A342723 a(n) is the number of convex 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.

0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 2, 1, 1, 3, 1, 2, 0, 0, 4, 3, 0, 0, 4, 7, 2, 0, 5, 2, 5, 0, 1, 0, 3, 4, 3, 4, 0, 6, 7, 3, 4, 0, 3, 4, 0, 0, 5, 0, 9, 10, 9, 3, 0, 5, 8, 0, 4, 0, 17, 4, 0, 9, 1, 19, 2, 0, 6, 2, 7, 0, 7, 7, 7, 23, 2, 8, 12, 0, 10, 0, 5, 0, 15, 27

Offset: 1

Author

Herbert Kociemba, Apr 25 2021

Keywords

• nonn

Comments

Without loss of generality we assume that a is the largest side length and that the diagonal e divides the convex quadrilateral into two triangles with sides a,b,e and c,d,e. The triangle inequality implies e > a-b and abs(e-c) < d < e+c.

Examples

			a(4)=1 is the smallest possible solution and is a rectangle with a=c=4, b=d=3, e=f=5 and area 12. a(24)=4 includes the smallest possible solution with all sides a,b,c,d different and a=24, b=20, c=15, d=7, e=20, f=25 and area 234. Furthermore there are three rectangles with a=24,b=7, a=24,b=10 and a=24,b=18.
		

Crossrefs

Cf. A340858 for trapezoids, A342720 and A342721 for concave integer quadrilaterals, A342722 for convex integer quadrilaterals with arbitrary area.

Programs

Mathematica

an={};
area[a_,b_,c_,d_,e_,f_]:=1/4 Sqrt[4e^2 f^2-(a^2+c^2-b^2-d^2)^2];
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*)
convexQ[{bx_,by_},{dx_,dy_},e_]:=If[(by-dy) e>by dx-bx dy>0,True,False]
(*define order on tuples*)
gQ[x_,y_]:=Module[{z=x-y,res=False},Do[If[z[[i]]>0,res=True;Break[],If[z[[i]]<0,Break[]]],{i,1,6}];res]
(*check if tuple is canonical*)
canonicalQ[{a_,b_,c_,d_,e_,f_}]:=Module[{x={a,b,c,d,e,f}},If[(gQ[{b,a,d,c,e,f},x]||gQ[{d,c,b,a,e,f},x]||gQ[{c,d,a,b,e,f},x]||gQ[{b,c,d,a,f,e},x]||gQ[{a,d,c,b,f,e},x]||gQ[{c,b,a,d,f,e},x]||gQ[{d,a,b,c,f,e},x]),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])&&(ar=area[a,b,c,d,e,f]; IntegerQ[ar])&&convexQ[pb,pd,e]&&canonicalQ[{a,b,c,d,e,f}],cnt++
(*;Print[{{a,b,c,d,e,f,ar},Graphics[Line[{pa,pb,pc,pd,pa}]]}]*)],{b,1,a},{e,a-b+1,a+b-1},{c,1,a},{d,Abs[e-c]+1,Min[a,e+c-1]}];
AppendTo[an,cnt],{a,1,85}]
an