Classroom Tips and Techniques: Maple Meets Marden's Theorem
Robert J. Lopez
Emeritus Professor of Mathematics and Maple Fellow
Maplesoft
Introduction
The statement of Marden's theorem in Table 1 is taken from [1] where Dan Kalman gives
insightful comments and an efficient geometric proof.
Let
be a third-degree polynomial with complex coefficients, and whose roots
,
and
are noncollinear points in the complex plane. Let
be the triangle with vertices
at
, and . There is a unique ellipse inscribed in
and tangent to the sides at their
midpoints. The foci of this ellipse are the roots of
.
Table 1
Dan Kalman's statement of Marden's theorem
It is quite natural in Maple to start with three complex points, interpolate them with a (monic)
cubic polynomial, and to find the zeros of its derivative. Then, from these two foci and the
midpoints of the sides of the triangle whose vertices are the original three points, it is possible to
obtain the equation of the inscribed ellipse. However, if the initial three points are taken as
arbitrary complex points, the expressions that arise become exceedingly large and cumbersome.
Kalman suggests the clever simplification of selecting the initial three points as
, and
, with the third point in the upper half-plane. There is no loss of generality by this device,
since Kalman shows that under a linear transformation capturing translation, rotation, and scaling,
Marden's theorem holds for a triple
if and only if it holds for the transformed triple.
After observing how cumbersome the general expressions can become, we readily adopt the
consequences of Kalman's insight.
In Example 1 we start with three fixed points and use Maple's geometry package to obtain the
inscribed ellipse. This is followed by an interactive animation that shows the dynamic dependence
of the triangle, its midpoints, the inscribed ellipse, and its foci.
In Example 2, we start with three points suggested by Kalman, the interpolating polynomial, and
the zeros of its derivative as the foci. The indeterminates
and
present grave problems for the
geometry package, so we obtain the equation of the inscribed ellipse by interpolating a general
quadratic through the midpoints of the sides of the triangle and having its center fall on the
midpoint of the line segment connecting the foci. Tangency at the midpoints is then readily shown.
Finally, for a constructive approach to Marden's theorem, we start with the triangle, and construct
the ellipse tangent to its sides at the midpoints. The coordinates of the foci are obtained after
transforming the ellipse to standard form. The foci so determined are shown to be the same as the
zeros of the derivative of the polynomial interpolating the vertices of the triangle.
Example 1: A Specific Case
Let the triple of complex numbers
The polynomial
be given by
that interpolates these three points is
or
and
is
The zeros of
, namely, the foci of the inscribed ellipse, are then
It now becomes convenient to use Maple's geometry package for constructing the circumscribed
ellipse. However, this requires that the vertices of the triangle, and the foci, be expressed as points
in the package.
Define the three interpolated vertices as
.
Define the foci
points
and
and
.
as the geometry
Obtain
, respectively the
midpoints of sides
of the triangle
formed by the three interpolated vertices
.
The coordinates of the three midpoints are
found, and assigned the respective names
.
Define
, the triangle formed by the vertices
}, as an object in the geometry
package.
The ellipse command in the geometry package will define an ellipse from the foci and the
distance from one focus to a point on the ellipse to the other focus. This distance is actually
the length of the major axis. We take as the point on the ellipse the midpoint of one side of
triangle .
Define (as an object in the geometry
package) the ellipse
to be the ellipse
whose foci are
and
, whose major
axis has length , and which contains the
point
.
Obtain the Cartesian equation of ellipse
, and assign this equation the name
.
Let
be the center of ellipse . Its
coordinates are displayed to the right. Note
the division by 3 for each coordinate. Careful
inspection of the coordinates of the points
, lead to the suspicion that center of
falls on the centroid of triangle .
This suspicion is confirmed by finding the
coordinates of the centroid of triangle .
Table 2
The ellipse of Marden's theorem obtained with Maple's geometry package
Figure 1, made with the draw command from the geometry package, shows the triangle
and
the ellipse . The foci
and
are shown as green dots, the center
as a red dot, and the
midpoints as blue dots. The vertices of
are drawn as black squares.
Figure 1
Circumscribed ellipse of Marden's theorem
That ellipse
touches the triangle at the midpoints of the sides is shown by the result of
substituting the coordinates of the midpoints into the Cartesian equation for . Ellipse
contains each midpoint since its equation is satisfied by each such point.
To show that ellipse
is tangent to the sides of the triangle at the midpoints, we need the
derivative along the ellipse. This is obtained by Maple's implicitdiff command.
Table 3 compares the slope of each side of triangle
and the slope of ellipse
at the midpoint
of the corresponding side.
Table 3
its sides
=
=
=
=
=
=
Verification that ellipse
is tangent to triangle
at the midpoints of
Apparently, taking as foci of an ellipse the zeros of the cubic polynomial interpolating the three
vertices of a triangle in the complex plane, and forcing the ellipse to pass through the midpoint of
one side of the triangle, is enough to determine an ellipse that is tangent to all three sides of the
triangle at the midpoints of the sides.
Interactive Graphics
Throughout the investigations detailed in this article, we tried to ignore the task of coding
interactive Figure 2. The lure was too strong, however, so we first tried a coding based on the
geometry package. Unfortunately, the resulting diagram responded too slowly to be useful. We
then coded a solution based on interpolation: we interpolated a general quadratic through the
midpoints of the triangle formed by three given points, using tangency to obtain the additional
conditions needed to determine the six coefficients in the general quadratic. But the general
quadratic is determined by just five coefficients - it is always possible to divide through by one of
the nonzero coefficients.
This distinction gives Maple's solve and fsolve commands some trouble. Since points are read
from a graph as floating-point numbers, Maple's solve command finds just the zero solution of six
equations in six unknowns that are not independent. Maple's fsolve command does not find any
solution. However, if the floats are converted to rational numbers, solve will obtain the solution of
the six equations in terms of one of the unknowns. By normalizing, we can then obtain the
equation of the ellipse, and its graph can be added to a plot of the relevant points and lines.
The code for Figure 2 is written as a module containing procedures that are invoked by the "click
and drag" fields in the plot component. Initialize this code (and the figure) by clicking the button
above Figure 2. It can be viewed by selecting Component Properties from the Context Menu for
the button. The code is similar to what was used for the Bézier Curves task template in the Curve
Fitting section of the task templates.
Figure 2 Interactive investigation of Marden's theorem.
Click to place three points. Third click generates the full
diagram of ellipse and triangle. Drag any vertex to resize the
triangle and the inscribed ellipse.
If the cursor on the graph does not default to the "click and drag" probe, select it from the toolbar.
Example 2: Using Kalman's Simplification
If all three vertices are taken as arbitrary complex points, there are six indeterminates in any
investigation of Marden's theorem. Kalman showed that it suffices to take the vertices
as the complex points
and
, thereby reducing the number of indeterminates to a
more tractable two. The interpolating polynomial
is then
or
Its derivative,
has the two zeros
the Cartesian coordinates of which are
Table 4 summarizes the ensuing calculations in the geometry package.
As we did earlier, define the three
interpolated vertices as
.
Define the foci
points
and
and
.
as the geometry
Obtain
, respectively the
midpoints of sides
of the triangle
formed by the three interpolated vertices
.
The coordinates of the three midpoints are
found, and assigned the respective names
.
The center of the circumscribed ellipse will
be at the midpoint of the segment joining the
foci. Its coordinates are again those of the
centroid of the triangle whose vertices are
.
The distance from one focus to a point on the ellipse to the other focus is
Unfortunately, this expression for the length of the major axis is too complex for the ellipse
command in the geometry package to handle. It is not possible to construct ellipse
as
we did in Table 2. We are forced to follow another path.
Table 4 Use of Maple's geometry package to find the ellipse of Marden's theorem in
terms of the parameters
and
The ellipse with foci
has the known center
. The coordinates of the center of the
conic governed by the quadratic equation
are
Since at least one of the six coefficients in the general quadratic must be nonzero, the ellipse in
Marden's theorem is determined by five constants. Interpolating the quadratic through the three
midpoints of the sides of the circumscribing triangle provides three conditions, and prescribing the
center provides two more. Hence, the solution of the five equations
is
so the ellipse is described by the equation
where we have divided through by the parameter
first obtain
implicitly as
. To parallel the verifications in Table 3, we
then construct Table 5.
=
=
=
=
=
=
Table 5 Verification that at the midpoints of the sides of the
enveloping triangle, the ellipse of Marden's theorem is tangent to
the triangle
Executing the command
generates (as a pop-up) the interactive plotting tool whose screen-capture is seen in Figure 3.
Figure 3
Interactive plot of the ellipse in Marden's theorem
This same tool can be launched from the Plot Builder by selecting "Interactive Plot with 2
parameters" as the plot type.
Marden's Theorem by Computation
By rotation and translation, but not scaling, the three vertices
of a triangle can be moved
to the points
in the plane. We next establish that the foci of an ellipse that is
tangent to the sides of the triangle at its midpoints are the zeros of the derivative of the polynomial
interpolating the vertices.
We start with the general quadratic equation
and the derivative
The midpoints and slopes of the sides of the triangle formed by the vertices
Table 6.
are listed in
Table 6 Midpoints (left column) and slopes (right column) of the sides of the triangle
whose vertices are
The six coefficients in are determined by the equations
whose solution is
leading to the quadratic
Using , we obtain the coordinates of the center as
in keeping with our earlier observation about the center of the ellipse being at the centroid of the
triangle. If we next translate coordinates with
the quadratic becomes
the template for which is
If the coordinates are rotated via
where
Setting
, the quadratic assumes the form
equal to
, we get
Unfortunately, Maple is unable to recognize that the coefficients of
simplified so that becomes
or
and
can be further
or
Since
, and
are positive, the coefficient of
Hence, the standard form for this ellipse is
is smaller than the coefficient of
.
where
and
Therefore we have
Since
, and since
for an ellipse, the denominator for
that
In order to show that the foci of the ellipse are the same as
is positive, so
and
(that is, the zeros of derivative of the interpolating cubic polynomial), we rotate and translate the
points
back to the
-coordinate system where the foci should then be
. The most difficult part of the rotation is expressing sine and
cosine of
in terms of radicals. For the sine we write
and make the substitutions
Expressing
in terms of
, and
, we have
The rotation and translation of the first coordinate of the first focus is then
The equivalence to the first coordinate of
is shown by the vanishing of the difference:
We leave three similar calculations to the reader.
Note on Hidden Code
Cells tinted in yellow contain hidden Maple input. This input can be seen by checking the "Show
input" box near the bottom of the Table Properties dialog, accessed either through the Table menu,
or the Context Menu. Figure 1 and Table 6 also contain hidden Maple input, but are not tinted.
References
[1] Dan Kalman, "An Elementary Proof of Marden's Theorem," American Mathematical
Monthly, Volume 115, Number 4, April 2008, pp 330-338.
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