Integral Co-ordinates

It can be clearly seen that for an isosceles right triangle of leg length $n$ , the number of points with integer coordinates is the triangular number $T_{n-2} = \displaystyle \sum_{k=1}^{n-1} k = \dfrac {(n-2)(n-1)}2$ . Therefore, for $n=2019$ , $T_{2017} = \dfrac {2017\times 2018}2 = \boxed{2035153}$ .

---

Alternative Solution: After solving a tougher problem Co-ordinates--2 , the following general solution for right triangle with $(0,0)$ , $(0,n)$ , and $(m,0)$ as vertices is obtained.

$\begin{aligned} N_\triangle (m,n) & = \frac {mn-m-n-\color{#3D99F6}\gcd(m,n)}2 + 1 & \small \color{#3D99F6} \text{where }\gcd(m,n) \text{ is the greatest common divisor of }m \text{ and }n. \end{aligned}$

For $m=n$ , we have $N_\triangle (n,n) = \dfrac {n^2 -3n}2 + 1 = \dfrac {(n-1)(n-2)}2$ . The same solution as above.

Let's directly do the general formula.

We will use two different methods to calculate the area of the triangle, one of them involves the number of points inside, so we can set both formulas equal to each other and solve this equation.

First, the area of the triangle is simply $A_1 = \frac 12 \cdot b \cdot h = \frac 12 n^2$ .

Now, there is another way to calculate this area that involves the number of points inside the triangle. By Pick's theorem , the area is $A_2 = I + \frac B2 -1$ where $I$ is the number of points inside the shape (it works for any shape without holes) and $B$ is the number of points on the boundary of the shape. We can find $B$ by counting the number of points on all three sides. There are $(n-1)$ on each side, plus the 3 corners, so the area is $A_2 = I + \frac {3+3\cdot (n-1)}{2}-1$

Since both ways calculate the same area, they must be equal.

$\begin{aligned} A_1 &= A_2 \\ \frac 12 n^2 &= I + \frac {3+3(n-1)}2-1 \\ n^2 &= 2I + 3n -2 \\ 2I & = n^2-3n+2 \\ I &= \frac 12n^2 - \frac 32 + 1 \\ &= \frac 12 (n-1)(n-2) \end{aligned}$

For the specific case of $n=2019$ , we can calculate $I$ as $\frac 12 (2019-1)(2019-2) = \boxed{2035153}$ .

Consider a square having vertices $(0,0) ,(0,n) ,(n, 0)$ and $(n, n)$

Cut it along the diagonal $(n, 0) ,(0,n)$ , We get our required triangle. Now, we need to find the total number of point inside this triangle. This triangle has a total of $\dfrac 12 (n-1)(n-1) = \dfrac 12 (n-1)^2$ number of points (integral) that lie inside it.

But actually when we observe we have counted the points that lie along the line $y=x$ twice. Hence, we need to remove that extra countings.

Observe that the hypotenuse and $y=x$ meet at $\begin{pmatrix} \dfrac n2 , \dfrac n2 \end{pmatrix}$ .There are a total of $(n-1)$ points between Origin and $\begin{pmatrix} \dfrac n2 , \dfrac n2 \end{pmatrix}$ .Hence, we need to remove a total of $\dfrac 12(n-1)$ number of points.

Hence, $N_n = \dfrac 12 (n-1)^2 - \dfrac 12 (n-1) = \dfrac 12(n-1)[(n-1) - 1] = \dfrac 12(n-1)(n-2)$

${\color{#20A900}{N_n = \dfrac 12 (n-1)(n-2)}}$

So for $n = 2019$

$N_{2019} = \dfrac 12(2019-1)(2019-2) = \boxed{2035153}$

Please correct me if I am wrong.