# Two sticks between two concentric circles

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 $CD$.

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 $(AP+DP)-(BP+CP)\ge\epsilon$?
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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
@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

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 $\delta$.