Wikipedia:Reference desk/Archives/Mathematics/2014 December 20

= December 20 =

Circumcircles of two triangles with a common edge
Hello!

I'm trying to write (for fun, not homework) a program that calculates the Delaunay triangulation of a set of points, and I have a small math/geometry question.

Lets say you have four points that make up a convex quadrilateral. Call these points p1, p2, p3 and p4. The points p1 and p4 are opposite each other, as are p2 and p3. Draw a line from p2 to p3 to split the quadrilateral into two triangles with one common edge (from p2 to p3). That is, one triangle is made up of (p1, p2, p3) and the other is (p2, p3, p4). The triangles don't intersect, since we've just split a convex quadrilateral in two.

My question is about the circumcircles of these triangles. In a valid Delaunay triangulation, none of the points of the triangulation lie on the inside of any of the circumcircles of the triangles. My question is this: if p4 is inside the circumcircle of the triangle (p1, p2, p3), does that automatically imply that p1 is inside the circumcircle of (p2, p3, p4)? That is, is it true that it is either the case that both of the points are inside the respective circumcircles, or that neither are? Or could it be that one point is inside the one of the circumcircles, but the other one isn't?

The reason I'm asking is that it'll help me to know how many points I need to test in various stages of the algorithm to make sure the triangulation is valid. I hope I made myself clear, but if anyone needs clarification, feel free to ask. 90.236.78.183 (talk) 11:37, 20 December 2014 (UTC)
 * The answer is yes. The location of p4 wrt to the circumcircle is determined by the following:
 * If P1 = (x1, y1), P2 = (x2, y2), P3 = (x3, y3) are oriented counterclockwise, then the point P4 = (x4, y4) lies inside/on/outside the circumcircle of P1, P2, P3 when the determinant
 * $$\begin{vmatrix}

x1^2+y1^2 & x1 & y1 & 1 \\ x2^2+y2^2 & x2 & y2 & 1 \\ x3^2+y3^2 & x3 & y3 & 1 \\ x4^2+y4^2 & x4 & y4 & 1 \end{vmatrix}$$
 * is positive/zero/negative resp.
 * You can get from "P4 inside the circumcircle of P1, P2, P3", to "P1 inside the circumcircle of P2, P4, P3" by an even permutation (leaving the determinant unchanged) so analytically the expressions are the same. If P1, P2, P3 are oriented clockwise then swap P2 and P3 and use the same idea.--RDBury (talk) 16:58, 20 December 2014 (UTC)

Trisecting the angle
Compass-and-straightedge construction notes that angles with certain values can be trisected, but not all possible angles. Is it possible to trisect a 270° angle with compass and straightedge? Given a real straightedge, you could simply place one of its sides adjacent to one of the angle's rays and use your compass to continue that ray beyond the vertex (do this twice, and you end up with four 90° angles), but it's been long enough since I did anything of this sort that I can't remember whether that kind of thing is allowed, and I'm not clear from the article. Nyttend (talk) 21:16, 20 December 2014 (UTC)


 * Yes, extending a line segment with the straight edge is allowed. (You only need the pencil from the compasses.)    D b f i r s   21:28, 20 December 2014 (UTC)
 * Thank you. What is the point of construction, anyway?  This was something I never figured out when I was in school; it always seemed to be "let's see what we can do", just an exercise in learning to use the compass and straightedge tools themselves.  Nyttend (talk) 21:30, 20 December 2014 (UTC)
 * There's some interesting mathematics mentioned in Compass-and-straightedge construction, but if you studied technical drawing you would probably learn several approximations and Neusis constructions to get round the "impossibilities".   D b f i r s   21:41, 20 December 2014 (UTC)
 * Well, there are exact constructions using a marked ruler as well. So it isn't really "impossible", it just isn't a ruler and compass construction.  As to "why", presumably there was a widespread belief that everything could be constructed just using unmarked rulers and compasses in the middle ages (this was all tied up with attitudes about religion as well, with its remnants still apparent in present-day Masonic obsession with trisecting the angle).  Of course, this was one of the Great Problems that led to the development of abstract algebra.   Sławomir Biały  (talk) 22:09, 20 December 2014 (UTC)
 * True, but the two marks are used as a "neusis" (i.e. an approximation because it has to be fitted "by eye"). The Greeks were fully aware of these methods.    D b f i r s   23:48, 20 December 2014 (UTC)
 * For purposes of this sort of discussion, you can trisect a 270° angle just by constructing a 90° angle. You don't actually have to use the 270° angle in a nontrivial way (or in fact in any way).
 * Contrast with trisecting a 60° angle. What the impossibility theorem says is that a 20° angle just cannot be constructed with compass and straightedge &mdash; whether you start with the 60° angle is beside the point.  --Trovatore (talk) 21:32, 20 December 2014 (UTC)
 * Constructing a 90° angle from scratch: Draw a line between two points. Draw the two circles centered at one of the points and going through the other (any overlapping circles with the same radius will actually work). Draw a line through the two intersection points of the circles. This line will be perpendicular to the original line. PrimeHunter (talk) 21:52, 20 December 2014 (UTC)

In answer to Nyttend's question "What is the point of construction, anyway?": basically it's about axiomatic systems. In math, you always want your proofs to be derived from from the fewest and simplest possible axioms; then they will apply as genearlly as possible. In describing mathematical constructions, things like this may be considered as axioms: That's not a complete list (I haven't talked about straight lines and their intersections), but you get the idea. Well, a compass-and-straightedge construction is one that can be carried out using the axioms I'm talking about here... and it's possible in any sort of geometry where the axioms work. And why do you care what you can construct? Because in proving a theorem, you may need to perform a construction as an intermediate step.
 * Given a point and a radius, a circle can be constructed with that point as center and having that radius.
 * Given two points, a straight line through those points can be constructed and prolonged indefinitely.
 * Given two circles, either they coincide completely, or they do not intersect, or each of their intersections determines a single point.

(Side note. You can actually use a simpler version of the first axiom: given any two distinct points, a circle can be constructed with the first point as center and passing through the second point. Instead of a compass as we known it, this corresponds to one where you aren't allowed to first set the radius, then move the compass and use it to draw a circle of that radius. This turns out to produce the same set of possible constructions as the other axiom, they just require extra steps.)

--65.94.50.4 (talk) 19:50, 21 December 2014 (UTC)


 * See Angle trisection for a more general discussion of methods of this construction. Archimedes was able to trisect the angle using a marked ruler.  There are other techniques for trisecting an angle.  Robert McClenon (talk) 01:56, 22 December 2014 (UTC)
 * By contrast, squaring the circle cannot be done using any finite set of tools in any finite number of steps. It isn't just impossible with Euclidean tools.  It really is impossible.  Robert McClenon (talk) 01:58, 22 December 2014 (UTC)
 * This last statement seems suspect. Squaring the circle can be done trivially with the tool "a ruler that has marks at 0, 1 and $$\pi$$", or similar. Perhaps something useful can be said if we restrict somehow the range of tools available? -- Meni Rosenfeld (talk) 13:25, 22 December 2014 (UTC)
 * An the $$\pi$$ mark can be made quite easily by rolling a marked circle upon a line. That's quite a simple tool to make - simpler and easier to make and use I think than the tools used for trisecting angles. Dmcq (talk) 13:31, 22 December 2014 (UTC)