💡Okay, so I have this grid that's 6 units wide and 5 units tall. I need to go from point A, which is at the bottom left, to point B, which is at the top right. Each block on the grid is 1 unit by 1 unit. The goal is to find out how many different paths I can take from A to B that are exactly 11 units long, without passing through the block at (3,2). First, I remember that in grid problems like this, the number of paths from one corner to another can be found using combinations. Since the grid is 6 units wide and 5 units tall, moving from A to B requires moving 6 units to the right and 5 units up. That means in total, I need to make 11 moves: 6 right (let's call them 'R') and 5 up (let's call them 'U').The number of different paths without any restrictions is the number of ways to arrange these moves. This is a combination problem where I choose 5 positions out of 11 for the 'U' moves (or equivalently, 6 positions for the 'R' moves). The formula for combinations is:[binom{n}{k} = frac{n!}{k!(n - k)!}]So, plugging in the numbers:[binom{11}{5} = frac{11!}{5!6!} = 462]That means there are 462 different paths from A to B without any restrictions.But now, I need to avoid passing through the block at (3,2). Hmm, okay. So, I need to subtract the number of paths that go through (3,2) from the total number of paths. To do this, I can break the problem into two parts: the number of paths from A to (3,2), and then from (3,2) to B. If I multiply these two numbers, I'll get the total number of paths that pass through (3,2). Then, subtracting this from the total number of paths will give me the number of valid paths that avoid (3,2).First, let's find the number of paths from A (0,0) to (3,2). To get from (0,0) to (3,2), I need to move 3 units to the right and 2 units up. So, the number of moves is 5 in total, with 3 'R's and 2 'U's. Using combinations again:[binom{5}{3} = frac{5!}{3!2!} = 10]So, there are 10 paths from A to (3,2).Next, I need to find the number of paths from (3,2) to B (6,5). From (3,2), I need to move 3 units to the right (since 6 - 3 = 3) and 3 units up (since 5 - 2 = 3). Wait, that's 3 'R's and 3 'U's, which is a total of 6 moves. So, the number of paths is:[binom{6}{3} = frac{6!}{3!3!} = 20]Wait, that doesn't seem right. Earlier, I thought it was 5 moves, but actually, from (3,2) to (6,5), it's 3 right and 3 up, which is 6 moves. So, the number of paths is 20, not 10 as I initially thought. Hmm, I need to correct that.So, the number of paths from (3,2) to B is 20. Therefore, the total number of paths passing through (3,2) is:[10 times 20 = 200]Wait, but earlier I thought it was 100. That was a mistake. I incorrectly calculated the number of paths from (3,2) to B as 10 instead of 20. So, the correct number of paths passing through (3,2) is 200.Therefore, the number of valid paths avoiding (3,2) is:[462 - 200 = 262]But wait, let me double-check my calculations. From (3,2) to (6,5), it's 3 right and 3 up, which is 6 moves. The number of combinations is indeed 20. So, 10 paths from A to (3,2), 20 paths from (3,2) to B, so 10*20=200 paths passing through (3,2). Subtracting that from the total 462 gives 262.But hold on, I think I made a mistake in the initial step. Let me recount. From (3,2) to (6,5), it's 3 right and 3 up, which is 6 moves. So, the number of paths is:[binom{6}{3} = 20]Yes, that's correct. So, 10*20=200. Therefore, 462-200=262.Wait, but in the initial problem statement, the grid is 6 units wide and 5 units tall. So, the coordinates go from (0,0) to (6,5). Therefore, the block at (3,2) is indeed a valid point on the grid.But let me visualize the grid. Starting at (0,0), moving to (6,5). The point (3,2) is somewhere in the middle. So, the number of paths passing through (3,2) is 10*20=200, as calculated.Therefore, the number of valid paths is 462-200=262.Wait, but earlier I thought it was 362. That was because I incorrectly calculated the number of paths from (3,2) to B as 10 instead of 20. So, the correct answer should be 262.But let me confirm once more. Total paths: 462. Paths through (3,2): 10*20=200. So, 462-200=262.Yes, that seems correct.But wait, another way to think about it is to use the principle of inclusion-exclusion. The total number of paths is 462. The number of paths passing through (3,2) is 10*20=200. So, subtracting 200 from 462 gives 262.Alternatively, I can think of it as the number of paths from A to B avoiding (3,2) is equal to the total number of paths minus the number of paths passing through (3,2).Yes, that makes sense.Therefore, the final answer should be 262.But wait, let me check if there's another way to calculate it. Maybe using dynamic programming or something else.Alternatively, I can think of the grid as a graph where each intersection is a node, and edges connect adjacent nodes. Then, the number of paths from A to B avoiding (3,2) can be calculated by subtracting the number of paths passing through (3,2) from the total number of paths.But that's essentially what I did earlier.Alternatively, I can calculate the number of paths from A to B without passing through (3,2) by considering all possible paths and subtracting those that go through (3,2).Yes, that's the same approach.Therefore, I think 262 is the correct answer.Wait, but in the initial problem statement, the user wrote:"Suppose we wish to walk from A to B via an 11-unit path, staying on the grid lines and avoiding passing through the block at (3,2). How many different paths can we take?"So, the path length is exactly 11 units. Since moving from (0,0) to (6,5) requires 6 right and 5 up moves, which is 11 moves, so that's consistent.Therefore, the total number of paths is 462, as calculated.Number of paths passing through (3,2): 10*20=200.Therefore, the number of valid paths is 462-200=262.Yes, that seems correct.But wait, let me check the calculation for the number of paths from (3,2) to (6,5). From (3,2), we need to move 3 right and 3 up. So, the number of paths is:[binom{6}{3} = 20]Yes, that's correct.Therefore, the number of paths passing through (3,2) is 10*20=200.Subtracting that from 462 gives 262.So, the final answer is 262.But wait, in the initial problem statement, the user wrote:"Suppose we wish to walk from A to B via an 11-unit path, staying on the grid lines and avoiding passing through the block at (3,2). How many different paths can we take?"So, the answer is 262.But in the initial thought process, I thought it was 362, but that was due to a miscalculation. After correcting that, it's 262.Therefore, the correct answer is 262.

💡First, I need to determine the combined speed of the two cyclists. Since they are moving in opposite directions, their speeds add up.Next, I'll calculate the total distance they cover together in 5 hours. Given that they are 50 km apart after this time, I can set up an equation to find the combined speed.Finally, by dividing the total distance by the time, I can find the combined speed and then determine the speed of each individual cyclist.

💡Alright, so I have this problem where I'm given a function ( f(x) = -frac{1}{2}x^2 + frac{13}{2} ). I need to find the interval ([a, b]) such that the minimum value of (f(x)) on this interval is (2a) and the maximum value is (2b). Hmm, okay, let's break this down step by step.First, I recall that quadratic functions have either a maximum or a minimum value, depending on the coefficient of (x^2). In this case, the coefficient is (-frac{1}{2}), which is negative, so the parabola opens downward. That means the function has a maximum point at its vertex and no minimum unless we restrict the domain to a specific interval.The vertex form of a quadratic function is ( f(x) = a(x - h)^2 + k ), where ((h, k)) is the vertex. Comparing this to my function, ( f(x) = -frac{1}{2}x^2 + frac{13}{2} ), I can see that the vertex is at ((0, frac{13}{2})). So, the maximum value of the function is (frac{13}{2}) at (x = 0).Now, since the function opens downward, as we move away from the vertex in either direction, the value of the function decreases. Therefore, on any interval ([a, b]), the maximum value will be at the vertex if the interval includes (x = 0), and the minimum value will be at one of the endpoints, depending on which side we're moving away from.But in this problem, the maximum value is given as (2b) and the minimum as (2a). So, I need to find (a) and (b) such that when I evaluate (f(x)) on ([a, b]), the maximum is (2b) and the minimum is (2a).Let me consider different cases based on where (a) and (b) are located relative to the vertex at (x = 0).**Case 1: Both (a) and (b) are on the right side of the vertex ((a geq 0))**In this case, since the function is decreasing for (x > 0), the maximum value on ([a, b]) would be at (x = a), and the minimum would be at (x = b).So, setting up the equations:- Maximum value: (f(a) = 2b)- Minimum value: (f(b) = 2a)Plugging into the function:1. (-frac{1}{2}a^2 + frac{13}{2} = 2b)2. (-frac{1}{2}b^2 + frac{13}{2} = 2a)Let me simplify these equations:From equation 1:(-frac{1}{2}a^2 + frac{13}{2} = 2b)Multiply both sides by 2:(-a^2 + 13 = 4b)So, (b = frac{13 - a^2}{4})From equation 2:(-frac{1}{2}b^2 + frac{13}{2} = 2a)Multiply both sides by 2:(-b^2 + 13 = 4a)So, (a = frac{13 - b^2}{4})Now, substitute (b) from equation 1 into equation 2:(a = frac{13 - left(frac{13 - a^2}{4}right)^2}{4})This looks complicated, but let's try to solve it step by step.First, compute (left(frac{13 - a^2}{4}right)^2):(left(frac{13 - a^2}{4}right)^2 = frac{(13 - a^2)^2}{16})So, substituting back:(a = frac{13 - frac{(13 - a^2)^2}{16}}{4})Multiply numerator and denominator by 16 to eliminate the fraction:(a = frac{16 times 13 - (13 - a^2)^2}{64})Simplify:(a = frac{208 - (169 - 26a^2 + a^4)}{64})(a = frac{208 - 169 + 26a^2 - a^4}{64})(a = frac{39 + 26a^2 - a^4}{64})Multiply both sides by 64:(64a = 39 + 26a^2 - a^4)Bring all terms to one side:(a^4 - 26a^2 + 64a - 39 = 0)This is a quartic equation, which might be difficult to solve directly. Maybe I can factor it or find rational roots.Let me try possible rational roots using the Rational Root Theorem. The possible roots are factors of 39 over factors of 1, so ±1, ±3, ±13, ±39.Testing (a = 1):(1 - 26 + 64 - 39 = 0)(1 - 26 = -25), (-25 + 64 = 39), (39 - 39 = 0)So, (a = 1) is a root.Therefore, we can factor out ((a - 1)):Using polynomial division or synthetic division:Divide (a^4 - 26a^2 + 64a - 39) by ((a - 1)).Using synthetic division:1 | 1 0 -26 64 -39 1 1 -25 39 ----------------- 1 1 -25 39 0So, the quartic factors as ((a - 1)(a^3 + a^2 - 25a + 39)).Now, let's factor the cubic (a^3 + a^2 - 25a + 39).Again, trying possible rational roots: ±1, ±3, ±13, ±39.Testing (a = 3):(27 + 9 - 75 + 39 = 0)(27 + 9 = 36), (36 - 75 = -39), (-39 + 39 = 0)So, (a = 3) is a root.Factor out ((a - 3)):Using synthetic division:3 | 1 1 -25 39 3 12 -39 ------------- 1 4 -13 0So, the cubic factors as ((a - 3)(a^2 + 4a - 13)).Now, the quartic is factored as:((a - 1)(a - 3)(a^2 + 4a - 13))Set each factor to zero:1. (a - 1 = 0) → (a = 1)2. (a - 3 = 0) → (a = 3)3. (a^2 + 4a - 13 = 0) Using quadratic formula: (a = frac{-4 pm sqrt{16 + 52}}{2} = frac{-4 pm sqrt{68}}{2} = frac{-4 pm 2sqrt{17}}{2} = -2 pm sqrt{17})So, the roots are (a = 1), (a = 3), (a = -2 + sqrt{17}), and (a = -2 - sqrt{17}).Now, since we're in Case 1 where (a geq 0), we can discard the negative roots (a = -2 - sqrt{17}) and (a = -2 + sqrt{17}) (since (sqrt{17} approx 4.123), so (-2 + 4.123 approx 2.123), which is positive, but let's check).Wait, actually, (a = -2 + sqrt{17}) is approximately (2.123), which is positive, so it's a valid root in this case.But let's see if these roots satisfy the original equations.First, (a = 1):From equation 1: (b = frac{13 - 1^2}{4} = frac{12}{4} = 3)So, interval ([1, 3]).Check if this satisfies the original conditions:- (f(1) = -frac{1}{2}(1)^2 + frac{13}{2} = -frac{1}{2} + frac{13}{2} = 6)- (f(3) = -frac{1}{2}(9) + frac{13}{2} = -frac{9}{2} + frac{13}{2} = 2)So, maximum value is 6, which should be (2b = 6) → (b = 3), and minimum value is 2, which should be (2a = 2) → (a = 1). Perfect.Next, (a = 3):From equation 1: (b = frac{13 - 9}{4} = frac{4}{4} = 1)But (b = 1) is less than (a = 3), which contradicts our assumption that (a leq b). So, this is not a valid solution.Next, (a = -2 + sqrt{17}):Compute (b = frac{13 - a^2}{4})First, compute (a^2):(a = -2 + sqrt{17})(a^2 = (-2 + sqrt{17})^2 = 4 - 4sqrt{17} + 17 = 21 - 4sqrt{17})So, (b = frac{13 - (21 - 4sqrt{17})}{4} = frac{13 - 21 + 4sqrt{17}}{4} = frac{-8 + 4sqrt{17}}{4} = -2 + sqrt{17})So, (a = -2 + sqrt{17}) and (b = -2 + sqrt{17}), which implies (a = b). But we need (a < b), so this is not a valid interval.Therefore, in Case 1, the only valid solution is ([1, 3]).**Case 2: Both (a) and (b) are on the left side of the vertex ((b leq 0))**In this case, since the function is increasing for (x < 0), the maximum value on ([a, b]) would be at (x = b), and the minimum would be at (x = a).Setting up the equations:- Maximum value: (f(b) = 2b)- Minimum value: (f(a) = 2a)Plugging into the function:1. (-frac{1}{2}b^2 + frac{13}{2} = 2b)2. (-frac{1}{2}a^2 + frac{13}{2} = 2a)These are similar to the previous equations, just swapping (a) and (b).From equation 1:(-frac{1}{2}b^2 + frac{13}{2} = 2b)Multiply by 2:(-b^2 + 13 = 4b)(b^2 + 4b - 13 = 0)Using quadratic formula:(b = frac{-4 pm sqrt{16 + 52}}{2} = frac{-4 pm sqrt{68}}{2} = frac{-4 pm 2sqrt{17}}{2} = -2 pm sqrt{17})Since (b leq 0), we take (b = -2 - sqrt{17}) (since (-2 + sqrt{17} approx 2.123 > 0)).Now, from equation 2:(-frac{1}{2}a^2 + frac{13}{2} = 2a)Multiply by 2:(-a^2 + 13 = 4a)(a^2 + 4a - 13 = 0)Using quadratic formula:(a = frac{-4 pm sqrt{16 + 52}}{2} = frac{-4 pm sqrt{68}}{2} = frac{-4 pm 2sqrt{17}}{2} = -2 pm sqrt{17})Again, since (a leq b leq 0), we take (a = -2 - sqrt{17}).So, the interval would be ([a, b] = [-2 - sqrt{17}, -2 - sqrt{17}]), but this implies (a = b), which is not valid as we need (a < b). Therefore, no solution exists in this case.**Case 3: (a < 0 < b) (Interval crosses the vertex)**In this case, the function increases from (a) to (0) and then decreases from (0) to (b). Therefore, the maximum value occurs at (x = 0), and the minimum value occurs at one of the endpoints, depending on which side is further from the vertex.Given that the maximum value is (2b) and the minimum is (2a), let's analyze:- Maximum value at (x = 0): (f(0) = frac{13}{2}), so (frac{13}{2} = 2b) → (b = frac{13}{4}).- Minimum value: We need to check both endpoints (x = a) and (x = b).Compute (f(a)) and (f(b)):- (f(a) = -frac{1}{2}a^2 + frac{13}{2})- (f(b) = -frac{1}{2}(frac{13}{4})^2 + frac{13}{2} = -frac{169}{32} + frac{208}{32} = frac{39}{32})Since (f(a)) is the minimum, we set (f(a) = 2a):(-frac{1}{2}a^2 + frac{13}{2} = 2a)Multiply by 2:(-a^2 + 13 = 4a)(a^2 + 4a - 13 = 0)Using quadratic formula:(a = frac{-4 pm sqrt{16 + 52}}{2} = frac{-4 pm sqrt{68}}{2} = frac{-4 pm 2sqrt{17}}{2} = -2 pm sqrt{17})Since (a < 0), we take (a = -2 - sqrt{17}).So, the interval is ([a, b] = [-2 - sqrt{17}, frac{13}{4}]).Let's verify:- (f(a) = -frac{1}{2}(-2 - sqrt{17})^2 + frac{13}{2})First, compute ((-2 - sqrt{17})^2 = 4 + 4sqrt{17} + 17 = 21 + 4sqrt{17})So, (f(a) = -frac{1}{2}(21 + 4sqrt{17}) + frac{13}{2} = -frac{21}{2} - 2sqrt{17} + frac{13}{2} = -4 - 2sqrt{17})But (2a = 2(-2 - sqrt{17}) = -4 - 2sqrt{17}), which matches.- (f(b) = frac{39}{32}), and (2b = 2(frac{13}{4}) = frac{13}{2}). Wait, this doesn't match because (frac{39}{32} neq frac{13}{2}). Hmm, that's a problem.Wait, I think I made a mistake here. The maximum value is at (x = 0), which is (frac{13}{2}), so (2b = frac{13}{2}) → (b = frac{13}{4}). That's correct.But the minimum value is at (x = a), which is (2a = f(a)). We computed (f(a) = -4 - 2sqrt{17}), which is indeed (2a) since (a = -2 - sqrt{17}). So, that part is correct.However, the value at (x = b) is (frac{39}{32}), which is greater than (2a) but less than (2b). Since (2b = frac{13}{2} = 6.5) and (frac{39}{32} approx 1.21875), which is greater than (2a approx -8.246). So, the minimum is indeed at (x = a), and the maximum is at (x = 0), which is correctly assigned to (2b).Therefore, this interval ([-2 - sqrt{17}, frac{13}{4}]) is valid.**Conclusion:**After analyzing all possible cases, we have two valid intervals:1. ([1, 3])2. ([-2 - sqrt{17}, frac{13}{4}])These are the intervals where the minimum value of (f(x)) is (2a) and the maximum value is (2b).

💡First, I need to determine the total area of the rectangular courtyard that needs to be paved. The courtyard is 30 meters long and 16.5 meters wide.Next, I'll calculate the area of the courtyard by multiplying its length by its width:30 m × 16.5 m = 495 m²Now, I know that 99 paving stones are required to cover this area. To find the area that each paving stone covers, I'll divide the total area of the courtyard by the number of paving stones:495 m² ÷ 99 = 5 m²Each paving stone has a length of 2.5 meters. To find the width of each paving stone, I'll use the area of each paving stone:Area = Length × Width5 m² = 2.5 m × WidthFinally, I'll solve for the width by dividing the area of each paving stone by its length:Width = 5 m² ÷ 2.5 m = 2 mTherefore, the width of each paving stone is 2 meters.

💡Alright, so I've got this problem about regular polyhedrons, specifically about drawing tangents to the midsphere through the midpoints of each edge. The goal is to prove two things: part (a) that the lines corresponding to the sides of one face intersect at one point, and part (b) that the lines corresponding to the edges from one vertex lie in one plane.First, I need to recall what a regular polyhedron is. It's a three-dimensional shape with all faces being congruent regular polygons, and the same number of faces meeting at each vertex. Examples include the tetrahedron, cube, octahedron, dodecahedron, and icosahedron.Now, the midsphere of a polyhedron is a sphere that is tangent to all of its edges. The center of the midsphere is the same as the centroid of the polyhedron, which is the point equidistant from all its vertices, edges, and faces. So, if we take the midpoint of each edge, the midsphere will pass through all these midpoints.The problem says to draw a tangent to the midsphere at each midpoint, and this tangent should be perpendicular to the edge. So, for each edge, we find its midpoint, and then at that point, we draw a line that's tangent to the midsphere and perpendicular to the edge.Let me visualize this. Imagine a cube. Each edge has a midpoint, and at each midpoint, we draw a line tangent to the midsphere, which is perpendicular to that edge. Since the cube is regular, all these lines should have some symmetrical properties.For part (a), we need to show that the lines corresponding to the sides of one face intersect at a single point. So, take one face of the cube, which is a square. Each side of this square has a midpoint, and at each midpoint, we've drawn a tangent line. These four tangent lines should all intersect at one point.Hmm, how can that be? Let me think about the geometry here. Each tangent line is perpendicular to the edge at its midpoint. So, in the case of a cube, each tangent line would be in a plane that's perpendicular to the edge and passes through its midpoint.Wait, but all these tangent lines are in different planes. How do they intersect at a single point? Maybe they all lie on a common sphere or something? Or perhaps they all pass through the center of the midsphere?Wait, no, the midsphere is tangent to the edges at their midpoints, so the center of the midsphere is equidistant from all edges. If I draw a tangent at each midpoint, the tangent line is perpendicular to the radius at that point. So, the tangent line is perpendicular to the line from the center of the midsphere to the midpoint of the edge.But the edge itself is also related to the center. Maybe there's some orthogonality here that causes the lines to converge.Let me try to think in terms of coordinates. Let's place the cube in a coordinate system with its center at the origin. Suppose the cube has edges of length 2, so the vertices are at (±1, ±1, ±1). The midpoints of the edges would then be at points like (1,1,0), (1,0,1), etc.The midsphere would have a radius equal to the distance from the center to the midpoint of an edge. For the cube, that distance is sqrt(1^2 + 1^2 + 0^2) = sqrt(2). Wait, no, the midpoint is at (1,1,0), so the distance from the origin is sqrt(1^2 + 1^2 + 0^2) = sqrt(2). So the radius of the midsphere is sqrt(2).Now, the tangent at the midpoint (1,1,0) would be perpendicular to the radius vector (1,1,0). So, the tangent line lies in a plane perpendicular to (1,1,0) and passes through (1,1,0). The equation of this plane would be x + y = 2, since the normal vector is (1,1,0).Similarly, for another midpoint, say (1,0,1), the tangent plane would be x + z = 2.Wait, but in 3D, these planes would intersect along a line, not a point. Hmm, maybe I'm confusing something here.Wait, no, the tangent line is not the entire plane, but just the line that's tangent to the sphere at that point. So, the tangent line lies on the tangent plane, but it's a one-dimensional line.So, for each edge midpoint, we have a tangent line lying on a tangent plane. These tangent lines are all different, but perhaps they intersect at a common point.Wait, in the cube case, maybe all these tangent lines intersect at the center of the cube? Let me check.Take the tangent line at (1,1,0). The line is perpendicular to (1,1,0), so its direction vector is (1,1,0). But wait, the tangent line is not along the direction of the radius, but perpendicular to it. So, the direction vector of the tangent line should be perpendicular to (1,1,0). So, possible direction vectors could be (1,-1,0) or (0,0,1), but actually, in 3D, there are infinitely many directions perpendicular to (1,1,0).Wait, maybe I need to parametrize the tangent line. The tangent line at point P on the sphere can be parametrized as P + t*v, where v is a direction vector perpendicular to OP, where O is the center.So, for point (1,1,0), the direction vector v must satisfy (1,1,0)·v = 0. So, v can be (1,-1,0) or (0,0,1), or any linear combination.But how do we determine the specific tangent line? The problem says it's the tangent line perpendicular to the edge. So, the edge has a certain direction, and the tangent line is perpendicular to that edge.Wait, in the cube, each edge is along the x, y, or z-axis. So, for example, the edge from (1,1,1) to (1,1,-1) is along the z-axis. Its midpoint is (1,1,0). The tangent line at this midpoint is supposed to be perpendicular to the edge, which is along the z-axis. So, the tangent line should lie in the plane perpendicular to the z-axis, which is the xy-plane.But the tangent line is also tangent to the midsphere at (1,1,0). So, in the xy-plane, the tangent line at (1,1,0) would be... Wait, in the xy-plane, the midsphere is a circle with radius sqrt(2), centered at the origin. The tangent at (1,1,0) would be the line x + y = 2, but that's a plane in 3D, but in the xy-plane, it's a line.Wait, no, in 3D, the tangent line is a line that touches the sphere only at (1,1,0) and is perpendicular to the radius. So, the direction vector of the tangent line must be perpendicular to the radius vector (1,1,0). So, direction vector v satisfies (1,1,0)·v = 0.Also, the tangent line is supposed to be perpendicular to the edge. The edge is along the z-axis, so its direction vector is (0,0,1). Therefore, the tangent line must be perpendicular to (0,0,1). So, the direction vector of the tangent line must satisfy (0,0,1)·v = 0, meaning v_z = 0.So, combining both conditions, the direction vector v must satisfy (1,1,0)·v = 0 and v_z = 0. So, v_x + v_y = 0 and v_z = 0. Therefore, v can be (1,-1,0) or any scalar multiple.So, the tangent line at (1,1,0) has direction vector (1,-1,0) and passes through (1,1,0). So, parametric equations are x = 1 + t, y = 1 - t, z = 0.Similarly, for another edge, say from (1,1,1) to (-1,1,1), which is along the x-axis. Its midpoint is (0,1,1). The tangent line here must be perpendicular to the edge, which is along the x-axis, so direction vector (1,0,0). Therefore, the tangent line must have direction vector perpendicular to (1,0,0), so v_x = 0. Also, it must be perpendicular to the radius vector (0,1,1). So, (0,1,1)·v = 0 => v_y + v_z = 0. Therefore, direction vector can be (0,1,-1).So, the tangent line at (0,1,1) is x = 0, y = 1 + t, z = 1 - t.Now, let's see if these two tangent lines intersect. The first line is x = 1 + t, y = 1 - t, z = 0. The second line is x = 0, y = 1 + s, z = 1 - s.To find an intersection, we need to solve:1 + t = 0 => t = -11 - t = 1 + s => 1 - (-1) = 1 + s => 2 = 1 + s => s = 1z = 0 vs z = 1 - s = 1 - 1 = 0. So, z matches.Therefore, the two lines intersect at (0, 2, 0). Wait, but (0,2,0) is not on the midsphere, which has radius sqrt(2). The distance from the origin is sqrt(0^2 + 2^2 + 0^2) = 2, which is greater than sqrt(2). So, it's outside the midsphere.Hmm, but the problem states that these lines are drawn through the midpoints, which are on the midsphere. So, the tangent lines are outside the midsphere, except at the midpoint.But in this case, the two tangent lines intersect at (0,2,0), which is outside the midsphere. So, does this mean that for the cube, the lines corresponding to the sides of one face intersect at a single point outside the midsphere?Wait, but in the cube, each face has four edges. So, we have four tangent lines, each at the midpoint of an edge of the face. Do all four lines intersect at the same point?Let's check another tangent line on the same face. Take the edge from (-1,1,0) to (-1,1,1). Its midpoint is (-1,1,0.5). The tangent line here must be perpendicular to the edge, which is along the z-axis, so direction vector (0,0,1). Therefore, the tangent line must have direction vector perpendicular to (0,0,1), so v_z = 0. Also, it must be perpendicular to the radius vector (-1,1,0.5). So, (-1,1,0.5)·v = 0 => -v_x + v_y + 0.5 v_z = 0. But since v_z = 0, we have -v_x + v_y = 0 => v_x = v_y.So, direction vector can be (1,1,0). Therefore, the tangent line is x = -1 + t, y = 1 + t, z = 0.5.Now, let's see if this line intersects the previous two lines at (0,2,0).From the first line: x = 1 + t, y = 1 - t, z = 0.From the second line: x = 0, y = 1 + s, z = 1 - s.From the third line: x = -1 + t, y = 1 + t, z = 0.5.Let's see if (0,2,0) is on the third line.Set x = 0: -1 + t = 0 => t = 1.Then y = 1 + 1 = 2, z = 0.5. But z is 0.5, not 0. So, (0,2,0) is not on the third line.Hmm, so the first two lines intersect at (0,2,0), but the third line doesn't pass through that point. So, maybe my initial assumption is wrong.Wait, maybe I made a mistake in calculating the direction vectors. Let me double-check.For the edge from (-1,1,0) to (-1,1,1), the midpoint is (-1,1,0.5). The edge is along the z-axis, so direction vector (0,0,1). The tangent line must be perpendicular to this edge, so its direction vector must satisfy (0,0,1)·v = 0 => v_z = 0.Also, the tangent line is tangent to the midsphere at (-1,1,0.5), so the direction vector must be perpendicular to the radius vector (-1,1,0.5). So, (-1,1,0.5)·v = 0 => -v_x + v_y + 0.5 v_z = 0. But since v_z = 0, we have -v_x + v_y = 0 => v_x = v_y.So, direction vector can be (1,1,0). Therefore, the tangent line is x = -1 + t, y = 1 + t, z = 0.5.Now, let's see if this line intersects the first line at (0,2,0).From the first line: x = 1 + t, y = 1 - t, z = 0.From the third line: x = -1 + s, y = 1 + s, z = 0.5.Set z equal: 0 ≠ 0.5, so they don't intersect.Wait, so maybe the four tangent lines on a face don't all intersect at a single point? But the problem says they should. So, perhaps my approach is wrong.Maybe I need to think about the reciprocal polyhedron or something like that. I remember that for regular polyhedrons, there's a dual polyhedron where vertices correspond to faces and vice versa. Maybe the lines we're drawing correspond to edges of the dual polyhedron, and their intersections correspond to vertices.In the case of the cube, the dual is the octahedron. So, maybe the lines we're drawing are edges of an octahedron, and their intersections are the vertices of the octahedron.But in the cube, the octahedron's vertices are at the centers of the cube's faces. So, for the cube's face, the center is at (0,0,0) for the face on the xy-plane, but wait, no, the center of the face is at (0,0,1) for the top face, etc.Wait, no, the dual of the cube is the octahedron, and its vertices are at the centers of the cube's faces. So, for the cube's face on the xy-plane at z=1, the center is (0,0,1). Similarly, the face on z=-1 has center (0,0,-1), and so on.So, if the lines we're drawing correspond to edges of the octahedron, then their intersections should be at the octahedron's vertices, which are at the centers of the cube's faces.But in my earlier calculation, the lines intersected at (0,2,0), which is not the center of the face. So, maybe I'm not drawing the lines correctly.Wait, perhaps I need to consider that the tangent lines are not just in the plane of the face, but also in some other orientation.Wait, the problem says to draw a tangent to the midsphere at the midpoint, perpendicular to the edge. So, the tangent line is in a plane that's perpendicular to the edge and passes through the midpoint.In the cube, for an edge along the z-axis, the tangent line lies in a plane perpendicular to the z-axis, which is the xy-plane. But the midsphere is a sphere, so the tangent line in the xy-plane at (1,1,0) is the line x + y = 2, but that's a plane in 3D. Wait, no, the tangent line is a line in that plane.Wait, maybe I'm confusing the tangent plane with the tangent line. The tangent line is a one-dimensional line that touches the sphere at one point and lies in the tangent plane.So, for the point (1,1,0), the tangent plane is x + y = 2, and the tangent line is a line in that plane that's also perpendicular to the edge.Wait, the edge is along the z-axis, so the tangent line must be perpendicular to the z-axis direction. So, in the tangent plane x + y = 2, the tangent line must have a direction vector with zero z-component.But also, the tangent line must be tangent to the sphere, so it must touch the sphere only at (1,1,0). So, in the plane x + y = 2, the sphere intersects as a circle, and the tangent line is tangent to that circle at (1,1,0).Wait, but in the plane x + y = 2, the sphere equation is x^2 + y^2 + z^2 = 2, but since x + y = 2, we can substitute y = 2 - x, so x^2 + (2 - x)^2 + z^2 = 2 => 2x^2 - 4x + 4 + z^2 = 2 => 2x^2 - 4x + z^2 + 2 = 0.This is a quadratic in x and z, which is a paraboloid, not a circle. Hmm, maybe my approach is wrong.Alternatively, perhaps I should use the fact that the tangent line is perpendicular to the radius. So, the direction vector of the tangent line is perpendicular to the radius vector (1,1,0). So, direction vector v satisfies (1,1,0)·v = 0.Also, the tangent line is perpendicular to the edge, which is along the z-axis, so direction vector v must satisfy (0,0,1)·v = 0 => v_z = 0.Therefore, v_x + v_y = 0 and v_z = 0. So, direction vector can be (1,-1,0).Thus, the tangent line is x = 1 + t, y = 1 - t, z = 0.Similarly, for another edge on the same face, say from (1,1,0) to (1,-1,0), which is along the y-axis. Its midpoint is (1,0,0). The tangent line here must be perpendicular to the y-axis, so direction vector (0,1,0). Also, it must be perpendicular to the radius vector (1,0,0). So, direction vector v must satisfy (1,0,0)·v = 0 => v_x = 0.Also, since it's perpendicular to the edge along y-axis, v must satisfy (0,1,0)·v = 0 => v_y = 0. Wait, that can't be, because if v_x = 0 and v_y = 0, then v_z can be anything, but the tangent line must lie in the plane perpendicular to the edge.Wait, maybe I'm getting confused. Let's think again.The edge is along the y-axis, from (1,1,0) to (1,-1,0). Its midpoint is (1,0,0). The tangent line must be perpendicular to this edge, which is along the y-axis, so the tangent line must lie in a plane perpendicular to the y-axis, which is the xz-plane.Also, the tangent line is tangent to the midsphere at (1,0,0). So, the direction vector of the tangent line must be perpendicular to the radius vector (1,0,0). So, (1,0,0)·v = 0 => v_x = 0.Therefore, direction vector v has v_x = 0. Also, since the tangent line lies in the xz-plane, v_y can be anything, but since it's in the xz-plane, v_y = 0? Wait, no, the xz-plane is defined by y=0, but the direction vector can have y-component.Wait, no, the direction vector lies in the xz-plane, so v_y can be non-zero, but the line itself is in the xz-plane. Wait, no, the line passes through (1,0,0) and lies in the xz-plane, so its direction vector must have v_y = 0.Wait, no, the line can have a direction vector with v_y ≠ 0, but since it's in the xz-plane, the line must satisfy y=0. So, the direction vector must have v_y = 0.Therefore, direction vector is (0,0,1) or (0,0,-1). But it must also be perpendicular to the radius vector (1,0,0). So, (1,0,0)·v = 0 => v_x = 0. So, direction vector can be (0,0,1).Thus, the tangent line is x = 1, y = 0, z = t.Now, let's see if this line intersects the previous tangent line at (0,2,0). The first line is x = 1 + t, y = 1 - t, z = 0. The second line is x = 1, y = 0, z = s.To find intersection, set x=1: 1 + t = 1 => t=0. Then y=1 - 0=1 vs y=0. Not equal. So, no intersection.Hmm, this is confusing. Maybe I'm not considering all the tangent lines correctly.Wait, perhaps I need to consider that for each edge, the tangent line is not just in the plane perpendicular to the edge, but also in some other orientation. Maybe I need to think about the reciprocal polyhedron more carefully.Alternatively, maybe I should use vector algebra to find the equations of these tangent lines and see if they intersect.Let me consider the cube again. Let's take one face, say the front face with vertices (1,1,1), (1,-1,1), (-1,-1,1), (-1,1,1). The midpoints of the edges are (1,0,1), (0,-1,1), (-1,0,1), (0,1,1).For each midpoint, we draw a tangent line to the midsphere, perpendicular to the edge.Let's take the midpoint (1,0,1). The edge here is from (1,1,1) to (1,-1,1), which is along the y-axis. So, the tangent line must be perpendicular to the y-axis, so its direction vector must satisfy (0,1,0)·v = 0 => v_y = 0.Also, the tangent line is tangent to the midsphere at (1,0,1), so the direction vector must be perpendicular to the radius vector (1,0,1). So, (1,0,1)·v = 0 => v_x + v_z = 0.Therefore, direction vector v satisfies v_y = 0 and v_x = -v_z. So, direction vector can be (1,0,-1).Thus, the tangent line is x = 1 + t, y = 0, z = 1 - t.Similarly, for the midpoint (0,-1,1), the edge is from (1,-1,1) to (-1,-1,1), along the x-axis. The tangent line must be perpendicular to the x-axis, so direction vector (1,0,0)·v = 0 => v_x = 0.Also, direction vector must be perpendicular to the radius vector (0,-1,1). So, (0,-1,1)·v = 0 => -v_y + v_z = 0 => v_z = v_y.Therefore, direction vector can be (0,1,1).Thus, the tangent line is x = 0, y = -1 + t, z = 1 + t.Now, let's see if these two lines intersect.First line: x = 1 + t, y = 0, z = 1 - t.Second line: x = 0, y = -1 + s, z = 1 + s.Set x equal: 1 + t = 0 => t = -1.Then, y = 0 vs y = -1 + s => 0 = -1 + s => s = 1.Check z: 1 - (-1) = 2 vs 1 + 1 = 2. So, z matches.Therefore, the two lines intersect at (0,0,2).Wait, that's interesting. So, the two tangent lines intersect at (0,0,2). Let's check another tangent line on the same face.Take the midpoint (-1,0,1). The edge is from (-1,1,1) to (-1,-1,1), along the y-axis. The tangent line must be perpendicular to the y-axis, so direction vector v_y = 0.Also, direction vector must be perpendicular to the radius vector (-1,0,1). So, (-1,0,1)·v = 0 => -v_x + v_z = 0 => v_z = v_x.Therefore, direction vector can be (1,0,1).Thus, the tangent line is x = -1 + t, y = 0, z = 1 + t.Now, let's see if this line intersects the previous two lines at (0,0,2).From the first line: x = 1 + t, y = 0, z = 1 - t.From the third line: x = -1 + s, y = 0, z = 1 + s.Set x equal: 1 + t = -1 + s => s = t + 2.Set z equal: 1 - t = 1 + s => 1 - t = 1 + t + 2 => -t = t + 2 => -2t = 2 => t = -1.Then, s = -1 + 2 = 1.Check x: 1 + (-1) = 0 vs -1 + 1 = 0. Good.Check y: 0 vs 0. Good.Check z: 1 - (-1) = 2 vs 1 + 1 = 2. Good.So, all three lines intersect at (0,0,2).Similarly, the fourth tangent line on the face would also intersect at (0,0,2). Let's confirm.Take the midpoint (0,1,1). The edge is from (-1,1,1) to (1,1,1), along the x-axis. The tangent line must be perpendicular to the x-axis, so direction vector v_x = 0.Also, direction vector must be perpendicular to the radius vector (0,1,1). So, (0,1,1)·v = 0 => v_y + v_z = 0.Therefore, direction vector can be (0,1,-1).Thus, the tangent line is x = 0, y = 1 + t, z = 1 - t.Now, check if this line passes through (0,0,2).From the line: x = 0, y = 1 + t, z = 1 - t.Set z = 2: 1 - t = 2 => t = -1.Then, y = 1 + (-1) = 0. So, yes, it passes through (0,0,2).Therefore, all four tangent lines corresponding to the sides of one face intersect at the point (0,0,2).So, in the cube, the lines corresponding to the sides of one face intersect at a single point outside the midsphere.Similarly, for other faces, the intersection points would be at (0,0,-2), (0,2,0), (0,-2,0), (2,0,0), (-2,0,0), etc., which are the vertices of an octahedron scaled by a factor of 2.Therefore, part (a) is proven for the cube.Now, for part (b), we need to show that the lines corresponding to the edges originating from one vertex lie in one plane.Take a vertex of the cube, say (1,1,1). It has three edges: along x, y, and z axes. The midpoints of these edges are (1,1,0), (1,0,1), and (0,1,1).We've already found the tangent lines at these midpoints:1. At (1,1,0): x = 1 + t, y = 1 - t, z = 0.2. At (1,0,1): x = 1 + t, y = 0, z = 1 - t.3. At (0,1,1): x = 0, y = 1 + t, z = 1 - t.We need to check if these three lines lie in the same plane.To find the plane, we can use the three points where the lines pass through.Wait, but each line is defined by a point and a direction. So, maybe we can find the plane equation by using three points.Alternatively, we can check if the three lines are coplanar.One way to check if three lines are coplanar is to see if the volume of the tetrahedron formed by any three points on the lines is zero.But since the lines are infinite, another approach is to find if there's a plane that contains all three lines.Let me try to find the equation of the plane.First, let's find three points on the three lines.From the first line: when t=0, we have (1,1,0).From the second line: when t=0, we have (1,0,1).From the third line: when t=0, we have (0,1,1).So, three points are A(1,1,0), B(1,0,1), and C(0,1,1).Let's find the equation of the plane passing through these three points.The general equation of a plane is ax + by + cz = d.Substitute A: a(1) + b(1) + c(0) = d => a + b = d.Substitute B: a(1) + b(0) + c(1) = d => a + c = d.Substitute C: a(0) + b(1) + c(1) = d => b + c = d.So, we have three equations:1. a + b = d2. a + c = d3. b + c = dSubtract equation 1 from equation 2: (a + c) - (a + b) = d - d => c - b = 0 => c = b.From equation 3: b + c = d => b + b = d => 2b = d.From equation 1: a + b = d => a + b = 2b => a = b.So, a = b = c, and d = 2b.Let's set b = 1, then a = 1, c = 1, d = 2.Thus, the plane equation is x + y + z = 2.Now, let's check if all three lines lie in this plane.First line: x = 1 + t, y = 1 - t, z = 0.Plug into plane equation: (1 + t) + (1 - t) + 0 = 2 => 2 = 2. True for all t.Second line: x = 1 + t, y = 0, z = 1 - t.Plug into plane equation: (1 + t) + 0 + (1 - t) = 2 => 2 = 2. True for all t.Third line: x = 0, y = 1 + t, z = 1 - t.Plug into plane equation: 0 + (1 + t) + (1 - t) = 2 => 2 = 2. True for all t.Therefore, all three lines lie in the plane x + y + z = 2.Thus, part (b) is also proven for the cube.This suggests that for any regular polyhedron, the lines corresponding to the edges from one vertex lie in a plane, and the lines corresponding to the sides of one face intersect at a point.I think the key idea here is that these tangent lines correspond to the edges of the dual polyhedron, and their intersections and coplanar properties are inherited from the dual's structure.So, in summary, for part (a), the lines intersect at a point which is the vertex of the dual polyhedron, and for part (b), the lines lie in a plane which is the face of the dual polyhedron.

💡Alright, so I've got this problem here: Let ( z ) be a complex number with ( |z + 1| > 2 ). I need to prove that ( |z^3 + 1| > 1 ). Hmm, okay. Let me try to break this down step by step.First, I know that ( |z + 1| > 2 ) means that the distance from the complex number ( z ) to the point ( -1 ) on the complex plane is greater than 2. So, ( z ) lies outside a circle centered at ( -1 ) with radius 2. That gives me a geometric interpretation, but I'm not sure yet how that helps with ( |z^3 + 1| ).Maybe I can express ( z ) in terms of another variable to simplify things. Let me set ( w = z + 1 ). Then, ( z = w - 1 ). The condition ( |w| > 2 ) holds, and I need to find ( |(w - 1)^3 + 1| ). Let's expand that expression:[(w - 1)^3 + 1 = w^3 - 3w^2 + 3w - 1 + 1 = w^3 - 3w^2 + 3w]So, ( |z^3 + 1| = |w^3 - 3w^2 + 3w| ). Hmm, that seems a bit complicated, but maybe I can factor it or find some relationship.Let me factor out a ( w ):[w^3 - 3w^2 + 3w = w(w^2 - 3w + 3)]So, ( |w(w^2 - 3w + 3)| = |w| cdot |w^2 - 3w + 3| ). Since ( |w| > 2 ), I know that ( |w| ) is at least something greater than 2. But I need to find a lower bound for the entire expression.Wait, maybe I can find the minimum value of ( |w^2 - 3w + 3| ) given that ( |w| > 2 ). If I can find a lower bound for ( |w^2 - 3w + 3| ), then multiplying it by ( |w| ) (which is greater than 2) might give me the desired result.Let me consider ( |w^2 - 3w + 3| ). I can write this as ( |w^2 - 3w + 3| ). Maybe I can use the triangle inequality here. The triangle inequality says that ( |a + b| geq ||a| - |b|| ). So, perhaps I can apply that to ( |w^2 - 3w + 3| ).Let me think: ( |w^2 - 3w + 3| geq ||w^2| - |3w - 3|| ). Hmm, not sure if that's helpful. Alternatively, maybe I can write it as ( |w^2 - 3w + 3| geq | |w^2| - |3w - 3| | ).But I'm not sure if that's the right approach. Maybe I can use the reverse triangle inequality or some other inequality.Alternatively, perhaps I can parametrize ( w ) as ( w = re^{itheta} ), since it's a complex number. Then, ( |w| = r > 2 ), and ( theta ) is the argument of ( w ).So, ( w = re^{itheta} ), then ( w^2 = r^2 e^{i2theta} ), and ( 3w = 3re^{itheta} ). So, ( w^2 - 3w + 3 = r^2 e^{i2theta} - 3r e^{itheta} + 3 ).Hmm, calculating the magnitude of this expression might be tricky. Maybe I can compute ( |w^2 - 3w + 3|^2 ) to make it easier, since ( |a + b|^2 = |a|^2 + |b|^2 + 2text{Re}(aoverline{b}) ).Let me try that:[|w^2 - 3w + 3|^2 = |r^2 e^{i2theta} - 3r e^{itheta} + 3|^2]Expanding this, it's equal to:[(r^2 e^{i2theta} - 3r e^{itheta} + 3)(r^2 e^{-i2theta} - 3r e^{-itheta} + 3)]Multiplying this out would be quite involved, but perhaps I can find a way to simplify it or find a lower bound.Alternatively, maybe I can consider the function ( f(w) = w^3 - 3w^2 + 3w ) and analyze its minimum modulus on the domain ( |w| > 2 ). Since ( f(w) = w(w^2 - 3w + 3) ), and ( |w| > 2 ), perhaps I can find the minimum of ( |f(w)| ) over this domain.Wait, maybe I can use the fact that ( |f(w)| geq | |w^3| - |3w^2| - |3w| | ) by the triangle inequality. Let's see:[|f(w)| = |w^3 - 3w^2 + 3w| geq ||w^3| - |3w^2| - |3w||]But that would give:[|f(w)| geq |r^3 - 3r^2 - 3r|]Hmm, but since ( r > 2 ), let's see what ( r^3 - 3r^2 - 3r ) is. For ( r = 2 ), it's ( 8 - 12 - 6 = -10 ). For ( r = 3 ), it's ( 27 - 27 - 9 = -9 ). Wait, but this is negative, so the absolute value would be positive, but I'm not sure if this helps because it's not clear how this relates to the actual modulus.Maybe another approach. Let's consider that ( |z^3 + 1| = |(z + 1)(z^2 - z + 1)| ). Wait, is that correct? Let me check:[(z + 1)(z^2 - z + 1) = z^3 - z^2 + z + z^2 - z + 1 = z^3 + 1]Yes, that's correct. So, ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| ). Since ( |z + 1| > 2 ), if I can show that ( |z^2 - z + 1| geq frac{1}{2} ), then multiplying by ( |z + 1| > 2 ) would give ( |z^3 + 1| > 1 ).So, let's try to find a lower bound for ( |z^2 - z + 1| ). Maybe I can use the reverse triangle inequality or some other inequality.Let me write ( z = x + yi ), where ( x ) and ( y ) are real numbers. Then, ( z^2 = (x + yi)^2 = x^2 - y^2 + 2xyi ), and ( -z = -x - yi ). So, ( z^2 - z + 1 = (x^2 - y^2 - x + 1) + (2xy - y)i ).The modulus squared is:[(x^2 - y^2 - x + 1)^2 + (2xy - y)^2]This looks complicated, but maybe I can find a way to simplify it or find a lower bound.Alternatively, perhaps I can use the fact that ( |z^2 - z + 1| geq | |z^2| - |z| - 1 | ) by the triangle inequality. Let's see:[|z^2 - z + 1| geq ||z^2| - |z| - 1|]But ( |z^2| = |z|^2 ), so:[|z^2 - z + 1| geq ||z|^2 - |z| - 1|]Hmm, but I don't know ( |z| ), only ( |z + 1| > 2 ). Maybe I can relate ( |z| ) to ( |z + 1| ).From the triangle inequality, ( |z| = |(z + 1) - 1| leq |z + 1| + 1 ). Since ( |z + 1| > 2 ), this gives ( |z| leq |z + 1| + 1 ), but that's an upper bound, not a lower bound. I need a lower bound for ( |z| ).Wait, maybe I can use the reverse triangle inequality: ( ||z + 1| - |1|| leq |z| ). So, ( |z| geq ||z + 1| - 1| ). Since ( |z + 1| > 2 ), this gives ( |z| geq 2 - 1 = 1 ). So, ( |z| geq 1 ).Okay, so ( |z| geq 1 ). Now, going back to ( |z^2 - z + 1| geq ||z|^2 - |z| - 1| ). Since ( |z| geq 1 ), let's denote ( r = |z| geq 1 ). Then, the expression becomes ( |r^2 - r - 1| ).We need to find the minimum of ( |r^2 - r - 1| ) for ( r geq 1 ). Let's analyze the function ( f(r) = r^2 - r - 1 ).The function ( f(r) ) is a quadratic in ( r ). Its derivative is ( f'(r) = 2r - 1 ), which is zero at ( r = 1/2 ). But since ( r geq 1 ), the minimum of ( f(r) ) occurs at ( r = 1 ):[f(1) = 1 - 1 - 1 = -1]As ( r ) increases beyond 1, ( f(r) ) increases because the derivative ( f'(r) = 2r - 1 ) is positive for ( r > 1/2 ). So, for ( r geq 1 ), ( f(r) ) is increasing, and the minimum value is ( -1 ) at ( r = 1 ).Therefore, ( |r^2 - r - 1| geq | -1 | = 1 ) for ( r geq 1 ). Wait, but that's not correct because ( f(r) ) is negative at ( r = 1 ) and becomes positive as ( r ) increases. So, actually, ( |r^2 - r - 1| ) is equal to ( |f(r)| ), which is 1 at ( r = 1 ) and increases as ( r ) moves away from 1.Wait, no, let me think again. At ( r = 1 ), ( f(r) = -1 ), so ( |f(r)| = 1 ). For ( r > 1 ), ( f(r) ) becomes positive and increases, so ( |f(r)| = f(r) ) which is greater than 1. Therefore, ( |r^2 - r - 1| geq 1 ) for ( r geq 1 ).Therefore, ( |z^2 - z + 1| geq 1 ). But wait, that's not possible because if ( |z| = 1 ), then ( |z^2 - z + 1| ) could be less than 1. Wait, maybe my approach is flawed.Wait, no, actually, when ( |z| = 1 ), ( z ) lies on the unit circle. Let's take ( z = 1 ). Then, ( |z^2 - z + 1| = |1 - 1 + 1| = 1 ). If ( z = -1 ), ( |z^2 - z + 1| = |1 + 1 + 1| = 3 ). If ( z = i ), ( |z^2 - z + 1| = |-1 - i + 1| = | -i | = 1 ). So, it seems that the minimum value is indeed 1 when ( |z| = 1 ).But in our case, ( |z| geq 1 ), so ( |z^2 - z + 1| geq 1 ). Therefore, ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| geq |z + 1| cdot 1 ). Since ( |z + 1| > 2 ), this would imply ( |z^3 + 1| > 2 ), which is stronger than what we need, ( |z^3 + 1| > 1 ).Wait, but the problem only asks to prove ( |z^3 + 1| > 1 ), so if my reasoning is correct, it's actually greater than 2. But maybe I made a mistake somewhere.Wait, let me double-check. I said that ( |z^2 - z + 1| geq 1 ) because ( |r^2 - r - 1| geq 1 ) for ( r geq 1 ). But actually, when ( r = 1 ), ( |r^2 - r - 1| = |-1| = 1 ), and for ( r > 1 ), ( r^2 - r - 1 ) is positive and increases, so ( |r^2 - r - 1| = r^2 - r - 1 ), which is greater than 1 for ( r > 1 ) because ( r^2 - r - 1 > 1 ) when ( r > phi ) (where ( phi ) is the golden ratio, approximately 1.618). Wait, actually, solving ( r^2 - r - 1 = 1 ) gives ( r^2 - r - 2 = 0 ), so ( r = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3] / 2 ). So, ( r = 2 ) or ( r = -1 ). Since ( r geq 1 ), the solution is ( r = 2 ). Therefore, for ( r geq 2 ), ( r^2 - r - 1 geq 1 ), so ( |z^2 - z + 1| geq 1 ).But in our case, ( |z + 1| > 2 ), which implies ( |z| geq | |z + 1| - 1 | = 2 - 1 = 1 ). So, ( |z| geq 1 ), but not necessarily ( |z| geq 2 ). Therefore, my previous conclusion that ( |z^2 - z + 1| geq 1 ) holds for ( |z| geq 1 ), but when ( |z| ) is between 1 and 2, ( |z^2 - z + 1| ) might be less than 1.Wait, let's test ( z = 1 ). Then, ( |z + 1| = 2 ), which is not greater than 2, so it's on the boundary. ( |z^3 + 1| = |1 + 1| = 2 ), which is greater than 1. If ( z ) approaches 1 from outside the circle ( |z + 1| > 2 ), say ( z = 1 + epsilon ) where ( epsilon ) is small, then ( |z + 1| = |2 + epsilon| > 2 ), and ( |z^3 + 1| ) approaches 2. So, in this case, ( |z^3 + 1| ) approaches 2, which is still greater than 1.Wait, but if ( |z| ) is between 1 and 2, say ( |z| = 1.5 ), then ( |z^2 - z + 1| ) might be less than 1. Let me check with ( z = 1.5 ). Then, ( z^2 = 2.25 ), ( z^2 - z + 1 = 2.25 - 1.5 + 1 = 1.75 ), so ( |z^2 - z + 1| = 1.75 ), which is greater than 1. Hmm, interesting.Wait, maybe my earlier conclusion was correct. Let me consider ( z ) such that ( |z| = r geq 1 ). Then, ( |z^2 - z + 1| geq | |z^2| - |z| - 1 | = |r^2 - r - 1| ). For ( r geq 1 ), ( r^2 - r - 1 ) is minimized at ( r = 1 ), giving ( -1 ), so ( |r^2 - r - 1| geq 1 ) for ( r geq 1 ). Therefore, ( |z^2 - z + 1| geq 1 ).Wait, but when ( r = 1 ), ( |z^2 - z + 1| = 1 ), but ( |z + 1| = |z - (-1)| ). If ( |z| = 1 ), then ( |z + 1| ) can be as small as 0 (if ( z = -1 )) or as large as 2 (if ( z = 1 )). But in our problem, ( |z + 1| > 2 ), so ( z ) cannot be on the unit circle because ( |z + 1| leq |z| + 1 leq 1 + 1 = 2 ) if ( |z| leq 1 ). Wait, that's not correct because ( |z + 1| ) can be greater than 2 even if ( |z| leq 1 ). For example, if ( z = 1 ), ( |z + 1| = 2 ), but if ( z ) is slightly more than 1, say ( z = 1 + epsilon ), then ( |z + 1| = 2 + epsilon > 2 ).Wait, no, if ( |z| leq 1 ), then ( |z + 1| leq |z| + 1 leq 2 ). So, to have ( |z + 1| > 2 ), ( |z| ) must be greater than 1. Because if ( |z| leq 1 ), then ( |z + 1| leq 2 ). Therefore, in our problem, ( |z| > 1 ).So, going back, since ( |z| > 1 ), and ( |z^2 - z + 1| geq 1 ), then ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| > 2 cdot 1 = 2 ). Therefore, ( |z^3 + 1| > 2 ), which is stronger than the required ( |z^3 + 1| > 1 ).But wait, the problem only asks to prove ( |z^3 + 1| > 1 ), so my conclusion is even stronger. However, I need to make sure that my reasoning is correct.Let me summarize:1. Given ( |z + 1| > 2 ), we set ( w = z + 1 ), so ( |w| > 2 ).2. Then, ( z = w - 1 ), and ( z^3 + 1 = (w - 1)^3 + 1 = w^3 - 3w^2 + 3w ).3. We factor this as ( w(w^2 - 3w + 3) ).4. We need to find ( |w(w^2 - 3w + 3)| = |w| cdot |w^2 - 3w + 3| ).5. Since ( |w| > 2 ), we need a lower bound for ( |w^2 - 3w + 3| ).6. Alternatively, we can express ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| ).7. Given ( |z + 1| > 2 ), we need to find a lower bound for ( |z^2 - z + 1| ).8. Using the reverse triangle inequality, we find that ( |z^2 - z + 1| geq 1 ) for ( |z| geq 1 ).9. Since ( |z + 1| > 2 ) implies ( |z| > 1 ), we have ( |z^2 - z + 1| geq 1 ).10. Therefore, ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| > 2 cdot 1 = 2 ), which is greater than 1.So, my conclusion is that ( |z^3 + 1| > 2 ), which certainly implies ( |z^3 + 1| > 1 ). Therefore, the original statement is proven.However, I'm a bit concerned because the problem only asks to show ( |z^3 + 1| > 1 ), and I've shown a stronger result. Maybe there's a simpler way to show just ( >1 ) without going through all this, but my approach seems valid.Alternatively, perhaps I can use the fact that ( |z^3 + 1| geq ||z|^3 - 1| ) by the reverse triangle inequality. Since ( |z + 1| > 2 ), and ( |z| geq | |z + 1| - 1 | = 1 ), but ( |z| ) could be just slightly above 1, making ( |z|^3 ) slightly above 1, so ( ||z|^3 - 1| ) could be close to 0. Therefore, this approach might not work.Wait, but earlier I found that ( |z^2 - z + 1| geq 1 ), so multiplying by ( |z + 1| > 2 ) gives a stronger result. So, perhaps my initial approach is the right way.Another thought: Maybe I can use the minimum modulus principle or consider the function ( f(z) = z^3 + 1 ) and analyze its behavior on the region ( |z + 1| > 2 ). But that might be more advanced than needed.Alternatively, perhaps I can consider specific cases or use geometric interpretations. For example, if ( z ) is real, then ( z + 1 > 2 ) implies ( z > 1 ). Then, ( z^3 + 1 > 1^3 + 1 = 2 ), so ( |z^3 + 1| > 2 ). Similarly, if ( z ) is purely imaginary, say ( z = yi ), then ( |z + 1| = sqrt{1 + y^2} > 2 ), so ( y^2 > 3 ). Then, ( z^3 + 1 = (yi)^3 + 1 = -y^3 i + 1 ), so ( |z^3 + 1| = sqrt{1 + y^6} ). Since ( y^2 > 3 ), ( y^6 > 27 ), so ( |z^3 + 1| > sqrt{1 + 27} = sqrt{28} > 5 ), which is again greater than 1.But these are just specific cases. I need a general proof.Wait, going back to the expression ( |z^3 + 1| = |z + 1| cdot |z^2 - z + 1| ). Since ( |z + 1| > 2 ), if I can show that ( |z^2 - z + 1| geq frac{1}{2} ), then multiplying by ( |z + 1| > 2 ) would give ( |z^3 + 1| > 1 ). But earlier, I found that ( |z^2 - z + 1| geq 1 ), which is stronger, so this would suffice.Alternatively, perhaps I can use the fact that ( |z^2 - z + 1| geq | |z|^2 - |z| - 1 | ), and since ( |z| > 1 ), ( |z|^2 - |z| - 1 ) is increasing for ( |z| > frac{1}{2} ), so the minimum occurs at ( |z| = 1 ), giving ( 1 - 1 - 1 = -1 ), so ( | |z|^2 - |z| - 1 | geq 1 ). Therefore, ( |z^2 - z + 1| geq 1 ).Thus, combining this with ( |z + 1| > 2 ), we get ( |z^3 + 1| > 2 cdot 1 = 2 ), which is greater than 1.Therefore, the proof is complete.