AMC 10 · 2018 · #16

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoremarea-trianglesinterval-arithmetic bound-inequality-then-enumerate ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
A right triangle has the right angle at B, with legs AB = 20 and BC = 21. Counting the two legs themselves, find how many segments from B to a point on the hypotenuse AC have a whole-number length.

Pick an answer.

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

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A picture pins down where B, the hypotenuse, and the foot of the altitude sit. The key is the extreme case: the shortest possible segment is the altitude from B, which fixes the smallest length. Then split the hypotenuse at the foot into two parts and count the integer lengths each part sweeps through.

1STEP 1

Find the hypotenuse

Right angle at B makes AC the hypotenuse, so AC = √(20² + 21²) = √841 = 29.

AC = √(20² + 21²) = √(841) = 29
2STEP 2

Shortest segment is the altitude

The shortest segment is the altitude h from B to AC; equating areas ½·20·21 = ½·29·h gives h = 420/29.

1/2(20)(21) = 1/2(29)h → h = 420/29
3STEP 3

Smallest whole-number length

Since 420/29 ≈ 14.48, no segment is that short, so the smallest possible whole-number length is 15.

h = 420/29 ≈ 14.48 → shortest integer length = 15
4STEP 4

Sweep each side of the foot

From A to the foot H the length falls through 20,19,18,17,16,15 (six); from H to C it rises through 15,16,17,18,19,20,21 (seven).

A → H: 20,19,18,17,16,15 H → C: 15,16,17,18,19,20,21
5STEP 5

Count without double counting

Different landing points mean the two sides share no segment, so 6 + 7 = 13 distinct segments — answer (D).

6 + 7 = 13
Answer
13
The lengths run from a minimum of about 14.48 up to a maximum of 21 (the longer leg). The only whole numbers in that range are 15 through 21. Length 21 is reached just once (it is the leg BC, the unique longest segment), while every length from 15 to 20 is reached twice — once on each side of the foot. That is 1 + 6*2 = 13, matching the count and confirming answer (D).
💡Key takeaway

The shortest reach from the corner is the straight-down drop, and from there the length climbs to each leg — count every whole number it passes on both sides.

  • Find the hypotenuse
  • Shortest segment is the altitude
  • Smallest whole-number length
  • Sweep each side of the foot
  • Count without double counting

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