Teaser 1958: Pent up
From The Sunday Times, 26th March 2000 [link]
The schoolchildren run around in a walled regular pentagonal playground, with sides of 20 metres and with an orange spot painted at its centre. When the whistle blows each child has to run from wherever they are to touch each of the five walls, returning each time to their starting point, and finishing back at the same point.
Brian is clever but lazy and notices that he can minimize the distance he has to run provided that his starting point is within a certain region. Therefore he has chalked the boundary of this region and he stays within in throughout playtime.
(a) How many sides does Brian’s region have?
(b) What is the shortest distance from the orange spot to Brian’s chalk line?
This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.
[teaser1958]
Hugh Casement and 
S2T2 
Jim Randell 9:28 am on 29 November 2020 Permalink |
I wrote a program to plot points with a path distance that is the same as that of the central point, and you get a plot like this:
This makes sense, as the shortest distance from a point to the line representing any particular side is a line through the point, perpendicular to the side. So for each side, if we sweep it towards the opposite vertex, we cover a “house shaped” region of the pentagon, that excludes two “wings” on each side. (The regions that overhang the base).
Sweeping each of the five sides towards the opposite vertex, the intersection of these five “house shaped” regions is the shaded decagon:
A line from the centre of one of the sides of this decagon, through the orange spot to the centre of the opposite side, is a minimal diameter of the decagon, and we see that this distance is the same as the length of one of the sides of the original pentagon. (It corresponds to the side swept towards the opposite vertex, in the position when it passes through the exact centre of the pentagon).
So the minimum distance from the centre to the perimeter of the decagon is half this distance.
Solution: (a) The region has 10 sides; (b) The shortest distance is 10 metres.
If the playground had been a regular 4-gon (square) or 3-gon (equilateral triangle), then every interior point would correspond to a minimal path.
For a 6-gon (hexagon) we get a smaller 6-gon inside as the minimal area, and for a 7-gon the area is the shape of a 14-gon.
In general for a k-gon we get an area in the shape of a smaller k-gon (when k is even) or 2k-gon (when k is odd).
The areas get smaller and smaller as k increases. In the limit we have a circular playground with a single point at the centre corresponding to the minimal path.
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Hugh Casement 1:23 pm on 29 November 2020 Permalink |
I think he chalks a line that runs (for example) from the north vertex to a point roughly two thirds of the way down the ESE wall, then to the midpoint of the S wall, to a point roughly a third of the way up the WSW wall, and back to the N vertex. He then stays on that line rather than within it. When the whistles blows he can run right round the line, clockwise or anticlockwise, touching two walls at the N vertex. His total distance is about 81.75 m.
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Jim Randell 2:39 pm on 29 November 2020 Permalink |
@Hugh: Except that after touching each wall you have to return to your starting point before you can set off to touch the next wall.
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Hugh Casement 3:17 pm on 29 November 2020 Permalink |
Oh, I misread it, thinking he just has to return to the start point at the end.
Sorry to introduce a red herring.
Anyway in my version I was wrong that the intermediate points are about a third of the way along the walls from the south: they’re precisely two thirds of the way.
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Hugh Casement 4:50 pm on 29 November 2020 Permalink |
Oops! Got it wrong again. Precisely one third, six and 2/3 metres.
I’m really not awake today. Feel free to delete all my posts on this subject.
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Hugh Casement 2:31 pm on 29 November 2020 Permalink |
Put t = tan(72°) = sqrt{5 + 2 sqrt(5)}.
Then if the midpoint of the south wall has coordinates (0,0) and the north vertex is at (0, 10t),
the lower right wall is y = t(x – 10) and the lower left wall is y = t(10 – x).
Then a bit of calculus — or some trial & error by program — shows that the intermediate points that he touches are at roughly (±12.060113, 6.340375) and the length of his path is about 81.75134 metres. If he starts at a point off the line (on either side) his overall path is longer.
Of course his chalk line could start at any vertex and go via the midpoint of the opposite side. He could draw five such overlapping quadrilaterals, and stick to any of them, swapping between them at will, so long as when the whistle blows he is not confused as to which line he is on.
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