AMC 10 · 2003 · #7
Grade 7 geometry-2dHow many non-congruent triangles with perimeter 7 have integer side lengths?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count the triangles whose three side lengths are whole numbers and add up to $7$, where two triangles that are the same shape and size (congruent) count only once.
Givens: Each triangle has three side lengths that are positive integers; The three side lengths add up to a perimeter of $7$; Triangles that are congruent (same three side lengths) are counted only once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: How many non-congruent integer-sided triangles have perimeter $7$
Understand
Restated: Count the triangles whose three side lengths are whole numbers and add up to $7$, where two triangles that are the same shape and size (congruent) count only once.
Givens: Each triangle has three side lengths that are positive integers; The three side lengths add up to a perimeter of $7$; Triangles that are congruent (same three side lengths) are counted only once; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #14 Extreme Principle, #3 Eliminate Possibilities
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 $a \le b \le c$ 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.
Execute — Answer: B
7.G.A.2 Step 1 Bound the longest side
- Write the sides in increasing order as $a \le b \le c$, so $c$ is the longest.
- The triangle inequality says the two shorter sides must together beat the longest: $a+b>c$.
- Since the perimeter is $7$, the two shorter sides add to $a+b=7-c$.
- Substituting gives $7-c>c$.
💡 The longest side is the one most likely to break the triangle, so testing it against the other two is the sharpest single check.
6.EE.B.8 Step 2 Pin the longest side to a single value
- Solve $7-c>c$: add $c$ to both sides to get $7>2c$, so $c<3.5$.
- Because $c$ is the longest of three sides summing to $7$, it must be at least a third of $7$, so $c \ge 3$.
- The only whole number with $3 \le c < 3.5$ is $c=3$.
- So every such triangle has longest side $3$, and the two shorter sides add to $a+b=7-3=4$.
💡 Squeezing the longest side between a lower and an upper bound leaves just one whole number, collapsing the whole search to one case.
7.G.A.2 Step 3 List the side combinations and keep the real triangles
- With $c=3$ fixed, list positive integers $a \le b \le 3$ that add to $a+b=4$: either $a=1,b=3$ giving sides $(1,3,3)$, or $a=2,b=2$ giving sides $(2,2,3)$.
- Check each with the triangle inequality: for $(1,3,3)$, $1+3=4>3$, so it works; for $(2,2,3)$, $2+2=4>3$, so it works too.
- Both are valid and they are different shapes, giving $2$ triangles, which is choice (B).
💡 Once the longest side is fixed, only a couple of ordered pairs remain, so a short honest list finishes the count.
7.G.A.2 Write the sides in increasing order as $a \le b \le c$, so $c$ is the longest. T 6.EE.B.8 Solve $7-c>c$: add $c$ to both sides to get $7>2c$, so $c<3.5$. Because $c$ is t 7.G.A.2 With $c=3$ fixed, list positive integers $a \le b \le 3$ that add to $a+b=4$: ei Review
Reasonableness: Sanity-check by ignoring the triangle rule and just listing integer sides $a \le b \le c$ 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).
Alternative: Skip the bounding step and brute-force every ordered triple with $a \le b \le c$ summing to $7$, then apply the triangle inequality to each: $(1,1,5)$ no, $(1,2,4)$ no, $(1,3,3)$ yes, $(2,2,3)$ yes. Counting the "yes" cases gives $2$, matching choice (B); the bounding method just reaches the same two cases faster by ruling out $c \ne 3$ up front.
CCSS standards used (min grade 7)
7.G.A.2Draw geometric shapes with given conditions including triangles (Applying the triangle inequality to decide which sets of three side lengths actually form a triangle.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Solving $7-c>c$ to get $c<3.5$ and combining it with $c \ge 3$ to pin $c=3$.)
⭐ 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.
⭐ 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.
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