POTW #14-16 A Distance Ratio
The Ratio Of Maximum To Minimum Distance Among Six Towns
John Snyder, FSA
December 6, 2013
17:30 EST
Problem
The smallest distance between any two of six towns is m miles. The largest distance between any two
of the towns is M miles. Show that M  m ³
3 . Assume that the land is flat.
Solution
A traditional proof would likely center around considering the shape of the convex hull containing the six
points. Since any convex hull can be tessellated into Delaunay triangles (a common technique in computational geometry), we would then focus on the knowledge that the critical triangle must always contain
an angle of at least 120°. Using this fact we could establish the required relationship by considering the
ratio of the maximum to minimum edge length of such a triangle.
It is more interesting to see if we can show that the required relationship is true by actually finding the
minimum ratio of M  m. To do this we’
ll first define 6 points in the plane.
In[1]:=
points = 88x1, y1<, 8x2, y2<, 8x3, y3<, 8x4, y4<, 8x5, y5<, 8x6, y6<<;
There are CH6, 2L = 15 combinations of these six points taken two at a time. We list these in the next cell.
In[2]:=
Out[2]=
pairs = Subsets@points, 82<D
888x1, y1<, 8x2, y2<<, 88x1, y1<, 8x3, y3<<, 88x1, y1<, 8x4, y4<<, 88x1, y1<, 8x5, y5<<,
88x1, y1<, 8x6, y6<<, 88x2, y2<, 8x3, y3<<, 88x2, y2<, 8x4, y4<<, 88x2, y2<, 8x5, y5<<,
88x2, y2<, 8x6, y6<<, 88x3, y3<, 8x4, y4<<, 88x3, y3<, 8x5, y5<<, 88x3, y3<, 8x6, y6<<,
88x4, y4<, 8x5, y5<<, 88x4, y4<, 8x6, y6<<, 88x5, y5<, 8x6, y6<<<
The 15 possible distances between the points are then as follows:
2
TownDistanceRatio1416.nb
In[3]:=
Out[3]=
d = EuclideanDistance žžž pairs  FullSimplify@ð, Element@Flatten@pointsD, RealsDD &
:
Hx1 - x2L2 + Hy1 - y2L2 ,
Hx1 - x3L2 + Hy1 - y3L2 ,
Hx1 - x4L2 + Hy1 - y4L2 ,
Hx1 - x5L2 + Hy1 - y5L2 ,
Hx1 - x6L2 + Hy1 - y6L2 ,
Hx2 - x3L2 + Hy2 - y3L2 ,
Hx2 - x4L2 + Hy2 - y4L2 ,
Hx2 - x5L2 + Hy2 - y5L2 ,
Hx2 - x6L2 + Hy2 - y6L2 ,
Hx3 - x4L2 + Hy3 - y4L2 ,
Hx3 - x5L2 + Hy3 - y5L2 ,
Hx3 - x6L2 + Hy3 - y6L2 ,
Hx4 - x5L2 + Hy4 - y5L2 ,
Hx4 - x6L2 + Hy4 - y6L2 ,
Hx5 - x6L2 + Hy5 - y6L2 >
Now we’
ll employ Mathematica’
s powerful global optimization function NMinimize. This function can
search using different methods, but we’
ll employ a random search with 210 = 1024 different initial seed
values. Using parallel kernels we can quickly solve the problem and select the solution giving the lowest
ratio of M  m. We do this in the next cell.
Max@dD
In[4]:=
sol = ParallelTableBNMinimizeB
, Flatten@pointsD,
Min@dD
Method ® 9"RandomSearch", "RandomSeed" ® RandomIntegerA91, 106 =E=F, 81024<F;
Min@
solP
All,
1TD
Out[5]=
1.90211
So we found a minimum ratio of M  m which is about 1.90211. The index number of this solution is:
In[6]:=
Out[6]=
p = Flatten@Position@solPAll, 1T, Min@solPAll, 1TDDDP1T
481
And the resulting 15 distances are:
In[7]:=
Out[7]=
d . solPp, 2T
80.617894, 0.617893, 0.617893, 0.617893, 0.617893, 0.726378, 1.1753,
1.1753, 0.726377, 1.1753, 0.726376, 1.1753, 0.726378, 0.726377, 1.1753<
Now let’
s plot the six points we found.
TownDistanceRatio1416.nb
In[8]:=
3
Graphics@8PointSize@LargeD, Red, Point@points . solPp, 2TD<,
Frame ® True, Background ® LightGrayD  Panel
0.2
0.0
-0.2
Out[8]=
-0.4
-0.6
-0.8
-0.2
0.0
0.2
0.4
0.6
0.8
It appears that the towns are arranged in a regular pentagon with five towns at the vertices and the sixth
town at the center of the circumscribing circle. If the edge length of the pentagon is set equal to one the
resulting distances are as shown in the following diagram.
4
TownDistanceRatio1416.nb
Using this arrangement the maximum distance between any two towns is:
5+
In[9]:=
5
max =
 FullSimplify
5-
5
1
I1 +
Out[9]=
5M
2
And the minimum distance between any two towns is:
-2
In[10]:=
min =
 FullSimplify
-5 +
5
1
I5 +
Out[10]=
5M
10
So the smallest possible ratio of M  m seems to be
solution.
1
2
J 5 + 5N » 1.90211 which matches the above
TownDistanceRatio1416.nb
max
In[11]:=
FullSimplifyB
F
min
1
I5 +
Out[11]=
5M
2
In[12]:=
Out[12]=
%  N
1.90211
Since the minimum ratio found using numerical optimization methods is greater than
reasonable to conclude that the ratio of M  n ³
3 . In fact,
3 , it seems
3 appears to be a rather poor lower
bound for the minimum ratio of M  m. The above analysis suggests that using
lower bound by about 9%.
3 understates the
3
 N
In[13]:=
1
2
Out[13]=
J5 +
5N
0.910593
I would conjecture that the true lower bound of the ratio of M  m is
proof of this fact would likely be very difficult.
1
2
J 5 + 5N , although a formal
5