Triangle dilemma
Probability
Level
2
Consider a regular 4 × 4 grid of sixteen points, as in this picture. How many (non-degenerate) triangles can be formed whose corners lie on the grid?
The answer is 516.
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1 solution
The number of ways of picking three points from the grid is 16C3 = 560. These three points will form a degenerate triangle (one with zero area) if and only if they all lie in a straight line. This straight line may be one of the rows, one of the columns, or one of the diagonals. The number of ways of picking three points, all from one row of the grid, is 4 × 4C3= 16 (pick the row, then pick 3 points within the row). Symmetrically, the number of ways of picking three points, all from the same column, is 16. Now let’s try the diagonals. There are three “bottom-left to top-right” diagonals with at least three points — the middle one has 4 points and the others have 3 each. The number of ways of picking three points, all from one of these diagonals, is 4C3+2×(3C3) = 6. The situation for “bottom-right to top-left” diagonals is identical and also gives 6 possibilities. So the total number of ways of picking a degenerate triangle is 2 × 16 + 2 × 6 = 44, and the number of ways of picking a non-degenerate triangle is 560 - 44 = 516.