Amber Truhanovitch November 29, 2007
POF Group-C
Problem Statement:
A
tetrahedral with edge length n is dipped in paint. Now we must go and find how
many faces of the small tetrahedron that make up the large tetrahedron; 0, 1,
2, 3, 4, have paint on them.
Process:
I started by
visualizing a tetrahedron and drawing out the top and base points. I did this
for the lengths n=1 to n=6. Then I built a model to try and look for a pattern.
The first step is to find out how many regular tetrahedrons there are in the
figure. After n=3 length I noticed that there was an octahedrons in the middle
of the figure. The down the level, n=3 there were three octahedrons figures.
Looking at the three D model these three octahedrons figures create an inverted
tetrahedron. So I can to a generalization that at the intersection of the three
octahedrons was an inverted tetrahedron. So on the next level, n=3, there where
six octahedrons and there were one intersecting vertices. This creates an
inverted tetrahedron. So for n=3 there are 6 regular tetrahedrons and one
inverted tetrahedron. But there is also a trick. What I am looking at is only
the base of n=3. That means that to find the total number of tetrahedrons for
the figure, you have to find the sum of the base plus the total amount of
tetrahedrons of that base from the figures above(just the total amount on the
base). So for example, n=3 has 7 total tetrahedrons on that base, plus n=2 that
has 3 total tetrahedrons on its base, and plus n=1 which has one total
tetrahedron on its base. So the n=3 figure has a total of 11 total tetrahedron.
The second way I did this was the total is the total on that base from the
total from the figure before it. So n=3 has 7 total on that base, and the total
for n=2 is 4 so 7 + 4 = 11. This process was done for n=1 up to n=7. I only did
this up to seven because this created a basis for a chart and a pattern to
form.
Step 1. Draw the n=1 tetrahedron. I visualized my
tetrahedron from above, so my drawings are of each tetrahedron. The colored
in circles are the top of the tetrahedrons and the open circles are the three
bases. The figure to the left shows a 3-d model so when the 1dimensional
model is produced you will understand the parts.
This is n=1. Again, the black dots are the middle point
at the top of the tetrahedron and the open circles are the base points. So
in an n=1 tetrahedron there are a total of, 1 tetrahedron. This n is equal to 2. There are three total tetrahedrons
and one octahedron in the middle. The three tetrahedrons are shaded in
blue. And the octahedron is in white. This is a view of just the base,
which means the top layers are cut off. So the total amount of tetrahedrons
are the base tetrahedrons (3) plus the total for the base one before (n=1)
so 3 +1=4 total in n=2 or the second way wish is the total for this base, 3
plus the total from the base before it, 1 so 3+1= 4
This is n=3. There are 6 regular upright tetrahedrons on this base.
Now there are three octahedrons. At the intersection of the three
octahedrons was an inverted tetrahedron. On level, n=3, there where six
octahedrons and there were one intersecting vertex. This creates an
inverted tetrahedron. So the total tetrahedrons on this level are 7. The total for a n=3 figure are 7 + (total
of base of n=2) + (total of base n=1) So 7+3+1= 11 total for n=3 or the second way wish is the total for this base, 7
plus the total from the base before it (n=2), 4 so 7+4= 11
This is n=4. The orange dots are the vertices of the
octahedron which create the inverted tetrahedrons. There are 10 regular tetrahedrons
and the vertices which create 3 inverted tetrahedrons. So a total of 13
tetrahedrons for the n=4 base. And the total amount of tetrahedrons for an
n=4 figure is 13 + (total of n=3 figure) 13 + 11= 24 tetrahedrons.
This is n=5. The orange dots are the vertices of the
octahedron which create the inverted tetrahedrons. There are 15 regular
tetrahedrons and the vertices which create6 inverted tetrahedrons. So a
total of 21 tetrahedrons for the n=5 base. And the total amount of
tetrahedrons for an n=5 figure is 21+ (total of n=4 figure) 21 + 24= 45 tetrahedrons.
This is n=6. The orange dots are the vertices of the
octahedron which create the inverted tetrahedrons. There are 21 regular
tetrahedrons and the vertices which create 10 inverted tetrahedrons. So a
total of 31 tetrahedrons for the n=6 base. And the total amount of
tetrahedrons for an n=6 figure is 31+ (total of n=5 figure) 31 + 45= 76 tetrahedrons.
This is n=7. The orange dots are the vertices of the
octahedron which create the inverted tetrahedrons. There are 28 regular
tetrahedrons and the vertices which create 15 inverted tetrahedrons. So a
total of 43 tetrahedrons for the n=7 base. And the total amount of
tetrahedrons for an n=7 figure is 43+ (total of n=6 figure) 43 + 76= 119 tetrahedrons.
From here I created a chart because well frankly I did not want to draw out any more triangles and count…. So here is my table I created for the total amount of tetrahedrons in a n length figure.(the arrows shows the method to make it more clear)
|
N length of figure |
# reg. Tetras in base |
# octahedrons in base |
# inverted tetras in base |
#total on base |
Total Tetras for N figure |
|
N=1 |
1 |
0 |
0 |
|
1 |
|
N=2 |
3 |
1 |
0 |
|
4 |
|
N=3 |
6 |
3 |
1 |
|
11 |
|
N=4 |
10 |
6 |
3 |
13 |
24 |
|
N=5 |
15 |
10 |
6 |
|
45 |
|
N=6 |
21 |
15 |
10 |
|
76 |
|
N=7 |
28 |
21 |
15 |
43 |
119 |
|
N=8 |
36 |
28 |
21 |
57 |
176 |
I found out a pattern in each of the columns to find out the next n= down. So for # reg tetras. Take the first (n=1) and add 2, (n=2)= 3, then take the second and add 3 (n=3) 3+3=6 third and add 4 (n=4) 6+4 = 10 so for n=8, its (n=7) +8 = 36 and for the total on base its total on base(n=8)= (2(n-1))+ (n-1total) so for n=8 the total amount of the base is…
(2(8-1))+ (8-1total)= (2(7))+ (7total)= 14+ 7total on base = 14 +43= 57
This expression is for the total number of tetrahedrons on base in the chart. (2(n-1))+ (n-1total)
And now I just found another equations, so that if you do not have a chart and you need to know how many are on that specific base you can easily… there is a pattern for the base column.
N values # base expression expression for any value of N Example
N=1 1 N2 (n2-(n-1)) (12-(1-1)= 1-0 =1
N=2 3 N2-1 (22-(2-1)=
4-1 =3
N=3 7 N2-2 (32-(3-1)=
9-2 =7
N=4 13 N2-3 (42-(4-1)=
16-3 =13
N=5 21 N2-4 (52-(5-1)= 25-4 =21
N=6 31 N2-5 (62-(6-1)= 36-5 =31
N=7 43 N2-6 (72-(7-1)= 49-6 =43
N=8 57 N2-7 (82-(8-1)= 64-7 =57
Ok so now I have completed this table… helps us a bit
Now even better, I used that chart made above, and found an expression for the total amount of tetrahedrons for the whole Nth figure. So first I wrote out the totals for the nth terms and found the differences until they were the same. I came to be the 3rd difference that they were all equal. So the equation is part of ax3+bx2+cx+d= total number in the figure.
So, x is equal to the Nth value. So it will look like an3+bn2+cn+d= total number in the figure.
The nest step was to go and find like values of a, b, c, and d. I did this by taking examples from the chart.
I used n=1 the total of 1, n=2 total of 4, n=3 total of 11, and n=4 total of 24. Four equations were needed because we had to find each variable which is four different equations.
SO using the values I created the equations…..
a+b+c+d=1
8 a+4 b+2 c+d=4, 27
a+9 b+3 c+d=11,
64 a+16 b+4 c+d=24
Then I used mathematica to solve and return
the values of a, b, c, and d.
In mathematica the code was: Solve[{ a+b+c+dŠ1, 8 a+4 b+2 c+dŠ4, 27 a+9 b+3 c+dŠ11,64 a+16 b+4, c+dŠ24 }]
And in return I received {{a®1/3,b®0,c®2/3,d®0}}
With this I can created the expression for
the total number of tetrahedrons per figure.
an3+bn2+cn+d= total number in the figure
a= (1/3) b=0 c=(2/3) d=0 1/3n3+0n2+2/3n+0= total number in the figure
simplified is: 1/3n3+2/3n= total number in the figure
simplified once more is: n3+2n = total number in the figure
3
I also tested this to make sure it was correct. I used n=6 the total is suppose to equal 76.
So 63+2(6) ?= 76 216+12 = 228 = 76
3 3 3
So this does work!!!
Now we have the total amount of tetrahedrons… now we need to find out what sides are painted and what sides aren’t.
I used my model that I created to help me visualize and count everything. I actually counted for n=8.
Here is one way to look at it 1 dimensionally.
N=1 this is the only tetrahedron with all four sides painted, because there are three sides and one bottom=4. And that means there are 0 for 3,2,1,0 faces painted
N=2 the top/tip tetrahedron no longer had paint
on the bottom because the base is now n=2. So it has 3 sides painted, the other
three have 3 sides painted also because they have the two sides and their base.
There is also and octahedron in the center that does not have paint because it
is counted as empty space.
3 3 2
N=3, there is still one top tetra that
has 3 sides painted, no, the second row, they only have two sides painted as
appose to three because that row is no longer the base. There will always be
four tetrahedrons with three sides painted because they are the tip and bases
of the figure. Now the two faces that
are painted, there is a total of six. It also has one with out any paint
because at the circle and the intersection of the octahedron is the inverted
triangle that is in the middle so it contains no paint.
3 N=4 is a bit different. It still has 0 tetras with 4
faces, 4 with 3 faces, but on 2 faces it contains 12. Which is 6 more than
n=3. For 1 face there are four tetrahedrons. The middle one on the 3rd
row on all 3 sides and there is one on the base (n=4) that is in the middle
and only the bottom is showing out that has paint on it. So that is equal
to three. And now there are three with out any paint on them. This is
because there are three total inverted tetras on the n=4 figure.
I did this same thing for all of the tetrahedron bases up to 8. It became reallllllly confusing one we got done to the lower levels, so I made a model that would help me. To help keep track of the faces with no paint, a good way to count was to count the number of inverted tetrahedrons on the figure and then the amount of regular tetrahedrons that also had no paint. These come from the larger levels. They are the tetrahedrons that where in the middle of the bottom /base and after adding another layer they were out of reach and away from the paint. To gain this information I also used the chart I made from above and I created a chart to help because patterns needed to be looked for also.
|
|
4 faces |
3 faces |
2 faces |
1face |
0 faces (regular) |
0 faces(inverted) |
0 faces (total) |
|
N=1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
|
N=2 |
0 |
4 |
0 |
0 |
0 |
0 |
0 |
|
N=3 |
0 |
4 |
6 |
0 |
0 |
1 |
1 |
|
N=4 |
0 |
4 |
12 |
4 |
0 |
4 |
4 |
|
N=5 |
0 |
4 |
18 |
12 |
1 |
10 |
11 |
|
N=6 |
0 |
4 |
24 |
24 |
4 |
20 |
24 |
|
N=7 |
0 |
4 |
30 |
40 |
10 |
35 |
45 |
|
N=8 |
0 |
4 |
36 |
60 |
20 |
56 |
76 |
Now my next step is to go and find expressions so that we can determine how many sides are painted for any given Nth value.
For so the exception of n=1, all 4 faces are 0 tetrahedron. n≠1 and obviously (because we can not have negative sides) n>1. 4faces=0tetrahedrons
For three faces, it also has the exception of 1. n≠1 and n>1 3 faces= 4 tetrahedrons.
For 2 faces with the exception on n=1, the values increase by six. So the first differential is 6 and they are constant, so there will be just n (no powers) and we will have to multiply by 6 to get the terms. But when I did this, every value was off by 12. So to fix this problem, if you take n-2 and then multiply it by six, you will obtain the total number of tetrahedrons with 2 faces for any given Nth value.
Examples: 6(n-2)
N=2 6(2-2)= 6(0)= 0
N=3 6(3-2)= 6(1)= 6
N=4 6(4-2)= 6(2)= 12
N=5 6(5-2)= 6(3)= 18
And so forth.
For 1 face with the exception on n=1, the values I found the second difference is constant which creates an2+bn+c= total 1 faces
And I used mathematica again to solve for the variables
Solve[{16 a+4
b+cŠ4,25 a+5
b+cŠ12, 36 a+6
b+cŠ24},{a,b,c}]
{{a®2,b®-10,c®12}}
So the equation comes out to be 2n2-10n+12= total 1 faces
For 0 faces, it’s the total o faced ones, and the differences are on the third difference, for mathamatica its an3+bn2+cn+d= total number in the figure
Solve[{ 27 a+9
b+3 c+dŠ1,64 a+16
b+4 c+dŠ4,125 a+25
b+5 c+dŠ11,216
a+36 b+6 c+dŠ24},{a,b,c,d}]
{{a®1/3,b®-2,c®14/3,d®-4}}
So the equation is
1/3n3-2n2+14/3n-4= total number in the figure
So all of the equations for all sides are…
|
4 face |
3 faces |
2 faces |
1 face |
0 face |
|
N ≠1 n>1. 4faces=0tetrahedrons |
N ≠1 and n>1 3 faces= 4 tetrahedrons. |
6(n-2) |
1/3n3-2n2+14/3n-4= total number in the figure |
1/3n3-2n2+14/3n-4= total number in the figure |
Solution:
A tetrahedral with edge length n is
dipped in paint. Now we must go and find how many faces of the small
tetrahedron that make up the large tetrahedron; 0, 1, 2, 3, 4, have paint on
them.
So to determine how many sides are painted the number of tetrahedrons had to be found in a given figure and then the next step to determine what sides are painted and what are not. But also, to do this problem you must understand what makes up the tetrahedrons.
Within the tetrahedron shape there are
octahedrons and inverted tetrahedrons. This can be seen by the three
octahedrons figures that create an inverted tetrahedron. This intersection of
the three octahedrons is an inverted tetrahedron.
So by following patters, you can see the different figures.
Below is a chart of the number of figures in the Nth tetrahedron.
|
N length of figure |
# reg. Tetras in base |
# octahedrons in base |
# inverted tetras in base |
#total on base |
Total Tetras for N figure |
|
N=1 |
1 |
0 |
0 |
|
1 |
|
N=2 |
3 |
1 |
0 |
|
4 |
|
N=3 |
6 |
3 |
1 |
|
11 |
|
N=4 |
10 |
6 |
3 |
13 |
24 |
|
N=5 |
15 |
10 |
6 |
|
45 |
|
N=6 |
21 |
15 |
10 |
|
76 |
|
N=7 |
28 |
21 |
15 |
43 |
119 |
|
N=8 |
36 |
28 |
21 |
57 |
176 |
There is a way to get any total value of any given Nth term, this is by the expression
n3+2n = total number in the figure (which was proved above)
3
Now, we must find out how many tetrahedrons have 0,1,2,3, and 4 painted sides on any given figure. This can be found by finding the first 7 or 8 n= tetrahedrons and finding there patterns. This is a chart of the n=1 – n=8 painted faces on the tetrahedrons.
|
|
4 faces |
3 faces |
2 faces |
1face |
0 faces (regular) |
0 faces(inverted) |
0 faces (total) |
|
N=1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
|
N=2 |
0 |
4 |
0 |
0 |
0 |
0 |
0 |
|
N=3 |
0 |
4 |
6 |
0 |
0 |
1 |
1 |
|
N=4 |
0 |
4 |
12 |
4 |
0 |
4 |
4 |
|
N=5 |
0 |
4 |
18 |
12 |
1 |
10 |
11 |
|
N=6 |
0 |
4 |
24 |
24 |
4 |
20 |
24 |
|
N=7 |
0 |
4 |
30 |
40 |
10 |
35 |
45 |
|
N=8 |
0 |
4 |
36 |
60 |
20 |
56 |
76 |
And by finding trends and differences expressions can be made to mind each individual number of tetrahedrons for each side. These expressions are charted below. (also proven above)
|
4 face |
3 faces |
2 faces |
1 face |
0 face |
|
N ≠1 n>1. 4faces=0tetrahedrons |
N ≠1 and n>1 3 faces= 4 tetrahedrons. |
6(n-2) |
1/3n3-2n2+14/3n-4= total number in the figure |
1/3n3-2n2+14/3n-4= total number in the figure |
So now with all of the equations, a long and hard question can be answered with a few algebraic calculations.
Generalizations:
While doing the tetrahedron problem, many generalizations need to be made and there are many that can be found.
To
start, a tetrahedron is a
solid contained by four plane faces; a triangular pyramid. An octahedron is any
polyhedron having eight plane faces.
In the tetrahedron, n is equal
to the number of tetrahedral bases. So n=2 means there are two tetrahedrons
across and two going up on the sides. It has three bases and one tip.
When putting tetrahedral figures
together, octahedrons are created. When these octahedrons interest they create
inverted tetrahedrons. So it can be
generalized that at the intersection of the three octahedrons was an inverted
tetrahedron. So for every three octahedrons there is an inverted tetrahedron,
but they only occur around the bases of tetrahedrons. So there must be three
octahedrons around a base ( a vertex) to create an inverted tetrahedron.
To find the total number of tetrahedrons for
the figure, you have to find the sum of the base plus the total amount of
tetrahedrons of that base from the figures above (just the total amount on the
base). Or the second way to do this was
to take the total on that base from the total from the figure before it. So for
any Nth value by using the expressions, (n2-(n-1))
will let you find the total amount of tetrahedrons on that base, and for the
total number of tetrahedrons in a figure,
n3+2n = total number in the figure is the expression.
3
The next step is to find out how many of the tetrahedrons have 0,1,2,3,or 4 painted faces.
When trying to find these values, more generalizations can be made.
To help keep track of the faces with no paint, a good way to count was to count the number of inverted tetrahedrons on the figure and then the amount of regular tetrahedrons that also had no paint. These come from the larger levels. They are the tetrahedrons that where in the middle of the bottom /base and after adding another layer they were out of reach and away from the paint.
So to find any amount of tetrahedron for a specific painted side the expressions below are used.
|
4 face |
3 faces |
2 faces |
1 face |
0 face |
|
N ≠1 n>1. 4faces=0tetrahedrons |
N ≠1 and n>1 3 faces= 4 tetrahedrons. |
6(n-2) |
1/3n3-2n2+14/3n-4 = total number in the figure |
1/3n3-2n2+14/3n-4 = total number in the figure |
Self-assessment:
I am very proud of myself on this POF. I found many patterns and proved many expression. I have never done that before and I actually get the problem. So I think I did very well and I am really happy. Also, I used mathematica to help me prove and solve some of the expressions and I could never do that before so I am pumped! J
And I built a small portion of the tetrahedrons to create a visual model to help me, and it worked well. J