From:tromp@cwi.nl (John Tromp)Date:Wed, 23 Oct 1996 17:06:04 GMTNewsgroups:sci.mathSubject:nice geometry puzzle

I was recently reminded of an old puzzle asking for a subdivision of a square into acute-angled triangles So there can be no angle of at least 90 degrees. In particular, all four corners of the square require bisecting at least. An additional challenge is to minimize the resulting number of triangles The answer appears to be eight, as witnessed by my PostScript signature. This fact may very well have been proven somewhere in the literature; let me know if you have a reference. I attacked the problem of how to make the angles as acute as possible. For an 8 triangle subdivision, this leads to an optimization problem involving two unknowns (x,y) (being one of the (symmetric) middle points in a square ranging from (-1,0) to (1,2)), and three necessarily equal angles. Application of the cosine law results in the simultaneous equations: x^2-x+y^2 (1-x)^2-2y+y^2 x sqrt((1-x)^2+y^2) ------------- = --------------------- = ------------------- sqrt(x^2+y^2) sqrt((1-x)^2+(2-y)^2) sqrt(x^2+(2-y)^2) With the help of expert Maple user Walter Lioen this gave me the desired solution (x,y) ~ (.13071391155196, .379831986484951) corresponding to an angle of about 85 degrees. Now I was wondering if a division in more triangles could achieve even more acute angles. I discovered that a division in 20 triangles can: First divide the square in two along a main diagonal. Then subdivide each of the two triangles into 10 smaller ones, by adding a small triangle in the middle (each of whose corners connects to one of the outer triangle), and adding one point on each side of the outer triangle, connecting to two corners of the inner one. My big question is: how do I go about optimizing the angles in such a subdivision? It seems to involve three instead of two parameters: If we position the outer triangle at (0,0),(1,0),(0,1), then the small inner triangle should be positioned at (w,w),(x,y),(y,x). The problem is I'm not sure which angles should up being equal when w,x,y are chosen so as to minimize the largest angle... By fixing some angles and optimizing a remaining single parameter I found a solution with all angles smaller than 84 degrees, but it seems something close to 80 degrees should be feasible. Any help is appreciated! regards, %!PS % -John Tromp (http://www.cwi.nl/~tromp/) 300 200 translate 2{-1 1 26 76 0 0 200 0 200 400 0 400 26 76 200 400 200 0 26 76 9 0 76 moveto{lineto}repeat scale}repeat stroke showpage

From:eppstein@ics.uci.edu (David Eppstein)Date:24 Oct 1996 14:17:50 -0700Newsgroups:sci.mathSubject:Re: nice geometry puzzle

In <tromp.846090364@news.cwi.nl> tromp@cwi.nl (John Tromp) writes: > I was recently reminded of an old puzzle asking for > a subdivision of a square into acute-angled triangles This was one of the problems in the Amer. Math. Monthly a number of years ago. Unfortunately I've lost the exact reference. > ...Now I was wondering if a division in more triangles could achieve > even more acute angles... I found a solution with all angles smaller > than 84 degrees, but it seems something close to 80 degrees should be > feasible. My paper "Provably good mesh generation" [J. Comp. Sys. Sci. 48 (1994) 384-409] has a triangulation of the unit square with angles at most roughly 80 degrees. It uses a total of 26 triangles, five along each side and six in the interior. I didn't try hard to optimize the angles in this triangulation and probably it could be improved a little. If you allow subdivisions in which triangles do not always meet edge to edge, a method of J. L. Gerver achieves 72 degrees for any polygon [The dissection of a polygon into nearly equilateral triangles, Geom. Ded. 16 (1984) 93-106]. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

To:sci.math@wormwood.ICS.UCI.EDU, tromp@cwi.nlSubject:nice geometry puzzleDate:Thu, 24 Oct 1996 17:37:10 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

John Tromp asked about subdivisions of a square into acute triangles, to which I responded with a pointer to 26-triangle solution having all angles at most 80, and a method for getting solutions in which triangles need not meet edge to edge, but with maximum angle 72. I now have a solution with 60 triangles, all of which do meet edge to edge, and in which the maximum angle is 72. Two opposite sides of the square are partitioned into thirds, each third is covered by a 72-54-54 isosceles triangle, and the remaining gaps on each side are filled with six 36-72-72 isosceles triangles. The space in the middle then forms three hexagons, each of which can be filled with 14 triangles (maybe fewer triangles per hexagon would also work). I've put a picture of this triangulation on my web site at http://www.ics.uci.edu/~eppstein/acute-square.gif. It almost works to do a similar solution in which the two opposite sides of the square are subdivided in half and then covered as before with isosceles triangles. The spaces between them look like they should be able to be filled with 12 more 36-72-72 triangles but there's just barely not enough room (the resulting rectangle has dimensions 1 x 1.013). Is 72 a hard minimum? Any better would require avoiding degree-five vertices inside the square, which seems to be difficult. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/

To:sci.math@wormwood.ICS.UCI.EDU, John.Tromp@cwi.nl, mfineman@ix.netcom.comSubject:nice geometry puzzleDate:Sat, 26 Oct 1996 10:44:36 -0700From:David Eppstein <eppstein@ICS.UCI.EDU>

Here's an even better solution to the problem of triangulating a square with all angles at most 72. Place as guides two congruent regular pentagons, meeting at a face, diagonally in the square (as large as possible, so that four of their corners touch the square's sides). Use as triangulation vertices these four points on the square's sides, the two centers of the pentagons (on one diagonal of the square), and the two points on the other diagonal where the pentagons share a corner. The triangulation has six 54-54-72 triangles (connecting each pentagon center to three pentagon edges), and eight 45-63-72 triangles (two at each corner of the square). Is fourteen triangles optimal? One can't do better than 72 degrees because (by some manipulations with Euler's formula) any triangulated square has (1) a vertex along a square edge where only two triangles meet, (2) a square corner covered by only one triangle, or (3) an interior vertex where at most five triangles meet. Cases (1) and (2) give at best 90 degrees, and case (3) gives at best 72. -- David Eppstein UC Irvine Dept. of Information & Computer Science eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/