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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewesta) 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.

Log in to reply

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.

Log in to reply