Categories

# Trapezoid Problem | AIME I, 1992 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Rationals.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on trapezoid.

## Trapezoid – AIME I, 1992

Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre 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.

• is 107
• is 164
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Trapezoid

Angle Bisectors

But try the problem first…

Source

AIME I, 1992, Question 9

Coordinate Geometry by Loney

## Try with Hints

First hint

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Second Hint

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

$$\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}$$

Final Step

solving we get 120y=(70)(92)

or, AP=y=$$\frac{161}{3}$$

or, 161+3=164.

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.