answer：Okay, so I have this problem where I need to find the conjugate of a complex number z, which is given as z = i divided by (1 + i). Hmm, complex numbers can be a bit tricky, especially when dealing with fractions like this. Let me try to break it down step by step.First, I remember that a complex number has a real part and an imaginary part, like a + bi, where a is the real part and b is the imaginary part. The conjugate of a complex number, denoted as (overline{z}), is where you change the sign of the imaginary part. So, if z is a + bi, then (overline{z}) is a - bi. That seems straightforward, but in this case, z is given as a fraction, so I need to simplify it first before finding its conjugate.Alright, z is equal to i over (1 + i). To simplify this, I think I need to get rid of the imaginary unit i in the denominator. I remember that one way to do this is by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of (1 + i) is (1 - i). So, if I multiply both the top and bottom by (1 - i), that should help me simplify the expression.Let me write that out:z = (frac{i}{1 + i}) × (frac{1 - i}{1 - i})Multiplying the numerators together and the denominators together, I get:z = (frac{i(1 - i)}{(1 + i)(1 - i)})Now, I need to carry out these multiplications. Starting with the numerator: i times (1 - i). Distributing i across the terms inside the parentheses:i × 1 = ii × (-i) = -i²So, the numerator becomes i - i². I remember that i² is equal to -1, so substituting that in:i - (-1) = i + 1So, the numerator simplifies to 1 + i.Now, moving on to the denominator: (1 + i)(1 - i). This is a difference of squares, so it should simplify nicely. Let me multiply it out:1 × 1 = 11 × (-i) = -ii × 1 = ii × (-i) = -i²Adding these together:1 - i + i - i²The -i and +i cancel each other out, leaving:1 - i²Again, since i² is -1, this becomes:1 - (-1) = 1 + 1 = 2So, the denominator simplifies to 2.Putting it all together, z is now:z = (frac{1 + i}{2})Which can be written as:z = (frac{1}{2} + frac{1}{2}i)Okay, so z is (frac{1}{2} + frac{1}{2}i). Now, to find the conjugate of z, (overline{z}), I just need to change the sign of the imaginary part. The real part remains the same, and the imaginary part, which is (frac{1}{2}i), becomes (-frac{1}{2}i).Therefore, the conjugate of z is:(overline{z} = frac{1}{2} - frac{1}{2}i)Looking back at the options provided:A: (frac{1}{2} + frac{1}{2}i)B: (1 + frac{1}{2}i)C: (1 - frac{1}{2}i)D: (frac{1}{2} - frac{1}{2}i)So, the correct answer should be option D.Wait, just to make sure I didn't make any mistakes. Let me double-check my steps.1. I started with z = i / (1 + i).2. Multiplied numerator and denominator by (1 - i), the conjugate of the denominator.3. Expanded the numerator: i(1 - i) = i - i² = i + 1.4. Expanded the denominator: (1 + i)(1 - i) = 1 - i² = 1 - (-1) = 2.5. So, z = (1 + i)/2 = 1/2 + (1/2)i.6. Conjugate of z is 1/2 - (1/2)i, which is option D.Everything seems to check out. I don't see any errors in my calculations. So, I'm confident that option D is the correct answer.

answer：Okay, so I have this triangle ABC where AB is 4 units, AC is 5 units, and the cosine of angle A is 1/10. I need to find the length of the angle bisector AD. Hmm, angle bisector... I remember there's a theorem about angle bisectors, maybe the Angle Bisector Theorem. Let me think.First, maybe I should sketch the triangle to visualize it better. So, triangle ABC with AB=4, AC=5, and angle A with cosine 1/10. Since cosine is positive, angle A must be acute. That makes sense.I think the first step is to find the length of BC. I can use the Law of Cosines for that. The Law of Cosines formula is:BC² = AB² + AC² - 2 * AB * AC * cos(angle A)Plugging in the values:BC² = 4² + 5² - 2 * 4 * 5 * (1/10)Calculating each term:4² is 16, 5² is 25, and 2 * 4 * 5 is 40. Then, 40 * (1/10) is 4.So, BC² = 16 + 25 - 4 = 37Therefore, BC = sqrt(37). Okay, so BC is the square root of 37. That's approximately 6.08 units, but I'll keep it as sqrt(37) for exactness.Now, since AD is the angle bisector of angle A, it splits BC into two segments, BD and DC, in the ratio of AB to AC. That's the Angle Bisector Theorem. So, BD/DC = AB/AC = 4/5.Let me denote BD as 4k and DC as 5k. Then, BD + DC = BC, which is sqrt(37). So:4k + 5k = sqrt(37)9k = sqrt(37)k = sqrt(37)/9Therefore, BD = 4k = 4*sqrt(37)/9 and DC = 5k = 5*sqrt(37)/9.Alright, so now I have BD and DC. How do I find AD? I think there's a formula for the length of an angle bisector. Let me recall. I believe it's called the Angle Bisector Length Formula.The formula for the length of an angle bisector from angle A is:AD = (2 * AB * AC * cos(angle A / 2)) / (AB + AC)Wait, is that correct? Hmm, I'm not entirely sure. Maybe I should derive it or look for another approach.Alternatively, I can use the Stewart's Theorem. Stewart's Theorem relates the lengths of the sides of the triangle to the length of a cevian (which AD is, since it's an angle bisector). The formula is:AB² * DC + AC² * BD = BC * (AD² + BD * DC)Let me write that down with the known values.AB is 4, AC is 5, BC is sqrt(37), BD is 4*sqrt(37)/9, DC is 5*sqrt(37)/9.So, plugging into Stewart's Theorem:4² * (5*sqrt(37)/9) + 5² * (4*sqrt(37)/9) = sqrt(37) * (AD² + (4*sqrt(37)/9)*(5*sqrt(37)/9))Calculating each term step by step.First, 4² is 16, multiplied by (5*sqrt(37)/9):16 * (5*sqrt(37)/9) = (80*sqrt(37))/9Next, 5² is 25, multiplied by (4*sqrt(37)/9):25 * (4*sqrt(37)/9) = (100*sqrt(37))/9Adding these two terms together:(80*sqrt(37))/9 + (100*sqrt(37))/9 = (180*sqrt(37))/9 = 20*sqrt(37)So, the left side of Stewart's Theorem is 20*sqrt(37).Now, the right side is sqrt(37) multiplied by (AD² + (4*sqrt(37)/9)*(5*sqrt(37)/9)).Let me compute the product (4*sqrt(37)/9)*(5*sqrt(37)/9):First, multiply the constants: 4*5 = 20Then, sqrt(37)*sqrt(37) = 37Denominator: 9*9 = 81So, the product is (20*37)/81 = 740/81Therefore, the right side becomes sqrt(37)*(AD² + 740/81)Putting it all together:20*sqrt(37) = sqrt(37)*(AD² + 740/81)Divide both sides by sqrt(37):20 = AD² + 740/81Now, solve for AD²:AD² = 20 - 740/81Convert 20 to 81 denominator:20 = 1620/81So, AD² = 1620/81 - 740/81 = (1620 - 740)/81 = 880/81Therefore, AD = sqrt(880/81) = sqrt(880)/9Simplify sqrt(880). Let's factor 880:880 = 16 * 55 = 16 * 5 * 11So, sqrt(880) = sqrt(16 * 55) = 4*sqrt(55)Thus, AD = 4*sqrt(55)/9Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting from Stewart's Theorem:AB² * DC + AC² * BD = BC*(AD² + BD*DC)Plugging in:16*(5*sqrt(37)/9) + 25*(4*sqrt(37)/9) = sqrt(37)*(AD² + (4*sqrt(37)/9)*(5*sqrt(37)/9))Left side:16*(5*sqrt(37)/9) = 80*sqrt(37)/925*(4*sqrt(37)/9) = 100*sqrt(37)/9Total left side: (80 + 100)*sqrt(37)/9 = 180*sqrt(37)/9 = 20*sqrt(37). That's correct.Right side:sqrt(37)*(AD² + (4*sqrt(37)/9)*(5*sqrt(37)/9))Compute BD*DC:(4*sqrt(37)/9)*(5*sqrt(37)/9) = (20*37)/81 = 740/81. Correct.So, right side is sqrt(37)*(AD² + 740/81)Thus, 20*sqrt(37) = sqrt(37)*(AD² + 740/81)Divide both sides by sqrt(37):20 = AD² + 740/81Convert 20 to 81 denominator:20 = 1620/81AD² = 1620/81 - 740/81 = 880/81AD = sqrt(880)/9Simplify sqrt(880):880 = 16*55, so sqrt(880) = 4*sqrt(55). Therefore, AD = 4*sqrt(55)/9.Hmm, that seems right. Let me think if there's another way to verify this.Alternatively, I can use the formula for the length of an angle bisector:AD = (2 * AB * AC * cos(angle A / 2)) / (AB + AC)But I don't know angle A / 2, but I know cos(angle A) = 1/10. Maybe I can find cos(angle A / 2) using the half-angle formula.The half-angle formula is:cos(angle A / 2) = sqrt[(1 + cos(angle A))/2]So, cos(angle A / 2) = sqrt[(1 + 1/10)/2] = sqrt[(11/10)/2] = sqrt(11/20) = sqrt(11)/sqrt(20) = sqrt(11)/(2*sqrt(5)) = sqrt(55)/10Wait, let me compute that step by step.(1 + 1/10) = 11/10Divide by 2: 11/20sqrt(11/20) = sqrt(11)/sqrt(20) = sqrt(11)/(2*sqrt(5)) = (sqrt(11)*sqrt(5))/(2*5) = sqrt(55)/10Yes, so cos(angle A / 2) = sqrt(55)/10Now, plug into the angle bisector formula:AD = (2 * AB * AC * cos(angle A / 2)) / (AB + AC)AB is 4, AC is 5, so:AD = (2 * 4 * 5 * (sqrt(55)/10)) / (4 + 5)Simplify numerator:2 * 4 * 5 = 4040 * (sqrt(55)/10) = 4 * sqrt(55)Denominator: 4 + 5 = 9So, AD = (4 * sqrt(55)) / 9Which matches the result from Stewart's Theorem. So, that's reassuring.Therefore, the length of angle bisector AD is 4*sqrt(55)/9.**Final Answer**The length of angle bisector overline{AD} is boxed{dfrac{4sqrt{55}}{9}}.

answer：Okay, so I have this problem about a parallelepiped being cut by three planes, resulting in eight hexahedrons. Each of these hexahedrons has all quadrilateral faces. It's given that a sphere can be circumscribed around one of these hexahedrons, and I need to prove that a sphere can be circumscribed around each of them. Hmm, interesting.First, let me try to visualize the setup. A parallelepiped is like a 3D figure formed by six parallelograms. When three planes cut it, each plane intersects two pairs of opposite faces and doesn't intersect the remaining two. So, each plane is slicing through the parallelepiped in a specific way, creating these hexahedrons. Each hexahedron has six faces, all of which are quadrilaterals.Now, it's given that one of these hexahedrons can have a sphere circumscribed around it. That means all its vertices lie on the surface of a sphere. I need to show that this property holds for all eight hexahedrons.I remember that in geometry, if a polyhedron is circumscribed around a sphere, it's called a circumscribed polyhedron, and all its vertices lie on a common sphere. For a hexahedron, which is a six-faced polyhedron, this would mean that all eight vertices lie on a sphere. Wait, actually, a hexahedron typically has six faces, but how many vertices does it have? Let me think. A cube, which is a type of hexahedron, has eight vertices. So, in this case, each hexahedron has eight vertices, all lying on a sphere if it's circumscribed.Given that one hexahedron is circumscribed, the others must also be. But how do I show that? Maybe there's some symmetry or property of the parallelepiped and the way the planes cut it that ensures this.Let me recall that a parallelepiped can be thought of as a three-dimensional figure formed by three pairs of parallel planes. When three additional planes cut it, each intersecting two pairs of opposite faces, they divide the parallelepiped into eight smaller hexahedrons. Each of these smaller hexahedrons is similar in some way because of the symmetry of the original parallelepiped.Wait, but the original parallelepiped isn't necessarily a cube or a rectangular prism; it's just a general parallelepiped, so the faces are parallelograms. That might complicate things a bit because the angles and edge lengths can vary.However, the key here might be that the cutting planes are arranged in such a way that each hexahedron is a "corner" piece of the original parallelepiped. So, each hexahedron is a smaller version, scaled down from the original, but maintaining the same structure.If one of these hexahedrons is circumscribed, meaning all its vertices lie on a sphere, then perhaps the others inherit this property due to the way the original parallelepiped is divided.Let me think about the properties of a circumscribed polyhedron. For a polyhedron to be circumscribed around a sphere, it must be tangent to the sphere at certain points, but in this case, it's the vertices that lie on the sphere, so it's more about the polyhedron being cyclic, meaning all its vertices lie on a sphere.In two dimensions, a quadrilateral is cyclic if and only if the sum of its opposite angles is 180 degrees. Maybe there's an analogous property in three dimensions for polyhedrons.But I'm not sure about that. Maybe I should think about the coordinates. If I assign coordinates to the vertices of the original parallelepiped, then the cutting planes can be defined by equations, and the vertices of each hexahedron can be determined. Then, if one hexahedron has all its vertices lying on a sphere, perhaps the others do as well because of the linear nature of the cuts.Wait, maybe it's about affine transformations. If the original parallelepiped can be transformed into a cube via an affine transformation, and the cutting planes correspond to planes in the cube, then the property of being circumscribed might be preserved under affine transformations.But affine transformations don't preserve spheres unless they're similarity transformations. So, that might not be the right approach.Alternatively, maybe I can use the fact that the hexahedrons are similar in some way. If one is circumscribed, then by symmetry, all must be.But I need a more rigorous approach.Let me consider the original parallelepiped. Let's denote it as ABCD A'B'C'D', where ABCD is the base and A'B'C'D' is the top face, with edges AA', BB', etc., connecting the corresponding vertices.Now, the three cutting planes each intersect two pairs of opposite faces. Let's say the first plane intersects the front and back faces, the second intersects the left and right faces, and the third intersects the top and bottom faces. Each plane cuts through the parallelepiped, creating intersections with the edges.These intersections divide each edge into segments, and the hexahedrons are formed by these divisions. Each hexahedron is bounded by parts of the original faces and the new faces created by the cutting planes.Since each cutting plane intersects two pairs of opposite faces, the resulting hexahedrons will each have three new faces from the cutting planes and three original faces from the parallelepiped.Now, if one of these hexahedrons is circumscribed, meaning all its vertices lie on a sphere, then perhaps the others must also be circumscribed because of the way the original parallelepiped is structured.Wait, maybe I can use the concept of dual polyhedrons or something related to symmetry.Alternatively, perhaps I can use the fact that if a polyhedron is circumscribed, then certain conditions on its edges or faces must hold, and these conditions are preserved across all eight hexahedrons due to the way they're formed.Let me think about the coordinates again. Suppose I place the original parallelepiped in a coordinate system with vertex A at the origin, and edges along the axes. Let’s denote the vectors defining the parallelepiped as **a**, **b**, and **c**. Then, any point inside the parallelepiped can be expressed as a linear combination of these vectors.The cutting planes can be defined by equations of the form α**a** + β**b** + γ**c** = d, where α, β, γ are constants. These planes intersect the edges of the parallelepiped at specific points, dividing the edges into segments.Each hexahedron will then have vertices that are combinations of these intersection points and the original vertices. If one hexahedron has all its vertices lying on a sphere, then the distances from the center of the sphere to each vertex must be equal.Given that the original parallelepiped is linear, and the cutting planes are linear, the positions of the vertices of each hexahedron are linear combinations of the original vertices. Therefore, if one hexahedron satisfies the condition of having all vertices equidistant from a center point, the others must as well because the transformations are linear and preserve the distances in a certain way.Wait, but linear transformations don't necessarily preserve distances unless they're isometries. So, maybe that's not the right path.Alternatively, perhaps I can use the fact that the hexahedrons are congruent. If the original parallelepiped is cut symmetrically by the three planes, then all eight hexahedrons are congruent, meaning they are identical in shape and size. Therefore, if one is circumscribed, all must be.But the problem doesn't specify that the cutting planes are symmetrically placed. It just says each plane intersects its own two pairs of opposite faces and doesn't intersect the remaining two. So, the cutting planes could be placed asymmetrically, leading to non-congruent hexahedrons.Hmm, that complicates things. So, maybe the hexahedrons aren't necessarily congruent, but they still must all be circumscribed.Wait, but the key is that the original parallelepiped is convex, and the cutting planes divide it into eight parts, each of which is a hexahedron. The fact that one is circumscribed might impose some condition on the others.Let me think about the properties of a circumscribed polyhedron. For a polyhedron to be circumscribed, it must have a circumsphere, meaning all its vertices lie on a sphere. In three dimensions, this is a non-trivial condition.But in this case, since the hexahedrons are formed by linear cuts, maybe the condition of one being circumscribed propagates to the others through some geometric constraints.Alternatively, perhaps I can use the concept of reciprocal figures or dual polyhedrons, but I'm not sure.Wait, maybe I can use the fact that the hexahedrons are related by translations or reflections. If the original parallelepiped is symmetric, then the hexahedrons might be images of each other under these symmetries, and thus if one is circumscribed, so are the others.But again, the problem doesn't specify that the cutting planes are symmetrically placed, so the hexahedrons might not be symmetric images of each other.Hmm, this is tricky. Maybe I need to think about the specific properties of the hexahedrons formed by the cutting planes.Each hexahedron has three original faces from the parallelepiped and three new faces from the cutting planes. All faces are quadrilaterals, which is given.If a sphere is circumscribed around one hexahedron, then all its vertices lie on a sphere. Let's denote this hexahedron as H1. Now, I need to show that all other hexahedrons H2, H3, ..., H8 also have all their vertices lying on some sphere.Perhaps I can consider the positions of the vertices relative to the original parallelepiped. Since the cutting planes are linear, the positions of the vertices of each hexahedron are determined by the intersections of these planes with the edges of the parallelepiped.If one hexahedron is circumscribed, then the distances from the center of the sphere to each of its vertices are equal. Given the linear nature of the cuts, maybe these distances impose conditions on the other hexahedrons, forcing their vertices to also lie on a sphere.Alternatively, maybe I can use the fact that the hexahedrons share edges and faces, so if one has a circumscribed sphere, the others must conform to this sphere as well.Wait, but the hexahedrons don't necessarily share the same center. Each might have its own center for the circumscribed sphere.Hmm, maybe I need to think about the coordinates more carefully. Let me assign coordinates to the original parallelepiped.Let’s say the original parallelepiped has vertices at points (0,0,0), (a,0,0), (0,b,0), (0,0,c), and their combinations. So, the eight vertices are all combinations of (0 or a, 0 or b, 0 or c).Now, the three cutting planes each intersect two pairs of opposite faces. Let's say the first plane intersects the faces at x = p, the second at y = q, and the third at z = r, where 0 < p < a, 0 < q < b, and 0 < r < c.These planes divide the parallelepiped into eight smaller hexahedrons, each occupying a corner of the original parallelepiped. Each hexahedron will have vertices at combinations of (0 or p, 0 or q, 0 or r), depending on the corner.Wait, actually, no. Each hexahedron will have vertices that are combinations of the original vertices and the intersection points on the edges.For example, one hexahedron will have vertices at (0,0,0), (p,0,0), (0,q,0), (0,0,r), (p,q,0), (p,0,r), (0,q,r), and (p,q,r). Similarly, the other hexahedrons will have vertices at the other corners, with coordinates involving a-p, b-q, c-r, etc.Now, if one of these hexahedrons, say the one with vertices at (0,0,0), (p,0,0), (0,q,0), (0,0,r), (p,q,0), (p,0,r), (0,q,r), and (p,q,r), is circumscribed, meaning all these eight points lie on a sphere.The equation of a sphere in 3D is (x - h)^2 + (y - k)^2 + (z - l)^2 = R^2, where (h,k,l) is the center and R is the radius.If all eight points lie on this sphere, then plugging each point into the equation must satisfy it.So, for (0,0,0): h^2 + k^2 + l^2 = R^2.For (p,0,0): (p - h)^2 + k^2 + l^2 = R^2.Similarly, for (0,q,0): h^2 + (q - k)^2 + l^2 = R^2.For (0,0,r): h^2 + k^2 + (r - l)^2 = R^2.And so on for the other points.Subtracting the first equation from the second, we get:(p - h)^2 - h^2 = 0 => p^2 - 2ph = 0 => p(p - 2h) = 0.Since p ≠ 0, we have h = p/2.Similarly, subtracting the first equation from the third:(q - k)^2 - k^2 = 0 => q^2 - 2qk = 0 => k = q/2.And subtracting the first equation from the fourth:(r - l)^2 - l^2 = 0 => r^2 - 2rl = 0 => l = r/2.So, the center of the sphere must be at (p/2, q/2, r/2).Now, plugging this back into the first equation:(p/2)^2 + (q/2)^2 + (r/2)^2 = R^2.So, R^2 = (p^2 + q^2 + r^2)/4.Now, let's check if the other points satisfy this equation.Take the point (p,q,0):(p - p/2)^2 + (q - q/2)^2 + (0 - r/2)^2 = (p/2)^2 + (q/2)^2 + (r/2)^2 = R^2. So, yes.Similarly, (p,0,r):(p/2)^2 + (0 - q/2)^2 + (r/2)^2 = R^2.And (0,q,r):(0 - p/2)^2 + (q/2)^2 + (r/2)^2 = R^2.Finally, (p,q,r):(p/2)^2 + (q/2)^2 + (r/2)^2 = R^2.So, all eight points lie on the sphere centered at (p/2, q/2, r/2) with radius sqrt(p^2 + q^2 + r^2)/2.Therefore, the hexahedron with vertices at (0,0,0), (p,0,0), (0,q,0), (0,0,r), etc., is indeed circumscribed by this sphere.Now, what about the other hexahedrons? Let's consider another hexahedron, say the one at the opposite corner, with vertices at (a,0,0), (a - p,0,0), (a, b - q,0), (a,0,c - r), (a - p, b - q,0), (a - p,0,c - r), (a, b - q,c - r), and (a - p, b - q,c - r).Wait, actually, the coordinates might be a bit different. Let me think.Each hexahedron is formed by the intersection of the three cutting planes. So, the first cutting plane at x = p divides the parallelepiped into two parts: one from x=0 to x=p, and the other from x=p to x=a. Similarly for y=q and z=r.Therefore, the eight hexahedrons are:1. (0,0,0) to (p,q,r)2. (p,0,0) to (a,q,r)3. (0,q,0) to (p,b,r)4. (0,0,r) to (p,q,c)5. (p,q,0) to (a,b,r)6. (p,0,r) to (a,q,c)7. (0,q,r) to (p,b,c)8. (p,q,r) to (a,b,c)Wait, actually, each hexahedron is defined by the intersection of the three half-spaces divided by the cutting planes. So, each hexahedron occupies a corner of the original parallelepiped, bounded by the cutting planes.Therefore, the eight hexahedrons are:1. x ≤ p, y ≤ q, z ≤ r2. x ≥ p, y ≤ q, z ≤ r3. x ≤ p, y ≥ q, z ≤ r4. x ≤ p, y ≤ q, z ≥ r5. x ≥ p, y ≥ q, z ≤ r6. x ≥ p, y ≤ q, z ≥ r7. x ≤ p, y ≥ q, z ≥ r8. x ≥ p, y ≥ q, z ≥ rEach of these has vertices that are combinations of the original vertices and the cutting planes.Now, if the first hexahedron (x ≤ p, y ≤ q, z ≤ r) is circumscribed, as we saw, it has a sphere centered at (p/2, q/2, r/2) with radius sqrt(p^2 + q^2 + r^2)/2.What about the second hexahedron (x ≥ p, y ≤ q, z ≤ r)? Its vertices are (p,0,0), (a,0,0), (p,q,0), (a,q,0), (p,0,r), (a,0,r), (p,q,r), (a,q,r).Let's check if these points lie on a sphere. Let's denote the center as (h,k,l). For the sphere to pass through all these points, the distances from (h,k,l) to each vertex must be equal.Let's compute the distances:1. From (p,0,0): (h - p)^2 + (k - 0)^2 + (l - 0)^2 = R^22. From (a,0,0): (h - a)^2 + k^2 + l^2 = R^23. From (p,q,0): (h - p)^2 + (k - q)^2 + l^2 = R^24. From (a,q,0): (h - a)^2 + (k - q)^2 + l^2 = R^25. From (p,0,r): (h - p)^2 + k^2 + (l - r)^2 = R^26. From (a,0,r): (h - a)^2 + k^2 + (l - r)^2 = R^27. From (p,q,r): (h - p)^2 + (k - q)^2 + (l - r)^2 = R^28. From (a,q,r): (h - a)^2 + (k - q)^2 + (l - r)^2 = R^2Now, let's subtract equation 1 from equation 2:(h - a)^2 - (h - p)^2 = 0Expanding:h^2 - 2ah + a^2 - (h^2 - 2ph + p^2) = 0Simplify:-2ah + a^2 + 2ph - p^2 = 0Factor:2h(p - a) + (a^2 - p^2) = 0Factor further:2h(p - a) + (a - p)(a + p) = 0Factor out (p - a):(p - a)(-2h + (a + p)) = 0Since p ≠ a, we have:-2h + a + p = 0 => h = (a + p)/2Similarly, subtract equation 1 from equation 3:(k - q)^2 - k^2 = 0 => q^2 - 2qk = 0 => k = q/2And subtract equation 1 from equation 5:(l - r)^2 - l^2 = 0 => r^2 - 2rl = 0 => l = r/2So, the center of the sphere for the second hexahedron must be at ((a + p)/2, q/2, r/2).Now, let's check if this center satisfies the distance to all points.Compute R^2 for the center ((a + p)/2, q/2, r/2):From (p,0,0):((a + p)/2 - p)^2 + (q/2 - 0)^2 + (r/2 - 0)^2= ((a - p)/2)^2 + (q/2)^2 + (r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Similarly, from (a,0,0):((a + p)/2 - a)^2 + (q/2 - 0)^2 + (r/2 - 0)^2= ((-a + p)/2)^2 + q^2/4 + r^2/4= (a - p)^2/4 + q^2/4 + r^2/4Same as above.From (p,q,0):((a + p)/2 - p)^2 + (q/2 - q)^2 + (r/2 - 0)^2= ((a - p)/2)^2 + (-q/2)^2 + (r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.From (a,q,0):((a + p)/2 - a)^2 + (q/2 - q)^2 + (r/2 - 0)^2= ((-a + p)/2)^2 + (-q/2)^2 + (r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.From (p,0,r):((a + p)/2 - p)^2 + (q/2 - 0)^2 + (r/2 - r)^2= ((a - p)/2)^2 + (q/2)^2 + (-r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.From (a,0,r):((a + p)/2 - a)^2 + (q/2 - 0)^2 + (r/2 - r)^2= ((-a + p)/2)^2 + (q/2)^2 + (-r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.From (p,q,r):((a + p)/2 - p)^2 + (q/2 - q)^2 + (r/2 - r)^2= ((a - p)/2)^2 + (-q/2)^2 + (-r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.From (a,q,r):((a + p)/2 - a)^2 + (q/2 - q)^2 + (r/2 - r)^2= ((-a + p)/2)^2 + (-q/2)^2 + (-r/2)^2= (a - p)^2/4 + q^2/4 + r^2/4Same.So, all points lie on the sphere centered at ((a + p)/2, q/2, r/2) with radius sqrt[(a - p)^2 + q^2 + r^2]/2.Wait, but this is different from the first hexahedron's sphere. The first hexahedron's sphere was centered at (p/2, q/2, r/2) with radius sqrt(p^2 + q^2 + r^2)/2.So, the second hexahedron has a different center and a different radius. But the problem states that a sphere can be circumscribed around one of the hexahedrons, not necessarily all. So, if the first hexahedron is circumscribed, the second one is also circumscribed, but with a different sphere.Wait, but the problem says "a sphere can be circumscribed around one of these hexahedrons. Prove that a sphere can be circumscribed around each of them." So, it doesn't specify that the same sphere, just that each has its own sphere.So, in this case, each hexahedron does have a circumscribed sphere, centered at the midpoint of the segment connecting the original vertex to the cutting plane intersection, with radius half the distance from that midpoint to the opposite vertex.Wait, but in the first hexahedron, the center was at (p/2, q/2, r/2), which is the midpoint between (0,0,0) and (p,q,r). Similarly, the second hexahedron's center was at ((a + p)/2, q/2, r/2), which is the midpoint between (a,0,0) and (p,q,r).So, in general, for each hexahedron, the center of its circumscribed sphere is the midpoint between the original vertex and the opposite vertex created by the cutting planes.Therefore, each hexahedron has a circumscribed sphere centered at the midpoint of the diagonal connecting its two opposite vertices.Since the original parallelepiped is convex and the cutting planes are linear, this property holds for all eight hexahedrons.Thus, if one hexahedron is circumscribed, all must be circumscribed because the same reasoning applies to each one individually.Therefore, the conclusion is that each of the eight hexahedrons can have a sphere circumscribed around it.

answer：First, I need to understand the ratio of work between Alice and Bob, which is 4:7. This means that for every 4 parts of work Alice does, Bob does 7 parts.Next, I'll determine the total number of parts by adding Alice's and Bob's parts together: 4 + 7 = 11 parts.The total area to be painted is 330 square feet. To find out how much area each part represents, I'll divide the total area by the total number of parts: 330 ÷ 11 = 30 square feet per part.Since Bob is responsible for 7 parts of the work, I'll multiply the area per part by the number of parts Bob does: 30 × 7 = 210 square feet.Therefore, Bob paints 210 square feet of the house.

answer：Okay, so I have this problem: For how many integer values of ( a ) does the equation ( x^2 + ax + 12a = 0 ) have integer solutions for ( x )? Hmm, let me try to figure this out step by step.First, I remember that if a quadratic equation ( x^2 + bx + c = 0 ) has integer solutions, then those solutions must be integers that multiply to ( c ) and add up to ( -b ). This is from Vieta's formulas, right? So in this case, the equation is ( x^2 + ax + 12a = 0 ). Let me denote the integer roots as ( m ) and ( n ). Then, according to Vieta's formulas, we have:1. ( m + n = -a )2. ( mn = 12a )Hmm, so I have these two equations. Maybe I can manipulate them to find something useful. Let me try substituting ( a ) from the first equation into the second one.From the first equation, ( a = -(m + n) ). Plugging this into the second equation:( mn = 12(- (m + n)) )( mn = -12m - 12n )Let me rearrange this equation:( mn + 12m + 12n = 0 )Hmm, this looks a bit messy. Maybe I can factor it somehow. I remember something called Simon's Favorite Factoring Trick, where you add a constant to both sides to make it factorable. Let me try that.If I add 144 to both sides:( mn + 12m + 12n + 144 = 144 )Now, let's see if this factors nicely. It looks like:( (m + 12)(n + 12) = 144 )Oh, nice! So now, ( (m + 12) ) and ( (n + 12) ) are integers that multiply to 144. That means they are pairs of divisors of 144. So, I need to find all pairs of integers ( (d, e) ) such that ( d times e = 144 ). Each pair will give me possible values for ( m + 12 ) and ( n + 12 ), which I can then use to find ( m ) and ( n ), and subsequently ( a ).Alright, let's list all the pairs of divisors of 144. Since 144 is positive, the pairs can be both positive and negative. Let me list them:Positive divisor pairs:1. (1, 144)2. (2, 72)3. (3, 48)4. (4, 36)5. (6, 24)6. (8, 18)7. (9, 16)8. (12, 12)Negative divisor pairs:1. (-1, -144)2. (-2, -72)3. (-3, -48)4. (-4, -36)5. (-6, -24)6. (-8, -18)7. (-9, -16)8. (-12, -12)Wait, but actually, each positive pair can also be paired with a negative pair, right? Because if ( d times e = 144 ), then ( (-d) times (-e) = 144 ) as well. So, for each positive pair, there is a corresponding negative pair. So, in total, how many pairs do we have?From the positive list, we have 8 pairs, and each has a corresponding negative pair, so that's another 8, making a total of 16 pairs. Hmm, but wait, actually, each positive pair is unique, and each negative pair is unique, so 8 positive and 8 negative, totaling 16.But hold on, let me make sure. For example, for the pair (1, 144), the corresponding negative pair is (-1, -144). Similarly, (2, 72) corresponds to (-2, -72), and so on. So, yes, 8 positive and 8 negative, making 16 pairs in total.Now, for each of these pairs, I can find ( m ) and ( n ) by subtracting 12 from each component. So, for each pair ( (d, e) ), we have:( m = d - 12 )( n = e - 12 )Then, using Vieta's formula, ( a = -(m + n) ). Let's compute ( a ) for each pair.Starting with the positive pairs:1. (1, 144): ( m = 1 - 12 = -11 ) ( n = 144 - 12 = 132 ) ( a = -(-11 + 132) = -(121) = -121 )2. (2, 72): ( m = 2 - 12 = -10 ) ( n = 72 - 12 = 60 ) ( a = -(-10 + 60) = -(50) = -50 )3. (3, 48): ( m = 3 - 12 = -9 ) ( n = 48 - 12 = 36 ) ( a = -(-9 + 36) = -(27) = -27 )4. (4, 36): ( m = 4 - 12 = -8 ) ( n = 36 - 12 = 24 ) ( a = -(-8 + 24) = -(16) = -16 )5. (6, 24): ( m = 6 - 12 = -6 ) ( n = 24 - 12 = 12 ) ( a = -(-6 + 12) = -(6) = -6 )6. (8, 18): ( m = 8 - 12 = -4 ) ( n = 18 - 12 = 6 ) ( a = -(-4 + 6) = -(2) = -2 )7. (9, 16): ( m = 9 - 12 = -3 ) ( n = 16 - 12 = 4 ) ( a = -(-3 + 4) = -(1) = -1 )8. (12, 12): ( m = 12 - 12 = 0 ) ( n = 12 - 12 = 0 ) ( a = -(0 + 0) = 0 )Now, moving on to the negative pairs:1. (-1, -144): ( m = -1 - 12 = -13 ) ( n = -144 - 12 = -156 ) ( a = -(-13 + (-156)) = -(-169) = 169 )Wait, hold on, that doesn't seem right. Let me double-check that calculation.Wait, ( m = -1 - 12 = -13 ) is correct, and ( n = -144 - 12 = -156 ) is also correct. Then, ( m + n = -13 + (-156) = -169 ), so ( a = -(-169) = 169 ). Yeah, that's correct.2. (-2, -72): ( m = -2 - 12 = -14 ) ( n = -72 - 12 = -84 ) ( a = -(-14 + (-84)) = -(-98) = 98 )3. (-3, -48): ( m = -3 - 12 = -15 ) ( n = -48 - 12 = -60 ) ( a = -(-15 + (-60)) = -(-75) = 75 )4. (-4, -36): ( m = -4 - 12 = -16 ) ( n = -36 - 12 = -48 ) ( a = -(-16 + (-48)) = -(-64) = 64 )5. (-6, -24): ( m = -6 - 12 = -18 ) ( n = -24 - 12 = -36 ) ( a = -(-18 + (-36)) = -(-54) = 54 )6. (-8, -18): ( m = -8 - 12 = -20 ) ( n = -18 - 12 = -30 ) ( a = -(-20 + (-30)) = -(-50) = 50 )7. (-9, -16): ( m = -9 - 12 = -21 ) ( n = -16 - 12 = -28 ) ( a = -(-21 + (-28)) = -(-49) = 49 )8. (-12, -12): ( m = -12 - 12 = -24 ) ( n = -12 - 12 = -24 ) ( a = -(-24 + (-24)) = -(-48) = 48 )Wait, hold on, let me make sure I didn't make a mistake in calculating ( a ) for the negative pairs.For (-1, -144):( m = -1 -12 = -13 )( n = -144 -12 = -156 )( m + n = -13 + (-156) = -169 )( a = -(-169) = 169 ). That's correct.Similarly, for (-2, -72):( m = -2 -12 = -14 )( n = -72 -12 = -84 )( m + n = -14 + (-84) = -98 )( a = -(-98) = 98 ). Correct.Continuing:(-3, -48):( m = -3 -12 = -15 )( n = -48 -12 = -60 )( m + n = -15 + (-60) = -75 )( a = -(-75) = 75 ). Correct.(-4, -36):( m = -4 -12 = -16 )( n = -36 -12 = -48 )( m + n = -16 + (-48) = -64 )( a = -(-64) = 64 ). Correct.(-6, -24):( m = -6 -12 = -18 )( n = -24 -12 = -36 )( m + n = -18 + (-36) = -54 )( a = -(-54) = 54 ). Correct.(-8, -18):( m = -8 -12 = -20 )( n = -18 -12 = -30 )( m + n = -20 + (-30) = -50 )( a = -(-50) = 50 ). Correct.(-9, -16):( m = -9 -12 = -21 )( n = -16 -12 = -28 )( m + n = -21 + (-28) = -49 )( a = -(-49) = 49 ). Correct.(-12, -12):( m = -12 -12 = -24 )( n = -12 -12 = -24 )( m + n = -24 + (-24) = -48 )( a = -(-48) = 48 ). Correct.Okay, so compiling all the values of ( a ) we got:From positive pairs:-121, -50, -27, -16, -6, -2, -1, 0From negative pairs:169, 98, 75, 64, 54, 50, 49, 48Wait, hold on, I notice that in the negative pairs, one of the ( a ) values is 50, which is the same as in the positive pairs? Wait, no, in the positive pairs, we have -50, and in the negative pairs, we have 50. So, they are different. So, all these ( a ) values are unique.Let me list them all:From positive pairs:-121, -50, -27, -16, -6, -2, -1, 0From negative pairs:169, 98, 75, 64, 54, 50, 49, 48So, that's 8 from positive and 8 from negative, totaling 16 distinct values.Wait, let me check if any of these are duplicates. Looking at the positive pairs, all are negative except 0, and the negative pairs are all positive. So, no overlaps. So, 16 distinct integer values of ( a ).Therefore, the number of integer values of ( a ) is 16.**Final Answer**The number of integer values of ( a ) is boxed{16}.

answer：Okay, so I have this problem here: I need to find the fifth-largest divisor of 1,209,600,000. The options are given as (A) 75,600,000; (B) 151,200,000; (C) 302,400,000; (D) 604,800,000; and (E) 1,209,600,000. The largest divisor is obviously the number itself, which is option E. But I need to figure out what the fifth-largest divisor is.First, I remember that to find the largest divisors of a number, it's helpful to look at its prime factorization. If I can break down 1,209,600,000 into its prime factors, I can then determine the divisors more easily.Let me start by factorizing 1,209,600,000. Hmm, 1,209,600,000 looks like a large number, but maybe I can break it down step by step. I notice that it ends with six zeros, so it's divisible by 10^6, which is 1,000,000. Let me divide 1,209,600,000 by 1,000,000 to simplify it. 1,209,600,000 ÷ 1,000,000 = 1,209.6Wait, that's not an integer. Maybe I made a mistake. Let me try dividing by 100,000 instead. 1,209,600,000 ÷ 100,000 = 12,096Okay, 12,096 is an integer. So, 1,209,600,000 = 12,096 × 100,000. Now, I can factorize 12,096 and 100,000 separately.Starting with 12,096. Let's divide by 2 repeatedly:12,096 ÷ 2 = 6,048 6,048 ÷ 2 = 3,024 3,024 ÷ 2 = 1,512 1,512 ÷ 2 = 756 756 ÷ 2 = 378 378 ÷ 2 = 189So, that's six divisions by 2. So, 12,096 = 2^6 × 189.Now, factorizing 189. Let's divide by 3:189 ÷ 3 = 63 63 ÷ 3 = 21 21 ÷ 3 = 7So, 189 = 3^3 × 7.Putting it all together, 12,096 = 2^6 × 3^3 × 7.Now, let's factorize 100,000. 100,000 is 10^5, which is (2×5)^5 = 2^5 × 5^5.So, combining both parts, 1,209,600,000 = 12,096 × 100,000 = (2^6 × 3^3 × 7) × (2^5 × 5^5) = 2^(6+5) × 3^3 × 5^5 × 7 = 2^11 × 3^3 × 5^5 × 7.Wait, hold on, earlier I thought it was 2^10, but now I have 2^11. Let me double-check my calculations.Wait, 12,096 ÷ 2 six times gives me 189, which is correct. 100,000 is 2^5 × 5^5. So, total 2s are 6 + 5 = 11. So, 2^11. Hmm, okay, so my initial factorization was wrong. It's 2^11 × 3^3 × 5^5 × 7.But wait, the original number is 1,209,600,000. Let me check if 2^11 × 3^3 × 5^5 × 7 equals that.Calculating 2^11 = 2048; 3^3 = 27; 5^5 = 3125; 7 is 7.Multiplying them together: 2048 × 27 = 55,296; 55,296 × 3125 = let's see, 55,296 × 1000 = 55,296,000; 55,296 × 2000 = 110,592,000; 55,296 × 125 = 6,912,000. So, adding up: 110,592,000 + 6,912,000 = 117,504,000. Then, 117,504,000 × 7 = 822,528,000. Wait, that's not 1,209,600,000. So, I must have messed up somewhere.Wait, maybe I made a mistake in the initial division. Let me try another approach.1,209,600,000. Let's divide by 1,000,000 first: 1,209,600,000 ÷ 1,000,000 = 1,209.6. Hmm, not an integer. Maybe divide by 100,000: 1,209,600,000 ÷ 100,000 = 12,096. That's correct. So, 12,096 × 100,000.But 12,096 × 100,000 = 1,209,600,000. So, 12,096 is correct.Wait, but when I factorized 12,096, I got 2^6 × 3^3 × 7. Let me check that.12,096 ÷ 2 = 6,048 6,048 ÷ 2 = 3,024 3,024 ÷ 2 = 1,512 1,512 ÷ 2 = 756 756 ÷ 2 = 378 378 ÷ 2 = 189 So, that's six 2s, correct. 189 ÷ 3 = 63 63 ÷ 3 = 21 21 ÷ 3 = 7 So, three 3s and one 7. So, 2^6 × 3^3 × 7.Then, 100,000 is 2^5 × 5^5.So, total is 2^(6+5) × 3^3 × 5^5 × 7 = 2^11 × 3^3 × 5^5 × 7.But when I multiplied it out, I got 822,528,000, which is less than 1,209,600,000. So, something's wrong.Wait, maybe I made a mistake in the multiplication. Let me try again.2^11 = 2048 3^3 = 27 5^5 = 3125 7 = 7So, 2048 × 27 = 55,296 55,296 × 3125: Let's do 55,296 × 1000 = 55,296,000 55,296 × 2000 = 110,592,000 55,296 × 125 = 6,912,000 So, 110,592,000 + 6,912,000 = 117,504,000 117,504,000 × 7 = 822,528,000Hmm, that's still not matching. Maybe my prime factorization is wrong.Wait, 1,209,600,000. Let me try another approach. Maybe it's 1,209.6 million. 1,209.6 million is 1,209,600,000.Alternatively, perhaps I can factorize 1,209,600,000 as 1,2096 × 10^5.Wait, 1,2096 is 12,096. So, same as before.Wait, maybe I should use a different method. Let me try dividing by 10 repeatedly.1,209,600,000 ÷ 10 = 120,960,000 120,960,000 ÷ 10 = 12,096,000 12,096,000 ÷ 10 = 1,209,600 1,209,600 ÷ 10 = 120,960 120,960 ÷ 10 = 12,096 12,096 ÷ 10 = 1,209.6Wait, that's not helpful because it's not an integer after five divisions. Hmm.Maybe I should factorize 1,209,600,000 as 1,2096 × 10^5, but 1,2096 is 12,096, which we already saw.Wait, maybe I made a mistake in the multiplication earlier. Let me try multiplying 2^11 × 3^3 × 5^5 × 7 again.2^11 = 2048 3^3 = 27 5^5 = 3125 7 = 7So, 2048 × 27 = 55,296 55,296 × 3125: Let's break it down.55,296 × 3125 = 55,296 × (3000 + 125) = 55,296 × 3000 + 55,296 × 12555,296 × 3000 = 165,888,000 55,296 × 125: Let's calculate 55,296 × 100 = 5,529,600; 55,296 × 25 = 1,382,400 So, 5,529,600 + 1,382,400 = 6,912,000 Adding to 165,888,000: 165,888,000 + 6,912,000 = 172,800,000 Now, multiply by 7: 172,800,000 × 7 = 1,209,600,000Ah, okay, so that works out. So, my prime factorization is correct: 2^11 × 3^3 × 5^5 × 7.Now, to find the divisors, especially the largest ones, I need to consider the exponents of the prime factors. The largest divisor is the number itself, which is 1,209,600,000.The next largest divisors would be obtained by dividing the number by the smallest prime factors first. Since 2 is the smallest prime factor, dividing by 2 will give the next largest divisor.So, the second largest divisor is 1,209,600,000 ÷ 2 = 604,800,000.Then, the third largest would be 604,800,000 ÷ 2 = 302,400,000.Fourth largest: 302,400,000 ÷ 2 = 151,200,000.Fifth largest: 151,200,000 ÷ 2 = 75,600,000.Wait, but let me think if this is always the case. Dividing by the smallest prime factor each time gives the next largest divisor? I think so because each time you're reducing the number by the smallest possible factor, thus keeping the result as large as possible.But let me verify. Suppose I have a number N = p1^a × p2^b × p3^c..., where p1 < p2 < p3... are primes. Then, the largest divisor is N. The next largest would be N ÷ p1, then N ÷ p1^2, and so on, until you can't divide by p1 anymore, then you move to p2.But in this case, since 2 is the smallest prime, and we have a high exponent of 2 (11), we can divide by 2 multiple times before moving to the next prime.So, starting from N:1. N = 2^11 × 3^3 × 5^5 × 7 2. N ÷ 2 = 2^10 × 3^3 × 5^5 × 7 3. N ÷ 2^2 = 2^9 × 3^3 × 5^5 × 7 4. N ÷ 2^3 = 2^8 × 3^3 × 5^5 × 7 5. N ÷ 2^4 = 2^7 × 3^3 × 5^5 × 7 6. N ÷ 2^5 = 2^6 × 3^3 × 5^5 × 7 7. N ÷ 2^6 = 2^5 × 3^3 × 5^5 × 7 8. N ÷ 2^7 = 2^4 × 3^3 × 5^5 × 7 9. N ÷ 2^8 = 2^3 × 3^3 × 5^5 × 7 10. N ÷ 2^9 = 2^2 × 3^3 × 5^5 × 7 11. N ÷ 2^10 = 2^1 × 3^3 × 5^5 × 7 12. N ÷ 2^11 = 3^3 × 5^5 × 7 After that, the next largest divisor would be N ÷ (2^11 × 3) = 3^2 × 5^5 × 7, which is smaller than N ÷ 2^10.So, the order of divisors from largest to smallest would be:1. N 2. N ÷ 2 3. N ÷ 2^2 4. N ÷ 2^3 5. N ÷ 2^4 6. N ÷ 2^5 7. N ÷ 2^6 8. N ÷ 2^7 9. N ÷ 2^8 10. N ÷ 2^9 11. N ÷ 2^10 12. N ÷ 2^11 13. N ÷ (2^11 × 3) ... and so on.Therefore, the fifth largest divisor is N ÷ 2^4.Calculating N ÷ 2^4:N = 1,209,600,000 2^4 = 16 1,209,600,000 ÷ 16 = 75,600,000.So, the fifth largest divisor is 75,600,000, which is option A.Wait, but let me make sure I didn't skip any larger divisors by only dividing by 2. For example, could there be a divisor larger than N ÷ 2^4 that isn't just N divided by a higher power of 2?For instance, what if I divide by 3 instead of 2 at some point? Would that give a larger divisor?Let's see. The next largest divisor after N ÷ 2^10 would be N ÷ (2^11 × 3), which is much smaller. Similarly, if I divide by 5 instead of 2, that would also result in a much smaller number. So, indeed, the largest divisors are obtained by dividing by the smallest prime factor as much as possible.Therefore, my initial conclusion seems correct. The fifth largest divisor is 75,600,000.