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We say that a finite set S\mathcal{S} of points in the plane is [i]balanced[/i] if, for any two different points AA and BB in S\mathcal{S}, there is a point CC in S\mathcal{S} such that AC=BCAC=BC. We say that S\mathcal{S} is [i]centre-free[/i] if for any three different points AA, BB and CC in S\mathcal{S}, there is no points PP in S\mathcal{S} such that PA=PB=PCPA=PB=PC.

(a) Show that for all integers n3n\ge 3, there exists a balanced set consisting of nn points.

(b) Determine all integers n3n\ge 3 for which there exists a balanced centre-free set consisting of nn points.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Consider a finite set S\mathcal{S} of points in the plane. The problem involves two specific definitions: a set is balanced if, for any two different points AA and BB in S\mathcal{S}, there is a point CC in S\mathcal{S} such that AC=BCAC = BC. Additionally, the set is centre-free if for any three different points AA, BB, and CC in S\mathcal{S}, there is no point PP in S\mathcal{S} such that PA=PB=PCPA = PB = PC.

### Part (a)

To show that for all integers n3n \geq 3, there exists a balanced set consisting of nn points, consider placing the points equally spaced on a circle. This configuration is symmetrical, and for any pair of points AA and BB, the perpendicular bisector of the segment ABAB will contain a point CC on the circle such that AC=BCAC = BC. This holds for any pair of points on the circle.

1. Place nn points on a circle such that each point is evenly spaced.
2. For any two points AA and BB, there exists a point CC on the circle because of the symmetrical nature of the circle (specifically, the point diametrically opposite to the midpoint of arc ABAB).

Thus, this configuration ensures that the set is balanced for any number of points n3n \geq 3.

### Part (b)

To determine all integers n3n \geq 3 for which there exists a balanced centre-free set consisting of nn points, analyze the geometric properties of balanced configurations.

1. Consider the property of being centre-free: if any three points AA, BB, and CC are considered, no fourth point PP can exist such that PA=PB=PCPA = PB = PC.
2. In a balanced configuration using an even number nn of points, symmetry can lead to points PP equidistant from any three others, which violates the centre-free condition.

By the above consideration, a balanced centre-free set cannot exist if nn is even, because having evenly spaced points on a circle for an even nn results in symmetry that allows for a central point equidistant from multiple others.

Thus, a balanced centre-free set consisting of nn points is only possible for:

All odd integers n3 \boxed{\text{All odd integers } n \geq 3}

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