Difference between revisions of "2000 AIME II Problems/Problem 10"
Mathcool2009 (talk | contribs) (→Solution) |
(→Solution) |
||
Line 12: | Line 12: | ||
Solving gives <math>r^2=\boxed{647}</math>. | Solving gives <math>r^2=\boxed{647}</math>. | ||
+ | == Solution 2== | ||
+ | Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on. | ||
+ | <cmath>A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}</cmath> | ||
+ | <math>r^2=\frac{A}{a+b+c+d} = \boxed{647}</math>. | ||
== See also == | == See also == |
Revision as of 12:40, 28 January 2019
Contents
[hide]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 .
Solution 2
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on. .
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.