2005 AIME I Problems/Problem 15

Revision as of 22:06, 16 January 2007 by JBL (talk | contribs) (Solution in progress -- all are welcome to finish.)

Problem

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

Solution

Let $E$ and $F$ be the points of tangency of the incircle with $BC$ and $AC$, respectively. Without loss of generality, let $AB > AC$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $DE^2 = 2m \cdot m = AF^2$, so $DE = AF$. Now, $CE$ and $CF$ are two tangents to a circle from the same point, so $CE = CF$ and thus $AC = AF + CF = DE + CE = CD = 10$. Then triangle $\triangle ACD$ is isosceles, This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also