Bit of a mystery Mathologer today with the title of the video not giving away much. Anyway it all starts with the quest for equilateral triangles in square grids and by the end of it we find ourselves once more in the realms of irrationality. This video contains some extra gorgeous visual proofs that hardly anybody seems to know about.
0:47 First puzzle
2:24 Second puzzle
3:50 Edward Lucas
4:41 Equilateral triangles
13:15 3d & 3rd puzzle
19:52 30 45 60
Here are links to/references of some of the things I mention in the video:
Joel Hamkin’s blog posts that inspired this video:
There is also a whole chapter about all this and much more related maths in his new book
Here is another really good article which includes a nice characterisation of the triangles that can be found in square grids plus a very good survey of relevant results:
Michael J. Beeson, Triangles with Vertices on Lattice Points, The American Mathematical Monthly 99 (1992), 243-252, https://www.jstor.org/stable/2325060?seq=1
Scherrer’s and Hadwinger’s articles:
Scherrer, Willy, Die Einlagerung eines regulären Vielecks in ein Gitter, Elemente der Mathematik 1 (1946), 97-98.
Hadwiger, Hugo Über die rationalen Hauptwinkel der Goniometrie, Elemente der Mathematik 1 (1946), 98-100.
Another, nice paper on rational (and algebraic) cosines
Here is a solution to the first puzzle (one way to find the general formula):
The music in this video is by Chris Haugen, Fresh Fallen Snow (playing in the video) and Morning Mandolin (for the credits)
A couple of remarks:
1. Probably the simplest way to deduce the sin and tan parts of the rational trig ratio theorem is to realise that they follow from the cos part via the trigonometric identities: sin(x)=cos(90-x) and tan^2(x) = (1-cos(2x))/(1+cos(2x)). Note that the second identity implies that if tan(x) is rational, then cos(2x) is rational (if tan(x)=c/d, then tan^2(x)=c^2/d^2=C/D and cos(2x)=(D-C)/(D+C)).
2. Bug report.
a) Here I redefine cos(120◦) = 1.
b) This transition to the good stuff I clearly did not think through properly.
It’s possible to make this work for all regular n-gons. There is only one complication that occurs for n’s that are of the form 2 * odd. For the corresponding regular n-gons, if you pick up the edges in the order that they appear around the n-gon and assemble them into a star, things close up into (n/2)-stars. For all other n, things work exactly as I showed in the video. Having said that you can also assemble the edges of one of the exceptions into stars. Have a look at this https://imgur.com/68A3fEe and you’ll get the idea. Anyway lots more nice side puzzles to be explored here if you are interested 🙂
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