2003 AIME II Problems/Problem 7
Contents
Problem
Find the area of rhombus given that the radii of the circles circumscribed around triangles
and
are
and
, respectively.
Solution
The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD and half of diagonal AC
. The length of the four sides of the rhombus is
.
The area of any triangle can be expressed as , where
,
, and
are the sides and
is the circumradius. Thus, the area of
is
. Also, the area of
is
. Setting these two expressions equal to each other and simplifying gives
. Substitution yields
and
, so the area of the rhombus is
.
Solution 2
Let . Let
. By the extended law of sines,
Since
,
, so
Hence
. Solving
,
. Thus
The height of the rhombus is
, so we want
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See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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