Aesthetically pleasing, somehow

Let the side of the larger equilateral triangle be $s$ and the side of the smaller equilateral triangle be $t$ , and label the diagram as follows:

Since the angle of the square is $90°$ and the angle of the equilateral triangle is $60°$ , by symmetry $\angle BAC = \frac{1}{2}(90° - 60°) = 15°$ , so $s = \frac{1}{\cos 15°} = \sqrt{6} - \sqrt{2}$ .

The inradius of the larger equilateral triangle is $r_s = \frac{1}{2\sqrt{3}}s = \frac{1}{2\sqrt{3}}(\sqrt{6} - \sqrt{2}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{6}$ .

Therefore, $a = 2$ , $b = 6$ , and $a + b = \boxed{8}$ .

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Extra credit:

The height of the larger equilateral triangle is $h_s = \frac{\sqrt{3}}{2}s$ and the height of the smaller equilateral triangle is $h_t = \frac{\sqrt{3}}{2}t$ . Since the triangles are placed along the diagonal of the unit square, $h_s + h_t = \sqrt{2}$ , or $\frac{\sqrt{3}}{2}s + \frac{\sqrt{3}}{2}t = \sqrt{2}$ , which rearranges to $\frac{t}{2} = \frac{\sqrt{6}}{3} - \frac{1}{2}s$ .

Substituting $s = \sqrt{6} - \sqrt{2}$ , we find that $\frac{t}{2} = \frac{\sqrt{6}}{3} - \frac{1}{2}(\sqrt{6} - \sqrt{2}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{6} = r_s$ .