Difference between revisions of "2003 AIME II Problems/Problem 7"

Revision as of 19:40, 4 July 2013

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 $20\cdot40/2=\boxed{400}$ .

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 == See also ==
 {{AIME box|year=2003|n=II|num-b=6|num-a=8}}
+{{MAA Notice}}

2003 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2003 AIME II Problems/Problem 7"

Revision as of 19:40, 4 July 2013

Problem

Solution

See also