APMO 2016

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

As in Solution 1, we introduce point $M^{\prime }$, and reduce the problem to proving $\frac{PR}{RN}=\frac{P{M}^{\prime }}{M^{\prime }A}$. 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 $$\begin{array}{c}\frac{FP}{NE}=\frac{P{M}^{\prime }}{M^{\prime }A}\end{array}\text{\hspace{1em}(1)}$$ This is a computation regarding the triangle $A{M}^{\prime }N$ and its excircle opposite $A$. Indeed, setting $a=M^{\prime }N$, $b=NA$, $c=M^{\prime }A$, $s=\frac{a+b+c}{2}$, $x=s-a$, $y=s-b$ and $z=s-c$, then $AE=AF=s$, $M^{\prime }F=z$ and $NE=y$. From $△OFP\sim △AFO$ 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 $$|A{M}^{\prime }N{|}^2=xyzs\text{(Heron’s formula) and}|A{M}^{\prime }N|=r_a(s-a),$$ we have $r_a^2=\frac{yzs}{x}$. Therefore, $FP=\frac{yz}{x}$. We can now write everything in (1) in terms of $x,y,z$. We conclude that we have to verify $$\frac{\frac{yz}{x}}{y}=\frac{z+\frac{yz}{x}}{x+y},$$ which is easily seen to be true.

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