Post B2VJqpH7U2J7SlPIv2 by 11011110@mathstodon.xyz
 (DIR) More posts by 11011110@mathstodon.xyz
 (DIR) Post #B2VJqjeIOxHU0uCjD6 by johncarlosbaez@mathstodon.xyz
       2024-02-15T21:59:57Z
       
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       Some geometry problems are easy to state but hard to solve!   For any triangle, can an ideal point-sized billiard ball bounce around inside in a 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐 trajectory - a path that repeats?The answer is "yes" for acute triangles, and this has been known since 1775.   It's also "yes" for right triangles.  But for obtuse triangles, nobody knows!In 2008, Richard Schwartz showed that the answer is "yes" for triangles with angles of 100° or less.   He broke the problem down into cases and checked each case with the help of a computer.   Then progress was stuck... until 2018, when Jacob Garber, Boyan Marinov, Kenneth Moore and George Tokarsky showed the answer is "yes" for triangles with angles of 112.3° or less.   Beyond that we're stuck.... except for triangles with all 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 angles (measured in degrees).  For them too the answer is "yes".   The picture here is fromGeorge Tokarsky, Jacob Garber, Boyan Marinov, Kenneth Moore, One hundred and twelve point three degree theorem, https://arxiv.org/abs/1808.06667and for more check out this article on Quanta:https://www.quantamagazine.org/the-mysterious-math-of-billiards-tables-20240215/
       
 (DIR) Post #B2VJqkoc457zdC2Tvk by lisyarus@mastodon.gamedev.place
       2024-02-15T22:20:31Z
       
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       @johncarlosbaez I'm probably misunderstanding something: how can all angles be rational if they should sum to pi?
       
 (DIR) Post #B2VJqmFwhwaO6FppAm by 11011110@mathstodon.xyz
       2026-01-20T08:18:32Z
       
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       @lisyarus @johncarlosbaez In this context "rational" means that the angles are rational multiples of pi (or equivalently rational numbers of degrees)
       
 (DIR) Post #B2VJqnGKxyUdDl1eQC by lisyarus@mastodon.gamedev.place
       2026-01-20T11:11:22Z
       
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       @11011110 @johncarlosbaez Oh I see, thank you!
       
 (DIR) Post #B2VJqoIV7PomQl2tQu by johncarlosbaez@mathstodon.xyz
       2026-01-20T17:15:39Z
       
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       I had written"Beyond that we're stuck.... except for triangles with all 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 angles (measured in degrees)."Of course this is an old Babylonian way of saying rational multiples of π, but I thought nonmathematicians would find it easier to understand.Impressive that this conversation is continuing after two years!  Thanks @11011110 - I hadn't seen @lisyarus's question.
       
 (DIR) Post #B2VJqpH7U2J7SlPIv2 by 11011110@mathstodon.xyz
       2026-01-21T07:03:52Z
       
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       @johncarlosbaez @lisyarus The explanation is that I rediscovered this old conversation while searching for material for a new Wikipedia article https://en.wikipedia.org/wiki/Triangular_billiardsIt needs more figures, and while playing around with some possibilities (which I haven't fleshed out yet) I ran into a curious phenomenon: You might expect the angles of most irrational triangles to have two independently irrational numbers. They can't have three because they sum to \(\pi\). But the first one I tried, the triangle with edge lengths 2,3,4, has only one, because its angles satisfy another unexpected relation: if \(\phi\) and \(\psi\) are the sharpest and second-sharpest angles, then \(3\phi+2\psi=\pi\). This appears to be closely related to the existence of periodic billiards paths perpendicular to the long side of this triangle.
       
 (DIR) Post #B2VJqq3Kams7sHxo48 by shironeko@fedi.tesaguri.club
       2026-01-21T08:35:28.672854Z
       
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       @11011110 @johncarlosbaez @lisyarus I love these open questions where at first glance you think "that doesn't sound too bad", then you read what cases are still unsolved and goes "oh".btw I was struck by the corner case (heh) where the trajectory simply ends, it seem more intuitive if the ball bounces back in some direction. I guess that makes the periodic path problem trivial.