Bogosiov I.I.

2011 numbers are written on the board. It turns out that the sum of any three of the written numbers is also a written number.

What is the smallest number of zeros that can be among these numbers?

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An example from 2009 zeros and numbers $1, -1$ satisfies the condition.

Assume the number of zeros is no more than 2008. Then on the board, there will be either three non-negative numbers, among which at least two are strictly positive, or three non-positive numbers, among which at least two are strictly negative. Suppose the first case holds: numbers $a$ and $b$ are positive, and $c$ is non-negative. We can assume that $a$ is the largest of all the numbers written. However, the number $a+b+c > a$ cannot be written. Contradiction.

## Answer

2009 zeros.

## Authors: Bogdanov I.I., Garber A.

Petya and Kolya each have two numbers written in their notebooks; initially, Petya has 1 and 2, and Kolya has 3 and 4. Every minute, Petya forms a quadratic polynomial $f(x)$ whose roots are the two numbers written in his notebook, and Kolya forms a quadratic polynomial $g(x)$ whose roots are the two numbers written in his notebook. If the equation $f(x) = g(x)$ has two distinct roots, one of the boys replaces his pair of numbers with these roots; otherwise, nothing happens. What could be the second number in Petya's notebook at the moment when the first became 5?

## Solution

We will write next to each pair of numbers the monic quadratic polynomial whose roots are the numbers in that pair. Let at some moment the boys have polynomials $p(x)$ and $q(x)$. Then they solved an equation of the form $a p(x) = b q(x)$, where $a, b$ are some different non-zero numbers. This means that the obtained numbers are the roots of the polynomial $a p(x) - b q(x)$. If one of the boys replaces his numbers with these roots, then next to them will be written a linear combination of the polynomials $p(x)$ and $q(x)$.

The initial two polynomials are $p_{0}(x) = (x-1)(x-2)$ and $q_{0}(x) = (x-3)(x-4)$. From the above, it follows that at each step, each boy has a polynomial of the form $r(x) = \alpha p_{0}(x) + \beta q_{0}(x)$, where $\alpha + \beta = 1$.

The parabolas $y = p_{0}(x)$ and $y = q_{0}(x)$ are obviously symmetric with respect to the line $x = 2.5$, so $p_{0}(2.5) = q_{0}(2.5) = 0.75$. Therefore, $r(2.5) = 0.75$.

If the number 5 is written on Petya's sheet, then $r(5) = 0$. Now the second root of the polynomial $r(x)$ can be easily found. For example, like this. After the substitution

$t = x - 2.5$ we get a monic polynomial with a constant term of 0.75 (this is its value at the point 0). One of its roots is $5 - 2.5 = 2.5$, by Vieta's formulas the second root is $0.75 : 2.5 = 0.3$. The second root of the polynomial $r(x)$ is $0.3 + 2.5 = 2.8$.

## Answer

2.8.

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