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2018 AMC 10A Problem 16

2018 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: October 28, 2026

Triangles: Area & Pythagorean · difficulty 3 of 5 · about 4 min · position P16-20

Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$ . Including $\overline{AB}$ and $\overline{BC}$ , how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$ ?

A) $5$
B) $8$
C) $12$
D) $13$
E) $15$

Answer

Key insight

Sliding along the hypotenuse, the segment shrinks from 20 to the height (about 14.5), then grows to 21, hitting each whole number once per side.

Solution

Picture a point $X$ sliding along the hypotenuse from $A$ to $C$ , and watch the length of $BX$ .

1. Find the hypotenuse. $AC = \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29$ .
2. Find the shortest segment. The shortest segment from $B$ to the hypotenuse is the one that meets it at a right angle, the height $BH$ . Compute it with area. Using the legs, the area is $\tfrac12 \cdot 20 \cdot 21 = 210$ . Using the hypotenuse as the base, the area is $\tfrac12 \cdot 29 \cdot BH$ . So
$\tfrac12 \cdot 29 \cdot BH = 210, \qquad BH = \frac{420}{29} \approx 14.48.$
3. Watch the length change. At $A$ , $BX = 20$ . As $X$ moves toward $H$ , $BX$ shrinks smoothly to about $14.48$ . As $X$ moves on from $H$ to $C$ , $BX$ grows smoothly to $21$ .
4. Count whole-number lengths on the first stretch ( $A$ to $H$ ). The length passes through every value from $20$ down to $14.48$ , so it hits $20, 19, 18, 17, 16, 15$ once each. That is $6$ segments (including $BA$ itself).
5. Count on the second stretch ( $H$ to $C$ ). The length passes through every value from $14.48$ up to $21$ , so it hits $15, 16, \dots, 21$ once each. That is $7$ segments (including $BC$ itself).
6. Nothing shorter. No segment is shorter than $14.48$ , so length $14$ or less never happens.

Total: $6 + 7 = 13$ .

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

Why this works

The distance from a fixed point to a point sliding along a line gets smaller until the right-angle spot, then gets larger. So each length between the shortest one and an end is reached exactly once on each side. The only real calculation is the height, which you can get from the area in two ways. Knowing that $20$ , $21$ , $29$ is a right-triangle triple saves a step.

The trap

Counting the lengths 15 through 21 only once (7 segments), forgetting that lengths 15 through 20 each appear twice, once on each side of the shortest segment.

Concise solution

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 $\boxed{\textbf{(D)}\ 13}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed