2023 AMC 10A Problem 11

1. Read the side lengths. The big square has area $3$, so its side is $\sqrt3$. The small square has area $2$, so its side is $\sqrt2$.
2. Name the legs. Call the shaded triangle's short leg $a$ and long leg $b$. Look at one side of the big square: it is made of the short leg of one triangle and the long leg of the next. So
$$a + b = \sqrt3.$$
3. The hypotenuse of each triangle is a side of the small square, so by the Pythagorean theorem
$$a^2 + b^2 = 2.$$
4. Get the product $ab$. Square the first equation: $(a+b)^2 = a^2 + 2ab + b^2 = 3$. We already know $a^2 + b^2 = 2$, so $2ab = 3 - 2 = 1$, and $ab = \frac12$.
5. Turn this into the ratio. We want $r = \frac{a}{b}$. Divide $a^2 + b^2 = 2$ by $ab = \frac12$:
$$\frac{a^2 + b^2}{ab} = \frac{a}{b} + \frac{b}{a} = \frac{2}{1/2} = 4.$$
So $r + \frac1r = 4$.
6. Solve for $r$. Multiply by $r$: $r^2 + 1 = 4r$, or $r^2 - 4r + 1 = 0$. The quadratic formula gives
$$r = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt3.$$
7. Pick the right one. The short leg divided by the long leg is less than $1$. Since $\sqrt3 \approx 1.73$, the value $2 + \sqrt3 \approx 3.73$ is too big, and $2 - \sqrt3 \approx 0.27$ is the answer.

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

Knowing the sum of two numbers and the sum of their squares tells you their product, because $(a+b)^2 = a^2 + 2ab + b^2$. Sum and product are enough to find anything that treats $a$ and $b$ fairly, including $\frac ab + \frac ba$. Going straight for the ratio avoids ever working out $a$ and $b$ themselves, which are ugly.

If you know a little trigonometry: call $x$ the small angle of the shaded triangle, opposite the short leg. The legs are $\sqrt2 \sin x$ and $\sqrt2 \cos x$, and they add to $\sqrt3$. Squaring, $2(1 + 2\sin x \cos x) = 3$, so $2 \sin x \cos x = \frac12$. That means $\sin 2x = \frac12$, so $2x = 30^\circ$ and $x = 15^\circ$. The ratio short over long is $\tan 15^\circ = 2 - \sqrt3$.