Question

In triangle $ABC,$ $AB = 13,$ $BC = 15,$ and $CA = 17.$ Point $D$ is on $\overline{AB},$ $E$ is on $\overline{BC},$ and $F$ is on $\overline{CA}.$ Let $AD = p \cdot AB,$ $BE = q \cdot BC,$ and $CF = r \cdot CA,$ where $p,$ $q,$ and $r$ are positive and satisfy $p + q + r = \frac{2}{3}$ and $p^2 + q^2 + r^2 = \frac{2}{5}.$ The ratio of the area of triangle $DEF$ to the area of triangle $ABC$ can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 61

Concepts:area ratio symmetry (algebra)algebraic manipulation

Difficulty rating: 2560

Solution:

Each corner triangle's area is a product of side fractions: $[ADF] = p(1-r)[ABC],$ $[BED] = q(1-p)[ABC],$ and $[CFE] = r(1-q)[ABC],$ using the formula $\frac{1}{2}xy\sin\theta$ on the shared angles. Subtracting, $\begin{aligned} \frac{[DEF]}{[ABC]} &= 1 - p(1-r) \\ &\quad {}- q(1-p) - r(1-q) \\ &= 1 - (p+q+r) \\ &\quad {}+ (pq+qr+rp). \end{aligned}$

From the given values, $pq + qr + rp = \frac{(2/3)^2 - 2/5}{2}$ $= \frac{4/9 - 2/5}{2}$ $= \frac{1}{45}.$

Therefore the ratio is $1 - \frac{2}{3} + \frac{1}{45} = \frac{16}{45},$ and $m + n = 16 + 45 = 61.$

2001 AIME I Problem 9

Answer: 61

Difficulty rating: 2560

Solution:

Problem 9 in Other Years