Triangles in the grid

There are ${25 \choose 3}$ or $\frac{25*24*23}{3*2*1}$ which is 2300 ways to choose 3 points in a 5 by 5 grid. However, there could be points in a straight line like this:

So, we must find how many such combinations there are on one line.

• For a line with 5 points, there are ${5 \choose 3}$ or $\frac{5*4*3}{3*2*1}$ or 10 total ways
• For a line with 4 points, there are ${4 \choose 3}$ or $\frac{4*3*2}{3*2*1}$ or 4 total ways
• For a line with 3 points, there are ${3 \choose 3}$ or $\frac{3*2*1}{3*2*1}$ or 1 way

Now, we must find the total number of each type of line.

For lines with length 5, there are 12 total lines (5 horizontal, 5 verticle, 2 diagonal)

For lines with length 4, there are a total of 4 llines (4 diagonal)

For lines with length 3, there are a total of 16 (4 through the center, 8 through the points directly horizontal or vertice from the center, 4 on the edges)

That gives us a total of $2300-12*10-4*4-16*1=\boxed{2148}$ total triangles

For three points to not form a triangle, they must be colinear, so we'll count how many ways there are to choose $3$ colinear points, and subtract that from the number of ways to choose $3$ points without any conditions. To do this, we'll consider every possible line in which the $3$ points can be:

In the red line, there are $\binom{5}{3}=10$ ways to pick $3$ points, and there are $10$ "similar" lines ( $5$ horizontal, and $5$ vertical)

In the green line, again, there are $\binom{5}{3}=10$ ways to pick $3$ points, and there are $2$ similar lines (each one is a diagonal)

For the blue line, there are $\binom{4}{3}=4$ ways, for the green line there is $\binom{3}{3}=1$ , and there are $4$ similar lines to each of those ( $2$ parallel to each diagonal of the square)

Finally, for the orange line, there is $\binom{3}{3}=1$ way to pick $3$ points, and there are $12$ of those ( $3$ with slope $2$ , $3$ with slope $-2$ , $3$ with slope $\frac{1}{2}$ and $3$ with slope $\frac{-1}{2}$ )

This makes our total

$10\cdot10+2\cdot10+4\cdot(4+1)+12\cdot1=152$

There are $\binom{25}{3}=2300$ ways to pick $3$ points in the $5\times5$ grid, so the number of triangles is $2300-152=2148$

There are ${25 \choose 3}=2300$ ways to choose 3 points. We need to subtract from this the number of ways the 3 points would be colinear. We will count these by the point that is between the other two. There are 6 possible types of points that might be between the other two; A: the center, B: a unit distance from the center, C: one diagonal distance from the center, D: the midpoint of an edge, E: between the midpoint of an edge and its corner, and F: a corner. If we use the lowercase of a point to denote the number of colinear triplets with that point between the other two, our answer can be written as $2300-a-4b-4c-4d-8e-4f$ . It does not take long to calculate that $f=0, e=3, d=4, c=10, b=13, a=20$ giving us the answer 2148.

I did this with brute force code after too much head-scratching. I wrote code with three nested FOR loops, one for each point of the triangle, each loop running through the possible 25 positions.

Inside the inner loop I calculated the area formed by every possible combination of three points in the grid and if non-zero, I added it to a counter.

Finally, I divided the total count by 6, as there are 6 possible orders that each valid triangle can be formed by three points. (ABC, ACB, BAC, BCA, CAB or CBA).

In order for a triangle to exist, it must be defined by three points. Thus, we can find the total number of triangles by finding the number of ways to choose three points out of the 25 total points. This is equal to $\binom{25}{3}\ = 2300$ .

But of course, this includes the many degenerate triangles that occur when the three chosen points are collinear. In order to find the actual number of triangles, we must subtract the many sets of three collinear points. In order to do this, we can use more counting methods:

First, note that there are ten lines of five points each, five horizontal lines and five vertical lines. Then, the two major diagonals are also lines with 5 points. In a single line of five points, there are $\binom{5}{3}$ ways of choosing 3 collinear points. In total, this is $12\cdot\binom{5}{3}\ = 120$

Now let's look at lines with four points. There are only four of them (they all have slopes $1$ or $-1$ ). Each line contains $\binom{4}{3}$ collinear dots of three. In total, this makes $4\cdot\binom{4}{3}\ = 16$

Finally, there are the lines of length three. Again, there are four of these with slopes $1$ or $-1$ . However, there are also many triples of points that make lines with slopes $\frac{1}{2}$ , $2$ , $\frac{-1}{2}$ , or $-2$ . In total, this adds 12 more lines or 16 total lines made of three dots. This means $16\cdot\binom{3}{3}\ = 16$

Altogether then, the total number of degenerate triangles made of three collinear points is $120 + 16 + 16 = 152$ .

Subtracting from the total number of ways to choose three points, $2300 - 152 =$ $\boxed{ 2148 }$

Thus, $2148$ triangles can be made by using three points on a $5 \times 5$ square grid.