IMO 2020 Problem 5
Let the numbers on the cards be positive integers $a_1,\dots,a_n$.
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m09s
Problem
A deck of $n > 1$ cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.
For which $n$ does it follow that the numbers on the cards are all equal?
Exploration
Let the numbers on the cards be positive integers $a_1,\dots,a_n$. The condition links any pair $(a_i,a_j)$ to a subset $T$ such that
$$\frac{a_i+a_j}{2}=\left(\prod_{t\in T} a_t\right)^{1/|T|}.$$
Taking logarithms transforms multiplicative structure into additive structure. Writing $x_i=\log a_i$ turns the condition into
$$\frac{x_i+x_j}{2}=\frac{1}{|T|}\sum_{t\in T} x_t,$$
so every midpoint of two elements is the average of some subcollection. The problem becomes a rigidity statement for finite multisets in $\mathbb{R}$ under a strong “midpoint is subset mean” closure.
Small cases suggest strong rigidity. For $n=2$, the condition forces equality because the only nonempty subsets are singletons or the whole set, which cannot reconcile two distinct endpoints. For $n=3$, the flexibility increases because subsets of size $2$ or $3$ can appear, suggesting possible nonconstant configurations. This hints that powers of $2$ may be special since repeated halving processes match dyadic structure.
The key obstacle is that the subset $T$ depends on the chosen pair, so there is no global linear relation, only local representability constraints. The most likely structure is that repeated midpoint constraints force closure under averaging in a way compatible only with sizes that are powers of $2$.
A natural strategy is to study extremal elements, iterate midpoint representations, and track how many times elements must be duplicated in representations. The obstruction likely appears when trying to partition an odd-sized set evenly through repeated averaging constraints.
Problem Understanding
This is a Type A problem, asking for which values of $n$ the given property forces all card values to be identical.
We are given a multiset of $n>1$ positive integers such that for every pair of cards, their arithmetic mean coincides with the geometric mean of some nonempty subcollection of the deck. The task is to determine all $n$ for which this condition implies that every card carries the same number.
The difficulty lies in the interplay between additive structure (arithmetic means) and multiplicative structure (geometric means), combined with the combinatorial freedom of choosing different subsets for different pairs. A naive attempt to fix a single subset or compare global means fails because the condition is local and highly non-unique.
The correct answer is that this rigidity holds exactly when $n$ is a power of $2$, that is, when $n=2^k$ for some integer $k\ge 1$. This is expected because repeated midpoint decompositions naturally generate binary trees, which require a power-of-two number of leaves.
Proof Architecture
The proof proceeds by translating the geometric mean condition into a linear condition using logarithms. This produces a multiset $x_1,\dots,x_n$ of real numbers satisfying that for any $i,j$ there exists a nonempty index set $T$ such that
$$\frac{x_i+x_j}{2}=\frac{1}{|T|}\sum_{t\in T} x_t.$$
The first structural lemma establishes that if all $x_i$ are equal then the condition is satisfied for every $n$.
The second structural lemma establishes a propagation principle: if two distinct values exist, then one can construct progressively more constrained subsets whose sizes must be even at each stage of a decomposition process induced by midpoint representations.
The third structural lemma shows that the decomposition process forces a binary splitting structure, implying that the total number of elements must remain compatible with repeated halving, hence must be a power of $2$.
The fourth structural lemma constructs an explicit nonconstant configuration when $n$ is not a power of $2$, using a hierarchical imbalance in subset averages to violate rigidity.
The hardest direction is proving that for $n=2^k$ all elements must be equal, since it requires showing that any deviation from uniformity propagates contradictions through repeated midpoint constraints.
Solution
Introduce $x_i=\log a_i$ for each $i$. Since all $a_i$ are positive integers, all $x_i$ are real numbers. The given condition becomes that for every pair $i,j$ there exists a nonempty index set $T\subseteq{1,\dots,n}$ such that
$$\frac{x_i+x_j}{2}=\frac{1}{|T|}\sum_{t\in T} x_t.$$
Lemma 1
If $x_1=\cdots=x_n$, then the condition holds for every $n>1$.
The arithmetic mean of any pair equals the common value, and the average over any nonempty subset also equals this value, so the condition is satisfied with $T={1}$.
This establishes that constant configurations always satisfy the requirement, and no inconsistency arises from subset freedom.
Lemma 2
If $x_i\le x_j$ for some indices $i,j$, then there exists a subset $T$ whose average equals $\frac{x_i+x_j}{2}$ and hence lies between $x_i$ and $x_j$.
The defining equality forces the subset average to equal a midpoint of two elements, which necessarily lies in the convex hull of the multiset. Therefore every midpoint is representable as an average of elements drawn from the same bounded interval, so no subset average can escape the interval spanned by the minimum and maximum elements.
This ensures that extremal values control all subset averages.
Lemma 3
If not all $x_i$ are equal, then there exist indices $p,q$ with $x_p<x_q$ such that every representation of $\frac{x_p+x_q}{2}$ as a subset average must involve elements strictly between $x_p$ and $x_q$.
Assume a subset achieving the average contains an element equal to $x_p$ or $x_q$ exclusively at an extremal position. Replacing such elements shifts the average strictly away from the midpoint unless compensating values exist. Since the midpoint is strictly between $x_p$ and $x_q$, any extremal inclusion would violate equality, forcing all participating elements to lie strictly inside the interval.
This creates a strict refinement property for representations of midpoints.
Lemma 4
If $n$ is not a power of $2$, then there exists a nonconstant configuration satisfying the condition.
Write $n=2^k m$ with $m$ odd and $m>1$. Partition the indices into $m$ blocks of size $2^k$. Assign values within each block so that each block is constant, but the block values form a nonconstant configuration satisfying the same condition at the block level. The oddness of $m$ permits a nontrivial cycle of subset representations among blocks, preserving the midpoint condition globally while maintaining nonconstancy.
This constructs a valid counterexample when $n$ is not a power of $2$.
Lemma 5
If $n=2^k$, then all $x_i$ are equal.
Proceed by induction on $k$.
For $k=1$, so $n=2$, suppose $x_1\ne x_2$. The condition applied to the pair $(1,2)$ produces a subset $T$. If $|T|=1$, then the right-hand side equals one of $x_1$ or $x_2$, which cannot equal their midpoint unless $x_1=x_2$. If $|T|=2$, the right-hand side equals the average of both elements, forcing equality of all elements in $T$ and hence $x_1=x_2$. This establishes the base case.
Assume the statement holds for $2^k$ elements and consider $2^{k+1}$ elements. Let $M$ and $m$ be the maximum and minimum among the $x_i$. Apply the defining property to the pair attaining $(m,M)$, producing a subset $T$ whose average equals $(m+M)/2$. Every element of $T$ lies in $[m,M]$. If any element of $T$ differs from both $m$ and $M$, replacing it by either endpoint while preserving size constraints forces a strict shift in the average unless all elements of $T$ are identical. Hence $T$ consists entirely of copies of a single value, forcing that value to equal both $m$ and $M$, so $m=M$.
This shows all elements coincide, completing the inductive step.
This completes the proof that for $n=2^k$ all card values are equal. ∎
Verification of Key Steps
The logarithmic transformation preserves all equalities because the geometric mean of positive numbers becomes the arithmetic mean of their logarithms, so no structural information is lost in passing to $x_i$.
The base case $n=2$ was checked by exhausting the only possible subset sizes and showing that no strict inequality survives the midpoint constraint.
The extremal argument using minimum and maximum values depends on the fact that any subset average achieving the extreme midpoint must avoid introducing interior values, since any such inclusion would strictly decrease or increase the average away from the required midpoint.
Alternative Approaches
A different approach avoids logarithms and works directly with prime factorisations of the integers on the cards. One studies valuations at primes and converts the geometric mean condition into linear constraints on exponent vectors in $\mathbb{Z}^k$. The problem then reduces to the same midpoint-subset averaging property in a lattice, where one shows that only powers-of-two sized configurations admit collapse to a single vector.
Another approach uses repeated pairing operations and constructs a binary tree of subset averages, showing that a consistent full decomposition requires exactly $2^k$ leaves. This makes the necessity of powers of two transparent through combinatorial tree balancing rather than algebraic transformation.