# What is the smallest number of $45^\circ-60^\circ-75^\circ$ triangles that a square can be divided into?

What is the smallest number of $45^\circ-60^\circ-75^\circ$ triangles that a square can be divided into?

The image below is a flawed example, from http://www.mathpuzzle.com/flawed456075.gif

Laczkovich gave a solution with many hundreds of triangles, but this was just an demonstration of existence, and not a minimal solution. ( Laczkovich, M. "Tilings of Polygons with Similar Triangles." Combinatorica 10, 281-306, 1990. )

I've offered a prize for this problem: In US dollars, (\$200-number of triangles). NEW: The prize is won, with a 50 triangle solution by Lew Baxter. - Your prize has an existentialist bias; there should also be a prize for proving that there's no solution with less than$100$triangles :-) – joriki Jun 11 '11 at 9:10 Perhaps it is also interesting to have lower bounds on the solution.. @EdPegg, I don't think people will be very interested in your prize money, they may end up having to pay you! – picakhu Jun 17 '11 at 18:20 @Ed Pegg How did Laczkovich packed the square with triangles? I'm very curious :D – Jineon Baek Sep 15 '11 at 5:00 @Peter I think 45-60-75 means that the triangle has 45, 60, 75 as its angles (in degree). – Jineon Baek Sep 20 '11 at 10:25 A copy of Laczkovich's paper can be found here: springerlink.com/content/p55415826m0j01w2/fulltext.pdf – Joel Reyes Noche Dec 7 '11 at 3:53 show 12 more comments ## 3 Answers I improved on Laczkovich's solution by using a different orientation of the 4 small central triangles, by choosing better parameters (x, y) and using fewer triangles for a total of 64 triangles. The original Laczkovich solution uses about 7 trillion triangles. Here's one with 50 triangles: - Can you provide some further details as to how you accomplished this? Regards. – Amzoti Jan 7 '13 at 22:25 add comment The following was posted by Ed Pegg as a suggested edit to Lew Baxter's answer, but was rejected for being too substantial a change. I thought it was useful information, so I reproduce it below. If you no longer want it to be posted here, Ed, leave a comment and I'll delete it. Exact points for the triangles are as follows, with$b=\sqrt3$: $$\{\{0,0\}, \{261+93b,0\}, \{522+186b,0\}, \{2709-489b,0\}, \{3492-210b,0\}, \{3890-140b,0\}, \{4288-70b,0\}, \{4686,0\}, \{252+9b,252+9b\}, \{513+102b,252+9b\}, \{774+195b,252+9b\}, \{3000-116b,492-94b\}, \{3398-46b,492-94b\}, \{3597-11b,492-94b\}, \{3796+24b,492-94b\}, \{4194+94b,492-94b\}, \{2262+25b,1230-235b\}, \{2859+130b,1230-235b\}, \{3456+235b,1230-235b\}, \{756+27b,756+27b\}, \{2214-423b,756+27b\}, \{1278+213b,756+27b\}, \{2736-237b,756+27b\}, \{1260+45b,1260+45b\}, \{1746-105b,1260+45b\}, \{2232-255b,1260+45b\}, \{1428+51b,1428+51b\}, \{1278+213b,2214-423b\}, \{1278+213b,1278+213b\}, \{1980+517b,2706-517b\}, \{0,1491+639b\}, \{1278+213b,3408-213b\}, \{0,4686\}\}$$ The triangles use points $$\{\{1,2,9\},\{2,9,10\},\{2,3,10\},\{3,10,11\},\{3,4,22\},\{4,22,23\},\{4,23,5\},\{5,12,13\},\{5,6,13\},\{6,13,15\},\{6,7,15\},\{7,15,16\},\{7,8,16\},\{9,11,20\},\{11,20,22\},\{12,17,18\},\{12,14,18\},\{14,18,19\},\{14,16,19\},\{20,21,24\},\{21,24,26\},\{21,26,23\},\{24,25,27\},\{25,27,28\},\{25,26,28\},\{27,28,29\},\{1,29,31\},\{29,31,32\},\{31,32,33\},\{17,19,30\},\{17,30,28\},\{28,30,32\}\}$$ Leading to the solution: - add comment I have no answer to the question, but here's a picture resulting from some initial attempts to understand the constraints that exist on any solution.$\qquad$This image was generated by considering what seemed to be the simplest possible configuration that might produce a tiling of a rectangle. Starting with the two “split pentagons” in the centre, the rest of the configuration is produced by triangulation. In this image, all the additional triangles are “forced”, and the configuration can be extended no further without violating the contraints of triangulation. If I had time, I'd move on to investigating the use of “split hexagons”. The forcing criterion is that triangulation requires every vertex to be surrounded either (a) by six$60^\circ$angles, three triangles being oriented one way and three the other, or else (b) by two$45^\circ$angles, two$60^\circ$angles and two$75^\circ\$ angles, the triangles in each pair being of opposite orientations.

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I don't understand why all the additional triangles are forced. –  Bob Hearn Oct 13 '11 at 16:35
@Bob: forcing criterion added –  David Bevan Dec 30 '11 at 16:59
Unfortunately, there is a solution with hundreds of triangles. See the Laczkovich paper. This forcing argument doesn't work, since you're proving a known solution is impossible. –  Ed Pegg Jan 5 '12 at 15:20
@EdPegg: Perhaps I didn’t explain it clearly enough; I only investigated a simple solution (as described in the paragraph under the image). All I’ve proved is that starting with two “split pentagons” and triangulating from then on (i.e. adding no additional vertices part way along a triangle’s edge) doesn’t work. –  David Bevan Jan 5 '12 at 16:27