$8$ ants are placed on the edges of the unit cube. Prove that there exists a pair of ants at a distance not exceeding $1$.
Problem
Source: II Caucasus Mathematical Olympiad
Tags: 3D geometry, geometric inequality, geometry, inequalities
21.03.2018 00:48
We see that the farthest away all the ants can be away from each other is if all of the atns are on the $8$ vertices. We see that if $2$ vertices are next to each other, the difference away from them is $1$ unit. Thus, proves this. Lol, I used no math...
21.03.2018 00:54
Pigeonhole Principle
21.03.2018 02:49
Can you elaborate?
21.03.2018 02:59
TheEvaluator wrote: We see that the farthest away all the ants can be away from each other is if all of the atns are on the $8$ vertices. We see that if $2$ vertices are next to each other, the difference away from them is $1$ unit. Thus, proves this. Lol, I used no math... Yes, you are right, thats not maths. Thats not a proof.
22.03.2018 10:22
Well, since theres still not a solution, I will post mine.$\newline$ For each ant ( denoted by $A_{i}$), there is a vertex $V_{i}$, endpoint of the edge that contains $A_{i}$, such that $d(A_{i},V_{i}) \leq \frac{1}{2} $ ( if $A_{i}$ is in the middle, just pick one ). $\newline$ Then, if $V_{i} = V_{j} = V$, $i \neq j$ $\Rightarrow d(A_{i},A_{j}) \leq d(A_{i},V) + d(V,A_{j}) \leq \frac{1}{2} + \frac{1}{2} = 1 $. Otherwise, all $V_{i}$ are diferents, let $d_{i} = d(A_{i},V_{i})$, WLOG $d_{1}= max (d_{i}).$ $\newline$ For some $i$, $V_{i} \neq V_{1}$ is an endpoint of the edge containing $A_{1}$ $\Rightarrow d(V_{i},A_{1}) = 1 - d_{1}$ $\Rightarrow d(A_{i},A_{1}) \leq d(A_{i},V_{i}) + d(V_{i},A_{1}) = d_{i} + (1 - d_{1}) \leq 1$ .