### 29th Putnam 1968

**Problem B3**

Given that a 60^{o} angle cannot be trisected with ruler and compass, prove that a 120^{o}/n angle cannot be trisected with ruler and compass for n = 1, 2, 3, ... .

**Solution**

This is a trap.

The obvious answer is that if we could trisect an angle 120^{o}/n, that would give us an angle 40^{o}/n and hence 60^{o}/3. But we are told we cannot trisect 60^{o}.

This does not work for a slightly subtle reason. We are given the angle 120^{o}/n. That *in itself* may help us to trisect 60^{o}. For example, it certainly allows us to trisect 60^{o} if n = 6.

So we have to use the standard Galois theory approach. It is not hard to show that a straight-edge allows us to get rational distances and compasses allow us to get any finite number of square roots. So we can construct distances which are in an extension K of the rationals of degree 2^{n} for some n. Constructing an angle is equivalent to constructing its cosine.

So if we are given an angle A, then we can construct distances which are in any extension K of degree 2^{n}over Q(cos A). Now if K is an extension of L of degree r and L is an extension of F of degree s, then K is an extension of degree rs of F. So suppose Q(cos A) has degree u and Q(cos(A/3 ) has degree v. If we can construct A/3 given A, then cos(A/3) must lie in an extension K of Q(cos A) of degree 2^{n}. But K is also an extension of Q(cos(A/3) ) so v/u must be a power of 2.

Finally, it is well-known that Q(cos(360/m) ) has degree φ(m) over Q, where φ is Euler's function (so that φ(m) is the number of 1, 2, ... m-1 relatively prime to m). So we have to show that φ(9n)/φ(3n) is not a power of 2. But φ(m) = m ∏(1 - 1/p), where the product is taken over all primes p dividing m. But 3n and 9n have the same prime divisors, so φ(9n) = 3 φ(3n), and 3 is not a power of 2.

29th Putnam 1968

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002