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Trapezoid Problem | AIME I, 1992 | Question 9

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

Check the Answer


Answer: is 164.

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

Trapezoid Problem

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.

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