2023 AMC 10A Problem 11

Let the shaded triangle have legs $a \le b$. Its hypotenuse is a side of the inner square, and its two legs together make up a side of the outer square (the four congruent triangles fit corner to corner). Therefore
$$a + b = \sqrt3, \qquad a^2 + b^2 = 2.$$

Squaring the first equation and subtracting the second:
$$2ab = (a+b)^2 - (a^2+b^2) = 3 - 2 = 1, \quad\text{so}\quad ab = \tfrac12.$$

We want $r = \frac{a}{b}$. Dividing $a^2 + b^2 = 2$ by $ab = \frac12$ gives
$$\frac{a}{b} + \frac{b}{a} = 4, \qquad\text{i.e.}\qquad r + \frac1r = 4.$$
So $r^2 - 4r + 1 = 0$ and $r = 2 \pm \sqrt3$. Since $a \le b$, we need $r \le 1$, so $r = 2 - \sqrt3$.

The answer is $\boxed{\textbf{(C)}\ 2-\sqrt3}$.

Two symmetric facts about the legs (their sum and their sum of squares) determine their product, and sum-plus-product is all you need for any symmetric quantity, including the ratio via $\frac ab + \frac ba = \frac{a^2+b^2}{ab}$. Solving for the ratio directly avoids computing $a$ and $b$, which are ugly nested radicals.

Let $\theta$ be the angle the inner square's side makes with the outer side. Then $\cos\theta + \sin\theta = \frac{\sqrt3}{\sqrt2}$, so squaring gives $\sin 2\theta = \frac12$ and $\theta = 15^\circ$. The ratio is $\tan 15^\circ = 2 - \sqrt3$.

• Solving the quadratic for a and b separately with messy nested radicals, or reporting the longer-to-shorter ratio 2 + sqrt 3.
• Setting $a + b$ equal to the inner side $\sqrt2$ instead of the outer side $\sqrt3$.