Difference between revisions of "2000 AIME II Problems/Problem 10"
m (→Solution 2) |
m (→Solution) |
||
| Line 2: | Line 2: | ||
A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[Perfect square|square]] of the [[radius]] of the circle. | A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[Perfect square|square]] of the [[radius]] of the circle. | ||
| − | == Solution == | + | == Solution 1== |
Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed. | Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed. | ||
| Line 12: | Line 12: | ||
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 (<math>a, b, c,</math> and <math>d</math> are the tangent lengths, not the side lengths). | 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). | ||
Revision as of 09:39, 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
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.