All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Unit lengths from movable points to unit triangles. (Posted on 2009-09-14)
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)

 No Solution Yet Submitted by Jer No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 intervals to 2 digits of precision (more to come) | Comment 5 of 11 |

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.

 Posted by Daniel on 2009-09-14 22:31:07

 Search: Search body:
Forums (0)