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Sphere is inscribed in a cube

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Sphere is inscribed in a cube

by crackgmat007 » Sat May 30, 2009 9:30 pm
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
a. 10( sqrt 3 – 1)

b. 5

c. 10( sqrt 2 – 1)

d. 5( sqrt 3 – 1)

e. 5( sqrt 2 – 1)

OA - D...pls expliain. tx
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by raleigh » Sun May 31, 2009 1:16 am
First, note that what you are looking for is the difference between [the distance between the center of the cube and a a vertex] and the radius of the sphere.

Distance formula in R^3 is sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2).

Make the center of the cube the origin. Then to get to a vertex you'd have to go 5 in each direction so the distance between the center of the cube and a vertex is sqrt(25 + 25 + 25) = sqrt(75) = 5sqrt(3).

The radius of the sphere is 5.

The difference is 5 sqrt(3) - 5 = 5(sqrt 3 - 1). The answer is D.

If you aren't comfortable with the distance formula for R^3, you can derive it from using the Pythagorean Theorem twice, basically.

Does this make sense? I just got in from a show and drank a lot of cheap beer tonight. Let me know if you need any clarification and I'll reply tomorrow.
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by Vemuri » Sun May 31, 2009 3:20 am
raleigh wrote: Distance formula in R^3 is sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2).

Make the center of the cube the origin. Then to get to a vertex you'd have to go 5 in each direction so the distance between the center of the cube and a vertex is sqrt(25 + 25 + 25) = sqrt(75) = 5sqrt(3).
Can you please explain the R^3 formula? Also, is there a simpler way to solve this problem?
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by ssmiles08 » Sun May 31, 2009 5:21 am
I did it the following way:

The largest possible distance within a cube is [a^2 + b^2 + c^2]^(1/2)

so the largest diagonal is (100 + 100 + 100)^1/2 = 10(3)^1/2

SInce the diameter of the circle is 10, we can subtract it from 10(3)^1/2.

10(3)^1/2 - 10 now has the the length of 2 distances from the vertex to the surface of the cube. To obtain 1 distance, we divide by 2.

5(3)^1/2 - 5
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by Vemuri » Sun May 31, 2009 5:24 am
ssmiles08 wrote:I did it the following way:

The largest possible distance within a cube is [a^2 + b^2 + c^2]^(1/2)

so the largest diagonal is (100 + 100 + 100)^1/2 = 10(3)^1/2

SInce the diameter of the circle is 10, we can subtract it from 10(3)^1/2.

10(3)^1/2 - 10 now has the the length of 2 distances from the vertex to the surface of the cube. To obtain 1 distance, we divide by 2.

5(3)^1/2 - 5
Cudos !!! That's the best GMAT solution according to me. Lets see if anyone can beat this :D
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by PAB2706 » Sun May 31, 2009 5:28 am
the diagonal of cube is sqrt3 *side

this diag will pass thru the center of the sphere...

sphere has radius 5

diag of the cube is 10sqrt3 half of this is 5sqrt3

so 5sqrt3-5 will give the distance.

i hope i am right.
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