1992 AIME Problems/Problem 9

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Problem

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 $AP^{}_{}=\frac mn$, where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$.

Solution

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See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

let AB be the base of the trapezoid and consider angles A and B. Let x equal the distance from A to P and let h equal the height of the trapezoid. Let r equal the radius of the circle.

then

1.  sinA= r/x = h/70   and  sinB= r/(92-x)  =  h/50 

let z be the distance along AB from A to where the perp from D meets AB

then h^2 +z^2 =70^2 and (73-z)^2 + h^2 =50^2 so h =(44710959)^.5 /146 now substitute this into 1 to get x= 11753/219 kevin raponi