HMMT 二月 2005 · TEAM2 赛 · 第 10 题

HMMT February 2005 — TEAM2 Round — Problem 10

[30] Let x , y , and z be three roots of unity. Prove that x + y + z is also a root of unity if and only if x + y = 0, y + z = 0, or z + x = 0. 1 Early Re-tile-ment [150] Let S = { s , . . . , s } be a finite set of integers, and define S + k = { s + k, . . . , s + k } . 0 n 0 n We say that S and T are equivalent , written S ∼ T , if T = S + k for some k . Given a (possibly infinite) set of integers A , we say that S tiles A if A can be partitioned into subsets equivalent to S . Such a partition is called a tiling of A by S .

解析

[30] Let x , y , and z be three roots of unity. Prove that x + y + z is also a root of unity if and only if x + y = 0, y + z = 0, or z + x = 0. Solution: Again, we consider the geometric picture. Arrange the vectors x , y , z , and − x − y − z so as to form a quadrilateral. If they are all roots of unity, they form a quadrilateral all of whose side lengths are 1. If the quadrilateral is degenerate, then two of the vectors sum to 0, which implies the result. But even if it is not degenerate, the quadrilateral must be a rhombus, and since opposite sides of a rhombus are parallel, this again implies that two of the four roots of unity sum to 0. Early Re-tile-ment [150] Let S = { s , . . . , s } be a finite set of integers, and define S + k = { s + k, . . . , s + k } . 0 n 0 n We say that S and T are equivalent , written S ∼ T , if T = S + k for some k . Given a (possibly infinite) set of integers A , we say that S tiles A if A can be partitioned into subsets equivalent to S . Such a partition is called a tiling of A by S . 3