2018 AMC 10A Problem 16

The hypotenuse is $AC = \sqrt{20^2 + 21^2} = \sqrt{841} = 29$. The shortest segment from $B$ to $\overline{AC}$ is the altitude $BH$, with
$$BH = \frac{AB \cdot BC}{AC} = \frac{420}{29} \approx 14.48.$$

Let a point $X$ travel from $A$ to $C$ along the hypotenuse. The length $BX$ starts at $BA = 20$, decreases steadily to $BH \approx 14.48$ at the foot $H$, then increases steadily to $BC = 21$.

• On the stretch from $A$ to $H$, $BX$ takes every value in $[14.48, 20]$ exactly once, so the integers $15, 16, 17, 18, 19, 20$ each occur once: $6$ segments.
• On the stretch from $H$ to $C$, $BX$ takes every value in $[14.48, 21]$ exactly once: the integers $15, \dots, 21$ occur once each: $7$ segments.

No segment has length $14$ or less, since $BH > 14$. Total: $6 + 7 = 13$.

The answer is $\boxed{\textbf{(D)}\ 13}$.

The distance from a fixed point to a moving point on a segment is a continuous function with a single minimum at the perpendicular foot, so it sweeps each value between the minimum and an endpoint exactly once on each side. Counting integer lengths becomes counting integers in two intervals; the only real computation is the altitude, which fixes the lower bound. Recognizing $20$-$21$-$29$ as a Pythagorean triple saves time.

• Counting integers from 15 to 21 once (7 values) and forgetting that lengths 15 through 20 occur twice, on both sides of the foot of the altitude.
• Rounding the altitude to $14$ and including length $14$, giving $15$, or excluding the legs themselves despite the problem saying to include them.