Let n≥2 be a positive integer and λ a positive real number. There are n fleas on a horizontal line, and we need to find the values of λ for which, given any point M and any initial positions of the fleas, there is a sequence of moves that can place all fleas to the right of M.
### Move Description:
A move consists of selecting two fleas located at points A and B (with A to the left of B), and moving the flea from A to a new point C such that ABBC=λ.
### Analysis:
- Assume the leftmost flea is initially at position x1 and the rightmost flea is at position xn.
- The goal is to transform the system such that all fleas are located at some position greater than M.
### Considerations:
1. Move Effect:
- If a flea initially at A jumps to C, then:
C=A+λ(B−A)=(1−λ)A+λB.
- This replaces A with a point closer to B (if λ>0).
2. Bounding Fleas to the Right:
- We need each flea to eventually move past M. Since fleas consecutively jump rightward, the greatest possible accumulation of fleas past M occurs when effective λ allows maximal stretching of intervals.
3. **Condition on λ:**
- Starting with fleas positioned in a finite interval covering x1 to xn, progressively applying transformations:
- If λ is too small, the rightward jumps might be insufficient to clear M in finite steps.
4. Sufficient Condition:
- Sufficiently large λ ensures that the accumulative forward motion possible among successive intervals exceeds the necessary coverage over distance xn−M.
- Analyzing proportion:
- For m iterations to push gaps from x1 through to beyond xn, having λ≥n−11 guarantees accumulative growth beyond necessary jumps.
### Conclusion:
With the above reasoning, we conclude that the values of λ that ensure an eventual placement of all fleas to the right of any point M, for any initial configuration of fleas, are:
λ≥n−11
This bound arises from ensuring that progressive cumulative extensions with each move can bridge the intervals ensuring encompassment reaches past any arbitrary point M.