Problem

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Tags: ratio, geometry, geometry unsolved



Let $T$ be an isosceles right triangle. Let $S$ be the circle such that the diļ¬€erence in the areas of $T \cup S$ and $T \cap S$ is the minimal. Prove that the centre of $S$ divides the altitude drawn on the hypotenuse of $T$ in the golden ratio (i.e., $\frac{(1 + \sqrt{5})}{2}$)