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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from number_theory import Qde # triangle_side = 3 x hexagon_side # # triangle side: t.(t + 1) / 2 # hexagon side: h.(h + 1) / 2 # # t.(t + 1) / 2 = 3.h.(h + 1) / 2 ==> # # (2.t + 1)^2 - 3.(2.h + 1)^2 == -2 # Hence solve the quadratic diophantine equation # generate solutions for 2.t + 1 and 2.h + 1 t = Qde(3, -2) for ttp1, hhp1 in t.solve(limit=4): if ttp1 % 2 and hhp1 % 2: t, h = ttp1 // 2, hhp1 // 2 t_side, h_side = t * (t + 1) // 2, h * (h + 1) // 2 # both sides must be even (and non zero) if t and h and not (t_side % 2 or h_side % 2): print(f"Triangle side = {t_side} mm, Hexagon side = {h_side} mm") |