Difference between revisions of "2013 AIME II Problems/Problem 5"
(→Problem 5) |
IMOJonathan (talk | contribs) |
||
Line 28: | Line 28: | ||
== Solution 3 == | == Solution 3 == | ||
We notice that <math>\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)=\sin(\angle{DAE})</math>. We can find <math>2\sin(\alpha)\cos(\alpha)</math>, to be <math>2(\frac{1}{\sqrt{28}})(\frac{3\sqrt{3}}{\sqrt{28}})=\frac{3\sqrt{3}}{14}</math>, so our answer is <math>3+3+14=\boxed{020}</math>. | We notice that <math>\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)=\sin(\angle{DAE})</math>. We can find <math>2\sin(\alpha)\cos(\alpha)</math>, to be <math>2(\frac{1}{\sqrt{28}})(\frac{3\sqrt{3}}{\sqrt{28}})=\frac{3\sqrt{3}}{14}</math>, so our answer is <math>3+3+14=\boxed{020}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Let A be the origin of the complex plane, B be <math>1+i\sqrt{3}</math>, and C be <math>2</math>. Also, WLOG, let D have a greater imaginary part than E. Then, D is <math>\frac{4}{3}+\frac{2i\sqrt{3}}{3}</math> and E is <math>\frac{5}{3}+\frac{i\sqrt{3}}{3}</math>. Then, <math>\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im(\frac{26+6i\sqrt{3}}{28}) = \frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=4|num-a=6}} | {{AIME box|year=2013|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:37, 8 October 2016
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
We notice that . We can find
, to be
, so our answer is
.
Solution 4
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,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.