Difference between revisions of "1995 AHSME Problems/Problem 30"
(→Solution) |
(→Solution) |
||
| Line 10: | Line 10: | ||
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>. | 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>. | ||
| − | 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\{ | + | 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\{2,3,4\}</math>. |
| − | There are | + | There are six cubes such that <math>x+y+z=2</math>: permutations of <math>(1,1,0)</math> and <math>(2,0,0)</math>. <br/> |
| + | Symmetrically, there are six cubes such that <math>x+y+z=4</math>. <br/> | ||
| + | Finally, there are seven cubes such that <math>x+y+z=3</math>: permutations of <math>(2,1,0)</math> and the central cube <math>(1,1,1)</math>. | ||
| + | |||
| + | That gives a total of <math>\boxed{19}</math> 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. | ||
==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 08:40, 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
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 
to 
. The midpoint of this diagonal is at 
. The plane that passes through this point and is orthogonal to the diagonal has the equation 
.
The unit cube with opposite corners at 
and 
is intersected by this plane if and only if 
. Therefore the cube is intersected by this plane if and only if 
.
There are six cubes such that 
: permutations of 
and 
.
Symmetrically, there are six cubes such that 
.
Finally, there are seven cubes such that 
: permutations of 
and the central cube 
.
That gives a total of 
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.
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 | ||
