2020 IMO Problems/Problem 5

Problem

A deck of 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 does it follow that the numbers on the cards are all equal?

Solution 1

Claim : For all n > 1, all numbers must be equal
Contradiction: Let us assume this is not true and for a certain n, there are k distinct positive integers which can be written in ascending order as follows :

z_k > z_k-1 > z_k-2 > … > z₁

Since z_k is the largest of the numbers, it has to be greater than 1. This implies that there will be a prime p₁ that divides z_k.

Now we know that the arithmetic mean of z_k and z_k-1 is more than z_k-1, thus the geometric mean which it is equivalent to must include the term z_k. Since we know the arithmetic mean is a rational number, this means the root of the products of the n numbers of the geometric mean must be positive, and this will also divide p.

We can use similar argument for the sum of z_k-1 and z_k-2 will be greater than z_k-2 and thus the geometric mean is once again divisible by p. We can repeat this all the way to z₁, which would mean that by dividing each term by p, we can reach another set of n cards which fulfil the conditions. We can keep applying this until z_k. However this is a contradiction.

Thus, for all n > 1, all the terms on the card will be equal.

QED

Video solution

https://www.youtube.com/watch?v=dTqwOoSfaAA [video covers all day 2 problems]

See Also

2020 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions