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Maximum Angle between Perpendicular Bisectors


Date: 06/11/99 at 06:09:03
From: Brecht Hommez
Subject: The maximum angle between two perpendicular bisectors

Hello Dr. Math,

First of all, congratulations on your excellent website. It has 
already been a great help to me (a Ph.D. student in high energy 
physics).

Right now, I have a very annoying problem and I wonder if there is a 
solution.

Suppose you have two circles with the same radius and they do not 
intersect.

Step 1:
Choose a point on each circle and consider the line segment formed by 
these two points. Then construct the perpendicular bisector of this 
line segment. 

Step 2:
Repeat step 1, but with different choices of points on each of the two 
circles.

Thus then we have two perpendicular bisectors.

Now, my question is: which four points do I have to choose (two points 
in step 1 and two points in step 2), to obtain the maximum angle 
between the two perpendicular bisectors?

I really hope that you can help me - after days of thinking about 
this, I couldn't find the answer.

Thanks,
Brecht Hommez


Date: 06/11/99 at 16:49:22
From: Doctor Peterson
Subject: Re: The maximum angle between two perpendicular bisectors

Hello, Brecht! I'm glad we've been helpful.

We can simplify the problem considerably by ignoring the perpendicular 
bisectors, because the angle between the perpendiculars is the same as 
the angle between the segments themselves. Then we can consider only 
the slope of each segment, ignoring details like where they intersect. 
This makes it fairly obvious that the maximum angle will occur when 
the two segments are tangent to the two circles, one sloped as far as 
possible to the left and the other as far as possible to the right:

               ***********
            ***           ***
         ***                 ***
        *                       *
       *                         *
      *                           *
     *              A              *
     *              +              *
     *         /    |    \ r       *
      *    /        |        \    *
     P3+           Q+------------+P1
        *           |     x     *
         ***        |        ***
          \ ***  d/2|     *** /
           \   ***********   /
            \       |       /
             \      |y     /sqrt((d/2)^2 - r^2)
              \     |     /
               \    |    /
                \   |   /
                 \  |  /
                  \ | /
                   \|/
                    *O
                   /|\
                  / | \
                 /  |  \
                /   |   \
               /    |    \
              /     |     \
             /      |      \
            /       |       \
           /   ***********   \
          / ***     |     *** \
         ***        |        ***
        *           |           *
     P2+            |            +P4
      *    \        |        /    *
     *         \    |    /         *
     *              +              *
     *              B              *
      *                           *
       *                         *
        *                       *
         ***                 ***
            ***           ***
               ***********

I don't know in what form you need to find the pairs of points, but we 
can see that the segments intersect at the midpoint of the segment 
connecting the centers of the circles, and the maximum angle will be

    2 arcsin(2r/d)

where r is the radius of the circles and d is the distance between 
their centers. The four points will be symmetrical about this point 
and about the connecting line.

If you need x and y coordinates of the points, and the circles are in 
something like the position I've shown, you can get them from these 
similar triangles:

Using APQ and AOP:

   PQ   OP     x    sqrt((d/2)^2 - r^2)
   -- = --    --- = -------------------
   PA   OA     r          d/2

Using POQ and AOP:

   OQ   OP             y            sqrt((d/2)^2 - r^2)
   -- = --    ------------------- = -------------------
   OP   OA    sqrt((d/2)^2 - r^2)         d/2


Let me know if I've misunderstood the problem or if there is more you 
need.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 06/14/99 at 05:45:38
From: Brecht Hommez
Subject: Re: The maximum angle between two perpendicular bisectors

Hello Dr Peterson,

Thanks for the quick answer to my problem, but ... I 'm still having 
trouble. In fact, the problem I sent to you was the derivative problem 
of the original problem. The solution of the derivative problem does 
not help me further with the original problem and therefore I think it 
may be wise to send the original problem to you:

Suppose you have three circles (call them circle 1, circle 2, and 
circle 3) with the same radius and they do not intersect. 

Step 1:
Choose a point on each of two different circles (e.g. circle 1 and 
circle 2) and consider the line segment formed by these two points. 
Then construct the perpendicular bisector of this line segment. 

Step 2:
Repeat step 1, but with two circles not the same as the circles used 
in step 1 (e.g. for step 2 you choose circle 2 and circle 3)

Step 3:
Thus we have two perpendicular bisectors. We calculate the point where 
they intersect. Let's call this point P.

We repeat these 3 steps for a lot of points on each circle and each of 
the 3 combinations of different circles in step 1 and step 2:

Combination 1:
   Step 1: circle 1 and circle 2
   Step 2: circle 2 and circle 3

Combination 2:
   Step 1: circle 1 and circle 2
   Step 2: circle 1 and circle 3

Combination 3:
   Step 1: circle 1 and circle 3
   Step 2: circle 2 and circle 3

I have written a computer program that calculates all these points P 
and they seem to lie on a polygon with 6 vertices.

Now my question is: Is there a way to calculate analytically the 
(x,y)-coordinates of the 6 vertices of this polygon?

Thanks for considering my question,
Brecht Hommez


Date: 06/15/99 at 09:04:32
From: Doctor Peterson
Subject: Re: The maximum angle between two perpendicular bisectors

Hello again, Brecht.

Your problem is an interesting one, and I have spent some time 
thinking about it. I do not have all the details worked out, but I can 
at least tell you the general idea, which may be all you need.

If we consider first only one pair of circles, I have convinced myself 
that the locus of the perpendicular bisectors of segments joining 
pairs of points on the circles is the interior of a hyperbola whose 
semiaxes are r (the radius of the circles) and sqrt(d^2 - r^2) (half 
the length of the tangents from my earlier answer; the angle I gave is 
the angle between the asymptotes). If the circles of radius r have 
their centers at (-d, 0) and (d, 0), the equation of this hyperbola is

    x^2      y^2
    --- - --------- = 1
    r^2   d^2 - r^2



The circles A and B are symmetrical about the midpoint O, through 
which I have drawn a secant CDEF; for reasons of symmetry I can say 
that the perpendicular bisectors of CE and DF are the extrema of the 
set of all perpendicular bisectors of segments connecting the two 
circles with the same slope. I have drawn the locus of these two 
bisectors in blue and green respectively; the locus of point P, in 
red, is one branch of the hyperbola which forms the envelope of these 
lines. It can be seen that this is a hyperbola, because the difference 
BP - AP is

    (BD + DP) - (FP - AF) = (DP - FP) + (BD + AF)
                          = 0 + 2r

which is constant, because P is on the bisector and DP = FP.

Now if we draw these hyperbolas for all three pairs of circles, we see 
that the locus of points of intersection of pairs of lines will be the 
union of the pairwise intersections of the hyperbolic regions. That 
is, the bisectors for one pair of circles may meet those for another 
pair of circles where their two hyperbolic regions intersect, which 
will look like a curved quadrilateral. If I understand your 
explanation correctly, I would expect the result to look like a curved 
hexagram (star) rather than a hexagon. Here is the intersection of two 
regions, one between two horizontal lines (approximating the branches 
of one hyperbola), and the other between slanted lines (representing 
another hyperbola):

           /
          /
         /
    ----+---------+
       /........./
      /........./
     /........./
    +---------+----
             /
            /
           /

and here is the union of all three such quadrilaterals:

           +
          /.\
         /...\
    +---+-----+---+
     \./.......\./
      +.........+
     /.\......./.\
    +---+-----+---+
         \.../
          \./
           +

It appears that you are getting instead the intersection of the three 
hyperbolic regions, which may mean I am slightly misunderstanding your 
intention; perhaps you are choosing one point on each of the three 
circles and finding the intersection of the three bisectors (that is, 
finding the locus of circumcenters of triangles with one vertex on 
each circle). Then your locus will be either the "curved hexagon," or 
at least a subset of it; apparently it's the whole thing:

           +
          / \
         /   \
    +---+-----+---+
     \ /.......\ /
      +.........+
     / \......./ \
    +---+-----+---+
         \   /
          \ /
           +

As you can see, since you have not an actual polygon but a slightly 
curved region, it will take more than the coordinates of the vertices 
to define it fully (though they would define a hexagon that would 
contain your region, and appears to be a good approximation). But what 
I have given you should be enough to find those points and to define 
the curves if needed.

Now that I have worked this out, I am very curious as to the possible 
application of it! What are the circles, and why is this region 
important?

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Euclidean Geometry
High School Euclidean/Plane Geometry

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