50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

Question

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?

Accepted Answer

Answer: a) necessary; b) not necessary.

a) We first write the condition, which implies that they eventually meet at one point:

2008 t = 2010 t + 1 - 3 m = 2009 t + 2 - 3 n

, where

m, n \in Z, t \in R

. We have:

t = 3 n - 2

or

2 t = 6 n - 4

. Moreover,

2 t = 3 m - 1 \Rightarrow 3 m - 1 = 6 n - 4 \Leftrightarrow m = 2 n - 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 = 4 t + 1 - 3 m = 2 t + 2 - 3 n

,

m, n \in Z, t \in R

. Then,

t = 3 n - 2

and

3 t = 3 m - 1

. So we obtain the following equation:

9 n - 6 = 3 m - 1

for integer

m, n

. This equation has no solutions, therefore, our runners will not meet at one point.