Prove a triangle is equilateral I have to prove that a triangle is equilateral given that there is a unit circle inscribed in the triangle and that the triangle's heights are integers.
I'm unsure on how to proceed with the question. My attempts at proving this end up in dead ends.
Edit: As requested, one of my dead ends (at least the one I think is closest to the right answer)
A unit circle inscribed in triangle ABC touches the it at the midpoints of line AB, line BC, and line AC (from now on I'll denote them by M, N, and P respectively). The height of the trangle is the line AN, which passes through the unit circle through its origin. Therefore, $$h = \overline{AN} = 2+x $$ where x is the measurement of the line between A and the top of the unit circle. X is also an integer since the question says the height has to be an integer.
This is where I reached a dead end
 A: Let $p$ and $A$ be the perimeter and the area of the triangle.
1) We have that
$$\frac{1}{r}=\frac{p}{2A}=\frac{a}{2A}+\frac{b}{2A}+\frac{c}{2A}=
\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}$$
where $a$, $b$, $c$ are the sides,  $h_a$, $h_b$, $h_c$ are the corresponding altitudes and $r$ is the radius of the inscribed circle. 
2) The triangle inequality $a<b+c$ implies
$$\frac{1}{h_a}=\frac{a}{2A}<\frac{b}{2A}+\frac{c}{2A}=\frac{1}{h_b}+\frac{1}{h_c}.$$
Hence, by 1) and 2), any altitude is longer than the diameter of the incircle:
$$\frac{1}{h_a}<\frac{1}{h_b}+\frac{1}{h_c}=r-\frac{1}{h_a}\quad
\Rightarrow\quad  h_a>2r.$$ 
Since $r=1$ then $h_a>2r=2$. Moreover $h_a\geq 3$ because $h_a$ is an integer.
In a similar way, we obtain $h_b\geq 3$, $h_c\geq 3$. 
By 1) we have that $h_a=h_b=h_c=3$ (otherwise $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}<1$) and finally
$$a=\frac{2A}{h_a}=\frac{2A}{3},\; b=\frac{2A}{h_b}=\frac{2A}{3},\;c=\frac{2A}{h_c}=\frac{2A}{3}$$
which means that the triangle is equilateral. 
