Number of integer triangles. 
Number of integer isosceles or equilateral triangle none of whose
  sides exceed $2c$ is?

I substituted $c =3$ checked, and got $27= 3 \times (3^2)$ i.e. $3c^2$ triplets. It gives the right answer($3c^2$) but how do I write a proper formal proof of this?
I tried it this way: We have three vacant places for three integers. First place can be filled in $2c$ ways, 2nd one in $2c$ ways, 3rd one in $2c$ way but it involves over-counting, is it possible to eliminate the extras? 
 A: We can define an isoceles triangle by an ordered pair $(b,n)$ where $b$ is the length of the base side and $n$ is the length of the two congruent sides.
The basic criteria are (basically the triangle inequalities)
i) $b > 0; n > 0$
ii) $b < 2n$
iii) (trivial and dumb) $n < b + n$ (which... well, duh....)
Any such pair is possible.
So with the additional requirement that $b \le 2c$ and $n \le 2c$, and $n,b \in \mathbb Z$ how many such ordered pairs may there be?
Well, our conditions are now:
i') $1 \le b\le 2c; 1\le n \le 2c$
ii') $b\le 2n-1$
iv) $n,b, c \in \mathbb Z$.
or in other words $1\le n \le 2c; 1 \le b \le \min(2n-1, 2c)$.
And ANY such pair of natural numbers will do.
So there are $\sum_{n=1}^{2c}(\sum_{b=1}^{\min(2c,2n-1)
} 1)=\sum_{n=1}^{2c} \min(2c,2n-1)$ such triangles.
$\sum_{n=1}^{2c} \min(2c,2n-1)=\sum_{n=1}^{c} \min(2c,2n-1))+ \sum_{n=c+1}^{2c} \min(2c,2n-1)$
$=\sum_{n=1}^{c}(2n-1)+ \sum_{n=c+1}^{2c}2c$
$= [2(\sum_{n=1}^cn)- c]+[c*2c]$
$= 2\frac {c(c+1)}2 - c + 2c^2$
$= 3c^2$
A: Let $b$ be the length of the base and $a$ be the length of the legs of such a triangle. Given $c\geq1$ we want
$$1\leq b\leq 2c, \quad {b\over2}<a\leq 2c\ .\tag{1}$$
If $b=2k-1$ with $1\leq k\leq c$ is odd the condition $(1)$ enforces $k\leq a\leq 2c$ and allows of $2c-(k-1)$ different integer values for $a$.
If $b=2k$ with $1\leq k\leq c$ is even then $(1)$ enforces $k+1\leq a\leq 2c$ and allows of $2c-k$ different integer values for $a$.
The total number $N$ of isosceles triangles having sides $\leq2c$ therefore is given by
$$N=\sum_{k=1}^c\bigl(2c-(k-1)\bigr)+\sum_{k=1}^c(2c-k)=4c^2-\sum_{k=1}^c(2k-1)=3c^2\ .$$
A: Seems like a recursive approach would work well.
The number of isoceles triangles with the two equal legs equal to $2c$, and the other shorter, is of course $2c{-}1$.
The number of isoceles triangles with the two equal legs equal to $2c{-}1$, and the other shorter, is $2c{-}2$.
The isosceles triangles with "base" of $2c$ and equal legs shorter is $c{-}1$, since the equal legs must be more than $c$ in length, and likewise with a base of $2c-1$ the count is also $c{-}1$ (since equal legs of $c$ will now qualify).
Adding the two equilateral triangles of sides $2c$ and $2c-1$, we get $3c{-}1$ triangles on the increment between $c{-}1$ and $c$.
So our cumulative value $t(c) = t(c-1) + 6c-3$, with $t(0)=0$ of course. Then the values $3,12,27,48,75,\ldots$ can be assessed to give $t(c) = 3c^2$.
