grobber wrote: Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent. Pretty tough to start. For clarity purposes, let A' = Z. Let E be the intersection of the incircle with AB and F the intersection of the incircle with AC. Let O be the intersection of AZ and the incircle. We want to prove that OM / MB * BZ / ZC * CN / NO = 1 (@). (By Ceva's, that implies BN, CM, and AZ are concurrent.) angle BOZ = angle BZM (incircle is tangent to BC at Z) and angle COZ = angle CZN. Thus, triangle BOZ ~ triangle BZM and triangle COZ ~ triangle CZN. We then have BO / BZ = BZ / BM and CO / CZ = CZ / CN. Substituting in @, we see that we must prove BO / BZ * OM / ON * CZ / CO = 1. Again, from similar triangles, we have BO / BZ = OZ / ZM and CO / CZ = OZ / ZN. Substituting again, we see that we must prove OM / ZM * NZ / NO = 1, or OM * ZN = ON * ZM. For this, we use inversion. Invert about point O with arbitrary radius. We have the following figure: The incircle I is inverted into a line that contains E', M', Z', N', and F' (in that order) and line BC is inverted into a circle that is tangent to I' at Z'. Line AB is inverted into a circle that is tangent to I' at E' and meets circle (BC)' at B' and O. Line AC is inverted into a circle that is tangent to I' at F' and meets circle (BC)' at C' and O. Finally, the ray Z'O goes through the other intersection of circles (BC)' and (AC)' at A'. Now, we claim that M'Z' = Z'N'. Z'E' = Z'O * Z'A' = Z'F'. M'E'^2 = M'B' * M'O = M'Z'^2 implies M'E' = M'Z'. Likewise, N'Z' = N'F'. Thus, Z'E' = Z'F' implies 2M'Z' = 2N'Z' implies M'Z' = Z'N'. By inversive distances, OM / OZ' = MZ / M'Z' and ON / OZ' = NZ / N'Z'. Because M'Z' = Z'N', we have MZ / MO = NZ / NO. Then OM * ZN = ON * ZM, as desired.

My solution: I did use some things I won't prove here. For example: Lemma Let ABCD be a cyclic quadrilateral. Assume the tangents at A and C to its circumcircle and BD are concurrent. Then the tangents at B and D to the circumcircle and AC are also concurrent. Let A', B', C' be the points where the incircle touches BC, CA, AB respectively. In order to prove that AA', BN, CM are concurrent me must prove (by Desargue's thm) that the intersection pts of the pairs of lines (MN, BC); (A'N, BP); (A'N, CP) are collinear. Let A'P \cap MN=X. It's well known that the line formed by the intersection pts of (A'N, BP) and (A'N, CP) is the polar of X with respect to the circle, so what we want is for MN \cap BC to be on the polar of X, which is equivalent to X being on the polar of BC \cap MN, so the tangents to the circle at A' and P and MN must be concurrent. What we've proved here is that this condition is equivalent to what we initially wanted to prove. I'll come back later. As you can see, it's going to take a while to explain everything.

grobber wrote: I did use some things I won't prove here. For example: Lemma Let ABCD be a cyclic quadrilateral. Assume the tangents at A and C to its circumcircle and BD are concurrent. Then the tangents at B and D to the circumcircle and AC are also concurrent. Let A', B', C' be the points where the incircle touches BC, CA, AB respectively. In order to prove that AA', BN, CM are concurrent me must prove (by Desargue's thm) that the intersection pts of the pairs of lines (MN, BC); (A'N, BP); (A'N, CP) are collinear. Let A'P \cap MN=X. It's well known that the line formed by the intersection pts of (A'N, BP) and (A'N, CP) is the polar of X with respect to the circle, so what we want is for MN \cap BC to be on the polar of X, which is equivalent to X being on the polar of BC \cap MN, so the tangents to the circle at A' and P and MN must be concurrent. What we've proved here is that this condition is equivalent to what we initially wanted to prove. I'll come back later. As you can see, it's going to take a while to explain everything. OK. Before, that, did you mean (A'N, BP) and (A'M, CP), or (A'N, BP) and (A'N, CP) as you stated? Because (A'N, CP) = N, in which case the line formed by the intersection pts of those two points is A'N. I am familiar with that lemma. Perhaps it can be posted as a problem on its own. Also, what is the "polar" of a point? =D Thanks!

I have solved this one before ( it's really a nice one): We name the intersection of MN and BC, D. Instead of showing that AA', BN and CM are concurrent, we show that (DBA'C) = -1. First we will prove that KP is tangent to the circle: K is on the polar of A, so its polar passes through A, it passes also through A' so AA' is the polar of K. So KP is the second tangent from K to the incircle. Now to finish the proof I will prove that P (DBA'C) = -1. We intersect the rays of this system with center P with the line MN, (let X = (AA', MN) so we have to prove that P(DMXN) = -1 so we have to prove that A'(DMXN)= -1.We intersect the rays with the incircle, so we have to prove that (A'MPN)= -1.So we have to prove that P(A'MPN) = -1 .We intersect the rays with BC ,so we have to prove that (KBA'C) = -1,and that is obvious because AA', BB' and CC' are concurrent.

hossein11652 wrote: We name the intersection of MN and BC, D. Instead of showing that AA, BN and CM are concurrent, we show that (DBAC) = -1. First we will prove that KP is tangent to the circle: K is on the polar of A, so its polar passes through A, it passes also through A so AA is the polar of K. So KP is the second tangent from K to the incircle. I think you have forgotten to say: Define K as the point of intersection of B'C' and BC. Hereby, B' and C' are the points where the incircle of ABC touches CA and AB. hossein11652 wrote: Now to finish the proof I will prove that P (DBAC) = -1. We intersect the rays of this system with center P with the line MN, (let X = (AA, MN) so we have to prove that P(DMXN) = -1 so we have to prove that A(DMXN)= -1.We intersect the rays with the incircle, so we have to prove that (AMPN)= -1.So we have to prove that P(AMPN) = -1 .We intersect the rays with BC ,so we have to prove that (KBAC) = -1,and that is obvious because AA, BB and CC are concurrent. Thanks, a very nice proof! Darij

well done! I think we should have more problems which can be solved by this method

sorry what does (ABCD) denote?

The term (ABCD) denotes the double ratio of four points A, B, C, D, i. e. (ABCD) = (AB / BC) / (AD / DC). The notion P (ABCD) means the same as (ABCD), but with the collateral signification that we project the points from the point P. Darij

I have found a merely philosophical solution to the problem. It uses nothing but the following lemma from projective geometry: Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle. Now, consider the projective transformation letting the incircle of triangle ABC fixed and mapping the triangle A'B'C' (formed by the points of tangency of the incircle with the sides BC, CA, AB) to an equilateral triangle. Then, the image of triangle ABC will be an equilateral triangle, too (since the tangents at the vertices of an equilateral triangle to its circumcircle form an equilateral triangle themselves), and everything follows from simple symmetry observations. This strategy helps killing several problems. Are we allowed to use it at the IMO? Darij

Are we allowed to use affine transformations, for example, to reduce a triangle geometry problem to the isosceles case? And when exactly is it "correct" to use this technique?

Yes one of of the contestantants solved this with affine transformations,and reduced the problem to isoceles triangle. He got the perfect mark

I think in Iran a lot of things are considered well - known. I saw some official solutions that used some hard to prove facts. Anyway, can somebody tell how much of projective tranformations can be used at IMO?

DusT wrote: I think in Iran a lot of things are considered well - known. I saw some official solutions that used some hard to prove facts. Anyway, can somebody tell how much of projective tranformations can be used at IMO? Any solution that uses well-known mathematical facts, even mathematicals facts that are way further the imo level (advanced calculus, etc.) are taken into consideration, but obivously you will never get partial credit for stating something like "this problem CAN be solved using advanced calculus easy, we are leaving the details for the reader ..." . You are more likely to receive partial credit if you state something as above, but instead of "advanced calculus" put a smart idea upon which a solution can be developed. In case of solved problems you will receive full-score if the mathematics you use are correct.

Hi, all. I have a solution to this one (hope it is different from Art of problem solving) : Let the incircle touch AB and AC at B', C'. Lemma : If $A_1, B_1, C_1$ be points lying on the incircle, then $AA_1, BB_1, CC_1$ are concurent if and only if $A'A_1, B'B_1, C'C_1$ are concurent. This one is easy to prove, just use Ceva trig. Therefore, $MC', NB', AP$ are concurent at $K$ since $BM, CN, AP$ are concurent at P. On the other hand, according to Pascal 's theroem for 6 points $M, N, B', C', B', C'$ we obtain $T$ lies on $AK$, where $T$ is the intersection of $C'N$ and $B'M$. Thus, T is on AA'. Applying our lemma again we finally get $AA', BN, CM$ are concurent since $A'P, C'N, B'M$ are concurent at $T$.

treegoner wrote: Let the incircle touch AB and AC at B', C'. Lemma : If $A_1, B_1, C_1$ be points lying on the incircle, then $AA_1, BB_1, CC_1$ are concurent if and only if $A'A_1, B'B_1, C'C_1$ are concurent. This one is easy to prove, just use Ceva trig. Are you sure about your Lemma? I think that if you have three points $A_1$, $B_1$, $C_1$ on the incircle such that the lines $A'A_1$, $B'B_1$, $C'C_1$ concur, then the lines $AA_1$, $BB_1$, $CC_1$ also concur, that's right, but the converse is not generally true, since you can call $A_2$ the second point intersection of the line $AA_1$ with the incircle, and then the line $AA_2$ is the same as the line $AA_1$, so that the lines $AA_2$, $BB_1$, $CC_1$ still concur, but the lines $A'A_2$, $B'B_1$, $C'C_1$ don't concur anymore. I think I know your error: You got an equation of the kind $\left( \frac{...}{...}\cdot \frac{...}{...}\cdot \frac{...}{...}\right) ^{2}=1$, and concluded that $\frac{...}{...}\cdot \frac{...}{...}\cdot \frac{...}{...}=1$ without caring for the sign. Actually, I myself have sometimes made this error... Darij

Thank you Darij. I did make a stupid mistake But luckily, this problem isn't influenced by that. OK. I'll rewrite my lemma Lemma Let $A_1, B_1, C_1$ be points lying on the incircle. Then If $A'A_1, B'B_1, C'C_1$ are concurent then $AA_1, BB_1, CC_1$ are concurent. The whole point of the problem is to prove $MC', AA', NB' $ are concurent (parallel is considered to be concurent). This one is easy since you can use Ceva to prove that (Don't worry, this one is true ) Then similarly you use the Pascal theorem and continue as I said in my solution In fact, my first solution is this one. But a stupid idea flew by my mind.

I'm asking someone for the proof of the following lemma: darij grinberg wrote: Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle. I just started playin with projective geometry so all help welcome. BTW. Do you know where I can find some useful theorems and lemmas for projective geometry with proofs ?

Omid Hatami wrote: Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the side $BC$ at $A^{\prime}$, and the line $AA^{\prime}$ meets the incircle again at a point $P$. Let the lines $CP$ and $BP$ meet the incircle of triangle $ABC$ again at $N$ and $M$, respectively. Prove that the lines $AA^{\prime}$, $BN$ and $CM$ are concurrent. This difficult and very nice problem is an instructive aplication of the harmonical quadrilateral and the its properties. I suggest to see the characterization of the harmonical quadrilateral from the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=43652 Proof. I"ll use only the properties of the harmonical quadrilateral $\mathrm{(\ h.q.\ )}$ Define the tangent $XX$ in the point $X\in w$ to the incircle $w\ .$ Denote the intersection $R\in PP\cap BC$ and the points $B'\in AC$ , $C'\in AB$ which belong to the incircle $w$ of the triangle $ABC\ .$ Recall the relations $A'B=BC'$ and $A'C=CB'\ .$ $1\blacktriangleright\ A\in C'C'\cap B'B'$ and $P\in AA'$ $\Longrightarrow$ $PB'A'C'\--\ \mathrm{h.q.}$ $\Longrightarrow$ $R\in B'C'$ and $\frac{PC'}{PB'}=\frac{A'C'}{A'B'}\ \ (1)\ .$ $2\blacktriangleright\{\begin{array}{c}B\in C'C'\cap A'A'\ ,\ M\in PB\Longrightarrow PC'MA'\--\ \mathrm{h.q.}\Longrightarrow T\in A'C'\ ,\ PC'\cdot MA'=PA'\cdot MC'=\frac{1}{2}\cdot PM\cdot A'C'\\\\ C\in BB'\cap A'A'\ ,\ N\in PC\Longrightarrow PB'NA'\--\ \mathrm{h.q.}\Longrightarrow S\in A'B'\ ,\ PB'\cdot NA'=PA'\cdot NB'=\frac{1}{2}\cdot PN\cdot A'B'\end{array}$ $\Longrightarrow$ $\frac{MP}{NP}=\frac{MC'}{A'C'}\cdot \frac{A'B'}{NB'}\ \ (2)\ .$ $3\blacktriangleright\{\begin{array}{c}\triangle CB'N\sim\triangle CPB'\Longrightarrow NC\cdot PB'=B'C\cdot B'N\\\\ \triangle BC'M\sim\triangle BPC'\Longrightarrow MB\cdot PC'=C'B\cdot C'M\end{array}$ $\Longrightarrow \frac{NC}{MB}=\frac{B'C}{C'B}\cdot\frac{B'N}{PB'}\cdot\frac{PC'}{C'M}$ and $(1)$ $\Longrightarrow$ $\frac{NC}{MB}=\frac{B'C}{C'B}\cdot\frac{B'N}{C'M}\cdot\frac{A'C'}{A'B'}\ \ (3)\ .$ Observe that $R\in B'C'\Longrightarrow$ $\frac{RB}{RC}=\frac{A'B}{A'C}\ \ (3)\ .$ From the product of the relations $(2)$ , $(3)$ , $(4)$ obtain $\frac{RB}{RC}\cdot \frac{NC}{NP}\cdot\frac{MP}{MB}=1\ ,$ i.e. the points $R$ , $M$ , $N$ are collinearly $\Longrightarrow$ the lines $BN$ , $CM$ , $AA'$ are concurrently.

Hello, this is just a small generalization of this question and an incomplete proof. Im sorry for not completing the proof; I've been trying this lemma for a while now (previously in connection with http://www.mathlinks.ro/Forum/viewtopic.php?t=113117 with no success ) , maybe someone who has seen it before or can solve it can help me with it, thanks. Let $ABCD$ be a quadrilateral circumscribed about a circle $\omega$. Let $P$ be any point on $AC$, and the segments $PB$ and $PD$ meet $\omega$ at $B'$ and $D'$. Then $DB', BD', AC$ are concurrent. (Note: this is also true if we take $B'$ and $D'$ as the intersections of $\omega$ with the lines but no the segments $PB$ and $PD$. ) As you can see, this becomes the topic problem when $C$ is on $\omega$, i.e. $\omega$ becomes the incircle of $\triangle ABD$, and $P$ is the other intersection point of $AC$ with $\omega$. Lemma: Let $ABCD$ be a quadrilateral circumscribed about a circle $\omega$. Let $P$ be any point on $AC$. Let the segments $PA, PB, PC, PD$ meet $\omega$ at $A', B', C', D'$ respectively. Then $A'B'C'D'$ is a harmonic quadrilateral. I don't know how to prove this, but I've tried it out in geometry software and it works. Let $K$ be the pole of $AC$ wrt $\omega$ and $E$ be the intersection of $AC$ and $BD$. By Ceva's Theorem, it suffices to show \[\frac{BE}{ED}\frac{DD'}{D'P}\frac{PB'}{B'B}=1----------(1)\] Since $K$ lies on the third diagonal of the complete quadrilateral $ABCD$ and the diagonals of the complete quadrilateral divide each other harmonically, we get $(K,E;B,D) =-1$ and: \[\frac{BE}{ED}\frac{BK}{KD}=-1-------(2)\] Combining (1) and (2), it suffices to show by Menelaus using $\triangle BPD$ that $B', D', K$ are collinear. But $K$ is the intersection of tangents to $\omega$ at $A'$ and $C'$, thus this is equivalent to showing that $A'B'C'D'$ is a harmonic quadrilateral, which is 'true' by the Lemma. Perhaps the lemma is no easier to prove than the question itself , but I've seen uses for this lemma in other problems so maybe it could be useful to know. Thanks!

Let $B'$ and $C'$ be the $B$- and $C$-intouch points respectively, and let $\overline{BC} \cap \overline{B'C'}=X$. Since $X$ lies on the polar of $A$, $A$ lies on the polar of $X$, which should include $A'$ as well, hence this polar is just $\overline{A'P}$ and $\overline{XP}$ is tangent to the incircle. Then, since $\overline{AA'},\overline{BB'},\overline{CC'}$ concur, $$-1=(X,A';B,C)\stackrel{P}{=}(P,A';M,N),$$hence $\overline{MN}$ passes through the intersection of the tangents to the incircle at $A'$ and $P$, which is just $X$. Therefore, $\overline{PA'},\overline{BN},\overline{CM}$ concur, since $(X,A';B,C)=-1$ (here we view $X$ as $\overline{BC} \cap \overline{MN}$), as desired. $\blacksquare$

yofro wrote: Megus wrote: I'm asking someone for the proof of the following lemma: darij grinberg wrote: Given a circle and a triangle inscribed in it, there always exists a projective transformation letting the circle fixed and mapping the triangle to an equilateral triangle. I have found a very nice proof of this lemma.

How can we be sure that $\omega$ is an ellipse and not some other conic? I am not too familiar with projective transformations but this does not seem immediately clear to me

@above Projective images of the circle under transformations in which a line not intersecting the circle is mapped into the line at infinity are ellipses. This is a fundamental fact of homographies. If you want a proof of even this fact you may consider the central projection definition of projective transformation. See this. Thanks for pointing this out though, I've updated my blog. If you want to see why the line is outside the circle, it's sort of obvious: it's the polar of the symmedian point. The symmedian point is clearly inside the circle so the result follows.

forgot that Ceva-Menelaus works in the other direction [asy][asy] unitsize(4cm); pair A = dir(120); pair B = dir(205); pair C = dir(335); pair I=incenter(A, B, C); pair AA=foot(I, B, C); pair BB=foot(I, C, A); pair CC=foot(I, A, B); pair X=extension(B, C, BB, CC); pair P = IP(Line(A, AA, 10), circumcircle(AA, BB, CC), 0); pair M = IP(Line(P, B, 10), circumcircle(AA, BB, CC), 1); pair N = IP(Line(P, C, 10), circumcircle(AA, BB, CC), 0); draw(A--B--C--cycle, heavygreen); draw(circumcircle(AA, BB, CC), heavygreen); draw(P--X--C, red); draw(X--N, red); draw(X--BB, red); draw(A--AA, red); draw(B--P--C, heavygreen); draw(B--N, red); draw(C--M, red); dot("$B$", B, SSW); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$I$", I, NE); dot("$A'$", AA, dir(AA)); dot("$B'$", BB, NE); dot("$C'$", CC, SW); dot("$X$", X, dir(X)); dot("$P$", P, dir(P)); dot("$N$", N,dir(N)); dot("$M$", M,NE); [/asy][/asy] Let $B'$ and $C'$ be the points at which the incircle meets $\overline{AC}$ and $\overline{AB}$, and $X$ be the intersection of the tangents to the circle at $P$ and $A'$. Then $A'P$ is the polar of $X$, so by La Hire's we find that $X$ lies on line $B'C'$ (the polar of $A$). Therefore, since $AA'$, $BB'$, and $CC'$ concur at the Gergonne point, by Ceva-Menelaus we get $$-1 = (X, A'; B, C) \overset{P}{=} (P, A'; M, N),$$so $PMA'N$ is harmonic and $X$ lies on line $MN$. Now, from Descargue's on $\triangle PBC$ and $\triangle A'NM$, it suffices to show that $X$, $\overline{PB} \cap \overline{A'N}$, and $\overline{PC} \cap \overline{A'M}$ are collinear; but this is true by Pascal's on $A'A'NPPM$. So, we are done.

Let $\ell$ be the tangent to the incircle at $P$ and $D=\ell\cap BC$. The incircle touches sides $CA$, $AB$ in points $E$, $F$, respectively. Note that line $A'P$ is the pole of $D$ in the incircle. Since $A\in A'P$ we have from La-Hire's theorem that $D$ lies on pole of $A$, which is line $EF$. So, $D,F,E$ are collinear. Since $AA',BE,CF$ are concurrent, we conclude from Ceva-Menelaus' configuration that $(D,A';B,C)=-1$. From perspectivity at $P$, we have that $PMA'N$ is a harmonic quadrilateral. So, lines $\ell$, $MN$ and $BC$ concur (At $D$), which means $D,M,N$ are collinear. Again from Ceva-Menelaus' configuration, we have that $BN,CM,AA'$ concur at some point. $\square$

Denote $B'$ and $C'$ as the touch points of the incircle with $AC$ and $AB$. Also, let $K=B'C'\cap BC$, $I$ be the incenter, and $I'$ be the foot of $I$ onto $AA'$. AS $AA'$, $BB'$, $CC'$ are concurrent, we can say: \[(KA';BC)=-1.\] Using the radical center on the incircle, $(II'A)$, $(B'II'C)$, we get that $II'$, $B'C'$, and $BC$ concur at $K$. Also, as $PB'A'C'$ is harmonic, $KP$ tangent to the incircle. Therefore: \[-1=(KA';BC)\overset{P}{=}(PA';MN),\]implying that $M,N,K$ are collinear. As a result, by Ceva-Menelaus, and $(KA';BC)=-1$, the desired line concur.

Define $B'$, $C'$ analogously. Let $D$ be the polar of $APA'$ with respect to the incircle, which lies on the tangents to the incircle at $P$ and $A'$. Then, since $A$ lies on the polar of $D$ it follows that $D$ lies on the polar of $A$, which is $B'C'$. Since $AA'$, $BB'$, $CC'$ concur at the Gergonne point, it follows that by Ceva-Menelaus that $(DA';BC) = -1$. As such, it follows that \[ (DA';BC) \overset{P}= (PA';MN) = -1 \]and thus $D$ also lies on $MN$. As such, by the converse of Ceva-Menelaus this implies that $AA'$, $BN$ and $CM$ concur.

Let $\omega$ be the incircle of $\triangle ABC$, and let $B'$ and $C'$ be the points of tangency of $\omega$ to $CA$ and $AB$, respectively. Also, define $X=CM\cap AA'$, $Y=BN\cap AA'$, and $G=AP\cap BB'\cap CC'$. Since $A$, $P$, and $A'$ are collinear, it follows that $G$ must exist, and is exactly the Gergonne point of $\triangle ABC$. Since $BA'$ and $BC'$ are both tangents to $\omega$ and $B$, $M$, and $P$ are collinear, it follows that $A'PC'M$ is a harmonic quadrilateral. Projecting from $C$ onto $AA'$, we find that \begin{align*} (PX;GA')&\overset{C}=(PM;C'A')\\ &=-1. \end{align*} By a symmetric argument, we also have that $A'NB'P$ is harmonic, so $(PY;GA')=-1$. Thus $(PX;GA')=(PY;GA')=1$, implying that $X=Y=AA'\cap BN\cap CM$. Hence $AA'$, $BN$, and $CM$ are concurrent, as desired. $\blacksquare$ - Jörg

Let $B'$ and $C'$ be defined analogously to $A'$. Let $T$ be the point on $BC$ such that $\angle TPI = 90$. It is well known that $TB'C'$ are collinear. Projecting through $P$ we find, \begin{align*} -1 = (TA', BC) \overset{P}{=} (PA', MN) \end{align*}so $PMA'N$ is harmonic. Then we claim that $T$, $M$ and $N$ are collinear. Let $G$ be the point such that $MG$ and $NG$ are tangents to the incircle. Clearly then $G$ lies on $AA'$ from harmonic $MA'NP$. Thus $G$ lies on the polar of $T$ . Then by La Hires we know that $T$ lies on the polar of $G$, or $MN$. Now we can conclude from $T$, $M$, $N$ collinear and $(TA', BC) = -1$ that in triangle $PBC$ the cevians $PA'$, $NB$ and $MC$ are concurrent.

Let $B'$ and $C'$ be the other intouch points. Since $PB'A'C'$ is harmonic, we have $B'C'$, $\overline{PP}$, and $\overline{A'A'}$ concur at $X$. By Ceva, $AA'$, $BB'$ and $CC'$ concur so by Ceva-Menelaus we have $(\overline{B'C'} \cap \overline{BC}, A'; C, B) = -1 \implies \overline{B'C'} \cap \overline{BC} = X$. Then notice that $(X, A'; C, B) \overset{P}= (P, A'; N, M) = -1$, so $PA'NM$ is a harmonic quadrilateral. This implies that $\overline{A'A'}$ and $\overline{MN}$ intersect at $X$, so by Ceva-Menelaus we have $BN$ and $CM$ and $AA'$ concurrent.

Very simple to be honest if you know basic projective geometry. Let $B'$, $C'$ be the tangency points of the incircle with $AC$ and $AB$ respectively. Now define $T=\overline{PP} \cap \overline{B'C'} \cap \overline{BC}$ (this exists since $(P,A';C',B')=-1)$. It is equivalent to show that $T$, $M$ and $N$ are collinear; and this is true since \[-1=(T,A';B,C) \overset{P}{=} (P,A';M,N)\]which implies our desired.

Let $B'$ and $C'$ be the $B$-intouch point and the $C$-intouch point respectively. Note that $(A',P;B',C')=-1$. Thus define $T$ as the intersection between $BC$, $B'C'$, and the tangent to the incircle at $P$. It is well-known that cevians $AA', BB', CC'$ of $\Delta ABC$ concur at a point, so $(A',T;B,C)=-1$. Since $(A',P,M,N) \stackrel{P}{=} (A',T;B,C)=-1$, $T$ lies on $MN$. So the cevians $PA',BN,CM$ of $\Delta PBC$ are concurrent.

Let $B'$ and $C'$ be other tangency points of incircle. We know $(A', B'; N, P)=-1$ and $(P,M;C',A')=-1$. Letting $X=C'N \cap PA',$ taking perspectivity at $X$ gives $$(P, B'X \cap incircle; C', A') \overset{X}{=} (A', B'; N, P) = -1=(P, M; C' A'),$$which implies $ B'X \cap incircle=M$ so $B', X, M$ are collinear. Since $(B', C'; P, A')=-1,$ taking perspectivity at $X$ gives $(M, N; A', P)=-1.$ This means $PP, A'A', MN$ are concurrent, say at $Y.$ Perspectivity at $P$ gives points $Y, B, A', C$ are harmonic. Thus, $PA', BN, CM$ are concurrent (follows from Ceva and Menelaus).

Define $B'$ and $C'$ analogously to $A'$. Note that $A'B'PC'$ is a harmonic quadrilateral, so $\overline{PP}$, $\overline{B'C'}$, and $\overline{A'A'}$ concur at a point $X$. Thus EGMO 9.11 gives \[=1 = (X,A';B,C) \overset{P}{=} (P,A',M,N).\] Hence, $X$ lies on $\overline{MN}$ as well, so the converse of EGMO 9.11 on $\triangle PBC$ proves the desired result. $\square$