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 $$\overline{HA}\cdot \overline{H{A}_1}=\overline{HB}\cdot \overline{H{B}_1}=\overline{HC}\cdot \overline{H{C}_1}=\frac{1}{2}{\mathcal{P}}_{H/(ABC)}.$$ Consider the inversion of center $H$ and the power equals to the value above, then we have $$\begin{array}{rl}& A\leftrightarrow A_1,B\leftrightarrow B_1,C\leftrightarrow C_1,\\ & (H{B}_1{C}_1)\leftrightarrow BC,(H{C}_1{A}_1)\leftrightarrow CA,(H{A}_1{B}_1)\leftrightarrow AB\end{array}$$ 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 $(H{B}_1{C}_1),(H{C}_1{A}_1),(H{A}_1{B}_1)$. Consider the homothety of center $H$, ratio $2$ which maps $d\to d^{\prime }$ and the circle $(H{B}_1{C}_1)$ to the circle of center $H$ and radius $HA$ (since $HA$ is the diameter of $(H{B}_1{C}_1)$). By the property of homothety, we have $d^{\prime }$ is also tangent to $(A,AH)$. Similarly to these circles $(B,BH),(C,CH)$ so the line $d^{\prime }$ satisfies the given condition. $\square$