2021 AMC 10B Problem 14

Step 1: find where the center is.

1. A chord's length depends only on how far it is from the center: the farther away, the shorter. So the two chords of length $38$ are the same distance from the center $O$.
2. That puts $O$ exactly halfway between those two chords. Since the lines are equally spaced, the two $38$-chords must be next to each other, and the $34$-chord is the third, outer line.
3. Call the spacing $d$. Then $O$ is $\frac{d}{2}$ from each $38$-chord and $\frac{d}{2} + d = \frac{3d}{2}$ from the $34$-chord.

Step 2: make right triangles.

4. The segment from $O$ straight to a chord (at a right angle) cuts the chord in half. Joining $O$ to an end of the chord makes a right triangle whose longest side is the radius $r$.
5. For a $38$-chord the legs are $19$ and $\frac{d}{2}$. For the $34$-chord the legs are $17$ and $\frac{3d}{2}$.
6. Both triangles have longest side $r$, so by the Pythagorean theorem
$$19^2 + \frac{d^2}{4} = r^2 = 17^2 + \frac{9d^2}{4}.$$

Step 3: solve.

1. $361 - 289 = \frac{9d^2}{4} - \frac{d^2}{4}$, which is $72 = 2d^2$.
2. So $d^2 = 36$ and $d = 6$.

The answer is $\boxed{\textbf{(B)}\ 6}$.

Half a chord, the distance to the center, and the radius always form a right triangle. Equal chords give you the center's position by symmetry. Writing $r^2$ two ways and setting them equal makes the unknown radius drop out, which is the standard move for parallel chords.