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2020 AMC 12B Problem 23

Problem 23 of 25HarderAlgebraLogic

How many integers n≥2n \ge 2 are there such that whenever z1,z_1, z2,z_2, …,\ldots, znz_n are complex numbers such that ∣z1∣=∣z2∣=⋯=∣zn∣=1 |z_1| = |z_2| = \cdots = |z_n| = 1 and z1+z2+⋯+zn=0, z_1 + z_2 + \cdots + z_n = 0, then the numbers z1,z_1, z2,z_2, …,\ldots, znz_n are equally spaced on the unit circle in the complex plane?

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Solution

For n=2,n = 2, z1+z2=0z_1 + z_2 = 0 forces z2=−z1,z_2 = -z_1, which is equally spaced. For n=3,n = 3, three unit vectors summing to zero must form an equilateral triangle, so they are equally spaced. For every even n≥4,n\ge4, choose n2\frac{n}{2} antipodal pairs at generic angles that do not form a regular nn-gon. For every odd n≥5,n\ge5, choose the vertices of an equilateral triangle together with n−32\frac{n-3}{2} generic antipodal pairs. Each construction has sum 00 but is not equally spaced. Hence only n=2n = 2 and n=3n = 3 work, giving 22 values. Thus, the correct answer is B.
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Tagged: roots of unity · complex number · counterexample

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