2003 AIME II Problems/Problem 7

Problem

Find the area of rhombus $ABCD$ given that the radii of the circles circumscribed around triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively.

Solution

The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$ . The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$ .

The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$ , where $a$ , $b$ , and $c$ are the sides and $R$ is the circumradius. Thus, the area of $\triangle ABD$ is $ab=2a(a^2+b^2)/(4\cdot12.5)$ . Also, the area of $\triangle ABC$ is $ab=2b(a^2+b^2)/(4\cdot25)$ . Setting these two expressions equal to each other and simplifying gives $b=2a$ . Substitution yields $a=10$ and $b=20$ , so the area of the rhombus is $20\cdot40/2=\boxed{400}$ .

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
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2003 AIME II Problems/Problem 7

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