If $G$ is the Gergonne point of $\triangle PQR$ and $A, D, E$ are the contact points of the incircle such that $\triangle PGE$ is equilateral. What is the ratio of area $\dfrac{\triangle PQR}{\triangle PGE}$?
My attempt:
Let's put this in $X$–$Y$ coordinates with $C$ as origin and line $PQ$ being parallel to the $X$-axis and $EC$ being the $Y$-axis ($C$ is the center of the incircle and radius $r$).
$\angle APQ = \angle PER = \alpha = 60^0$; Say, $PG = EG = PE = 1$ unit
Coordinates of points: $E(0,-r), P(-1,-r), C(0,0)$.
The equation of the line $PR$: $y + r = m(x+1)$
As line $PR$ is tangent to the circle, the perpendicular from center $C$ will be equal to the radius.
$r = \dfrac{r – m}{\sqrt{m^2+1}}$ or, $m = \dfrac{2r}{1-r^2}$
So, the equation of the line $PR$: $y + r = \dfrac{2r}{1-r^2} (x+1)$ …(1)
Equation of line $ER$: $y + r = -x \tan\alpha$ …(2)
Equating (1) and (2), we get the intersection point $R (x_0,y_0)$
[$\dfrac{-2r}{2r+(1-r^2)\tan \alpha}, \dfrac{2r\tan \alpha}{2r+(1-r^2)\tan \alpha} – r$], where $\tan \alpha = \tan 60^0 = \sqrt3$
Equation of line $QR$:
$y – y_0 = m_0 (x – x_0)$
Again equating perpendicular distance from the center and the radius
$r = \dfrac{m_0x_0 – y_0}{\sqrt{m_0^2+1}}$
or, $(x_0^2-r^2)m_0^2 – 2x_0y_0m_0 + (y_0^2 – r^2) = 0$
As there are two lines passing through $(x_0,y_0)$ that are tangent to the circle ($PR$ and $QR$), the quadratic equation will give us two values of $m_0$, one of them being $m_0 = \dfrac{2r}{1-r^2}$ as we have established before.
This is how far I reached. I tried to replace values of ($x_0, y_0$) in the quadratic equation and we already know one of the roots but it is becoming very tedious to find the slope of line $QR$.
Once I know the slope, I can find the point $A$ and as we know the equation of line $PA$, I should be able to find the value of $r$ as well as coordinates of points $P, Q, R$. That will lead to us finding the area of $\triangle PQR$ and the ratio to $\triangle PGE$ (area of $\triangle PGE = \dfrac {\sqrt3}{4}$ units).
Any suggestions on how to simplify the solution? Are there any known properties of gergonne point that I am missing which makes the solution simpler?
Best Answer
$\begin{array}{} λ(C,r) & ω(P,1) & C=(0,0) \\ P=(-1,-r) & E=(0,-r) \\ \text{ΔPGE is equilateral} & PG=EG=EP=1 & G=(\frac{-1}{2},\frac{\sqrt{3}}{2}-r) \\ \end{array}$
$\begin{array}{} λ∩ω=D & PG∩λ=A & DG∩PE=Q \\ AQ⟂AC & m_{AQ}·m_{AC}=-1 & PD∩EG=R \\ x^2+y^2=r^2 & (x+1)^2+(y+r)^2=1 & x+r·y+r^2=0 \\ \end{array}$
$\begin{array}{} y=\frac{-(x+r^2)}{r} & D=\left( \frac{-2r^2}{r^2+1},\frac{r(1-r^2)}{r^2+1} \right) \end{array}$
$\begin{array}{} PG∩λ=A & \left| \begin{array}{} x & y & 1 \\ -1 & -r & 1 \\ \frac{-1}{2} & \frac{\sqrt{3}}{2}-r & 1 \\ \end{array} \right|=0 & y=x·\sqrt{3}+\sqrt{3}-r & x^2+y^2=r^2 \end{array}$
$\begin{array}{} x_{A}=\frac{1}{4}\left( \sqrt{3r^2+2\sqrt{3}r-3}+\sqrt{3}r-3 \right) & y_{A}=x_{A}·\sqrt{3+\sqrt{3}-r} \end{array}$
$\begin{array}{} x_{Q}=\frac{2r}{\sqrt{3}-r} & y_{Q}=-r \end{array}$
$\begin{array}{} m_{AQ}·m_{AC}=-1 & ⇒ & -\sqrt{3r^2+2r\sqrt{3}-3}+3r^2-9+(\sqrt{3r^2+2r\sqrt{3}-3})\sqrt{3}r)=0 & ⇒ \\ r=\frac{7\sqrt{3}}{9}≈1.3471506 \\ \end{array}$
$ratio=\frac{ΔPQR}{ΔPGE}=\frac{\frac{1}{2}\left| \begin{array}{} -1 & \frac{7\sqrt{3}}{9} & 1 \\ 7 & \frac{-7\sqrt{3}}{9} & 1 \\ \frac{-21}{10} & \frac{119\sqrt{3}}{90} & 1 \\ \end{array} \right| }{\frac{\sqrt{3}}{4}}=\frac{168}{5}=33.6$