# 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 1

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.

## Solution 2

Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\sqrt{3}$, the altitude of the top vertex of the newly placed cube is $3\sqrt{3}$. The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$, so that the height of the plane is $\frac{3\sqrt{3}}{2}$.

By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.

We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$. The base of the pyramid is an equilateral triangle with side length of $\sqrt{3^2+3^2}=3\sqrt{2}$. The height of the triangle is $\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}$. The distance of the center of the triangle to the vertex is $\frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}$. Therefore, $h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}$. The altitude of the pyramid at the bottom of the cube is also $h$. The altitude in the middle is $3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}$.

The altitude of the vertex at the top is $3\sqrt{3}$. The altitude of the second highest $3$ vertices are all $2\sqrt{3}$. The altitude of the third highest $3$ vertices are all $\sqrt{3}$. The altitude of the bottom-most vertex is $0$.

By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\sqrt{3}$. The altitude of the second highest $3$ vertices are all $\frac{2\sqrt{3}}{3}$. The altitude of the third highest $3$ vertices are all $\frac{\sqrt{3}}{3}$. The altitude of the bottom-most vertex is $0$.

The length of the space diagonal of a unit cube is $\sqrt{3}$. The highest vertex of the bottom-most unit cube has an altitude of $\sqrt{3}$. As $\sqrt{3} < \frac{3\sqrt{3}}{2}$, therefore, the plane will not pass through the unit cube at the bottom.

For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}$. As $\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}$, therefore, the plane will not pass through the next $3$ unit cubes.

For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}$. As $\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}$, therefore, the plane will pass through the next $3$ unit cubes.

So at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.

Thus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\boxed{\textbf{(D) } 19}$ unit cubes.

## See also

 1995 AHSME (Problems • Answer Key • Resources) Preceded byProblem 29 Followed byFinal Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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