Let $ T_{\infty} $ be the infinity point on $ AE. $ From $ OP $ $ \parallel $ $ EF $ we get the second tangent from $ P $ to $ \odot (O) $ is parallel to $ AE, $ so from the converse of Brianchon's theorem for the degenerate hexagon $ MFPT_{\infty}EN $ we conclude that $ MN $ is tangent to $ \omega. $

Let $X$ be the second tangency point from $N$ to $\omega$. Let $GE$ be a diameter of $\omega$. Let $Z = PG \cap NE \cap MR$ now (at infinity). [asy][asy]size(5cm); defaultpen(fontsize(9pt)); pair E = dir(140); pair F = dir(40); pair A = 2*E*F/(E+F); pair O = origin; pair X = dir(100); pair N = 2*E*X/(E+X); pair M = extension(A, F, N, X); pair G = -E; pair P = 2*F*G/(F+G); pair R = extension(N, P, E, F); filldraw(unitcircle, palecyan, blue); draw(E--A--P, blue); draw(O--P, mediumcyan); draw(E--F, mediumcyan); draw(M--R, red); draw(A--E, red); draw(G--P, dashed+red); draw(N--P, blue); draw(E--G, dotted+red); draw(N--X, blue); draw(X--M, dotted+blue); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$A$", A, dir(A)); dot("$O$", O, dir(225)); dot("$X$", X, dir(X)); dot("$N$", N, dir(N)); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$P$", P, dir(P)); dot("$R$", R, dir(250)); [/asy][/asy] Take a projective transform which sends $EXFG$ to a rectangle centered at $R$. Then $M = PF \cap NX \cap RZ$. Thus $MNX$ collinear, done.

Should i actually post this? I feel embarrassed to post this with the existence of the above two solutions Redefine $M$ to be the point on $AB$ so that $MN$ is tangent to $\omega$, then we want to prove $RM\parallel AC$. Let $X=AO\cap EF$, use Menelaus theorem, then $\frac{AP}{PF}\cdot \frac{FR}{RE}\cdot \frac{EN}{NA}=1$. So we want to prove $$\frac{PR}{RN}=\frac{PM}{MA}$$$$\iff \frac{AP}{PF}\cdot \frac{PM}{MA}\cdot \frac{EN}{NA}=1$$But $OP\parallel EF$, so this is equivalent to $$\frac{AO}{OX}\cdot \frac{PM}{MA}\cdot\frac{EN}{NA}=1$$Let us view $\omega$ to be the excircle of $\triangle AMN$, then we get the following problem: Given $\triangle ABC$, if $A$-excircle with center $O$ is tangent to $AB, AC$ at $M, N$, and $AO\cap MN=X$ then prove $\frac{AO}{OX}=\frac{AB}{BM}\cdot\frac{AC}{CN}$. Upon homothety mapping excircle to incircle, and use the standard lengths on incircle and excircle, we get the following problem: Given $\triangle ABC$, if incircle with center $I$ is tangent to $BC, CA, AB$ at $D, E, F$, and $AO\cap EF=M$ then prove $\frac{AI}{IM}=\frac{AB}{BD}\cdot \frac{AC}{CD}$. Now we tackle this problem using trigo, this is standard incircle configuration, so it shouldn't be hard. Here we go: Denote $\alpha=\frac{\angle A}{2}, \beta=\frac{\angle A}{2}, \gamma=\frac{\angle C}{2}$ First compute that $$\frac{AI}{IM}=(\frac{AF}{FM})^2=\frac{1}{\sin^2 \angle FAI}=\frac{1}{\sin^2 \alpha}$$Next, let's compute $\frac{AB}{BD}$. We have $$\frac{AB}{BD}=\frac{AB}{BI}\cdot \frac{BI}{BD}=\frac{\sin \angle AIB}{\sin \angle BAI}\cdot \frac{1}{\cos\angle IBD}=\frac{\cos \gamma}{\cos \beta}\cdot \frac{1}{\sin \alpha}$$Then likewise $\frac{AC}{CD}=\frac{\cos \beta}{\cos \gamma}\cdot \frac{1}{\sin \alpha}$. So this easily verifies $\frac{AI}{IM}=\frac{1}{\sin^2 \alpha}=\frac{AB}{BD}\cdot \frac{AC}{CD}$. Done. To above two solutions: What is your motivation to brianchon/projective transformation idea? I am curious to know how you came out with it.

navi_09220114 wrote: To above two solutions: What is your motivation to brianchon/projective transformation idea? I am curious to know how you came out with it. This problem falls into a class of problems which I like to call "purely projective", id est it can be phrased entirely in terms of incidence (concurrency, tangency, et al) and at most one circle. In any problem of this form, one can generally take a projective transformation to destroy the entire problem like I've done. This method is discussed in more detail in Section 9.7 "Projective Transformations" of my textbook, and briefly in Prasolov.

My solution, found after the test, relies upon no synthetic observation and just on trigonometry, although very ugly(to type it out). Anyways yours are really nice solutions!

Let $X$ be the second tangency point from $N$ to $\omega$ and $Y$ be the reflection of $F$ over $OP.$ Clearly $Y$ lies on $\omega$ and $FY$ is the diameter of $\omega,$ moreover $Y$ is the second tangency point form $P$ to $\omega.$ Let $R^*$ be the intersection of $\overline{XY}$ and $\overline{EF}.$ It's easy to see that $N,P,R^*$ lies on the polar of $EX\cap YF,$ hence we conclude that $R^*\equiv R.$ Therefore $\angle RXF=\tfrac{1}{2}\angle A=\tfrac{1}{2}\angle MRF$ (here $\angle RMF=\angle A$ follows by parallel condition), which amounts to $M$ is the center of $\odot(XRF)\implies MX=MF\implies MX$ is tangent to $\omega.$ The statement follows. $\square$ [asy][asy] size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair O=D("O",origin,S); pair E=D("E",dir(150),dir(150)); pair F=D("F",dir(30),dir(-100)*1.5); pair A=D("A",IP(L(-E*dir(90)+E,E,5,5),L(-F*dir(-90)+F,F,5,5)),N); pair P=D("P",IP(L(F-E,O,5,5),L(A,F,5,5)),dir(45)); pair R=D("R",6*F/11+5*E/11,dir(-130)); pair N=D("N",extension(P,R,A,E),dir(135)); pair X=D("X",IP(CP(N,E),unitcircle),dir(60)); pair M=D("M",extension(N,X,A,F),dir(45)); pair Y=D("Y",extension(X,R,E,O),dir(-60)); D(unitcircle); D(A--E--F--cycle); D(N--M,blue+dashed); D(X--Y--E,red); D(O--P--F); D(M--R); D(CP(M,R),dotted); D(X--O--F); D(Y--P,dashed); } b(); pathflag=false; b(); [/asy][/asy]

v_Enhance wrote: Let $X$ be the second tangency point from $N$ to $\omega$. Also, let $GF$ be a diameter of $\omega$. Let $Z = PG \cap NF \cap MR$ now. Take a projective transform which sends $EXFG$ to a rectangle centered at $R$. Then $M = PE \cap NX \cap RZ$. Thus $MNX$ collinear, done. Please elaborate on this - what do you mean by "a projective transform"? Do you have links (aside from your textbook)? Could you also elaborate on your method and how does it "destroy" the "purely-projective" problems? I have prasolov - which chapter is this? thanks a lot

TelvCohl wrote: Let $ T_{\infty} $ be the infinity point on $ AE. $ From $ OP $ $ \parallel $ $ EF $ we get the second tangent from $ P $ to $ \odot (O) $ is parallel to $ AE, $ so from the converse of Brianchon's theorem for the degenerate hexagon $ MFPT_{\infty}EN $ we conclude that $ MN $ is tangent to $ \omega. $ Would you mind explaining why does this work? Brianchon's theorem (the other direction) used for a degenerate circumscribed hexagon in which three consecutive vertices lie on a line isn't always true, as noted here. It is possible that the converse doesn't have any such special cases, but a short Google search doesn't confirm or refute anything about the matter. For example, the author of this handout claims the degenerate version is true, but only uses it when one of the three collinear points is the contact point of the conic section, like it is used in this solution.

I've found something about this issue: Quote: A geometrical theorem can be used in a degenerate case if either its proof still functions in this case, or one can deduce the degenerate case from the generic case by a limiting argument. Our application of the Brianchon theorem to the hexagon AX'BCZD did not match any of these two conditions; thus, it was not legitimate. Hence, there is no wonder the resulting assertion was wrong. But: Darij Grinberg wrote: In other words: The hexagon formed by the lines a1; b1; c1; d1; e1; f1 may be degenerated, but if two adjacent sides lie on one line, then the vertex where these sides meet must be the point of tangency of this line with the circle, and not just an arbitrary point on this line. If the proof of the converse is analogous, then the proof works because the tangent points are the vertices of the degenerate hexagon, not just any points on the sides.

Here is a boring solution: Let the tangent to $\omega$ at $M$ different from $MF$ meet $AE$ at $N'$. We wish to prove that $N', R$, and $P$ are collinear. By Menelaus' Theorem in $\triangle AEF$ it suffices to prove that $$\frac{AF}{FP} \cdot \frac{FR}{RE} \cdot \frac{EN'}{N'A} = -1$$ Now, set $MN' = a$, $N'A = b$, $AM = c$, $s = \frac{a + b + c}{2}$ and $\angle FOP = \alpha$. Then $$\frac{EN'}{N'A} = \frac{s - b}{b}$$ And $$\frac{FR}{RE} = \frac{FM}{MA} = \frac{s - c}{c}$$ Now, notice that $PO^2 = PF \cdot PA$, and hence $\frac{AP}{PF} = -\frac{PO^2}{PF^2} = -\frac{1}{\sin^2 \alpha}$. Putting all of this together, we wish to prove $$\sin^2 \alpha = \frac{(s - b)(s - c)}{bc}$$ Which upon noticing that $\angle MAO = \alpha = \frac{\angle MAN'}{2}$, is a trigonometric identity depending only on triangle $MAN'$, which is easy to prove using the Law of Cosines and standard manipulation.

์Newton 's theorem

Just an observation: Almost each year recently has a purely projective APMO problem!

v_Enhance wrote: Let $X$ be the second tangency point from $N$ to $\omega$. Let $GE$ be a diameter of $\omega$. Let $Z = PG \cap NE \cap MR$ now (at infinity). [asy][asy]size(5cm); defaultpen(fontsize(9pt)); pair E = dir(140); pair F = dir(40); pair A = 2*E*F/(E+F); pair O = origin; pair X = dir(100); pair N = 2*E*X/(E+X); pair M = extension(A, F, N, X); pair G = -E; pair P = 2*F*G/(F+G); pair R = extension(N, P, E, F); filldraw(unitcircle, palecyan, blue); draw(E--A--P, blue); draw(O--P, mediumcyan); draw(E--F, mediumcyan); draw(M--R, red); draw(A--E, red); draw(G--P, dashed+red); draw(N--P, blue); draw(E--G, dotted+red); draw(N--X, blue); draw(X--M, dotted+blue); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$A$", A, dir(A)); dot("$O$", O, dir(225)); dot("$X$", X, dir(X)); dot("$N$", N, dir(N)); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$P$", P, dir(P)); dot("$R$", R, dir(250)); [/asy][/asy] Take a projective transform which sends $EXFG$ to a rectangle centered at $R$. Then $M = PF \cap NX \cap RZ$. Thus $MNX$ collinear, done. I wanted to ask. If we observe the diagram when it is a rectangle, we have $F = P = G$ ane $N = E$. But you said that $M = PF \cap NX \cap RZ$. I understand that $PF \cap RZ = M$. But how come either $PF \cap NX $ or $NX \cap RZ$ is the point at infinity? They are not parallel, are they?

IndoMathXdZ wrote: If we observe the diagram when it is a rectangle, we have $F = P = G$ ane $N = E$. This seems wrong. Here is the post-transformed diagram. [asy][asy] pair E = dir(220); pair F = -E; pair X = dir(140); pair G = -X; pair O = origin; pair N = 2*E*X/(E+X); pair P = 2*F*G/(F+G); pair M = extension(P, F, N, X); pair R = extension(N, P, E, F); pair T = extension(N, E, P, G); filldraw(unitcircle, invisible, blue); draw(M--P, blue); draw(O--P, heavycyan); draw(E--F, heavycyan); draw(M--T, red); draw(N--T--P, red); draw(N--P, blue); draw(E--G, dotted+red); draw(N--X, blue); draw(X--M, dotted+blue); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$X$", X, dir(X)); dot("$G$", G, dir(G)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$R$", R, dir(315)); dot("$T$", T, dir(T)); /* TSQ Source: E = dir 220 F = -E X = dir 140 G = -X O := origin N = 2*E*X/(E+X) P = 2*F*G/(F+G) M = extension P F N X R = extension N P E F R315 T = extension N E P G unitcircle 0.1 lightcyan / blue M--P blue O--P heavycyan E--F heavycyan M--T red N--T--P red N--P blue E--G dotted red N--X blue X--M dotted blue */ [/asy][/asy]

v_Enhance wrote: IndoMathXdZ wrote: If we observe the diagram when it is a rectangle, we have $F = P = G$ ane $N = E$. This seems wrong. Here is the post-transformed diagram. [asy][asy] pair E = dir(220); pair F = -E; pair X = dir(140); pair G = -X; pair O = origin; pair N = 2*E*X/(E+X); pair P = 2*F*G/(F+G); pair M = extension(P, F, N, X); pair R = extension(N, P, E, F); pair T = extension(N, E, P, G); filldraw(unitcircle, invisible, blue); draw(M--P, blue); draw(O--P, heavycyan); draw(E--F, heavycyan); draw(M--T, red); draw(N--T--P, red); draw(N--P, blue); draw(E--G, dotted+red); draw(N--X, blue); draw(X--M, dotted+blue); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$X$", X, dir(X)); dot("$G$", G, dir(G)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$R$", R, dir(315)); dot("$T$", T, dir(T)); /* TSQ Source: E = dir 220 F = -E X = dir 140 G = -X O := origin N = 2*E*X/(E+X) P = 2*F*G/(F+G) M = extension P F N X R = extension N P E F R315 T = extension N E P G unitcircle 0.1 lightcyan / blue M--P blue O--P heavycyan E--F heavycyan M--T red N--T--P red N--P blue E--G dotted red N--X blue X--M dotted blue */ [/asy][/asy] Thanks a lot for your help

Here is my solution Let $MX$ be the second tangent through $M$ and let it intersect ray $AC$ at point $X$ We prove that $P$,$R$ and $X$ are collinear For convenience we denote $AM$ by $c$ ,$AN$ by $b$ and $MN$ by $a$ It needs to be proven that $$AP\div PF\times FR\div RE\times EX\div XA=1$$$$FR\div RE=FM\div MA=(s-c)\div c$$and $$EX\div XA=(s-b)\div b$$Since OP is parallel to FE it follows that $AO\perp OP$ Also $OF\perp AP$ Therefore $$AP\div PF=(AP\times AF)\div (PF\times AF)=AO^2\div OF^2=(s^2+r_A^2)\div r_A^2$$Now it needs to be proven that $$((s^2+r_A^2)\div s^2)\times (s-b)(s-c)\div bc=1$$Using $(s-b)(s-c)=r\times r_A$ and $(s-b)+(s-c)=(a)$ we get the desired result. Comment:I guess this problem is hard to prove using elementary techniques like angle chasing and radical axis and easy using bash, inversion,projective geometry,etc.

MonsterS wrote: ์Newton 's theorem Redefine $M$ s.t. $M$ lies on $FF$ and $MN$ is tangent to $\omega$. Let $G$ be the antipode of $E$. Then $GP$ is tangent to $\omega$. Let $L$ be the point at infinity along $EE$ and $GG$. It suffices to prove that $M, R, L$ are collinear, which follows from Newton's Theorem in $LNMP$.

Here's a nice solution using moving points. shinichiman wrote: Let $AB$ and $AC$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $AC$ at $E$ and ray $AB$ at $F$. Let $R$ be a point on segment $EF$. The line through $O$ parallel to $EF$ intersects line $AB$ at $P$. Let $N$ be the intersection of lines $PR$ and $AC$, and let $M$ be the intersection of line $AB$ and the line through $R$ parallel to $AC$. Prove that line $MN$ is tangent to $\omega$. Move $R$ on the line $EF$ and fix everything else. Clearly the degrees of $N$ and $M$ are both $1.$ Hence degree of the line $NM$ is at most $1+1=2.$ So the degree of the pole of $NM$ is $2$ which hence moves on a conic, and so by the projective dual of this $NM$ is tangent to a fixed conic $\mathcal{C}$. To show $\mathcal{C} \equiv \omega,$ we need to check this for $5$ positions of $R,$ since a conic is uniquely determined by $5$ points. Take $R=E,F,\infty_{EF}$ and the result is trivial here. Take $R \in EF$ such that $PR \parallel AC$ and the result is again true here. For the last position, take $N$ such that the line parallel to $EF$ through $N$ is tangent to $\omega.$ With some work, we can show that this works. So we are done. $\blacksquare$

Draw the second tangent to $\omega$ through $P$ which touches $\omega$ at $Y$. Now if $\angle FAE=\theta\implies\angle AFE=90^\circ-\frac{\theta}{2}=\angle APO\implies\angle APY=180^\circ-\theta\implies PY\|AE\|MR$. So let $PY,MR,AE$ concur at the point at infinity $(P_{\infty})$. So, $M,R,P_{\infty}$ are collinear. Now let $X$ be a point on $\omega$ such that $NX$ is a tangent to $\omega$. So, it suffices to prove that $MX$ is also a tangent to $\omega$. Now let $X'$ be a point on $\omega$ such that $MX'$ is a tangent to $\omega$. So, by Pascal's Theorem on $FFX'X'EY$ and $EEYYFX'$ we get that $M,YX'\cap EF,P_{\infty}$ are collinear. So, $YX'\cap EF\equiv R\equiv EF\cap XY\implies X'\equiv X$. So, $MX$ is also a tangent to $\omega$. Hence, $MN$ is tangent to $\omega$. $\blacksquare$.

A solution without branchion, projective transformation and polars lemma:

return to our problem: redifine it as Let $N$ a point on $AE$ , $X \neq E$ a point on $\omega$ such that $NX$ is tangent to $\omega$ $M = NX \cap AF $ , $R = PN \cap EF$ prove that $MR || AE$ solution :

and we are done

Here's a synthetic solution I found (which incidentally doesn't use projective geometry): First, we claim the following: Let $ABC$ be a triangle with $DEF$ being the intouch triangle, and $I_A$ the $A-$excenter. Let $E', F'$ be the $A-$mixtilinear excircle touchpoints with $AC, AB$ respectively. Then the intersection of the line through $B$ which is parallel to $AC$ and the line $CF'$ is the reflection of $E$ in the midpoint of $BC$ and lies on the $A-$excircle touchchord. For the proof, note that by $\sqrt{bc}$ inversion at $A$ and a flip over the angle bisector, since $F'$ is mapped to $E$, the lines $BE, CF'$ are parallel. Taking a reflection in the midpoint of $BC$, the first part of our claim is proved. Now, since the $A-$intouch chord and the $A-$extouch chord meet $BC$ at isotomic points, they are reflections in the midpoint of $BC$ since they are parallel, proving the second part of the claim. Coming back to the main problem, let the tangent from $N$ to $\omega$ meet $AF$ at $X$. Applying our claim to $AXN$ as the reference triangle, we see that $XR \parallel AN$ (since $OP$ coincides with the $A$-mixtilinear excircle touchchord), so $X = M$, as needed.

Problem is about excircle but nothing prevents it to be about incircle, so for the sake of variety I'll reformulate it as follows. Restated wrote: Incircle of $ABC$ with center $I$ tangent to sides $BC, AC$ at $D,E$ resp. $DE$ meet line through $A$ parallel to $BC$ at $P$. $BP$ meet $AC$ at $X$. Prove that $IX \parallel DE$.

By Brianchon on $DBAEX\infty_{BC}$ we see that tangent from $X$ to incircle is parallel to $BC$. So it follows that $\angle IXC = \angle DEC$.

dame dame

My first homography problem! I probably messed up somewhere. Special thanks to Elliott Liu and Jeffrey Chen for helping me work through my debilitating confusion. Let $G$ be the second tangency point from $P$ to $\omega$. Then, since $FE\parallel PO$ we have that $\angle (EF,FG)= \angle (OP,FG) = 90$. Thus, by Thale's $EG$ is a diameter. Denote with $Q$ the point at infinity defined by $AE\cap MR \cap PG$. Now by Theorem 2.4, the special case of conic mappings, we may take the interior point $R$ and map it to the center of $\omega$ while maintaining collinearity and the cross ratio. In our new diagram, $R$ is the center and $EF$ is a diameter. Note that since $N,R,P$ are still collinear and $N,P$'s tangents are parallel, we have that $R$ is the midpoint of $NP$. Similar argument holds for $Q,M$. Thus, after taking a homothety with factor $-1$ at $R$, we have that $PQ$ is tangent to $\omega$ at $G$ implies that $MN$ is tangent to $\omega$ and we're done. $\blacksquare$.

This was pretty easy to trig-length bash. Here's a sketch. We prove $MR||AN$, and assume tangency. Do Menelaus for triangle $AEF$ and $P-R-N$. If we let $<APO=x$, then $AP/EP=cos(x)$ and $<MAN=180-2x$. Use Thales in $AEF$ to see we want $ER/RF=EM/MA=MX/MA$ ($X$ is the tangency point). Also for the Menelaus expression, we use $FN/AN=XN/AN$. So now we have everything in the Menelaus expression in terms of the sides of $AMN$ and $cos(x)$. Use LoC is triangle $AMN$ to finish (we have the lengths $MX$ and $NX$ because we have an excircle for triangle $AMN$).

My solution by moving points. Suppose the tangent from $N$ to $\omega$ touches $\omega$ at $X\neq E$. We have to prove that the points $M, N, X$ are collinear. Let calculate the degrees of the points (here degrees are at most). Let $R$ be the moving point on line $EF$ and $\omega , E, F, A$ are fixed. Therefore $P$ is also fixed. Now deg(N)=1 just because $N$ is the intersection of $AE, PR$. $AE$ has degree $0$ and $RP$ has $1$. So deg(N)=1. By the similar process deg(M)=1. Now have to calculate the degree of the point $X$. As $R$ varies on the line $EF$, the point varies on the circle $\omega$. Therefore deg(X)=2. So we have to check for $1+1+2+1=5$ cases that $M, N, X$ collinear. Before checking $5$ cases, lets reduce the degrees. (i) when $F=P$ then easily $N=X=E$. (ii) when $A$ lies on the circle then everything is same. Easily checked. (iii) when $R=P$ then same as previous. reducing $3$ degrees we have to show only $2$ cases. when $R=E$ then collinear and when $R=F$ then also collinear. Remark: Please anybody can fix error then inform kindly.

Let the tangent from $P$ to $\omega$ intersect $\omega$ at $G.$ Claim: $PG \parallel MR \parallel AE.$ Proof: $\angle{AFE}=90-\frac{\angle{A}}{2}=\angle{APO}=\angle{GPO} \implies \angle{GPB}=\angle{CAB}.$ $\Box$ Now, use the converse of Brianchon's Theorem on degenerate hexagon $MFPP_{\infty}EN$ gives us the desired. $\blacksquare$

Move $R$ on $EF.$ Let distinct from $AB$ tangent to $\omega$ through $M$ meet $AC$ by $M';$ since angle $\angle MOM'=\frac{\pi-\angle BAC}{2}$ is fixed, mapping $M\mapsto M'$ is composition of rotation and central projection, so it's a homography. Also composition of central projections $M\stackrel{\infty_{AC}}{\mapsto} R\stackrel{P}{\mapsto} N$ is a homography, so to prove $M'=N$ it's suffice to check it for three cases; trivial when $R$ coincides with $E,F.$ When $R=\infty_{AC}$ we see $M=\infty_{AB}, N=OP\cap AC,$ and $MN$ is tangent to $\omega$ by symmetry wrt $O.$

Let tangent at N to $\omega$ meet it at K and meet AF at M'. we'll prove M' is M. we can prove M'R || AC. NE/NQ = NR/NP so we will prove AM'/AP = NE/NQ. Let PO meet $\omega$ at Q. Let perpendicular to M' meet PQ at S. we will prove ASP,NOQ and ASM',NOE are similar. ( I found triangle ENQ interesting so my idea was to find a point S such that ASP,NOQ and ASM',NOE are similar and then we can easily prove AM'/AP = NE/NQ.) Claim1 : OEN and SM'A are similar. Proof : ∠OEN = 90 = ∠SM'A and ∠NOE = (180 - KOF - FOP - EOQ)/2 = 90 - ∠M'OF - ∠FOP = ∠AOM' = ∠ASM'. Claim2 : ONQ and SAP are cyclic. ∠NOQ = ∠NOE + ∠EOQ = ∠AOM' + ∠FOP = ∠ASM' + ∠M'SP = ∠ASP and ∠ONQ = 90 - ∠NOE = 90 - ∠ASM' = ∠SAP. Now by similarities we have NE/AM' = NQ/AP so NE/NQ = AM'/AP. we're Done.

Here's a different solution using Moving Points Move $M$ on $(AB)$ and let the tangent from $M$ to $\omega$ (other than $MB$ ) intersect line $AC$ at $N_1$ $, $ $N_2=PR\cap AC$ Since $M\to R\to N_2$ all are projective $\implies M\to N_2$ is projective Let $D$ be the tangency point of $MN_1$ with $\omega$ and denote by $X'$ the image of $X$ under an inversion w.r.t to $\omega$ Since $M\to M'\to D\to N_1'\to N_1$ all are projective hence $M\to N_1$ is also projective Therefore to show $N_1=N_2$, it suffices to check three values of $M.$ $\bullet$ $M=A\implies R=N_1=E$ so $N_2=E \implies N_1=N_2$ $\bullet$ $M=F\implies R=F,N_1=A$ so $N_2=A \implies N_1=N_2$ $\bullet $$M=\infty_{AB}\implies R=\infty_{EF},MN_1\parallel AB$ So $D$ is the reflection of $F$ w.r.t to $O$ Since $R=\infty_{EF}$ then $PR\parallel EF$ so to show$N_1=N_2$ it's enough to check that $ON_1\parallel EF$ but note that : $\angle{DON_1}=\angle{DEN_1}=\angle{EDN_1}=\angle{DFE}$ $ $ $ $ $ $ $\implies ON_1\parallel EF $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$ \blacksquare$

Define $K$ and $L$, distinct from $E$ and $F$, such that $NK$ and $PL$ are tangent to $\omega$. $\textcolor{blue}{\textbf{Claim 1:}}$ Points $K$, $R$, and $L$ are collinear. This is equivalent to proving $KL \cap EF$ lies on $NP$. By Brocard on $EKFL$, we know $KL \cap EF$ lies on the polar of $FL \cap EK$. However, we know $N$ and $P$ also lie on this polar by La Hire's. ${\color{blue} \Box}$ $\textcolor{blue}{\textbf{Claim 2:}}$ $M$ is the circumcenter of $\triangle FRK$. We first note $L$ and $F$ are reflections over $OP$ by symmetry, so \[FL \perp PO \parallel EF \implies EL \text{ is a diameter of } \omega.\] Since $\triangle FMR \sim \triangle FAE$ by the given parallel condition, we know $FM = MR$ and \[\angle FKR = \angle FEL = \frac{180 - \angle FOE}{2} = \frac{\angle A}{2} = \frac{\angle FAE}{2} = \frac{\angle FMR}{2}.~{\color{blue} \Box}\] Thus $MF = MK$, so $MKN$ is tangent to $\omega$. $\blacksquare$ [asy][asy] size(270); pair A, B, C, O, E, F, P, R, N, M, K, L; A = dir(95); B = dir(210); C = dir(320); O = incenter(A, B, C); E = foot(O, A, C); F = foot(O, A, B); P = extension(O, O+F-E, A, B); R = .2E + .8F; N = extension(A, C, P, R); M = extension(A, B, R, A+R-E); K = foot(O, M, N); L = 2*foot(F, P, O)-F; draw(C--A--B^^E--F^^K--L); draw(M--R^^P--N--K^^O--P--L); draw(M--K^^L--extension(L, F, E, K)--E, dashed); draw(incircle(A, B, C)); dot("$A$", A, dir(90)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$E$", E, dir(0)); dot("$F$", F, dir(180)); dot("$K$", K, dir(90)); dot("$L$", L, dir(270)); dot("$O$", O, dir(270)); dot("$P$", P, dir(180)); dot("$M$", M, .6dir(140)); dot("$N$", N, dir(0)); dot("$R$", R, dir(305)); dot(extension(L, F, E, K)); [/asy][/asy]

APMO vibes

Notice $M \to R \to N$ is projective. When $M = E, A$ the result is obvious. To show $M = \infty_{AC}$ works, it suffices to prove that $P$ is the intersection of $AB$ and the reflection of $AC$ through $O$, since in this case $N = P$. This follows immediately since $OP$ is the external angle bisector of $\angle MOE$, finishing. Now since a projective map is defined by three points and it is well known that mapping from a line to a second line via tangents in a circle tangent to both lines is projective, we are done.

Animate $R$ with degree $1$, so $M$ has degree $1$, $\overline{OM}$ has degree $1$, and thus the reflection of the fixed $\overline{AF}$ over $\overline{OM}$—also the other tangent from $M$—has degree $2$ (by a tangent addition coordinate calculation). Its intersection $N'$ with $\overline{AE}$ has degree $1$ by strong Zack's lemma since when $R=E$ the reflection coincides with $\overline{AE}$. It suffices to show that $P,R,N'$ are collinear, which is now a degree $2$ condition, so we check $3$ cases. When $R=F$ we have $N'=A$ and the collinearity is obvious When $R=\infty_{\overline{EF}}$ we have $M=\infty_{\overline{AF}}$ and the reflected line is the other line parallel to $\overline{AF}$ tangent to $\omega$, so $N'$ is the reflection of $P$ over $O$, implying the desired collinearity. When $R$ is chosen such that $M=P$, the reflected line is the other line parallel to $\overline{AE}$ tangent to $\omega$, so $N'=\infty_{\overline{AE}}$ and the collinearity follows. This finishes the problem. $\blacksquare$

By replacing $N$ with a phantom point defined according to a tangency (so that $\omega$ is the $A$-excircle of $\triangle AMN$) extraverting, and relabeling, it suffices to prove the following. Quote: Let $ABC$ be a triangle with incircle $\omega$, incenter $I$, and intouch triangle $DEF$. Let $P$ be the point on $\overline{AB}$ such that $\overline{AI} \perp \overline{IP}$. Prove that $\overline{EF}$, $\overline{CP}$, and the line $\ell$ through $B$ parallel to $\overline{AC}$ concur. We use barycentric coordinates. Note that the point at infinity along $\overline{AC}$ has coordinates $(1:0:-1)$, and thus $\ell$ has equation $x+z=0$. We have $E=(s-c:0:s-a)$ and $F=(s-b:s-a:0)$, so $\overline{EF}$ has equation $(s-a)x-(s-b)y-(s-c)z=0$, so to find $\overline{EF} \cap \ell$ we need to solve $$(s-a)x+(s-c)x=bx=(s-b)y,$$hence the intersection $X$ has coordinates $(s-b:b:b-s)$. Thus $P':=\overline{CX} \cap \overline{AB}=(s-b:b:0)$. It suffices to show $\overline{AI} \perp \overline{IP'}$; the relevant displacement vectors are $(b+c:-b:-c)$ and $(c-b:b:-c)$, and we want to show \begin{align*} 0&=a^2(bc-bc)+b^2(-c(b+c)-c(c-b))+c^2(b(b+c)-b(c-b))\\ &=-b^3c-b^2c^2-b^2c^2+b^3c+b^2c^2+bc^3-bc^3+b^2c^2=0 \end{align*}so we're done. $\blacksquare$

By replacing $N$ with a phantom point defined according to a tangency (so that $\omega$ is the $A$-excircle of $\triangle AMN$) extraverting, and relabeling, it suffices to prove the following. Quote: Let $ABC$ be a triangle with incircle $\omega$, incenter $I$, and intouch triangle $DEF$. Let $P$ be the point on $\overline{AB}$ such that $\overline{AI} \perp \overline{IP}$. Prove that $\overline{EF}$, $\overline{CP}$, and the line $\ell$ through $B$ parallel to $\overline{AC}$ concur. Let $X=\ell \cap \overline{EF}$; we use Menelaus on $\triangle AEF$ to show that $X,P,C$ are collinear. In what proceeds all ratios are unsigned. We have $\frac{AC}{CE}=\frac{b}{s-c}$. Since $\triangle BFX \sim \triangle AFE$, we have $\frac{EX}{XF}=\frac{AB}{BF}=\frac{c}{s-b}$. Finally, let $T=\overline{AI} \cap \overline{EF}$; since $\overline{EF} \perp \overline{IP}$, we have $$\frac{FP}{PA}=1-\frac{AT}{AI}=1-\frac{\cos \frac{\angle A}{2}}{\sec \frac{\angle A}{2}}=1-\cos^2 \frac{\angle A}{2}=\sin^2 \frac{\angle A}{2}=\frac{1-\cos \angle A}{2}=\frac{a^2-b^2-c^2+2bc}{4bc}=\frac{(a-b+c)(a+b-c)}{4bc}=\frac{(s-b)(s-c)}{bc}.$$Clearly none of $X,P,C$ lie on the segment sides of $\triangle AEF$, so since $\frac{b}{s-c} \cdot \frac{c}{s-b} \cdot \frac{(s-b)(s-c)}{bc}=1$ we conclude. $\blacksquare$

Here goes homography: Let $G$ denote the second tangent from point $P$. Note that: $PG \parallel MR \parallel AE$ (say they are concurrent at $T$, thus $T$ being point at infinity on line $AE$ right now). Let $NX$ denote the second tangent of point $N$. Thus, there exists an homography which maps $EFGE$ to a rectangle. Thus: by converse of Brianchon's theorem on $MFPTEN$, $NM$ must be tangent to $\omega$ at $X$.