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Difference between revisions of "1995 AHSME Problems/Problem 30"

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==Solution==
 
==Solution==
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Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
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Now consider the diagonal from <math>(0,0,0)</math> to <math>(3,3,3)</math>. The midpoint of this diagonal is at <math>\left(\frac 32,\frac 32,\frac 32\right)</math>. The plane that passes through this point and is orthogonal to the diagonal has the equation <math>x+y+z=\frac 92</math>.
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The unit cube with opposite corners at <math>(x,y,z)</math> and <math>(x+1,y+1,z+1)</math> is intersected by this plane if and only if <math>x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3</math>. Therefore the cube is intersected by this plane if and only if <math>x+y+z\in\{1,2,3,4\}</math>.
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There are three cubes with <math>x+y+z=1</math> and six cubes with <math>x+y+z=2</math>. The cases <math>x+y+z\in\{3,4\}</math> are symmetric. Therefore there are <math>2(3+6)=\boxed{18}</math> intersected cubes.
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1995|num-b=29|after=Final Question}}
 
{{AHSME box|year=1995|num-b=29|after=Final Question}}

Revision as of 08:34, 7 January 2009

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 $x+y+z\in\{1,2,3,4\}$.

There are three cubes with $x+y+z=1$ and six cubes with $x+y+z=2$. The cases $x+y+z\in\{3,4\}$ are symmetric. Therefore there are $2(3+6)=\boxed{18}$ intersected cubes.

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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