From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Triangulation Problem Date: 29 Mar 1995 23:20:54 GMT A few days ago I wrote a truly uninspired answer to a previous post, to which I would like to add some elegance and further information. The problem stated by Ash Morgan was as follows: given a triangle in the plane and three angles which sum to 2pi, find the point inside the triangle meeting these conditions: when the new point is connected to the vertices by line segments, the angles formed between the segments are the prescribed three angles. I answered the question as first presented (which was in terms of coordinates) by describing two quartic curves, each of which contained the points making one of the three interior angles correct. One need only intersect the quartics by eliminating y (say) and then solving for x. Yuk. I had Maple do the elimination and found that x had to satisfy a 1-variable quartic, which was a truly horrifying polynomial in the coordinates of the triangle and the cosines of the angles. However, Maple also found that, in each example I tested, the Galois group was D4 of order 8. This implied a ruler-and-compass solution ought to be possible. Anders Linner recognized the problem as an extension of the simpler case in which all three angles were to be 2pi/3. The solution he mentioned for that case extends simply to cover this case too. Here's the key fact I neglected: if a chord is drawn across a circle, all the triangles drawn with that chord as base and with the last vertex on the circle form the same angle at that vertex. (Actually the angle is replaced by its supplement when the triangle is drawn on the other side of the chord.) Thus the quartic curve I described above is nothing but the union of two circles (of the same radii) containing two of the original triangle's vertices. (My mistake was to ask Maple to factor the quartic, forgetting that it would only do this if the constituent curves could be given with coefficients which were _rational_ functions of the original cosines.) So here is the solution of the problem. Given a triangle ABC in the plane and three angles a, b, c which sum to 2pi, do the following. Bisect the angle a and draw an isoceles triangle ABD opposite ABC which has angles ABD and BAD both equal to a/2. Draw the circle S passing through A, B, and D (recall this is done by intersecting the perpendicular bisectors of AD and BD to find the center E of S; the radius is the distance SE of course). Then any point X on S which lies inside the triangle may be shown to make an angle AXB equal to a. Similarly we draw a circle S' for which AC is a chord, such that any point X on S' inside the triangle makes the angle AXC equal to b. These circles intersect at A and at one more point Z. That point Z is the one we seek. This approach also simplifies the less elegant coordinate approach. If I haven't messed up while typing, the following formulas should work. Assuming coordinates of A=(0,0) and B=(d,e), let a2 = tan(a/2) ; then D = (1/2)( (d,e) +- a2.(-e,d) ) (the sign chosen to make D point roughly away from C). If a1=tan(a) then the center E of the circle S is similarly E = (1/2)( (d,e) -+ (1/a1).(-e,d) ). The circle passing through the origin and centered at (h,k) is x^2+y^2 -2h.x -2k.y = 0, so S is described by the equation x^2+y^2 - (d+-e/a1)x - (e-+d/a1)y = 0. Likewise if C=(c,0), we let b1=tan(b) and find the equation of the points making an angle b with A and C to be x^2+y^2 - cx + (c/b1)y = 0. Assuming c>0 and e>0, we then have two equations which simplify first to (c-(d+e/a1))x + (-c/b1 - (e-d/a1))y=0 and x^2+y^2 - cx + (c/b1)y = 0, but which allow us to solve for x in terms of y to get x=(c/b1 + (e-d/a1))/(c-(d+e/a1)) . y and then an equation in y alone: y=0 or y= - [ - c.(c/b1 + (e-d/a1))/(c-(d+e/a1)) + (c/b1) ] / [((c/b1 + (e-d/a1))/(c-(d+e/a1)))^2 + 1] which may then be substituted into the formula giving x. dave