Teaser 3203: Circuit maker
From The Sunday Times, 11th February 2024 [link] [link]
The racing chief tasked his designer for a new circuit. To create a 100:1 scale plan of the circuit the designer cut paper circles of 10cm radius into 12 equal sectors, then removed the top segments to create isosceles triangles with the two equal sides being 10cm. He arranged them on his desk to make a closed loop with no overlaps. Adjacent pieces always share the entire 10cm edge without overlap, but the short edges remain open.
The chief gave the following requirements. The start-finish line must be located on a straight section of at least 140m in length. Two circular viewing towers of 19m diameter must fit into the internal area, with one at either end of the circuit.
The designer minimised the internal area. How many triangles did he use?
[teaser3203]
Tony Smith, 
Hugo, 
S2T2 
Jim Randell 5:06 pm on 9 February 2024 Permalink |
I am assuming we place triangles together so that the track runs though the long edges of the triangles, and the short edges form the edge of the track (with barriers).
The smaller sides of the triangles would correspond to ≈5.18m (= 1 unit). So we would need 28 of them to allow a straight 140m long (27 is slightly too short). And a length of 4 of them would allow the viewing towers to fit in a central rectangle.
So this would require 2 × (28 + 4) = 64 triangles around the inside, and another 2 × (27 + 3) + 4 × 4 = 76 to complete a simple circuit. Making 140 triangles in total, and the enclosed area is 28 × 4 = 112 sq units (≈ 3001 sq m).
But is this simple design minimal?
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Jim Randell 5:18 pm on 9 February 2024 Permalink |
No, it isn’t.
I have found another design that allows the viewing towers to fit in a smaller central area (110.42 sq units ≈ 2959 sq m) (and also uses fewer triangles).
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Jim Randell 3:47 pm on 10 February 2024 Permalink |
And now I’ve found another design that I’m fairly convinced must have a minimal enclosed area for the viewing towers.
The total area enclosed is 623.2 sq m, and 567.1 sq m is taken up by the viewing towers, leaving just over 56 sq m of unused area.
However this does lead to a family of designs with the same enclosed area, but different numbers of triangles. Maybe we want the smallest number of triangles made from a whole number of circles?
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Mark Valentine 6:25 pm on 10 February 2024 Permalink |
Interesting. Can’t visualise how there are multiple triangle numbers for the minimal area.
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Jim Randell 9:26 am on 12 February 2024 Permalink |
I thought I would share a couple of the answers I have found (as links for now):
This is the smallest enclosed area I found (623 sq m), but the area is split into two parts and the track can be extended without altering the area, so I don’t think this is what the setter had in mind. [ link ]
And here is the smallest area I have found with a single enclosed area (about 820 sq m – which I calculated numerically). [ link ]
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Brian Gladman 2:03 pm on 12 February 2024 Permalink |
@Jim, Your first answer was my second published answer on the Sunday Times Teaser discussion group and was ruled inadmissible by Mark himself because the base sides of the triangular tiles must remain open.
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Jim Randell 3:21 pm on 12 February 2024 Permalink |
@Brian: I thought the method of constructing the circuit could have been explained better, and I wasn’t really sure what “the short edges remain open” was meant to mean in this context, although after a while I think I worked out how the track was to be constructed.
But as this first layout leads to multiple possible answers for the puzzle I thought it probably wasn’t what the setter was looking for, so I looked for a solution where the enclosed area was a simple polygon, and came up with the second layout. (Although I followed a systematic approach for this layout, I don’t know for sure that the enclosed area is minimal).
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Brian Gladman 4:37 pm on 12 February 2024 Permalink |
I got a bit bored with drawing layouts so I thought this morning about the vertical displacement in half wrapping a tower which is 1 + 2s + 2c where s and c are sin(30) and cos(30).
For the other half of the (partial) tower wrapping we want combinations of s ‘s and c’s that fall just short of cancelling out the vertical displacement on the other side.
The two smallest results are 4s + 2c == 0 and 2s + 3c = 0.13. The last is your second layout which hence seems the best that can be done if 0 is ruled out.
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Jim Randell 8:48 am on 18 February 2024 Permalink |
Here are more complete notes on the solution(s) I found:
I found a design that uses 180 triangles, and has hardly any unused enclosed space.
The enclosed space is 623.21 sq m, but 567.06 sq m of this is taken up by the viewing towers, leaving just 56.15 sq m of unused space.
The “spine” of the track uses two rows of 28 triangles, which gives a distance of 144.94m between the centre lines of the outside pair, and this is enough to fit the required straight.
Although if we are allowed to extend the straight slightly into the corners, we could reduce this to two rows of 27 triangles (to give a total of 176 triangles), with a distance of 139.76m between the centre lines of the outside pair. But as we are not trying to minimise the number of triangles, and the length of the straight is an absolute minimum the total of 180 triangles is a better fit to the requirements.
And if the designer used a whole number of circles to make the triangles, then 15 circles would give him 180 triangles (but we can expand the design by adding 4 triangles on the spine to use any larger multiple of 4 triangles while enclosing the same area).
A variation on this puzzle is where there is a single connected area in the middle of the track, and the solution above does not provide this. (And this may be what the setter of the puzzle originally had in mind – the requirement that the “short edges remain open” is apparently meant to restrict a triangle from being zero distance away from another triangle along the short edge).
And we can modify the solution above to give the following layout:
This uses 168 triangles (which is 14 circles) and gives an enclosed area of 820.13 sq m (unused area = 253.07 sq m). Although I don’t know if this is the smallest possible contiguous internal area, but if we were to extend the long straights the enclosed area would increase.
Solution: The track was constructed using 168 triangles.
The program I used to plot the circuits, and to calculate the internal area is here [@replit].
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Hugo 11:18 am on 18 February 2024 Permalink |
It is not usual for racing cars to go round sharp corners.
Unrealistic, even in Puzzleland. One of the worst Teasers in recent times.
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Tony Smith 8:50 am on 19 February 2024 Permalink |
Station Hairpin, Monaco Grand Prix.
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