# Critical Line – Volume 10

Welcome back to another year of puzzles and ponderings! In this month’s installment: the Kakeya Needle Problem and Numbered Cones.

Suppose you have a line segment of unit length (a needle with nil width) that sits on a plane. You can twist and translate the segment across the plane and it will sweep over a certain area. The goal is to twist and translate the line segment without removing it from the plane so that it makes an entire $$360^{\circ}$$ rotation. For instance, if you rotate the segment around its midpoint then it will sweep out the area of a circle, with area $$\frac{\pi}4$$.

The Kakeya Needle Problem asks: what is the smallest area required to continuously rotate a needle $$360^{\circ}$$?

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Surprisingly, the answer is: there is no minimum.

For any value $$\varepsilon > 0$$, we can find a set with area less than $$\varepsilon$$ with the desired property.

The construction is a bit involved to go through rigorously in this article, but there are three key observations. Firstly, you can rotate a line segment $$360^{\circ}$$ within an equilateral triangle of height $$1$$. Secondly, you can translate a unit line segment from one place to another using an arbitrarily small area by shifting the needle along a line, rotating it slightly then bringing it back as depicted below.

Finally, you can cut a triangle into $$2^n$$ pieces, then overlap pairs of piece until the total area covered is only $$\frac{1}n$$ of the original area, as illustrated below. This enables you to move the needle around continuously $$360^{\circ}$$ with arbitrarily small area, by choosing a suitably large $$n$$.

You might ask – is there a set of area zero which has the Kakeya property? The answer to this question is in fact, no. So there really is no minimum. Proof of the later is slightly more accessible and can be found here.

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Now for the main event…

**An Orderly Stroll**

Along a straight line of unit length I have placed $$n$$ equally spaced cones (at $$\frac{1}{n+1}, \frac{2}{n+1}, \dots, \frac{n}{n+1}$$) and I have numbered them from $$1$$ to $$n$$. You start at position $$0$$ and walk to cone labelled $$1$$, then to the cone labelled $$2$$ and so on until you have collected all the cones, at which point you walk to point $$1$$. Given the cones were labelled at random from one of the $$n$$! different permutations, what is the expected distance that you have to walk?

For your chance to win the coveted $50 book voucher – send your solution to Puzzles@actuaries.asn.au

## The Critical Line Volume 9 solution

By Chris Ebbs (christine.j.ebbs@gmail.com) and Jevon Fulbrook (jevon01@gmail.com)

- gin
- rum
- ale
- beer
- wine
- sake
- port
- mead
- rosé
- Moet
- ouzo
- grog

This month’s winner is Stephen Woods, as the first person to send in a correct solution. Congratulations!

CPD: Actuaries Institute Members can claim two CPD points for every hour of reading articles on Actuaries Digital.

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