Argentine National Olympiad 2016

The values in question are all integers $r$ greater than $1$.

We start with a general observation about two particles $P_1$ and $P_2$ moving on $AB$ by the given rules, with different constant speeds $v_1$ and $v_2$, $v_1>v_2$. Suppose that they are at the same point $Q$ of $AB$ at a certain moment $t$. There are two possibilities for the distances $v_{1}t$ and $v_{2}t$ the particles have traveled until that moment. If $P_1$ and $P_2$ are moving in the same direction when they simultaneously reach $Q$, then the integer parts of $v_{1}t$ and $v_{2}t$ have the same parity, and their fractional parts are equal. Hence $v_{1}t-v_{2}t$ is an even positive integer. And if $P_1$ and $P_2$ are moving in opposite directions when they meet at $Q$, then the integer parts of $v_{1}t$ and $v_{2}t$ have different parity, and the sum of their fractional parts is $1$. Therefore $v_{1}t+v_{2}t$ is an even positive integer.

Now let the rational $r>1$ have the stated property, for any number $n+1$, $n\ge 1$, of particles with speeds $1,r,r^2,\dots ,r^n$. Let $t$ be a moment when all of them are at the same point. Then all of them are at the same point of $AB$ at instant $t$. Apply the observation to the first and the last particle, with speeds $v_1=r^n$ and $v_2=1$. We infer that $(r^n-1)t$ or $(r^n+1)t$ is an integer. Because $r$ is rational, $t$ is rational too. Write $r$ and $t$ as irreducible fractions: $r=\frac{a}{b}$, $t=\frac{c}{d}$. Then $(r^n\pm 1)t=\frac{(a^n\pm b^n)c}{b^{n}d}$. Since $a^n\pm b^n$ and $b^n$ are coprime, it follows that $b^n$ divides $c$. Moreover, the latter holds for each $n>1$ by hypothesis. This is possible only if $b=1$, that is, if $r>1$ is an integer.

Conversely, every integer $r>1$ is a solution. Let $r\ge 3$ be odd and $n\ge 1$ arbitrary. Then all $n+1$ particles with speeds $1,r,r^2,\dots ,r^n$ will be at the midpoint of $AB$ at $t=\frac{1}{2}$. Indeed, $r^k\cdot \frac{1}{2}$ has fractional part $\frac{1}{2}$ for each $k=0,1,2,\dots ,n$ since $r^k$ is odd.

Let $r=2m$, $m\ge 1$, be even and $n\ge 1$ arbitrary. Then at the moment $t=\frac{2m}{2m+1}$ all $n+1$ particles with speeds $1,r,r^2,\dots ,r^n$ will be at the point $Q$ at distance $\frac{2m}{2m+1}$ from $A$. It is enough to prove that for each $k=0,1,2,\dots$ the next equality holds:

$$(2m{)}^k\frac{2m}{2m+1}=\{\begin{array}{ll}2q+\frac{2m}{2m+1}& \text{con }q=0,1,2,\dots \text{ si }k\ge 0\text{ es par;}\\ 2q+1+\frac{1}{2m+1}& \text{con }q=0,1,2,\dots \text{ si }k\ge 1\text{ es impar.}\end{array}$$

Indeed, these relations mean that, for $k$ even, the particle with speed $r^k$ will be moving from $A$ towards $B$ at the moment $t=\frac{2m}{2m+1}$, and it will be at distance $\frac{2m}{2m+1}$ from $A$, that is, at point $Q$.

For $k$ odd, the particle with speed $r^k$ will be moving from $B$ towards $A$ at the moment $t=\frac{2m}{2m+1}$, and it will be at distance $\frac{1}{2m+1}$ from $B$, hence at point $Q$ again.

So it remains to prove the displayed equalities, which we do by induction in $k$. The case $k=0$ is obvious. Proceed to the inductive step $k\to k+1$. The induction hypothesis yields

$$k\text{ odd: }(2m{)}^{k+1}\frac{2m}{2m+1}=2m(2q+1)+\frac{2m}{2m+1}=2q^{\prime }+\frac{2m}{2m+1},q^{\prime }=0,1,2,\dots ;$$ $$k\text{ even: }(2m{)}^{k+1}\frac{2m}{2m+1}=4mq+\frac{4m^2}{2m+1}=4mq+2m-1+\frac{1}{2m+1}=2q^{\prime }+1+\frac{1}{2m+1},q^{\prime }=0,1,2,\dots$$

This completes the induction and the solution.