2013 AIME II Problems/Problem 5

Problem

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 1

$[asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$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, $a + b + c = \boxed{020}$ .

Solution 2

We find that, as before, $AE = \sqrt{7}$ , and also the area of $\Delta DAE$ is 1/3 the area of $\Delta ABC$ . Thus, using the area formula, $1/2 * 7 * \sin(\angle EAD) = 3\sqrt{3}/4$ , and $\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}.$

Solution 3

We notice that $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)=\sin(\angle{DAE})$ . We can find $2\sin(\alpha)\cos(\alpha)$ , to be $2(\frac{1}{\sqrt{28}})(\frac{3\sqrt{3}}{\sqrt{28}})=\frac{3\sqrt{3}}{14}$ , so our answer is $3+3+14=\boxed{020}$ .

Solution 4

Let A be the origin of the complex plane, B be $1+i\sqrt{3}$ , and C be $2$ . Also, WLOG, let D have a greater imaginary part than E. Then, D is $\frac{4}{3}+\frac{2i\sqrt{3}}{3}$ and E is $\frac{5}{3}+\frac{i\sqrt{3}}{3}$ . Then, $\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) = \frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

Solution 5

Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\cos(\angle DAE)$ or $AE=\sqrt(7)$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\cos(\angle DAE)=(1-7-7)/-2(/sqrt{7})(/sqrt{7})=13/14$ $\sin^2(\angle DAE)=1-cos^2(\angle DAE)$ $\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{169}$ So $\sin(\angle DAE)=\frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$

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