2015 AMC 12B Problems/Problem 10

Problem

How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?

$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$

Solution

Listing all possible triangle side lengths satisfying the constraints, we find the following:

$\text{2-3-4}\\ \text{2-4-5}\\ \text{2-5-6}\\ \text{3-4-6}\\ \text{3-5-6}$

Thus the answer is $\fbox{\textbf{(C)}\; 5}$ .

More explanation:

Since we want non-congruent triangles that are not isosceles or equilateral, we can just list side lengths $a-b-c$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$

There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$

When $a=2$ then, similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives $3$ triangles ( $2-3-4, 2-4-5, 2-5-6$ ) within our perimeter constraint.

When $a=2$ then $c$ can be $b+1$ or $b+2$ , which gives $3-4-5, 3-4-6, 3-5-6$ . Note that $3-4-5$ is a right triangle, so we get rid of it and we get $2$ triangles.

All in all this gives us 5 triangles.

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2015 AMC 12B Problems/Problem 10

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