1230, notes 3

1 Appendix to notes 2, on Hyperbolic

geometry:

The axioms of hyperbolic geometry are axioms 1-4 of Euclid, plusan alternative to axiom 5:

Axiom 5-h: Given a line l and a point p not on l, there are at leasttwo distinct lines which contain p and do not intersect l.

Theorem: Assuming Euclid’s first four axioms, and axiom 5-h, andgiven l and p not on l, there are infinitely many lines containingp and not intersecting l .

Proof ?

Comment on the “extendability” axiom.

The wording in Euclid is a bit unclear. Recall that I stated it asfollows:

A line can be extended forever.

This is basically what Euclid wrote. Hilbert was more precise,and adopted an axiom also used by Archimedes:

Axiom: If AB and CD are any segments, then there exists anumber n such that n copies of CD constructed contiguouslyfrom A along the ray AB willl pass beyond the point B.

A model of hyperbolic geometry: Consider an open disk, say ofradius 1, in the earlier model of Euclidean geometry. Thus, let

D ={(α, β) | α2 + β2 < 1

}

Definition: A “point” is a pair (α, β) in this set.

Definition: A “line”is any diameter of this disk, or the set of pointson a circle which intersects D and which meets the boundary ofD in two places and at right angles to the boundary of D at bothof these places. Notice that a diameter can be considered part ofsuch a circle of infinite radius.

However, there a lot to be said before we can claim that this modelsatisfies Euclid’s first four axioms. In particular, work is requiredto define what “congruent” line segments are. It is necessaryto define some sort of distance between points which is differentfrom Euclidean distance. With this definition, it turns out thata line segment near the boundary can look very short to us, andyet according to the definition of distance, be congruent (samelength) as a segment near the middle of the disk which looks alot longer to us. It is in this sense that a “line” can be extendedindefinitely (to “infinity”). This will be discussed in more detail iftime permits later.

1.1 Euclidean solid geometry

Planes are mentioned in some of the 1st 20 definitions in chapter1.

They don’t appear again until chapter 11, where there are new de-finitions (e.g. "solid angle", definition 11), but no new postulates.(Hilbert included several axioms about planes.)

2 Euclid, Book 13 "Platonic Solids”

What are they?

Definitions:

Although Platonic solids are three dimensional objects, we startwith a definition in R2.

1. polygon. (class exercise)

As an example of a polygon we have, for example, triangles,squares, etc. Please note that I am talking here about the bound-ary and everything inside. For example, the set

S = {(x, y) | 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1}

is a polygon (called the “unit square”). Its boundary is{(x, y) | (x, y) ∈ S and either x = 0, x = 1,

y = 0 or y = 1o9r

}

but this set is not itself a polygon, at least not as a polygon isusually defined.

Definitions:

1. polygon

A polygon in R2 is a closed bounded subset of R2 with the fol-lowing properties:

(a) It has a non-empty interior.

(b) Its boundary consists of a finite set of line segments.

(c) Each of these segments intersects exactly two of the others,one at each of its endpoints. There are no other intersectionsbetween these segments.

(d) The boundary is connected.

A polygon in R3 is a subset of a plane which is a polygon in thatplane.

What “unusual”polygons are there with this definition?

2. A set in Rn, is “convex” if each intersection of this set with a(straight) line is either empty, a single point, or a segment of thatline.

3. A polygon is regular if it is convex, each edge is congruentto each other edge and all the angles between adjacent edges areequal.

convex, regular

non-convex irregular

Now we move to three dimensional objects.

4. A polyhedron is a connected closed bounded region inR3 whoseboundary consists of a finite set of polygons. These polygons arecalled “faces” of the polyhedron. Polyhedra can also be convexor non-convex.

convex polyhedron | nonconvex polyhedron

Not a polyhedron

But, all definitions are arbitrary, and authors often differ, accord-ing to the setting in which they are working. The definition I gaveprobably allows for some weird looking polyhedra. Some authorsdon’t even attempt to define polyhedron, but simply define “con-vex polyhedron”, which would just add the conditon that the set isconvex to what I have above, but would very much simplify whatis allowed.

Test of geometric intuition: Start with a cube.

Looking down from the top, cut vertical slices (parallel to thevertical edges) along the four slanted lines, thus cutting off thefour vertical edges.

Make the same cuts on a second side,

oriented as shown.

Do this again.

Do this again.

What is the result?

Definition: A Platonic solid, or “regular polyhedron”, is aconvex polyhedron such that all faces are congruent to eachother, as are all solid angles, and each face is a regularpolygon.

Plato: “The first will be the simplest and smallest construc-tion,and its element is that triangle which has its hypotenuse twicethe lesser side. When two such triangles are joined at the diago-nal, and this is repeated three times, and the triangles rest theirdiagonals and shorter sides on the same point as a centre, a singleequilateral triangle is formed out of six triangles; and four equi-lateral triangles, if put together, make out of every three planeangles one solid angle∗, being that which is nearest to the mostobtuse of plane angles; and out of the combination of these fourangles arises the first solid form which distributes into equal andsimilar parts the whole circle in which it is inscribed.”∗See Elements, chapter 11 for defintion of "solid angle"

He is describing a tetrahedron, in which three equilateral trianglesmeet at each vertex:

sum of angles at vertex: 3× 60o = 180 < 360.

Other possibilities:

Four triangular faces meet at a vertex. This gives an octahedron

sum of angles at vertex = 4× 60 = 240 < 360.

Five triangular faces meet at a vertex, giving an icosahedron (20faces):

sum of angles at vertex = 5× 60 = 300 < 360

Can six triangular faces meet at a vertex? Angle sum = 6 ×60 = 360, so the surface would be flat and we don’t get a threedimensional solid. (See Euclid, Book 11, Propositions 20,21.)

How about square faces? The only possibility is a cube — threefaces meet at a vertex.

sum of angles at vertex= 3× 90 = 270..

Four faces meeting at a point would give a sum of 360

There is one with pentagonal faces, three meeting at a vertex.This gives a dodecahedron (12 faces)

sum of angles at vertex: 3× 108 = 324 < 360.

That’s all there can be (five), because any others would give toolarge an angle sum ( ≥ 360 )

Exactly what has been proved here?

Theorem: There are no more than five platonic solids.

Theorem: There are no more than five platonic solids.

But how do we know there are any at all?

The greatest idea of Greek mathematicians: Assertions such asthe existence of any of the platonic solids need to be “proved”.

The greatest idea of Greek mathematicians: Assertions such asthe existence of any of the platonic solids need to be “proved”.

For example, do the Platonic solids exist without the parallel pos-tulate? (This may be the universe in which we live!)

The greatest idea of Greek mathematics: Assertions such as theexistence of any of the platonic solids need to be “proved”.

Earlier example: The "Pythagorean theorem". The result wasdiscovered many times, in many places, but as far as we know,without proof before Pythagoras.

Construction of a cube. Proposition 15, Book 13.

To construct a cube, and enclose it in a sphere.

Let the diameter AB of the given sphere be laid on,† and let it havebeen cut at C such that AC is double CB.‡ And Let CD have beendrawn from C at right angles to AB. And let the semi-circle ADBhave been drawn on AB. And let DB have been joined. And letthe square EFGH, having its side equal to DB, be laid out.§ Andlet EK, FL, GM, and HN have been drawn from points E, F, G,and H, respctively, at right angles to the plane of square EFGH.¶

And let EK, FL, GM, and HN, equal to one of EF, FG, GH, andHE, have been cut off from EK, FL, GM, and HN, respectively.And let KL, LM, MN, and NK have been joined. Thus a cubecontained by six equal squares has been constructed.

Question: Did this require the parallel postulate?†Book 1, proposition 2‡prop 6.10§proposition 46, book 1¶prop. 11.12

Definition: Let P be a convex polyhedron. Then the “dual” ofP is the polyhedron with vertices at the centers of the faces of P,and edges connecting the centers of any two faces with a commonedge.

Proposition: If P has v vertices, f faces, and e edges, then thedual of P has f vertices, e edges, and v faces.

Construction of an Octahedron: Book 11, prop 14 or:

The polyhedron whose vertices are the centers of the faces ofa cube is an octahedron. (The "dual" of the cube".) Also,the polyhedron whose vertices are the centers of the faces of anoctahedron is a cube.)

Construction of a tetrahedron: Proposition 13, Book 13. (pg.519 in Fitzpatrick edition, 37 lines) or: Prove that an appropriateset of 4 vertices of a cube form a tetrahedron.

Construction of a Dodecahedron: Proposition 17. (pg. 530,88 lines)

Construction of an icosahedron: Prop 16, (pg. 512. 85 lines)

“Woodworker’s proof” that a dodecahedron exists

In the slicing method we leave, at the end, one line segment oneach of the original faces of the cube. We assume that these areof equal length and symmetrically placed.

Consider three of the cut planes as shown:

Here we are assuming that each of the cutting planes goes from acenter line of one face to a point a distance d2 from the center lineof another face, with orientation as shown. We need to calculated in terms of the length s of a side of the cube. In fact, d will thelength of an edge of our dodecahedron.

We calculate the intersection of the three planes (yellow dot). Thisis a straight forward problem in linear algebra. After some furtheralgebra it turns out that in order for that point to be the distanced from the three other points we must have(

sd− s2)2+ s4 + (s− d)4 = d2 (4s− 2d)2 (1)

If we can find a positive solution to this equation which is lessthan s then we will have proven that a dodecahedron exists. Infact, some algebra leads to four solutions,

d =1 +√5

2s

(golden mean, ≈ 1.618 s,

but we want d < s

)

d =1−√5

2s (< 0)

d =3 +√5

2s (> 1)

d =3−√5

2s,

only feasible value,= 2− golden mean≈ 0.382 s

d =3−√5

2s

Here is one of the pentagons (yellow dots) :

The golden mean can be constructed by methods of Euclid (straightedge and compass; book 2, proposition 11), and so in essence this

method can be said to be Euclidean. But algebra makes it easierto prove this.

The Icosahedron is the dual of the dodecahedron, and vice-versa.

Better method for constructing a dodecahedron: Start with theicosahedron, using a construction due to Luca Pacioli, 1509. (SeeStillwell, page 21-22)

First define the "golden ratio" as usual:

1

x=

x

1− x

x =1 +√5

2= 1.618.....

1. Construct a rectangle with this ratio of the longer side to theshorter. (Book 2, proposition 11, plus book 6, prop 30.)

2. Construct three perpendicular rectangles of this size as shown(Book 11, prop. 11)

Prove that ABC is an equilateral triangle.

3 Beyond the Platonic Solids

4 The Archimedian solids

Faces are all regular polyhedra, but they are not all the samepolyhedron.

Archimedes proved that there are exactly 13 of these.

4.1 Euler’s formula

e = v + f − 2.

Applies to any “simply connected”polyhedron. Really a theoremin "topology"!

Plausibility argument (??): Consider a polyhedron with only fourvertices. Then this is a tetrahedron (possibly irregular). Showthat if one vertex is added, then another edge is added also, sov+ f − e is unchanged. Add an edge. This adds a face, so againv + f − e is unchanged. Check its value for a tetrahedron

Proofs: There are many. For example, see page 471 of Stillwell.

(There are many formulas due to Euler. The best known are thisone and eiπ = −1.)

This can be used to prove (again) that there are only 5 regularpolyhedra.