| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1013.1 | I doubt it. | AKQJ10::YARBROUGH | I prefer Pi | Fri Jan 13 1989 15:06 | 8 |
| It's a little tricky to come up with a configuration of three circles that
intersect in exactly four points in which ANY spherical triangles are
formed, unless two of the circles intersect in two points, and the third
circle is tangent to both the others, in which case the triangles formed
have three concave or three convex sides, so the angles sum in general to
less or more than 180. That makes the statement false.
Lynn Yarbrough
|
| 1013.2 | Ooops! Let's try again | HPSCAD::HERMAN | | Fri Jan 13 1989 16:37 | 12 |
|
The problem as stated is incorrect but not for the reasons stated in .1;
There are plenty of such configurations.
Correct statement:
Given three circles on the surface of a sphere (not necessarily great circles)
which intersect at exactly four unique points ONE OF WHICH IS COMMON TO ALL
THREE CIRCES, show that the sum of the angles of one of the four curved
triangles formed by this configuration is 180 degrees.
-Franklin
|
| 1013.3 | a diagram | KOBAL::GILBERT | Ownership Obligates | Mon Jan 16 1989 12:14 | 22 |
| Let's see... The three circles should look something like:
--------
/ \
/ \
/ \
| |
--+ C |
/ |\ |
| A \| /
--*- /
\ / / \ \ /
+-- -+------
| B |
\ /
----
At the point where all three circles intersect (*), there are three lenses
and three triangles.
|
| 1013.4 | Missing a triangle? | HPSCAD::HERMAN | | Tue Jan 17 1989 21:05 | 29 |
| Regarding .3, the author has correctly drawn a member of
one of the three families of configurations in question.
WRT the same drawing, Triangles A, B and C do NOT have the required
property. However this triangle does (remember, these circles are on
the sphere):
--------
/ \
/ \
/ \
| |
--+ |
/ |
| * /
/
\ /
+ +------
| |
\ /
----
-Franklin
|
| 1013.5 | conformal map to canonical form | CTCADM::ROTH | If you plant ice you'll harvest wind | Wed Jan 18 1989 11:35 | 14 |
| The situation is where three planes in general position intersect at a
point, say P, lying on the surface of a sphere. The other three points
of intersection are given by the sphere and two of the planes in all
three ways.
There exists a unique circle passing though the latter three points.
Now think of the Riemann sphere and conformally map this fourth
circle to a great circle. Now it is clear that there exists
a triangle with sides summing to 180 degrees. Its vertices will lie
on the great circle and it will be on the opposite side of the
great circle from the image of P.
- Jim
|
| 1013.6 | | CTCADM::ROTH | If you plant ice you'll harvest wind | Fri Jan 20 1989 07:04 | 20 |
| The answer in .5 is actually a bit clumsy on thinking about it.
It just seemed easier to visualize that way.
A much nicer way is to just add a fourth circle to the set so that there
are 4 circles passing through 3 out of 4 points in 4 ways.
Now label the angles the circular arcs make with each other - there are
only 6 possibilities, each angle being repeated 4 times. To see this,
consider just two intersecting circles - the angle of intersection is
repeated 4 times.
Now there will be 4 triangles whose angles sum to 180 degrees. Each
one will have its edges made up of 3 of the circles and its vertices
incident on the fourth. The common point of intersection of the 3 circles
will lie "outside" the fourth circle, and the interior angles will
always appear consecutively around that outer point - hence 180 degrees.
It has a symmetry that's a lot like something from projective geometry.
- Jim
|
| 1013.7 | Why Great Circle Reduction? | HPSCAD::HERMAN | | Sat Jan 21 1989 20:41 | 20 |
|
Regarding .5
> Now think of the Riemann sphere and conformally map this fourth
> circle to a great circle.
No problem.
> Now it is clear that there exists
> a triangle with sides summing to 180 degrees.
Since the image of the three original circles under the conformal mapping
map to new circles, you have reduced the problem to where the other three
points of intersection lie on great circle. Its not at all clear to me why
this reduction gives the result. Please explain further.
-Franklin
|
| 1013.8 | Pro-ject, my good man, pro-ject! | NOEDGE::HERMAN | | Tue Aug 07 1990 00:07 | 25 |
|
Mr. Roth was very close and since its been a while, here goes:
The idea for this problem was inspired by the crummy hand rendered
drawing of stereographic projection on the 1st couple of pages of
the book:
Fractals Everywhere
by Michael Barnsley
The solution to the problem follows from the mentioned drawing:
Rotating if necessary, use the point of common intersection of
the three circles as the north pole for stereographically projecting
the sphere onto the x-y plane. Since stereographic projection
o carries circles on the sphere which contain the point
of projection onto lines in the plane (and conversely)
o is conformal (preserves angles)
the three circles are transformed into three lines which intersect
in a triangle.
-Franklin
|