Sunday Times Teaser 2485
by Nick MacKinnon
Published: 9 May 2010 (link)
Jack and Kay [1] have inherited a circular field, with North Gate (N) at the northernmost point and East Gate (E), South Gate (S) and West Gate (W) at the appropriate points.
Jack’s share of the field is one hectare in area. For each point P in his share, three of the angles NPE, EPS, SPW and WPN are acute.
For each point in Katy’s [1] share, however, fewer than three of those angles are acute.
How far is it between North Gate and East Gate?
[1] There is a Sunday Times typographic error here – the intended name is is not known.
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This diagram of the field illustrates the areas involved in this puzzle.
The angle subtended by the diameter of a circle at any point on the circle is 90 degrees. This angle is larger for points inside the circle and smaller for those outside.
Consider the four semi-circles on the chords between the North, East, South and West gates on the circular field. The angles subtended by one of these chords will be acute for points outside ‘its’ circle and obtuse otherwise. So the points where three of these angles are acute are those that lie outside three of these four semi-circles, that is, points in the green area.
We can calculate this area as the area of the field (\(\pi r^2\)) minus eight times the area of the orange coloured segment. And the orange area is one quarter of a circle of radius \(r/\sqrt{2}\) minus the area of a triangle of the width and height equal to \(r/\sqrt{2}\). So the green area is:
\[\pi r^2 – 8(\pi r^2/8 – r^2/4) = 2r^2\]
Hence \(2r^2 = 10000\) square metres, which gives \(r\) as \(100/\sqrt{2}\) metres. Hence the distance between the north and east gates (\(\sqrt{2}r\)) is 100 metres.