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High School POW: the cube
Posted:
Jul 17, 1997 11:29 AM
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Given a painted cube n units on a side that is cut into smaller cubes 1
unit on a side, how many smaller cubes will have 3 painted surfaces, 2, 1,
0 painted surfaces?
answer: 8 cubes 3 painted surfaces (corners)
12(n-2) cubes 2 pntd srfc (edges)
6(n-2)^2 cbs 1 ps (interior exposed surfaces)
(n-2)^3 cbs 0 ps (interior
We approached the problem individually at first, then we began to compare
notes. We first agreed that N could not be 1 because that cube would have 6
painted sides
Some started making a table for cubes of increasing dimension: 2*2*2,
3*3*3, 4*4*4, etc.They counted cubes in each of the four categories for
each size cube, filled in the table and looked for a pattern.
Others took an algebraic approach:
if n was some number not equal to 1, the cube would have 8 corners with 3
ptd fcs;
each of the 12 edges would minus the corners would have (n-2) cubes with
ptd fcs;
each of the 6 exposed surfaces (minus corners and edges) would have (n-2)^2
cubes with ptd fcs;
all the inner unpainted cubes would for a cube with side (n-2) for a total
of (n-2)^3 small cubes.
The general form for the total num,ber of cubes works out to be: n^3 - 4n^2
+ 18n - 28
We plugged values from the table into the algebraic formula and found that
both methods produced the same results for several different cubes. We
assumed the others would correspond,also.
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