Let $P$ and $Q$ be two distinct points in the plane. Let us denote by $m(PQ)$ the segment bisector of $PQ$. Let $S$ be a finite subset of the plane, with more than one element, that satisfies the following properties: (i) If $P$ and $Q$ are in $S$, then $m(PQ)$ intersects $S$. (ii) If $P_1Q_1, P_2Q_2, P_3Q_3$ are three diferent segments such that its endpoints are points of $S$, then, there is non point in $S$ such that it intersects the three lines $m(P_1Q_1)$, $m(P_2Q_2)$, and $m(P_3Q_3)$. Find the number of points that $S$ may contain.
Problem
Source: Spanish Communities
Tags: pigeonhole principle, geometry, circumcircle, rhombus, perpendicular bisector, combinatorics proposed, combinatorics
daniel73
20.09.2007 14:32
No solution after over a year? Let us see if this one works... Assume that $ S$ contains exactly $ n$ points. We may then define $ \binom{n}{2}=\frac{n(n-1)}{2}$ segments. Assume $ n > 5$. Then, there are more than $ 2n$ segments, or more than $ 2n$ perpendicular bisectors, passing through the $ n$ points of $ S$. By Dirichlet's pigeonhole principle, there is at least one point such that 3 perpendicular bisectors pass through it, contradiction. Or $ n\leq5$. For $ n = 5$, take $ S$ as the set of vertices of a regular pentagon. Through each vertex pass exactly two perpendicular bisectors, those of the opposite side and the diagonal parallel to the latter. For $ n = 3$, take $ S$ as the set of vertices of an equilateral triangle. Through each vertex pass exactly one perpendicular bisector, that of the opposite side. So $ S$ may contain 3 or 5 points. $ S$ may not contain 2 points, since there is no point left in $ S$, for the perpendicular bisector of the segment they determine, to intersect. Assume a set $ S$ may be found with exactly 4 points, and call them $ A,B,C,D$. Assume wlog that the perpendicular bisector of $ AB$ passes through $ C$. Then, $ ABC$ is isosceles in $ C$. Assume that $ ABC$ is not equilateral, then the perpendicular bisectors of $ BC$, $ CA$ cannot pass through $ A$, $ B$, respectively, or they both pass through $ D$, which is then the circumcenter of $ ABC$, and the perpendicular bisector of $ AB$ passes also through it, contradiction. $ ABC$ is then equilateral. Consider now the perpendicular bisector of $ CD$. Since it passes through $ A$ wlog, $ CDA$ is also equilateral, or $ ABCD$ is a rhombus with lesser angle $ \pi/3$ and greater angle $ 2\pi/3$, or the perpendicular bisector of $ BD$ passes through both $ A$ and $ C$. So, as before, $ ABD$ and $ BCD$ are equilateral, which is absurd. So $ S$ may contain only 3 or 5 points.