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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

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

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
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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