Diamond Tiling

[bonanova]

A regular hexagon is sliced into congruent equilateral triangles and then tiled with diamond shapes made of two triangles in three different orientations. Prove that every tiling of the hexagon contains equal numbers of red, green and blue diamonds.

[Buddyboy3000]

7 hours ago, bonanova said:

A regular hexagon is sliced into congruent equilateral triangles and then tiled with diamond shapes made of two triangles in three different orientations. Prove that every tiling of the hexagon contains equal numbers of red, green and blue diamonds.

First, I need to see if it is theoretically possible. Counting all of the triangles, I got 96. Divided by 2 for all of the diamonds is 48. 48 divided by 3 is 16. So, 16 of each diamond needs to be used. I think it would be best to start placing them at the edges. I am able to do the outer edge evenly placing one type of diamond on two sides adjacent. With this, 7 of each diamond is used. Doc1.docx (Link leads to image of outer edge). Next, I can eliminate that layer and use the next. The next will have 6 less spots for diamonds to be put in. So, instead of 7, 5 will be placed onto the two sides. Then it's 3, and 1. So, each number of diamonds used is 7+5+3+1. This equals 16, the number of each diamond.

To conclude, there can be an even number of red, green, and blue diamonds in the hexagon.

[bonanova]

Is a tiling possible that does not have equal representation of colors?

Edit: OP asks:
Prove that every tiling of the hexagon contains equal numbers of red, green and blue diamonds.

Edited December 15, 2015 by bonanova
Clarify OP

[bonanova]

Clue:

Spoiler

4 x 4 x 4.

 

[bonanova]

It's a very simple result. Buddyboy3000 was close ...

[bonanova]

Oh my. Such a pretty puzzle and still no proof?
Final clue:

Spoiler

Print it out and color it.

 

[plasmid]

Spoiler

Consider the top row of triangles in the hexagon. It has four triangles in a V and five triangles in an A orientation. Consider the four triangles in a V. They obviously can't be part of the blue shape since the other triangle that would contribute to the blue shape isn't present in the hexagon, so they must be part of a red or green shape. Those four red/green shapes that occupy the Vs must also occupy four of the A triangles in the top row, so only one of the A triangles could be involved in forming a blue shape.

Then consider the second row of triangles, with five triangles in a V and six triangles in an A orientation. Since there could be no more than one blue shape in the preceding row of triangles, only one of the five V triangles in the second row could be involved in a blue shape and the other four must be involved in red or green shapes. And therefore at most two of the six triangles in an A orientation can be part of a blue shape. This process continues for each subsequent row. So the first row has no V triangles in a blue shape and (at most) one A triangle in a blue shape, the second row has at most one V triangle and two A triangles in a blue shape, etc. Also note that a symmetric argument works on the bottom row.

Putting everything together, the maximum total number of triangles involved in forming blue shapes would be

Row 1: 1 triangle (an A)

Row 2: 3 triangles (a V and two As)

Row 3: 5 triangles (two Vs and three As)

Row 4: 7 triangles

Rows 5-8: mirror image of rows 1-4

And the total is 32 small triangles involved in forming blue shapes, which is 1/3 of the total hexagon. Symmetric arguments apply to the red shape if you consider the upper left and lower right borders like we considered the top and bottom borders for the blue shape, and if you consider the upper right and lower left borders for the green shape. So each of the three colors can occupy no more than 1/3 of the hexagon.

 

 

[plainglazed]

Spoiler for may be at least a partial proof along the lines of the formal proof

Spoiler

Any hexagon can be constructed of three large diamond shapes congruent to the red, green and blue unit sized shapes.  Each of those shapes can only be constructed in their entirety while within their bounds from unit size shapes of the same color.  But can visualize if a different color is placed within a large diamond shape you have to break the boundary of that large diamond in at least two (an even number of) places with half diamonds in order to fill that large diamond.  Consequently...oh bother - so maybe just an observation.

 

[bonanova]

This is one of my fav puzzles. If you want one final clue:

Spoiler

It's not a hexagon.

Otherwise, read on:
It's from Peter Winkler's book "Mathematical Puzzles, A Connoisseur's Collection."
Its answer constitutes a "proof without words."

Spoiler

 

[Logophobic]

For further information regarding this puzzle and its origin, see http://gurmeet.net/puzzles/tiling-with-calissons/

" Source: On the back cover of Mathematical Puzzles: A Connoisseur's Collection(163 pages, 2003) by Peter Winkler. Originally, the problem appeared in "The Problem of Calissons" by G David and C Tomei, American Mathematical Monthly, Vol 96(5), May 1989, p. 429 - 431."

 

[bonanova]

Words can be added to the picture to create a proof. The notion, suggested by one of the writers quoted above, that an optical illusion turns the proof itself into an illusion is specious. The proof would go as follows.

Spoiler

Draw nxn square grids on the three inside walls of a cube corner and create an orthographic projection along the cube diagonal. Painting each wall a different color gives a trivial diamond tiling of the puzzle's hexagon with n² tiles of each type. We note (1) that a tiling cannot be achieved with arbitrary placement of tiles and (2) that all legal tilings can be achieved through modification of the trivial tiling: Pack into the cube corner any number, from 1 to n³, of 1x1x1 cubes whose visible faces are colored according to their orientation. Each cube obscures three diamonds, one of each color, and replaces them with three others, also one of each color. The observance of equal numbers of colors is thus preserved for every possible tiling.