AIME 2002 II · 第 13 题

Solution 1

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 $m+n=\boxed{901}$. 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 $X$.

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 :)

Solution 3

Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$, we can see that the mass of $P$ is $26$, hence we know that $\frac{BP}{PE}=\frac{10}{3}$, now we can find that $\frac{PQ}{AE}=\frac{10}{13}$ which implies $PQ=\frac{30}{13}$, it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$, so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$. ~bluesoul

Solution 4(Ceva & Menelaus)

Construct $\overline{CP}$ and extend it to line $\overline{AB}$ at point $F$.

Using Ceva's Theorem on triangle $ABC$ and point $P$, we get

$$\frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}=1$$ Thus, $\frac{AF}{BF}=\frac{6}{5}$. Then, using this info, we apply Menelaus on triangle $ACF$ and line $BE$, obtaining

$$\frac{AE}{BE} \times \frac{CP}{FP} \times \frac{FB}{AB}=1$$ Simplifying and substituting, we find that $\frac{CP}{FP}=\frac{11}{15}$. Alternatively, $\frac{FP}{CF}=\frac{15}{26}$, which is also the ratio between the heights of the desired triangles. Finishing, $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ achieving the final answer of $\boxed{901}$. ~faliure167