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A293260
Number of adventitious quadrangles (convex, noncyclic, not kite) such that Pi/n is the largest number that divides all the angles.
0
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, 436
OFFSET
1,10
COMMENTS
"All the angles" in the title means any angle formed by 3 vertices. There are 8 nonoverlapping angles in total.
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.
Cyclic quadrilaterals have properties a=d, b=g, c=f, e=h, thus making the adventitious case trivial.
Kites have properties a=b, c=h, d=g, e=f, thus making the adventitious case trivial.
Some properties:
1. b+c = f+g := x, d+e = h+a := y, x+y = Pi.
2. sin(a)sin(c)sin(e)sin(g) = sin(b)sin(d)sin(f)sin(h).
3. In an adventitious quadrangle, swapping angles (b,c) with (f,g) or (a,h) with (d,e) gives another adventitious quadrangle.
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.
Perhaps the most famous case is Langley's problem (where n=18).
EXAMPLE
a(8) = 1 because there is one quadrangle where all angles are divisible by 180/8 = 22.5 degrees.
a=90, b=45, c=22.5, d=45, e=67.5, f=45, g=22.5, h=22.5.
a(10) = 2 (180/10 = 18):
72 54 18 36 72 36 36 36
108 36 18 54 72 36 18 18
a(12) = 12 (180/12 = 18):
75 30 45 45 60 60 15 30
75 60 15 45 60 30 45 30
90 30 30 45 75 45 15 30
90 45 15 45 75 30 30 30
90 45 30 45 60 60 15 15
90 45 30 75 30 45 30 15
90 60 15 45 60 45 30 15
105 30 15 30 105 30 15 30
105 30 30 75 45 45 15 15
105 45 15 30 90 45 15 15
105 45 15 75 45 30 30 15
120 30 15 60 75 30 15 15
MATHEMATICA
Remove[f];
f[n_Integer] := Do[
If[A == B < n/2 - C, Continue[]]; (* if A == B then C >= H *)
If[A == B == n/2 - C || C == D == n/2 - B, Continue[]]; (* remove kite *)
F = n/\[Pi] ArcTan[(Sin[d] Sin[a + b])/(Sin[a] Sin[c] Sin[e]) -
Cot[e], 1] /. Thread[{a, b, c, d, e} -> \[Pi]/n {A, B, C, D, E}];
F = Round[F, 10^-6];
If[A < F, Continue[]];
If[GCD[A, B, C, D, E, F] != 1, Continue[]];
If[A == E && B < F, Continue[]]; (* if A == E then B >= F* )
If[A == F && B < E, Continue[]]; (* if A == F then B >= E* )
{A, B, C, D, E, F, B + C - F, D + E - A} // Sow;
, {A, n/4 // Ceiling, n - 3}
, {B, Max[1, n - 3 A + 2], Min[A, n - A - 2]}(* B <= A and C < A and H < A *)
, {C, Max[1, n - 2 A - B + 1], Min[A - 1, n - A - B - 1]}(* C < A and H < A *)
, {D, n - A - B - C, A - 1}(* D < A and E <= A *)
, {E, {n - B - C - D}}
] // Reap // Last // If[# == {}, {}, # // Last] &;
Table[f[n] // Length, {n, 30}]
(* 180/n f[n] /. n -> 18 // TableForm *)
CROSSREFS
Sequence in context: A075533 A355783 A053814 * A095238 A167345 A292496
KEYWORD
nonn,more
AUTHOR
Albert Lau, Oct 04 2017
STATUS
approved