Vietnamese Mathematical Olympiad

a) Let $I$, $X$ and $Y$ be the midpoints of $BC$, $BE$ and $CF$, respectively. Let $IX$, $IY$ intersect $AC$, $AB$ at $S$, $T$ respectively. Since $IB$ is tangent to $(ABG)$, we have
$$IB^2 = IG \cdot IA = IC^2,$$
so $IC$ is tangent to $(AGC)$ as well. Since $I$, $Y$ are midpoints of $BC$, $CF$, we have $IY$ is parallel to $BF$, then $\angle IYG = \angle BFG = \angle BAG$, and thus $A, T, Y$ and $G$ are concyclic, so $IY \cdot IT = IG \cdot IA$.

Similarly,
$$IS \cdot IX = IG \cdot IA = IY \cdot IT,$$
implying that $A, T, S, Y, X$ and $G$ lie on the same circle.

Hence, $\angle AXY = \angle AGY = \angle IGC = \angle ACI = \angle ABF = \angle ABM$.
Therefore, $A, M, B$ and $X$ lie on the same circle. Similarly, $A, Y, C$ and $N$ lie on the same circle. Thus, we have $\angle AMX = \angle GBD$,
and $\angle NAY = \angle GCD$. So we have
$$\begin{align*} \angle MAN &= \angle AMX + \angle NAY - \angle YAX \\ &= \angle GBD + \angle GCD + \angle BGC - 180^\circ \\ &= 180^\circ - \angle BDC, \end{align*}$$
implying that $A, D, M$ and $N$ lie on the same circle.

b) For a triangle $ABC$, define the transformation namely $\gamma_{A,\triangle ABC} = T \circ I$, which is the union of the inversion $I$ of center $A$, power $AB \cdot AC$, and the reflection $T$ through the angle bisector of $\angle ABC$. First, let prove the following lemma

Lemma. Let $ABC$ be a triangle and $(I)$ be a circle through $B, C$. Assume that $\gamma_{A,\triangle ABC}$ transforms $(I)$ into $(J)$, then $AI, AJ$ are isogonal with respect to $\angle ABC$.

Proof. We have the inversion with center $A$ and power $AB \cdot AC$ transforms $(I)$ into $(J')$, and $A, I, J$ are collinear due to the property of the inversion. Then after the reflection through the bisector of $\angle BAC$, $J'$ transforms to $J$. This implies that $AI, AJ$ are isogonal with respect to angle $BAC$. $\square$

Back to our main problem, let $O_a, O_b$ and $O_c$ be the centers of $(HEF)$, $(HKF)$ and $(HKE)$. Note that $O_a, O_b, O_c$ are the mutual intersections of the perpendicular bisectors of $HK, HE, HF$, so the perpendicular bisectors of $HK, HE$ and $HF$ are precisely the lines $O_bO_c, O_cO_a$ and $O_aO_b$. By Desargue's theorem, $R, P$ and $Q$ are collinear if and only if $AO_a, BO_b$ and $CO_c$ are concurrent. So we will show that indeed $AO_a, BO_b$ and $CO_c$ concur.

Let $W$ be the center of the circle $\omega \equiv (KEF)$. Now let's consider the transformation $\gamma_{A,\triangle ABC}$. Since $\angle BAD = \angle KAC$ and $\angle ABD = 180^\circ - \angle C = \angle AGC$, we have $\triangle ABD \sim \triangle AGC$, and thus
$$AB \cdot AC = AG \cdot AD,$$
and $AG, AD$ are isogonal with respect to $\angle BAC$. So $\gamma_{A,\triangle ABC}$ turns $D$ to $G$ and vice versa, and hence it transforms $(BHC)$ to $(BKC)$ and vice versa. Since $H \in (BHC)$, $K \in (BKC)$ and $AH, AK$ are isogonal with respect to $\angle BAC$, the transformation will transform $H$ to $K$ and vice versa. Now, we have
$$\angle ABF - \angle ACB = \angle KGC = \angle AEC,$$
and similarly $\angle ACE = \angle AFB$, then $\triangle AFB \sim \triangle ACE$, thus
$$AE \cdot AF = AB \cdot AC.$$
Also, since $\angle FAB = \angle BAC = \angle CAE$, $AE$, $AF$ are isogonal with respect to $\angle BAC$. Hence $\gamma_{A,\Delta ABC}$ turns $E, F$ to each other and hence it turns $(HEF)$ and $(KEF)$ to each other. By the lemma, we have $AO_a$ and $AW$ are isogonal with respect to angle $BAC$.

Using this similar argument, we can show that $\gamma_{B,\Delta ABC}$ turns $H, E$ to each other, by noting that $\triangle AHB \sim \triangle ECB$ so $BA \cdot BC = BH \cdot BE$, and $\angle ABH = \angle BAG = \angle GBC$. It also turns $F, G$ to each other since
$$\angle ABF = \angle ACB = \angle FBC$$
and $\triangle ABF \sim \triangle KBC$, so $BK \cdot BF = BA \cdot BC$. Thus, this transformation turns $(HFK)$ and $(EFK)$ to each other, and by the lemma, $BO_b, BW$ are isogonal with respect to $\angle ABC$. Similarly, we can show that $CO_c, CW$ are isogonal with respect to $\angle ACB$. Hence, $AO_a, BO_b$ and $CO_c$ will concur on the isogonal conjugate of $W$ with respect to triangle $ABC$. The proof is completed.
$\square$