All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Scrutiny (Posted on 2023-05-01) Difficulty: 2 of 5
A puzzle by V. Dubrovsky, from Quantum, January-February 1992:

In a certain planetary system, no two planets are separated by the same distance. On each planet sits an astronomer who observes the planet closest to hers.

Prove that if the total number of planets is odd, there must be a planet that no one is observing.

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
proof by contradiction | Comment 5 of 6 |
Assume that in an odd-numbered collection of planets every planet is observed. 

Say there are P planets and consider the two that are nearest each other.

Astronomers on those planets observe each other.

If an astronomer from any other planet observes either of those two, then at most P-3 observers are left to observe P-2 planets, leaving one unobserved which would violate the opening assumption.

So we can disregard those two planets that observe each other and are observed by no others.  

However the same argument applies to the two nearest remaining planets out of the group of P-2.  We can continually eliminate pairs until we're left with a single planet.  There will be no one left to observe it and that contradicts the assumption that all planets are observed.  

  Posted by xdog on 2023-05-02 10:14:55
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information