Jump to content
BrainDen.com - Brain Teasers
  • 0
Sign in to follow this  
bonanova

Tiling a hexagon

Question

A regular hexagon is divided into 2n equilateral triangles.
Pairing triangles that share an edge produces diamond shapes with three distinct orientations, as shown.

diamonds.jpg

Prove that any n-diamond tiling of the hexagon will use the three types in equal numbers.

Share this post


Link to post
Share on other sites

4 answers to this question

  • 0
9 hours ago, bonanova said:

True, and that is a proof by construction, getting you a point.

There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?

I think I might have it.

We can construct a stylised 3D cube out of the three diamonds in the shape of a 2D hexagon.  In order to complete the cube/hexagon, we require one of each diamond orientation.  For a cube of length n, there will be n^2 number of diamonds of each orientation. 

I hope that makes sense.  I wish I could scan and upload my drawing.  Once I realised that it's a stylised cube for n=1, it took me a bit to actually draw out n=2.  Once I did, though, it's easy to see that as one side of the cube grows, the others must (obviously) grow as well in order to maintain the shape of the regular hexagon (and cube).

Share this post


Link to post
Share on other sites
  • 1

First, we'll look at each "ring" of hexagons.  The centre hexagon contains 6 diamonds in 2 sets of 3.  In order to not overlap diamonds, choosing the first one forces the orientation of the other two to be different.  For each ring around the centre hexagon, the same is true.  Once a single orientation is chosen, the orientation of all of the diamonds in that ring are chosen.  Additionally, two adjacent edges will share an orientation before the orientation is forced to change (if you want me to prove why the orientation is forced to change, I can't come up with anything other than "you've run out of room").  This pattern continues for all rings, regardless of how many you add.

Share this post


Link to post
Share on other sites
  • 0
Spoiler

The symmetry of course is 6-fold, so 2 is a red herring, and we can say, instead:

A regular hexagon is divided into 6p equilateral triangles.
Paring triangles that share an edge produces diamond shapes with three distinct orientations, as shown.
Prove that any
tiling of the hexagon will use exactly p diamonds of each type.

 

Share this post


Link to post
Share on other sites
  • 0
3 hours ago, Molly Mae said:
Spoiler

First, we'll look at each "ring" of hexagons.  The centre hexagon contains 6 diamonds in 2 sets of 3.  In order to not overlap diamonds, choosing the first one forces the orientation of the other two to be different.  For each ring around the centre hexagon, the same is true. etc. (bn)

 

True, and that is a proof by construction, getting you a point.

There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this  

  • Recently Browsing   0 members

    No registered users viewing this page.

×