🧠:<think>Okay, so I've got this problem here about two curves, C₁ and C₂, and I need to find the minimum value of S(k), which is the sum of the areas enclosed by these curves and the lines x=0 and x=π. The curves are given by y = sin(2x) for C₁ and y = k cos(x) for C₂, with k between 0 and 2. Alright, first things first, I need to visualize what's going on here. So, C₁ is a sine curve with a period of π, which means it completes one full cycle from 0 to π. It'll start at 0, go up to 1 at π/4, back down to 0 at π/2, down to -1 at 3π/4, and back to 0 at π. On the other hand, C₂ is a cosine curve scaled by k. Since k is between 0 and 2, this curve will oscillate between k and -k, but because we're only considering from 0 to π, it'll start at k, go down to 0 at π/2, and then to -k at π.Now, the problem mentions four parts enclosed by these curves and the lines x=0 and x=π. So, I imagine the curves intersecting each other a couple of times between 0 and π, creating four distinct regions whose areas we need to sum up to get S(k). Then, we need to find the value of k that minimizes this total area.To approach this, I think I need to find the points where C₁ and C₂ intersect. That is, solve sin(2x) = k cos(x). Let me write that down:sin(2x) = k cos(x)Using the double-angle identity for sine, sin(2x) = 2 sin(x) cos(x), so substituting that in:2 sin(x) cos(x) = k cos(x)Assuming cos(x) ≠ 0 (since at x=π/2, cos(x)=0, which is another intersection point), I can divide both sides by cos(x):2 sin(x) = kSo, sin(x) = k/2Therefore, the solutions for x are:x = arcsin(k/2) and x = π - arcsin(k/2)These will be the x-coordinates where the two curves intersect. Let's denote these as x₁ = arcsin(k/2) and x₃ = π - arcsin(k/2). Also, we know that at x=π/2, both curves pass through zero, so that's another intersection point, x₂ = π/2.So now, I have three intersection points: x₁, x₂, and x₃. Between these points, the curves will cross each other, creating four regions:1. From x=0 to x=x₁2. From x=x₁ to x=x₂3. From x=x₂ to x=x₃4. From x=x₃ to x=πFor each of these intervals, I need to figure out which curve is on top and which is on the bottom to set up the integrals correctly.Let me consider each interval:1. From x=0 to x=x₁: At x=0, sin(2x)=0 and k cos(x)=k. So, C₂ is above C₁ here. Since x₁ is the first intersection point, C₂ remains above C₁ until x=x₁.2. From x=x₁ to x=x₂: After x₁, sin(2x) becomes greater than k cos(x). This is because at x just above x₁, sin(2x) is increasing while k cos(x) is decreasing. So, C₁ is above C₂ in this interval.3. From x=x₂ to x=x₃: At x=π/2, both curves are zero. Just to the right of π/2, sin(2x) becomes negative, while k cos(x) is also negative. But let's check which one is larger in magnitude. Since sin(2x) at x just above π/2 is sin(2*(π/2 + ε)) = sin(π + 2ε) = -sin(2ε), which is approximately -2ε. Meanwhile, k cos(π/2 + ε) = -k sin(ε) ≈ -k ε. Since k is between 0 and 2, depending on k, either could be larger. Wait, but actually, let's evaluate the behavior more carefully. Wait, maybe it's better to consider specific points. For example, at x=3π/4, sin(2x)=sin(3π/2)=-1, and k cos(3π/4)=-k√2/2. So, which one is larger? Since sin(2x) is -1 and k cos(x) is -k√2/2. So, sin(2x) is more negative, meaning that C₂ is above C₁ here because it's less negative. So, from x=x₂ to x=x₃, C₂ is above C₁.4. From x=x₃ to x=π: After x=x₃, sin(2x) becomes greater than k cos(x) again. Let me check at x=π, sin(2π)=0 and k cos(π)=-k. So, sin(2x) is zero, and k cos(x) is negative, so C₁ is above C₂.Wait, that seems a bit confusing. Maybe I should plot these curves mentally or sketch them roughly.- C₁ starts at 0, goes up to 1 at π/4, back to 0 at π/2, down to -1 at 3π/4, and back to 0 at π.- C₂ starts at k, goes down to 0 at π/2, and then to -k at π.So, depending on the value of k, the points where they intersect can vary.But since we have x₁ = arcsin(k/2) and x₃ = π - arcsin(k/2), these will always be symmetric around π/2.So, in terms of areas:1. From 0 to x₁: C₂ is above C₁, so the area is ∫(k cos x - sin 2x) dx from 0 to x₁.2. From x₁ to π/2: C₁ is above C₂, so the area is ∫(sin 2x - k cos x) dx from x₁ to π/2.3. From π/2 to x₃: C₂ is above C₁, so the area is ∫(k cos x - sin 2x) dx from π/2 to x₃.4. From x₃ to π: C₁ is above C₂, so the area is ∫(sin 2x - k cos x) dx from x₃ to π.So, S(k) is the sum of these four areas.Now, to compute S(k), I need to set up these integrals and then add them up.Let's compute each integral one by one.First, let's compute the integral from 0 to x₁: ∫(k cos x - sin 2x) dx.The antiderivative of k cos x is k sin x, and the antiderivative of sin 2x is (-1/2) cos 2x.So, the integral becomes:[k sin x - (-1/2 cos 2x)] from 0 to x₁= [k sin x + (1/2) cos 2x] from 0 to x₁= [k sin x₁ + (1/2) cos 2x₁] - [k sin 0 + (1/2) cos 0]= k sin x₁ + (1/2) cos 2x₁ - 0 - (1/2)(1)= k sin x₁ + (1/2)(cos 2x₁ - 1)Similarly, the integral from x₁ to π/2: ∫(sin 2x - k cos x) dx.Antiderivative of sin 2x is (-1/2) cos 2x, and antiderivative of k cos x is k sin x.So, the integral becomes:[(-1/2 cos 2x) - k sin x] from x₁ to π/2= [(-1/2 cos π) - k sin(π/2)] - [(-1/2 cos 2x₁) - k sin x₁]= [(-1/2)(-1) - k(1)] - [(-1/2 cos 2x₁) - k sin x₁]= (1/2 - k) - (-1/2 cos 2x₁ - k sin x₁)= 1/2 - k + (1/2 cos 2x₁ + k sin x₁)Now, the integral from π/2 to x₃: ∫(k cos x - sin 2x) dx.Again, antiderivative of k cos x is k sin x, antiderivative of sin 2x is (-1/2) cos 2x.So, the integral becomes:[k sin x - (-1/2 cos 2x)] from π/2 to x₃= [k sin x + (1/2) cos 2x] from π/2 to x₃= [k sin x₃ + (1/2) cos 2x₃] - [k sin(π/2) + (1/2) cos π]= k sin x₃ + (1/2) cos 2x₃ - [k(1) + (1/2)(-1)]= k sin x₃ + (1/2) cos 2x₃ - k + (1/2)= (k sin x₃ - k) + (1/2)(cos 2x₃ + 1)Similarly, the integral from x₃ to π: ∫(sin 2x - k cos x) dx.Antiderivative is (-1/2 cos 2x) - k sin x.So, the integral becomes:[(-1/2 cos 2x) - k sin x] from x₃ to π= [(-1/2 cos 2π) - k sin π] - [(-1/2 cos 2x₃) - k sin x₃]= [(-1/2)(1) - k(0)] - [(-1/2 cos 2x₃) - k sin x₃]= (-1/2) - [ -1/2 cos 2x₃ - k sin x₃ ]= -1/2 + (1/2 cos 2x₃ + k sin x₃)Now, let's sum up all these four integrals to get S(k):S(k) = [k sin x₁ + (1/2)(cos 2x₁ - 1)] + [1/2 - k + (1/2 cos 2x₁ + k sin x₁)] + [(k sin x₃ - k) + (1/2)(cos 2x₃ + 1)] + [-1/2 + (1/2 cos 2x₃ + k sin x₃)]Let me simplify term by term.First, expand each bracket:1. First integral: k sin x₁ + (1/2 cos 2x₁ - 1/2)2. Second integral: 1/2 - k + (1/2 cos 2x₁ + k sin x₁)3. Third integral: k sin x₃ - k + (1/2 cos 2x₃ + 1/2)4. Fourth integral: -1/2 + (1/2 cos 2x₃ + k sin x₃)Now, let's combine all these terms:= [k sin x₁ + (1/2 cos 2x₁ - 1/2)] + [1/2 - k + (1/2 cos 2x₁ + k sin x₁)] + [k sin x₃ - k + (1/2 cos 2x₃ + 1/2)] + [-1/2 + (1/2 cos 2x₃ + k sin x₃)]Let me group like terms:Terms with sin x₁:k sin x₁ + k sin x₁ = 2 k sin x₁Terms with cos 2x₁:1/2 cos 2x₁ + 1/2 cos 2x₁ = cos 2x₁Constant terms:-1/2 + 1/2 - k - k + 1/2 -1/2Let me compute that:-1/2 + 1/2 = 0Then, -k - k = -2kThen, +1/2 -1/2 = 0So, total constant terms: -2kTerms with sin x₃:k sin x₃ + k sin x₃ = 2 k sin x₃Terms with cos 2x₃:1/2 cos 2x₃ + 1/2 cos 2x₃ = cos 2x₃So, putting it all together:S(k) = 2 k sin x₁ + cos 2x₁ - 2k + 2 k sin x₃ + cos 2x₃Now, let's recall that x₁ = arcsin(k/2) and x₃ = π - arcsin(k/2).So, sin x₃ = sin(π - x₁) = sin x₁.Similarly, cos 2x₃ = cos(2π - 2x₁) = cos(2x₁), because cosine is even and cos(2π - θ) = cos θ.Therefore, cos 2x₃ = cos 2x₁Similarly, sin x₃ = sin x₁So, substituting these into S(k):S(k) = 2 k sin x₁ + cos 2x₁ - 2k + 2 k sin x₁ + cos 2x₁Combine like terms:2 k sin x₁ + 2 k sin x₁ = 4 k sin x₁cos 2x₁ + cos 2x₁ = 2 cos 2x₁So, S(k) = 4 k sin x₁ + 2 cos 2x₁ - 2kNow, let's express everything in terms of k.We know that x₁ = arcsin(k/2), so sin x₁ = k/2.Also, cos 2x₁ = 1 - 2 sin²x₁ = 1 - 2*(k²/4) = 1 - k²/2So, substituting back:S(k) = 4k*(k/2) + 2*(1 - k²/2) - 2kSimplify:4k*(k/2) = 2k²2*(1 - k²/2) = 2 - k²So, S(k) = 2k² + 2 - k² - 2k = k² + 2 - 2kSo, S(k) simplifies down to k² - 2k + 2.Wait, that's interesting. So, S(k) = k² - 2k + 2.Now, to find the minimum value of S(k), we can treat this as a quadratic function in k.The quadratic is S(k) = k² - 2k + 2.The general form of a quadratic is ax² + bx + c, and its minimum occurs at k = -b/(2a).Here, a=1, b=-2, so the minimum occurs at k = -(-2)/(2*1) = 2/2 = 1.Therefore, the minimum value of S(k) is S(1) = (1)^2 - 2*(1) + 2 = 1 - 2 + 2 = 1.Wait, so the minimum value is 1?But the initial curves are y = sin(2x) and y = k cos x. When k=1, C₂ is y = cos x.Let me verify if this makes sense.At k=1, the curves are y = sin(2x) and y = cos x.They intersect at x where sin(2x) = cos x.Which is 2 sin x cos x = cos x.So, either cos x = 0, which gives x=π/2, or 2 sin x = 1, so sin x = 1/2, so x=π/6 and 5π/6.So, x₁=π/6, x₂=π/2, x₃=5π/6.Then, computing the areas:From 0 to π/6: cos x - sin 2xFrom π/6 to π/2: sin 2x - cos xFrom π/2 to 5π/6: cos x - sin 2xFrom 5π/6 to π: sin 2x - cos xComputing each integral:First integral: ∫₀^{π/6} (cos x - sin 2x) dx= [sin x + (1/2) cos 2x] from 0 to π/6= [sin(π/6) + (1/2) cos(π/3)] - [sin 0 + (1/2) cos 0]= [1/2 + (1/2)(1/2)] - [0 + (1/2)(1)]= [1/2 + 1/4] - 1/2= 3/4 - 1/2 = 1/4Second integral: ∫_{π/6}^{π/2} (sin 2x - cos x) dx= [(-1/2 cos 2x) - sin x] from π/6 to π/2= [(-1/2 cos π) - sin(π/2)] - [(-1/2 cos(π/3)) - sin(π/6)]= [(-1/2)(-1) - 1] - [(-1/2)(1/2) - 1/2]= [1/2 - 1] - [-1/4 - 1/2]= (-1/2) - (-3/4)= (-1/2) + 3/4 = 1/4Third integral: ∫_{π/2}^{5π/6} (cos x - sin 2x) dx= [sin x + (1/2) cos 2x] from π/2 to 5π/6= [sin(5π/6) + (1/2) cos(5π/3)] - [sin(π/2) + (1/2) cos π]= [1/2 + (1/2)(1/2)] - [1 + (1/2)(-1)]= [1/2 + 1/4] - [1 - 1/2]= 3/4 - 1/2 = 1/4Fourth integral: ∫_{5π/6}^{π} (sin 2x - cos x) dx= [(-1/2 cos 2x) - sin x] from 5π/6 to π= [(-1/2 cos 2π) - sin π] - [(-1/2 cos(5π/3)) - sin(5π/6)]= [(-1/2)(1) - 0] - [(-1/2)(1/2) - 1/2]= (-1/2) - (-1/4 - 1/2)= (-1/2) - (-3/4)= (-1/2) + 3/4 = 1/4So, each of the four areas is 1/4, so S(1) = 4*(1/4) = 1.Therefore, the minimum value of S(k) is indeed 1, occurring at k=1.Wait, but in the initial thought process, I thought the answer might be around -1.0577, but that seems incorrect because S(k) is a sum of areas, which should be positive. So, getting a positive minimum makes sense.But wait, the initial problem statement says "the sum of the areas of four parts enclosed by C₁, C₂ and two lines x=0, x=π." So, areas are always positive, hence S(k) must be positive.But in my earlier simplification, I got S(k) = k² - 2k + 2, which is a quadratic opening upwards, with its minimum at k=1, giving S(1)=1.So, that's consistent.Wait, but in the initial problem, the user wrote "the sum of the areas of four parts enclosed by C₁, C₂ and two lines x=0, x=π." So, if S(k) is always positive, and the minimum is 1, then 1 is the answer.But looking back at the initial problem, the user wrote "the minimum value of S(k)." Since S(k) is a sum of areas, it's always positive, so the minimum would be the smallest positive value it can take, which is 1.But wait, maybe I made a mistake in the simplification.Wait, let's go back to the expression for S(k):S(k) = 4k sin x₁ + 2 cos 2x₁ - 2k.But x₁ = arcsin(k/2), so sin x₁ = k/2, cos 2x₁ = 1 - 2 sin²x₁ = 1 - (k²)/2.So, substituting:S(k) = 4k*(k/2) + 2*(1 - k²/2) - 2k= 2k² + 2 - k² - 2k= k² - 2k + 2Yes, that's correct.So, S(k) = k² - 2k + 2.This is a quadratic function in k, which can be written as S(k) = (k - 1)^2 + 1.Since (k - 1)^2 is always non-negative, the minimum value occurs when (k - 1)^2 = 0, i.e., k=1, giving S(k)=1.Therefore, the minimum value of S(k) is 1.But wait, in the initial problem, the user wrote "Find the minimum value of S(k)." So, the answer should be 1.But in the initial problem statement, the user also wrote "the sum of the areas of four parts enclosed by C₁, C₂ and two lines x=0, x=π." So, if S(k) is 1 when k=1, that's the minimum.But let me double-check with k=0.5, for example.If k=0.5, then x₁ = arcsin(0.25) ≈ 0.2527 radians.Compute S(k):S(k) = (0.5)^2 - 2*(0.5) + 2 = 0.25 - 1 + 2 = 1.25, which is larger than 1.Similarly, for k=2, S(k)=4 - 4 + 2=2, which is larger than 1.For k=0, S(k)=0 -0 +2=2.So, indeed, the minimum is at k=1, with S(k)=1.Therefore, the minimum value of S(k) is 1.</think>

🧠:<think>Okay, so I have this geometry problem here, and I need to prove that point Q is the circumcenter of triangle CDE. Let me try to visualize the setup first. There are two parallel lines, l₁ and l₂, and a circle ω that's tangent to both of them. Then there's another circle ω₁ that's tangent to l₁ at point A and externally tangent to ω at point C. Similarly, there's a third circle ω₂ that's tangent to l₂ at point B, externally tangent to ω at point D, and also externally tangent to ω₁ at point E. The lines AD and BC intersect at point Q, and I need to show that Q is the circumcenter of triangle CDE.Alright, let me break this down. First, since ω is tangent to both l₁ and l₂, which are parallel, ω must be sandwiched between these two lines. The distance between l₁ and l₂ should be twice the radius of ω, right? Because the circle is tangent to both lines, so the distance from the center of ω to each line is the radius, and since the lines are parallel, the total distance between them is twice the radius.Now, ω₁ is tangent to l₁ at A and externally tangent to ω at C. So, ω₁ is sitting above l₁, touching it at A, and also touching ω from the outside. Similarly, ω₂ is sitting below l₂, touching it at B, and also touching ω from the outside and ω₁ from the outside at E. So, ω₁ and ω₂ are both externally tangent to ω and to each other.I need to find the point Q where AD and BC intersect and show that it's the circumcenter of triangle CDE. The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle, so maybe I can show that Q lies on the perpendicular bisectors of CD, DE, and EC.Let me think about the properties of tangent circles. If two circles are externally tangent, the line connecting their centers passes through the point of tangency. So, for example, the center of ω and the center of ω₁ lie on the line connecting their centers, which passes through point C. Similarly, the centers of ω and ω₂ lie on the line connecting their centers, passing through D, and the centers of ω₁ and ω₂ lie on the line connecting their centers, passing through E.Maybe I can assign coordinates to these points to make things more concrete. Let's set up a coordinate system where l₁ is the x-axis and l₂ is the line y = 2r, where r is the radius of ω. Then, the center of ω, let's call it O, is at (0, r). Now, ω₁ is tangent to l₁ at A, so A must be a point on the x-axis. Let's say A is at (a, 0). The center of ω₁, let's call it O₁, must be at (a, r₁), where r₁ is the radius of ω₁. Since ω₁ is externally tangent to ω, the distance between their centers should be equal to the sum of their radii. So, the distance between O and O₁ is sqrt((a - 0)^2 + (r₁ - r)^2) = r + r₁.Squaring both sides, we get a² + (r₁ - r)² = (r + r₁)². Expanding both sides: a² + r₁² - 2r r₁ + r² = r² + 2r r₁ + r₁². Simplifying, a² - 2r r₁ = 2r r₁, so a² = 4r r₁. Therefore, a = 2√(r r₁). So, point A is at (2√(r r₁), 0).Similarly, for ω₂, which is tangent to l₂ at B. Let's denote the center of ω₂ as O₂, at (b, 2r - r₂), where r₂ is the radius of ω₂. The distance between O and O₂ is sqrt((b - 0)^2 + (2r - r₂ - r)^2) = sqrt(b² + (r - r₂)^2). Since ω₂ is externally tangent to ω, this distance should be equal to r + r₂. So, sqrt(b² + (r - r₂)^2) = r + r₂.Squaring both sides: b² + (r - r₂)^2 = (r + r₂)^2. Expanding: b² + r² - 2r r₂ + r₂² = r² + 2r r₂ + r₂². Simplifying, b² - 2r r₂ = 2r r₂, so b² = 4r r₂. Therefore, b = 2√(r r₂). So, point B is at (2√(r r₂), 2r).Next, ω₁ and ω₂ are externally tangent to each other at E. So, the distance between their centers O₁ and O₂ should be equal to r₁ + r₂. The coordinates of O₁ are (2√(r r₁), r₁) and O₂ are (2√(r r₂), 2r - r₂). The distance between O₁ and O₂ is sqrt[(2√(r r₂) - 2√(r r₁))² + (2r - r₂ - r₁)^2] = r₁ + r₂.That's a bit complicated, but maybe we can find a relationship between r₁ and r₂. Let me denote x = √(r r₁) and y = √(r r₂). Then, the x-coordinates of O₁ and O₂ become 2x and 2y, respectively, and the y-coordinates are r₁ and 2r - r₂.Wait, let's express r₁ and r₂ in terms of x and y. From earlier, we have x² = r r₁, so r₁ = x² / r. Similarly, y² = r r₂, so r₂ = y² / r.Now, the distance between O₁ and O₂ is sqrt[(2y - 2x)² + (2r - (y² / r) - (x² / r))²] = (x² / r) + (y² / r).Simplify the expression inside the square root:First, (2y - 2x)² = 4(y - x)².Second, (2r - (y² / r + x² / r)) = (2r² - x² - y²)/r. So, squaring this gives (2r² - x² - y²)² / r².Therefore, the distance squared is 4(y - x)² + (2r² - x² - y²)² / r², and this should equal (x² + y²)/r.Wait, that seems messy. Maybe there's another way. Since ω₁ and ω₂ are externally tangent, the line connecting their centers passes through E. So, perhaps E lies on the line O₁O₂.Alternatively, maybe instead of coordinates, I should think about homothety or inversion. Homothety is a scaling transformation that can map one circle to another, preserving tangency points. Since all the circles are tangent to ω, maybe there's a homothety center at the point where the tangents meet.Wait, the lines AD and BC intersect at Q. Maybe Q is the homothety center that maps ω₁ to ω₂ or something like that.Alternatively, perhaps considering radical axes. The radical axis of two circles is the locus of points with equal power concerning both circles. Since Q lies on AD and BC, which are radical axes of certain pairs of circles, maybe it has equal power concerning all three circles, making it the circumcenter.Let me think about the radical axis properties. The radical axis of ω and ω₁ is the line perpendicular to the line connecting their centers, passing through the points with equal power. Similarly for the radical axis of ω and ω₂. The intersection of these radical axes would be the radical center, which has equal power concerning all three circles. But the radical center is the center of the circle orthogonal to all three circles, which might not necessarily be the circumcenter.Wait, but the circumcenter is the intersection of the perpendicular bisectors. Maybe Q lies on the perpendicular bisectors of CD, DE, and EC. If I can show that QC = QD = QE, then Q is equidistant from C, D, and E, making it the circumcenter.So, how can I show that QC = QD = QE?Let me consider the power of point Q with respect to ω, ω₁, and ω₂. Since Q lies on AD, which is tangent to ω at C, the power of Q with respect to ω is QC². Similarly, since Q lies on BC, which is tangent to ω at D, the power of Q with respect to ω is QD². Therefore, QC² = QD², so QC = QD.Similarly, since Q lies on AD, which is tangent to ω₁ at C, the power of Q with respect to ω₁ is QC². And since Q lies on BC, which is tangent to ω₂ at D, the power of Q with respect to ω₂ is QD². But wait, that might not directly give me QE.Alternatively, since Q is the intersection of AD and BC, and these lines are common tangents to the circles, the power of Q with respect to each circle is equal to the square of the length of the tangent from Q to that circle.So, power of Q w.r. to ω is QC² = QD², as AD and BC are both tangents from Q to ω.Similarly, power of Q w.r. to ω₁ is QA² (since A is the tangency point on l₁), but also, since Q is on BC, which is tangent to ω₂ at D, maybe there's another relation.Wait, actually, ω₁ is tangent to ω at C, so the radical axis of ω and ω₁ is the common tangent at C. Similarly, the radical axis of ω and ω₂ is the common tangent at D. Therefore, the radical center of ω, ω₁, and ω₂ is the point where their radical axes meet, which is the intersection of the common tangents at C and D.But in this case, the radical axes are the common tangents at C and D, which are lines AD and BC, respectively. So, their intersection Q is the radical center, meaning it has equal power concerning all three circles.Therefore, the power of Q concerning ω, ω₁, and ω₂ is equal. So, QC² = QD² = QE², which implies QC = QD = QE. Hence, Q is equidistant from C, D, and E, meaning it's the circumcenter of triangle CDE.Wait, but I need to be careful here. The radical center has equal power concerning all three circles, which means the distances from Q to each circle are equal in terms of tangent lengths. However, the circumcenter requires that the distances from Q to the points C, D, and E are equal. But since C, D, and E are points on the circles, not necessarily the centers, does equal power imply equal distances?Hmm, not exactly. The power of a point concerning a circle is equal to the square of the tangent length from the point to the circle. So, if Q has equal power concerning ω, ω₁, and ω₂, then the lengths of the tangents from Q to each circle are equal. But the points C, D, and E are points of tangency on the circles, so the tangent lengths from Q to each circle would be QC, QD, and QE, respectively. Therefore, if QC² = QD² = QE², then QC = QD = QE, meaning Q is equidistant from C, D, and E, which makes it the circumcenter.Therefore, Q is the circumcenter of triangle CDE.I think that makes sense. Let me recap:1. Q lies on AD and BC, which are common tangents to ω, ω₁, and ω₂.2. The radical axes of the pairs (ω, ω₁), (ω, ω₂), and (ω₁, ω₂) are the common tangents at C, D, and E, respectively.3. The intersection of these radical axes is the radical center, which is Q.4. The radical center has equal power concerning all three circles, meaning QC² = QD² = QE².5. Therefore, Q is equidistant from C, D, and E, making it the circumcenter of triangle CDE.Yeah, that seems solid. I think that's the proof.</think>

🧠:<think>First, express 625 as a power of 5:[625 = 5^4]Rewrite the expression (625^{log_2 250}) as:[(5^4)^{log_2 250}]Apply the power of a power rule:[(5^4)^{log_2 250} = 5^{4 cdot log_2 250}]Simplify the exponent using the property of logarithms:[4 cdot log_2 250 = log_2 (250^4)]Rewrite the expression using the exponential and logarithmic relationship:[2^{log_2 (250^4)} = 250^4]Take the fourth root:[(250^4)^{frac{1}{4}} = 250]The final answer is (boxed{B}).</think>

🧠:<think>Okay, so I have this expression to evaluate:10 cdot 73 log _{3}left[(3+1)left(3^{2}+1right)left(3^{4}+1right) cdotsleft(3^{64}+1right)+frac{1}{2}right]+log _{3} 2Wow, that's a mouthful. Let me break it down step by step.First, I notice that the expression inside the logarithm is a product of terms like (3^{2^k} + 1) starting from k=0 up to some point. Let me see how many terms there are. The exponents are 1, 2, 4, ..., 64. So that's 1, 2, 4, 8, 16, 32, 64. That's 7 terms in total.So the product is (3+1)(3^2 + 1)(3^4 + 1)...(3^{64} + 1). Then, there's an additional frac{1}{2} being added to this product.The entire expression is inside a logarithm base 3, multiplied by 10*73, and then we add another log base 3 of 2.Hmm. Maybe I can simplify the product first. I remember that products of terms like (x - 1)(x + 1) = x^2 - 1. So maybe something similar can be applied here.Let me denote the product as P:P = (3+1)(3^2 + 1)(3^4 + 1)...(3^{64} + 1)If I multiply P by (3 - 1), which is 2, I get:(3 - 1)P = (3 - 1)(3 + 1)(3^2 + 1)(3^4 + 1)...(3^{64} + 1)Using the difference of squares, (3 - 1)(3 + 1) = 3^2 - 1 = 9 - 1 = 8.So now, (3^2 - 1)(3^2 + 1) = 3^4 - 1 = 81 - 1 = 80.Continuing this pattern:(3^4 - 1)(3^4 + 1) = 3^8 - 1And so on, each time doubling the exponent.So, after multiplying by (3 - 1), the product telescopes:(3 - 1)P = 3^{128} - 1Because each step doubles the exponent until we reach 64, which is 2^6, so the next exponent is 128.Therefore, solving for P:P = frac{3^{128} - 1}{2}So, the product P is equal to frac{3^{128} - 1}{2}.Now, going back to the original expression inside the logarithm, it's P plus frac{1}{2}:P + frac{1}{2} = frac{3^{128} - 1}{2} + frac{1}{2} = frac{3^{128} - 1 + 1}{2} = frac{3^{128}}{2}So, the argument of the logarithm simplifies to frac{3^{128}}{2}.Now, let's rewrite the entire expression:10 cdot 73 log_{3}left(frac{3^{128}}{2}right) + log_{3} 2I can split the logarithm into two parts:log_{3}left(frac{3^{128}}{2}right) = log_{3}(3^{128}) - log_{3}(2) = 128 - log_{3}(2)So, substituting back:10 cdot 73 cdot (128 - log_{3}(2)) + log_{3}(2)Let me compute 10 multiplied by 73 first:10 * 73 = 730.So now, the expression is:730 cdot (128 - log_{3}(2)) + log_{3}(2)Distribute the 730:730 cdot 128 - 730 cdot log_{3}(2) + log_{3}(2)Combine like terms:The terms with log_{3}(2):-730 cdot log_{3}(2) + log_{3}(2) = (-730 + 1) cdot log_{3}(2) = -729 cdot log_{3}(2)So now, the expression is:730 cdot 128 - 729 cdot log_{3}(2)Hmm, that seems a bit complicated. Let me see if I can simplify it further.Wait, maybe I made a mistake in the earlier steps. Let me verify.We had:log_{3}left(frac{3^{128}}{2}right) = 128 - log_{3}(2)So, 10*73*(128 - log3(2)) + log3(2)Which is 730*128 - 730*log3(2) + log3(2)That's correct.So, factoring log3(2):730*128 + (-730 + 1)*log3(2) = 730*128 - 729*log3(2)Hmm, that still leaves us with a term involving log3(2). But the answer choices are all integers: 32, 64, 128, or none of the above.Wait, maybe I made a miscalculation earlier.Let me go back.We had P = (3+1)(3^2 +1)...(3^64 +1) = (3^{128} -1)/2So, P + 1/2 = (3^{128} -1)/2 + 1/2 = (3^{128} -1 +1)/2 = 3^{128}/2So, that's correct.Then, log3(3^{128}/2) = log3(3^{128}) - log3(2) = 128 - log3(2)So, 10*73*(128 - log3(2)) + log3(2) = 730*(128 - log3(2)) + log3(2)Expanding: 730*128 - 730*log3(2) + log3(2) = 730*128 - 729*log3(2)Hmm, so unless log3(2) is a rational number that cancels out, this might not be an integer.Wait, log3(2) is irrational, so unless -729*log3(2) is equal to something, but I don't think so.Wait, maybe I made a mistake in the initial product.Wait, let's re-examine the product.We have P = (3+1)(3^2 +1)...(3^{64} +1). When we multiply by (3 -1), we get 3^{128} -1.Therefore, P = (3^{128} -1)/ (3 -1) = (3^{128} -1)/2So, P + 1/2 = (3^{128} -1)/2 + 1/2 = (3^{128})/2So, that's correct.So, log3(P +1/2) = log3(3^{128}/2) = 128 - log3(2)So, then 10*73*(128 - log3(2)) + log3(2) = 730*128 - 730*log3(2) + log3(2) = 730*128 - 729*log3(2)Hmm, so unless 729*log3(2) is an integer, which it is not, because log3(2) is irrational, this expression is not an integer. Therefore, the value is not among the options 32, 64, 128. So, the answer would be D) None of the above.Wait, but let me double-check. Maybe I missed something in the initial steps.Wait, let's compute 730*128:730 * 100 = 73,000730 * 28 = 20,440So, total is 73,000 + 20,440 = 93,440So, 730*128 = 93,440Then, 729*log3(2). Since log3(2) is approximately 0.6309, so 729*0.6309 ≈ 729*0.63 ≈ 458.67So, the expression is approximately 93,440 - 458.67 ≈ 92,981.33But the answer choices are 32, 64, 128. So, clearly, the value is not among those. Therefore, the answer is D) None of the above.Wait, but the initial problem had 10*73, which is 730, but maybe I misread the problem. Let me check:Yes, it's 10 multiplied by 73, so 730.Alternatively, maybe I misapplied the logarithm properties.Wait, let's see:log3(P +1/2) = log3(3^{128}/2) = log3(3^{128}) - log3(2) = 128 - log3(2)So, 10*73*(128 - log3(2)) + log3(2) = 730*(128 - log3(2)) + log3(2)Which is 730*128 - 730*log3(2) + log3(2) = 730*128 - (730 -1)*log3(2) = 730*128 - 729*log3(2)Yes, that's correct.So, unless I'm missing some simplification, this seems like the answer is not an integer, so D) None of the above.But wait, maybe I made a mistake in the product P. Let me re-examine that.We have P = (3+1)(3^2 +1)(3^4 +1)...(3^{64} +1)Multiply both sides by (3 -1):(3 -1)P = (3 -1)(3 +1)(3^2 +1)...(3^{64} +1) = (3^2 -1)(3^2 +1)(3^4 +1)...(3^{64} +1) = (3^4 -1)(3^4 +1)...(3^{64} +1) = ... = 3^{128} -1So, P = (3^{128} -1)/ (3 -1) = (3^{128} -1)/2So, P + 1/2 = (3^{128} -1)/2 + 1/2 = (3^{128})/2So, that's correct.Therefore, log3(3^{128}/2) = 128 - log3(2)Therefore, the expression is 730*(128 - log3(2)) + log3(2) = 730*128 - 729*log3(2)Which is approximately 93,440 - 458.67 ≈ 92,981.33, which is not an integer, so D) None of the above.Wait, but the answer choices are 32, 64, 128, or none. So, the answer is D.But just to be thorough, maybe I misapplied the logarithm property.Wait, log3(a/b) = log3(a) - log3(b). Yes, that's correct.So, log3(3^{128}/2) = 128 - log3(2). Correct.So, the expression is 730*(128 - log3(2)) + log3(2) = 730*128 - 729*log3(2)Which is 93,440 - 729*log3(2). Since log3(2) is irrational, this is not an integer. Therefore, the answer is D) None of the above.</think>