How well do you know your special right triangles?

Let AC = OG = BZ = x. If we calculate the length of a corresponding side on all 3 triangles in terms of x, we will be able to determine their scale factors, since the ratio of one pair of sides in two triangles is equal to the ratios of their perimeters.

Now the shortest side on a triangle is the one opposite the smallest angle, so the largest triangle must be the one for which x is opposite the smallest angle: triangle CAT. Also, the smallest triangle must be the one for which x is opposite the largest angle: triangle DOG.

I will choose to calculate AT and OG in terms of x (since they correspond.) This is easy to do, since we can draw the altitude CC' two special right triangles.

Using 30-60-90 right triangle ratios, the altitude CC' on triangle CAT has length $\frac{x \sqrt{3}}{2}$ , and the distance AC' is $\frac{x}{2}$ . Now, we can use the 45-45-90 ratio to determine that C'T is $\frac{x \sqrt{3}}{2}$ to match the altitude. Thus the length AT is $\frac{x (1 + \sqrt{3})}{2}$ .

On DOG, the distance OG is just x.

The ratio from AT:OG is thus $\frac{1 + 1\sqrt{3}}{2}$ . Therefore a = b = 1, c = 3, and d = 2.

1 + 1 + 3 + 2 = 7.

For completeness, I'll compute the length UZ:

On BUZ, draw the altitude BB'. First use the 45-45-90 triangle to determine that the distance B'Z is $\frac{x}{\sqrt{2}}$ and the altitude BB' is also $\frac{x}{\sqrt{2}}$ . You can next use the 30-60-90 right triangle to determine that B'U is $\frac{x}{\sqrt{2} \sqrt {3}}$ which can be rewritten as $\frac{1}{\sqrt{6}}$ . Note that $\frac{1}{\sqrt{6}} +\frac{1}{\sqrt{2}}$ is smaller than $\frac{x (1 + \sqrt{3})}{2}$ .