Problem 1337

Let's break down the problem into two parts and solve each part step-by-step.

### Part (a)
We need to find the condition for $\triangle AMN$ to have maximum area.

1. Angle Chasing:
We start by analyzing the angles. Given that $M$ and $N$ are points on the circumcircles of $\triangle ABH$ and $\triangle ACH$ respectively, we have:
$$\angle AMN \equiv \angle AMH + \angle HMC \equiv \angle ABH + \angle HCA \equiv 2 \angle BAC \pmod{\pi}$$
This implies that $\triangle AMN$ is isosceles with $AM = AN$.

2. Maximizing the Area:
The area of $\triangle AMN$ is maximized when $AM$ is maximized. Since $M$ and $N$ lie on the circumcircles of $\triangle ABH$ and $\triangle ACH$, the maximum distance $AM$ can achieve is the diameter of the circumcircle of $\triangle ABC$, which is $2R$ (where $R$ is the circumradius of $\triangle ABC$).

3. **Condition for Maximum $AM$:**
For $AM$ to be $2R$, the line $\Delta$ must be perpendicular to $AH$ at $H$. This is because the diameter of the circumcircle subtends a right angle to any point on the circle.

Thus, the condition for $\triangle AMN$ to have maximum area is that $\Delta$ is perpendicular to $AH$ at $H$.

### Part (b)
We need to prove that point $P$ moves on a fixed circle.

1. Perpendicular Lines:
Let $d_1$ be the line through $M$ perpendicular to $DB$, and $d_2$ be the line through $N$ perpendicular to $DC$. These lines intersect at point $P$.

2. Angle Chasing:
From part (a), we have:
$$\angle AMN \equiv 2 \angle BAC \equiv \angle DBN \equiv \angle PMN \pmod{\pi}$$
This implies that $P$ lies on the circle with diameter $AN$.

3. Reflection and Fixed Circle:
Draw a line $d$ through $A$ parallel to $CD$. This line $d$ is the perpendicular bisector of $NP$. Hence, $P$ is the reflection of $N$ through $d$. Since $N$ lies on the circumcircle of $\triangle AHC$, the reflection of this circle through $d$ is also a circle.

Therefore, $P$ lies on a fixed circle, which is the reflection of the circumcircle of $\triangle AHC$ through the line $d$.

$\blacksquare$