2000 AIME II Problems/Problem 10

Problem

A circle is inscribed in quadrilateral $ABCD$ , tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ , $PB=26$ , $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.

Solution

Call the center of the circle O. By drawing the lines from O tangent to the sides and from O to the vertices of the quadrilateral, eight congruent right triangles are formed. Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$ , or $(arctan(19/r)+arctan(26/r))+(arctan(37/r)+arctan(23/r))=180$ . Take the tan of both sides and use the identity for tan(A+B) to get $(tan(arctan(19/r)+arctan(26/r))+tan(arctan(37/r)+arctan(23/r)))/n=0$ , so $tan(arctan(19/r)+arctan(26/r))+tan(arctan(37/r)+arctan(23/r))=0$ . Use the identity for tan(A+B) again to get $(45/r)/(1-(19\cdot26/r^2)+(60/r)/(1-37\cdot23/r^2)=0$ . Solving gives $r^2=647$ .

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2000 AIME II Problems/Problem 10

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