Cube / Octahedron Surface Area

Each of the blue triangles is an isosceles right triangle with legs of $\frac{1}{2}$ , a hypotenuse of $\frac{\sqrt{2}}{2}$ , and an area of $B = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

Each of the yellow triangles is an equilateral triangle with sides of $\frac{\sqrt{2}}{2}$ and an area of $Y = \frac{\sqrt{3}}{4} \cdot (\frac{\sqrt{2}}{2})^2 = \frac{\sqrt{3}}{8}$ .

There are $24$ blue and $24$ yellow triangles, so the surface area is $S = 24B + 24Y = 24 \cdot \frac{1}{8} + 24 \cdot \frac{\sqrt{3}}{8} = 3 + 3\sqrt{3}$ .

Therefore, $a = b = c = 3$ and $a + b + c = \boxed{9}$ .