How to find the number of integer coordinates in the interior of triangle?

Question

How to find the number of integer coordinates in the interior of the triangle with vertices $(0,0)$ $(0,21)$ $(21,0)$?

Answer 1

Hint.

This is an isocele rectangle triangle. So it is easy to find the number of points of the associated square. Namely $22 \times 22 =484$ points.

Then you have to substract the points that are on the boundaries of the square, namely $2(22+20)=84$ points. And also the ones lying in the interior of the diagonal, that is additional $20$ points. Remains $380$ points.

Divided by two for the triangle $190$ points.

Answer 2

Using Pick Theorem:: $\displaystyle A = i+\frac{b}{2}-1$

Where $A$ is Area of $\triangle$ formed by the points $(0,0),(21,0),(0,21)$

So $\displaystyle A = \frac{1}{2}\times 21 \times 21 = \frac{441}{2}$ and $i=$ Interior ordered pair within a $\triangle.$

and $b = $Boundry points, Which is equal to $ = 21+21+20+1 = 63$

Here $21$ points along $X$ axis and $21$ points along $Y$ axis and $20$ points

on boundry and $1$ point is origin.

So Put into formula, We get $\displaystyle \frac{441}{2} = i+\left(\frac{63}{2}\right)-1$

So we get $i = 190$ Integer ordered pairs which lie exactly interior of $\triangle $

Answer 3

The number of points on $y$ when $x = 1$ is $19$. The number of points on $y$ when $x = 2$ is $18$ and so on till $x = 19$, whete $y$ has only $1$ point.

So, the total is $1+2+3+...+18+19= (19 ×20)/2=190$