💡Alright, so I have this problem about a cone. The vertex is S, and there are some things given: the cosine of the angle between two generatrices SA and SB is sqrt(15)/4, the angle between SA and the base is 60 degrees, and the area of triangle SAB is 2. I need to find the volume of the cone.Okay, let's start by understanding what's given. A cone has a vertex S, and SA and SB are generatrices, which means they are the slant heights from the vertex to the base. The angle between SA and SB has a cosine of sqrt(15)/4. Also, the angle between SA and the base is 60 degrees. The area of triangle SAB is 2.First, I think I need to find the length of the generatrix SA. Since the area of triangle SAB is given, and I know the angle between SA and SB, maybe I can use the formula for the area of a triangle involving two sides and the sine of the included angle.The formula is: Area = (1/2)*ab*sin(theta), where a and b are the lengths of the sides, and theta is the included angle.In this case, both SA and SB are generatrices, so they should be equal in length because it's a cone. Let's denote the length of SA as 'l'. So, the area of triangle SAB is (1/2)*l*l*sin(theta) = 2.We know that cos(theta) is sqrt(15)/4, so we can find sin(theta) using the identity sin^2(theta) + cos^2(theta) = 1.Calculating sin(theta):sin(theta) = sqrt(1 - cos^2(theta)) = sqrt(1 - (sqrt(15)/4)^2) = sqrt(1 - 15/16) = sqrt(1/16) = 1/4.So, sin(theta) is 1/4.Now, plug this into the area formula:(1/2)*l^2*(1/4) = 2Simplify:(1/8)*l^2 = 2Multiply both sides by 8:l^2 = 16Take square root:l = 4So, the length of each generatrix SA and SB is 4 units.Next, I need to relate this to the cone's dimensions. The angle between SA and the base is 60 degrees. In a cone, the generatrix makes an angle with the base, which is related to the height and the radius of the cone.Let me visualize the cone: the vertex S is connected to the base, which is a circle. The generatrix SA is slant height, and the angle between SA and the base is 60 degrees. So, if I consider the right triangle formed by the height (h), radius (r), and generatrix (l), the angle between SA and the base is 60 degrees.In this right triangle, the angle at the base is 60 degrees, so the adjacent side is the radius, and the hypotenuse is the generatrix (l = 4). Therefore, cos(60 degrees) = adjacent/hypotenuse = r/l.We know cos(60 degrees) is 0.5, so:0.5 = r / 4Multiply both sides by 4:r = 2So, the radius of the base is 2 units.Now, to find the height (h), we can use the Pythagorean theorem:l^2 = r^2 + h^2We know l = 4 and r = 2, so:4^2 = 2^2 + h^216 = 4 + h^2Subtract 4:h^2 = 12Take square root:h = 2*sqrt(3)So, the height of the cone is 2*sqrt(3) units.Now, to find the volume of the cone, the formula is:Volume = (1/3)*pi*r^2*hPlugging in the values:Volume = (1/3)*pi*(2)^2*(2*sqrt(3)) = (1/3)*pi*4*(2*sqrt(3)) = (1/3)*pi*8*sqrt(3) = (8*sqrt(3)/3)*piSo, the volume of the cone is (8*sqrt(3)/3)*pi.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Calculated sin(theta) correctly: sqrt(1 - (sqrt(15)/4)^2) = 1/4.2. Used area formula: (1/2)*l^2*(1/4) = 2, leading to l = 4. That seems correct.3. Angle between SA and base is 60 degrees, so cos(60) = r/l => r = 2. That makes sense.4. Pythagorean theorem: l^2 = r^2 + h^2 => 16 = 4 + h^2 => h^2 = 12 => h = 2*sqrt(3). Correct.5. Volume formula: (1/3)*pi*r^2*h = (1/3)*pi*4*2*sqrt(3) = (8*sqrt(3)/3)*pi. Yes, that's right.I think all steps are correct. So, the volume should be (8*sqrt(3)/3)*pi.**Final Answer**The volume of the cone is boxed{dfrac{8sqrt{3}}{3}pi}.

💡Okay, so I need to figure out how many ways to distribute 7 volunteer positions among 3 schools. Each school has to get at least one position, and the allocations need to be distinct. Hmm, let me break this down step by step.First, I know that distributing positions among schools is a combinatorial problem. Since each school must get at least one position, I can't have any school getting zero positions. Also, the allocations need to be distinct, which probably means that each school gets a different number of positions.So, I need to find all the possible ways to partition 7 positions into 3 distinct positive integers. That is, I need to find all sets of three different numbers that add up to 7. Let me list them out.Starting with the smallest possible number, which is 1. If one school gets 1 position, then the remaining 6 positions need to be split between the other two schools, each getting at least one position, and the numbers need to be distinct.So, let's see:1. If one school gets 1, another gets 2, then the third must get 4 (since 1 + 2 + 4 = 7).2. If one school gets 1, another gets 3, then the third must get 3. But wait, that's not allowed because the allocations need to be distinct. So, 1, 3, 3 is invalid.3. If one school gets 1, another gets 4, then the third must get 2. But that's the same as the first case, just reordered.4. If one school gets 1, another gets 5, then the third must get 1. Again, duplicates, which isn't allowed.So, the only valid partition when starting with 1 is 1, 2, 4.Now, let's check if there are other partitions without starting with 1.If I start with 2:1. If one school gets 2, another gets 3, then the third must get 2. Again, duplicates, so invalid.2. If one school gets 2, another gets 4, then the third must get 1. That's the same as the first case, just reordered.If I start with 3:1. If one school gets 3, another gets 4, then the third must get 0, which is invalid because each school must get at least one position.So, it seems like the only valid partition is 1, 2, 4.Now, since the allocations are distinct, the order matters. That is, assigning 1 to School A, 2 to School B, and 4 to School C is different from assigning 1 to School B, 2 to School A, and 4 to School C, and so on.So, how many ways can we assign these numbers to the schools? Since there are 3 schools and 3 distinct numbers, it's a permutation problem.The number of permutations of 3 distinct items is 3! = 6.Therefore, there are 6 ways to distribute the 7 volunteer positions among the 3 schools, ensuring each school gets at least one position and the allocations are distinct.

💡Okay, so I've got this problem about Henrique and his two urns. Let me try to wrap my head around it step by step. First, there are two urns. The first one has only black balls, and the second one has only white balls. That's clear. Now, Henrique takes a certain number of balls from the first urn and moves them to the second urn. Then, he takes the same number of balls from the second urn and moves them back to the first urn. After all that, the question is asking whether the number of white balls in the first urn is greater than, less than, or equal to the number of black balls in the second urn.Hmm, okay. Let me break this down. Let's say the first urn has B black balls, and the second urn has W white balls. Now, Henrique takes, say, k balls from the first urn and moves them to the second urn. So, after this first transfer, the first urn will have B - k black balls left, and the second urn will have W + k balls, which includes the original W white balls plus k black balls.Now, the second step is where it gets a bit tricky. Henrique takes k balls from the second urn and moves them back to the first urn. But here's the thing: the second urn now has both white and black balls. So, when he takes k balls from it, some of them might be white and some might be black. Let me think about this. Suppose that out of the k balls he takes from the second urn, p are black and q are white. So, p + q = k. That means, after this second transfer, the first urn will have B - k + p black balls and q white balls. Meanwhile, the second urn will have W + k - k = W white balls and k - p black balls left.Wait, so the number of white balls in the first urn is q, and the number of black balls in the second urn is k - p. But since p + q = k, that means q = k - p. So, the number of white balls in the first urn is equal to the number of black balls in the second urn.Is that right? Let me double-check. If I start with B black balls and W white balls, move k black balls to the second urn, then move k balls back, which are a mix of black and white, the number of white balls that end up in the first urn should be the same as the number of black balls that stayed in the second urn. Yeah, that makes sense. Because every white ball that moves to the first urn corresponds to a black ball that didn't move back. So, they should balance out. Therefore, the number of white balls in the first urn is equal to the number of black balls in the second urn.I think that's the answer. It seems a bit counterintuitive at first, but when you break it down step by step, it makes sense.

💡Okay, so I have this problem about order numbers in arrays. Let me try to understand it step by step. First, the problem defines an "order" in an array as a pair of elements where the earlier element is smaller than the later one. For example, in the array (2, 4, 3, 1), the orders are "2, 4" and "2, 3" because 2 is less than both 4 and 3, which come after it. So, the order number is just the total count of such pairs. In this case, it's 2.Now, the problem gives me an array (a₁, a₂, a₃, a₄, a₅) with a distinct positive number elements, and it tells me that its order number is 4. I need to find the order number of the reversed array, which is (a₅, a₄, a₃, a₂, a₁). The options are A:7, B:6, C:5, D:4.Hmm, okay. So, first, I need to figure out what the original array looks like, or at least understand its structure, given that it has an order number of 4. Then, I can reverse it and compute the order number for the reversed array.Let me recall that in any array, the maximum possible order number is the number of pairs, which for n elements is n(n-1)/2. For n=5, that would be 10. So, the original array has 4 orders, which is relatively low. That suggests that the array is mostly decreasing, but not entirely. Because if it were entirely decreasing, the order number would be 0, right? Since no element is smaller than the one after it.Wait, no. Actually, if the array is entirely increasing, the order number would be maximum, which is 10 for n=5. If it's entirely decreasing, the order number would be 0 because no element is smaller than the one after it. So, in this case, the original array has 4 orders, which is somewhere between 0 and 10. So, it's not too increasing or too decreasing.I need to figure out how the orders are distributed. Maybe it's helpful to think about the concept of inversions. Wait, in computer science, an inversion is a pair of elements where the earlier element is greater than the later one. So, the number of inversions is related to how "out of order" the array is. But in this problem, the order number is the number of increasing pairs, which is the opposite of inversions.So, if the order number is 4, that means there are 4 increasing pairs, and the number of inversions would be 10 - 4 = 6. So, the original array has 6 inversions.But maybe I don't need to think in terms of inversions. Let me think about how the orders are formed. Each order is a pair (i_p, i_q) where p < q and i_p < i_q. So, for each element, I can count how many elements after it are larger than it. The sum of these counts across all elements is the order number.So, for the array (a₁, a₂, a₃, a₄, a₅), the order number is 4. That means the total number of such increasing pairs is 4.I need to figure out how the elements are arranged. Let's denote the elements in increasing order as b₁ < b₂ < b₃ < b₄ < b₅. So, each a_i is one of these b's, but in some permutation.Given that the order number is 4, which is relatively low, the array is mostly decreasing, but with a few increasing steps.Let me think about possible permutations of 5 elements with exactly 4 increasing pairs.One approach is to consider that in a permutation, the number of increasing pairs can be calculated by looking at the number of ascents. An ascent is a position where a_i < a_{i+1}. The number of increasing pairs is related to the number of ascents, but not exactly the same because each ascent can contribute multiple increasing pairs.Wait, actually, the number of increasing pairs is the sum over each element of the number of elements after it that are larger. So, for example, if the array is (1, 3, 2, 5, 4), then:- 1 has 3 elements after it that are larger: 3, 2, 5, 4 (but wait, 2 is smaller than 1, so actually, 3 and 5 are larger, so 2 elements.- 3 has 1 element after it that is larger: 5- 2 has 1 element after it that is larger: 5- 5 has 0 elements after it that are larger- 4 has 0 elements after it that are largerSo, total increasing pairs: 2 + 1 + 1 = 4. So, this array has an order number of 4.So, the array (1, 3, 2, 5, 4) has order number 4. Now, if we reverse it, we get (4, 5, 2, 3, 1). Let's compute its order number.Compute the number of increasing pairs:- 4: elements after it are 5, 2, 3, 1. How many are larger than 4? 5 is larger, so 1.- 5: elements after it are 2, 3, 1. None are larger than 5.- 2: elements after it are 3, 1. 3 is larger, so 1.- 3: elements after it are 1. None are larger.- 1: no elements after it.Total increasing pairs: 1 + 0 + 1 + 0 + 0 = 2. Wait, that's only 2. But the options don't have 2. Hmm, maybe I made a mistake.Wait, let's check the original array again. (1, 3, 2, 5, 4). The increasing pairs are:1 < 3, 1 < 2, 1 < 5, 1 < 4, 3 < 5, 2 < 5. Wait, that's 6 increasing pairs. But earlier, I thought it was 4. Hmm, I must have miscalculated.Wait, no. The definition is that an "order" is a pair where p < q and a_p < a_q. So, for the array (1, 3, 2, 5, 4):- 1 < 3: yes- 1 < 2: yes- 1 < 5: yes- 1 < 4: yes- 3 < 2: no- 3 < 5: yes- 3 < 4: yes- 2 < 5: yes- 2 < 4: yes- 5 < 4: noWait, so actually, the increasing pairs are:(1,3), (1,2), (1,5), (1,4), (3,5), (3,4), (2,5), (2,4). That's 8 increasing pairs. But the problem says the order number is 4. So, I must have misunderstood the definition.Wait, the problem says: "if i_p < i_q when p < q, then 'the order of i_p and i_q' is considered an 'order' of the array." So, each such pair is an order. So, the total number is the count of all such pairs.But in the example given, (2,4,3,1) has order number 2. Let's check:- 2 < 4: yes- 2 < 3: yes- 2 < 1: no- 4 < 3: no- 4 < 1: no- 3 < 1: noSo, only two increasing pairs: (2,4) and (2,3). So, the order number is 2.Wait, so in that example, only the first element has two increasing pairs. The other elements don't contribute any. So, in the original array, the order number is 4, which means there are 4 such increasing pairs.So, going back to my previous example, (1,3,2,5,4) has more than 4 increasing pairs, so it's not a good example.I need to find an array of 5 elements with exactly 4 increasing pairs.Let me try to construct such an array.Let's think about the minimal number of increasing pairs. If the array is almost decreasing, except for a few places.Suppose the array is (3,4,2,5,1). Let's count the increasing pairs:- 3 < 4: yes- 3 < 2: no- 3 < 5: yes- 3 < 1: no- 4 < 2: no- 4 < 5: yes- 4 < 1: no- 2 < 5: yes- 2 < 1: no- 5 < 1: noSo, increasing pairs are (3,4), (3,5), (4,5), (2,5). That's 4. So, this array has order number 4.Now, if I reverse this array, I get (1,5,2,4,3). Let's compute its order number.Compute the increasing pairs:- 1 < 5: yes- 1 < 2: yes- 1 < 4: yes- 1 < 3: yes- 5 < 2: no- 5 < 4: no- 5 < 3: no- 2 < 4: yes- 2 < 3: yes- 4 < 3: noSo, increasing pairs are (1,5), (1,2), (1,4), (1,3), (2,4), (2,3). That's 6 increasing pairs. So, the order number is 6.Therefore, the answer should be B:6.Wait, but let me double-check with another example to make sure.Suppose the original array is (2,3,1,4,5). Let's count the increasing pairs:- 2 < 3: yes- 2 < 1: no- 2 < 4: yes- 2 < 5: yes- 3 < 1: no- 3 < 4: yes- 3 < 5: yes- 1 < 4: yes- 1 < 5: yes- 4 < 5: yesWait, that's way more than 4. So, that's not a good example.Wait, maybe another approach. Let's think about the number of increasing pairs as the sum over each element of the number of elements after it that are larger.So, for the original array, the total is 4.Let me denote the array as a₁, a₂, a₃, a₄, a₅.We have:Number of elements after a₁ that are larger: let's say x₁Number of elements after a₂ that are larger: x₂Similarly, x₃, x₄, x₅.But x₅ is always 0 because there are no elements after a₅.So, x₁ + x₂ + x₃ + x₄ = 4.We need to find a permutation where this sum is 4.One way is to have x₁=2, x₂=1, x₃=1, x₄=0.So, a₁ has 2 elements after it that are larger, a₂ has 1, a₃ has 1, a₄ has 0.So, let's construct such an array.Let's assign the smallest element to a₁, then the next smallest to a position where it can have 2 larger elements after it.Wait, maybe it's better to think in terms of ranks.Let me assign ranks to the elements. Let's say the elements are 1,2,3,4,5 in some order.We need to arrange them such that the sum of the number of larger elements after each position is 4.So, for example:If a₁=1, then x₁=4 (since all elements after it are larger). But we need x₁ + x₂ + x₃ + x₄ =4, so if x₁=4, the rest must be 0, which would mean the array is (1,5,4,3,2). Let's check:- 1 <5,1<4,1<3,1<2: 4- 5 <4,5<3,5<2: 0- 4 <3,4<2: 0- 3 <2:0- 2:0Total:4. So, this array has order number 4.Now, reversing it gives (2,3,4,5,1). Let's compute its order number.Compute the increasing pairs:- 2 <3: yes- 2 <4: yes- 2 <5: yes- 2 <1: no- 3 <4: yes- 3 <5: yes- 3 <1: no- 4 <5: yes- 4 <1: no- 5 <1: noSo, increasing pairs: (2,3), (2,4), (2,5), (3,4), (3,5), (4,5). That's 6. So, order number is 6.Another example: original array (1,5,4,3,2), reversed is (2,3,4,5,1). Order number 6.Another example: original array (3,4,2,5,1). Reversed is (1,5,2,4,3). Order number 6 as computed earlier.So, it seems consistent that reversing an array with order number 4 results in an array with order number 6.Therefore, the answer is B:6.

💡Okay, so I have this geometry problem here. It says that in a convex quadrilateral ABCD, M and N are the midpoints of BC and AD respectively. We're given that AM equals BN and DM equals CN. We need to prove that AC equals BD. Hmm, interesting. Let me try to visualize this.First, let me draw a convex quadrilateral ABCD. So, A, B, C, D are the vertices in order, making sure that all interior angles are less than 180 degrees. M is the midpoint of BC, so I'll mark that point. Similarly, N is the midpoint of AD, so I'll mark that too.Now, given that AM equals BN. So, the length from A to M is the same as from B to N. Also, DM equals CN, meaning the length from D to M is the same as from C to N. I need to show that the diagonals AC and BD are equal in length.Hmm, maybe coordinate geometry can help here. Let me assign coordinates to the points. Let's set point A at (0, 0) for simplicity. Let me denote point B as (2b, 0) so that the midpoint calculations might be easier. Then, point D can be at (2d, 2e), and point C at (2c, 2f). I'm doubling the coordinates to make the midpoints have integer coordinates, which might simplify things.So, coordinates:- A: (0, 0)- B: (2b, 0)- D: (2d, 2e)- C: (2c, 2f)Now, M is the midpoint of BC. So, coordinates of M would be the average of B and C:M_x = (2b + 2c)/2 = b + cM_y = (0 + 2f)/2 = fSo, M is (b + c, f)Similarly, N is the midpoint of AD:N_x = (0 + 2d)/2 = dN_y = (0 + 2e)/2 = eSo, N is (d, e)Now, let's compute the lengths AM, BN, DM, and CN.First, AM is the distance from A(0,0) to M(b + c, f):AM² = (b + c - 0)² + (f - 0)² = (b + c)² + f²BN is the distance from B(2b, 0) to N(d, e):BN² = (d - 2b)² + (e - 0)² = (d - 2b)² + e²Given that AM = BN, so their squares are equal:(b + c)² + f² = (d - 2b)² + e² ...(1)Similarly, DM is the distance from D(2d, 2e) to M(b + c, f):DM² = (b + c - 2d)² + (f - 2e)²CN is the distance from C(2c, 2f) to N(d, e):CN² = (d - 2c)² + (e - 2f)²Given that DM = CN, so their squares are equal:(b + c - 2d)² + (f - 2e)² = (d - 2c)² + (e - 2f)² ...(2)Now, I have two equations (1) and (2) with variables b, c, d, e, f. I need to manipulate these equations to find a relationship between the diagonals AC and BD.First, let's expand equation (1):(b + c)² + f² = (d - 2b)² + e²Expanding both sides:b² + 2bc + c² + f² = d² - 4bd + 4b² + e²Bring all terms to the left:b² + 2bc + c² + f² - d² + 4bd - 4b² - e² = 0Simplify:-3b² + 2bc + c² + f² - d² + 4bd - e² = 0 ...(1a)Now, let's expand equation (2):(b + c - 2d)² + (f - 2e)² = (d - 2c)² + (e - 2f)²Expanding each term:First term: (b + c - 2d)² = b² + c² + 4d² + 2bc - 4bd - 4cdSecond term: (f - 2e)² = f² + 4e² - 4efThird term: (d - 2c)² = d² + 4c² - 4cdFourth term: (e - 2f)² = e² + 4f² - 4efSo, putting it all together:(b² + c² + 4d² + 2bc - 4bd - 4cd) + (f² + 4e² - 4ef) = (d² + 4c² - 4cd) + (e² + 4f² - 4ef)Simplify both sides:Left side: b² + c² + 4d² + 2bc - 4bd - 4cd + f² + 4e² - 4efRight side: d² + 4c² - 4cd + e² + 4f² - 4efSubtract right side from left side:b² + c² + 4d² + 2bc - 4bd - 4cd + f² + 4e² - 4ef - d² - 4c² + 4cd - e² - 4f² + 4ef = 0Simplify term by term:b² + (c² - 4c²) + (4d² - d²) + 2bc + (-4bd) + (-4cd + 4cd) + (f² - 4f²) + (4e² - e²) + (-4ef + 4ef) = 0Which simplifies to:b² - 3c² + 3d² + 2bc - 4bd - 3f² + 3e² = 0 ...(2a)Now, I have equations (1a) and (2a):(1a): -3b² + 2bc + c² + f² - d² + 4bd - e² = 0(2a): b² - 3c² + 3d² + 2bc - 4bd - 3f² + 3e² = 0Hmm, these are two equations with variables b, c, d, e, f. Maybe I can add them or manipulate them to find a relationship.Let me write them again:Equation (1a):-3b² + 2bc + c² + f² - d² + 4bd - e² = 0Equation (2a):b² - 3c² + 3d² + 2bc - 4bd - 3f² + 3e² = 0Let me add (1a) and (2a):(-3b² + b²) + (2bc + 2bc) + (c² - 3c²) + (f² - 3f²) + (-d² + 3d²) + (4bd - 4bd) + (-e² + 3e²) = 0 + 0Simplify each term:-2b² + 4bc - 2c² - 2f² + 2d² + 0 + 2e² = 0Divide both sides by 2:-b² + 2bc - c² - f² + d² + e² = 0Rearranged:d² + e² - b² - c² - f² + 2bc = 0 ...(3)Hmm, not sure what to do with this yet. Maybe I can find expressions for AC and BD in terms of these variables.Diagonal AC is from A(0,0) to C(2c, 2f). So, AC² = (2c - 0)² + (2f - 0)² = 4c² + 4f²Diagonal BD is from B(2b, 0) to D(2d, 2e). So, BD² = (2d - 2b)² + (2e - 0)² = 4(d - b)² + 4e²We need to show that AC² = BD², which would mean:4c² + 4f² = 4(d - b)² + 4e²Divide both sides by 4:c² + f² = (d - b)² + e²So, if I can show that c² + f² = (d - b)² + e², then AC = BD.Looking back at equation (3):d² + e² - b² - c² - f² + 2bc = 0Let me rearrange equation (3):d² + e² = b² + c² + f² - 2bcSo, (d - b)² = d² - 2bd + b²From equation (3), d² = b² + c² + f² - 2bc - e²Wait, maybe I can substitute d² from equation (3) into (d - b)².Wait, let's compute (d - b)² + e²:(d - b)² + e² = d² - 2bd + b² + e²From equation (3), d² + e² = b² + c² + f² - 2bcSo, substitute d² + e²:(d - b)² + e² = (b² + c² + f² - 2bc) - 2bd + b²Wait, that seems messy. Maybe another approach.Wait, from equation (3):d² + e² = b² + c² + f² - 2bcSo, (d - b)² + e² = d² - 2bd + b² + e² = (d² + e²) - 2bd + b² = (b² + c² + f² - 2bc) - 2bd + b² = 2b² + c² + f² - 2bc - 2bdHmm, not sure. Maybe I need to find another relation.Wait, let's go back to equation (1a):-3b² + 2bc + c² + f² - d² + 4bd - e² = 0From equation (3), we have d² + e² = b² + c² + f² - 2bcSo, let's substitute d² + e² into equation (1a):-3b² + 2bc + c² + f² - (d² + e²) + 4bd = 0But d² + e² = b² + c² + f² - 2bc, so:-3b² + 2bc + c² + f² - (b² + c² + f² - 2bc) + 4bd = 0Simplify:-3b² + 2bc + c² + f² - b² - c² - f² + 2bc + 4bd = 0Combine like terms:(-3b² - b²) + (2bc + 2bc) + (c² - c²) + (f² - f²) + 4bd = 0Which simplifies to:-4b² + 4bc + 4bd = 0Divide both sides by 4:-b² + bc + bd = 0Factor:b(-b + c + d) = 0So, either b = 0 or -b + c + d = 0If b = 0, then point B would be at (0,0), which is the same as point A, which can't be in a convex quadrilateral. So, we discard b = 0.Thus, -b + c + d = 0 => c + d = b ...(4)Okay, that's a useful relation. Now, let's use this in equation (3):From equation (3):d² + e² = b² + c² + f² - 2bcBut from equation (4), c = b - d. Let's substitute c = b - d into equation (3):d² + e² = b² + (b - d)² + f² - 2b(b - d)Expand (b - d)²:= b² - 2bd + d²So, substitute:d² + e² = b² + (b² - 2bd + d²) + f² - 2b² + 2bdSimplify:d² + e² = b² + b² - 2bd + d² + f² - 2b² + 2bdCombine like terms:d² + e² = (b² + b² - 2b²) + (-2bd + 2bd) + d² + f²Which simplifies to:d² + e² = d² + f²Subtract d² from both sides:e² = f²So, e = ±fBut since we're dealing with lengths, e and f are real numbers, so e = f or e = -f. However, in a convex quadrilateral, the coordinates are such that points are arranged in order without crossing, so e and f would have the same sign. Thus, e = f.So, e = f.Now, let's recall that we need to show AC² = BD², which is 4c² + 4f² = 4(d - b)² + 4e²Divide both sides by 4:c² + f² = (d - b)² + e²But we have e = f, so:c² + f² = (d - b)² + f²Subtract f² from both sides:c² = (d - b)²Take square roots:c = ±(d - b)But from equation (4), c = b - d. So, c = -(d - b). Therefore, c = - (d - b) => c = b - d, which is consistent.Thus, c² = (d - b)², so indeed, c² + f² = (d - b)² + f²Therefore, AC² = BD², so AC = BD.Hence, we've proven that AC equals BD.

💡Okay, so I have this problem where Derek is filling a 10x10 grid with 50 ones and 50 twos. Then he takes the product of the numbers in each row and each column, sums all those products up, and gets an integer N. I need to find the minimum possible value of N.Alright, let me break this down. First, the grid is 10x10, so there are 100 cells in total. He's putting 50 ones and 50 twos in there. So, each cell is either a 1 or a 2, with exactly half of them being 1s and the other half being 2s.Now, for each row, he takes the product of the numbers in that row. Since each row has 10 numbers, and each number is either 1 or 2, the product of each row will be 2 raised to the number of 2s in that row. Similarly, for each column, the product will be 2 raised to the number of 2s in that column.So, if I denote the number of 2s in the i-th row as a_i, then the product for that row is 2^{a_i}. Similarly, for the j-th column, if the number of 2s is b_j, the product is 2^{b_j}. Since there are 10 rows and 10 columns, Derek will have 10 row products and 10 column products. He then sums all these 20 products to get N. So, N is the sum of 2^{a_1} + 2^{a_2} + ... + 2^{a_{10}} + 2^{b_1} + 2^{b_2} + ... + 2^{b_{10}}}.Our goal is to arrange the 50 twos and 50 ones in the grid such that this sum N is minimized.Hmm, okay. So, to minimize N, we need to arrange the twos in such a way that the sum of these 2^{a_i} and 2^{b_j} is as small as possible.I remember that for products, having numbers as close as possible tends to minimize or maximize certain expressions. Maybe something similar applies here.Let me think about the properties of exponents. Since 2^x is an increasing function, the larger the exponent, the larger the value. So, to minimize the sum, we want the exponents to be as small as possible. But we have a fixed total number of twos, which is 50.So, if we spread out the twos as evenly as possible across the rows and columns, we might be able to minimize the sum. Because if some rows or columns have a lot of twos, their products will be much larger, which would increase N.So, the key idea is to distribute the 50 twos as evenly as possible across the 10 rows and 10 columns.Let me calculate how many twos each row and each column should ideally have. Since there are 10 rows and 50 twos, each row should have 5 twos. Similarly, each column should have 5 twos. Because 10 rows times 5 twos per row is 50, and similarly for columns.So, if each row has exactly 5 twos and each column has exactly 5 twos, then the product for each row is 2^5, and the product for each column is also 2^5. Therefore, each row product is 32, and each column product is 32.Then, the sum N would be 10*32 (for the rows) plus 10*32 (for the columns), which is 320 + 320 = 640.Is this the minimum possible value? Let me check if this is achievable.Yes, if we arrange the grid such that each row and each column has exactly 5 twos, then this configuration is possible. For example, a checkerboard pattern where every other cell is a 2 would result in each row and column having exactly 5 twos. So, this is achievable.But wait, let me make sure that this is indeed the minimum. Suppose we have some rows with more than 5 twos and some with fewer. Would that make the sum N smaller?Let me consider two rows. Suppose one row has 6 twos and another has 4 twos. The products would be 2^6 = 64 and 2^4 = 16. The sum for these two rows would be 64 + 16 = 80. If both rows had 5 twos, their products would be 32 each, so the sum would be 64. So, 64 is less than 80. Therefore, having uneven distribution increases the sum.Similarly, for columns, the same logic applies. So, to minimize the sum, we need to have as uniform a distribution as possible.Therefore, having each row and each column with exactly 5 twos gives the minimal sum.Hence, the minimal N is 640.Wait, but let me think again. Is there a way to have some rows with fewer twos and some with more, but somehow the total sum ends up being smaller? Maybe because the products are multiplicative, perhaps some combination could lead to a lower total?Let me test with a smaller grid. Suppose we have a 2x2 grid with 2 ones and 2 twos. What's the minimal sum of row products and column products?If we arrange it as:1 22 1Then, row products are 1*2=2 and 2*1=2. Column products are 1*2=2 and 2*1=2. So, total sum is 2+2+2+2=8.Alternatively, if we arrange it as:1 12 2Then, row products are 1*1=1 and 2*2=4. Column products are 1*2=2 and 1*2=2. Total sum is 1+4+2+2=9, which is higher.Another arrangement:1 21 2Row products: 1*2=2 and 1*2=2. Column products: 1*1=1 and 2*2=4. Total sum: 2+2+1+4=9, which is also higher.So, in the 2x2 case, the minimal sum is achieved when each row and column has an equal number of twos, which is 1 per row and column.Similarly, in the 10x10 case, distributing the twos as evenly as possible across rows and columns minimizes the sum.Therefore, my initial conclusion seems correct. The minimal N is 640.