AMC 10 · 2015 · #2

Grade 7 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
polygon-inequalityperimeterinterval-arithmetic bound-inequality-then-enumerate ↑ Prerequisites: polygon-inequalityperimeter
📏 Medium solution 💡 2 insights
Problem
Two sides of a triangle are known and the third is free. Find the perimeter that is impossible.

Pick an answer.

(A)
52
(B)
57
(C)
62
(D)
67
(E)
72
How to solve
Strategy Extreme Principle

The whole problem is a question about a boundary — where does the set of possible perimeters stop — so Tool #14 (Extreme Principle) is the one that decides it. Pushing the two known sides to their two extreme positions, folded flat open and flat closed, pins down both ends of the range at once and shows why each end is excluded rather than included. Tool #4 (Introduce a Variable) names the third side x so that the perimeter and the third side become the same unknown wearing two hats. Tool #1 (Draw a Diagram) supplies the hinge picture that makes the extremes visible and, just as importantly, shows that the in-between values are all actually reached. Tool #3 (Eliminate Possibilities) then does the cheap final pass: four choices sit inside the range, one does not.

1STEP 1

Tie the perimeter to the third side

The perimeter and the third side move together.

P=20+15+x=35+x ⟺ x=P-35
2STEP 2

Swing the two sides on a hinge

Swinging the two sides sweeps a whole range.

gap at 0° = 20-15 = 5, gap at 180° = 20+15 = 35
3STEP 3

The two ends are not triangles

The two ends are flat, not triangles.

x=5 or x=35 → corners collinear, zero area → 5 ＜ x ＜ 35
4STEP 4

Shift the range to perimeters

Shifting gives the perimeter range.

5 ＜ x ＜ 35 → 5+35 ＜ x+35 ＜ 35+35 → 40 ＜ P ＜ 70
5STEP 5

Test the five numbers

Only 72 falls outside it, choice (E).

52,57,62,67 ∈ (40,70) but 72 ＞ 70 → (E)
Answer
72
The rejected perimeter can be checked on its own terms with no reference to the range at all: 72 forces a third side of 37, and 20+15=35 ＜ 37, so the two known sides cannot meet. Each accepted perimeter can be verified the same direct way — for 52 the sides are 20,15,17 and all three checks pass, since 15+17=32 ＞ 20, 20+17=37 ＞ 15, and 20+15=35 ＞ 17. The two ways to go wrong here are both about care rather than cleverness. First, checking only the inequality x ＜ 20+15 and forgetting x ＞ 20-15 gives an upper bound with no lower bound; that still happens to reject 72, but it would fail on a version of this problem whose odd choice was too small. Second, treating the bounds as inclusive would admit 70; that is genuinely wrong, though it does not change the answer here because 70 is not offered. Note also that the answer does not depend on the third side being a whole number, which is right, because the problem never says it is.
💡Key takeaway

Two sides joined at a hinge can only span a gap between their difference and their sum, so the perimeter can be anything strictly between twice the longer side and twice the two sides added together.

  • Tie the perimeter to the third side
  • Swing the two sides on a hinge
  • The two ends are not triangles
  • Shift the range to perimeters
  • Test the five numbers

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