redacted... i will post a solution once I solve the problem. Edit: I solved the problem but it is identical to the solution below.

bora_olmez wrote: Does anybody know if this is real (i.e, actually Iran TST)? Do you have any reason to doubt it?

[asy][asy] import graph; size(13.455854810467093cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.6980415915099805, xmax = 2.029885813723566, ymin = -2.110469671334513, ymax = 1.3126540963980047; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (-0.6430689346649243,0.7658082953774523), B = (-0.8526372509201448,-0.5225033189782989), C = (0.8434498568353982,-0.5372079104074569), M = (0.008669388951461646,0.9999624201414813), D = (-4.1493674322957075,-0.4939216086713019), Y = (-0.5737265065554792,-0.9776385912138741), Z = (0.47276323725564023,-1.1192089758389103), P = (-0.016361180134094235,-1.887164755769925), F = (1.651842103706722,-0.5442164406005033); draw(A--B--C--cycle, linewidth(0.8) + qqffff); draw(arc(Z,0.12240013472225979,-154.0032446489424,-122.4937707182029)--(0.47276323725564023,-1.1192089758389103)--cycle, linewidth(2.) + ffxfqq); draw(arc(A,0.12240013472225979,-72.74583691171233,-41.2363629809728)--(-0.6430689346649243,0.7658082953774523)--cycle, linewidth(2.) + ffxfqq); draw(arc(Z,0.12240013472225979,385.99675535105763,417.5062292817971)--(0.47276323725564023,-1.1192089758389103)--cycle, linewidth(2.) + ffxfqq); draw(arc((-1.3082395641083786,-0.5185533768798192),0.12240013472225979,327.99380044907593,359.5032743798155)--(-1.3082395641083786,-0.5185533768798192)--cycle, linewidth(2.) + ffxfqq); draw(arc(Y,0.12240013472225979,301.5003194778338,327.99380044907593)--(-0.5737265065554792,-0.9776385912138741)--cycle, linewidth(2.) + qqwuqq); draw(arc(Y,0.12240013472225979,121.50031947783386,147.99380044907596)--(-0.5737265065554792,-0.9776385912138741)--cycle, linewidth(2.) + qqwuqq); draw(arc(F,0.12240013472225979,179.5032743798155,205.9967553510576)--(1.651842103706722,-0.5442164406005033)--cycle, linewidth(2.) + qqwuqq); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(0.8)); draw(A--B, linewidth(0.8) + qqffff); draw(B--C, linewidth(0.8) + qqffff); draw(C--A, linewidth(0.8) + qqffff); draw(D--M, linewidth(0.8)); draw(B--D, linewidth(0.8)); draw(D--Z, linewidth(0.8)); draw(P--B, linewidth(0.8)); draw(P--C, linewidth(0.8)); draw(circle((0.40365722923891034,-0.06566250349155542), 1.3367794700595546), linewidth(0.8)); draw(circle((-0.23009406180570016,-0.2624979990028388), 1.1081345047760305), linewidth(0.8)); draw((-1.3082395641083786,-0.5185533768798192)--(0.012158769469482744,-1.3438284972384043), linewidth(0.8)); draw(F--C, linewidth(0.8)); draw(F--(0.012158769469482744,-1.3438284972384043), linewidth(0.8)); draw(A--Y, linewidth(0.8)); draw((0.012158769469482744,-1.3438284972384043)--A, linewidth(0.8)); draw(A--Z, linewidth(0.8)); draw(A--(-1.3082395641083786,-0.5185533768798192), linewidth(0.8)); draw(A--F, linewidth(0.8)); draw(F--Y, linewidth(0.8)); draw(Z--(-1.3082395641083786,-0.5185533768798192), linewidth(0.8)); draw(P--(0.012158769469482744,-1.3438284972384043), linewidth(0.8)); draw(A--P, linewidth(0.8)); /* dots and labels */ dot(A,linewidth(3.pt) + dotstyle); label("$A$", (-0.6873971771106012,0.8148935485274835), NE * labelscalefactor); dot(B,linewidth(3.pt) + dotstyle); label("$B$", (-0.9281174420643787,-0.6171880277229499), NE * labelscalefactor); dot(C,linewidth(3.pt) + dotstyle); label("$C$", (0.8793245473343241,-0.6335080456859178), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("$M$", (0.018443599787763592,1.0433738000090342), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); label("$D$", (-4.135000971787585,-0.47030786605623875), NE * labelscalefactor); dot(Y,linewidth(3.pt) + dotstyle); label("$Y$", (-0.6588371456754072,-1.106788566611987), NE * labelscalefactor); dot(Z,linewidth(3.pt) + dotstyle); label("$Z$", (0.5080441386768028,-1.1924686609175685), NE * labelscalefactor); dot(P,linewidth(3.pt) + dotstyle); label("$P$", (0.03068361325998957,-1.930949473741866), NE * labelscalefactor); dot((0.012158769469482744,-1.3438284972384043),linewidth(3.pt) + dotstyle); label("$N$", (0.03884362224147356,-1.449508943834313), NE * labelscalefactor); dot((-1.3082395641083786,-0.5185533768798192),linewidth(3.pt) + dotstyle); label("$E$", (-1.385077945027482,-0.600868009759982), NE * labelscalefactor); dot(F,linewidth(3.pt) + dotstyle); label("$F$", (1.6830854320104967,-0.5804679873062721), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $NY$ intersect line $BC$ at $E$. Note that \[\angle NZY+\angle NYZ=2\angle A -\angle A \implies ZNY=180^{\circ}-A \]so the point $E$ is on the circle $CZN$. By the similar argument we can show that $F=NZ \cap BC$ is on the circle $BYN$. on the other hand it is trivial that $\triangle BEY \sim \triangle CFZ$. now by law of sines we have: \[\frac{sin(PNZ)}{sin(PNY)}=\frac{PZ\times sin(NZP)}{PY \times sin(NYP)}\]so we have to prove that: \[\frac{PY}{PZ}=\frac{sin(NZP)}{sin(NYP)} \quad (1)\]also by Menelaus's theorem we have that: \begin{align*} \frac{PY}{PZ} &= \frac{CD}{BD} \times \frac{PY}{ZC} \\ &= \frac{AC}{AB} \times \frac{BY}{CZ} \end{align*}now we have that: \begin{align*} \frac{AC}{AB} \times \frac{BY}{CZ}&=\frac{sin(CZE)}{sin(BYE)} \\ \iff \frac{AC}{AB}\times \frac{BY}{sin(BEY)}\times \frac{sin(BYE)}{CZ}&=1 \\ \iff \frac{AC}{AB}\times \frac{BE}{CZ}&=1 \\ \iff \frac{AC}{AB}\times \frac{AB}{AC}&=1 \end{align*}so we are done.

iman007 wrote: \begin{align*} \iff \frac{AC}{AB}\times \frac{BY}{sin(BEY)}\times \frac{sin(BYE)}{CZ}&=1 \\ \iff \frac{AC}{AB}\times \frac{BE}{CZ}&=1 \\ \iff \frac{AC}{AB}\times \frac{AB}{AC}&=1 \end{align*} Can you elaborate on this part?

Maretangens wrote: Can you elaborate on this part? just apply law of sines in triangle $\triangle BYE$ also note that $\triangle ABE \sim ACZ$

Let $NY,NZ \cap BC$ at $U,V$ respectively. Since $\measuredangle ANU = \measuredangle ANY = \measuredangle ABY = \measuredangle ACB = \measuredangle ACU$. So, $A,C,Z,U,N$ are concyclic. Similary to $A,B,Y,V,N$ are concyclic. Let $M = DN \cap BC$. By DDIT $\square BCYZ$ and project by point $N$, $\{B,V\},\{C,U\},\{M,X\}$ are recipocal pairs. Therefore, $\odot(BVA),\odot(CUA),\odot(MXA)$ are coaxial. So, $A,M,X,N$ are concyclic. So, $\angle MNA = \angle AXM = \frac{|\angle B|-|\angle C|}{2}$. But $\angle AXY = \angle C$ and $\angle AXZ = B$. So, $\angle MXY = \angle MXZ = \frac{|\angle B|+|\angle C|}{2}$ [asy][asy] import graph; import geometry; size(500); pair A,B,C,R,X,B0,C0,Y,Z,U,V; A =dir(130); B = dir(200); C = dir(-20); R = dir(90); X = extension(A,R,B,C); B0 = foot(B,(0,0),(-1,0)); C0 = foot(C,(0,0),(1,0)); pair conjB = 2*B0-B; pair conjC = 2*C0-C; pair D = 2/(conjB+conjC); Y = 0.4*D+0.6*B; Z = extension(X,Y,C,D); pair[] N = intersectionpoints(circumcircle(A,C,Z),circumcircle(A,B,Y)); U = extension(B,C,Y,N[1]); V = extension(N[1],Z,B,C); pair M = extension(D,N[1],B,C); fill(B--C--Z--Y--cycle,0.1*orange+0.9*white); fill(X--B--D--Z--cycle,0.1*green+0.9*white); fill(A--B--C--cycle,0.1*cyan+0.9*white); draw(unitcircle); draw(line(B,C)); draw(B--D); draw(C--D); draw(X--1.5*Z-0.5*Y,red); draw(circumcircle(A,C,Z),blue); draw(circumcircle(A,B,Y),blue); draw(D--M,green); draw(N[1]--U,magenta); draw(N[1]--V,magenta); draw(A--B--C--cycle); draw(circumcircle(A,M,X),red); dot("$A$",A,dir(100)); dot("$B$",B,dir(220)); dot("$C$",C,dir(-40)); dot("$D$",D,dir(D)); dot("$X$",X,dir(220)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$U$",U,dir(220)); dot("$V$",V,dir(-40)); dot("$N$",N[1],dir(N[1])); dot("$M$",M,dir(240)); [/asy][/asy]

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Menelaus theorem Oson masalaku bu Kim nimalar qib tashlagansila

Really Nice problem Let $BC$ meet $AYB$ at $K$ and $AZC$ at $S$. Claim1 : $ASY$ and $AZK$, $ASB$ and $AZC$, $ABY$ and $ACK$, $BSY$ and $CZK$ are similar. Proof : $\angle AYS = \angle AKN = \angle AKZ$ and $\angle ASY = \angle ASN = \angle AZK$ $\implies$ $ASY$ and $AZK$ are similar. $\angle SAB = \angle SAZ - \angle BAZ = \angle SCZ - \angle BAZ = \angle BAC - \angle BAZ = \angle ZAC$ and $\angle ASB = \angle ASC = \angle AZC$ $\implies$ $ASB$ and $AZC$ are similar. we prove similarity of $ABY$ and $ACK$ with same approach and Note that proving all similarities proves that $BSY$ and $CZK$ are similar as well. Claim2 : $N,Y,S$ and $N,Z,K$ are collinear. Proof : $\angle ZNS = \angle 180 - \angle ZAS = \angle 180 - \angle BAC = \angle ABC + \angle ACB = \angle ANZ + \angle ANY = \angle ZNY$ so $N,Y,S$ are collinear. we prove the other one with same approach. Now note that $\frac{\sin{DNY}}{\sin{DNZ}} = \frac{DY}{DZ} . \frac{\sin{DYN}}{\sin{DZN}}$ and $\frac{\sin{DYN}}{\sin{DZN}} = \frac{\sin{BYS}}{\sin{CZK}} = \frac{\sin{BYS}}{\sin{BSY}}$ so we need to prove $\frac{DZ}{DY} = \frac{\sin{BYS}}{\sin{BSY}}$. Note that with Menelaus on $BCD$ we have $\frac{DZ}{DY} = \frac{BX}{CX} . \frac{ZC}{BY} = \frac{AB}{AC} . \frac{ZC}{BY} = \frac{BS}{CZ} . \frac{ZC}{BY} = \frac{BS}{BY} = \frac{\sin{BYS}}{\sin{BSY}}$. we're Done.

Why $$\frac{CD}{BD}=\frac{AC}{BC}$$?

Figure of problem !

Moving points trivializes the problem? Define $W$ as the midpoint of arc $BC$ not containing $A$. $Y$ is a moving point on line $l_b$, hence has degree $1$, and evidently $$Y \rightarrow YD \rightarrow ZD \rightarrow Z \Leftrightarrow Y \rightarrow Z$$Therefore, $Z$ has degree $1$.Evidently, $(AYB),AZC$ have degree $1$. Hence, $N$ has at most $d(AYB)+d(AZC)=2$ degree. Notice that if $Y$ approaches $B$, then $(AYB)=(AZC)$. Hence, $N$ has degree $2-1=1$. Therefore it is enough to prove the problem for $1+1=2$ cases. \begin{align*} & \bullet Y = D \qquad \text{We have $Y=N=Z=D$ which is clearly true.}\\ &\bullet Y=B \qquad \text{ $Z=C ,\ N=W$. Since $D$ is intersection of tangent lines from $B,C$, by definition of $W$ it is true.} \end{align*}Hence we are done. Remark: Well, I realized #$4$ used the same thing, but i am confused why $l \neq BC$, I think it is indeed possible

Iora wrote: Moving points trivializes the problem? Define $W$ as the midpoint of arc $BC$ not containing $A$. $Y$ is a moving point on line $l_b$, hence has degree $1$, and evidently $$Y \rightarrow YD \rightarrow ZD \rightarrow Z \Leftrightarrow Y \rightarrow Z$$Therefore, $Z$ has degree $1$.Evidently, $(AYB),AZC$ have degree $1$. Hence, $N$ has at most $d(AYB)+d(AZC)=2$ degree. Notice that if $Y$ approaches $B$, then $(AYB)=(AZC)$. Hence, $N$ has degree $2-1=1$. Therefore it is enough to prove the problem for $1+1=2$ cases. \begin{align*} & \bullet Y = D \qquad \text{We have $Y=N=Z=D$ which is clearly true.}\\ &\bullet Y=B \qquad \text{ $Z=C ,\ N=W$. Since $D$ is intersection of tangent lines from $B,C$, by definition of $W$ it is true.} \end{align*}Hence we are done. Remark: Well, I realized #$4$ used the same thing, but i am confused why $l \neq BC$, I think it is indeed possible $N$ doesn't even have degree $1$