All the sides of a triangle are integers. If the perimeter of the triangle is 16, then how many different possible triangles are there? (Assume that the triangle is non-degenerate. Two triangles are considered the same if they are congruent.)

Answer 1

To find the number of different triangles with integer sides and a perimeter of 16, we denote the sides of the triangle as \(a\), \(b\), and \(c\), where \(a \leq b \leq c\). The triangle inequality states that the sum of the lengths of any two sides must be greater than the length of the third side. In this scenario, we also have the condition:

\[ a + b + c = 16 \]

To apply the triangle inequality, we focus on the most restrictive form:

\[ a + b > c \]

Given the perimeter condition, we can express \(c\) in terms of \(a\) and \(b\):

\[ c = 16 - a - b \]

Thus, the triangle inequality can be rewritten as:

\[ a + b > 16 - a - b \]

Simplifying this, we obtain:

\[ 2(a + b) > 16 \implies a + b > 8 \]

Furthermore, since \(a \leq b \leq c\) implies that \(c\) is the largest side, we also know:

\[ c \geq b \implies 16 - a - b \geq b \implies 16 - a \geq 2b \implies 16 \geq a + 2b \]

We now have two inequalities to satisfy:

1. \(a + b > 8\)
2. \(a + 2b \leq 16\)

We can express \(b\) in terms of \(a\):

From \(a + b > 8\):

\[ b > 8 - a \]

From \(a + 2b \leq 16\):

\[ 2b \leq 16 - a \implies b \leq \frac{16 - a}{2} \]

Now we combine these results for \(b\):

\[ 8 - a < b \leq \frac{16 - a}{2} \]

Next, we set bounds on \(b\).

Starting with the first part:

1. Rearranging gives \(b > 8 - a\) and writing \(b\) as \(b = k\), we have: \[ k > 8 - a \]
2. The second part gives: \[ k \leq \frac{16 - a}{2} \]

To find integer solutions for \(a\) and \(b\) (and \(c\)), we will explore integer values for \(a\) from 1 to \(7\) (since \(a + b + c = 16\) and \(c\) must be the largest side).

Iterating over possible values of \(a\):

For \(a = 1\):

• \(b > 7\), \(b \leq \frac{15}{2}\) → no valid \(b\).

For \(a = 2\):

• \(b > 6\), \(b \leq 7\) → \(b = 7\) → \(c = 7\) → 1 triangle: \((2, 7, 7)\).

For \(a = 3\):

• \(b > 5\), \(b \leq 6.5\) → \(b = 6\) → \(c = 7\) → 1 triangle: \((3, 6, 7)\).

For \(a = 4\):

• \(b > 4\), \(b \leq 6\) → \(b = 5, 6\) → Triangles:
  • \((4, 5, 7)\)
  • \((4, 6, 6)\) → 2 triangles.

For \(a = 5\):

• \(b > 3\), \(b \leq 5.5\) → \(b = 5\) → \(c = 6\) → 1 triangle: \((5, 5, 6)\).

For \(a = 6\):

• \(b > 2\), \(b \leq 4\) → no valid \(b\).

For \(a = 7\):

• \(b > 1\), \(b\) must be 7: no valid triangle possible (as \(b = 7\) leads \(c\) to be 2).

Summarizing, we find the unique non-congruent triangles:

1. \((2, 7, 7)\)
2. \((3, 6, 7)\)
3. \((4, 5, 7)\)
4. \((4, 6, 6)\)
5. \((5, 5, 6)\)

In total, there are \(5\) unique triangles. Thus, the answer is:

\[ \boxed{5} \]