Let be a positive integer. In the plane, there are pairwise disjoint disks , with radii in order. For every , a point is marked in . Let be an arbitrary point in the plane. Prove that:
(Note: here the disks are assumed to contain their boundaries.)
Taiwan IMO Selection Camp, 2021
Solution
We will make use of the following lemma.
Lemma. Let be disjoint disks in the plane with radii . Let be a point in , and let be an arbitrary point. Then there exist indices and such that .
Proof. Let be the center of . Consider six rays (if , then the ray may be assumed to have an arbitrary direction). These rays partition the plane into six angles (one of which may be non-convex) whose measures sum up to ; hence one of the angles, say , has measure at most . Then cannot be the unique largest side in (possibly degenerate) triangle , so, without loss of generality, . Therefore, , as desired.
Now we prove the required inequality by induction on . The base case is trivial. For the induction step, apply the Lemma to the six largest disks, in order to find indices and such that and . Removing from the configuration and applying the inductive hypothesis, we get
Adding up this inequality with we establish the induction step.