I think this problem, in slightly different form was given in Shortlist 99: http://www.kalva.demon.co.uk/short/soln/sh99g8.html Strange that it appeared in a Chinese TST then.

epitomy01 wrote: I think this problem, in slightly different form was given in Shortlist 99: http://www.kalva.demon.co.uk/short/soln/sh99g8.html Strange that it appeared in a Chinese TST then. actually it's harder than the one that was given in shortlist.and wants more things.

notice that $ \frac {PN}{PM} = \frac {NI_1}{MI_2} = \frac {NI}{MI}$ (cause $ NI = NI_1,MI = MI_2$) hence the circumcircle of $ \triangle AIP$ is apollonius circle and devide $ MN$ in a harmonic ratio. now let DI intersect circumcircle of $ ABC$ at $ P'$ and $ A'$ (Where A' and A are on the same side of MN) cause the line DA is reflection of line DA' wrt MN and D is midpoint of MN hence $ AA'||MN$ hence $ \angle NP'A' = \angle ANM = \angle IND$ hence $ DN^2 = DI.DP'$ but $ DN^2$ is power of D wrt circumcircle of AIP hence P' lies on circumcircle of AIP also hence $ P\equiv P'$

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I have used another idea. If $ IP$ passes through the mid-point of $ MN$, then $ \bigtriangleup INP$ and $ \bigtriangleup IMP$ have the same area, which is equivalent to $ \frac{PB}{PC}= \frac{sin^2 \frac{C}{2}}{sin^2 \frac{B}{2}}$ But I don't know how to find this ratio easily, so I try trigonometry: As $ \bigtriangleup PNM$ and $ \bigtriangleup PI_1I_2$ are similar, we have $ \frac{PN}{PM}= \frac{NI_1}{MI_2}= \frac{sin \frac{C}{2}}{sin \frac{B}{2}}$ Therefore if we let $ \angle PAB=\alpha$, $ \angle PAC=\beta$ We have $ \frac{sin \frac{C}{2}}{sin \frac{B}{2}}= \frac{PN}{PM}= \frac{sin( \frac{C}{2}+ \alpha)}{sin( \frac{B}{2}+ \beta)}$ Also $ \alpha+ \beta = \angle A$ And $ \frac{PB}{PC}= \frac{sin \alpha}{sin \beta}$ Thus for $ \alpha+ \beta = \angle A$ and $ \frac{sin \frac{C}{2}}{sin \frac{B}{2}}= \frac{sin( \frac{C}{2}+ \alpha)}{sin( \frac{B}{2}+ \beta)}$ We hope to show $ \frac{sin \alpha}{sin \beta} = \frac{sin^2 \frac{C}{2}}{sin^2 \frac{B}{2}}$ To prove this, I have constructed another right-angled triangle A very long solution

I have a short solution. It seems that I use similar ideas to Amir.S. I prefer a different notation. Problem: Let $ \triangle ABC$ have circumcircle $ \omega$ and incenter $ I$. $ BI\cap\omega=\{B,N\}$; $ CI\cap \omega=\{C,P\}$; $ M$ is the midpoint of $ NP$. $ G$ is an arbitrary point on arc $ BC$ not containing $ A$. $ I_1;I_2$ are the incenters of $ ABG;ACG$. The circumcircle of $ I_1I_2G$ intersects $ \omega$ again at $ T$ (if they are tangent, $ T=G$). Prove that $ P,I,T$ are collinear. Proof: Lemma 1: $ \triangle TI_1I_2\sim\triangle TPN$ and $ \triangle TPI_1\sim\triangle TNI_2$. Proof: note $ G,I_1,P$; $ G,I_2,N$ are triples of collinear points. From the fact that $ NPTG$ and $ I_1I_2TG$ are cyclic, we get the first similarity. Then we get the second similarity. Lemma 2: Let $ AI$ intersect the circumcircle of $ ABC$ at $ A'$. Then $ A'I=A'B=A'C$. Proof: angle chase. Lemma 3: $ B,C$ are fixed points. The locus of points, $ P$, such that $ BP: PC=k\ne 1$ is a circle. Proof: well-known (coordinates? there are other proofs). Lemma 4: $ ABCD$ with $ AB\parallel CD$. $ E=AC\cap BD$; $ F=BC\cap AD$, $ M,N$ are midpoints of $ AB$, $ CD$. Then $ M,N,E,F$ are collinear. Proof: well-known. Using lemma 1 and 2, $ TP: TN=PI_1: NI_2=AP: AN$ which is constant. By lemma 3, there are only two points that lie on $ \omega$ and have that constant ratio since two distinct circles intersect at at most two points. Thus $ T$ is fixed and unique. Now choose $ G=T$ and define incenters $ I_1'$ and $ I_2'$. Then $ C(GI_1'I_2')$ must be be tangent to $ \omega$ (if we let there be a second intersection of those circles, we get a contradiction since $ T$ is unique). It follows that $ I_1'I_2'\parallel NP$. Then we are done by lemma 4 using trapezoid $ NPI_1I_2$.

Let me propose another problem Let $ ABC$ be a triangle inscribed $ (O)$ and $ P$ be a point lies on small arc $ BC$, let $ I_1,I_2$ be A-excenter of triangle $ APB,APC$ prove that circumcircle of $ PI_1I_2$ always pass through a fix point.

Hint

encyclopedia wrote: Let me propose another problem Let $ ABC$ be a triangle inscribed $ (O)$ and $ P$ be a point lies on small arc $ BC$, let $ I_1,I_2$ be A-excenter of triangle $ APB,APC$ prove that circumcircle of $ PI_1I_2$ always pass through a fix point.

Hint

dear encyclopedia Could you draw a figure?I can't see your result

Hi dear plane geometry, maybe I has a typo I denote midpoint of $ MN$ as the same point $ P$, sorry, now I let it be point $ K$.

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PROOF: 1.It is easy to prove A,B,Ia1,Ia are concyclic ∠IaBP=90-∠Ia1BIa-1/2∠ABP=90-1/2∠PAC-1/2∠ABP=1/2(∠ACP-∠PAC) Notice N is the midpoint of big arc APC so∠NBP=1/2(∠ACP-∠PAC) this indicates that Ia,N,B are collinear analogously,Ia,M,C are collinear 2.From 1 we can acquire NIaMQ is a parallelogram Denote R,S are midpoints of IaQ,IaA , respectively Ia,R,Q are collinear From the original problem we know that AQ⊥IA And we have MN//AQ => MN⊥IA => MN is the perpendicular bisector of AIa 3.form 2 we have MN is the perpendicular bisector of AIa But we already know that M is on perpendicular bisector of AB => M is the circumcenter of ⊙A,B,Ia1,Ia analogously, N is the circumcenter of ⊙A,C,Ia2,Ia 4.from 3 , ∠Ia2PM=∠Ia2PC-∠CPM=90-1/2∠B-1/2(∠A-∠B)=90-1/2∠A ∠MQN=∠A-1/2(∠A-∠C)-1/2(∠A-∠B)=90-1/2∠A N,P,Ia2 are collinear analogously, M,P,Ia1 are collinear 5.Denote T=⊙O∩IaQ MT/MIa1=MT/MA=MT/NQ NT/Nia2=NT/NA=NT/MQ Notice we have MT/NQ=RT/NR=RT/RM=NT/MQ ∠TNP=∠TMP => △NTIa2∽△TMIa1 P,T,Ia1,Ia2 are concyclic Thus we are done!

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Fang-jh wrote: Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear. It is not difficult to see that $ NB = NI_1 = NI = NA$ and $ MC = MI_2 = MI = MA$. Since $ \angle I_1PN = \angle I_2PM$ and $ \angle NI_1P = \angle PI_2M$,we have $ \triangle PI_1N$ and $ \triangle PI_2M$ are congruent. =>$ \frac {PM}{PN} = \frac {I_2M}{I_2N} = \frac {AM}{AN} = \frac {sin\frac {B}{2}}{sin\frac {C}{2}}$ =>$ PM sin \frac {C}{2} = PNsin\frac {B}{2}$ Since $ \triangle PIN$ and $ \triangle PCM$ are congruent,we have $ \angle IPN = \angle CPM = \frac {B}{2}$.Likewise,$ \angle IPM = \frac {C}{2}$. $ D,I,P$ are collinear <=> $ [IPN] = [IPM]$ <=> $ PMsinIPN = PNsinCPM$ <=> $ PM sin \frac {C}{2} = PN sin \frac {B}{2}$,which is proven above.$ Q.E.D$ This is too nice a problem!

Basically, after disposing of all the non-sense fluff about the incenter and the mid-point of the arc $BAC$ it is just this http://www.artofproblemsolving.com/community/u243741h19777p5682901

Note this result - https://artofproblemsolving.com/community/c6h19777p29800996 If $T$ is Mixtilinear Incircles touch point and Mixtilinear Incircles touch $AB,AC$ at $K_1,K_2$ and $AT$ interest Mixtilinear Incircles at $X$ $$-1=(T,X;K_1,K_2)\stackrel{T}{=}(T,A;N,M) \implies P=T$$ Note if $L$ is midpoint of arc $BC$ contains $A$ then $\overline{P-I-L}$ If $Y=AI \cap (ABC)$ then $I$ is orthocenter of $\triangle YMN$ and $L$ is antipode of $Y$ so $D= IL \cap MN$ is midpoint of $MN$ $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.569cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.866804980004579, xmax = 32.960363534416956, ymin = -10.432859072921945, ymax = 8.97488583667342; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); 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