Ellipse in the triangle

Solution 1:

Here's a geometric construction for the ellipse.

Reflect lines $PQ$ and $PR$ about centre $S$ to $AP'$ and $DP'$, which are also tangent to the ellipse. Hence the ellipse is inscribed into pentagon $ABCDP$.

There is a simple construction to find tangency points (see here for a proof): if diagonals $PC$ and $DA$ of the pentagon meet at $H$, line $BH$ intersects $PD$ at tangency point $E=(3,0)$. One can just repeat this construction to find the other four tangency points and plug their coordinates into the generic equation of a conic, to get its coefficients.

Otherwise, reflect $E$ about $S$ to find a diameter $EF$. Conjugate diameter $MN$ is parallel to $PD$ and $MS^2=PS^2-ES^2$ (because $P$ lies on the director circle). Once we have two conjugate diameters we can construct the axes of the ellipse as explained here.

Solution 2:

A very different answer, inspired by a property which is well described in pages 139-142 of the book "Mathematical Plums" by G.D. Chakerian, edited by Ross Honsberger, available as a Google book.

Fig. 1: Desmos graphical representation._

The construction we are going to give is based on barycentric coordinates of center $S$:

$$\left(\dfrac{area(SBC)}{area(ABC)},\dfrac{area(ASC)}{area(ABC)},\dfrac{area(ABS)}{area(ABC)}\right)=(0.35, 0.25, 0.40) \approx \underbrace{(7, 5, 8)}_{\text{non normalized}}$$

Let us build a triangle $PRT$ with sides $(7, 5, 8)$ with common side $PR$ with the initial triangle, in order to have simpler computations.

The construction that we are going to do is based on two principles:

• the incenter $I$ of triangle $PRT$ has the same barycentric coordinates as $S$ w.r.t. triangle $PRQ$.

• there exists a linear correspondence mapping triangle $PRT$ onto triangle $PRQ$.

The coordinates of $T$ are

$$(x,y)=(4,4 \sqrt{3}).$$

They are found by solving the following system:

$$\begin{cases}x^2+y^2&=&8^2\\(x-5)^2+y^2&=&7^2\end{cases}$$

Now, consider the following linear transform :

$$A=\begin{pmatrix}1&1\\0&\sqrt{3}\end{pmatrix} \ \ \ \text{with} \ \ \ A^{-1}=\begin{pmatrix}1&-1/\sqrt{3}\\0&1/\sqrt{3}\end{pmatrix}$$

mapping triangle $PRQ$ onto triangle $PRT$.

As linear mappings preserve barycentric coordinates, center $I$ of the incircle of $PRT$ is the image of $S$ (the center point of the ellipse) by $A$, i.e., has coordinates:

$$\begin{pmatrix}1&1\\0&\sqrt{3}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}3\\ \sqrt{3}\end{pmatrix}$$

Therefore the incircle has radius $r=\sqrt{3}$ giving it the following parametric equations:

$$\begin{cases}x(t)&=&3+\sqrt{3}\cos(t)\\y(t)&=&\sqrt{3}+\sqrt{3}\cos(t)\end{cases}\tag{1}$$

Coming back to triangle $PRQ$ by applying $A^{-1}$ to (1) ; finally, we obtain the following equations for the ellipse:

$$\begin{cases}X(t)&=&2(1+\cos(t))\\Y(t)&=&1+\frac12(-\cos(t)+\sqrt{3}\sin(t))\end{cases}\tag{2}$$

Important: the following rich article in Forum Geometricum has a cousin approach with many results/remarks, based in particular on what they call the perspector point which is the image by $A^{-1}$ of the Gergonne point of triangle $PRT$.

Remark: $S$ was "a priori" eligible as the center of a certain inscribed ellipse because it is inside the midpoint triangle of triangle $PQR$.

Here is a GeoGebra construction (many thanks to @Toby Mak who made it).