, 2016

Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $AB$ at point $M'$. It suffices to show that $M'R \parallel AC$, since this would yield $M' = M$.
Suppose that the line $PO$ intersects $AC$ at $Q$ and the circumcircle of $AM'O$ at $Y$, respectively. Then
$$\angle AYM' = \angle AOM' = 90^\circ - \angle M'OP.$$
By angle chasing we have $\angle EOQ = \angle FOP = 90^\circ - \angle AOF = \angle M'AO = \angle M'YP$ and by symmetry $\angle EQO = \angle M'PY$. Therefore $\triangle M'YP \sim \triangle EOQ$.
On the other hand, we have
$$\begin{aligned} \angle M'OP &= \angle M'OF + \angle FOP = \frac{1}{2}(\angle FOX + \angle FOP + \angle EOQ) = \\ &= \frac{1}{2}\left(\frac{180^\circ - \angle XOE}{2}\right) = 90^\circ - \frac{\angle XOE}{2} \end{aligned}$$
Since we know that $\angle AYM'$ and $\angle M'OP$ are complementary this implies
$$\angle AYM' = \frac{\angle XOE}{2} = \angle NOE$$
Therefore, $\angle AYM'$ and $\angle NOE$ are congruent angles, and this means that $A$ and $N$ are corresponding points in the similarity of triangles $\triangle M'YP$ and $\triangle EOQ$. It follows that
$$\frac{AM'}{M'P} = \frac{NE}{EQ} = \frac{NR}{RP}$$
We conclude that $M'R \parallel AC$, as desired.

As in Solution 1, we introduce point $M'$, and reduce the problem to proving $\frac{PR}{RN} = \frac{PM'}{M'A}$. Menelaus theorem in triangle $ANP$ with transversal line $FRE$ yields
$$\frac{PR}{RN} \cdot \frac{NE}{EA} \cdot \frac{AF}{FP} = 1.$$
Since $AF = EA$, we have $\frac{FP}{NE} = \frac{PR}{RN}$, so that it suffices to prove
$$\begin{equation*} \frac{FP}{NE} = \frac{PM'}{M'A} \tag{1} \end{equation*}$$
This is a computation regarding the triangle $AM'N$ and its excircle opposite $A$. Indeed, setting $a = M'N$, $b = NA$, $c = M'A$, $s = \frac{a+b+c}{2}$, $x = s-a$, $y = s-b$ and $z = s-c$, then $AE = AF = s$, $M'F = z$ and $NE = y$. From $\triangle OFP \sim \triangle AFO$ we have $FP = \frac{r_a^2}{s}$, where $r_a = OF$ is the exradius opposite $A$. Combining the following two standard formulas for the area of a triangle
$$|AM'N|^2 = x y z s \quad \text{(Heron's formula) and} \quad |AM'N| = r_a(s-a),$$
we have $r_a^2 = \frac{y z s}{x}$. Therefore, $FP = \frac{y z}{x}$. We can now write everything in (1) in terms of $x, y, z$. We conclude that we have to verify
$$\frac{\frac{y z}{x}}{y} = \frac{z + \frac{y z}{x}}{x + y},$$
which is easily seen to be true.

As in Solution 1, we introduce point $M'$. Let the line through $M'$ and parallel to $AN$ intersect $EF$ at $R'$. Let $P'$ be the intersection of lines $NR'$ and $AM$. It suffices to show that $P'O \parallel FE$, since this would yield $P = P'$, and then $R = R'$ and $M = M'$. Hence it is enough to prove that
$$\begin{equation*} \frac{AF}{FP'} = \frac{AD}{DO}, \tag{2} \end{equation*}$$
where $D$ is the intersection of $AO$ and $EF$. Once again, this reduces to a computation regarding the triangle $AM'N$ and its excircle opposite $A$.
Let $u = P'F$ and $x, y, z, s$ as in Solution 2a. Note that since $AE = AF$ and $M'R' \parallel AE$, we have $M'R' = M'F = z$. Since $M'R' \parallel AN$, we have $\frac{P'M'}{P'A} = \frac{M'R'}{NA}$, that is,
$$\frac{u + z}{u + x + y + z} = \frac{z}{x + z}$$
From this last equation we obtain $u = \frac{y z}{x}$. Hence $\frac{AF}{FP'} = \frac{x s}{y z}$. Also, as in Solution 2a, we have $r_a^2 = \frac{y z s}{x}$.
Finally, using similar triangles $ODF, FDA$ and $OFA$, and the above equalities, we have
$$\frac{AD}{DO} = \frac{AD}{DF} \cdot \frac{DF}{DO} = \frac{AF}{OF} \cdot \frac{AF}{OF} = \frac{s^2}{r_a^2} = \frac{s^2}{\frac{y z s}{x}} = \frac{x s}{y z} = \frac{AF}{FP'},$$
as required.