Finding the number of Triangles

The longest side must be of length at least $\lceil\frac{200}{3}\rceil=67$ , and length less than $\frac{200}{2}=100$ .

When the longest side length is $l$ and the shortest is $s$ , we must have $200-l-s\le l$ and $s\le 200-s-l$ . That is, $200-2l\le s\le 100-\lceil \frac{l}{2}\rceil$ . Then given $l$ , there are $100-\lceil\frac{l}{2}\rceil-(200-2l-1)=2l-\lceil\frac{l}{2}\rceil-99$ possible triangles with perimeter $200$ (up to congruence).

Thus, the answer is $\displaystyle\sum_{l=67}^{99}(2l-\lceil\frac{l}{2}\rceil-99)=2\left(\frac{99\cdot 100}{2}-\frac{66\cdot 67}{2}\right)-\left(2\sum_{k=34}^{49}k+50\right)-99\cdot 33$

$\displaystyle=5478-50-2\left(\frac{49\cdot 50}{2}-\frac{33\cdot 34}{2}\right)-3267=\boxed{833}$ .

The proof of the formula is lengthy. (But I encourage you to simplify the derivation :)) If the perimeter is even, use the formula #Triangles=flr(n^2/48) If the perimeter is odd , use the formula #Triangles=floor((n+3)^2/48) Any other solutions are encouraged.

nice problem, i like this one

Ans: 833 This can be calculated by using Alcuin's sequence.

http://oeis.org/search?q=Alcuin%27s+sequence&sort=&language=&go=Search