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>. | ||
== Solution 2== | == Solution 2== | ||
− | Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on. | + | Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on (<math>a, b, c,</math> and <math>d</math> are the tangent lengths, not the side lengths). |
<cmath>A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}</cmath> | <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>. | <math>r^2=\frac{A}{a+b+c+d} = \boxed{647}</math>. |
Revision as of 20:15, 8 April 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 ( 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.