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# Problem 1! IMO 2015

We say that a finite set $$\mathcal{S}$$ of points in the plane is balanced if, for any two different points A and B in $$\mathcal{S}$$, there is a point C in $$\mathcal{S}$$ such that $$AC=BC$$. We say that $$\mathcal{S}$$ is centre-free if for any three different points A, B and C in $$\mathcal{S}$$, there is no points P in $$\mathcal{S}$$ such that $$PA=PB=PC$$

(a) Show that for all integers $$n\geq 3$$, there exists a balanced set consisting of $$n$$ points.

(b) Determine all integers $$n\geq 3$$ for which there exists a balanced centre-free set consisting of points.

##### This is part of the set IMO 2015

Note by Sualeh Asif
2 years, 6 months ago

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a) Consider the set of points $$O, P_1, P_2, \ldots P_n$$ such that $$P_kO=r$$ for all $$1\le k\le n$$; and for all $$P_i$$, there exists a $$P_j$$ such that $$P_iP_j=r$$. It is easy to construct this for any $$n\ge 3$$.

b) I claim that only odd integers $$n\ge 3$$ satisfy that it can be center-free. For all odd integers $$n\ge 3$$, a working example is just the vertices of a regular $$n$$-gon. It remains to prove that it is not possible for even numbers $$n\ge 3$$.

Note that there are $$\dbinom{n}{2} = \dfrac{n(n-1)}{2}$$ edges total, and $$n$$ points, meaning that on average, each point lies on the perpendicular bisector of $$\dfrac{n-1}{2}$$ edges. When $$n$$ is even, this means that there exists a point $$P$$ that lies on the perpendicular bisector of $$\dfrac{n}{2}$$ edges. Consider the edges that have $$P$$ on their perpendicular bisector. At least two edges share a point, because if no edge shared a point there can be at most $$\dfrac{n}{2}-1$$ edges as there are only $$n-1$$ available points. But then we're done; since two edges share a point, and $$P$$ is a perpendicular bisector of both these edges, then the distance between $$P$$ to the three points that make up these two edges is the same, contradiction.

- 2 years, 6 months ago

We can prove (b) using double counting(refer to Daniel's solution for the rest), here's the basic outline:

Our object being counted will be $$(P, \{ A,B\})$$ satisfying $$PA=PB$$, the number of which will be denoted by $$k$$. On one hand, every combination of two points(i.e. $$\{A,B\}$$) has at least one $$P$$ since the set is balanced; this gives $$k\ge C(n,2)=\frac {n(n-1)}{2}$$. On the other hand, every point has at most $$\lfloor \frac {n-1}{2}\rfloor$$ combinations of two points (If it has more, then by PHP there exists two combinations that share a point, contradicting the centre-free condition); this gives $$k\le n\lfloor \frac {n-1}{2}\rfloor$$. Together we have:

$$\frac {n(n-1)}{2}\le n\lfloor \frac {n-1}{2}\rfloor$$

Clearly this inequality only holds when $$n$$ is odd.

- 2 years, 6 months ago