Finding the area enclosed by the locus of centroids of equilateral triangles inscribed in an ellipse.

Question

Let a and b be the lengths of the semi-major and semi-minor axes of an ellipse, respectively.

How to find the area enclosed by the locus of the centroids of equilateral triangles inscribed in the ellipse.

Answer. How to answer this question using calculus,trigonometry?

Answer 1

Since a circle of center $G(x_G,\,y_G)$ and radius $R > 0$ can be parameterized as:

$$ (x,\,y) := (x_G,\,y_G) + R\left(\cos u,\,\sin u\right) $$

it follows that the vertices of an equilateral triangle of centroid $G$ can be parameterized as:

$$ (x_n,\,y_n) := (x_G,\,y_G) + R\left(\cos u_n,\,\sin u_n\right), $$

where $u_n = u + \frac{2\,n\,\pi}{3}$, with $u \in [0,\,2\pi)$ and $n = 0,\,1,\,2$.

So, imposing that these vertices belong to an ellipse of Cartesian equation:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

with $a,\,b > 0$ the lengths of the semi-axes, the following system of equations is obtained:

$$ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1 \; \; \; \land \; \; \; \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1 \; \; \; \land \; \; \; \frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} = 1 $$

in the unknowns $x_G,\,y_G,\,R$, whose four solutions are:

$$ -a \le x_G \le a\,; \; \; \; y_G = \pm \frac{b}{a}\sqrt{a^2 - x_G^2}\,; \; \; \; R = 0\,; $$

or:

$$ \tiny \begin{aligned} & x_G = \pm \frac{-a^4\left(\sin u_0 - \sin u_1\right)\left(\sin u_0 - \sin u_2\right)\left(\sin u_1 - \sin u_2\right)-a^2b^2\left(\left(\sin u_0 - \sin u_1\right)\cos^2 u_2 - \left(\sin u_0 - \sin u_2\right)\cos^2 u_1 + \left(\sin u_1 - \sin u_2\right)\cos^2 u_0\right)}{\sqrt{\left(a^2\left(\sin u_0 - \sin u_1\right)^2+b^2\left(\cos u_0 - \cos u_1\right)^2\right)\left(a^2\left(\sin u_0 - \sin u_2\right)^2+b^2\left(\cos u_0 - \cos u_2\right)^2\right)\left(a^2\left(\sin u_1 - \sin u_2\right)^2+b^2\left(\cos u_1 - \cos u_2\right)^2\right)}} \;; \\ & . \\ & y_G = \pm \frac{b^4\left(\cos u_0 - \cos u_1\right)\left(\cos u_0 - \cos u_2\right)\left(\cos u_1 - \cos u_2\right)-a^2b^2\left(\left(\sin^2 u_0 - \sin^2 u_1\right)\cos u_2 - \left(\sin^2 u_0 - \sin^2 u_2\right)\cos u_1 + \left(\sin^2 u_1 - \sin^2 u_2\right)\cos u_0\right)}{\sqrt{\left(a^2\left(\sin u_0 - \sin u_1\right)^2+b^2\left(\cos u_0 - \cos u_1\right)^2\right)\left(a^2\left(\sin u_0 - \sin u_2\right)^2+b^2\left(\cos u_0 - \cos u_2\right)^2\right)\left(a^2\left(\sin u_1 - \sin u_2\right)^2+b^2\left(\cos u_1 - \cos u_2\right)^2\right)}} \;; \\ & . \\ & R = \pm \frac{2a^2b^2\left(\left(\sin u_0 - \sin u_1\right)\cos u_2 - \left(\sin u_0 - \sin u_2\right)\cos u_1 + \left(\sin u_1 - \sin u_2\right)\cos u_0\right)}{\sqrt{\left(a^2\left(\sin u_0 - \sin u_1\right)^2+b^2\left(\cos u_0 - \cos u_1\right)^2\right)\left(a^2\left(\sin u_0 - \sin u_2\right)^2+b^2\left(\cos u_0 - \cos u_2\right)^2\right)\left(a^2\left(\sin u_1 - \sin u_2\right)^2+b^2\left(\cos u_1 - \cos u_2\right)^2\right)}} \;; \end{aligned} $$

which, simplified according to the above reports, offers the only desired solution:

$$ \begin{aligned} & x_G = \frac{\sqrt{2}\,a^2\left(a^2-b^2\right)\cos(3u)}{\sqrt{\left(a^2+b^2\right)\left(a^4+14a^2b^2+b^4\right)+\left(a^2-b^2\right)^3\cos(6u)}} \;; \\ & y_G = \frac{\sqrt{2}\,b^2\left(a^2-b^2\right)\sin(3u)}{\sqrt{\left(a^2+b^2\right)\left(a^4+14a^2b^2+b^4\right)+\left(a^2-b^2\right)^3\cos(6u)}} \;; \\ & R = \frac{4\sqrt{2}\,a^2b^2}{\sqrt{\left(a^2+b^2\right)\left(a^4+14a^2b^2+b^4\right)+\left(a^2-b^2\right)^3\cos(6u)}} \;. \end{aligned} $$

Since:

$$ \frac{x_G^2}{\left(x_G | u=0\right)^2} + \frac{y_G^2}{\left(y_G | u=\frac{\pi}{2}\right)^2} = 1 $$

i.e.

$$ \frac{x_G^2}{\left(\frac{a^2-b^2}{a^2+3b^2}\,a\right)^2} + \frac{y_G^2}{\left(-\frac{a^2-b^2}{3a^2+b^2}\,b\right)^2} = 1 $$

is an identity for each $u \in [0,\,2\pi)$, we can answer the question of the topic:

The locus of centroids of equilateral triangles inscribed in an ellipse of semiaxis $a,\,b > 0$ is a concentric ellipse of semiaxis $\frac{\left|a^2-b^2\right|}{a^2+3b^2}\,a,\,\frac{\left|a^2-b^2\right|}{3a^2+b^2}\,b$. So, since the first ellipse has area $\pi\,a\,b$, the second ellipse has area $\frac{\left(a^2-b^2\right)^2}{\left(a^2+3b^2\right)\left(3a^2+b^2\right)}\,\pi\,a\,b$.

Furthermore, by slightly manipulating the last relationship, we obtain:

$$ \frac{\left(\frac{a^2+3b^2}{a^2-b^2}\,x_G\right)^2}{a^2} + \frac{\left(-\frac{3a^2+b^2}{a^2-b^2}\,y_G\right)^2}{b^2} = 1 $$

from which the coordinates of the fourth point $P(x_P,\,y_P)$ are highlighted where the circle circumscribed to the equilateral triangle intersects the ellipse of semi-axes $a,\,b\,$:

$$ x_P = \frac{a^2+3b^2}{a^2-b^2}\,x_G\,, \; \; \; \; \; \; y_P = -\frac{3a^2+b^2}{a^2-b^2}\,y_G\,. $$

Then, to complete the work, compacting everything in the following way:

$$ \begin{aligned} & R(u) := \sqrt{\frac{32a^4b^4}{\left(a^2+b^2\right)\left(a^4+14a^2b^2+b^4\right)+\left(a^2-b^2\right)^3\,\cos(6u)}} \;; \\ & G(u) := R(u)\left(\frac{a^2-b^2}{4b^2}\,\cos(3u),\;\frac{a^2-b^2}{4a^2}\,\sin(3u)\right); \\ & V(u,\,v) := G(u) + R(u)\left(\cos\left(u + \frac{2\pi}{3}\,v\right),\;\sin\left(u + \frac{2\pi}{3}\,v\right)\right); \\ & P(u) := R(u)\left(\frac{a^2+3b^2}{4b^2}\,\cos(3u),\;-\frac{3a^2+b^2}{4a^2}\,\sin(3u)\right); \end{aligned} $$

with $u \in [0,\,2\pi)$ and $v = 0,\,1,\,2$, this is easily implemented in Wolfram Mathematica 12.0:

{a, b} = {2, 1};

ellipse1 = {a Cos[u], b Sin[u]};
ellipse2 = Abs[a^2 - b^2] {1 / (a^2 + 3 b^2), 1 / (3 a^2 + b^2)} ellipse1;
plot1 = ParametricPlot[{ellipse1, ellipse2}, {u, 0, 2π}, PlotStyle -> {Blue, Red}];

R[u_] := Sqrt[32 a^4 b^4 / ((a^2 + b^2) (a^4 + 14 a^2 b^2 + b^4) + (a^2 - b^2)^3 Cos[6 u])]
G[u_] := R[u] {(a^2 - b^2) Cos[3 u] / (4 b^2), (a^2 - b^2) Sin[3 u] / (4 a^2)}
V[u_, v_] := G[u] + R[u] {Cos[u + 2π/3 v], Sin[u + 2π/3 v]}
P[u_] := R[u] {(a^2 + 3 b^2) Cos[3 u] / (4 b^2), -(3 a^2 + b^2) Sin[3 u] / (4 a^2)}

frames = Table[{xG, yG} = G[u];
               {x1, y1} = V[u, 0];
               {x2, y2} = V[u, 1];
               {x3, y3} = V[u, 2];
               {xP, yP} = P[u];

               list1 = {{x1, y1}, {x2, y2}, {x3, y3}, {x1, y1}};
               list2 = {{{xG, yG}, {x1, y1}}, {{x2, y2}, {x3, y3}}, {{xP, yP}}};
               plot2 = ParametricPlot[G[u] + R[u] {Cos[v], Sin[v]}, {v, 0, 2π}, 
                                      PlotStyle -> {Black, Thin}];

               plot3 = Graphics[{Black, Thin, Line[list1]}];
               plot4 = ListPlot[list2, PlotStyle -> {Green, Yellow, Magenta}];
               Magnify[Show[{plot1, plot2, plot3, plot4}, 
                            PlotRange -> {{-2.0, 2.0}, {-1.5, 1.5}}], 2],

              {u, 0, 2π, 0.1}];

Export["image.gif", frames, "AnimationRepetitions" -> ∞, "DisplayDurations" -> 1];