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

Since we want non-congruent triangles that are neither isosceles nor 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 $a=2$, 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 $a=3$, $c$ can be $b+1$ or $b+2$, which gives triangles $(3,4,5), (3,4,6), (3,5,6)$. Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.

All in all, this gives us $3+2 = \boxed{\textbf{(C)}\; 5}$ triangles.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png