2001 AIME I Problems/Problem 7

Revision as of 00:36, 19 January 2008 by Dgreenb801 (talk | contribs) (Solution)

Problem

Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

By Heron's formula, the area of the whole triangle is $21\sqrt{1311}/2$. Since the area of a triangle is the inradius times the semiperimeter, the inradius is $\sqrt{1311}/6$. The ratio of the heights of triangles ADE and ABC is equal to the ratio between sides DE and BC. Thus, we have $(21\sqrt{1311}/40-\sqrt{1311}/6)/(21\sqrt{1311}/40)=x/20$. Solving for x gives x=$860/63$, so the answer is $923$.

See also

2001 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions