Maths Olympiad Prep

Library / China Mathematical Competition /72 of 136

China Mathematical Competition

Algebra Difficulty 4.9 AIME Prove it China

Let complex number z=(a+cos⁡θ)+(2a−sin⁡θ)iz = (a + \cos \theta) + (2a - \sin \theta)i. If ∣z∣≤2|z| \le 2 for any θ∈R\theta \in \mathbb{R}, then the range of real number aa is ______.

Solution

By the definition given above, we have, for any θ∈R\theta \in \mathbb{R},
∣z∣≤2⇔(a+cos⁡θ)2+(2a−sin⁡θ)2≤4⇔2a(cos⁡θ−2sin⁡θ)≤3−5a2⇔−25asin⁡(θ−φ)≤3−5a2⇒25∣a∣≤3−5a2⇒∣a∣≤55. \begin{align*} |z| \le 2 & \Leftrightarrow (a + \cos \theta)^2 + (2a - \sin \theta)^2 \le 4 \\ & \Leftrightarrow 2a(\cos \theta - 2\sin \theta) \le 3 - 5a^2 \\ & \Leftrightarrow -2\sqrt{5}a \sin(\theta - \varphi) \le 3 - 5a^2 \\ & \Rightarrow 2\sqrt{5} |a| \le 3 - 5a^2 \\ & \Rightarrow |a| \le \frac{\sqrt{5}}{5}. \end{align*}
So the range of aa is [−55,55]\left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right].

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.