A293260 - OEIS

%I #15 Oct 23 2017 21:13:21

%S 0,0,0,0,0,0,0,1,0,2,0,12,0,9,0,14,0,75,0,26,0,35,0,110,0,54,0,57,0,

%T 436

%N Number of adventitious quadrangles (convex, noncyclic, not kite) such that Pi/n is the largest number that divides all the angles.

%C "All the angles" in the title means any angle formed by 3 vertices. There are 8 nonoverlapping angles in total.

%C Consider convex quadrilateral ABCD. Let a,b,c,d,e,f,g,h be the angles ABD,DBC,BCA,ACD,CDB,BDA,DAC,CAB, respectively. A quadrangle is adventitious if all these angles are rational multiples of Pi.

%C Cyclic quadrilaterals have properties a=d, b=g, c=f, e=h, thus making the adventitious case trivial.

%C Kites have properties a=b, c=h, d=g, e=f, thus making the adventitious case trivial.

%C Some properties:

%C   1. b+c = f+g := x, d+e = h+a := y, x+y = Pi.

%C   2. sin(a)sin(c)sin(e)sin(g) = sin(b)sin(d)sin(f)sin(h).

%C   3. In an adventitious quadrangle, swapping angles (b,c) with (f,g) or (a,h) with (d,e) gives another adventitious quadrangle.

%C From empirical observation, it seems that no adventitious quadrangles exist for odd numbers n. For example, take n=9: 180 degrees/9 = 20 degrees, and forming a quadrangle in which all angles are multiples of 20 degrees is impossible (proven by brute force). It seems to hold for all odd numbers n.

%C Perhaps the most famous case is Langley's problem (where n=18).

%H Kevin S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath734/kmath734.htm">Adventitious Solutions</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Langley%E2%80%99s_Adventitious_Angles">Langley’s Adventitious Angles</a>

%e a(8) = 1 because there is one quadrangle where all angles are divisible by 180/8 = 22.5 degrees.

%e   a=90, b=45, c=22.5, d=45, e=67.5, f=45, g=22.5, h=22.5.

%e a(10) = 2 (180/10 = 18):

%e    72  54  18  36  72  36  36  36

%e   108  36  18  54  72  36  18  18

%e a(12) = 12 (180/12 = 18):

%e    75  30  45  45  60  60  15  30

%e    75  60  15  45  60  30  45  30

%e    90  30  30  45  75  45  15  30

%e    90  45  15  45  75  30  30  30

%e    90  45  30  45  60  60  15  15

%e    90  45  30  75  30  45  30  15

%e    90  60  15  45  60  45  30  15

%e   105  30  15  30 105  30  15  30

%e   105  30  30  75  45  45  15  15

%e   105  45  15  30  90  45  15  15

%e   105  45  15  75  45  30  30  15

%e   120  30  15  60  75  30  15  15

%t Remove[f];

%t f[n_Integer] := Do[

%t       If[A == B < n/2 - C, Continue[]];(* if A == B then C >= H *)

%t       If[A == B == n/2 - C || C == D == n/2 - B, Continue[]];(* remove kite *)

%t       F = n/\[Pi] ArcTan[(Sin[d] Sin[a + b])/(Sin[a] Sin[c] Sin[e]) -

%t            Cot[e], 1] /. Thread[{a, b, c, d, e} -> \[Pi]/n {A, B, C, D, E}];

%t       F = Round[F, 10^-6];

%t       If[A < F, Continue[]];

%t       If[GCD[A, B, C, D, E, F] != 1, Continue[]];

%t       If[A == E && B < F, Continue[]];(* if A == E then B >= F* )

%t       If[A == F && B < E, Continue[]];(* if A == F then B >= E* )

%t       {A, B, C, D, E, F, B + C - F, D + E - A} // Sow;

%t       , {A, n/4 // Ceiling, n - 3}

%t       , {B, Max[1, n - 3 A + 2], Min[A, n - A - 2]}(* B <= A and C < A and H < A *)

%t       , {C, Max[1, n - 2 A - B + 1], Min[A - 1, n - A - B - 1]}(* C < A and H < A *)

%t       , {D, n - A - B - C, A - 1}(* D < A and E <= A *)

%t       , {E, {n - B - C - D}}

%t       ] // Reap // Last // If[# == {}, {}, # // Last] &;

%t Table[f[n] // Length, {n, 30}]

%t (* 180/n f[n] /. n -> 18 // TableForm *)

%K nonn,more

%O 1,10

%A _Albert Lau_, Oct 04 2017