Question

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 $m + n.$

Answer

Answer: 901

Concepts:mass points homothety area ratio

Difficulty rating: 2990

Solution:

Assign masses $5$ at $A,$ $6$ at $B,$ and $15$ at $C.$ Then $E$ balances $\overline{AC}$ ( $5 \cdot 3 = 15 \cdot 1$ ) and $D$ balances $\overline{BC}$ ( $6 \cdot 5 = 15 \cdot 2$ ), so the cevians $\overline{AD}$ and $\overline{BE}$ meet at the center of mass $P,$ of total mass $26.$ Extending $\overline{CP}$ to meet $\overline{AB}$ at $F,$ the mass at $F$ is $5 + 6 = 11,$ so on segment $CF$ we get $CP : PF = 11 : 15,$ that is, $\frac{FP}{FC} = \frac{15}{26}.$

The homothety centered at $F$ with ratio $\frac{15}{26}$ sends $C$ to $P$ and maps line $AB$ to itself; it carries line $CA$ to the parallel line through $P$ — which is line $PQ$ — and line $CB$ to line $PR.$ Hence it maps triangle $CAB$ onto triangle $PQR,$ and $\frac{[PQR]}{[ABC]} = \left(\frac{15}{26}\right)^2 = \frac{225}{676}.$

Since $\gcd(225, 676) = 1,$ the answer is $m + n = 225 + 676 = 901.$

2002 AIME II Problem 13

Answer: 901

Difficulty rating: 2990

Solution:

Problem 13 in Other Years