Mathematics of Billiards

Igor Rivin

St AndrewsOctober 9 2015

-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”

Was referring to billiards (but could well

be referring to Mathematics)

However, let us remember W. C.

Fields:

-W. C. Fields

“I spent half my money on booze, women, and gambling. The other half I

wasted.”

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.

Simplify!

• Everything is linear!• Collisions all elastic!• And, to make it even simpler, move to

dimension 1!

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.

Question:

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

Fundamental trick: forget the colors!

Two dimensions

• Billiards obey Fermat’s principle:• Angle of incidence is equal the angle of

reflection!

Why?

• Distance minimization!

Remember, in the one-dimensional case, the

dynamics is purely periodic. What about dimension 2?

Hard question: does a billiard have ANY

closed orbit?

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).

Cases

• Polygons - does not work, since longest trajectory might go through vertex.

• Acute-angled triangle: YES!• Proof (Fagnano):

Fagnano construction

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

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!

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.

Other polygons

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

What does a random orbit look

like?