Question

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 $DE = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 923

Concepts:similarity incircle, incenter, and inradius

Difficulty rating: 2390

Solution:

Since $\overline{DE} \parallel \overline{BC},$ triangles $ADE$ and $ABC$ are similar, and the ratio equals the ratio of their heights from $A.$ The line $DE$ passes through the incenter, which sits at height $r$ (the inradius) above $BC,$ so the ratio is $\frac{h - r}{h} = 1 - \frac{r}{h},$ where $h$ is the height from $A$ to $\overline{BC}.$

If $K$ is the area and $s = \frac{21 + 22 + 20}{2} = \frac{63}{2}$ the semiperimeter, then $r = \frac{K}{s}$ and $h = \frac{2K}{20},$ so $\frac{r}{h} = \frac{20}{2s} = \frac{20}{63}.$

Therefore $DE = 20\left(1 - \frac{20}{63}\right)$ $= 20 \cdot \frac{43}{63}$ $= \frac{860}{63},$ which is in lowest terms, and $m + n = 860 + 63 = 923.$

2001 AIME I Problem 7

Answer: 923

Difficulty rating: 2390

Solution:

Problem 7 in Other Years