answer：First, I need to calculate the volume of the cylinder. The formula for the volume of a cylinder is ( V_{text{cylinder}} = pi r_{text{cyl}}^2 h_{text{cyl}} ). Given that the radius of the cylinder is 10 cm and the height is 30 cm, the volume is:[V_{text{cylinder}} = pi (10)^2 (30) = 3000pi text{ cubic cm}]Next, I'll calculate the volume of one cone. The formula for the volume of a cone is ( V_{text{cone}} = frac{1}{3} pi r_{text{cone}}^2 h_{text{cone}} ). With a radius of 10 cm and a height of 15 cm, the volume of one cone is:[V_{text{cone}} = frac{1}{3} pi (10)^2 (15) = 500pi text{ cubic cm}]Since there are two congruent cones, the total volume occupied by the cones is:[2V_{text{cone}} = 2 times 500pi = 1000pi text{ cubic cm}]Finally, to find the volume of the cylinder that is not occupied by the cones, I'll subtract the total volume of the cones from the volume of the cylinder:[V_{text{unoccupied}} = V_{text{cylinder}} - 2V_{text{cone}} = 3000pi - 1000pi = 2000pi text{ cubic cm}]

answer：First, I need to calculate the length of an arc in a circle with a radius of 2 units and a central angle of 300 degrees.I recall that the formula for arc length is:[text{Arc Length} = theta times r]where (theta) is the central angle in radians and (r) is the radius of the circle.Since the given angle is in degrees, I need to convert it to radians. To do this, I use the conversion factor:[180^circ = pi text{ radians}]So, 300 degrees in radians is:[300^circ times frac{pi}{180^circ} = frac{5pi}{3} text{ radians}]Now, I can plug the values into the arc length formula:[text{Arc Length} = frac{5pi}{3} times 2 = frac{10pi}{3}]Therefore, the length of the arc is (frac{10pi}{3}) units.

answer：Okay, so I have this problem about two skew lines, l and m. On line l, there are three points A, B, and C such that AB equals BC. From each of these points, we drop perpendiculars to line m, and the lengths of these perpendiculars are given: AD is sqrt(15), BE is 7/2, and CF is sqrt(10). I need to find the distance between the two skew lines l and m.First, I remember that the distance between two skew lines is the length of the common perpendicular between them. So, if I can find the length of this common perpendicular, that should be the answer.Since AB equals BC, points A, B, and C are equally spaced on line l. Let me denote the distance AB as 'd', so BC is also 'd'. So, the total distance from A to C is 2d.Now, from each of these points, we drop perpendiculars to line m. The lengths of these perpendiculars are given. I think these lengths vary because the points A, B, and C are at different positions along line l, which is skew to m.I recall that in such cases, the lengths of the perpendiculars from points on one skew line to another can be related to the distance between the lines and the angle between them. Maybe I can use some trigonometric relationships here.Let me denote the distance between the two lines as 'h'. Since the common perpendicular is the shortest distance, h is the minimum distance between l and m.Now, considering the perpendiculars AD, BE, and CF, these are not necessarily equal to h because they are not the common perpendiculars. Instead, they are the distances from points A, B, and C to line m, which can be expressed in terms of h and the angles between the lines.Wait, maybe I can model this with coordinates. Let me try to set up a coordinate system where line l is along the x-axis, and line m is somewhere in 3D space. Since l and m are skew, they don't intersect and aren't parallel.Let me assign coordinates to points A, B, and C on line l. Let's say point B is at the origin (0, 0, 0). Then, since AB = BC, point A can be at (-d, 0, 0) and point C at (d, 0, 0). So, AB = BC = d.Now, line m is another skew line. Let me denote a point on line m as D, E, or F, which are the feet of the perpendiculars from A, B, and C respectively. So, AD, BE, and CF are perpendiculars from A, B, and C to line m.Since AD, BE, and CF are perpendiculars, their lengths are the distances from points A, B, and C to line m. So, AD = sqrt(15), BE = 7/2, and CF = sqrt(10).I think the key here is to realize that these distances can be related to the distance between the lines and the direction of the lines. Maybe if I can express the coordinates of D, E, and F in terms of the coordinates of A, B, and C, I can find the common perpendicular.Alternatively, I remember that for skew lines, the distance can be found using the formula involving the cross product of their direction vectors and the vector connecting a point on each line. But in this case, I don't have the direction vectors or specific points on both lines, so maybe that's not directly applicable.Wait, perhaps I can use the fact that the distances from A, B, and C to line m form a quadratic relationship because of the equal spacing of A, B, and C on line l. Since AB = BC, the distances from A, B, and C to m should follow a quadratic progression.Let me denote the distance from a general point on line l to line m as a function of the position along l. Let's say the distance squared from a point at position x on l to m is a quadratic function: f(x) = ax² + bx + c.Given that at x = -d, f(-d) = (sqrt(15))² = 15; at x = 0, f(0) = (7/2)² = 49/4; and at x = d, f(d) = (sqrt(10))² = 10.So, we have three equations:1. f(-d) = a(-d)² + b(-d) + c = ad² - bd + c = 152. f(0) = a(0)² + b(0) + c = c = 49/43. f(d) = a(d)² + b(d) + c = ad² + bd + c = 10From equation 2, we know c = 49/4. Plugging this into equations 1 and 3:1. ad² - bd + 49/4 = 152. ad² + bd + 49/4 = 10Let me subtract equation 1 from equation 2:(ad² + bd + 49/4) - (ad² - bd + 49/4) = 10 - 15Simplify:2bd = -5So, bd = -5/2Now, let me add equations 1 and 2:(ad² - bd + 49/4) + (ad² + bd + 49/4) = 15 + 10Simplify:2ad² + 98/4 = 252ad² + 24.5 = 252ad² = 0.5ad² = 0.25So, a = 0.25 / d²Now, from bd = -5/2, we have b = (-5/2) / dSo, now we have expressions for a and b in terms of d.But we need another equation to find d. Wait, maybe we can use the fact that the minimum distance between the lines occurs at some point, and the function f(x) should have a minimum. The minimum occurs where the derivative is zero.The function f(x) = ax² + bx + c. The derivative f’(x) = 2ax + b. Setting this equal to zero:2ax + b = 0x = -b/(2a)Plugging in the values of a and b:x = -[(-5/2)/d] / [2*(0.25/d²)]Simplify numerator: (5/2)/dDenominator: 0.5/d²So, x = (5/(2d)) / (0.5/d²) = (5/(2d)) * (2d²/1) = 5dWait, that can't be right because x is supposed to be the position along line l where the minimum distance occurs. But if x = 5d, that would be far beyond point C, which is at x = d. Maybe I made a mistake in the calculation.Let me double-check:f’(x) = 2ax + b = 0So, x = -b/(2a)We have a = 0.25/d² and b = (-5/2)/dSo, x = -[(-5/2)/d] / [2*(0.25/d²)] = (5/(2d)) / (0.5/d²) = (5/(2d)) * (2d²/1) = 5dHmm, same result. That suggests that the minimum distance occurs at x = 5d, which is beyond point C. But that doesn't make sense because the points A, B, and C are at x = -d, 0, d. Maybe the minimum distance isn't necessarily among these points.Wait, but the distance function f(x) is a quadratic, so it's a parabola. If the coefficient a is positive, it opens upwards, meaning the minimum is at x = 5d. If a is negative, it opens downwards, meaning the maximum is at x = 5d. But in our case, a = 0.25/d², which is positive, so the parabola opens upwards, and the minimum is at x = 5d.But since our points are at x = -d, 0, d, the minimum distance is beyond x = d, so the minimum distance isn't achieved at any of these points. Therefore, the distance between the lines is the minimum value of f(x), which occurs at x = 5d.But wait, the distance between the lines is the minimum distance, which is the value of f(x) at x = 5d. Let me compute that.f(5d) = a*(5d)² + b*(5d) + c = a*25d² + b*5d + cWe know a = 0.25/d², b = (-5/2)/d, c = 49/4So,f(5d) = (0.25/d²)*25d² + [(-5/2)/d]*5d + 49/4Simplify:= 0.25*25 + (-5/2)*5 + 49/4= 6.25 + (-12.5) + 12.25= (6.25 + 12.25) - 12.5= 18.5 - 12.5= 6So, the minimum distance squared is 6, so the distance is sqrt(6).Wait, that seems to make sense. So, the distance between the lines is sqrt(6).But let me verify this because I might have made a mistake in the earlier steps.We set up f(x) = ax² + bx + c, with f(-d)=15, f(0)=49/4, f(d)=10. Solved for a, b, c in terms of d, found that the minimum occurs at x=5d, and then computed f(5d)=6, so the distance is sqrt(6).Alternatively, maybe there's another approach. I remember that for three equally spaced points on a line, the distances to another skew line can be used to find the common perpendicular. Maybe using the formula for the distance between skew lines involving the lengths of the perpendiculars from three points.Wait, I think there's a formula that relates the distances from three equally spaced points to the other line, and from that, you can find the common perpendicular. Let me recall.I think the formula is something like the common perpendicular squared is equal to the middle distance squared minus the average of the squares of the other two distances. Or maybe it's a different combination.Wait, let me think. If we have three points equally spaced on line l, and we drop perpendiculars to line m, the distances from these points to m can be used to find the common perpendicular.Let me denote the distances as AD = sqrt(15), BE = 7/2, CF = sqrt(10). So, AD² = 15, BE² = 49/4, CF² = 10.Since AB = BC, the spacing is equal. Let me denote AB = BC = d.Now, the distances from A, B, C to m are AD, BE, CF respectively. The common perpendicular distance h can be found using these.I think the formula is h² = BE² - (AD² + CF²)/2Let me test this:h² = (49/4) - (15 + 10)/2 = 49/4 - 25/2 = 49/4 - 50/4 = -1/4That can't be right because h² can't be negative.Hmm, maybe the formula is different. Maybe h² = (AD² + CF²)/2 - BE²So,h² = (15 + 10)/2 - 49/4 = 25/2 - 49/4 = 50/4 - 49/4 = 1/4So, h = 1/2. But that contradicts the earlier result of sqrt(6). So, which one is correct?Wait, maybe the formula is h² = (AD² + CF²)/2 - BE², but only if the points are equally spaced and the distances are in a certain configuration.But in our case, the points are equally spaced, so maybe this formula applies.But then h² = (15 + 10)/2 - (49/4) = 25/2 - 49/4 = 50/4 - 49/4 = 1/4, so h = 1/2.But earlier, using the quadratic function approach, I got h = sqrt(6). These two results are conflicting.I must have made a mistake somewhere. Let me check both approaches.In the quadratic function approach, I assumed f(x) = ax² + bx + c, with f(-d)=15, f(0)=49/4, f(d)=10. Then solved for a, b, c in terms of d, found that the minimum occurs at x=5d, and then computed f(5d)=6, so h= sqrt(6).In the formula approach, I used h² = (AD² + CF²)/2 - BE², which gave h=1/2.These two results can't both be correct. So, I need to figure out which approach is correct.Wait, maybe the formula I used is incorrect. Let me think about the geometry.If we have three equally spaced points on line l, and we drop perpendiculars to line m, the distances from these points to m can be related to the common perpendicular and the angle between the lines.Let me denote the angle between the lines l and m as θ. Then, the distance from a point on l to m can be expressed as h / sinθ, but I'm not sure.Wait, actually, the distance from a point on l to m is the length of the perpendicular from that point to m. If we consider the common perpendicular h, then the distances from points A, B, C to m can be expressed in terms of h and the positions of A, B, C relative to the common perpendicular.Let me denote the position of B as the midpoint between A and C, so AB = BC = d. Let me denote the common perpendicular as MN, with M on l and N on m. Then, the distance from B to m is BE = 7/2, which is the length of the perpendicular from B to m.Similarly, the distances from A and C to m are AD = sqrt(15) and CF = sqrt(10).Now, if I consider the points A, B, C on l, and their projections D, E, F on m, the distances AD, BE, CF can be related to the common perpendicular h and the positions of A, B, C.I think the distances AD, BE, CF can be expressed as sqrt(h² + x²), where x is the distance along l from the projection of the common perpendicular.Wait, let me try to model this.Let me denote the projection of the common perpendicular MN on line l as point B. So, the common perpendicular is from B to m, which is BE = 7/2. Wait, but BE is given as 7/2, which is the distance from B to m. So, if BE is the common perpendicular, then h = 7/2. But that contradicts the earlier results.Wait, no, because BE is the distance from B to m, which is not necessarily the common perpendicular. The common perpendicular is the shortest distance between the lines, which might not pass through B.Wait, actually, the common perpendicular is the shortest distance between the lines, so it must be the minimum of AD, BE, CF. But in our case, AD = sqrt(15) ≈ 3.87, BE = 3.5, CF = sqrt(10) ≈ 3.16. So, the minimum is CF ≈ 3.16, which is less than BE and AD. So, the common perpendicular is CF? No, because CF is just the distance from C to m, not necessarily the common perpendicular.Wait, no, the common perpendicular is the shortest distance between the two lines, which is the minimum distance between any two points on l and m. So, it's possible that the common perpendicular is less than all of AD, BE, CF.Wait, but in our case, the distances from A, B, C to m are sqrt(15), 7/2, sqrt(10). So, the minimum among these is sqrt(10) ≈ 3.16. But is that the common perpendicular? Not necessarily, because the common perpendicular could be even shorter if there's a point on l closer to m than C.Wait, but we don't have information about other points on l. So, maybe the common perpendicular is indeed the minimum of these three distances, which is sqrt(10). But that contradicts the earlier quadratic approach where h = sqrt(6) ≈ 2.45, which is less than sqrt(10).Hmm, I'm confused now. Let me try to think differently.I remember that for skew lines, the distance can be found using the formula:Distance = |(P2 - P1) · (v1 × v2)| / |v1 × v2|where P1 and P2 are points on each line, and v1 and v2 are their direction vectors.But in this problem, we don't have the direction vectors or specific points on both lines, so maybe we can't use this formula directly.Alternatively, maybe we can use the fact that the distances from A, B, C to m form a quadratic function, as I did earlier, and then find the minimum of that function, which would be the common perpendicular.In that approach, I found that the minimum distance squared is 6, so the distance is sqrt(6). That seems plausible.But then why does the formula h² = (AD² + CF²)/2 - BE² give a different result? Maybe that formula is incorrect or applies under different conditions.Wait, let me check the formula again. I think the correct formula when you have three equally spaced points on one skew line and their distances to the other line is:h² = (AD² + CF²)/2 - BE²But in our case, this gives h² = (15 + 10)/2 - (49/4) = 25/2 - 49/4 = 50/4 - 49/4 = 1/4, so h = 1/2.But this contradicts the quadratic approach. So, which one is correct?Wait, maybe the formula is h² = (AD² + CF²)/2 - BE², but only when the points are equally spaced and the distances are in a certain configuration. Maybe I need to verify this formula.Alternatively, perhaps the formula is h² = BE² - (AD² + CF²)/2, but that would give a negative result, which is impossible.Wait, let me think geometrically. If we have three equally spaced points on line l, and we drop perpendiculars to line m, the distances from these points to m can be used to find the common perpendicular.Let me denote the common perpendicular as h. Then, the distances from A, B, C to m can be expressed in terms of h and the positions of A, B, C relative to the common perpendicular.Let me denote the position of B as the midpoint between A and C, so AB = BC = d. Let me denote the common perpendicular as MN, with M on l and N on m. Then, the distance from B to m is BE = 7/2, which is the length of the perpendicular from B to m.Wait, but if MN is the common perpendicular, then the distance from B to m is not necessarily equal to h unless B is the point where the common perpendicular meets l. But in reality, the common perpendicular could be from some other point on l, not necessarily B.So, let me denote the common perpendicular as from point P on l to point Q on m, with PQ = h. Then, the distances from A, B, C to m can be expressed in terms of h and the distances from A, B, C to P.Let me denote the distance from A to P as x, from B to P as y, and from C to P as z. Since AB = BC = d, and assuming P is between A and C, we have x = y + d and z = y - d. But this might complicate things.Alternatively, let me consider the coordinates again. Let me set up a coordinate system where line l is along the x-axis, and the common perpendicular is along the z-axis. Let me denote the coordinates of A, B, C on l as (-d, 0, 0), (0, 0, 0), (d, 0, 0). Then, line m is somewhere in 3D space, and the common perpendicular from l to m is along the z-axis, so the coordinates of the foot of the common perpendicular on l is (p, 0, 0), and on m is (q, r, h), where h is the distance between the lines.Wait, this might be getting too complicated. Maybe I should stick with the quadratic function approach.In that approach, I found that the minimum distance squared is 6, so h = sqrt(6). Let me see if that makes sense.Given that AD² = 15, BE² = 49/4 = 12.25, CF² = 10.If h² = 6, then h = sqrt(6) ≈ 2.45.Now, the distances from A, B, C to m are sqrt(15) ≈ 3.87, 3.5, sqrt(10) ≈ 3.16. So, the minimum distance is indeed less than all of these, which makes sense because the common perpendicular is the shortest distance.So, the quadratic function approach seems correct, and the formula I used earlier might be incorrect or misapplied.Therefore, I think the distance between the lines l and m is sqrt(6).But wait, let me double-check the quadratic function approach.We set f(x) = ax² + bx + c, with f(-d)=15, f(0)=49/4, f(d)=10.Solved for a, b, c in terms of d:a = 0.25/d²b = (-5/2)/dc = 49/4Then, the minimum occurs at x = -b/(2a) = (5/(2d)) / (0.5/d²) = 5d.Then, f(5d) = a*(25d²) + b*(5d) + c = 0.25*25 + (-5/2)*5 + 49/4 = 6.25 - 12.5 + 12.25 = 6.So, f(5d) = 6, so h = sqrt(6).Yes, that seems correct.Therefore, the distance between lines l and m is sqrt(6).

answer：Alright, so I have this problem where I need to compute a sum involving a function t(x) that has some missing coefficients. The function is given as:t(x) = cos(5x) + * cos(4x) + * cos(3x) + * cos(2x) + *² cos(x) + *The stars (*) represent missing coefficients, and the math student claims that he can compute the sum without knowing these coefficients. I need to figure out if he's right.First, let me understand what exactly is being asked. The sum to compute is:t(0) - t(π/5) + t(2π/5) - t(3π/5) + ... + t(8π/5) - t(9π/5)So, it's an alternating sum of t evaluated at multiples of π/5, starting from 0 up to 9π/5.Given that t(x) is a combination of cosines with different frequencies and some missing coefficients, I need to see if the sum can be evaluated without knowing the specific values of these coefficients.Let me write out the function t(x) explicitly, replacing the stars with variables to represent the missing coefficients:t(x) = cos(5x) + a cos(4x) + b cos(3x) + c cos(2x) + d² cos(x) + eHere, a, b, c, d, and e are the missing coefficients. The problem is to compute the sum:S = t(0) - t(π/5) + t(2π/5) - t(3π/5) + t(4π/5) - t(5π/5) + t(6π/5) - t(7π/5) + t(8π/5) - t(9π/5)So, S is an alternating sum of t evaluated at these specific points.Since t(x) is a linear combination of cosines and a constant term, I can split the sum S into separate sums for each term in t(x):S = [cos(0) - cos(π) + cos(2π) - cos(3π) + cos(4π) - cos(5π) + cos(6π) - cos(7π) + cos(8π) - cos(9π)] + a[cos(0) - cos(4π/5) + cos(8π/5) - cos(12π/5) + cos(16π/5) - cos(20π/5) + cos(24π/5) - cos(28π/5) + cos(32π/5) - cos(36π/5)] + b[cos(0) - cos(3π/5) + cos(6π/5) - cos(9π/5) + cos(12π/5) - cos(15π/5) + cos(18π/5) - cos(21π/5) + cos(24π/5) - cos(27π/5)] + c[cos(0) - cos(2π/5) + cos(4π/5) - cos(6π/5) + cos(8π/5) - cos(10π/5) + cos(12π/5) - cos(14π/5) + cos(16π/5) - cos(18π/5)] + d²[cos(0) - cos(π/5) + cos(2π/5) - cos(3π/5) + cos(4π/5) - cos(5π/5) + cos(6π/5) - cos(7π/5) + cos(8π/5) - cos(9π/5)] + e[1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1]Now, let's analyze each of these sums one by one.Starting with the constant term e:e[1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1] = e[0] = 0So, the constant term contributes nothing to the sum.Next, let's look at the term involving cos(5x):[cos(0) - cos(π) + cos(2π) - cos(3π) + cos(4π) - cos(5π) + cos(6π) - cos(7π) + cos(8π) - cos(9π)]But cos(kπ) = (-1)^k, so:= [1 - (-1) + 1 - (-1) + 1 - (-1) + 1 - (-1) + 1 - (-1)]= [1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1]= 10So, the sum for cos(5x) is 10.Now, let's consider the term involving cos(4x):a[cos(0) - cos(4π/5) + cos(8π/5) - cos(12π/5) + cos(16π/5) - cos(20π/5) + cos(24π/5) - cos(28π/5) + cos(32π/5) - cos(36π/5)]Simplify the angles modulo 2π:cos(0) = 1cos(4π/5) = cos(4π/5)cos(8π/5) = cos(8π/5)cos(12π/5) = cos(12π/5 - 2π) = cos(2π/5)cos(16π/5) = cos(16π/5 - 2π*3) = cos(16π/5 - 6π/5) = cos(10π/5) = cos(2π) = 1cos(20π/5) = cos(4π) = 1cos(24π/5) = cos(24π/5 - 2π*2) = cos(24π/5 - 10π/5) = cos(14π/5) = cos(14π/5 - 2π) = cos(4π/5)cos(28π/5) = cos(28π/5 - 2π*2) = cos(28π/5 - 10π/5) = cos(18π/5) = cos(18π/5 - 2π*1) = cos(8π/5)cos(32π/5) = cos(32π/5 - 2π*3) = cos(32π/5 - 10π/5) = cos(22π/5) = cos(22π/5 - 2π*2) = cos(2π/5)cos(36π/5) = cos(36π/5 - 2π*3) = cos(36π/5 - 10π/5) = cos(26π/5) = cos(26π/5 - 2π*2) = cos(6π/5) = cos(6π/5 - 2π) = cos(-4π/5) = cos(4π/5)So, substituting back:a[1 - cos(4π/5) + cos(8π/5) - cos(2π/5) + 1 - 1 + cos(4π/5) - cos(8π/5) + cos(2π/5) - cos(4π/5)]Simplify:= a[1 - cos(4π/5) + cos(8π/5) - cos(2π/5) + 1 - 1 + cos(4π/5) - cos(8π/5) + cos(2π/5) - cos(4π/5)]Combine like terms:= a[1 + ( - cos(4π/5) + cos(4π/5) - cos(4π/5)) + (cos(8π/5) - cos(8π/5)) + (- cos(2π/5) + cos(2π/5)) + 1 - 1]Simplify further:= a[1 - cos(4π/5) + 0 + 0 + 0]But wait, that doesn't seem right. Let me check my simplification again.Looking back:= a[1 - cos(4π/5) + cos(8π/5) - cos(2π/5) + 1 - 1 + cos(4π/5) - cos(8π/5) + cos(2π/5) - cos(4π/5)]Grouping similar terms:= a[1 + 1 - 1] + a[- cos(4π/5) + cos(4π/5) - cos(4π/5)] + a[cos(8π/5) - cos(8π/5)] + a[- cos(2π/5) + cos(2π/5)]Simplify:= a[1] + a[- cos(4π/5)] + a[0] + a[0]So, = a[1 - cos(4π/5)]But cos(4π/5) is equal to cos(π - π/5) = -cos(π/5). So,= a[1 - (-cos(π/5))] = a[1 + cos(π/5)]But I need to compute this sum for the term involving cos(4x). However, I realize that this approach might not be the most efficient. Maybe there's a better way to handle these sums.I recall that for sums of cosines at equally spaced angles, there's a formula involving the Dirichlet kernel or using complex exponentials. Maybe I can use that.Let me consider the general sum:Sum_{k=0}^{n-1} (-1)^k cos(kθ)This can be expressed as the real part of the sum:Sum_{k=0}^{n-1} (-1)^k e^{ikθ}Which is a geometric series with ratio -e^{iθ}.The sum is:Sum_{k=0}^{n-1} (-1)^k e^{ikθ} = (1 - (-e^{iθ})^n) / (1 + e^{iθ})Taking the real part:Re[(1 - (-e^{iθ})^n) / (1 + e^{iθ})]This might be complicated, but perhaps for specific values of θ and n, it simplifies.In our case, for each term in t(x), we have different frequencies (5x, 4x, 3x, 2x, x), and we're evaluating at x = kπ/5 for k = 0 to 9.So, for each frequency m (where m = 1 to 5), the sum becomes:Sum_{k=0}^{9} (-1)^k cos(m * kπ/5)This is similar to the general sum I mentioned earlier, with θ = mπ/5 and n = 10.So, applying the formula:Sum_{k=0}^{9} (-1)^k cos(m * kπ/5) = Re[Sum_{k=0}^{9} (-1)^k e^{i m k π/5}]= Re[(1 - (-e^{i m π/5})^{10}) / (1 + e^{i m π/5})]Simplify the numerator:1 - (-e^{i m π/5})^{10} = 1 - (-1)^{10} e^{i m π/5 * 10} = 1 - (1) e^{i 2mπ} = 1 - 1 = 0Wait, that would make the entire sum zero. But that can't be right because earlier, for m=5, the sum was 10.Hmm, perhaps I made a mistake in applying the formula.Wait, when m=5, the frequency is 5, so mπ/5 = π. So, the sum becomes:Sum_{k=0}^{9} (-1)^k cos(kπ) = Sum_{k=0}^{9} (-1)^k (-1)^k = Sum_{k=0}^{9} 1 = 10Which matches what I got earlier.But for other m's, let's see:For m=1:Sum_{k=0}^{9} (-1)^k cos(kπ/5)Using the formula:= Re[(1 - (-e^{i π/5})^{10}) / (1 + e^{i π/5})]= Re[(1 - (-1)^{10} e^{i 2π}) / (1 + e^{i π/5})]= Re[(1 - 1 * 1) / (1 + e^{i π/5})] = Re[0 / (1 + e^{i π/5})] = 0Similarly, for m=2:Sum_{k=0}^{9} (-1)^k cos(2kπ/5)= Re[(1 - (-e^{i 2π/5})^{10}) / (1 + e^{i 2π/5})]= Re[(1 - (-1)^{10} e^{i 4π}) / (1 + e^{i 2π/5})]= Re[(1 - 1 * 1) / (1 + e^{i 2π/5})] = 0Same for m=3 and m=4:Sum_{k=0}^{9} (-1)^k cos(3kπ/5) = 0Sum_{k=0}^{9} (-1)^k cos(4kπ/5) = 0Only for m=5, the sum is non-zero and equals 10.Therefore, going back to our original sum S:S = [Sum for cos(5x)] + a[Sum for cos(4x)] + b[Sum for cos(3x)] + c[Sum for cos(2x)] + d²[Sum for cos(x)] + e[Sum for constant]From the above analysis:Sum for cos(5x) = 10Sum for cos(4x) = 0Sum for cos(3x) = 0Sum for cos(2x) = 0Sum for cos(x) = 0Sum for constant = 0Therefore, S = 10 + a*0 + b*0 + c*0 + d²*0 + e*0 = 10So, regardless of the values of a, b, c, d, and e, the sum S is always 10.Therefore, the math student is correct; he can compute the sum without knowing the missing coefficients.

answer：Okay, so I have this problem about an acute triangle ABC. The problem says that the altitudes AX and BY intersect at H. I know that in a triangle, the point where all three altitudes intersect is called the orthocenter. So, H is the orthocenter of triangle ABC.The given angles are angle BAC, which is 50 degrees, and angle ABC, which is 85 degrees. I need to find the measure of angle CHX. Hmm, I'm not exactly sure where point X is located. Since AX is an altitude, I think X must be the foot of the altitude from A to BC. Similarly, BY is an altitude, so Y must be the foot of the altitude from B to AC.Let me try to visualize the triangle. Let me sketch triangle ABC with angle BAC = 50°, angle ABC = 85°, so angle BCA must be 180° - 50° - 85° = 45°. So, angle C is 45°. That's a nice angle, maybe it will help later.Since H is the orthocenter, it lies inside the triangle because the triangle is acute. Now, I need to find angle CHX. Let me think about point X. Since AX is the altitude from A to BC, X is on BC. So, CHX is the angle at point H between points C and X.Wait, I need to make sure I understand the notation. Angle CHX is the angle at H between points C and X. So, it's the angle formed at H by the lines HC and HX.I think I need to find some relationships between the angles in the triangle and the angles involving the orthocenter. Maybe I can use some properties of the orthocenter or cyclic quadrilaterals.Let me recall that in a triangle, the orthocenter, along with the vertices, forms several cyclic quadrilaterals. For example, the feet of the altitudes and the orthocenter lie on a circle called the nine-point circle. But I'm not sure if that's directly helpful here.Alternatively, maybe I can use the fact that in triangle ABC, the angles at the orthocenter relate to the original angles of the triangle. I remember that the angles at the orthocenter can be calculated using the original angles of the triangle.Wait, let me think step by step.First, I can find all the angles of triangle ABC:- Angle at A: 50°- Angle at B: 85°- Angle at C: 45°Now, since H is the orthocenter, the altitudes from each vertex intersect at H.So, AX is the altitude from A to BC, meeting BC at X.BY is the altitude from B to AC, meeting AC at Y.And there's also the altitude from C to AB, which would meet AB at some point, say Z.But in this problem, we're only concerned with points X and Y, and the orthocenter H.I need to find angle CHX. So, let's focus on triangle CHX.To find angle CHX, I might need to know some other angles or sides in that triangle.Alternatively, maybe I can find some similar triangles or use the properties of cyclic quadrilaterals.Wait, since AX and BY are altitudes, they are perpendicular to BC and AC respectively.So, angle AXB is 90°, and angle BYC is 90°.Also, since H is the orthocenter, it lies on both AX and BY.So, HX is a segment from H to X on BC, and HY is a segment from H to Y on AC.I think I can use the fact that in triangle ABC, the orthocenter H creates several right angles with the sides.Wait, maybe I can consider triangle HXC.In triangle HXC, I know that angle HXC is 90° because AX is perpendicular to BC, and H lies on AX.Wait, no, HX is part of AX, but HX itself is not necessarily perpendicular to BC unless H coincides with X, which it doesn't because H is the orthocenter inside the triangle.Wait, maybe I need to think differently.Let me consider the quadrilateral formed by points A, H, X, and the foot of another altitude. Hmm, not sure.Alternatively, maybe I can use the fact that in triangle ABC, the angles at the orthocenter relate to the original angles.I recall that the angles at the orthocenter can be calculated as 180° minus the original angles of the triangle.Wait, let me think. In triangle ABC, the angles at the orthocenter H are equal to 180° minus the angles at the opposite vertices.So, angle BHC is equal to 180° - angle BAC.Similarly, angle AHB is equal to 180° - angle ACB.Wait, let me verify that.Yes, in a triangle, the measure of the angle at the orthocenter opposite to a vertex is equal to 180° minus the measure of the angle at that vertex.So, angle BHC = 180° - angle BAC = 180° - 50° = 130°.Similarly, angle AHB = 180° - angle ACB = 180° - 45° = 135°.And angle AHC = 180° - angle ABC = 180° - 85° = 95°.Wait, but I'm not sure if that's directly helpful for angle CHX.Alternatively, maybe I can consider triangle CHX and find some angles there.Since AX is an altitude, angle AXB is 90°, and H lies on AX, so HX is part of AX.Similarly, since BY is an altitude, angle BYC is 90°, and H lies on BY.Wait, maybe I can consider triangle HXC.In triangle HXC, I know that angle HXC is part of the altitude AX, which is perpendicular to BC.Wait, no, HX is part of AX, but HX itself is not necessarily perpendicular to BC unless H is at X, which it isn't.Wait, maybe I can use the fact that in triangle ABC, the orthocenter H creates several right triangles.Wait, let me think about the coordinates. Maybe assigning coordinates to the triangle might help.Let me place triangle ABC in a coordinate system.Let me set point A at (0,0), point B at (c,0), and point C somewhere in the plane.But maybe it's better to use trigonometric relationships.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H has certain properties.Wait, another approach: since H is the orthocenter, the angles at H can be related to the original angles.Wait, I think I can use the fact that in triangle ABC, the angles at H are equal to 180° minus the angles at the opposite vertices.So, angle BHC = 180° - angle BAC = 130°, as I thought earlier.Similarly, angle AHB = 180° - angle ACB = 135°, and angle AHC = 180° - angle ABC = 95°.But I'm not sure how that helps with angle CHX.Wait, maybe I can consider triangle CHX and find some angles.Since AX is an altitude, angle AXB is 90°, so triangle AXB is right-angled at X.Similarly, triangle BYC is right-angled at Y.Since H is the orthocenter, it lies on both AX and BY.So, in triangle AXB, H is a point on AX, and in triangle BYC, H is a point on BY.Wait, maybe I can use similar triangles.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several cyclic quadrilaterals.For example, quadrilateral HXBC is cyclic because angles HXB and HCB are both 90°, making them subtend the same arc.Wait, is that true?Wait, angle HXB is 90° because AX is an altitude, and angle HCB is 90° because the altitude from C would be perpendicular to AB.Wait, but H is the orthocenter, so the altitude from C is CH, which is perpendicular to AB.Wait, so angle HCB is 90°, but angle HXB is also 90°, so points H, X, B, and C lie on a circle because they all subtend right angles over the same segment.Wait, no, actually, in a circle, if two points subtend right angles over the same segment, they lie on the circle with that segment as the diameter.But in this case, H and X both subtend right angles over BC.Wait, no, H is the orthocenter, so the altitude from C is CH, which is perpendicular to AB, not BC.Wait, maybe I'm getting confused.Let me clarify:- AX is the altitude from A to BC, so AX is perpendicular to BC, meeting at X.- BY is the altitude from B to AC, so BY is perpendicular to AC, meeting at Y.- The altitude from C would be CZ, which is perpendicular to AB, meeting at Z.So, H is the intersection of AX, BY, and CZ.Therefore, H is inside the triangle, and each altitude is perpendicular to the opposite side.So, in this case, angle AXB is 90°, angle BYC is 90°, and angle CZB is 90°.Now, considering quadrilateral HXBC, we have:- Angle HXB is 90° (since AX is perpendicular to BC)- Angle HCB is 90° (since CZ is perpendicular to AB, but H is on CZ, so angle HCB is 90°)Wait, no, angle HCB is not necessarily 90°, because H is on CZ, which is perpendicular to AB, but angle HCB is the angle at C between BC and CH.Wait, since CZ is perpendicular to AB, and H lies on CZ, then angle ZCB is 90°, but angle HCB is not necessarily 90° unless H coincides with Z, which it doesn't.Wait, maybe I need to think differently.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several similar triangles.Wait, let me consider triangle HXC.In triangle HXC, I know that HX is part of the altitude AX, which is perpendicular to BC.So, in triangle HXC, angle HXC is 90°, because AX is perpendicular to BC.Wait, no, HX is part of AX, but HX itself is not necessarily perpendicular to BC unless H is at X, which it isn't.Wait, no, AX is perpendicular to BC, so the entire line AX is perpendicular to BC, meaning that any point on AX, including H, will have the property that the line from that point to BC is perpendicular.Wait, no, that's not correct. The entire line AX is perpendicular to BC, so any point on AX, when connected to BC, will form a right angle with BC.Wait, but H is on AX, so the line from H to BC is HX, which is part of AX, so HX is perpendicular to BC.Wait, that makes sense. So, HX is perpendicular to BC, meaning that angle HXC is 90°.Wait, but earlier I thought that HX is part of AX, which is perpendicular to BC, so HX is indeed perpendicular to BC, making angle HXC 90°.Wait, but that contradicts my earlier thought that HX is not necessarily perpendicular to BC.Wait, no, actually, since AX is perpendicular to BC, and H is on AX, then HX is a segment of AX, so HX is also perpendicular to BC.Therefore, angle HXC is 90°.Wait, that seems correct.So, in triangle HXC, angle at X is 90°, angle at C is angle BCA, which is 45°, so angle at H, which is angle CHX, can be found using the triangle angle sum.Wait, but wait, triangle HXC has angles at H, X, and C.We know angle at X is 90°, angle at C is 45°, so angle at H is 180° - 90° - 45° = 45°.Wait, that seems too straightforward. So, angle CHX is 45°.Wait, but let me double-check.In triangle HXC:- Angle at X: 90° (since HX is perpendicular to BC)- Angle at C: 45° (angle BCA)Therefore, angle at H: 180° - 90° - 45° = 45°.So, angle CHX is 45°.Wait, but I'm not sure if that's correct because I might be misapplying the triangle angle sum.Wait, let me think again.In triangle HXC, angle at X is 90°, angle at C is 45°, so angle at H is 45°, making triangle HXC an isosceles right triangle.Wait, but is that the case?Wait, in triangle ABC, angle at C is 45°, and HX is perpendicular to BC, so triangle HXC has angle at X of 90°, angle at C of 45°, so angle at H is 45°, making triangle HXC a 45-45-90 triangle.Therefore, angle CHX is 45°.Wait, that seems correct.Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several right angles, and using the properties of cyclic quadrilaterals.But I think the approach I took is correct.So, angle CHX is 45°.

answer：To determine the probability that the product of two distinct numbers from the set is a multiple of 126, we first factorize 126 into its prime components: (126 = 2 times 3^2 times 7). This means the product must include at least one factor of 2, two factors of 3, and one factor of 7.We analyze the given set ({3, 7, 14, 18, 21, 27, 35, 42, 48, 54, 63}) to identify numbers that contribute to these factors:- **Multiples of 7**: 7, 14, 21, 35, 42, 63- **Multiples of 3**: 3, 18, 21, 27, 42, 48, 54, 63- **Factors of 2**: 14, 48Next, we enumerate pairs that can provide the necessary factors:1. **14** (provides 2 and 7) can pair with 21, 27, 42, 54, or 63 (each provides at least one 3).2. **21** (provides 3 and 7) can pair with 48 (provides 2).3. **42** (provides 2, 3, and 7) can pair with 48 (provides an additional 2).4. **63** (provides 3 and 7) can pair with 14 or 48 (to get the necessary 2).Summarizing the viable pairs:- 14 pairs with 5 numbers.- 21 pairs with 1 number.- 42 pairs with 1 number.- 63 pairs with 2 numbers.This gives a total of (5 + 1 + 1 + 2 = 9) successful pairs. The total number of possible pairs is (binom{11}{2} = 55).Therefore, the probability is (frac{9}{55}).