Integral Co-ordinates -2

Consider a right triangle of vertices $(0,0)$ , $(0,n)$ , and $(m,0)$ . I am using $m=20$ and $n=15$ for convenience. If we consider the rectangle twice the size of the triangle with vertices $(0,0)$ , $(0,n)$ , $(m,0)$ , and $(m,n)$ . Then we can find the number of points with integer coordinates in the triangle as:

$N_\triangle = \dfrac {N_{\text{rectangle}} - N_{\text{4 sides}}-N_{\text{diagonal-2}}}2$

where

• $N_{\text{rectangle}} = (m+1)(n+1)$ is the total number of points with integer coordinates along the four sides and within the rectangle.
• $N_{\text{4 sides}} = 2(m+n)$ is the number of points with integer coordinates along the four sides of the rectangle.
• $N_{\text{diagonal - 2}} = \gcd(m,n)-1$ is the number of points with integer coordinates along a diagonal or the hypotenuse of the rectangle but excluding the two at $(0,n)$ and $(m,0)$ , where $\gcd(m,n)$ is the greatest common divisor of $m$ and $n$ .

Therefore,

$\begin{aligned} N_\triangle & = \frac {(m+1)(n+1) - 2(m+n)-\gcd(m,n)+1}2 \\ & = \frac {mn - m-n - \gcd(m,n)}2 + 1 \end{aligned}$

For $m=720$ and $n=360$ , $N_\triangle = \dfrac {720\times 360 - 720-360-360}2 + 1 = \boxed{128881}$ .

Using Pick's Theorem as used by @Henry U in my previous simpler version of this problem.

The Area of a Polygon, ${\color{#20A900}{A= I + \dfrac B2 -1 }}$

Here, $I$ is the total number of interior points inside the polygon.

$B$ is the total number of points on the boundary of the polygon.

Here, we can simply tell that area of our triangle having $(0,0),(0,n)$ and $(m, 0)$ as vertices is $A'= \dfrac 12 mn$

Now, we need to find $B$ we have $n$ points on one side and $m$ points on another side and origin and also $k$ points on the hypotenuse.

We get $k = gcd(-n, m) -1$ For detailed explanation visit .

Now we have, $A=A'$

$\dfrac 12 mn = I + \dfrac 12( 1 + m +n + gcd(-n, m)-1) -1$

We get $I = \dfrac 12 mn - \dfrac 12 (m + n + gcd(-n, m)) + 1$

Finally, $\boxed{{\color{#20A900}{N_{(n, m) } = I = \dfrac 12 ( mn -m -n -gcd(-n, m)) + 1 }}}$

Hence, for $n =360 , m = 720$

$N_{(360,720)} = \dfrac 12( 720\times 360 - 720 - 360 - gcd(-360,720)) + 1 = \boxed{128881}$

I might be seriously wrong, please do correct me if so.