how many possible acute triangles with perimeter given

Question

How many possible acute triangles exist with perimeter 18? All sides are positive integers. The triangle (7,7,4) is the same as (4,7,7). I need the work in a way that a geometry 9th grade student would be able to come up with.

Answer 1

Since the triangle is required to be an acute, recall that by a corollary to the Pythagorean Theorem we have that:

$$a^2 + b^2 > c^2$$

Where $a$, $b$, and $c$ are the given side lengths of the triangle.

Also, note that we have by the Triangle Inequality that: $a + b > c$, and is respective with all three of the sides, and the constraint that $a + b + c = 18$, or that the perimeter is $18$.

The only way to find all of the cases is through case-work, given the constraints i have mentioned above. So keep guessing numbers and checking them.

Answer 2

Let's assume that $c \geq a \land c \geq b$; this forms the first constraint:

$$ c \geq \frac{18}{3} \implies c \geq 6 $$

A valid triangle requires $a + b > c$; substituting the perimeter constraint gives us:

$$ a + b > c \implies (18 - c) > c \implies c < 9 $$

You can work out the values for $a$ and $b$ from the three possible values of $c$ and the following inequality between $a$ and $b$:

$$ a = (18 - c) - b\\ 18 - 2 \cdot c \leq a \leq c $$

• $(6, 6, 6)$
• $(4, 7, 7)$
• $(5, 6, 7)$
• $(2, 8, 8)$
• $(3, 7, 8)$
• $(4, 6, 8)$
• $(5, 5, 8)$

These findings must also satisfy the condition of an acute triangle; from the Pythagorean theorem's converse, we learn that:

If $a^2 + b^2 > c^2$, then the triangle is acute.

After substituting $c$ for $18 - a - b$:

$$ a^2 + b^2 > (18 - a - b)^2\\ \implies a + b - \frac{a\cdot b}{18} > \frac{18^2}{2\cdot18}\\ \implies a + b - \frac{a\cdot b}{18} > 9 $$

This eliminates $(4, 6, 8)$, $(3, 7, 8)$ and $(5, 5, 8)$.