2013 AIME II Problems/Problem 5: Difference between revisions
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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) = \sqrt{1-\sin(\angle EAM)^2}=\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>. | 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) = \sqrt{1-\sin(\angle EAM)^2}=\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>. | ||
== Solution 2 == | |||
We find that, as before, <math>AE = \sqrt{7}</math>, and also the area of <math>\Delta DAE</math> is 1/3 the area of <math>\Delta ABC</math>. Thus, using the area formula, <math>1/2 * 7 * \sin(\angle EAD) = 3\sqrt{3}/4</math>, and <math>\sin(\angle EAD) = \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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:13, 12 February 2014
Problem 5
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, 
.
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, 
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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