💡Okay, so I have this problem about an ellipse and a parabola. Let me try to figure it out step by step. First, part (1) asks for the equation of the ellipse. The ellipse is given by (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0). The point (P(2,3)) lies on this ellipse, so plugging that in, I get:[frac{2^2}{a^2} + frac{3^2}{b^2} = 1 implies frac{4}{a^2} + frac{9}{b^2} = 1]That's one equation. Now, the right focus (F) of the ellipse is also the focus of the parabola (C_2: y^2 = 2px). The point ((2,4)) is on this parabola, so plugging that in:[4^2 = 2p cdot 2 implies 16 = 4p implies p = 4]So the parabola is (y^2 = 8x). The focus of a parabola (y^2 = 2px) is at ((p/2, 0)). Wait, no, actually, for (y^2 = 4ax), the focus is at ((a,0)). So in this case, (2p = 8) implies (p = 4), so the focus is at ((p/2, 0)) which is ((2, 0)). So the right focus (F) of the ellipse is ((2, 0)).For an ellipse, the distance from the center to each focus is (c), where (c^2 = a^2 - b^2). Since the focus is at ((2, 0)), the center of the ellipse must be at the origin because the foci are on the x-axis. So (c = 2), which gives:[c^2 = a^2 - b^2 implies 4 = a^2 - b^2]So now I have two equations:1. (frac{4}{a^2} + frac{9}{b^2} = 1)2. (a^2 - b^2 = 4)I need to solve these simultaneously. Let me denote (a^2 = A) and (b^2 = B) for simplicity. Then:1. (frac{4}{A} + frac{9}{B} = 1)2. (A - B = 4)From equation 2, (A = B + 4). Substitute into equation 1:[frac{4}{B + 4} + frac{9}{B} = 1]Let me find a common denominator:[frac{4B + 9(B + 4)}{B(B + 4)} = 1 implies frac{4B + 9B + 36}{B(B + 4)} = 1 implies frac{13B + 36}{B(B + 4)} = 1]Multiply both sides by (B(B + 4)):[13B + 36 = B^2 + 4B]Bring all terms to one side:[B^2 + 4B - 13B - 36 = 0 implies B^2 - 9B - 36 = 0]Solving this quadratic equation:[B = frac{9 pm sqrt{81 + 144}}{2} = frac{9 pm sqrt{225}}{2} = frac{9 pm 15}{2}]So, (B = frac{24}{2} = 12) or (B = frac{-6}{2} = -3). Since (B = b^2) must be positive, (B = 12). Then (A = B + 4 = 16). Therefore, (a^2 = 16) and (b^2 = 12). So the equation of the ellipse is:[frac{x^2}{16} + frac{y^2}{12} = 1]Okay, that seems solid. Now, moving on to part (2). This seems more complicated. We have a line (l) passing through the right focus (F(2,0)) of the ellipse, intersecting the ellipse at points (A) and (B). We need to find the maximum value of the product (t) of the slopes of the lines containing the sides of triangle (ABP). First, let me parse this. The triangle is (ABP), so the sides are (AB), (BP), and (PA). The slopes of these sides are (m_{AB}), (m_{BP}), and (m_{PA}). The product (t) is (m_{AB} cdot m_{BP} cdot m_{PA}). Wait, but the problem says "the product of the slopes of the lines containing the sides of triangle (ABP)". So each side is a line, so each has a slope. So yes, three slopes multiplied together. But in the solution provided earlier, it seems like they considered two slopes, not three. Hmm, maybe I need to double-check.Wait, the triangle has three sides, each with a slope, so the product (t) should be the product of these three slopes. But in the solution, they only considered two slopes. Maybe I need to check the problem statement again.Wait, the problem says: "the product of the slopes of the lines containing the sides of triangle (ABP)". So, each side is a line, so each has a slope. So, three slopes. But in the solution, they considered (t = k cdot frac{y_1 - 3}{x_1 - 2} cdot frac{y_2 - 3}{x_2 - 2}). That is, they took the slope of line (l) (which is (k)) and the slopes from (A) to (P) and (B) to (P). So, that's three slopes: (k), (m_{AP}), (m_{BP}). So, (t = k cdot m_{AP} cdot m_{BP}). But the sides of triangle (ABP) are (AB), (BP), and (PA). So, the slopes are (m_{AB}), (m_{BP}), and (m_{PA}). So, is the product (t = m_{AB} cdot m_{BP} cdot m_{PA})? Wait, but in the solution, they took (k) as the slope of line (l), which is the same as (m_{AB}), since (l) is the line passing through (A) and (B). So, (m_{AB} = k). Then, (m_{AP}) is the slope from (A) to (P), which is (frac{y_1 - 3}{x_1 - 2}), and similarly for (m_{BP}). So, yes, (t = m_{AB} cdot m_{AP} cdot m_{BP}). So, that's correct. So, the product is (k cdot frac{y_1 - 3}{x_1 - 2} cdot frac{y_2 - 3}{x_2 - 2}). Now, the solution parametrizes the line (l) as (y = k(x - 2)), which makes sense since it passes through (F(2,0)). Then, they substitute this into the ellipse equation to find points (A) and (B). Let me try to follow that. Substitute (y = k(x - 2)) into (frac{x^2}{16} + frac{y^2}{12} = 1):[frac{x^2}{16} + frac{k^2(x - 2)^2}{12} = 1]Multiply both sides by 48 to eliminate denominators:[3x^2 + 4k^2(x^2 - 4x + 4) = 48]Expand:[3x^2 + 4k^2x^2 - 16k^2x + 16k^2 = 48]Combine like terms:[(3 + 4k^2)x^2 - 16k^2x + (16k^2 - 48) = 0]This is a quadratic in (x). Let me denote this as:[Ax^2 + Bx + C = 0]Where:- (A = 3 + 4k^2)- (B = -16k^2)- (C = 16k^2 - 48)The solutions to this quadratic will give the x-coordinates of points (A) and (B). Let me denote them as (x_1) and (x_2). Using Vieta's formula:- (x_1 + x_2 = -B/A = 16k^2 / (3 + 4k^2))- (x_1 x_2 = C/A = (16k^2 - 48)/(3 + 4k^2))Similarly, the y-coordinates are (y_1 = k(x_1 - 2)) and (y_2 = k(x_2 - 2)). So, (y_1 + y_2 = k(x_1 + x_2 - 4) = k(16k^2 / (3 + 4k^2) - 4)). Let me compute that:[y_1 + y_2 = kleft( frac{16k^2 - 4(3 + 4k^2)}{3 + 4k^2} right) = kleft( frac{16k^2 - 12 - 16k^2}{3 + 4k^2} right) = kleft( frac{-12}{3 + 4k^2} right) = frac{-12k}{3 + 4k^2}]Similarly, (y_1 y_2 = k^2(x_1 - 2)(x_2 - 2)). Let me compute that:First, expand ((x_1 - 2)(x_2 - 2)):[x_1 x_2 - 2(x_1 + x_2) + 4]Substitute from Vieta's:[frac{16k^2 - 48}{3 + 4k^2} - 2 cdot frac{16k^2}{3 + 4k^2} + 4 = frac{16k^2 - 48 - 32k^2 + 4(3 + 4k^2)}{3 + 4k^2}]Simplify numerator:[16k^2 - 48 - 32k^2 + 12 + 16k^2 = (16k^2 - 32k^2 + 16k^2) + (-48 + 12) = 0k^2 - 36 = -36]So, ((x_1 - 2)(x_2 - 2) = frac{-36}{3 + 4k^2}). Therefore, (y_1 y_2 = k^2 cdot frac{-36}{3 + 4k^2} = frac{-36k^2}{3 + 4k^2}).Wait, but in the solution, they have (y_1 y_2 = - frac{36k^2}{4k^2 + 3}). That's the same as what I have here, just written differently. So, that's correct.Now, going back to the expression for (t):[t = k cdot frac{y_1 - 3}{x_1 - 2} cdot frac{y_2 - 3}{x_2 - 2}]Let me compute (frac{y_1 - 3}{x_1 - 2}) and (frac{y_2 - 3}{x_2 - 2}). First, note that (y_1 = k(x_1 - 2)), so:[frac{y_1 - 3}{x_1 - 2} = frac{k(x_1 - 2) - 3}{x_1 - 2} = k - frac{3}{x_1 - 2}]Similarly,[frac{y_2 - 3}{x_2 - 2} = k - frac{3}{x_2 - 2}]So, the product becomes:[t = k cdot left(k - frac{3}{x_1 - 2}right) cdot left(k - frac{3}{x_2 - 2}right)]Let me denote (u = frac{1}{x_1 - 2}) and (v = frac{1}{x_2 - 2}). Then,[t = k cdot (k - 3u) cdot (k - 3v) = k left[ k^2 - 3k(u + v) + 9uv right]]So, I need expressions for (u + v) and (uv). First, (u + v = frac{1}{x_1 - 2} + frac{1}{x_2 - 2}). Let me compute this:[u + v = frac{x_2 - 2 + x_1 - 2}{(x_1 - 2)(x_2 - 2)} = frac{(x_1 + x_2) - 4}{(x_1 - 2)(x_2 - 2)}]We already have (x_1 + x_2 = frac{16k^2}{3 + 4k^2}) and ((x_1 - 2)(x_2 - 2) = frac{-36}{3 + 4k^2}). So,[u + v = frac{frac{16k^2}{3 + 4k^2} - 4}{frac{-36}{3 + 4k^2}} = frac{frac{16k^2 - 4(3 + 4k^2)}{3 + 4k^2}}{frac{-36}{3 + 4k^2}} = frac{16k^2 - 12 - 16k^2}{-36} = frac{-12}{-36} = frac{1}{3}]Similarly, (uv = frac{1}{(x_1 - 2)(x_2 - 2)} = frac{3 + 4k^2}{-36}). So,[uv = frac{3 + 4k^2}{-36} = -frac{3 + 4k^2}{36}]Now, substitute back into (t):[t = k left[ k^2 - 3k cdot frac{1}{3} + 9 cdot left(-frac{3 + 4k^2}{36}right) right] = k left[ k^2 - k - frac{9(3 + 4k^2)}{36} right]]Simplify the terms inside the brackets:First, (9 cdot frac{3 + 4k^2}{36} = frac{27 + 36k^2}{36} = frac{3 + 4k^2}{4}).So,[t = k left[ k^2 - k - frac{3 + 4k^2}{4} right] = k left[ frac{4k^2 - 4k - 3 - 4k^2}{4} right] = k left[ frac{-4k - 3}{4} right] = k cdot left( -k - frac{3}{4} right)]So,[t = -k^2 - frac{3}{4}k]Wait, that's a quadratic in (k). To find its maximum, since the coefficient of (k^2) is negative, the parabola opens downward, so the maximum is at the vertex.The vertex occurs at (k = -frac{b}{2a}), where (a = -1), (b = -frac{3}{4}). So,[k = -frac{-frac{3}{4}}{2 cdot -1} = frac{frac{3}{4}}{-2} = -frac{3}{8}]So, the maximum value of (t) is:[t = -left(-frac{3}{8}right)^2 - frac{3}{4} cdot left(-frac{3}{8}right) = -frac{9}{64} + frac{9}{32} = -frac{9}{64} + frac{18}{64} = frac{9}{64}]So, the maximum value of (t) is (frac{9}{64}).Wait, but in the solution, they wrote (t = -k^2 - frac{3}{4}k = -(k + frac{3}{8})^2 + frac{9}{64}). That's another way to write it, completing the square. So, the maximum occurs when the square term is zero, i.e., when (k = -frac{3}{8}), giving (t = frac{9}{64}).So, that seems correct.But let me check if I made any mistakes in my calculations. First, when I computed (u + v), I had:[u + v = frac{(x_1 + x_2) - 4}{(x_1 - 2)(x_2 - 2)} = frac{frac{16k^2}{3 + 4k^2} - 4}{frac{-36}{3 + 4k^2}} = frac{frac{16k^2 - 12 - 16k^2}{3 + 4k^2}}{frac{-36}{3 + 4k^2}} = frac{-12}{-36} = frac{1}{3}]That seems correct.Then, (uv = frac{1}{(x_1 - 2)(x_2 - 2)} = frac{3 + 4k^2}{-36}). So,[uv = -frac{3 + 4k^2}{36}]Yes, that's correct.Then, substituting into (t):[t = k left[ k^2 - 3k(u + v) + 9uv right] = k left[ k^2 - 3k cdot frac{1}{3} + 9 cdot left(-frac{3 + 4k^2}{36}right) right]]Simplify:[k^2 - k - frac{9(3 + 4k^2)}{36} = k^2 - k - frac{3 + 4k^2}{4}]Convert (k^2) to (frac{4k^2}{4}):[frac{4k^2}{4} - k - frac{3 + 4k^2}{4} = frac{4k^2 - 4k - 3 - 4k^2}{4} = frac{-4k - 3}{4}]So,[t = k cdot frac{-4k - 3}{4} = frac{-4k^2 - 3k}{4} = -k^2 - frac{3}{4}k]Yes, that's correct.Then, completing the square:[t = -k^2 - frac{3}{4}k = -left(k^2 + frac{3}{4}kright)]Complete the square inside the brackets:[k^2 + frac{3}{4}k = left(k + frac{3}{8}right)^2 - left(frac{3}{8}right)^2 = left(k + frac{3}{8}right)^2 - frac{9}{64}]So,[t = -left( left(k + frac{3}{8}right)^2 - frac{9}{64} right) = -left(k + frac{3}{8}right)^2 + frac{9}{64}]Therefore, the maximum value of (t) is (frac{9}{64}), achieved when (k = -frac{3}{8}).So, the solution seems correct. But let me think if there's another way to approach this problem, maybe using parametric equations or something else, but I think the method used is solid. Another thought: when the line (l) is vertical, i.e., (k) is undefined, but in that case, the line would be (x = 2), which intersects the ellipse at (x=2). Plugging into the ellipse equation:[frac{4}{16} + frac{y^2}{12} = 1 implies frac{1}{4} + frac{y^2}{12} = 1 implies frac{y^2}{12} = frac{3}{4} implies y^2 = 9 implies y = pm 3]So, points (A(2,3)) and (B(2,-3)). But point (P) is also ((2,3)), so triangle (ABP) would have points (A(2,3)), (B(2,-3)), and (P(2,3)). But (A) and (P) are the same point, so the triangle collapses into a line. Therefore, the slopes would be undefined or zero, but in this case, the product (t) would be zero or undefined. Since the problem specifies a line passing through (F) intersecting the ellipse at two points, I think (x=2) is a valid case, but it results in (t=0), which is less than (frac{9}{64}). So, the maximum is indeed (frac{9}{64}).Another edge case: when the line (l) is horizontal, i.e., (k=0). Then, the line is (y=0), which is the x-axis. It intersects the ellipse at points where (y=0):[frac{x^2}{16} = 1 implies x = pm 4]So, points (A(4,0)) and (B(-4,0)). Then, triangle (ABP) has points (A(4,0)), (B(-4,0)), and (P(2,3)). The slopes are:- (m_{AB}): Since (A) and (B) are on the x-axis, the slope is 0.- (m_{AP}): Slope from (A(4,0)) to (P(2,3)): (frac{3 - 0}{2 - 4} = frac{3}{-2} = -frac{3}{2})- (m_{BP}): Slope from (B(-4,0)) to (P(2,3)): (frac{3 - 0}{2 - (-4)} = frac{3}{6} = frac{1}{2})So, the product (t = 0 cdot (-frac{3}{2}) cdot frac{1}{2} = 0). So, again, (t=0), which is less than (frac{9}{64}).Therefore, the maximum value of (t) is indeed (frac{9}{64}).I think that's thorough enough. I don't see any mistakes in the reasoning, so I agree with the solution.

💡Okay, so I need to convert the polar coordinate (5, 5π/4) to rectangular coordinates. Hmm, I remember that polar coordinates are given in the form (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis. Rectangular coordinates, on the other hand, are given as (x, y). I think the formulas to convert from polar to rectangular coordinates involve cosine and sine. Let me try to recall them. I believe x is equal to r multiplied by the cosine of θ, and y is equal to r multiplied by the sine of θ. So, x = r cos θ and y = r sin θ. Alright, so in this case, r is 5 and θ is 5π/4. I need to calculate cos(5π/4) and sin(5π/4). Let me think about where 5π/4 is on the unit circle. I know that π radians is 180 degrees, so 5π/4 is π + π/4, which is 225 degrees. That places it in the third quadrant of the unit circle, where both cosine and sine are negative. Now, cos(π/4) is √2/2, and sin(π/4) is also √2/2. Since 5π/4 is in the third quadrant, both cosine and sine will be negative. So, cos(5π/4) should be -√2/2 and sin(5π/4) should be -√2/2. Let me double-check that. Yes, in the third quadrant, both x and y are negative, so cosine and sine are negative. So, cos(5π/4) = -√2/2 and sin(5π/4) = -√2/2. Now, plugging these into the formulas for x and y. First, x = r cos θ = 5 * (-√2/2). Multiplying 5 by -√2/2 gives x = -5√2/2. Similarly, y = r sin θ = 5 * (-√2/2). Multiplying 5 by -√2/2 gives y = -5√2/2. So, putting it all together, the rectangular coordinates should be (-5√2/2, -5√2/2). Wait, let me make sure I didn't mix up the signs or the values. Since 5π/4 is in the third quadrant, both x and y should be negative, which they are. The cosine and sine values are both -√2/2, which seems correct. Multiplying by r = 5 gives the coordinates as negative numbers, which makes sense. I think that's it. The rectangular coordinates are (-5√2/2, -5√2/2).

💡Okay, so I have this problem about oblique projection and how it affects the area of a triangle when drawing a perspective diagram. The question is asking how many times the area of the original triangle is compared to the area of the perspective diagram. The options are A: 1/2, B: 2√2, C: 2, D: √2/4. Hmm, I need to figure this out step by step.First, I remember that oblique projection is a type of parallel projection where the projectors are not perpendicular to the projection plane. This means that shapes can appear distorted in the projection. In particular, I think in oblique projection, one axis is usually scaled differently than the others. I think in cavalier projection, the scaling factor is 1, which doesn't change lengths, but in cabinet projection, it's 0.5, which halves the lengths along that axis.Wait, so maybe in this case, since it's an oblique projection, they might be using a specific scaling factor. The problem mentions a triangle, so maybe the base is kept the same, and the height is scaled? Or is it the other way around?Let me try to visualize. If I have a triangle, and I'm projecting it using oblique projection, I need to decide which sides are aligned with the projection axes. The solution mentioned taking one side as the x-axis and the height as the y-axis. So, the base remains unchanged, and the height is scaled.In oblique projection, typically, one axis (like the x-axis) remains the same, and the other axis (like the y-axis) is scaled. The scaling factor can vary, but in cabinet projection, it's usually 0.5, which is half. So, if the height is along the y-axis, its length would be halved in the projection.But wait, the solution said something about sin45°. Hmm, maybe the projection isn't just scaling but also involves an angle. I think in oblique projection, the projectors make an angle with the projection plane. If it's a 45° angle, then the scaling factor might be related to sin45°, which is √2/2.So, if the height is scaled by sin45°, which is √2/2, but since it's oblique projection, maybe it's scaled by 1/2 of that? Wait, let me think. If the height is scaled by sin(theta), where theta is the angle between the projectors and the projection plane, then if theta is 45°, sin(theta) is √2/2. So, the height becomes √2/2 times the original height.But the solution mentioned it's halved and then multiplied by sin45°, so maybe it's (1/2)*sin45°, which would be (1/2)*(√2/2) = √2/4. So, the height in the projection is √2/4 times the original height.So, the area of the triangle is (base * height)/2. In the original triangle, the area is (b * h)/2. In the projected triangle, the base remains b, and the height becomes (√2/4)h. So, the projected area is (b * (√2/4)h)/2 = (b * h)/2 * (√2/4) = original area * √2/4.Therefore, the area of the perspective diagram is √2/4 times the original area. So, the original area is the perspective area multiplied by 4/√2, which simplifies to 2√2. So, the original area is 2√2 times the perspective area.Wait, let me double-check. If the projected height is √2/4 times the original, then the area scales by √2/4. So, original area = projected area / (√2/4) = projected area * 4/√2 = 2√2 * projected area. Yeah, that makes sense.So, the answer should be B: 2√2.I think that's it. I considered the scaling of the height due to the oblique projection, calculated the effect on the area, and found the factor by which the original area is larger than the projected area. It seems consistent with the solution provided earlier.**Final Answer**boxed{B}

💡Okay, so I have this problem about coloring strips in the Cartesian plane and proving something about rectangles. Let me try to understand it step by step.First, the problem defines strips ( S_n = {(x, y) mid n leq x < n + 1} ) for every integer ( n ). So, each strip is a vertical band that's one unit wide, starting at ( x = n ) and going up to ( x = n + 1 ). These strips are colored either red or blue. Now, the goal is to prove that for any two distinct positive integers ( a ) and ( b ), there exists a rectangle with side lengths ( a ) and ( b ) such that all four vertices of the rectangle are the same color. Hmm, okay. So, I need to show that no matter how you color these strips red or blue, you can always find a rectangle of size ( a times b ) where all four corners are either red or blue. Let me think about how to approach this. Maybe I can use some kind of pigeonhole principle or consider the periodicity of the coloring. Since the strips are colored either red or blue, the coloring is periodic in some sense, but since it's arbitrary, I can't assume a specific pattern.Wait, maybe I can assume the contrary and reach a contradiction. Let's try that.Assume that there is no such rectangle with all vertices of the same color. That means, for every possible rectangle of size ( a times b ), the four vertices are not all red or all blue. So, in every such rectangle, there must be at least two red and two blue vertices.But how can I use this assumption to find a contradiction?Let's consider the strips ( S_n ). Each strip is colored either red or blue. Let's fix a starting strip, say ( S_0 ). Without loss of generality, let's assume ( S_0 ) is red. Now, if I look at the strip ( S_a ), which is ( a ) units to the right of ( S_0 ), what can we say about its color?If ( S_a ) were also red, then we could form a rectangle with vertices at ( (0, y) ), ( (a, y) ), ( (0, y + b) ), and ( (a, y + b) ). Since both ( S_0 ) and ( S_a ) are red, and if the vertical distance ( b ) doesn't affect the color (since strips are vertical), then all four vertices would be red, which contradicts our assumption. Therefore, ( S_a ) must be blue.Similarly, if ( S_a ) is blue, then ( S_{2a} ) must be red, because otherwise, if ( S_{2a} ) were blue, we could form a rectangle from ( S_a ) to ( S_{2a} ) with height ( b ), and all vertices would be blue, which is again a contradiction. So, the coloring alternates every ( a ) units.But wait, this alternation is only along the x-axis. What about the y-axis? The problem doesn't specify anything about horizontal strips, only vertical ones. So, the coloring is only dependent on the x-coordinate, not the y-coordinate. That means, for any point ( (x, y) ), its color is determined solely by the strip ( S_n ) where ( n leq x < n + 1 ).So, if ( S_0 ) is red, ( S_a ) is blue, ( S_{2a} ) is red, ( S_{3a} ) is blue, and so on. Similarly, if we consider ( b ), then ( S_b ) must be blue, ( S_{2b} ) must be red, ( S_{3b} ) must be blue, etc.But here's the problem: if ( a ) and ( b ) are distinct positive integers, they might have some common multiple. Let's say the least common multiple (LCM) of ( a ) and ( b ) is ( L ). Then, ( L ) is a multiple of both ( a ) and ( b ). So, considering the strip ( S_L ), according to the alternation from ( a ), it should be red if ( L ) is a multiple of ( 2a ), but since ( L ) is also a multiple of ( b ), according to the alternation from ( b ), it should be blue if ( L ) is a multiple of ( 2b ). But wait, ( L ) is the LCM of ( a ) and ( b ), so it's the smallest number that is a multiple of both. Therefore, depending on whether ( L ) is even or odd in terms of multiples of ( a ) and ( b ), we might have a contradiction.Let me formalize this. Suppose ( L = k cdot a = m cdot b ) for some integers ( k ) and ( m ). Then, from the alternation along ( a ), ( S_L ) should be red if ( k ) is even and blue if ( k ) is odd. Similarly, from the alternation along ( b ), ( S_L ) should be red if ( m ) is even and blue if ( m ) is odd.But since ( L ) is the LCM, ( k = L / a ) and ( m = L / b ). If ( a ) and ( b ) are coprime, then ( L = a cdot b ), so ( k = b ) and ( m = a ). If ( a ) and ( b ) are not coprime, ( L ) is still the product divided by their GCD.In any case, ( k ) and ( m ) are integers. Now, if ( k ) is even and ( m ) is odd, or vice versa, then ( S_L ) would have conflicting color assignments, which is impossible. Therefore, there must be a contradiction in our initial assumption that no such rectangle exists.Wait, but what if ( k ) and ( m ) have the same parity? Then, ( S_L ) would have consistent color assignments. Hmm, so maybe my earlier reasoning isn't sufficient.Let me think again. If ( S_0 ) is red, then ( S_a ) is blue, ( S_{2a} ) is red, etc. Similarly, ( S_b ) is blue, ( S_{2b} ) is red, etc. Now, if ( L ) is a multiple of both ( a ) and ( b ), then ( S_L ) must be both red and blue depending on the parity of ( L/a ) and ( L/b ). But ( L/a = b / gcd(a, b) ) and ( L/b = a / gcd(a, b) ). So, if ( gcd(a, b) ) is 1, then ( L = a cdot b ), and ( L/a = b ), ( L/b = a ). So, if ( b ) is even, then ( S_L ) is red if ( L/a ) is even, but ( L/b = a ), so if ( a ) is even, it's blue. Wait, this is getting confusing.Maybe I need to consider specific cases. Let's say ( a = 2 ) and ( b = 3 ). Then, ( L = 6 ). So, ( S_6 ) should be red if ( 6/2 = 3 ) is odd, so blue. Similarly, ( 6/3 = 2 ) is even, so red. Contradiction! So, ( S_6 ) cannot be both blue and red. Therefore, our assumption that no such rectangle exists is false.Similarly, if ( a = 3 ) and ( b = 5 ), ( L = 15 ). ( 15/3 = 5 ) is odd, so ( S_{15} ) should be blue. ( 15/5 = 3 ) is odd, so ( S_{15} ) should be blue as well. Wait, no contradiction here. Hmm, so maybe my earlier reasoning only works when ( L/a ) and ( L/b ) have different parities.Wait, in the case of ( a = 2 ) and ( b = 3 ), ( L = 6 ), ( 6/2 = 3 ) (odd) and ( 6/3 = 2 ) (even). So, one is odd, the other is even, leading to a contradiction. But in the case of ( a = 3 ) and ( b = 5 ), both ( L/a = 5 ) and ( L/b = 3 ) are odd, so no contradiction. Hmm, so maybe my approach only works when ( a ) and ( b ) are such that ( L/a ) and ( L/b ) have different parities.But the problem states that ( a ) and ( b ) are distinct positive integers. So, they can be any distinct positive integers, not necessarily coprime or anything. So, in some cases, the contradiction arises, and in others, it doesn't. Therefore, my earlier approach might not cover all cases.Wait, maybe I need to consider more than just the LCM. Maybe I need to look at the coloring pattern more carefully. Let's think about the coloring as a function of ( n ). Each strip ( S_n ) is colored red or blue, so we can represent this as a function ( f: mathbb{Z} rightarrow {R, B} ).Now, if we assume that no rectangle of size ( a times b ) has all four vertices of the same color, then for any ( n ), the colors of ( S_n ), ( S_{n+a} ), ( S_{n+b} ), and ( S_{n+a+b} ) must not all be the same. But this seems a bit vague. Maybe I can model this as a graph or use some combinatorial argument. Alternatively, perhaps I can use the infinite pigeonhole principle, considering that there are infinitely many strips, so some pattern must repeat.Wait, another idea: if we fix the vertical distance ( b ), then for each strip ( S_n ), the strip ( S_{n+b} ) must be a different color if ( S_n ) is the same color as ( S_{n+a} ). Hmm, not sure.Alternatively, maybe I can consider the coloring as a periodic function with period ( a ) or ( b ), but since the coloring is arbitrary, it's not necessarily periodic.Wait, going back to the initial assumption: if no such rectangle exists, then for every ( n ), the colors of ( S_n ) and ( S_{n+a} ) must alternate in a way that prevents four corners from being the same color. But if I can show that this alternation leads to a contradiction when considering both ( a ) and ( b ), then the assumption is false.Let me try to formalize this. Suppose ( S_0 ) is red. Then, as before, ( S_a ) must be blue, ( S_{2a} ) must be red, etc. Similarly, ( S_b ) must be blue, ( S_{2b} ) must be red, etc. Now, consider the strip ( S_{a + b} ). If ( S_{a + b} ) is red, then looking at the rectangle from ( S_0 ) to ( S_{a + b} ) with height ( b ), the vertices would be ( (0, y) ), ( (a + b, y) ), ( (0, y + b) ), ( (a + b, y + b) ). Since ( S_0 ) is red and ( S_{a + b} ) is red, the top vertices would be in ( S_0 ) and ( S_{a + b} ), which are both red, but the bottom vertices would be in ( S_b ) and ( S_{a + 2b} ). Wait, ( S_b ) is blue, and ( S_{a + 2b} ) would be... Let's see, ( S_{a + 2b} ) is ( S_{a} + 2b ). Since ( S_a ) is blue, ( S_{a + 2b} ) would be blue if ( 2b ) is even, but ( 2b ) is even, so ( S_{a + 2b} ) is blue. Therefore, the bottom vertices are blue, and the top vertices are red, so the rectangle has two red and two blue vertices, which is okay. But if ( S_{a + b} ) is blue, then the top vertices are blue, and the bottom vertices would be ( S_b ) (blue) and ( S_{a + 2b} ) (blue). So, all four vertices would be blue, which contradicts our assumption. Therefore, ( S_{a + b} ) cannot be blue, so it must be red.But wait, if ( S_{a + b} ) is red, then looking at the rectangle from ( S_a ) to ( S_{a + b} ) with height ( b ), the vertices would be ( (a, y) ), ( (a + b, y) ), ( (a, y + b) ), ( (a + b, y + b) ). ( S_a ) is blue, ( S_{a + b} ) is red, ( S_{a + b} ) is red, and ( S_{a + 2b} ) is blue. So, the vertices are blue, red, red, blue. Again, two of each color, which is okay.Hmm, so maybe this approach isn't leading to a contradiction. Maybe I need to consider more strips or a different configuration.Wait, another idea: if we consider the coloring as a function ( f(n) ) where ( f(n) ) is the color of ( S_n ), then our assumption implies that ( f(n) neq f(n + a) ) or ( f(n) neq f(n + b) ) for all ( n ). But I'm not sure.Alternatively, maybe I can use the fact that the coloring is periodic with period ( 2a ) and ( 2b ), leading to a contradiction if ( 2a ) and ( 2b ) are not commensurate.Wait, perhaps I need to consider the coloring function ( f(n) ) and its periodicity. If ( f(n) ) is periodic with period ( d ), then ( d ) must divide both ( 2a ) and ( 2b ). But since ( a ) and ( b ) are distinct, their periods might not align, leading to a contradiction.Alternatively, maybe I can use the infinite Ramsey theorem, which states that in any infinite coloring, there exists an infinite monochromatic subset. But I'm not sure how to apply that here.Wait, going back to the initial idea with the LCM. Let me try again. Suppose ( L = text{lcm}(a, b) ). Then, ( L ) is a multiple of both ( a ) and ( b ). So, ( L = k cdot a = m cdot b ) for some integers ( k ) and ( m ).From the alternation along ( a ), ( S_L ) must be red if ( k ) is even and blue if ( k ) is odd. Similarly, from the alternation along ( b ), ( S_L ) must be red if ( m ) is even and blue if ( m ) is odd.But since ( L ) is the LCM, ( k = L / a = b / gcd(a, b) ) and ( m = L / b = a / gcd(a, b) ). Now, if ( gcd(a, b) ) is 1, then ( k = b ) and ( m = a ). So, if ( b ) is even, ( k ) is even, so ( S_L ) is red. If ( a ) is even, ( m ) is even, so ( S_L ) is red. If both ( a ) and ( b ) are odd, then ( k ) and ( m ) are both odd, so ( S_L ) would be blue from both perspectives, which is consistent. Wait, but if ( a ) and ( b ) are both odd, then ( L = a cdot b ) is odd, so ( S_L ) would be blue if starting from red at ( S_0 ). But if ( a ) and ( b ) are both even, then ( L ) is even, so ( S_L ) would be red if starting from red at ( S_0 ).Wait, maybe I'm overcomplicating this. The key point is that if ( a ) and ( b ) are such that ( L/a ) and ( L/b ) have different parities, then ( S_L ) would have conflicting color assignments, leading to a contradiction. If they have the same parity, then no contradiction arises, but perhaps another approach is needed.Wait, but in the problem statement, ( a ) and ( b ) are distinct positive integers. So, they could be both even, both odd, or one even and one odd. If they are both even, then ( L ) is even, so ( S_L ) would be red if starting from red at ( S_0 ). If they are both odd, ( L ) is odd, so ( S_L ) would be blue. If one is even and the other is odd, then ( L ) is even if ( a ) is even and ( b ) is odd, or vice versa. Wait, no, ( L ) is the LCM, so if one is even and the other is odd, ( L ) would be even.Wait, let's take an example where ( a = 2 ) and ( b = 3 ). Then, ( L = 6 ). From ( a )'s perspective, ( L/a = 3 ), which is odd, so ( S_6 ) should be blue. From ( b )'s perspective, ( L/b = 2 ), which is even, so ( S_6 ) should be red. Contradiction! Therefore, ( S_6 ) cannot be both blue and red, so our assumption that no such rectangle exists is false.Similarly, if ( a = 3 ) and ( b = 5 ), ( L = 15 ). ( L/a = 5 ) (odd), so ( S_{15} ) should be blue. ( L/b = 3 ) (odd), so ( S_{15} ) should be blue. No contradiction here. Hmm, so in this case, the contradiction doesn't arise. So, my earlier approach only works when ( L/a ) and ( L/b ) have different parities.But the problem states that ( a ) and ( b ) are distinct positive integers, so they could be both even, both odd, or one even and one odd. Therefore, in some cases, the contradiction arises, and in others, it doesn't. So, my approach isn't sufficient for all cases.Wait, maybe I need to consider more than just the LCM. Maybe I need to look at the coloring pattern over multiple periods or consider the interaction between ( a ) and ( b ) in a different way.Another idea: if we fix the vertical distance ( b ), then for each strip ( S_n ), the strip ( S_{n + b} ) must be a different color if ( S_n ) is the same color as ( S_{n + a} ). But I'm not sure.Alternatively, perhaps I can use the fact that the coloring is periodic with period ( 2a ) and ( 2b ), leading to a contradiction if ( 2a ) and ( 2b ) are not commensurate.Wait, maybe I can consider the coloring as a function ( f(n) ) where ( f(n) ) is the color of ( S_n ). Then, our assumption implies that ( f(n) neq f(n + a) ) or ( f(n) neq f(n + b) ) for all ( n ). But I'm not sure how to use this.Wait, another approach: consider the coloring as a sequence of red and blue strips. Since the coloring is arbitrary, it's an infinite sequence. Now, if we look at the sequence modulo ( a ) and modulo ( b ), we might find some periodicity or pattern that leads to a contradiction.Alternatively, maybe I can use the infinite pigeonhole principle. Since there are infinitely many strips, and only two colors, some pattern must repeat infinitely often, leading to a monochromatic rectangle.Wait, let's think about it this way: for each strip ( S_n ), consider the color of ( S_n ) and ( S_{n + a} ). If they are the same color, then we can form a rectangle with height ( b ) that has two vertices of that color. But we need all four vertices to be the same color.Wait, maybe I need to consider pairs of strips. For each ( n ), look at the pair ( (S_n, S_{n + a}) ). There are only two possible colorings for each pair: both red, both blue, or one red and one blue. Since there are infinitely many pairs, by the pigeonhole principle, some pair must repeat infinitely often. If a pair repeats, then we can find two such pairs separated by ( b ) units, leading to a rectangle with all four vertices of the same color.Wait, that sounds promising. Let me try to formalize this.Consider the coloring of each pair ( (S_n, S_{n + a}) ). Each pair can be in one of three states: both red, both blue, or one red and one blue. Since there are infinitely many such pairs, by the pigeonhole principle, at least one of these states must occur infinitely often.Case 1: There are infinitely many ( n ) such that ( S_n ) and ( S_{n + a} ) are both red. Then, among these, consider the vertical distance ( b ). If we can find two such ( n ) and ( n' ) such that ( n' = n + k cdot b ) for some integer ( k ), then the rectangle with vertices at ( (n, y) ), ( (n + a, y) ), ( (n, y + b) ), ( (n + a, y + b) ) would have all four vertices red, which is what we want.But wait, how do we ensure that such ( n ) and ( n' ) exist? Since there are infinitely many ( n ) with ( S_n ) and ( S_{n + a} ) both red, we can apply the pigeonhole principle again on the vertical shifts. Specifically, for each such ( n ), consider the color of ( S_{n + b} ) and ( S_{n + a + b} ). If either ( S_{n + b} ) or ( S_{n + a + b} ) is red, then we can form a rectangle. If not, then both are blue, but then we can consider the next shift.Wait, this is getting a bit tangled. Maybe a better way is to consider that since there are infinitely many ( n ) with ( S_n ) and ( S_{n + a} ) both red, we can find two such ( n ) and ( n' ) such that ( n' = n + k cdot b ) for some ( k ), leading to a rectangle with all four vertices red.Alternatively, if Case 1 doesn't hold, then Case 2 or Case 3 must. Case 2: infinitely many ( n ) with ( S_n ) and ( S_{n + a} ) both blue. Similar reasoning applies; we can find a rectangle with all four vertices blue.Case 3: infinitely many ( n ) with ( S_n ) and ( S_{n + a} ) of different colors. Then, for these ( n ), ( S_n ) is red and ( S_{n + a} ) is blue, or vice versa. Now, consider the vertical shifts by ( b ). For each such ( n ), look at ( S_{n + b} ) and ( S_{n + a + b} ). If ( S_{n + b} ) is red and ( S_{n + a + b} ) is blue, or vice versa, then we can form a rectangle with two red and two blue vertices, which is okay. But if for some ( n ), ( S_{n + b} ) and ( S_{n + a + b} ) are both red or both blue, then we can form a rectangle with all four vertices of the same color.But wait, if there are infinitely many ( n ) where ( S_n ) and ( S_{n + a} ) are of different colors, then among these, there must be some ( n ) where ( S_{n + b} ) and ( S_{n + a + b} ) are both red or both blue, leading to a monochromatic rectangle.Therefore, in all cases, we can find a rectangle with all four vertices of the same color, contradicting our initial assumption.So, putting it all together, by assuming that no such rectangle exists, we derived a contradiction by considering the periodicity and the pigeonhole principle on the coloring of the strips. Therefore, our assumption must be false, and such a rectangle must exist.

💡Okay, so I have this problem where I need to find the shortest distance between two points, R and S, which lie on two different lines in 3D space. The lines are given in parametric form, so I can write R and S in terms of parameters u and v, respectively. First, let me write down the parametric equations for both lines to get a clear picture.For point R on the first line:- The starting point is (2, -3, 5).- The direction vector is (3, -1, 2).So, R can be expressed as:R = (2 + 3u, -3 - u, 5 + 2u)Similarly, for point S on the second line:- The starting point is (1, 2, 0).- The direction vector is (1, 3, -1).So, S can be expressed as:S = (1 + v, 2 + 3v, 0 - v) = (1 + v, 2 + 3v, -v)Now, I need to find the shortest distance between R and S. The distance between two points in 3D space is given by the distance formula:Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]But since I want to minimize this distance, it's often easier to work with the square of the distance to avoid dealing with the square root, which complicates differentiation. So, I'll define RS² as the square of the distance between R and S.Let me compute RS² step by step.First, compute the differences in each coordinate:Δx = (2 + 3u) - (1 + v) = 2 + 3u - 1 - v = 1 + 3u - vΔy = (-3 - u) - (2 + 3v) = -3 - u - 2 - 3v = -5 - u - 3vΔz = (5 + 2u) - (-v) = 5 + 2u + vNow, square each of these differences:(Δx)² = (1 + 3u - v)²(Δy)² = (-5 - u - 3v)²(Δz)² = (5 + 2u + v)²So, RS² = (1 + 3u - v)² + (-5 - u - 3v)² + (5 + 2u + v)²Now, I need to expand each of these squared terms.Let's start with (1 + 3u - v)²:= (1)^2 + (3u)^2 + (-v)^2 + 2*(1)*(3u) + 2*(1)*(-v) + 2*(3u)*(-v)= 1 + 9u² + v² + 6u - 2v - 6uvNext, (-5 - u - 3v)²:= (-5)^2 + (-u)^2 + (-3v)^2 + 2*(-5)*(-u) + 2*(-5)*(-3v) + 2*(-u)*(-3v)= 25 + u² + 9v² + 10u + 30v + 6uvLastly, (5 + 2u + v)²:= (5)^2 + (2u)^2 + (v)^2 + 2*(5)*(2u) + 2*(5)*(v) + 2*(2u)*(v)= 25 + 4u² + v² + 20u + 10v + 4uvNow, let's add all these expanded terms together to get RS².First, list all the terms:From (1 + 3u - v)²:1 + 9u² + v² + 6u - 2v - 6uvFrom (-5 - u - 3v)²:25 + u² + 9v² + 10u + 30v + 6uvFrom (5 + 2u + v)²:25 + 4u² + v² + 20u + 10v + 4uvNow, combine like terms:Constants:1 + 25 + 25 = 51u² terms:9u² + u² + 4u² = 14u²v² terms:v² + 9v² + v² = 11v²uv terms:-6uv + 6uv + 4uv = 4uvu terms:6u + 10u + 20u = 36uv terms:-2v + 30v + 10v = 38vSo, putting it all together, RS² = 14u² + 11v² + 4uv + 36u + 38v + 51Wait, hold on, that doesn't seem to match the initial expansion. Let me double-check my addition.Wait, in the initial expansion, I think I might have made a mistake in combining the terms. Let me go back.Wait, when I expanded (1 + 3u - v)², I got 1 + 9u² + v² + 6u - 2v - 6uvFrom (-5 - u - 3v)², I got 25 + u² + 9v² + 10u + 30v + 6uvFrom (5 + 2u + v)², I got 25 + 4u² + v² + 20u + 10v + 4uvNow, adding them up:Constants: 1 + 25 + 25 = 51u² terms: 9u² + u² + 4u² = 14u²v² terms: v² + 9v² + v² = 11v²uv terms: -6uv + 6uv + 4uv = 4uvu terms: 6u + 10u + 20u = 36uv terms: -2v + 30v + 10v = 38vSo, RS² = 14u² + 11v² + 4uv + 36u + 38v + 51Wait, but in the initial problem statement, the user had a different expansion, which resulted in 35u² + 54uv + 14v² + 32u + 8v + 51. So, I must have made a mistake in my calculation.Let me check each expansion again.First, (1 + 3u - v)²:= (1)^2 + (3u)^2 + (-v)^2 + 2*(1)*(3u) + 2*(1)*(-v) + 2*(3u)*(-v)= 1 + 9u² + v² + 6u - 2v - 6uvThat seems correct.Next, (-5 - u - 3v)²:= (-5)^2 + (-u)^2 + (-3v)^2 + 2*(-5)*(-u) + 2*(-5)*(-3v) + 2*(-u)*(-3v)= 25 + u² + 9v² + 10u + 30v + 6uvThat also seems correct.Lastly, (5 + 2u + v)²:= (5)^2 + (2u)^2 + (v)^2 + 2*(5)*(2u) + 2*(5)*(v) + 2*(2u)*(v)= 25 + 4u² + v² + 20u + 10v + 4uvThat also seems correct.Now, adding all terms:Constants: 1 + 25 + 25 = 51u²: 9 + 1 + 4 = 14v²: 1 + 9 + 1 = 11uv: -6 + 6 + 4 = 4u: 6 + 10 + 20 = 36v: -2 + 30 + 10 = 38So, RS² = 14u² + 11v² + 4uv + 36u + 38v + 51But the user had 35u² + 54uv + 14v² + 32u + 8v + 51. Hmm, that's different. Maybe I made a mistake in the initial expansion.Wait, let me check the user's expansion:They had:= (3u - v + 1)^2 + (-u - 3v - 5)^2 + (2u + v + 5)^2.Which expands to:9u² - 6uv + v² + 1 + 6u - 2v + u² + 9v² + 30uv + 25 + 25u² + 4v² + 20uv + 20u + 10v + 25.Then combined to 35u² + 54uv + 14v² + 32u + 8v + 51.Wait, but in my calculation, I have 14u² + 11v² + 4uv + 36u + 38v + 51.So, there's a discrepancy here. Maybe the user made a mistake in their expansion, or I did.Wait, let's see:In the user's expansion, they have:(3u - v + 1)^2 = 9u² + v² + 1 - 6uv + 6u - 2v(-u - 3v -5)^2 = u² + 9v² + 25 + 6uv + 10u + 30v(2u + v +5)^2 = 4u² + v² + 25 + 4uv + 20u + 10vAdding these together:9u² + v² + 1 -6uv +6u -2v + u² +9v² +25 +6uv +10u +30v +4u² +v² +25 +4uv +20u +10vNow, combine like terms:u²: 9 +1 +4 =14v²:1 +9 +1=11uv: -6 +6 +4=4u:6 +10 +20=36v:-2 +30 +10=38Constants:1 +25 +25=51So, RS²=14u² +11v² +4uv +36u +38v +51Which matches my calculation, not the user's. So, the user must have made a mistake in their expansion.Therefore, I think my expansion is correct.So, RS²=14u² +11v² +4uv +36u +38v +51Now, to find the minimum distance, I need to find the values of u and v that minimize RS².To do this, I can take partial derivatives with respect to u and v, set them equal to zero, and solve the resulting system of equations.First, compute the partial derivative of RS² with respect to u:∂(RS²)/∂u = 28u + 4v + 36Similarly, compute the partial derivative with respect to v:∂(RS²)/∂v = 22v + 4u + 38Set both partial derivatives equal to zero:28u + 4v + 36 = 0 ...(1)4u + 22v + 38 = 0 ...(2)Now, we have a system of two linear equations:28u + 4v = -364u + 22v = -38Let me write them as:28u + 4v = -36 ...(1)4u + 22v = -38 ...(2)To solve this system, I can use the method of elimination or substitution. Let me use elimination.First, let's simplify equation (1) by dividing all terms by 4:7u + v = -9 ...(1a)Similarly, equation (2) can be simplified by dividing by 2:2u + 11v = -19 ...(2a)Now, from equation (1a), we can express v in terms of u:v = -9 -7uNow, substitute this expression for v into equation (2a):2u + 11*(-9 -7u) = -19Compute:2u -99 -77u = -19Combine like terms:(2u -77u) + (-99) = -19-75u -99 = -19Now, add 99 to both sides:-75u = -19 +99-75u =80Divide both sides by -75:u =80 / (-75) = -16/15So, u= -16/15Now, substitute u back into equation (1a) to find v:v = -9 -7*(-16/15) = -9 + 112/15Convert -9 to fifteenths:-9 = -135/15So, v= -135/15 +112/15= (-135 +112)/15= (-23)/15So, v= -23/15Now, we have u= -16/15 and v= -23/15Now, let's compute RS² using these values.But before that, let me verify if these values satisfy both equations.From equation (1a):7u + v =7*(-16/15) + (-23/15)= (-112/15) + (-23/15)= (-135)/15= -9, which matches.From equation (2a):2u +11v=2*(-16/15) +11*(-23/15)= (-32/15) + (-253/15)= (-285)/15= -19, which also matches.So, the values are correct.Now, let's compute RS².RS²=14u² +11v² +4uv +36u +38v +51First, compute each term:u= -16/15, v= -23/15Compute u²:u²= (256)/(225)v²= (529)/(225)uv= (368)/(225)Now, compute each term:14u²=14*(256/225)=3584/22511v²=11*(529/225)=5819/2254uv=4*(368/225)=1472/22536u=36*(-16/15)= -576/15= -38.438v=38*(-23/15)= -874/15≈-58.266751=51Now, let's compute each term as fractions to avoid decimal approximations.36u=36*(-16/15)= (-576)/15= (-192)/538v=38*(-23/15)= (-874)/15So, RS²=3584/225 +5819/225 +1472/225 + (-192)/5 + (-874)/15 +51First, combine the fractions with denominator 225:3584 +5819 +1472= 3584+5819=9403+1472=10875So, 10875/225=48.333...=48 and 1/3Now, the other terms:-192/5= -38.4-874/15≈-58.266751=51Now, convert all to fractions with denominator 15 to combine:48 and 1/3= 145/3=725/15-38.4= -384/10= -192/5= -576/15-58.2667= -874/1551=765/15Now, RS²=725/15 -576/15 -874/15 +765/15Combine numerators:725 -576 -874 +765= (725 +765) - (576 +874)=1490 -1450=40So, RS²=40/15=8/3≈2.6667Wait, that can't be right because RS² should be a positive number, but 8/3 is approximately 2.6667, which is positive, but let's verify the calculations.Wait, I think I made a mistake in converting 48 and 1/3 to fifteenths.48 and 1/3=48 +1/3= (48*3 +1)/3=145/3145/3= (145*5)/15=725/15Similarly, -192/5= (-192*3)/15= -576/15-874/15 remains as is.51=51*15/15=765/15So, RS²=725/15 -576/15 -874/15 +765/15Now, compute numerator:725 -576=149149 -874= -725-725 +765=40So, RS²=40/15=8/3Therefore, RS= sqrt(8/3)= (2*sqrt(6))/3≈1.63299Wait, but in the initial problem, the user had RS²=10, leading to RS=√10≈3.1623. But according to my calculations, RS²=8/3, so RS=2√6/3≈1.63299.This suggests that either I made a mistake in my calculations, or the user's initial expansion was incorrect.Wait, let me double-check the RS² calculation.RS²=14u² +11v² +4uv +36u +38v +51u= -16/15, v= -23/15Compute each term:14u²=14*(256/225)=3584/22511v²=11*(529/225)=5819/2254uv=4*(368/225)=1472/22536u=36*(-16/15)= -576/15= -38.438v=38*(-23/15)= -874/15≈-58.266751=51Now, let's compute RS²:3584/225 +5819/225 +1472/225 + (-576/15) + (-874/15) +51First, sum the fractions with denominator 225:3584 +5819 +1472=10875So, 10875/225=48.333...=48 and 1/3Now, sum the fractions with denominator 15:-576/15 -874/15= (-576 -874)/15= (-1450)/15= -96.666...Now, add the constants:48.333... -96.666... +51Compute:48.333 -96.666= -48.333-48.333 +51=2.666...Which is 8/3.So, RS²=8/3, so RS=√(8/3)=2√6/3≈1.63299But the user had RS²=10, leading to RS=√10≈3.1623.This suggests that either the user made a mistake in their expansion, or I did.Wait, let me check the initial parametrization.Wait, in the user's initial problem, they wrote R as (3u +2, -u -3, 2u +5), which is correct.Similarly, S as (v +1, 3v +2, -v), which is also correct.Then, they computed RS² as ((3u +2) - (v +1))² + ((-u -3) - (3v +2))² + ((2u +5) - (-v))²Which is correct.Then, they expanded each term:(3u -v +1)^2 + (-u -3v -5)^2 + (2u +v +5)^2Which is correct.Then, they expanded each term:=9u² -6uv +v² +1 +6u -2v +u² +9v² +30uv +25 +25u² +4v² +20uv +20u +10v +25Wait, let's check this expansion.First term: (3u -v +1)^2=9u² +v² +1 -6uv +6u -2vSecond term: (-u -3v -5)^2= u² +9v² +25 +6uv +10u +30vThird term: (2u +v +5)^2=4u² +v² +25 +4uv +20u +10vNow, adding all terms:9u² +v² +1 -6uv +6u -2v +u² +9v² +25 +6uv +10u +30v +4u² +v² +25 +4uv +20u +10vNow, combine like terms:u²:9 +1 +4=14v²:1 +9 +1=11uv:-6 +6 +4=4u:6 +10 +20=36v:-2 +30 +10=38Constants:1 +25 +25=51So, RS²=14u² +11v² +4uv +36u +38v +51Which is what I had.But the user had:=9u² -6uv +v² +1 +6u -2v +u² +9v² +30uv +25 +25u² +4v² +20uv +20u +10v +25Which seems to have an error in the coefficients.Wait, in the user's expansion, they have:=9u² -6uv +v² +1 +6u -2v +u² +9v² +30uv +25 +25u² +4v² +20uv +20u +10v +25Wait, the third term is (2u +v +5)^2, which expands to 4u² +v² +25 +4uv +20u +10vBut in the user's expansion, they have 25u², which is incorrect. It should be 4u².So, the user made a mistake in expanding (2u +v +5)^2 as 25u² instead of 4u².That explains the discrepancy.Therefore, my expansion is correct, and the user's initial expansion was wrong.So, proceeding with my correct expansion, RS²=14u² +11v² +4uv +36u +38v +51We found u= -16/15 and v= -23/15So, RS²=8/3, so RS=2√6/3But let me verify this by plugging u and v back into the distance formula.Compute R and S with u= -16/15 and v= -23/15Compute R:R= (2 +3u, -3 -u,5 +2u)= (2 +3*(-16/15), -3 -(-16/15),5 +2*(-16/15))Compute each coordinate:x=2 + (-48/15)=2 - 16/5= (10/5 -16/5)= -6/5y= -3 +16/15= (-45/15 +16/15)= -29/15z=5 + (-32/15)= (75/15 -32/15)=43/15So, R=(-6/5, -29/15,43/15)Compute S:S=(1 +v,2 +3v, -v)= (1 + (-23/15),2 +3*(-23/15), -(-23/15))Compute each coordinate:x=1 -23/15= (15/15 -23/15)= -8/15y=2 -69/15= (30/15 -69/15)= -39/15= -13/5z=23/15So, S=(-8/15, -13/5,23/15)Now, compute RS vector:RS= S - R= [(-8/15 - (-6/5)), (-13/5 - (-29/15)), (23/15 -43/15)]Compute each component:x: -8/15 +6/5= -8/15 +18/15=10/15=2/3y: -13/5 +29/15= (-39/15 +29/15)= -10/15= -2/3z:23/15 -43/15= -20/15= -4/3So, RS vector=(2/3, -2/3, -4/3)Now, compute the magnitude of RS:|RS|=sqrt[(2/3)^2 + (-2/3)^2 + (-4/3)^2]=sqrt[(4/9)+(4/9)+(16/9)]=sqrt[24/9]=sqrt[8/3]=2√6/3Which matches our earlier result.Therefore, the minimal distance RS is 2√6/3.But wait, the user had RS²=10, leading to RS=√10≈3.1623, but according to my calculations, it's 2√6/3≈1.63299.So, there's a significant difference. Therefore, I think the user made a mistake in their initial expansion, leading them to an incorrect answer.Therefore, the correct minimal distance is 2√6/3.But let me check once more.Alternatively, another method to find the minimal distance between two skew lines is to use the formula:Distance= |(P2 - P1) · (d1 × d2)| / |d1 × d2|Where P1 and P2 are points on the two lines, and d1 and d2 are their direction vectors.Let me try this method to confirm.Given:Line 1: P1=(2,-3,5), direction vector d1=(3,-1,2)Line 2: P2=(1,2,0), direction vector d2=(1,3,-1)Compute vector P2 - P1= (1-2,2-(-3),0-5)=(-1,5,-5)Compute cross product d1 × d2:|i   j   k||3   -1   2||1   3   -1|= i[(-1)(-1) - (2)(3)] - j[(3)(-1) - (2)(1)] + k[(3)(3) - (-1)(1)]= i[1 -6] - j[-3 -2] + k[9 +1]= (-5)i - (-5)j +10k= (-5,5,10)Compute |d1 × d2|=sqrt[(-5)^2 +5^2 +10^2]=sqrt[25 +25 +100]=sqrt[150]=5√6Now, compute (P2 - P1) · (d1 × d2)= (-1,5,-5) · (-5,5,10)= (-1)*(-5) +5*5 + (-5)*10=5 +25 -50= -20Take absolute value: | -20 | =20Therefore, Distance=20 / (5√6)=4/√6= (4√6)/6= (2√6)/3Which matches our previous result.Therefore, the minimal distance is indeed 2√6/3.So, the user's initial answer of √10 was incorrect due to an error in expansion.Therefore, the correct minimal distance is 2√6/3.

💡First, I need to calculate the distance Amanda and her friends traveled during each segment of their trip.For the first part, they traveled at an average speed of 40 miles per hour for 1 hour and 30 minutes. Converting 1 hour and 30 minutes to hours gives 1.5 hours. Multiplying the speed by the time gives the distance for this segment.Next, after the first stop, they continued at an average speed of 50 miles per hour for 1 hour. Multiplying these values directly provides the distance for the second segment.After the second stop, they traveled at an average speed of 30 miles per hour for 2 hours and 15 minutes. Converting 2 hours and 15 minutes to hours results in 2.25 hours. Multiplying the speed by this time gives the distance for the final segment.Finally, I will add up the distances from all three segments to find the total distance Amanda and her friends covered to reach Super Fun-tastic Land.