Undergraduate talk on billiards at St Andrews, 9 Oct 2015

1. Mathematics of
Billiards
Igor Rivin

2. St Andrews
October 9 2015

4. -Herbert Spencer
“A certain dexterity in games of skill argues a
well-balanced mind, but such such dexterity
as you have shown is evidence, I fear, of a
misspent youth”

5. Was referring to billiards
(but could well be
referring to Mathematics)

6. However, let us
remember W. C. Fields:

7. -W. C. Fields
“I spent half my money on booze, women, and
gambling. The other half I wasted.”

8. Notice that “real” billiards are
• Very complex
• Highly nonlinear (the ball spins, the table has
friction)
• So, hard. In Mathematics, we avoid hard things
as best we can.

10. Simplify!
• Everything is linear!
• Collisions all elastic!
• And, to make it even simpler, move to dimension
1!

12. How?
• We have a circular skating rink.
• We have some beginning skaters, who are so
bad that:
• They have to hug the edge of the rink.
• They can only all go at speed one.
• when they collide, they bounce off elastically.

13. Question:
• If they start in a certain configuration (positions
and directions), will they eventually return to that
configuration?

15. Fundamental trick:
forget the colors!

16. Two dimensions
• Billiards obey Fermat’s principle:
• Angle of incidence is equal the angle of reflection!

18. Why?
• Distance minimization!

20. Remember, in the one-dimensional
case, the dynamics is purely
periodic. What about dimension 2?

21. Hard question: does a
billiard have ANY closed
orbit?

22. Cases
• Smooth boundary - YES, with any number of
reflections!
• Proof: variational, consider the closed polygonal
path with N vertices and maximal length (this is a
theorem of Steinhaus).

23. Cases
• Polygons - does not work, since longest trajectory
might go through vertex.
• Acute-angled triangle: YES!
• Proof (Fagnano):

24. Fagnano construction
• Q, R, P are bases of altitudes
(so QRP is the “pedal triangle”)

25. Fagnano construction
• Proof: BPOR has two (opposite) right angles, so
is inscribed. APR and ABQ are supported on the
same arc, so equal. Similarly, so are APQ and
ACR. Why are ABQ and ACR equal? Because
both complement BAC to 90 degrees!

26. What about other triangles?
• Used to be OPEN for ALL obtuse triangles,
however, some cases have been recently done
for SOME (not very) obtuse angled triangles by
Pat Hooper and Rich Schwartz.

27. Other polygons
• Angles rational multiples of Pi: yes. Otherwise,
completely open.

28. What does a random
orbit look like?