Difference between revisions of "2003 AIME II Problems/Problem 7"
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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 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 <math>a</math>, <math>b</math>, and <math>c</math> 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 <math>b=2a</math>. Substitution yields <math>a=10</math> and <math>b=20</math>, so the area of the rhombus is <math>20\cdot40/2=400</math>. | + | The area of any triangle can be expressed as <math>\frac{a\cdot b\cdot c}{4R}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> 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 <math>b=2a</math>. Substitution yields <math>a=10</math> and <math>b=20</math>, so the area of the rhombus is <math>20\cdot40/2=\boxed{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 15:31, 19 February 2012
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 , , and 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 . 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 |