💡Alright, so I've got this problem about polygonal distance, which is defined as the sum of the absolute differences in the x-coordinates and y-coordinates of two points. It's like measuring distance on a grid where you can only move horizontally or vertically, not diagonally. Cool, I think I get that.Part (1) is straightforward: Given points A(1,0) and B(2,3), find the polygonal distance ρ(A,B). So, I just need to plug into the formula. Let me write that down:ρ(A,B) = |x2 - x1| + |y2 - y1|Plugging in the values:ρ(A,B) = |2 - 1| + |3 - 0| = |1| + |3| = 1 + 3 = 4Okay, that seems simple enough. So, ρ(A,B) is 4. I think that's it for part (1). Moving on to part (2).Part (2) says: Given point A(1,0), point B is a moving point on the line l: x - √2 y + 2 = 0. Find the minimum value of ρ(A,B).Hmm, so I need to find the point B on the line l such that the polygonal distance from A to B is minimized. Polygonal distance is the Manhattan distance, right? So, it's like the shortest path on a grid. But here, B is constrained to lie on a specific line.I remember that for Manhattan distance, the closest point on a line might not be the same as the closest point in Euclidean distance. So, I need to think about how to minimize |x - 1| + |y - 0|, given that x - √2 y + 2 = 0.Let me express x in terms of y from the line equation:x = √2 y - 2So, substituting into the distance formula:ρ(A,B) = |√2 y - 2 - 1| + |y - 0| = |√2 y - 3| + |y|So, now I have ρ(A,B) as a function of y: f(y) = |√2 y - 3| + |y|I need to find the y that minimizes this function. Since it's a piecewise linear function, the minimum will occur at a point where the slope changes, which is at the points where the expressions inside the absolute values change sign.So, let's find the critical points:1. √2 y - 3 = 0 => y = 3 / √2 ≈ 2.1212. y = 0So, the critical points are at y = 0 and y = 3 / √2.Now, let's analyze the function f(y) in different intervals:1. For y < 0: f(y) = -(√2 y - 3) - y = -√2 y + 3 - y = (-√2 - 1)y + 3 This is a linear function with a negative slope, so it's decreasing as y increases.2. For 0 ≤ y < 3 / √2: f(y) = -(√2 y - 3) + y = -√2 y + 3 + y = (-√2 + 1)y + 3 The slope here is (-√2 + 1). Since √2 ≈ 1.414, so (-1.414 + 1) ≈ -0.414, which is negative. So, the function is still decreasing in this interval.3. For y ≥ 3 / √2: f(y) = (√2 y - 3) + y = √2 y - 3 + y = (√2 + 1)y - 3 The slope here is (√2 + 1) ≈ 2.414, which is positive. So, the function starts increasing after y = 3 / √2.Therefore, the function f(y) decreases until y = 3 / √2 and then increases. So, the minimum occurs at y = 3 / √2.Wait, but hold on. If the function is decreasing up to y = 3 / √2 and then increasing, the minimum should be at y = 3 / √2. But let me double-check.Wait, actually, for y < 0, the function is decreasing as y increases, so the minimum in that interval would be at y approaching 0 from the left. But since y can't be less than 0 in this context (because the line x - √2 y + 2 = 0 can have negative y values, but we need to see if y can be negative.Wait, actually, y can be any real number because the line extends infinitely in both directions. So, for y < 0, the function is decreasing as y increases, but as y approaches negative infinity, f(y) tends to positive infinity because of the absolute values. So, the function must have a minimum somewhere.But according to the analysis, the function decreases until y = 3 / √2, which is approximately 2.121, and then increases. So, the minimum is at y = 3 / √2.But wait, let me think again. If I consider y approaching negative infinity, f(y) = |√2 y - 3| + |y| ≈ |√2 y| + |y| = (√2 + 1)|y|, which goes to infinity. Similarly, as y approaches positive infinity, f(y) ≈ (√2 + 1)y, which also goes to infinity. So, the function must have a minimum somewhere in between.But according to the critical points, the function changes slope at y = 0 and y = 3 / √2. So, the function is decreasing on (-∞, 3 / √2) and increasing on (3 / √2, ∞). Wait, no, that can't be because at y = 0, the slope changes from (-√2 - 1) to (-√2 + 1). Both slopes are negative, so the function is decreasing throughout y < 3 / √2.Wait, that doesn't make sense because if the function is decreasing up to y = 3 / √2 and then increasing, then the minimum is at y = 3 / √2.But let me plot the function or think of specific values.At y = 0: f(y) = |0 - 3| + |0| = 3 + 0 = 3At y = 3 / √2 ≈ 2.121: f(y) = |√2*(3 / √2) - 3| + |3 / √2| = |3 - 3| + 3 / √2 = 0 + 3 / √2 ≈ 2.121At y = 1: f(y) = |√2*1 - 3| + |1| ≈ |1.414 - 3| + 1 ≈ 1.586 + 1 ≈ 2.586At y = 2: f(y) = |√2*2 - 3| + |2| ≈ |2.828 - 3| + 2 ≈ 0.172 + 2 ≈ 2.172At y = 3 / √2 ≈ 2.121: f(y) ≈ 2.121At y = 3: f(y) = |√2*3 - 3| + |3| ≈ |4.242 - 3| + 3 ≈ 1.242 + 3 ≈ 4.242So, from these values, it seems that the function decreases until y ≈ 2.121 and then increases. So, the minimum is indeed at y = 3 / √2.Therefore, the minimum ρ(A,B) is 3 / √2. But wait, let me calculate it properly.At y = 3 / √2:x = √2 y - 2 = √2*(3 / √2) - 2 = 3 - 2 = 1So, point B is (1, 3 / √2). Then, ρ(A,B) = |1 - 1| + |3 / √2 - 0| = 0 + 3 / √2 = 3 / √2 ≈ 2.121But wait, earlier when I plugged in y = 3 / √2, I thought f(y) was 3 / √2, but actually, it's |√2 y - 3| + |y| = |3 - 3| + 3 / √2 = 0 + 3 / √2 = 3 / √2.But wait, is that correct? Because when y = 3 / √2, x = 1, so point B is (1, 3 / √2). Then, ρ(A,B) is |1 - 1| + |3 / √2 - 0| = 0 + 3 / √2.But wait, is there a point B on the line l that gives a smaller ρ(A,B)? Because when I took y = 0, I got ρ(A,B) = 3, which is larger than 3 / √2 ≈ 2.121.Wait, but is 3 / √2 the minimum? Let me see.Alternatively, maybe I should consider reflecting point A across the line l and then find the intersection point.Wait, in Manhattan distance, the reflection method might not work as it does in Euclidean distance. Hmm.Alternatively, maybe I can parameterize the line l and express ρ(A,B) in terms of a parameter, then find its minimum.Let me try that.The line l is x - √2 y + 2 = 0, which can be written as x = √2 y - 2.So, any point B on l can be written as (√2 t - 2, t), where t is a real number.Then, ρ(A,B) = |√2 t - 2 - 1| + |t - 0| = |√2 t - 3| + |t|So, f(t) = |√2 t - 3| + |t|Now, to minimize f(t), we can consider different cases based on the value of t.Case 1: t < 0In this case, |t| = -t, and |√2 t - 3| = -(√2 t - 3) = -√2 t + 3So, f(t) = (-√2 t + 3) + (-t) = (-√2 - 1)t + 3This is a linear function with a negative slope since (-√2 - 1) is negative. So, as t increases, f(t) decreases. Therefore, the minimum in this interval would be at t approaching 0 from the left.Case 2: 0 ≤ t < 3 / √2Here, |t| = t, and |√2 t - 3| = -(√2 t - 3) = -√2 t + 3So, f(t) = (-√2 t + 3) + t = (-√2 + 1)t + 3The slope here is (-√2 + 1) ≈ (-1.414 + 1) ≈ -0.414, which is still negative. So, f(t) is decreasing in this interval as well.Case 3: t ≥ 3 / √2Here, |t| = t, and |√2 t - 3| = √2 t - 3So, f(t) = (√2 t - 3) + t = (√2 + 1)t - 3The slope here is (√2 + 1) ≈ 2.414, which is positive. So, f(t) is increasing in this interval.Therefore, the function f(t) decreases until t = 3 / √2 and then increases. Hence, the minimum occurs at t = 3 / √2.So, the minimum ρ(A,B) is f(3 / √2) = |√2*(3 / √2) - 3| + |3 / √2| = |3 - 3| + 3 / √2 = 0 + 3 / √2 = 3 / √2But 3 / √2 can be rationalized as (3√2)/2 ≈ 2.121So, the minimum ρ(A,B) is (3√2)/2.Wait, but earlier when I plugged in y = 3 / √2, I got point B as (1, 3 / √2). Let me verify the coordinates.Given x = √2 y - 2, so if y = 3 / √2, then x = √2*(3 / √2) - 2 = 3 - 2 = 1. So, yes, point B is (1, 3 / √2).Then, ρ(A,B) = |1 - 1| + |3 / √2 - 0| = 0 + 3 / √2 = 3 / √2 = (3√2)/2.Okay, that makes sense.So, the minimum ρ(A,B) is (3√2)/2.Moving on to part (3):For two different points A(x1, y1) and B(x2, y2) on the plane, does there exist a point C(x,y) that satisfies:① ρ(A,C) + ρ(C,B) = ρ(A,B);② ρ(A,C) = ρ(C,B).If it exists, find all points that satisfy the conditions; if not, provide a proof.Hmm, so we need to find a point C such that the sum of the polygonal distances from A to C and C to B equals the polygonal distance from A to B, and also that the polygonal distances from A to C and C to B are equal.First, let's recall that in polygonal distance, the triangle inequality holds, meaning ρ(A,C) + ρ(C,B) ≥ ρ(A,B). So, equality holds if and only if C lies on the "polygonal path" from A to B, meaning that C is on the line segment connecting A and B in the Manhattan metric.But in this case, we also have the condition that ρ(A,C) = ρ(C,B). So, C must be the midpoint of the polygonal path from A to B.But wait, in Manhattan distance, the midpoint isn't necessarily the same as the Euclidean midpoint. Hmm.Wait, let's think about it. For two points A and B, the set of points C such that ρ(A,C) + ρ(C,B) = ρ(A,B) is the set of points lying on the Manhattan path from A to B. This path consists of moving horizontally and then vertically, or vertically and then horizontally, from A to B.So, for example, if A is (x1, y1) and B is (x2, y2), the Manhattan path from A to B can be either:1. Move horizontally from A to (x2, y1), then vertically to B.2. Move vertically from A to (x1, y2), then horizontally to B.So, the set of points C that satisfy ρ(A,C) + ρ(C,B) = ρ(A,B) are all the points on these two paths.Now, among these points, we need to find those where ρ(A,C) = ρ(C,B).So, essentially, we're looking for points C on the Manhattan path from A to B such that C is equidistant from A and B in terms of polygonal distance.Let me denote S = ρ(A,B). Then, we need ρ(A,C) = ρ(C,B) = S/2.So, does such a point C exist?Well, in the Manhattan path, the total distance is S = |x2 - x1| + |y2 - y1|.If S is even, then S/2 is an integer, and perhaps such a point exists. But if S is odd, then S/2 is not an integer, but in the plane, coordinates can be real numbers, so maybe it still exists.Wait, but in the Manhattan path, the movement is along the grid, so the coordinates of C must lie on the grid lines between A and B.Wait, but in the plane, coordinates can be any real numbers, so even if S is odd, we can still have a point C such that ρ(A,C) = ρ(C,B) = S/2.But let's think about it more carefully.Suppose we have two points A(x1, y1) and B(x2, y2). Let's assume without loss of generality that x2 ≥ x1 and y2 ≥ y1. So, the Manhattan path from A to B is moving right to (x2, y1), then up to B.Now, let's parameterize the path. Let me define a parameter t, where t ranges from 0 to 1.From A to (x2, y1), t goes from 0 to t1, and from (x2, y1) to B, t goes from t1 to 1.Wait, actually, it's better to think in terms of the horizontal and vertical distances.Let me denote dx = x2 - x1 and dy = y2 - y1.So, the total Manhattan distance S = dx + dy.We need to find a point C such that ρ(A,C) = ρ(C,B) = S/2.So, let's consider moving from A towards B along the Manhattan path.Case 1: Moving horizontally first.Suppose we move a distance t along the x-axis from A, reaching point C1(t) = (x1 + t, y1), where t ranges from 0 to dx.Then, the remaining distance to B is S - t = (dx + dy) - t.But we need ρ(A,C1) = t, and ρ(C1,B) = (dx - t) + dy.We want t = (dx - t) + dy.So, t = dx - t + dy => 2t = dx + dy => t = (dx + dy)/2.But t must be ≤ dx, because we're only moving horizontally.So, if (dx + dy)/2 ≤ dx, then t = (dx + dy)/2 is valid.Otherwise, we have to move horizontally to (x2, y1), then move vertically.Similarly, if (dx + dy)/2 > dx, then we move horizontally to (x2, y1), then move vertically a distance t' = (dx + dy)/2 - dx = (dy - dx)/2.But since dy = y2 - y1 ≥ 0, if dy ≥ dx, then t' is non-negative.So, in this case, point C would be (x2, y1 + t') = (x2, y1 + (dy - dx)/2).Similarly, if we first move vertically, then horizontally.So, in general, such a point C exists if and only if the total distance S is even? Wait, no, because even if S is odd, we can still have a point C with real coordinates.Wait, no, the coordinates can be real numbers, so even if S is odd, S/2 is just a real number, so we can still have such a point C.But wait, in the Manhattan path, the movement is along the grid, so the coordinates of C must lie on the grid lines. But in the plane, coordinates can be any real numbers, so even if S is odd, we can still have a point C with real coordinates such that ρ(A,C) = ρ(C,B) = S/2.Wait, but in the Manhattan path, the movement is along the grid, so the coordinates of C must lie on the grid lines. But in the plane, coordinates can be any real numbers, so even if S is odd, we can still have a point C with real coordinates such that ρ(A,C) = ρ(C,B) = S/2.Wait, maybe I'm overcomplicating.Let me think of it algebraically.We need to find C(x,y) such that:1. |x - x1| + |y - y1| + |x2 - x| + |y2 - y| = |x2 - x1| + |y2 - y1|2. |x - x1| + |y - y1| = |x2 - x| + |y2 - y|From condition 1, we have the triangle inequality holding with equality, which implies that C lies on the Manhattan path from A to B.From condition 2, we have that the distance from A to C equals the distance from C to B.So, combining these, C must lie on the Manhattan path and be equidistant from A and B.In the Manhattan path, the total distance is S = |x2 - x1| + |y2 - y1|.So, C must be at a distance of S/2 from both A and B along the path.Therefore, such a point C exists if and only if S is even? Wait, no, because S can be any positive real number, and S/2 is just half of that.But in the Manhattan path, the coordinates of C must satisfy certain conditions.Wait, let's consider that in the Manhattan path, moving from A to B, we can either move horizontally first or vertically first.Suppose we move horizontally first:C = (x1 + t, y1), where t is between 0 and dx.Then, ρ(A,C) = t.ρ(C,B) = (dx - t) + dy.We need t = (dx - t) + dy => 2t = dx + dy => t = (dx + dy)/2.But t must be ≤ dx, so (dx + dy)/2 ≤ dx => dy ≤ dx.Similarly, if dy > dx, then t = (dx + dy)/2 > dx, which is not possible in this case. So, in that case, we have to move horizontally to (x2, y1), then move vertically.So, after moving horizontally to (x2, y1), the remaining distance to B is dy.So, the total distance is S = dx + dy.We need to find a point C such that ρ(A,C) = ρ(C,B) = S/2.If S is even, then S/2 is an integer, but in the plane, coordinates can be real numbers, so it's not necessary for S to be even.Wait, no, S is just a real number, so S/2 is also a real number.So, regardless of whether S is even or odd, we can find such a point C.Therefore, such a point C exists.But wait, let me think about it more carefully.Suppose A is (0,0) and B is (1,1). Then, S = 1 + 1 = 2.So, S/2 = 1.Now, the Manhattan path from A to B can be either right then up, or up then right.If we move right to (1,0), then up to (1,1). So, the point C must be at a distance of 1 from A and 1 from B.So, moving right from A, after distance 1, we reach (1,0). Then, moving up from (1,0) to (1,1), which is another distance of 1.So, the point C is (1,0), which is at distance 1 from A and 1 from B.Similarly, if we move up first to (0,1), then right to (1,1), the point C would be (0,1).Wait, but in this case, there are two points C: (1,0) and (0,1).But wait, ρ(A,C) = 1 and ρ(C,B) = 1 for both points.So, in this case, there are two points C that satisfy the conditions.Similarly, if A is (0,0) and B is (2,2), then S = 4, S/2 = 2.Moving right to (2,0), then up to (2,2). The point C would be (2,0), which is at distance 2 from A and 2 from B.Alternatively, moving up to (0,2), then right to (2,2). The point C would be (0,2).But wait, in this case, there are two points C: (2,0) and (0,2).But wait, what if A is (0,0) and B is (3,1). Then, S = 3 + 1 = 4.So, S/2 = 2.Now, moving right to (3,0), then up to (3,1). The total distance is 4.We need to find a point C such that ρ(A,C) = 2 and ρ(C,B) = 2.So, moving right from A, after distance 2, we reach (2,0). Then, the remaining distance to B is 2: from (2,0) to (3,0) is 1, and from (3,0) to (3,1) is 1, so total 2.So, point C is (2,0).Alternatively, moving up first: from A(0,0) to (0,1), then right to (3,1). The total distance is 4.We need to find a point C such that ρ(A,C) = 2 and ρ(C,B) = 2.Moving up from A, after distance 2, we reach (0,2). But wait, B is at (3,1), so moving up to (0,2) is beyond B's y-coordinate. So, that's not possible.Wait, no, because the Manhattan path is either right then up or up then right.If we move up first, we go from (0,0) to (0,1), then right to (3,1). So, the total distance is 1 (up) + 3 (right) = 4.So, to find C such that ρ(A,C) = 2 and ρ(C,B) = 2.Moving up from A, after distance 1, we reach (0,1). Then, moving right from (0,1), we need to cover 2 units to reach C.So, from (0,1), moving right 2 units, we reach (2,1). Then, from (2,1) to B(3,1) is 1 unit.So, ρ(A,C) = 1 (up) + 2 (right) = 3, which is more than 2.Wait, that doesn't work.Alternatively, moving right first: from A(0,0) to (2,0), then up to (2,1). Then, ρ(A,C) = 2 (right) + 1 (up) = 3, which is more than 2.Wait, this is confusing.Wait, maybe I need to think differently.We need to find a point C such that ρ(A,C) = ρ(C,B) = 2.So, starting from A(0,0), moving towards B(3,1), we can move right and up in such a way that the total distance from A to C is 2, and from C to B is also 2.So, let's denote C as (x,y).Then, |x - 0| + |y - 0| = 2, and |3 - x| + |1 - y| = 2.So, we have:x + y = 2,and(3 - x) + (1 - y) = 2 => 4 - x - y = 2 => x + y = 2.So, both equations reduce to x + y = 2.Therefore, any point C on the line x + y = 2 that lies on the Manhattan path from A to B.But the Manhattan path from A(0,0) to B(3,1) can be either right then up or up then right.If we move right then up:From (0,0) to (3,0), then up to (3,1).The line x + y = 2 intersects this path at (2,0) and (3, -1), but (3, -1) is not on the path.So, the intersection is at (2,0).Similarly, if we move up then right:From (0,0) to (0,1), then right to (3,1).The line x + y = 2 intersects this path at (1,1) and (0,2), but (0,2) is not on the path.So, the intersection is at (1,1).But wait, (1,1) is on the path from (0,1) to (3,1), but moving right from (0,1), we reach (1,1) after moving 1 unit right.So, ρ(A,C) = 1 (up) + 1 (right) = 2,and ρ(C,B) = 2 (right) + 0 (up) = 2.So, point C is (1,1).Similarly, point (2,0) is on the path from (0,0) to (3,0), then up to (3,1).So, ρ(A,C) = 2 (right),and ρ(C,B) = 1 (right) + 1 (up) = 2.So, both points (1,1) and (2,0) satisfy the conditions.Therefore, in this case, there are two points C that satisfy the conditions.Wait, but in the previous example where A was (0,0) and B was (1,1), we had two points C: (1,0) and (0,1). Similarly, here, we have two points.So, in general, for two points A and B, there can be two points C that satisfy the conditions: one on the horizontal segment and one on the vertical segment of the Manhattan path.But wait, in the case where A and B are aligned either horizontally or vertically, what happens?Suppose A is (0,0) and B is (4,0). So, S = 4.We need to find C such that ρ(A,C) = ρ(C,B) = 2.So, moving along the x-axis, C would be at (2,0).Similarly, if A is (0,0) and B is (0,4), then C would be at (0,2).So, in these cases, there's only one point C.Therefore, in general, if A and B are aligned either horizontally or vertically, there's only one point C that satisfies the conditions. Otherwise, there are two points.So, to answer part (3):Yes, such a point C exists. If A and B are aligned either horizontally or vertically, there is exactly one such point C, which is the midpoint along the line connecting A and B. If A and B are not aligned either horizontally or vertically, there are two such points C, one on the horizontal segment and one on the vertical segment of the Manhattan path from A to B.Therefore, the points C that satisfy the conditions are the midpoints along the Manhattan path from A to B.But wait, in the case where A and B are not aligned, we have two midpoints: one on the horizontal path and one on the vertical path.So, in general, there are either one or two points C depending on the alignment of A and B.But the question says "for two different points A and B on the plane", so it's general.Therefore, the answer is that such points C exist, and they are the midpoints along the Manhattan path from A to B. Depending on the orientation of A and B, there may be one or two such points.But to be precise, in the case where A and B are not aligned horizontally or vertically, there are two points C: one on the horizontal segment and one on the vertical segment of the Manhattan path. If A and B are aligned either horizontally or vertically, there is exactly one point C, which is the midpoint along that line.So, to summarize:Yes, such points C exist. They are the midpoints along the Manhattan path from A to B. If A and B are aligned either horizontally or vertically, there is exactly one such point C. If A and B are not aligned, there are two such points C.Therefore, the answer is that such points C exist, and they are the midpoints along the Manhattan path from A to B. Depending on the alignment of A and B, there may be one or two such points.But wait, in the case where A and B are not aligned, we have two points C: one on the horizontal path and one on the vertical path. So, the set of points C is the union of these midpoints.Therefore, the final answer is that such points C exist, and they are the midpoints along the Manhattan path from A to B. If A and B are aligned either horizontally or vertically, there is exactly one such point C. If A and B are not aligned, there are two such points C.But the question asks: "does there exist a point C(x,y) that satisfies... If it exists, find all points that satisfy the conditions; if it does not exist, please provide a proof."So, the answer is yes, such points exist, and they are the midpoints along the Manhattan path from A to B. Depending on the alignment of A and B, there may be one or two such points.Therefore, the final answer is:Yes, such points C exist. They are the midpoints along the Manhattan path from A to B. If A and B are aligned either horizontally or vertically, there is exactly one such point C. If A and B are not aligned, there are two such points C.But to express this more formally, we can say that C is the midpoint of the line segment AB in the Manhattan metric, which can be either the horizontal midpoint or the vertical midpoint, depending on the path taken.Therefore, the points C are given by:If moving horizontally first:C = (x1 + (dx + dy)/2, y1) if (dx + dy)/2 ≤ dx,otherwise, C = (x2, y1 + (dx + dy)/2 - dx).Similarly, if moving vertically first:C = (x1, y1 + (dx + dy)/2) if (dx + dy)/2 ≤ dy,otherwise, C = (x1 + (dx + dy)/2 - dy, y2).But this might be more detailed than necessary.Alternatively, we can express C as:C = (x1 + t, y1) where t = (dx + dy)/2, if t ≤ dx,or C = (x2, y1 + (dx + dy)/2 - dx).Similarly, C = (x1, y1 + t) where t = (dx + dy)/2, if t ≤ dy,or C = (x1 + (dx + dy)/2 - dy, y2).But perhaps a better way is to express C in terms of the coordinates.Let me define:Let dx = |x2 - x1|,dy = |y2 - y1|.Then, the total distance S = dx + dy.We need to find C such that ρ(A,C) = ρ(C,B) = S/2.Case 1: Moving horizontally first.If S/2 ≤ dx,then C = (x1 + (S/2), y1).Otherwise,C = (x2, y1 + (S/2 - dx)).Case 2: Moving vertically first.If S/2 ≤ dy,then C = (x1, y1 + (S/2)).Otherwise,C = (x1 + (S/2 - dy), y2).Therefore, depending on the relative sizes of dx and dy, we have different expressions for C.But in general, there are two possible points C, one on each path (horizontal then vertical, vertical then horizontal), unless one of the paths requires moving beyond the other coordinate, in which case only one point exists.Wait, no, actually, regardless of the relative sizes, both points exist, but they might coincide if A and B are aligned.Wait, no, if A and B are aligned horizontally or vertically, then both points coincide at the midpoint.Otherwise, they are two distinct points.Therefore, in general, there are two points C unless A and B are aligned, in which case there is only one.Therefore, the answer is:Yes, such points C exist. They are the midpoints along the Manhattan path from A to B. Specifically, if A and B are not aligned either horizontally or vertically, there are two such points C: one on the horizontal segment and one on the vertical segment of the Manhattan path. If A and B are aligned either horizontally or vertically, there is exactly one such point C, which is the midpoint along that line.Therefore, the points C are given by:C1 = (x1 + (dx + dy)/2, y1) if (dx + dy)/2 ≤ dx,otherwise, C1 = (x2, y1 + (dx + dy)/2 - dx).C2 = (x1, y1 + (dx + dy)/2) if (dx + dy)/2 ≤ dy,otherwise, C2 = (x1 + (dx + dy)/2 - dy, y2).But since dx and dy are non-negative, and S = dx + dy,if (dx + dy)/2 ≤ dx => dy ≤ dx,and if (dx + dy)/2 ≤ dy => dx ≤ dy.Therefore, if dx ≥ dy, then C1 is (x1 + (dx + dy)/2, y1),and C2 is (x1, y1 + (dx + dy)/2 - dx) = (x1, y1 + (dy - dx)/2).But wait, if dx ≥ dy, then (dy - dx)/2 is negative, which would place C2 below A, which is not on the Manhattan path.Wait, no, because in the vertical path, we move up first, then right.So, if dx ≥ dy, then moving vertically first, after moving dy units up, we still need to move (dx - dy) units right.But we need to find C such that ρ(A,C) = S/2.So, if we move vertically first, after moving t units up, we have ρ(A,C) = t,and then we need to move (S/2 - t) units right.But since we can't move beyond B, we have to ensure that t ≤ dy and (S/2 - t) ≤ dx.Wait, this is getting too convoluted.Perhaps a better approach is to consider that for any two points A and B, there are two points C that satisfy the conditions: one on the horizontal path and one on the vertical path, unless A and B are aligned, in which case they coincide.Therefore, the answer is:Yes, such points C exist. They are the midpoints along the Manhattan path from A to B. If A and B are aligned either horizontally or vertically, there is exactly one such point C. If A and B are not aligned, there are two such points C.Therefore, the final answer is:Yes, such points C exist. They are the midpoints along the Manhattan path from A to B. Specifically, if A and B are not aligned either horizontally or vertically, there are two such points C: one on the horizontal segment and one on the vertical segment of the Manhattan path. If A and B are aligned either horizontally or vertically, there is exactly one such point C, which is the midpoint along that line.So, in conclusion, the answer is that such points C exist, and they are the midpoints along the Manhattan path from A to B. Depending on the alignment of A and B, there may be one or two such points.

💡To determine the largest prime factor of 5292, I will start by dividing the number by the smallest prime, which is 2.5292 divided by 2 equals 2646. Since 2646 is still even, I will divide by 2 again, resulting in 1323.Next, I will check for divisibility by 3. The sum of the digits of 1323 is 9, which is divisible by 3. Dividing 1323 by 3 gives 441.I will continue factoring 441 by dividing it by 3 again, obtaining 147. Repeating this step, I divide 147 by 3 to get 49.Finally, I recognize that 49 is 7 squared. Therefore, the prime factors of 5292 are 2, 3, and 7, with 7 being the largest prime factor.

💡First, I need to simplify the given expression:(√1.21) / (√0.81) + (√0.81) / (√x)I know that √1.21 is 1.1 and √0.81 is 0.9. Substituting these values into the expression gives:1.1 / 0.9 + 0.9 / √xSimplifying the first fraction, 1.1 divided by 0.9 equals approximately 1.222222222222222.Now, the expression is:1.222222222222222 + 0.9 / √xGiven that the value of this expression is 2.507936507936508, I set up the equation:1.222222222222222 + 0.9 / √x = 2.507936507936508Subtracting 1.222222222222222 from both sides:0.9 / √x = 1.285714285714286To solve for √x, I divide 0.9 by 1.285714285714286:√x ≈ 0.7Finally, squaring both sides to find x:x = (0.7)^2 = 0.49Therefore, the value of x is 0.49.

💡First, I need to determine the discount pattern based on the number of pizzas purchased. The given information shows that for 1 pizza, there is 0% discount; for 2 pizzas, there is a 4% discount; and for 3 pizzas, there is an 8% discount.I observe that the discount increases by 4% for each additional pizza. This means that the multiple by which the discount increases for every additional pizza is 4%.To confirm this, I can calculate the total cost for 3 pizzas with the given discount. Each pizza costs 10, so without any discount, 3 pizzas would cost 3 * 10 = 30. With an 8% discount, the total cost becomes 30 - (8% of 30) = 30 - 2.40 = 27.60.Since the man paid 37, which is more than 27.60, there might be additional costs or taxes involved. However, based on the discount pattern provided, the multiple by which the discount increases for each additional pizza is clearly 4%.

💡First, I recognize that the initial ratio of boarders to day scholars is 7 to 16. This means that for every 7 boarders, there are 16 day scholars.Let’s denote the initial number of boarders as (7x) and the initial number of day scholars as (16x), where (x) is a common factor.After 80 new boarders join the school, the total number of boarders becomes (7x + 80). The problem states that the new ratio of boarders to day scholars is 1 to 2. This means that for every boarder, there are two day scholars.I can set up the equation based on the new ratio:[frac{7x + 80}{16x} = frac{1}{2}]To solve for (x), I'll cross-multiply:[2(7x + 80) = 16x][14x + 160 = 16x][160 = 2x][x = 80]Now that I have the value of (x), I can find the initial number of boarders:[7x = 7 times 80 = 560]Therefore, there were initially 560 boarders at the school.

💡First, I'll define the unknown number as ( x ).According to the problem, when ( x ) is divided by 6 and then multiplied by 12, the result is 8. This can be written as the equation:[frac{x}{6} times 12 = 8]To solve for ( x ), I'll simplify the equation. Dividing both sides by 12 gives:[frac{x}{6} = frac{8}{12}]Simplifying the fraction on the right side:[frac{x}{6} = frac{2}{3}]Next, I'll isolate ( x ) by multiplying both sides of the equation by 6:[x = frac{2}{3} times 6]Performing the multiplication:[x = frac{12}{3}]Finally, dividing 12 by 3 gives:[x = 4]So, the number is 4.