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Difference between revisions of "1995 AHSME Problems/Problem 30"
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We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be <math>h</math>. The side length of the base of the pyramid is <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. <math>h = \sqrt{3^2-(\frac{ \sqrt{3} }{2} \cdot \frac23 \cdot \sqrt{2})^2} = \sqrt{9-6}=\sqrt{3}</math>. The altitude of the pyramid at the bottom of the cube is also <math>h</math>. The altitude in the middle is <math>3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}</math>.
 
We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be <math>h</math>. The side length of the base of the pyramid is <math>\sqrt{1^2+1^2}=\sqrt{2}</math>. <math>h = \sqrt{3^2-(\frac{ \sqrt{3} }{2} \cdot \frac23 \cdot \sqrt{2})^2} = \sqrt{9-6}=\sqrt{3}</math>. The altitude of the pyramid at the bottom of the cube is also <math>h</math>. The altitude in the middle is <math>3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}</math>.
  
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The altitude from the vertex at the top to the bottom is <math>3\sqrt{3}</math>. The altitude from the next <math>3</math> vertices to the bottom are all <math>2\sqrt{3}</math>. The altitude from the next <math>3</math> vertices to the bottom are all <math>\sqrt{3}</math>. The altitude from the bottom-most vertex to the bottom is <math>0</math>.
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The length of the space diagonal of a unit cube is <math>\sqrt{3}</math>. <math>\sqrt{3} \< \frac{3\sqrt{3}}{2}</math>, therefore, the plane that is perpendicular will not pass through the unit cube at the top.
  
 
==See also==
 
==See also==

Revision as of 04:14, 30 September 2023

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 newly placed cube is $3\sqrt{3}$. The plane that is perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$.

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 side length of the base of the pyramid is $\sqrt{1^2+1^2}=\sqrt{2}$. $h = \sqrt{3^2-(\frac{ \sqrt{3} }{2} \cdot \frac23 \cdot \sqrt{2})^2} = \sqrt{9-6}=\sqrt{3}$. 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 from the vertex at the top to the bottom is $3\sqrt{3}$. The altitude from the next $3$ vertices to the bottom are all $2\sqrt{3}$. The altitude from the next $3$ vertices to the bottom are all $\sqrt{3}$. The altitude from the bottom-most vertex to the bottom is $0$.

The length of the space diagonal of a unit cube is $\sqrt{3}$. $\sqrt{3} \< \frac{3\sqrt{3}}{2}$ (Error compiling LaTeX. Unknown error_msg), therefore, the plane that is perpendicular will not pass through the unit cube at the top.

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Final 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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