2000 AIME II Problems/Problem 10: Difference between revisions
| Line 13: | Line 13: | ||
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
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 | ||
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