VMO

a) Let the line passing through $A$ and parallel to $BC$ meet $(O)$ at $A_0$. We have $A{A}_{0}CB$ is an isosceles trapezoid so $A_{0}C=AB=2F{H}_a$. We also have $\angle F{H}_{a}B=\angle FB{H}_a=\angle A_{0}CB$ then $F{H}_a\parallel A_{0}C$. But $GC=2GC$, it follows that $G$, $A_0$ and $H_a$ are collinear.



Let $S$ be the midpoint of $A_0{H}_a$ then we have $SE\parallel A{A}_0$ so $\angle AES=\angle A_{0}AC=\angle A_{0}XC$, which implies $SECX$ is a cyclic quadrilateral.

Let $T$ be the midpoint of $B{H}_a$ then we have $ST\parallel B{A}_0$. We obtain $\angle XST=\angle X{A}_{0}B=\angle XCB$ and it follows that $XCST$ is a cyclic quadrilateral.

From the above, we have $X,C,E,S$ and $T$ lie on the same circle, thus the circumcircle of triangle $XEC$ passes through the midpoint of $B{H}_a$.

b) Firstly, we will prove the following lemma.

Lemma. Given triangle $ABC$ with circumcircle $(O)$ and $P,Q$ are two arbitrary points. Lines $PA,PB$ and $PC$ intersect $(O)$ at the second points $P_a,P_b$ and $P_c$. Lines $QA,QB$ and $QC$ intersect $(O)$ at the second points $Q_a,Q_b$ and $Q_c$. Lines $P_a{Q}_a,P_b{Q}_b$ and $P_c{Q}_c$ intersect $BC,CA$ and $AB$ at $D,E$ and $F$ respectively. Then the points $D,E$ and $F$ are collinear.

Proof. Using the property of cyclic quadrilaterals, we have $$\frac{DB}{DC}=\frac{P_{a}B}{P_{a}C}\cdot \frac{Q_{a}B}{Q_{a}C}.$$ From there, we obtain that $$\begin{array}{rl}\prod \frac{DB}{DC}& =\prod \left(\frac{P_{a}B}{P_{a}C}\cdot \frac{Q_{a}B}{Q_{a}C}\right)=\prod \frac{P_{a}B}{P_{a}C}\cdot \prod \frac{Q_{a}B}{Q_{a}C}\\ & =\prod \frac{\sin PAB}{\sin PAC}\cdot \prod \frac{\sin QAB}{\sin QAC}=1.\end{array}$$ Using Menelaus' theorem, we deduce that three points $D,E$ and $F$ are collinear. $\square$

Back to the problem, according to part a), $A{A}_0\parallel BC$ so $XA$ and $X{A}_0$ are isogonal with respect to $\angle BXC$. Let $U$ be the intersection of $AX$ and $BC$. We have $$\frac{UB}{UC}\cdot \frac{H_{a}B}{H_{a}C}=\frac{X{B}^2}{X{C}^2}.(1)$$ It is also well-known that $$\frac{UB}{UC}=\frac{BA}{CA}\cdot \frac{BX}{XC}.$$ Therefore, $$\frac{X{B}^2}{X{C}^2}=\frac{U{B}^2}{U{C}^2}÷\frac{A{B}^2}{A{C}^2}.$$ Combining with (1), we get $$\frac{UB}{UC}=\frac{U{B}^2}{U{C}^2}÷\frac{A{B}^2}{A{C}^2}÷\frac{H_{a}B}{H_{a}C}.$$ Hence, $$\frac{UB}{UC}=\frac{A{B}^2}{A{C}^2}\cdot \frac{H_{a}B}{H_{a}C}.$$ Let $V$ be the intersection of $BY$ and $CA$, $W$ be the intersection of $CZ$ and $AB$ then we have similar equality. Thus, $$\prod \frac{UB}{UC}=\prod \left(\frac{A{B}^2}{A{C}^2}\cdot \frac{H_{a}B}{H_{a}C}\right)=1.$$ It follows that $AX$, $BY$, and $CZ$ are concurrent at $I$.

Let $A_1$, $B_1$ and $C_1$ be the symmetric points of $A$, $B$ and $C$, respectively, through the midpoints of $BC$, $CA$ and $AB$. By simple calculation, we observe that $A_{1}X\parallel DM$ and $\frac{G{A}_1}{GD}=3$.

Hence the homothety with center $G$ in ratio $3$ turns the line $DM$ into a straight line $A_{1}X$. Similarly, we can prove that this homothety also transforms $EN$ and $FP$ to $B_{1}Y$ and $C_{1}Z$.

From here, we just need to prove that the lines $A_{1}X$, $B_{1}Y$ and $C_{1}Z$ are concurrent. Consider triangle $XYZ$ and two points $I$ and $G$, we have $XI$, $YI$ and $ZI$ intersect $(XYZ)$ at $A$, $B$ and $C$; $XG$, $YG$ and $ZG$ cut $(XYZ)$ at $A_0$, $B_0$ and $C_0$. Applying the lemma, we have $YZ$, $ZX$ and $XY$ cut $A{A}_0$, $B{B}_0$ and $C{C}_0$ in three collinear points. Applying Desargues' theorem, we conclude that $A_{1}X$, $B_{1}Y$ and $C_{1}Z$ are concurrent. $\square$