Team Selection Tests for BMO, 2018

1) Since $AM$ passes through the midpoint of the segment $BE$ and $AP \parallel BE$, we can see that
$$(AP, AM, AB, AC) = -1 \text{ or } (MP, MA, MD, ME) = -1.$$
Suppose that $AM$ cuts $(I)$ at the second point $T$ then $MDTE$ is a harmonic quadrilateral, then $(Mx, MT, MD, ME) = -1$ with $Mx$ as the tangent line of $(I)$.
From these results, we get $Mx \equiv MP$ which implies that $PM$ is tangent to $(I)$ or $\angle IMP = 90^\circ$. Similarly, we also have $\angle INP = 90^\circ$. Hence, if we denote $K$ as the midpoint of $DE$ then $IK \perp DE$ and $K \in (MIP), (NIQ)$. Since $A \in IK$ as the radical axis of these two circles, the result follows.

2) Denote $A_1, B_1, C_1$ as the projections from $A, B, C$ to the opposite side, respectively then
$$\overline{HA} \cdot \overline{HA_1} = \overline{HB} \cdot \overline{HB_1} = \overline{HC} \cdot \overline{HC_1} = \frac{1}{2} \mathscr{P}_{H/(ABC)}.$$
Consider the inversion of center $H$ and the power equals to the value above, then we have
$$\begin{aligned} & A \leftrightarrow A_1, \quad B \leftrightarrow B_1, \quad C \leftrightarrow C_1, \\ & (HB_1C_1) \leftrightarrow BC, \quad (HC_1A_1) \leftrightarrow CA, \quad (HA_1B_1) \leftrightarrow AB \end{aligned}$$
The image of $(I)$ should be some line $d$ since $H \in (I)$. Because $BC, CA, AB$ are tangent to $(I)$ then $d$ is also tangent to $(HB_1C_1), (HC_1A_1), (HA_1B_1)$.
Consider the homothety of center $H$, ratio $2$ which maps $d \rightarrow d'$ and the circle $(HB_1C_1)$ to the circle of center $H$ and radius $HA$ (since $HA$ is the diameter of $(HB_1C_1)$). By the property of homothety, we have $d'$ is also tangent to $(A, AH)$. Similarly to these circles $(B, BH), (C, CH)$ so the line $d'$ satisfies the given condition. $\square$