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

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== Solution ==
 
== Solution ==
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The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD <math>a</math> and half of diagonal AC <math>b</math>. The length of the four sides of the rhombus is <math>\sqrt{a^2+b^2}</math>. The area of any triangle can be expressed as <math>\frac{a\cdot b\cdot c}{4R}</math>, where a, b, and c are the sides and R is the circumradius. Thus, the area of triangle ABD is <math>ab=2a(a^2+b^2)/(4\cdot12.5)</math>. Also, the area of triangle ABC is <math>ab=2b(a^2+b^2)/(4\cdot25)</math>. 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 <math>20\cdot40/2=400</math>.
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The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD <math>a</math> and half of diagonal AC <math>b</math>. The length of the four sides of the rhombus is <math>\sqrt{a^2+b^2}</math>. The area of any triangle can be expressed as <math>\frac{a\cdot b\cdot c}{4R}</math>, where a, b, and c are the sides and <math>R</math> is the circumradius. Thus, the area of <math>\triangle ABD</math> is <math>ab=2a(a^2+b^2)/(4\cdot12.5)</math>. Also, the area of <math>\triangle ABC</math> is <math>ab=2b(a^2+b^2)/(4\cdot25)</math>. Setting these two expressions equal to each other and simplifying gives b=2a. Substitution yields <math>a=10</math> and <math>b=20</math>, so the area of the rhombus is <math>20\cdot40/2=400</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2003|n=II|num-b=6|num-a=8}}

Revision as of 22:37, 10 March 2010

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 $20\cdot40/2=400$.

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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