A geometry problem by anweshan bor

The number of integral points in the interior of a triangle with vertices $\large (0,0) , (0,n), (n,0)$ is given by $\large \Sigma (n-2) = \frac{(n-2)(n-2+1)}{2} = \frac{(n-2)(n-1)}{2}$ .

In the given question $\large n=21$ , therefore answer would be $\large \Sigma (21-2) = \Sigma (19) = \frac{(19)(20)}{2} = 190$ .

In place of a triangle if it was a square we would have 20 * 20 =400 without boundary. So the triangle as we have would have 200. But know the diagonal 20 are also on the boundary. 20/2=10 of them are contributed by each triangle.

We have only one triangle so it will have 200 - 10 =190.