Source: IMO-2020-P3
Category: Existence and Construction

Problem:
There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n$. Each pebble is coloured in one of $n$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles such that the total weights of both piles are the same, and each pile contains two pebbles of each colour.

Solution:
We begin by pairing the weights such that the sum of each pair is constant:
\[ 1+4n = 2+(4n-1) = 3+(4n-2) = \dots = 2n+(2n+1) = 4n+1. \]
Place all four pebbles of the same colour into a single box, yielding $n$ boxes in total. For each $k=1,2,\dots,2n$, we connect pebble $k$ and pebble $4n+1-k$ with a piece of string. This creates $2n$ strings, each connecting a pair of pebbles whose weights sum to $4n+1$.

To solve the problem, it suffices to paint each string either blue or green such that each box has exactly two blue strings and two green strings attached to its pebbles (where a string connecting two pebbles within the same box counts twice for that box). If we can achieve this, we can assign the pairs of pebbles connected by blue strings to the first pile and those connected by green strings to the second pile. Since each box has two string ends of each colour, each pile will contain exactly two pebbles of each colour. Furthermore, since there are exactly $n$ blue strings and $n$ green strings, each pile will consist of $n$ pairs of pebbles, giving both piles an equal total weight of $n(4n+1)$.

We can rephrase the problem in terms of graph theory by viewing the $n$ boxes as vertices and the $2n$ strings as edges. This forms a $4$-regular multigraph on $n$ vertices, where self-loops are allowed and contribute $2$ to the degree of a vertex. We wish to colour the edges blue and green such that each vertex is incident to exactly two blue edges and two green edges.

Since every vertex in this multigraph has an even degree of $4$, each connected component of the graph admits an Eulerian circuit. Consider a connected component with $k$ vertices. The sum of the degrees of the vertices in this component is $4k$, so it contains exactly $2k$ edges. Because the Eulerian circuit for this component has an even length ($2k$ edges), we may colour the edges along the circuit in alternating colours: green and blue.

This alternating colouring ensures that every time the circuit passes through a vertex (entering via one edge and leaving via the next), the two edges involved have different colours. Since each vertex has a degree of $4$, the Eulerian circuit passes through each vertex exactly twice. If the vertex has no self-loops, it is visited on two separate occasions, with each visit contributing one blue and one green edge, yielding exactly two blue and two green edges in total. If a vertex has a self-loop, the self-loop is traversed as a single edge of one colour, contributing two ends of that colour to the vertex; the alternating nature of the circuit ensures that the edges immediately preceding and following the self-loop are of the opposite colour, providing two ends of the other colour. A vertex with two self-loops simply receives one blue self-loop and one green self-loop. Thus, every vertex receives exactly two blue and two green ends. This yields the desired arrangement, completing the proof.


Grading guidelines:
(Partial) 1. Used the pairing 1+4n=2+(4n-1)=...=4n+1, and reduced the task to selecting exactly n of these 2n pairs so that each colour appears exactly twice among the selected pebbles. 2. Reformulated the problem as a 4-regular multigraph on n vertices (colours as vertices; each complementary-weight pair as an edge, allowing loops), and stated the goal as a 2-edge-colouring with exactly two edges of each colour incident to every vertex. 3. Proved (or correctly invoked) that each connected component of an even-degree graph has an Euler tour, with the intention to alternate colours along such a tour. (Almost) 1. Carried out the Euler-circuit alternating-colouring argument to obtain two edges of each colour at every vertex, but did not correctly justify the handling of self-loops (counting two incidences) or multiple edges in the incidence count. 2. Completed the graph-colouring construction, but made an arithmetic/logic slip in translating back to pebbles (e.g. did not explicitly conclude both piles have weight n(4n+1), or miscounted the number of pairs/pebbles per pile while the colouring condition is correct). 3. Solution is almost complete, but made minor mistakes which are not negligible.