Geometry #5

by sjaelee, Aug 23, 2011, 11:53 PM

Quote:
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a+b\sqrt{c}$, where a,b,c are postive integers and are not divisible by the square of any prime. Find a+b+c.

Source: AIME

Let $R=$large radius and $r=$ small radius and let $d=$ parrallel distance between centers. Then

$(R+r)^2-(R-r)^2=d^2$ by Pythagorean Theorem.

This means $d=4Rr$. Now, we consider the triangle between center of the ocatagon and the vertices of the octagon. The two equal angles are $135/2$, so the other is $45$. The sides are:

$d,d,2r$. Using the law of cosines,

$2(d)^2-2(d)(d)\cos 45=(2r)^2$, so

$2d^2-\sqrt{2}d^2=4r^2$

$(2-\sqrt{2})d^2=4r^2$, and $d=4Rr$, so

$(8-4\sqrt{2})Rr=4r^2$. Dividing by $4r$,

$(2-\sqrt{2})R=r$. We know $r=100$, so

$(2-\sqrt{2})R=100$. Multiplying $2+\sqrt{2}$,

$2R=200+100\sqrt{2}$, so $R=100+50\sqrt{2}$.

Thus, $+b+c=100+50+2=152$.

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3 Comments

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Someone esle got the same answer as me with a different method, so pretty sure this is right?

by sjaelee, Aug 23, 2011, 11:53 PM

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Isn't this really just a 2D problem?

by djmathman, Aug 24, 2011, 1:40 AM

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As with most aime 3d probes, it is

by sjaelee, Aug 24, 2011, 6:09 AM

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