2003 AIME II Problems/Problem 7

Revision as of 21:05, 17 February 2010 by Mermaid (talk | contribs) (Solution)

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 $a\cdot b\cdot c/4R$, 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 $20\cdot40/2=400$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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