Download Math 3329-Uniform Geometries — Lecture 04 1. Constructions with

Math 3329-Uniform Geometries — Lecture 04
1. Constructions with straightedge and compass
Traditional geometry often deals with constructions with straightedge and compass. You can see this in
Euclid’s First Postulate, “To draw a straight line from any point to any point,” and his third, “To describe
a circle with any centre and distance.” With this, we allow ourselves only the ability to do the following.
Given two points A and B, we can draw a straight line through the two points with the straightedge, and
given a third point C, we can draw a circle with center C and radius equal to the distance between A and
B with the compass. It should be emphasized that the straightedge is not a ruler, and so measuring lengths
with it is against the rules. I don’t see this as a practical approach, but like driving with your left foot, it is
exciting, and it will instill a greater appreciation for life.
In some sense, Euclid’s Elements is as a very methodical description of the things you can do with straightedge
and compass constructions. It actually makes more sense to think of it this way, as opposed to thinking of
it as an axiom system. The first three postulates tell you how you can construct geometric figures, draw a
line through any two points with the straightedge, extend a line you already have with the straight edge,
and draw circles with the compass. The last two postulates give some basis for interpreting what the figures
represent and that the things you see in a figure always behave the same way.
Euclid’s Elements consist of thirteen Books, and these contain 432 Propositions. The propositions are
basically theorems that tell you that a certain kind of figure can be constructed (the proof tells you how)
or some fact about a particular kind of figure. We’ll look at some of these propositions to get some sort of
feeling for Euclid’s work.
Euclid’s Proposition I (from Book I) states [Euclid, p 241]
On a given finite straight line to construct an equilateral triangle.
Here Euclid is saying that if you have a line segment (finite straight line), then you can construct an
equilateral triangle (a triangle with three equal-length sides) with this segment as one of the sides. Euclid’s
proof goes something like this. Let’s say our segment has endpoints A and B, and we’ll call the segment
AB. We then draw two circles each with radius AB, one with center at A and one with center at B. The
circles will have two points of intersection, C and C 0 . Both 4ABC and 4ABC 0 are equilateral triangles,
since AC, AC 0 , BC, and BC 0 are all radii of one of these two circles. See Figure 1.
C
A
B
Figure 1. Given segment AB, we can construct an equilateral triangle 4ABC.
For some reason, Euclid used a collapsing compass. He could put one end at a point (the center) and
the drawing end at another point, and then he could draw the circle. Once he picked it up, however, the
length of the radius was lost. We’re obviously not talking about a real-world compass, and the motivations
here are probably that Euclid was trying to start with the most basic assumptions possible. In his second
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proposition, Euclid shows that a collapsing compass is equivalent to a non-collapsing one. This is of no
concern to me. I’m interested in how Euclid’s big geometric ideas got us to where we are today.
Let’s skip up to Proposition 4 [Euclid, p 247]
If two triangles have the two sides equal to two sides respectively, and have the angles
contained by the equal straight lines equal, they will also have the base equal to the base,
the triangle will be equal to the triangle, and the remaining angles will be equal to the
remaining angles respectively, namely those which the equal sides subtend.
This proposition illustrates one of Euclid’s logical flaws as an axiom system. Euclid starts the Elements
with some basic assumptions, and then seems to prove the propositions from these assumptions. The proof
he gives for Proposition 4, however, says little more than “it’s true, because it’s obviously true.” If you
read the statement carefully, you may recognize this as the side-angle-side criterion for the congruence of
triangles or SAS . One of Hilbert’s fixes is to assume SAS as an axiom. Geometrically, SAS tells us a
couple of things. One is that a triangle only has three degrees of freedom. In other words, designating a
angle and the two adjacent sides completely determines the triangle (the lengths of all its sides, the measures
of its angles, and its area). It also expresses the uniformity of the Euclidean plane: the geometry of a triangle
is the same no matter where it is.
Before we go on, let’s make clear what we mean by congruent triangles.
Basic Principle 1. Intuitively, two triangles are congruent, if you can pick one up, and place it on the
other so that they coincide exactly. Corresponding sides will be the same length, corresponding angles will
have the same measure, and the areas will be the same.
1.1. Quiz.
–1– Is there a SSSS criterion of congruence for quadrilaterals? In other words, suppose we have two
quadrilaterals ABCD and A0 B 0 C 0 D0 , and AB = A0 B 0 , BC = B 0 C 0 , CD = C 0 D0 , and DA = D0 A0 .
Must the two quadrilaterals be congruent?
–2– Is there a AAAA criterion of congruence?
–3– Consider a quadrilateral with two sides of length 3 and two sides of length 2. It also has two right
angles. Draw two non-congruent quadrilaterals with these properties.
Getting back to the SAS criterion, in addition to telling us whether two triangles are congruent, it also tells
us how to copy a triangle. If we were given a triangle 4ABC, we could copy the angle ∠A, and then mark
off lengths AB and AC on the two sides of the angle. At this point, there is only one way to put in the third
side.
Let’s investigate how an SSA criterion might work.
C
A
B
Figure 2. Here’s a triangle 4ABC.
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1.2. Quiz.
–1– Copy the triangle in Figure 2 as best you can.
–2– Now copy the angle at ∠A as best you can, and extend the sides out longer than you’ll need (maybe
twice as long as they are in 4ABC). Label the vertex A0 . We’ve got ∗ ∗ A, right?
–3– As best you can, mark of a distance AC to get a point C 0 . Now we have ∗SA.
–4– From C 0 , we need to add a side C 0 B 0 that is the same length as CB. Draw a circle centered at C 0
with radius equal to CB. The circle represents all the segments of length CB with one endpoint at
C 0 . We also want B 0 to be on the other side of ∠A.
–5– How many triangles do you see that match SSA with 4ABC?
–6– Can there be an SSA criterion of congruence?
Proposition 11 tells us how we can construct a perpendicular. For example, suppose that we have
a line l and a point P on it. To construct a line through P that is perpendicular to l, we would do the
following. Draw a circle (with any radius) centered at P . This would give us two points A and B. See Figure
3.
A
P
B
Figure 3. Given a point P on a line, we draw a circle centered at P to find points A and B.
Now we do what we did in Proposition 1, and draw circles centered at A and B, both with radii AB. Then
we draw the perpendicular through the two points where the two circles intersect. Let’s call the two points
C and D. See Figure 4. The triangle 4ABC is an equilateral triangle, which means that the three sides
have the same length. It’s probably obvious to you that the three angles must be equal also (i.e., that the
triangle is also equiangular), but it’s not terribly easy to prove this (actually it’s technically impossible)
within Euclid’s postulate system. That’s much of what the previous ten propositions are trying to establish.
In addition, we will find that similarly obvious “facts” are not necessarily true. We’ll skip over this, and once
we accept that the three angles of an equilateral triangle are equal, it easily follows that ∠AP C = ∠BP C,
and so they must be right angles (see Euclid’s Definition 10). Note that if we had started with the points
A and B, then the line or segment CD gives us the point P and a perpendicular bisector. (Which
proposition is that?).
C
A
P
B
D
Figure 4. As in Proposition 1, we can then find the perpendicular through P .
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Proposition 23 tells us how to construct the copy of an angle. Suppose we have an angle with vertex
A and a line l with A0 on it. These are the lines in black in Figure 5. We want to construct another line
through A0 creating the copied angle. We start by drawing any circle with A at its center, which produces
points B and C on the two sides of the original angle. Next, draw a circle with the same length radius with
A0 at its center (ours is not a collapsing compass). This will give us a point B 0 on l. Now, draw a circle
with center at B 0 with radius equal to BC. The circles with centers at A0 and B 0 will intersect at a point
C 0 (there are two choices). The line through A0 and C 0 gives us a copy of the original angle. We know
this because 4ABC and 4A0 B 0 C 0 are congruent triangles. Again, this isn’t as easy to prove as you might
expect. It is easy to see that the corresponding sides of the two triangles are equal, but it takes some work
to show that the corresponding angles are also equal. We will skip over this part too.
C0
C
A
B
A0
B0
Figure 5. We can copy an angle onto a line at a given point.
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References
[Bonola]
Roberto Bonola (1955). Non-Euclidean Geometry (H.S. Carslaw, Trans.). Dover Publications, New York. (Original
translation, 1912, and original work published in 1906.)
[Descartes] Rene Descartes (1954). The Geometry of Rene Descartes (D.E. Smith and M.L. Latham, Trans.). Dover Publications, New York. (Original translation, 1925, and original work published in 1637.)
[Euclid]
Euclid (1956). The Thirteen Books of Euclid’s Elements (2nd Ed., Vol. 1, T.L. Heath, Trans.). Dover Publications,
New York. (Original work published n.d.)
[Eves]
Howard Eves (1990). An Introduction to the History of Mathematics (6th Ed.). Harcourt Brace Jovanovich, Orlando, FL.
[Federico] P.J. Federico (1982). Descartes on Polyhdra: A study of the De Solidorum Elementis. Springer-Verlag, New York.
[Henderson] David W. Henderson (2001). Experiencing Geometry: In Euclidean, Spherical, and Hyperbolic Spaces 2nd Ed.
Prentice Hall, Upper Saddle River, NJ.
[Henle]
Michael Henle (2001). Modern Geometries: Non-Euclidean, Projective, and Discrete 2nd Ed. Prentice Hall, Upper
Saddle River, NJ.
[Hilbert]
David Hilbert (1971). Foundations of Geometry (2nd Ed., L. Unger, Trans.). Open Court, La Salle, IL. (10th
German edition published in 1968.)
[Hilbert2]
D. Hilbert and S. Cohn-Vossen (1956). Geometry and the Imagination (P. Nemenyi, Trans.). Chelsea, New York.
(Original work, Anschauliche Geometrie, published in 1932.)
[Motz]
Lloyd Motz and Jefferson Hane Weaver (1993). The Story of Mathematics. Avon Books, New York.
[Weeks]
Jeffrey R. Weeks (1985). The Shape of Space. Marcel Dekker, New York.