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Difference between revisions of "2000 AIME II Problems/Problem 10"
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== Solution ==
 
== 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, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(arctan(19/r)+arctan(26/r))+(arctan(37/r)+arctan(23/r))=180</math>. Take the tan of both sides and use the identity for tan(A+B) to get <math>(tan(arctan(19/r)+arctan(26/r))+tan(arctan(37/r)+arctan(23/r)))/n=0</math>, so <math>tan(arctan(19/r)+arctan(26/r))+tan(arctan(37/r)+arctan(23/r))=0</math>. Use the identity for tan(A+B) again to get <math>(45/r)/(1-(19\cdot26/r^2)+(60/r)/(1-37\cdot23/r^2)=0</math>. Solving gives <math>r^2=647</math>.
+
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.
 +
 
 +
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>.
 +
 
 +
Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get <math>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0</math>.
 +
 
 +
Use the identity for <math>\tan(A+B)</math> again to get <math>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}}=0</math>.
 +
 
 +
Solving gives <math>r^2=\boxed{647}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2000|n=II|num-b=9|num-a=11}}

Revision as of 03:27, 15 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 $r^2=\boxed{647}$.

See also

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
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