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2002 AIME II Problems/Problem 13

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Problem

In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Let $X$ be the intersection of $\overline{CP}$ and $\overline{AB}$ .

Since $\overline{PQ} \parallel \overline{CA}$ and $\overline{PR} \parallel \overline{CB}$ , $\angle CAB = \angle PQR$ and $\angle CBA = \angle PRQ$ . So $\Delta ABC \sim \Delta QRP$ , and thus, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2$ .

Using mass points:

WLOG, let $W_C=15$ .

Then:

$W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5$ .

$W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6$ .

$W_X=W_A+W_B=5+6=11$ .

$W_P=W_C+W_X=15+11=26$ .

Thus, $\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}$ . Therefore, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}$ , and $m+n=\boxed{901}$ .

See also

2002 AIME II (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

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