# 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 1

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 $15+95+8=\boxed{118}$.

## Solution 2 (Vectors)

We know the radii to $A$,$B$, and $C$ form a triangular pyramid $OABC$. We know the lengths of the edges $OA = OB = OC = 20$. First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$. Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$, and $z$ occupy the other dimension, with the origin as $C$. We look at vectors $OA$ and $OC$. Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$, hence it is $7$ units along the $x$ axis. Hence for vector $OC$, in form $$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$, since $A$ is $9$ along $x$, and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$. We know both vector magnitudes are $20$. Solving for $h$ yields $\frac{15\sqrt{95} }{8}$, so Answer = $\boxed{118}$.