Normat 59:2, 67–81 (2011) 67
Bisecting lines of convex regions
Ulf Persson
Matematiska Institutionen
Chalmers Tekniska Högskola och
Göteborgs Universitet
ulfp@chalmers.se
Introduction
Let us to fix ideas consider compact convex domains in the plane (although much
of what is going to be said will also hold in a more general case) and furthermore
let us call a line bisecting if it divides into two parts of equal area.
It is easy to see that there are plenty of bisecting lines. In fact if we fix the
direction of lines there will be a unique line among them which is bisecting. We
simply translate a given line and see what is say on the left of it in . The area of
which will vary monotonically and continuously as we move to the right, starting
with zero and ending up with the area of . Somewhere in between we will have
area
1
2
µ(). In fact all parallel lines can be written under the form ax + by = c
fixing (a, b) ”=0. And we can define
c
= {(x, y) œ :ax + by < c}. The function
µ(c)=µ(
c
) is strictly increasing and continuous varying from 0 to µ(
c
).
A similar argument shows that through each point there goes a bisecting line. We
simply rotate the line through the point. It will divide into a left and right part
say. When we have rotated a half turn, those two parts have been switched. So if
we consider the difference in area, which likewise varies c ontinuously its value have
changed signs, thus somewhere between it has to have been zero.
68 Ulf Persson Normat 2/2011
Note that if the point lies outside there will be a unique bisecting line through
it. On the other hand if it lies inside there may be many bisecting lines through it.
A natural question to ask is whether all the bisecting lines go through a single
point. This is obviously the case for a circle, or more generally an ellipse or a
rectangle. Medians are bisecting lines for triangles, and medians intersect in a
single point. Such a common point to all the bisecting lines would be a natural
candidate for being the true center of the re gion. However, most regions do not
have a center in fact we can easily establish the following.
Proposition: A point p is the center of a convex region iff it is on the midpoint
of each segment L fl (a so called bisecting segment) where L is a bisecting line.
PROOF: Consider a radius vector through the point p going to the boundary of
. Those are parametrized by the angles (◊) they make with the x≠axis. We can
then define a function l(◊) given by their lengths. As the radius vector rotates in
the positive direction (◊ increases) we can define the sets
+
◊
,
≠
◊
of the points in
ahead or behind the line. If we let µ(◊)=µ(
+
◊
) we see that
dµ(◊)
d◊
=
1
2
(l(◊) ≠ l(◊ + fi))
Thus the function µ is constant iff l(◊)=l(◊ + fi)
Thus a region with a center in this sense, would be invariant under reflection in
that center. In other words if the center is translated to the origin, the region would
be invariant under (x, y) ‘æ (≠x, ≠y). Clearly most regions do not have points with
this property, in particular triangles do not, because the only candidate for such a
point would be its center of gravity (the intersection of any two of its medians) and
for such it is easy to check that it is not the case. But what do we have instead?
So let us fix a bisecting line L and consider all other bisecting lines M.They
will all intersect L in a varying point LM which always will lie inside from a
remark made above. When M approaches L one will expect a limiting intersection
point, but aprioriit is not entirely clear that this will not depend on whether we
approach in the positive or negative direction. But it is actually easy to see that
the self-intersection will be the midpoint of the bisecting segment.
Proposition:The s elf-intersection point of a bisecting segment is its mid-point.
Normat 2/2011 Ulf Persson 69
PROOF:
rs
L
L’
P
∆θ
∆θ
Consider two bisecting segments L, L
Õ
which are close and intersect in P .
They cut the region into four parts,
two and two non-adjecent. The non-
adjecent parts have the same areas,
in particular the two areas defined by
the acute angle. Those areas will be
better and better approximated by
1
2
r
2
◊ and
1
2
s
2
◊ as ◊ æ 0 from
which we conclude that in the limit
as L
Õ
approaches L we get r = s
The total image of all the intersection points will make up a connected s egm ent
ˆ
L of L contained in the interior of . As we vary L there will be a region traced
out inside and whose boundary will be made up of the boundary points of the
ˆ
L. That region we will call the bisecting image and denote by B() (supressing
if the context make it obvious). The set B itself will be sub-divided in various
subregions B
k
k>0 in which through each point there will pass 2k +1 bisecting
lines. (In fact the whole plane minus B would make up a natural B
0
). If we talk
about 2k +1 distinct lines, those subsets B
k
will be open. There boundaries will
consist of points through which two of the bisections coalesce. One speaks about
ramification. As we cross such a boundary the number of bisecting lines either drop
or increase by an amount of two. The bisecting lines that stay distinct survive the
crossing, while the two that are made to coincide do not. On the other hand going
from the other direction, they will appear ’from nowhere’. The boundary points will
make up a curve, consisting of all the mid-points of bisecting segments. As those
are naturally parametrized by their directions we obtain a naturally parametrized
curve which we will denote by ˆB() and refer to as the bisecting locus, although
in general we do not exp ec t it to coincide with the topological boundary, even if
it will contain it. What kind of region B are we talking about? To find out let us
make an experiment and try to plot it in a very simple case, namely the triangular
region made up of the three inequalities x, y > 0 and x + y<1. But before that
it would be convenient to make a digression on duality.
The space of lines
A line can be given by an equation ax + by +c =0where not all the coefficients are
zero. The uninitiated reader may be puzzled by what is meant by the line c =0but
that will subsequently be explained. The main point to note is that the coefficients
(a, b, c) are not determined by the line, and multiple (ta, tb, tc) will do as well.
One says that (a, b, c) are homogenous co-ordinates. One formal way of presenting
this is to consider the space S = R
3
\{(0, 0, 0)}≥where two points are identified
iff they are multiples of each other in the sense above, i.e. (a, b, c) ≥ (ta, tb, tc)
with t ”=0. Geometrically we can think of this as the space of all lines in R
3
through the origin. Or in other words all the possible directions. Some of those
70 Ulf Persson Normat 2/2011
lines can be parametrized by points on the hyperplane H given by c =1simply by
considering the intersection of the line with H. Not all lines have intersection with
H, those that do, form a subset isomorphic with R
2
. The lines parallel to H form
a set isomorphic with a circle with antipodal points identified, which is actually
topologically a circle, and constitute what is called the line at infinity. One can
think of those as all the directions as given by lines in R
2
through the origin, and
be made up of equivalence classes of parallel lines, which can be thought of as the
intersections at infinity. By the use of perspective and parallel lines converging to a
vanishing point at the horizon, this notion can be given a very tangible meaning. In
fact one has a natural notion of line in S, namely given by the planes through the
origin. Two distinct lines will always meet at a point, as two distinct planes through
the origin will always meet in a line. The set S which we have defined is called the
real projective plane and denoted by RP
2
.Thelinec =0with co-ordinates (0, 0,c)
(or if you prefer normalized to (0, 0, 1)) corresponds to the line at infinity.
Now how s hould we represent the projective plane in a nice way? One way is to
normalize the co-ordinates such that a
2
+ b
2
+ c
2
=1by choosing t =
1
Ô
a
2
+b
2
+c
2
.
We then get a sphere with antipodal points identified. To represent it in the plane
we can consider a so called orthogonal projection which will map one hemisphere
to a circle. The boundary of the circle will have anti-podal points identified. The
picture we will get is the following.
Note that in this way we get a natural metric on the space of lines, by considering
them as antipodal points on a sphere. Each line in RP
2
will naturally have length
fi as they can be represented by half great circles with the end points (lying at
the line at infinity) identified. Now the projective plane can be thought in many
ways as a compactification of the plane, simply by choosing a distinguished line
and call that the line at infinity. In the picture above, the line at infinity will be
the boundary circle antipodally identified.
In this conte xt the notion of duality arises naturally. A line in R
2
gives rise to
a point in our S by construction. On the other hand a point P =(x
0
,y
0
) in R
2
gives rise to a line in S namely all the lines (a, b, c) going through the point P .The
co-ordinates for those lines satisfy the linear condition ax
0
+b
0
+c =0which is the
equation of a line. This sets up a principle of duality when extended to RP
2
and its
dual space RP
2ú
, two distinct points gives rise to a unique line, two distinct lines
gives rise to a unique point. It can all be elegantly expressed via the innerproduct
<x,y,z,a,b,c>= ax + by + cz.FixingA =(a, b, c) œ RP
2ú
we get a line via
0=<X,A>= ax + by = cz inside RP
2
, on the other hand fixing X œ RP
2
and
consider A as variable, we get a line in RP
2ú
.
Normat 2/2011 Ulf Persson 71
We may also consider another way of normalizing. Any line (except the line at infi-
nity) can be given in the form x cos ◊+y sin ◊ = c (corresponding to (cos ◊, sin ◊, ≠c).
We simply let ◊ be the angle the normal of the line makes with the x-axis
θ
θ
ψ
c
R
(X,Y)
In this way we can represent the space of lines by points (◊, c) œ S
1
◊R in a cylinder
with antipodal points (◊, c) and (◊ + fi, ≠c) identified. Note that the line at infinity
will not be covered by this parametrization. Now the lines through the point (X, Y )
gives rise to a sinusoidal curve on the cylinder, namely by c = R sin(◊ ≠ Â)
In fact this way of representing lines is given by the map projection obtained by
projecting from the center of the sphere to a c ylinder wrapped around the equator.
We could also note in passing that if we consider directed lines, i.e. lines with
an arrow, then they will form an S
2
the universal double cover of RP
2
.Inthe
cylindrical representation the arrow will be positively oriented with respect to the
normal
Finally we may note that we ac tually can make a canonical identification between
a space and its dual given the unit circle. This allows us to associate to each point
in R
2
alineinR
2
in the following way. If L does not go through the origin it can
be represented in a unique way as x cos ◊ + y sin ◊ = R with R>0 (note that in
our original notation R = ≠c). Similarly any point different from the origin can
be written uniquely in polar form (R cos ◊, R sin ◊) with R. Now associate to each
point (◊, R) the line (◊, 1/R). The points on the unit circle will be associated to the
corresponding tangent lines, while the origin will correspond to the line at infinity.
The whole thing can also be run backwards, associating a line to a point. If the
line intersects the circle, the corresponding tangents will intersect in a point which
will set up the duality. Finally one may consider the map (◊, R) ‘æ (◊, 1/R) which
is nothing but inversion in the unit circle.
Now to return to our convex regions . We may consider all the lines L which do
not intersect they form a subset of R
2
. The boundary of that subset will consists
of lines that intersect the boundary of in one point, if it is an extremal point, or
along a line-segment, and subdivide the plane into two parts, one of which contains
. Such lines are called supporting lines, and through each point outside there
will be exactly two such supporting lines. While through a point inside there
will be no supporting lines at all. The subset
ˆ
of lines intersecting will form a
convex set in S called the dual of . The bisecting lines will form a curve inside
ˆ
which we will refer to as the bisecting locus. The corresponding mid-points will
form a curve inside itself, a curve which we already have denoted by ˆB().
Those two curves will be in so called duality, meaning that either of them is gioven
72 Ulf Persson Normat 2/2011
by the other by associating their tangents. That indeed the bisecting segment is
tangent to the bisecting locus will be shown later.
The Triangle
Now let us explicitly identify all the bise cting lines of . It is clear that we will
have four cases depending on the slope k of the line,
In each of the four cases it is straightforward to compute the area of the sliced off
triangle and determine the value of c for which its area becomes
1
4
. By normalizing
to b Ø 0 we get the following table (where – = arctan
1
2
)
k Ø 1 ≠c = a +
1
2
a(a ≠ b)
3fi
4
Æ ◊ Æ fi cos ◊ +
1
2
1+
Ô
2cos(2◊ +
fi
4
)
≠
1
2
Æ k Æ 1 ≠c = b ≠
1
2
b(b ≠ a)
fi
2
≠ – Æ ◊ Æ
3fi
4
sin ◊ ≠
1
2
1 ≠
Ô
2 sin(2◊ +
fi
4
)
≠2 Æ k Æ≠
1
2
≠c =
1
2
ab – Æ ◊ Æ
fi
2
≠ –
1
2
Ô
sin 2◊
k Æ≠2 ≠c = a ≠
1
2
a(a ≠ b) 0 Æ ◊ Æ – cos ◊ ≠
1
2
1+
Ô
2cos(2◊ +
fi
4
)
We can now draw ’all’ the bisecting lines and get the following picture.
In the middle we get an amoeba-like
blob with three cusps. It is tempting
to guess that the corresponding cu-
spidal tangents are the three medi-
ans. The following picture seems to
bear it out.
Let us now fix a simple bisecting line such as the median x = y and see what is
going on along it.
Normat 2/2011 Ulf Persson 73
We see that the function has a single minimum and
a single maximum. Thus through each point on the
segment bounded by the extremal values there will
be three bisecting lines going through.
We can compute the locus of the points which
are given by bisectors intersecting themselves. If
the lines are parametrized by the upper part of
the circle, via C(◊)=c(cos ◊, sin ◊) (those func-
tions are given explicitly in table 1) then the points
will be parametrized by
x =
-
-
-
-
C(◊)sin◊
C
Õ
(◊) cos ◊
-
-
-
-
y =
-
-
-
-
cos ◊C(◊)
≠sin ◊C
Õ
(◊)
-
-
-
-
We will now make the necessary computations for the four cases and present below
x =1+
cos ◊+
Ô
2cos(◊+
fi
4
)
2
Ô
1+
Ô
2cos(2◊+
fi
4
)
y =
sin ׭
Ô
2 sin(◊+
fi
4
)
2
Ô
1+
Ô
2cos(2◊+
fi
4
)
x =
≠cos ◊+
Ô
2 sin(◊+
fi
4
)
2
Ô
1≠
Ô
2 sin(2◊+
fi
4
)
y =1+
≠sin ◊+
Ô
2cos(◊+
fi
4
)
2
Ô
1≠
Ô
2 sin(2◊+
fi
4
)
x =
sin ◊
2
Ô
sin 2◊
y =
cos ◊
2
Ô
sin 2◊
x =1+
≠cos ◊≠
Ô
2cos(◊+
fi
4
)
2
Ô
1+
Ô
2cos(2◊+
fi
4
)
y =
≠sin ◊+
Ô
2 sin(◊+
fi
4
)
2
Ô
1+
Ô
2cos(2◊+
fi
4
)
If we use those formulas to map those points they will trace out the boundary ˆB.
Those arcs involved are actually arcs of hyperbolas. One c ase
is simple, namely the third, when it is easy to see that the
points satisfy xy =
1
8
. The others we can easily work out,
if not by making a linear transformation that transforms
our triangle into an equilateral one placed at the its center
at the origin. It is easy to see that all the objects we have
defined so far will transform in the same linear way, as a
bisecting line will be carried to a bisec ting line under any
linear transformation.
Before we will look at the general case of the bisecting locus we will make a slight
digression on curvature and singularities.
Curvature and Singularities
Given a curve C(t) it bends in general, meaning that the directions of the tangents
vary. We talk about the total curvature between two points on it as being given by
the angle made by the two tangents
74 Ulf Persson Normat 2/2011
θ
P
Q
Having the notion of a total curvature, we get the local by differentiation. I.e. we
let the two points P, Q approach each other and divide the total curvature with the
distance between and getting a limit. The angle made by the tangents is the same
as being given by the normals. If the points are close the lengths of the two normals
from their intersection point to the tangency points will likewise be close. In fact
chose a third point R be tween, and consider the unique circle going through the
three points P, Q, R. Its center will be given by the intersection of the two midpoints
normals to the segments PR and RQ. Hence there will be an approximating circle,
formed by three points coming together in analogy with two approaching points
defining a tangent line. The limit of the approximating circles will be called the
circle of curvature. Its center will be called the center of curvature, and its radius r
the radius of curvature, while 1/r will be denoted the curvature. This fits well with
the original definition, because the arc of an approximate circle will approximate
the length of the arc between the two points. The former being r◊ we are done.
The locus traced by the center of curvatures is interesting and can be thought of as
the dual of the curve in the dual space of lines given by the normals of the curve.
The circle of curvature is the circle who has the best fit of all the circles to the
curve. In fact, as seen by the construction, a curvature circle touches the curve in
three coinciding points. Would we compare the Taylor expansion of the circular arc
with that of the curve they would coincide up to and including the second order
term. If we use instead all quadric curves, we have instead of three parameter, five
parameters, and would be able to get approximations that also included fourth
order terms.
To compute the curvature is easy in the case of an horizontal tangent. We can
look at the case of the parabola y = ax
2
. The tangent line at (x
0
,y
0
) is given by
y ≠ y
0
=2ax
0
(x ≠ x
0
) hence that of the normal will be y ≠ y
0
= ≠
1
2ax
0
(x ≠ x
0
)
and its intersection with the y-axis
1
2a
+ ax
2
0
and as x
0
æ 0 we see that the limit
will be
1
2a
which both gives the position of the center, and the radius of the circle.
Thus the curvature is given by 2a. An alternative is to consider a circle of radius r
tangent to the x-axis at the origin. It has the equation x
2
+(y ≠ r)
2
= r
2
and we
can solve locally for y getting
y = r ≠
r
2
≠ x
2
= r(1 ≠
Ú
1 ≠
x
2
r
2
)=
x
2
2r
+ ...
Comparing with the parabolic arc y = ax
2
we find that it will have the same
curvature as the circle if a =
1
2r
, i.e. we get curvature 2a (and radius of curvature
Normat 2/2011 Ulf Persson 75
1
2a
). If f would be a function with the same Taylor expansion as the parabola, we
would have a =
1
2
f
ÕÕ
(0) and thus curvature is given by f
ÕÕ
(0).
Now the c urvature is intimately related to the second derivative which tells you
how fast the curve bends, or rather how far it is from being a straight line. However
the relation is not straightforward.
θ
θ
Rotating the picture to the left the new co-ordinates X, Y will be given in terms
of the old as X = x/ cos ◊ and Y = y cos ◊.Ify = –x
2
we get Y = y cos ◊ =
–
x
2
cos
2
◊
cos
2
◊ cos ◊ = cos
3
◊–X
2
. As tan ◊ = f
Õ
(x
0
) we get cos ◊ =
1
Ô
1+(f
Õ
(x
0
))
2
and
hence the curvature in general is given by
f
ÕÕ
(x
0
)
(1+(f
Õ
(x
0
))
2
)
3
2
.
Now let us consider a piece of curve which is given in polar-coordinates by r(◊)
a function which tends to zero as ◊ æ 0. Clearly the origin will be a point on
the curve, but what type of point? Assume that lim
◊ æ0
r(◊)=A>0.Wethen
get (x
Õ
(0),y
Õ
(0)) = (A, 0) thus the curve is tangent to the x-axis. Writing x ≥
A◊ cos ◊, y ≥ A◊ sin ◊ we can for small values of ◊ write y = A◊◊ =
1
A
(A◊)
2
thus
locally y =
1
A
x
2
+ ... i.e. the curve approaches its tangent quadratically.
However if A would abruptly change sign as ◊ becomes negative, then the curve
would back up on itself and stay on the right half plane. We will have occasion to
return to this phenomenon later.
Finally what happens if lim
◊ æ0
r◊
◊
2
=0. The same
argument as before yields y = A◊
2
sin ◊ = A◊
3
=
1
Ô
A
(A◊
2
)
3
2
=
1
Ô
A
x
3
2
. This is the cuspidal singu-
larity given by the polynomial equation Ay
2
= x
3
.
The two branches approach the tangent much
slower than in the quadratic case. In fact it is a bit
hard to get an intuitive approach of how it actually
approaches zero.
76 Ulf Persson Normat 2/2011
The picture to the left shows a comparison betwe-
en a cuspidal approach to the tangent and a regu-
lar quadratic. While the curvature of the parabola
approaches a finite non-zero value as we go to the
origin, for the cuspidal curve it goes to infinity,
which is however quite hard to see from a picture.
Finally we will discuss how one can interpolate between two tangents with given
tangent-points. In other words to find a convex arc that joins P
0
and P
1
below,
while being tangent to the corresponding lines at those points.
P
0
P
1
Write the lines as x cos ◊
0
+ y sin ◊
0
= c
0
and x cos ◊
1
+ y sin ◊
1
= c
1
respectively,
and let us give the points as P
0
= c
0
(cos ◊
0
, sin ◊
0
)+r
0
(≠sin ◊
0
, cos ◊
0
) and P
1
=
c
1
(cos ◊
1
, sin ◊
1
)+r
1
(≠sin ◊
1
, cos ◊
1
). It is then natural to consider the curve C(◊)=
c(◊)(cos ◊, sin ◊)+r(◊)(≠sin ◊, cos ◊) with the obvious boundary conditions c(◊
i
)=
c
i
,r(◊
i
)=r
i
. Taking the derivative of C we get (≠(c + r
Õ
)sin◊ +(c
Õ
≠r) cos ◊, (c +
r
Õ
) cos ◊ +(c
Õ
≠ r)sin◊) from which we derive the condition c
Õ
= r. Now we can
take any function r that interpolates between r
0
and r
1
which will determine c up
to an additive constant. If r is merely linear we will not have enough freedom to
fit c but if we take r quadratic it will always work out uniquely. We may also use
the freedom in choosing the origin of our presentation. A shift to (A, B) will result
in ˜c(◊)=c(◊) ≠ A cos ◊ ≠ B sin ◊, ˜r(◊)=r(◊)+A sin ◊ ≠ B cos ◊.Inthiswayby
appropriate choices of A, B we may arrange to get a linear choice for ˜r. In fact a
linear choice is given by r(◊)=
r
0
+r
1
2
+K(◊ ≠
◊
0
+◊
1
2
) from which follows that c(◊)=
k +
r
0
+r
1
2
◊ +
K
2
(◊ ≠
◊
0
+◊
1
2
)
2
and hence that we need c(◊
1
) ≠c(◊
0
)=
r
0
+r
1
2
(◊
1
≠◊
2
)
which can easily be effected given the freedom of (A, B).
In the picture below we have done this for two choices of P
0
. Note that what
we are interpolating are the tangent rays, not the tangents. In the smooth case the
angular change is given by the acute angle in the intersection, while in the alter-
native case with a cusp, we have an angular change that is greater corresponding
to the obtuse angle.
Normat 2/2011 Ulf Persson 77
P
0
P
1
P’
0
We may remark that the curvature of C(◊) is easily computed to be (c + r
Õ
) and
that the locus of centers of curvature is given by ≠r
Õ
(cos ◊, sin ◊)+r(≠sin ◊, cos ◊)
The bisecting locus
We would now like to study the bisecting locus more carefully. The sequence of
pictures below, suitably interpreted will give the key to many of its properties.
O
P
E’
1
E
1
E
2
E’
2
r
ψ
1
ψ
2
∆θ
0
P
Q
h
∆θ
∆θ/2
So let us fix one bisecting segment L = E
1
E
2
which will make the angles Â
1
,Â
2
respectively with the boundary ˆ of the region and let us denote the lengths
of OE
1
and OE
2
with r. Now take a neighboring bisecting segment L
Õ
, making
78 Ulf Persson Normat 2/2011
the angle ◊ with L. Its length up to first order will be given by the sum of the
lengths of OE
Õ
1
= r(1 + ◊ tan Â
1
) and OE
Õ
2
= r(1 + ◊ tan Â
2
). If we take the
midpoint of the segment we will not in general land in O but in P and the length
of OP is
1
2
r(tan Â
1
≠ tan Â
2
). However this is not the entire story, as the areas of
the triangles OE
1
E
Õ
1
and OE
2
E
Õ
2
are not the same, as they would b e if L
Õ
was a
bisecting segment. We will have to make a parallel shift of L
Õ
and in this picture
to the right. The excess of the first over the second is given by the difference of
the areas of the top and bottom parts respectively i.e.
1
2
r
2
(tan Â
1
≠ tan Â
2
)(◊)
2
.
Thus if we want to shift with an amount h we will get
h(r(1 + tan Â
1
)+r(1 + tan(Â
2
)=
1
2
r
2
(tan Â
1
≠ tan Â
2
)(◊)
2
.
From this it follows that h =
1
4
r(tan Â
1
≠ tan Â
2
)(◊)
2
. As h is of se cond order in
◊ this shift will not affect the length of L
Õ
to the first order. The length k of QP
will be given by k sin ◊ = h thus to be given by k =
1
4
r(tan Â
1
≠ tan Â
2
)◊.The
point P on L
Õ
will hence be moved that amount further up (in our case). Finally
if we draw the segment PO this will have length
1
2
r(tan Â
1
≠tan Â
2
)◊ and make
the angle
◊
2
with L.
From this we can draw a number of conclusions.
Conclusion If Â
1
”= Â
2
then O is a smooth point of ˆB and the bisecting line L is
tangent to ˆB at O. Furthermore it lies on the concave side of the curve. The speed
of the natural parametrization of ˆB is given by
1
2
r|(tan Â
1
≠ tan Â
2
)| Finally the
curvature at the point O is given by
2
r
(tan Â
1
≠ tan Â
2
)
≠1
.
PROOF: By construction as P approaches O the angle it makes with the line L
goes to zero, this line will hence be the tangent. As ◊ switches sign the p oint P
will still stay at the same side. We note also that the point Q on the tangent line L
sits inside B because the latter is made up of such intersection points. The formula
for the speed is immediate. As for the curvature we refer to a previous section for
the following approach.
If we intersect the parabola with a line of slope tan(◊/2) through the origin the
distance to the other intersection point will be of the order
1
2a
. Comparing with
what we have above we conclude that the radius of curvature will be
1
2
r(tan Â
1
≠
tan Â
2
) hence the formula.
We also note that if the end point belongs to a corner, the length of the radius
vector will so to speak switch signs, and thus we will get two tangent branches.
It is easy to work out the conditions for this, using the methods above. It has to
do with the acuteness of the corner, and how slope at the opposing point. Precise
statements are safely left to the interested reader.
We note that the curve ˆB cannot enclose a convex re gion B as in that case for
each direction ◊ there will be two tangents, but each direction only contains one
bisector. In fact ˆB will consist of a number of concave arcs joined together by
singularities, and have the property that no two tangents are parallel. We cannot
build an enclosing curve with just two concave arcs, at least three is needed as in
our example of the triangle. But more than three will not work either if we insist
on a simple curve, as the following picture explains.
Normat 2/2011 Ulf Persson 79
One should think of the points AB etc as
being joined by a concave arc tangent at the
appropriate lines. The total curvature will then
be given by “ etc. The total contribution – +
— + “ is then seen as fi the maximum allow-
ed. If a concave arc is split up into two parts
joined by a cusplike singularity, the total cur-
vature will increase.
A more complicated curve is given by the following example of a regular pentagon
Finally we note that bisecting segments are not the only chords that can be as-
sociated to a convex region. As we have already noted, for each direction there are
two supporting lines. They touch the boundary at two points that c an be joined.
That s egm ent may or may not be a bisecting one but if it is then it corresponds to
a critical one, i.e. when the direction of the boundary coincide at opposite points,
and conversely every critical bisecting segment occurs in this way. Furthermore
between two supporting lines there will be a bisecting one, as well as one which lies
half-way between, and which will sometimes also be a bisecting segment. However,
if we trace the self-intersection points there will sometimes be jumps, when the
boundary of the region has segments and corners.
Now let us consider the singularities that occur on the curves.
Singularities
We assume that the bisecting segment is vertical and orthogonal to the boundary
ˆ at its endpoints, and those arcs are given locally as r(1 ≠ ax
2
+ ...) and
r(1 ≠bx
2
+ ...). Close bisectors making a small angle ◊ will have radii of lengths
r(1 + –◊
2
) and r(1 + —◊
2
) respectively. How do we compute those lengths? For
that purpose we look at the picture below.
80 Ulf Persson Normat 2/2011
O
P
ax
2
r
r(1+αθ
2
+...)
θ
Making the ’Ansatz’ above we get
x = r(1 + –◊
2
+ ...)sin◊
and the equation
cos ◊r(1 + –◊
2
+ .....)=r ≠ a(sin ◊r(1 +
–◊
2
+ ...))
2
or (1 ≠
◊
2
2
+ ...)(1 + –◊
2
+ ...)=
1 ≠ a◊
2
r(1 + ...)
from which follows (– ≠
1
2
)=≠ar,thus– =0is
equivalent with a =
1
2r
i.e. the circle of radius r is
the circle of curvature to the parabolic arc.
In order to compute the shift h as in the previous
exercise, we need to be able to compute areas of the
shaded region (which in this case is the defect to
make up a circular segment). Those are given by a
simple integration
s
”◊
0
r
2
–◊
2
d◊ =
1
3
r
2
–◊
3
.
As before we need to have 2rh =
1
3
r
2
–◊
3
≠
1
3
r
2
–◊
3
the shift will be given by
h/ sin ◊ =
1
6
r
2
(– ≠—)◊
2
) while the shift from the extended lengths will be given
by
1
2
r(– ≠ —)◊
2
. The important thing is that the length of the radius vector of
angle ◊ will no be quadratic and not linear in the angle. This corresponds to an
approach given by r(◊) ≥ ◊
2
in polar co-ordinates, thus x ≥ ◊
2
and y ≥ ◊
3
.In
other words a cusp y
2
= x
3
. Note however that in the case of our triangle, the
singularties will be so called tacnodal, due to the corners of ˆ and correspond to
two tangent branches.
The Symmetrizer
We may translate the bisecting segments such that their midpoints land on the
origin. What we get is a star-shaped region and which is symmetric with respect
to reflec tion in the center. Let us give it a name -the symmetrizer S() of .In
the case of our triangle we get the following non-convex region
Normat 2/2011 Ulf Persson 81
What is the area of S()? There is a natural map from S
ú
= S() \{(0, 0)}
to \ ˆB which ’blows up’ the origin to the curve ˆB. Any point p œ S
ú
lies
on a unique ray through the origin, which corresponds to a unique point P œ ˆB
which defines a translation T which carries (0, 0) to P .Thendefine(p)=T (p).
Intuitively this map is area-preserving. Locally it can be well-approximated by so
called ’shearing maps’. Those are maps that for some choice of co-ordinates can be
written under the form (x, y) æ (x, y + h(x, y)) for some function h which need not
even be continuous, let alone differentiable, only me asurable n general. What that
map does is to translate s mall squares without overlapping or te aring. In the case
h is differentiable it is straightforward to compute its Jacobian and check that it is
identically equal to one. In our case we can also write down explicitly by using
a parametrization „
0
,„
1
of ˆB by writing down
(x, y) ‘æ (x + „
0
(arctan
y
x
),y+ „
1
(arctan
y
x
))
The Jacobian will be given by the determinant
-
-
-
-
-
1+„
Õ
0
(arctan
y
x
)
≠y
x
2
+y
2
„
Õ
1
(arctan
y
x
)
x
x
2
+y
2
„
Õ
0
(arctan
y
x
)
≠y
x
2
+y
2
1+„
Õ
1
(arctan
y
x
)
x
x
2
+y
2
-
-
-
-
-
=1+
≠y„
Õ
0
+ x„
Õ
1
x
2
+ y
2
=1
as the bisecting curve is tangent to the defining line.
Note that the map defined has a strange property as it approaches the origin.
Although the quotient |(p) ≠ (q)|/|p ≠ q| goes to infinity as p, q approaches the
origin, the map is nevertheless area-preserving.
Now the following should be obvious, where µ denotes area.
Proposition µ(S()) = µ() + 2
q
k
kµ(B
k
())
PROOF: We simply remark that on the inverse image
≠1
(B
k
) the map is 2k+1 : 1.