by Bill Graham
The key to solving this problem is realizing that all face, corner, and edge pieces will
have paint on them. The remainder will not. These unpainted cubes form another, smaller
cuboid that is exactly 2 cm smaller in all dimensions than the original cuboid.
The number of cubes in the original cuboid is given by
A × B × C
where A, B and C are the dimensions of the original cuboid.
The smaller cuboid will be 2 cm smaller in each dimension or
(A-2) × (B-2) × (C-2)
The requirement of one-half the number of cubes can be expressed as follows.
A × B × C = 2 × (A-2) × (B-2) × (C-2)
Before throwing your hands up in panic at this complex equation, look again.
You can simplify things by noting that it really states that the product of
three rational fractions is one-half.
(A-2)/A × (B-2)/B × (C-2)/C = 1/2
If you enjoy playing with numbers, you'll like this one.
One case would be that A cancels with B-2, and B cancels C-2 so that the
final result is (A-2)/C = 1/2, where A = B-2, and B = C-2. Manipulate these
facts just a little more to get
(C-6)/C = 1/2 or
C = 12
The resulting cuboid is 8×10×12 with an volume of 960 and an
unpainted volume of 480.
It turns out that this is the smallest among twenty solutions.
5×13×132 = 8580
5×14×72 = 5040
5×15×52 = 3900
5×16×42 = 3360
5×17×36 = 3060
5×18×32 = 2880
5×20×27 = 2700
5×22×24 = 2640
6×9×56 = 3024
6×10×32 = 1920
6×11×24 = 1584
6×12×20 = 1440
6×14×16 = 1344
7×7×100 = 4900
7×8×30 = 1680
7×9×20 = 1260
7×10×16 = 1120
8×8×18 = 1152
8×9×14 = 1008
8×10×12 = 960
Now, take a look at a ratio of 9:1 or the interior volume being 1/10 of the total.
A × B × C = 10 × (A-2) × (B-2) × (C-2)
Try for a solution with two equal sides first.
Let A = B.
B × B × C = 10 × (B-2) × (B-2) × (C-2)
Start with B = 3 and progress to find an integral result for C.
9C = 10(C-2)
C = 20
Lucked out on the first try. You'll find only two solutions.
3×3×20 = 180
3×4×5 = 60
Vol 1, No 6 Answers
