- 18 Aug '11 19:43 / 1 editYou have a bunch of sections of fence (however many you need), all the same length, all straight and impossible to bend. Your goal is to build two closed fences, one containing the other, such that the inner fence uses a greater number of sections than the outer one.

What is the minimum number of sections that you need to achieve your goal?

(note: a section of fence can have an arbitrarily small, but positive, thickness) - 18 Aug '11 22:38

One thing you didn't specify is whether the fence sections have to be touching end-to-end. In other words can a fence section have a part that is dangling, not enclosing anything, while another section pokes it in its side. If so, I have constructed a toothpick example that I think is the minimum,*Originally posted by Anthem***You have a bunch of sections of fence (however many you need), all the same length, all straight and impossible to bend. Your goal is to build two closed fences, one containing the other, such that the inner fence uses a greater number of sections than the outer one.**

What is the minimum number of sections that you need to achieve your goal?

(note: a section of fence can have an arbitrarily small, but positive, thickness)a rhombus of 4 enclosing a weird arrangement of 5but it's hard to describe and impossible to really prove that I did it because of the olives or maybe it was the gin. - 19 Aug '11 06:37How about, making the outer fence a regular pentagon ABCDE.

For inner fencing,

Place a section MN that touches AB at M and BC at N.

Place a section NO that touches BC at N and CD at O.

Place a section OP that touches CD at O and DE at P.

Place a section PQ that touches DE at P and EA at Q.

Now, place sections QR and RM (in V shape) to complete the inner fencing (MNOPQR).

Assuming thickness is very small (epsilon) the points for the inner fencing can be chosen arbitrarily inside the pentagon. - 19 Aug '11 11:09It may not be the minimal solution but I reckon 14 pieces is do-able without the outer fence and the inner fence touching. By a 2x1 rectangle on the outside and sort of two deformed w-shapes in the middle to ensure both shapes are closed loops.

smartarin: I think your idea assumes is right however it rests on the assumption that you can fit a square of lengths n into a pentagon of lengths n, which I think you can. So that would make your answer better than mine.