IMO 1989 SL 7

Origin: FIN

Problem

Show that any two points lying inside a regular n-gon E can be joined by two circular arcs lying inside E and meeting at an angle of at least
1 −2 n  \pi.

Solution

Assume that P1 and P2 are points inside E, and that the line P1P2 inter- sects the perimeter of E at Q1 and Q2. If we prove the statement for Q1 and Q2, we are done, since these arcs can be mapped homothetically to join P1 and P2. Let V1, V2 be two vertices of E. Then applying two homotheties to the inscribed circle of E one can find two arcs (one of them may be a side of E) joining these two points, both tangent to the sides of E that meet at V1 and V2. If A is any point of the side V2V3, two homotheties with center V1 take the arcs joining V1 to V2 and V3 into arcs joining V1 to A; their angle of incidence at A remains (1 −2/n)\pi. Next, for two arbitrary points Q1 and Q2 on two different sides V1V2 and V3V4, we join V1 and V2 to Q2 with pairs of arcs that meet at Q2 and have an angle of incidence (1 −2/n) \pi. The two arcs that meet the line Q1Q2 again outside E meet at Q2 at an angle greater than or equal to (1 −2/n) \pi. Two homotheties with center Q2 carry these arcs to ones meeting also at Q1 with the same angle of incidence.