Problem 10924
(American Mathematical Monthly, Vol.109, February 2002) Proposed by A. J. Sasane (Netherlands).
A regular polygon of2001 sides is inscribed in a circle of unit radius. Prove that its side and all its diagonals have irrational lengths.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will prove a more general statement:
In a regular polygon of N ≥ 2 sides and inscribed in a circle of unit radius, the side and all the diagonals have irrational lengths iffN is an odd integer.
First recall that for all integers n ≥ 1 and for all θ ∈ R
cos nθ + i sin nθ = (cos θ + i sin θ)n =
n
X
k=0
n k
ik(sin θ)k(cos θ)n−k.
After taking the imaginery part, we have
sin nθ =
n
X
k=1, kodd
n k
(−1)k−12 (sin θ)k(cos θ)n−k.
If n is odd then the exponents n−k are even and
(cos θ)n−k = (cos2θ)n−k2 = (1 − sin2θ)n−k2 .
Hence, for all odd integers n ≥ 1, there is a polynomial Qn(x) ∈ Z[x] of degree n−1, of the following form
Qn(x) = (−1)n−12
1 +n
2
xn−1+ . . . + n, such that
sin nθ = sin θ · Qn(sin θ) for all odd integers n ≥ 1. (1) After this key remark, we start the proof of our statement.
If N is even then the diagonal between two opposite vertices has length 2 ∈ Q. On the other hand, if N is odd then the side and the diagonals have lengths: 2 sin(πm/N ) for m = 1, . . . , N − 1. Assume that one of these numbers is rational, then we will reach a contradiction.
Since 1 ≤ m ≤ N − 1, there are an odd prime p and some integer e ≥ 1 such that pedivides N , pe−1 divides m, but pe does not divide m. Therefore N/peis an odd integer and by (1)
sin π m′ p
= sin N pe·π m
N
= sinπ m N
· QpeN
sinπ m N
∈ Q
where m′ = m/pe−1. Since MCD(m′, p) = 1, if k ∈ Z then there are two integers s0 and t0 such that s = s0+ j p and t = t0− j m′ solve the following linear diophantine equation for all j ∈ Z:
s m′+ t p = k.
We can always pick j such that s is an odd positive integer. Then, by (1) sin π k
p
= sin π(s m′+ t p) p
= (−1)tsin π m′ p
· Qs
sin π m′ p
∈ Q.
Varying k = ±1, . . . , ±p−12 , we obtain p − 1 distinct rational numbers which are the zeroes of the polynomial
Qp(x) = (−1)p−12
1 +p
2
xp−1+ . . . + p.
The polynomial Qphas integer coefficients and therefore every rational zero has the following prop- erty: the numerator divides p whereas the denominator divides a = 1 + p(p − 1)/2. It follows that
S =
sin π k
p
: k = ±1, . . . , ±p − 1 2
⊂ T =
±1 d, ±p
d : d | a
∩ (−1, 1).
Hence
cos π 2p
= sin π
p·(p − 1) 2
= max S ≤ max T ≤ p
p + 1 = 1 − 1 p + 1, and therefore
1
p + 1 ≤ 1 − cos π 2p
= 2 sin2 π 4p
< 2 π 4p
2
< 2 p2. This inequality never holds for p ≥ 3 which is a contradiction.
Note that, if N = 2001 = 3 · 23 · 29 then we can get a contradiction in a different way:
if p = 23 then a = 254 = 2 · 127 and T = {±12, ±1271 , ±2541 ±12723, ±25423}, if p = 29 then a = 407 = 11 · 37 and T = {±111, ±371, ±4071 , ±2937, ±40729}.
In both cases the number of elements of T is much less than p − 1. The remaining case p = 3 is easily solved because sin(π/3) =√
3/2 6∈ Q.
