Maths Olympiad Prep

Library / /90 of 121

Geometry Difficulty 6.6 National Olympiad Prove it India

Problem:
Let A1B1C1D1E1A_{1} B_{1} C_{1} D_{1} E_{1} be a regular pentagon. For 2≤n≤112 \leq n \leq 11, let AnBnCnDnEnA_{n} B_{n} C_{n} D_{n} E_{n} be the pentagon whose vertices are the midpoints of the sides of the pentagon An−1Bn−1Cn−1Dn−1En−1A_{n-1} B_{n-1} C_{n-1} D_{n-1} E_{n-1}. All the 5 vertices of each of the 11 pentagons are arbitrarily coloured red or blue. Prove that four points among these 55 points have the same colour and form the vertices of a cyclic quadrilateral.

Solution

Solution:
We first observe that all the eleven pentagons are regular. Moreover, there are 5 fixed directions and all the 55 sides are in one of these directions. If we consider any two sides which are parallel, they are the parallel sides of an isosceles trapezium, which is cyclic.

If we consider any pentagon, its two adjacent vertices have the same colour. Consider all such 11 sides whose end points are of the same colour. These are in 5 fixed directions. By pigeon-hole principle, there are 3 sides which are in the same directions and therefore parallel to each other. Among these three sides, two must have end points having one colour (again by P−H\mathrm{P}-\mathrm{H} principle). Thus there are two parallel sides among the 55 and the end points of these have one fixed colour. But these two sides are parallel sides of an isosceles trapezium. Hence the four end points are concyclic.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.