Axiverse 366 Report post Posted April 8, 2007 How do you find the minimum bounding circle of a triangle? I was reading an email response on mathworld, but it said the minimum bounding circle was the circumcircle which seems obviously wrong for near colinear triangles. So I know that it's either a circumcircle, or a circle defined by the longest line, but I don't know the limits for these. 0 Share this post Link to post Share on other sites
JohnBolton 1372 Report post Posted April 8, 2007 If the triangle is obtuse, then the diameter of the bounding circle is the longest edge. Otherwise, the bounding circle is the circumcircle. 0 Share this post Link to post Share on other sites
Raghar 96 Report post Posted April 8, 2007 Quote:Original post by AxiverseHow do you find the minimum bounding circle of a triangle? I was reading an email response on mathworld, but it said the minimum bounding circle was the circumcircle which seems obviously wrong for near colinear triangles. So I know that it's either a circumcircle, or a circle defined by the longest line, but I don't know the limits for these.R = a/2a^2 = 2*b^2b = cNow get to the paper, and try to derive this. It's easy 15 minutes to 5 hours exercise. (and you'd practice trigonometry on a simple problem.) I however more often seen a problem "find a sphere that is bounding volume for n-points", so I'm unsure how much often you'd use the above identities. 0 Share this post Link to post Share on other sites
Zipster 2393 Report post Posted April 8, 2007 To supplement JohnBolton's answer, you can easily check if a triangle is obtuse by finding the midpoint of the longest edge, and seeing if the distance from this midpoint to the third point (not on the edge), is less than the half-length of this longest edge, which indicates obtuseness. This is a process based on inscribed angle properties that's a lot better than literally finding and checking each angle of the triangle. 0 Share this post Link to post Share on other sites
Nathan Baum 1027 Report post Posted April 8, 2007 Surelya, b, c = vertices of the triangle (1)origin = (a + b + c) / 3 (2)radius = max (|origin - a| , |origin - b|, |origin - c|) (3)? 0 Share this post Link to post Share on other sites
Zipster 2393 Report post Posted April 8, 2007 Quote:Original post by Nathan BaumSurely[...]?Only in the case of an acute triangle. Imagine an obtuse triangle that's very long and thin. We can let the diameter of the circle be the longest edge (such that the center is on the edge), and the third point would still be inside the circle somewhere. If we instead chose to center the circle at the barycenter, the center is further away from the endpoints of the longest edge than it was when it was directly on the edge, contradicting the fact that it's the smallest such bounding circle.I admit it's a slippery problem though. You can only imagine how crazy things get when you have an arbitrary number of points in an arbitrary dimension :) 0 Share this post Link to post Share on other sites