2003 AIME II Problems/Problem 7: Difference between revisions
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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 08:56, 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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