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Joined: 02 Sep 2009
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Re M2508
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16 Sep 2014, 00:23
Official Solution:A sphere is inscribed in a cube with an edge of \(x\) centimeters. In terms of x what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? A. \(x(\sqrt{3}1)\) B. \(\frac{x}{2}\) C. \(x(\sqrt{2}1)\) D. \(\frac{x}{2}(\sqrt{3}1)\) E. \(\frac{x}{2}(\sqrt{2}1)\) Say \(x=10\) centimeters. Then, since a sphere is inscribed in cube then the edge of the cube equals to the diameter of a sphere, thus \(Diameter=10\). Next, diagonal of a cube equals to \(Diagonal=\sqrt{10^2+10^2+10^2}=10\sqrt{3}\). Now half of Diagonal minus Diameter is gap between the vertex of a cube and the surface of the sphere, thus \(gap=\frac{DiagonalDiameter}{2}=\frac{10*\sqrt{3}10}{2}=5(\sqrt{3}1)\). Since \(x=10\) then \(5(\sqrt{3}1)=\frac{x}{2}(\sqrt{3}  1)\). Answer: D
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Re: M2508
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08 Oct 2014, 04:50
Cube length = x cm
therefore, diagonal of the cube  a = x*sqrt(3) diameter of the sphere inscribed  b = x * sqrt(2)
distance left is a  b. this distance is the distance between both parts which are left, either side of the diameter of the sphere. so the distance between the point is half the distance. hence, 1/2 * (ab)



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Re: M2508
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08 Oct 2014, 13:39
Could someone please explain why E is not correct?
To me, here we are dealing with a right triangle with dimensions xxxsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)1).



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08 Oct 2014, 13:56



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Re: M2508
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08 Oct 2014, 14:16
Bunuel wrote: temoalta wrote: Could someone please explain why E is not correct?
To me, here we are dealing with a right triangle with dimensions xxxsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)1). We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face. I do understand we have a right triangle. I guess I don't see what is wrong with my logic.



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08 Oct 2014, 14:25



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Re: M2508
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09 Oct 2014, 07:03
Isn't that formula used to calculate the greatest distance? The question is asking for the shortest possible distance. What is the correct triangle we should we use here?



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09 Oct 2014, 07:16



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09 Oct 2014, 07:27
I understand now. I thought that vertices were the edges of the cube. Thanks.



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04 Mar 2015, 00:32
I think this question is good and helpful. Could anyone explain the above with the help of a diagram for better understanding?



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30 Dec 2015, 18:59
I think this is a highquality question and I agree with explanation.



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25 Mar 2016, 20:35
The sphere and cube arrangement will be as shown in figure. 1/2(DiagonalDiameter) = 1/2 (a sqrt3 a)= a/2 (sqrt31)
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Sphere.png [ 9.32 KiB  Viewed 6915 times ]



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18 Jan 2017, 08:43
I think this is a highquality question and I agree with explanation.



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Re: M2508
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22 May 2018, 07:56
"half of Diagonal minus Diameter is gap between the vertex of a cube and the surface of the sphere" Shouldn't it be 0.5 (Diagonal)  Radius of the sphere As the diagonal will have two segments which are not part of the diameter?
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Re: M2508
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09 Jun 2018, 08:58
temoalta wrote: Bunuel wrote: temoalta wrote: Could someone please explain why E is not correct?
To me, here we are dealing with a right triangle with dimensions xxxsqrt(2). x is the diameter of the sphere. If I subtract xsqrt(2)x I would get the distance between two of the vertices and the sphere. Then I divide that by 2 to get just one side, and so it would be x/2(sqrt(2)1). We are dealing with a right triangle where the hypotenuse is the diagonal of the cube, and the legs are the edge and the diagonal of a face. I do understand we have a right triangle. I guess I don't see what is wrong with my logic. Why it is a 30:60:90 right triangle and not a 45:45:90 right triangle is because of the 3D plane. A normal square would have a diagonal of \(x\sqrt{2}\), but since this is a cube the diagonal is \(x\sqrt{3}\).










