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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.  
  
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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) = \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>.
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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>.
  
 
==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 22:46, 3 June 2013

Problem 5

In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$.

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]

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 $M$ be the midpoint of $\overline{DE}$. Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$, $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$. Since the triangle $\Delta AME$ is right, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$. Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$. So we can use the double angle formula for sine, $\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$. Therefore, $a + b + c = \boxed{020}$.

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions
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