Project of the Month WINNERS, November 1996

[Geometry Problem of the Week]

We have two winners this month and five (!) honorable mention. This is a
neat problem because I knew that a lot of people would get it right, and I
would get to concentrate on the explanations. That's exactly how it
worked out.

Here's how I scored the solutions this month. There were four parts to
the problem, and a point was given for each right equation, and a point
for each good explanation. There was also a point given for pointing out
any special cases or exceptions, and another point could be earned by
writing something exceptional.

Five people scored a straight 8 points - they got all four equations and
provided good explanations. These were the folks who won honorable
mention. It is a thin line between honorable mention and the winners, but
the winners did go just one step further.

Both the winners pointed out not only the four equations and provided good
explanations, but they also pointed out that this only works if n is
greater than 1. A small point, but it's important to look at all the
parts of a problem and see if there is anything special going on with any
of the cases.

There were a number of people who figured out the number of cubes of each
sort for different size cubes, and then made a table and looked for
patterns and figured out formulas for the different parts. That's okay,
and most of those folks got the right answers, but this is a problem that
can be attacked from a visual standpoint. There are always 8 cubes with
three painted sides not because the tables of numbers said so, but because
every cube has eight vertices.

Same thing goes for the other parts. The number of cubes with no painted
faces is a little cube in the middle. That makes more sense from a visual
standpoint than the way some people solved this part - they took n^3,
which is the size of the whole thing, and subtracted their answers for the
other parts. That answer is no less right, but it doesn't describe what's
going on if you're simply looking at a cube. Just something to think
about when you are solving problems like this.

Here are the winners, along with some of my comments. The honorable
mentions will follow in a separate message, also with comments, along with
a list of the students who also got all the right equations, but didn't
provide as much, if any, explanation.

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WINNER, NOVEMBER PROJECT OF THE MONTH
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Michael Mandel
Germantown Academy, Fort Washington, Pennsylvania

From: Mand...@aol.com

Michael Mandel
Mrs. Carver's Geometry Honors
Germantown Academy
Fort Washington, Pa
11-26-96

In this cube where all sides are painted and then is divided into unit cubes
all of my calculations are all for cubes more than one unit on each side.
Every cube has exactly eight corners, and each corner is painted on three
sides because it is the intersection of three of the sides of the cube. To
find out the number of unit cubes painted on two sides in a cube of sides of
length "x", you must multiply (x - 2) by 12. It must be x - 2 because on
each edge the two ends are both corners and therefore painted on three sides
and don't count towards the two sided cubes. It must be multiplied by 12
because there are four edges on the upper square, four on the bottom square
and four connecting the two. It is only the edges because they are the
intersection of two sides. The number of cubes painted on one side can be
found by multiplying (x - 2) by (x - 2) by 6. It must be (x - 2) squared
because each side is x long and the sides on the edges have paint on more
than one side. The edges are on both sides of the face so it must be (x - 2)
squared. It will be multiplied by 6 because there are 6 faces of the cube.
The number of sides with no paint on them can be found by cubing (x - 2).
This is because on both ends of the unpainted cube there are painted unit
cubes, so you cannot count them in the unpainted number. This holds true for
all three dimensions and therefore it is (x - 2) cubed. In a cube measuring
1 unit cube on each side there is one cube painted on six sides and no other
cubes painted on any different number of sides.

MY COMMENTS:

Michael did a nice job of explaining where the equations come from, and
does it from a very visual perspective. He also points out that the
equations only work if n is greater than 1. The one thing he might have
done to improve his solution is to break it up visually - it is hard to
read a huge blocks of text, and it's a good idea to at least break it into
paragraphs.

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WINNER, NOVEMBER PROJECT OF THE MONTH
***********************************************

Jeffrey Chang
BN High School, somewhere...

From: Hai-Feng Chang <bnh...@mail.erols.com>

Name: Jeffrey Chang
Email Address: bnh...@mail.erols.com
School:
Grade: 8

It is taken that n is integral, because otherwise the cube would be
exceedingly difficult, if not impossible, to divide in the aforesaid way.

a) No matter how long the side, only the corner unit cubes will have
three sides of paint on them, because this is the most number of corners
revealed for any unit cube in a cube with side larger than 1. There are
always eight corner unit cubes (in cubes with sides larger than 1), and
so the formula is f(n) = 8, where f(n) --> 3 painted sides, unless n = 1,
where f(n) = 0. (There is only one cube for a unit cube itself, with 6
painted sides).

b) The edges of the cube, not including the corners, will have two of
their sides painted, those cubes having only two sides revealed. There
are twelve edges for each cube. However, each edge of n includes the two
corner cubes, which must be subtracted because of their painting of three
sides. Thus, g(n) = 12(n-2), where n is larger than 1 (for reasons
explained already in a)).

c) The center unit cubes, of each face, will only have one surface
painted, because this is the only one revealed. These center cubes do
not include those of the edge, or the corners. Thus, there are n-2 by
n-2 of these, for each face, and six faces of the large cube. h(n) = 6 *
(n-2)^2, where n is larger than 1 (for reasons explained already in a)).

d) The remaining cubes are not painted, as the most exposed unit cube in
a cubical structure would only be the corner, revealed on three sides.
These three, two, and one sides have already been covered, and so the
remainder are unpainted unit cubes. There are n^3 total unit cubes
present, and so the number of these unpainted unit cubes is j(n) = n^3 -
6 * (n-2)^2 - 12(n-2) - 8, where n is necessarily larger than 1 (for the
above explained reasons, in a)).

MY COMMENTS:

Jeffrey explains each part of the problem very well from a visual
viewpoint, and it's written very clearly. Breaking the solution up into
parts makes it a lot easier to read, and you can refer back to the other
parts as necessary.