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Prismoids

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July 20, 2013

Abstract

This is the papers abstract . . .

1 Introduction

2 Proof 1

(here goes pictures)Now, our quadrilateral reduces down to two right triangles of hypotenuse

r. Converting these into a system of equations gives

b cos 1 = a cos 2

and

1+2 = .To solve this system, we substitute variables:x = cos 1, 1 = cos

-1 x

y = cos 2, 2 = cos-1 y

c = a/b.

Plugging into our system and condensing into one equation givescos-1 x + cos-1 y = cos-1 cy + cos-1 y = .Solving this equation gives

y =

sin2

c2 2c cos + 1 . (1)

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Since r = by

and c = ab

= coscos

, our equation transforms into

r =cos

( coscos

)2 2 coscos

cos + 1

sin, (2)

Which finally becomes

r =

cos2 2coscoscos + cos2

sin. (3)

So, now, because h =

2 r2, we have our relation for the height of aprismatoid:

h =

1 cos

2 2coscoscos + cos2sin2

. (4)

3 ORourke

(another picture showing that ORourkes statement reduces to smallest di-hedral)

Therefore, the total travel distance is just (1 + cos). However, because

is just hsin( ) = hsin, the travel distance reduces to h(1+cos)

sin . Asthe height is constant for every side on a prismoid and because (1+cos)sin

ismonotone decreasing between 0 and , the travel distance is maximized when is as small as possible; ORourkes statement reduces down to placing thetop on the side with the smallest dihedral angle will always form a validunfolding.

In this paper, we will prove that ORourkes statement is correct; everyprismoid can be unfolded in a volcano unfolding with the top placed on theside with the smallest dihedral angle. For brevity, we will hereon call the sidewith the smallest dihedral angle A and the measure of this angle . First,we notice that if the supporting line of the outward edge of side A does notcross any part of the rest of the unfolding, the fact that the unfolding is validfollows. Therefore, our approach will be to show that the supporting line ofA will not cross the rest of the unfolding.

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4 Proof 2

First, we notice that at the corners between A and the prismoids base, theprismoid is locally a tetrahedron. We will now prove the first lemma:

lemma 1: small dihedral - large corner angle. [use citation]

Therefore, because the relevant dihedrals remain the same after cuttingthe prismoid down to a tetrahedron and that the dihedral of the side adjacentto A is necessarily greater than As, we have that the corner angle of A atthat point is larger than that of As adjacent side. So, because of this fact,we can also show that As corner angle is larger than /2. So, it necessarilyfollows that the adjacent sides are below the supporting line.

Now, we will need to consider the nonadjacent sides. To evaluate thisproblem, we will start by defining the side we want to show is below thesupporting line as B. Now, we will take the solid prismoid and cut throughit along the plane that is perpendicular to the bases of the prismoid andgoes through the points C and D marked above. The resulting figure thatcontains A and B is also a prismoid, and because A and B are the same withrespect to their location in space and dihedrals, our problem remains thesame. Therefore, showing our conjecture works for sides separated by oneedge will show that our conjecture works for sides separated by any number

of edges, which is to say our conjecture is true.

To analyze the problem using our cut prismoid, we will need to calculatesome variables. First, we have the trapezoidal height, H, of side A as sin,where is the edge length of A and is the interior corner angle of A. Simi-larly, we have the veritcal rise of B, V, as sin, where is the edge lengthof B and is the angle of rotation given by (i); the sum being overall the exterior angles of the base from As corner to Bs. As we are dealingwith a simplified case in which there are only two angles, 1 and 2(markedabove), between A and B, simplifies to 2 + 1 + 2. We will also needthe vertical travel distance, T, that side B declines by with respect to A.Defining the edge between A and B as S and the angle between A and Sas 1, we get that T is equal to Ssin1. Given these values for a prismoidunfolding, it is easy to see that the supporting line will not be crossed iff

H V T. pictures are needed throughout

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Therefore, plugging in our values for H, V, and T, we get

sin sin( 2 + 1 + 2) Ssin1 (5)Now, we will need a value for S. To do this, we will need to analyze

the trapezoid between A and B, which we will call N. We will now definethe corner angle of N at point C as and the corner angle at D as . Inaddition, because N is adjacent to both A and B, we also know the measure oftwo of its edges: the edge bordering A is of length , and the edge borderingB has length . This gives us the following information about N. [picturehere]

However, this is not sufficient information to obtain a value for S. This

is because N is essentially two triangles plus a rectangle in the middle, andthere is not enough information to uniquely determine the rectangles base.This means that S could be as large as we wanted without changing any ofits angles or lengths. The best we can do is obtain a lower bound for S bycalculating the length of the base of both of the smaller triangles that borderthe rectangle. The base length of the triangle of the left is equal to cos,while the base of the one on the right is equal to cos. This gives a lowerbound for S of cos + cos. Luckily, replacing S with its lower boundsimply represents the worst case scenario; it must work for the conjectureto be true and its correctness implies that the entire conjecture is true as

well. Our condition for a lack of intersection is nowsin sin(+ 1 + 2) (cos + cos)sin1. (6)

Taking another look at N reveals that because its height is a constant, itshould not matter if we measure it using and or with and . Thismeans that we have

sin = sin (7)

or

= sin

sin(8)

Plugging into (6) and simplifying gives

sin + sin1cos

sin sin(+ 1 + 2) sin1cos

sin 0 (9)

One possible approach from here is to use that , , , ,and2 are in-dependent from 1 to take the derivative of the above inequality with respect

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to 1, then setting equal to 0 to find the minimum value of the above. Then,

we can show that the minimum value is always greater than 0. Doing theabove gives

f(1min) =

c

1 + a2 2acosb, if cosb > a

c sin(b), if cosb a (10)

witha = sin(cot + cot) (11)

b = + 2 (12)

and

c = sinsin

sin. (13)

Ill do more work on the problem later today.

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