how many possible acute triangles with perimeter given 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.
 A: 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.
A: 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 $(5, 5, 8)$.
