1987 AIME Problem 9

Attempt Problem 9 of the 1987 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1987 AIME solutions, or check the answer key.

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9.

Triangle

ABC

has a right angle at

B

and contains a point

P

for which

PA=10,

PB=6,

and

\angle APB=\angle BPC=\angle CPA.

Find

PC.

Answer

Answer: 33

Concepts:vector right triangle

Difficulty rating: 2380

Small hint:

The three equal angles around

P

are each

120^\circ

Big hint:

Use vectors from

P

and translate the right angle at

B

into a dot product

Solution:

Let

\mathbf a,

\mathbf b,

and

\mathbf c

be the vectors from

P

to

A,

B,

C,

and put

|\mathbf c|=x.

Their pairwise angles are

120^\circ,

so

\mathbf a\cdot\mathbf b=-30,

\mathbf b\cdot\mathbf c=-3x,

and

\mathbf a\cdot\mathbf c=-5x.

Since

AB\perp BC,

(\mathbf a-\mathbf b)\cdot(\mathbf c-\mathbf b)=0.

Expanding gives

-5x+30+3x+36=0,

so

x=33.