Question

Trapezoid $ABCD$ has sides $AB=92,$ $BC=50,$ $CD=19,$ and $AD=70,$ with $AB$ parallel to $CD.$ A circle with center $P$ 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 $m+n.$

Answer

Answer: 164

Concepts:coordinate geometry trapezoid tangent circles

Difficulty rating: 2230

Small Hint:

Put $AB$ on the $x$ -axis and compare the distances from $P$ to the two legs

Big Hint:

The common trapezoid height cancels, leaving an equation involving $AP$ and $PB$ divided by the leg lengths

Solution:

Put $A=(0,0),$ $B=(92,0),$ and let the height of the trapezoid be $h.$ If $P=(p,0),$ its perpendicular distances to legs $AD$ and $BC$ are $\frac{hp}{70}$ and $\frac{h(92-p)}{50},$ respectively. Tangency to both legs makes these equal, so $\frac p{70}=\frac{92-p}{50}.$ Hence $120p=6440$ and $AP=p=\frac{161}{3}.$ Therefore $m+n=161+3=164.$

1992 AIME Problem 9

Answer: 164

Difficulty rating: 2230

Small Hint:

Big Hint:

Solution:

Problem 9 in Other Years