🧠:Okay, so I need to prove that a ball shot from a corner of a rectangular billiard table with dimensions p by 2q, where both p and q are odd numbers, will definitely end up in one of the middle pockets. The table has pockets at each corner and in the middle of each longer side (since the longer sides are length 2q). The ball is hit at a 45-degree angle to the sides. First, let me visualize the table. The table is longer along the 2q side, so the middle pockets are on the sides of length 2q. Since p and q are both odd, that might have something to do with the reflections and how the ball travels. When a ball is shot at 45 degrees on a rectangular table, it reflects off the sides in such a way that the angle of incidence equals the angle of reflection. Instead of tracking the actual reflections, a common technique in billiard problems is to "unfold" the table by reflecting it across its sides each time the ball would hit a wall. This way, the ball's path becomes a straight line in this grid of reflected tables, and we can analyze where it would end up based on the least common multiples or some periodic pattern.Given the 45-degree angle, the ball will travel in a diagonal path, moving one unit in the x-direction and one unit in the y-direction for each unit of distance. But since the table is p by 2q, the actual distances might be different. Wait, but the table's dimensions are given as p and 2q. Since the ball is moving at 45 degrees, the number of reflections it undergoes before returning to a corner (or a pocket) should relate to the least common multiple (LCM) of the table's dimensions. However, since we're dealing with reflections, maybe the LCM of p and 2q? But since p and q are odd, 2q is even. The LCM of an odd and even number. Hmm.Wait, but when you unfold the table, the ball's path is a straight line in this extended grid. So, the ball will end up in a pocket when the coordinates of the extended grid correspond to a corner or a middle pocket. But since the original pockets are at the corners and midpoints of the longer sides, maybe the middle pockets are at (p, q) and (0, q) if we consider the table from (0,0) to (p, 2q). Wait, the middle of the longer sides (which are length 2q) would be at (0, q) and (p, q). So, starting from a corner, say (0,0), the ball is shot at 45 degrees. Let's think about unfolding the table.Each time the ball hits a side, we can mirror the table, so the ball continues straight. So, after some number of reflections, the ball's path will reach a point that is equivalent to a pocket. Since the ball is moving diagonally, the number of times it reflects horizontally and vertically will depend on the table's dimensions.Alternatively, the key idea is that the number of reflections needed for the ball to reach a pocket is determined by the least common multiple of the table's length and width, but scaled by the direction of movement. However, since the ball is moving at 45 degrees, the horizontal and vertical distances traveled are the same. Therefore, the ball will have traveled a distance that is a multiple of both p and 2q. Wait, but because it's moving diagonally, the distance until it reaches a corner would be LCM(p, 2q), but since it's moving in both directions, perhaps the number of times it reflects is LCM(p, 2q)/p in the horizontal direction and LCM(p, 2q)/(2q) in the vertical direction. But since LCM(p, 2q) is the smallest number that's a multiple of both p and 2q, then LCM(p, 2q) = (p*2q)/GCD(p, 2q). Since p and q are both odd, GCD(p, 2q) = GCD(p, q). Because 2 and p are coprime (p is odd), so GCD(p, 2q) = GCD(p, q). Let’s denote d = GCD(p, q). Then LCM(p, 2q) = (p*2q)/d.Therefore, the number of horizontal reflections would be (2q)/d - 1? Wait, maybe not. Let me think again. When you unfold the table, the number of tables in the horizontal direction would be LCM(p, 2q)/p, and similarly in the vertical direction LCM(p, 2q)/2q. But since the ball is moving at 45 degrees, the number of tables traversed in each direction should be the same. Wait, but LCM(p, 2q)/p and LCM(p, 2q)/2q should be integers. Let’s see. For example, if p = 3 and q = 1, then 2q = 2. LCM(3, 2) = 6. Then LCM/p = 6/3 = 2, LCM/(2q) = 6/2 = 3. But 2 and 3 are different, which would mean that the path crosses 2 tables horizontally and 3 vertically? That seems contradictory. Wait, maybe this approach isn't correct.Alternatively, perhaps we can model the coordinates of the ball after reflections. Let’s use coordinate system where the table is from (0, 0) to (p, 2q). The ball starts at (0, 0) and moves with direction (1, 1). Each time it hits a wall, it reflects, which can be modeled by flipping the direction in the corresponding axis. But instead of changing direction, we can think of continuing the straight line in an unfolded grid of mirrored tables. So, the ball's trajectory is a straight line with slope 1 in this grid. The pockets are located at the corners (0,0), (0,2q), (p,0), (p,2q), and the middle pockets at (0, q), (p, q).So, the ball will end up in a pocket when the coordinates (k*p, m*2q) or (k*p, q + m*2q) for integers k, m. Since the ball is moving along the line y = x. So, we need to find integers t such that the coordinates (t, t) satisfy either t = k*p and t = m*2q (for a corner pocket), or t = k*p and t = q + m*2q (for a middle pocket).Therefore, solving t = k*p = m*2q would give t as a common multiple of p and 2q. The least such t is LCM(p, 2q). Similarly, solving t = k*p = q + m*2q. Let's analyze both possibilities.First, for the corner pocket: t must be a multiple of both p and 2q. Since p and 2q are both integers, LCM(p, 2q) is the smallest t where this happens. For the middle pocket: t must be a multiple of p and equal to q plus a multiple of 2q. So, t = k*p = q + m*2q. Then, k*p = q*(1 + 2m). Since p and q are coprime? Wait, no. Since p and q are both odd, GCD(p, q) is some odd integer d. Let d = GCD(p, q). Then, p = d*p', q = d*q', where GCD(p', q') = 1. Then, k*d*p' = d*q'*(1 + 2m). Cancel d: k*p' = q'*(1 + 2m). Since GCD(p', q') = 1, p' divides (1 + 2m). Therefore, 1 + 2m = n*p' for some integer n. Then, m = (n*p' - 1)/2. Since p' is odd (since p is odd, d is odd, so p' is odd), then n*p' is odd, so n*p' - 1 is even, so m is integer if n is integer. Therefore, there exist integers n and m such that this holds. Then, t = k*p = q*(1 + 2m) = d*q'*(1 + 2m) = d*q'*n*p' = d*n*p'*q' = n*p*q'. Wait, since p = d*p' and q = d*q', then p*q' = d*p'*q' = q*p'. Hmm, not sure. Let me see.Alternatively, given that p and 2q are the dimensions, and the ball is moving along (t, t), so we need to find the smallest t where t is a multiple of p and either t is a multiple of 2q or t = q mod 2q.Wait, maybe modular arithmetic can help here. Let’s see. For the ball to end up in a corner, t must satisfy t ≡ 0 mod p and t ≡ 0 mod 2q. For the middle pocket, t must satisfy t ≡ 0 mod p and t ≡ q mod 2q. So, the problem reduces to solving these two congruences.Case 1: t ≡ 0 mod p and t ≡ 0 mod 2q. Then t = LCM(p, 2q).Case 2: t ≡ 0 mod p and t ≡ q mod 2q. We need to see if such a t exists. If such a t exists, then the ball will reach the middle pocket at (t, t) which is (k*p, k*p) = (k*p, q + m*2q). So, need k*p ≡ q mod 2q. That is, k*p ≡ q mod 2q. Since p and 2q may not be coprime, we need to check if q is divisible by GCD(p, 2q). Let d = GCD(p, 2q). Since p is odd and 2q is even, d must be GCD(p, q). Because GCD(p, 2q) = GCD(p, q) as p is odd. Let d = GCD(p, q). Then, for the equation k*p ≡ q mod 2q, divide both sides by d: k*(p/d) ≡ (q/d) mod (2q/d). Since GCD(p/d, 2q/d) = GCD(p/d, 2*(q/d))). But since p/d and q/d are coprime (as d is GCD(p, q)), GCD(p/d, 2*(q/d)) is GCD(p/d, 2). But p/d is odd (since p is odd and d is a divisor of p), so GCD(p/d, 2) = 1. Therefore, the equation k*(p/d) ≡ (q/d) mod (2q/d) has a solution because p/d and 2q/d are coprime. Hence, there exists an integer k satisfying this congruence. Therefore, such a t exists. Therefore, there are two possible t values where the ball could end up in a pocket: one at LCM(p, 2q) (a corner) and another at some smaller t (a middle pocket). We need to determine which one comes first, i.e., which t is smaller. Since LCM(p, 2q) is a common multiple, but the t for the middle pocket is LCM(p, 2q)/2 if possible? Wait, maybe not. Let me think with an example. Let’s take p = 3, q = 1 (so 2q = 2). Then LCM(3, 2) = 6. For the middle pocket, we need t ≡ 0 mod 3 and t ≡ 1 mod 2. Solving t ≡ 0 mod 3, t ≡ 1 mod 2. The smallest such t is 3. So t = 3, which is smaller than 6. So in this case, the ball would reach the middle pocket first. Similarly, take p = 5, q = 3. Then 2q = 6. LCM(5,6) = 30. For the middle pocket: solve t ≡ 0 mod 5 and t ≡ 3 mod 6. Find t such that t = 5k and 5k ≡ 3 mod 6. 5k ≡ -k ≡ 3 mod 6 => -k ≡ 3 mod 6 => k ≡ -3 mod 6 => k ≡ 3 mod 6. So k = 6m + 3. Then t = 5*(6m + 3) = 30m + 15. The smallest t is 15. Which is less than 30. So the ball would reach the middle pocket at t = 15. So in both cases, the middle pocket is reached at t = LCM(p, 2q)/2. Wait, in the first example, LCM(3,2)=6, 6/2=3. In the second example, LCM(5,6)=30, 30/2=15. So it seems that t = LCM(p, 2q)/2. Is this a general result?Yes, because when p and 2q are such that LCM(p, 2q) = L, then solving t ≡ 0 mod p and t ≡ q mod 2q would result in t = L/2. Because L is the LCM, so L = k*p = m*2q. If we take half of that, L/2 = (k/2)*p = (m/2)*2q = m*q. But k and m must be integers. Wait, but if L is LCM(p, 2q), then L is even since 2q is even and p is odd. Therefore, L must be even because it's a multiple of 2q. So L/2 is an integer. Then, L/2 must be a multiple of p? Wait, L is the least common multiple, so L = LCM(p, 2q). Then, L/2 = LCM(p, 2q)/2. Since p is odd and 2q is even, LCM(p, 2q) = LCM(p, q) if q is even? Wait, no. Let me think again. LCM(p, 2q) is equal to (p*2q)/GCD(p, 2q). Since p is odd, GCD(p, 2q) = GCD(p, q). So LCM(p, 2q) = (2p q)/d where d = GCD(p, q). Therefore, LCM(p, 2q)/2 = (p q)/d. Since d = GCD(p, q), then (p q)/d = LCM(p, q). So LCM(p, 2q)/2 = LCM(p, q). But LCM(p, q) is the LCM of two odd numbers. So in this case, LCM(p, q) must be odd. But LCM(p, 2q)/2 is equal to LCM(p, q). Wait, let's verify with the examples. For p=3, q=1: LCM(3,2)=6, LCM(3,1)=3. So 6/2=3. Yes. For p=5, q=3: LCM(5,6)=30, LCM(5,3)=15. 30/2=15. Correct. So in general, LCM(p, 2q)/2 = LCM(p, q). Therefore, t = LCM(p, q). But LCM(p, q) is a multiple of p, so t = LCM(p, q) = k*p. Also, LCM(p, q) is a multiple of q, so LCM(p, q) = m*q. Therefore, when the ball travels t = LCM(p, q), it has moved k*p horizontally and m*q vertically. But since the ball is moving at 45 degrees, it's moving (t, t). Wait, but in the unfolded grid, the horizontal direction is p and vertical is 2q. Wait, maybe I need to be careful here.Wait, in the coordinate system where the original table is (0,0) to (p, 2q). The middle pockets are at (0, q) and (p, q). So, when the ball is at (t, t), in the unfolded grid, this corresponds to (t mod p, t mod 2q). If t is a multiple of p, then t mod p = 0. If t ≡ q mod 2q, then t mod 2q = q. Therefore, (0, q) or (p, q). But starting from (0,0), the ball is moving in the positive x and y direction. So when t = LCM(p, q), which is LCM(p, 2q)/2, the coordinates would be (LCM(p, q), LCM(p, q)). But LCM(p, q) mod p = 0, and LCM(p, q) mod 2q. Since LCM(p, q) is a multiple of q, so LCM(p, q) = q * k. Therefore, LCM(p, q) mod 2q = q*k mod 2q. If k is even, then LCM(p, q) mod 2q = 0; if k is odd, LCM(p, q) mod 2q = q. But since LCM(p, q) is LCM(p, q). Let’s note that since p and q are both odd, their LCM is also odd. Because the product p*q is odd, and LCM(p, q) divides p*q. So LCM(p, q) must be odd. Therefore, LCM(p, q) is odd. Then, LCM(p, q) = q*k, where k = LCM(p, q)/q = p / GCD(p, q). Since p and q are both odd, GCD(p, q) is odd, so k = p / GCD(p, q) is also odd. Therefore, k is odd, so LCM(p, q) mod 2q = q*k mod 2q = q*(2m + 1) mod 2q = q + 2q*m mod 2q = q. Therefore, LCM(p, q) mod 2q = q. Therefore, in coordinates, (LCM(p, q) mod p, LCM(p, q) mod 2q) = (0, q). Hence, the ball reaches (0, q), but since the original table is from (0,0) to (p, 2q), (0, q) is a middle pocket on the left side. But we started at (0,0), moving right and up. Wait, but how does the unfolding work? If the ball travels t = LCM(p, q), which is equal to LCM(p, 2q)/2, then in the unfolded grid, it's at (LCM(p, q), LCM(p, q)) = (k*p, q + m*2q). Therefore, reflecting the table m times vertically. But the x-coordinate is k*p, meaning it's gone through k tables horizontally, and y-coordinate is q + m*2q, meaning it's gone through m tables vertically and half a table more. But when you unfold the table, each reflection over the vertical sides (length p) would place the ball in a new table copy either to the right or left, and each reflection over the horizontal sides (length 2q) would place the ball in a new table copy above or below. Since the ball is moving northeast at 45 degrees, each time it hits a vertical wall (x = 0 or x = p), it reflects, which in the unfolded grid would be continuing into the adjacent table. Similarly for horizontal walls. But when the ball reaches (k*p, q + m*2q), which is a middle pocket, how does this correspond in the original table? The x-coordinate k*p is a multiple of p, so that's at x = 0 or x = p in some reflected table. The y-coordinate q + m*2q. Since the original table's vertical length is 2q, each reflection over the horizontal walls adds or subtracts 2q. Therefore, the y-coordinate modulo 2q gives the position in the original table. q + m*2q modulo 2q is q, so the y-coordinate is q. Therefore, the position in the original table would be (0, q) or (p, q) depending on whether k is even or odd. Wait, since the ball starts at (0,0), moving right and up. Each time it crosses p in the x-direction, it's equivalent to reflecting the table. If k is even, then x = 0; if k is odd, x = p. Wait, but in the example with p=3, q=1: LCM(3,1)=3. So t=3. Then, coordinates (3,3). 3 mod 3=0, 3 mod 2=1. So (0,1). Which is the middle pocket on the left. But the ball started at (0,0), moving right and up. How does it reach (0,1)? That seems like it went through some reflections. Wait, perhaps when the ball travels to (3,3) in the unfolded grid, it's actually moving through 3 tables to the right and 1.5 tables up. But since you can't have half a table, maybe the vertical direction has an odd number of reflections. Let me think again. The vertical length is 2q, so each table copy vertically is 2q. If the ball goes up 3 units when q=1 (so 2q=2), starting at (0,0). So moving to (3,3). The vertical position 3 is equivalent to 3 mod 2*1=1. So 3 mod 2 =1. So in the original table, it's at y=1. The horizontal position 3 mod 3=0. So (0,1). But how does the ball get there? Starting at (0,0), moving right and up. It would first hit the right wall at (3,3) after traveling 3 units? Wait, no. Wait, in reality, the table is 3 units long (p=3) and 2 units wide (2q=2). So the ball is moving at 45 degrees, so after moving 1 unit right and 1 unit up, it's at (1,1). Then another unit to (2,2). Then another unit would hit the corner (3,3), but the table's width is only 2. Wait, so actually, the ball would hit the top wall before reaching x=3. Let me trace the actual path.Starting at (0,0). Moving to (1,1), which is within the table. Then to (2,2), which is at the top edge (since the table's height is 2q=2). So at (2,2), it hits the top middle pocket? Wait, the middle pockets are at (0, q) and (p, q). When q=1, the middle pockets are at (0,1) and (3,1). But the table's height is 2, so the middle of the vertical sides (which are length 2q=2) would be at y=1. Wait, but the vertical sides are length 2q=2, so the middle is at (0,1) and (3,1). But the ball is at (2,2), which is a corner? Wait, no. The table is size p x 2q = 3x2. So the four corners are at (0,0), (3,0), (0,2), (3,2). The middle pockets are at (0,1), (3,1). So when the ball reaches (2,2), that's on the top side (y=2), but not a corner. Wait, (2,2) isn't a pocket. Then it should reflect. Wait, perhaps my earlier approach is oversimplified. Maybe the unfolding method needs more careful consideration. Let's actually simulate the path.Starting at (0,0), moving at 45 degrees. The table is 3 units long (x from 0 to 3) and 2 units high (y from 0 to 2). The ball moves right and up. First, it would hit the right wall at x=3 when y=3, but the table's height is only 2. So it actually hits the top wall first. The collision point is when y=2, which occurs at x=2. So the ball travels from (0,0) to (2,2), reflects off the top wall. Since the reflection reverses the y-component, so the new direction is right and down. Then it goes to (3,1), which is a middle pocket. So in this case, it reaches the middle pocket at (3,1) after two reflections: first off the top, then off the right. Wait, but according to the coordinates in the unfolded grid, the total distance traveled would be sqrt((3)^2 + (1)^2) but since it's moving at 45 degrees, maybe not. Wait, no. In the unfolded grid, reflecting the table over the top wall would place a mirrored table above. But instead of reflecting the direction, we can think of continuing straight. So from (0,0), moving to (2,2), then into the mirrored table, continuing to (3,3). But in reality, the ball in the original table after hitting (2,2) would go to (3,1). So in the unfolded grid, the straight path would go through multiple reflections. But perhaps my earlier conclusion that t=LCM(p, q) leads to the middle pocket is correct, as in the example, t=3, which is LCM(3,1)=3. So in coordinates (3,3) in the grid, which corresponds to (0,1) in the original table. But in reality, the ball ends up at (3,1). Hmm, this discrepancy suggests maybe my coordinate system was flipped. Alternatively, perhaps the horizontal direction is p and vertical is 2q, so when unfolding, moving right corresponds to adding p each time, and moving up corresponds to adding 2q. But the ball is moving in (1,1) direction, so each unit step in x and y. Wait, confusion arises between the grid of tables and the actual coordinates. Maybe a better approach is to use the least common multiple approach.The key point is that the ball will have traveled a number of horizontal periods (p) and vertical periods (2q) such that the number of horizontal and vertical traversals are integers. Because the slope is 1, the horizontal and vertical distances must be equal. Therefore, the distance traveled must be a common multiple of p and 2q. The least common multiple LCM(p, 2q) is the first time this happens. However, since we have middle pockets at q, maybe the ball reaches q in the vertical direction before reaching 2q. But how does the horizontal and vertical movement interact? Let’s think in terms of reflections. Each time the ball reflects off a vertical wall (left or right), it inverts the x-direction. Each time it reflects off a horizontal wall (top or bottom), it inverts the y-direction. Since the ball starts at (0,0) moving NE (right and up). Let’s model the reflections. Let’s consider the general case.Let p and q be odd. The table is width p, height 2q. The ball starts at (0,0), moving at 45 degrees. Let’s track the ball’s position after each reflection.But tracking each reflection might get complicated. Instead, use the method of images: reflect the table each time the ball would hit a wall, creating a grid of mirrored tables. Then the ball travels in a straight line through this grid. The destination pocket is where this line intersects a corner or a middle pocket.Given the ball is moving along the line y = x. We need to find the smallest positive integer t such that (t, t) is either a corner or a middle pocket in this grid.A corner in the grid occurs at (kp, m*2q) for integers k, m. A middle pocket occurs at (kp, q + m*2q). So, we need to solve t = kp and t = m*2q (for corner) or t = kp and t = q + m*2q (for middle pocket).Therefore, the ball will fall into a pocket when t is the smallest positive integer satisfying either:1. t = kp = m*2q (corner pocket), or2. t = kp = q + m*2q (middle pocket).We need to show that the smallest such t satisfies the second condition (middle pocket) rather than the first (corner pocket).Since p and q are both odd, 2q is even. Let’s compute LCM(p, 2q). As p is odd and 2q is even, LCM(p, 2q) = p*2q / GCD(p, 2q). Since p is odd, GCD(p, 2q) = GCD(p, q). Let’s denote d = GCD(p, q). Therefore, LCM(p, 2q) = (2pq)/d.Similarly, consider the middle pocket condition: t = kp = q + m*2q. Rearranging, kp - 2q m = q. This is a linear Diophantine equation in variables k and m. A solution exists if and only if GCD(p, 2q) divides q. Let’s check: GCD(p, 2q) = d, and since d divides p and q (because d = GCD(p, q)), d divides q. Therefore, the equation has solutions.Therefore, the smallest positive solution t for the middle pocket is t = LCM(p, 2q)/2. Wait, but LCM(p, 2q) = (2pq)/d. Then LCM(p, 2q)/2 = pq/d = LCM(p, q). Since LCM(p, q) = pq/d. Since LCM(p, q) is less than LCM(p, 2q) when q < 2q, which it always is. Therefore, the ball will reach the middle pocket at t = LCM(p, q), which is earlier than t = LCM(p, 2q). Therefore, the ball falls into the middle pocket before reaching the corner pocket.Therefore, since LCM(p, q) < LCM(p, 2q), the ball will reach the middle pocket first.But we need to confirm that LCM(p, q) is indeed a solution to t = kp = q + m*2q. Let’s verify:Given t = LCM(p, q), which is a multiple of p, so t = kp. Also, LCM(p, q) is a multiple of q, so LCM(p, q) = mq. Therefore, we need to check if mq = q + 2q*m', which simplifies to m = 1 + 2m'. Since m and m' are integers, this requires that m is odd. Since LCM(p, q) = mq, and since LCM(p, q)/q = p / GCD(p, q). Since p and q are both odd, GCD(p, q) is odd, so p / GCD(p, q) is an integer. Let’s denote p' = p / GCD(p, q) and q' = q / GCD(p, q), which are coprime. Then LCM(p, q) = p' q' GCD(p, q) = p' q. Therefore, LCM(p, q)/q = p', which is an integer. Therefore, m = p'. Since p is odd and p' is a divisor of p, p' is also odd. Therefore, m = p' is odd, so m = 1 + 2m' where m' = (m - 1)/2. Therefore, such m' exists. Therefore, LCM(p, q) = q + 2q*m', so t = LCM(p, q) satisfies t = q + 2q*m', which means the ball reaches the middle pocket.Therefore, since LCM(p, q) < LCM(p, 2q) (because LCM(p, 2q) = 2 LCM(p, q) when q is even, but in our case q is odd, so LCM(p, 2q) = LCM(p, 2q) = 2 LCM(p, q) if p and q are coprime, otherwise it's LCM(p, 2q) = 2 LCM(p, q / GCD(p, 2q))). Wait, perhaps not exactly, but in general, LCM(p, 2q) is twice LCM(p, q) if q is coprime with 2, which it is since q is odd. Wait, if p and q are coprime, then LCM(p, 2q) = 2pq. LCM(p, q) = pq. So LCM(p, 2q) = 2 LCM(p, q). Therefore, in that case, LCM(p, q) = pq, LCM(p, 2q) = 2pq. So LCM(p, q) is indeed half of LCM(p, 2q), hence LCM(p, q) < LCM(p, 2q). Therefore, the ball reaches the middle pocket at t = LCM(p, q), which is sooner than the corner pocket at t = LCM(p, 2q). Hence, the ball must fall into a middle pocket.But we need to ensure that when the ball reaches t = LCM(p, q), it actually lands exactly on a middle pocket and not somewhere else. Given that in the unfolded grid, the coordinates (LCM(p, q), LCM(p, q)) correspond to (0, q) in the original table, considering the reflections. Wait, coordinates in the grid are (t, t). When you fold the grid back into the original table, the x-coordinate is t mod p and the y-coordinate is t mod 2q. Since t = LCM(p, q) = kp, so t mod p = 0. Then t mod 2q. Since t = LCM(p, q) = mq (as LCM(p, q) is a multiple of q). But since q is odd, mq mod 2q. If m is even, mq mod 2q = 0. If m is odd, mq mod 2q = q. But since LCM(p, q) = kp = mq, and since p and q are both odd, m must be such that mq = kp. Let’s take m = p / GCD(p, q) and k = q / GCD(p, q). Then mq = (p / d) q and kp = (q / d) p, where d = GCD(p, q). Since (p/d) and (q/d) are coprime, and both are integers. Since p and q are both odd, (p/d) and (q/d) are also odd. Therefore, m = p/d is odd, so mq mod 2q = q*(p/d) mod 2q. Since p/d is odd, q*(p/d) = q + q*(p/d - 1). Since p/d - 1 is even (because p/d is odd), then q*(p/d) = q + 2q*( (p/d - 1)/2 ). Therefore, q*(p/d) mod 2q = q. Hence, t mod 2q = q. Therefore, the coordinates in the original table are (0, q), which is a middle pocket. However, the ball started at (0,0) moving to the right and up. So in the unfolded grid, the ball reaches (kp, q + m*2q). The number of horizontal reflections is k-1, and vertical reflections is m. But since the ball is at (kp, q + m*2q), which is a middle pocket, this corresponds to having crossed k tables horizontally and m tables vertically plus half a table. But in the original table, after k reflections on the x-axis and m reflections on the y-axis, the ball's position would be at (kp mod p, (q + m*2q) mod 2q) = (0, q). But depending on whether k is even or odd, the x-coordinate could be 0 or p. Wait, no. Since kp mod p is always 0, regardless of k. Similarly, (q + m*2q) mod 2q is q. So the position is either (0, q) or (p, q) depending on the direction after reflections. Wait, actually, when unfolding, each horizontal reflection alternates the direction. If k is even, the ball is on the original table or an even number of reflections, so x = 0. If k is odd, it's on a reflected table, so x = p. Similarly, vertical reflections: if m is even, y remains in the original vertical direction, but since we have q + m*2q, which is q when m=0, q + 2q when m=1, etc. Wait, (q + m*2q) mod 2q is q when m is even, and q + 2q mod 2q = q when m is odd. Wait, no. (q + m*2q) mod 2q = q for any m. Therefore, regardless of m, the y-coordinate is q. But the x-coordinate, if k is even, is 0; if k is odd, is p. Therefore, the ball ends up at either (0, q) or (p, q). But since the ball is moving in the positive x direction, starting from (0,0), and the first time it reaches x = kp is when k=1, so x = p, but wait, t = LCM(p, q). Let’s see with the example where p=3, q=1, LCM(3,1)=3. Then the ball is at (3,3). In the original table, x=3 mod 3=0, y=3 mod 2=1. So (0,1). But the ball started at (0,0), moving right and up. How does it end up at (0,1)? That seems like it wrapped around horizontally but ended up on the left side. Wait, perhaps the number of horizontal reflections is k=1, meaning it went to the right table once, but since LCM(p, q)=3, which is equal to p=3, so k=1. Then, the x-coordinate is p, but mirrored, so in the original table, it's p mod p=0. But mirrored tables mean that after an odd number of reflections, the direction flips. So in reality, the ball would have exited the right side of the table and entered the next table to the right, but since we are considering the grid, it's equivalent to wrapping around. However, pockets are only at the corners and midpoints of the original table. Therefore, in this case, the ball would have left the original table and entered the reflected table, but the pocket in the reflected table's (0, q) is equivalent to the original table's (p, q). Wait, maybe not. Alternatively, perhaps the middle pockets in the reflected tables are considered, but the problem states that the pockets are only at the original table's corners and midpoints. Therefore, the ball must enter a pocket in the original table. Therefore, even though in the grid it seems like the ball is at (0, q) in some reflected table, but in reality, the ball cannot exit the original table except through one of its own pockets. Therefore, my earlier reasoning might be flawed.Let me think differently. Instead of unfolding indefinitely, the ball travels through the original table, reflecting each time it hits a wall. We need to count the number of times it reflects horizontally and vertically before reaching a pocket.Since the ball is moving at 45 degrees, the number of horizontal and vertical reflections will be related to the dimensions p and 2q. Because it's moving diagonally, the ball will reflect alternately off horizontal and vertical walls. Let’s denote that after each reflection, the direction changes. But since it's moving at 45 degrees, the pattern of reflections can be periodic. The key is to find after how many reflections the ball returns to a corner or a midpoint.Alternatively, consider the parity of the number of reflections. Since p and q are odd, after traveling p units horizontally, the ball would have reflected off the vertical walls (p-1) times (since each reflection occurs after traveling a unit distance p). Wait, no. Wait, if the table is length p, then traveling from one end to the other is p units. But the ball is moving at 45 degrees, so each unit of distance corresponds to sqrt(2) units in x and y. Wait, perhaps I need to parametrize by time or distance.Alternatively, parametrize the movement. Let’s let the ball move with velocity (1,1), so position at time t is (t, t). The table has dimensions p x 2q. The ball reflects off the walls, reversing the respective component of velocity upon collision.But instead, use the method where we reflect the table and keep the velocity constant. Then, the ball travels in a straight line through the reflected grid. The first pocket it encounters will be either a corner or a middle pocket.In the grid of reflected tables, the pockets are located at every (kp, m*2q) for corners and (kp, q + m*2q) for middle pockets, where k and m are integers. The ball's path is the line y = x. We need to find the smallest t > 0 where (t, t) coincides with one of these pocket positions.Thus, solving t = kp and t = m*2q (corner) or t = kp and t = q + m*2q (middle).For the corner, t must be a multiple of both p and 2q, so t = LCM(p, 2q).For the middle pocket, t must be a multiple of p and equal to q plus a multiple of 2q. So, t = kp = q + m*2q. We need to find the smallest such t.Since p and 2q may not be coprime, we need to solve for k and m in integers. Let’s rearrange the equation:kp - 2q m = q.This is a linear Diophantine equation of the form ax + by = c, where a = p, b = -2q, c = q.A solution exists if and only if GCD(p, 2q) divides q. Since GCD(p, 2q) = GCD(p, q) (as p is odd), and GCD(p, q) divides q, yes, solutions exist.The general solution can be found once a particular solution is known. Let’s find the minimal positive solution t.Let’s denote d = GCD(p, q), so p = d p', q = d q', where GCD(p', q') = 1.Then, the equation becomes:k d p' - 2 d q' m = d q'Divide both sides by d:k p' - 2 q' m = q'So, k p' - 2 m q' = q'This simplifies to:k p' = q' (1 + 2 m)Since GCD(p', q') = 1, p' divides (1 + 2m). Therefore, 1 + 2m = p' n for some integer n.Thus, m = (p' n - 1)/2.Similarly, k = q' n.Since m must be an integer, (p' n - 1) must be even. Since p' is odd (as p is odd and p' divides p), p' n is odd if n is odd, even if n is even. Therefore, p' n - 1 is even if and only if n is odd. Let’s set n = 2 t + 1 for integer t ≥ 0.Wait, but maybe a better approach is to find the minimal n such that (p' n - 1) is even. Since p' is odd, p' n is odd when n is odd, even when n is even. Therefore, p' n -1 is even when n is odd. Therefore, the minimal positive n is 1.Therefore, set n = 1:m = (p' * 1 - 1)/2 = (p' - 1)/2k = q' * 1 = q'Therefore, the minimal solution is:k = q'm = (p' - 1)/2Therefore, the minimal t is:t = k p = q' * d p' = q' pSince p = d p', q = d q', so q' p = q' * d p' = p * q' = p * (q / d) = (p q)/d = LCM(p, q).Therefore, the minimal t is LCM(p, q), as previously concluded.Thus, the ball reaches the middle pocket at t = LCM(p, q), which is less than t = LCM(p, 2q) since LCM(p, 2q) = 2 LCM(p, q) when p and q are coprime. If they are not coprime, since d = GCD(p, q), LCM(p, 2q) = LCM(p, 2q) = LCM(d p', 2 d q') = d LCM(p', 2 q'). Since p' and q' are coprime, LCM(p', 2 q') is 2 p' q' if p' and 2 are coprime (i.e., p' is odd, which it is). Therefore, LCM(p, 2q) = d * 2 p' q' = 2 d p' q' = 2 LCM(p, q). Therefore, LCM(p, 2q) = 2 LCM(p, q). Therefore, LCM(p, q) is indeed half of LCM(p, 2q), hence smaller. Therefore, the ball reaches the middle pocket first.Therefore, the ball will fall into a middle pocket before reaching a corner pocket, proving the required result.Another way to see it is considering the parity. Since both p and q are odd, LCM(p, q) is odd, and LCM(p, 2q) is even. The middle pockets are located at q, which is half of 2q. Since LCM(p, q) is odd, when the ball travels LCM(p, q) units, it has made an odd number of horizontal traversals (each of length p) and an odd number of vertical traversals (each of length q). Therefore, the final position in terms of reflections would be at (p, q), a middle pocket. Whereas LCM(p, 2q) is even, leading to a corner pocket, but since LCM(p, q) comes first, the middle pocket is reached earlier.Therefore, in all cases where p and q are odd, the ball shot at 45 degrees from the corner will end up in a middle pocket.

✅:To prove that a ball shot from a corner of a rectangular billiard table of size ( p times 2q ) (with ( p ) and ( q ) odd) at a 45° angle will fall into a middle pocket, we analyze the ball's trajectory using reflections and number theory.Key Steps:1. Unfolding the Table: Instead of tracking reflections, "unfold" the table into a grid of mirrored tables. The ball's path becomes a straight line in this grid.2. Pocket Positions: Corners are at ((kp, m cdot 2q)) and middle pockets at ((kp, q + m cdot 2q)) for integers ( k, m ).3. Path Analysis: The ball travels along the line ( y = x ). We find the smallest ( t > 0 ) such that ( (t, t) ) aligns with a pocket: - Corner Pocket: Solve ( t = text{LCM}(p, 2q) ). - Middle Pocket: Solve ( t = kp = q + m cdot 2q ).4. Diophantine Equation: The equation ( kp - 2qm = q ) has solutions since ( gcd(p, 2q) = gcd(p, q) ) divides ( q ).5. Least Common Multiples: - ( text{LCM}(p, q) ) (middle pocket) is smaller than ( text{LCM}(p, 2q) ) (corner pocket) because ( text{LCM}(p, 2q) = 2 cdot text{LCM}(p, q) ).6. Parity Consideration: Since ( p ) and ( q ) are odd, ( text{LCM}(p, q) ) is odd, ensuring the ball reaches the middle pocket (at ( q ), midpoint of ( 2q )) before the corner.Conclusion: The ball first reaches a middle pocket at ( t = text{LCM}(p, q) ), proving it will fall into one of the middle pockets.boxed{text{The ball will fall into a middle pocket.}}

🔑:: 1. Consider the billiard table: We start by considering a rectangular billiard table of dimensions ( p times 2q ) where ( p ) and ( q ) are odd numbers. The table has corner pockets and pockets in the middle of each side of length ( 2q ).2. Set up the coordinates: - Place the lower left corner of the table at the origin ((0,0)) in the coordinate plane. - The corners of the table are at ((0, 0)), ((p, 0)), ((0, 2q)), and ((p, 2q)). - The middle pockets on the longer sides are at ((frac{p}{2}, 0)), ((frac{p}{2}, 2q)).3. Path of the ball: - The ball is shot at an angle of (45^{circ}) to the side from a corner, say ((0,0)). - At (45^{circ}), the ball's path can be represented by a line ( y = x ) (more precisely, a ray starting from ((0,0))). 4. Extend the billiard table by reflection: - Imagine reflecting the table across its boundaries. This can be visualized by considering a grid where each cell is ( p times 2q ). - Within this grid, the ball will continue along the line ( y = x ) straight through multiple reflected images of the table until it returns to the original grid.5. Analyze the trajectory: - Because we are dealing with reflections, the ball's actual path through multiple cells will follow the main diagonal until it intersects one of the pocket locations. - Whenever the main diagonal intersects a lattice point ((m, m)) where both coordinates are multiples, ( m ) must be divisible by both ( p ) and ( 2q ).6. Finding the intersection point: - The smallest such intersection where this occurs is at least common multiple of ( p ) and ( 2q ), denoted ( m = operatorname{LCM}(p, 2q) ). - Therefore, the ball will first hit a lattice point at coordinates ( (m, m) ).7. Determining the pocket: - Since ( p ) and ( q ) are odd, ( operatorname{LCM}(p, 2q)) is also an odd number. - The pocket locations within the table are such that: if ( m ) is odd, the engagement with the central pocket happens when ( y = x = m ), guaranteeing that the ball will pass through one of the middle pockets.8. Conclusion:The exact trajectory analyzed shows that the ball, shot from any corner at a (45^{circ}) angle, will have to pass through one of the middle pockets along its path.[boxed{text{The ball will necessarily fall into one of the middle pockets.}}]

🧠:Okay, let me try to figure out this tire swapping problem. So, the car has front and rear tires that wear out at different rates. The front tires last 25,000 km, and the rear ones last 15,000 km. We need to determine when to swap them so that all four tires wear out at the same time. Hmm, okay.First, I need to understand why they wear out at different rates. Maybe the front tires handle more weight or have more friction due to steering? But the problem doesn't specify, so I guess the exact reason isn't important here. The key is that they have different lifespans if not swapped.The goal is to make all four tires wear out simultaneously. That means after swapping them at some optimal point, both the front and rear tires should be completely worn at the same total distance. Let me think about how to model this.Let's denote the distance at which we should swap the tires as 'x' kilometers. So, the car will drive 'x' km with the original setup, then swap the front and rear tires, and continue driving until all tires wear out. The total distance covered when all tires are worn should be the same for all four.Each tire's wear rate can be considered as a fraction of their total lifespan per kilometer. For the front tires, their wear rate is 1/25,000 per km, and for the rear tires, it's 1/15,000 per km. When we swap them, each tire will have been used in both the front and rear positions, accumulating wear from both periods.So, let's break it down. Suppose we swap the tires after x km. Then, each original front tire would have been used for x km on the front and then (total distance - x) km on the rear. Similarly, each original rear tire would have been used for x km on the rear and then (total distance - x) km on the front. The total wear on each tire should equal 1 (or 100%, meaning fully worn).Therefore, for one of the original front tires:Wear from front position: x / 25,000Wear from rear position: (D - x) / 15,000Total wear: x/25,000 + (D - x)/15,000 = 1Similarly, for one of the original rear tires:Wear from rear position: x / 15,000Wear from front position: (D - x) / 25,000Total wear: x/15,000 + (D - x)/25,000 = 1So, we have two equations here, but actually, since all four tires need to wear out at the same total distance D, these equations should both equal 1. Let me check if they can be solved together.Wait, but each equation represents a different tire. Let me write them down properly.For original front tires (which are moved to rear after x km):x/25,000 + (D - x)/15,000 = 1 ...(1)For original rear tires (which are moved to front after x km):x/15,000 + (D - x)/25,000 = 1 ...(2)Now, we have two equations with two variables: x and D. Our goal is to find x (the swap distance) such that D (the total distance when all tires are worn) is the same for both equations.Let me solve both equations for D and set them equal to each other.From equation (1):Multiply both sides by 75,000 (LCM of 25,000 and 15,000) to eliminate denominators:3x + 5(D - x) = 75,0003x + 5D -5x = 75,000-2x +5D =75,000 ...(1a)From equation (2):Multiply both sides by 75,000:5x +3(D -x) =75,0005x +3D -3x =75,0002x +3D =75,000 ...(2a)Now, we have two equations:-2x +5D =75,000 ...(1a)2x +3D =75,000 ...(2a)Let me add these two equations to eliminate x:(-2x +5D) + (2x +3D) =75,000 +75,0008D =150,000D=150,000 /8= 18,750 kmSo, the total distance when all tires wear out is 18,750 km. Now, substitute D back into one of the equations to find x.Using equation (2a):2x +3*18,750 =75,0002x +56,250=75,0002x=75,000 -56,250=18,750x=18,750 /2=9,375 kmTherefore, the swap should occur at 9,375 km. Let me verify this with both equations.Check equation (1a):-2*(9,375) +5*(18,750)= -18,750 +93,750=75,000 ✓Check equation (2a):2*(9,375) +3*(18,750)=18,750 +56,250=75,000 ✓Also, check the original equations:For original front tire:9,375/25,000 + (18,750 -9,375)/15,000 = 0.375 + 0.625 =1 ✓For original rear tire:9,375/15,000 + (18,750 -9,375)/25,000 =0.625 +0.375=1 ✓Looks good. So swapping at 9,375 km allows all tires to wear out at 18,750 km.Wait, but let me think again. The problem says "the front tires wear out after 25,000 km, and the rear after 15,000 km." So if you don't swap, front tires last 25k, rear 15k. But if you swap once at x km, then each tire spends x km in one position and (D -x) km in the other. The total D is 18,750 km, which is less than both 25k and 15k. That seems counterintuitive. If we swap, shouldn't the total distance be longer than 15k? Wait, but actually, by swapping, we balance the wear so that all tires are used until their combined wear reaches 100%.But let me think in terms of tire usage. Each tire is used in both positions. So the front tires are used for x km front and then D -x km rear. The rear tires are used for x km rear and then D -x km front. So the total wear on each tire is the sum of their front and rear wear rates.But since front and rear positions have different wear rates, swapping allows distributing the wear more evenly.Another way to approach this: think of the total wear per tire.Total front wear needed for the car: For the entire distance D, the front position requires 2 tires. So total front wear is 2 * D. Similarly, total rear wear is 2 * D. But each tire can contribute some front wear and some rear wear.Each of the four tires can contribute some amount to front and rear wear. Let me model this.Suppose we have four tires: A, B, C, D.Initially, A and B are front, C and D are rear.After swapping at x km, A and B move to rear, and C and D move to front.Total distance D.So, each tire (A and B) spends x km front and (D -x) km rear.Each tire (C and D) spends x km rear and (D -x) km front.Total front wear: For the first x km, front tires are A and B. For the remaining (D -x) km, front tires are C and D.Total front wear is x*2 (since two tires) * (1/25,000) + (D -x)*2*(1/25,000). Wait, no. Wait, the wear per tire is per km. Wait, maybe this approach is more complicated.Alternatively, think of the total wear required for front and rear positions.Each kilometer driven, the front tires wear at 1/25,000 per tire, so total front wear per km is 2*(1/25,000). Similarly, rear wear per km is 2*(1/15,000).The total front wear over D km is D*(2/25,000). Similarly, total rear wear is D*(2/15,000).But we have four tires, each can contribute to front and rear wear. Each tire can be used in front for some distance and rear for some distance. Let the total front wear per tire be t_f and rear wear per tire be t_r. Since all four tires are the same, each tire's total wear is t_f + t_r =1.Total front wear from all tires: 4*t_f = D*(2/25,000). Similarly, total rear wear from all tires:4*t_r = D*(2/15,000).But each tire's t_f is the time it spent on front divided by 25,000, and t_r is time on rear divided by 15,000.Wait, perhaps this is a better way. Let me consider that each tire's time on front is s, and time on rear is t. Then:For each tire: s/25,000 + t/15,000 =1.But all tires have the same s and t because they are swapped symmetrically.Total front position usage: 2*D = sum of s for all four tires. But each tire is on front for s, so 4*s =2D => s= D/2.Similarly, total rear position usage: 2*D = sum of t for all four tires => 4*t=2D => t=D/2.Wait, that can't be. If each tire is on front for D/2 km and rear for D/2 km, then:Each tire's wear is (D/2)/25,000 + (D/2)/15,000 = D/(2*25,000) + D/(2*15,000) = D*(1/50,000 +1/30,000)= D*(3/150,000 +5/150,000)=D*(8/150,000)= D/18,750.Set this equal to 1 (full wear):D/18,750 =1 => D=18,750 km.Ah, that's the same D as before! So regardless of when you swap, as long as each tire spends half the total distance on front and half on rear, the total wear is D/18,750. Wait, but that seems contradictory. If you swap once at x km, then each original front tire spends x km front and (D -x) km rear, and each original rear tire spends x km rear and (D -x) km front. But in this case, if D=18,750, then each tire's total front + rear is 18,750 km, but the wear depends on the position.Wait, but according to this, if each tire is on front for D/2 and rear for D/2, then wear is (D/2)/25,000 + (D/2)/15,000. Setting this equal to 1 gives D=18,750. So that's the total distance. But how does the swapping time x affect this?In the earlier approach, we found that swapping at x=9,375 km (which is exactly half of D=18,750 km) gives each tire half the distance on each position. So swapping at halfway allows each tire to spend half the time on front and half on rear. That makes sense. So the optimal swap is at half the total distance, which is 9,375 km. Then drive another 9,375 km, totaling 18,750 km.But wait, if you swap at x=9,375 km, then each original front tire is on front for 9,375 km and rear for 9,375 km. Each original rear tire is on rear for 9,375 km and front for 9,375 km. So each tire's wear is:Original front: 9,375/25,000 +9,375/15,000 =0.375 +0.625=1.0Original rear:9,375/15,000 +9,375/25,000=0.625 +0.375=1.0Exactly. So swapping at half the total distance ensures equal wear.But how did we get x=9,375 km from the equations, which is exactly D/2. So in this case, swapping at halfway point. Therefore, the answer is to swap the tires after 9,375 km, and then all tires will wear out at 18,750 km.But the question is asking "when should the tires be swapped". So the answer is after 9,375 km. Let me check the calculation again.Total distance D=18,750 km. Swap at x=9,375 km. Yes. So if you drive 9,375 km, swap front and rear, then drive another 9,375 km, all tires will be worn out.Alternatively, if you drive more than one swap, maybe multiple swaps, but the problem doesn't specify that. It just asks when to swap them so that they wear out at the same time. So a single swap is sufficient.Therefore, the answer is swap at 9,375 km.

✅:To determine when to swap the front and rear tires so they wear out simultaneously, consider the following steps:1. Define Variables: - Let ( x ) be the distance (in km) before swapping. - Let ( D ) be the total distance when all tires are worn out.2. Wear Equations: - Original Front Tires: After swapping, they spend ( x ) km on front and ( D - x ) km on rear. [ frac{x}{25,000} + frac{D - x}{15,000} = 1 ] - Original Rear Tires: After swapping, they spend ( x ) km on rear and ( D - x ) km on front. [ frac{x}{15,000} + frac{D - x}{25,000} = 1 ]3. Solve the System of Equations: - Multiply both equations by 75,000 (LCM of 25,000 and 15,000) to eliminate denominators: - First equation becomes: ( 3x + 5(D - x) = 75,000 Rightarrow -2x + 5D = 75,000 ) - Second equation becomes: ( 5x + 3(D - x) = 75,000 Rightarrow 2x + 3D = 75,000 ) - Add the two equations to eliminate ( x ): [ (-2x + 5D) + (2x + 3D) = 75,000 + 75,000 Rightarrow 8D = 150,000 Rightarrow D = 18,750 text{ km} ] - Substitute ( D = 18,750 ) back into one of the equations to find ( x ): [ 2x + 3(18,750) = 75,000 Rightarrow 2x = 18,750 Rightarrow x = 9,375 text{ km} ]4. Verification: - Original Front Tires: ( frac{9,375}{25,000} + frac{9,375}{15,000} = 0.375 + 0.625 = 1 ) - Original Rear Tires: ( frac{9,375}{15,000} + frac{9,375}{25,000} = 0.625 + 0.375 = 1 )Final Answer: The tires should be swapped after 9,375 km, allowing all four tires to wear out simultaneously at a total distance of 18,750 km.

🔑:Given:- Front tires wear out after (25{,}000 , text{km})- Rear tires wear out after (15{,}000 , text{km})We need to determine the distance (x) at which to swap the tires so that all tires wear out simultaneously.1. Set up the Equations: Let's define ( x ) as the distance driven before the tires are swapped, and ( y ) as the additional distance driven after swapping the tires. Then: - Front tires wear out after (25{,}000 , text{km}): [ x + y ] - Rear tires wear out after (15{,}000 , text{km}): [ x + y ]2. Determine the Remaining Distances: After ( x ) km, the front tires have worn: [ frac{x}{25{,}000} ] and the rear tires have worn: [ frac{x}{15{,}000} ]3. Swapping Positions: After swapping, the front tires become rear tires and vice versa. Let (d) be the additional distance after the swapping, then the cumulative distance after ( d ) km that both front and rear tires will have worn ( 1 ): - Distance left to wear for front tires after swapping: [ (25{,}000)(1 - frac{x}{25{,}000}) = 25{,}000 - x ] - Distance left to wear for rear tires after swapping: [ (15{,}000)(1 - frac{x}{15{,}000}) = 15{,}000 - x ]4. Equating the Worn Distance: For the tires to wear out simultaneously: [ 25{,}000 - x = d times frac{25{,}000}{15{,}000} ] Rearranging to determine ( x ) and ( d ) to meet the wear-out distance, solve: [ d = 15{,}000 - x ] Substitute ( d ) back: [ 25{,}000 - x = frac{25{,}000}{15{,}000} (15{,}000 - x) ]5. Solve for ( x ): [ 25{,}000 - x = frac{25{,}000}{15{,}000} cdot (15{,}000 - x) ] Simplify: [ 25{,}000 - x = frac{25{,}000 cdot 15{,}000}{15{,}000} - frac{25{,}000}{15{,}000} cdot x ] [ 25{,}000 - x = 25{,}000 - frac{25{,}000}{15{,}000} cdot x ] [ 25{,}000 - x = 25{,}000 - frac{25{,}000}{15{,}000} x ] Combine terms to solve ( x ): [ 25{,}000 - 25{,}000 = left(1 - frac{25{,}000}{15{,}000}right)x ] Simplify and solve for common factor: [ x = frac {25{,}000 cdot 15{,}000}{25{,}000 + 25{,}000} = 9,375 , text{km} ]So, the tires need to be swapped after:[boxed{9{,}375 , text{km}}]

🧠:Okay, so I have this problem about dividing 1990 people into subsets with certain conditions. Let me try to parse through each part step by step. First, the problem states that the subsets must be non-intersecting, meaning each person is in exactly one subset. The conditions are:(a) No one in a subset knows all the others in the subset. So, in any subset, there's no person who is acquainted with everyone else in that subset. That means for every person in a subset, there's at least one person they don't know.(b) Among any three persons in a subset, there are at least two who do not know each other. This seems like a restriction on the density of the acquaintance graph. In graph theory terms, this would mean that the graph doesn't contain a triangle (three mutual acquaintances), but actually, it's stronger. Because even if there's a triangle, condition (a) already says no one can know everyone else. But condition (b) is saying that in any trio, there's at least one pair that doesn't know each other. So, the graph is triangle-free? Wait, no. A triangle is three people all knowing each other. If we have three people where two don't know each other, then that trio satisfies the condition. So actually, condition (b) is that the graph does not contain a complete tripartite subgraph of three nodes. In other words, the graph has no triangles? Wait, no. If there is a triangle (three people all knowing each other), then in that trio, all pairs know each other. So condition (b) would be violated because among those three, there are not at least two who do not know each other. Therefore, condition (b) actually implies that the graph is triangle-free. So the acquaintance graph within each subset is triangle-free.But wait, let me confirm. If in every trio of people in the subset, at least two do not know each other. So, in any three people, you can't have all three knowing each other, which is a triangle. So yes, the graph is triangle-free. But actually, even more: it's not just triangle-free, because in a triangle-free graph, you can have three people where two know each other and the third doesn't know one of them. But here, the condition is that in any three people, at least two do not know each other. So that is a stronger condition. For example, a cycle of three nodes (a triangle) would violate this, but a cycle of four nodes (a square) would satisfy triangle-free but not necessarily satisfy this condition. Wait, let's take three people in a square. A square is four people where each is connected to two others. If I take any three people in the square, then in the square, each person is connected to two others. So, in the three people, how many connections are there? If the three people are consecutive in the square, say A-B-C-D, and we take A, B, C. Then A knows B, B knows C, and A doesn't know C. So in this trio, two people (A and C) don't know each other. So that satisfies condition (b). Similarly, if we take A, B, D. A knows B and D? Wait, no, in a square, A is connected to B and D. If we take A, B, D: A knows B and D, B knows A and C, D knows A and C. So trio A, B, D: A knows B and D, B knows A, D knows A. So in this trio, B and D don't know each other. So again, two people don't know each other. So in the square graph, any three people would have at least two who don't know each other. So actually, the square graph satisfies condition (b). But the square graph is not triangle-free? Wait, no. A triangle is three mutual edges. The square graph doesn't have any triangles, so it is triangle-free. Wait, but the square graph is a cycle of four nodes, which is triangle-free. So maybe condition (b) is equivalent to being triangle-free. Wait, no. Wait, suppose we have a graph that is a pentagon. It's a cycle of five nodes. Then, any three nodes in the pentagon will have at least two who are not adjacent, hence not knowing each other. So that also satisfies condition (b). However, if you have a graph that has a triangle, then in that triangle, all three know each other, so condition (b) is violated. Therefore, condition (b) is equivalent to the graph being triangle-free. Wait, but in the problem statement, it's "among any three persons in a subset, there are always at least two who do not know each other." So in other words, there is no set of three people where all three know each other. That is exactly the definition of triangle-free. So yes, condition (b) is equivalent to the graph being triangle-free. Therefore, each subset's acquaintance graph is triangle-free.Then condition (c): For any two persons in a subset who do not know each other, there is exactly one person in the same subset knowing both of them. So if two people are non-adjacent, there is exactly one common acquaintance. In graph theory terms, this is a strongly regular graph. Specifically, a strongly regular graph with parameters (v, k, λ, μ), where v is the number of vertices, k the degree, λ the number of common neighbors for adjacent pairs, and μ the number of common neighbors for non-adjacent pairs. In this case, condition (c) says that for any non-adjacent pair, there is exactly one common neighbor. So μ = 1. Also, condition (b) says the graph is triangle-free, which would mean that adjacent pairs have no common neighbors. Because if two adjacent vertices had a common neighbor, that would form a triangle. Therefore, λ = 0. So we have a strongly regular graph with parameters (v, k, 0, 1). Such graphs are known as triangle-free strongly regular graphs with μ=1.Moreover, condition (a) says that no one knows all others, so in each subset, every person has degree at most v-2. But in a regular graph (if it is regular), the degree k would be less than v-1. Wait, but if the graph is strongly regular with λ=0 and μ=1, then certain conditions must be met. Let me recall the properties of strongly regular graphs. For a strongly regular graph with parameters (v, k, 0, 1), the eigenvalues can be calculated, and there are certain feasibility conditions.First, the basic equations for strongly regular graphs. The number of vertices v, each vertex has degree k. For any two adjacent vertices, they have λ common neighbors (here λ=0), and for any two non-adjacent vertices, they have μ common neighbors (here μ=1). The eigenvalues of such a graph are given by:θ = [ (1/2)( -1 + sqrt(4k - 3) ) ]andτ = [ (1/2)( -1 - sqrt(4k - 3) ) ]But I might need to check the standard formulas. Alternatively, the eigenvalues can be calculated using the equation:s^2 + (1 - k)s + (μ - k) = 0 for non-main eigenvalues. Wait, maybe I should use the standard relations.For a strongly regular graph, the following equation holds:k(k - λ - 1) = (v - k - 1)μPlugging in λ=0 and μ=1:k(k - 1) = (v - k - 1)(1)So:k^2 - k = v - k - 1Therefore:k^2 = v - 1Thus, v = k^2 + 1So each subset must have v = k^2 + 1 vertices, where each vertex has degree k.Additionally, another condition for strongly regular graphs is that the eigenvalues must be integers. The eigenvalues are given by:θ = [1/2][(λ - μ) + sqrt((λ - μ)^2 + 4(k - μ))]Wait, let me recall the correct formula. The eigenvalues of a strongly regular graph are:θ = (1/2)[-1 + sqrt(4k - 3)]andτ = (1/2)[-1 - sqrt(4k - 3)]Wait, that seems too specific. Alternatively, perhaps I need to use the standard formula.For a strongly regular graph with parameters (v, k, λ, μ), the eigenvalues other than k are:[ (-λ - μ ± sqrt( (λ - μ)^2 + 4(k - μ) )) / 2 ]In our case, λ=0, μ=1, so:[ (-0 - 1 ± sqrt( (0 - 1)^2 + 4(k - 1) )) / 2 ]= [ (-1 ± sqrt(1 + 4k - 4)) / 2 ]= [ (-1 ± sqrt(4k - 3)) / 2 ]Therefore, the eigenvalues are [ (-1 + sqrt(4k - 3) ) / 2 ] and [ (-1 - sqrt(4k - 3) ) / 2 ]For these to be integers, sqrt(4k - 3) must be an integer. Let me denote m = sqrt(4k - 3), which must be an integer. Then 4k - 3 = m^2 => k = (m^2 + 3)/4Since k must be an integer (as it's the degree of each vertex), (m^2 + 3) must be divisible by 4. Therefore, m^2 ≡ 1 mod 4. Since squares modulo 4 are 0 or 1, so m must be odd. Let m = 2t + 1 for some integer t. Then m^2 = 4t^2 + 4t + 1. Then k = (4t^2 + 4t + 1 + 3)/4 = (4t^2 + 4t + 4)/4 = t^2 + t + 1. Therefore, k = t^2 + t + 1 for some integer t ≥ 0. Then, since v = k^2 + 1, substituting k:v = (t^2 + t + 1)^2 + 1= t^4 + 2t^3 + 3t^2 + 2t + 2But we also have that the multiplicities of the eigenvalues must be integers. The multiplicities can be calculated as follows. Let the eigenvalues be θ and τ, with multiplicities f and g. Then:f + g + 1 = vAnd the multiplicities satisfy:f = [ ( -τ - 1 )( v - 1 - k ) ] / ( θ - τ )But perhaps this is getting too complicated. Alternatively, for a strongly regular graph, the multiplicities of the eigenvalues are given by:m_1 = [ (v - 1) - k + sqrt( (v - 1 - k)^2 + 4(n - k - 1) ) ) / 2 ]Wait, maybe I need to check a standard reference.Alternatively, let's recall that for a strongly regular graph, the eigenvalues other than k have multiplicities:f = (1/2)[ (v - 1) - (2k + (v - 1)(λ - μ)) / sqrt( (λ - μ)^2 + 4(k - μ) ) ) ]But maybe this is not helpful.Alternatively, since we have v = k^2 + 1, and k = t^2 + t + 1, perhaps for specific values of t, we can get feasible parameters.Let me try small values of t:For t = 0:k = 0 + 0 + 1 = 1v = 1^2 + 1 = 2So a graph with 2 vertices, each of degree 1. That's just an edge. But in this case, condition (a): no one knows all others. But in a subset of two people, each person knows the other. So each person would know the only other person, violating condition (a). Therefore, t=0 is invalid.t=1:k = 1 + 1 + 1 = 3v = 3^2 + 1 = 10So a graph with 10 vertices, each of degree 3, triangle-free, and every non-adjacent pair has exactly one common neighbor. This is known as the Petersen graph. Yes, the Petersen graph is a well-known example of a strongly regular graph with parameters (10, 3, 0, 1). So that fits. The Petersen graph satisfies all the given conditions: it's triangle-free, every pair of non-adjacent vertices has exactly one common neighbor, and no vertex is connected to all others (since each has degree 3 in a 10-vertex graph). So this works.t=2:k = 4 + 2 + 1 = 7v = 7^2 + 1 = 50So a graph with 50 vertices, each of degree 7, triangle-free, and every non-adjacent pair has exactly one common neighbor. Is such a graph known? I'm not sure, but perhaps it exists. The question is, does such a strongly regular graph exist for t=2? The eigenvalues would be:sqrt(4k - 3) = sqrt(28 - 3) = sqrt(25) = 5So the eigenvalues would be [ (-1 + 5)/2 ] = 2 and [ (-1 - 5)/2 ] = -3.Then the multiplicities can be calculated. Let me recall that for a strongly regular graph, the multiplicities of the eigenvalues satisfy certain equations. Let m_1 and m_2 be the multiplicities of the eigenvalues θ and τ.For our case, θ = 2 and τ = -3.The multiplicities can be found using the equations:1 + m_1 + m_2 = vandk + m_1θ + m_2τ = 0Substituting the values:1 + m_1 + m_2 = 50and7 + 2m_1 - 3m_2 = 0So we have:m_1 + m_2 = 492m_1 - 3m_2 = -7Solving these equations:From the first equation: m_1 = 49 - m_2Substitute into the second equation:2(49 - m_2) - 3m_2 = -798 - 2m_2 - 3m_2 = -798 - 5m_2 = -7-5m_2 = -105m_2 = 21Then m_1 = 49 - 21 = 28So the multiplicities are 28 and 21. These are integers, so the eigenvalues are integers, which is a good sign. However, another condition for strongly regular graphs is that the multiplicities must also be integers, which they are here. Therefore, such a graph might exist. However, I don't know if a (50, 7, 0, 1) strongly regular graph is known. The existence isn't immediately clear to me. Similarly, for higher t, we can get more parameters, but the existence is questionable.But given that the problem is asking for the maximum possible number of subsets, we need to find the minimal size of each subset such that the total number of subsets is maximized. Since the total number of people is 1990, if each subset has size v, then the number of subsets is 1990 / v, so to maximize the number of subsets, we need to minimize v. The minimal possible v is when t=1, which gives v=10 (the Petersen graph), but let's check if smaller v exists.Wait, when t=0, we saw that v=2, but it doesn't satisfy condition (a). For t=1, v=10, which works. For t=2, v=50. So the possible subset sizes are 10, 50, etc. Therefore, the minimal subset size is 10, so the maximum number of subsets would be 1990 / 10 = 199. But we need to check if 10 divides 1990. 1990 divided by 10 is 199, so yes. But we have to verify that the graph with v=10 actually satisfies all the given conditions.Wait, the Petersen graph is a well-known example that satisfies all three conditions. Let's check:(a) No one knows all others. In the Petersen graph, each vertex has degree 3, and there are 10 vertices, so no one knows all others. So condition (a) is satisfied.(b) Among any three persons, at least two do not know each other. Since the Petersen graph is triangle-free, there are no three mutual acquaintances. In fact, it's even more restrictive—its girth is 5, meaning the shortest cycle is length 5, so it has no triangles or quadrilaterals. So condition (b) is satisfied.(c) For any two persons who do not know each other, there is exactly one person who knows both. In the Petersen graph, any two non-adjacent vertices have exactly one common neighbor. Yes, that's a property of the Petersen graph. Therefore, condition (c) is satisfied.Therefore, subsets of size 10 satisfy all the conditions. If we can divide the 1990 people into subsets of size 10, then the maximum number of subsets would be 199. But we need to make sure that there are no smaller subsets possible. For instance, if there exists a graph with v=5 that satisfies the conditions, then 1990 / 5 = 398 would be larger. But does such a graph exist?Let me check for t=1/2? Wait, t must be an integer because m was an integer (sqrt(4k -3) = m). So t has to be integer. So the next possible subset size after 10 is 50, which would give 1990 /50 = 39.8, which is not an integer. 1990 divided by 50 is 39.8, so we can't have 50-vertex subsets. Similarly, the next one would be for t=3:k = t^2 + t + 1 = 9 + 3 +1 = 13v = 13^2 +1 = 170170 divides 1990? 1990 /170 = 11.7..., not an integer. So not possible.Therefore, the only feasible subset size that divides 1990 is 10, giving 199 subsets. Therefore, the maximum possible number of subsets is 199.But wait, let me confirm that there are no other possible strongly regular graphs with different parameters that could have smaller subset sizes. For example, is there a strongly regular graph with parameters different from the ones we considered that could satisfy conditions (a), (b), (c)?Let me think. For a graph to satisfy conditions (a), (b), (c):- (a) No vertex is connected to all others: so degree k ≤ v - 2.- (b) Triangle-free: so no edges between neighbors.- (c) Any two non-adjacent vertices have exactly one common neighbor.So such a graph is a strongly regular graph with λ=0, μ=1, and k ≤ v - 2. The only known such graphs are the Petersen graph (10,3,0,1) and maybe some others. For example, the Petersen graph is the unique such graph with those parameters. Are there others?Another example is the complement of the Petersen graph, but the complement would have parameters (10,6,3,4), which doesn't fit our case. There is also the 5-cycle, but in the 5-cycle, each vertex has degree 2. Let's check if the 5-cycle satisfies the conditions.For the 5-cycle:(a) Each person knows 2 others, so no one knows all others. Satisfies (a).(b) Any three persons: in a 5-cycle, take any three consecutive vertices. Each is connected to the next, but the first and third are not connected. So among three consecutive vertices, two do not know each other. Similarly, any three vertices in the cycle will have at least two non-adjacent. So condition (b) is satisfied.(c) For any two non-adjacent vertices, how many common neighbors do they have? In the 5-cycle, two non-adjacent vertices are distance 2 apart. Each has two neighbors. For example, in a 5-cycle labeled A-B-C-D-E-A, the non-adjacent pairs are (A,C), (B,D), (C,E), (D,A), (E,B). Take A and C: their neighbors are B and E (for A) and B and D (for C). Common neighbor is B. So exactly one common neighbor. Similarly, A and D: A is connected to B and E; D is connected to C and E. Common neighbor is E. Similarly, all non-adjacent pairs have exactly one common neighbor. Therefore, the 5-cycle satisfies condition (c).Therefore, the 5-cycle is another example of a graph satisfying all three conditions. Wait, but the 5-cycle has parameters (5,2,0,1). So in terms of strongly regular graph parameters, (5,2,0,1). Let me check if this fits the earlier equation.From before, we had the equation k(k - 1) = (v - k -1)μ. Plugging in v=5, k=2, μ=1:Left side: 2*1 = 2Right side: (5 - 2 -1)*1 = 2*1 = 2So yes, it satisfies the equation. Also, the eigenvalues would be sqrt(4k -3) = sqrt(8 -3) = sqrt(5), which is not an integer. Wait, but earlier we had a different way of calculating eigenvalues. Wait, no, the formula for the eigenvalues when λ=0, μ=1 is [ (-1 ± sqrt(4k -3))/2 ]So for k=2, sqrt(4*2 -3) = sqrt(5), so the eigenvalues are (-1 ± sqrt(5))/2, which are not integers. However, in the 5-cycle, the eigenvalues are 2*cos(2πj/5) for j=0,1,2,3,4, which are 2, 2*cos(72°), 2*cos(144°), etc., which are not integers. Therefore, the 5-cycle is a strongly regular graph with non-integer eigenvalues. But the earlier condition about eigenvalues being integers might not hold for all cases. Wait, but in the 5-cycle, the strongly regular graph parameters (5,2,0,1) do not require integer eigenvalues. So perhaps the integrality condition is only for certain cases.But this suggests that the 5-cycle is a valid graph satisfying all the given conditions, and it has v=5. Then, if subsets of size 5 are allowed, then 1990 /5 = 398 subsets, which is much larger than 199.But wait, this contradicts our earlier conclusion. So there must be an error in my reasoning.Wait, so the 5-cycle graph: let's verify the conditions again.(a) No one knows everyone else. Each person knows 2 others in the subset of 5, so yes, condition (a) is satisfied.(b) Among any three persons, at least two do not know each other. Let's take three consecutive vertices in the cycle: A, B, C. A knows B, B knows C, but A doesn't know C. So in this trio, A and C don't know each other. Similarly, any trio will have at least two who don't know each other. So condition (b) is satisfied.(c) For any two persons who don't know each other, there is exactly one person who knows both. Let's take two non-adjacent vertices. In the 5-cycle, two vertices are non-adjacent if they are two steps apart. For example, A and C. Their common neighbor is B. Similarly, A and D: their common neighbor is E. Similarly for others. So yes, each non-adjacent pair has exactly one common neighbor. Therefore, condition (c) is satisfied.Therefore, the 5-cycle is indeed a valid subset. So that's a graph with v=5, k=2, which fits into the equation k(k -1) = (v -k -1)μ => 2*1 = (5 -2 -1)*1 => 2=2. So it works.But then why didn't this come up in the previous analysis where we set m = sqrt(4k -3) and got k=(m^2 +3)/4? Because in this case, k=2, so 4k -3 = 5, so m=sqrt(5), which is not an integer. However, the eigenvalues don't have to be integers for a strongly regular graph. The integrality condition is only necessary if the graph is conference graphs (a type of strongly regular graph) or in certain cases. So maybe the 5-cycle is an exception. So in this case, even though the eigenvalues are not integers, the graph is still strongly regular.Therefore, there exist strongly regular graphs with parameters (v, k, 0,1) where v=5, k=2, which is the 5-cycle. So this graph is a valid subset. Similarly, are there other cycles? Let's check 4-cycle.In a 4-cycle (square), each vertex has degree 2. Check the conditions:(a) No one knows everyone else. Correct, each knows two out of three others. Wait, in a subset of 4, each person knows two others, so there are two they don't know. So condition (a) is satisfied.(b) Among any three persons, at least two do not know each other. Take three vertices in the 4-cycle: for example, A, B, C. A knows B and D, B knows A and C, C knows B and D. So in trio A, B, C: A doesn't know C. Similarly, trio A, B, D: A knows B and D, B knows A, D knows A. So B and D don't know each other. So yes, any three have at least two who don't know each other. Wait, but in a 4-cycle, if you take three vertices, two are adjacent and one is not. But in the 4-cycle, is there a trio where two don't know each other? Let's see. Take A, B, C: A knows B, B knows C, A doesn't know C. So two who don't know each other. Similarly, trio A, C, D: A knows D, D knows C, A doesn't know C. So again, two who don't know each other. So condition (b) is satisfied.(c) For any two persons who don't know each other, there is exactly one person who knows both. In the 4-cycle, non-adjacent pairs are A-C and B-D. Let's take A and C. They are not adjacent. Their common neighbors: A is connected to B and D; C is connected to B and D. So common neighbors are B and D. Wait, two common neighbors. But condition (c) requires exactly one. Therefore, the 4-cycle does not satisfy condition (c). Therefore, the 4-cycle is invalid.Similarly, the 3-cycle is a triangle, which violates condition (b). So the 5-cycle works, the 4-cycle does not. What about the 6-cycle? Let's check.In a 6-cycle, each vertex has degree 2. Check condition (c): For two non-adjacent vertices. In a 6-cycle, non-adjacent vertices can be distance 2 or 3 apart. For example, in a cycle A-B-C-D-E-F-A:- A and C are distance 2 apart. Common neighbors: B and F. So two common neighbors. That violates condition (c).Similarly, A and D (distance 3) have no common neighbors. Wait, A is connected to B and F; D is connected to C and E. So no common neighbors. Therefore, non-adjacent pairs can have 0 or 2 common neighbors. Therefore, the 6-cycle does not satisfy condition (c).Therefore, the 5-cycle is unique in that non-adjacent pairs have exactly one common neighbor. Let's confirm:In a 5-cycle, two non-adjacent vertices are distance 2 apart. Each has two neighbors. For example, A and C in the 5-cycle A-B-C-D-E-A. A is connected to B and E; C is connected to B and D. Common neighbor is B. Exactly one common neighbor. Similarly, A and D: A connected to B and E; D connected to C and E. Common neighbor is E. Exactly one. Similarly for others. So yes, all non-adjacent pairs have exactly one common neighbor. Therefore, the 5-cycle satisfies all three conditions.Therefore, subset size 5 is possible. So then, why in our earlier analysis using strongly regular graphs, we only found the Petersen graph (v=10) and larger. But the 5-cycle is also a strongly regular graph with (5,2,0,1). So the problem is that the earlier approach required eigenvalues to be integers, which is not the case here. Therefore, perhaps there are other strongly regular graphs with parameters (v, k, 0, 1) that do not satisfy the integrality conditions but still exist. The 5-cycle is such an example.Therefore, subset sizes can be 5 or 10. If subset size 5 is allowed, then the maximum number of subsets would be 1990 /5 = 398. But wait, we need to check if such subsets can coexist. The problem states that the entire set of 1990 is divided into non-intersecting subsets, each satisfying (a), (b), (c). So if we can use subsets of size 5, that would be better (more subsets). But does a 5-cycle graph satisfy all three conditions? We checked that it does. Then why didn't the earlier analysis with the equation k^2 = v -1 account for this?Because for the 5-cycle, k=2, v=5. Then k^2 =4, v -1 =4. So 4=4, so it does satisfy k^2 = v -1. Wait, but earlier, when I derived v =k^2 +1, that came from the equation k(k -1) = (v -k -1)μ. But in this case, μ=1, so substituting:k(k -1) = (v -k -1)(1)So k^2 -k = v -k -1Therefore, k^2 = v -1Thus, v =k^2 +1For the 5-cycle, k=2, so v=5=4 +1, yes. So the 5-cycle does fit into this equation. Therefore, the earlier analysis was correct, but I missed that t can be such that m is sqrt(4k -3). In the 5-cycle case:k=2, so 4k -3 =5, so m= sqrt(5). But m needs to be integer for the eigenvalues to be integers. However, the 5-cycle graph still exists even though its eigenvalues are not integers. Therefore, the equation v =k^2 +1 is necessary for a strongly regular graph with λ=0, μ=1, but the integrality of eigenvalues is not a strict requirement for existence, although it is a requirement for certain types of strongly regular graphs, like conference graphs.Therefore, the 5-cycle is a valid example with v=5, k=2, and other parameters. Therefore, subset size 5 is possible. Then, subset size 10 (Petersen graph) is another. So now the question is, are there other subset sizes?Let's see, the next case would be k=3, so v =3^2 +1=10, which is the Petersen graph. Then k=2 gives v=5. Next, k=1 would give v=2, but that doesn't satisfy condition (a). For k=4, v=17, but we can check if such a graph exists. Let's see if a (17,4,0,1) strongly regular graph exists.Using the equation:k(k -1) = (v -k -1)*μ4*3 = (17 -4 -1)*1 =>12=12. So the equation holds.Then eigenvalues would be:sqrt(4k -3) = sqrt(16 -3)=sqrt(13), non-integer. But does such a graph exist? I don't know. There is a known strongly regular graph with parameters (16,6,2,2), but (17,4,0,1) is the complement of the (17,8,3,4) strongly regular graph. However, I don't think such a graph is known to exist. The existence of strongly regular graphs with non-integer eigenvalues is questionable. In fact, strongly regular graphs with such parameters are called "pseudo" graphs if they don't satisfy the integrality conditions, and their existence is often ruled out by the Krein conditions or the absolute bound. The Krein conditions state that for a strongly regular graph, (θ + 1)(k + θ + 2θτ) ≤ (k + θ)(τ + 1)^2, and similarly for τ. These conditions must be satisfied for a strongly regular graph to exist. If the eigenvalues are not integers, it's more complicated, but generally, non-integer eigenvalues can still satisfy Krein conditions.However, rather than getting bogged down in existence, which might be beyond my current knowledge, let's focus on known examples. The 5-cycle and the Petersen graph are known to exist. The 5-cycle has parameters (5,2,0,1), and the Petersen graph has (10,3,0,1). Are there others? For example, the 6-vertex graph: does a 6-vertex graph satisfy the conditions? Let's check.If v=6, then k^2 =v -1=5, so k= sqrt(5), which is not integer. Therefore, no such graph. So subset size 6 is invalid.Similarly, for v=5, it works. For v=10, it works. For v=17, if it exists, it would work, but since 17 doesn't divide 1990 (1990 divided by 17 is ~117), but 1990 /17 = 117.05, not integer. So even if such a graph exists, we can't use it here.Similarly, v=50: 1990 /50=39.8, not integer. So the only subset sizes that divide 1990 are 5 and 10. 1990 divided by 5 is 398, which is an integer. 1990 divided by 10 is 199, which is also an integer. So both are possible. But we need to check if the 5-cycle and the Petersen graph are the only possible subset types, or if other graphs with different v could also be combined.However, the problem asks for the maximum possible number of subsets. To maximize the number of subsets, we need to minimize the size of each subset. The smallest possible subset size is 5 (the 5-cycle), allowing for 398 subsets. The next is 10 (Petersen graph), allowing 199 subsets. Therefore, 398 is larger. But we need to verify if all subsets can indeed be of size 5.But wait, there is a problem. The total number of people is 1990. If we partition into subsets of size 5, we need 1990 /5 = 398 subsets. Since 5 divides 1990 (1990=5*398), this is possible. So if subsets of size 5 are allowed, then the maximum number of subsets is 398. However, the question is, are there any constraints that prevent using subsets of size 5?Wait, the problem statement says that the set is divided into non-intersecting subsets satisfying (a), (b), (c). The 5-cycle satisfies all three conditions, so there shouldn't be any problem. Therefore, the maximum number of subsets would be 398.But this contradicts the initial analysis where I thought the answer was 199. So where is the mistake?Ah, the key is that the 5-cycle graph satisfies all three conditions, and allows for smaller subsets. Therefore, the maximum number of subsets is indeed 398. However, I need to confirm this with the problem constraints.Wait, let me verify again the conditions for the 5-cycle:(a) No one knows all others. In a 5-cycle, each person knows 2 others, so yes, condition satisfied.(b) Any three people have at least two who don't know each other. As analyzed earlier, yes.(c) Any two who don't know each other have exactly one common acquaintance. Yes, as each non-adjacent pair has exactly one common neighbor.Therefore, the 5-cycle is a valid subset. Therefore, subsets of size 5 are allowed, and since 5 divides 1990, the maximum number of subsets is 398.But why does the problem part (i) ask to prove that within each subset, every person has the same number of acquaintances? In the 5-cycle, each person has exactly 2 acquaintances, which is the same for everyone. In the Petersen graph, each person has 3 acquaintances. So in both cases, the graph is regular. Therefore, part (i) is satisfied.Therefore, for part (ii), the maximum number of subsets is 1990 divided by the minimal subset size. The minimal subset size is 5, so 1990 /5=398.But wait, let's check whether there exists a strongly regular graph with parameters (v, k, 0,1) for v=5 and v=10. Since both exist, and these are the only known ones, and for larger v, either they don't exist or don't divide 1990, then 398 is the answer.However, I need to make sure that there are no other subset sizes. For example, the 5-cycle is size 5, the Petersen graph size 10. What about the complement of the Petersen graph? It has parameters (10,6,3,4), which doesn't fit our conditions. So no.Another example: the Clebsch graph, which is a (16,5,0,2) strongly regular graph. Doesn't fit our μ=1 condition. So no.The Shrikhande graph is (16,6,2,2). Not applicable.So the only known graphs with μ=1 and λ=0 are the 5-cycle and the Petersen graph. Therefore, the minimal subset sizes are 5 and 10. Since 5 is smaller, and 5 divides 1990, the maximum number of subsets is 398.But wait, I need to check some references in my mind. The 5-cycle is indeed a valid example, and the problem doesn't restrict subset sizes beyond the given conditions. Therefore, the answer should be 398.However, I recall that in some problems, such as in the theory of block designs or friendship graphs, the friendship theorem states that if every pair of vertices has exactly one common neighbor, then the graph is a friendship graph, which consists of triangles sharing a common vertex. However, in our case, the graph is triangle-free, which would conflict with the friendship theorem. Wait, the friendship theorem says that if any two vertices have exactly one common neighbor, then the graph is a friendship graph. But the friendship graph is not triangle-free. In our case, we have a triangle-free graph where any two non-adjacent vertices have exactly one common neighbor. Adjacent vertices have no common neighbors (due to triangle-free). So this is different from the friendship theorem's hypothesis.In the friendship theorem, the condition is that any two vertices (whether adjacent or not) have exactly one common neighbor. In our problem, it's only for any two non-adjacent vertices. Therefore, the friendship theorem does not apply here.Therefore, our graphs are allowed to have adjacent pairs with zero common neighbors (due to triangle-free) and non-adjacent pairs with exactly one common neighbor. Such graphs are known as "strongly regular graphs" with λ=0, μ=1.The 5-cycle and the Petersen graph are two examples. It is known that there are only a few such graphs. According to some references, the 5-cycle, the Petersen graph, the Hoffman-Singleton graph, and possibly the Higman-Sims graph are examples, but the Hoffman-Singleton graph has parameters (50, 22, 0, 6), so μ=6, which doesn't fit. Wait, no. Wait, the Hoffman-Singleton graph is a (50, 22, 0, 6) strongly regular graph, so μ=6. Not relevant here.Wait, maybe the only known triangle-free strongly regular graphs with μ=1 are the 5-cycle and the Petersen graph. If that's the case, then the minimal subset sizes are 5 and 10. Therefore, using 5 gives the maximum number of subsets.But to confirm, let's check for v=5 and v=10.For v=5, 1990 divided by 5 is 398.For v=10, it's 199.If there exists a graph with v=15, k=4, then 1990 /15=132.666, not integer. But anyway, such a graph's existence is doubtful.Therefore, the conclusion is that the maximum number of subsets is 398, achieved by dividing the set into 398 subsets of size 5, each arranged as a 5-cycle, satisfying all the given conditions.But wait, let me check once more. In the 5-cycle, the number of edges is 5, each person has degree 2. The total number of edges in the subset is (5*2)/2=5. In terms of the conditions:(a) Satisfied.(b) Satisfied.(c) Satisfied.Yes. Therefore, 5-cycles are valid. Therefore, answer is 398.But I need to verify part (i), which asks to prove that within each subset, every person has the same number of acquaintances. In the 5-cycle, each person has 2 acquaintances, which is regular. In the Petersen graph, each has 3, which is regular. So if all subsets are either 5-cycles or Petersen graphs, then within each subset, everyone has the same number of acquaintances. Therefore, part (i) is proven.Therefore, the answer to part (ii) is 398.However, the problem might be expecting the answer 199, assuming the Petersen graph is the minimal subset. But given that the 5-cycle satisfies all conditions, the correct answer should be 398. But to ensure that, I need to make sure there are no other constraints.Wait, another thought: in condition (c), for any two persons in the subset who do not know each other, there is exactly one person in the same subset knowing both of them.In the 5-cycle, take two non-adjacent people: they have exactly one common neighbor. In the 5-cycle, this is true. So yes, condition (c) is satisfied.Therefore, the 5-cycle is valid, and thus the maximum number is 398.But perhaps the problem assumes that the graph is regular, which the 5-cycle is, so part (i) is satisfied.Therefore, my final answer is:(i) Proven by noting that each subset is a strongly regular graph (specifically a 5-cycle or Petersen graph), which are regular.(ii) The maximum number of subsets is 398.However, to cross-verify, let me recall that in the literature, such a graph (triangle-free, every two non-adjacent vertices have exactly one common neighbor) is known as a friendship graph minus the central node or something else? Wait, no. The friendship graph is different.Alternatively, these graphs are known as the polygons (5-cycle) and the Petersen graph. According to some references, the 5-cycle and the Petersen graph are the only triangle-free strongly regular graphs with μ=1. If that is the case, then the minimal subset size is 5, leading to 398 subsets.Therefore, unless there's a mistake in my analysis, the answer should be 398.But let me check another angle. Suppose all subsets are 5-cycles. Then, the total number of edges in the entire graph is 1990/5 *5=1990 edges. But each person is in a 5-cycle with degree 2, so total edges per subset:5. Total edges:398*5=1990. But since each edge is counted twice in the total degree (each edge contributes to two people), wait no. Total number of edges in the entire graph would be 1990 people *2 acquaintances /2=1990 edges. Yes, that's correct.But the problem doesn't mention anything about the entire graph, only about the subsets. The entire graph is divided into subsets, each of which is a graph with the given properties. The entire graph is the disjoint union of these subset graphs. Therefore, the entire graph's structure is a union of cycles of length 5 or Petersen graphs, etc. But the problem doesn't impose any conditions on the entire graph, only on each subset.Therefore, the answer is 398.But I have a doubt because in some sources, the only triangle-free strongly regular graphs are the 5-cycle, Petersen graph, and maybe a few others. But none of the others would divide 1990.Therefore, I think the correct answer is 398.Final Answer(i) boxed{text{Every person within each subset has the same number of acquaintances.}} (ii) boxed{398}

✅:The problem involves dividing a set of 1990 persons into non-intersecting subsets satisfying specific conditions. The key steps to solve the problem are as follows:1. Understanding the Conditions: - (a) No one in a subset knows all others. - (b) Among any three persons in a subset, at least two do not know each other (triangle-free). - (c) For any two persons who do not know each other, there is exactly one common acquaintance.2. Graph Theory Interpretation: - The conditions describe a strongly regular graph with parameters ((v, k, 0, 1)), where (v) is the number of vertices, (k) is the degree, (lambda = 0) (triangle-free), and (mu = 1) (exactly one common neighbor for non-adjacent pairs).3. Feasible Graphs: - The 5-cycle (pentagon) and the Petersen graph are known examples of such graphs. The 5-cycle has parameters ((5, 2, 0, 1)) and the Petersen graph has parameters ((10, 3, 0, 1)).4. Verification: - Both graphs satisfy all conditions (a), (b), and (c). The 5-cycle is a regular graph where each vertex has 2 acquaintances, and the Petersen graph is a regular graph where each vertex has 3 acquaintances.5. Maximizing the Number of Subsets: - The minimal subset size is 5 (5-cycle), leading to the maximum number of subsets. The total number of subsets is (1990 / 5 = 398).6. Conclusion: - Part (i) is proven by noting that each subset is a regular graph (either 5-cycle or Petersen graph). - Part (ii) is determined by using the smallest subset size (5), leading to the maximum number of subsets. Final Answer(i) boxed{text{Every person within each subset has the same number of acquaintances.}} (ii) boxed{398}

🔑:# Part (i):1. Consider one subset ( S ) satisfying the given conditions: Let's take ( S ) as a subset of persons. We need to show that within ( S ), every person has the same number of acquaintances.2. Identify a person with the maximum number of acquaintances: Let ( x in S ) be a person who knows the maximum number of persons in ( S ). Suppose ( x ) knows ( x_1, x_2, ldots, x_n ).3. Explore relationships between acquaintances: By condition (b), no subset of three people contains a triangle of acquaintance. Thus, ( x_i ) and ( x_j ) do not know each other if ( i neq j ), because then ( x, x_i ), and ( x_j ) would violate condition (b).4. Define sets ( N_i ) of persons who know ( x_i ) but not ( x ): For each ( x_i ) (where ( 1 leq i leq n )), let ( N_i ) be the set of persons in ( S ) who know ( x_i ) but not ( x ). Note that if ( i neq j ), then ( N_i ) and ( N_j ) have no person in common. If there were a common person, then there would be more than one person knowing both ( x_i ) and ( x_j ), which would contradict condition (c).5. Assume ( N_1 ) is non-empty and explore the acquaintances of a member of ( N_1 ): Assume ( N_1 ) is not empty. Let ( y_1 in N_1 ). By condition (c), for each ( k > 1 ), there is exactly one person ( y_k ) in ( N_k ) who knows ( y_1 ). This implies that ( y_1 ) knows ( n ) persons, namely ( x_1, y_2, ldots, y_n ).6. Conclude knowledge network consistency: Since ( n ) is the maximum number of acquaintances a person can have in ( S ), ( y_1 ) can only know exactly ( n ) persons. By the same reasoning, each person in ( N_i ) (for ( i = 1, 2, ldots, n )) also knows exactly ( n ) persons in ( S ).7. By symmetry, generalize the acquaintance structure: Replacing ( y_1 ) with ( x ) in our argument, we see that also each ( x_i ) knows exactly ( n ) persons.8. Sum up relationships based on condition (c): Every person in ( S ) other than ( x, x_1, ldots, x_n ) must be in some ( N_j ). Therefore, every person in ( S ) knows exactly ( n ) persons in ( S ), confirming that everyone in ( S ) has the same number of acquaintances.[blacksquare]# Part (ii):1. Objective to maximize subset count: To maximize the number of subsets, we have to minimize the size of each group following the given constraints.2. Identify the smallest viable subset structure: A subset where each person knows exactly 2 people would minimize the group size. This configuration is satisfied by arranging 5 persons in a cycle, where each person knows the two persons adjacent to them.3. Calculate the maximal number of subsets: The smallest subset satisfying the conditions contains exactly 5 persons. Hence, the maximum number of such subsets is given by: [ text{Maximum number of subsets} = frac{1990}{5} = 398 ][boxed{398}]

🧠:Okay, so I have this problem here. Let me read it again. In triangle ABC, AC equals AB equals 25, and BC is 40. D is a point on BC. From D, perpendiculars are drawn to meet AC at E and AB at F. The question is asking for DE + DF, and the options are given. Alright.First, let me visualize the triangle. ABC is an isosceles triangle because AB equals AC, both 25 units. The base BC is 40 units. So, ABC is an isosceles triangle with AB = AC = 25 and BC = 40. Point D is somewhere on BC. From D, we drop perpendiculars to AC and AB, meeting them at E and F respectively. Then, we need to find the sum DE + DF. Hmm. Interesting.Since ABC is isosceles with AB = AC, the triangle is symmetric with respect to the altitude from A to BC. Let me maybe draw that altitude. If I let H be the foot of the altitude from A to BC, then BH = HC = 20 because BC is 40. Then, using the Pythagorean theorem, the height AH can be calculated. Let me compute that.In triangle ABH, AB is 25, BH is 20. So, AH squared plus BH squared equals AB squared. So, AH^2 + 20^2 = 25^2. AH^2 = 625 - 400 = 225. Therefore, AH is 15. So the height of the triangle is 15 units. Got that.Now, point D is somewhere on BC. Let me assign coordinates to the points to make this problem easier. Let's place the triangle in a coordinate system. Let me put point B at (-20, 0), point C at (20, 0), and point A at (0, 15). Wait, because BC is 40 units, so from -20 to 20 on the x-axis, that's 40 units. The altitude is 15 units, so point A is at (0,15). That makes sense.So, coordinates:- B: (-20, 0)- C: (20, 0)- A: (0, 15)Now, point D is on BC. Let's parametrize point D. Let me let D be at (x, 0), where x ranges from -20 to 20. Wait, but BC is from B (-20,0) to C (20,0). Wait, but BC is 40 units long. So, yes, from x = -20 to x = 20. But actually, the problem says BC is 40, so perhaps the coordinates are from (0,0) to (40,0). Wait, maybe I should adjust that. Wait, perhaps I made a mistake in coordinate placement. Let me check.Wait, if BC is 40 units, then if we set B at (0,0) and C at (40,0), then the base BC is along the x-axis from 0 to 40. Then, point A is somewhere above the x-axis. Since AB = AC = 25, we can find the coordinates of A. Let me recast the coordinates for clarity.Let’s place point B at (0,0), point C at (40,0). Then, point A will be somewhere above the x-axis such that AB = AC = 25. Let me compute the coordinates of A. The x-coordinate of A will be at the midpoint of BC since the triangle is isosceles. So, midpoint of BC is at (20, 0). Therefore, point A is at (20, h), where h is the height. Then, the distance from A to B is 25. So, using the distance formula:Distance from A(20, h) to B(0,0) is sqrt((20 - 0)^2 + (h - 0)^2) = sqrt(400 + h^2) = 25. Therefore:sqrt(400 + h^2) = 25Squaring both sides: 400 + h^2 = 625h^2 = 225h = 15Therefore, coordinates:- B: (0, 0)- C: (40, 0)- A: (20, 15)This might be a better coordinate system. So, BC is from (0,0) to (40,0), and A is at (20,15). Then, point D is on BC, so coordinates of D can be parametrized as (t, 0), where t ranges from 0 to 40. From D(t,0), we need to draw perpendiculars to AC and AB, meeting them at E and F respectively.We need to find DE + DF. Hmm. Let's see.First, let's find equations for lines AB and AC.Line AB connects points A(20,15) and B(0,0). The slope of AB is (15 - 0)/(20 - 0) = 15/20 = 3/4. So, the equation of AB is y = (3/4)x.Line AC connects points A(20,15) and C(40,0). The slope of AC is (0 - 15)/(40 - 20) = (-15)/20 = -3/4. So, the equation of AC is y - 15 = (-3/4)(x - 20), which simplifies to y = (-3/4)x + 15 + 15 = y = (-3/4)x + 30. Wait, let's check that:Slope is -3/4. So, using point A(20,15):y - 15 = (-3/4)(x - 20)y = (-3/4)x + ( -3/4 * -20 ) + 15= (-3/4)x + 15 + 15Wait, no. Wait, (-3/4)*(x - 20) is (-3/4)x + 15. Then, adding 15 gives y = (-3/4)x + 15 + 15 = (-3/4)x + 30. Wait, that can't be. Let me verify with point C(40,0):Plugging x = 40 into the equation y = (-3/4)x + 30: y = (-3/4)*40 + 30 = -30 + 30 = 0. Correct. So equation of AC is y = (-3/4)x + 30.Okay, so equations:AB: y = (3/4)xAC: y = (-3/4)x + 30Now, from point D(t, 0), we need to draw perpendiculars to AB and AC. Let's first find the perpendicular from D to AB, which meets AB at F, and the perpendicular from D to AC, which meets AC at E.First, let's find DF, the length from D(t,0) to F on AB, which is the foot of the perpendicular from D to AB.Similarly, DE is the length from D(t,0) to E on AC, the foot of the perpendicular from D to AC.We need to compute DF and DE as functions of t, then add them together, and see if the sum is constant (since the answer is one of the options, implying it doesn't depend on t). Wait, that might be the case. If DE + DF is constant regardless of where D is on BC, then the answer is that constant value. Let's check.Alternatively, maybe it's not constant, but in this problem, the answer is one of the given options, so perhaps it's a constant.But let's proceed step by step.First, let's find DF. DF is the distance from D(t,0) to line AB. The formula for the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2).First, let's write equation AB in standard form. AB: y = (3/4)x. So, rearranged: (3/4)x - y = 0. Multiply both sides by 4: 3x - 4y = 0. So, standard form is 3x - 4y = 0.Therefore, distance from D(t,0) to AB is |3*t - 4*0 + 0| / sqrt(3^2 + (-4)^2) = |3t| / 5. Therefore, DF = 3t / 5.Wait, but t is between 0 and 40? Wait, in our coordinate system, D is on BC from (0,0) to (40,0), so t ranges from 0 to 40. Therefore, DF = 3t /5.Wait, but let me verify that. Because DF is the distance from D(t,0) to line AB. Let's compute that.Alternatively, since DF is the foot of the perpendicular from D to AB, maybe we can compute coordinates of F and then compute the distance DF.Alternatively, since the formula gives DF = |3t - 4*0| / 5 = 3t/5. But that would imply DF increases as t increases, which seems odd if DE + DF is constant.Wait, perhaps the formula is correct. Let me check with a specific point. Let's take D at B, which is (0,0). Then DF is the distance from B to AB. But B is on AB, so DF should be zero. Plugging t=0 into 3t/5 gives 0. Correct. Similarly, if D is at midpoint (20,0), then DF would be 3*20/5 = 12. If D is at C(40,0), then DF would be 3*40/5 = 24. But wait, point C is not on AB. Wait, but DF is the distance from D to AB. So, when D is at C(40,0), DF is the distance from C to AB, which would be a fixed value. Let me compute that.Using the distance formula from C(40,0) to AB: 3*40 -4*0 | /5 = 120/5 =24. Correct. So DF at D=C is 24. Similarly, if D is at B, DF is 0, which makes sense. So DF= 3t/5.Similarly, let's compute DE, the distance from D(t,0) to AC.First, equation of AC is y = (-3/4)x + 30. Let's write that in standard form. Multiply both sides by 4: 4y = -3x + 120 → 3x + 4y - 120 = 0.Therefore, standard form is 3x + 4y -120 =0.So, the distance from D(t,0) to AC is |3*t + 4*0 -120| / sqrt(3^2 +4^2) = |3t -120| /5.Therefore, DE = |3t -120| /5.Since t ranges from 0 to 40, 3t -120 ranges from -120 to 0. Therefore, |3t -120| = 120 -3t.Therefore, DE = (120 -3t)/5 = 24 - (3t)/5.Therefore, DE + DF = [24 - (3t)/5] + [3t/5] = 24. So DE + DF =24. Therefore, the answer is 24, which is option C.Wait, that's interesting. So regardless of where D is on BC, DE + DF is always 24. That's a constant. Therefore, the correct answer is 24, option C.But let me verify this with a specific example. Let's take D at B(0,0). Then DF is 0 (since D is on AB), and DE is the distance from B to AC. Let's compute that. Using the distance formula from B(0,0) to AC: |3*0 +4*0 -120| /5 = | -120 | /5 =24. So DE=24, DF=0. Sum=24.Similarly, take D at C(40,0). Then DE is 0 (since D is on AC), and DF is the distance from C to AB, which we calculated as 24. So DE=0, DF=24. Sum=24.Take D at midpoint (20,0). Then DF=3*20/5=12, DE=24 -12=12. So DE + DF=24. Yep, works. So regardless of where D is on BC, DE + DF=24. Therefore, the answer is 24, option C.But wait, let me check with another point. Let's take D at (10,0). Then DF=3*10/5=6. DE=24 - 3*10/5=24-6=18. 6+18=24. Correct. So DE + DF=24. Seems consistent.Therefore, the answer is 24, which is option C. So the answer is C.Final Answerboxed{C}

✅:Given (triangle ABC) with (AC = AB = 25) and (BC = 40). Point (D) is chosen on (BC), and perpendiculars are drawn from (D) to (AC) at (E) and to (AB) at (F). We need to find (DE + DF).1. Coordinates Setup: - Place (B) at ((0, 0)), (C) at ((40, 0)), and (A) at ((20, 15)) using the midpoint and height calculated from the Pythagorean theorem. - The height (AH) from (A) to (BC) is 15, calculated as follows: [ AH^2 + 20^2 = 25^2 implies AH^2 = 225 implies AH = 15 ]2. Equations of Lines: - Line (AB): (y = frac{3}{4}x) - Line (AC): (y = -frac{3}{4}x + 30)3. Distance Calculations: - Distance from (D(t, 0)) to line (AB) (standard form (3x - 4y = 0)): [ DF = frac{|3t - 4 cdot 0|}{5} = frac{3t}{5} ] - Distance from (D(t, 0)) to line (AC) (standard form (3x + 4y - 120 = 0)): [ DE = frac{|3t + 4 cdot 0 - 120|}{5} = frac{|3t - 120|}{5} = frac{120 - 3t}{5} ]4. Sum of Distances: - Calculate (DE + DF): [ DE + DF = frac{120 - 3t}{5} + frac{3t}{5} = frac{120}{5} = 24 ]Thus, (DE + DF) is constant and equal to 24 regardless of the position of (D) on (BC).[boxed{C}]

🔑:1. We start with identifying the essential dimensions of triangle triangle ABC. - Given: AC = AB = 25 and BC = 40. - triangle ABC is an isosceles triangle with AC = AB.2. We draw an altitude from vertex A perpendicular to the base BC, and let the foot of the perpendicular be M. Since triangle ABC is isosceles, point M is the midpoint of BC. - Therefore, BM = CM = frac{BC}{2} = frac{40}{2} = 20.3. To find AM (the altitude), we apply the Pythagorean theorem in triangle ACM: [ AM^2 + CM^2 = AC^2 ] [ AM^2 + 20^2 = 25^2 ] [ AM^2 + 400 = 625 ] [ AM^2 = 225 ] [ AM = sqrt{225} = 15 ]4. We now consider the total area of triangle ABC. Using the base BC and height AM: [ text{Area of } triangle ABC = frac{1}{2} times BC times AM = frac{1}{2} times 40 times 15 = 300 ]5. We focus on the point D on BC, and draw perpendiculars DE and DF from D to AC and AB respectively.6. Denote by x the altitude (perpendicular) from D to AC (i.e., DE) and by y the altitude from D to AB (i.e., DF). The areas of triangles triangle ADE and triangle ADF can now be considered.7. The area of triangle ABC can also be calculated by summing the areas of triangle ADE, triangle ADF, and triangle DEF: - text{Area of } triangle ADE = frac{1}{2} times AC times DE = frac{1}{2} times 25 times x - text{Area of } triangle ADF = frac{1}{2} times AB times DF = frac{1}{2} times 25 times y - Hence total area is: [ text{Area of } triangle ABC = frac{1}{2} times 25 times DE + frac{1}{2} times 25 times DF = frac{25}{2}(DE + DF) ]8. Equating the two expressions for the area of triangle ABC: [ 300 = frac{25}{2}(DE + DF) ] Therefore: [ DE + DF = frac{2 times 300}{25} = frac{600}{25} = 24 ] Conclusion:[boxed{24}]

🧠:Okay, let's see. I need to find the sum of f(1) to f(100) given these functional equations. Hmm, first, let's try to understand what these equations are telling me about the function f(x). The first equation is f(1 + x) = f(3 - x) for any real number x. That seems like some kind of symmetry. If I let y = 1 + x, then x = y - 1, so substituting back in, the equation becomes f(y) = f(3 - (y - 1)) = f(4 - y). So this means that f(y) = f(4 - y) for all real y. So the function is symmetric about the line x = 2. Because reflecting y over 2 gives 4 - y. So, for example, f(1) = f(3), f(0) = f(4), f(2 + t) = f(2 - t), right?Then the second equation is f(2 + x) = -f(1 - x). Let me see if I can relate this to the first equation or find another symmetry. Let's substitute y = 2 + x, so x = y - 2. Then the equation becomes f(y) = -f(1 - (y - 2)) = -f(3 - y). So f(y) = -f(3 - y). That's interesting. So combining this with the first equation, maybe we can find some periodicity or relations between the function values.Wait, from the first equation, we have f(y) = f(4 - y). From the second equation, we have f(y) = -f(3 - y). Let me see if I can combine these two. Let's take f(4 - y) from the first equation and substitute into the second. Since f(y) = f(4 - y), and f(y) = -f(3 - y), so f(4 - y) = -f(3 - y). Let me write that as f(4 - y) = -f(3 - y). Let me replace y with another variable to make it clearer. Let z = 4 - y, so y = 4 - z. Then the equation becomes f(z) = -f(3 - (4 - z)) = -f(z - 1). So f(z) = -f(z - 1). That's a recursive relation. So f(z) = -f(z - 1). Which means f(z) + f(z - 1) = 0. So this tells me that each function value is the negative of the previous one. So if I can find the value at some point, the next one is just the negative. Let's check this. For example, if z = 2, then f(2) = -f(1). Similarly, f(3) = -f(2) = --f(1) = f(1). Then f(4) = -f(3) = -f(1). Wait, but from the first equation, f(4) = f(0). So f(0) = -f(3). But f(3) = f(1) from the first equation. So f(0) = -f(1). Hmm, interesting. Let's see if this holds with the other relations.Alternatively, maybe there's a period here. If f(z) = -f(z - 1), then f(z + 1) = -f(z). So applying this twice: f(z + 2) = -f(z + 1) = --f(z) = f(z). So period 2? So f(z + 2) = f(z). So the function is periodic with period 2. So every 2 steps, the function repeats. If that's the case, then the function has period 2.Wait, let's verify this. If f(z + 2) = f(z), then the function repeats every 2 units. Let's check with the previous relations. From the first equation, f(1 + x) = f(3 - x). If the function has period 2, then f(3 - x) = f(1 - x), which would mean f(1 + x) = f(1 - x), which is a symmetry about x = 1. But earlier we had symmetry about x = 2. Wait, maybe there's a combination here. But if the function is periodic with period 2, then f(4 - y) = f(4 - y - 2*1) = f(2 - y), which would relate to the second equation.Wait, perhaps combining both symmetries. Let's step back. If the function is periodic with period 2, then f(z + 2) = f(z). And from the recursive relation f(z) = -f(z - 1), we can see that f(z + 2) = f(z) as derived before. So the periodicity is 2. Therefore, the function alternates between two values. Let's suppose that f(1) = a, then f(2) = -f(1) = -a, f(3) = -f(2) = a, f(4) = -f(3) = -a, and so on. So the function alternates between a and -a every integer step. But also, from the first equation, f(1) = f(3) = f(5) = ..., and f(2) = f(4) = f(6) = ..., which aligns with the periodicity of 2. So the function alternates between a and -a starting from f(1) = a.But wait, let's check if this is consistent with the other equation. The second equation was f(2 + x) = -f(1 - x). Let's plug x = 0. Then f(2) = -f(1). Which is exactly what our previous conclusion was. If x = 1, then f(3) = -f(0). From our earlier, f(3) = a, and f(0) = -a. So indeed, a = -(-a) = a, which holds. If x = -1, then f(1) = -f(2). Which again is f(1) = -f(2) => a = -(-a) = a, which is consistent. So this seems to check out.Therefore, the function has period 2, so f(n) = a when n is odd, and f(n) = -a when n is even. But we need to check if there are any additional constraints. Let's check with the first equation. For example, take x = 1. Then f(1 + 1) = f(3 - 1) => f(2) = f(2). Which is trivially true. If x = 2, f(3) = f(1). Which is true because f(3) = a and f(1) = a. Similarly, x = 0.5, f(1.5) = f(2.5). Since the function is periodic with period 2, f(1.5) = f(1.5 + 2k) for integer k, but this is for all real numbers. Wait, but if the function is periodic with period 2, then yes, f(1.5) = f(1.5 + 2k). So that works. Similarly, the second equation, f(2 + x) = -f(1 - x). Let's take x = 0.5, so f(2.5) = -f(0.5). But f(2.5) = f(0.5 + 2) = f(0.5) due to periodicity. So f(0.5) = -f(0.5), which implies f(0.5) = 0. Wait, this is important. If the function is periodic with period 2, then for any x, f(x) = -f(x - 1). So f(x) + f(x - 1) = 0. But in the real numbers, this is possible only if for any x, f(x) = -f(x - 1). But in that case, if x is an integer, then f(n) alternates between a and -a. But for non-integer x, this might impose additional constraints.Wait, let's take x = 0.5. Then f(0.5) = -f(-0.5). But since the function is periodic with period 2, f(-0.5) = f(-0.5 + 2) = f(1.5). Then f(0.5) = -f(1.5). But from the first symmetry, f(1.5) = f(4 - 1.5) = f(2.5). But f(2.5) = f(0.5) by periodicity. So f(0.5) = -f(1.5) = -f(2.5) = -f(0.5). Therefore, f(0.5) = -f(0.5), which implies f(0.5) = 0. Similarly, for any x = n + 0.5, where n is an integer, f(n + 0.5) = -f(n - 0.5). But by periodicity, f(n + 0.5) = f(0.5), and f(n - 0.5) = f(-0.5) = f(1.5) = f(0.5). Wait, this seems to lead to f(0.5) = -f(0.5), hence f(0.5) = 0. Therefore, all half-integers must be zero?But wait, if f(x) is periodic with period 2 and f(x) = -f(x - 1), then for any x, f(x) = -f(x - 1). Let's apply this recursively. Then f(x) = -f(x - 1) = --f(x - 2) = f(x - 2). Which gives f(x) = f(x - 2), which is the periodicity. So this is consistent. But also, for x = 0.5, we have f(0.5) = -f(-0.5). But f(-0.5) = f(-0.5 + 2) = f(1.5) = -f(0.5) from the relation f(1.5) = -f(0.5). Therefore, f(0.5) = -(-f(0.5)) = f(0.5), which is a tautology. Wait, that doesn't lead to anything. Hmm, maybe I need to check again.Wait, using f(x) = -f(x - 1), let's set x = 0.5: f(0.5) = -f(-0.5). Then using periodicity, f(-0.5) = f(-0.5 + 2) = f(1.5). Then f(0.5) = -f(1.5). But f(1.5) = -f(0.5) from the same relation: f(1.5) = -f(0.5). Therefore, f(0.5) = -(-f(0.5)) = f(0.5). Which doesn't give new information. So maybe the function can have arbitrary values at certain points? Wait, but this seems like an infinite loop. Unless f(0.5) = 0. Wait, if we go back to the second equation. Let's take x = 0.5. Then f(2 + 0.5) = -f(1 - 0.5) => f(2.5) = -f(0.5). But f(2.5) = f(0.5) due to periodicity. So f(0.5) = -f(0.5) => 2f(0.5) = 0 => f(0.5) = 0. Ah, there we go. So using the second equation directly, we find that f(2.5) = -f(0.5). But since the function is periodic with period 2, f(2.5) = f(0.5). Therefore, combining these gives f(0.5) = -f(0.5) => f(0.5) = 0. Similarly, for any x, if we take x such that 2 + x = y, then 1 - x = 1 - (y - 2) = 3 - y. So f(y) = -f(3 - y). But from the first equation, f(y) = f(4 - y). So substituting f(y) = f(4 - y) into the second equation gives f(4 - y) = -f(3 - y). Then replacing y with 4 - z, gives f(z) = -f(z - 1). So this is consistent with our previous result.Therefore, the function satisfies f(z) = -f(z - 1), which implies f(z) + f(z - 1) = 0. So for integer values, this alternates between a and -a. But for half-integer values, like 0.5, 1.5, etc., the function is zero. For example, f(0.5) = 0, f(1.5) = -f(0.5) = 0, f(2.5) = -f(1.5) = 0, etc. So all half-integers are zero. But since the problem is asking about the sum from f(1) to f(100), which are all integers, the half-integers don't affect this sum. So, considering integer arguments, the function alternates between a and -a. So f(1) = a, f(2) = -a, f(3) = a, f(4) = -a, and so on. Therefore, the sum from f(1) to f(100) is (a - a + a - a + ...) for 100 terms. Let's see. Each pair (f(1) + f(2)) = a - a = 0, similarly (f(3) + f(4)) = 0, etc. Since 100 is even, there are 50 such pairs, each summing to zero. Therefore, the total sum is 0. But wait, let me double-check. If we have 100 terms starting at f(1) up to f(100), that's 100 terms. Since it's an even number, each pair cancels. So yes, the sum is 0. But let me check with smaller numbers. For example, sum from f(1) to f(2): a - a = 0. Sum from f(1) to f(4): a - a + a - a = 0. So it holds. Hence, for 100 terms, it's 50 pairs each summing to zero. Therefore, the answer should be 0. But let's make sure that there isn't any other condition that could affect this. For example, if a was zero, then all function values would be zero. But the problem doesn't state that the function is identically zero. However, even if a is non-zero, the sum still cancels out. Therefore, regardless of the value of a, the sum is zero. Therefore, the final answer should be 0.Final Answerboxed{0}

✅:Given a function ( f(x) ) with the domain ( mathbf{R} ), we know that for any real number ( x ), it holds that ( f(1+x) = f(3-x) ) and ( f(2+x) = -f(1-x) ). We need to find the value of ( f(1) + f(2) + cdots + f(100) ).First, we analyze the given functional equations:1. ( f(1 + x) = f(3 - x) ) implies that ( f(y) = f(4 - y) ) for all ( y ), indicating symmetry about ( x = 2 ).2. ( f(2 + x) = -f(1 - x) ) implies that ( f(y) = -f(3 - y) ) for all ( y ).Combining these two results, we substitute ( f(4 - y) ) from the first equation into the second equation:[ f(4 - y) = -f(3 - y) ]Letting ( z = 4 - y ), we get:[ f(z) = -f(z - 1) ]This recursive relation shows that ( f(z) = -f(z - 1) ), leading to:[ f(z) + f(z - 1) = 0 ]Applying this recursively, we find that the function ( f ) is periodic with period 2:[ f(z + 2) = f(z) ]For integer values, this periodicity implies that ( f(n) ) alternates between two values ( a ) and ( -a ):- ( f(1) = a )- ( f(2) = -a )- ( f(3) = a )- ( f(4) = -a )- and so on.Summing the values from ( f(1) ) to ( f(100) ), we observe that each pair ( (f(1) + f(2), f(3) + f(4), ldots) ) cancels out:[ (a - a) + (a - a) + cdots + (a - a) = 0 ]Since there are 100 terms, forming 50 pairs, each pair sums to zero. Therefore, the total sum is:[ boxed{0} ]

🔑:1. Given Conditions: - Function ( f(x) ) is defined over all real numbers, (mathbb{R}). - Two functional equations: 1. ( f(1 + x) = f(3 - x) quad text{for any real number } x ). 2. ( f(2 + x) = -f(1 - x) quad text{for any real number } x ).2. Investigate the First Equation: - For ( x = 2 ), we get: [ f(1 + 2) = f(3 - 2) implies f(3) = f(1) ] - For ( x = 0 ), we have: [ f(1 + 0) = f(3 - 0) implies f(1) = f(3) ] Since ( f(1 + x) = f(3 - x) ) holds for all ( x ), changing ( x to x - 2 ): [ f(1 + (x - 2)) = f(3 - (x - 2)) implies f(x - 1) = f(5 - x) ] Again, putting ( x= 4 ), we get: [ f(1 + 4) = f(3 - 4) implies f(5) = f(-1) ]3. Investigate the Second Equation: - For ( x = -2 ), we get: [ f(2 - 2) = -f(1 + 2) implies f(0) = -f(3) implies f(0) = -f(1) ] Since ( f(2 + x) = -f(1 - x) ), for ( x = 1 ), we have: [ f(2 + 1) = -f(1 - 1) implies f(3) = -f(0) ] And since we already have ( f(0) = -f(1) ) and ( f(3) = f(1) ): [ f(3) = -(-f(1)) implies f(3) = f(1) ]4. Periodicity Check: - Combining both ( f(3) = f(1) ) and ( f(x + 4) = f(x) ), we get: - For ( x + 4 = x ), substituting ( x + 2 ) in the periodic equation: [ f(2 + x) = -f(1 - x) ] On further substitution: [ x = 4k ] ( T = 4 ) gives periodicity. Observing the periodic pattern: - We have ( f(x + 2) = -f(x) ) - Hence ( f(x+4) = f(x) )5. Summation Across 100 Terms: - Since ( f(x) ) is periodic with ( T = 4 ), we observe: ( f(1) + f(2) + f(3) + f(4) = 0 ) As: [ f(1) + f(3) = 0 quad text{and} quad f(2) + f(4) = 0 implies -f(1) - f(2) implies 0 ] - Since ( f(1) + f(2) + ldots + f(100) ): - Each group of four sums to zero. - There are twenty-five groups. Therefore, ( f(1) + f(2) + cdots + f(100) = 0 ). Conclusion:[boxed{0}]

🧠:Okay, so I need to prove that for a sequence of non-negative numbers (a_1, a_2, ldots) that satisfies the condition (a_{n+m} leq a_n + a_m) for any natural numbers (n) and (m), the inequality (a_n leq m a_1 + left(frac{n}{m} - 1right) a_m) holds for any (n geq m). Hmm, let's break this down step by step.First, the given condition (a_{n+m} leq a_n + a_m) reminds me of subadditive functions or sequences. Subadditive sequences have properties where the value at the sum of two indices is less than or equal to the sum of the values at those indices. So maybe some theorems or properties related to subadditive sequences can be applied here. But I don't recall the exact theorems, so I need to work it out from scratch.The goal is to show that for any (n geq m), (a_n) is bounded above by (m a_1 + left(frac{n}{m} - 1right) a_m). Let's parse this expression. The term (frac{n}{m}) suggests that we might need to divide (n) into chunks of size (m) or something similar. Since (n geq m), maybe we can express (n) as a multiple of (m) plus some remainder, and then apply the given subadditive condition repeatedly.Let me try to write (n) as (k m + r), where (k) is a non-negative integer and (0 leq r < m). However, since (n geq m), (k) must be at least 1 when (r = 0), but if (r > 0), (k) can be 1 or more. Wait, but depending on how (n) relates to (m), this division might vary. Alternatively, maybe I can use induction on (n) or relate (a_n) to (a_m) and (a_1) through recursive applications of the inequality.Alternatively, if we fix an (m), then for any (n geq m), perhaps we can write (n = q m + r) with (q) being the quotient and (r) the remainder, then apply the subadditive property (q) times to break (a_n) into (q) terms of (a_m) and one term of (a_r). Then, since (r < m), (a_r) can be related to (a_1) through (r) applications of the subadditive condition. Let's explore this.Suppose (n = q m + r), where (0 leq r < m). Then, by the subadditive property:(a_n = a_{q m + r} leq a_{q m} + a_r).Then, applying the subadditive property repeatedly on (a_{q m}):(a_{q m} leq a_{(q-1)m} + a_m leq a_{(q-2)m} + 2 a_m leq ldots leq q a_m).So, (a_{q m} leq q a_m). Therefore, combining this with the previous inequality:(a_n leq q a_m + a_r).Now, since (r < m), and all terms are non-negative, maybe we can bound (a_r) in terms of (a_1). Since (a_r leq r a_1), by applying the subadditive condition (r-1) times: (a_r leq a_{r-1 + 1} leq a_{r-1} + a_1), and by induction, (a_r leq r a_1). So, if that's the case, then:(a_n leq q a_m + r a_1).But (q = lfloor frac{n}{m} rfloor - 1) if (r neq 0), or (q = frac{n}{m} - 1) if (r = 0). Wait, let's check. If (n = q m + r), then (q = lfloor frac{n}{m} rfloor) when (0 leq r < m). But actually, if (n = q m + r), then (q = lfloor frac{n}{m} rfloor) when (0 < r < m), and (q = frac{n}{m}) when (r = 0). So, depending on whether (n) is a multiple of (m) or not.But in the inequality we need to prove, the coefficient of (a_m) is (frac{n}{m} - 1), which is not necessarily an integer. Hmm, so perhaps there's a way to express this bound using real coefficients rather than integer ones. Alternatively, maybe we can use some form of division where we allow real numbers? But since (n) and (m) are integers, that might complicate things.Wait, maybe instead of expressing (n) as (q m + r), we can approach this by induction. Let me try induction on (n). Suppose that for some (n geq m), the inequality holds. Then, we need to show that it holds for (n + 1). But induction might be tricky here because the inequality involves both (n) and (m), so maybe we need a different approach.Alternatively, fix (m) and consider varying (n). Let's think of (m) as a fixed integer, and (n) as any integer greater than or equal to (m). The inequality to prove is linear in (n), which suggests that it's related to an upper bound growing linearly with (n), which makes sense for subadditive sequences because of the subadditive property leading to linear growth.Another thought: the given inequality resembles the form of the conclusion from Fekete's lemma, which states that for a subadditive sequence, the limit of (a_n / n) exists and is equal to the infimum of (a_n / n). But I'm not sure if that's directly applicable here since we need a specific inequality, not just the existence of a limit.Wait, maybe using the idea behind Fekete's lemma. If we can express (a_n) in terms of multiples of (m) and then bound the remainder. Let me try again the approach of writing (n = k m + r), where (k) is an integer and (0 leq r < m). Then:(a_n = a_{k m + r} leq a_{k m} + a_r leq k a_m + a_r).Since (r < m), we can express (a_r) as (a_{r}) where (r leq m - 1). Then, by applying the subadditive property repeatedly on (a_r), we get (a_r leq r a_1). Therefore:(a_n leq k a_m + r a_1).Now, note that (k = lfloor frac{n}{m} rfloor), and (r = n - k m). Therefore:(a_n leq lfloor frac{n}{m} rfloor a_m + (n - lfloor frac{n}{m} rfloor m) a_1).But the inequality we need to prove is (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m). Let me compare these two expressions.Our derived upper bound is (lfloor frac{n}{m} rfloor a_m + (n - lfloor frac{n}{m} rfloor m) a_1), while the target inequality is (m a_1 + left( frac{n}{m} - 1 right) a_m). These look different. Let's check if one implies the other or if there's a way to manipulate one into the other.Let me take the target inequality: (m a_1 + left( frac{n}{m} - 1 right) a_m). Let's rewrite that as (frac{n}{m} a_m - a_m + m a_1 = frac{n}{m} a_m + m a_1 - a_m). Hmm.Compare this to our derived expression (lfloor frac{n}{m} rfloor a_m + (n - lfloor frac{n}{m} rfloor m) a_1). Let's denote (k = lfloor frac{n}{m} rfloor), so (n = k m + r), (0 leq r < m). Then our expression is (k a_m + r a_1). The target is (frac{n}{m} a_m - a_m + m a_1 = left( k + frac{r}{m} right) a_m - a_m + m a_1 = (k - 1) a_m + frac{r}{m} a_m + m a_1).So we have two expressions:1. (k a_m + r a_1)2. ((k - 1) a_m + frac{r}{m} a_m + m a_1)We need to show that (k a_m + r a_1 leq (k - 1) a_m + frac{r}{m} a_m + m a_1). Let's subtract the left side from the right side:((k - 1) a_m + frac{r}{m} a_m + m a_1 - k a_m - r a_1 = -a_m + frac{r}{m} a_m + m a_1 - r a_1)Simplify:(-a_m left(1 - frac{r}{m}right) + a_1 (m - r))So the difference is (-a_m left(frac{m - r}{m}right) + a_1 (m - r) = (m - r)left( -frac{a_m}{m} + a_1 right)).Therefore, for our derived expression to be less than or equal to the target inequality, we need:((m - r)left( -frac{a_m}{m} + a_1 right) geq 0)Since (m - r geq 0) (because (r < m)), this inequality holds if (-frac{a_m}{m} + a_1 geq 0), i.e., (a_1 geq frac{a_m}{m}).But is this necessarily true? We don't know. The sequence is non-negative and subadditive, but can we guarantee that (a_1 geq frac{a_m}{m})?Wait, actually, let's check for a simple case. Suppose (m = 2). Then (a_2 leq 2 a_1) by subadditivity. Therefore, (frac{a_2}{2} leq a_1), so (a_1 geq frac{a_m}{m}) when (m = 2). Similarly, for general (m), since (a_m leq m a_1) by applying the subadditive condition (m-1) times:(a_m = a_{1 + (m-1)} leq a_1 + a_{m-1} leq a_1 + a_1 + a_{m-2} leq ldots leq m a_1).Therefore, (a_m leq m a_1), which implies that (a_1 geq frac{a_m}{m}). Therefore, the term (-frac{a_m}{m} + a_1 geq 0). Therefore, the difference ((m - r)left( -frac{a_m}{m} + a_1 right) geq 0), which implies that:(k a_m + r a_1 leq (k - 1) a_m + frac{r}{m} a_m + m a_1).Hence, our derived upper bound (k a_m + r a_1) is less than or equal to the target expression (m a_1 + left( frac{n}{m} - 1 right) a_m). Therefore, combining these results:(a_n leq k a_m + r a_1 leq m a_1 + left( frac{n}{m} - 1 right) a_m).Therefore, the inequality holds. Wait, let me verify this again.Given (k = lfloor frac{n}{m} rfloor), which is the integer part of (frac{n}{m}), and (r = n - k m). Then, (frac{n}{m} = k + frac{r}{m}). So the target expression is (m a_1 + left( k + frac{r}{m} - 1 right) a_m = m a_1 + (k - 1) a_m + frac{r}{m} a_m). As we showed earlier, the difference between this and our derived bound (k a_m + r a_1) is ((m - r)( a_1 - frac{a_m}{m} )), which is non-negative because (a_1 geq frac{a_m}{m}) and (m - r geq 0). Therefore, indeed:(k a_m + r a_1 leq m a_1 + (k - 1) a_m + frac{r}{m} a_m = m a_1 + left( frac{n}{m} - 1 right) a_m).Therefore, the inequality (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m) holds for all (n geq m).Let me check with an example to see if this makes sense. Let's take (m = 2), so the inequality becomes (a_n leq 2 a_1 + left( frac{n}{2} - 1 right) a_2) for (n geq 2). Let's test this with the sequence (a_n = n a_1), which is additive and hence subadditive. Then, (a_2 = 2 a_1). Plugging into the inequality:Right-hand side (RHS) = (2 a_1 + left( frac{n}{2} - 1 right) 2 a_1 = 2 a_1 + (n - 2) a_1 = n a_1), which matches (a_n = n a_1). So equality holds here.Another example: take a subadditive sequence where (a_n) is constant, say (a_n = c) for all (n). Then the subadditive condition (a_{n + m} leq a_n + a_m) becomes (c leq 2 c), which holds since (c geq 0). Then the inequality becomes (c leq m c + left( frac{n}{m} - 1 right) c = c left( m + frac{n}{m} - 1 right)). For (n geq m), let's take (n = m). Then RHS is (c (m + 1 - 1) = c m). So the inequality becomes (c leq c m), which holds since (m geq 1) and (c geq 0). If (n > m), say (n = 2m), then RHS is (c (m + 2 - 1) = c (m + 1)). The inequality is (c leq c (m + 1)), which also holds. So this example works.What if the sequence is not linear? Suppose (a_n = lfloor frac{n}{2} rfloor). Let's check subadditivity. For any (n, m), (a_{n + m} = lfloor frac{n + m}{2} rfloor leq lfloor frac{n}{2} rfloor + lfloor frac{m}{2} rfloor + 1). But actually, is this sequence subadditive? Let's see:(a_{n + m} = lfloor frac{n + m}{2} rfloor), and (a_n + a_m = lfloor frac{n}{2} rfloor + lfloor frac{m}{2} rfloor). The difference is that (lfloor frac{n + m}{2} rfloor leq lfloor frac{n}{2} rfloor + lfloor frac{m}{2} rfloor + 1), which is true because the left side is either equal to the sum or one more. But we need (a_{n + m} leq a_n + a_m). Let's test with (n = 1), (m = 1): (a_2 = 1), (a_1 + a_1 = 0 + 0 = 0). Wait, that's not true. So this sequence isn't subadditive. Let me pick another example.Suppose (a_n) is defined as (a_1 = 1), (a_n = 1) for all (n). Then (a_{n + m} = 1 leq 1 + 1 = 2), so it is subadditive. Then the inequality for (n geq m) is (1 leq m times 1 + (frac{n}{m} - 1) times 1 = m + frac{n}{m} - 1). So (1 leq m + frac{n}{m} - 1). Since (n geq m), (frac{n}{m} geq 1), so (m + frac{n}{m} - 1 geq m + 1 - 1 = m geq 1), which holds. So equality isn't necessarily achieved here, but the inequality holds.Another example: Let (a_1 = 1), (a_2 = 1), and (a_n = a_{n-1} + 1) for (n geq 3). Wait, is this subadditive? Let's check (a_{n + m} leq a_n + a_m). For (n, m geq 3), (a_{n + m} = (n + m - 1)), while (a_n + a_m = (n - 1) + (m - 1) = n + m - 2). So (n + m - 1 leq n + m - 2), which is false. So this isn't subadditive. Hmm, maybe choose a different example.Take the Fibonacci sequence, but with non-negative terms. Wait, Fibonacci is additive, but maybe not the best example. Alternatively, take (a_n = log(n+1)). But log(n+1) is subadditive? For (n, m), (log(n + m + 1) leq log(n + 1) + log(m + 1)). Not sure. Let me check for n = m = 1: (log(3) leq 2 log(2)). Since (log 3 ≈ 1.0986), and (2 log 2 ≈ 1.3863). So yes. For n=2, m=2: (log(5) ≈ 1.6094 leq 2 log(3) ≈ 2.1972). Seems to hold, but not sure if it's always true. Anyway, perhaps this is getting too sidetracked.Back to the proof. The key steps are:1. Express (n) as (k m + r), where (0 leq r < m).2. Use subadditivity to break down (a_n leq k a_m + r a_1).3. Show that this expression is less than or equal to (m a_1 + left( frac{n}{m} - 1 right) a_m) using the fact that (a_m leq m a_1).Therefore, the inequality holds. Another way to think about it is that by choosing to split (n) into (m) and 1s, we can bound the total using a combination of (a_m) terms and (a_1) terms, and the given inequality optimizes this split to give the desired bound.Alternatively, perhaps using mathematical induction on (k) where (n = k m). Wait, but the original problem states (n geq m), so induction over (n) starting from (m). Let's try that.Base case: (n = m). Then the inequality becomes (a_m leq m a_1 + left( frac{m}{m} - 1 right) a_m = m a_1 + 0 = m a_1). But we already know from the subadditive condition that (a_m leq m a_1), as we can split (m) into (1 + 1 + ldots + 1) (m times) and apply the subadditive property m-1 times. So the base case holds.Assume that for some (n geq m), the inequality holds: (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m).Now, consider (n + 1). We need to show that (a_{n+1} leq m a_1 + left( frac{n + 1}{m} - 1 right) a_m).But how can we relate (a_{n + 1}) to (a_n)? Using the subadditive condition: (a_{n + 1} = a_{n + 1} leq a_n + a_1). By the induction hypothesis, (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m). Therefore:(a_{n + 1} leq m a_1 + left( frac{n}{m} - 1 right) a_m + a_1 = (m + 1) a_1 + left( frac{n}{m} - 1 right) a_m).But the desired upper bound is (m a_1 + left( frac{n + 1}{m} - 1 right) a_m = m a_1 + left( frac{n}{m} + frac{1}{m} - 1 right) a_m).Comparing the two:Inductive step gives: ((m + 1) a_1 + left( frac{n}{m} - 1 right) a_m).Desired: (m a_1 + left( frac{n}{m} - 1 + frac{1}{m} right) a_m).So the difference is:((m + 1) a_1 + left( frac{n}{m} - 1 right) a_m - left[ m a_1 + left( frac{n}{m} - 1 + frac{1}{m} right) a_m right] = a_1 - frac{1}{m} a_m).For the induction step to hold, we need (a_1 - frac{1}{m} a_m leq 0), i.e., (a_1 leq frac{a_m}{m}). But earlier, we established that (a_m leq m a_1), which implies (frac{a_m}{m} leq a_1). So (a_1 geq frac{a_m}{m}), which would mean (a_1 - frac{a_m}{m} geq 0). Therefore, the difference is non-negative, meaning that the inductive step gives an upper bound that is larger than the desired one. Therefore, the induction step doesn't directly work because we have:(a_{n + 1} leq text{something} geq text{desired bound}).Therefore, induction might not be the right approach here. This suggests that the initial approach using division into (k m + r) is more promising, as it directly relates to the structure of the problem.Alternatively, think of the inequality (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m) as a linear combination of (a_1) and (a_m). Since (a_n) is built up from smaller terms, using the subadditive property, we can distribute the "weight" of (n) between (m) and 1. The coefficient of (a_m) is approximately (frac{n}{m}), but adjusted by subtracting 1, and the coefficient of (a_1) is (m), which accounts for the remainder when distributing (n) into chunks of (m).Another perspective: Let's consider dividing (n) into (k) chunks of size (m) and a remainder. The total number of chunks would be roughly (frac{n}{m}), but since we can't have a fraction of a chunk, we use (k = lfloor frac{n}{m} rfloor). However, the given inequality uses a real number coefficient (frac{n}{m} - 1), which suggests that instead of taking the floor, we can allow a fractional coefficient but subtract 1 to adjust. Then, by utilizing the bound on (a_r leq r a_1) and the relation (a_m leq m a_1), we can combine these to get the desired linear combination.In summary, the key steps are:1. Decompose (n) into (k m + r), with (0 leq r < m).2. Apply the subadditive property to get (a_n leq k a_m + r a_1).3. Use the fact that (k = frac{n - r}{m}), so (k leq frac{n}{m} - frac{r}{m}). Wait, no. (k = lfloor frac{n}{m} rfloor = frac{n - r}{m}).4. Then substitute (k = frac{n - r}{m}) into the inequality (a_n leq k a_m + r a_1), getting (a_n leq frac{n - r}{m} a_m + r a_1).5. Rearranging, (a_n leq frac{n}{m} a_m - frac{r}{m} a_m + r a_1 = frac{n}{m} a_m + r (a_1 - frac{a_m}{m})).6. Since (a_1 geq frac{a_m}{m}), the term (r (a_1 - frac{a_m}{m})) is non-negative. Therefore, the maximum value of this expression occurs when (r) is as large as possible, which is (r = m - 1).7. But to get an upper bound independent of (r), note that (r (a_1 - frac{a_m}{m}) leq (m - 1)(a_1 - frac{a_m}{m})).8. However, this seems complicated. Instead, since we need to show (a_n leq m a_1 + (frac{n}{m} - 1)a_m), let's consider:From step 4, (a_n leq frac{n - r}{m} a_m + r a_1). Let's rewrite this as:(a_n leq frac{n}{m} a_m - frac{r}{m} a_m + r a_1 = frac{n}{m} a_m + r(a_1 - frac{a_m}{m})).Since (a_1 geq frac{a_m}{m}), (a_1 - frac{a_m}{m} geq 0). Therefore, the term (r(a_1 - frac{a_m}{m}) leq m(a_1 - frac{a_m}{m})), because (r < m). Hence,(a_n leq frac{n}{m} a_m + m(a_1 - frac{a_m}{m}) = frac{n}{m} a_m + m a_1 - a_m = m a_1 + (frac{n}{m} - 1)a_m).Which is exactly the inequality we wanted to prove. Therefore, this completes the proof.So, to recap:By expressing (n) as (k m + r) with (0 leq r < m), using subadditivity to bound (a_n) by (k a_m + r a_1), and then using the relationship (a_m leq m a_1) to further bound the expression, we arrive at the desired inequality.I think this covers all the necessary steps. Let me just write a concise version of the proof to ensure clarity.Step-by-Step Explanation and Proof:Given a non-negative sequence (a_1, a_2, ldots) satisfying (a_{n + m} leq a_n + a_m) for all (m, n in mathbb{N}), we aim to prove that for any (n geq m),[a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m.]Proof:1. Express (n) in terms of (m): For any (n geq m), divide (n) by (m) to get (n = k m + r), where (k) is a non-negative integer and (0 leq r < m). Here, (k = leftlfloor frac{n}{m} rightrfloor) and (r = n - k m).2. Apply subadditivity recursively: Using the subadditive property (a_{n + m} leq a_n + a_m), we decompose (a_n) as follows: [ a_n = a_{k m + r} leq a_{k m} + a_r. ] Applying the subadditive property (k - 1) more times on (a_{k m}): [ a_{k m} leq a_{(k - 1)m} + a_m leq a_{(k - 2)m} + 2 a_m leq ldots leq k a_m. ] Therefore, [ a_n leq k a_m + a_r. ]3. Bound the remainder term (a_r): Since (r < m) and the sequence is non-negative, apply the subadditive property (r) times to (a_r): [ a_r leq r a_1. ] Substituting this into the previous inequality: [ a_n leq k a_m + r a_1. ]4. Relate (k) and (r) to (n): From (n = k m + r), we have (k = frac{n - r}{m}). Substituting (k) into the inequality: [ a_n leq frac{n - r}{m} a_m + r a_1 = frac{n}{m} a_m - frac{r}{m} a_m + r a_1. ] Factor out (r) from the last two terms: [ a_n leq frac{n}{m} a_m + r left( a_1 - frac{a_m}{m} right). ]5. Utilize the bound (a_m leq m a_1): By applying the subadditive property (m - 1) times on (a_m): [ a_m leq m a_1 implies frac{a_m}{m} leq a_1 implies a_1 - frac{a_m}{m} geq 0. ] Since (r geq 0), the term (r left( a_1 - frac{a_m}{m} right)) is non-negative. Therefore, the maximum value of this term occurs when (r) is maximized (i.e., (r = m - 1)): [ r left( a_1 - frac{a_m}{m} right) leq (m - 1) left( a_1 - frac{a_m}{m} right). ] However, to establish an upper bound independent of (r), note that: [ r left( a_1 - frac{a_m}{m} right) leq m left( a_1 - frac{a_m}{m} right), ] since (r < m). Substituting this into the inequality for (a_n): [ a_n leq frac{n}{m} a_m + m left( a_1 - frac{a_m}{m} right) = frac{n}{m} a_m + m a_1 - a_m. ] Simplifying the right-hand side: [ a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m. ]Conclusion:For any non-negative subadditive sequence (a_1, a_2, ldots) and integers (n geq m), the inequality (a_n leq m a_1 + left( frac{n}{m} - 1 right) a_m) holds. This result leverages the decomposition of (n) into multiples of (m) and a remainder, coupled with the subadditive property and the relationship between (a_m) and (a_1).

✅:To prove that for a sequence (a_1, a_2, ldots) of non-negative numbers satisfying (a_{n+m} leq a_n + a_m) for all (m, n in mathbb{N}), the inequality (a_n leq m a_1 + left(frac{n}{m} - 1right) a_m) holds for any (n geq m), follow these steps:1. Express (n) in Terms of (m): Write (n = k m + r) where (k = leftlfloor frac{n}{m} rightrfloor) and (0 leq r < m). Here, (k) is the integer quotient and (r) is the remainder.2. Apply Subadditivity Recursively: Using the subadditive property, decompose (a_n): [ a_n leq a_{k m} + a_r leq k a_m + a_r. ] Here, (a_{k m} leq k a_m) by applying the subadditive condition (k-1) times.3. Bound the Remainder Term (a_r): Since (r < m), apply the subadditive property (r) times to get: [ a_r leq r a_1. ]4. Combine the Results: Substitute (k = frac{n - r}{m}) into the inequality: [ a_n leq frac{n - r}{m} a_m + r a_1 = frac{n}{m} a_m - frac{r}{m} a_m + r a_1. ] Factor out (r): [ a_n leq frac{n}{m} a_m + rleft(a_1 - frac{a_m}{m}right). ]5. Utilize (a_m leq m a_1): Since (a_m leq m a_1), we have (a_1 - frac{a_m}{m} geq 0). The term (rleft(a_1 - frac{a_m}{m}right)) is maximized when (r = m - 1), leading to: [ a_n leq frac{n}{m} a_m + mleft(a_1 - frac{a_m}{m}right) = m a_1 + left(frac{n}{m} - 1right)a_m. ]Final Conclusion:For any (n geq m),[a_n leq m a_1 + left(frac{n}{m} - 1right)a_m.]This inequality follows from decomposing (n) into multiples of (m) and applying the subadditive property, leveraging the relationship between (a_m) and (a_1).(boxed{a_n leq m a_1 + left(frac{n}{m} - 1right) a_m})

🔑:1. Let ( n = mq + r ), where ( q in mathbb{N} ) and ( 0 leq r < m ). 2. Using the given condition ( a_{n+m} leq a_{n} + a_{m} ), we get: [ a_{n} = a_{mq+r} leq a_{mq} + a_{r} ]3. By applying the given inequality recursively on ( a_{mq} ): [ a_{mq} leq q a_{m} ] Combining these results, we obtain: [ a_{n} leq q a_{m} + a_{r} ]4. Express ( q ) and ( r ) in terms of ( n ) and ( m ): [ q = leftlfloor frac{n}{m} rightrfloor quad text{and} quad r = n - mq ] Since ( 0 leq r < m ), it follows that: [ n - mq = r implies q = frac{n-r}{m} ] 5. Substitute ( q = frac{n-r}{m}) back into the inequality: [ a_{n} leq left( frac{n-r}{m} right) a_{m} + a_{r} ]6. Now we rewrite ( a_{r} ) in terms of ( a_{1} ) since ( 0 leq r < m ): [ a_{r} leq r a_{1} ]7. Substituting the upper bound for ( a_{r} ): [ a_{n} leq left( frac{n - r}{m} right) a_{m} + r a_{1} ]8. Simplifying the expression further: [ a_{n} leq left( frac{n}{m} - frac{r}{m} right) a_{m} + r a_{1} ]9. Notice that ( 0 leq frac{r}{m} < 1 ), thus: [ a_{n} leq left( frac{n}{m} - 1 right) a_{m} + a_{m} + r a_{1} ]10. As ( r < m ), we obtain: [ r a_{1} leq m a_{1} ]11. Hence: [ a_{n} leq left( frac{n}{m} - 1 right) a_{m} + a_{m} + m a_{1} ]12. We combine like terms to get the final form: [ a_{n} leq m a_{1} + left( frac{n}{m} - 1 right) a_{m} ] Conclusion:[boxed{a_n leq m a_1 + left( frac{n}{m} - 1 right) a_{m}}]