Number of integral coordinates in a given region.

Question

The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices $(0,0) ,(0, 41$) and $(41,0)$ , is:

(1) 901 (2) 861 (3) 820 (4) 780.

I tried to find out the number of points manually, but it didn't look like the optimal method. Thanks in advance.

Answer 1

Manually seems fine to me. First draw the triangle.

Consider the integral points on the line $x=1$. The point on the hypotenuse is $(1,40)$. The number of points inside the triangle are $39$, that is, $\{1,2\dots,39\}$.

Similarly number of integral points on $x=2$ are $38$, that is $\{1,2,\dots,38\}$.

So basically you have to sum up $39+38+\dots+1 = \frac{40.39}{2}=780.$

Answer 2

The problem can be approached by finding the area of the triangle in two ways:

First, as $\displaystyle A = \frac{1}{2}(41^2) = \frac{1681}{2}$

Second, by an application of Pick's theorem $\displaystyle A = i + \frac{b}{2} - 1$

Now boundary points $b$ can be calculated by considering that there is a single origin point $(0,0)$ common to both cathetuses, $40$ points unique to each of the two cathetuses, and $42$ points along the hypotenuse (inclusive of the end points where it intersects with the cathetuses).

So $\displaystyle b = 1+ 2(40)+42 = 123$, giving $\displaystyle A = i + \frac{121}{2}$

Equating the two expressions for $A$, we get the number of interior points $i = 780$.