20th Putnam 1959

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Problem A2

Let ω3 = 1, ω ≠ 1. Show that z1, z2, -ωz1 - ω2z2 are the vertices of an equilateral triangle.

 

Solution

Let A be the point z1, B the point z2. Then z2 - z1 represents the vector from A to B. Now -ω2 has unit length and makes an angle ±π/3 with the positive real axis, so multiplying z2 - z1 by it rotates AB through an angle π/3 (clockwise or counterclockwise). Adding the result to z2 - z1 gives a point C such that AC is at an angle π/3 to AB. In other words ABC is equilateral. It is easily checked that C is then z1 - ω2(z2 - z1) = -ωz1 - ω2z2 (since 1 + ω + ω2= 0).

 


 

20th Putnam 1959

© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002