2017 AMC 10A Problem 21

Both squares cut off small right triangles that have the same shape as the $3$-$4$-$5$ triangle. Matching shapes means matching ratios of sides, and that is enough to find each square.

Part 1: the square in the corner ($x$).

1. Stand the triangle with the leg of length $4$ along the bottom and the leg of length $3$ going up the left side. The square sits in the bottom-left corner, and its top-right corner touches the hypotenuse.
2. Above the square is a small right triangle. Its bottom side is the top of the square, length $x$. Its height is the rest of the left leg, $3 - x$.
3. Its top side is flat like the big triangle's bottom, and it shares the top corner, so it has the same shape as the big triangle. In the big triangle, bottom $\div$ height $= 4 \div 3$. So
$$\frac{x}{3 - x} = \frac43.$$
4. Cross-multiply: $3x = 4(3 - x) = 12 - 4x$, so $7x = 12$ and $x = \dfrac{12}{7}$.

Part 2: the square on the hypotenuse ($y$).

5. Now the square lies along the hypotenuse, and its two top corners touch the two legs. Along the hypotenuse, from one end to the other, you pass: a small corner triangle, the square's bottom side $y$, and another small corner triangle.
6. Each corner triangle has a right angle (the square's side stands straight up from the hypotenuse) and shares a corner angle with the big triangle. So each has the $3$-$4$-$5$ shape. In each one, the square's side $y$ is the leg standing up, and the other leg lies along the hypotenuse.
7. Corner where the leg of length $4$ meets the hypotenuse. In the big triangle, the side across from this corner is the short leg $3$ and the side next to it is $4$. In the small triangle, $y$ is across from this corner, so $y$ plays the role of $3$. The piece along the hypotenuse plays the role of $4$: it is $\tfrac43 y$.
8. Corner where the leg of length $3$ meets the hypotenuse. Here the roles swap: $y$ plays the role of $4$, and the piece along the hypotenuse is $\tfrac34 y$.
9. The three pieces fill the hypotenuse of length $5$:
$$\tfrac43 y + y + \tfrac34 y = 5.$$
Over twelfths: $\tfrac{16}{12}y + \tfrac{12}{12}y + \tfrac{9}{12}y = \tfrac{37}{12}y = 5$, so $y = \dfrac{60}{37}$.

Part 3: the ratio.

1. $$\frac{x}{y} = \frac{12}{7} \div \frac{60}{37} = \frac{12}{7} \cdot \frac{37}{60} = \frac{37}{35}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{37}{35}}$.

A square tucked inside a right triangle always leaves small right triangles that share an angle with the big one, so they are all the same shape. Write one side of the big triangle as a sum of pieces, express each piece using the square's side and the $3 : 4$ ratio, and solve one equation. The only care needed is matching each small triangle's legs to the right sides of the big triangle.

For the square on the hypotenuse, use the triangle on top instead. The height of the big triangle measured from the hypotenuse is $\frac{3 \cdot 4}{5} = \frac{12}{5}$ (twice the area divided by the base). The triangle above the square has base $y$ (the square's top side, parallel to the hypotenuse) and height $\frac{12}{5} - y$, and it has the same shape as the big triangle. So base $\div$ height matches:
$$\frac{y}{\frac{12}{5} - y} = \frac{5}{\frac{12}{5}}.$$
Cross-multiplying gives $\tfrac{12}{5}y = 12 - 5y$, so $\tfrac{37}{5}y = 12$ and $y = \frac{60}{37}$, as before.