The Special Point in a Triangle 2

Solution using areal coordinates:

Let $A=(1,0,0),B=(0,1,0),C=(0,0,1),O=(l,m,n)$ then we have: $D=(0,m,n),E=(l,0,n),F=(l,m,0)$

The line $AD$ has equation $yn-mz=0$ and $EF$ has equation $-xnm+ynl+zlm=0$ . It follows that $X=(2l,m,n)$ is on both lines. Symettrically, we have $Y=(l,2m,n),Z=(l,m,2n)$ .

The line $XY$ has equation $-mnx-nly+3mlz=0$ and $CF$ has equation $-xm+yl=0$ . Again, $P=(3l,3m,2n)$ is on both of these lines. By symmetry, $Q=(3l,2m,3n)$ .

It follows then that $R=(0,3m,2n),S=(0,2m,3n)$ so $\frac{BR}{RC}=\frac{2n}{3m}$ .

It is known that the circumcentre has coordinates $(sin(2A),sin(2B),sin(2C))$ and the orthocentre: $(tan(A),tan(B),tan(C))$

So $\frac{2 sin(2C)}{3 sin(2B)}=\frac{4 \sqrt{3}}{9} \Rightarrow \frac{sin(2C)}{sin(2B)}=\frac{2 \sqrt{3}}{3} \Rightarrow \frac{sin(C) cos(C)}{sin(B) cos(B)}=\frac{2 \sqrt{3}}{3}$

And: $\frac{2 tan(C)}{3 tan(B)}=\frac{2 \sqrt{3}}{3} \Rightarrow \frac{cos(B)sin(C)}{sin(B)cos(C)}=\sqrt{3}$

Multiplying these two gives: $(\frac{sin(C)}{sin(B)})^2=\frac{6}{3}=2 \Rightarrow \frac{sin(C)}{sin(B)}=\sqrt{2}$

Putting this back into the first expression gives $\frac{cos(C)}{cos(B)}=\frac{\sqrt{6}}{3}$

Combining $cos(C)=\frac{\sqrt{6}}{3} cos(B) \Rightarrow (cos(C))^2=\frac{2}{3} (cos(B))^2$ and $sin(C)=\sqrt{2} sin(B) \Rightarrow (sin(C))^2=2 (sin(B))^2$ gives:

$(cos(C))^2+(sin(C))^2=1=2 (sin(B))^2+\frac{2}{3} (cos(B))^2=\frac{4}{3} (sin(B))^2+\frac{2}{3} (cos(B))^2+(sin(B))^2)=\frac{4}{3} (sin(B))^2+ \frac{2}{3}$

$\Rightarrow \frac{1}{3}=\frac{4}{3} (sin(B))^2 \Rightarrow (sin(B))^2=\frac{1}{4} \Rightarrow sin(B)=\frac{1}{2} \Rightarrow \angle B=30^{\circ}$ as $ABC$ is acute.

$sin(C)=\sqrt{2} sin(B) \Rightarrow sin(C)=\frac{\sqrt{2}}{2} \Rightarrow \angle C=45^{\circ}$

So $\frac{4 \times \angle C}{\angle B}=\frac{180}{30}=6$ .

I wonder if someone can come up with a more elegant solution.