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

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Solving gives <math>r^2=\boxed{647}</math>.
 
Solving gives <math>r^2=\boxed{647}</math>.
 +
 +
Note: the equation may seem nasty at first, but once you cancel the <math>r</math>s and other factors, you are just left with <math>r^2</math>. That gives us <math>647</math> quite easily
  
 
== Solution 2==
 
== Solution 2==

Revision as of 11:07, 8 October 2020

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 1

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, four pairs of 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 $r^2=\boxed{647}$.

Note: the equation may seem nasty at first, but once you cancel the $r$s and other factors, you are just left with $r^2$. That gives us $647$ quite easily

Solution 2

Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ($a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}\] $r^2=\frac{A}{a+b+c+d} = \boxed{647}$.

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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