1989 AIME Problem 15

Question

Point

P

is inside triangle

ABC.

Line segments

APD,

BPE,

and

CPF

are drawn with

D

on

BC,

E

on

AC,

and

F

on

AB

(see the figure). Given that

AP=6,

BP=9,

PD=6,

PE=3,

and

CF=20,

find the area of triangle

ABC.

Answer

Answer: 108

Concepts:area ratio triangle area vector

Difficulty rating: 3060

Small hint:

Use the two known cevian ratios to find the barycentric weights of

A

and

B

at

P

Big hint:

Place

P

at the origin; the resulting vector relation determines the angle between

PA

and

PB

Solution:

Since

AP=PD,

the barycentric weight of

A

at

P

is

\frac{1}{2}.

Since

BP:PE=3:1,

the weight of

B

is

\frac{1}{4},

so the weight of

C

is also

\frac{1}{4}.

Along

CF,

this means

\frac{PF}{CF}=\frac{1}{4},

hence

PF=5

and

CP=15.

Place

P

at the origin and denote the position vectors

A,

B,

and

C

by the same letters. The barycentric relation is

2A+B+C=0.

Thus

C=-2A-B.

Using

|A|=6,

|B|=9,

and

|C|=15,

\begin{aligned}225&=|2A+B|^2\\&=4(36)+81+4A\mathbin{\cdot}B,\end{aligned}

so

A\mathbin{\cdot}B=0.

Therefore

PA\perp PB.

Expanding the cross product gives

(B-A)\mathbin{\times}(C-A)=4(A\mathbin{\times}B).

Consequently,

\begin{aligned}{}[ABC]&=2|A\mathbin{\times}B|\\&=2(6)(9)=108.\end{aligned}