2003 AIME II Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
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>. | 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 21:37, 10 March 2010
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 a, b, and c 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 b=2a. Substitution yields 
and 
, so the area of the rhombus is 
.
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 | ||
