Solution: Day 4, problem 3
Obviously the thickness and radius of the pie play no role, all the action taking place along the circumference,
which we can identify with T = R/Z. Hence we can restate the problem more mathematically as follows: Given two
numbers a, b strictly between 0 and 1, show that the piecewise linear map f: T to T defined by x |--> a - x + b if
xis in I = Z + [0,a] and x |--> x + b if x is not in I, has finite order. We first show that for any x in T there
is a positive integer n = n(x) with fn(x) = x, and then show that n(x) is bounded. (This proves the
assertion of the problem since then the least common multiple of all the n(x)'s is a period for f.)
We distinguish two cases. If x + Z b is disjoint from I (which can only happen if b is rational, since
otherwise x + Z b is dense in R/Z), then fn(x) = x + nb for all n and therefore fN(x) = x where
N is the denominator of b. If not, then there are integers n1 > 0 and n2 > = 0 such
that x - n1b and x + n2b belong to I. Choose them minimal. Then x belongs to an orbit
of f of length 2(n1 + n2), consisting of the points x + nb (n = -n1 + 1, -n1 + 2, ... ,n2)
followed by the points a - x + nb (n = -n2 + 1, -n2 + 2, ... , n1). It remains only to show that
the integer N = n1 + n2 is bounded. This integer has the property that there is an element y of I
(namely, x - n1b) such that y + Nb is in I but y + nb is not in I for 0 < n < N.
If b is rational, then clearly N is bounded by the denominator of b. If not, then there is a
positive integer M such that Mb is congruent modulo 1 to a number d between 0 and a.
Then the sequence y + Mb, y + 2Mb, ... advances in steps of d < |I|, so must meet I after at
most [(1 - a)/d] steps, giving for N an upper bound M[(1 - a)/d] which is independent of x.
Return to the problems.
Don Zagier 1996