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%I #15 Apr 16 2021 00:13:31
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,2,0,0,2,0,0,0,3,1,1,0,0,1,3,0,0,0,2,
%T 1,0,6,0,4,4,2,1,0,0,1,0,0,6,0,2,8,6,2,0,1,2,0,2,0,9,0,0,2,0,13,1,0,4,
%U 0,3,0,3,5,10,11
%N a(n) is the number of concave integer quadrilaterals (up to congruence) with integer side lengths a,b,c,d with n=Max(a,b,c,d), integer diagonals e,f and integer area.
%C Without loss of generality we assume that a is the largest side length and that the diagonal e divides the concave quadrilateral into two triangles with sides a,b,e and c,d,e. Then e < a is a necessary condition for concavity. The triangle inequality further implies e > a-b and abs(e-c) < d < e+c.
%e a(66)=1 because the only concave integer quadrilateral with longest edge length 66 and integer area has sides a=66, b=55, c=12, d=65, diagonals e=55, f=65 and area 1650.
%t an={};
%t area[a_,b_,c_,d_,e_,f_]:=(1/4)Sqrt[(4e^2 f^2-(a^2+c^2-b^2-d^2)^2)]
%t he[a_,b_,e_]:=(1/(2 e))Sqrt[(-((a-b-e) (a+b-e) (a-b+e) (a+b+e)))];
%t paX[e_]:={e,0} (*vertex A coordinate*)
%t pbX[a_,b_,e_]:={(-a^2+b^2+e^2)/(2 e),he[a,b,e]}(*vertex B coordinate*)
%t pc={0,0};(*vertex C coordinate*)
%t pdX[c_,d_,e_]:={(c^2-d^2+e^2)/(2 e),-he[c,d,e]}(*vertex D coordinate*)
%t concaveQ[{bx_,by_},{dx_,dy_},e_]:=If[by dx-bx dy<0||by dx-bx dy>(by-dy) e,True,False]
%t gQ[x_,y_]:=Module[{z=x-y,res=False},Do[If[z[[i]]>0,res=True;Break[],If[z[[i]]<0,Break[]]],{i,1,4}];res]
%t canonicalQ[{a_,b_,c_,d_}]:=Module[{m={a,b,c,d}},If[(gQ[{b,a,d,c},m]||gQ[{d,c,b,a},m]||gQ[{c,d,a,b},m]),False,True]]
%t Do[cnt=0;
%t Do[pa=paX[e];pb=pbX[a,b,e];pd=pdX[c,d,e];
%t If[(f=Sqrt[(pb-pd).(pb-pd)];IntegerQ[f])&&(ar=area[a,b,c,d,e,f]; IntegerQ[ar])&&concaveQ[pb,pd,e]&&canonicalQ[{a,b,c,d}],cnt++
%t (*;Print[{{a,b,c,d,e,f,ar},Graphics[Line[{pa,pb,pc,pd,pa}]]}]*)],
%t {b,1,a},{e,a-b+1,a-1},{c,1,a},{d,Abs[e-c]+1,Min[a,e+c-1]}];
%t AppendTo[an,cnt],{a,1,75}
%t ]
%t an
%Y Cf. A340858 for trapezoids, A342720 for concave integer quadrilaterals with arbitrary area.
%K nonn
%O 1,17
%A _Herbert Kociemba_, Mar 19 2021
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