The points of this 3-by-3 grid are equally spaced horizontally and vertically. How many different sets of three points of this grid can be the three vertices of a triangle?

Guest May 22, 2022

#1**0 **

We can choose 3 distinct points to form a triangle, but there are some cases that the 3 points do not form a triangle (for example the 3 points are all on the same diagonal/row/column). We need to subtract those cases from the number of ways to choose 3 distinct points from 9.

We can count the number of cases of collinear points by hand.

All on same row: 3 cases

All on same column: 3 cases

All on same diagonal: 2 cases

Total: 8 cases that do not form a triangle. We subtract that from 9C3:

\(\text{Number of ways} = \binom{9}3 - 8 = 76\)

MaxWong May 23, 2022