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M25-08

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M25-08  [#permalink]

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New post 16 Sep 2014, 01:23
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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. \(\frac{x}{2}(\sqrt{2}-1)\)

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Re M25-08  [#permalink]

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New post 16 Sep 2014, 01:23
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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 \(gap=\frac{Diagonal-Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)\).

Since \(x=10\) then \(5(\sqrt{3}-1)=\frac{x}{2}(\sqrt{3} - 1)\).


Answer: D
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Re: M25-08  [#permalink]

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New post 08 Oct 2014, 05:50
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 * (a-b)
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Re: M25-08  [#permalink]

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New post 08 Oct 2014, 14:39
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(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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Re: M25-08  [#permalink]

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New post 08 Oct 2014, 14:56
temoalta wrote:
Could someone please explain why E is not correct?

To me, here we are dealing with a right triangle with dimensions x-x-xsqrt(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.
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Re: M25-08  [#permalink]

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New post 08 Oct 2014, 15:16
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 x-x-xsqrt(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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Re: M25-08  [#permalink]

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New post 08 Oct 2014, 15:25
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 x-x-xsqrt(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.


You are considering wrong right triangle.
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Re: M25-08  [#permalink]

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New post 09 Oct 2014, 08:03
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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New post 09 Oct 2014, 08:16
temoalta wrote:
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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New post 09 Oct 2014, 08:27
I understand now. I thought that vertices were the edges of the cube. Thanks.
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New post 04 Mar 2015, 01:32
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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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New post 04 Mar 2015, 03:52
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New post 30 Dec 2015, 19:59
I think this is a high-quality question and I agree with explanation.
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New post 25 Mar 2016, 21:35
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The sphere and cube arrangement will be as shown in figure.

1/2(Diagonal-Diameter) = 1/2 (a sqrt3 -a)= a/2 (sqrt3-1)
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New post 18 Jan 2017, 09:43
I think this is a high-quality question and I agree with explanation.
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Re: M25-08  [#permalink]

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New post 22 May 2018, 08:56
"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: M25-08  [#permalink]

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New post 09 Jun 2018, 09:58
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 x-x-xsqrt(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 3-D plane. A normal square would have a diagonal of \(x\sqrt{2}\), but since this is a cube the diagonal is \(x\sqrt{3}\).
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Re: M25-08 &nbs [#permalink] 09 Jun 2018, 09:58
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