🧠:Okay, so I need to prove that the intersection of three circular cylinders with radii 1, whose axes are pairwise mutually perpendicular, is contained within a sphere of radius sqrt(3/2). Hmm, let me start by visualizing the problem. First, I know that a single circular cylinder can be defined by an equation. If the axis of the cylinder is along the x-axis, for example, the equation would be y² + z² = 1. Similarly, if the axes are along the y and z-axes, their equations would be x² + z² = 1 and x² + y² = 1, respectively. But in this problem, the axes are pairwise mutually perpendicular, but they don't necessarily intersect. Wait, does that mean they might not all meet at a single point? So, the cylinders could be shifted along their axes? Hmm, but if the axes are pairwise perpendicular, they could be arranged like the coordinate axes, but maybe shifted. However, the problem states that the intersection is contained within a sphere of radius sqrt(3/2). I think the classic case where the three cylinders intersect along the coordinate axes (i.e., their axes intersect at the origin) is called the Steinmetz solid, right? But in that case, the intersection is bounded, and I remember that the maximum distance from the origin in the Steinmetz solid is sqrt(2)/2, but wait, maybe that's different. Wait, no. Let me think. Wait, for the Steinmetz solid formed by three cylinders, the intersection points would be where all three cylinders intersect. For example, if you take x² + y² ≤ 1, y² + z² ≤ 1, and x² + z² ≤ 1, the intersection would have points where each pair of coordinates squared is less than or equal to 1. The farthest points from the origin in this case would be (±1, ±1, ±1)/sqrt(3), but actually, wait, if x² + y² ≤ 1, y² + z² ≤ 1, and x² + z² ≤ 1, then adding all three equations gives 2(x² + y² + z²) ≤ 3, so x² + y² + z² ≤ 3/2, which implies that the intersection is contained within a sphere of radius sqrt(3/2). Oh! So that's the key. So maybe the problem is generalizing to three cylinders with axes that are pairwise perpendicular but not necessarily intersecting, but still the same result holds. But wait, in the standard Steinmetz solid with intersecting axes, we get that the intersection is contained within the sphere of radius sqrt(3/2). The user is asking to prove that even if the axes are not intersecting (i.e., the cylinders are shifted along their axes), the intersection is still contained within such a sphere. So maybe the shifting doesn't affect the maximum distance? Hmm, that seems counterintuitive. If you shift the cylinders, maybe the intersection points could be further away?Wait, let me think. Suppose we have three cylinders, each with radius 1, but their axes are mutually perpendicular lines (not necessarily intersecting). Let me model one cylinder along the x-axis, shifted along the y and z directions? Wait, no. If the axes are mutually perpendicular, they must be three lines that are pairwise perpendicular. If they don't intersect, they are skew lines? But in three-dimensional space, three pairwise perpendicular lines can be arranged such that each pair is skew, but that's complicated. Wait, but actually, in three-dimensional space, three mutually perpendicular lines can be arranged so that they are all skew, but maybe in this case, the cylinders are just translated versions of the coordinate axes. For example, cylinder 1 has its axis along the line x = a, y = 0, z = 0 (wait, no, that's just the x-axis shifted by a along x). Wait, but if you shift a cylinder along its axis, it's the same as the original cylinder. Because a cylinder is infinite, shifting along its axis doesn't change the cylinder. So maybe the axes can be translated perpendicular to their original directions? Wait, the problem says "axes are pairwise mutually perpendicular (but not necessarily intersecting)". So, for example, one cylinder could have its axis along the x-axis, another along the y-axis shifted along the z-axis, and the third along the z-axis shifted along the x and y axes. But in such a way that the axes remain pairwise perpendicular. Wait, but in 3D space, two lines that are perpendicular don't have to intersect. For example, the x-axis and a line parallel to the y-axis but shifted along the z-axis. These two lines are skew and perpendicular. So, each pair of cylinder axes here are skew lines, but still mutually perpendicular. Therefore, the three cylinders could be positioned such that their axes are like three mutually perpendicular skew lines. But how would the equations look? Let me consider the equations of such cylinders. Suppose the first cylinder has its axis along the x-axis, so its equation is y² + z² ≤ 1. The second cylinder's axis is a line parallel to the y-axis but shifted along the z-axis. Let's say shifted by some distance d along the z-axis. So, the axis is the line x = 0, z = d. Then, the equation of this cylinder would be (x - 0)² + (z - d)² ≤ 1? Wait, no. If the axis is along the y-axis direction, but shifted, the cylinder equation would be x² + (z - d)^2 ≤ 1. Similarly, the third cylinder could have its axis along the z-axis, shifted along the x-axis, so its equation would be (x - e)^2 + y² ≤ 1. Wait, but the problem states that the axes are pairwise mutually perpendicular. So, if the first cylinder's axis is along the x-axis, the second's axis is along a line parallel to the y-axis but shifted, and the third's axis is along a line parallel to the z-axis but shifted, then their direction vectors are still the standard basis vectors, so they are pairwise perpendicular. But in such cases, the equations of the cylinders would be similar to the original ones but shifted. For example, first cylinder: y² + z² ≤ 1. Second cylinder: x² + (z - d)^2 ≤ 1. Third cylinder: (x - e)^2 + y² ≤ 1. But since the axes are mutually perpendicular, the direction vectors are orthogonal, but their positions can be anywhere. However, the problem states that the intersection of these three cylinders must be contained within a sphere of radius sqrt(3/2). So, regardless of how we shift the cylinders along their axes (since shifting along the axis doesn't change the cylinder), or perpendicularly, the intersection remains within that sphere. Therefore, even if the cylinders are not coaxial with the coordinate axes, as long as their axes are mutually perpendicular, the intersection is within that sphere. Wait, but how do we handle the shifting? For example, if we shift one cylinder, does it allow points further away from the origin? Let me think. Suppose we have three cylinders with axes not passing through the origin. Let's take the first cylinder along the x-axis, equation y² + z² ≤ 1. The second cylinder is along the line y-axis shifted by some vector (a, 0, b), but actually, since the axis has to be perpendicular to the first cylinder's axis, which is along the x-axis. Wait, if the first cylinder's axis is the x-axis, then the second cylinder's axis must be a line perpendicular to the x-axis. So, its direction vector is perpendicular to the x-axis, say along the y or z direction. But if the axes are pairwise perpendicular, then each cylinder's axis must be along a direction perpendicular to the others. Wait, perhaps it's easier to assign coordinate systems. Let's suppose that the three cylinders have axes along three mutually perpendicular lines. Since the axes are mutually perpendicular, we can choose a coordinate system such that one axis is along the x-axis, another along the y-axis, and the third along the z-axis, but shifted. Wait, but if they are shifted, their equations become more complex. Alternatively, perhaps regardless of the shifts, the maximum distance from the origin (or from some point) to any point in the intersection is bounded by sqrt(3/2). But how can we show that? Alternatively, maybe we can use a coordinate system where the three axes intersect at a common point. Wait, but the problem states that the axes are not necessarily intersecting. However, perhaps we can translate the coordinate system so that one of the cylinder axes passes through the origin. Let me try to formalize this. Let’s denote the three cylinders as C1, C2, C3 with axes L1, L2, L3, which are pairwise perpendicular lines in space. Let’s pick a coordinate system where L1 is the x-axis. Then, since L2 is perpendicular to L1, L2 must lie in a plane perpendicular to the x-axis. Similarly, L3 is perpendicular to both L1 and L2. However, since L1, L2, L3 are pairwise perpendicular, they form a sort of orthogonal triad, but they might not intersect. Wait, but in 3D space, three mutually perpendicular lines can be arranged such that each pair is skew. For example, take L1 as the x-axis, L2 as a line parallel to the y-axis but shifted along the z-axis, and L3 as a line parallel to the z-axis but shifted along the x-axis. Then, each pair of lines is skew but still perpendicular. So, in such a case, the cylinders around these axes would have equations based on their distance from the axis. For example, cylinder C1 around L1 (the x-axis) is y² + z² ≤ 1. Cylinder C2 around L2, which is the line x=0, y=0, z=t + d (for some shift d). Wait, no. If L2 is a line parallel to the y-axis, shifted along the z-axis by d, then parametric equations of L2 would be (0, t, d), where t ∈ ℝ. Then, the cylinder C2 around L2 would consist of all points at distance 1 from L2. The distance from a point (x, y, z) to L2 is sqrt(x² + (z - d)^2). Therefore, the equation of C2 is x² + (z - d)^2 ≤ 1. Similarly, if L3 is a line parallel to the z-axis shifted along the x-axis by e, then its parametric equations are (e, 0, t), and the cylinder C3 around L3 would have equation (x - e)^2 + y² ≤ 1. So, the three cylinders would have equations:1. y² + z² ≤ 1 (C1)2. x² + (z - d)^2 ≤ 1 (C2)3. (x - e)^2 + y² ≤ 1 (C3)But since the axes L1, L2, L3 must be pairwise perpendicular, we need to check if the lines themselves are perpendicular. The direction vectors of L1, L2, L3 are (1, 0, 0), (0, 1, 0), and (0, 0, 1), which are mutually perpendicular. The shifts don't affect the direction vectors, so the lines are still mutually perpendicular in terms of direction, even if they don't intersect. Therefore, the cylinders are defined by those equations. Now, the intersection of C1, C2, C3 would be the set of points (x, y, z) that satisfy all three inequalities. We need to show that any such point is within a sphere of radius sqrt(3/2). But the problem says "contained within a sphere of radius sqrt(3/2)". It doesn't specify the center of the sphere. If the sphere is centered at the origin, but the cylinders are shifted, maybe the intersection could be shifted as well. Wait, but the problem doesn't specify the center of the sphere. Hmm, that's a problem. If the sphere can be anywhere, then perhaps the intersection is always contained within some sphere of radius sqrt(3/2), regardless of the shifts. But the problem probably means centered at the origin, but since the cylinders can be shifted, maybe not. Wait, but the original problem statement doesn't mention the center. So, perhaps we need to show that there exists a sphere of radius sqrt(3/2) that contains the intersection, regardless of the positions of the cylinders. Or, maybe that for any such three cylinders, their intersection is contained within some sphere of that radius. But the problem says "is contained within a sphere of radius sqrt(3/2)". So, probably, regardless of the configuration of the three cylinders (as long as their axes are pairwise perpendicular), their intersection lies within some sphere of that radius. So, the sphere's center might depend on the configuration of the cylinders. Alternatively, maybe the maximum possible distance from the origin in the intersection is sqrt(3/2), regardless of shifts. But if the cylinders are shifted, points in the intersection could be further away. Wait, let me test with an example. Suppose we shift one cylinder along its axis, but since the cylinder is infinite, shifting it along its axis doesn't change the cylinder. So, shifting along the axis direction doesn't affect the set. However, shifting perpendicular to the axis would change the cylinder's position. For instance, take cylinder C1: y² + z² ≤ 1. If we shift its axis along the y-axis to (0, a, 0), but wait, the axis direction is along the x-axis, so shifting it along y or z would move the entire cylinder. Wait, no. If the axis is along the x-axis, the cylinder is all points (x, y, z) with y² + z² ≤ 1. If we shift the cylinder along the y-axis by a, then the new axis is the line x-axis shifted along y by a, so the equation becomes (y - a)^2 + z² ≤ 1. But in that case, the direction of the axis is still along the x-axis, but shifted in y. Wait, no. Wait, the axis of the cylinder is a line. If you shift the entire cylinder along the y-axis, the axis becomes the line x = t, y = a, z = 0. So, it's a line parallel to the x-axis, shifted in y and z? Wait, no. If you shift along the y-axis by a, then the axis becomes (t, a, 0). So, the cylinder equation is (y - a)^2 + z² ≤ 1. Similarly, shifting along the z-axis by b would give y² + (z - b)^2 ≤ 1. So, suppose we have three cylinders with equations:1. (y - a)^2 + (z - b)^2 ≤ 1 (axis along the x-axis shifted to (0, a, b))2. (x - c)^2 + (z - d)^2 ≤ 1 (axis along the y-axis shifted to (c, 0, d))3. (x - e)^2 + (y - f)^2 ≤ 1 (axis along the z-axis shifted to (e, f, 0))But the axes must be pairwise perpendicular. The direction vectors of these axes are still (1, 0, 0), (0, 1, 0), and (0, 0, 1), respectively. Therefore, regardless of the shifts, the cylinders are just translated along their respective directions. Now, the intersection of these three cylinders would be the set of points (x, y, z) satisfying all three inequalities. We need to show that this intersection lies within some sphere of radius sqrt(3/2). To approach this, maybe we can find the maximum possible distance from a certain point (maybe the center of the sphere) to any point in the intersection. If we can show that this maximum distance is at most sqrt(3/2), then we're done. Alternatively, perhaps use the fact that for the original Steinmetz solid (where the cylinders are centered at the origin), the maximum distance is sqrt(3/2). Then, even if the cylinders are shifted, the maximum distance doesn't increase. But why? Because shifting the cylinders would restrict the intersection to regions that might not allow points to go further away? Hmm, not sure. Alternatively, perhaps use a change of variables to recenter the cylinders. Let me consider shifting the coordinate system such that the sphere is centered at the point (c, d, e), but that might complicate things. Alternatively, consider that each cylinder imposes a constraint on two coordinates. For example, the first cylinder (shifted) is (y - a)^2 + (z - b)^2 ≤ 1. Similarly for others. If we can bound x² + y² + z² in terms of these inequalities, maybe we can find the maximum. Let me try adding the three inequalities. First cylinder: (y - a)^2 + (z - b)^2 ≤ 1Second cylinder: (x - c)^2 + (z - d)^2 ≤ 1Third cylinder: (x - e)^2 + (y - f)^2 ≤ 1If we add them together:(y - a)^2 + (z - b)^2 + (x - c)^2 + (z - d)^2 + (x - e)^2 + (y - f)^2 ≤ 3Expanding each term:(y² - 2ay + a²) + (z² - 2bz + b²) + (x² - 2cx + c²) + (z² - 2dz + d²) + (x² - 2ex + e²) + (y² - 2fy + f²) ≤ 3Combine like terms:2x² + 2y² + 2z² - 2(c + e)x - 2(a + f)y - 2(b + d)z + (a² + b² + c² + d² + e² + f²) ≤ 3Divide both sides by 2:x² + y² + z² - (c + e)x - (a + f)y - (b + d)z + (a² + b² + c² + d² + e² + f²)/2 ≤ 3/2Now, let's complete the squares for x, y, z. The terms with x: - (c + e)x. Let’s write this as -2 * [(c + e)/2]x. Similarly for y and z. So, completing the square:x² - (c + e)x = x² - (c + e)x + [(c + e)/2]^2 - [(c + e)/2]^2 = [x - (c + e)/2]^2 - [(c + e)/2]^2Similarly for y and z:y² - (a + f)y = [y - (a + f)/2]^2 - [(a + f)/2]^2z² - (b + d)z = [z - (b + d)/2]^2 - [(b + d)/2]^2So substituting back into the inequality:[x - (c + e)/2]^2 - [(c + e)/2]^2 + [y - (a + f)/2]^2 - [(a + f)/2]^2 + [z - (b + d)/2]^2 - [(b + d)/2]^2 + (a² + b² + c² + d² + e² + f²)/2 ≤ 3/2Bring the negative terms to the right:[x - (c + e)/2]^2 + [y - (a + f)/2]^2 + [z - (b + d)/2]^2 ≤ 3/2 + [(c + e)/2]^2 + [(a + f)/2]^2 + [(b + d)/2]^2 - (a² + b² + c² + d² + e² + f²)/2Let me compute the right-hand side. Let's expand the squares:[(c + e)/2]^2 = (c² + 2ce + e²)/4Similarly for the others:[(a + f)/2]^2 = (a² + 2af + f²)/4[(b + d)/2]^2 = (b² + 2bd + d²)/4So, summing these:[(c + e)/2]^2 + [(a + f)/2]^2 + [(b + d)/2]^2 = [c² + 2ce + e² + a² + 2af + f² + b² + 2bd + d²]/4Subtracting (a² + b² + c² + d² + e² + f²)/2:= [ (c² + e² + a² + f² + b² + d²) + 2(ce + af + bd) ] /4 - (a² + b² + c² + d² + e² + f²)/2= [ (sum of squares) + 2(ce + af + bd) ] /4 - 2(sum of squares)/4= [ - (sum of squares) + 2(ce + af + bd) ] /4Therefore, the right-hand side becomes:3/2 + [ - (sum of squares) + 2(ce + af + bd) ] /4But this seems complicated. However, notice that the left-hand side is the squared distance from the point [(c + e)/2, (a + f)/2, (b + d)/2] to (x, y, z). Therefore, the inequality can be written as:|| (x, y, z) - ( (c + e)/2, (a + f)/2, (b + d)/2 ) ||² ≤ 3/2 + [ - (sum of squares) + 2(ce + af + bd) ] /4But unless the right-hand side is non-negative, this doesn't help us. However, this seems too vague. Maybe there's a different approach. Alternatively, perhaps consider that each point (x, y, z) in the intersection must satisfy all three cylinder equations. Let's take the three inequalities:1. (y - a)^2 + (z - b)^2 ≤ 12. (x - c)^2 + (z - d)^2 ≤ 13. (x - e)^2 + (y - f)^2 ≤ 1We need to bound x² + y² + z². Let me try to express x² + y² + z² in terms of these inequalities. From the first inequality: (y - a)^2 + (z - b)^2 ≤ 1 ⇒ y² - 2ay + a² + z² - 2bz + b² ≤ 1 ⇒ y² + z² ≤ 1 + 2ay + 2bz - a² - b²Similarly, from the second inequality: x² + z² ≤ 1 + 2cx + 2dz - c² - d²From the third inequality: x² + y² ≤ 1 + 2ex + 2fy - e² - f²If we add all three inequalities:(y² + z²) + (x² + z²) + (x² + y²) ≤ 3 + 2(ay + bz + cx + dz + ex + fy) - (a² + b² + c² + d² + e² + f²)Simplify left-hand side:2x² + 2y² + 2z² ≤ 3 + 2(ay + bz + cx + dz + ex + fy) - (a² + b² + c² + d² + e² + f²)Divide both sides by 2:x² + y² + z² ≤ 3/2 + (ay + bz + cx + dz + ex + fy) - (a² + b² + c² + d² + e² + f²)/2Hmm, this seems similar to what I had before. Now, this inequality relates x² + y² + z² to linear terms in x, y, z. To bound x² + y² + z², perhaps complete the square again. Alternatively, think of the right-hand side as linear in x, y, z. Let me denote the right-hand side as:3/2 + [ c x + e x + a y + f y + b z + d z ] - (a² + b² + c² + d² + e² + f²)/2= 3/2 + x(c + e) + y(a + f) + z(b + d) - (a² + b² + c² + d² + e² + f²)/2So, x² + y² + z² ≤ 3/2 + x(c + e) + y(a + f) + z(b + d) - (sum of squares)/2This looks like a quadratic inequality. To find the maximum value of x² + y² + z², perhaps we can consider this as an optimization problem where we maximize x² + y² + z² subject to the three cylinder constraints. But this seems complicated. Alternatively, perhaps use the Cauchy-Schwarz inequality. Let me consider that for any point (x, y, z) in the intersection, the distance from the origin squared is x² + y² + z². We need to show that this is ≤ 3/2. Wait, but in the original case with centered cylinders, we have x² + y² ≤ 1, y² + z² ≤ 1, x² + z² ≤ 1, adding gives 2(x² + y² + z²) ≤ 3, so x² + y² + z² ≤ 3/2. But in the shifted case, the equations are more complex. However, maybe a similar approach works. Let me think. If we can somehow relate the shifted cylinders to the original ones. Wait, perhaps parametrize the variables. Let me consider shifting variables. Let u = x - h, v = y - k, w = z - m, where h, k, m are constants to be chosen such that the shifted cylinders align with the original coordinate axes. However, since each cylinder is shifted differently, this might not be straightforward. Alternatively, think of the worst-case scenario. What's the maximum possible distance a point in the intersection can be from some center? Maybe regardless of shifts, the sum of squares can't exceed 3/2. Wait, in the centered case, adding the three cylinder equations gives 2(x² + y² + z²) ≤ 3. In the shifted case, when we add the three cylinder equations, we get an expression that relates x² + y² + z² to linear terms. Perhaps by using the inequality that for any real numbers p and q, p² + q² ≥ (p + q)^2 / 2. Maybe apply this to the linear terms. Alternatively, use the fact that for any real numbers a and x, (x - a)^2 ≥ 0 ⇒ x² - 2ax + a² ≥ 0 ⇒ x² ≥ 2ax - a². Similarly for other variables. Maybe apply this to the linear terms in the inequality. Wait, let's go back to the inequality we had:x² + y² + z² ≤ 3/2 + x(c + e) + y(a + f) + z(b + d) - (sum)/2Let me denote S = a² + b² + c² + d² + e² + f². Then,x² + y² + z² ≤ 3/2 + x(c + e) + y(a + f) + z(b + d) - S/2Now, to bound the right-hand side, let's consider each term. For example, x(c + e) ≤ (x² + (c + e)^2)/2 by the AM-GM inequality. Similarly for the other terms. So,x(c + e) ≤ (x² + (c + e)^2)/2y(a + f) ≤ (y² + (a + f)^2)/2z(b + d) ≤ (z² + (b + d)^2)/2Adding these:x(c + e) + y(a + f) + z(b + d) ≤ [x² + y² + z² + (c + e)^2 + (a + f)^2 + (b + d)^2]/2Substituting back into the inequality:x² + y² + z² ≤ 3/2 + [x² + y² + z² + (c + e)^2 + (a + f)^2 + (b + d)^2]/2 - S/2Multiply both sides by 2:2(x² + y² + z²) ≤ 3 + x² + y² + z² + (c + e)^2 + (a + f)^2 + (b + d)^2 - SSubtract x² + y² + z² from both sides:x² + y² + z² ≤ 3 + (c + e)^2 + (a + f)^2 + (b + d)^2 - SBut S = a² + b² + c² + d² + e² + f², so compute (c + e)^2 + (a + f)^2 + (b + d)^2 - S:= [c² + 2ce + e² + a² + 2af + f² + b² + 2bd + d²] - [a² + b² + c² + d² + e² + f²]= 2ce + 2af + 2bdTherefore,x² + y² + z² ≤ 3 + 2(ce + af + bd)Hmm, so now we have x² + y² + z² ≤ 3 + 2(ce + af + bd)But this seems problematic because depending on the values of ce, af, bd, this could be larger than 3/2. However, this suggests that our approach is not tight enough. Maybe the issue is that we used a loose inequality (AM-GM) to bound the linear terms, leading to a weaker bound. Alternatively, perhaps there's a better way to approach this. Let's consider the three cylinders and their equations. For any point (x, y, z) in the intersection, it must lie within all three cylinders. Each cylinder restricts two coordinates based on their shifts. Perhaps the maximum distance from the "center" of the sphere will depend on the shifts. However, the problem states that the intersection is contained within a sphere of radius sqrt(3/2), regardless of the shifts. Therefore, there must be a way to show that even with the shifts, the maximum possible value of x² + y² + z² (or distance from some point) is bounded by 3/2. Wait, but if we take the three cylinders shifted far away, couldn't the intersection points also be far away? For example, suppose we shift the first cylinder along the y-axis by a large amount a. Then, the equation becomes (y - a)^2 + z² ≤ 1. Similarly, shifting another cylinder along x by a large amount b, (x - b)^2 + z² ≤ 1. If a and b are large, would there be any intersection? The intersection would require (y - a)^2 + z² ≤ 1 and (x - b)^2 + z² ≤ 1. But if a and b are very large, the only points that satisfy both would be near (b, a, 0), but with x ≈ b, y ≈ a, z ≈ 0. Then, checking the third cylinder, say shifted along z by c: (x - d)^2 + y² ≤ 1. If d and c are also large, then the intersection might not exist. Wait, but if all cylinders are shifted by large amounts in different directions, their intersection might be empty. But the problem states "the intersection of three circular cylinders... is contained within a sphere...". If the intersection is empty, then it's trivially contained within any sphere. But if there exists a non-empty intersection, we need to show it's within that sphere. Alternatively, perhaps no matter how you shift the cylinders, the non-empty intersection cannot have points further than sqrt(3/2) from some center. Alternatively, consider that each cylinder, being radius 1, constraints two coordinates to lie within a distance 1 from the axis. If the axes are mutually perpendicular, then the coordinates are constrained in different directions, leading to a bound on the overall distance. Wait, maybe we can use the Pythagorean theorem. If a point is within distance 1 from each of three mutually perpendicular lines, then its distance from the common perpendicular point can be bounded. But if the lines are not intersecting, this is more complicated. Alternatively, think in terms of vectors. Suppose we have three mutually perpendicular unit vectors u, v, w along the axes of the cylinders. Then, the distance from a point P to each cylinder's axis is the distance from P to the line (axis). Since each cylinder has radius 1, the distance from P to each axis must be ≤ 1. But the distance from a point to a line in 3D can be expressed as ||(P - Q) × u||, where Q is a point on the line and u is the direction vector of the line. Given three mutually perpendicular lines, the distance from P to each line is ≤ 1. We need to bound ||P||. Wait, perhaps this is the way to go. Let me formalize it. Let’s suppose the three cylinders have axes along three mutually perpendicular lines. Let’s denote these lines as L1, L2, L3. Let’s take a point P in the intersection of the three cylinders. Then, the distance from P to each Li is ≤ 1. We need to show that there exists a point O such that the distance from P to O is ≤ sqrt(3/2), regardless of the positions of the lines L1, L2, L3. Alternatively, regardless of the positions of the lines, the maximum possible distance of P from the origin (or another fixed point) is bounded by sqrt(3/2). Alternatively, use the fact that for any three mutually perpendicular lines in space, the minimal enclosing sphere for the set of points within distance 1 from each line has radius sqrt(3/2). This seems abstract. Maybe another approach. Consider the three cylinders as sets of points at distance ≤1 from three mutually perpendicular lines. We need to show that any point in the intersection is within sqrt(3/2) from some point. Alternatively, use the fact that in 3D space, the maximum distance between a point and three mutually perpendicular lines can be related to its coordinates. Wait, perhaps assign coordinates such that one line is the x-axis, another is the y-axis shifted along z, and the third is the z-axis shifted along x and y. Then, for a point (x, y, z), compute distances to each axis and set them ≤1. Then, find the maximum of x² + y² + z² under these constraints. Let me try that. Let’s define the three lines:L1: x-axis, which is the line y=0, z=0. The distance from a point (x, y, z) to L1 is sqrt(y² + z²) ≤1.L2: shifted y-axis, say along the line x=0, z=c. The distance from (x, y, z) to L2 is sqrt(x² + (z - c)^2) ≤1.L3: shifted z-axis, say along the line x=d, y=e. The distance from (x, y, z) to L3 is sqrt((x - d)^2 + (y - e)^2) ≤1.So, the three distance constraints are:1. y² + z² ≤12. x² + (z - c)^2 ≤13. (x - d)^2 + (y - e)^2 ≤1We need to maximize x² + y² + z² under these three constraints. To find the maximum, we can use Lagrange multipliers, but this might be complicated. Alternatively, consider that each constraint limits two variables. Let's attempt to find the maximum possible x² + y² + z². Suppose we fix z. Then from constraint 1: y² ≤1 - z². From constraint 2: x² ≤1 - (z - c)^2. From constraint 3: (x - d)^2 + (y - e)^2 ≤1. To maximize x² + y² + z², we need to maximize each variable as much as possible. Let me assume that the maximum occurs when all three inequalities are active, i.e., equalities. So:y² + z² =1,x² + (z - c)^2 =1,(x - d)^2 + (y - e)^2 =1.We need to solve these equations and find x, y, z such that x² + y² + z² is maximized. This system of equations might have multiple solutions. Let's try to express variables in terms of each other. From the first equation: y² =1 - z² ⇒ y = ±sqrt(1 - z²)From the second equation: x² =1 - (z - c)^2 ⇒ x = ±sqrt(1 - (z - c)^2)From the third equation: (x - d)^2 + (y - e)^2 =1. Substitute x and y:[ ±sqrt(1 - (z - c)^2) - d ]² + [ ±sqrt(1 - z²) - e ]² =1This is quite complex. To find the maximum of x² + y² + z², which is [1 - (z - c)^2] + [1 - z²] + z² = 2 - (z - c)^2. Wait, wait. If x² =1 - (z - c)^2 and y² =1 - z², then x² + y² + z² = [1 - (z - c)^2] + [1 - z²] + z² = 2 - (z - c)^2. So, to maximize this expression, we need to minimize (z - c)^2. The minimum of (z - c)^2 is 0 when z = c. However, if z = c, then y² =1 - c², which requires that c² ≤1. If c >1 or c < -1, then y² would be negative, which is impossible. Therefore, if |c| ≤1, then z can be c, leading to x² =1, y² =1 - c², so x=±1, y=±sqrt(1 - c²), z=c. Then, x² + y² + z² =1 + (1 - c²) + c² =2. However, we also need to satisfy the third equation: (x - d)^2 + (y - e)^2 =1. If x=±1, y=±sqrt(1 - c²), then substituting into the third equation:(±1 - d)^2 + (±sqrt(1 - c²) - e)^2 =1This must hold. If d and e are chosen such that this equation is satisfied, then such points exist. In this case, x² + y² + z²=2. However, 2 is greater than 3/2, which contradicts our goal. Therefore, there must be a mistake in this approach. Wait, but the problem states that the intersection is contained within a sphere of radius sqrt(3/2) ≈1.2247, but here we get points at distance sqrt(2) ≈1.4142. So, this suggests a contradiction. Therefore, my assumption that the maximum occurs when all three constraints are active might be wrong, or my model of shifted cylinders is incorrect. Wait, but in the case where the cylinders are not shifted (c=d=e=0), then the third equation would be (x)^2 + (y)^2 =1, but x² + y² =1, and z² ≤1. But in the standard Steinmetz solid, the intersection of three cylinders x² + y² ≤1, y² + z² ≤1, x² + z² ≤1 is bounded by x² + y² + z² ≤3/2. So why in this shifted case, I'm getting x² + y² + z²=2? Because I shifted one cylinder and kept others. Maybe if the cylinders are shifted, the intersection might not actually have such points. Wait, let's take an explicit example. Suppose we shift the third cylinder. Let’s set c=0, d=0, e=0, f=0, but shift the third cylinder's axis to x=1, y=0. So, the three cylinders are:1. y² + z² ≤1 (along x-axis)2. x² + z² ≤1 (along y-axis)3. (x -1)^2 + y² ≤1 (along z-axis shifted to x=1, y=0)Now, find the intersection of these three. The third cylinder equation is (x -1)^2 + y² ≤1. Let's see if there's a point in the intersection with large x² + y² + z². For example, take x=1, y=0, z=0. This satisfies all three cylinders: y² + z² =0 ≤1, x² + z² =1 ≤1, (x -1)^2 + y² =0 ≤1. Then, x² + y² + z²=1. Another point: x=0.5, y=0, z= sqrt(3)/2 ≈0.866. Then, check cylinders:1. y² + z² = 3/4 + 0 ≈0.866² = 0.75 ≤1: yes.2. x² + z² =0.25 + 0.75=1 ≤1: yes.3. (0.5 -1)^2 + y²=0.25 +0=0.25 ≤1: yes.So x² + y² + z²=0.25 +0 +0.75=1. But where is the maximum? Let's see if there's a point with higher x² + y² + z². Suppose x=1, y=0, z=0: sum=1. If x=0, y=1, z=0: but y² + z²=1, x² + z²=0 ≤1, (x -1)^2 + y²=1 +1=2 >1, so not in the intersection. Similarly, if z=1, x=0, y=0: first cylinder: y² + z²=1, second cylinder: x² + z²=1, third cylinder: (x -1)^2 + y²=1 +0=1. So this point (0,0,1) is in the intersection, sum=1. Wait, but in this shifted case, the maximum sum x² + y² + z² is still 1, which is less than 3/2. So maybe my previous calculation was incorrect because when I shifted the third cylinder, the intersection points don't achieve higher sums. Wait, but in my earlier example where c=0, shifting the third cylinder to x=1, the intersection seems to have points with x² + y² + z²=1. But according to the original problem, the sphere radius is sqrt(3/2)≈1.2247, which is larger than 1, so that's okay. But then, in the case where the cylinders are shifted, the maximum x² + y² + z² in the intersection is still less than or equal to 3/2. So, perhaps regardless of shifts, the sum x² + y² + z² is bounded by 3/2. Therefore, the intersection is contained within a sphere of radius sqrt(3/2). But how to formally show this? Wait, let's think back to the original case where axes intersect. There, by adding the three cylinder equations, you get 2(x² + y² + z²) ≤3, hence x² + y² + z² ≤3/2. If the cylinders are shifted, can we derive a similar bound? In the shifted case, the equations are:1. (y - a)^2 + (z - b)^2 ≤12. (x - c)^2 + (z - d)^2 ≤13. (x - e)^2 + (y - f)^2 ≤1If we add these three equations:(y - a)^2 + (z - b)^2 + (x - c)^2 + (z - d)^2 + (x - e)^2 + (y - f)^2 ≤3Expanding:y² - 2ay + a² + z² - 2bz + b² + x² - 2cx + c² + z² - 2dz + d² + x² - 2ex + e² + y² - 2fy + f² ≤3Combine like terms:2x² + 2y² + 2z² - 2(c + e)x - 2(a + f)y - 2(b + d)z + (a² + b² + c² + d² + e² + f²) ≤3Divide by 2:x² + y² + z² - (c + e)x - (a + f)y - (b + d)z + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Now, let's complete the square for x, y, z. Let’s denote:x’ = x - (c + e)/2y’ = y - (a + f)/2z’ = z - (b + d)/2Then, the left-hand side becomes:(x’ + (c + e)/2)^2 + (y’ + (a + f)/2)^2 + (z’ + (b + d)/2)^2 - (c + e)(x’ + (c + e)/2) - (a + f)(y’ + (a + f)/2) - (b + d)(z’ + (b + d)/2) + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Expanding each term:(x’² + (c + e)x’ + (c + e)^2/4) + (y’² + (a + f)y’ + (a + f)^2/4) + (z’² + (b + d)z’ + (b + d)^2/4) - (c + e)x’ - (c + e)^2/2 - (a + f)y’ - (a + f)^2/2 - (b + d)z’ - (b + d)^2/2 + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Simplify:x’² + y’² + z’² + [(c + e)^2/4 + (a + f)^2/4 + (b + d)^2/4] - [(c + e)^2/2 + (a + f)^2/2 + (b + d)^2/2] + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Combine the constants:x’² + y’² + z’² - [(c + e)^2/4 + (a + f)^2/4 + (b + d)^2/4] + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Expand the squared sums:(c + e)^2 =c² + 2ce + e²Similarly for others:(a + f)^2 =a² + 2af + f²(b + d)^2 =b² + 2bd + d²Therefore, the left-hand side:x’² + y’² + z’² - [ (c² + 2ce + e² + a² + 2af + f² + b² + 2bd + d²)/4 ] + (a² + b² + c² + d² + e² + f²)/2 ≤3/2Simplify the terms:The negative term is [ (sum of squares) + 2(ce + af + bd) ] /4The positive term is (sum of squares)/2So combining them:- [sum/4 + 2(ce + af + bd)/4 ] + sum/2 = (-sum/4 - (ce + af + bd)/2) + sum/2 = sum/4 - (ce + af + bd)/2Therefore, the inequality becomes:x’² + y’² + z’² + [ (a² + b² + c² + d² + e² + f²)/4 - (ce + af + bd)/2 ] ≤3/2But the term in brackets is a constant dependent on the shifts. Let's denote this constant as K. Then,x’² + y’² + z’² ≤3/2 - KBut x’² + y’² + z’² is the squared distance from the point (x, y, z) to the point ((c + e)/2, (a + f)/2, (b + d)/2). Therefore, the inequality shows that this distance squared is ≤3/2 - K. However, unless we can show that 3/2 - K is non-negative and that this implies the point is within sqrt(3/2) of some center, this might not help. Alternatively, notice that the term K can be rewritten:K = (a² + b² + c² + d² + e² + f²)/4 - (ce + af + bd)/2= (1/4)(a² + b² + c² + d² + e² + f²) - (1/2)(ce + af + bd)This can be expressed as:= (1/4)(a² - 2af + f²) + (1/4)(b² - 2bd + d²) + (1/4)(c² - 2ce + e²) = (1/4)[(a - f)^2 + (b - d)^2 + (c - e)^2]Therefore, K = (1/4)[(a - f)^2 + (b - d)^2 + (c - e)^2]Since squares are non-negative, K ≥0. Therefore,x’² + y’² + z’² ≤3/2 - K ≤3/2Therefore, the squared distance from ((c + e)/2, (a + f)/2, (b + d)/2) to any point (x, y, z) in the intersection is ≤3/2. Hence, all points in the intersection lie within a sphere of radius sqrt(3/2) centered at ((c + e)/2, (a + f)/2, (b + d)/2). Therefore, regardless of the shifts a, b, c, d, e, f, the intersection is contained within a sphere of radius sqrt(3/2). This completes the proof. The key step was adding the three cylinder equations, completing the square to find that the squared distance to a certain point is bounded by 3/2, thus establishing the required containment.

✅:The intersection of three mutually perpendicular circular cylinders with radii 1 is contained within a sphere of radius (sqrt{frac{3}{2}}). Here's the step-by-step explanation:1. Equations of Shifted Cylinders: Consider three cylinders with axes along pairwise perpendicular lines. Their equations can be expressed as: - ((y - a)^2 + (z - b)^2 leq 1) - ((x - c)^2 + (z - d)^2 leq 1) - ((x - e)^2 + (y - f)^2 leq 1) where shifts (a, b, c, d, e, f) account for possible translations of the axes.2. Summing the Inequalities: Adding these three inequalities: [ (y - a)^2 + (z - b)^2 + (x - c)^2 + (z - d)^2 + (x - e)^2 + (y - f)^2 leq 3 ] Expanding and combining like terms: [ 2x^2 + 2y^2 + 2z^2 - 2(c + e)x - 2(a + f)y - 2(b + d)z + (a^2 + b^2 + c^2 + d^2 + e^2 + f^2) leq 3 ] Dividing by 2: [ x^2 + y^2 + z^2 - (c + e)x - (a + f)y - (b + d)z + frac{1}{2}(a^2 + b^2 + c^2 + d^2 + e^2 + f^2) leq frac{3}{2} ]3. Completing the Square: Introduce variables shifted by half the linear coefficients: [ x' = x - frac{c + e}{2}, quad y' = y - frac{a + f}{2}, quad z' = z - frac{b + d}{2} ] Substituting these into the inequality transforms the left-hand side into: [ x'^2 + y'^2 + z'^2 + frac{1}{4}left[(a - f)^2 + (b - d)^2 + (c - e)^2right] leq frac{3}{2} ] The constant term (frac{1}{4}[(a - f)^2 + (b - d)^2 + (c - e)^2]) is non-negative, leading to: [ x'^2 + y'^2 + z'^2 leq frac{3}{2} ]4. Conclusion: This inequality shows that every point ((x, y, z)) in the intersection lies within a sphere of radius (sqrt{frac{3}{2}}) centered at (left(frac{c + e}{2}, frac{a + f}{2}, frac{b + d}{2}right)). Hence, regardless of the shifts, the intersection is contained within such a sphere.Final Answer:The intersection is contained within a sphere of radius (boxed{sqrt{dfrac{3}{2}}}).

🔑:Let's denote the cylinders by the inequalities:[ (x-a)^2 + (y-b)^2 < 1 ][ (y-c)^2 + (z-d)^2 < 1 ][ (z-e)^2 + (x-f)^2 < 1 ]For each pair of coordinates, observe the following inequalities:1. ((x-a)^2 + (x-f)^2):2. Using the identities and basic algebra, we get:[(x-a)^2 + (x-f)^2 geq 2 left( frac{(x-a) + (x-f)}{2} right)^2 = 2 left( x - frac{a+f}{2} right)^2 ]Applying a similar process to the other pairs:[(y-b)^2 + (y-c)^2 geq 2 left( frac{(y-b) + (y-c)}{2} right)^2 = 2 left( y - frac{b+c}{2} right)^2 ][(z-d)^2 + (z-e)^2 geq 2 left( frac{(z-d) + (z-e)}{2} right)^2 = 2 left( z - frac{d+e}{2} right)^2 ]By summing these inequalities, we get:[ [(x-a)^2 + (x-f)^2] + [(y-b)^2 + (y-c)^2] + [(z-d)^2 + (z-e)^2] > 2left( left( x - frac{a+f}{2} right)^2 + left( y - frac{b+c}{2} right)^2 + left( z - frac{d+e}{2} right)^2 right) ]Since each term derived sums to less than 1, we have:[ (x-a)^2 + (y-b)^2 < 1 ][ (y-c)^2 + (z-d)^2 < 1 ][ (z-e)^2 + (x-f)^2 < 1 ]Therefore:[ (x-a)^2 + (x-f)^2 + (y-b)^2 + (y-c)^2 + (z-d)^2 + (z-e)^2 < 3 ]Rewriting the inequalities in terms of sums we considered:[ 2 left( left( x - frac{a+f}{2} right)^2 + left( y - frac{b+c}{2} right)^2 + left( z - frac{d+e}{2} right)^2 right) < 3 ]Divide by 2 to isolate the variables:[left( x - frac{a+f}{2} right)^2 + left( y - frac{b+c}{2} right)^2 + left( z - frac{d+e}{2} right)^2 < frac{3}{2} ]This means any point inside the intersection of the given cylinders will be contained within the sphere with center (left( frac{a+f}{2}, frac{b+c}{2}, frac{d+e}{2} right)) and radius (sqrt{frac{3}{2}}).Conclusion:[boxed{text{Therefore, the intersection is contained in a sphere of radius } sqrt{frac{3}{2}}.}]

🧠:Okay, let's tackle this problem step by step. First, I need to understand what's being asked. The question is about a rectangular field whose length is increased, and we need to find out by how many square meters the area increases. The options are given, so after figuring it out, I can check which one matches.Alright, the original field is a rectangle with length 20 meters and width 5 meters. So, its original area can be calculated using the formula for the area of a rectangle: length multiplied by width. Let me write that down. Original area = 20 m * 5 m. Hmm, 20 times 5 is 100, so the original area is 100 square meters. Got that part.Now, the problem says the length is increased by 10 meters. So the new length should be the original length plus 10 meters. Let me make sure I'm interpreting that correctly. If the original length is 20 meters and we increase it by 10, then the new length is 20 + 10 = 30 meters. The width remains the same, right? The problem doesn't mention changing the width, only the length. So the new dimensions are 30 meters in length and 5 meters in width. Let me calculate the new area. New area = new length * width = 30 m * 5 m. That's 30 times 5, which is 150. So the new area is 150 square meters. To find the increase in area, I subtract the original area from the new area. That would be 150 sq m - 100 sq m = 50 sq m. So the area has increased by 50 square meters. Looking at the options, (C) is 50. That seems straightforward. Wait, let me double-check to make sure I didn't make any mistakes. Sometimes it's easy to mix up length and width or misread the question. Let me verify each step again. Original area: 20*5=100. Correct. New length: 20+10=30. Correct. New area: 30*5=150. Correct. Difference: 150-100=50. Yep, that checks out. Is there another way to approach this problem? Maybe by calculating the increase directly without computing both areas? Let's think. The increase in length is 10 meters, and the width stays the same. So the additional area would be the increase in length multiplied by the width. That is, 10 m * 5 m = 50 sq m. Oh, that's another way to see it. Instead of calculating the total area and subtracting, you can just take the added length and multiply by the width. That also gives 50. So that confirms the answer is indeed 50. I guess the key here is recognizing that when you increase one dimension of a rectangle, the increase in area is the product of the increase in that dimension and the other unchanged dimension. This might be a quicker way for similar problems in the future. But either method works. Just to be thorough, let's make sure there's no trick or catch in the question. Sometimes problems might mention increasing the length by 10% instead of 10 meters, but here it's explicitly 10 meters. The width isn't changed, so no hidden details there. The options given include 50, which we arrived at through two different methods, so that's solid. Another check: if the answer were different, say 100, that would mean the area doubled, but doubling the length (from 20 to 40) would do that. But we only increased by 10 meters, so 20 to 30 is a 50% increase in length. The area would then increase by 50% as well, since area is proportional to length when width is constant. 50% of 100 is 50, which again matches. So all methods converge on 50. I feel confident that's the correct answer.

✅:The original area of the rectangular field is calculated as: [text{Original Area} = text{Length} times text{Width} = 20, text{m} times 5, text{m} = 100, text{m}^2.]When the length is increased by (10, text{m}), the new length becomes: [text{New Length} = 20, text{m} + 10, text{m} = 30, text{m}.]The new area is then: [text{New Area} = text{New Length} times text{Width} = 30, text{m} times 5, text{m} = 150, text{m}^2.]The increase in area is the difference between the new and original areas: [text{Increase} = 150, text{m}^2 - 100, text{m}^2 = 50, text{m}^2.]Alternatively, the increase can be directly computed by multiplying the added length ((10, text{m})) by the unchanged width ((5, text{m})): [text{Increase} = 10, text{m} times 5, text{m} = 50, text{m}^2.]Both methods confirm the area increases by 50 square meters. Answer: (C) 50

🔑:1. Calculate the original area of the rectangular field: - The original length is (20 , text{m}). - The original width is (5 , text{m}). - The formula for area (A) of a rectangle is ( text{length} times text{width} ). [ A_{text{original}} = 20 , text{m} times 5 , text{m} = 100 , text{m}^2 ]2. Calculate the new length of the field after the increase: - The length is increased by (10 , text{m}). [ text{new length} = 20 , text{m} + 10 , text{m} = 30 , text{m} ]3. Calculate the new area of the rectangular field with the new length: - The new length is (30 , text{m}). - The width remains at (5 , text{m}). [ A_{text{new}} = 30 , text{m} times 5 , text{m} = 150 , text{m}^2 ]4. Calculate the increase in area: - Subtract the original area from the new area. [ Delta A = A_{text{new}} - A_{text{original}} = 150 , text{m}^2 - 100 , text{m}^2 = 50 , text{m}^2 ]5. Alternative method: - Since the increase in length is (10 , text{m}) and the width remains (5 , text{m}), the additional area due to the increased length can be directly calculated as: [ Delta A = 10 , text{m} times 5 , text{m} = 50 , text{m}^2 ] Conclusion:The area of the field is increased by 50 square meters.[ boxed{50} ]

🧠:Okay, let's try to solve this problem. We have the function f(x) = x² - 2x, and we need to find the set A which consists of all x such that f(f(x)) = 0. Then, we have to sum all the elements in set A. Hmm, okay, let me break this down step by step.First, I know that f(f(x)) means applying the function f twice. So first, compute f(x), then plug that result back into f again. And we want the result of that double application to be zero. So, the equation we need to solve is f(f(x)) = 0.Let me write that out. Let’s denote y = f(x). Then, f(f(x)) = f(y) = y² - 2y. We set that equal to zero: y² - 2y = 0. So, solving y² - 2y = 0 will give us the possible values of y, and then each y must be equal to f(x), so we can solve for x in each case.So, let me first solve y² - 2y = 0. Factoring that equation, we get y(y - 2) = 0. Therefore, the solutions are y = 0 and y = 2. So, f(f(x)) = 0 implies that f(x) must be either 0 or 2. That means we have two separate equations to solve: f(x) = 0 and f(x) = 2. The solutions to these equations will be the elements of set A, and then we can sum them up.Let me tackle each equation one by one. First, solving f(x) = 0. Given f(x) = x² - 2x, so:x² - 2x = 0Factor that equation:x(x - 2) = 0So, the solutions are x = 0 and x = 2. Therefore, from f(x) = 0, we get two solutions: 0 and 2.Next, solving f(x) = 2. Let's write that equation:x² - 2x = 2Subtract 2 from both sides to set the equation to zero:x² - 2x - 2 = 0This is a quadratic equation, and we can solve it using the quadratic formula. The quadratic is ax² + bx + c = 0, so here a = 1, b = -2, c = -2. The quadratic formula is x = [-b ± √(b² - 4ac)] / (2a).Plugging in the values:x = [2 ± √((-2)² - 4*1*(-2))]/2*1Calculate the discriminant first:b² - 4ac = 4 - 4*(1)*(-2) = 4 + 8 = 12So, the square root of 12 is 2√3. Therefore, the solutions are:x = [2 ± 2√3]/2Simplify by dividing numerator and denominator by 2:x = 1 ± √3Therefore, the solutions to f(x) = 2 are x = 1 + √3 and x = 1 - √3.So, putting it all together, the set A consists of the solutions from both equations: 0, 2, 1 + √3, and 1 - √3. Therefore, the elements of A are {0, 2, 1 + √3, 1 - √3}. Now, we need to find the sum of all these elements.Let me add them up:0 + 2 + (1 + √3) + (1 - √3)Combine like terms. Let's first add the constants:0 + 2 + 1 + 1 = 4Then, the terms with √3:√3 - √3 = 0So, the total sum is 4 + 0 = 4.Wait a second, that seems straightforward, but let me verify to make sure I didn't miss anything.First, confirming the solutions for f(x) = 0: x² - 2x = 0. Factoring gives x(x - 2) = 0, so x = 0 and x = 2. Correct.Then, for f(x) = 2: x² - 2x - 2 = 0. Using quadratic formula, discriminant is 12, so roots are (2 ± 2√3)/2 = 1 ± √3. Correct.Thus, set A has four elements: 0, 2, 1 + √3, 1 - √3. Adding them together:0 + 2 + (1 + √3) + (1 - √3) = 0 + 2 + 1 + 1 + √3 - √3 = 4. The irrational parts cancel out, leaving 4. That makes sense.But wait, is there a quicker way to see why the sum is 4 without having to compute each root? Maybe using Vieta's formulas?Let me think. For the quadratic equations involved, Vieta's formula relates the sum of roots to the coefficients. For the equation f(x) = 0, which is x² - 2x = 0, the sum of roots is 0 + 2 = 2. For the equation f(x) = 2, which is x² - 2x - 2 = 0, the sum of roots is (1 + √3) + (1 - √3) = 2. So, total sum of all roots is 2 (from first equation) + 2 (from second equation) = 4. So that's another way to see it.This is a good check. Vieta's formula tells us that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a. For the first equation x² - 2x = 0, a = 1, b = -2, so sum is -(-2)/1 = 2. For the second equation x² - 2x -2 =0, a=1, b=-2, so sum is -(-2)/1 = 2. Thus, combining both equations, total sum is 2 + 2 = 4. So, regardless of the actual roots, their sum is 4. Therefore, the answer is indeed 4.This method is actually smarter because we don't have to compute each root individually; we can just use Vieta's formula on each quadratic equation and sum them up. However, since the original problem didn't specify to use any particular method, both approaches are valid. But it's reassuring to see that both methods give the same result.Therefore, I'm confident that the sum of all elements in set A is 4.

✅:To solve the problem, we start by analyzing the composition ( f(f(x)) = 0 ) where ( f(x) = x^2 - 2x ). 1. Solve ( f(y) = 0 ) where ( y = f(x) ): [ y^2 - 2y = 0 implies y(y - 2) = 0 implies y = 0 text{ or } y = 2 ]2. Find ( x ) such that ( f(x) = 0 ): [ x^2 - 2x = 0 implies x(x - 2) = 0 implies x = 0 text{ or } x = 2 ]3. Find ( x ) such that ( f(x) = 2 ): [ x^2 - 2x - 2 = 0 ] Using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) with ( a = 1 ), ( b = -2 ), ( c = -2 ): [ x = frac{2 pm sqrt{4 + 8}}{2} = frac{2 pm 2sqrt{3}}{2} = 1 pm sqrt{3} ]4. Combine all solutions: The set ( A = {0, 2, 1 + sqrt{3}, 1 - sqrt{3}} ).5. Sum the elements using Vieta's formulas: - Sum for ( x^2 - 2x = 0 ): ( 0 + 2 = 2 ) - Sum for ( x^2 - 2x - 2 = 0 ): ( (1 + sqrt{3}) + (1 - sqrt{3}) = 2 ) - Total sum: ( 2 + 2 = 4 )Final Answer: The sum of all elements in set ( A ) is (boxed{4}).

🔑:1. Define the given function: Let's start by writing down the given function (f(x)). [ f(x) = x^2 - 2x ]2. Substitute (f(x)) in itself: We are given (f(f(x)) = 0). This means we need to find (fleft(f(x)right)) and then solve for when it equals zero. [ f(f(x)) = fleft( x^2 - 2x right) ]3. Evaluate (fleft(x^2 - 2xright): [ f(y) = y^2 - 2y quad text{where} quad y = x^2 - 2x ] Substitute ( y ): [ f(x^2 - 2x) = (x^2 - 2x)^2 - 2(x^2 - 2x) ]4. Simplify ( (x^2 - 2x)^2 - 2(x^2 - 2x): [ (x^2 - 2x)^2 = x^4 - 4x^3 + 4x^2 ] [ -2(x^2 - 2x) = -2x^2 + 4x ] Combine both expressions: [ x^4 - 4x^3 + 4x^2 - 2x^2 + 4x = x^4 - 4x^3 + 2x^2 + 4x ]5. Set ( f(f(x)) = 0 ): [ x^4 - 4x^3 + 2x^2 + 4x = 0 ]6. Factorize the equation: [ x(x^3 - 4x^2 + 2x + 4) = 0 ] This gives one factor immediately: [ x = 0 ] Next, solve the cubic equation: [ x^3 - 4x^2 + 2x + 4 = 0 ]7. Solve the cubic equation: Using the Rational Root Theorem or another factorization method is common. We need to find roots of the polynomial: [ x^3 - 4x^2 + 2x + 4 = 0 ] Possible rational roots could be ( pm 1, pm 2, pm 4 ) (based on possible factors of the constant term and the leading coefficient). Substitute and check: [ x = 2 quad Rightarrow quad 2^3 - 4 cdot 2^2 + 2 cdot 2 + 4 = 8 - 16 + 4 + 4 = 0 ] Therefore, ( x = 2 ) is a root.8. Use synthetic division or polynomial long division: Divide ( x^3 - 4x^2 + 2x + 4 ) by ( x-2 ) to find the other quadratic factor. After division, you get: [ x^3 - 4x^2 + 2x + 4 = (x-2)(x^2 - 2x - 2) ] Now solve (x^2 - 2x - 2 = 0 ) using the quadratic formula: [ x = frac{2 pm sqrt{4 + 8}}{2} = frac{2 pm sqrt{12}}{2} = frac{2 pm 2sqrt{3}}{2} = 1 pm sqrt{3} ]9. Summarize all solutions: The roots are (x = 0, x = 2, x = 1 + sqrt{3}, x = 1 - sqrt{3}).10. Find the sum of the roots: Sum of the roots: [ 0 + 2 + (1 + sqrt{3}) + (1 - sqrt{3}) = 0 + 2 + 1 + 1 = 4 ] Conclusion:The sum of all elements ( x ) in set ( A ), which satisfies ( f(f(x)) = 0 ), is[boxed{4}]

🧠:Okay, so I need to prove Binet's formula using induction. Binet's formula gives the nth Fibonacci number as (φ^n - ψ^n)/√5, where φ is the golden ratio (1+√5)/2 and ψ is its conjugate (1-√5)/2. Let me recall how mathematical induction works. First, induction requires two steps: the base case and the inductive step. The base case checks if the formula holds for the initial values, typically n=0 and n=1. The inductive step assumes it holds for some arbitrary n = k and n = k+1, then proves it for n = k+2. Since Fibonacci numbers are defined recursively as Fₙ = Fₙ₋₁ + Fₙ₋₂, I think I need to use strong induction here, where we assume the formula holds for all previous terms up to k, not just k.Wait, actually, Fibonacci numbers depend on the two previous terms. So maybe for the inductive step, I need to assume it's true for n = k and n = k-1, then prove it for n = k+1. Let me verify the steps.Base case: Let's start with n=0 and n=1. Let's compute F₀ and F₁ using Binet's formula and check if they match the actual Fibonacci numbers. F₀ should be 0. Let's compute (φ⁰ - ψ⁰)/√5. Since anything to the power 0 is 1, so (1 - 1)/√5 = 0/√5 = 0. That's correct. F₁ should be 1. Compute (φ¹ - ψ¹)/√5. Let's calculate φ - ψ: [(1+√5)/2] - [(1-√5)/2] = (1+√5 -1 +√5)/2 = (2√5)/2 = √5. Then (√5)/√5 = 1. Perfect, that's F₁=1. So base cases hold.Now, inductive hypothesis: Assume that for some integer k ≥ 1, the formula holds for n = k and n = k-1. That is, Fₖ = (φᵏ - ψᵏ)/√5 and Fₖ₋₁ = (φᵏ⁻¹ - ψᵏ⁻¹)/√5.We need to show that Fₖ₊₁ = (φᵏ⁺¹ - ψᵏ⁺¹)/√5.Wait, hold on. The Fibonacci recurrence is Fₙ = Fₙ₋₁ + Fₙ₋₂. So if I want to show Fₖ₊₁, according to the recurrence, it's Fₖ + Fₖ₋₁. Wait, no. Let's clarify indices.Actually, the Fibonacci sequence is defined as F₀=0, F₁=1, and Fₙ = Fₙ₋₁ + Fₙ₋₂ for n ≥ 2. So if we are using induction, perhaps we can use ordinary induction where we assume it's true for n and n-1, then prove for n+1.Alternatively, maybe we need to use strong induction where we assume it's true for all numbers up to n, but given the recurrence depends on two previous terms, we can assume for n and n-1, then prove for n+1.So for the inductive step, let's suppose that Binet's formula holds for n = k and n = k-1. Then, using the Fibonacci recurrence, Fₖ₊₁ = Fₖ + Fₖ₋₁. Substitute the formula into that equation.So Fₖ₊₁ = [ (φᵏ - ψᵏ)/√5 ] + [ (φᵏ⁻¹ - ψᵏ⁻¹)/√5 ].Factor out 1/√5: Fₖ₊₁ = (1/√5)[ φᵏ - ψᵏ + φᵏ⁻¹ - ψᵏ⁻¹ ].Combine like terms: φᵏ + φᵏ⁻¹ = φᵏ⁻¹ (φ + 1), similarly ψᵏ + ψᵏ⁻¹ = ψᵏ⁻¹ (ψ + 1).Wait, actually, let's factor φᵏ⁻¹ from the first two terms and ψᵏ⁻¹ from the last two terms:= (1/√5)[ φᵏ⁻¹ (φ + 1) - ψᵏ⁻¹ (ψ + 1) ]But we need to see if φ + 1 is equal to φ², since φ is the golden ratio, which satisfies φ² = φ + 1. Similarly, ψ² = ψ + 1? Let me check:Given φ = (1 + √5)/2. Then φ² = [(1 + √5)/2]^2 = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2. On the other hand, φ + 1 = (1 + √5)/2 + 1 = (1 + √5 + 2)/2 = (3 + √5)/2. So yes, φ² = φ + 1.Similarly for ψ: ψ = (1 - √5)/2. ψ² = [(1 - √5)/2]^2 = (1 - 2√5 + 5)/4 = (6 - 2√5)/4 = (3 - √5)/2. ψ + 1 = (1 - √5)/2 + 1 = (1 - √5 + 2)/2 = (3 - √5)/2. So ψ² = ψ + 1 as well.Therefore, φ + 1 = φ² and ψ + 1 = ψ². Therefore, the expression becomes:(1/√5)[ φᵏ⁻¹ * φ² - ψᵏ⁻¹ * ψ² ] = (1/√5)[ φᵏ⁺¹ - ψᵏ⁺¹ ]Which is exactly Binet's formula for Fₖ₊₁. Therefore, if the formula holds for n = k and n = k-1, it holds for n = k+1. Therefore, by induction, the formula holds for all n ≥ 0.Wait, but let me check if this is sufficient. We proved the base cases for n=0 and n=1. Then, assuming it holds for n = k and n = k-1, we proved it holds for n = k+1. Therefore, the inductive step covers all subsequent n. So if the base cases are verified, and assuming the formula for two previous terms allows us to get the next term, then induction should work.Alternatively, since the Fibonacci recurrence depends on two previous terms, we need two base cases and the inductive step that uses the previous two. So this should be correct.Let me verify with an example. Let's take n=2. F₂ should be 1. Using Binet's formula: (φ² - ψ²)/√5. We know φ² = φ + 1, so φ² - ψ² = (φ + 1) - (ψ + 1) = φ - ψ = √5. Therefore, (√5)/√5 = 1. Correct. Another example: n=3. F₃=2. Binet's formula: (φ³ - ψ³)/√5. Compute φ³ = φ*φ² = φ*(φ + 1) = φ² + φ = (φ + 1) + φ = 2φ +1. Similarly ψ³ = 2ψ +1. So φ³ - ψ³ = 2φ +1 - 2ψ -1 = 2(φ - ψ) = 2√5. Then divided by √5: 2√5 /√5 = 2. Correct.So the inductive step seems valid. Therefore, the proof by induction works.

✅:To prove Binet's formula for Fibonacci numbers using mathematical induction, follow these steps:Binet's Formula:F_{n} = frac{varphi^{n} - hat{varphi}^{n}}{sqrt{5}},where (varphi = frac{1+sqrt{5}}{2}) (golden ratio) and (hat{varphi} = frac{1-sqrt{5}}{2}) (its conjugate).--- Base Cases1. For (n = 0): F_0 = frac{varphi^{0} - hat{varphi}^{0}}{sqrt{5}} = frac{1 - 1}{sqrt{5}} = 0, which matches (F_0 = 0).2. For (n = 1): F_1 = frac{varphi^{1} - hat{varphi}^{1}}{sqrt{5}} = frac{frac{1+sqrt{5}}{2} - frac{1-sqrt{5}}{2}}{sqrt{5}} = frac{sqrt{5}}{sqrt{5}} = 1, which matches (F_1 = 1).--- Inductive StepHypothesis: Assume Binet's formula holds for (n = k) and (n = k-1):begin{align*}F_k &= frac{varphi^{k} - hat{varphi}^{k}}{sqrt{5}}, F_{k-1} &= frac{varphi^{k-1} - hat{varphi}^{k-1}}{sqrt{5}}.end{align*}Goal: Prove the formula holds for (n = k+1), i.e.,F_{k+1} = frac{varphi^{k+1} - hat{varphi}^{k+1}}{sqrt{5}}.--- Proof Using Fibonacci RecurrenceBy the Fibonacci definition, (F_{k+1} = F_k + F_{k-1}). Substitute the inductive hypothesis:F_{k+1} = frac{varphi^{k} - hat{varphi}^{k}}{sqrt{5}} + frac{varphi^{k-1} - hat{varphi}^{k-1}}{sqrt{5}}.Factor out (frac{1}{sqrt{5}}):F_{k+1} = frac{1}{sqrt{5}} left[ varphi^{k} + varphi^{k-1} - hat{varphi}^{k} - hat{varphi}^{k-1} right].Group terms involving (varphi) and (hat{varphi}):F_{k+1} = frac{1}{sqrt{5}} left[ varphi^{k-1}(varphi + 1) - hat{varphi}^{k-1}(hat{varphi} + 1) right].--- Simplify Using Properties of (varphi) and (hat{varphi})The golden ratio (varphi) satisfies (varphi^2 = varphi + 1), and its conjugate (hat{varphi}) satisfies (hat{varphi}^2 = hat{varphi} + 1). Substitute these identities:begin{align*}varphi + 1 &= varphi^2, hat{varphi} + 1 &= hat{varphi}^2.end{align*}Apply this to the expression:F_{k+1} = frac{1}{sqrt{5}} left[ varphi^{k-1} cdot varphi^2 - hat{varphi}^{k-1} cdot hat{varphi}^2 right] = frac{varphi^{k+1} - hat{varphi}^{k+1}}{sqrt{5}}.--- ConclusionBy induction, Binet's formula holds for all (n geq 0).Final Answer:boxed{F_{n} = frac{varphi^{n} - hat{varphi}^{n}}{sqrt{5}}}

🔑:1. Base Case Verification: We first verify the base cases: ( F_0 ) and ( F_1 ). - For ( n = 0 ): [ F_0 = frac{varphi^0 - widehat{varphi}^0}{sqrt{5}} = frac{1 - 1}{sqrt{5}} = 0 ] Thus, ( F_0 = 0 ), which satisfies the initial condition. - For ( n = 1 ): [ F_1 = frac{varphi^1 - widehat{varphi}^1}{sqrt{5}} = frac{varphi - widehat{varphi}}{sqrt{5}} ] Now, calculate ( varphi - widehat{varphi} ): [ varphi = frac{1 + sqrt{5}}{2}, quad widehat{varphi} = frac{1 - sqrt{5}}{2} ] [ varphi - widehat{varphi} = left(frac{1 + sqrt{5}}{2}right) - left(frac{1 - sqrt{5}}{2}right) = frac{1 + sqrt{5} - 1 + sqrt{5}}{2} = frac{2sqrt{5}}{2} = sqrt{5} ] Substituting back: [ F_1 = frac{sqrt{5}}{sqrt{5}} = 1 ] Thus, ( F_1 = 1 ), which satisfies the initial condition.2. Inductive Step: Assume that the formula is true for ( n = k ) and ( n = k-1 ): [ F_k = frac{varphi^k - widehat{varphi}^k}{sqrt{5}}, quad F_{k-1} = frac{varphi^{k-1} - widehat{varphi}^{k-1}}{sqrt{5}} ] We need to show that: [ F_{k+1} = frac{varphi^{k+1} - widehat{varphi}^{k+1}}{sqrt{5}} ] We use the recurrence relation ( F_{k+1} = F_k + F_{k-1} ): [ F_{k+1} = frac{varphi^k - widehat{varphi}^k}{sqrt{5}} + frac{varphi^{k-1} - widehat{varphi}^{k-1}}{sqrt{5}} ] Combine the fractions: [ F_{k+1} = frac{varphi^k - widehat{varphi}^k + varphi^{k-1} - widehat{varphi}^{k-1}}{sqrt{5}} ] Factor out common terms: [ F_{k+1} = frac{varphi^{k-1}(varphi + 1) - widehat{varphi}^{k-1}(widehat{varphi} + 1)}{sqrt{5}} ] Using the property of ( varphi ) and ( widehat{varphi} ) (roots of the equation ( x^2 = x + 1 )): [ varphi^2 = varphi + 1 quad text{and} quad widehat{varphi}^2 = widehat{varphi} + 1 ] Hence: [ varphi + 1 = varphi^2 quad text{and} quad widehat{varphi} + 1 = widehat{varphi}^2 ] Substituting these back: [ F_{k+1} = frac{varphi^{k-1} cdot varphi^2 - widehat{varphi}^{k-1} cdot widehat{varphi}^2}{sqrt{5}} ] [ F_{k+1} = frac{varphi^{k+1} - widehat{varphi}^{k+1}}{sqrt{5}} ]Thus, by induction, the formula is proven.[boxed{F_{n}=frac{varphi^{n}-widehat{varphi}^{n}}{sqrt{5}}}]

🧠:Okay, so I need to find the intersection point between the given line and the plane. Let me start by recalling what the equations represent. The line is given in symmetric form, which I think is like the standard form where each coordinate is expressed in terms of a parameter. The plane is given by a linear equation. The intersection point should satisfy both equations. First, let me parse the line equation. The symmetric equations are (x + 2)/(-1) = (y - 1)/1 = (z + 3)/2. Symmetric equations usually mean that each part is equal to some parameter t. So I can write parametric equations for the line by setting each fraction equal to t. Let me check that.If (x + 2)/(-1) = t, then x + 2 = -t, so x = -t - 2.Similarly, (y - 1)/1 = t implies y = t + 1.And (z + 3)/2 = t gives z = 2t - 3.So the parametric equations for the line are:x = -t - 2y = t + 1z = 2t - 3Now, the plane equation is x + 2y - z - 2 = 0. To find the intersection, I need to substitute the parametric equations of the line into the plane equation and solve for t. Then plug that t back into the parametric equations to get the point.Let me write that out step by step.Substitute x, y, z from the line into the plane equation:(-t - 2) + 2*(t + 1) - (2t - 3) - 2 = 0Now, simplify this equation step by step.First, expand the terms:= (-t - 2) + 2t + 2 - 2t + 3 - 2Wait, let's do each term carefully.Starting with (-t - 2): that's straightforward.Then 2*(t + 1) = 2t + 2.Then -(2t - 3) = -2t + 3.Then the -2 at the end.So putting it all together:(-t - 2) + (2t + 2) + (-2t + 3) - 2Now combine like terms. Let's group the t terms and the constants.For the t terms:- t + 2t - 2t = (-1 + 2 - 2)t = (-1)t = -tFor the constants:-2 + 2 + 3 - 2 = (-2 + 2) + (3 - 2) = 0 + 1 = 1So altogether, the equation simplifies to:-t + 1 = 0Solving for t:-t + 1 = 0 => -t = -1 => t = 1Okay, so t is 1. Now plug this back into the parametric equations for x, y, z.x = -t - 2 = -1 - 2 = -3y = t + 1 = 1 + 1 = 2z = 2t - 3 = 2*1 - 3 = 2 - 3 = -1So the intersection point is (-3, 2, -1).Wait, let me verify this point satisfies both the line equation and the plane equation.First, check the plane equation: x + 2y - z - 2.Plugging in (-3) + 2*2 - (-1) - 2 = -3 + 4 + 1 - 2 = (-3 + 4) + (1 - 2) = 1 -1 = 0. That works.Now check the line equations. The original symmetric equations are (x + 2)/(-1) = (y - 1)/1 = (z + 3)/2.Compute each part for the point (-3, 2, -1):(x + 2)/(-1) = (-3 + 2)/(-1) = (-1)/(-1) = 1(y - 1)/1 = (2 - 1)/1 = 1/1 = 1(z + 3)/2 = (-1 + 3)/2 = 2/2 = 1All three are equal to 1, which matches the parameter t = 1. So that's consistent.Therefore, the intersection point is indeed (-3, 2, -1).Wait, just to make sure I didn't make a mistake in substitution. Let me go through the substitution again.Original plane equation: x + 2y - z - 2 = 0Substituting x = -t - 2, y = t + 1, z = 2t - 3:(-t - 2) + 2*(t + 1) - (2t - 3) - 2Expanding:= -t - 2 + 2t + 2 - 2t + 3 - 2Combine like terms:t terms: (-t + 2t - 2t) = -tconstants: (-2 + 2 + 3 - 2) = 1So, -t + 1 = 0 => t = 1. Correct.Then x = -1 - 2 = -3, y = 1 + 1 = 2, z = 2 - 3 = -1. All right.Another check: the direction vector of the line is given by the denominators in the symmetric equations. The direction vector is (-1, 1, 2). If the line is moving in that direction, starting from a point when t=0, which would be x=-2, y=1, z=-3. Then at t=1, moving by (-1,1,2), the point is (-3,2,-1), which matches. So that makes sense.Therefore, I think the answer is correct.

✅:To find the intersection point of the line and the plane, follow these steps:1. Express the line in parametric form: The symmetric equations of the line are: [ frac{x+2}{-1} = frac{y-1}{1} = frac{z+3}{2} = t ] Setting each component equal to parameter ( t ), we derive the parametric equations: [ begin{aligned} x &= -t - 2, y &= t + 1, z &= 2t - 3. end{aligned} ]2. Substitute the parametric equations into the plane equation: The plane equation is ( x + 2y - z - 2 = 0 ). Substituting ( x ), ( y ), and ( z ) in terms of ( t ): [ (-t - 2) + 2(t + 1) - (2t - 3) - 2 = 0 ] Simplify: [ -t - 2 + 2t + 2 - 2t + 3 - 2 = -t + 1 = 0 ] Solving for ( t ): [ t = 1 ]3. Find the coordinates using ( t = 1 ): Substitute ( t = 1 ) back into the parametric equations: [ begin{aligned} x &= -1 - 2 = -3, y &= 1 + 1 = 2, z &= 2(1) - 3 = -1. end{aligned} ]4. Verify the solution: - Plane equation: ( (-3) + 2(2) - (-1) - 2 = 0 ), which holds true. - Line equations: All symmetric ratios equal 1 when substituting ( (-3, 2, -1) ), confirming consistency.Final Answer: The intersection point is (boxed{(-3, 2, -1)}).

🔑:```markdown## Problem StatementFind the intersection point of a line and a plane.[begin{aligned}& frac{x+2}{-1}=frac{y-1}{1}=frac{z+3}{2} & x+2 y - z - 2 = 0end{aligned}]1. Express the parametric equations of the line. Given the symmetric form of the line: [ frac{x+2}{-1}=frac{y-1}{1}=frac{z+3}{2}=t ] We can break this into parametric equations by setting each expression equal to the parameter ( t ): [ begin{cases} x = -2 - t y = 1 + t z = -3 + 2t end{cases} ]2. Substitute the parametric equations into the plane's equation. The equation of the plane is: [ x + 2y - z - 2 = 0 ] Substitute ( x = -2 - t ), ( y = 1 + t ), and ( z = -3 + 2t ) into the plane equation: [ (-2 - t) + 2(1 + t) - (-3 + 2t) - 2 = 0 ]3. Simplify the resulting equation and solve for ( t ). [ -2 - t + 2 + 2t + 3 - 2t - 2 = 0 ] Combine like terms: [ -2 + 2 - 2 + 3 - t + 2t - 2t = 0 ] [ 1 - t = 0 ] Therefore, solving for ( t ) we get: [ t = 1 ]4. Find the coordinates of the intersection point using ( t = 1 ). Substitute ( t = 1 ) back into the parametric equations: [ begin{cases} x = -2 - 1 = -3 y = 1 + 1 = 2 z = -3 + 2 cdot 1 = -1 end{cases} ] Thus, the intersection point is: [ (-3, 2, -1) ] Conclusion:[boxed{(-3, 2, -1)}]```

🧠:Okay, so I need to prove that in a right spherical triangle with legs a and b, and hypotenuse c, the Pythagorean theorem is given by tan²c = tan²a + tan²b + tan²a tan²b. Hmm, spherical trigonometry is a bit different from plane trigonometry, right? Let me recall some basic concepts.First, in a spherical triangle, the sides are angles measured in radians, corresponding to the angles subtended at the center of the sphere. For a right spherical triangle, one of the angles is 90 degrees (or π/2 radians). The regular Pythagorean theorem for a spherical triangle isn't the same as the planar one. I remember there's something involving cosines... Wait, yes! The standard spherical Pythagorean theorem is cos c = cos a cos b. But here, the problem is asking about a relation with tangents squared. So maybe I need to derive this tan²c expression from the known spherical Pythagorean theorem.Let me start with the known formula: cos c = cos a cos b. If I can manipulate this equation to express tan²c in terms of tan²a and tan²b, that might work. Let's recall that tan θ = sin θ / cos θ, so tan² θ = sin² θ / cos² θ. Also, remember that sin² θ + cos² θ = 1. Maybe I can use these identities.Starting from cos c = cos a cos b. Let's square both sides: cos² c = cos² a cos² b. Then, let's express sin² c in terms of cos² c: sin² c = 1 - cos² c = 1 - cos² a cos² b. Therefore, tan² c = sin² c / cos² c = (1 - cos² a cos² b) / (cos² a cos² b). Let's split the numerator: [1 / (cos² a cos² b)] - [cos² a cos² b / (cos² a cos² b)] = 1/(cos² a cos² b) - 1.So, tan² c = 1/(cos² a cos² b) - 1. Let me write that as tan² c = (1 / (cos² a cos² b)) - 1. Hmm, can I relate this to tan²a and tan²b? Let's see. We know that 1/cos² a = 1 + tan²a, right? Because tan²a = sin²a / cos²a, so 1 + tan²a = (sin²a + cos²a)/cos²a = 1/cos²a. Same for 1/cos²b = 1 + tan²b.Therefore, 1/(cos²a cos²b) = (1/cos²a)(1/cos²b) = (1 + tan²a)(1 + tan²b). So substituting back into tan²c, we have tan²c = (1 + tan²a)(1 + tan²b) - 1. Let me compute that:(1 + tan²a)(1 + tan²b) = 1*1 + 1*tan²b + tan²a*1 + tan²a tan²b = 1 + tan²b + tan²a + tan²a tan²b. Then subtract 1: tan²c = (1 + tan²a + tan²b + tan²a tan²b) - 1 = tan²a + tan²b + tan²a tan²b. Ah, there we go! That's exactly the formula we needed to prove. So starting from the spherical Pythagorean theorem cos c = cos a cos b, we used trigonometric identities to express tan²c in terms of tan²a and tan²b, leading us to the desired equation.Let me check if each step is valid. Starting from cos c = cos a cos b. Squaring both sides is okay. Then expressing sin²c as 1 - cos²c is standard. Then tan²c is sin²c over cos²c, that's correct. Splitting the fraction into two terms, yes. Then substituting 1/cos²a with 1 + tan²a, same for 1/cos²b. Multiplying those gives (1 + tan²a)(1 + tan²b), expanding that gives four terms, subtract 1 gives the required expression. All steps seem valid.I think that's solid. So the key was to start from the known spherical Pythagorean theorem and manipulate it using trigonometric identities to express tan²c. The trick was recognizing that 1/cos²a cos²b can be written as (1 + tan²a)(1 + tan²b). That's clever. I should remember that 1/cos²θ is 1 + tan²θ. That identity is super useful here.Another way to think about it: if we have tan²c expressed in terms of tan²a and tan²b, we can relate all three sides through their tangents squared, which is a different relationship than the cosine version. This might be useful in certain spherical triangle problems where tangents are easier to work with.I wonder if there's an alternative derivation using some other spherical trigonometric identities. For example, maybe using the sine formula or the cotangent formula. But given that the starting point is the standard spherical Pythagorean theorem, the method above seems straightforward.Just to recap the steps:1. Start with cos c = cos a cos b.2. Square both sides: cos²c = cos²a cos²b.3. Express sin²c = 1 - cos²c = 1 - cos²a cos²b.4. Then tan²c = sin²c / cos²c = (1 - cos²a cos²b)/(cos²a cos²b) = [1/(cos²a cos²b)] - 1.5. Recognize 1/(cos²a cos²b) as (1/cos²a)(1/cos²b) = (1 + tan²a)(1 + tan²b).6. Expand that product and subtract 1 to get tan²c = tan²a + tan²b + tan²a tan²b.Yes, each step follows logically. I don't see any gaps or mistakes here. So this should be a valid proof for the given identity.

✅:To prove the Pythagorean theorem for a right spherical triangle with legs ( a, b ) and hypotenuse ( c ), expressed as:[tan^{2} c = tan^{2} a + tan^{2} b + tan^{2} a tan^{2} b,]we start with the fundamental spherical Pythagorean theorem:[cos c = cos a cos b.]Step 1: Square both sides to eliminate the cosine:[cos^{2} c = cos^{2} a cos^{2} b.]Step 2: Use the Pythagorean identity (sin^{2} c = 1 - cos^{2} c) to express (tan^{2} c):[tan^{2} c = frac{sin^{2} c}{cos^{2} c} = frac{1 - cos^{2} a cos^{2} b}{cos^{2} a cos^{2} b}.]Step 3: Split the fraction into two terms:[tan^{2} c = frac{1}{cos^{2} a cos^{2} b} - 1.]Step 4: Recognize that (frac{1}{cos^{2} theta} = 1 + tan^{2} theta) for any angle (theta). Substitute this identity for (a) and (b):[frac{1}{cos^{2} a cos^{2} b} = (1 + tan^{2} a)(1 + tan^{2} b).]Step 5: Expand the product and simplify:[(1 + tan^{2} a)(1 + tan^{2} b) = 1 + tan^{2} a + tan^{2} b + tan^{2} a tan^{2} b.]Step 6: Substitute back into the expression for (tan^{2} c) and subtract 1:[tan^{2} c = left(1 + tan^{2} a + tan^{2} b + tan^{2} a tan^{2} bright) - 1 = tan^{2} a + tan^{2} b + tan^{2} a tan^{2} b.]Thus, the Pythagorean theorem for a right spherical triangle in terms of tangents is proven:[boxed{tan^{2} c = tan^{2} a + tan^{2} b + tan^{2} a tan^{2} b}.]

🔑:1. Start with the spherical cosine rule for sides, which states that for any spherical triangle with sides a, b, and c: [ cos c = cos a cos b + sin a sin b cos C ] For a right spherical triangle where C = frac{pi}{2}, we have: [ cos left( frac{pi}{2} right) = 0 quad Rightarrow quad cos c = cos a cos b ]2. Square both sides of the equation: [ (cos c)^2 = (cos a cos b)^2 ] which simplifies to: [ cos^2 c = cos^2 a cos^2 b ]3. Recognize the trigonometric identity related to the tangent function: [ cos^2 x = frac{1}{1 + tan^2 x} ] Apply this identity to cos^2 a, cos^2 b, and cos^2 c: [ cos^2 a = frac{1}{1 + tan^2 a}, quad cos^2 b = frac{1}{1 + tan^2 b}, quad cos^2 c = frac{1}{1 + tan^2 c} ]4. Substitute the expressions for cos^2 a, cos^2 b, and cos^2 c into the equation cos^2 c = cos^2 a cos^2 b: [ frac{1}{1 + tan^2 c} = left( frac{1}{1 + tan^2 a} right) left( frac{1}{1 + tan^2 b} right) ]5. Simplify the right-hand side: [ frac{1}{1 + tan^2 c} = frac{1}{(1 + tan^2 a)(1 + tan^2 b)} ]6. To separate the tangents, multiply both sides of the equation by (1 + tan^2 c)(1 + tan^2 a)(1 + tan^2 b) to get rid of the fractions: [ 1 = (1 + tan^2 c) ]7. Expand and simplify the right-hand side: [ 1 + tan^2 c = 1 + tan^2 a + tan^2 b + tan^2 a tan^2 b ]8. Rearrange the equation to isolate tan^2 c: [ tan^2 c = tan^2 a + tan^2 b + tan^2 a tan^2 b ] Conclusion:[boxed{operatorname{tg}^{2} c= operatorname{tg}^{2} a+ operatorname{tg}^{2} b+ operatorname{tg}^{2} a operatorname{tg}^{2} b}]