Let S be a cube. Compute the number of planes which pass through at least 3 vertices of S

Guest Jul 2, 2021

#1**0 **

Since any three points determine a plane, the answer is $\binom{8}{3} = \boxed{56}$.

Guest Jul 2, 2021

#4**0 **

We can break up the planes based on how many vertices of the bottom face they pass through.

$1$ plane passes through exactly $4$ vertices, and $0$ pass through exactly $3$ vertices.

The planes passing through $2$ vertices of the bottom face either pass through an edge or a diagonal of the bottom face. Each edge on the bottom face has $2$ valid planes passing through it and each diagonal has $3$ valid planes, for a total of $4 \cdot 2 + 2 \cdot 3 = 14$.

The valid planes passing through exactly $1$ vertex of the bottom face must intersect the top face on a specific diagonal, so there are $4$ such planes.

There is $1$ plane passing through $0$ vertices, so this makes a total of $1 + 14 + 4 + 1 =$ **20** total planes.

Awesomeguy Jul 2, 2021