| Postulate 1
To draw a straight line from any point to any point. Postulate 2 To produce a finite straight line continuously in a straight line. Postulate 3 To describe a circle with any center and radius. |
The first says that it is always possible to join two points with a
straight line segment; the second says that any such segment can be extended
to any length one pleases; and the third asserts that a circle can be drawn
of any size with center at any point. The ground rules established
by these postulates indicate that geometry was to be performed with just
two tools: a straightedge for drawing line segments, and a compass for
drawing circles. It is important to note that the straightedge was
not
a ruler; no marks were allowed on the straightedge.
An example of the sort of problem that was considered
is given by Proposition
i.9 ("To bisect a given rectilineal angle"). The procedure for
bisecting an angle described here should be well known to any modern student
of high school geometry.
Book
4 of the Elements included the methods of construction with
straightedge and compass (which are called Euclidean tools) of each of
the follwing regular polygons: the equilateral triangle, the square, the
pentagon, the hexagon, the decagon (10 sides) and the 15-gon. There
is no doubt that Euclid and his contemporaries could also have perfomed
the constructions of regular polygons having 8, or 12 , or 16 sides, as
well as many others. But there were obvious omissions: How was one
to construct a regular polygon with, say, 7 or 9 or 11 sides? Take,
for instance, the case of the nine-sided polygon. This polygon has
an interior angle of measure 140 degrees = (7/9)p,
so if this construction could be carried out, it would also then be possible
to construct an angle of measure (1/9)p = 20
degrees. But since the equilateral triangle can be constructed, so
can an angle of measure 60 degrees. Thus, the construction of the
9-gon would make possible the trisection of a 60-degree angle.
None of the Greek geometers was able to successfully determine a procedure
for constructing the regular 9-gon. More generally, the trisection
problem (to trisect an arbitrary angle with the Euclidean tools) also remained
unsolved for many centuries. It was not until the nineteenth century
that it was found that this problem is fundamentally impossible:
with Euclidean tools alone, there is no way to devise a construction process
that trisects a general angle. (But here's a neat origami
solution to the problem.)
Nonetheless, many did solve the trisection problem
by employing additional tools besides straightedge and compass. One
way to do this is to assume the existence of a special curve that can be
used in the construction together with the lines and circles drawn by the
straightedge and compass. Any such curve is called a trisectrix.
This lab will walk you through one of these trisection procedures that
makes use of one type of trisectrix curve.
A second "golden age" of geometry began in the
seventeenth century in the wake of Descartes' introduction of algebraic
methods into geometric analysis. Among the large number of curves
studied with these new techniques was a family of curves called roulettes.
Imagine a curve C rolling without slipping along a second fixed
curve C0. If we somehow attach a point P
to the rolling curve, it traces out a third curve as C rolls along
C0.
The path of P is the roulette of C with respect to C0
with pole P. For instance, the cycloid is the
roulette of a circle rolling on a line whose pole is a point on the circle
(see p. 225 of your textbook, and recall exercises 4.1.9 and 4.1.10(a)).
More interestingly, roulette curves are precisely what makes a Spirograph®
set so fun to play with.
An important class of roulette curves are the trochoids,
formed when the rolling and fixed curves C and C0
are both circles. The word trochoid is derived from the Greek trochos,
meaning "wheel". You are encouraged to experiment with the SpiroGraph
applet found at the link provided here, designed by David Little at
UC San Diego. It allows you to draw trochoids of two types: epitrochoids,
formed when the rolling circle C rolls on the outside of the fixed
circle C0, and hypotrochoids, for which C
rolls on the inside of C0. One can adjust the radii
of the two circles--Little denotes as R the radius of C0
and r the radius of C--and the distance p from the
pole P to the center Q of C to obtain a number of
interesting curves. Near the bottom of his webpage, Little provides
some suggestions for values of these parameters which will produce nice
curves.
In the special case that the pole P lies
on the rolling circle C (that is, p = r), we call
the epitrochoid an epicycloid and the hypotrochoid a hypocycloid.
The picture on p. 228 of the text illustrates the generation of an epicycloid.
When p < r, then P lies inside C as it rolls
and the curve is said to be curtate; when p > r, P
lies outside C and the curve is called prolate.
x(t)=(R+r)cos(t) - p*cos((R+r)/r*t)
y(t)=(R+r)sin(t) - p*sin((R+r)/r*t)
We assume that at time t = 0 the center Q of C lies on the positive x-axis, as does P, with P a distance p to the left of Q. Explain how these equations arise by giving parametric equations first for Q as C rolls around C0, then for P as it turns around Q (assuming Q fixed). The composition of these two motions generates the epitrochoid.
x(t)=(R-r)cos(t) + p*cos((R-r)/r*t)
y(t)=(R-r)sin(t) - p*sin((R-r)/r*t)
Again, we assume that at time t = 0 the center Q of C lies on the positive x-axis, as does P, with P a distance p to the right of Q. Explain how these equations are derived.
x(q)=
4cosq*cos(3q)
y(q)=
4sinq*cos(3q)
The last result above shows that the trifolium has polar equation r = 4 cos(3q). This curve can be used to trisect angles. Here is the procedure:
Let us first illustrate the procedure by drawing a Maple plot that shows the trisection of the angle a of measure 1 radian.
- On a set of coordinate axes, plot the trifolium, on which
- an angle a is drawn with the positive x-axis at the origin O (the upper arm of the angle is drawn with a straightedge); this is the angle we wish to trisect.
- Add to this picture the circle of radius 1/2 centered at (1/2, 0). This can be done by bisecting the segment between the origin and the point of intersection A of the x-axis and the trifolium, the point (1, 0). This locates the center C of the circle. Now with the compass draw the circle at that center that passes through the origin. Let B be the point on the circle where the upper arm of the angle crosses it. Therefore, the angle we wish to trisect is angle AOB.
- With the point of the compass at O, open it to the point B and mark the point T on the right loop of the trifolium so that OT = OB.
- It turns out that the ray OT will trisect the angle; that is, angle AOT is precisely one-third angle AOB!