Given A=(a,0), B=(0,0), and C=(0,a)
Let f(a)=the total number of unit equilateral triangles XYZ that can be formed such that the lengths AX, BY, and CZ are all 1 unit.
Give a piecewise definition by intervals for f(a)
I used the following Mathematica code:
Dst[x1_,y1_,x2_,y2_]:=Sqrt[(x2-x1)^2+(y2-y1)^2];
Sols[a_]:=Length[FindInstance[{Dst[x1,x2,a,0]ƒú1,Dst[y1,y2,0,0]ƒú1,Dst[z1,z2,0,a]ƒú1,Dst[x1,x2,y1,y2]ƒú1,Dst[x1,x2,z1,z2]ƒú1,Dst[y1,y2,z1,z2]ƒú1},{x1,x2,y1,y2,z1,z2},Reals,100]];
Trans[a_]:=(
s=Sols[a];
stp=0.1;
x=a;
For[d=1,d„T5,d++,
x+=stp;
While[Sols[x]ƒús,
x+=stp;
];
x=x-stp;
stp=stp/10;
];
Return[{x,Sols[x+10*stp]}];
);
x=0;
t=0;
s=1;
prec=2;
xp=0;
While[s
„j0 ,
s=s-1;
stp=0.1;
For[i=1,i
„Tprec,i++,
x=x+stp;
While[Sols[x]
ƒús,
x=x+stp;
];
x=x-stp;
stp=stp/10;
];
Print["on [",xp,",",x,") F(a)=",s];
s=Sols[x+10*stp];
xp=x;
x=(Floor[10*x]+1)/10;
If[s
„j0,s=s+1];
];
which then gives the intervals as
on [0,0.49) F(a)=0
on [0.49,0.51) F(a)=4
on [0.51,0.63) F(a)=6
on [0.63,0.7) F(a)=8
on [0.7,1.06) F(a)=12
on [1.06,1.15) F(a)=8
on [1.15,1.41) F(a)=6
on [1.41,1.93) F(a)=2
on [1.93,2.) F(a)=4
on [2,Infinity) F(a)=0
I am now going to bump the precision up to 6 digits and let it run (could be a while). Hopefully this greater precission will assist us in determining possible exact values for the intervals.
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Posted by Daniel
on 2009-09-14 22:31:07 |
intervals to 2 digits of precision (more to come)
