Teaser 3021: Field for thought
From The Sunday Times, 16th August 2020 [link]
Farmer Giles had a rectangular field bordered by four fences that was 55 hectares in size. He divided the field into three by planting two hedges, from the mid-point of two fences to two corners of the field. He then planted two more hedges, from the mid-point of two fences to two corners of the field. All four hedges were straight, each starting at a different fence and finishing at a different corner.
What was the area of the largest field when the four hedges had been planted?
[teaser3021]
S2T2 
Jim Randell 10:24 pm on 14 August 2020 Permalink |
The x and y co-ordinates of the intersections of the hedges helpfully divide the edges of the rectangle into equal divisions.
The area of the central quadrilateral is then easily calculated.
Solution: The central field has an area of 11 hectares.
In the first diagram, each of the four coloured triangular areas has an area of xy/5, so the central region also has an area of xy/5.
Another way to think of it is that there are five quadrilaterals each identical to the central one, but four of them have had bits sliced off them, and then the sliced off bits have been repositioned to make the original rectangle.
This gives a handy practical construction achievable using folding and straight edge to divide a rectangle into 5 equal strips (or a 5×5 grid of identical smaller rectangles).
This is a bit like Routh’s Theorem (see: Enigma 1313, Enigma 320, Enigma 1076, Teaser 2865) but for a rectangle instead of a triangle.
In general, if the the lines are drawn from a corner to a point a fraction f along a side we can determine the area of the central quadrilateral (as a fraction of the overall parallelogram) as:
In the case of f = 1/2 we get:
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