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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 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

Check the Answer


But try the problem first…

Answer: is 164.

Source
Suggested Reading

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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