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# M25-08

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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16 Sep 2014, 00:23
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Difficulty:

65% (hard)

Question Stats:

55% (01:10) correct 45% (01:17) wrong based on 457 sessions

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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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16 Sep 2014, 00:23
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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 $$5(\sqrt{3}-1)=\frac{x}{2}(\sqrt{3} - 1)$$.

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Schools: Anderson '17
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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 * (a-b)
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Joined: 08 Oct 2014
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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 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).
Math Expert
Joined: 02 Sep 2009
Posts: 52294

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08 Oct 2014, 13:56
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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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 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.
Math Expert
Joined: 02 Sep 2009
Posts: 52294

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08 Oct 2014, 14:25
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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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
Intern
Joined: 08 Oct 2014
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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
1
I think this question is good and helpful.
Could anyone explain the above with the help of a diagram for better understanding?
Math Expert
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Posts: 52294

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04 Mar 2015, 02:52
schak2rhyme wrote:
I think this question is good and helpful.
Could anyone explain the above with the help of a diagram for better understanding?

Check the links in my post above.
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Joined: 20 Dec 2015
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30 Dec 2015, 18:59
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 24 May 2013
Posts: 79

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25 Mar 2016, 20:35
2
The sphere and cube arrangement will be as shown in figure.

1/2(Diagonal-Diameter) = 1/2 (a sqrt3 -a)= a/2 (sqrt3-1)
Attachments

Sphere.png [ 9.32 KiB | Viewed 6915 times ]

Intern
Joined: 12 May 2016
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18 Jan 2017, 08:43
I think this is a high-quality question and I agree with explanation.
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Posts: 192
Location: India
Concentration: General Management, Healthcare
GMAT 1: 640 Q40 V38
GMAT 2: 680 Q48 V35
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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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Chasing the dragon

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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 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 $$x\sqrt{3}$$.
Re: M25-08 &nbs [#permalink] 09 Jun 2018, 08:58
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# M25-08

Moderators: chetan2u, Bunuel

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