2023 AMC 10B Problem 3

To compare the circles, we need their sizes. The triangles give us those sizes through one key fact.

1. The key fact. If a right triangle has all three corners on a circle, its longest side (the hypotenuse) goes straight through the center. In other words, the hypotenuse is a diameter. (The right angle "looks at" half the circle, and half a circle is cut off by a diameter.)
2. Read off the diameters. The $3$-$4$-$5$ triangle has hypotenuse $5$, so circle $A$ has diameter $5$ and radius $\frac{5}{2}$. The $5$-$12$-$13$ triangle has hypotenuse $13$, so circle $B$ has diameter $13$ and radius $\frac{13}{2}$.
3. Compare the areas. Area is $\pi r^2$:
$$\frac{\pi \left(\frac{5}{2}\right)^2}{\pi \left(\frac{13}{2}\right)^2} = \frac{\frac{25}{4}}{\frac{169}{4}} = \frac{25}{169}.$$
The $\pi$ and the $\frac{1}{4}$ cancel, so you could simply square $\frac{5}{13}$.

The answer is $\boxed{\textbf{(D)}\ \frac{25}{169}}$.

"A right triangle inside a circle" should make you think "the hypotenuse is a diameter" right away; the two shorter sides do not matter after that. And when one circle is a scaled copy of another, the area ratio is the square of the length ratio, so you never need to multiply out $\pi$.

An inscribed angle of $90^\circ$ subtends a semicircle, so the hypotenuse of each right triangle is a diameter of its circle. Circle $A$ has diameter $5$ and circle $B$ has diameter $13$.

Area is proportional to the square of the radius (or of the diameter), so
$$\frac{\text{area of } A}{\text{area of } B} = \left(\frac{5}{13}\right)^2 = \frac{25}{169}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{25}{169}}$.