Vietnamese Mathematical Olympiad

a) Firstly, note that when $\ell$ changes, the angles $\angle AMN$ and $\angle ANM$ both remain unchanged, so triangle $AMN$ is always self-congruent. Draw $AK$ perpendicular to $MN$ ($K \in MN$), then $AK \le AH$. Therefore, the area of triangle $AMN$ attains maximal value when $AH$ is the altitude or $MN \perp AH$. Thus, when $\Delta \perp AH$, the area of triangle $AMN$ is largest.

b) Let the line passing through $A$ and parallel to $BD$ intersect (ABM) at $E$ and the line passing through $A$ and parallel to $CD$ intersect (ACH) at $F$. Notice that the radius of the circles ($O$), (ABH) and (ACH) are equal, hence $\angle ADB = \angle AEB$, which implies $\angle ABD = \angle BAE$ (because $BD \parallel AE$). Hence, $\angle BAD = \angle ABE$, it follows that $AD \parallel BE$. Thus $ADBE$ is a parallelogram.

Similarly, $ADCF$ is a parallelogram. Therefore, triangle $AEF$ is the image of triangle $DBC$ through the translation $T_v$ with vector $\vec{v} = \vec{DA}$. Denote $O_1, H_1$ to be the circumcenter and orthocenter of triangle $AEF$, respectively. It is well known that
$$\begin{aligned} \overrightarrow{OH} &= \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} \\ \overrightarrow{O_1H_1} &= \overrightarrow{O_1A} + \overrightarrow{O_1E} + \overrightarrow{O_1F}, \end{aligned}$$
However, $\overrightarrow{O_1H_1} = \overrightarrow{O_1O} + \overrightarrow{OH_1}$, $\overrightarrow{O_1A} = \overrightarrow{O_1O} + \overrightarrow{OA}$, and $\overrightarrow{O_1E} = \overrightarrow{OB}$, $\overrightarrow{O_1F} = \overrightarrow{OC}$ (image via translation forward $T_v$). Hence
$$\overrightarrow{O_1O} + \overrightarrow{OH_1} = \overrightarrow{O_1O} + \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC},$$
so $\overrightarrow{OH_1} = \overrightarrow{OH}$. Thus, $H_1 \equiv H$.

Now we will prove that the intersection $P$ of $d_1$ and $d_2$ always lies on the circle ($AEF$). First, since $AE \parallel BD$ and $AF \parallel CD$, we have $MP \perp AE$ and $NP \perp AF$. Let $P_1$ be the point of symmetry of $M$ through $AE$, $P_2$ is the symmetry point of $N$ over $AF$. Since the circles ($AEF$), ($ABH$) and ($ACH$) are equal (because they are equal to the same circle ($O$)) so $P_1, P_2$ are all on the circle ($AEF$). Note that the Steiner lines of $P_1$ and $P_2$ with respect to the triangle $AEF$ are coincident (that is the line $\Delta$). According to the property of the Steiner line, we deduce that $P_1 \equiv P_2 \equiv P$ so $P$ always lies on the fixed circle ($AEF$). $\square$