HMMT 十一月 2022 · 冲刺赛 · 第 29 题

HMMT November 2022 — Guts Round — Problem 29

[15] Consider the set S of all complex numbers z with nonnegative real and imaginary part such that 2 | z + 2 | ≤ | z | . Across all z ∈ S , compute the minimum possible value of tan θ , where θ is the angle formed between z and the real axis.

解析

[15] Consider the set S of all complex numbers z with nonnegative real and imaginary part such that 2 | z + 2 | ≤ | z | . Across all z ∈ S , compute the minimum possible value of tan θ , where θ is the angle formed between z and the real axis. Proposed by: Vidur Jasuja √ Answer: 7 Solution: Let z = a + bi . Then, 2 2 2 2 z + 2 = ( a − b + 2) + 2 ab · i. 2 2 2 2 2 2 2 Recall the identity ( a − b ) + (2 ab ) = ( a + b ) , so we have 2 2 2 2 2 2 2 | z + 2 | = ( a + b ) + 4( a − b ) + 4 2 2 2 2 2 2 2 Thus, z ∈ S if and only if ( a + b ) + 4( a − b ) + 4 ≤ a + b . 2 2 2 2 2 2 2 Suppose ( a − b ) = c ( a + b ). Note that | c | < 1 . Let a + b = x. Then, x + (4 c − 1) x + 4 ≤ 0 . This 3 inequality must have real solutions, forcing − 4 < 4 c − 1 < 4, implying that − 1 ≤ c ≤ − . The smaller 4 2 b 3 2 2 the magnitude of c , the smaller the ratio , so then c = − . This gives us that a : b = 1 : 7, so then 2 a 4 √ √ 1 i 7 tan θ ≥ 7 . This value of c achieves exactly one solution, at x = 2, in particular z = + . 2 2