A342720 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) and integer diagonals e,f.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 3, 1, 2, 2, 3, 1, 4, 2, 4, 2, 5, 3, 7, 1, 2, 4, 3, 13, 7, 20, 12, 5, 3, 7, 10, 3, 8, 2, 14, 12, 10, 15, 17, 8, 11, 10, 20, 13, 15, 10, 45, 9, 18, 25, 46, 38, 18, 2, 25, 20, 30, 18, 32, 17, 32, 43

Offset: 1

Author

Herbert Kociemba, Mar 19 2021

Keywords

• nonn

Comments

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.

Examples

			a(15)=1 because the only concave integer quadrilateral with longest edge length 15 has a=15, b=13, c=13, d=15 and diagonals e=4 and f=24. a(20)=3 because there are three solutions (a,b,c,d,e,f): (20,13,15,18,9,26), (20,13,13,20,11,24) and {20,15,15,20,7,24}.
		

Crossrefs

Cf. A340858 for trapezoids, A342721 for concave integer quadrilaterals with integer area.

Programs

Mathematica

an={};
he[a_,b_,e_]:=1/(2 e) Sqrt[(-((a-b-e) (a+b-e) (a-b+e) (a+b+e)))]
paX[e_]:={e,0} (*vertex A coordinate*)
pbX[a_,b_,e_]:={(-a^2+b^2+e^2)/(2 e),he[a,b,e]}(*vertex B coordinate*)
pc={0,0};(*vertex C coordinate*)
pdX[c_,d_,e_]:={(c^2-d^2+e^2)/(2 e),-he[c,d,e]}(*vertex D coordinate*)
concaveQ[{bx_,by_},{dx_,dy_},e_]:=If[by dx-bx dy<0||by dx-bx dy>(by-dy) e,True,False]
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]
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]]
Do[cnt=0;
Do[pa=paX[e];pb=pbX[a,b,e];pd=pdX[c,d,e];
If[(f=Sqrt[(pb-pd).(pb-pd)];IntegerQ[f])&&concaveQ[pb,pd,e]&&canonicalQ[{a,b,c,d}],cnt++
(*;Print[{{a,b,c,d,e,f},Graphics[Line[{pa,pb,pc,pd,pa}]]}]*)],
{b,1,a},{e,a-b+1,a-1},{c,1,a},{d,Abs[e-c]+1,Min[a,e+c-1]}];
AppendTo[an,cnt],
{a,1,75}
]
an