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Sunday Times Teaser 3054 – Discs A Go-Go

by Brian Gladman on April 2, 2021

by Howard Williams

Published Sunday April 04 2021 (link)

My kitchen floor is tiled with identically-sized equilateral triangle tiles while the floor of the bathroom is tiled with identically-sized regular hexagon tiles, the tiles being less than 1m across. In both cases the gaps between tiles are negligible. After much experimenting I found that a circular disc dropped at random onto either the kitchen or bathroom floor had exactly the same (non-zero) chance of landing on just one tile.

The length of each side of the triangular tiles and the length of each side of the hexagon tiles are both even triangular numbers of mm (ie, of the form 1+2+3+…).

What are the lengths of the sides of the triangular and hexagonal tiles?

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  1. Brian Gladman permalink

    The diagrams above illustrate that if the disc is to fall entirely within a single hexagonal (triangular) tile, it centre has to fall within the inner green areas shown. Hence the probability of this happening is simply the ratio of the areas of these inner and outer shapes.

    If the radius of the circular disc is \(r\) and the sides of the regular hexagon and equilateral triangle are \(h\) and \(t\) respectively, inspection of the small lower right triangles in each shape shows that the side lengths of the inner hexagon and the triangle are respectively \(h\,-\, 2r/\sqrt{3}\)  and \(t\,-\,2\sqrt{3}r\).   With \(p_h\) and \(p_t\) as the probabilities of the disc falling entirely within a tile, these become:\[p_h = \left(\frac{h \;- 2r/\sqrt{3}}{h}\right)^2=\left(1-\frac{2r}{\sqrt{3}h}\right)^2\]
    \[p_t = \left(\frac{t\; – 2\sqrt{3}r}{t}\right)^2=\left(1-\frac{2\sqrt{3}r}{t}\right)^2\]
    Equating these shows that equal probabilities requires that \(t=3h\). (see [1])
    Since the sides are triangular numbers, let the hexagon’s and the triangle’s numbers be \(h(h+1)/2\) and \(t(t+1)/2\) respectively (this is a a change of notation from that above). We then have: \[t(t+1) = 3h(h+1)\] which, after rearrangement, can be put in the form:\[(2t+1)^2\; -\; 3(2h+1)^2=-2\]This is a quadratic Diophantine equation for which there are standard techniques for solution which can be used to show that, if we can find one solution, an infinite series of solutions can be produced using two recursions:\[t_{n+1}=2 t_{n}+3h_{n}+2\]\[h_{n+1}=t_{n}+2h_{n}+1\] Since the equation has the solution \((t, h) = (0, 0)\) the sequence of solutions can easily be generated \[(t,h) = (0,0),(2,1),(9,5),(35,20),(132,76), …\] from which the possible side lengths are \[(0,0),(3,1),(45,15),(630,210),(8778,2926), …\]The solution is hence equilateral triangle tiles with sides of 630 mm and regular hexagon tiles with sides of 210 mm.

    [1] More generally for regular polyhedra with \(m\) and \(n\) sides \((s)\) equal probabilities requires that \[\tan(180/m)/s_m = \tan(180/n)/s_n\]Here is a programmed solution using the quadratic Diophantine equation solver in my number theory library (available here).

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