Classroom Tips and Techniques: Maple Meets Marden's Theorem. Robert J. Lopez Emeritus Professor of Mathematics and Maple Fellow Maplesoft

Transcription

1 Introduction Classroom Tips and Techniques: Maple Meets Marden's Theorem Robert J. Lopez Emeritus Professor of Mathematics and Maple Fellow Maplesoft 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

2 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 be given by The polynomial that interpolates these three points is or and is The zeros of, namely, the foci of the inscribed ellipse, are then

3 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 and as the geometry points and. 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.

4 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.

5 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

6 of the corresponding side. = = = = = = Table 3 Verification that ellipse is tangent to triangle at the midpoints of its sides 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

7 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

8 or Its derivative, has the two zeros the Cartesian coordinates of which are Table 4 summarizes the ensuing calculations in the geometry package.

9 As we did earlier, define the three interpolated vertices as. Define the foci and as the geometry points and. 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

10 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

11 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. = = = =

12 = = 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

13 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

14 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

15 the template for which is If the coordinates are rotated via where, the quadratic assumes the form Setting equal to, we get Unfortunately, Maple is unable to recognize that the coefficients of and can be further simplified so that becomes or

16 or Since, and are positive, the coefficient of is smaller than the coefficient of. Hence, the standard form for this ellipse is where and Therefore we have Since, and since for an ellipse, the denominator for is positive, so that In order to show that the foci of the ellipse are the same as

17 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

18 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.

19 References [1] Dan Kalman, "An Elementary Proof of Marden's Theorem," American Mathematical Monthly, Volume 115, Number 4, April 2008, pp Legal Notice: Maplesoft, a division of Waterloo Maple Inc Maplesoft and Maple are trademarks of Waterloo Maple Inc. This application may contain errors and Maplesoft is not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact Maplesoft for permission if you wish to use this application in for-profit activities.