Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB…

Question

Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices, and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to:

Accepted Answer

Triangles (one point from 3 of the 4 sides, choosing which side to exclude):
$$\alpha = 5\cdot7\cdot6 + 5\cdot7\cdot9 + 5\cdot6\cdot9 + 7\cdot6\cdot9 = 210+315+270+378 = 1173$$

Quadrilaterals (one point from each of the 4 sides):
$$\beta = 5\times7\times6\times9 = 1890$$

$$\beta-\alpha = 1890-1173 = 717$$

Solutions · 1