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Sunday Times Teaser 2549 – A Round Trip

by Brian Gladman

Published: 31 July 2011 (link)

I own a circular field with six trees on its perimeter. One day I started at one tree and walked straight to the next, continuing in this way around the perimeter from tree to tree until I returned to my starting point. In this way I found that the distances between consecutive trees were 12, 12, 19, 19, 33 and 33 metres.

What is the diameter of the field?

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    The layout on the left of the diagram shows the configuration of the triangles formed by the tree positions described in the teaser. But since the lengths of the paths between the trees come in pairs, we can rearrange the triangles as shown on the right where the three different triagles in combination each occupy one half of the circle.

    From this rearrangement we can see that the sum of the central angles of the three different triangles is 180 degrees, which means that the sum of their half angles is 90 degrees (i.e. \(\pi/2\) radians).

    For a chord of length \(l\) and a circle of diameter \(d\), we can evaluate the half angle as \(\sin^{-1}(l/d)\). With the three chords \(a\), \(b\) and \(c\) as shown, we have: \[\sin^{-1}\left(\frac{a}{d}\right)+\sin^{-1}\left(\frac{b}{d}\right)+\sin^{-1}\left(\frac{c}{d}\right)=\frac{\pi}{2}\] This can be expressed as: \[\cos\left(\sin^{-1}\left(\frac{a}{d}\right)+\sin^{-1}\left(\frac{b}{d}\right)\right)=\cos\left(\frac{\pi}{2}-\sin^{-1}\left(\frac{c}{d}\right)\right)\]Expanding the cosines now gives \[\sqrt{1-\left(\frac{a}{d}\right)^2} \sqrt{1-\left(\frac{b}{d}\right)^2}-\frac{ab}{d^2}=\frac{c}{d}\]and hence: \[\left(d^2-a^2\right)\left(d^2-b^2\right)=\left(ab+cd\right)\]This can now be expanded and simplified to give a cubic equation for \(d\):\[d^3-\left(a^2+b^2+c^2\right)d-2abc=0\] Inserting the values for \(a\), \(b\) and \(c\) into this equation now gives:\[d^3-1549d-15048=0\] We can guess that the diameter is likely to be an integer, in which case it will be a factor of \(15048=2^3.3^2.11.19\). And from this it does not take long to identify \(d=44\) as the solution (the other solutions, \(-22\pm\sqrt{142}\), are negative).

    In fact the cubic for \(d\) can be solved analytically to give:\[d=2\sqrt{\frac{a^2+b^2+c^2}{3}}\cos\left(\frac{1}{3}\cos^{-1}\left(\frac{abc}{\left((a^2+b^2+c^2)/3\right)^{3/2}}\right)\right)\]

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