Difference between revisions of "2000 AIME II Problems/Problem 10"
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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>. | 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 | + | 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>. | Solving gives <math>r^2=\boxed{647}</math>. | ||
Revision as of 16:52, 14 July 2014
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
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 
.
Use the identity for again to get 
.
Solving gives 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
