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Difference between revisions of "2000 AIME II Problems/Problem 12"

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So the final answer is <math>15+95+8=\boxed{118}</math>
 
So the final answer is <math>15+95+8=\boxed{118}</math>
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== See also ==
 
 
{{AIME box|year=2000|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2000|n=II|num-b=11|num-a=13}}

Revision as of 19:35, 18 March 2008

Problem

The points $A$, $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$. It is given that $AB=13$, $BC=14$, $CA=15$, and that the distance from $O$ to triangle $ABC$ is $\frac{m\sqrt{n}}k$, where $m$, $n$, and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$.

Solution

Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$. By the Pythagorean Theorem on triangles $\triangle OAD$, $\triangle OBD$ and $\triangle OCD$ we get:

$DA^2=DB^2=DC^2=20^2-OD^2$

It follows that $DA=DB=DC$, so $D$ is the circumcenter of $\triangle ABC$.

By Heron's Formula the area of $\triangle ABC$ is:

$K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = \sqrt{(21)(8)(7)(6)} = \sqrt{(3\cdot 7)(2^3)(7)(2\cdot 3)} =\sqrt{2^4\cdot 3^2\cdot 7^2} = 2^2\cdot 3\cdot 7 = 84$

Now the circumradius of $\triangle ABC$ is:

$R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}$

Thus by the Pythagorean Theorem again,

$OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \sqrt{\frac{160^2-65^2}{8^2}} = \frac{\sqrt{(160-65)(160+65)}}{8} = \frac{\sqrt{(95)(225)}}{8} = \frac{15\sqrt{95}}{8}$.

So the final answer is $15+95+8=\boxed{118}$

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions
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