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., $\frac{(1 + \sqrt{5})}{2}$)