Difference between revisions of "2013 AIME II Problems/Problem 5"
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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. | 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 <math>M</math> be the midpoint of <math>\overline{DE}</math>. Then <math>\Delta MCA</math> is a 30-60-90 triangle with <math>MC = \dfrac{3}{2}</math>, <math>AC = 3</math> and <math>AM = \dfrac{3\sqrt{3}}{2}</math>. Since the triangle <math>\Delta AME</math> is right, then we can find the length of <math>\overline{AE}</math> by pythagorean theorem, <math>AE = \sqrt{7}</math>. Therefore, since <math>\Delta AME</math> is a right triangle, we can easily find <math>\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}</math> and <math>\cos(\angle EAM) = \dfrac{3\sqrt{3}}{2\sqrt{7}}</math>. So we can use the double angle formula for sine, <math>\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = | + | Let <math>M</math> be the midpoint of <math>\overline{DE}</math>. Then <math>\Delta MCA</math> is a 30-60-90 triangle with <math>MC = \dfrac{3}{2}</math>, <math>AC = 3</math> and <math>AM = \dfrac{3\sqrt{3}}{2}</math>. Since the triangle <math>\Delta AME</math> is right, then we can find the length of <math>\overline{AE}</math> by pythagorean theorem, <math>AE = \sqrt{7}</math>. Therefore, since <math>\Delta AME</math> is a right triangle, we can easily find <math>\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}</math> and <math>\cos(\angle EAM) = \dfrac{3\sqrt{3}}{2\sqrt{7}}</math>. So we can use the double angle formula for sine, <math>\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{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}} | ||
Revision as of 15:36, 6 April 2013
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
![[asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, S); label("$M$", M, S); draw(A--D); draw(A--E); draw(A--M);[/asy]](http://latex.artofproblemsolving.com/f/9/f/f9f9726a27a1fdee751794de373a0f07ceab353d.png)
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, 
.
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 | ||
