# 1995 AHSME Problems/Problem 30

## Problem

A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is $\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$

## Solution

Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.

Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$. The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$. The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$.

The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$. Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$.

There are six cubes such that $x+y+z=2$: permutations of $(1,1,0)$ and $(2,0,0)$.
Symmetrically, there are six cubes such that $x+y+z=4$.
Finally, there are seven cubes such that $x+y+z=3$: permutations of $(2,1,0)$ and the central cube $(1,1,1)$.

That gives a total of $\boxed{19}$ intersected cubes.

Note that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.

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