Difference between revisions of "2013 AIME II Problems/Problem 5"

Revision as of 23:37, 7 October 2016

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, $a + b + c = \boxed{020}$

@@ Line 30: / Line 30: @@
 == 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>
+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\left(\frac{26+6i\sqrt{3}}{28}\right) =  \frac{3\sqrt{3}}{14}</math>. Therefore, <math>a + b + c = \boxed{020}</math>
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=4|num-a=6}}
 {{MAA Notice}}

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