2003 AIME II Problems/Problem 7
Problem
Find the area of rhombus given that the circumradii of 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 .
Solution 2
Let . Let . By the extended law of sines, Since , , so Hence . Solving , . Thus The height of the rhombus is , so we want
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Video Solution by Sal Khan
https://www.youtube.com/watch?v=jpKjXtywTlQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=16 - AMBRIGGS
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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