Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's start with two concentric circles of radii $r<R$. Then we put two sticks inside the outer circle while avoiding the inner circle, say $AB$ and $CD$.

Two sticks between the two circles

Then we compare the length of inner part $AP+DP$ with the exterior part $BP+CP$. It seem that there exist a lower bound for the difference: $(AP+DP)-(BP+CP)\ge\epsilon$ for all such sticks with a uniform constant $\epsilon>0$.

I don't if this is true. (For specific reason we can use $r=1$ and $R=2$).
If it is, could you give any hint how to prove it? Thanks!


Edit: Now I know there is no uniform lower bound since the distance goes to zero if

  1. the point $C$ moves close to $A$;

  2. the lengths of both sticks goes to zero: $AB\to0$ and $CD\to0$.

The second case can be viewed as a special case of the first one. So a reasonable modification of my question is:

  • Will there exist $\epsilon=\epsilon(\delta)>0$ such that for if the distance of the two ends $AC\ge\delta$, then the difference $(AP+DP)-(BP+CP)\ge\epsilon$?
share|improve this question
1  
a) Do I understand correctly that the two sticks must be placed such that they intersect in a point $P$? b) In what sense is the constant $\delta$ uniform? c) What keeps us from making $BP+CP$ arbitrarily small by choosing $P$ arbitrarily close to the outer circle, and $AP+DP$ supremal by choosing sticks of maximal length, nearly touching the inner circle? –  joriki May 2 '13 at 7:31
1  
@joriki at least for c) I can answer you. The process you describe would maximize the difference, and as such not prove anything about the existence of a lower bound, because you're moving in the wrong direction. However, moving the two lines closer and closer should make the difference closer and closer to $0$. –  Arthur May 2 '13 at 8:07
    
Thank you! Now I see such a lower bound can't exist. –  Pengfei May 2 '13 at 13:46

1 Answer 1

up vote 0 down vote accepted

Your original proposition about a global lower bound is not true. You can move all points arbitrarily close together while still maintaining the required properties. Thus all distances will become arbitrarily small as well, and a lower bound on their differences can not exist.

Your updated question appears to be true. Given fixed $A,B,C$ the minimal value of $(AP+DP)−(BP+CP)$ is obtained (although this remains to be proved for now) by setting $D=B$. Conversely for given $A,C,D$ the minimal value corresponds to $B=D$. In this case you have $BP=DP=0$, and one can show $AP>CP$ for $AC=\delta>0$. For fixed $A,C$ the lowest value is obtained if $AB$ touches the inner circle. So you can use that configuration to compute $\epsilon$ from $\delta$.

Screenshot from Cinderella

All of the above was obtained from experiments, so this should give you an idea as to what the steps of a proof would be, but it is not a proof so far, and it might even be wrong for some non-obvious reason.

share|improve this answer
    
Thank you! I just modified the question. –  Pengfei May 2 '13 at 14:06
    
@Pengfei: Updating a question in such a way that it renders a formerly correct answer now incorrect is not exactly the best style, in my opinion. I updated my answer accordingly nevertheless. –  MvG May 2 '13 at 14:36
    
Thank you! I faced the same dilemma before. There might be some trivial conditions I didn't realize when posting questions. Should I leave this question as it is and post the modified version in a new post? –  Pengfei May 2 '13 at 14:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.