SOAR Homework Six
These homework problems are meant to expand your understanding of what
goes on during class. Any you turn in will be graded and returned to
you. Answers may or may not be posted on the web, depending on demand.
- Make a model of hyperbolic space, as follows. Print out the page of a tiling of
the Euclidean plane by equilateral triangles. Each vertex is
surrounded by six triangles. At each of these vertices, cut a slit
and add a seventh triangle. (This means that the sheet of triangles
can no longer lie flat. This is the whole point: the hyperbolic
plane is not flat like the Euclidean plane.) (See also
http://members.tripod.com/professor_tom/hyperbolic/, where the
same sort of construction is done with 7-sided polygons. You could
try this, instead, if you like.)
This problem shows that great circles on the sphere correspond to
circles and lines (``generalized circles'') in the Euclidean plane.
We're going to define a map from the sphere x2 +
y2 + z2 = 1 to the Euclidean plane as follows:
draw a line from the north pole (0,0,1) through the point (x,y,z) on
the sphere as in the figure.
This will intersect the xy-plane in a point
(x0,y0,0) (provided the point (x,y,z)
is not the north pole!). (See the pdf file for the pictures.)
It might also help to view this picture from the side of the
xy-plane, as seen below. (See the pdf file for the pictures.)
A great circle on the sphere x2 + y2 + z2 = 1 is the
intersection of the sphere with a plane Ax+By+Cz=0 (this is the
equation of a plane through the origin - the center of the
sphere). We begin by assuming that this great circle does
not pass through the north pole (0,0,1), which means that
C ¹ 0. We can assume that C=1 (just divide through by C and
re-name the resulting variables). For the next parts, therefore,
assume that Ax + By + z = 0.
Use this second figure and some similar triangles to show that
Use similar triangles to show that
Hint: Look at the projection of this situation into the
three planes: the xy-plane, the xz-plane, and the yz-plane.
From part (a), conclude that
x02 + y02 = ||
= 1 + ||
- Using the fact that Ax + By + z = 0, show that the equation
x02 + y02 = 1 + [ 2z/(1-z)] from part (c) simplifies to
x02 + 2 A x0 + y02 + 2 B y0 = 1. |
The equation from the previous part is a circle. Find the
center and radius of this circle. (Your answer will involve A
Now assume that the great circle on the sphere passes through the
north pole (0,0,1). In terms of the equation Ax+By+Cz=0, this
means that C=0, so Ax+By=0. For this last part, therefore,
assume that Ax + By = 0.
- Using the fact that Ax + By = 0, show that the equation from
part (b) simplifies to A x0 + B y0 = 0. This is a line.
At the beginning of this problem, I called circles and lines
``generalized circles'' on the plane. We've shown that great
circles on the sphere correspond to circles in the plane and
certain lines in the plane. What lines are these? That is,
what lines in the plane correspond to great circles on the
In class we saw that the Cantor set C, obtained from the interval
[0,1] by removing the middle third of each interval repeatedly,
has the following properties:
Now consider a revised Cantor set, call it K. Obtain K from the
unit interval [0,1] by removing the middle fifth (rather
than third) of each interval repeatedly. Does K have the same
two properties (I) and (II) as C? Explain why your answer is
There are no intervals in C. That is, given two points a and
b in C, there is a point x not in C between a and
b (that is, with a < x < b).
The length of C is zero. (Well, we saw that the total length of
the removed intervals was 1, so that there was no length left
over for C.)
These problems are also available as a pdf file.
Please email Peter
if you are interested in answers or solutions for the web. Thanks.
SOAR Winter 2003 Course Homepage