# Difference between revisions of "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 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). $$A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}$$ $r^2=\frac{A}{a+b+c+d}^2 = \boxed{647}$.

## Solution 3 (Smart algebra to make 2 less annoying)

Using the formulas established in solution 2, one notices: $$r^2=\frac{A^2}{a+b+c+d}$$ $$r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}$$ $$r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}$$ $$r^2=\boxed{647}$$

which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand.