2013 AIME II Problems/Problem 5
Contents
Problem
In equilateral let points
and
trisect
. Then
can be expressed in the form
, where
and
are relatively prime positive integers, and
is an integer that is not divisible by the square of any prime. Find
.
Solution 1
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let be the midpoint of
. Then
is a 30-60-90 triangle with
,
and
. Since the triangle
is right, then we can find the length of
by pythagorean theorem,
. Therefore, since
is a right triangle, we can easily find
and
. So we can use the double angle formula for sine,
. Therefore,
.
Solution 2
We find that, as before, , and also the area of
is 1/3 the area of
. Thus, using the area formula,
, and
. Therefore,
Solution 3
Let A be the origin of the complex plane, B be , and C be
. Also, WLOG, let D have a greater imaginary part than E. Then, D is
and E is
. Then,
. Therefore,
Solution 4
Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, or
The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines,
Since
Then
So
. Therefore,
Solution 5 (Veectors)
Setting up a convinient coordinate system, we let be at point
,
be at point
, and
be at point
. Then
and
will be at points
and
. Then
. From here, we see that $\sin(\DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$ (Error compiling LaTeX. Unknown error_msg)
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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