2000 AIME II Problems/Problem 10
Contents
Problem
A circle is inscribed in quadrilateral 
, tangent to 
at 
and to 
at 
. Given that 
, 
, 
, and 
, find the square of the radius of the circle.
Solution 1
Call the center of the circle 
. By drawing the lines from tangent to the sides and from to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, 
, or 
.
Take the 
of both sides and use the identity for 
to get ![\[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]](http://latex.artofproblemsolving.com/1/0/d/10d199ea13199b92aa7ee186b88f17c3facdac3f.png)
Use the identity for again to get ![\[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]](http://latex.artofproblemsolving.com/9/d/2/9d29e7ed99258e5a4ffe3dce264d0d35eb5d2362.png)
Solving gives 
.
Note: the equation may seem nasty at first, but once you cancel the 
s and other factors, you are just left with 
. That gives us 
quite easily.
Solution 2
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on (
and 
are the tangent lengths, not the side lengths).
![\[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\]](http://latex.artofproblemsolving.com/f/1/e/f1ea5edc88f82cfc369d50547923150dfa497d4d.png)
.
Solution 3 (Smart algebra to make 2 less annoying)
Using the formulas established in solution 2, one notices:
![\[r^2=\frac{A^2}{a+b+c+d}\]](http://latex.artofproblemsolving.com/6/d/e/6ded0cd3ee8e3f293b4d8b16e1e0472bb619c1f3.png)
![\[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\]](http://latex.artofproblemsolving.com/d/9/6/d96fe4f12c44152b152ac43bbf722e3d562beaf9.png)
![\[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\]](http://latex.artofproblemsolving.com/b/b/9/bb9bb28e5952a3a9f2a1d47418a8057875c33bf9.png)
![\[r^2=\boxed{647}\]](http://latex.artofproblemsolving.com/9/c/9/9c9171f3e7c73b6c8155565038345618f8c788a3.png)
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.
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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