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.
Integers
Trapezoid
Angle Bisectors
Answer: is 164.
AIME I, 1992, Question 9
Coordinate Geometry by Loney
Let AP=y or, PB=92-y
extending AD and BC to meet at Y
and YP bisects angle AYB
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
\(\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}\)
solving we get 120y=(70)(92)
or, AP=y=\(\frac{161}{3}\)
or, 161+3=164.
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