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BdMO Online Forum • View topic - IMO 2015 - Problem 1

## IMO 2015 - Problem 1

Discussion on International Mathematical Olympiad (IMO)

### IMO 2015 - Problem 1

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 point $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

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

Proposed by Netherlands.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.
Nirjhor

Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: IMO 2015 - Problem 1

a) Consider any circle with it’s center as one point A. If n is odd, take any $\frac {n-1}{2}$ different equilateral triangles $A_i$ with one vertice A, other two on the circle; such that any two $A_i$ do not share vertices except A. If n is even, consider similar configuration of $\frac {n}{2}$ triangles, but now exactly one pair $A_i,A_j [i \neq j]$ share two vertices including A. A and all the vertices of the$A_i$’s constitute the desired balanced set.
b) First note that for all odd $n = 2m+1[m \in N]$, there exists a balanced center-free set of n-points. We take any regular n-gon. For any two vertices X,Y of it; there exists exactly one another vertice Z so that XZ = XY. For,if we label the points clockwise as $X = A_1,A_2,…,A_n$; the perpendicular from X to edge $A_m,A_{m+1}$ is a line of symmetry for the n-gon; while the distances $XA_i$ is strictly increasing for $1 \leq I \leq m$. So if the vertices constitute a balanced set, it is center-free. To note it is balanced, pick any vertices X,Y. XY naturally partitions the n-gon into two part not containing X and Y. One of them contains odd number of vertices, their central vertice lies on the perpendicular bisector of XY.

I claim that n has to be odd to exist such a center-free n-point configuration S. Assume that exists. Here’s some terminology: any point in the set S is ‘dot’ and for P,Q distinct dots, segment PQ is an ‘edge’. For P,Q,R distinct dots; if RP = RQ,we say P is R-companion of Q.

Choose any two different dots A,B; I claim that there exists exactly one dot C so that B lies on the perpendicular bisector of AC. Let S\{A} = K. The (n-1) distinct edges AX, with X varying over K, do not share a common dot in S on their perpendicular bisector; though for each AX there have to be at least one dot in S on the bisector. |S| = (n-1), so this edges AX has a unique dot of K in the perpendicular bisector of it; and also each dot in K appear in the perpendicular bisector of exactly one such edge AX.

Choose any point P in S. Let S\{P} = T. By the previous claim, for each Q in T, there exists exactly exactly one point R in T[other than Q, of course] so that R is a P-companion of Q. Note also that being P-companion is a mutual relation. So the points in T pair up under this relation, with nobody pair itself. It follows that |T| is even, so n is odd.

Fm Jakaria

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Joined: Thu Feb 28, 2013 11:49 pm