AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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



Angle Bisectors

Check the Answer

Answer: is 164.

AIME I, 1992, Question 9

Coordinate Geometry by Loney

Try with Hints

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Trapezoid Problem

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


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

or, AP=y=\(\frac{161}{3}\)

or, 161+3=164.

Subscribe to Cheenta at Youtube