Maths Olympiad Prep

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Number theory Difficulty 5.7 AIME, harder Prove it

Let α\alpha be an irrational number with 0<α<10<\alpha<1, and draw a circle in the plane whose circumference has length 1. Given any integer n3n \geq 3, define a sequence of points P1,P2,,PnP_{1}, P_{2}, \ldots, P_{n} as follows. First select any point P1P_{1} on the circle, and for 2kn2 \leq k \leq n define PkP_{k} as the point on the circle for which the length of arcPk1Pk\operatorname{arc} P_{k-1} P_{k} is α\alpha, when travelling counterclockwise around the circle from Pk1P_{k-1} to PkP_{k}. Suppose that PaP_{a} and PbP_{b} are the nearest adjacent points on either side of PnP_{n}. Prove that a+bna+b \leq n.

Solution

No points coincide since α\alpha is irrational. Assume for contradiction that n<a+b<2nn<a+b<2 n. Then it follows that
PnPa+bnPaPb \overline{P_{n} P_{a+b-n}} \| \overline{P_{a} P_{b}}
as shown below. !
This is an obvious contradiction since then Pa+bnP_{a+b-n} is contained in the arc PaPb^\widehat{P_{a} P_{b}} of the circle through PnP_{n}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.