Shortest Path off the Sides of an Equilateral Triangle

The shortest path between two points $D$ and $E$ is one taken by a beam of light. Treat sides of the equilateral triangle as mirrors where the beam of light reflects off. We can find the image light path by reflecting the triangle about the mirrors.

Starting from $D(5.3)$ , we find the final image of $E$ is at $E'\left(-\frac {1+\sqrt 3}2, \frac {3+11\sqrt 3}2 \right)$ . Therefore the shortest length of the path is $\sqrt{\left(5+\frac {1+\sqrt 3}2 \right)^2 + \left( 3-\frac {3+11\sqrt 3}2 \right)^2} = \sqrt{124-11\sqrt 3}$ , and $a+b+c= 124+11+3 = \boxed{138}$ .

The above figure shows how to find the shortest path by reflecting the end point (point $E$ ) about the sides in reverse order, so first about side $AB$ and this generates point $E'$ . Then we reflect point $E'$ about side $BC$ generating point $E"$ , and finally we reflect $E"$ about $AC$ generating point $E'''$ . The distance $DE'''$ is the shortest distance. The coordinates of $E''' = (-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}, \dfrac{11 \sqrt{3}}{2} + \dfrac{3}{2} )$ , and this makes the distance $\overline{DE'''} = \sqrt{ 124 - 11 \sqrt{3}}$ . Hence the answer is $124 + 11 + 3 = \boxed{138}$