Let me check this example again. Original square: sum(a² + b²) = 2s². After splitting into four smaller squares each of side s/2, each small square contributes ( (s/2)^2 + (s/2)^2 ) = s²/2. Four such squares: 4*(s²/2) = 2s². So total sum is same as original. So in this case, splitting into four squares doesn't change the sum. But according to previous reasoning, splitting should increase the sum. So where is the mistake?Wait, the key is that when you split a rectangle into smaller rectangles, the sum of (a_i² + b_i²) can sometimes stay the same or increase. In the case of splitting a square into four smaller squares, the total sum remains the same. Therefore, equality can be achieved even with non-trivial partitions. Hmm, so my previous conclusion that equality only holds for the trivial partition is incorrect.Wait, this is a problem. So in the example of splitting the square into four equal smaller squares, the sum remains 2s². Therefore, the equality is achieved here as well. Therefore, the inequality is not strict for non-trivial partitions, but there exist non-trivial partitions where equality holds.So, this suggests that the inequality sum(a_i² + b_i²) ≥ 2s² holds, and equality occurs when all the rectangles are squares whose circumscribed circles have the same total area as the original square's circumscribed circle.Wait, but in the case of four smaller squares, each with side s/2, their circumscribed circles would each have radius sqrt( (s/2)^2 + (s/2)^2 ) / 2 = sqrt(2)*(s/2)/2 = s/(2*sqrt(2)). So the area of each small circle is π*(s/(2*sqrt(2)))^2 = π*s²/(8). There are four such circles, so total area is 4*(π*s²/8) = π*s²/2, which is equal to the area of the circumscribed circle around the original square. So indeed, equality holds here.Therefore, the inequality is not strict; there are non-trivial partitions where equality is achieved. Therefore, the correct statement is that the sum is not less than the area of the original circle, and it can be equal even when partitioned into smaller squares.Therefore, the key idea is that for each rectangle, a_i² + b_i² ≥ 2a_i b_i, with equality when a_i = b_i, i.e., when the rectangle is a square. Therefore, sum(a_i² + b_i²) ≥ 2 sum(a_i b_i) = 2s². If all the rectangles are squares, then each term a_i² + b_i² = 2a_i b_i, so the sum becomes exactly 2 sum(a_i b_i) = 2s². Hence, equality holds when all the rectangles are squares. Otherwise, the sum would be greater.Therefore, the conclusion is that regardless of how you partition the square into rectangles, the sum of the areas of their circumscribed circles is at least the area of the circumscribed circle around the original square. Equality holds if and only if all the rectangles are squares (including the trivial case of a single square).But wait, in the example of four smaller squares, they are squares, so equality holds. But if we partition into non-square rectangles, then the sum would be greater. For example, splitting the square into two equal rectangles: each rectangle is s by s/2. Then each has a_i² + b_i² = s² + (s²/4) = 5s²/4. Sum over two rectangles: 5s²/2, which is greater than 2s². So indeed, non-square rectangles lead to a larger sum.Therefore, the proof using the inequality a_i² + b_i² ≥ 2a_i b_i for each rectangle, summing up, and recognizing that sum(a_i b_i) = s² is valid. Hence, the result follows.But let me check another aspect. What if the square is partitioned in a more complex way, like a spiral or something? Does the sum still hold? But since the sum is purely based on each rectangle's side lengths, and regardless of their arrangement, as long as the total area is s², the inequality applies. Because each a_i and b_i are positive real numbers, and the Cauchy-Schwarz or AM-QM inequality applies regardless of their spatial arrangement.Therefore, the key is that the mathematical inequality holds for any partition, because it's a per-rectangle inequality that is then summed over all rectangles. The geometrical arrangement doesn't affect the sum, only the side lengths. Hence, the proof is solid.Therefore, the final answer is that the sum of the areas of the circumscribed circles around each rectangle is indeed not less than the area of the circumscribed circle around the original square, proven by applying the AM-QM inequality to each rectangle and summing up the results.

✅:To prove that the sum of the areas of the circles circumscribed around each rectangle formed by cutting a square is not less than the area of the circle circumscribed around the original square, follow these steps:1. Circumscribed Circle Area Formula: For a rectangle with side lengths (a) and (b), the circumscribed circle has a radius equal to half the diagonal: (frac{sqrt{a^2 + b^2}}{2}). The area of this circle is: [ text{Area} = pi left(frac{sqrt{a^2 + b^2}}{2}right)^2 = frac{pi}{4}(a^2 + b^2). ]2. Original Square's Circumscribed Circle: For a square of side length (s), its diagonal is (ssqrt{2}). The radius of its circumscribed circle is (frac{ssqrt{2}}{2}), and its area is: [ text{Area} = pi left(frac{ssqrt{2}}{2}right)^2 = frac{pi s^2}{2}. ]3. Sum of Areas for Partitioned Rectangles: Let the square be partitioned into (n) rectangles with side lengths ((a_i, b_i)). The total area of their circumscribed circles is: [ sum_{i=1}^n frac{pi}{4}(a_i^2 + b_i^2). ]4. Key Inequality: For each rectangle, by the Arithmetic Mean-Quadratic Mean (AM-QM) inequality: [ a_i^2 + b_i^2 geq 2a_i b_i. ] Summing over all rectangles: [ sum_{i=1}^n (a_i^2 + b_i^2) geq 2 sum_{i=1}^n a_i b_i. ] Since (sum_{i=1}^n a_i b_i = s^2) (total area of the square), we get: [ sum_{i=1}^n (a_i^2 + b_i^2) geq 2s^2. ]5. Final Comparison: Multiply the inequality by (frac{pi}{4}): [ sum_{i=1}^n frac{pi}{4}(a_i^2 + b_i^2) geq frac{pi}{4} cdot 2s^2 = frac{pi s^2}{2}. ] The left side is the total area of the circumscribed circles around the rectangles, and the right side is the area of the original square's circumscribed circle.Conclusion: The sum of the areas of the circumscribed circles around all rectangles formed by partitioning the square is always at least equal to the area of the circumscribed circle around the original square. Equality occurs if and only if all partitioned rectangles are squares themselves.[boxed{text{The sum of the areas of the circumscribed circles around the rectangles is not less than the area of the circumscribed circle around the square.}}]

🔑:1. Let ( s ) be the area of the original square and ( s_1, s_2, ldots, s_n ) be the areas of the constituent rectangles that make up the square.2. Similarly, let ( S ) be the area of the circle circumscribed around the original square, and ( S_1, S_2, ldots, S_n ) be the areas of the circles circumscribed around each of the rectangles.3. Consider one of the rectangles, ( k )-th rectangle, with sides ( a ) and ( b ). The area of the circle circumscribed around this rectangle is given by the formula: [ S_k = frac{1}{4} pi (a^2 + b^2) ]4. The area of this rectangle itself is: [ s_k = ab ]5. Now, to relate the areas of the circles and the rectangles, begin by noting the comparison: [ pi s_k = pi ab ] To establish an upper bound for ( pi ab ), we can employ the arithmetic mean – geometric mean (AM-GM) inequality: [ frac{a^2 + b^2}{2} geq ab implies a^2 + b^2 geq 2ab ] Given that ( ab = s_k ), we get: [ pi s_k = pi ab leq pi frac{a^2 + b^2}{2} ]6. Returning to the area of the circle circumscribed around the rectangle, we multiply our inequality by ( 2 ): [ pi ab leq frac{pi}{2} (a^2 + b^2) = 2 left( frac{1}{4} pi (a^2 + b^2) right) = 2 S_k ] Hence: [ pi s_k leq 2 S_k ]7. Summing this inequality over all the rectangles, we get: [ pi (s_1 + s_2 + cdots + s_n) leq 2 (S_1 + S_2 + cdots + S_n) ]8. Observing properties of the entire structure, the total area of the original square ( s ) is the sum of the areas of its constituent rectangles: [ s = s_1 + s_2 + cdots + s_n ] Taking the circles into account, we know from the formula for the area of the circle circumscribed around a square of side length ( s ): [ S = frac{1}{4} pi (2s) = pi s ]9. Therefore, substituting ( pi s ) with ( 2S ): [ 2 S = pi s = pi (s_1 + s_2 + cdots + s_n) leq 2 (S_1 + S_2 + cdots + S_n) ]Conclusion:[boxed{2S leq 2 left( S_1 + S_2 + cdots + S_n right)}]Thus, we have shown that the sum of the areas of the circles circumscribed around each of the rectangles is at least as large as the area of the circle circumscribed around the original square. blacksquare

🧠:Okay, let's tackle these two problems where I need to determine if the numbers N are prime or composite. Both parts (a) and (b) involve products of four consecutive numbers plus one. Hmm, that structure seems familiar. Maybe there's a pattern or formula that applies here?Starting with part (a): N = 2011 × 2012 × 2013 × 2014 + 1. Let me first compute what this product looks like. But wait, multiplying four consecutive numbers directly would be tedious and probably unnecessary. Maybe there's a smarter way to analyze this.I recall that sometimes expressions like n(n+1)(n+2)(n+3) can be rewritten in terms of squares or other polynomials. Let me see if there's a way to express this product plus one as a perfect square or something similar, which might help in factoring.Let's denote the four consecutive numbers as n, n+1, n+2, n+3. Then the product is n(n+1)(n+2)(n+3). Let me try to pair them: n(n+3) and (n+1)(n+2). Calculating these:n(n+3) = n² + 3n(n+1)(n+2) = n² + 3n + 2So the product becomes (n² + 3n)(n² + 3n + 2) = (n² + 3n)² + 2(n² + 3n). Then adding 1 gives:(n² + 3n)² + 2(n² + 3n) + 1 = (n² + 3n + 1)²Wait, that's a perfect square! Let me check with smaller numbers to verify this pattern. Suppose n = 1:1×2×3×4 +1 = 24 +1 =25 =5². Now, (1² +3×1 +1)² = (1 +3 +1)² =5². Correct.Another test: n=2: 2×3×4×5 +1=120 +1=121=11². Compute (2² +3×2 +1)=4+6+1=11. Squared is 121. Correct again.So the formula seems to hold. Therefore, in general, n(n+1)(n+2)(n+3) +1 = (n² + 3n +1)². That's a perfect square. Therefore, N in part (a) is a perfect square. Since it's a square greater than 1, it must be composite.Wait, but let's confirm with the given numbers. For part (a), n=2011. Then:n² +3n +1 =2011² +3×2011 +1. Let's compute 2011² first. 2011×2011. Let's compute 2000²=4,000,000. Then 11²=121. Then cross terms: 2×2000×11=44,000. So 2011²= (2000+11)²=2000² +2×2000×11 +11²=4,000,000 +44,000 +121=4,044,121. Then add 3×2011=6033. So total is 4,044,121 +6,033 +1=4,050,155. Therefore, N = (4,050,155)². Hence, N is a composite number as it's a square.For part (b): N=2012×2013×2014×2015 +1. Let's apply the same approach. Let n=2012. Then the product is n(n+1)(n+2)(n+3) +1, which according to our previous formula, should be (n² +3n +1)². Therefore, N would be a perfect square and composite.Wait, but let me verify with the actual numbers. Let's compute n² +3n +1 where n=2012.n² =2012². Let's calculate that:2012² = (2000 +12)² =2000² +2×2000×12 +12²=4,000,000 +48,000 +144=4,048,144. Then 3n=3×2012=6,036. So n² +3n +1=4,048,144 +6,036 +1=4,054,181. Therefore, N=(4,054,181)², so it's a perfect square and composite.But wait, both parts (a) and (b) result in perfect squares? That seems surprising. Let me check with another example to ensure I didn't make a mistake. Take n=3:3×4×5×6 +1=360 +1=361=19². Compute n² +3n +1=9 +9 +1=19. Correct. So the formula holds. Therefore, indeed, the product of four consecutive integers plus one is the square of (n² +3n +1). Therefore, both N in (a) and (b) are perfect squares, hence composite.But wait, the problem is presented as two separate parts, so maybe I need to check each individually, but according to the formula, both should be composite. However, maybe I need to verify if the square is a prime. But since the square is of a number greater than 1, the square must be composite. For example, 25=5×5, which is composite.Therefore, both N in (a) and (b) are composite numbers. But let me just confirm with part (a) again. If N=(4050155)^2, then it's 4050155 squared, which is clearly composite. Similarly for part (b), 4054181 squared is composite.Wait, but maybe the number inside the square could be prime? For example, if (n² +3n +1) is prime, then N would be a square of a prime, hence composite. However, even if the number inside is prime, the square is still composite. So regardless of whether (n² +3n +1) is prime or composite, N is composite.But wait, let's check if (n² +3n +1) is prime. For part (a), n=2011: 2011² +3×2011 +1. Let's compute this as 2011*(2011 +3) +1 =2011*2014 +1. Wait, 2011 is a prime number, right? Let me check. Yes, 2011 is a prime. Then 2011*2014 +1=2011*(2*19*53) +1. Wait, 2014=2×19×53. Therefore, 2011×2014 is divisible by 2011, 2,19,53. Then adding 1 would make it not divisible by any of these. So 2011×2014 +1=4050155. Is 4050155 prime? Well, since 4050155 is equal to 2011² +3×2011 +1, which is a quadratic. But I don't know if it's prime. However, regardless, even if it were prime, N would be its square, which is composite. But in reality, 4050155 is likely composite. For instance, 4050155 ends with a 5, so it's divisible by 5. Let me check: 4,050,155 ÷5=810,031. So 4050155=5×810,031. Hence, it's composite, so N is 5²×810,031², which is composite. Similarly for part (b), let's check if 4,054,181 is prime. It ends with a 1, so not divisible by 2 or 5. Let's check divisibility by 3: 4+0+5+4+1+8+1=23, which is not divisible by 3. 7? Let's do the division: 4054181 ÷7. 7×579,168=4,054,176. Subtract: 4,054,181 -4,054,176=5. Not divisible by 7. 11? Alternating sum: (4 +5 +8) - (0 +4 +1)=17 -5=12, not divisible by 11. 13? Let's try 4054181 ÷13. 13×311,860=4,054,180. Subtract: 1. Not divisible by 13. 17? This could take a while. But even if 4,054,181 is prime, N would still be composite as a square. However, given the size, it's more likely composite. But regardless, the question is whether N is prime or composite, and since N is a square greater than 1, it's composite in both cases.So the answer for both (a) and (b) is composite.

✅:Both numbers ( N ) are composite. Explanation:For both parts (a) and (b), ( N ) is expressed as the product of four consecutive integers plus 1. This structure can be generalized as ( N = n(n+1)(n+2)(n+3) + 1 ). By algebraic manipulation, this simplifies to a perfect square:[n(n+1)(n+2)(n+3) + 1 = (n^2 + 3n + 1)^2]Verification:- Part (a): Let ( n = 2011 ). Then: [ N = (2011^2 + 3 cdot 2011 + 1)^2 = 4,!050,!155^2 ] Since ( N ) is a perfect square greater than 1, it is composite.- Part (b): Let ( n = 2012 ). Then: [ N = (2012^2 + 3 cdot 2012 + 1)^2 = 4,!054,!181^2 ] Similarly, ( N ) is a perfect square and thus composite.Conclusion: Both ( N ) in (a) and (b) are composite numbers.[boxed{text{Both } N text{ are composite}}]

🔑:: Is the given number N prime or composite? Consider the following cases:(a) N=2011 cdot 2012 cdot 2013 cdot 2014+1We will first consider whether N is composite.1. Observe that the numbers ( 2011, 2012, 2013, ) and ( 2014 ) are consecutive integers. We want to know whether the product of four consecutive integers plus one, i.e.,[ N = 2011 cdot 2012 cdot 2013 cdot 2014 + 1, ]is composite. We notice a pattern from the problem 7.1 (as referenced) which implies that in general:[ N = (k-1)k(k+1)(k+2) + 1, ]is composite when ( k ) is positive integer. 2. For ( k = 2012 ):[ N = (2012-1) cdot 2012 cdot (2012+1) cdot (2012+2) + 1 = 2011 cdot 2012 cdot 2013 cdot 2014 + 1, ]3. By the properties derived from problem 7.1, ( N ) is composite.Conclusion for (a): ( N ) is composite.(b) N=2012 cdot 2013 cdot 2014 cdot 2015+1We will next check whether this ( N ) is composite.1. Following a general approach similar to part (a): For any natural number ( n ),[ N = (n+1)(n+2)n(n+3) + 1. ]2. We can simplify this product: Let ( M = n^2 + 3n ). Then:[ N = (M)(M+2) + 1 = M^2 + 2M + 1 = (M+1)^2. ]3. Substituting back, ( M = n^2 + 3n ):[ N = (n^2 + 3n + 1)^2. ]4. Since ( N ) is a perfect square in form ( N = (k+1)^2 ) where ( k = n^2 + 3n ), ( N ) is a composite number (a perfect square is never prime unless it is 1).In our case:For ( n = 2012 ), we have:[ N = 2012 cdot 2013 cdot 2014 cdot 2015 + 1 = (2012^2 + 3 cdot 2012 + 1)^2. ]Since it is a perfect square, it is composite.Conclusion for (b): ( N ) is composite.Therefore, the final boxed answers are:[ boxed{text{a) Composite,} quad text{b) Composite.}}]

🧠:Alright, let's tackle this geometry problem. It's about an acute-angled triangle (ABC) where the altitude (AD), the median (BE), and the angle bisector (CF) intersect at a common point (O). We need to find the measure of angle (C) given that (OE = 2 OC). Hmm, okay, let's start by drawing a diagram to visualize the problem. First, in an acute-angled triangle, all altitudes, medians, and angle bisectors will lie inside the triangle. The point (O) where these three lines intersect is crucial here. Since (BE) is a median, it must connect vertex (B) to the midpoint of (AC). The altitude (AD) drops from (A) perpendicular to (BC), and the angle bisector (CF) splits angle (C) into two equal parts. Given that (OE = 2 OC), I need to relate the lengths of these segments to the properties of the triangle. Since (O) is the intersection point, maybe we can use mass point geometry, coordinate geometry, or trigonometric relations. Let me consider coordinate geometry as a possible approach because it allows for precise calculations.Let's set up a coordinate system. Let me place point (C) at the origin ((0, 0)) for simplicity. Then, let’s denote point (B) as ((b, 0)) and point (A) as ((a, h)), where (h) is the height from (A) to (BC). However, since (AD) is the altitude, (D) must lie on (BC), so the coordinates of (D) would be ((a, 0)), but wait, no—if (AD) is the altitude from (A) to (BC), then (D) is the foot of the perpendicular from (A) to (BC). If (BC) is along the x-axis from ((0,0)) to ((c, 0)), then the coordinates might need adjustment. Maybe I should define the coordinates more carefully.Alternatively, perhaps using barycentric coordinates with respect to triangle (ABC) could be helpful since we're dealing with medians, altitudes, and angle bisectors. But barycentric coordinates might complicate things. Let me try a different approach.Let me recall that in a triangle, the centroid, orthocenter, incenter, and circumcenter have specific relations. However, in this problem, the point (O) is the intersection of an altitude, a median, and an angle bisector. Typically, the centroid is the intersection of medians, the orthocenter is the intersection of altitudes, the incenter is the intersection of angle bisectors. So unless the triangle is equilateral, these points are different. However, here (O) is a common point for an altitude, median, and angle bisector. So unless the triangle has some special properties, this point is unique. Therefore, perhaps angle (C) is 60 degrees? Wait, but we need to confirm that.Given that (OE = 2 OC), where (E) is the midpoint of (AC) (since (BE) is the median), and (O) lies somewhere inside the triangle. Let's denote some variables. Let’s let (OC = x), so (OE = 2x). Therefore, the distance from (O) to (C) is (x), and from (O) to (E) is (2x). Since (E) is the midpoint of (AC), maybe we can use coordinate geometry here.Let’s assign coordinates again. Let me place point (C) at ((0, 0)), point (B) at ((2b, 0)) so that the midpoint (E) of (AC) can be calculated easily. Let’s let point (A) be at ((0, 2a)), so the midpoint (E) of (AC) would be at ((0, a)). Wait, but then (BE) is the median from (B) to (E). Hmm, point (E) would be the midpoint of (AC), so if (A) is at ((0, 2a)) and (C) is at ((0,0)), then (E) is at ((0, a)). Then the median (BE) connects (B) at ((2b, 0)) to (E) at ((0, a)). The altitude (AD) from (A) to (BC)—since (BC) is from ((2b, 0)) to ((0,0)), the line (BC) is along the x-axis. The altitude from (A) at ((0, 2a)) to (BC) is vertical, so (D) would be at ((0, 0)), but that's point (C). Wait, that can't be. Because in the problem, (AD) is an altitude, so if (AD) is from (A) to (BC), then (D) is on (BC). But if (A) is at ((0, 2a)) and (BC) is from ((0,0)) to ((2b, 0)), then the altitude from (A) to (BC) is the vertical line dropping from (A) to ((0,0)), which is point (C). But in the problem statement, (AD) is the altitude, so (D) would be (C) in this coordinate system. But then the altitude (AD) coincides with the angle bisector (CF) if (F) is somewhere else. Wait, this seems conflicting. Maybe my coordinate system is not ideal.Let me try a different coordinate setup. Let's place point (C) at ((0, 0)), point (B) at ((c, 0)), and point (A) at ((d, e)). Then, the median (BE) would connect (B) to the midpoint (E) of (AC), which is ((d/2, e/2)). The altitude (AD) from (A) to (BC) would have (D) on (BC). The line (BC) is the x-axis from ((0,0)) to ((c, 0)). The altitude from (A) is perpendicular to (BC), so since (BC) is horizontal, the altitude is vertical. Therefore, the foot (D) is ((d, 0)). But since (D) is on (BC), which is from ((0,0)) to ((c, 0)), we must have (0 leq d leq c). The angle bisector (CF) from (C) to point (F) on (AB). The angle bisector divides the angle at (C) into two equal parts. The coordinates of (F) can be found using the angle bisector theorem, which states that (AF/FB = AC/BC). Wait, the angle bisector theorem: in triangle (ABC), if (CF) is the angle bisector of angle (C), then (AF/FB = AC/BC). So if we let (AF = k) and (FB = m), then (k/m = AC/BC). Given all these, we need to find the coordinates of point (O) where altitude (AD), median (BE), and angle bisector (CF) intersect. Then, using the given condition (OE = 2 OC), find angle (C).This seems quite involved. Let's proceed step by step.First, define coordinates:- Let (C = (0, 0))- Let (B = (c, 0))- Let (A = (d, e))Then, midpoint (E) of (AC) is ((d/2, e/2)).Altitude (AD) is the vertical line from (A) to (D = (d, 0)).Median (BE) connects (B = (c, 0)) to (E = (d/2, e/2)). The equation of median (BE) can be parametrized. Similarly, angle bisector (CF) goes from (C = (0, 0)) to point (F) on (AB). Let's find coordinates of (F).Using angle bisector theorem:(AF / FB = AC / BC)Compute (AC): distance from (A) to (C) is (sqrt{d^2 + e^2})Compute (BC): distance from (B) to (C) is (c)Thus, (AF / FB = sqrt{d^2 + e^2} / c)Let’s parametrize point (F) on (AB). Coordinates of (A) are ((d, e)), coordinates of (B) are ((c, 0)). Let’s use the ratio (AF/FB = k/m = sqrt{d^2 + e^2}/c). Therefore, coordinates of (F) can be written as:(F_x = frac{m cdot d + k cdot c}{k + m})(F_y = frac{m cdot e + k cdot 0}{k + m})But since (AF/FB = k/m = sqrt{d^2 + e^2}/c), let’s set (k = sqrt{d^2 + e^2}) and (m = c). Then:(F_x = frac{c cdot d + sqrt{d^2 + e^2} cdot c}{c + sqrt{d^2 + e^2}})Wait, no. Wait, in the section formula, if (AF/FB = k/m), then (F) divides (AB) in the ratio (k:m). So coordinates of (F) are:(F_x = frac{m cdot d + k cdot c}{k + m})(F_y = frac{m cdot e + k cdot 0}{k + m})Given (k/m = AC / BC = sqrt{d^2 + e^2}/c), so (k = sqrt{d^2 + e^2}), (m = c). Hence:(F_x = frac{c cdot d + sqrt{d^2 + e^2} cdot c}{c + sqrt{d^2 + e^2}} = frac{c(d + sqrt{d^2 + e^2})}{c + sqrt{d^2 + e^2}} = c cdot frac{d + sqrt{d^2 + e^2}}{c + sqrt{d^2 + e^2}})Similarly,(F_y = frac{c cdot e + sqrt{d^2 + e^2} cdot 0}{c + sqrt{d^2 + e^2}} = frac{c e}{c + sqrt{d^2 + e^2}})Therefore, the coordinates of (F) are (left( frac{c(d + sqrt{d^2 + e^2})}{c + sqrt{d^2 + e^2}}, frac{c e}{c + sqrt{d^2 + e^2}} right))Now, the angle bisector (CF) is the line from (C = (0,0)) to (F). Let's find the equation of (CF). The direction vector from (C) to (F) is ((F_x, F_y)), so parametric equations are (x = t F_x), (y = t F_y), where (t in [0, 1]).Now, we need to find the point (O) where altitude (AD), median (BE), and angle bisector (CF) intersect. Let's find equations for these lines and find their intersection.First, altitude (AD) is the vertical line (x = d).Median (BE) connects (B = (c, 0)) to (E = (d/2, e/2)). Let's find its parametric equation. The vector from (B) to (E) is ((d/2 - c, e/2 - 0) = (d/2 - c, e/2)). So parametric equations can be written as:(x = c + (d/2 - c)s)(y = 0 + (e/2)s), where (s in [0, 1])Alternatively, in slope-intercept form, let's compute the equation of BE.The slope of BE is ((e/2 - 0)/(d/2 - c) = (e/2)/(d/2 - c) = e/(d - 2c)So the equation is (y - 0 = [e/(d - 2c)](x - c)), which simplifies to (y = [e/(d - 2c)](x - c))Since altitude (AD) is (x = d), plugging (x = d) into the equation of BE gives:(y = [e/(d - 2c)](d - c))Therefore, the coordinates of (O) from the intersection of BE and AD are ((d, [e/(d - 2c)](d - c)))Now, this point (O) must also lie on the angle bisector (CF). The parametric equations for CF are (x = t F_x), (y = t F_y), so we need to find (t) such that:(d = t F_x)and([e/(d - 2c)](d - c) = t F_y)From the first equation, (t = d / F_x). Substitute into the second equation:([e/(d - 2c)](d - c) = (d / F_x) F_y)Substitute (F_x) and (F_y):([e/(d - 2c)](d - c) = left( d / left( frac{c(d + sqrt{d^2 + e^2})}{c + sqrt{d^2 + e^2}} right) right) cdot left( frac{c e}{c + sqrt{d^2 + e^2}} right))Simplify the right-hand side:First, (d / F_x = d / left( frac{c(d + sqrt{d^2 + e^2})}{c + sqrt{d^2 + e^2}} right) = d cdot frac{c + sqrt{d^2 + e^2}}{c(d + sqrt{d^2 + e^2})} ) = frac{d(c + sqrt{d^2 + e^2})}{c(d + sqrt{d^2 + e^2})})Then multiply by (F_y = frac{c e}{c + sqrt{d^2 + e^2}}):So overall RHS:(frac{d(c + sqrt{d^2 + e^2})}{c(d + sqrt{d^2 + e^2})} cdot frac{c e}{c + sqrt{d^2 + e^2}} = frac{d e}{d + sqrt{d^2 + e^2}})Therefore, the equation becomes:(frac{e(d - c)}{d - 2c} = frac{d e}{d + sqrt{d^2 + e^2}})We can cancel (e) from both sides (assuming (e neq 0), which it is since it's an altitude):(frac{d - c}{d - 2c} = frac{d}{d + sqrt{d^2 + e^2}})Cross-multiplying:((d - c)(d + sqrt{d^2 + e^2}) = d(d - 2c))Let’s expand the left-hand side:(d(d) + d sqrt{d^2 + e^2} - c d - c sqrt{d^2 + e^2})= (d^2 + d sqrt{d^2 + e^2} - c d - c sqrt{d^2 + e^2})Right-hand side is:(d^2 - 2 c d)Therefore, set equal:(d^2 + d sqrt{d^2 + e^2} - c d - c sqrt{d^2 + e^2} = d^2 - 2 c d)Subtract (d^2) from both sides:(d sqrt{d^2 + e^2} - c d - c sqrt{d^2 + e^2} = -2 c d)Bring all terms to the left:(d sqrt{d^2 + e^2} - c d - c sqrt{d^2 + e^2} + 2 c d = 0)Simplify:(d sqrt{d^2 + e^2} + c d - c sqrt{d^2 + e^2} = 0)Factor terms:(sqrt{d^2 + e^2}(d - c) + c d = 0)Hmm, this equation relates (c), (d), and (e). Let’s try to rearrange it:(sqrt{d^2 + e^2}(d - c) = -c d)But the left side is a product of a square root (which is non-negative) and ((d - c)). The right side is (-c d). Since the triangle is acute-angled, all coordinates should be positive. Let me check the signs.Assuming (c > 0), (d > 0), (e > 0). Then:If (d - c) is positive, the left side is positive, but the right side is negative, which can't be. If (d - c) is negative, then left side is negative (since (sqrt{d^2 + e^2} > 0)), and right side is (-c d), which is negative. So possible.Thus, (d - c < 0), so (d < c). Therefore, we can write:(sqrt{d^2 + e^2}(c - d) = c d)Divide both sides by (c):(sqrt{d^2 + e^2}(1 - d/c) = d)Let’s let (k = d/c), where (k < 1). Then:(sqrt{(k c)^2 + e^2}(1 - k) = k c)Square both sides to eliminate the square root:((k^2 c^2 + e^2)(1 - k)^2 = k^2 c^2)Expand the left side:( (k^2 c^2 + e^2)(1 - 2k + k^2) = k^2 c^2 )Multiply out:(k^2 c^2 (1 - 2k + k^2) + e^2 (1 - 2k + k^2) = k^2 c^2)Expand term by term:First term: (k^2 c^2 - 2k^3 c^2 + k^4 c^2)Second term: (e^2 - 2k e^2 + k^2 e^2)So combining:(k^2 c^2 - 2k^3 c^2 + k^4 c^2 + e^2 - 2k e^2 + k^2 e^2 = k^2 c^2)Subtract (k^2 c^2) from both sides:(-2k^3 c^2 + k^4 c^2 + e^2 - 2k e^2 + k^2 e^2 = 0)Rearrange:(k^4 c^2 - 2k^3 c^2 + k^2 e^2 - 2k e^2 + e^2 = 0)Factor terms where possible:Let’s factor (e^2) terms:(e^2(k^2 - 2k + 1) = e^2(k - 1)^2)And (c^2) terms:(c^2(k^4 - 2k^3) = c^2 k^3(k - 2))Wait, but combining all together:(k^4 c^2 - 2k^3 c^2 + k^2 e^2 - 2k e^2 + e^2 = 0)Hmm, perhaps we can group terms:(k^4 c^2 - 2k^3 c^2 + k^2 e^2 + (-2k e^2 + e^2))Alternatively, factor terms with (c^2) and (e^2):(c^2 k^2(k^2 - 2k) + e^2(k^2 - 2k + 1) = 0)Note that (k^2 - 2k + 1 = (k - 1)^2), so:(c^2 k^2(k^2 - 2k) + e^2(k - 1)^2 = 0)But since (c), (e), (k) are positive (with (k <1)), and ((k - 1)^2) is positive, both terms are positive (since (k^2 - 2k = k(k - 2)); but (k <1), so (k - 2 <0), so (k(k - 2)) is negative. Therefore, (c^2 k^2(k^2 - 2k)) is negative, and (e^2(k -1)^2) is positive. So the equation is a negative term plus a positive term equals zero. Let's write it as:( - c^2 k^2(2k - k^2) + e^2(1 - k)^2 = 0 )So,( e^2(1 - k)^2 = c^2 k^2(2k - k^2) )Take square roots? Hmm, perhaps not. Let's express (e^2) in terms of (c^2):( e^2 = frac{c^2 k^2(2k - k^2)}{(1 - k)^2} )But (e^2) must be positive, so the numerator and denominator must have the same sign. Since (k <1), denominator ((1 - k)^2) is positive. The numerator (k^2(2k - k^2)) must also be positive. So (2k - k^2 > 0), which implies (k(2 - k) > 0). Since (k >0) and (k <1), (2 -k >1 >0), so the numerator is positive. Therefore, this is valid.Now, recall that in triangle (ABC), angle (C) is at the origin, so coordinates are (C(0,0)), (B(c,0)), (A(d,e)). The angle at (C) is the angle between vectors (CB) and (CA). Vector (CB) is ((c, 0)), vector (CA) is ((d, e)). The angle (C) can be found via the dot product formula:(cos angle C = frac{CB cdot CA}{|CB||CA|})Compute the dot product:(CB cdot CA = c cdot d + 0 cdot e = c d)The magnitudes:(|CB| = c)(|CA| = sqrt{d^2 + e^2})Therefore,(cos angle C = frac{c d}{c sqrt{d^2 + e^2}} = frac{d}{sqrt{d^2 + e^2}})So, angle (C = arccosleft( frac{d}{sqrt{d^2 + e^2}} right))But from our previous equation, we have (e^2 = frac{c^2 k^2(2k - k^2)}{(1 - k)^2}), where (k = d/c). Let’s substitute (d = k c), so (e^2 = frac{c^2 k^2(2k - k^2)}{(1 - k)^2}). Therefore,(sqrt{d^2 + e^2} = sqrt{k^2 c^2 + frac{c^2 k^2(2k - k^2)}{(1 - k)^2}} = c k sqrt{1 + frac{2k - k^2}{(1 - k)^2}})Simplify the term inside the square root:(1 + frac{2k - k^2}{(1 - k)^2} = frac{(1 - k)^2 + 2k - k^2}{(1 - k)^2})Expand ((1 - k)^2 = 1 - 2k + k^2), so numerator:(1 - 2k + k^2 + 2k - k^2 = 1)Therefore,(sqrt{d^2 + e^2} = c k sqrt{frac{1}{(1 - k)^2}} = frac{c k}{1 - k})Therefore,(cos angle C = frac{d}{sqrt{d^2 + e^2}} = frac{k c}{(c k)/(1 - k)} = 1 - k)So angle (C = arccos(1 - k))But we need to find angle (C) in terms of the given condition (OE = 2 OC). Let's recall that (O) is the intersection point, and we need to relate (OE) and (OC). From earlier, we have coordinates of (O) as ((d, [e/(d - 2c)](d - c))). Let's compute the distances (OE) and (OC).First, coordinates of (E) are ((d/2, e/2)). Coordinates of (O) are ((d, y_O)), where (y_O = [e/(d - 2c)](d - c)). Coordinates of (C) are ((0, 0)).Compute (OC):Distance from (O) to (C) is (sqrt{(d - 0)^2 + (y_O - 0)^2} = sqrt{d^2 + y_O^2})Compute (OE):Distance from (O) to (E) is (sqrt{(d - d/2)^2 + (y_O - e/2)^2} = sqrt{(d/2)^2 + (y_O - e/2)^2})Given that (OE = 2 OC), so:(sqrt{(d/2)^2 + (y_O - e/2)^2} = 2 sqrt{d^2 + y_O^2})Square both sides:((d/2)^2 + (y_O - e/2)^2 = 4(d^2 + y_O^2))Expand:(d^2/4 + y_O^2 - e y_O + e^2/4 = 4 d^2 + 4 y_O^2)Bring all terms to the left:(d^2/4 + y_O^2 - e y_O + e^2/4 - 4 d^2 - 4 y_O^2 = 0)Simplify:(-15 d^2 /4 - 3 y_O^2 - e y_O + e^2 /4 = 0)Multiply both sides by -4:(15 d^2 + 12 y_O^2 + 4 e y_O - e^2 = 0)Hmm, this seems complicated. But perhaps we can substitute (y_O = [e/(d - 2c)](d - c)) into this equation. Let's do that.Let (y_O = frac{e(d - c)}{d - 2c})Then,Substitute into the equation:(15 d^2 + 12 left( frac{e(d - c)}{d - 2c} right)^2 + 4 e cdot frac{e(d - c)}{d - 2c} - e^2 = 0)Let’s simplify term by term:First term: (15 d^2)Second term: (12 cdot frac{e^2 (d - c)^2}{(d - 2c)^2})Third term: (4 e cdot frac{e(d - c)}{d - 2c} = frac{4 e^2 (d - c)}{d - 2c})Fourth term: (-e^2)Therefore, equation becomes:(15 d^2 + frac{12 e^2 (d - c)^2}{(d - 2c)^2} + frac{4 e^2 (d - c)}{d - 2c} - e^2 = 0)Let’s factor (e^2) terms:(15 d^2 + e^2 left( frac{12 (d - c)^2}{(d - 2c)^2} + frac{4 (d - c)}{d - 2c} - 1 right) = 0)From earlier, we have an expression for (e^2) in terms of (c) and (d) (from the angle bisector intersection condition):Recall that (e^2 = frac{c^2 k^2(2k - k^2)}{(1 - k)^2}) where (k = d/c). Let's substitute (k = d/c) into this. Let (k = d/c), so (d = k c). Then (e^2 = frac{c^2 (k c / c)^2 (2k - (k c /c)^2)}{(1 - k)^2} = frac{c^2 k^2 (2k - k^2)}{(1 - k)^2}). So,(e^2 = frac{c^2 k^2 (2k - k^2)}{(1 - k)^2})Therefore, substitute into the equation:(15 (k c)^2 + frac{c^2 k^2 (2k - k^2)}{(1 - k)^2} left( frac{12 (k c - c)^2}{(k c - 2c)^2} + frac{4 (k c - c)}{k c - 2c} - 1 right) = 0)Simplify each term inside the big bracket:First term: (frac{12 (k c - c)^2}{(k c - 2c)^2} = frac{12 c^2 (k - 1)^2}{c^2 (k - 2)^2} = frac{12 (1 - k)^2}{(2 - k)^2})Second term: (frac{4 (k c - c)}{k c - 2c} = frac{4 c (k - 1)}{c (k - 2)} = frac{4 (k - 1)}{k - 2} = frac{4 (1 - k)}{2 - k})Third term: (-1)Thus, the equation becomes:(15 k^2 c^2 + frac{c^2 k^2 (2k - k^2)}{(1 - k)^2} left( frac{12 (1 - k)^2}{(2 - k)^2} + frac{4 (1 - k)}{2 - k} - 1 right) = 0)Factor out (c^2):(c^2 left[ 15 k^2 + frac{k^2 (2k - k^2)}{(1 - k)^2} left( frac{12 (1 - k)^2}{(2 - k)^2} + frac{4 (1 - k)}{2 - k} - 1 right) right] = 0)Since (c^2 neq 0), the expression inside the brackets must be zero:(15 k^2 + frac{k^2 (2k - k^2)}{(1 - k)^2} left( frac{12 (1 - k)^2}{(2 - k)^2} + frac{4 (1 - k)}{2 - k} - 1 right) = 0)Let’s compute the term inside the big parenthesis:Let’s denote (A = frac{12 (1 - k)^2}{(2 - k)^2} + frac{4 (1 - k)}{2 - k} - 1)Compute (A):First term: (12 frac{(1 - k)^2}{(2 - k)^2})Second term: (4 frac{(1 - k)}{(2 - k)})Third term: (-1)Let’s compute this step by step. Let’s set (t = 1 - k), then (2 - k = 1 + t). Let’s see if substitution helps.Alternatively, find a common denominator for the terms. The common denominator would be ((2 - k)^2).Thus,First term: (12 (1 - k)^2 / (2 - k)^2)Second term: (4 (1 - k)(2 - k) / (2 - k)^2)Third term: (- (2 - k)^2 / (2 - k)^2)Therefore,(A = [12(1 - k)^2 + 4(1 - k)(2 - k) - (2 - k)^2] / (2 - k)^2)Compute numerator:Expand each term:12(1 - k)^2 = 12(1 - 2k + k^2) = 12 - 24k + 12k^24(1 - k)(2 - k) = 4[(2 - k - 2k + k^2)] = 4(2 - 3k + k^2) = 8 - 12k + 4k^2-(2 - k)^2 = -(4 - 4k + k^2) = -4 + 4k - k^2Sum all terms:12 -24k +12k^2 +8 -12k +4k^2 -4 +4k -k^2Combine like terms:Constants: 12 +8 -4 =16Linear terms: -24k -12k +4k = -32kQuadratic terms:12k^2 +4k^2 -k^2 =15k^2So numerator: 15k^2 -32k +16Therefore,(A = (15k^2 -32k +16)/(2 -k)^2)Therefore, the equation becomes:(15k^2 + frac{k^2 (2k -k^2)}{(1 -k)^2} cdot frac{15k^2 -32k +16}{(2 -k)^2} =0)Let’s write this as:(15k^2 + frac{k^2 (2k -k^2)(15k^2 -32k +16)}{(1 -k)^2 (2 -k)^2} =0)This is a complicated equation in terms of (k). Let’s denote:(15k^2 + frac{k^2 (2k -k^2)(15k^2 -32k +16)}{(1 -k)^2 (2 -k)^2} =0)Multiply both sides by ((1 -k)^2 (2 -k)^2) to eliminate the denominator:(15k^2 (1 -k)^2 (2 -k)^2 + k^2 (2k -k^2)(15k^2 -32k +16) =0)Factor out (k^2):(k^2 [15(1 -k)^2 (2 -k)^2 + (2k -k^2)(15k^2 -32k +16)] =0)Since (k neq 0), we have:(15(1 -k)^2 (2 -k)^2 + (2k -k^2)(15k^2 -32k +16) =0)Expand the first term:First, compute ((1 -k)^2 (2 -k)^2). Let’s expand ((1 -k)(2 -k)) first:((1 -k)(2 -k) = 2 -k -2k +k^2 =2 -3k +k^2)Then square it:((2 -3k +k^2)^2 =4 -12k +13k^2 -6k^3 +k^4)Wait, let me compute it properly:Let’s compute ((a + b + c)^2) where a=2, b=-3k, c=k^2.Wait, actually, the square of ( (2 -3k +k^2) ):= (2)^2 + (-3k)^2 + (k^2)^2 + 2*(2*(-3k) + 2*(k^2) + (-3k)*(k^2))=4 +9k^2 +k^4 +2*(-6k +2k^2 -3k^3)=4 +9k^2 +k^4 -12k +4k^2 -6k^3Combine like terms:=4 -12k + (9k^2 +4k^2) + (-6k^3) +k^4=4 -12k +13k^2 -6k^3 +k^4Therefore,15*(4 -12k +13k^2 -6k^3 +k^4) =60 -180k +195k^2 -90k^3 +15k^4Now, compute the second term: (2k -k^2)(15k^2 -32k +16)First, expand 2k*(15k^2 -32k +16) =30k^3 -64k^2 +32kThen, expand -k^2*(15k^2 -32k +16) =-15k^4 +32k^3 -16k^2Combine:30k^3 -64k^2 +32k -15k^4 +32k^3 -16k^2 =-15k^4 + (30k^3 +32k^3) + (-64k^2 -16k^2) +32k= -15k^4 +62k^3 -80k^2 +32kTherefore, the entire expression becomes:60 -180k +195k^2 -90k^3 +15k^4 -15k^4 +62k^3 -80k^2 +32k =0Combine like terms:-15k^4 +15k^4 =0-90k^3 +62k^3 =-28k^3195k^2 -80k^2=115k^2-180k +32k=-148k+60So the equation simplifies to:-28k^3 +115k^2 -148k +60=0Multiply both sides by -1:28k^3 -115k^2 +148k -60=0Now, we need to solve this cubic equation:28k³ -115k² +148k -60=0Let’s attempt to find rational roots using Rational Root Theorem. Possible roots are factors of 60 divided by factors of 28. Possible candidates: ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60 divided by 1,2,4,7,14,28.So possible rational roots include: 1, 1/2, 1/4, 1/7, 2, 3, 3/2, 5, 5/2, etc. Let’s test k=1:28(1) -115(1) +148(1) -60=28-115+148-60= (28+148)+( -115-60)=176 -175=1≠0k=2: 28*8 -115*4 +148*2 -60=224 -460 +296 -60= (224+296)-(460+60)=520-520=0. Hey, k=2 is a root!Therefore, we can factor out (k-2):Using polynomial division or synthetic division.Divide 28k³ -115k² +148k -60 by (k -2).Using synthetic division:2 | 28 -115 148 -60 56 -118 60 28 -59 30 0So the quotient is 28k² -59k +30. Therefore,28k³ -115k² +148k -60=(k-2)(28k² -59k +30)Now, factor 28k² -59k +30. Let's compute discriminant D=59² -4*28*30=3481 -3360=121=11²Roots: k=(59±11)/56So k=(59+11)/56=70/56=5/4=1.25, and k=(59-11)/56=48/56=6/7≈0.857Therefore, the roots are k=2, k=5/4, k=6/7But recall that in our problem, (k = d/c <1) because (d <c). So k=2 and k=5/4 are greater than 1, which would contradict (k <1). Hence, the only feasible root is k=6/7≈0.857.Therefore, k=6/7.Now, recall that angle (C = arccos(1 -k)). Since k=6/7,1 -k=1 -6/7=1/7Thus,angle (C = arccos(1/7))But the problem asks to find angle (C). However, we need to check if this is correct and if there are any calculation errors.Let’s verify step by step:1. We derived the equation relating k and ended up with k=6/7.2. Then, angle C= arccos(1 -k)=arccos(1/7). Let’s compute arccos(1/7) in degrees. 1/7≈0.1429, so angle≈81.79 degrees. But let's verify if this is correct with the given condition OE=2OC.But wait, the problem states that in an acute-angled triangle. If angle C is approximately 81.79 degrees, then the triangle is acute-angled as all angles would be less than 90 degrees. But we need to confirm whether this angle satisfies OE=2OC.Alternatively, perhaps there is a miscalculation. Let's check the steps again.We had:After substituting all conditions, we arrived at angle C= arccos(1 -k), where k=6/7. So 1 -k=1/7, so angle C≈81.79 degrees. But does this satisfy OE=2OC?Alternatively, let's check with coordinate system.Given k=6/7, which is d/c=6/7. Let’s set c=7, so d=6. Then, from the earlier equation, e²= c² k²(2k -k²)/(1 -k)^2Compute e²:c=7, k=6/7:e²=7²*(6/7)^2*(2*(6/7) - (6/7)^2)/(1 -6/7)^2Compute step by step:e²=49*(36/49)*(12/7 -36/49)/(1/7)^2Simplify:=49*(36/49)*( (84/49 -36/49) )/(1/49)=36*(48/49)/(1/49)=36*48=1728Thus, e²=1728, so e=√1728=12√12=12*2√3=24√3. Wait, √1728=√(144*12)=12√12=12*2√3=24√3. Yes.Therefore, coordinates:C(0,0), B(7,0), A(6,24√3)Check if this satisfies the given condition OE=2 OC.Compute point O as the intersection of altitude AD, median BE, and angle bisector CF.First, altitude AD is vertical line x=6, from A(6,24√3) to D(6,0).Median BE connects B(7,0) to E, midpoint of AC. Coordinates of E: midpoint of A(6,24√3) and C(0,0) is (3,12√3).Equation of BE: from (7,0) to (3,12√3). The slope is (12√3 -0)/(3 -7)=12√3 / (-4)= -3√3.Equation: y -0= -3√3(x -7)So y= -3√3 x +21√3Intersection with AD (x=6):y= -3√3*6 +21√3= -18√3 +21√3=3√3Thus, O is at (6,3√3)Compute OC: distance from O(6,3√3) to C(0,0)=√(6² + (3√3)^2)=√(36 +27)=√63=3√7Compute OE: distance from O(6,3√3) to E(3,12√3)=√[(6-3)^2 + (3√3 -12√3)^2]=√[9 + (-9√3)^2]=√[9 +243]=√252=√(36*7)=6√7Indeed, OE=6√7 and OC=3√7, so OE=2 OC. This satisfies the given condition.Therefore, angle C= arccos(1/7). But the problem asks to find angle C. However, 1/7 is not a standard cosine value. Is there a calculation mistake?Wait, angle C was defined as arccos(d / sqrt(d² + e²))= arccos(k c / sqrt( (k c)^2 + e² )). Earlier, we found that sqrt(d² + e²)=c k / (1 -k). Therefore, d / sqrt(d² + e²)= (k c)/ (c k / (1 -k))=1 -k. Therefore, angle C= arccos(1 -k). Given that k=6/7, then angle C= arccos(1 -6/7)= arccos(1/7). But 1/7 is approximately 0.1429, which corresponds to an angle of approximately 81.79 degrees. However, the problem might expect an exact value. Is arccos(1/7) a special angle? Probably not. Let me check my steps again to see if there's a miscalculation.Wait, but when we set up coordinate system with C at (0,0), B at (c,0), A at (d,e), we derived angle C= arccos(1 -k) where k= d/c. However, in the case where k=6/7, angle C= arccos(1/7). But in our specific coordinate example with c=7, d=6, e=24√3, we can compute angle C using coordinates.Vector CB is from C(0,0) to B(7,0): (7,0)Vector CA is from C(0,0) to A(6,24√3): (6,24√3)The angle at C is the angle between these two vectors.Dot product: 7*6 +0*24√3=42|CB|=7|CA|=sqrt(6² + (24√3)^2)=sqrt(36 + 1728)=sqrt(1764)=42Therefore, cos(angle C)= dot product/( |CB| |CA| )=42/(7*42)=1/7Therefore, angle C= arccos(1/7)≈81.79 degrees. However, the problem asks to find angle C. Since 1/7 is not a standard angle, perhaps the answer is 60 degrees? But in our example, it's approximately 81.79 degrees, which contradicts.Wait, maybe there was an error in assuming coordinate system. Let me check again.Wait, in our coordinate system, we found that with k=6/7, the given condition OE=2 OC is satisfied, and angle C is arccos(1/7). However, the problem states that the triangle is acute-angled. In our example, angle C≈81.79 degrees, and other angles can be computed.Compute other angles:Coordinates: A(6,24√3), B(7,0), C(0,0)Compute sides:AB: distance between A(6,24√3) and B(7,0)=sqrt( (7-6)^2 + (0 -24√3)^2 )=sqrt(1 + 1728)=sqrt(1729)BC:7AC:42Check angles:Using Law of Cosines:Angle at A: cos(angle A)= (AB² + AC² - BC²)/(2 AB * AC)= (1729 + 42² -7²)/(2*sqrt(1729)*42)Compute 42²=1764, 7²=49= (1729 +1764 -49)/(2*42*sqrt(1729))= (1729+1715)/(84 sqrt(1729))=3444/(84 sqrt(1729))=41/sqrt(1729)Which is a small angle, acute.Angle at B: Similarly, using Law of Cosines:cos(angle B)= (AB² + BC² - AC²)/(2 AB * BC)= (1729 +49 -1764)/(2*sqrt(1729)*7)= (14)/(14 sqrt(1729))=1/sqrt(1729). Thus, angle B≈ close to 90 degrees but slightly less, so acute.Thus, all angles are acute. Therefore, angle C≈81.79 degrees is acceptable. However, the problem likely expects an exact value. Since 1/7 is not a standard cosine value, perhaps there's an error in the approach.Alternatively, maybe in the problem, angle C is 60 degrees, and when you compute, OE=2 OC.Wait, let's test if angle C=60 degrees satisfies the condition.Assume angle C=60 degrees. Then, cos(angle C)=0.5=1 -k, so k=1 -0.5=0.5. So d/c=0.5, so d=0.5c.Then, e²= c² k²(2k -k²)/(1 -k)^2= c²*(0.25)*(1 -0.25)/(0.5)^2= c²*0.25*0.75/0.25= c²*0.75Thus, e= (c * sqrt(3))/2. So coordinates:C(0,0), B(c,0), A(0.5c, (sqrt(3)/2)c)This is an equilateral triangle? Wait, no. If angle C=60 degrees, and coordinates of A are (0.5c, (sqrt(3)/2)c), then distance AC= sqrt( (0.5c)^2 + ( (sqrt(3)/2)c )^2 )=sqrt(0.25c² +0.75c²)=sqrt(c²)=c. So AC=c, BC=c, angle C=60 degrees. So it's an equilateral triangle. But in an equilateral triangle, all medians, altitudes, and angle bisectors coincide. So point O would be the centroid, orthocenter, incenter, and circumcenter. In that case, OC= 1/3 of the median, and OE= distance from centroid to midpoint, which is also 1/3 of the median? Wait, in an equilateral triangle, the centroid divides the median in a 2:1 ratio. So if OE is from centroid to midpoint, which would be 1/3 the length. But in the problem, OE=2 OC. In an equilateral triangle, if OC is the distance from centroid to vertex C, then OC=2/3 of the median length. While OE, distance from centroid to midpoint E, is 1/3 of the median length. Therefore, OE=1/3, OC=2/3, so OE=0.5 OC, which is the opposite of the given condition OE=2 OC. Therefore, angle C=60 degrees does not satisfy the problem's condition.Thus, our previous result of angle C= arccos(1/7)≈81.79 degrees is the solution. However, the problem might expect an answer in terms of inverse trigonometric functions, but generally, competition problems like this often have nice angles. Hence, perhaps there's a miscalculation or a different approach.Wait, but according to our detailed coordinate analysis, the answer is angle C= arccos(1/7). However, this is approximately 81.79 degrees, which is not a standard angle. Maybe the problem is designed to have an answer of 60 degrees, but our calculation shows otherwise. Alternatively, perhaps there's a synthetic geometry solution without coordinates.Let me try a synthetic approach.Given that in triangle ABC, altitude AD, median BE, and angle bisector CF concur at O, with OE=2 OC. Need to find angle C.Let’s recall some properties:1. In a triangle, the centroid divides the median in ratio 2:1. However, here O is not necessarily the centroid because it's also on an altitude and angle bisector.2. Since O is on the angle bisector CF, by the angle bisector theorem, the ratio of distances from O to the sides is equal to the ratio of the adjacent sides. But since O is also on the median BE and altitude AD, perhaps we can use mass point geometry or Ceva's theorem.Let’s consider using Ceva's theorem. For concurrency of AD, BE, CF, Ceva's condition must hold.But AD is an altitude, BE is a median, CF is an angle bisector. Ceva's Theorem states that for concurrent lines:(AF/FB) * (BD/DC) * (CE/EA) =1Given that BE is the median, E is the midpoint of AC, so CE=EA=1/2 AC. Therefore, CE/EA=1.BD/DC: Since AD is an altitude, D is the foot from A to BC. BD/DC= ?AF/FB: CF is the angle bisector, so by angle bisector theorem, AF/FB=AC/BC.Therefore, Ceva's condition gives:(AF/FB)*(BD/DC)*(CE/EA)=(AC/BC)*(BD/DC)*1=1Therefore,(AC/BC)*(BD/DC)=1 => BD/DC=BC/ACBut BD + DC=BC. Let’s denote BD= x, DC= y. Then x + y= BC, and x/y= BC/AC => x= (BC/AC) yTherefore,(BC/AC)y + y = BC => y( BC/AC +1 )=BC => y= BC / ( BC/AC +1 )= (BC * AC)/(BC + AC)Therefore,DC= y= (BC * AC)/(BC + AC)Similarly, BD= (BC/AC) y= (BC/AC)*(BC * AC)/(BC + AC)= BC²/(BC + AC)But how does this help in finding angle C?We also know that O is the intersection point, and OE=2 OC.Let’s consider the positions of O on the median BE and on the angle bisector CF.Since E is the midpoint of AC, and BE is a median, in mass point terms, the centroid divides the median in a 2:1 ratio. But here, O is not the centroid because it also lies on the altitude and angle bisector. However, given OE=2 OC, perhaps mass point geometry can be applied.Let’s assign masses to the points such that the masses are inversely proportional to the lengths. Let’s let mass at C be m, then mass at E should be such that the ratio OE/OC=2/1, so the mass at C times OC equals mass at E times OE. Therefore, m * OC = mass_E * OE => mass_E= m * (OC/OE)= m*(1/2). Therefore, mass at E is m/2.But E is the midpoint of AC, so masses at A and C must be equal to make E the center. Since E is the midpoint, masses at A and C must be equal. Let mass at A be equal to mass at C=m. Therefore, mass at E is m +m=2m. But previously we got mass at E=m/2. Contradiction.Wait, perhaps mass point approach here is tricky due to the different cevians intersecting. Alternatively, use coordinate geometry again but in a normalized system.Alternatively, let’s consider the ratios. Since OE=2 OC, and E is the midpoint of AC, which is a key point. Maybe use vectors.Let me consider vector approach. Let’s place point C at the origin, vector c=0. Let’s denote vector b=CB, vector a=CA. Then, midpoint E of AC is (a +0)/2= a/2. The median BE connects point B (vector b) to E (a/2). The altitude AD from A must be perpendicular to BC. Since BC is vector b -0= b, the direction of BC is vector b. The altitude from A to BC is perpendicular to vector b. The angle bisector CF: since it's the angle bisector of angle C, it divides AB in the ratio of AC to CB. By the angle bisector theorem, AF/FB= AC/CB= |a|/|b|.The point O is the intersection of the three cevians. Let’s express O in terms of parameters along each cevian.First, parametrize median BE: any point on BE can be expressed as b + t(a/2 -b), where t∈[0,1].Parametrize altitude AD: since AD is from A (a) to D on BC. Since D is the foot of the altitude from A to BC, we can find D using projection. The vector AD is perpendicular to BC, so (a -d) ⋅ b=0. Let’s write D as scalar multiple of b: d= s b, since BC is from C(0) to B(b). Then, the vector AD= a -d= a -s b. This must be perpendicular to BC (vector b), so:(a -s b) ⋅ b=0 => a ⋅ b -s |b|²=0 => s= (a ⋅ b)/|b|²Thus, point D= s b= (a ⋅ b /|b|² ) bThus, parametric form of altitude AD: from A(a) to D(s b), can be written as a + u (s b -a), u∈[0,1]Similarly, parametrize angle bisector CF: from C(0) to F on AB. By angle bisector theorem, AF/FB= |a|/|b|. Therefore, point F divides AB in ratio |a| : |b|. Thus, vector F= (|b| a + |a| b )/(|a| + |b|)Therefore, parametrization of CF: t F, t∈[0,1]Now, O is the intersection of BE, AD, and CF.Let’s find O on BE and CF.Let’s express O as a point on BE: O= b + t(a/2 -b)= (1 -t)b + t a/2Also, O is on CF: O= s F= s (|b| a + |a| b )/(|a| + |b|)Therefore, equate the two expressions:(1 -t)b + t a/2 = s (|b| a + |a| b )/(|a| + |b| )This is a vector equation. Equate coefficients of a and b:Coefficient of a:t/2 = s |b| / (|a| + |b| )Coefficient of b:(1 -t) = s |a| / (|a| + |b| )Let’s denote |a|=AC= p, |b|=BC= q. Then:From coefficient of a:t/2 = s q / (p + q )From coefficient of b:1 -t = s p / (p + q )We can solve for s and t.From the first equation: s= (t/2)(p + q)/qSubstitute into the second equation:1 -t = [ (t/2)(p + q)/q ] * p / (p + q )Simplify:1 -t = (t/2)(p/q)Thus,1 -t = (t p)/(2 q )Rearrange:1= t( p/(2q) +1 )Therefore,t=1/(1 + p/(2q ))= 2q/(2q +p )Then,s= (t/2)(p + q )/q= ( (2q/(2q +p ))/2 )(p + q)/q= ( q/(2q +p ) )(p + q)/q= (p + q)/(2q +p )Now, we have parameters t and s. Now, also, O lies on altitude AD. Let’s use that to find more relations.Parametrize AD: O= a + u (D -a )=a + u ( (a ⋅ b /|b|² ) b -a )=a(1 -u ) + u ( (a ⋅ b /|b|² ) b )But we also have O expressed as (1 -t )b + t a/2. Therefore, equate the two expressions:a(1 -u ) + u ( (a ⋅ b /|b|² ) b ) = (1 -t )b + t a/2Again, equate coefficients of a and b:Coefficient of a:1 -u = t/2Coefficient of b:u (a ⋅ b /|b|² ) = 1 -tFrom coefficient of a: u=1 - t/2Substitute into coefficient of b:(1 - t/2)(a ⋅ b /|b|² )=1 -tMultiply both sides by |b|²:(1 - t/2)(a ⋅ b )= (1 -t )|b|²From earlier, we have t=2q/(2q +p )Substitute t into this equation:First, compute (1 -t/2 ):=1 - (2q/(2q +p )) /2=1 - q/(2q +p )=(2q +p -q )/(2q +p )=(q +p )/(2q +p )Compute (1 -t ):=1 -2q/(2q +p )=(2q +p -2q )/(2q +p )=p/(2q +p )Therefore,Left side: (q +p )/(2q +p ) * (a ⋅ b )Right side: p/(2q +p ) * |b|²Multiply both sides by (2q +p ):(q +p )(a ⋅ b )=p |b|²But a ⋅ b= |a||b| cosθ, where θ is angle C.Therefore,(q +p ) |a||b| cosθ= p |b|²Divide both sides by |b|:(q +p ) |a| cosθ= p |b|But |a|=AC=p, |b|=BC=q. Therefore:(q +p ) p cosθ= p qDivide both sides by p:(q +p ) cosθ= qThus,cosθ= q / (p +q )But θ is angle C, which we need to find.Therefore,cos(angle C)= BC / (AC + BC )But this seems unusual. Let me check this derivation again.From the equation:(q +p ) p cosθ= p qDivide both sides by p:(q +p ) cosθ= qThus,cosθ= q/(p +q )But angle C is angle at vertex C, between sides CB and CA. So, by the Law of Cosines:cos(angle C)= (a² + b² - c² )/(2ab )Wait, but according to this result, cos(angle C)= BC/(AC + BC ). This doesn't align with the standard formula. Therefore, there must be a miscalculation.Wait, in the derivation, we used vectors where point C is at the origin, vector a=CA, vector b=CB.Then, a ⋅ b= |a||b|cosθ, which is correct.But from the equation:(q +p ) |a||b| cosθ= p |b|²Here, p= |a|, q= |b|Therefore,(p +q ) p q cosθ= p q²Divide both sides by p q:(p +q ) cosθ= qThus,cosθ= q/(p +q )Therefore, angle C= arccos( q/(p +q ) )But in our coordinate example earlier, we had AC=42, BC=7, so p=42, q=7, so cosθ=7/(42+7)=7/49=1/7, which matches our previous result. Therefore, this is correct.Thus, angle C= arccos( BC/(AC + BC) )But the problem states that OE=2 OC. In our coordinate system, we found that this leads to angle C= arccos(1/7 )However, the problem wants us to find angle C given that OE=2 OC. Since the expression involves BC and AC, perhaps there's a relation between sides that can be expressed in terms of angle C.From the result cos(angle C)= BC/(AC + BC ), we can write:AC + BC= BC / cos(angle C )Let’s denote AC= b, BC= a, angle C= γ. Then:cosγ= a/(a +b )Using the Law of Sines:a/sinA = b/sinB = c/sinγBut this might not directly help. However, since the problem involves a median and an angle bisector, perhaps we can find relations using the given condition.Alternatively, since we know angle C= arccos( a/(a +b ) ), but we need to relate this to the given condition OE=2 OC.But from our earlier analysis, this relation angle C= arccos( BC/(AC + BC) ) is a direct consequence of OE=2 OC. Therefore, angle C is determined uniquely by this relation. However, unless BC/(AC + BC ) corresponds to a standard cosine value, angle C is a general angle.But in our specific example with BC=7, AC=42, angle C= arccos(1/7 ). Therefore, the answer is arccos(1/7 ), which is approximately 81.79 degrees. However, the problem likely expects an exact value, which suggests that there might be a special case where this ratio results in a standard angle, but our analysis shows it's not the case. Alternatively, perhaps in the problem, the given ratio OE=2 OC leads to angle C=60 degrees, but as we saw earlier, in an equilateral triangle, OE=0.5 OC, which does not satisfy the given condition. Therefore, the answer must be arccos(1/7 ), which is approximately 81.79 degrees. However, since the problem is likely expecting an exact answer, and given that 1/7 is a fraction, the answer is angle C= arccos(1/7 ). But maybe there is a different approach leading to a standard angle.Wait, let me revisit the cubic equation we solved earlier. We found k=6/7, leading to angle C= arccos(1/7 ). But what if there's a miscalculation in solving the cubic equation?We had the cubic equation 28k³ -115k² +148k -60=0, which had roots k=2, k=5/4, k=6/7. We discarded k=2 and k=5/4 since k<1, and took k=6/7. This led us to angle C= arccos(1/7 ). If there's no error in solving the cubic, then this is the correct result.Thus, after careful analysis through coordinate geometry, vector approach, and Ceva's theorem, we arrive at angle C= arccos(1/7 ). However, converting this into degrees is approximately 81.79 degrees, which isn't a standard angle. Therefore, the answer is likely expressed as arccos(1/7 ), but since the problem might expect an exact value, and given that in many Olympiad problems, such answers are acceptable, I will proceed with that.But wait, the problem states that the triangle is acute-angled. Let’s confirm that angle C≈81.79 degrees is acute. Yes, since it's less than 90 degrees. The other angles: angle A and angle B can be computed but are not necessary here.Therefore, the final answer is angle C= arccos(1/7 ), but perhaps in fractional terms of π, but since 1/7 doesn't correspond to a standard fraction of π, we leave it as arccos(1/7 ). However, the problem might expect the answer in degrees, but without a calculator, we can't compute the exact degree measure. However, in mathematical problems, especially competitions, answers in terms of inverse trigonometric functions are acceptable.But wait, in the original problem statement, is there any information we haven't used? Let me check again.The problem states that in an acute-angled triangle ABC, the altitude AD, median BE, and angle bisector CF intersect at point O. Find angle C if OE=2 OC.In our solution, angle C= arccos(1/7 ). But maybe there's a different approach using trigonometric identities or properties of the incenter, centroid, or orthocenter. However, since O is the intersection of an altitude, median, and angle bisector, it's a unique point not corresponding to the standard centers.Alternatively, consider using trigonometric Ceva's theorem.Trigonometric Ceva's theorem states that for concurrent cevians, the following holds:[sin(angle OCB)/sin(angle OCA)] * [sin(angle OAC)/sin(angle OAB)] * [sin(angle OBC)/sin(angle OBA)] =1But this might complicate things further.Alternatively, since O lies on the angle bisector CF, then the ratio of distances from O to the sides is proportional to the adjacent sides. However, combining this with the other conditions is non-trivial.Given the time I've spent and the consistent result from multiple approaches (coordinate, vector, Ceva), I believe the answer is indeed angle C= arccos(1/7 ). Therefore, the measure of angle C is arccos(1/7 ), which can be written in boxed notation as:boxed{arccosleft(dfrac{1}{7}right)}However, since the problem might expect an answer in degrees or in a different form, but given the lack of a standard angle, this is the precise answer.

✅:To solve the problem, we start by analyzing the given conditions in an acute-angled triangle (ABC) where the altitude (AD), median (BE), and angle bisector (CF) intersect at point (O). We need to find (angle C) given (OE = 2 OC).1. Coordinate Setup and Key Points: - Place (C) at the origin ((0, 0)), (B) at ((c, 0)), and (A) at ((d, e)). - The midpoint (E) of (AC) is ((d/2, e/2)). - The altitude (AD) is vertical from (A) to (D = (d, 0)). - The median (BE) connects (B) to (E). - The angle bisector (CF) divides (angle C) and intersects (AB) at (F) such that (AF/FB = AC/BC).2. Equations of Lines: - The median (BE) has the equation (y = frac{e}{d - 2c}(x - c)). - The altitude (AD) is (x = d). - The angle bisector (CF) is parametrized as (x = tF_x), (y = tF_y).3. Intersection Point (O): - Solving the intersection of (BE) and (AD) gives (O = (d, frac{e(d - c)}{d - 2c})). - This point (O) must also lie on (CF), leading to equations involving (d), (e), and (c).4. Distance Condition (OE = 2 OC): - Calculate distances (OE) and (OC) using coordinates of (O), (E), and (C). - Set up the equation (sqrt{(d/2)^2 + (y_O - e/2)^2} = 2 sqrt{d^2 + y_O^2}) and solve for the parameters.5. Solving for (k = d/c): - Introduce (k = d/c) and solve the resulting cubic equation (28k^3 - 115k^2 + 148k - 60 = 0). - The feasible root (k = 6/7) leads to (cos angle C = 1 - k = 1/7).6. Conclusion: - The angle (C) is found using (arccos(1/7)).Thus, the measure of (angle C) is (boxed{arccosleft(dfrac{1}{7}right)}).

🔑:We are given an acute-angled triangle ABC with the altitude AD, median BE, and angle bisector CF intersecting at a point O. We need to find the measure of angle C under the condition OE = 2 OC.1. Assign Values and Simplify the Problem: - Assume AC = 2a and let angle C = 2gamma. - Then, CE = a (because E is the midpoint of AC).2. Express CD and Relations for OC and OE: - Since AD is the altitude, use the cosine relation in triangle ADC: [ CD = AC cos(2gamma) = 2a cos(2gamma) ] - Since O lies on AD, F, and E: [ OC = CD sec(gamma) = 2a cos(2gamma) sec(gamma) ] - Given OE = 2 OC: [ OE = 4a cos(2gamma) sec(gamma) ]3. Apply the Law of Cosines in triangle EOC: - From the cosine law for triangle EOC: [ EO^2 = CO^2 + CE^2 - 2 cdot CO cdot CE cdot cos(angle EOC) ] - Substitute EO and CO: [ (4a cos(2gamma) sec(gamma))^2 = (2a cos(2gamma) sec(gamma))^2 + a^2 - 2 cdot (2a cos(2gamma) sec(gamma)) cdot a cdot cos(gamma) ]4. Simplify and Solve for cos(2gamma): - Expand and simplify: [ 16a^2 cos^2(2gamma) sec^2(gamma) = 4a^2 cos^2(2gamma) sec^2(gamma) + a^2 - 4a^2 cos(2gamma) cos(gamma) ] [ 16 cos^2(2gamma) sec^2(gamma) = 4 cos^2(2gamma) sec^2(gamma) + 1 - 4 cos(2gamma) cos^2(gamma) ] [ 12 cos^2(2gamma) = 1 - 4 cos(2gamma) cos^2(gamma) ]5. Eliminate and Solve: [ 12 cos^2(2gamma) = cos^2(gamma) - 4 cos(2gamma) cos^2(gamma) ] [ cos^2(gamma) + 2 cos(2gamma) cos^2(gamma) = 1 ]6. Isolate cos(2gamma): [ 12 cos^2(2gamma) = (1 + cos(2gamma))(1 - 4cos^2(gamma)) ] Given that 1 + cos(2gamma) = 2 cos^2(gamma) [ 24 cos^2(2gamma) = 1 - 4 cos(2gamma) ] 7. Find the Value of cos(2gamma): [ 24 cos^2(2gamma) + 4 cos(2gamma) - 1 = 0 ] [ tan(2theta)*((2.007 m/s)*(t)) = x(t) ]boxed{angle C = arcos(1/7)}

🧠:Okay, let's tackle this problem step by step. First, I need to understand what is being asked. The question is about determining how many of the given statements correctly conclude the negation of "p and q" (∼(p ∧ q)). Alternatively, the same problem is presented in another form, asking about the negation of "both p and q are true." So essentially, both versions are the same, just phrased differently. Let me recall some basic logic principles. The negation of a conjunction "p and q" (p ∧ q) is logically equivalent to "not p or not q" (∼p ∨ ∼q) by De Morgan's laws. This means that for the negation to hold, at least one of p or q must be false. Therefore, the correct negation should cover all scenarios where either p is false, q is false, or both are false. Now, looking at the four statements provided:1. p ∧ q2. p ∧ ∼q3. ∼p ∧ q4. ∼p ∧ ∼qEach of these is a conjunction. The original statement to negate is "p ∧ q." The negation, as we established, is "∼p ∨ ∼q," which is a disjunction. However, the given statements are all conjunctions. So, none of them individually capture the full negation because each one specifies a particular case where one is true and the other is false, or both are false. But the actual negation includes all those cases together. Wait a second. The question is asking which of these statements are correct in concluding ∼(p ∧ q). So, if any of these four statements is true, does that necessarily mean that ∼(p ∧ q) is true? Let me think. If, for example, statement 2 (p ∧ ∼q) is true, then in that case, since q is false, the conjunction p ∧ q would be false. Therefore, ∼(p ∧ q) would be true. Similarly, if statement 3 (∼p ∧ q) is true, then p is false, so again p ∧ q is false, making ∼(p ∧ q) true. Statement 4 (∼p ∧ ∼q) being true also makes p ∧ q false. However, statement 1 (p ∧ q) being true would make ∼(p ∧ q) false. Therefore, statements 2, 3, and 4, if true, would each individually lead to the conclusion that ∼(p ∧ q) is true. Statement 1, on the other hand, would contradict the conclusion. But wait, the question is phrased as "how many of these statements are correct in concluding ∼(p ∧ q)." So, if a statement can be a premise that leads to the conclusion ∼(p ∧ q), then statements 2, 3, and 4 would each imply ∼(p ∧ q). However, statement 1 would not; in fact, it would negate the conclusion. But here's a confusion point: Is the question asking which of these statements, when taken as a conclusion, is entailed by ∼(p ∧ q), or is it asking which of these statements, if true, would make ∼(p ∧ q) true? Looking back at the original problem: "How many of these statements are correct in concluding ∼(p ∧ q)." The wording is a bit ambiguous, but in logic, "concluding" usually refers to if the statement itself can serve as a conclusion for some premise. But in this context, it's more likely asking which of these statements, if true, would correctly lead to the conclusion that ∼(p ∧ q) is true. Alternatively, the problem could be interpreted as: if you have the conclusion ∼(p ∧ q), which of these statements would be valid ways to conclude that. But given the alternative phrasing in the second part, which is about negating "both p and q are true," it seems that the question is about which of these propositions correctly represent the negation. Wait, the second part says: "How many of these propositions are correct in negating 'both p and q are true'?" So the first part is about concluding ∼(p ∧ q), and the second is about the negation of "both p and q are true." Since these are the same, the answer should be consistent. But the negation of "both p and q are true" is "at least one of p or q is false," which is equivalent to ∼p ∨ ∼q. However, each of the given propositions in the problem are specific cases: 1. Both p and q are true. (p ∧ q)2. p is true and q is false. (p ∧ ∼q)3. p is false and q is true. (∼p ∧ q)4. Both p and q are false. (∼p ∧ ∼q)The negation of "both p and q are true" is not any single one of these, but rather the disjunction of 2, 3, and 4. So, the correct negation is the combination of cases where at least one is false. Therefore, none of the individual propositions 1-4 alone can be the correct negation, unless we consider all of 2,3,4 together. However, the question is asking how many of these propositions are correct in negating the original statement. But here's the key: In logic, the negation of a statement is another statement that is true whenever the original is false. So, the original statement is "both p and q are true" (p ∧ q). Its negation is "it is not the case that both p and q are true," which is equivalent to "at least one is false." Therefore, the negation is a single statement: ∼p ∨ ∼q. However, the options given are four different propositions. The question is, how many of these four propositions are correct negations? But each of propositions 2, 3, 4 individually make the original statement false. For example, if proposition 2 is true (p ∧ ∼q), then the original statement (p ∧ q) is false. Similarly, proposition 3 being true (∼p ∧ q) makes the original statement false, and proposition 4 (∼p ∧ ∼q) also makes the original false. However, proposition 1 (p ∧ q) being true would make the original statement true, so it's not a negation. But the negation of "both p and q are true" is not just any one of these, but the disjunction of 2,3,4. So, in other words, the correct negation is "either 2, 3, or 4 is true." However, the question is asking how many of the propositions 1-4 are correct negations. Since propositions 2,3,4 each individually serve as sufficient conditions for the negation (i.e., if any of them is true, the negation holds), but none of them alone are equivalent to the negation. The negation is the disjunction of these three. Therefore, if the question is interpreted as "how many of these propositions, when asserted, correctly negate 'both p and q are true'," then each of 2,3,4 do negate the original statement. Because if you can show that p is true and q is false, then certainly "both p and q are true" is false. Similarly for the others. However, in logic, the full negation would require covering all possibilities where at least one is false, but each of 2,3,4 is a specific case of that. Wait, but in standard logic, the negation of a statement is another statement that has the opposite truth value in all cases. So the negation of p ∧ q is ∼p ∨ ∼q. The propositions 2,3,4 are each specific instances where ∼p ∨ ∼q is true. However, none of them alone are equivalent to the negation. The negation is the disjunction of 2,3,4. Therefore, individually, none of them are the negation, but together they form the negation. But the question is asking how many of these propositions are correct in negating “both p and q are true”. If "correct in negating" means "are equivalent to the negation," then none of them are, because each is a conjunction, and the negation is a disjunction. However, if "correct in negating" means "if this proposition is true, then the original statement is false," then propositions 2,3,4 each individually suffice to make the original statement false. In that case, each of them is a correct way to negate the original statement, but they are not equivalent to the negation. This is a crucial distinction. Let me clarify:- The negation of "both p and q are true" is "at least one is false" (∼p ∨ ∼q). - However, the given options are four different scenarios:1. Both true (p ∧ q) – this is the original statement, not the negation.2. p true and q false (p ∧ ∼q) – this is one specific case where the negation holds.3. p false and q true (∼p ∧ q) – another specific case.4. Both false (∼p ∧ ∼q) – third specific case.Each of 2,3,4 is a subset of the full negation (which includes all three). So, if any of 2,3,4 is true, then the negation ∼(p ∧ q) is true. However, the problem is likely presented in the context of understanding that the negation is not just one of these cases, but all of them together. Therefore, in standard logic, the correct negation is the disjunction of 2,3,4, not any one individually. But the question is phrased as: "how many of these propositions are correct in negating..." So if someone claims that "p ∧ ∼q" is a negation of "p ∧ q", is that correct? From a logical equivalence standpoint, no, because "p ∧ ∼q" is not equivalent to "∼(p ∧ q)". However, "p ∧ ∼q" does imply "∼(p ∧ q)". Similarly, "∼p ∧ q" and "∼p ∧ ∼q" each imply "∼(p ∧ q)". Therefore, each of 2,3,4 are sufficient conditions for the negation, but not equivalent. But the question is about "negating" the original statement. In logic, to negate a statement means to provide a statement that is equivalent to its negation. Therefore, unless the proposition is logically equivalent to ∼(p ∧ q), it's not a correct negation. Since propositions 2,3,4 are each more specific than the negation (they are all subsets of the negation), they are not equivalent. Therefore, none of them individually are correct negations. Only the disjunction of 2,3,4 is equivalent. However, in some contexts, especially in natural language, people might consider any situation that contradicts the original statement as a "negation," but technically in formal logic, the negation must cover all possibilities where the original is false. Given that, the answer depends on the interpretation. If the question is asking for the number of propositions that are logically equivalent to the negation, the answer is 0, because none of them are equivalent. However, if it's asking how many propositions, if true, would make the original statement false (i.e., serve as sufficient conditions for the negation), then the answer is 3 (statements 2,3,4). But looking back at the original problem, in the first version, it's phrased as "how many of these statements are correct in concluding ∼(p ∧ q)". So, if we interpret "correct in concluding" as "if this statement is true, then ∼(p ∧ q) must be true," then statements 2,3,4 are correct, because each of them being true necessarily makes ∼(p ∧ q) true. Statement 1 being true would make ∼(p ∧ q) false, so it's not correct. Therefore, the answer would be 3. However, in the second version, it's phrased as "how many of these propositions are correct in negating 'both p and q are true'." If "negating" here is meant to find the logical equivalence, then none are correct. But if it's sufficient to show that the proposition makes the original statement false, then again 3. But the options given are (A) 0, (B) 1, (C) 2, (D) 3, (E) 4. This is a bit confusing because different interpretations lead to different answers. But given the context of a multiple-choice question, likely from a logic textbook or similar, we need to recall that the negation of a conjunction is a disjunction of the negations. The question is trying to test whether the student understands that the negation isn't just one of the cases but the combination. However, the propositions given are individual cases. Wait, but in the first version, the problem is about "concluding ∼(p ∧ q)". So if we have premises that are p ∧ q, p ∧ ∼q, etc., and we want to see which of these premises can lead to the conclusion ∼(p ∧ q). But wait, if the premise is p ∧ q, then the conclusion ∼(p ∧ q) would be false. If the premise is p ∧ ∼q, then the conclusion ∼(p ∧ q) is true. Similarly for ∼p ∧ q and ∼p ∧ ∼q. So, statements 2,3,4, when assumed as premises, do lead to the conclusion ∼(p ∧ q). Therefore, as premises, they are correct in concluding the negation. However, in formal logic, when we talk about a statement being a conclusion, we usually talk about logical consequence. So, if S is a statement, and C is a conclusion, then C is a logical consequence of S if whenever S is true, C is also true. Applying that here:- If S is p ∧ q, then C is ∼(p ∧ q). But p ∧ q being true makes C false. Therefore, S does not lead to C.- If S is p ∧ ∼q, then whenever S is true, p is true and q is false, so p ∧ q is false, hence C is true. Therefore, S leads to C.- Similarly for ∼p ∧ q and ∼p ∧ ∼q.Therefore, statements 2,3,4 each individually entail the conclusion ∼(p ∧ q). Statement 1 does not. Therefore, three of the statements are correct in concluding ∼(p ∧ q). However, another angle: the problem might be asking which of these statements are equivalent to ∼(p ∧ q). In that case, none, because ∼(p ∧ q) is equivalent to ∼p ∨ ∼q, which is not equivalent to any of the given conjunctions. But the question says "correct in concluding," which is different from equivalence. If "correct in concluding" means that the statement entails the conclusion, then 2,3,4 are correct. But in logic, if A entails B, then A is a sufficient condition for B. So, if you have A, you can conclude B. Therefore, in that sense, each of 2,3,4 are sufficient to conclude ∼(p ∧ q). Therefore, the answer should be 3. However, another possible interpretation is that the question is asking which of these statements are correct as negations, i.e., logically equivalent to the negation. In that case, since the negation is a disjunction and the options are conjunctions, none of them are equivalent. Therefore, the answer would be 0. But this is conflicting. To resolve this, let's check the exact wording again. First version: "How many of these statements are correct in concluding ∼(p ∧ q)?" The key term is "in concluding." If you have the statement as a premise, does it allow you to conclude ∼(p ∧ q)? Yes, for statements 2,3,4. Because if any of them is true, then ∼(p ∧ q) must be true. So they are correct premises to conclude the negation. Therefore, the answer is 3, option D. However, another possible angle: perhaps the question is asking which of these statements are conclusions that can be derived from ∼(p ∧ q). In other words, if we have ∼(p ∧ q), which of these statements can we conclude? In that case, since ∼(p ∧ q) is equivalent to ∼p ∨ ∼q, we cannot conclude any of the specific cases (2,3,4) because the disjunction doesn't specify which one is true. Therefore, none of them can be concluded from ∼(p ∧ q) alone. But the question is phrased as "correct in concluding ∼(p ∧ q)", which is the opposite direction. Given the wording "correct in concluding ∼(p ∧ q)", it's more about whether the statement can be used to conclude ∼(p ∧ q), not the other way around. Therefore, if the statement is a premise that allows us to conclude ∼(p ∧ q), then 2,3,4 work. Therefore, the answer should be 3, option D. But wait, looking at the alternative phrasing: "How many of these propositions are correct in negating 'both p and q are true'?" If "negating" here means "serving as the negation," which in logic requires equivalence, then none are correct because the negation is a disjunction. However, if "negating" is used more loosely to mean "contradicts" the original statement, then propositions 2,3,4 each contradict the original statement. But in formal terms, the negation of a proposition is another proposition that is true exactly when the original is false. The original is false in three cases: 2,3,4. The negation is the disjunction of these three. Therefore, each of 2,3,4 individually are not the full negation, but each is a part of the negation. However, sometimes people make the mistake of thinking that to negate a conjunction, you have to specify one of the cases, but actually, the proper negation is the disjunction. Therefore, a common trick question is to present these four options and see if people think 2,3,4 are all correct negations. But in reality, none of them are correct because they are not equivalent. This is a classic logic question. The negation of "both p and q are true" is "at least one is false," which is not the same as "p is true and q is false," "p is false and q is true," or "both are false." Each of those is a separate case, but the negation includes all three. Therefore, individually, none of them are correct negations; only their disjunction is. Therefore, the correct answer should be 0. But this contradicts the previous reasoning. Wait, but the key here is the difference between logical equivalence and implication. If the question is about equivalence, then none are correct. If it's about implying the negation, then 2,3,4 are correct. The problem says "correct in concluding ∼(p ∧ q)." In logical terms, if we say that a statement A is correct in concluding B, it usually means that A entails B (i.e., A → B is a tautology). So, in this case, if the statement (e.g., p ∧ ∼q) entails ∼(p ∧ q), then it is correct. Let's verify:1. p ∧ q → ∼(p ∧ q): This would be false, because if p ∧ q is true, then ∼(p ∧ q) is false.2. p ∧ ∼q → ∼(p ∧ q): If p ∧ ∼q is true, then p ∧ q is false, so this implication holds.3. ∼p ∧ q → ∼(p ∧ q): Similarly, this holds.4. ∼p ∧ ∼q → ∼(p ∧ q): This also holds.Therefore, statements 2,3,4 each entail the conclusion ∼(p ∧ q), so they are "correct in concluding" it. Statement 1 does not. Therefore, the answer is 3, option D. However, in the alternative phrasing, asking about negating “both p and q are true”, if we take the term "negating" as forming the logical negation (i.e., equivalent statement), then none are correct. But if "negating" is taken to mean "contradicts" or "refutes" the original statement, then 2,3,4 are correct. But in formal logic, the negation must be equivalent. Therefore, the answer would be 0. This creates confusion. Looking at the problem again, since it's presented in two phrasings but the same options, we need to consider what is being tested here. This is a common question in logic exams where students are asked about the negation of a conjunction, and the options are the different possible cases. The correct answer is that none of the individual cases are equivalent to the negation, hence the answer is 0. However, many students mistakenly choose 3, thinking that each case contradicts the original. To clarify, the negation of "both p and q are true" is "it is not the case that both are true," which can be due to any of the three scenarios: p true q false, p false q true, both false. However, the negation itself is the disjunction of these three, not any one individually. Therefore, none of the given propositions correctly represent the negation on their own. But wait, the question is not asking for equivalence, but rather how many are correct in negating the original. If by "correct in negating" they mean "are equivalent to the negation," then answer is 0. If it means "if this proposition is true, then the original is false," then answer is 3. Given that the original statements in the first version are presented as statements that conclude ∼(p ∧ q), which would mean that each statement is a premise leading to that conclusion. Since 2,3,4 do lead to the conclusion, answer is 3. However, given the alternative phrasing about negating, which is more about equivalence, answer is 0. But the options are the same for both phrasings. This is a bit of a ambiguity, but given that it's a common type of question, and considering that in logic, the negation is a single statement that is logically equivalent, the answer is likely 0. Because none of the given propositions are equivalent to ∼(p ∧ q). However, if the question is from a source that considers any of the three cases as valid ways to negate the original (even though they are not equivalent), then the answer would be 3. To resolve this, perhaps looking at the exact terminology used. The first question says "correct in concluding ∼(p ∧ q)". If "concluding" is in the sense that if you assume the statement is true, then you can conclude ∼(p ∧ q), then 2,3,4 are correct. However, in formal logic, when you talk about a conclusion, the validity depends on the form of the argument. If the argument is "Statement, therefore ∼(p ∧ q)", then for each of 2,3,4, the argument would be valid. Therefore, those three statements are correct in the sense that they can be used to validly conclude ∼(p ∧ q). Therefore, the answer should be 3, option D. But I recall that in some textbooks, particularly introductory ones, they might phrase the question as "which of the following are ways to negate the statement," expecting students to recognize that all three cases are part of the negation, but since they are separate options, choosing all three. However, if the options are separate, the correct answer is 3. Wait, but the answer options are (A) 0, (B)1, etc. So if the answer is 3, then option D. Alternatively, if the question is from the perspective that the negation must cover all cases where the original is false, and since none of the options individually do that, the answer is 0. This is confusing, but given the way the question is phrased—asking "how many of these statements are correct in concluding"—it's likely testing the understanding that any of the three cases (2,3,4) being true would allow you to conclude the negation. Therefore, the answer is 3. However, another perspective is that all four statements are possible scenarios, and the negation of "both p and q are true" is the disjunction of 2,3,4. But since each of these is a separate statement, and the question asks how many of them are correct negations, the answer should be 3, because each one individually contradicts the original. But in formal logic, the negation is not any of these individually, but their disjunction. So technically, none of them are correct as negations because they are not equivalent. However, if the question is using "negating" in a non-technical sense, meaning "contradicts," then 3. Given the problem is presented in two forms, one about "concluding" and one about "negating," and given that in both cases the options are the same, it's likely expecting the answer 3. However, I need to check for any possible errors in this reasoning. Wait, let's take a concrete example. Suppose the original statement is "It is raining and cold." The negation is "It is not raining or not cold," which includes "raining and not cold," "not raining and cold," and "not raining and not cold." If someone claims that the negation is "raining and not cold," that is a part of the negation but not the entire negation. Therefore, in formal terms, it's not equivalent. However, if the question is asking which of the given statements contradict the original, then all three (2,3,4) do. But the term "negating" in formal logic requires equivalence. Therefore, the answer should be 0. However, this might not be the case in all contexts. Given that this is a multiple-choice question with options 0 to 4, and considering common misconceptions, it's possible that the intended answer is 3, but the technically correct answer is 0. Wait, let's verify with truth tables. Let's list all possible scenarios:p | q | p ∧ q | ∼(p ∧ q) | p ∧ ∼q | ∼p ∧ q | ∼p ∧ ∼qT | T | T | F | F | F | FT | F | F | T | T | F | FF | T | F | T | F | T | FF | F | F | T | F | F | TNow, the negation of p ∧ q is ∼(p ∧ q), which is true in the last three rows. Each of the statements 2,3,4 correspond to these three rows. However, each of them is only true in one specific row where ∼(p ∧ q) is true. Therefore, each of them is a subset of the negation. To be equivalent, the statement would have to be true in exactly the same rows as ∼(p ∧ q). Since none of 2,3,4 cover all three rows where ∼(p ∧ q) is true, none are equivalent. Therefore, the answer is 0. However, if the question is asking how many of these statements are true when ∼(p ∧ q) is true, then in each row where ∼(p ∧ q) is true, exactly one of 2,3,4 is true. Therefore, in that sense, there are three statements that can be true when the negation holds, but they are mutually exclusive. But the question is not about when ∼(p ∧ q) is true, but rather how many of the given statements correctly conclude ∼(p ∧ q). If "correctly conclude" means that whenever the statement is true, ∼(p ∧ q) is also true, then yes, statements 2,3,4 have this property. Therefore, they are valid in concluding the negation. In logic, an argument is valid if the conclusion follows necessarily from the premises. So, if the premises are p ∧ ∼q, then the conclusion ∼(p ∧ q) necessarily follows. Similarly for the others. Therefore, each of these three statements can serve as valid premises to conclude ∼(p ∧ q). Therefore, the answer should be 3, option D. But there's a catch here. If the question is asking how many of these statements are equivalent to the negation, then the answer is 0. If it's asking how many can be used as premises to validly conclude the negation, then the answer is 3. Given that the question says "correct in concluding," it's more about the validity of the argument from the statement to the conclusion. Therefore, since each of 2,3,4 validly imply the conclusion, the answer is 3. However, another angle: sometimes in logic, "concluding" can mean that the conclusion is logically equivalent. But that's not standard terminology. Usually, "concluding" just means that the conclusion follows validly from the premises. Therefore, based on standard logical terminology, the answer should be 3. But I want to double-check with a reliable source. According to logical consequence definitions, if a statement A entails statement B, then B is a logical consequence of A. So, if we have A = p ∧ ∼q, and B = ∼(p ∧ q), then A entails B. Similarly for the others. Therefore, each of 2,3,4 are correct in the sense that they entail the conclusion. Therefore, the answer is 3, option D. However, I'm still conflicted because in some contexts, particularly in teaching logic, this question is used to trick students into thinking that the negation is one of the cases, but the correct answer is that none are, because the negation is the disjunction. But given the phrasing "correct in concluding," which implies validity rather than equivalence, the answer should be 3. Given that both phrasings of the question are present, and considering that the answer options include 3, which is a common distractor, but in reality, the technically correct answer is 0 because none are equivalent. However, based on the wording about "concluding," which is about implication rather than equivalence, the answer is 3. But wait, let's think about how this question would be graded. If it's a question from a textbook or a professor, they might expect students to recognize that each of the three statements (2,3,4) individually contradict the original statement, hence can be used to conclude the negation. Therefore, the answer would be 3. Alternatively, if they are testing knowledge that the negation must be a disjunction and not a conjunction, then answer is 0. This is really ambiguous. To resolve, perhaps look at the original problem's source or similar problems. For example, in the book "How to Prove It" by Velleman, similar exercises clarify that the negation of "P and Q" is "not P or not Q," and that individual cases are not the negation but part of it. In standard multiple-choice logic questions, when asked for the negation of a conjunction, if the options are the disjunction and various conjunctions, the disjunction is correct. However, if the options are conjunctions, the answer is none. But here, the question is not asking for the negation directly, but how many of the given statements are correct in concluding the negation. If we take "correct in concluding" to mean that the statement logically implies the conclusion, then 2,3,4 do, so answer is 3. Therefore, considering the terminology and standard logical definitions, the answer is D) 3.

✅:The negation of ( p wedge q ) is ( sim p vee sim q ), which includes all cases where at least one proposition is false. The given statements (2, 3, 4) represent specific instances of this disjunction (i.e., ( p wedge sim q ), ( sim p wedge q ), ( sim p wedge sim q )). While none of these individual conjunctions are logically equivalent to the full negation ( sim(p wedge q) ), each one implies ( sim(p wedge q) ) when true. For example:- If ( p wedge sim q ) is true, ( p wedge q ) must be false, making ( sim(p wedge q) ) true.- Similarly, ( sim p wedge q ) and ( sim p wedge sim q ) being true also ensure ( sim(p wedge q) ) is true.Thus, statements 2, 3, and 4 are sufficient conditions for concluding ( sim(p wedge q) ). Statement 1 (( p wedge q )) directly contradicts the conclusion and is invalid. Answer: (D) 3

🔑:# Part (I)We need to determine for which statements (1)-(4) we can correctly deduce sim(p wedge q), where - sim p means the negation of p - wedge means "and"To do this, let's first use De Morgan's laws, which state:[ sim(p wedge q) = (sim p) vee (sim q) ]However, we need to check the correctness of the statements against sim(p wedge q) = sim p wedge sim q. Let’s analyze each:1. ( p wedge q ): Both p and q are true. - sim(p wedge q) = sim(text{True} wedge text{True}) = text{False}2. ( p wedge sim q ): p is true and q is false. - sim(p wedge q) = sim(text{True} wedge text{False}) = text{True}, and it is considered as sim p wedge sim q, which results in sim(text{True}) wedge sim(text{False}) = text{False} wedge text{True} = text{False}3. ( sim p wedge q ): p is false and q is true. - sim(p wedge q) = sim(text{False} wedge text{True}) = text{True}4. ( sim p wedge sim q ): Both p and q are false. - sim(p wedge q) = sim(text{False} wedge text{False}) = text{False}Evaluations:- Case (1) yields false equivalence.- Case (2), (3), and (4) are true equivalences.# Conclusion: - Statements (2), (3), and (4) are correct deducing sim(p wedge q), i.e., there are 3 correct statements.# Part (II)We need to find out how many of the expressions deny "both p and q are true."Evaluating each:1. ( p text{ and } q ) both true. - Does not deny it. 2. ( p ) true and ( q ) false. - Denies it. Only one true. 3. ( p ) false and ( q ) true. - Denies it. Only one true.4. Both ( p ) and ( q ) false. - Denies it. Both false.# Conclusion:- Cases (2), (3), and (4) deny "both p and q are true," so there are 3 correct denials. Therefore, the correct choice for both parts is:boxed{D}

🧠:Okay, let's try to tackle this problem step by step. First, I need to understand what's given and what's being asked. We have the function ( S = sin^{2} x + sin^{2} y + left(8 sin^{2} t + 2right) sin x sin y ), where ( x ) and ( y ) are non-negative real numbers such that ( x + y = t ), and ( t ) is in the interval ([0, pi]). The task is to find the maximum value of ( g(t) = M(t) - N(t) ), where ( M(t) ) is the maximum of ( S ) and ( N(t) ) is the minimum of ( S ) for each ( t ). Then, we need to determine the value of ( t ) where this maximum of ( g(t) ) occurs.Alright, so first, for each ( t ), we need to find the maximum and minimum values of ( S ) given ( x + y = t ). Then, subtract the minimum from the maximum to get ( g(t) ), and finally find the maximum of ( g(t) ) over all ( t ) in ([0, pi]).Let me start by analyzing the function ( S ). It's a function of ( x ) and ( y ), but since ( x + y = t ), we can express one variable in terms of the other. Let me set ( y = t - x ). Then, ( S ) becomes a function of ( x ) alone, with ( x ) ranging from 0 to ( t ).So substituting ( y = t - x ), we have:( S(x) = sin^{2} x + sin^{2}(t - x) + left(8 sin^{2} t + 2right) sin x sin(t - x) )Our goal is to find the maximum and minimum of ( S(x) ) for ( x in [0, t] ), for each ( t in [0, pi] ).Let me first consider if there's a way to simplify ( S(x) ). Perhaps using trigonometric identities to combine terms.First, note that ( sin^2 x + sin^2 y ). There's an identity for the sum of squares of sines. Alternatively, maybe express each ( sin^2 ) term using the double-angle formula:( sin^2 theta = frac{1 - cos 2theta}{2} )Similarly, the product ( sin x sin y ) can be expressed using a product-to-sum identity:( sin x sin y = frac{cos(x - y) - cos(x + y)}{2} )But let me see if applying these identities would help.Alternatively, since ( y = t - x ), perhaps we can express ( sin y = sin(t - x) ), which can be expanded as ( sin t cos x - cos t sin x ).But before diving into that, maybe consider substituting ( x = u ) and ( y = t - u ), so the problem becomes a single-variable optimization over ( u in [0, t] ). Let's stick with ( x ) as the variable.So, let's write ( S(x) ) as:( S(x) = sin^2 x + sin^2(t - x) + (8 sin^2 t + 2) sin x sin(t - x) )Let me compute each term separately.First, ( sin^2 x + sin^2(t - x) ). Let's compute that:Using the identity ( sin^2 a = frac{1 - cos 2a}{2} ), so:( sin^2 x = frac{1 - cos 2x}{2} )( sin^2(t - x) = frac{1 - cos 2(t - x)}{2} )So adding them together:( frac{1 - cos 2x}{2} + frac{1 - cos 2(t - x)}{2} = frac{2 - cos 2x - cos 2(t - x)}{2} )Simplify the cosines:( cos 2x + cos 2(t - x) ). Hmm, perhaps using another identity here. The sum of cosines:( cos A + cos B = 2 cos frac{A + B}{2} cos frac{A - B}{2} )Let me apply this identity to ( cos 2x + cos 2(t - x) ):Let ( A = 2x ), ( B = 2(t - x) ). Then,( cos A + cos B = 2 cos frac{A + B}{2} cos frac{A - B}{2} )Compute ( frac{A + B}{2} = frac{2x + 2t - 2x}{2} = frac{2t}{2} = t )Compute ( frac{A - B}{2} = frac{2x - (2t - 2x)}{2} = frac{2x - 2t + 2x}{2} = frac{4x - 2t}{2} = 2x - t )Therefore,( cos 2x + cos 2(t - x) = 2 cos t cos(2x - t) )So substituting back into the expression:( sin^2 x + sin^2(t - x) = frac{2 - 2 cos t cos(2x - t)}{2} = 1 - cos t cos(2x - t) )Okay, that's a nice simplification. Now, moving on to the cross term:( (8 sin^2 t + 2) sin x sin(t - x) )Let me first compute ( sin x sin(t - x) ). Again, using the product-to-sum identity:( sin a sin b = frac{cos(a - b) - cos(a + b)}{2} )Here, ( a = x ), ( b = t - x ), so:( sin x sin(t - x) = frac{cos(x - (t - x)) - cos(x + (t - x))}{2} = frac{cos(2x - t) - cos t}{2} )Therefore, the cross term becomes:( (8 sin^2 t + 2) cdot frac{cos(2x - t) - cos t}{2} = frac{8 sin^2 t + 2}{2} (cos(2x - t) - cos t) )Simplify the coefficient:( frac{8 sin^2 t + 2}{2} = 4 sin^2 t + 1 )So, the cross term is:( (4 sin^2 t + 1)(cos(2x - t) - cos t) )Putting it all together, the entire expression for ( S(x) ):( S(x) = 1 - cos t cos(2x - t) + (4 sin^2 t + 1)(cos(2x - t) - cos t) )Let me expand this:First, distribute the terms:( S(x) = 1 - cos t cos(2x - t) + (4 sin^2 t + 1)cos(2x - t) - (4 sin^2 t + 1)cos t )Combine like terms:The terms involving ( cos(2x - t) ):( [ - cos t + 4 sin^2 t + 1 ] cos(2x - t) )The constant terms (with respect to x):( 1 - (4 sin^2 t + 1)cos t )So, simplifying:Let me compute the coefficient of ( cos(2x - t) ):( - cos t + 4 sin^2 t + 1 )Note that ( sin^2 t = 1 - cos^2 t ), so substituting:( - cos t + 4(1 - cos^2 t) + 1 = - cos t + 4 - 4 cos^2 t + 1 = (-4 cos^2 t) - cos t + 5 )So, the coefficient is ( -4 cos^2 t - cos t + 5 )Thus, the expression becomes:( S(x) = (-4 cos^2 t - cos t + 5) cos(2x - t) + 1 - (4 sin^2 t + 1)cos t )Now, let's compute the constant term:( 1 - (4 sin^2 t + 1)cos t = 1 - 4 sin^2 t cos t - cos t )Again, note that ( sin^2 t = 1 - cos^2 t ):So, substituting:( 1 - 4(1 - cos^2 t)cos t - cos t = 1 - 4 cos t + 4 cos^3 t - cos t = 1 - 5 cos t + 4 cos^3 t )Therefore, the entire expression for ( S(x) ):( S(x) = (-4 cos^2 t - cos t + 5) cos(2x - t) + 1 - 5 cos t + 4 cos^3 t )Combine the constant terms:Wait, the expression is:( S(x) = [ -4 cos^2 t - cos t + 5 ] cos(2x - t) + [1 - 5 cos t + 4 cos^3 t ] )This seems complicated, but perhaps we can write ( S(x) ) as:( S(x) = A(t) cos(2x - t) + B(t) )Where:( A(t) = -4 cos^2 t - cos t + 5 )( B(t) = 1 - 5 cos t + 4 cos^3 t )Now, since ( S(x) ) is expressed as ( A(t) cos(2x - t) + B(t) ), the maximum and minimum of ( S(x) ) will depend on the coefficient ( A(t) ).Because ( cos(2x - t) ) oscillates between -1 and 1, so:- If ( A(t) > 0 ), then the maximum of ( S(x) ) is ( A(t) cdot 1 + B(t) ) and the minimum is ( A(t) cdot (-1) + B(t) )- If ( A(t) < 0 ), then the maximum of ( S(x) ) is ( A(t) cdot (-1) + B(t) ) and the minimum is ( A(t) cdot 1 + B(t) )- If ( A(t) = 0 ), then ( S(x) = B(t) ), so maximum and minimum are both ( B(t) )Therefore, ( M(t) - N(t) = |2 A(t)| ), since if ( A(t) neq 0 ), the difference is ( (A(t) + B(t)) - (-A(t) + B(t)) = 2 A(t) ) if ( A(t) > 0 ), or ( (-A(t) + B(t)) - (A(t) + B(t)) = -2 A(t) ) if ( A(t) < 0 ). So in either case, the difference is ( 2 |A(t)| ). If ( A(t) = 0 ), then the difference is 0.Therefore, ( g(t) = 2 |A(t)| ), where ( A(t) = -4 cos^2 t - cos t + 5 ).Therefore, our task reduces to finding the maximum of ( 2 | -4 cos^2 t - cos t + 5 | ) over ( t in [0, pi] ).Wait, but before accepting this conclusion, let me verify. The reasoning is that since ( S(x) = A(t) cos(2x - t) + B(t) ), the maximum and minimum of ( S(x) ) are ( B(t) pm |A(t)| ), hence the difference is ( 2 |A(t)| ). But we need to confirm that the variable substitution and the range of ( 2x - t ) allows the cosine term to attain both 1 and -1.Wait, the variable ( x ) ranges from 0 to ( t ). Therefore, ( 2x - t ) ranges from ( -t ) to ( t ). The cosine function is even, so ( cos(2x - t) ) can attain all values between ( cos(-t) ) and ( cos(t) ). Wait, but the range is actually over ( 2x - t in [-t, t] ). However, depending on ( t ), the angle ( 2x - t ) may or may not reach angles where cosine is 1 or -1.But the maximum of ( cos theta ) is 1, which occurs at ( theta = 0 ), and minimum is -1 at ( theta = pi ). However, if ( 2x - t ) can reach 0 or ( pi ) within the interval ( x in [0, t] ), then the cosine term can reach 1 or -1. Let's check:For ( 2x - t = 0 ), solving for x: ( x = t/2 ). Since ( x in [0, t] ), t/2 is always within the interval, so the maximum of 1 is attainable.For ( 2x - t = pi ), solving for x: ( x = (t + pi)/2 ). But x must be ≤ t, so (t + π)/2 ≤ t ⇒ t + π ≤ 2t ⇒ π ≤ t. However, t ∈ [0, π], so this is only possible when t = π. Then x = (π + π)/2 = π, which is allowed. So when t = π, x = π, but then y = π - π = 0. So in that case, 2x - t = 2π - π = π. Therefore, for t = π, the cosine term can reach -1.But for t < π, (t + π)/2 > t ⇒ since t < π, (t + π)/2 > t ⇒ π > t ⇒ (t + π)/2 > t ⇒ π > t, which is true, but x must be ≤ t, so x = (t + π)/2 is only possible if (t + π)/2 ≤ t ⇒ t + π ≤ 2t ⇒ π ≤ t. But t ∈ [0, π], so equality holds only when t = π. Therefore, for t < π, the angle ( 2x - t ) cannot reach π, so the minimum value of the cosine term would be ( cos(t) ), since when x = 0, ( 2x - t = -t ), so cos(-t) = cos t, and as x increases, the angle goes from -t to t. Therefore, the minimum of cos(2x - t) would be cos t if t ∈ [0, π/2], because cos is decreasing on [0, π], so if the angle ranges from -t to t, which is symmetric around 0, the minimum would be cos t (since for t ≤ π/2, t ≤ π/2, so the angle ranges from -t to t, which is within [-π/2, π/2], where cos is positive and decreasing). Wait, but when t > π/2, then the angle 2x - t can reach up to t, which is greater than π/2, so cos(t) would be negative. Wait, let's think.Wait, for t ∈ [0, π], when x ranges from 0 to t, then 2x - t ranges from -t to t. So the angle θ = 2x - t ∈ [-t, t]. The cosine function is even, so the minimum of cos θ over θ ∈ [-t, t] is cos t when t ∈ [0, π], since cos is decreasing on [0, π], so the minimum occurs at θ = ±t (since cos(-t) = cos t). Wait, no. Wait, cos θ is symmetric about θ = 0. So over the interval [-t, t], the minimum value of cos θ is cos t if t ∈ [0, π], because as θ increases from -t to t, the cosine decreases from cos(-t) = cos t to cos t (since it's symmetric). Wait, actually, cos θ is maximum at θ = 0 and decreases as |θ| increases. Therefore, over the interval [-t, t], the minimum value of cos θ is cos t (since t is in [0, π], so the farthest points from 0 are ±t, where the cosine is cos t). Therefore, the minimum of cos θ over θ ∈ [-t, t] is cos t, and the maximum is 1 (attained at θ = 0). Therefore, regardless of t ∈ [0, π], the cosine term cos(2x - t) ranges from cos t to 1. Therefore, the maximum value of cos(2x - t) is 1 and the minimum is cos t.Wait, this contradicts my earlier thought. Wait, for example, take t = π. Then θ ranges from -π to π, but cos θ ranges from -1 (at θ = π) to 1 (at θ = 0). So in this case, the minimum is -1. But according to the previous reasoning, if t = π, then the minimum is cos π = -1, which matches. If t = π/2, then θ ranges from -π/2 to π/2, so the minimum is cos π/2 = 0, but the actual minimum of cos θ over [-π/2, π/2] is 0 (at θ = ±π/2). Wait, but cos π/2 is 0, which is the minimum in that interval.Wait, actually, if θ ∈ [-t, t], then the minimum of cos θ is cos t when t ∈ [0, π], because cos is decreasing on [0, π], so the minimum occurs at θ = ±t, which gives cos t. However, when t > π/2, cos t becomes negative. For example, t = 2π/3. Then θ ranges from -2π/3 to 2π/3. The minimum of cos θ here is cos(2π/3) = -1/2. But cos θ at θ = 2π/3 is -1/2, which is indeed the minimum in that interval. Similarly, at t = π, the minimum is cos π = -1. So yes, regardless of t ∈ [0, π], the minimum of cos θ over θ ∈ [-t, t] is cos t, and the maximum is 1. Therefore, the previous conclusion is correct.Therefore, in our expression for S(x):( S(x) = A(t) cos(2x - t) + B(t) ), and since cos(2x - t) ∈ [cos t, 1], then:If A(t) > 0, then:Maximum S = A(t)*1 + B(t)Minimum S = A(t)*cos t + B(t)Thus, g(t) = M(t) - N(t) = A(t)(1 - cos t)If A(t) < 0, then:Maximum S = A(t)*cos t + B(t)Minimum S = A(t)*1 + B(t)Thus, g(t) = M(t) - N(t) = A(t)(cos t - 1) = |A(t)| (1 - cos t)Wait, but A(t) could be positive or negative. Let me think.If A(t) > 0, then since cos(2x - t) is in [cos t, 1], the maximum occurs at cos(2x - t) = 1, giving S = A(t)*1 + B(t), and the minimum at cos(2x - t) = cos t, giving S = A(t)*cos t + B(t). Therefore, g(t) = A(t)(1 - cos t)If A(t) < 0, then since A(t) is negative, multiplying by cos(2x - t) which is in [cos t, 1], the maximum of S occurs at the minimum of cos(2x - t), which is cos t, so S = A(t)*cos t + B(t), and the minimum of S occurs at cos(2x - t) = 1, so S = A(t)*1 + B(t). Therefore, g(t) = (A(t) cos t + B(t)) - (A(t) + B(t)) = A(t)(cos t - 1) = -A(t)(1 - cos t). Since A(t) is negative, this becomes |A(t)|(1 - cos t).If A(t) = 0, then S(x) = B(t), so g(t) = 0.Therefore, in all cases, g(t) = |A(t)| (1 - cos t). So our previous conclusion that g(t) = 2 |A(t)| was incorrect. The correct expression is g(t) = |A(t)| (1 - cos t). Therefore, my earlier step had an error.Wait, that was a crucial mistake. Let me verify this again.Given that S(x) = A(t) * cos(2x - t) + B(t), and cos(2x - t) ∈ [cos t, 1], then:If A(t) > 0:- Maximum S = A(t)*1 + B(t)- Minimum S = A(t)*cos t + B(t)So difference is A(t)*(1 - cos t)If A(t) < 0:- Maximum S = A(t)*cos t + B(t)- Minimum S = A(t)*1 + B(t)Difference is A(t)*(cos t - 1) = |A(t)|*(1 - cos t) since A(t) is negative.Therefore, regardless of the sign of A(t), g(t) = |A(t)|*(1 - cos t)Therefore, the correct expression for g(t) is |A(t)|*(1 - cos t), where A(t) = -4 cos² t - cos t + 5.Therefore, we need to maximize g(t) = | -4 cos² t - cos t + 5 | * (1 - cos t) over t ∈ [0, π].So, let's first write A(t) as:A(t) = -4 cos² t - cos t + 5Let me denote u = cos t, so that u ∈ [-1, 1] since t ∈ [0, π]. Then,A(t) = -4 u² - u + 5Therefore, g(t) becomes | -4 u² - u + 5 | * (1 - u )Our task is to maximize this expression over u ∈ [-1, 1].So, first, let's analyze the function h(u) = | -4 u² - u + 5 | * (1 - u ) for u ∈ [-1, 1].First, let's consider the quadratic inside the absolute value:f(u) = -4 u² - u + 5We can write this as f(u) = -4u² - u + 5. Let's find its roots to determine where it is positive or negative.Find roots of f(u) = 0:-4u² - u + 5 = 0Multiply both sides by -1:4u² + u - 5 = 0Using quadratic formula:u = [ -1 ± sqrt(1 + 80) ] / (8) = [ -1 ± sqrt(81) ] / 8 = [ -1 ± 9 ] / 8Thus, the roots are:u = ( -1 + 9 ) / 8 = 8 / 8 = 1u = ( -1 - 9 ) / 8 = -10 / 8 = -5/4Since the quadratic coefficient is positive (4), the parabola opens upwards. Therefore, f(u) = -4u² - u + 5 is positive between the roots u = -5/4 and u = 1. However, since u ∈ [-1, 1], the interval where f(u) ≥ 0 is u ∈ [-1, 1], because the roots are at u = -5/4 (outside our interval) and u = 1. Therefore, in the interval u ∈ [-1, 1], f(u) starts at u = -1:f(-1) = -4(1) - (-1) + 5 = -4 + 1 + 5 = 2At u = 1:f(1) = -4(1) -1 + 5 = -4 -1 +5 = 0Since the parabola opens upward (after multiplying by -1), the original function f(u) is a downward-opening parabola. Therefore, f(u) is decreasing from u = -infty to the vertex, then increasing. Wait, f(u) = -4u² - u +5 is a quadratic with a maximum at u = -b/(2a) where a = -4, b = -1. So vertex at u = -(-1)/(2*(-4)) = 1/( -8 ) = -1/8.Therefore, f(u) has a maximum at u = -1/8. Let's compute f(-1/8):f(-1/8) = -4*(1/64) - (-1/8) +5 = -1/16 + 1/8 +5 = ( -1 + 2 ) /16 +5 = 1/16 +5 = 5.0625So f(u) reaches maximum at u = -1/8, then decreases to f(1) = 0.But since u ∈ [-1,1], f(u) starts at f(-1) = 2, increases to f(-1/8) ≈5.0625, then decreases to f(1)=0. Therefore, in the interval u ∈ [-1,1], f(u) is always positive except at u=1 where it is zero. Therefore, |f(u)| = f(u) for u ∈ [-1,1). At u=1, |f(1)| = 0.Therefore, h(u) = f(u)*(1 - u) for u ∈ [-1,1).So, h(u) = (-4u² -u +5)(1 - u) for u ∈ [-1,1)At u=1, h(1)=0*0=0.Therefore, we can express h(u) as (-4u² - u +5)(1 - u) over u ∈ [-1,1].Now, we need to find the maximum of h(u) over u ∈ [-1,1].Let me expand h(u):h(u) = (-4u² - u +5)(1 - u) = (-4u²)(1 - u) - u(1 - u) +5(1 - u)Compute each term:First term: -4u²(1 - u) = -4u² +4u³Second term: -u(1 - u) = -u + u²Third term:5(1 - u) =5 -5uCombine all terms:-4u² +4u³ -u + u² +5 -5uCombine like terms:4u³ + (-4u² + u²) + (-u -5u) +5Simplify:4u³ -3u² -6u +5Therefore, h(u) = 4u³ -3u² -6u +5But wait, this seems incorrect. Wait, let's check:Wait, expanding (-4u² - u +5)(1 - u):Multiply each term:-4u² * 1 = -4u²-4u² * (-u) = +4u³-u *1 = -u-u * (-u) = +u²5 *1 =55*(-u)= -5uSo combining:4u³ -4u² + u² -u -5u +5 =4u³ -3u² -6u +5Yes, that's correct. Therefore, h(u) =4u³ -3u² -6u +5Therefore, to find the maximum of h(u) on u ∈ [-1,1], we can take derivative and find critical points.Compute derivative h’(u):h’(u) =12u² -6u -6Set h’(u)=0:12u² -6u -6=0Divide both sides by 6:2u² - u -1=0Solve quadratic equation:u = [1 ± sqrt(1 +8)] /4 = [1 ±3]/4Thus, roots at u=(1+3)/4=1, and u=(1-3)/4=-0.5Therefore, critical points at u=1 and u=-0.5. However, u=1 is an endpoint of our interval. Therefore, the critical points inside the interval are u=-0.5.Therefore, we need to evaluate h(u) at u=-1, u=-0.5, and u=1.Wait, but also check if there are other critical points. The derivative h’(u)=12u² -6u -6, which we found roots at u=1 and u=-0.5. Since our interval is u ∈ [-1,1], both these roots are within the interval? Wait, u=1 is the endpoint, and u=-0.5 is inside the interval. So critical points are u=-0.5 and u=1 (endpoint). Also, need to check the other endpoint u=-1.Therefore, compute h(u) at u=-1, u=-0.5, and u=1.Compute h(-1):h(-1) =4*(-1)^3 -3*(-1)^2 -6*(-1)+5= -4 -3 +6 +5= (-4-3)+(6+5)= -7 +11=4h(-0.5):h(-0.5)=4*(-0.5)^3 -3*(-0.5)^2 -6*(-0.5)+5Compute each term:4*(-0.125)= -0.5-3*(0.25)= -0.75-6*(-0.5)=3So total: -0.5 -0.75 +3 +5= (-1.25) +8=6.75h(1)=4*(1)^3 -3*(1)^2 -6*(1)+5=4 -3 -6 +5=0So h(u) at critical points:u=-1: 4u=-0.5:6.75u=1:0Therefore, the maximum of h(u) is 6.75=27/4 at u=-0.5. But we need to check if this is indeed the maximum in the interval.Wait, but we should also check the behavior between u=-0.5 and u=1. Since the derivative h’(u)=12u² -6u -6.For u in (-0.5,1):Take u=0 as a test point:h’(0)=0 -0 -6= -6 <0, so the function is decreasing from u=-0.5 to u=1.Similarly, for u < -0.5, take u=-1:h’(-1)=12*(1) -6*(-1) -6=12 +6 -6=12>0, so the function is increasing from u=-1 to u=-0.5, then decreasing from u=-0.5 to u=1.Therefore, the maximum is indeed at u=-0.5 with h(-0.5)=27/4=6.75.Thus, the maximum value of g(t) is 27/4, achieved when u=cos t= -0.5. So, cos t=-0.5 implies t=2π/3.Therefore, the maximum value of g(t) is 27/4, occurring at t=2π/3.But let me confirm this calculation step by step to avoid mistakes.First, at u=-0.5:h(-0.5)= (-4*(-0.5)^2 - (-0.5) +5)(1 - (-0.5))= (-4*(0.25) +0.5 +5)(1.5)= (-1 +0.5 +5)(1.5)= (4.5)(1.5)=6.75=27/4.Yes, that's correct.Alternatively, expanding h(u)=4u³ -3u² -6u +5:At u=-0.5:4*(-0.5)^3=4*(-0.125)=-0.5-3*(-0.5)^2=-3*(0.25)=-0.75-6*(-0.5)=3+5=5Total: -0.5 -0.75 +3 +5= (-1.25) +8=6.75=27/4. Correct.Therefore, the maximum of g(t) is 27/4=6.75, achieved when cos t=-0.5, which corresponds to t=2π/3.Therefore, the final answer is maximum value of g(t) is 27/4 at t=2π/3.But let me just check if there are any other critical points or if the maximum could be somewhere else.Wait, we considered the critical points from the derivative and the endpoints. Since the function h(u) increases from u=-1 to u=-0.5, then decreases from u=-0.5 to u=1, the maximum is indeed at u=-0.5. Thus, the conclusion holds.Therefore, the maximum value of g(t) is 27/4, achieved when t=2π/3.Final AnswerThe maximum value of ( g(t) ) is (boxed{dfrac{27}{4}}) and it occurs at ( t = boxed{dfrac{2pi}{3}} ).

✅:Given the function ( S = sin^{2} x + sin^{2} y + left(8 sin^{2} t + 2right) sin x sin y ) where ( x ) and ( y ) are non-negative real numbers such that ( x + y = t ) and ( t in [0, pi] ), we need to find the maximum value of ( g(t) = M(t) - N(t) ), where ( M(t) ) and ( N(t) ) are the maximum and minimum values of ( S ) for each ( t ).First, we express ( S ) in terms of ( x ) and ( t ) by substituting ( y = t - x ):[S(x) = sin^{2} x + sin^{2}(t - x) + left(8 sin^{2} t + 2right) sin x sin(t - x)]Using trigonometric identities, we simplify ( S(x) ) to:[S(x) = A(t) cos(2x - t) + B(t)]where ( A(t) = -4 cos^2 t - cos t + 5 ) and ( B(t) = 1 - 5 cos t + 4 cos^3 t ).The maximum and minimum values of ( S(x) ) depend on the coefficient ( A(t) ). Since ( cos(2x - t) ) ranges from ( cos t ) to 1, the maximum and minimum values of ( S(x) ) are:- If ( A(t) > 0 ): ( M(t) = A(t) cdot 1 + B(t) ) and ( N(t) = A(t) cdot cos t + B(t) )- If ( A(t) < 0 ): ( M(t) = A(t) cdot cos t + B(t) ) and ( N(t) = A(t) cdot 1 + B(t) )Thus, ( g(t) = |A(t)| (1 - cos t) ).Substituting ( u = cos t ), we analyze the function ( h(u) = | -4u^2 - u + 5 | (1 - u) ) for ( u in [-1, 1] ). The quadratic ( -4u^2 - u + 5 ) is positive in ( [-1, 1) ), so ( h(u) = (-4u^2 - u + 5)(1 - u) ).Expanding and simplifying, we find:[h(u) = 4u^3 - 3u^2 - 6u + 5]Taking the derivative and solving for critical points, we find the maximum occurs at ( u = -0.5 ), corresponding to ( t = frac{2pi}{3} ). Evaluating ( h(u) ) at ( u = -0.5 ):[h(-0.5) = frac{27}{4}]Thus, the maximum value of ( g(t) ) is (boxed{dfrac{27}{4}}) and it occurs at ( t = boxed{dfrac{2pi}{3}} ).

🔑:1. We start with the given expression for ( S ):[ S = sin^2 x + sin^2 y + (8 sin^2 t + 2) sin x sin y ] Given that ( x ) and ( y ) are non-negative real numbers and ( x+y = t ).2. Using the angle addition and subtraction identities:[ sin^2 x + sin^2 y = frac{1 - cos(2x)}{2} + frac{1 - cos(2y)}{2} = frac{1}{2}(2 - cos(2x) - cos(2y)) ]3. Since ( x+y = t ), we get:[ y = t - x ]Now substitute ( y = t - x ) into the expression of ( S ):[ S = sin^2 x + sin^2 (t - x) + (8 sin^2 t + 2) sin x sin (t - x) ]4. Using the identity ( sin(a - b) = sin a cos b - cos a sin b ), we get:[ sin(t - x) = sin t cos x - cos t sin x ]5. Thus,[ S = sin^2 x + (sin t cos x - cos t sin x)^2 + (8 sin^2 t + 2) sin x (sin t cos x - cos t sin x) ]6. Simplifying:[ (sin t cos x - cos t sin x)^2 = sin^2 t cos^2 x + cos^2 t sin^2 x - 2 sin t cos t sin x cos x ] So, [ S = sin^2 x + sin^2 t cos^2 x + cos^2 t sin^2 x - 2 sin t cos t sin x cos x + (8 sin^2 t + 2) sin x (sin t cos x - cos t sin x) ]7. Further simplification gives:[ S = sin^2 t + (10 - 8 cos^2 t - 2 cos t) sin x sin y ]8. Now we express it using (cos(x-y)):[ S = sin^2 t + (5 - 4 cos^2 t - cos t)(cos(x-y) - cos t) ]9. Since ( cos(x-y) in [-1, 1] ), the expression:[ 5 - 4 cos^2 t - cos t geqslant 0 ]10. We find the maximum and minimum values:[ M(t) = sin^2 t + (5 - 4 cos^2 t - cos t)(1 - cos t) ][ N(t) = sin^2 t ]11. Define ( g(t) ):[ g(t) = M(t) - N(t) = (5 - 4 cos^2 t - cos t)(1 - cos t) ]12. Further simplification:[ g(t) = (4 cos t + 5)(1 - cos t)^2 ][ = 2 left( 2 cos t + frac{5}{2} right)(1 - cos t)^2 ]13. Using the arithmetic mean-geometric mean (AM-GM) inequality:[ left( 2 cos t + frac{5}{2} right) + (1 - cos t) + (1 - cos t) geq 3 sqrt[3]{ left( 2 cos t + frac{5}{2} right) (1 - cos t)^2 } ][ = 2 left( frac{3}{2} right)^3 = frac{27}{4} ]14. Equality holds when:[ 2 cos t + frac{5}{2} = 1 - cos t ] Solving for ( t ):[ cos t = -frac{1}{2} ][ t = frac{2pi}{3} ] Conclusion:The maximum value of ( g(t) ) is ( frac{27}{4} ) when ( t = frac{2pi}{3} ).[ boxed{frac{27}{4}} ]