Difference between revisions of "2000 AIME II Problems/Problem 10"
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Solving gives <math>r^2=\boxed{647}</math>. | Solving gives <math>r^2=\boxed{647}</math>. | ||
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+ | Note: the equation may seem nasty at first, but once you cancel the <math>r</math>s and other factors, you are just left with <math>r^2</math>. That gives us <math>647</math> quite easily | ||
== Solution 2== | == Solution 2== |
Revision as of 11:07, 8 October 2020
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
Use the identity for again to get
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).
.
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.