Complex numbers kill it easily

bigant146 wrote: In an acute traingle $ABC$ with $AB< BC$ let $BH_b$ be its altitude, and let $O$ be the circumcenter. A line through $H_b$ parallel to $CO$ meets $BO$ at $X$. Prove that $X$ and the midpoints of $AB$ and $AC$ are collinear. My solution is that First, label the midpoint of $AB M$. Because $BC$ and $BH_b$ is isogonal $\angle MH_bB=\angle MbH_b=\angle OBC=\angle OCB$ So we could conclude that $\angle AMH_b=180-\angle BOC=\angle H_bXB$ And $(MBXH_b)$ is concyclic!! $\angle MXB=\angle MH_bB=\angle OBC$ So $MX$ parallel to $BC$, proved.

Let $M $ midpoint of $ AC$ and $L=OX\cap AC$ we want to prove that X is on the $A$-midline $\angle MXO =\angle OBC =\frac{\pi}{2}-A$, since $\angle H_bXO= \angle XOC =\pi -2\angle A$ then it s suffice to show that $XM$ is the bisector of $\angle LXH_b$ or $ \frac{XH_b}{XL}=\frac{MH_b}{ML}\ :$ Thales yields $\frac{XH_b}{XL}=\frac{OC}{OL}=\frac{OB}{OL}\ \ \overset{\text{by Thales}}{=}\ \ \frac{MH_b}{ML}\ \square$ RH HAS

With areals the line through the midpoint of $AB,AC$ has equation $x=y+z$. Let $BO$ intersect this line at $X'$ then: $$X'=\left (a^2 S_A,a^2 S_A-c^2 S_C,c^2S_C \right)$$It remains to check $H_B,X'$ and the infinity point on $CO$ are colinear: $$H_B X' \parallel CO \Leftrightarrow 0=\begin{vmatrix} a^2 S_A & a^2 S_A-c^2 S_C & c^2S_C \\ S_C & 0 & S_A \\ a^2 S_A & b^2 S_B & -a^2 S_A-b^2 S_B \end{vmatrix}= \begin{vmatrix} 0 & a^2 S_A-c^2 S_C-b^2 S_B & \sum_{cyc} a^2 S_A \\ S_C & 0 & S_A \\ a^2 S_A & b^2 S_B & -a^2 S_A-b^2 S_B \end{vmatrix}$$$$ \Leftrightarrow S_{BC} \sum_{cyc} a^2 S_A=(b^2S_B+c^2S_C-a^2 S_A)(a^2S_A+S_{BC}) \Leftrightarrow 2S_{BC}=b^2S_B+c^2S_C-a^2 S_A$$Which can be expanded out an checked to be correct so $X=X'$ as desired.

With complex set $\odot ABC$ to the unit circle and $A=a,B=b,C=\frac{1}{b}$. Let $X'=x$ be the intersection point of $BO$ and the $A-$midline: $$XBO \text{ colinear} \Leftrightarrow x=b^2 \overline{x}$$Take homothey factor $2$ about $A$ taking $X' \rightarrow 2x-a$ is on $BC$ so: $$b+\frac{1}{b}=2x-a+2 \overline{x}-\frac{1}{a}=2x-a+2 \frac{x}{b^2}-\frac{1}{a} \implies x=\frac{1}{2} \left (\frac{b^2-a^2}{a(1+b^2)} \right)$$As $H_B=\frac{1}{2} \left (a+b+\frac{b-a}{b^2} \right)$ it remains to check $H_B X' \parallel CO$: $$2(H_B-X)=\frac{b-a}{b^2}-\frac{b^2-a^2}{a(1+b^2)}=\frac{(b-a)(a-b^3)}{ab^2(1+b^2)}$$$$\frac{2(H_B-X)}{C-O}=\frac{\frac{(b-a)(a-b^3)}{ab^2(1+b^2)}}{\frac{1}{b}}=\frac{(b-a)(a-b^3)}{ab(1+b^2)}$$$$\overline{\left(\frac{(b-a)(a-b^3)}{ab(1+b^2)} \right)}=\frac{\left(\frac{1}{b}-\frac{1}{a} \right) \left (\frac{1}{a}-\frac{1}{b^3} \right)}{\frac{1}{ab} \left(1+\frac{1}{b^2} \right)}=\frac{(b-a)(a-b^3)}{ab(1+b^2)}$$So $\frac{2(H_B-X)}{C-O} \in \mathbb{R} \implies H_B X' \parallel CO \implies X=X'$ as desired.

This is also Romania 2018. JTST 3. They offer 3 different solutions! (For Juniors)

Why everyone is bashing this cute problem? $\text{Solution:}$ Let $M$ be the midpoint of$AB$ , $N=XM\cap BH_b$ and connect $MH_b$. We wish to prove that $MX\parallel BC$. $\angle BMH_b=2\angle A$ And $\angle H_bXB=\angle XOC=\angle OBC +\angle OCB=180^\circ -2\angle A$ So, $MH_bXB$ is cyclic quadrilateral. $\text{Final chase:}$ $\angle MXB=\angle MH_bB=90^\circ -\angle A$ And, $\angle XBC=\angle OBC= 90^\circ -\angle A$ $\therefore MX\parallel BC$. $\blacksquare$