## Teaser 2549: A round trip

**From The Sunday Times, 31st 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?

[teaser2549]

## Jim Randell 8:19 am

on8 April 2019 Permalink |I assumed the trees were visited by taking a direct straight line between them, rather than walking along an arc of the field boundary.

We can cut the field (like a pizza) into sectors from the trees to the centre point. The field can then be rearranged (like a pizza) with the sectors in any order we like, and retain its circular shape.

In particular we can rearrange the circle into two semi-circles each involving a 12, 19, and 33 sector.

If we draw the quadrilateral formed from the trees in one of these semi-circles we get:

The length of the lines are:

AB = a, BC = b, CD = c, AD = x, AC = y, BD = z.As the triangles ABD and ACD are inscribed in a semi-circle they are right-angled:

so:

and multiplying these together:

Also, ABCD is a cyclic quadrilateral, so by Ptolomy’s Theorem [ link ]:

squaring both sides:

equating both expressions for

y²z²we get:dividing by

x, and simplifying:We can look for a solution to this equation numerically:

Solution:The diameter of the field is 44 metres.If there is an integer solution

k, then the cubic polynomial is divisible by(x – k), sokwill be a smallish divisor of the constant term in the polynomial, in this case2abc.In fact using:

a = 12, b = 19, c = 33we get the following polynomial:which can be factorised as:

Giving a positive root of 44, and negative roots of –22 ± √(142).

There are only 5 divisors to be checked in the required range.

LikeLike

## Jim Randell 11:28 pm

on8 April 2019 Permalink |Here is an alternative approach that uses a bit less analysis, and works with any viable collection of distances.

For a sector with a distance of

din a circle with radiusrwe can use the Cosine Rule [ link ] to calculate the angle at the centre of the circle:We then just need to find a value for

r, which makes the sum of the angles for all sectors achieve a value of 360°.Run:[ @repl.it ]LikeLike