Are w1 and w2 tangent to w external and internal?

Xblade124 wrote: Are w1 and w2 tangent to w external and internal? $\omega_1$ and $\omega_2$ pass through P which is inside $\omega $. So they only can be tangent internally.(assume ABCD is convex)

Let the $O$ be the center of $\omega$. Let $\mathbb{I}$ be the negative inversion centered at $P$ which preserves $\omega$. Denote by $Z'$ the image of $Z$ under $\mathbb{I}$. Clearly $A=C'$ and $B=D'$. Note that $DPCE_1$ and $APBF_1$ are cyclic and inversely similar (1). Let $E'C$ intersect $\omega$ the second time at $C_1$ and let $BF'$ intersect $\omega$ the second time at $B_1$. $$\angle{DC_1E'}=\angle{PBC} \ \ \text{and} \ \ \angle{DE'C_1}=\angle{CPB} \implies \triangle{E'C_1D} \sim \triangle{PBC}\implies \frac{E'D}{PC}=\frac{E'C_1}{PB} \ \ \ (2)$$$$\angle{AB_1F'}=\angle{PCB} \ \ \text{and} \ \ \angle{AF'B_1}=\angle{CPB} \implies \triangle{F'B_1A} \sim \triangle{PCB}\implies \frac{F'A}{PB}=\frac{F'B_1}{PC} \ \ \ (3)$$$$\frac{BF'}{E'C}\stackrel{\text{(1)}}{=}\frac{AB}{DC}\stackrel{\text{(1)}}{=}\left(\frac{PB}{PC}\right)^2\frac{PC}{PB}\stackrel{\text{(1)}}{=}\left(\frac{PB}{PC}\right)^2\frac{DE'}{AF'}=\frac{PB\frac{E'D}{PC}}{PC\frac{F'A}{PB}}\stackrel{\text{(2),(3)}}{=}\frac{E'C_1}{B_1F'} \implies \text{pow}(F',\omega)=\text{pow}(E',\omega) \implies OE'=OF'$$Now we clearly have that the tangent to $\omega$ at $X'$ is parallel to $E'F'$ from inversion, but combining this with the fact from above we have that $OX'$ is the perpendicular bisector of $E'F'$, which means that $X'E'=X'F' \implies$ $$\frac{EX}{XF}\stackrel{\text{from inversion}}{=}\frac{\frac{E'X'}{t}PE\cdot PX}{\frac{F'X'}{t}PF\cdot PX} =\frac{PE}{PF}\stackrel{\text{from symmetry}}{=}\frac{EY}{YF}\implies \frac {EX}{EY}=\frac {FX}{FY} \ \ \ \blacksquare$$

Etemadi wrote: Consider quadrilateral $ABCD $ inscribed in circle $\omega $. $P\equiv AC\cap BD$. $E$, $F$ lie on sides $AB$, $CD$ respectively such that $\hat {APE}=\hat {DPF} $. Circles $\omega_1$, $\omega_2$ are tangent to $\omega$ at $X $, $Y $ respectively and also both tangent to the circumcircle of $\triangle PEF $ at $P $. Prove that: $$\frac {EX}{EY}=\frac {FX}{FY} $$Proposed by Ali Zamani We make the stronger claim that $X,Y$ lie on the $P$-appollonian circle of $\triangle PEF$. Invert at point $P$. Now we get the new set-up: Inverted Iran TST 2018/3/2/6 wrote: Let $ABCD$ be a cyclic quadrilateral circumscribed in circle $\omega$ with center $O$ and let $P=\overline{AC} \cap \overline{BD}$. Let $E,F$ be points on $\odot(PAB)$ and $\odot(PCD)$ with $\angle APE=\angle DPF$. Let $X,Y$ be points on $\omega$ such that tangents to $\omega$ at $X,Y$ are parallel to line $\overline{EF}$. Then prove that $\overline{XY}$ is the perpendicular bisector of line $\overline{EF}$. It suffices to show $OE=OF$ since the chord cut by $\overline{EF}$ on $\omega$ has $\overline{XY}$ as its perpendicular bisector. Now we have two methods to conclude from here: (a) By Brokard's Theorem, $O$ is the inverse of the Miquel point of $ABCD$ that lies on the $P$-appollonian circle in $\triangle PEF$, in the original problem. Hence $OE=OF$. (b) Notice that $APBE$ and $DPCF$ are similar quadrilaterals. Let $\overline{PF}$ meet $\odot(APB)$ again at $E'$ and $\overline{PE}$ meet $\odot(CPD)$ again at $F'$. Note that $\tfrac{PE}{PE'}=\tfrac{PF}{PF'}$ proving $EFE'F'$ is cyclic. Now $AEE'B$ is an isosceles trapezoid hence $OE=OE'$. Similarly, $OF=OF'$ hence $O$ is the center of $\odot(EFE'F')$ proving $OE=OF$. Either way, we're done! $\blacksquare$

anantmudgal09 wrote: (b) Notice that $APBE$ and $DPCF$ are similar quadrilaterals. Let $\overline{PF}$ meet $\odot(APB)$ again at $E'$ and $\overline{PE}$ meet $\odot(CPD)$ again at $F'$. Note that $\tfrac{PE}{PE'}=\tfrac{PF}{PF'}$ proving $EFE'F'$ is cyclic. Now $AEE'B$ is an isosceles trapezoid hence $OE=OE'$. Similarly, $OF=OF'$ hence $O$ is the center of $\odot(EFE'F')$ proving $OE=OF$. You missed some special cases: $AB\parallel CD$ or $\angle APE=\angle BPE$.

Where is Iran 2018 TST other problems

Here are the set of problems (Iran TST) from 2017. https://artofproblemsolving.com/community/c435323_2017_iran_team_selection_test

Been trying to avoid inversion, so here is a solution without inversion. By Radical Axis, the tangent in $X$ at the circles $\omega_1,\omega$,the tangent in $Y$ at the circles $\omega_2,\omega$ and the tangent in $P$ at the circles $\omega_1,\omega_2$ are concurrent, name the concurrency point $S$. We would like $EF$ to pass through $S$ too, because that would mean $\frac{SF}{SE}=\frac{FX^2}{XE^2}=\frac{FY^2}{YE^2}$ implying the desired result. Let the tangent in $P$ at $\omega'$=the circumcircle of $\triangle PEF$ intersect $\omega$ in $P_1,P_2$. Let $EF$ intersect $P_1P_2$ in $U$. Let $EF$ intersect $\omega$ at $E_1,E_2$. Let $EF$ intersect $AC,BD$ at $E_2,F_2$(the order on the line $EF$ will be $F_1,F,F_2,E_2.E,E_1$). $UP$ is tangent to $PEF$, obviously. Since $\angle E_2PE=\angle FPU$ and $\angle E_2PE=\angle F_2PF$, we get $\angle F_2E_2P=\angle F_2PU$, so $PU$ is tangent to the circumcircle of the triangle $PE_2F_2$. By Power of Point, we have $UP^2=UF_2\cdot UE_2=UF\cdot UE$. But by Desargues Involution Theorem, we have $(E_1,F_1),(E_2,F_2),(E,F)$ are pairs of involution. Since $UF_2\cdot UE_2=UF\cdot UE$, the involution is an inversion with center $U$. So $UF_2\cdot UE_2=UF\cdot UE=UF_1\cdot UE_1$. By Power of Point, $UP_1\cdot UP_2=UF_1\cdot UE_1=UE\cdot UF$. So $P_1,P_2,E,F$ are concyclic. Let $l$ be the radical axis of $\omega'$ and $\omega$. Now by radical axis for the circles $\omega,\omega',(P_1FEP_2)$, we have $P_1P_2\cap EF\cap l=U$. But again by radical axis for $\omega,\omega_1,\omega'$ we have $P_1P_2\cap l\cap SX=U$. So $S,U$ coincide, so $EF$ passes through $S$. Hence we are done.

A NEW SOLUTION Suppose tangents at $X,Y,P$ concur at $S$, then the powers of $S$ to $\omega$ and the 'point circle' $P$ are equal. Note that $EA \cdot EB / EP^2 = FC \cdot FD / FP^2$, which implies $E,F$ lie on the same circle (named $\gamma$) that is coaxial with $\omega$ and the 'point circle' $P$. Now, suppose $EF$ intersect the tangent at $P$ at $T$, then the power of $T$ to $\gamma$ and the 'point circle' $P$ are equal. Since $\omega$, $\gamma$, and the 'point circle' $P$ are coaxial, we have $S=T$ (intersection of the shared axis and the tangent at $P$). Done.

Did anyone bash this problem?

Is there a way to solve this problem using barycentric coordinates or complex numbers?

I don't think this has been posted before? inverted wrote: Let $ABCD$ be a cyclic quadrilateral, $F',E'$ points on $(APB), (CPD)$ respectively with $|angle F'PB = \angle E'PC$. Let $X',Y'$ be points on $(ABCD)$ such that diameter $X'Y'$ is perpendicular to $E'F'$. Prove that $\frac {E'X'}{E'Y'}=\frac {F'X'}{F'Y'} $. Proof: Remove the $'$s for convenience. Let $E_1 = PE \cap (ABF), F_1 = PF \cap (CDE)$. Then $EE_1FF_1$ is cyclic, from $ABCD$ cyclic (consider $\Psi$, the composition of the homothety around $P$ + reflection over angle bisector of $\angle APD$). Form $\angle AOE_1 = \angle FPB$ it follows that $FE_1 || AB$. $AB \perp$ the line through $O$ and the center of $(ABF')$, hence so does $F'E_1$, therefore $O$ lies on the perpendicular bisector of $FE_1$. Likewise $O$ lies on the $\perp $bisector of $EF_1$, so $O$ is the center of $(EFE_1F_1)$, hence $XY$ bisects $EF$, from which the desired ratio relation follows.

Consider the inversion at $P$ which swaps $A$ and $C$. For any object $Z$ denote its image by $Z'$. Let $G=E'$ and $I=F'$. Let $O$ be the center of $(ABCD)$. Let $K=X'$ and $J=Y'$. Now notice that $\omega_1'$ becomes the tangent to $(ABCD)$ at $K$. Since $\omega_1$ is tangent to $(PEF)$, the tangent at $K$ is parallel to $IG$. This implies $OK\perp IG$. Similarly $OJ\perp IG$. Now by inversion distance formula it suffices to show \begin{align} \frac{KI}{KG}=\frac{JI}{JG} \end{align} CLAIM. $OI=OG$ Proof. Let $(PAB)=\Gamma_1$ and $(PCD)=\Gamma_2$. Suppose $PG$ meet $\Gamma_1$ at $L\neq P$ and $IP$ meet $\Gamma_2$ at $M\neq P$. Now since $\angle AIP=\angle ABP=\angle DCP=\angle DGP$ and $\angle IPA=\angle GPD$ we have $\triangle IAP\sim\triangle GDP$. Therefore $$\angle IBA=\angle IPA=\angle DPG=\angle LPB=\angle LAB$$which implies that $IABL$ is an isoceles trapezoid. Since $OA=OB$ we have $OI=OL$. By symmetry $OG=OM$. Moreover $$\frac{IP}{PG}=\frac{AP}{PD}=\frac{PB}{PC}=\frac{PL}{PM}$$hence $I,L,M,G$ are conyclic. This shows that $O$ is the circumcenter of $(ILGM)$ and hence $OI=OG$ as desired. Now the claim together with the fact that $OK\perp IG$ implies that $KJ$ is the perpendicular bisector of $IG$. Hence both sides of $(1)$ equals 1 and the proof is completed.