97
$\begingroup$

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

math puzzle 45-60-75 triangles

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.

  • 0
    @Ed: I hope you don't mind: I added the image to your post! (I realize you need a minimal amount of rep to do so: Here we refer to "rep" as in "reputation" ;p2011-06-11
  • 0
    Is it correct to assume that you don't know what the minimum is? If so, I think this should be tagged "open-problem". Welcome to math.SE!2011-06-11
  • 2
    Interesting optical illusion: I could swear the two parts of the top-left-to-bottom-right diagonal don't match up, but they do. Also, it looks more as if they do if you tilt your head $45°$ (in either direction).2011-06-11
  • 7
    Your prize has an existentialist bias; there should also be a prize for proving that there's no solution with less than $100$ triangles :-)2011-06-11
  • 1
    Yes, I don't know the minimum. It's an open problem. The Laczkovich solution has close to 400 triangles, and when I was trying to follow his paper to draw it, I started thinking "there has to be a more elegant solution than this."2011-06-11
  • 26
    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!2011-06-17
  • 0
    In my opinion sinus theorem and Heron's formula should be used in order to answer this question2011-09-14
  • 1
    @Ed Pegg How did Laczkovich packed the square with triangles? I'm very curious :D2011-09-15
  • 1
    @Peter I think 45-60-75 means that the triangle has 45, 60, 75 as its angles (in degree).2011-09-20
  • 0
    @Jineon Baek Thankyou for the clarification.2011-09-20
  • 0
    I gave up on drawing the solution from Laczkovich's paper. It's a complex, technically-explained answer. Also, I've doubled the cash prize.2011-10-03
  • 0
    Hmm, I'd almost think the prize formula is backwards: it would be much harder to find and prove a 150-triangle minimum, e.g., than 30.2011-10-03
  • 1
    Do you have a reference for Laczkovich's solution? The only pages I can find in Google are copies of this problem.2011-10-05
  • 2
    Added the reference. Wow... top unanswered question now.2011-10-05
  • 1
    The interesting thing is that it would be trivial to prove that a concrete given arrangement is correct.2011-10-08
  • 5
    A copy of Laczkovich's paper can be found here: http://www.springerlink.com/content/p55415826m0j01w2/fulltext.pdf2011-12-07
  • 0
    @Ed Pegg: [Your suggested edit](http://math.stackexchange.com/review/suggested-edits/43223) on Lew Baxter's answer was rejected as it "changes too much in the original post". You could post your own answer containing the point coordinates you added, which was useful information.2013-02-02
  • 2
    I wonder, if there exist true asymetric solutions (not divided by the square-diagonal into two right-angled structures). This might also lower the number of triangles needed for a solution because it gives an additional degree of freedom for the structure, i think.2017-05-21

5 Answers 5

2

I found a nearly perfect solution with 28 triangles for the square. Since the error is very small, this could be used for a jigsaw-puzzle. here are 3 versions with different locations of the flawed triangles:

Sol#28 flawed

21

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.

tiling with 64 triangles

Here's one with 50 triangles:

enter image description here

  • 11
    Can you provide some further details as to how you accomplished this? Regards.2013-01-07
19

I found a minor improvement to Lew Baxter's solution. There are only 46 triangles needed to tile a square:

                          This is my design 

45-60-75 x 46 @ 1 x 1

Actually i tried to find an optimal solution with a minimum number of tiles by creating a database with about 26.000 unique rhomboids & trapezoids consisting of 2-15 triangles. I searched trough various promising setups (where the variable width/height-ratio of one element defines a second and you just have to look, if it's in the database,too) but nothing showed up. So this 46-tiles solution was in some sense just a by-product. As there probably exist some more complex combinations of triangles which i was not able to include, an even smaller solution could be possible.

with b = $\sqrt3$ the points have the coordinates:
{{4686, 0}, {4686, 6 (582 - 35 b)}, {4686, 4089 - 105 b}, {4686, 4686}, {4194 + 94 b, 3000 - 116 b}, {141 (28 + b), 3351 + 36 b}, {4194 + 94 b, -11 (-327 + b)}, {141 (28 + b), 141 (28 + b)}, {3456 + 235 b, 2262 + 25 b}, {3456 + 235 b, 2859 + 130 b}, {3456 + 235 b, 3456 + 235 b}, {3426 - 45 b, 45 (28 + b)}, {3426 - 45 b, 3 (582 - 35 b)}, {3426 - 45 b, 3 (744 - 85 b)}, {3258 - 51 b, 51 (28 + b)}, {2472 + 423 b, 213 (6 + b)}, {-213 (-16 + b), 213 (6 + b)}, {2754 - 69 b, 2754 - 69 b}, {-639 (-5 + b), 0}, {213 (6 + b), 213 (6 + b)}, {0, 0}, {4686, 15 (87 + 31 b)}, {3930 - 27 b, 2736 - 237 b}, {3930 - 27 b, 213 (6 + b)}, {0, 4686}, {6 (582 - 35 b), 4686}, {4089 - 105 b, 4686}, {3000 - 116 b, 4194 + 94 b}, {3351 + 36 b, 141 (28 + b)}, {-11 (-327 + b), 4194 + 94 b}, {2262 + 25 b, 3456 + 235 b}, {2859 + 130 b, 3456 + 235 b}, {45 (28 + b), 3426 - 45 b}, {3 (582 - 35 b), 3426 - 45 b}, {3 (744 - 85 b), 3426 - 45 b}, {51 (28 + b), 3258 - 51 b}, {213 (6 + b), 2472 + 423 b}, {213 (6 + b), -213 (-16 + b)}, {0, -639 (-5 + b)}, {15 (87 + 31 b), 4686}, {2736 - 237 b, 3930 - 27 b}, {213 (6 + b), 3930 - 27 b}}

which build the 46 triangles with pointnumbers:
{{6, 5, 2}, {3, 2, 6}, {8, 7, 3}, {4, 3, 8}, {9, 10, 5}, {5, 6, 10}, {10, 11, 7}, {7, 8, 11}, {12, 15, 13}, {13, 15, 16}, {14, 13, 16}, {17, 15, 16}, {1, 19, 17}, {19, 17, 20}, {21, 20, 19}, {11, 18, 9}, {18, 9, 16}, {20, 16, 18}, {1, 22, 12}, {2, 23, 22}, {22, 24, 23}, {23, 14, 24}, {24, 12, 14}, {4, 27, 8}, {8, 30, 27}, {30, 8, 11}, {32, 11, 30}, {11, 18, 31} , {27, 26, 29} , {28, 29, 32}, {29, 28, 26}, {31, 32, 28}, {26, 41, 40}, {40, 42, 41}, {18, 31, 37}, {20, 37, 18}, {41, 35, 42}, {35, 34, 37}, {38, 36, 37}, {34, 36, 37}, {33, 36, 34}, {42, 33, 35}, {25, 40, 33}, {25, 39, 38}, {39, 38, 20}, {21, 20, 39}}

10

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:

Full square solution

2

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.

$\qquad$ 45-60-75

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.

  • 2
    I don't understand why all the additional triangles are forced.2011-10-13
  • 1
    @Bob: forcing criterion added2011-12-30
  • 0
    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.2012-01-05
  • 1
    @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.2012-01-05