1992 AIME Problem 9

Question

Trapezoid

ABCD

has sides

AB=92,

BC=50,

CD=19,

and

AD=70,

with

AB

parallel to

CD.

A circle with center

P

on

AB

is drawn tangent to

BC

and

AD.

Given that

AP=\frac{m}{n},

where

m

and

n

are relatively prime positive integers, find

m+n.

Answer

Put

AB

on the

x

-axis and compare the distances from

P

to the two legs

The common trapezoid height cancels, leaving an equation involving

AP

and

PB

divided by the leg lengths

Put

A=(0,0),

B=(92,0),

and let the height of the trapezoid be

h.

If

P=(p,0),

its perpendicular distances to legs

AD

and

BC

are

\frac{hp}{70}

and

\frac{h(92-p)}{50},

respectively. Tangency to both legs makes these equal, so

\frac p{70}=\frac{92-p}{50}.

Hence

120p=6440

and

AP=p=\frac{161}{3}.

Therefore

m+n=161+3=164.

Problem 9 in other years