Points arranged in array...!

image

We proceed by negation. First we select $3$ points from the $5 \times 4=20$ points. This can be done in $\binom{20}{3}$ ways. From this we will subtract the degenerate cases, that is, the line segments.

There are four types of degenerate triangles.

• Horizontal

• Vertical

• Diagonal

• With slope $\pm 1/2$ (for example, the line segment formed by joining A4, C3, and E2).

Number of horizontal degenerate triangles $=4 \cdot \binom{5}{3}$

Number of vertical degenerate triangles $=5 \cdot \binom{4}{3}$

Number of diagonal degenerate triangles $=4 \cdot \left(\binom{4}{3}+1\right)$

For this, count their number in the triangle with vertices (E1 , B4, E4) and multiply by four as there are four such triangles.

Number of degenerate triangles with slope $\pm 1/2$ $=4$

Therefore, number of non-degenerate triangles $\begin{aligned} &=\binom{20}{3}-\left[4 \cdot \binom{5}{3}+5 \cdot \binom{4}{3}+4 \cdot \binom{4}{3}+4\cdot 1+4\right]\\ &=\boxed{1056} \end{aligned}$

Here's a python implementation of this problem:

def find(A):
    result = []
    for i in A:
        for j in A:
            for k in A:
                if valid(i, j, k):
                    result.append([i,j,k])
    return len(result) / 6   # divide by 6 since each triangle is counted 3! times

def valid(a, b, c):
    if a == b or a == c or b == c:
        return False
    if a[0] == b[0] == c[0]:  # three points are on the same horizontal line
        return False
    elif b[0] == a[0] or c[0] == b[0] or a[0] == c[0]:  # to avoid division by 0 in the next step
        return True
    # three points are on the same line
    if (b[1] - a[1])/(b[0] - a[0]) == (c[1] - b[1])/(c[0] - b[0]):
        return False
    return True

def generate(): # generate all 20 coordinates
    result = []
    for i in range(4):
        for j in range(5):
            result += [(i,j)]
    return result

A = generate()
print(find(A))