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>. |
== 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 20:57, 31 January 2008
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 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, 
, or 
. Take the tan of both sides and use the identity for tan(A+B) to get 
, so 
. Use the identity for tan(A+B) 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 | ||
