IMC 1996 · Day 2 · P10

Assume the contrary – there is an arc $A \subset C$ with length $l(A) = \dfrac{\pi}{2}$ such that $A \subset B \setminus \Gamma$. Without loss of generality we may assume that the ends of $A$ are $M = (1/\sqrt{2}, 1/\sqrt{2})$, $N = (1/\sqrt{2}, -1/\sqrt{2})$. $A$ is compact and $\Gamma$ is closed. From $A \cap \Gamma = \emptyset$ we get $\delta > 0$ such that $\operatorname{dist}(x,y) > \delta$ for every $x \in A$, $y \in \Gamma$.

Given $\varepsilon > 0$ with $E_\varepsilon$ we denote the ellipse with boundary: $$\frac{x^2}{(1+\varepsilon)^2} + \frac{y^2}{b^2} = 1,$$ such that $M, N \in E_\varepsilon$. Since $M \in E_\varepsilon$ we get $$b^2 = \frac{(1+\varepsilon)^2}{2(1+\varepsilon)^2 - 1}.$$ Then we have $$\operatorname{area} E_\varepsilon = \pi \frac{(1+\varepsilon)^2}{\sqrt{2(1+\varepsilon)^2 - 1}} > \pi = \operatorname{area} D.$$ In view of the hypotheses, $E_\varepsilon \setminus B \ne \emptyset$ for every $\varepsilon > 0$. Let $S = \{ (x,y) \in \mathbb{R}^2 : |x| > |y| \}$. From

$E_\varepsilon \setminus S \subset D \subset B$ it follows that $E_\varepsilon \setminus B \subset S$. Taking $\varepsilon < \delta$ we get that $$\emptyset \ne E_\varepsilon \setminus B \subset E_\varepsilon \cap S \subset D_{1+\varepsilon} \cap S \subset B$$ – a contradiction (we use the notation $D_t = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \le t^2 \}$).

Remark. The ellipse with maximal area is well known as John's ellipse. Any coincidence with the President of the Jury is accidental.