AMC 10 · 2003 · #7

Grade 7 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
polygon-inequalitysystematic-enumeration bound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: polygon-inequality
📏 Medium solution 💡 2 insights
Problem
A triangle has three side lengths that are whole numbers and a perimeter of 7, and two congruent (same shape and size) triangles count as one. How many such triangles are there?

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question asks "how many" whole-number triangles fit a fixed perimeter, which is a finite counting task — tool #2 (Make a Systematic List) is built for exactly this, and writing sides in order a ≤ b ≤ c makes the list complete with no repeats. Tool #14 (Extreme Principle) tames the search first: the single longest side c is the tightest constraint, and the triangle inequality caps how big it can be, which pins c to one value before any listing starts. Tool #3 (Eliminate Possibilities) does the final filtering: after listing the integer side combinations, throw out any that fail the triangle inequality, and count what survives.

1STEP 1

Bound the longest side

Order the sides a ≤ b ≤ c. The triangle inequality needs a+b ＞ c, and a+b=7-c, so 7-c ＞ c.

a+b ＞ c → 7-c ＞ c
2STEP 2

Pin the longest side to a single value

Solving 7 ＞ 2c gives c ＜ 3.5, and c is at least a third of 7, so c=3 and the other two sides sum to 4.

7-c ＞ c → c ＜ 3.5, c ≥ 7/3 → c=3, a+b=4
3STEP 3

List the side combinations and keep the real triangles

Pairs summing to 4 with a ≤ b ≤ 3 give (1,3,3) and (2,2,3); both pass the inequality, so there are 2 triangles.

(1,3,3): 1+3 ＞ 3 ✓ (2,2,3): 2+2 ＞ 3 ✓ → 2 triangles (B)
Answer
2
Sanity-check by ignoring the triangle rule and just listing integer sides a ≤ b ≤ c with a+b+c=7: (1,1,5),(1,2,4),(1,3,3),(2,2,3) — four combinations. The first two fail the triangle inequality (1+1 ＜ 5 and 1+2 ＜ 4: too flat to close up), leaving exactly the two found, (1,3,3) and (2,2,3). This confirms 2 and shows why the answer is not the larger 4 (choice D).
💡Key takeaway

For integer-sided triangles with a fixed perimeter, first bound the longest side with the triangle inequality, then list the few remaining side combinations that fit.

  • Bound the longest side
  • Pin the longest side to a single value
  • List the side combinations and keep the real triangles

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