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

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== Problem ==
 
== Problem ==
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A circle is inscribed in quadrilateral <math>ABCD</math>, tangent to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the square of the radius of the circle.
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A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[square]] of the [[radius]] of the circle.
  
 
== Solution ==
 
== Solution ==
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Call the center of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, eight congruent right triangles are formed.
+
Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, eight congruent [[right triangle]]s are formed.
  
 
Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180</math>.
 
Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180</math>.

Revision as of 08:10, 20 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)
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Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions
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