Download Mike Williams MATH 4800 Spring 12 The Pyramid

Mike Williams
MATH 4800
Spring 12
The Pyramid/Tetrahedron Problem
The Problem
Attach a regular tetrahedron (made of equilateral triangles) to a single face of a regular square
pyramid. How many Faces does the resulting object have?
The Solution
Without thinking about it, one would say that the number of faces follows the general formula
for connecting 2 three dimensional polygons, # faces = #faces object 1  #facesobject 2 −2. Using this
form, the solution would be 5 + 4 – 2 = 7. However in this particular situation, with these two regular
polygons an interesting phenomenon occurs. Two of the Faces on the Tetrahedron and Two of the faces
on the square pyramid end up being parallel with each other. So the actual result is 2 less than the
general formula would predict.
Uniqueness
Clearly, any problem that deals with the number of faces on a geometric object is unique.
Either it has 4 faces or 5 or 6 et cetera. This particular problem has one solution, that solution is 5
faces. In the following paragraph, I will discuss two separate ways to get the equivalent answers when
solving for the angle needed between the base and any one face, so the answer is unique, the solution
method is not unique.
Extreme Cases
Now we get into interesting concepts, what are the specific conditions in which the phenomena,
described in the solution, occurs. By placing a tetrahedron onto a triangular face of a pyramid, at what
conditions does the tetrahedron not add any faces to the pyramid. Many interesting cases emerge if we
take the square pyramid and just change the shape of the base. Taking a minimalistic approach,
decreasing the number of sides of the pyramid by one, does this face alignment phenomena occur when
a tetrahedron is placed onto a second tetrahedron? The answer is no. The angle between the two
tetrahedra is obtuse, exactly 150 degrees (when using equilateral triangles). The more acute the Apex
of the tetrahedron base, the larger the angle is between the faces of the two tetrahedra. The only way
for the phenomena to occur on a triangular base is if the apex angle of the base is 0 degrees. Assuming
one tetrahedron (the one acting as the pyramid) does not change, i.e it is a regular polygon base, the one
being placed onto the pyramid, the height of the base would have to go to infinity, to produce a 0
degree Apex. Since this is clearly asinine, this phenomena does not occur in any pyramid with a base
that is less than four sides. When dealing with regular polygons, this means a square base.
Going the opposite direction, adding sides to the base, also brings up interesting cases. Without
a loss in generality, for all future discussion the side length of each regular polygon base will be 1 and
the tetrahedron that will be attached to the pyramid will have an equilateral base with side length of 1.
When examining a pentagonal base for the pyramid, what are the conditions in which attaching a
tetrahedron adds no sides to the pyramid?
To answer that one must first examine why it occurs for the square base. The square base
pyramid has equilateral triangle faces that make up the pyramid. The dihedral angle, the angle between
the two planes that meet at the apex of the tetrahedron base is defined in the general case, see below.
cos a −cos bcos c 
A = cos−1
sinb sin c
Where a is the apex angle of the base and angles b and c are the base angles

of the triangles on the faces. For a

3 . regular tetrahedron (which
2
produces the phenomenon when the base has for sides) all three angles a, b,
and c are 60 degrees. Therefore the dihedral angle is approximately 70.5
degrees.
By taking a cross section of the pyramid/tetrahedron system,
specifically a plane that bisects the dihedral angle described above, see fig.1,
the required base angle for the pyramid is easily found. Since the plane was
Fig. 1:
setup as a bisector of the dihedral
A
, angle, θ is clearly and the plane is orthogonal to the plane of
2
the base of the pyramid as shown.
In the figure, the darker line depicts the boundary of the
pyramid/tetrahedron object. So by constructing another line orthogonal to the base on the edge of the
object (as opposed to the middle) one can easily see by the alternate interior angles theorem in
euclidean geometry, both angles marked as θ are congruent. This means that the angle φ is exactly 90 –
θ, or approximately 54.75 degrees. Now, this result can be found using simple trigonometry, as all the
faces are regular polygons, cos  = 1 . using the base angle of 60 degrees one can find the
3
height of the equilateral triangle to be The angle between the base and the face is easily found next
using the cosine function, This yields a theta of approximately 54.75 degree as well.
So, what was the point of all that other junk, if the answer can be easily found using trig? Well,
when adding more sides to the base, the faces are no longer equilateral triangles. How do I know that
this phenomena even occurs when adding more sides to the base? The quick
answer, because I made a model of it, the long answer is the following
paragraph. To expand this problem to include any regular polygon base, the
first thing was to determine if the phenomena even occurred. Well, as I said, I
made a model as a proof of concept, then I made the following conjecture.
The phenomena occurs when the angle between the face and the base added
Fig. 2:
to half the dihedral angle of the tetrahedron add up to 90 degrees. The first thing I checked was to see
if equilateral triangles worked for the pentagonal pyramid. Clearly, the apothem of the pentagonal base
is larger than the apothem of the square base, therefore, a pentagonal pyramid made of equilateral
triangles would be shorter than one with a square base. Meaning of course that
the angle between the base and any one face would be less than the that of a
square base pyramid. Since I am assuming a side length of 1, the only other
option is to make the faces of isosceles triangles, see figure 2. One can clearly
see, as is the case in general with an isosceles triangle, that the height of
Fig. 3:
s
tan  and the angle between the face
2
a
and the base, φ, is defined by the cosine function,  = cos−1
. Where a is the apothem of the
h1
the triangular face is defined by the tangent function, h 1 =
 
polygon defined by the following equation, where s is the side length and n is the number of sides.
s
a=

2tan
n
This phenomena is entirely dependent on the height of the pyramid, and because the pyramid height is

entirely dependent on the base angle of the isosceles triangles, the two planes will be in alignment at
specific base angle values, dependent on the number of sides the base has. Thus concludes the
discussion on the requirements for the pyramid.
The requirements for the tetrahedron are similar to the pyramid. First, using regular polygon
bases, the base of the tetrahedron is a regular (equilateral) triangle. Second the faces of the tetrahedron
must be the same shape as the faces of the pyramid. These two requirements describe the tetrahedron
exactly. Using the dihedral angle formula from above we can see that the angle between any two faces
on the tetrahedron is:
A = cos−1

cos

− cos 2 
3
sin 2 

Using this equation for the dihedral angle, A, and the equation for the angle between the base and the
face of the pyramid, φ, and the conjecture,  
A 
= . Notice that both φ and A are in terms of θ,
2
2
so using numerical methods via Maple to solve for theta based on the side length and the number of
sides of the base, see Appendix 1 for specific details.
Now for the results of that Maple work. As a check method to make sure the maple program
worked, I input side length of 1 and 4 sides. This yielded a base angle of 60 degrees. The rest of the
results from the Maple program I wrote (detailed in Appendix A) are in the Table below.
Input (side length, Isosceles base Pyramid height Dihedral angle, A Base/face angle, φ
# base edges)
angle
(1,4)
60
2
70.5
54.75
2
(1,5)
68.7
1.08
64.9
57.54
(1,6)
73.22
1.41
62.96
58.51
(1,7)
76.07
1.73
62.02
58.99
(1,8)
78.05
2.03
61.47
59.26
(1,9)
79.52
2.33
61.13
59.44
(1,10)
80.65
2.62
60.89
59.55
(1,25)
86.38
6.84
60.13
59.94
(1,50)
88.2
13.76
60.03
59.98
(1 , 100)
89.1
27.55
60.01
60
Extrapolating these results to an infinite sided base, the base angles of the sides go to 90 degrees, and
the height of the pyramid goes to infinity. Thus on a macro scale, the pyramid becomes a cone and the
angle between the base and the faces goes to 60 degrees. As long as the conjecture continues to hold,
the dihedral angle of the tetrahedron goes to 60 degrees, for all intents and purposes, it becomes an
infinitely tall triangular prism, although the sides are still triangular. That attaches to the cone in two
places such that both faces of the prism are tangent to the cone.
Specialization and Generalization
Specializing this problem into 2 dimensions brings up some other interesting characteristics. To
specialize this into less dimensions, one must examine what the phenomenon looks like in less
dimensions. In three dimensions putting these two objects together does not add or take away any
faces, a two dimensional piece of the object. So in two dimensions, one would look at an object that
does not add or take away any extra sides, the one dimensional piece of the two dimensional object.
One would start by going through all the different two dimensional regular polygons, analogous
to the pyramid in three dimensions, and attaching a trapezoid, analogous to the tetrahedron, starting
with a regular triangle. Without loss in generality, all future two dimensional discussion of regular
polygons will have a side length of 1. Putting requirements on the trapezoid, the longest side will have
a length of 1, and will attach to the polygon. With these restrictions on the
polygons being used, a triangle does not work as the angle between the sides
of the triangle and the sides of the trapezoid is inherently less than 180
degrees. However removing the stipulation that the largest side must attach to
the polygon, a triangle does indeed work given the base angles of the trapezoid
are both equal to the base angles of the triangle.
Fig. 4:
With the given stipulations, a square does not work either, as a trapezoid is has two acute
angles, see figure 4, the sides will never be parallel to the square sides unless the base angles of the
trapezoid were 90 degrees, making it a square. As with the triangle problem, removing the restrictions
on the trapezoid allow for a solution. Instead of forcing the longest side to
attach to the regular polygon one could place the parallel parts of the
trapezoid parallel to the sides of the square, see figure 5. This allows for all
stipulations to occur, mainly, a trapezoid was placed against a regular polygon
and the resulting polygon had equal numbers of sides as the original polygon.
Fig. 5:
All other regular polygons, five or more sided, the stipulations made on
the trapezoid do indeed work. To determine the base angles of the trapezoid, one must first know the
interior angles of the polygon they are using. The interior angles of a regular polygon are dependent
only on their number of sides, of which the general form is:
Interior Angle =
 n−2∗180
n
The trapezoid is then constructed so that the interior angle of the polygon added to the base angle of the
trapezoid add up to 180 degrees. This ensures that the edges of the trapezoid and the edges of the
polygon line up. Because of the stipulations on the side length, the height of the trapezoid is bounded.
The lower (open) bound on the height is obviously 0, the upper bound is slightly more difficult. As the
length of the upper base of the trapezoid goes to 0, it becomes an isosceles triangle. So the upper
(open) bound is: h = s tan base angle .
2
The generalizations are imbedded in the two previous sections, generalizing the threedimensional problem and generalizing the two dimensional problem.
Analogy
The thing I was thinking when solving this problem, by extrapolating the three dimensional
pyramids, was Optics. More specifically, light moving through different media. The solution method
of using the angles between the base and the faces and the dihedral
angle forcing those to line up reminds me of the following optics
problem.
A beam of light travels through air, n 1 = 1, at an angle 40
degrees below horizontal until it strikes a medium with an index of
refraction, n 2 = 1.5 . If this layer is 15 cm thick and there is another
Fig. 6:
n
=
2,
layer of liquid, 3
under the first layer with a light source emitting light towards the top, at what
angle off horizontal does the bottom light source have to emit light so that the two beams of light run
parallel through the second medium?
The solution uses a physical principle known as Snell's Law and requires use of angle
measurements and requires them to line up. First off, Snell's Law says, n 1 sin 1  = n 2 sin  2 where
1 and 2 are the angle off of a line orthogonal to the surface of the medium, See figure 6. The
solution is fairly simple if you understand optics, seeing as light bends towards the vertical when the
index of refraction increases and bends away from the vertical if the index of refraction decreases. The
other part is noticing that the light strikes the same angle on both the edges of the second medium by
the alternate interior angles theorem of geometry. With these two pieces of knowledge and Snell's Law
one can easily solve for the angle needed in the tertiary medium, which is 22.5 degrees off the vertical,
so 77.5 degrees off horizontal is the angle needed to produce the secondary parallel beam.
A second Analogy comes from placing two of the Sand Crawler shapes together in a specific
way. By place the two objects square bases together so that one has the tetrahedra oriented to the sides
and the other has the tetrahedra oriented top and bottom one can create a larger tetrahedron with
equilateral sides. The reason this occurs is actually fairly simple in concept (not so simple in math).
The dihedral angle between the two faces that make up the top section of the sand crawler shape is 70.5
degrees. This is defined above in the extreme cases section.
The two tetrahedron bases make special angles with the base of the pyramid. While the faces of
the pyramid make 54.75 degree angles with the base, each of the tetrahedra's bases make exactly 180 –
54.75 degrees. By expanding the planes that these two faces constitute to see where they intersect, one
sees that the angle between them is exactly 70.5 degrees. So orienting them in the manner described
above, one can see clearly why these two shape can create a tetrahedron.
Interpretation and Representation
In the previous discussion, the problem has been defined in geometric terms. The solutions
have been expressed in geometric (counting the number of faces), and trigonometric terms. The
methods used to expand the three dimensional pyramid base to infinitely sided based used both
algebraic methods and technological methods.
One interesting hurdle while using the maple software was solving for the base angle of the
isosceles triangle that makes up the faces of both the pyramid and the tetrahedron for a base with five
or more sides. The problem encountered was with the floating point solve, fsolve, function in the
Maple program. Specifically, it was kicking out a negative solution for the base angle of the triangle,
and therefore, a negative pyramid height. I spent the better portion of a day going back through the
equations making sure they were accurate and that they were correct for the problem I was trying to
solve. It was not until later that I realized the equations I was using would give two solutions as there
was a cosine squared and sine squared term in one of the equations.
Unfortunately, fsolve was not returning both solutions, it was just returning the first solution.
Now I should have realized right out of the gate that a solution to the equation (and arccosine function)
that returned a value of -1.8 was not possible as the principle value of the arccosine function is bounded
between 0 and π. After some time working with Maple settings I finally gave up and dropped the
floating point solve for the general solve function. When this kicked back the results I immediately
noticed the error. The solve kicked out two solutions, the answer I was looking for, and the answer
minus π. Finally I was able to get Maple to solve the functions for me, and I was able to find
meaningful results. As with everything, using the tool to make it easier, I wrote the program(s) that are
detailed in Appendix A.
Duplicate the Solution
There are several ways to duplicate the solutions, going back to three dimensional pyramids and
tetrahedra, the easiest way to duplicate the solution is to place more tetrahedra onto the pyramid so that
it still has the same number of faces. For this to occur, the faces used to create the phenomena
described in the original problem must remain undisturbed. This is done by examining the number of
faces that are not used in this phenomena.
For a four sided base, three of the four sides are used in the first phenomena, so, by placing a
second tetrahedron on the side opposite the first, it produces the same phenomenon and does not add
any faces to the pyramid. The resulting shape is very similar to the sand crawler that the Jawa's in
George Lucas' 1977 Blockbuster film “Star Wars” use.
In general, the number of tetrahedra that can be placed on an n sided pyramid is exactly the
floor of n over 2. This is obvious in even numbered bases, for example, base 4 gives 2 possible
n
# Tetrahedra =
2
tetrahedron mounting places, base 6 gives 3 tetrahedron mounting places. The odd numbered bases are
⌊⌋
why the floor function is needed. In a five sided base, there are 2 places that one could place another
tetrahedron after the first, however, by choosing one of those two places, the other is no longer a viable
place for a tetrahedron. Thus there are only two mountable tetrahedra for a 5 sided base. A seven sided
base is similar, after two placements of the tetrahedra, there will be a choice of two places right next to
each other. By choosing one, the other becomes useless.
Homework
1. Given an angle between the base and a face of the pyramid of 70, is it possible for a tetrahedron
to line up with each face. (Comprehension)
2. What is the range of values for the height of a Trapezoid attached to a regular Octagon with a
side length of 1? (Application)
3. Are there any other 3-D objects that when combined the object faces line up? (Synthesis)