# Difference between revisions of "2002 AIME II Problems/Problem 13"

## 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 1

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

$[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("A",A,WSW); label("B",B,ESE); label("C",C,NNW); label("D",D,NE); label("E",E,WNW); label("X",X,SSE); label("P",P,NNE); label("Q",Q,SSW); label("R",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]$

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 $m+n=\boxed{901}$. Note we can just use mass points to get \left( \frac{15}{26} \right)^2= \frac{225}{676}$which is \boxed{901}$.

## Solution 2

First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$.

$[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("A",A,WSW); label("B",B,ESE); label("C",C,NNW); label("D",D,NE); label("E",E,WNW); label("X",X,SSE); label("P",P,NNE); label("Q",Q,SSW); label("R",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]$

We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$

By Ceva's, $$3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}$$ That means that $$\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}$$

Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that

$$W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}$$

Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$. Therefore,

$$\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}$$

Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$. Therefore,

$$\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}$$

Because $\triangle{PQR}$ is similar to $\triangle{CAB}$, $\angle{C}=\angle{P}$.

As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$.

Therefore, $$\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}$$

• Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)