AMC 10 · 2019 · #10

Grade 8 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

pythagorean-theoremcoordinate-geometryarea-trianglesperimeterpolygon-inequality identify-subproblemscasework ↑ Prerequisites: pythagorean-theoremarea-trianglescoordinate-geometry

📏 Medium solution 💡 3 insights

Problem

Points A and B are 10 units apart in a plane. Count all points C such that △ ABC has perimeter exactly 50 and area exactly 100.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Plant AB on a coordinate axis (Tool #1 + #9 makes the picture concrete). Split the requirements into (a) area → height of C above AB, and (b) perimeter → sum of slanted sides AC + BC (Tool #7 sub-questions). Then test whether ANY C at the required height can give the required slant sum — try the best (closest) candidate first (Tool #6); if even that fails, no C works.

1STEP 1

Put A = (-5, 0), B = (5, 0) so AB lies on the x-axis, and let C = (x, y).

A = (-5, 0), B = (5, 0), C = (x, y)

2STEP 2

Area: with AB (length 10) as base, area = 5|y| = 100 gives |y| = 20, so C lies on y = 20 or y = -20.

5|y| = 100 → |y| = 20

3STEP 3

Perimeter: 10 + |AC| + |BC| = 50, so |AC| + |BC| = 40 — the two slanted sides must total 40.

|AC| + |BC| = 50 - 10 = 40

4STEP 4

Test the symmetric point C = (0, 20): Pythagoras gives |AC| = |BC| = √(425) = 5√(17), so the sum is 10√(17).

|AC| + |BC| = 10√(17)

5STEP 5

Estimate √(17) ≈ 4.123, so 10√(17) ≈ 41.23 — already more than 40.

10√(17) ≈ 41.23 ＞ 40

6STEP 6

Reflecting A across y = 20 with the triangle inequality shows C = (0, 20) is the minimum, so |AC| + |BC| ≥ 10√(17) ＞ 40 for all such C.

min_C on y=20(|AC|+|BC|) = 10√(17) ＞ 40

7STEP 7

By symmetry the same fails on y = -20, so no C meets both conditions — the count is 0, choice (A).

Count of valid C = 0 → (A)

Answer

Quick gut check: a triangle with base 10 and area 100 has height 20 — it's a very TALL skinny triangle. The shortest slant from a point 20 up to either endpoint of a base 10 wide is already at least √(5² + 20²) ≈ 20.6. Two of those slants sum to at least ≈ 41.2, plus the base 10 gives perimeter at least ≈ 51.2 ＞ 50. There's no room — answer (A) 0 is consistent.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean thinking you already know: area =100 forces C to sit 20 above (or below) AB. Even the closest such C gives slants summing to 10√(17)≈ 41.2 ＞ 40, so the perimeter 50 is impossible — 0 points.

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