Difference between revisions of "2000 AIME II Problems/Problem 10"

Revision as of 19:35, 18 March 2008

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 $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$ .

Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get $\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0$ .

Use the identity for $\tan(A+B)$ again to get $\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}}=0$ .

Solving gives $r^2=\boxed{647}$ .

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

@@ Line 13: / Line 13: @@
 Solving gives <math>r^2=\boxed{647}</math>.
-== See also ==
 {{AIME box|year=2000|n=II|num-b=9|num-a=11}}

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Difference between revisions of "2000 AIME II Problems/Problem 10"

Revision as of 19:35, 18 March 2008

Problem

Solution