Difference between revisions of "2003 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Find the area of rhombus <math>ABCD</math> given that the circumradii | + | Find the area of rhombus <math>ABCD</math> given that the circumradii of triangles <math>ABD</math> and <math>ACD</math> are <math>12.5</math> and <math>25</math>, respectively. |
== Solution == | == Solution == |
Latest revision as of 09:56, 16 September 2022
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
~yofro
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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