2021 AMC 10B Problem 14

Let the spacing between adjacent lines be $d$ and the radius be $r$. Chords of equal length are the same distance from the center, so the center $O$ lies on the line midway between the two chords of length $38$, at distance $\frac{d}{2}$ from each. The chords of length $38$ must therefore be the two adjacent ones, and the $34$-chord is the third line, at distance $\frac{d}{2} + d = \frac{3d}{2}$ from $O$.

Drop perpendiculars from $O$ to a $38$-chord and to the $34$-chord; each bisects its chord, giving right triangles with legs $19$ and $\frac{d}{2}$, and $17$ and $\frac{3d}{2}$, both with hypotenuse $r$:
$$19^2 + \frac{d^2}{4} = 17^2 + \frac{9d^2}{4}.$$
So $361 - 289 = 2d^2$, i.e. $72 = 2d^2$ and $d = 6$.

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

Chord length is determined by distance from the center ($\text{half-chord}^2 + \text{distance}^2 = r^2$), so two equal chords locate the center by symmetry, and a third chord's length becomes an equation in the unknown spacing. Whenever parallel chords appear, write $r^2$ two ways and subtract; the unknown $r$ disappears.

• Placing the center on the middle line or on one of the 38-chords, instead of halfway between the two equal chords.
• Using the full chord lengths $38$ and $34$ instead of the half-chords $19$ and $17$ in the Pythagorean relation.