Re: Using half-planes to determine point location
- From: Hans-Bernhard Broeker <broeker@xxxxxxxxxxxxxxxxxxxxx>
- Date: 2 Apr 2006 11:40:53 GMT
garrett.j.young@xxxxxxxxx wrote:
[Problem] Connect the vertices of a pentagon with five lines. This
forms 10 triangles.
This already fails to make a lot of sense. There are *10* lines
connecting the vertices of a pentagon --- the 5 edges forming the
pentagon itself, and 5 diagonals, a.k.a. the "pentagram". And they
form not just 10 triangles, but 15 triangles, plus one pentagon, 5
angles, and 5 "wedges" (like angles, but with 3 vertices near the
apex). And that's before one begins to count edges, rays and
intersection vertices as separate regions "formed" by this
arrangement.
Determine which triangles contain that point using a
half-plane test method.
[Difficulty]
How the results of the half-plane test uniquely relate to the specific
triangles that contain the point
A triangle is the intersection of three half-planes. Each line cuts
the plane into two half-planes. Now: how is "intersection" expressed
algebraically?
If you limit yourself to the interior of the pentagon, you have 5
diagonals splitting the pentagon into two regions each. So a point's
position relative to the diagonals is given by 5 testable signs, plus
5 known signs from the edges of the pentagon. Each triangle is defined
by 3 of these 10 signs having particular values.
--
Hans-Bernhard Broeker (broeker@xxxxxxxxxxxxxxxxxxxxx)
Even if all the snow were burnt, ashes would remain.
.
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