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, 2010

Number theory Difficulty 7.1 National olympiad, round 2 Prove it Ukraine

There are three runners in different vertices of an equilateral triangle with side $1$ : First, Second and Third. They start moving simultaneously in the same direction (Second in First's direction, Third in Second's direction, First in Third's direction). Is it necessary that they all meet in one point at the same time, if:

a) First, Second and Third have velocity $2008$ , $2009$ and $2010$ respectively?

b) They are moving with distinct natural velocities?

Solution

Answer: a) necessary; b) not necessary.

a) We first write the condition, which implies that they eventually meet at one point: $2008t = 2010t + 1 - 3m = 2009t + 2 - 3n$ , where $m, n \in \mathbb{Z}, t \in \mathbb{R}$ . We have: $t = 3n - 2$ or $2t = 6n - 4$ . Moreover, $2t = 3m - 1 \Rightarrow 3m - 1 = 6n - 4 \Leftrightarrow m = 2n - 1$ , from that, we can easily find solutions, for instance, if $m = n = 1$ then $t = 1$ . If $t = 1$ then, indeed, First will run $2008$ , Second $2009$ , Third $2010$ , hence, they will meet at one point.

b) Let us suppose that First, Second and Third have velocities $1$ , $2$ and $4$ respectively. Then, the condition that they meet at one point can be rewritten as follows: $t = 4t + 1 - 3m = 2t + 2 - 3n$ , $m, n \in \mathbb{Z}, t \in \mathbb{R}$ . Then, $t = 3n - 2$ and $3t = 3m - 1$ . So we obtain the following equation: $9n - 6 = 3m - 1$ for integer $m, n$ . This equation has no solutions, therefore, our runners will not meet at one point.

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