14 May 2013
Section B
Lambert Wang
Cutting Corners
Painting (again)
There is a tetrahedron with an integer sidelength n. The tetrahedron is dipped in paint and then left
to dry. Four sets of n minus one cuts are made parallel to each face such that they are each equally spaced.
The resulting polyhedra all have a sidlength of one. How many of the resulting figures have paint on four,
three, two, one, or no faces?
Wierd Shapes
A tetrahedron is a four sided regular polyhedron. What that means is it is a triangular pyramid with
congruent faces. When cuts are made parallel to the faces, more tetrahedra with sidelength 1 are made. If n
is greater than two, in addition to tetrahedra, ocrahedra, eight sided polyhedra, are also made. This is
because of truncation, an operaton that increases the number of faces on a polyhedron through cutting off its
corners. In this case, when a tetrahedron is truncated fully, four smaller tetrahedra are made at the vertices
while one ocrahedra is produced from the center of the original tetrahedron.
With n equal to one, there is one tetrahedron and no octahedea. With n equal to two, there are four
tetrahedra and one octahedra. When a tetrahedron is cut, forms layer are formed with numbers of tetrahedra
represented by triangular numbers. For example, the first layer has one tetrahedron. Layer two has three,
layer three has six, layer four has ten, etc.
The total number of cut tetrahedra from a tetrahedron with a sidelength of n is modeled by:
â
n
k=1
n2
+n
2
=
1
6
Hn + 2L Hn + 1L n
Similarly, the number of octahedra in the first layer is none, layer two is one, layer three is three,
layer four is six.
The total number of cut octahedra from a tetrahedron with a sidelength of n is modeled by the same
equation as tetrahedra except n is n minus one:
â
n
k=1
Hn - 1L2 + Hn - 1L
2
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=
1
6
Hn + 1L n Hn - 1L
page 1 of 4
14 May 2013
Section B
Lambert Wang
Oh my god, why is the POW prompt written in Comic Sans?
Now that we have the basics of the problem done, it’s time to look at the paint.
Tetrahedra 4 Sides of Paint
When n equals one, there is one tetrahedron with four sides painted. Boring...
1
Tetrahedra 3 Sides of Paint
When n greater than one, the only tetrahedra with four painted sides are the tetrahedra at the vertices. Since there are four vertices, there are four tetrahedra with four painted sides when n is greater than
one.
4
Two Faced Tetrahedra
The tetrahedra with two faces painted after being cut are on the edges of the original tetrahedron.
Since there are n tetrahedra on each edge, and two those are vertices, there are n minus two painted tetrahedra on each edge. There are six edges so the number of two faced tetrahedra for n greater than one is modeled by:
6 Hn - 2L
Tetrahedra 1 Side Painted
The tetrahedra with one painted side are on the centers of the faces of the original tetrahedron. The
number of tetrahedra on each face is simply a triangular number. Since for n equal to three, there are no
tetrahedra with zero painted faces, we have to use n minus three for n in the equation for triangular numbers.
Also, since there are four faces, we simple multiply the expression by four:
4
Hn - 3L2 + Hn - 3L
2
= 2 Hn - 2L Hn - 3L
Vanilla Tetrahedra
The number of tetrahedra with no painted faces is simply the total number of tetrahedra minus the
number of tetrahedra with three, two, and one painted side:
1
Hn + 2L Hn + 1L n - H4L - H6 Hn - 2LL - H2 Hn - 3L Hn - 2LL =
6
1
I-24 + 26 n - 9 n2 + n3 M
6
Another way to look at the unpainted tetrahedra is to see that all the unpainted tetrahedra are in the
center of the original tetrahedra, blocked from the paint by one layer on each of the four sides. So, we can
use the equation for the total tetrahedra but use n minus four instead of n.
1
1
Hn - 4L + 2L HHn - 4L + 1L Hn - 4L = Hn - 2L Hn - 3L Hn - 4L
6
6
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page 2 of 4
14 May 2013
Section B
Lambert Wang
Octahedra? I hardly know ‘er!
The expressions for octahedra are similar to the expressions for tetrahedra except that instead of n,
we use n minus one.
Octahedra 4 Sides of Paint
For n equal to two:
1
Octahedra 3 Sides of Paint
For n greater than two:
4
Two Faced Octahedra
For n greater than two:
6 Hn - 3L
For n greater than two:
2 Hn - 3L Hn - 4L
Octahedra 1 Side Painted
Vanilla Octahedra
For n greater than two:
1
6
Hn - 3L Hn - 4L Hn - 5L
Why is it that simple?
When you’re cutting the original tetrahedra, you don’t start making octahedra until n is greater than
one. Once you start cutting octahedra, you have none in layer one, one in layer two, three in layer three, six
in layer four, etc. We know the number of tetrahedra share a similar model to the number of octahedra, but
is it the same for paint?
Since at least four sides must be truncated for an octahedron to appear, all octahedra must have four
or fewer sides painted. Four sides painted only occurs when n is two. There are three painted sides on the
octahedra directly adjecent to the tetrahedra on the vertices, so therefore, for n greater than two, there are
four tetrahedra with three painted sides. The octahedra on the edges, but not on the corners, have two
painted sides. There are one less than n octahedra on each edge. For unpainted octahedra, there is a layer of
octahedra on each of the four sides separating the center octahedra from the paint, so it is the same expression as the total number of octahedra only n minus four instead of n.
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page 3 of 4
14 May 2013
Section B
Lambert Wang
Ravioli Ravioli give me the Formuoli (Solution)
Here are all the formulas put into a table.
Tetrahedra
Total
n
4 sides
3 sides
2 sides
1 side
n=1
n>1
n>1
n>1
no sides n > 1
1
6
1
6
Hn + 2L Hn + 1L HnL
1
4
6 Hn - 2L
2 Hn - 2L Hn - 3L
Octahedra
1
6
n
n=2
n>2
n>2
n>2
Hn - 2L Hn - 3L Hn - 4L n > 2
1
6
Hn + 1L Hn - 1L HnL
1
4
6 Hn - 3L
2 Hn - 3L Hn - 4L
Hn - 3L Hn - 4L Hn - 5L
Generalizations
I could no make any further generalizations on the problem.
Self-Assessment
This POW wasn’t particularly more difficult than the last. I was able to produce expressions fairly
quickly, however, it took a fair about of time to explain how I came to those solutions. This was a very fun
and interesting POW. I feel as if my ability to construct geometric models has improved.
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