China Western Mathematical Olympiad

Question

Four frogs are positioned at four points on a straight line such that the distance between any two neighboring points is one unit of length. Suppose that every frog can jump to its corresponding point of reflection, by taking any one of the other three frogs as the reference point. Prove that there is no case where the distances between any two neighboring points, where the frogs stay, are all equal to 2008 units of length. (Posed by Liu Shixiong)

Accepted Answer

Without loss of generality, we may think of the initial positioning of the four frogs as being on the real number line at points

1

,

2

,

3

, and

4

. Further, it can be established that the frogs at odd number positions remain at odd number positions after each jump, and likewise for frogs at even number positions. Thus, no matter after how many jumps, there are two frogs remaining at odd number positions while the other two frogs remain at even number positions.

Therefore, in order that the distances between any two neighboring frogs are all equal to

2008

, all the frogs need to stay at points which are either all odd or all even, which is contrary to the actual situation. Hence, the proposition is proven.