Question

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 $a + b + c.$

Answer

Answer: 20

Concepts:law of cosines triangle area equilateral triangle

Difficulty rating: 2330

Solution:

Scale so the side length is $6,$ with $BD = DE = EC = 2.$ In triangle $AEC,$ the law of cosines gives $\begin{aligned} AE^2 &= 6^2 + 2^2 - 2 \cdot 6 \cdot 2 \cos 60^\circ \\ &= 28, \end{aligned}$ so $AE = 2\sqrt{7},$ and $AD = 2\sqrt{7}$ by symmetry.

Since $DE$ is one third of $BC$ and triangles $ADE$ and $ABC$ share the apex $A,$ we get $[ADE] = \frac{1}{3}[ABC]$ $= \frac{1}{3} \cdot \frac{36\sqrt{3}}{4}$ $= 3\sqrt{3}.$ On the other hand $[ADE] = \frac{1}{2} \cdot AD \cdot AE$ $\cdot \sin(\angle DAE)$ $= 14\sin(\angle DAE).$

Therefore $\sin(\angle DAE) = \frac{3\sqrt{3}}{14},$ and $a + b + c = 3 + 3 + 14 = 20.$

2013 AIME II Problem 5

Answer: 20

Difficulty rating: 2330

Solution:

Problem 5 in Other Years