No. of isosceles triangles possible of integer sides with sides $\leq n$ Prove that the no. of isosceles triangles with integer sides, no sides exceeding $n$ is $\frac{1}{4}(3n^2+1)$ or $\frac{3}{4}(n^2)$ according as n is odd or even, n is any integer.
How to do it? I found that under these conditions no. of triangles possible may be
${n\choose 2}$
 A: Fix the length of the two equal sides, say $k$. In how many ways can you choose the length of the base $b(k)$? Obviously $b(k) \ge 1$ and, for the triangle inequality, $b(k) < 2k$. But, since no side can exceed $n$, $b(k) \le n$. Putting these things together, we conclude that:


*

*when $2k - 1 \le n$, that is, when $\displaystyle k \le \left\lfloor\frac{n+ 1}{2}\right\rfloor$, $b(k)$ can range from $1$ to $2k - 1$ inclusive;

*when $2k - 1 > n$, which means $k > \left\lfloor\frac{n + 1}{2}\right\rfloor$, $b(k)$ can range from $1$ to $n$ inclusive.


Hence the total number of isosceles triangles is $$\sum_{k = 1}^n b(k) = \sum_{k = 1}^{\left\lfloor\frac{n + 1}{2}\right\rfloor} (2k - 1) + \sum_{k = \left\lfloor\frac{n + 1}{2}\right\rfloor + 1}^n n = \left\lfloor\frac{n + 1}{2}\right\rfloor^2 + n\left(n - \left\lfloor\frac{n + 1}{2}\right\rfloor\right)$$
Now when $n$ is odd $\displaystyle \left\lfloor\frac{n + 1}{2}\right\rfloor = \frac{n + 1}{2}$, so the expression above reduces to $$\left(\frac{n + 1}{2}\right)^2 + n\left(n - \frac{n + 1}{2}\right) = \frac{n^2 + 2n + 1 + 4n^2 - 2n^2 - 2n}{4} = \boxed{\frac{3n^2 + 1}{4}}$$ while, when $n$ is even, $\displaystyle \left\lfloor\frac{n + 1}{2}\right\rfloor = \frac{n}{2}$ and the formula is $$\left(\frac{n}{2}\right)^2 + n\left(n - \frac{n}{2}\right) = \frac{n^2 + 4n^2 - 2n^2}{4} = \boxed{\frac{3n^2}{4}}$$
A: ${n \choose 2}$ is the number of triplets $(k,k, m)$.  But not all triplets can be triangles.  To be a triangle i) $m < k + k = 2k$ and ii) $k < k + m$.  (i) is a essential, ii) is trivially redundant).
So we need to find all possible triplets $(k,k,m)$ where $k,m \le n$ and $m < 2k$.  As $m$ is an integer, that means $m \le 2k-1$ and $m \le n$.
That is $\sum_{k=1}^n\sum_{m=1}^{\min (n, 2k - 1)}1=\sum_{k=1}^n{\min (n, 2k - 1)}$
$\sum_{k= 1; k \le \frac n2} (\min (n, 2k - 1)) + \sum_{k> \frac n2}^n(\min(n,2k-1)) =$
$\sum_{k=1; k \le \frac n2} (2k-1)  + \sum_{k>\frac n2}^n n$
Case 1:  $n$ is even
$= [\sum_{k=1}^{\frac n2}(2k-1)] + [\sum_{k=\frac n2 + 1}^n n]$
$=[(\frac n2)^2] + [\frac n2*n]$
$=\frac 34n^2$.
Case 2: $n$ is odd
$= [\sum_{k=1}^{\frac {n-1}2}(2k-1)] + [\sum_{k=\frac {n+1}2 }^n n]$
$=[(\frac {n-1}2)^2] + [\frac {n+1}2*n]$
$=\frac {(n^2 - 2n + 1)+ (2n^2 + 2n)}4 = \frac 14(3n^2 + 1)$
.....
Example $n = 10$.
Now we need to find all possible $(k,k,m)$ so that $k,m \le 10$ and $m < 2k$.
If $k =1 $ then $m$ may be $1$.  Total: $1$ way.
If $k=2$ then $m$ may be $1,2,3$.  Total: $3$ + $1$ = $4$ ways.
If $k=3$ then $m$ may be $1...5$. Total: $5+3+1=9$ ways.
If $k=4$ then $m$ may be $1... 7$.  Total: $7+5+3+1=16$ ways.
If $k = 5$ then $m$ may be $1....9$. Total: $9 + 7+ ..... + 1 = 25$ ways.
If $k = 6$ then $m$ may be $1....10$.  Total: $10 + 25 = 35$ ways.
If $k = 7$ then $m$ may be $1..... 10$.  Total: $10 + 10 + 25 = 45$ ways.
If $k = 8 .... 10$ then $m$ may be $1.... 10$.  Total $10 + 10 + ..... + 10 + 25$ ways.
Total number of ways:  $1 + 3 + 5+ 7 + +9 + 10+10+10 + 10+10 = 75$
For $n$ even we will have
$(1 + 3 + 5 + ..... + (2*(\frac n2)-1) + (n + n + n + .....n) = $
$(\frac n2)^2 + \frac n2 * n = \frac 34 n^2$.
If $n$ is odd we will have.
$(1 + 3 + 5 + ..... + (2 * (\frac {n-1}2) - 1) + (n+n+n .... +n) = $
$(\frac {n-1{2}^2} + \frac {n+1} 2*n = \frac 14(3n^2 + 1))$.
A: Hint: Break it up into cases based on the repeated side, and don't forget about equilateral triangles or the triangle inequality!
A: First note that the number of isosceles triangles with equal lengths $j$ and base lengths $k$ is the same as the number of right angled triangles with hypothenuse $j$ and one fixed side equal to $k/2$, drop a perpendicular bisector from the top of the triangle to see this. So for $j \in \{1,...,n\}$we just need to count the number $s_j$ of $k$'s with $1 \leq k \leq n$ for which
$$j^2=(k/2)^2+s^2$$
has a positive solution $s \in \mathbb{R}$, and add all these $s_j$'s . The above equation has a solution if and only if $k<2j$. This gives 
$$s_j=\min\{2j-1, n\},$$
which is $n$ if $j\geq \frac{n+1}{2}$, and $2j-1$ else. So for $n$ odd we get for the total number
$$\sum_{j=1}^n s_j=\sum_{j=1}^{\frac{n-1}{2}} (2j-1) +\sum_{j=\frac{n+1}{2}}^n n,$$
and for $n$ even we get a total number
$$\sum_{j=1}^n s_j = \sum_{j=1}^{\frac{n}{2}} (2j-1) +\sum_{j=\frac{n+2}{2}}^n n$$
which I'm sure you can both evaluate using gauss' formula for the sum of consecutive numbers starting from $1$. 
