$\begingroup$Find number of non-congruent parallelogram of perimeter $40$ with integer side lengths and at least one diagonal of integer length.
My working:
Let sides be $a\le b$ and one diagonal of integer length be $c$
Now we have $b-a<c<a+b$ $\implies 20-2a<c<20$ $\implies$ number of values of $c$ for given $a$ is $2a-1\implies $ total choices are $\sum_{a=1}^{10}(2a-1)=100$
But here there may be some repetition due to both diagonal becomes integer!
Length of other diagonal will be $\sqrt{2a^2+2b^2-c^2}.$
Now should I check all cases ?
$\endgroup$1 Answer
$\begingroup$A parallelogram can be thought of as two separate triangles. Therefore, we only need to consider one triangle, with smaller sides $a$ and $b$, with some other side $c$. (WLOG $a \leqslant b$)
We have that $a+b=20$. By triangle inequality theorem, $c<20$. This becomes a much easier casework question.
We see that this easily boils down to $1 + 3 + 5 + . . . 19$. This comes from solving for the number of integral values of $c$ for the cases when $(a= 1, b = 19), (a=2, b=18) . . .$ etc. This is equal to $10^2 = 100$. Thus, there are a $100$ total paralellograms.
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