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Romanian Mathematical Olympiad - District Round

Algebra Difficulty 5.8 AIME, harder Prove it Romania

Let $a, b \in \mathbb{R}$ and $z \in \mathbb{C} \setminus \mathbb{R}$ such that $|a - b| = |a + b - 2z|$ .

a. Prove that there exists an unique real number $x$ which satisfies $|z - a|^x + |\bar{z} - b|^x = |a - b|^x$ .

b. Find all real numbers $x$ such that $|z - a|^x + |\bar{z} - b|^x \le |a - b|^x$ .

Solution

a. Set $u = z - a$ , $v = z - b$ . The relation gives $|v - u| = |u + v|$ where $u, v, u + v \in \mathbb{C} \setminus \mathbb{R}$ , so $u, v, u + v \neq 0$ . Thus $|u + v|^2 = |u|^2 + |v|^2$ . Since $|v| = |\bar{v}|$ , the equation is written successively $|u|^x + |v|^x = (\sqrt{|u|^2 + |v|^2})^x$ and then $\left(\frac{|u|}{\sqrt{|u|^2+|v|^2}}\right)^x + \left(\frac{|v|}{\sqrt{|u|^2+|v|^2}}\right)^x = 1$ .

The function $f: \mathbb{R} \to \mathbb{R}$ , $f(x) = \left( \frac{|u|}{\sqrt{|u|^2+|v|^2}} \right)^x + \left( \frac{|v|}{\sqrt{|u|^2+|v|^2}} \right)^x$ is strictly decreasing, hence $x = 2$ is the only solution.

b. The solution is $[2, +\infty)$ , for $f$ is strictly decreasing.

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