Let $ABC$ an acute triangle and $O$ its circumcircle ($O$ is in the interior of the triangle). Suppose the circumradius is equal to 1 and consider $D_1$, $D_2$ and $D_3$ the discs of radius 1 centered in $A, B$ and $C$ respectively. Prove that the only common point of the three discs is $O$.
My trial was considering a complex plane and the coordinates of the points $A(a),B(b),C(c)$ and the origin $O(0)$. Then $|a|=|b|=|c|=1$ and suppose by contradiction that there would exist a point $z \neq 0$ such that $|z-a|\leq 1, |z-b| \leq 1$ and $|z-c|\leq 1$.
There's a simple geometric solution : a point lying on the circles $D_1, D_2$ belongs to the mediator of the segment $[AB]$. Likewise, a point lying on the circles $D_1, D_3$ belongs to the mediator of the segment $[AC]$ etc.
Hence, a point lying on all three circles must be $O$ (the intersection of mediators).