Difference between revisions of "2003 AIME II Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | The | + | 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 <math>\sqrt{a^2+b^2}</math>. The area of any triangle can be expressed as the abc/4R, 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>. |
== 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 21:02, 17 February 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 a and half of diagonal AC b. The length of the four sides of the rhombus is 
. The area of any triangle can be expressed as the abc/4R, where a,b, and c are the sides and R is the circumradius. Thus, the area of triangle ABD is 
. Also, the area of triangle ABC is 
. 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 
.
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 | ||
