2007 AIME II Problem 9

Question

Rectangle

ABCD

is given with

AB = 63

and

BC = 448.

Points

E

and

F

lie on

\overline{AD}

and

\overline{BC}

respectively, such that

AE = CF = 84.

The inscribed circle of triangle

BEF

is tangent to

\overline{EF}

at point

P,

and the inscribed circle of triangle

DEF

is tangent to

\overline{EF}

at point

Q.

Find

PQ.

Answer

Answer: 259

Concepts:incircle, incenter, and inradius Pythagorean Theorem

Difficulty rating: 2650

Small hint:

Compute

BE,

BF,

and

EF

with the Pythagorean theorem; triangles

BEF

and

DFE

turn out congruent

Big hint:

The tangency point splits

EF

by tangent lengths:

EP = s - BF

and

FQ = s - DE,

where

s

is the common semiperimeter. Then

PQ = EF - EP - FQ.

Solution:

Place

A = (0, 0),

B = (63, 0),

C = (63, 448),

D = (0, 448),

so

E = (0, 84)

and

F = (63, 364).

Then

BE = DF = \sqrt{63^2 + 84^2}

= 21\sqrt{3^2 + 4^2} = 105,

BF = DE = 448 - 84 = 364,

and

EF = \sqrt{63^2 + 280^2}

= 7\sqrt{9^2 + 40^2} = 287.

In particular triangles

BEF

and

DFE

are congruent, with common semiperimeter

s = \frac{105 + 364 + 287}{2} = 378.

In any triangle, the distance from a vertex to the incircle’s tangency points on its two sides is the semiperimeter minus the opposite side. In triangle

BEF,

EP = s - BF

= 378 - 364 = 14;

in triangle

DEF,

FQ = s - DE

= 378 - 364 = 14.

Therefore

PQ = EF - EP - FQ

= 287 - 14 - 14 = 259.