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

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

Integers

Trapezoid

Angle Bisectors

But try the problem first...

Answer: is 164.

Source

Suggested Reading

AIME I, 1992, Question 9

Coordinate Geometry by Loney

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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