# 1994 Question 2

Okay so there’s a formula for this that states that given a perimeter of 2n, the number of integer non-congruent triangles is \lfloor {\frac{n^2}{12}} \rfloor

For odd perimeters there’s a slightly different formula.
This formula is quite hard to prove so I’ll do this problem in a different way.

Solution:
Given sides a, b, c such that a \geq b \geq c WLOG, we have:
665 \leq a \leq 996 (from the triangle inequality) and b+c=1994-a.

Now we have that:
\frac{1994-a}{2} \leq b \leq a from the previously installed inequality

So let’s solve for b.

In this interval, b has a- \lceil {\frac {1994-a}{2}} \rceil +1 solutions (with a fixed a).

Now we can try to sum over all possible values of a.
We’ll compute this sum for even values 2n=a and odd values 2n-1=a.

\sum_{a=665}^{996} = a- \lceil {\frac {1994-a}{2}} \rceil +1.

Splitting into two cases as aforementioned we get:
\sum_{n=333}^{498}=2n- \lceil {\frac {1994-2n}{2}} \rceil +1

and

\sum_{n=333}^{498}=(2n-1)- \lceil {\frac {1994-(2n-1)}{2}} \rceil +1

Now since the summations have the same index, we can combine them:

\sum_{n=333}^{498}=6n-1994

Now we can simply compute this to get 82834 possible triangles.

(Note that the mentioned formula should also give this answer)

QED.

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Fantastic!!!