tri-ellipse_construction.ggb |
A circle is the collection of all points that are a fixed distance from a focus (center). An ellipse is the collection of points, the sum of whose distances to two fixed foci is constant. These shapes are covered extensively in many high school classes. A tri-ellipse is the collection of points, the sum of whose distances to three fixed foci is constant. As we have seen, these shapes can take some unfamiliar forms . What else can we understand about these shapes?
It is possible to use geometric software to perform an asymmetric construction of a tri-ellipse. Let the foci of the tri-ellipse be points $A, B, \text{ and } C$ and the distance-sum be $d$. Write $d$ as $d=e+(d-e)$ for any $0\leq e \leq d.$ Draw the ellipse with foci $A \text{ and } B$ and distance-sum $e.$ Draw the circle with center $C$ and radius $d-e.$ The intersection(s) of the ellipse with the circle lie on the tri-ellipse. The tri-ellipse is then the locus of such intersection points as $e$ varies from $0$ to $d.$ Experiment with the applet below to create your own tri-ellipses.
You can also download the Geogebra file here if you want to look into the nuts and bolts.
Additionally, the tri-ellipse can be approached from an algebraic perspective. If the foci $A, B, \text{ and } C$ have coordinates $(x_A, y_A), (x_B, y_B), \text{ and } (x_C, y_C),$ then the tri-ellipse is represented by the equation
$$\sum_{L=A, B, C} \sqrt{(x-x_L)^2+(y-y_L)^2}=d.$$
Squaring three times, combined with some algebraic manipulations turns this equation into a degree-eight polynomial in two variables. Perhaps there are some algebraic techniques to bring to bear to further elucidate properties of the tri-ellipse.
Open Questions
There is a parametrization of an ellipse given by $x = r \cos \theta\text{ and } y = r\sin\theta.$ Is there a similar parametrization for a tri-ellipse in terms of an angle? This parametrization would be centered at the point found in the last post.
0 Comments
I am now prepared to answer question 5 from my previoius post: "How can one construct the 120°-point of a given triangle?" Recall that the 120°-point of a triangle is the point so that if the segments from that point to the three vertices of the triangle are drawn, three 120° angles are formed. Construction Use special right triangles to construct a circle through points A and B with center F so that $m\angle AFB = 120^{\circ}.$ Notice that for any point G on $\overarc{AB}$, $m\angle AGB=120^{\circ}.$ Repeat this for each pair of points of the triangle $\Delta ABC.$ The point where the three circles intersect is the 120°-point of the triangle. I've created an interactive Geogebra Worksheet that lets you play with the construction yourself. In the course of my research, I've discovered another interpretation for the significance of point P. P, in addition to being the point that minimizes the distance sum $AP + BP + CP$ and being the 120°-point of $\Delta ABC$ is also the point of equal tension for points A, B, and C. In other words, if three people, one standing at each of A, B, and C were to pull with equal tension on three lengths of rope connected at a single point, then that point would stabilize to point P. Open Questions
I created this image as part of my coursework for an online course about Geometer's Sketchpad – a computer program designed for high school math students. The assignment was intended to help the user gain proficiency with the parametric coloring functionality of the program. Here, the picture is colored based on the sum of the distances from the variable point $P$ to three fixed vertices $A, B,$ and $C.$ The smaller the sum $PA + PB + PC,$ the more violet the color of the point, the larger the sum, the redder the color. I was struck by the image's beauty. Even the blotchy nature of the picture, which is due to limitations of the program, seemed endearing. While the assignment for my course was only an exercise in the functionality of the computer program, it raised two interesting mathematical questions.
Notice that here we are considering the sum of the distances from a point $P$ to three other points $A, B,$ and $C,$ which I will call foci. We can rephrase the questions above without referencing color as, "Describe the set of all $P$ for which the sum $PA + PB + PC$ is a fixed value $d$" and "For which point $P$ is the same sum minimized?" If we consider diagrams with only one or two foci, the questions have well-established answers. With one focus, the contours form circles and the "center" point is the center of the circle. With two foci, the contours form ellipses, and the "center" points form the segment connecting the foci. Investigation My cousin was able to develop a computer program to help investigate this question. Specifically, the program allows the user to select the location of the three foci and which contour to draw. Below are a few screenshots of the program. A few observations are apparent from the images.
Hypothesis This hypothesis attempts to answer the second question posed above. We give the point $P$ that minimizes the sum $PA + PB + PC$ the name tri-center of the points $A, B,$ and $C.$ Case 1: The three foci form a triangle $\Delta ABC$ all of whose angles measure less than $120^{\circ}.$ In this case the point $P$ is the "120°-point" of the triangle as illustrated in the image. The 120°-point of a triangle $\Delta ABC$ is the point $P$ such that $m\angle APC = m \angle BPC = m \angle CPA=120^{\circ}.$ Case 2: The three foci form a triangle $\Delta ABC$ one of whose angles measures greater than $120^{\circ}.$ In this case, the point $P$ is located at the vertex of this largest angle. Proof I present an outline of a proof of the above hypothesis for the case when the three foci form an isosceles triangle with non-base vertex measuring less than 120°. $\Delta ABC$ is an isosceles triangle with $m \angle A < $ 120° We can assume that the tri-center of the triangle lies on the segment $\overline{AM},$ which bisects segment $\overline{BC},$ due to symmetry. Let $P$ be the 120°-point of the triangle and $P'$ be any point on $\overline{AM}.$ We show that $AP' + BP' + CP'$ is minimized when $P'=P$. To aid our proof, we note that minimizing $AP' + BP' + CP'$ is equivalent to minimizing $BP' + CP' - P'M$ and divide our proof into two cases. Let $x=PM, x\sqrt{3} = BM = CM, 2x=PB = PC.$ In addition, let $y=BP' = CP'$ and $z=PP'.$ Case 1: $P'$ is above $P.$ To show our point, we need to show that $3x < 2y-x-z,$ which is equivalent to $z < 2y - 4x.$ By the Pythagorean Theorem, $z=\sqrt{y^2-3x^2}-x.$ We show the result by reversing the steps of $$\sqrt{y^2-3x^2}-x<2y-4x$$ $$\sqrt{y^2-3x^2}<2y-3x$$ $$y^2-3x^2<4y^2-12xy+9x^2$$ $$0<3y^2-12xy+12x^2$$ $$0<3(y-2x)^2,$$ which is certainly true, because $2y-3x$ is positive. Questions
|
About Me
I started this blog to share my transformation from math nerd to math nerd who loves to share math with young people. I teach high school in Hanoi, Vietnam. Your comments are always welcome. Archives
May 2021
|