2003 AMC 10A Problem 7

P1-10~2 minPrint

Triangles: Area & Pythagorean Basic Counting Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

How many non-congruent triangles with perimeter $7$ have integer side lengths?

Show solution without answering

Key insight

The longest side must be shorter than the other two together, so it is less than half of 7. That makes it 3, and the others add to 4.

Solution

Three lengths make a triangle only if the two shorter ones together are longer than the longest one. The longest side is the one that can cause trouble, so pin it down first.

1. The longest side has to be less than the other two together. All three add to $7$ , so the longest side is less than half of $7$ , which is $3.5$ . As a whole number, it is at most $3$ .
2. It cannot be smaller than $3$ : three sides of at most $2$ add to at most $2 + 2 + 2 = 6$ , which is not $7$ . So the longest side is exactly $3$ .
3. The other two sides add to $7 - 3 = 4$ , and neither is more than $3$ . The options are $1 + 3$ and $2 + 2$ .
4. Check both triangles:
- Sides $1, 3, 3$ : $1 + 3 = 4 > 3$ . Works.
- Sides $2, 2, 3$ : $2 + 2 = 4 > 3$ . Works.
5. For comparison, splits like $1, 1, 5$ or $1, 2, 4$ fail, because $1 + 1 < 5$ and $1 + 2 < 4$ : the short sides cannot reach.

There are exactly $2$ triangles.

The answer is $\boxed{\textbf{(B)}\ 2}$ .

Why this works

For whole-number triangles with a fixed perimeter, the only rule to check is the triangle inequality, and it is easiest to check on the longest side: that side must be less than half the perimeter. Fixing the longest side first shrinks the list to a couple of cases you can check by hand.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let the sides be integers $a \le b \le c$ with $a + b + c = 7$ . The triangle inequality $a + b > c$ is equivalent to $c < \tfrac{7}{2}$ , so $c \le 3$ . Since three sides of at most $3$ must total $7$ , and $2 + 2 + 2 = 6$ is too small, the longest side is exactly $c = 3$ .

Then $a + b = 4$ with $a \le b \le 3$ : either $(a, b) = (1, 3)$ or $(2, 2)$ .

Both triples $(1, 3, 3)$ and $(2, 2, 3)$ satisfy $a + b > c$ , so there are exactly $2$ non-congruent triangles.

The answer is $\boxed{\textbf{(B)}\ 2}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)