2000 AIME II Problems/Problem 12

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 (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles):

$K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84$

From $R = \frac{abc}{4K}$ , we know that 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,

$OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.$

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2000 AIME II Problems/Problem 12

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