🧠:Okay, let's try to tackle this geometry problem. It's about a right triangle with squares constructed on the legs, and projections of certain points onto the extension of the hypotenuse. The claims are that the distances from these projections to the vertices of the hypotenuse are equal, and the sum of the lengths of the segments EE' and FF' equals the length of the hypotenuse BC. Hmm, this seems a bit abstract. Let me start by visualizing the problem.First, we have a right triangle ABC with the right angle at A. So, angle A is 90 degrees. Then, squares are constructed on the legs AB and AC. The outermost points of these squares are labeled E and F. I need to make sure I understand where E and F are located. If we construct a square on AB, which is one of the legs, then E is probably the vertex of the square that's not on the triangle. Similarly for F on the other leg AC. Let me sketch this mentally.Let’s denote the triangle with right angle at A, so sides AB and AC are the legs, and BC is the hypotenuse. Constructing squares on AB and AC would mean that on AB, the square is either extending outward from the triangle or inward. Since E and F are outermost points, they must be on the squares constructed externally. So, for the square on AB, starting from point A, moving along AB, then turning 90 degrees away from the triangle to construct the square. Similarly for AC. Therefore, point E would be the corner of the square on AB that's not on AB, and similarly F for AC.Now, the projections of E and F onto the extension of the hypotenuse BC are E' and F'. Projection here likely means orthogonal projection. So, we need to drop perpendiculars from E and F onto the line that extends BC, and the feet of these perpendiculars are E' and F'. The first claim is that E'B equals CF', and the second is that EE' + FF' equals BC.Let me try to break this down step by step. Maybe coordinate geometry would help here. Assign coordinates to the triangle to make calculations easier.Let's place point A at the origin (0,0). Let’s let AB be along the x-axis and AC along the y-axis. Suppose AB has length c and AC has length b. Then coordinates would be:- A: (0, 0)- B: (c, 0)- C: (0, b)Then BC is the hypotenuse from (c, 0) to (0, b). The equation of BC can be found. The slope of BC is (b - 0)/(0 - c) = -b/c, so the equation is y = (-b/c)x + b.Now, constructing squares on AB and AC. Let's start with the square on AB. AB goes from (0,0) to (c,0). To construct the square outward from the triangle, we can go from B (c,0) up or down. Since the square is constructed on AB, which is a leg, and we need the outermost point. If we construct the square outside the triangle, then from AB, moving upwards (since the triangle is in the first quadrant, constructing the square outward would be in the positive y direction from AB). Wait, AB is along the x-axis from (0,0) to (c,0). To construct a square on AB, the square can be constructed either above or below AB. Since the problem mentions "outermost points", I think it's the direction away from the triangle. Since the triangle is in the first quadrant with right angle at A, the square on AB would be constructed above AB (if we consider the usual orientation). Similarly, the square on AC would be constructed to the right of AC.Wait, perhaps another way. Let's be precise. The square on AB: AB is the base. Starting at A (0,0), moving to B (c,0). The square can be constructed by moving upward from AB. So the next vertex after B would be (c, c), but wait, AB has length c, so moving up c units? Wait, no. The side length of the square is equal to the length of AB, which is c. So, if AB is along the x-axis from (0,0) to (c,0), then the square constructed on AB would have vertices at A (0,0), B (c,0), (c, c), and (0, c). But then the outermost point E would be (c, c). Similarly, the square on AC: AC is from (0,0) to (0, b). The square constructed on AC would go to the right. So the vertices would be A (0,0), C (0, b), (-b, b), and (-b, 0). Wait, but that's to the left. Hmm, maybe the direction matters. Wait, if the square is constructed on AC as a side, then starting at A (0,0) up to C (0, b), then turning right (since it's a square) would be in the positive x direction. Wait, but from AC, which is vertical, constructing the square to the right would mean moving in the positive x direction. So from C (0, b), moving right along the x-axis for length b. So the next vertex would be (b, b), then down to (b, 0), then back to A? Wait, but that would overlap with the square on AB. Hmm, perhaps not. Wait, maybe the squares are constructed externally, so they don't overlap with the triangle. Let me think again.Alternatively, maybe the squares are built on the legs extended. For AB, which is from A to B (c,0), constructing the square outward would mean extending AB beyond B and constructing the square. Wait, but the problem says "constructed on the legs", so the squares have the legs as one of their sides. Therefore, each square shares the entire side with the leg. So for AB, the square has AB as one side, and is constructed outside the triangle. Since the triangle is right-angled at A, the squares on AB and AC would be constructed in such a way that they are on the outside of the triangle. So, for AB, starting at A (0,0) to B (c,0). The square constructed on AB would extend downward, because upward would be towards the triangle. Wait, the triangle is in the first quadrant with A at (0,0), B at (c,0), and C at (0, b). So, constructing the square on AB outside the triangle would be below AB, but since AB is along the x-axis from (0,0) to (c,0), constructing the square below AB would have negative y coordinates. But the problem mentions "outermost points", which might refer to the direction away from the triangle. Alternatively, maybe the squares are constructed in the plane of the triangle. Wait, this is getting confusing. Maybe I should assign coordinates to E and F based on the squares.Wait, let's take AB as the base from (0,0) to (c,0). To construct a square on AB, outside the triangle. If the triangle is in the first quadrant, then outside could be either above or below. But since the square is constructed on the leg, which is part of the triangle, maybe the square is constructed in such a way that it's not overlapping with the triangle. So if AB is along the x-axis, then constructing the square above AB (in the positive y direction) would be outside the triangle. Wait, but the triangle is already in the first quadrant, with AC going up the y-axis. If we construct the square on AB upwards, then the square would be in the first quadrant, but not overlapping with the triangle because the triangle's other leg is along the y-axis. Hmm, actually, if AB is from (0,0) to (c,0), and we construct the square above AB, the square would have vertices at (0,0), (c,0), (c, c), (0, c). But point (0, c) would be somewhere along the y-axis above A, but since AC has length b, unless c = b, which is not necessarily the case. Wait, this might not be the right approach.Alternatively, perhaps the squares are constructed on each leg such that each square shares the entire leg as one of its sides. So for AB, the square would have AB as its base, and extend outward perpendicular to AB. Since AB is horizontal, the square would extend either upwards or downwards. Similarly, AC is vertical, so the square on AC would extend either left or right. To determine the direction, "outermost" points E and F. If we consider the triangle in standard position, then the outermost direction for AB would be upwards (if the triangle is considered as lying below the squares), or maybe the direction away from the hypotenuse. This is a bit ambiguous. Wait, maybe it's better to define coordinates for E and F.Let me try again. Let's place the right-angled triangle at A(0,0), B(c,0), and C(0,b). Now, constructing a square on AB. The square should have AB as one of its sides. Let's construct it outward from the triangle. If the triangle is in the first quadrant, then outward from the triangle along AB would mean constructing the square in the direction away from the y-axis. Since AB is along the x-axis from (0,0) to (c,0), constructing the square outward (away from the triangle) would be in the negative y direction. Wait, but that's towards the negative y-axis, which is outside the first quadrant. Alternatively, maybe outward is the direction that's not towards the interior of the triangle. Since the triangle is in the first quadrant, the square on AB could be constructed above AB (positive y direction) or below (negative y). Similarly for AC.But the problem mentions "outermost" points E and F. If we construct the square on AB above AB, then the outermost point would be the farthest from AB, which would be the upper vertex of the square. Similarly, for AC, constructing the square to the right of AC, the outermost point would be the rightmost vertex. Wait, maybe that's the case. Let's assume that.So for AB, constructing the square above AB (positive y direction). The square has vertices at A(0,0), B(c,0), (c, c), and (0, c). So the outermost point E is (c, c). Wait, but the length of AB is c, so the square should have sides of length c. But if AB is length c, then moving up c units from B would get to (c, c). Similarly, from A(0,0) moving up c units would get to (0, c). Then connecting those gives the square. But in this case, the square on AB would extend into the first quadrant, but if the triangle has vertex C at (0, b), unless b = c, the square doesn't interfere with the triangle.Similarly, constructing the square on AC. AC is from A(0,0) to C(0, b). Constructing the square to the right of AC (positive x direction). The square would have vertices at A(0,0), C(0, b), (b, b), and (b, 0). So the outermost point F is (b, b). But wait, the length of AC is b, so the square on AC would have sides of length b. So starting at A(0,0), moving up to C(0, b), then right to (b, b), then down to (b, 0), and back to A. But then the square on AC would overlap with the square on AB if b = c, but in general, if b ≠ c, they don't overlap.But in this configuration, point E is (c, c) and point F is (b, b). Now, we need to project E and F onto the extension of the hypotenuse BC. Let's find the equation of BC first. As before, points B(c,0) and C(0, b). The slope of BC is (b - 0)/(0 - c) = -b/c. So the equation is y = (-b/c)x + b.Now, we need to find the projections E' and F' of E and F onto the extension of BC. Projection means the orthogonal projection, so we need to drop perpendiculars from E and F onto the line BC (or its extension) and find where they land.First, let's find E'. E is (c, c). The line BC has slope -b/c, so the perpendicular to BC will have slope c/b. The line perpendicular to BC passing through E(c, c) is y - c = (c/b)(x - c). We need to find the intersection point E' between this perpendicular and the line BC.The line BC is y = (-b/c)x + b. The perpendicular from E is y = (c/b)(x - c) + c. Let's set them equal:(-b/c)x + b = (c/b)(x - c) + cMultiply both sides by bc to eliminate denominators:-b^2 x + b^2 c = c^2 (x - c) + b c^2Wait, let's compute step by step:Left side: (-b/c)x + bRight side: (c/b)(x - c) + cMultiply both sides by bc:Left side: -b^2 x + b^2 cRight side: c^2 (x - c) + b c^2Wait, expanding right side:(c^2)x - c^3 + b c^2So:-b^2 x + b^2 c = c^2 x - c^3 + b c^2Bring all terms to left side:-b^2 x + b^2 c - c^2 x + c^3 - b c^2 = 0Factor x terms:x(-b^2 - c^2) + (b^2 c + c^3 - b c^2) = 0Factor out c in the constants:x(- (b^2 + c^2)) + c(b^2 + c^2 - b c) = 0Hmm, this seems complicated. Let me check my calculations again.Starting over:Equation of BC: y = (-b/c)x + bEquation of perpendicular from E(c, c): y = (c/b)(x - c) + cSet equal:(-b/c)x + b = (c/b)x - (c^2)/b + cMultiply all terms by bc to eliminate denominators:- b^2 x + b^2 c = c^2 x - c^3 + b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c + c^3 - b c^2 = 0Factor x: -x(b^2 + c^2) + c(b^2 + c^2 - b c) = 0Thus:x = [c(b^2 + c^2 - b c)] / (b^2 + c^2)Similarly, once x is found, substitute back into BC's equation to find y.But this seems messy. Maybe there's a smarter way. Alternatively, maybe using vectors or parametric equations.Alternatively, since projection involves some formula. The formula for the projection of a point (x0, y0) onto the line ax + by + c = 0 is:x' = (b(bx0 - ay0) - ac) / (a^2 + b^2)y' = (a(-bx0 + ay0) - bc) / (a^2 + b^2)But maybe it's better to use parametric coordinates.Alternatively, let me parametrize the line BC. Let’s say a point on BC can be written as B + t*(C - B) = (c,0) + t*(-c, b), where t is a real number. So parametric equations:x = c - c ty = 0 + b tWe need to find t such that the vector from E(c, c) to (x, y) is perpendicular to BC. The direction vector of BC is (-c, b). So the vector from E to (x, y) is (x - c, y - c) = (-c t - c, b t - c). This vector must be perpendicular to (-c, b), so their dot product is zero:(-c t - c)(-c) + (b t - c)(b) = 0Compute:(c t + c)c + (b t - c)b = 0c^2 t + c^2 + b^2 t - b c = 0t(c^2 + b^2) + c^2 - b c = 0Therefore:t = (b c - c^2) / (b^2 + c^2)So t = c(b - c)/(b^2 + c^2)Then the coordinates of E' are:x = c - c t = c - c*(c(b - c)/(b^2 + c^2)) = c - (c^2(b - c))/(b^2 + c^2)Similarly,y = b t = b*(c(b - c)/(b^2 + c^2)) = (b c (b - c))/(b^2 + c^2)Simplify x:x = c*( (b^2 + c^2) - c(b - c) ) / (b^2 + c^2 )= c*(b^2 + c^2 - c b + c^2 ) / (b^2 + c^2 )= c*(b^2 - c b + 2 c^2 ) / (b^2 + c^2 )Wait, this seems complicated. Maybe I made a miscalculation.Wait, x = c - (c^2 (b - c))/(b^2 + c^2 )= c*( (b^2 + c^2 ) - c(b - c) ) / (b^2 + c^2 )= c*( b^2 + c^2 - c b + c^2 ) / (b^2 + c^2 )= c*( b^2 - c b + 2 c^2 ) / (b^2 + c^2 )Hmm. Similarly for y:y = (b c (b - c))/(b^2 + c^2 )Not sure if this helps. Let me compute E'B and see if it relates to CF'.Point B is at (c, 0). E' is at ( x, y ). The distance E'B is sqrt( (x - c)^2 + (y - 0)^2 )But since E' is on the line BC, which is parametrized as t going from 0 to 1 for the segment BC. However, since E' is the projection, t could be outside [0,1], meaning E' is on the extension of BC beyond B or C.Wait, in our parametrization earlier, t = 0 corresponds to B (c,0) and t = 1 corresponds to C (0,b). So if t is negative, we are extending beyond B, and if t > 1, beyond C.In our calculation, t = c(b - c)/(b^2 + c^2). Depending on the values of b and c, t could be positive or negative. For example, if b > c, then t is positive, but numerator is c(b - c), denominator positive. If b < c, t is negative.But in any case, E' is the projection of E onto BC (or its extension). Similarly for F'.Now, let's compute E'B. Since E' is on the line BC, which is parametrized from B to C, the distance E'B can be calculated as the absolute value of the parameter t multiplied by the length of BC. Wait, in the parametrization earlier, the direction vector is (-c, b), so each unit of t corresponds to moving by (-c, b). But the length of BC is sqrt(c^2 + b^2). Therefore, the distance from B to E' would be |t| * sqrt(c^2 + b^2). But t was calculated as t = c(b - c)/(b^2 + c^2). Therefore, E'B = |t| * sqrt(b^2 + c^2) = |c(b - c)/(b^2 + c^2)| * sqrt(b^2 + c^2) = |c(b - c)| / sqrt(b^2 + c^2 )Similarly, we need to compute CF'. Let's perform similar steps for point F.Point F is the outermost point of the square on AC. If AC is from A(0,0) to C(0, b), constructing the square outward. If outward is to the right, then the square on AC would have vertices at A(0,0), C(0, b), (b, b), (b, 0). Therefore, point F is (b, b). Wait, but the length of AC is b, so the square has sides of length b. So moving from C(0, b) to the right b units gives (b, b). Then down to (b, 0), then left to (0,0). So yes, F is (b, b).Now, projecting F(b, b) onto the extension of BC. Let's find F'.Using the same method as before. The line BC is y = (-b/c)x + b. The perpendicular from F(b, b) to BC will have slope c/b, same as before. So the equation of the perpendicular is y - b = (c/b)(x - b)Set this equal to BC's equation:(-b/c)x + b = (c/b)(x - b) + bSimplify:(-b/c)x + b = (c/b)x - (c^2)/b + bSubtract b from both sides:(-b/c)x = (c/b)x - (c^2)/bMultiply all terms by bc to eliminate denominators:- b^2 x = c^2 x - c^3Bring all terms to left:- b^2 x - c^2 x + c^3 = 0Factor x:- x(b^2 + c^2 ) + c^3 = 0Thus:x = c^3 / (b^2 + c^2 )Then substitute back into BC's equation to find y:y = (-b/c)(c^3 / (b^2 + c^2 )) + b = (-b c^2 / (b^2 + c^2 )) + b = b - (b c^2)/(b^2 + c^2 ) = b ( (b^2 + c^2 ) - c^2 ) / (b^2 + c^2 ) = b^3 / (b^2 + c^2 )Therefore, F' has coordinates ( c^3 / (b^2 + c^2 ), b^3 / (b^2 + c^2 ) )Now, compute CF'. Point C is at (0, b). The distance CF' is sqrt( (c^3 / (b^2 + c^2 ) - 0 )^2 + ( b^3 / (b^2 + c^2 ) - b )^2 )Simplify:First component: c^3 / (b^2 + c^2 )Second component: (b^3 - b(b^2 + c^2 )) / (b^2 + c^2 ) = (b^3 - b^3 - b c^2 ) / (b^2 + c^2 ) = (-b c^2 ) / (b^2 + c^2 )Therefore, CF' = sqrt( [c^3 / (b^2 + c^2 )]^2 + [ -b c^2 / (b^2 + c^2 ) ]^2 )Factor out [c^2 / (b^2 + c^2 )]^2:sqrt( c^2 (c^2 + b^2 ) / (b^2 + c^2 )^2 )Wait, let's compute:[c^3 / D]^2 + [ -b c^2 / D ]^2 = c^6 / D^2 + b^2 c^4 / D^2 = c^4 (c^2 + b^2 ) / D^2, where D = b^2 + c^2So CF' = sqrt( c^4 (b^2 + c^2 ) / (b^2 + c^2 )^2 ) = sqrt( c^4 / (b^2 + c^2 )) = c^2 / sqrt(b^2 + c^2 )Similarly, earlier for E'B, we had E'B = |c(b - c)| / sqrt(b^2 + c^2 )Wait, but according to the problem statement, E'B = CF'So if E'B = |c(b - c)| / sqrt(b^2 + c^2 ) and CF' = c^2 / sqrt(b^2 + c^2 ), then setting them equal:|c(b - c)| = c^2Assuming all lengths are positive, we can drop the absolute value by considering cases.Case 1: b >= c, so |c(b - c)| = c(b - c). Then c(b - c) = c^2 => b - c = c => b = 2cCase 2: b < c, so |c(b - c)| = c(c - b). Then c(c - b) = c^2 => c - b = c => b = 0, which is impossible since b is the length of AC.Therefore, only possible if b = 2c. But this contradicts the generality of the problem, which should hold for any right triangle. Therefore, there must be a mistake in my calculations.Wait, this suggests that the equality E'B = CF' only holds when b = 2c, but the problem states it as a general fact. Hence, my approach must have an error.Let me go back and check.First, coordinates:A(0,0), B(c,0), C(0,b)Square on AB: constructed externally, which direction? If AB is along x-axis, constructing the square above AB (assuming external means away from the triangle which is in the first quadrant). So square on AB would have vertices at A(0,0), B(c,0), (c, c), (0, c). Thus, outermost point E is (c, c).Square on AC: constructed externally. AC is along y-axis. So square on AC would have vertices at A(0,0), C(0, b), (-b, b), (-b, 0). Therefore, outermost point F is (-b, b). Wait, this is different from my previous assumption. If we construct the square externally on AC, it would extend to the left (negative x direction), making F at (-b, b). Whereas before I thought it extended to the right, but that might be conflicting with the square on AB.Ah, this is the mistake! The squares are constructed on the legs, but the direction of construction depends on the orientation. If the square on AB is constructed upwards (positive y), then the square on AC should be constructed to the left (negative x) to be external. Therefore, point E is (c, c) and point F is (-b, b). This makes sense because constructing the squares outside the triangle, which is in the first quadrant.Therefore, my previous coordinates for F were wrong. I assumed F was (b, b), but actually, it should be (-b, b). That changes things.Let me recast everything with F at (-b, b).First, recalculate projection of E(c, c) onto BC:Line BC: y = (-b/c)x + bPerpendicular from E(c, c): slope c/bEquation: y - c = (c/b)(x - c)Intersection with BC:(-b/c)x + b = (c/b)x - (c^2)/b + cMultiply both sides by bc:- b^2 x + b^2 c = c^2 x - c^3 + b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c + c^3 - b c^2 = 0Factor x: -x(b^2 + c^2 ) + c(b^2 + c^2 - b c) = 0Thus, x = c(b^2 + c^2 - b c)/(b^2 + c^2 )Then y = (-b/c)x + b = (-b/c)*[c(b^2 + c^2 - b c)/(b^2 + c^2 )] + b = -b*(b^2 + c^2 - b c)/(b^2 + c^2 ) + b = [ -b(b^2 + c^2 - b c) + b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ -b^3 - b c^2 + b^2 c + b^3 + b c^2 ] / (b^2 + c^2 ) = (b^2 c ) / (b^2 + c^2 )So E' is at ( c(b^2 + c^2 - b c)/(b^2 + c^2 ), b^2 c / (b^2 + c^2 ) )Now, compute E'B. Point B is at (c, 0). Distance E'B:sqrt( [ c(b^2 + c^2 - b c)/(b^2 + c^2 ) - c ]^2 + [ b^2 c / (b^2 + c^2 ) - 0 ]^2 )Simplify x-component:c [ (b^2 + c^2 - b c ) - (b^2 + c^2 ) ] / (b^2 + c^2 ) = c [ -b c ] / (b^2 + c^2 ) = -b c^2 / (b^2 + c^2 )Absolute value squared: (b c^2 / (b^2 + c^2 ))^2Y-component squared: (b^2 c / (b^2 + c^2 ))^2Thus, E'B^2 = (b^2 c^4 + b^4 c^2 ) / (b^2 + c^2 )^2 = (b^2 c^2 (c^2 + b^2 )) / (b^2 + c^2 )^2 ) = (b^2 c^2 ) / (b^2 + c^2 )Therefore, E'B = (b c ) / sqrt(b^2 + c^2 )Now, compute CF'. Point F is (-b, b). Projection F' onto BC.Equation of BC: y = (-b/c)x + bSlope of BC: -b/c, so slope of perpendicular is c/bEquation of perpendicular from F(-b, b): y - b = (c/b)(x + b )Set equal to BC's equation:(-b/c)x + b = (c/b)(x + b ) + b - bWait, set equal:(-b/c)x + b = (c/b)(x + b )Multiply both sides by bc:- b^2 x + b^2 c = c^2 (x + b )Expand right side: c^2 x + b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c - b c^2 = 0Factor x: -x(b^2 + c^2 ) + b c (b - c ) = 0Thus,x = [ b c (b - c ) ] / (b^2 + c^2 )Then y = (-b/c)x + b = (-b/c)*(b c (b - c ) / (b^2 + c^2 )) + b = -b^2 (b - c ) / (b^2 + c^2 ) + b = [ -b^3 + b^2 c + b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ -b^3 + b^2 c + b^3 + b c^2 ] / (b^2 + c^2 ) = (b^2 c + b c^2 ) / (b^2 + c^2 ) = b c (b + c ) / (b^2 + c^2 )Thus, F' is at ( [ b c (b - c ) ] / (b^2 + c^2 ), b c (b + c ) / (b^2 + c^2 ) )Now, compute CF'. Point C is at (0, b). Distance CF':sqrt( [ b c (b - c ) / (b^2 + c^2 ) - 0 ]^2 + [ b c (b + c ) / (b^2 + c^2 ) - b ]^2 )Simplify x-component: b c (b - c ) / (b^2 + c^2 )Y-component: [ b c (b + c ) - b (b^2 + c^2 ) ] / (b^2 + c^2 ) = [ b^2 c + b c^2 - b^3 - b c^2 ] / (b^2 + c^2 ) = [ -b^3 + b^2 c ] / (b^2 + c^2 ) = -b^2 (b - c ) / (b^2 + c^2 )Therefore, CF'^2 = [ b c (b - c ) / (b^2 + c^2 ) ]^2 + [ -b^2 (b - c ) / (b^2 + c^2 ) ]^2Factor out [ (b - c )^2 / (b^2 + c^2 )^2 ]:= (b - c )^2 / (b^2 + c^2 )^2 [ b^2 c^2 + b^4 ]= (b - c )^2 (b^2 (c^2 + b^2 )) / (b^2 + c^2 )^2 )= (b^2 (b - c )^2 (b^2 + c^2 )) / (b^2 + c^2 )^2 )= b^2 (b - c )^2 / (b^2 + c^2 )Thus, CF' = b |b - c | / sqrt(b^2 + c^2 )But from earlier, E'B = (b c ) / sqrt(b^2 + c^2 )For E'B to equal CF', we need:(b c ) / sqrt(b^2 + c^2 ) = b |b - c | / sqrt(b^2 + c^2 )Cancel denominators and b (assuming b ≠ 0):c = |b - c |Which gives two cases:1. b - c ≥ 0: c = b - c → 2c = b → b = 2c2. b - c < 0: c = -(b - c ) → c = -b + c → 0 = -b → b = 0, which is impossible.Hence, E'B = CF' only if b = 2c. But the problem states this as a general property for any right triangle. This suggests a contradiction, meaning my calculations are still wrong.Wait, this can't be. The problem must hold generally, so either my coordinates for E and F are incorrect, or my projections are miscalculated.Re-examining the construction of squares on the legs. Maybe the squares are constructed differently. For instance, on AB, instead of constructing the square upwards (positive y) or downwards (negative y), perhaps it's constructed in another direction. Similarly for AC.Wait, perhaps "outermost" refers to the direction away from the hypotenuse BC. To determine that, we need to see the orientation.In triangle ABC with right angle at A, hypotenuse BC. The square on AB should be constructed outward from the triangle, which would be in the direction away from BC. Similarly for AC.To figure out the direction, consider the normal vector pointing outward from the triangle. For AB, which is along the x-axis from A(0,0) to B(c,0), the outward normal would be in the positive y direction (if the triangle is considered as the base AB with height AC upwards). Similarly, for AC, the outward normal would be in the positive x direction. Wait, but if the square on AC is constructed in the positive x direction, then point F would be at (b, b), but that might not be outermost with respect to BC.Alternatively, perhaps "outermost" is determined relative to the hypotenuse BC. The point E is constructed such that it's the farthest from BC among the square's vertices. Similarly for F.Alternatively, maybe the squares are built on the legs extended beyond A or B/C. For example, the square on AB could be constructed by extending AB beyond B and building the square there. But the problem says "constructed on the legs", which are the segments AB and AC, so the squares must have AB and AC as their sides.Given the confusion in coordinates, perhaps a different approach is better, such as using vectors or similarity.Let me try vector approach.Let’s denote vectors with their position vectors. Let’s place A at the origin, so vector A = (0,0). Let vector AB = c i, and vector AC = b j. Then vector BC = -c i + b j.The square on AB will have vertices A, B, B + (AB rotated 90 degrees), and A + (AB rotated 90 degrees). Rotating AB (which is c i) 90 degrees counterclockwise gives c j. So the square on AB is A(0,0), B(c,0), B + c j = (c,0) + (0,c) = (c,c), and A + c j = (0,c). Therefore, the outermost point E is (c,c).Similarly, the square on AC. Rotating AC (b j) 90 degrees clockwise (to be outward from the triangle) gives b i. So the square on AC is A(0,0), C(0,b), C + b i = (0,b) + (b,0) = (b,b), and A + b i = (b,0). Therefore, the outermost point F is (b,b). Wait, but this brings us back to the earlier problem where projecting F(b,b) led to a contradiction unless b=2c.But in this case, if squares are constructed by rotating the legs 90 degrees, then the outer points E and F are (c,c) and (b,b). But then the projections would still lead to the previous results where equality only holds for specific b and c.Alternatively, maybe the rotation is in the other direction. For the square on AB, rotating AB 90 degrees clockwise would give -c j. So the square would be A(0,0), B(c,0), B - c j = (c, -c), and A - c j = (0,-c). Then outermost point E is (c, -c). Similarly, for AC, rotating 90 degrees counterclockwise gives -b i, so square is A(0,0), C(0,b), C - b i = (-b, b), and A - b i = (-b, 0). Then outermost point F is (-b, b).Ah, this might resolve the earlier issue. If E is (c, -c) and F is (-b, b), then projecting these points onto BC's extension might give different results.Let’s recalculate with E(c, -c) and F(-b, b).First, projection of E(c, -c) onto BC.Line BC: from B(c,0) to C(0,b), equation y = (-b/c)x + b.Slope of BC: -b/c, so perpendicular slope is c/b.Equation of perpendicular from E(c, -c): y + c = (c/b)(x - c )Intersect with BC:(-b/c)x + b = (c/b)(x - c ) - cMultiply both sides by bc:- b^2 x + b^2 c = c^2 (x - c ) - b c^2Expand right side: c^2 x - c^3 - b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c + c^3 + b c^2 = 0Factor x: -x(b^2 + c^2 ) + c(b^2 + c^2 + b c ) = 0Thus,x = c(b^2 + c^2 + b c ) / (b^2 + c^2 )Hmm, this seems different.Wait, perhaps I made an error in calculation. Let's go step by step.Equation of perpendicular from E(c, -c): y + c = (c/b)(x - c )Thus, y = (c/b)x - (c^2)/b - cIntersection with BC: y = (-b/c)x + bSet equal:(c/b)x - (c^2)/b - c = (-b/c)x + bMultiply all terms by bc:c^2 x - c^3 - b c^2 = -b^2 x + b^2 cBring all terms to left:c^2 x + b^2 x - c^3 - b c^2 - b^2 c = 0Factor x: x(c^2 + b^2 ) - c^3 - b c^2 - b^2 c = 0Thus,x = (c^3 + b c^2 + b^2 c ) / (b^2 + c^2 ) = c(c^2 + b c + b^2 ) / (b^2 + c^2 )This seems more complicated. Substitute back into BC's equation to find y:y = (-b/c)x + b = (-b/c)*[ c(c^2 + b c + b^2 ) / (b^2 + c^2 ) ] + b = -b(c^2 + b c + b^2 ) / (b^2 + c^2 ) + b = [ -b(c^2 + b c + b^2 ) + b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ -b c^2 - b^2 c - b^3 + b^3 + b c^2 ] / (b^2 + c^2 ) = (-b^2 c ) / (b^2 + c^2 )Thus, E' is at ( c(c^2 + b c + b^2 ) / (b^2 + c^2 ), -b^2 c / (b^2 + c^2 ) )Now, compute E'B. Point B is at (c,0).Distance E'B = sqrt( [ c(c^2 + b c + b^2 ) / (b^2 + c^2 ) - c ]^2 + [ -b^2 c / (b^2 + c^2 ) - 0 ]^2 )Simplify x-component:c [ (c^2 + b c + b^2 - (b^2 + c^2 )) / (b^2 + c^2 ) ] = c [ b c / (b^2 + c^2 ) ] = b c^2 / (b^2 + c^2 )Y-component squared: ( -b^2 c / (b^2 + c^2 ) )^2 = b^4 c^2 / (b^2 + c^2 )^2Thus, E'B^2 = (b^2 c^4 + b^4 c^2 ) / (b^2 + c^2 )^2 = b^2 c^2 (c^2 + b^2 ) / (b^2 + c^2 )^2 = b^2 c^2 / (b^2 + c^2 )Therefore, E'B = (b c ) / sqrt(b^2 + c^2 )Similarly, projecting F(-b, b) onto BC.Perpendicular from F(-b, b) to BC.Slope of BC: -b/c, so perpendicular slope is c/b.Equation of perpendicular: y - b = (c/b)(x + b )Intersect with BC:(-b/c)x + b = (c/b)(x + b ) + b - bWait, set equal:(-b/c)x + b = (c/b)x + (c/b)*bSimplify:(-b/c)x + b = (c/b)x + cMultiply all terms by bc:- b^2 x + b^2 c = c^2 x + b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c - b c^2 = 0Factor x: -x(b^2 + c^2 ) + b c (b - c ) = 0Thus,x = [ b c (b - c ) ] / (b^2 + c^2 )Then y = (-b/c)x + b = (-b/c)*(b c (b - c ) / (b^2 + c^2 )) + b = -b^2 (b - c ) / (b^2 + c^2 ) + b = [ -b^3 + b^2 c + b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ -b^3 + b^2 c + b^3 + b c^2 ] / (b^2 + c^2 ) = (b^2 c + b c^2 ) / (b^2 + c^2 ) = b c (b + c ) / (b^2 + c^2 )Thus, F' is at ( b c (b - c ) / (b^2 + c^2 ), b c (b + c ) / (b^2 + c^2 ) )Compute CF'. Point C is at (0, b).Distance CF' = sqrt( [ b c (b - c ) / (b^2 + c^2 ) - 0 ]^2 + [ b c (b + c ) / (b^2 + c^2 ) - b ]^2 )X-component squared: [ b c (b - c ) / (b^2 + c^2 ) ]^2Y-component: [ b c (b + c ) - b (b^2 + c^2 ) ] / (b^2 + c^2 ) = [ b^2 c + b c^2 - b^3 - b c^2 ] / (b^2 + c^2 ) = ( -b^3 + b^2 c ) / (b^2 + c^2 ) = -b^2 (b - c ) / (b^2 + c^2 )Thus, y-component squared: [ -b^2 (b - c ) / (b^2 + c^2 ) ]^2 = b^4 (b - c )^2 / (b^2 + c^2 )^2Thus, CF'^2 = [ b^2 c^2 (b - c )^2 + b^4 (b - c )^2 ] / (b^2 + c^2 )^2 = b^2 (b - c )^2 (c^2 + b^2 ) / (b^2 + c^2 )^2 = b^2 (b - c )^2 / (b^2 + c^2 )Therefore, CF' = b |b - c | / sqrt(b^2 + c^2 )So comparing E'B and CF':E'B = (b c ) / sqrt(b^2 + c^2 )CF' = b |b - c | / sqrt(b^2 + c^2 )For these to be equal:(b c ) = b |b - c | → c = |b - c |Assuming b ≠ 0.Case 1: b > c → c = b - c → 2c = b → b = 2cCase 2: b ≤ c → c = c - b → b = 0, impossible.Therefore, equality holds only when b = 2c. But the problem statement doesn’t specify any relation between b and c, so this suggests a mistake in the problem interpretation or in the construction.This implies that either the problem has a specific condition where b = 2c, or my construction of the squares is incorrect. Since the problem states it generally, I must have misconstructed the squares.Wait, going back to the problem statement: "squares are constructed on the legs, with their outermost points labeled as E and F". The key is "outermost" — which probably means the vertices of the squares that are furthest from the hypotenuse BC.To find the outermost points, we need to determine which vertices of the squares are furthest from BC. Depending on the orientation, this could vary.Alternatively, perhaps "outermost" refers to the direction away from the triangle. Given the right triangle at A, squares constructed on AB and AC would have their outermost points in directions perpendicular to AB and AC, away from the triangle.If AB is along the x-axis and AC along the y-axis, then the square on AB would extend in the positive y direction, and the square on AC would extend in the positive x direction, making E at (c, c) and F at (b, b). Wait, but earlier this led to a contradiction unless b = 2c. Alternatively, if squares are constructed in the negative y and negative x directions, points E(c, -c) and F(-b, b). But projecting those might give different results.Alternatively, maybe the squares are constructed such that E and F are on the same side relative to BC. However, this is getting too vague.Alternatively, let's consider specific values to test. Let’s take a 3-4-5 triangle for concreteness. Let’s set AB = 3, AC = 4, then BC = 5.So coordinates: A(0,0), B(3,0), C(0,4)Square on AB: constructed outward. If outward is positive y, then E is (3,3). Square on AC: constructed outward in positive x direction, so F is (4,4).Project E(3,3) onto BC. Line BC: from (3,0) to (0,4). Equation: y = (-4/3)x + 4.Perpendicular from E(3,3) has slope 3/4. Equation: y - 3 = (3/4)(x - 3)Intersection: (-4/3)x + 4 = (3/4)x - 9/4 + 3Simplify:-4/3 x + 4 = 3/4 x - 9/4 + 12/4-4/3 x + 4 = 3/4 x + 3/4Multiply all terms by 12:-16x + 48 = 9x + 9-25x = -39 → x = 39/25 = 1.56y = (-4/3)(39/25) + 4 = (-156/75) + 300/75 = 144/75 = 48/25 = 1.92So E’ is (1.56, 1.92)Distance E’B: from (3,0) to (1.56,1.92):sqrt( (3 - 1.56)^2 + (0 - 1.92)^2 ) = sqrt( (1.44)^2 + (-1.92)^2 ) = sqrt( 2.0736 + 3.6864 ) = sqrt(5.76) = 2.4Now compute CF’:Point F is (4,4). Project onto BC.Perpendicular from F(4,4) to BC: slope 3/4. Equation: y - 4 = (3/4)(x - 4)Intersection with BC:(-4/3)x + 4 = (3/4)x - 3 + 4Simplify:-4/3 x + 4 = 3/4 x + 1Multiply all terms by 12:-16x + 48 = 9x + 12-25x = -36 → x = 36/25 = 1.44y = (-4/3)(36/25) + 4 = (-144/75) + 300/75 = 156/75 = 52/25 = 2.08F’ is (1.44, 2.08)Distance CF’: from C(0,4) to (1.44,2.08):sqrt( (1.44)^2 + (2.08 - 4)^2 ) = sqrt( 2.0736 + 3.6864 ) = sqrt(5.76) = 2.4So in this specific 3-4-5 triangle, E’B = CF’ = 2.4, and BC = 5. Now check EE’ + FF’:EE’ is distance from E(3,3) to E’(1.56,1.92):sqrt( (3 - 1.56)^2 + (3 - 1.92)^2 ) = sqrt( (1.44)^2 + (1.08)^2 ) = sqrt(2.0736 + 1.1664) = sqrt(3.24) = 1.8FF’ is distance from F(4,4) to F’(1.44,2.08):sqrt( (4 - 1.44)^2 + (4 - 2.08)^2 ) = sqrt( (2.56)^2 + (1.92)^2 ) = sqrt(6.5536 + 3.6864) = sqrt(10.24) = 3.2Sum EE’ + FF’ = 1.8 + 3.2 = 5, which equals BC = 5. So in this case, both conditions hold.But in this example, b = 4 and c = 3, so b = 4, c = 3, which doesn't satisfy b = 2c. Therefore, my previous conclusion that b must be 2c was incorrect. The error was in the general symbolic calculation, but in the numeric example, it works. Hence, there must be a mistake in the symbolic approach.Re-examining the symbolic expressions:We had E'B = (b c ) / sqrt(b^2 + c^2 )CF' = b |b - c | / sqrt(b^2 + c^2 )Wait, in the numeric example, b = 4, c = 3:E’B = (4*3)/5 = 12/5 = 2.4CF' = 4*|4 - 3| /5 = 4*1/5 = 0.8 → Wait, this contradicts the numeric result where CF’ was 2.4. Clearly, my symbolic expression for CF’ is wrong.Wait, in the numeric example, CF’ was 2.4, which is equal to (4*3)/5 = 12/5 = 2.4. But according to the earlier symbolic expression, CF’ = b |b - c | / sqrt(b^2 + c^2 ). For b=4, c=3: 4*1/5=0.8, which is wrong. Hence, there must be an error in the symbolic calculation of CF’.Looking back at the calculation of CF’ when F is at (4,4) in the numeric example:Projection of F(4,4) onto BC:We found F’ at (1.44, 2.08). Then CF’ is the distance from C(0,4) to (1.44,2.08):sqrt(1.44^2 + (2.08 - 4)^2 ) = sqrt( 2.0736 + 3.6864 ) = sqrt(5.76) = 2.4. Which is equal to (b c ) / sqrt(b^2 + c^2 ) = (4*3)/5 = 2.4. So CF’ should also be equal to (b c ) / sqrt(b^2 + c^2 ). Hence, my earlier symbolic expression for CF’ was incorrect. Let me re-express CF’.In the numeric example, CF’ = 2.4 = (4*3)/5 = same as E’B. Hence, in general, CF’ = (b c ) / sqrt(b^2 + c^2 ), same as E’B. Therefore, the earlier symbolic derivation must have an error.Re-examizing the CF’ calculation when F is at (b,b):Wait, in the numeric example, F was at (4,4), but according to the square on AC, which is of length 4, the square should be built on AC. If AC is from (0,0) to (0,4), then the square constructed outward (positive x direction) would have vertices at A(0,0), C(0,4), (4,4), (4,0). So the outermost point F is (4,4). Projection of F onto BC gave CF’ = 2.4.Symbolically, if F is at (b,b), then projecting onto BC:Line BC: y = (-b/c)x + bPerpendicular from F(b,b) has slope c/b. Equation: y - b = (c/b)(x - b )Intersection with BC:(-b/c)x + b = (c/b)x - (c^2)/b + bMultiply both sides by bc:- b^2 x + b^2 c = c^2 x - c^3 + b c^2Bring all terms to left:- b^2 x - c^2 x + b^2 c + c^3 - b c^2 = 0Factor x:- x(b^2 + c^2 ) + c(b^2 + c^2 - b c ) = 0Thus,x = c(b^2 + c^2 - b c ) / (b^2 + c^2 )Then y = (-b/c)x + b = (-b/c)*[ c(b^2 + c^2 - b c ) / (b^2 + c^2 ) ] + b = -b(b^2 + c^2 - b c ) / (b^2 + c^2 ) + b = [ -b^3 - b c^2 + b^2 c + b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ -b^3 - b c^2 + b^2 c + b^3 + b c^2 ] / (b^2 + c^2 ) = (b^2 c ) / (b^2 + c^2 )Thus, F’ is at ( c(b^2 + c^2 - b c ) / (b^2 + c^2 ), b^2 c / (b^2 + c^2 ) )Distance CF’ is from C(0,b) to F’:sqrt( [ c(b^2 + c^2 - b c ) / (b^2 + c^2 ) - 0 ]^2 + [ b^2 c / (b^2 + c^2 ) - b ]^2 )X-component: c(b^2 + c^2 - b c ) / (b^2 + c^2 )Y-component: [ b^2 c - b(b^2 + c^2 ) ] / (b^2 + c^2 ) = [ b^2 c - b^3 - b c^2 ] / (b^2 + c^2 ) = -b(b^2 + c^2 - b c ) / (b^2 + c^2 )Thus, CF’^2 = [ c(b^2 + c^2 - b c ) / (b^2 + c^2 ) ]^2 + [ -b(b^2 + c^2 - b c ) / (b^2 + c^2 ) ]^2 = (b^2 + c^2 - b c )^2 (c^2 + b^2 ) / (b^2 + c^2 )^2 ) = (b^2 + c^2 - b c )^2 / (b^2 + c^2 )Thus, CF’ = |b^2 + c^2 - b c | / sqrt(b^2 + c^2 )But in the numeric example, b=4, c=3:b^2 + c^2 - b c = 16 + 9 - 12 = 13. |13| /5 = 13/5 = 2.6, which does not match the numeric result of 2.4. Contradiction. Therefore, there's a mistake.Wait, but earlier in the numeric example, with F at (4,4), projection gave CF’=2.4. However, according to this formula, it should be 13/5=2.6. Hence, inconsistency.This suggests that the error is in the assumption of the location of point F. If in the numeric example, F is at (4,4), which is the square on AC extended in the positive x direction, but when projecting, we get CF’=2.4 which equals (b c)/sqrt(b^2 + c^2 ). However, according to the symbolic calculation, CF’ is |b^2 + c^2 - b c | / sqrt(b^2 + c^2 ). For b=4, c=3, this gives 13/5=2.6, which doesn't match.Therefore, the error must be in the projection calculation when F is at (b,b).Wait, let's recalculate the projection for the numeric example with F(4,4):Perpendicular from F(4,4) to BC (y = -4/3 x + 4) has slope 3/4. Equation: y - 4 = (3/4)(x -4 )Intersection with BC:-4/3 x + 4 = 3/4 x - 3 +4 → -4/3 x +4 = 3/4 x +1Multiply by 12:-16x +48 =9x +12 → -25x= -36 → x= 36/25=1.44y= -4/3*(36/25)+4= -48/25+100/25=52/25=2.08F’=(1.44,2.08)CF’= sqrt( (1.44-0)^2 + (2.08-4)^2 )= sqrt(2.0736 + 3.6864)= sqrt(5.76)=2.4=12/5=(4*3)/5= bc/sqrt(b² +c²). Hence, CF’= bc/sqrt(b² +c² ) in the numeric case.But according to the symbolic calculation when F is at (b,b), CF’= sqrt( (c(b² +c² - bc)/D )² + ( -b(b² +c² - bc)/D )² ), where D= b² +c². Which is sqrt( (b² +c² - bc )² (c² +b²)/D² )= |b² +c² - bc | / sqrt(b² +c² )But in numeric case, b=4, c=3: |16 +9 -12| /5=13/5=2.6≠2.4. Hence, contradiction. Therefore, the mistake is in assuming F is at (b,b). In reality, when constructing the square on AC outward, the direction might not be positive x. Let's re-express.If AC is from A(0,0) to C(0,b), constructing the square outward (away from the triangle) would require rotating AC 90 degrees clockwise (if AB is along x-axis and AC along y-axis, outward normal to AC would be to the left, negative x direction). So the square on AC would have vertices A(0,0), C(0,b), (-b, b), (-b,0). Hence, outermost point F is (-b, b). But projecting F(-b, b) onto BC.Let's recalculate the numeric example with F at (-4,4).Project F(-4,4) onto BC.Equation of BC: y = -4/3 x +4.Perpendicular slope: 3/4.Equation of perpendicular from F(-4,4): y -4 = (3/4)(x +4 )Intersection with BC:-4/3 x +4 = (3/4)x +3 +4Wait, set equal:-4/3 x +4 = (3/4)x +3 +4 -4 (wait, no):Perpendicular equation: y = (3/4)x +3 +4?Wait, no. From point (-4,4), perpendicular equation:y -4 = (3/4)(x +4 )So y = (3/4)x + 3 +4 = (3/4)x +7Intersection with BC: y = -4/3 x +4Set equal:(3/4)x +7 = -4/3 x +4Multiply all terms by 12 to eliminate denominators:9x +84 = -16x +4825x = -36 → x = -36/25 = -1.44y = -4/3*(-1.44)+4 = 1.92 +4 =5.92Thus, F’=(-1.44,5.92)Distance CF’ from C(0,4) to F’(-1.44,5.92):sqrt( (-1.44)^2 + (5.92 -4)^2 )= sqrt(2.0736 +3.6864)=sqrt(5.76)=2.4Thus, CF’=2.4= same as E’B. Hence, when F is correctly placed at (-b,b), CF’ equals (b c ) / sqrt(b² +c² )Therefore, the correct coordinates for F are (-b,b), not (b,b). This resolves the contradiction. Hence, in symbolic terms:Projection of F(-b,b) onto BC gives CF’= (b c ) / sqrt(b² +c² ) = E’BSimilarly, the sum EE’ + FF’:In the numeric example, EE’ is from E(3,3) to E’(1.56,1.92): 1.8FF’ is from F(-4,4) to F’(-1.44,5.92): distance sqrt( (-4 +1.44)^2 + (4 -5.92)^2 )= sqrt( (-2.56)^2 + (-1.92)^2 )= sqrt(6.5536 +3.6864)= sqrt(10.24)=3.2Sum 1.8+3.2=5=BC=5. Hence, holds.Thus, the key was that F is at (-b,b), not (b,b). Therefore, correcting the earlier analysis:When constructing the square on AC outward, the direction is negative x-axis, resulting in F at (-b,b).Re-doing the symbolic projection for F(-b,b):Line BC: y = (-b/c)x + bPerpendicular from F(-b,b): slope c/bEquation: y -b = (c/b)(x +b )Intersection with BC:(-b/c)x + b = (c/b)(x +b ) +b -b → (-b/c)x +b = (c/b)x + cMultiply by bc:- b² x + b² c = c² x + b c²Bring all terms to left:- b² x - c² x + b² c - b c² =0Factor x: -x(b² +c² ) +b c (b -c )=0Thus,x= [b c (b -c ) ] / (b² +c² )Then y= (-b/c)x +b= (-b/c)(b c (b -c ) / (b² +c² )) +b= -b² (b -c ) / (b² +c² ) +b= [ -b³ +b² c +b(b² +c² ) ] / (b² +c² )= [ -b³ +b² c +b³ +b c² ] / (b² +c² )= (b² c +b c² ) / (b² +c² )= bc (b +c ) / (b² +c² )Thus, F’ is at ( b c (b -c ) / (b² +c² ), bc (b +c ) / (b² +c² ) )Distance CF’ from C(0,b) to F’:sqrt( [ b c (b -c ) / (b² +c² ) ]² + [ bc (b +c ) / (b² +c² ) -b ]² )Compute x-component squared: [ b c (b -c ) / D ]², where D = b² +c²Y-component: [ bc (b +c ) -b D ] / D = [ bc (b +c ) -b (b² +c² ) ] / D= [ b² c +b c² -b³ -b c² ] / D= [ -b³ +b² c ] / D= -b² (b -c ) / DThus, y-component squared: [ -b² (b -c ) / D ]²= b⁴ (b -c )² / D²Thus, CF’²= [ b² c² (b -c )² +b⁴ (b -c )² ] / D²= b² (b -c )² (c² +b² ) / D²= b² (b -c )² / DTherefore, CF’= |b (b -c )| / sqrt(D )But from earlier numeric example, CF’= (b c ) / sqrt(D )This suggests a contradiction. Wait, in the numeric example, CF’ was 2.4=12/5, which is (4*3)/5=12/5. But according to this formula, CF’= |4*(4-3)| /5=4/5=0.8, which is wrong. Hence, the error must be elsewhere.Wait, but in the numeric example, when F is at (-4,4), the projection F’ is at (-1.44,5.92). The x-coordinate is x= -1.44= -36/25, which is [ b c (b -c ) ] / D= [4*3*(4-3)]/(16+9)=12*1/25=12/25=0.48, but in reality x was -36/25. This inconsistency indicates a miscalculation.Wait, in the symbolic calculation, x= [b c (b -c ) ] / D, but in the numeric case, x= [4*3*(4-3)] / (16+9)=12/25=0.48, but actual x was -36/25=-1.44. This is a discrepancy.Therefore, there's a mistake in the projection calculation when F is at (-b,b).Let me recalculate the projection of F(-b,b) onto BC.Parametrize BC: from B(c,0) to C(0,b), direction vector (-c,b). A point on BC can be written as B + t*(C - B) = (c - c t, 0 + b t), t ∈ R.The projection of F(-b,b) onto BC is the point E’ on BC such that vector FE’ is perpendicular to BC.Vector FE’ = (c - c t +b, b t -b )This must be perpendicular to BC's direction vector (-c,b):Dot product = (c -c t +b)(-c) + (b t -b)(b) =0Expand:- c(c -c t +b) + b(b t -b )=0- c² +c² t -b c +b² t -b²=0Group terms:t(c² +b² ) -c² -b c -b²=0Thus,t= (c² +b c +b² ) / (b² +c² )Thus, the parameter t= [c² +b c +b² ] / [b² +c² ]Thus, coordinates of F’:x= c -c t= c -c*(c² +b c +b² ) / (b² +c² )= [c(b² +c² ) -c(c² +b c +b² ) ] / (b² +c² )= [c b² +c³ -c³ -b c² -c b² ] / (b² +c² )= [ -b c² ] / (b² +c² )y= b t= b*(c² +b c +b² ) / (b² +c² )= [b c² +b² c +b³ ] / (b² +c² )Thus, F’= ( -b c² / (b² +c² ), b(c² +b c +b² ) / (b² +c² ) )Now, distance CF’ from C(0,b) to F’:x-component: -b c² / (b² +c² )y-component: [b(c² +b c +b² ) / (b² +c² ) -b ]= [b(c² +b c +b² ) -b(b² +c² ) ] / (b² +c² )= [b c² +b² c +b³ -b³ -b c² ] / (b² +c² )= b² c / (b² +c² )Thus, CF’= sqrt( [ -b c² / (b² +c² ) ]² + [ b² c / (b² +c² ) ]² )= sqrt( b² c⁴ +b⁴ c² ) / (b² +c² )= (b c ) sqrt(c² +b² ) / (b² +c² )= (b c ) / sqrt(b² +c² )Thus, CF’= (b c ) / sqrt(b² +c² ), which matches E’B and the numeric example.Therefore, the previous symbolic calculation was incorrect due to miscalculating the coordinates of F’. The correct CF’ is (b c ) / sqrt(b² +c² ), same as E’B. Hence, E’B=CF’ in general.Now, the second part: EE’ + FF’=BC.EE’ is the distance from E(c,c) to E’(c(b² +c² -b c ) / D, b² c / D ), where D=b² +c²Compute EE’:x-component: c - c(b² +c² -b c ) / D= c[ D - (b² +c² -b c ) ] / D= c[ b c ] / D= b c² / Dy-component: c - b² c / D= c(D -b² ) / D= c c² / D= c³ / DThus, EE’= sqrt( (b c² / D )² + (c³ / D )² )= sqrt( b² c⁴ +c⁶ ) / D= c² sqrt(b² +c² ) / D= c² / sqrt(b² +c² )Similarly, FF’ is the distance from F(-b,b) to F’( -b c² / D, b(c² +b c +b² ) / D )Compute FF’:x-component: -b - (-b c² / D )= -b +b c² / D= -b(D -c² ) / D= -b(b² ) / D= -b³ / D (absolute value: b³ / D )y-component: b - b(c² +b c +b² ) / D= b(D -c² -b c -b² ) / D= b(-b c ) / D= -b² c / D (absolute value: b² c / D )Thus, FF’= sqrt( (b³ / D )² + (b² c / D )² )= sqrt( b⁶ +b⁴ c² ) / D= b² sqrt(b² +c² ) / D= b² / sqrt(b² +c² )Thus, EE’ + FF’= c² / sqrt(b² +c² ) +b² / sqrt(b² +c² )= (b² +c² ) / sqrt(b² +c² )= sqrt(b² +c² )= BCHence, EE’ + FF’=BC.Therefore, both statements hold generally.Final Answerboxed{E'B = CF'} quad text{and} quad boxed{EE' + FF' = BC}

✅:Given a right triangle (ABC) with ( angle A = 90^circ ), squares are constructed on the legs (AB) and (AC) with their outermost points labeled as (E) and (F). The projections of these points on the extension of the hypotenuse (BC) are denoted as (E') and (F'). We need to show that (E'B = CF') and (EE' + FF' = BC). Key Steps and Analysis:1. Coordinates and Projections: - Place (A) at the origin ((0,0)), (B) at ((c,0)), and (C) at ((0,b)). - The squares on (AB) and (AC) are constructed outward from the triangle. The outermost points (E) and (F) are ((c, c)) and ((-b, b)) respectively.2. Projections onto Hypotenuse (BC): - The equation of (BC) is (y = -frac{b}{c}x + b). - The perpendicular from (E(c, c)) to (BC) intersects at (E'), and the perpendicular from (F(-b, b)) to (BC) intersects at (F').3. Calculating Projection Coordinates: - For (E(c, c)): - The projection (E') is found by solving the intersection of the perpendicular from (E) to (BC). - Coordinates of (E') are (left( frac{c(b^2 + c^2 - bc)}{b^2 + c^2}, frac{b^2 c}{b^2 + c^2} right)). - For (F(-b, b)): - The projection (F') is found by solving the intersection of the perpendicular from (F) to (BC). - Coordinates of (F') are (left( -frac{b c^2}{b^2 + c^2}, frac{b(c^2 + bc + b^2)}{b^2 + c^2} right)).4. Distances (E'B) and (CF'): - Calculate distances using coordinates: - (E'B = frac{bc}{sqrt{b^2 + c^2}}) - (CF' = frac{bc}{sqrt{b^2 + c^2}}) - Hence, (E'B = CF').5. Sum of Distances (EE') and (FF'): - Calculate (EE') and (FF'): - (EE' = frac{c^2}{sqrt{b^2 + c^2}}) - (FF' = frac{b^2}{sqrt{b^2 + c^2}}) - Summing these distances: - (EE' + FF' = frac{c^2 + b^2}{sqrt{b^2 + c^2}} = sqrt{b^2 + c^2} = BC). Final Answer:[boxed{E'B = CF'} quad text{and} quad boxed{EE' + FF' = BC}]

🔑:To solve the problem, we need to show and explain the steps that lead us to the conclusion based on the given conditions and geometric properties.1. Assign the notation and draw the auxiliary points: - Let ( Delta ABC ) be a right triangle with ( angle A = 90^circ ). - Construct squares on the legs ( AB ) and ( AC ) of the triangle. - Let ( E ) and ( F ) be the farthest points of these squares from ( A ) along ( AB ) and ( AC ), respectively.2. Understand the properties and perform projections: - We project points ( E ) and ( F ) perpendicularly onto the hypotenuse ( BC ) extended beyond ( C ): - Let ( E' ) be the projection of ( E ). - Let ( F' ) be the projection of ( F ).3. Use congruent triangles to relate segments: - Let ( A' ) be the foot of the altitude drawn from ( A ) to ( BC ). - Consider the similar triangles that emerge: - In ( triangle ABA' ) and ( triangle BEE' ), the triangles are congruent by the AA criterion because ( triangle ABA' ) and ( triangle BEE' ) share ( angle A ) and both have right angles. - Similarly, in ( triangle ACA' ) and ( triangle CFF' ), the triangles are congruent by the same AA criterion. 4. Establish segment equalities from congruence: - From the congruence of ( triangle ABA' cong triangle BEE' ), we get: [ BE' = AA' ] - From the congruence of ( triangle ACA' cong triangle CFF' ), we get: [ CF' = AA' ]5. Summarize and substitute in the given relation: - Therefore, ( BE' = CF' = AA' ). - We are given that the sum of the segments from the projections of ( E ) and ( F ) onto ( BC ) extended should equal the length of ( BC ): [ EE' + FF' = BA' + CA' ]6. Express the conclusion using right triangle properties: - Since ( BC ) is the hypotenuse of the right triangle ( Delta ABC ) and by definition, it is equal to: [ BC = BA' + CA' ] Combining everything, we have: [ EE' + FF' = BC ] In conclusion, using the properties of right triangles and congruent triangles, we have shown that ( BE' = CF' ) and ( EE' + FF' = BC ):[ boxed{BC}]

🧠:Okay, let's try to figure out this problem step by step. So, we have two cleaning vehicles, A and B. Vehicle A can clean the road alone in 10 hours, and Vehicle B takes 15 hours. They start from opposite cities, East and West, moving towards each other. When they meet, Vehicle A has cleaned 12 kilometers more than Vehicle B. We need to find the distance between the two cities.First, I need to understand the problem clearly. Both vehicles start cleaning the roads towards each other. The time they take to meet each other would be the same for both, right? Because they start at the same time and meet at some point. So the time taken until they meet is equal for both A and B.Now, Vehicle A's speed: If it can clean the entire road in 10 hours, then its speed is total distance divided by 10. Similarly, Vehicle B's speed is total distance divided by 15. Wait, but here's a thing: the total distance is what we need to find. Let's denote the total distance as D kilometers. Then, Vehicle A's speed would be D/10 km per hour, and Vehicle B's speed would be D/15 km per hour.But hold on, that might not be the right way to model their speeds. Let me think again. If Vehicle A takes 10 hours to clean the entire road alone, then its speed is D/10 km/h. Similarly, Vehicle B's speed is D/15 km/h. Yes, that makes sense because speed equals distance divided by time. So their speeds are D/10 and D/15 respectively.Now, when they start moving towards each other, their relative speed is the sum of their individual speeds. So, the combined speed would be D/10 + D/15. Let's compute that. To add these fractions, find a common denominator, which is 30. So, D/10 is 3D/30 and D/15 is 2D/30. Adding them gives 5D/30, which simplifies to D/6. So their combined speed is D/6 km per hour.Since they are moving towards each other, the time it takes for them to meet can be found by dividing the total distance D by their combined speed. So, time = D / (D/6) = 6 hours. Wait, that's interesting. The time taken until they meet is 6 hours, regardless of the distance? Because D cancels out. So no matter what D is, they meet after 6 hours? That seems counterintuitive, but mathematically, it checks out. Let me verify.If Vehicle A's speed is D/10 km/h, in 6 hours, it would cover (D/10)*6 = (6D)/10 = (3D)/5 km. Similarly, Vehicle B's speed is D/15 km/h, so in 6 hours, it would cover (D/15)*6 = (6D)/15 = (2D)/5 km. Adding those two distances together: (3D)/5 + (2D)/5 = D, which matches the total distance. So, yes, they meet after 6 hours. That makes sense.Now, according to the problem, when they meet, Vehicle A has cleaned 12 kilometers more than Vehicle B. From the calculations above, the distance cleaned by A is (3D)/5, and by B is (2D)/5. The difference between these distances is (3D)/5 - (2D)/5 = D/5. According to the problem, this difference is 12 km. So, D/5 = 12. Solving for D gives D = 12 * 5 = 60. So the total distance is 60 kilometers.Wait, let me check this again. Vehicle A cleans 3D/5, Vehicle B cleans 2D/5. Difference is D/5 = 12. Therefore, D = 60. That seems straightforward. Let me see if there's another way to approach this problem to confirm.Alternatively, maybe using relative speeds or different variables. Let's denote the total distance as D. Let the speed of A be v_A = D/10, speed of B be v_B = D/15. When moving towards each other, their relative speed is v_A + v_B = D/10 + D/15 = (3D + 2D)/30 = 5D/30 = D/6. Time to meet is D / (D/6) = 6 hours. In 6 hours, A travels v_A * 6 = (D/10)*6 = 3D/5. B travels v_B *6 = (D/15)*6 = 2D/5. The difference is 3D/5 - 2D/5 = D/5. Set D/5 = 12 => D = 60. Same result.Alternatively, perhaps setting up equations with variables for speeds. Let me try that.Suppose the total distance is D. Let speed of A be v_A, speed of B be v_B. Given that A takes 10 hours to clean D, so v_A = D / 10. Similarly, v_B = D / 15. When they move towards each other, time until meeting is t. In time t, distance covered by A is v_A * t, distance covered by B is v_B * t. These two distances should add up to D, so:v_A * t + v_B * t = DWe also know that v_A * t - v_B * t = 12 kmSo, we have two equations:1) (v_A + v_B) * t = D2) (v_A - v_B) * t = 12From equation 1, since v_A = D/10 and v_B = D/15, substituting:(D/10 + D/15) * t = DSimplify the left side: (3D/30 + 2D/30) * t = D => (5D/30) * t = D => (D/6) * t = DDivide both sides by D: (1/6) * t = 1 => t = 6 hours.Then, equation 2:(v_A - v_B) * t = 12Substitute v_A and v_B:(D/10 - D/15) * 6 = 12Compute D/10 - D/15: (3D - 2D)/30 = D/30Multiply by 6: (D/30)*6 = D/5Set equal to 12: D/5 = 12 => D = 60Same answer. So, all approaches lead to D = 60 km. Therefore, the distance between East City and West City is 60 kilometers.But let me just think again if there's any possibility of misunderstanding the problem. The problem states that when they meet, Vehicle A has cleaned 12 km more than Vehicle B. So the difference in their cleaned distances is 12 km. Our calculations show that this difference is D/5, so D must be 60. Yes.Alternatively, maybe someone could think that the time taken is different? But no, since they start at the same time and move towards each other until they meet, the time elapsed must be the same for both. So, their meeting time is the same. Therefore, the distance each covers is speed multiplied by that time. The difference between these distances is 12 km, leading us to the equation (v_A - v_B) * t = 12. As we found, this gives D = 60.Another sanity check: If D is 60 km, then Vehicle A's speed is 60/10 = 6 km/h, Vehicle B's speed is 60/15 = 4 km/h. Combined speed is 10 km/h, so time to meet is 60 / 10 = 6 hours. In 6 hours, A travels 6*6=36 km, B travels 4*6=24 km. The difference is 36-24=12 km, which matches the problem statement. Perfect, that checks out.So, after all these steps and checks, I'm confident the answer is 60 kilometers.

✅:The distance between East City and West City is determined by analyzing the speeds of the cleaning vehicles and their meeting point. Here's the step-by-step solution:1. Define Variables: - Let ( D ) = total distance between the cities (km). - Vehicle A's speed: ( frac{D}{10} ) km/h (cleans ( D ) km in 10 hours). - Vehicle B's speed: ( frac{D}{15} ) km/h (cleans ( D ) km in 15 hours).2. Relative Speed and Meeting Time: - Combined speed: ( frac{D}{10} + frac{D}{15} = frac{3D + 2D}{30} = frac{D}{6} ) km/h. - Time to meet: ( frac{D}{frac{D}{6}} = 6 ) hours (since distance ( D ) cancels out).3. Distances Cleaned: - Distance cleaned by A: ( frac{D}{10} times 6 = frac{3D}{5} ) km. - Distance cleaned by B: ( frac{D}{15} times 6 = frac{2D}{5} ) km. - Difference: ( frac{3D}{5} - frac{2D}{5} = frac{D}{5} ).4. Solve for ( D ): - Given ( frac{D}{5} = 12 ) km → ( D = 12 times 5 = 60 ) km.Verification:- Speeds: A = ( 6 ) km/h, B = ( 4 ) km/h.- Meeting time: ( 6 ) hours → A cleans ( 36 ) km, B cleans ( 24 ) km.- Difference: ( 36 - 24 = 12 ) km (matches problem statement).Final Answer: The cities are (boxed{60}) kilometers apart.

🔑:1. Determine the Speeds: First, we need to determine the speeds of the two vehicles (甲车 and 乙车). Our vehicles, 甲车, and 乙车 take 10 hours and 15 hours respectively to clean a specific length. - 甲车's speed: [ text{Speed of 甲车} = frac{1}{10} text{kilometers/hour} ] - 乙车's speed: [ text{Speed of 乙车} = frac{1}{15} text{kilometers/hour} ]2. Combined Speed: Next, we find the combined speed of both vehicles. Since they are moving towards each other, their combined speeds add up. [ text{Combined speed} = frac{1}{10} + frac{1}{15} ] To add these, we need a common denominator: [ frac{1}{10} = frac{3}{30}, quad frac{1}{15} = frac{2}{30} ] [ text{Combined speed} = frac{3}{30} + frac{2}{30} = frac{5}{30} = frac{1}{6} text{kilometers/hour} ]3. Time to Meet: Given the combined speed, we need to find the time it takes for the two vehicles to meet. This is given by the reciprocal of their combined speed: [ text{Time to meet} = frac{1}{frac{1}{6}} = 6 text{hours} ]4. Distance Cleaned by Each Vehicle: We know that when they meet, 甲车 has cleaned 12 kilometers more than 乙车. We can set up an equation to represent this scenario. - Let distance cleaned by 甲车 be ( D_甲 ) - Let distance cleaned by 乙车 be ( D_乙 ) From their relative cleaning speeds in 6 hours: [ D_甲 = frac{1}{10} times 6 = 0.6 text{kilometers} ] [ D_乙 = frac{1}{15} times 6 = 0.4 text{kilometers} ]5. Establish the 12-kilometer Difference: Given the problem's condition: [ D_甲 - D_乙 = 12 text{kilometers} ] Substituting the distances: [ 6 times (frac{1}{10} - frac{1}{15}) = 12 ]6. Calculate the Difference in Speeds: Simplified: [ frac{1}{10} - frac{1}{15} = frac{3}{30} - frac{2}{30} = frac{1}{30} text{kilometers/hour} ] Applying into the difference: [ D_甲 - D_乙 = 6 times left(frac{1}{10} - frac{1}{15}right) ] [ 6 times frac{1}{30} = 12 ] Therefore, [ 1 times 6 = 6 times frac{1}{5} ] [ 12 times 5 = 60 ] Therefore, the distance between the two cities is:[boxed{60 text{ kilometers}}]

🧠:Okay, let's see. I have a triangle with a circumradius of 4 cm. One of its interior angles is 60 degrees. The difference between the two sides that enclose this angle is 4 cm. I need to find the area of this triangle. Hmm, let me start by recalling some triangle properties and formulas that might be useful here.First, the circumradius (R) of a triangle is related to its sides and angles. I remember the formula that relates the sides of a triangle to the circumradius: a / sin A = b / sin B = c / sin C = 2R. So, each side divided by the sine of its opposite angle equals twice the circumradius. Since one of the angles is 60 degrees, maybe I can use that to express the sides in terms of the other angles?Let me denote the triangle as triangle ABC, where angle A is 60 degrees. The sides opposite to angles A, B, and C are a, b, and c respectively. Given that R is 4 cm, then 2R = 8 cm. So, a / sin A = 8 cm. Since angle A is 60 degrees, sin 60° is √3/2. Therefore, a = 8 * (√3/2) = 4√3 cm. So, side a is 4√3 cm.Now, the problem states that the difference between the two sides enclosing the 60-degree angle is 4 cm. The sides enclosing angle A are sides b and c. So, |b - c| = 4 cm. Let's note that down: |b - c| = 4. Since the problem doesn't specify which side is longer, we might have to consider both possibilities, but maybe the triangle's other angles can help determine that.We also need to find the area of the triangle. The area can be calculated in several ways. One common formula is (1/2)ab sin C, where a and b are two sides and C is the included angle. In this case, if we can find sides b and c, then the area would be (1/2)*b*c*sin 60°, since angle A is 60 degrees. Alternatively, using Heron's formula if we know all three sides, but maybe the first approach is better here.But first, we need to find the lengths of sides b and c. We know that |b - c| = 4. Also, from the Law of Sines, we can express b and c in terms of their opposite angles. Let's denote angles B and C. Since the sum of angles in a triangle is 180 degrees, angle B + angle C = 120 degrees.Using the Law of Sines: b / sin B = 8 and c / sin C = 8. Therefore, b = 8 sin B and c = 8 sin C. Since angle B + angle C = 120°, maybe we can let angle B = θ and angle C = 120° - θ. Then, b = 8 sin θ and c = 8 sin (120° - θ). Then, the difference |b - c| = |8 sin θ - 8 sin (120° - θ)| = 8 |sin θ - sin (120° - θ)| = 4 cm. So, 8 |sin θ - sin (120° - θ)| = 4, which simplifies to |sin θ - sin (120° - θ)| = 0.5.Now, let's compute sin (120° - θ). Using the sine subtraction formula: sin(120° - θ) = sin 120° cos θ - cos 120° sin θ. Sin 120° is √3/2, cos 120° is -1/2. So, sin(120° - θ) = (√3/2) cos θ - (-1/2) sin θ = (√3/2) cos θ + (1/2) sin θ.Therefore, sin θ - sin(120° - θ) = sin θ - [ (√3/2) cos θ + (1/2) sin θ ] = sin θ - (√3/2) cos θ - (1/2) sin θ = (1 - 1/2) sin θ - (√3/2) cos θ = (1/2) sin θ - (√3/2) cos θ.So, the absolute value of (1/2 sin θ - √3/2 cos θ) equals 0.5. Let me denote that:| (1/2 sin θ - √3/2 cos θ ) | = 0.5Notice that (1/2 sin θ - √3/2 cos θ) can be written as sin(θ - 60°). Because sin(A - B) = sin A cos B - cos A sin B. So, sin θ cos 60° - cos θ sin 60° = (1/2) sin θ - (√3/2) cos θ, which matches the expression above. So, |sin(θ - 60°)| = 0.5Therefore, sin(θ - 60°) = ±0.5So, θ - 60° = 30° + 360°n or θ - 60° = 150° + 360°n, where n is an integer. But since θ is an angle in a triangle, it must be between 0° and 180°, so we can ignore the 360° multiples. Therefore:Case 1: θ - 60° = 30° → θ = 90°Case 2: θ - 60° = -30° → θ = 30°Case 3: θ - 60° = 150° → θ = 210°, which is not possible because θ must be less than 180°Case 4: θ - 60° = -150° → θ = -90°, which is also not possible.So, the only valid solutions are θ = 90° or θ = 30°. Let's check these two cases.Case 1: θ = 90°Then angle C = 120° - 90° = 30°Therefore, sides would be:b = 8 sin 90° = 8 * 1 = 8 cmc = 8 sin 30° = 8 * 0.5 = 4 cmThus, the difference b - c = 8 - 4 = 4 cm. That matches the given condition. So, this case is valid.Case 2: θ = 30°Then angle C = 120° - 30° = 90°Therefore, sides would be:b = 8 sin 30° = 8 * 0.5 = 4 cmc = 8 sin 90° = 8 * 1 = 8 cmThus, the difference c - b = 8 - 4 = 4 cm. Since the absolute value is considered, this also satisfies the condition. So, this case is also valid.Therefore, the two possible triangles have sides (a=4√3, b=8, c=4) or (a=4√3, b=4, c=8). However, we should check if both triangles are valid. Let's check the triangle sides.In the first case, sides are 4√3, 8, and 4. Let's check the triangle inequality.4√3 ≈ 6.928 cm.So, sides are approximately 6.928, 8, and 4.6.928 + 8 > 4: 14.928 > 4 ✔️6.928 + 4 > 8: 10.928 > 8 ✔️8 + 4 > 6.928: 12 > 6.928 ✔️Similarly, in the second case, sides are 4√3, 4, and 8.Same as the first case, since 4 and 8 are swapped. The triangle inequalities still hold. So both triangles are valid. But since the sides enclosing the 60-degree angle are b and c, which in one case are 8 and 4, and in the other case 4 and 8. The difference is 4 cm in both cases. So both are valid solutions.Therefore, there are two possible triangles, but since the problem asks for the area, we need to check if both triangles have the same area.Calculating the area for the first case: (1/2)*b*c*sin A = (1/2)*8*4*sin 60° = 16*(√3/2) = 8√3 cm².Similarly, for the second case, it's (1/2)*4*8*sin 60° = same result, 8√3 cm². So, regardless of the order of b and c, the area is the same. Therefore, the area is 8√3 cm².Wait, but let me verify this. Alternatively, maybe I can compute the area using another formula to check.Alternatively, using Heron's formula. Let's compute the semi-perimeter (s) and then the area.In the first case, sides are 4√3, 8, 4.s = (4√3 + 8 + 4)/2 = (12 + 4√3)/2 = 6 + 2√3.Area = √[s(s - a)(s - b)(s - c)] = √[ (6 + 2√3)(6 + 2√3 - 4√3)(6 + 2√3 - 8)(6 + 2√3 - 4) ]Simplify each term:s - a = 6 + 2√3 - 4√3 = 6 - 2√3s - b = 6 + 2√3 - 8 = -2 + 2√3s - c = 6 + 2√3 - 4 = 2 + 2√3So, Area = √[ (6 + 2√3)(6 - 2√3)(-2 + 2√3)(2 + 2√3) ]First, compute (6 + 2√3)(6 - 2√3) = 6² - (2√3)² = 36 - 12 = 24Then, compute (-2 + 2√3)(2 + 2√3) = (-2)(2) + (-2)(2√3) + 2√3*2 + 2√3*2√3 = -4 -4√3 + 4√3 + 4*3 = -4 + 12 = 8Therefore, Area = √(24 * 8) = √192 = √(64*3) = 8√3 cm². So that checks out.Similarly, in the second case, sides are 4√3, 4, 8. It's the same set of sides, just reordered, so Heron's formula would give the same result. Therefore, the area is indeed 8√3 cm².But wait, the problem says "the difference between the two sides that enclose this angle is 4 cm". The two sides enclosing the 60-degree angle are b and c, which we found to be 8 and 4, so their difference is 4 cm. So, everything seems consistent.Alternatively, could there be another approach? For example, using the formula that relates the area with the circumradius: Area = (a*b*c)/(4R). Let's see if that gives the same answer.We have sides a=4√3, b=8, c=4. Then, a*b*c = 4√3 *8*4 = 128√3. Divided by 4R: 128√3 / (4*4) = 128√3 /16 = 8√3 cm². Yep, same result. So that's another confirmation.Therefore, the area of the triangle is 8√3 cm².But wait, let me just make sure I didn't make any mistakes in the trigonometric steps earlier. When I converted the expression (1/2 sin θ - √3/2 cos θ) into sin(θ - 60°), that's correct because sin(A - B) = sin A cos B - cos A sin B. So if A = θ and B = 60°, then sin(θ - 60°) = sin θ cos 60° - cos θ sin 60° = (1/2) sin θ - (√3/2) cos θ. So that step is correct. Therefore, |sin(θ - 60°)| = 0.5, leading to θ - 60° = ±30°, hence θ = 90° or 30°, which is correct.Therefore, the conclusion holds. The area is 8√3 cm².Final AnswerThe area of the triangle is boxed{8sqrt{3}} square centimeters.

✅:Given a triangle with a circumradius of (4 mathrm{~cm}) and one interior angle of (60^circ), where the difference between the two sides enclosing this angle is (4 mathrm{~cm}), we need to find the area of the triangle.1. Using the Law of Sines: - The circumradius (R = 4 mathrm{~cm}) implies (2R = 8 mathrm{~cm}). - For the side (a) opposite the (60^circ) angle: (a = 8 sin 60^circ = 8 times frac{sqrt{3}}{2} = 4sqrt{3} mathrm{~cm}).2. Setting up the sides enclosing the (60^circ) angle: - Let the sides be (b) and (c) with (|b - c| = 4 mathrm{~cm}). - Using the Law of Sines: (b = 8 sin B) and (c = 8 sin C), where (B + C = 120^circ).3. Solving for angles (B) and (C): - Let (B = theta) and (C = 120^circ - theta). - The difference (|8 sin theta - 8 sin (120^circ - theta)| = 4 mathrm{~cm}) simplifies to (|sin theta - sin (120^circ - theta)| = 0.5). - Using trigonometric identities, this simplifies to (| sin (theta - 60^circ) | = 0.5), leading to (theta = 90^circ) or (theta = 30^circ).4. Calculating the sides: - For (theta = 90^circ): (b = 8 sin 90^circ = 8 mathrm{~cm}) and (c = 8 sin 30^circ = 4 mathrm{~cm}). - For (theta = 30^circ): (b = 8 sin 30^circ = 4 mathrm{~cm}) and (c = 8 sin 90^circ = 8 mathrm{~cm}).5. Calculating the area: - The area using the formula (frac{1}{2}bc sin 60^circ): [ text{Area} = frac{1}{2} times 8 times 4 times frac{sqrt{3}}{2} = 8sqrt{3} mathrm{~cm}^2 ] - Verification using Heron's formula and the formula (frac{abc}{4R}) confirms the area.Thus, the area of the triangle is (boxed{8sqrt{3}}) square centimeters.

🔑:Consider the triangle ( triangle ABC ) with ( angle A = 60^circ ), ( b - c = 4 ), and circumradius ( R = 4 ) cm. To find the area of the triangle, we will proceed through the following steps:1. Using the Law of Cosines: From the Law of Cosines, we have: [ a^2 = b^2 + c^2 - 2bc cos A ] Given ( cos 60^circ = frac{1}{2} ), we can substitute cosine's value into the equation: [ a^2 = b^2 + c^2 - 2bc cdot frac{1}{2} = b^2 + c^2 - bc ]2. Using the given side difference ( b - c = 4 ): Knowing ( (b - c)^2 ), we can also express: [ (b - c)^2 = b^2 - 2bc + c^2 = 16 ]3. Combine the equations: We have: [ (b - c)^2 = 16 quad text{and} quad a^2 = b^2 + c^2 - bc ] Substituting ( (b - c)^2 + bc ) into ( a^2 ): [ a^2 = (b^2 + c^2) - bc = 16 + bc ]4. Determine the side ( a ): Using the relationship between ( a ), ( R ) and ( angle A ): [ a = 2R sin angle A ] Since ( R = 4 ) cm and ( sin 60^circ = frac{sqrt{3}}{2} ): [ a = 2 times 4 times frac{sqrt{3}}{2} = 4 sqrt{3} ]5. Calculate ( bc ) from ( a^2 ): Substitute ( a = 4 sqrt{3} ) into the ( a^2 ) expression: [ a^2 = (4 sqrt{3})^2 = 48 ] Thus: [ 48 = 16 + bc implies bc = 48 - 16 = 32 ]6. Determine the area ( S ) of ( triangle ABC ): Using the formula for the area of a triangle: [ S = frac{1}{2} bc sin A ] Substituting ( bc = 32 ) and ( sin 60^circ = frac{sqrt{3}}{2} ): [ S = frac{1}{2} times 32 times frac{sqrt{3}}{2} = 8 sqrt{3} ]Conclusion:The area of the triangle is:[boxed{8 sqrt{3} text{ cm}^2}]

🧠:Okay, let's see. So the problem is about Xiaoming and Xiaoliang exchanging stamps. I need to find the total number of stamps they have together. Hmm, let's break it down step by step.First, the problem says Xiaoming exchanges two stamps each with a face value of 1 yuan 6 jiao for Xiaoliang's stamps with a face value of 2 jiao each. Wait, exchanging according to face value. So the total value of the stamps Xiaoming gives should equal the total value of the stamps he receives from Xiaoliang. Let me confirm that. Yeah, exchanging according to face value means the total value exchanged is equal on both sides. So, Xiaoming gives two stamps each worth 1.6 yuan (since 1 yuan is 10 jiao, so 1 yuan 6 jiao is 16 jiao or 1.6 yuan). So each stamp is 1.6 yuan, so two stamps would be 3.2 yuan total. He exchanges this for Xiaoliang's stamps that are 2 jiao each. 2 jiao is 0.2 yuan. So the number of stamps Xiaoliang gives to Xiaoming would be total value divided by 0.2. So 3.2 / 0.2 = 16 stamps. So Xiaoming gives 2 stamps and receives 16 stamps from Xiaoliang.Before the exchange, Xiaoliang's number of stamps was 5 times Xiaoming's. After the exchange, Xiaoliang's stamps are 3 times Xiaoming's. We need to find the total number of stamps they have.Let me assign variables. Let's let M be the number of stamps Xiaoming had initially, and L be the number Xiaoliang had initially. Then, before the exchange, L = 5M. That's given.After the exchange, Xiaoming gives away 2 stamps and receives 16, so his new number is M - 2 + 16 = M + 14. Xiaoliang gives away 16 stamps and receives 2, so his new number is L - 16 + 2 = L - 14.After the exchange, Xiaoliang's stamps are 3 times Xiaoming's. So:L - 14 = 3*(M + 14)We already know that L = 5M. So substitute L with 5M in the equation:5M - 14 = 3*(M + 14)Let me solve that:5M - 14 = 3M + 42Subtract 3M from both sides:2M - 14 = 42Add 14 to both sides:2M = 56Divide by 2:M = 28So Xiaoming initially had 28 stamps. Then Xiaoliang had 5*28 = 140 stamps.Total stamps before exchange: 28 + 140 = 168.After exchange, Xiaoming has 28 + 14 = 42, Xiaoliang has 140 -14 = 126. 126 is indeed 3 times 42. So that checks out.Therefore, the total number of stamps they have is 168. Wait, but the problem says "Find the total number of stamps they have." Is it before or after the exchange? Hmm. The problem says "Find the total number of stamps they have." Usually, unless specified, I think it refers to the total they have in the end, but maybe it's asking for the total before the exchange. Wait, let me check.Wait, the problem says "Find the total number of stamps they have." Since the exchange is part of the problem's scenario, but the total number of stamps might not change because they are just exchanging, not creating or destroying stamps. Let's verify. Xiaoming gives 2 stamps and receives 16, so net gain of 14. Xiaoliang gives 16 and receives 2, so net loss of 14. So total stamps remain the same. So total before and after is the same. Therefore, whether it's before or after, the total is 168.So the answer is 168. Let me double-check the calculations.Initial M =28, L=140. After exchange, M becomes 28 -2 +16 =42. L becomes 140 -16 +2=126. 42 +126=168. Yep, same as before. So total is 168.

✅:To solve the problem, we start by defining variables for the initial number of stamps each person has. Let ( M ) be the number of stamps Xiaoming had initially, and ( L ) be Xiaoliang's initial count. Before the exchange, Xiaoliang had 5 times as many stamps as Xiaoming: [L = 5M]During the exchange, Xiaoming trades two stamps worth 1.6 yuan each (total value: ( 2 times 1.6 = 3.2 ) yuan) for Xiaoliang's stamps worth 0.2 yuan each. The number of 0.2 yuan stamps Xiaoliang gives is: [frac{3.2}{0.2} = 16 text{ stamps}.]After the exchange:- Xiaoming has ( M - 2 + 16 = M + 14 ) stamps.- Xiaoliang has ( L - 16 + 2 = L - 14 ) stamps.The problem states that after the exchange, Xiaoliang's stamps are 3 times Xiaoming's: [L - 14 = 3(M + 14)]Substituting ( L = 5M ) into the equation: [5M - 14 = 3(M + 14)]Solving for ( M ): [5M - 14 = 3M + 42 2M = 56 M = 28]Thus, Xiaoming initially had 28 stamps, and Xiaoliang had: [L = 5 times 28 = 140 text{ stamps}.]Total stamps before/after exchange: [28 + 140 = 168]Verification:- After exchange: Xiaoming has ( 28 + 14 = 42 ), Xiaoliang has ( 140 - 14 = 126 ).- ( 126 = 3 times 42 ), confirming correctness.Final Answer: [boxed{168}]

🔑:Let's break this problem down step by step.1. Exchange Calculation: - Xiao Ming exchanged two stamps of face value 1 yuan 6 jiao each for Xiao Liang's stamps of face value 2 jiao each. - First, convert the face value into the same denomination (jiao): (1 text{ yuan} = 10 text{ jiao}). - Therefore, (1 text{ yuan} 6 text{ jiao} = 10 text{ jiao} + 6 text{ jiao} = 16 text{ jiao}).2. Total face value exchanged: - Xiao Ming used two stamps, each of 16 jiao. - Total face value exchanged by Xiao Ming: (2 times 16 text{ jiao} = 32 text{ jiao}).3. Equivalent in Xiao Liang's stamps: - Since each of Xiao Liang's stamps is worth 2 jiao, the number of Xiao Liang's stamps equivalent to 32 jiao is: (frac{32 text{ jiao}}{2 text{ jiao/stamp}} = 16 text{ stamps}).4. Changes in the number of stamps: - After the exchange, Xiao Ming received 16 stamps from Xiao Liang, and he gave away 2 stamps to Xiao Liang. - Net increase in stamps for Xiao Ming: (16 - 2 = 14 text{ stamps}).5. Initial relationship between Xiao Liang's and Xiao Ming's stamps: - Before the exchange, Xiao Liang had five times as many stamps as Xiao Ming. - Let (N_m) be the number of stamps Xiao Ming initially had. - Therefore, Xiao Liang initially had (5N_m) stamps.6. Post-exchange relationship: - After the exchange, Xiao Ming had (N_m + 14) stamps. - Xiao Liang gave away 16 stamps and received 2, so they had (5N_m - 14) stamps left. - According to the problem, Xiao Liang's post-exchange number of stamps is three times that of Xiao Ming's post-exchange number of stamps: [ 5N_m - 14 = 3(N_m + 14) ]7. Solve for (N_m): [ 5N_m - 14 = 3N_m + 42 ] [ 5N_m - 3N_m = 42 + 14 ] [ 2N_m = 56 ] [ N_m = 28 ]8. Total number of stamps: - Xiao Ming initially had 28 stamps. - Xiao Liang initially had (5 times 28 = 140) stamps. - Together, initially, they had (28 + 140 = 168) stamps.Conclusion:[boxed{168}]

🧠:Alright, so I have these five volume of revolution problems to solve. Let me tackle them one by one. I need to remember the methods for calculating volumes when rotating around different axes. The main techniques I know are the disk method and the washer method. Sometimes, maybe the shell method would be more appropriate, especially if rotating around a vertical or horizontal line that's not the axis itself. Let me start with the first problem.Problem 1: ( y^{2} = 2px ), ( x = a ) around the ( Ox )-axis.Okay, so first, I need to visualize the region bounded by the curve ( y^2 = 2px ) and the line ( x = a ). The curve ( y^2 = 2px ) is a parabola that opens to the right, with vertex at the origin. The line ( x = a ) is a vertical line. So the region we're looking at is from the vertex of the parabola up to ( x = a ). When rotated around the x-axis, this should form a solid like a paraboloid.To find the volume, since we're rotating around the x-axis, and the cross-sections perpendicular to the x-axis are disks. The radius of each disk is given by the y-value of the function. Since ( y^2 = 2px ), solving for y gives ( y = sqrt{2px} ). But since the parabola is symmetric around the x-axis, the radius is actually from ( y = -sqrt{2px} ) to ( y = sqrt{2px} ), so the radius R is ( sqrt{2px} ).The volume using the disk method is ( pi int_{0}^{a} [R(x)]^2 dx ). Plugging in R(x):( V = pi int_{0}^{a} (sqrt{2px})^2 dx = pi int_{0}^{a} 2px dx )Calculating the integral:( 2ppi int_{0}^{a} x dx = 2ppi [ frac{x^2}{2} ]_{0}^{a} = 2ppi (frac{a^2}{2} - 0) = ppi a^2 )Wait, but let me check. The parabola ( y^2 = 2px ) is standard. When rotated around the x-axis, each cross-sectional area is a circle with area ( pi y^2 ), which is exactly ( pi (2px) ). So integrating from 0 to a:( pi int_{0}^{a} 2px dx ). Which is the same as above, so the result is ( pi p a^2 ). That seems right. So problem 1's answer is ( pi p a^2 ).Problem 2: ( frac{x^{2}}{a^{2}} + frac{y^{2}}{b^{2}} = 1 ) around the ( Oy )-axis.This is the equation of an ellipse centered at the origin. Rotating around the y-axis. So when rotating around the y-axis, we can use either the disk/washer method or the shell method. Let's see. If we use the disk method, we need to express x in terms of y. Solving for x:( frac{x^2}{a^2} = 1 - frac{y^2}{b^2} )( x^2 = a^2 (1 - frac{y^2}{b^2}) )So x = ±a sqrt(1 - y²/b²). The cross-sectional radius at a given y is x, so the radius is a sqrt(1 - y²/b²). However, since we're rotating around the y-axis, each disk's radius is x, and the volume is ( pi int_{-b}^{b} [x(y)]^2 dy ). But since the ellipse is symmetric about the x-axis, we can integrate from 0 to b and double it. Let me verify.But actually, integrating from -b to b would cover the entire ellipse. The formula using the disk method would be:( V = pi int_{-b}^{b} [x(y)]^2 dy )Plugging in x(y):( V = pi int_{-b}^{b} a^2 (1 - frac{y^2}{b^2}) dy )Since the integrand is even (symmetric about y=0), we can do 2 times integral from 0 to b:( V = 2pi a^2 int_{0}^{b} (1 - frac{y^2}{b^2}) dy )Calculate the integral:First term: ( int_{0}^{b} 1 dy = b )Second term: ( int_{0}^{b} frac{y^2}{b^2} dy = frac{1}{b^2} cdot frac{y^3}{3} bigg|_{0}^{b} = frac{1}{b^2} cdot frac{b^3}{3} = frac{b}{3} )So combining:( 2pi a^2 (b - frac{b}{3}) = 2pi a^2 (frac{2b}{3}) = frac{4}{3}pi a^2 b )Wait, but that seems familiar. The volume of an ellipsoid. When you rotate an ellipse around one of its axes, the volume is ( frac{4}{3}pi a b^2 ) if rotated around the x-axis, and ( frac{4}{3}pi a^2 b ) if rotated around the y-axis. So here, rotating around the y-axis, the answer is ( frac{4}{3}pi a^2 b ). That seems correct.Alternatively, if using the shell method, but since we’re rotating around the y-axis, and the ellipse is expressed as x in terms of y, disk method is straightforward here. So problem 2's answer is ( frac{4}{3}pi a^2 b ).Problem 3: ( 2y = x^{2} ), ( 2x + 2y - 3 = 0 ) around the ( Ox )-axis.First, find the region bounded by these two curves. Let me rewrite the equations:1. ( y = frac{x^2}{2} ): a parabola opening upwards.2. ( 2x + 2y - 3 = 0 ) => ( y = frac{3 - 2x}{2} = frac{3}{2} - x ): a straight line with slope -1.To find the points of intersection, set ( frac{x^2}{2} = frac{3}{2} - x )Multiply both sides by 2:( x^2 = 3 - 2x )Bring all terms to left:( x^2 + 2x - 3 = 0 )Factor: Looking for two numbers multiplying to -3 and adding to 2. That would be 3 and -1.So, (x + 3)(x - 1) = 0 → x = -3 or x = 1.So the curves intersect at x = -3 and x = 1. Let me find the corresponding y-values.For x = -3: y = (-3)^2 / 2 = 9/2 = 4.5, and from the line: y = 3/2 - (-3) = 3/2 + 3 = 4.5. So point (-3, 4.5).For x = 1: y = 1/2, and from the line: y = 3/2 - 1 = 1/2. So point (1, 1/2).Therefore, the region bounded by these curves is between x = -3 and x = 1, with the top boundary being the line y = (3 - 2x)/2 and the bottom boundary being the parabola y = x²/2.When rotating around the x-axis, the volume can be computed using the washer method. The outer radius R(x) is the distance from the x-axis to the line, and the inner radius r(x) is the distance to the parabola.So, R(x) = (3 - 2x)/2 and r(x) = x²/2.The volume is ( pi int_{-3}^{1} [R(x)^2 - r(x)^2] dx )Compute this integral.First, expand R(x)^2 and r(x)^2.( R(x) = frac{3 - 2x}{2} )( R(x)^2 = frac{(3 - 2x)^2}{4} = frac{9 - 12x + 4x^2}{4} )( r(x)^2 = left( frac{x^2}{2} right)^2 = frac{x^4}{4} )So the integrand becomes:( frac{9 - 12x + 4x^2}{4} - frac{x^4}{4} = frac{9 - 12x + 4x^2 - x^4}{4} )Therefore, the integral is:( pi int_{-3}^{1} frac{9 - 12x + 4x^2 - x^4}{4} dx = frac{pi}{4} int_{-3}^{1} (9 - 12x + 4x^2 - x^4) dx )Let me compute each term separately.Integral of 9 dx = 9xIntegral of -12x dx = -6x²Integral of 4x² dx = (4/3)x³Integral of -x^4 dx = - (1/5)x^5So putting it all together:( frac{pi}{4} [9x - 6x^2 + frac{4}{3}x^3 - frac{1}{5}x^5 ] ) evaluated from x = -3 to x = 1.Compute at upper limit x=1:9(1) -6(1)^2 + (4/3)(1)^3 - (1/5)(1)^5 = 9 - 6 + 4/3 - 1/5Compute each:9 -6 = 34/3 ≈ 1.333..., 1/5 = 0.2, so 3 + 1.333... - 0.2 ≈ 4.133...But let's do exact fractions:3 = 15/5, 4/3 = 20/15, 1/5 = 3/15Wait, perhaps better to get common denominators.Alternatively, convert all to fifteenths:3 = 45/154/3 = 20/151/5 = 3/15So total: 45/15 - 3/15 + 20/15 = (45 - 3 + 20)/15 = 62/15Wait, wait. Wait, the expression is:9(1) = 9-6(1)^2 = -6(4/3)(1)^3 = 4/3- (1/5)(1)^5 = -1/5So sum: 9 -6 + 4/3 -1/5Convert all to fifteenths:9 = 135/15-6 = -90/154/3 = 20/15-1/5 = -3/15So total: 135 -90 +20 -3 = (135-90) + (20-3) = 45 +17 = 62 → 62/15So upper limit is 62/15.Lower limit x = -3:9(-3) -6(-3)^2 + (4/3)(-3)^3 - (1/5)(-3)^5Compute each term:9*(-3) = -27-6*(-3)^2 = -6*9 = -54(4/3)*(-3)^3 = (4/3)*(-27) = -36-(1/5)*(-3)^5 = -(1/5)*(-243) = 243/5So adding them up:-27 -54 -36 + 243/5First, sum the integers: -27 -54 -36 = -117Convert 243/5 to decimal if needed, but let's keep as fractions.So total is -117 + 243/5. Convert -117 to fifths: -117 = -585/5.So total: (-585/5 + 243/5) = (-585 +243)/5 = (-342)/5 = -342/5.Therefore, the lower limit is -342/5.Subtract lower limit from upper limit:[62/15 - (-342/5)] = 62/15 + 342/5 = convert 342/5 to fifteenths: 342/5 = 1026/15.So total: 62/15 + 1026/15 = (62 + 1026)/15 = 1088/15.Multiply by π/4:V = (π/4)*(1088/15) = π * 1088 / 60 = Simplify 1088/60.Divide numerator and denominator by 4: 272/15.So V = (272/15)π ≈ 18.133...πBut let me check the calculation again for possible errors. Because integrating from -3 to 1, which is a bit messy.Wait, let me verify the integral computation step by step.First, the antiderivative:F(x) = 9x -6x² + (4/3)x³ - (1/5)x^5At x =1:F(1) = 9*1 -6*1 + (4/3)*1 - (1/5)*1 = 9 -6 + 4/3 -1/5 = 3 + 4/3 -1/5Convert to fifteenths:3 = 45/15, 4/3 = 20/15, 1/5 = 3/15So 45/15 +20/15 -3/15 = 62/15. That's correct.At x = -3:F(-3) = 9*(-3) -6*(-3)^2 + (4/3)*(-3)^3 - (1/5)*(-3)^5= -27 -6*9 + (4/3)*(-27) - (1/5)*(-243)= -27 -54 -36 + 243/5= (-27 -54 -36) + 243/5= -117 + 48.6Wait, 243/5 is 48.6, so -117 +48.6 = -68.4. But in fractions:-117 +243/5 = -585/5 +243/5 = (-585 +243)/5 = (-342)/5 = -68.4. So that's -342/5.Then F(1) - F(-3) = 62/15 - (-342/5) = 62/15 +342/5 = as before, 62/15 +1026/15 =1088/15.So V = (π/4)*(1088/15) = π*(1088)/60 = 272π/15. So the volume is ( frac{272}{15}pi ). Hmm, that seems a bit complicated. Let me check if there was a miscalculation in the setup.Wait, the curves are between x = -3 and x =1. The line is above the parabola in this interval. So using the washer method is correct. The outer radius is the line, inner radius the parabola. So the integrand is (R^2 - r^2). The algebra seems okay.Alternatively, maybe I can compute the integral in a different way. Let's see.Alternatively, could I have made a mistake in expanding R(x)^2?Original R(x) = (3 -2x)/2, so squared is (9 -12x +4x²)/4. Yes, that's correct. Then r(x)^2 is x^4/4. So the integrand is (9 -12x +4x² -x^4)/4. Correct.So the integral is from -3 to1 of that. Then splitting into four terms:9/4 -3x +x² - (x^4)/4.Wait, no:Wait, (9 -12x +4x² -x^4)/4 is 9/4 - 3x + x² - (x^4)/4.So integrating term by term:Integral of 9/4 dx = (9/4)xIntegral of -3x dx = -(3/2)x²Integral of x² dx = (1/3)x³Integral of -x^4/4 dx = - (1/20)x^5Then evaluating from -3 to1.At x=1:(9/4)(1) - (3/2)(1)^2 + (1/3)(1)^3 - (1/20)(1)^5= 9/4 -3/2 +1/3 -1/20Convert to 60 denominator:9/4 =135/60, -3/2 = -90/60, 1/3=20/60, -1/20= -3/60Sum: 135 -90 +20 -3 = 62 → 62/60 = 31/30 ≈1.033...At x=-3:(9/4)(-3) - (3/2)(-3)^2 + (1/3)(-3)^3 - (1/20)(-3)^5= -27/4 - (3/2)(9) + (1/3)(-27) - (1/20)(-243)= -27/4 -27/2 -9 + 243/20Convert all to twentieths:-27/4 = -135/20-27/2 = -270/20-9 = -180/20243/20 is 243/20Sum: -135 -270 -180 +243 = (-585 +243)/20 = -342/20 = -171/10So the difference is:Upper limit: 31/30 - (-171/10) = 31/30 + 171/10 = convert 171/10 to 513/30So 31/30 +513/30 =544/30 =272/15Multiply by π (since the integral is (272/15)π)Wait, but in the previous step, we split the integral as 9/4 -3x +x² -x^4/4, but actually, the integral was divided by 4. Wait no, wait:Wait, when we expanded the integrand:(R(x)^2 - r(x)^2) = (9 -12x +4x² -x^4)/4. So each term is divided by 4. Therefore, the integral becomes:Integral of [9/4 -3x +x² -x^4/4] dxSo integrating term by term:9/4 x - (3/2)x² + (1/3)x³ - (1/20)x^5 evaluated from -3 to1.Which is exactly what we did. Then the difference was 272/15 π. So the calculation seems correct. Therefore, the answer is 272π/15.Problem 4: ( x = a cos^{3} t ), ( y = a sin^{3} t ) around the ( Ox )-axis.This is a parametric equation, which is an astroid. The parametric equations are x = a cos³t, y = a sin³t. Rotating around the x-axis. To find the volume, we can use the method for parametric curves. The formula for the volume when rotating around the x-axis is ( pi int_{t_1}^{t_2} [y(t)]^2 cdot frac{dx}{dt} dt ). We need to determine the limits of t that trace the entire curve once.Since it's an astroid, it's a closed curve. The parameter t ranges from 0 to 2π to complete the curve. However, due to symmetry, we might compute from 0 to π/2 and multiply by 4, but let's check.Alternatively, since the entire curve is traced as t goes from 0 to 2π, but due to the symmetry, maybe integrating from 0 to π/2 and multiplying by 4? Wait, but let's see.Alternatively, perhaps using the formula for the volume generated by a parametric curve. When rotating around the x-axis, the volume is:( V = pi int_{a}^{b} [y(x)]^2 dx ). But since it's parametric, we can write dx = (dx/dt) dt, so the integral becomes:( V = pi int_{t_1}^{t_2} [y(t)]^2 cdot frac{dx}{dt} dt )So need to compute this integral.First, compute y(t)^2: [a sin³t]^2 = a² sin⁶tCompute dx/dt: derivative of a cos³t with respect to t is 3a cos²t (-sin t) = -3a cos²t sin tTherefore, the integrand is:( pi cdot a² sin⁶t cdot (-3a cos²t sin t) dt )But we need to check the limits. When rotating around the x-axis, does the parametric curve enclose a region? The astroid is a closed curve, so when rotated around the x-axis, the solid formed is like a sort of curved diamond shape rotated around the x-axis. However, since it's a closed loop, we have to be careful about how the volume is computed.Alternatively, maybe we can use the theorem of Pappus: the volume is the area of the shape multiplied by the distance traveled by its centroid. But I need to remember the formula. The volume is equal to the area A times the circumference traveled by the centroid, which is 2π times the centroid's y-coordinate. Wait, no, when rotating around the x-axis, the centroid (x̄, ȳ) would have the volume as A * 2πȳ. Wait, no, actually, Pappus's theorem states that the volume is equal to the product of the area and the distance traveled by the centroid. So if rotating around the x-axis, the centroid's y-coordinate is ȳ, so the distance traveled is 2πȳ. Therefore, Volume = A * 2πȳ.But maybe that's more complex here because we need to compute the area and the centroid. Let's see if that's feasible.First, the astroid parametric equations: x = a cos³t, y = a sin³t. The area enclosed by the astroid can be computed using the formula for parametric curves:A = ∫ y dx (integrated over the closed curve). So:A = ∫_{0}^{2π} y(t) dx/dt dtBut since the curve is traced once from t=0 to t=π/2, and due to symmetry, we can compute from 0 to π/2 and multiply by 4.Wait, actually, no. Let's compute the area:A = 4 ∫_{0}^{π/2} y(t) dx/dt dtBecause the astroid has four symmetrical quadrants.Compute dx/dt = -3a cos²t sin tSo:A = 4 ∫_{0}^{π/2} a sin³t * (-3a cos²t sin t) dt= 4*(-3a²) ∫_{0}^{π/2} sin⁴t cos²t dtBut area should be positive, so take absolute value:A = 12a² ∫_{0}^{π/2} sin⁴t cos²t dtThis integral can be computed using beta functions or using reduction formulas. Alternatively, using substitution. Let me recall that ∫ sin^n t cos^m t dt from 0 to π/2 is equal to (Γ((n+1)/2)Γ((m+1)/2))/(2Γ((n+m+2)/2))). For even exponents, we can use the formula for beta functions.Here, n=4, m=2.So, Γ((4+1)/2)=Γ(5/2)= (3/2)(1/2)√π = (3/4)√πΓ((2+1)/2)=Γ(3/2)= (1/2)√πΓ((4+2+2)/2)=Γ(8/2)=Γ(4)=3!=6Therefore, the integral is (Γ(5/2)Γ(3/2))/(2Γ(4)) ) = [(3/4 √π)(1/2 √π)] / (2*6) )= [ (3/8 π) ] / 12 = (3π/8)/12 = π/32. Wait, that seems conflicting. Wait, maybe my substitution is off.Alternatively, using the formula:∫_{0}^{π/2} sin^n t cos^m t dt = [ (n-1)!!(m-1)!! ) / (n + m)!! ) ] * π/2 if n and m are even,otherwise different. For n=4, m=2, both even.So:= [ (3!!)(1!!) ) / (6!!) ] * π/2But 3!! = 3*1=3, 1!!=1, 6!!=6*4*2=48So:= (3*1 /48 ) * π/2 = (3/48)(π/2) = (1/16)(π/2) = π/32Therefore, the integral is π/32. Therefore, area A=12a²*(π/32)= (12/32)a²π= (3/8)a²π.But I recall that the area of an astroid is known to be (3/8)πa². So that's correct.Now, using Pappus's theorem, we need the centroid ȳ. Since the astroid is symmetric about both axes, the centroid is at (0,0) if it's centered at the origin. Wait, but the parametric equations are x = a cos³t, y = a sin³t. The centroid would actually be at the origin? Wait, no. Wait, no, the centroid of the area is at the origin due to symmetry. Wait, but the astroid is symmetric in all four quadrants, so the centroid (x̄, ȳ) is (0,0). But then according to Pappus's theorem, Volume = A * 2π * ȳ = 0. That can't be right. That would suggest the volume is zero, which is impossible. So there's a mistake here.Wait, actually, no. The astroid is a 2D shape; when rotated around the x-axis, the generated solid is a 3D shape. But Pappus's theorem says that the volume is equal to the area of the 2D shape multiplied by the distance traveled by its centroid. However, if the centroid is on the axis of rotation (the x-axis here, since ȳ=0), then the distance traveled is zero, hence the volume is zero. But that's not correct. Therefore, my assumption that the centroid is at (0,0) is wrong.Wait, no. Wait, the astroid is symmetric about the x-axis and y-axis. So the centroid (x̄, ȳ) is indeed (0,0). Therefore, rotating the astroid around the x-axis, which is an axis of symmetry, would result in the centroid traveling a distance of 2π * ȳ = 0. Therefore, Pappus's theorem gives zero, which is wrong. So clearly, something's wrong here. Wait, but Pappus's theorem applies to surfaces and solids of revolution. Maybe I misapplied the theorem here. Wait, no. Wait, the theorem says that the volume of the solid of revolution generated by rotating a plane figure about an external axis is equal to the product of the area of the figure and the distance traveled by its centroid. However, if the centroid is on the axis of rotation, then the distance traveled is zero, hence the volume is zero. But that's not the case here. Because rotating the astroid around the x-axis, the entire area is off the axis (except the points on the x-axis). Wait, but the centroid is at the origin. Wait, the astroid has points on the x-axis (when y=0) and on the y-axis (when x=0). But as a whole, the centroid is at (0,0). So according to Pappus, rotating around the x-axis would produce a volume equal to the area times the distance traveled by the centroid. Since the centroid is on the x-axis (ȳ=0), the distance traveled is 2π*0=0. Therefore, Volume = 0, which is not correct. Therefore, there must be a misunderstanding.Wait, perhaps I made a mistake in using Pappus's theorem here. Wait, the astroid is a closed curve, but when rotated around the x-axis, it creates a solid which is like a surface of revolution. However, if the original figure is a loop, then the volume generated is the same as if you rotate the area enclosed by the loop. But in this case, the astroid is the boundary, and when rotated, it creates a surface, but perhaps the volume inside the surface? But if the centroid is on the axis, then the volume is zero? That can't be. So perhaps my application of Pappus's theorem is incorrect here because the figure is being rotated about an axis that it's symmetric around, but in reality, the volume is not zero. Therefore, maybe Pappus's theorem is not applicable here, or I made a mistake in computing the centroid.Alternatively, maybe the issue is that the astroid is a parametric curve, not a lamina (a filled-in area). If we consider the astroid as a curve, not an area, then rotating it would form a surface, but the theorem applies to areas. So if we want the volume enclosed by the surface of revolution, we need to compute it via integration. So perhaps the problem is that the parametric equations given are for the boundary, and we need to compute the volume generated by rotating the region bounded by the astroid around the x-axis. But an astroid is a closed curve, so the region bounded by it is the interior. Therefore, perhaps integrating using the disk/washer method for parametric equations.Alternatively, perhaps using the formula for the volume generated by a parametric curve when rotated about the x-axis:( V = pi int_{t_1}^{t_2} [y(t)]^2 cdot frac{dx}{dt} dt )But we need to be careful here. If the parametric equations trace the entire boundary of the region, then integrating this would give the volume. However, because the astroid is a closed curve, we need to ensure that the integral accounts for the entire area.Given the parametric equations x = a cos³t, y = a sin³t, which trace the astroid once as t goes from 0 to 2π. However, due to symmetry, perhaps integrating from 0 to π/2 and multiplying by 4.But let's proceed step by step.Compute V = π ∫ [y(t)]^2 dx/dt dt over the interval where the curve is above the x-axis. Wait, but since the curve is symmetric, maybe we can integrate from 0 to π/2 and multiply by 4, then account for the lower half?Wait, no, when rotating around the x-axis, the volume is generated by the entire region above and below the x-axis. However, in parametric form, y(t) is given as a sin³t, which is positive when t is in 0 to π, and negative when t is in π to 2π. So integrating from 0 to 2π would cover the entire rotation.But let's compute the integral:V = π ∫_{0}^{2π} [a sin³t]^2 * (-3a cos²t sin t) dt= π * a² * (-3a) ∫_{0}^{2π} sin⁶t * cos²t * sin t dt= -3π a³ ∫_{0}^{2π} sin⁷t cos²t dtBut sin⁷t cos²t is an odd function? Wait, no. Wait, the integrand is sin⁷t cos²t. Let's check the parity.If we substitute t → -t, sin(-t) = -sin t, cos(-t) = cos t. So sin⁷(-t) cos²(-t) = (-sin t)^7 cos²t = -sin⁷t cos²t. So the integrand is odd. But integrating over a symmetric interval from -π to π would give zero. But we're integrating from 0 to 2π, which is not symmetric. However, integrating over 0 to 2π, the integral of an odd function over a full period is zero? Wait, but sin⁷t cos²t over 0 to 2π. Wait, the function is periodic with period 2π. Let's check:The integral over 0 to 2π of sin⁷t cos²t dt.However, sin^7 t cos^2 t is a product of odd and even functions? Wait, cos^2 t is even, sin^7 t is odd. The product is odd * even = odd. But integrating an odd function over a full period (symmetric around 0) would give zero. But the interval here is 0 to 2π, which is not symmetric. However, the integral over 0 to 2π is the same as integrating over -π to π, shifted by π. But sin^7(t) cos²(t) has period π, maybe?Wait, this is getting too complicated. Let's compute the integral directly.But considering that sin⁷t cos²t can be written as sin t (1 - cos²t)^3 cos²t. But perhaps expanding it would help.Alternatively, use substitution. Let’s note that over the interval 0 to 2π, the integral of sin^m t cos^n t dt can be evaluated using beta functions if we use substitution. However, given that the exponents are both even and odd, it might be complex. Alternatively, consider that integrating from 0 to 2π, and due to periodicity, we can compute over 0 to π/2 and multiply by 4.Wait, let's try that. Split the integral into four parts: 0 to π/2, π/2 to π, π to 3π/2, 3π/2 to 2π. Due to symmetry, each quadrant integral may be the same or have specific relations.But let me make a substitution. Let u = t - π/2, etc., but this might not help.Alternatively, use the substitution z = t - π. Then when t ranges from 0 to 2π, z ranges from -π to π. The integrand sin^7(t)cos^2(t) becomes sin^7(z + π)cos^2(z + π) = (-sin z)^7 (cos(z + π))^2 = (-sin z)^7 (-cos z)^2 = -sin^7 z cos^2 z. Therefore, the integral becomes ∫_{-π}^{π} -sin^7 z cos^2 z dz. Since this is an odd function, the integral over symmetric limits is zero. Therefore, the integral from 0 to 2π is zero.Wait, but that would mean the volume is zero, which is impossible. Therefore, there must be a mistake in the setup.Wait, the original integral for the volume is:V = π ∫_{0}^{2π} [y(t)]^2 dx/dt dtBut if this integral equals zero, that can't be. Therefore, perhaps the error is in the limits of integration or in the parametrization.Wait, but when rotating a closed curve around an axis, the parametric integral must account for the entire enclosed area. However, the problem is that when using the formula ( V = pi int [y(t)]^2 dx ), for a parametric curve, we are essentially using the disk method where y(t) is the radius and dx is the infinitesimal thickness. However, if the curve is closed, when you integrate around the entire curve, the contributions above and below the axis might cancel out, leading to zero. Which is not the case. Therefore, perhaps this formula is only applicable for simple, non-closed curves where y(t) remains non-negative and the curve is traversed once from left to right.Alternatively, maybe we need to compute the volume using the surface of revolution generated by the upper half of the astroid, and then double it, considering symmetry.The astroid has four cusps. The upper half (where y ≥ 0) is from t = 0 to t = π. But actually, in the parametrization, when t is from 0 to π, y(t) = a sin³t is non-negative. Then from t = π to 2π, y(t) is non-positive. However, since we are squaring y(t), [y(t)]^2 is positive in both cases. But dx/dt is -3a cos²t sin t. So let's analyze the integrand:[y(t)]^2 dx/dt = [a² sin⁶t][-3a cos²t sin t] = -3a³ sin⁷t cos²tThus, the integrand is -3a³ sin⁷t cos²t.The integral over 0 to 2π of this would be zero, as we saw before, due to symmetry. Therefore, this approach gives zero, which is incorrect. Therefore, this method isn't suitable here. What's the issue?Ah! I think I realize the problem. The formula ( V = pi int [y(t)]^2 dx ) is applicable when the curve is a function y(x), i.e., single-valued, and you are moving from left to right along the x-axis. However, the astroid is a closed curve, so when t goes from 0 to 2π, the curve is traced clockwise (since dx/dt is negative when t is in 0 to π/2, where x decreases as t increases). Therefore, integrating from t=0 to t=2π would actually subtract the volume below the x-axis. Wait, but [y(t)]^2 is always positive, so the integral accumulates the volume both above and below the x-axis. But since dx/dt is negative in some regions, it might subtract in those parts.Alternatively, maybe we need to take the absolute value of dx/dt? No, because in the parametric integral, dx is just dx/dt dt, which can be positive or negative. But when computing the volume, the disk method requires that you move along the x-axis from left to right, stacking the disks. However, if the parametrization causes x to move back and forth, then integrating without considering the direction would lead to cancellation.Therefore, for closed curves, the standard disk/washer method may not apply directly. Instead, we need to parametrise only the upper half of the curve where y is positive and integrate from x leftmost to rightmost, then double it (if symmetric). But in this case, the astroid is symmetric about the x-axis, so the volume generated by rotating the upper half around the x-axis would be half the total volume. Wait, no. Because when you rotate the full astroid around the x-axis, the volume is generated by both the upper and lower halves, but since it's symmetric, the total volume would be twice the volume generated by the upper half.But how to parametrise the upper half? The upper half of the astroid corresponds to y ≥ 0, which in parametric terms is t from 0 to π. But x(t) in this interval goes from a (at t=0) to -a (at t=π). So x decreases from a to -a. If we use this parametrisation, we would integrate from t=0 to t=π, but since x is decreasing, dx/dt is negative. So when using the formula, integrating from t=0 to t=π would give a negative contribution. Therefore, to compute the volume, we should take the absolute value of dx/dt? Or integrate from leftmost to rightmost x.Alternatively, split the integral into two parts: from x=a to x=-a (upper half) and x=-a to x=a (lower half). But this might complicate things.Alternatively, express y as a function of x for the upper half and integrate. Let's try that.The parametric equations are x = a cos³t, y = a sin³t. To express y in terms of x, solve for t.From x = a cos³t, we get cos t = (x/a)^{1/3}, so t = arccos((x/a)^{1/3}). Then y = a sin³t. Using the identity sin²t + cos²t =1:sin t = sqrt(1 - cos²t) = sqrt(1 - (x/a)^{2/3})Therefore, y = a [sqrt(1 - (x/a)^{2/3})]^3 = a [1 - (x/a)^{2/3}]^{3/2}Therefore, for the upper half, y = a [1 - (x/a)^{2/3}]^{3/2}But integrating this from x=-a to x=a might be complex. Let's make a substitution.Let me change variable: let u = x/a, so x = a u, dx = a du. Then y = a [1 - u^{2/3}]^{3/2}The volume when rotating around the x-axis is:V = π ∫_{-a}^{a} [y(x)]^2 dx = π ∫_{-a}^{a} [a^2 (1 - (x/a)^{2/3})^3 ] dx= π a² ∫_{-a}^{a} [1 - (x/a)^{2/3}]^3 dxAgain, since the integrand is even (symmetric about x=0), we can compute from 0 to a and double it:V = 2π a² ∫_{0}^{a} [1 - (x/a)^{2/3}]^3 dxLet’s substitute u = (x/a)^{2/3}. Let’s see. Let’s set t = (x/a)^{2/3}, then x = a t^{3/2}, dx = a*(3/2) t^{1/2} dtBut perhaps another substitution. Let’s set u = (x/a)^{2/3}, so x = a u^{3/2}, dx = a*(3/2) u^{1/2} duWhen x=0, u=0; x=a, u=1.So the integral becomes:∫_{0}^{a} [1 - (x/a)^{2/3}]^3 dx = a*(3/2) ∫_{0}^{1} [1 - u]^3 u^{1/2} du= (3a/2) ∫_{0}^{1} (1 - u)^3 u^{1/2} duThis is a beta function integral. Recall that ∫_{0}^{1} u^{c-1} (1 - u)^{d-1} du = B(c, d) = Γ(c)Γ(d)/Γ(c+d)Here, c = 3/2 (since u^{1/2} = u^{3/2 -1}), and d = 4 (since (1 - u)^3 = (1 - u)^{4 -1})So B(3/2, 4) = Γ(3/2)Γ(4)/Γ(11/2)Compute each gamma function:Γ(3/2) = (1/2)√πΓ(4) = 3! = 6Γ(11/2) = (9/2)(7/2)(5/2)(3/2)(1/2)√π = (945/32)√πTherefore, B(3/2,4) = ( (1/2)√π * 6 ) / (945/32 √π ) ) = (3√π) / (945/32 √π ) = 3 / (945/32) = (3*32)/945 = 96/945 = 32/315Therefore, the integral becomes:(3a/2) * (32/315) = (3a/2)(32/315) = (96a)/630 = (16a)/105So the volume V is:2π a² * (16a/105) = 32π a³ / 105Thus, the volume is ( frac{32}{105}pi a^3 ).Wait, but let me check this result. Let's verify with another method.Alternatively, recall that the astroid is a hypocycloid with four cusps. The volume generated by rotating it around the x-axis is known. If I recall correctly, the volume should be ( frac{32}{105}pi a^3 ). Yes, this seems familiar. So that's correct.Problem 5: ( y = 4 - x^{2} ), ( y = 0 ) around the line ( x = 3 ).This is a rotated region around a vertical line x=3. The curves are y=4 -x² (a downward parabola) and y=0 (the x-axis). The region bounded by these is the area under the parabola from x=-2 to x=2, since y=4 -x² intersects y=0 at x=±2.When rotating around x=3, which is a vertical line to the right of the region (since the parabola is between x=-2 and 2, and x=3 is at x=3). To compute this volume, the shell method might be appropriate, as the axis of rotation is vertical, and we can integrate with respect to x.The shell method formula is ( 2pi int_{a}^{b} (radius)(height) dx ). The radius of each shell is the horizontal distance from x to x=3, which is (3 - x). The height of each shell is y = 4 -x². The limits of integration are from x=-2 to x=2.Therefore, the volume V is:V = 2π ∫_{-2}^{2} (3 - x)(4 -x²) dxFirst, expand the integrand:(3 -x)(4 -x²) = 12 -3x² -4x +x³So:V = 2π ∫_{-2}^{2} (12 -3x² -4x +x³) dxIntegrate term by term:Integral of 12 dx = 12xIntegral of -3x² dx = -x³Integral of -4x dx = -2x²Integral of x³ dx = (1/4)x^4So the antiderivative is:12x -x³ -2x² + (1/4)x^4Evaluate from x=-2 to x=2.Compute at x=2:12*(2) - (2)^3 -2*(2)^2 + (1/4)*(2)^4= 24 -8 -8 + (1/4)*16= 24 -8 -8 +4 = 12Compute at x=-2:12*(-2) - (-2)^3 -2*(-2)^2 + (1/4)*(-2)^4= -24 - (-8) -2*(4) + (1/4)*16= -24 +8 -8 +4 = (-24 -8 +8 +4) = -20Subtract lower limit from upper limit:12 - (-20) = 32Multiply by 2π:V = 2π *32 = 64πWait, that seems straightforward. Let me check the calculations again.At x=2:12*2 =24- (2)^3 = -8-2*(2)^2 = -8(1/4)*(2)^4 = (1/4)*16=4Sum: 24 -8 -8 +4 = 12. Correct.At x=-2:12*(-2)= -24- (-2)^3 = -(-8)=8-2*(-2)^2 = -2*4= -8(1/4)*(-2)^4= (1/4)*16=4Sum: -24 +8 -8 +4= (-24 -8) + (8 +4)= -32 +12= -20. Correct.Difference: 12 - (-20)=32. Multiply by 2π: 64π. So the volume is 64π.Alternatively, since the integrand has odd and even functions. Let's check:The integrand is 12 -3x² -4x +x³.The integral from -2 to 2 of -4x dx is zero, because it's an odd function.Similarly, the integral of x³ dx from -2 to2 is zero, since x³ is odd.Therefore, the integral reduces to:∫_{-2}^{2} (12 -3x²) dxWhich is even function, so 2 times ∫_{0}^{2} (12 -3x²) dxCompute:2 [12x -x³] from 0 to2 = 2[ (24 -8) -0 ]= 2*16=32Hence, same result. Then V=2π*32=64π. Correct.Therefore, the answer is 64π.Summary of Solutions:1. ( pi p a^2 )2. ( frac{4}{3}pi a^2 b )3. ( frac{272}{15}pi )4. ( frac{32}{105}pi a^3 )5. ( 64pi )

✅:1) Volume for ( y^{2} = 2px ), ( x = a ) around the ( Ox )-axis Using the disk method: [V = pi int_{0}^{a} 2px , dx = pi p a^2] Answer: (boxed{pi p a^2})2) Volume for ( frac{x^{2}}{a^{2}} + frac{y^{2}}{b^{2}} = 1 ) around the ( Oy )-axis Using the disk method for an ellipse: [V = frac{4}{3}pi a^2 b] Answer: (boxed{frac{4}{3}pi a^2 b})3) Volume for ( 2y = x^{2} ), ( 2x + 2y - 3 = 0 ) around the ( Ox )-axis Using the washer method between intersection points ( x = -3 ) and ( x = 1 ): [V = frac{272}{15}pi] Answer: (boxed{frac{272}{15}pi})4) Volume for ( x = a cos^{3} t ), ( y = a sin^{3} t ) around the ( Ox )-axis Using parametric integration for the astroid: [V = frac{32}{105}pi a^3] Answer: (boxed{frac{32}{105}pi a^3})5) Volume for ( y = 4 - x^{2} ), ( y = 0 ) around the line ( x = 3 ) Using the shell method: [V = 64pi] Answer: (boxed{64pi})

🔑:1) The problem requires us to find the volume of the solid formed by rotating the region bounded by the parabola (y^2 = 2px) and the line (x = a) around the (Ox) axis. To solve this, we use the formula for the volume of a solid of revolution around the (Ox) axis: [ V = pi int_{x_1}^{x_2} y^2 , dx ] Given (y^2 = 2px), we can substitute (y^2) in the integral: [ V = pi int_{0}^{a} 2px , dx ] Evaluate the integral: [ V = pi int_{0}^{a} 2px , dx = pi p int_{0}^{a} 2x , dx ] Integrate (2x) with respect to (x): [ V = pi p left[ x^2 right]_0^a = pi p left[ a^2 - 0 right] = pi p a^2 ] Conclusion: The volume of the solid is: [ boxed{pi p a^2} ]2) For the solid formed by rotating the ellipse described by (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) around the (Oy) axis, we find the volume using the formula: [ V = pi int_{y_1}^{y_2} x^2 , dy ] From the ellipse equation, solve for (x^2): [ x^2 = a^2 left(1 - frac{y^2}{b^2}right) ] Substitute (x^2) in the integral: [ V = pi a^2 int_{-b}^{b} left(1 - frac{y^2}{b^2}right) , dy ] Evaluate the integral: [ V = pi a^2 left[ y - frac{y^3}{3b^2} right]_{-b}^b ] Compute the values at the bounds: [ V = pi a^2 left( 2b - frac{2b^3}{3b^2}right) = frac{4}{3} pi a^2 b ] Conclusion: The volume of the solid is: [ boxed{frac{4}{3} pi a^2 b} ]3) For the region bounded by the parabola (2y = x^2) and the line (2x + 2y - 3 = 0) rotated around the (Ox) axis, we set up the volume integral. The equation will be given in suitable bounds and each expression handled similarly per integration rules.4) For the volume of the solid obtained by rotating the figure given by the parametric equations (x = a cos^3 t), (y = a sin^3 t) around the (Ox) axis, we transform the integral accordingly. Using the appropriate transformation ( y = a sin^3 t ) and converting to the new variable (t): The transformed integral then requires evaluation of trigonometric identities and integration.5) Finally, for the volume of the solid formed by rotating the area bounded by (y = 4 - x^2) and (y = 0) around the line (x = 3): The approach involves determining cross-sections and computing the resulting integral based on transformation of coordinates and resulting integration. Conclusion for all parts: Intermediate steps detailed, evaluations made clear to ensure precise volumes are found for all parts.

🧠:Okay, so I need to prove two things here. First, that for points A and B on the ellipse x²/a² + y²/b² = 1, with OA perpendicular to OB, the sum 1/|OA|² + 1/|OB|² equals 1/a² + 1/b². Second, that the line AB is tangent to the circle x² + y² = (a²b²)/(a² + b²). Let me start with the first part.First, I need to recall that if two vectors are perpendicular, their dot product is zero. Since OA and OB are perpendicular, the vectors from the origin to A and B must satisfy this condition. Let me denote point A as (x₁, y₁) and point B as (x₂, y₂). Since they're on the ellipse, they satisfy x₁²/a² + y₁²/b² = 1 and similarly for (x₂, y₂). Also, OA perpendicular to OB implies x₁x₂ + y₁y₂ = 0.Now, |OA|² is x₁² + y₁², and |OB|² is x₂² + y₂². So, the expression we need to compute is 1/(x₁² + y₁²) + 1/(x₂² + y₂²). We need to show that this equals 1/a² + 1/b².Hmm, how can I relate this? Maybe parametrize the points A and B on the ellipse. Since OA and OB are perpendicular, perhaps there's a parameter θ such that A is (a cosθ, b sinθ) and B is (-a sinθ, b cosθ) or something like that. Wait, but if OA is (a cosθ, b sinθ), then OB would need to be perpendicular. Let's check: the dot product of (a cosθ, b sinθ) and (x₂, y₂) should be zero. So a cosθ x₂ + b sinθ y₂ = 0. But how does B relate to θ? Maybe another parameter? Maybe B is related to θ + π/2?Alternatively, maybe use parametric coordinates for perpendicular points on an ellipse. Wait, I remember that for an ellipse, if two points are such that their radii are perpendicular, there might be a specific parametrization. Let me think.Alternatively, use coordinates in terms of slopes. Suppose OA has a slope m, then OB will have slope -1/m since they're perpendicular. Then, express points A and B in terms of m and substitute into the ellipse equation. But that might get complicated. Let's see.Suppose OA is along a line y = m x. Then point A is where this line intersects the ellipse. Solving y = m x and x²/a² + y²/b² = 1. Substitute y = m x into the ellipse equation: x²/a² + (m²x²)/b² = 1 => x² (1/a² + m²/b²) = 1 => x² = 1/(1/a² + m²/b²) => x = ±1/√(1/a² + m²/b²). Then y = m x. So point A is (1/√(1/a² + m²/b²), m/√(1/a² + m²/b²)).Similarly, OB has slope -1/m, so line y = (-1/m)x intersects the ellipse at point B. Similarly, substituting y = (-1/m)x into the ellipse equation: x²/a² + ( (1/m²)x² )/b² = 1 => x² (1/a² + 1/(b² m²)) = 1 => x² = 1/(1/a² + 1/(b² m²)) => x = ±1/√(1/a² + 1/(b² m²)), and y = (-1/m)x. So point B is (1/√(1/a² + 1/(b² m²)), -1/(m √(1/a² + 1/(b² m²)))).Now compute |OA|² and |OB|². For point A:|OA|² = x₁² + y₁² = [1/(1/a² + m²/b²)] + [m²/(1/a² + m²/b²)] = (1 + m²)/(1/a² + m²/b²) = [ (1 + m²) ] / [ (b² + a² m²)/a² b² ) ] = (1 + m²) * (a² b²) / (b² + a² m² )Similarly, |OB|² = x₂² + y₂² = [1/(1/a² + 1/(b² m²))] + [1/(m² (1/a² + 1/(b² m²)) )] = [1 + 1/m² ] / [1/a² + 1/(b² m² ) ]Let me compute that denominator first: 1/a² + 1/(b² m² ) = (b² m² + a² ) / (a² b² m² )So |OB|² = [ (1 + 1/m² ) ] / [ (b² m² + a² ) / (a² b² m² ) ] = ( (m² + 1)/m² ) * (a² b² m² ) / (a² + b² m² ) ) = ( (m² + 1) ) * (a² b² ) / (a² + b² m² )Therefore, 1/|OA|² + 1/|OB|² = [ (b² + a² m² ) / (a² b² (1 + m² )) ] + [ (a² + b² m² ) / (a² b² (1 + m² )) ]Wait, let's compute 1/|OA|²:1/|OA|² = (b² + a² m² ) / (a² b² (1 + m² ) )Similarly, 1/|OB|² = (a² + b² m² ) / (a² b² (1 + m² ) )Therefore, adding them together:[ (b² + a² m² ) + (a² + b² m² ) ] / (a² b² (1 + m² )) = [ (a² + b² ) + (a² m² + b² m² ) ] / (a² b² (1 + m² )) = [ (a² + b² )(1 + m² ) ] / (a² b² (1 + m² )) = (a² + b² ) / (a² b² ) = 1/a² + 1/b²Yes! So that works out. Therefore, the first part is proved by parametrizing the points A and B in terms of slope m and showing that 1/|OA|² + 1/|OB|² simplifies to 1/a² + 1/b² regardless of m. Nice.Now, the second part: proving that line AB is tangent to the circle x² + y² = (a² b² )/(a² + b² ). Let's see. So we need to show that the distance from the origin to line AB is equal to the radius of the circle, which is sqrt( (a² b² )/(a² + b² ) ). Since the circle is centered at the origin, the distance from the origin to the tangent line AB must equal the radius.First, let's find the equation of line AB. Points A and B were parametrized above in terms of m. Let me recall the coordinates of A and B:Point A: ( 1 / sqrt(1/a² + m²/b² ), m / sqrt(1/a² + m²/b² ) )Point B: ( 1 / sqrt(1/a² + 1/(b² m² ) ), -1/(m sqrt(1/a² + 1/(b² m² ) )) )But this looks complicated. Maybe there's a better parametrization. Alternatively, since OA and OB are perpendicular, we can use parametric angles. Let me try that.Suppose point A is (a cosθ, b sinθ), which is a standard parametrization for the ellipse. Then, since OA and OB are perpendicular, point B should be such that the vector OB is perpendicular to OA. The condition for perpendicularity is (a cosθ)(x) + (b sinθ)(y) = 0, where (x, y) are the coordinates of B. But B is also on the ellipse, so x²/a² + y²/b² = 1.So we can parametrize B as (-b sinθ, a cosθ) scaled appropriately? Wait, if OA is (a cosθ, b sinθ), then a perpendicular vector would be (-b sinθ, a cosθ), but we need to check if that lies on the ellipse.Let's check if (-b sinθ, a cosθ) satisfies x²/a² + y²/b² = 1. Compute x²/a² + y²/b² = (b² sin²θ)/a² + (a² cos²θ)/b². Unless a = b, this is not equal to 1. So that point is not on the ellipse. So that approach might not work.Alternatively, we can find B by rotating point A by 90 degrees, but scaled appropriately to lie on the ellipse. Hmm. Alternatively, use parametric coordinates with angles θ and θ + π/2.Wait, but the ellipse isn't a circle, so rotating by 90 degrees doesn't preserve the ellipse. So that complicates things.Alternatively, consider that for any point A on the ellipse, there's a unique point B such that OA is perpendicular to OB. Let me think of solving the equations for B given A.Given point A (a cosθ, b sinθ), then point B (x, y) must satisfy x²/a² + y²/b² = 1 and a cosθ * x + b sinθ * y = 0.So we can solve these two equations for x and y. Let me write that down.From the perpendicularity condition: a cosθ x + b sinθ y = 0 => y = - (a cosθ / (b sinθ)) x.Substitute into the ellipse equation:x²/a² + [ (a² cos²θ / b² sin²θ ) x² ] / b² = 1 => x²/a² + (a² cos²θ / b⁴ sin²θ ) x² = 1 => x² [ 1/a² + a² cos²θ / (b⁴ sin²θ ) ] = 1.This seems messy. Let's factor out x²:x² [ (b⁴ sin²θ + a^4 cos²θ ) / (a² b⁴ sin²θ ) ] = 1 => x² = (a² b⁴ sin²θ ) / (b⁴ sin²θ + a^4 cos²θ )Similarly, y = - (a cosθ / (b sinθ )) x = - (a cosθ / (b sinθ )) * sqrt( (a² b⁴ sin²θ ) / (b⁴ sin²θ + a^4 cos²θ ) )This is getting too complicated. Maybe there's a better approach.Wait, earlier when I used the slope m, I found expressions for points A and B. Maybe using those coordinates to find the equation of line AB and then compute its distance from the origin.Let me try that. From the previous part, points A and B are:A: ( 1 / sqrt(1/a² + m²/b² ), m / sqrt(1/a² + m²/b² ) )B: ( 1 / sqrt(1/a² + 1/(b² m² ) ), -1/(m sqrt(1/a² + 1/(b² m² ) )) )Let me denote denominator1 = sqrt(1/a² + m²/b² ) and denominator2 = sqrt(1/a² + 1/(b² m² ) )So A is (1/denominator1, m/denominator1 )B is (1/denominator2, -1/(m denominator2 ) )Now, find the equation of line AB. Let's use the two-point form.First, compute the slope of AB. The slope is (y_B - y_A)/(x_B - x_A )Compute y_B - y_A = [ -1/(m denominator2 ) ] - [ m / denominator1 ]x_B - x_A = [1 / denominator2 ] - [1 / denominator1 ]This seems complicated. Maybe instead, find the equation of the line in terms of coordinates.Alternatively, use the formula for the distance from the origin to the line AB. The formula is |Ax + By + C| / sqrt(A² + B²), but since the line passes through A and B, which are points not at the origin, we can write the line equation as follows.Let me write the equation of AB using determinant formula:| x y 1 || x_A y_A 1 | = 0| x_B y_B 1 |Expanding the determinant:x(y_A - y_B) + y(x_B - x_A) + (x_A y_B - x_B y_A) = 0So the line equation is (y_A - y_B)x + (x_B - x_A)y + (x_A y_B - x_B y_A) = 0The distance from the origin (0,0) to this line is |0 + 0 + (x_A y_B - x_B y_A)| / sqrt( (y_A - y_B)^2 + (x_B - x_A)^2 )So we need to compute |x_A y_B - x_B y_A| / sqrt( (y_A - y_B)^2 + (x_B - x_A)^2 ) and show that it equals sqrt( a² b² / (a² + b² ) )Alternatively, square both sides to simplify:( x_A y_B - x_B y_A )² / [ (y_A - y_B )² + (x_B - x_A )² ] = a² b² / (a² + b² )So let's compute numerator and denominator.First, compute x_A y_B - x_B y_A:x_A = 1 / denominator1y_B = -1/(m denominator2 )x_B = 1 / denominator2y_A = m / denominator1Thus,x_A y_B - x_B y_A = (1 / denominator1)( -1/(m denominator2 ) ) - (1 / denominator2 )( m / denominator1 )= -1/( m denominator1 denominator2 ) - m / ( denominator1 denominator2 )= [ -1/m - m ] / ( denominator1 denominator2 )= [ - (1 + m² ) / m ] / ( denominator1 denominator2 )So numerator squared is [ (1 + m² )² / m² ] / ( denominator1² denominator2² )Denominator of the distance expression is (y_A - y_B )² + (x_B - x_A )².Compute y_A - y_B = m / denominator1 - ( -1/(m denominator2 ) ) = m / denominator1 + 1/(m denominator2 )Similarly, x_B - x_A = 1 / denominator2 - 1 / denominator1So the denominator is [ m / denominator1 + 1/(m denominator2 ) ]² + [ 1 / denominator2 - 1 / denominator1 ]²This looks complicated, but maybe we can find a relationship between denominator1 and denominator2.From earlier, denominator1 = sqrt( 1/a² + m² / b² ) = sqrt( (b² + a² m² ) / (a² b² ) ) = sqrt( b² + a² m² ) / (a b )Similarly, denominator2 = sqrt( 1/a² + 1/(b² m² ) ) = sqrt( (b² m² + a² ) / (a² b² m² ) ) = sqrt( a² + b² m² ) / (a b m )So denominator1 = sqrt( b² + a² m² ) / (a b )Denominator2 = sqrt( a² + b² m² ) / (a b m )Note that denominator1 and denominator2 are related. Let's denote D = sqrt( a² + b² m² )Then denominator1 = sqrt( b² + a² m² ) / (a b )Wait, but D = denominator2 * (a b m )Wait, D = sqrt( a² + b² m² )So denominator2 = D / (a b m )Similarly, denominator1 = sqrt( b² + a² m² ). Hmm, unless m is 1, these are different.Wait, actually, denominator1 is sqrt(1/a² + m² / b² )^{-1} or wait, no:Wait, earlier, denominator1 was sqrt(1/a² + m² / b² ), but when we calculated |OA|², we had:denominator1 = sqrt( (1/a² + m² / b² ) )Wait, but in the coordinates of point A, we had x = 1 / sqrt(1/a² + m² / b² )Wait, so denominator1 is sqrt(1/a² + m² / b² )Similarly, denominator2 is sqrt(1/a² + 1/(b² m² ) )But 1/a² + m² / b² = (b² + a² m² ) / (a² b² )So denominator1 = sqrt( (b² + a² m² ) / (a² b² ) ) = sqrt(b² + a² m² ) / (a b )Similarly, denominator2 = sqrt( (1/a² + 1/(b² m² ) ) ) = sqrt( (b² m² + a² ) / (a² b² m² ) ) = sqrt(a² + b² m² ) / (a b m )Therefore, denominator1 = sqrt(b² + a² m² ) / (a b )Denominator2 = sqrt(a² + b² m² ) / (a b m )Therefore, note that denominator1 * denominator2 = [ sqrt(b² + a² m² ) / (a b ) ] * [ sqrt(a² + b² m² ) / (a b m ) ] = sqrt( (b² + a² m² )(a² + b² m² ) ) / (a² b² m )But this seems complicated. Let's see:Let me compute denominator1 * denominator2:= [ sqrt(b² + a² m² ) * sqrt(a² + b² m² ) ] / (a² b² m )But the product inside the square root is (b² + a² m² )(a² + b² m² )= a² b² + b^4 m² + a^4 m² + a² b² m^4= a² b² (1 + m^4 ) + m² (a^4 + b^4 )Hmm, not sure if that helps.Alternatively, let's compute numerator and denominator in the distance squared expression.Numerator squared: [ (1 + m² )² / m² ] / ( denominator1² denominator2² )Denominator1² = (b² + a² m² ) / (a² b² )Denominator2² = (a² + b² m² ) / (a² b² m² )Therefore, denominator1² denominator2² = [ (b² + a² m² )(a² + b² m² ) ] / (a^4 b^4 m² )Thus, numerator squared / denominator1² denominator2² = [ (1 + m² )² / m² ] / [ (b² + a² m² )(a² + b² m² ) / (a^4 b^4 m² ) ] = [ (1 + m² )² / m² ] * [ a^4 b^4 m² / ( (b² + a² m² )(a² + b² m² ) ) ] = (1 + m² )² a^4 b^4 / [ (b² + a² m² )(a² + b² m² ) ]Now, the denominator of the distance squared is (y_A - y_B )² + (x_B - x_A )². Let's compute this.First, compute y_A - y_B = m / denominator1 - ( -1/(m denominator2 ) ) = m / denominator1 + 1/(m denominator2 )Similarly, x_B - x_A = 1 / denominator2 - 1 / denominator1So, let's compute these terms:Compute y_A - y_B:= m / [ sqrt( (b² + a² m² ) / (a² b² ) ) ] + 1/( m * [ sqrt( (a² + b² m² ) / (a² b² m² ) ) ] )Simplify denominators:sqrt( (b² + a² m² ) / (a² b² ) ) = sqrt(b² + a² m² ) / (a b )Similarly, sqrt( (a² + b² m² ) / (a² b² m² ) ) = sqrt(a² + b² m² ) / (a b m )Therefore,y_A - y_B = m / ( sqrt(b² + a² m² ) / (a b ) ) + 1/( m * sqrt(a² + b² m² ) / (a b m ) )= m * (a b ) / sqrt(b² + a² m² ) + (a b m ) / ( m sqrt(a² + b² m² ) )= (a b m ) / sqrt(b² + a² m² ) + (a b ) / sqrt(a² + b² m² )Similarly, x_B - x_A = 1 / [ sqrt( (a² + b² m² ) / (a² b² m² ) ) ] - 1 / [ sqrt( (b² + a² m² ) / (a² b² ) ) ]= sqrt( (a² + b² m² ) / (a² b² m² ) )^{-1} - sqrt( (b² + a² m² ) / (a² b² ) )^{-1}Wait, actually:x_B = 1 / denominator2 = 1 / [ sqrt(1/a² + 1/(b² m² ) ) ] = 1 / [ sqrt( (b² m² + a² ) / (a² b² m² ) ) ] = sqrt( a² b² m² / (a² + b² m² ) ) = (a b m ) / sqrt(a² + b² m² )Similarly, x_A = 1 / denominator1 = 1 / sqrt(1/a² + m² / b² ) = 1 / sqrt( (b² + a² m² ) / (a² b² ) ) = (a b ) / sqrt(b² + a² m² )Therefore, x_B - x_A = (a b m ) / sqrt(a² + b² m² ) - (a b ) / sqrt(b² + a² m² )Similarly, y_A - y_B = (a b m ) / sqrt(b² + a² m² ) + (a b ) / sqrt(a² + b² m² )Therefore, the denominator (y_A - y_B )² + (x_B - x_A )² can be written as:[ (a b m ) / sqrt(b² + a² m² ) + (a b ) / sqrt(a² + b² m² ) ]² + [ (a b m ) / sqrt(a² + b² m² ) - (a b ) / sqrt(b² + a² m² ) ]²Factor out (a b )²:= (a b )² [ ( m / sqrt(b² + a² m² ) + 1 / sqrt(a² + b² m² ) )² + ( m / sqrt(a² + b² m² ) - 1 / sqrt(b² + a² m² ) )² ]Let me compute the terms inside the brackets:Let me denote S = sqrt(b² + a² m² ), T = sqrt(a² + b² m² )Then the expression becomes:[ ( m / S + 1 / T )² + ( m / T - 1 / S )² ]Expand the squares:First term: m² / S² + 2 m / (S T ) + 1 / T²Second term: m² / T² - 2 m / (S T ) + 1 / S²Adding them together:m² / S² + 2 m / (S T ) + 1 / T² + m² / T² - 2 m / (S T ) + 1 / S²Simplify:(m² / S² + 1 / S² ) + (1 / T² + m² / T² ) + (2 m / (S T ) - 2 m / (S T ) )Which simplifies to:( (m² + 1 ) / S² ) + ( (1 + m² ) / T² )Factor out (1 + m² ):(1 + m² )( 1 / S² + 1 / T² )Therefore, the denominator is (a b )² (1 + m² )( 1 / S² + 1 / T² )Recall that S = sqrt(b² + a² m² ), so S² = b² + a² m²Similarly, T² = a² + b² m²Therefore, 1/S² + 1/T² = 1/(b² + a² m² ) + 1/(a² + b² m² )Combine these fractions:= (a² + b² m² + b² + a² m² ) / [ (b² + a² m² )(a² + b² m² ) ]= (a² + b² + a² m² + b² m² ) / [ (b² + a² m² )(a² + b² m² ) ]= ( (a² + b² ) + m² (a² + b² ) ) / [ (b² + a² m² )(a² + b² m² ) ]= ( (a² + b² )(1 + m² ) ) / [ (b² + a² m² )(a² + b² m² ) ]Therefore, the denominator becomes:(a b )² (1 + m² ) * ( (a² + b² )(1 + m² ) ) / [ (b² + a² m² )(a² + b² m² ) ]= (a b )² (1 + m² )² (a² + b² ) / [ (b² + a² m² )(a² + b² m² ) ]Now, going back to the distance squared expression:Numerator squared / denominator1² denominator2² divided by denominator:Wait, the distance squared is:[ (1 + m² )² a^4 b^4 / ( (b² + a² m² )(a² + b² m² ) ) ] / [ (a b )² (1 + m² )² (a² + b² ) / ( (b² + a² m² )(a² + b² m² ) ) ) ]Simplify numerator divided by denominator:= [ (1 + m² )² a^4 b^4 / ( (b² + a² m² )(a² + b² m² ) ) ] * [ (b² + a² m² )(a² + b² m² ) ) / ( (a b )² (1 + m² )² (a² + b² ) ) ]Cancel out terms:(1 + m² )² cancels, (b² + a² m² )(a² + b² m² ) cancels, a^4 b^4 / (a² b² ) = a² b²Thus, distance squared = a² b² / (a² + b² )Therefore, distance = sqrt( a² b² / (a² + b² ) ) = (a b ) / sqrt(a² + b² )But the radius of the circle is sqrt( a² b² / (a² + b² ) ), which is the same as (a b ) / sqrt(a² + b² ). Hence, the distance from the origin to line AB is equal to the radius of the circle, which means the line AB is tangent to the circle.Therefore, both statements are proved.Final AnswerThe two properties are proved as shown, and thus the final results are encapsulated in the boxed expressions:For the first property: boxed{frac{1}{|OA|^{2}}+frac{1}{|OB|^{2}}=frac{1}{a^{2}}+frac{1}{b^{2}}},And for the second property, the line (AB) is indeed tangent to the specified circle.

✅:Given that (O) is the coordinate origin, and there are two points (A) and (B) on the ellipse (frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1) such that (OA perp OB), we need to prove two properties:1. (frac{1}{|OA|^{2}}+frac{1}{|OB|^{2}}=frac{1}{a^{2}}+frac{1}{b^{2}}).2. The line (AB) is tangent to the circle (x^{2}+y^{2}=frac{a^{2}b^{2}}{a^{2}+b^{2}}). Proof of the First Property1. Parametrization of Points (A) and (B): - Let point (A) be ((a cos theta, b sin theta)) on the ellipse. - Since (OA perp OB), point (B) must satisfy the condition (a cos theta cdot x + b sin theta cdot y = 0) and lie on the ellipse.2. Using Slopes for Parametrization: - Suppose the slope of (OA) is (m), then the coordinates of (A) are (left(frac{1}{sqrt{frac{1}{a^2} + frac{m^2}{b^2}}}, frac{m}{sqrt{frac{1}{a^2} + frac{m^2}{b^2}}}right)). - The slope of (OB) is (-frac{1}{m}), leading to coordinates of (B) as (left(frac{1}{sqrt{frac{1}{a^2} + frac{1}{b^2 m^2}}}, frac{-1/m}{sqrt{frac{1}{a^2} + frac{1}{b^2 m^2}}}right)).3. Calculating (|OA|^2) and (|OB|^2): - (|OA|^2 = frac{1 + m^2}{frac{1}{a^2} + frac{m^2}{b^2}}) - (|OB|^2 = frac{1 + frac{1}{m^2}}{frac{1}{a^2} + frac{1}{b^2 m^2}})4. Simplifying the Sum: - (frac{1}{|OA|^2} + frac{1}{|OB|^2} = frac{frac{1}{a^2} + frac{m^2}{b^2}}{1 + m^2} + frac{frac{1}{a^2} + frac{1}{b^2 m^2}}{1 + frac{1}{m^2}}) - Simplifying both terms results in (frac{1}{a^2} + frac{1}{b^2}). Proof of the Second Property1. Equation of Line (AB): - Using the coordinates of (A) and (B), derive the equation of line (AB) using the determinant formula.2. Distance from Origin to Line (AB): - The distance from the origin to line (AB) is calculated as (frac{|x_A y_B - x_B y_A|}{sqrt{(y_A - y_B)^2 + (x_B - x_A)^2}}).3. Simplifying the Distance: - After simplification, the distance is found to be (frac{ab}{sqrt{a^2 + b^2}}), which matches the radius of the circle (sqrt{frac{a^2 b^2}{a^2 + b^2}}). Final AnswerFor the first property:[boxed{frac{1}{|OA|^{2}}+frac{1}{|OB|^{2}}=frac{1}{a^{2}}+frac{1}{b^{2}}}]For the second property, the line (AB) is tangent to the circle (x^{2}+y^{2}=frac{a^{2}b^{2}}{a^{2}+b^{2}}).

🔑:1. Set Up Coordinates: Let O be the origin (0, 0), and consider two points A and B on the ellipse frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1 such that OA perp OB. We set |OA|=r_{1} and |OB|=r_{2}.2. Parameterization: Suppose angle xOA = theta, then given OA perp OB, we have angle xOB = frac{pi}{2} + theta. Thus, we can express the coordinates of A and B as: [ A left( r_{1} cos theta, r_{1} sin theta right), quad B left( -r_{2} sin theta, r_{2} cos theta right) ]3. Points on the Ellipse: Substituting the coordinates of A and B into the ellipse equation frac{x^2}{a^2} + frac{y^2}{b^2} = 1 gives: [ begin{cases} frac{r_{1}^{2} cos ^{2} theta}{a^{2}} + frac{r_{1}^{2} sin ^{2} theta}{b^{2}} = 1 frac{r_{2}^{2} sin ^{2} theta}{a^{2}} + frac{r_{2}^{2} cos ^{2} theta}{b^{2}} = 1 end{cases} ]4. Simplify Equations: Dividing both equations by r_{1}^{2} and r_{2}^{2} respectively: [ begin{cases} frac{cos ^{2} theta}{a^{2}} + frac{sin ^{2} theta}{b^{2}} = frac{1}{r_{1}^{2}} frac{sin ^{2} theta}{a^{2}} + frac{cos ^{2} theta}{b^{2}} = frac{1}{r_{2}^{2}} end{cases} ]5. Add the Equations: Adding the two equations together: [ frac{cos ^{2} theta}{a^{2}} + frac{sin ^{2} theta}{b^{2}} + frac{sin ^{2} theta}{a^{2}} + frac{cos ^{2} theta}{b^{2}} = frac{1}{r_{1}^{2}} + frac{1}{r_{2}^{2}} ] Simplify using the identity cos^2 theta + sin^2 theta = 1: [ frac{cos ^{2} theta + sin ^{2} theta}{a^{2}} + frac{sin ^{2} theta + cos ^{2} theta}{b^{2}} = frac{1}{r_{1}^{2}} + frac{1}{r_{2}^{2}} ] Thus: [ frac{1}{a^{2}} + frac{1}{b^{2}} = frac{1}{r_{1}^{2}} + frac{1}{r_{2}^{2}} ]6. Perpendicular Distance from O to AB: Considering point H where OH perp AB, we use the formula for the perpendicular distance from a point to a line: [ OH = frac{2S_{triangle AOB}}{AB} ] Here, the area S_{triangle AOB} can be expressed using the determinants of the coordinates of A and B: [ S_{triangle AOB} = frac{1}{2} left| r_{1} cos theta cdot r_{2} cos theta - (-r_{2} sin theta) cdot r_{1} sin theta right| = frac{1}{2} left| r_{1} r_{2} (cos^2 theta + sin^2 theta) right| = frac{1}{2} r_{1} r_{2} ] So: [ OH = frac{|OA||OB|}{|AB|} = sqrt{frac{1}{|OA|^{2}} + frac{1}{|OB|^{2}}} = sqrt{frac{1}{r_{1}^{2}} + frac{1}{r_{2}^{2}}} = sqrt{frac{1}{a^{2}} + frac{1}{b^{2}}} ]7. Radius of the Circle: The radius of the circle that intersects the line AB is determined by the expression we found for OH. Setting it equal to the circle's radius gives: [ OH = frac{ab}{sqrt{a^{2}+b^{2}}} ]Hence, we conclude that the line AB is tangent to the circle x^{2} + y^{2} = frac{a^{2} b^{2}}{a^{2} + b^{2}}.[boxed{frac{1}{|O A|^{2}} + frac{1}{|O B|^{2}} = frac{1}{a^{2}} + frac{1}{b^{2}}}]