Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Triangles $AOE,AOF$ are equilateral because $AE=OE=OA,\ AF=OF=OA$. The condition $\angle AOB<120^{\circ}\iff \angle AOC>60^{\circ}$ simply means that $F$ (assume $F$ is on the same side of $OA$ as $C$) lies on the arc $AC$ which doesn't contain $E$, so $CA$ is the internal bisector of $\angle ECF$ because $A$ is the midpoint of the arc $EF$ (without the condition, $AC$ would be the external bisector). This means that $J$ lies on the bisector of $\angle ECF$. On the other hand, we have $AD\|OJ$ and $OD\|AJ$ because they're both $\perp AB$, so $J$ is the reflection of $D$ in the midpoint of $OA$. This means $\angle EJF=\angle EDF=120^{\circ}$, and since $\angle ECF=\frac{\angle EOF}2=60^{\circ}$ and $J$ lies on the internal bisector of $\angle ECF$, it means that $J$ is the incenter (the only other position on $AC$ from which $EF$ is seen under an angle of $120^{\circ}$ is $A$, and it's clear that $J\ne A$).

orl wrote: The circle $ S$ has centre $ O$, and $ BC$ is a diameter of $ S$. Let $ A$ be a point of $ S$ such that $ \angle AOB < 120{{}^\circ}$. Let $ D$ be the midpoint of the arc $ AB$ which does not contain $ C$. The line through $ O$ parallel to $ DA$ meets the line $ AC$ at $ I$. The perpendicular bisector of $ OA$ meets $ S$ at $ E$ and at $ F$. Prove that $ I$ is the incentre of the triangle $ CEF.$ Consider the complex plane with origin $ O$. Wlog assume that $ S$ is the unitcircle. For uppercase letter points, let the lowercase letter denote their corresponding complex number. (Except for $ I$, which will has the complex number $ z_I$) $ D$ is the midpoint of the arc $ AB$, ie. $ \angle BOD = \angle DOA \iff \frac {b}{d} = \frac {d}{a} \iff a = d^2$. (Since $ |a| = |b| = |d| = 1|$) I will prove that $ z_I = d^2 - d$. Obviously $ OI \parallel AD$ since $ z_I - o = a - d = d^2 - d$. $ \frac {z_I - a}{c - a} = \frac {d}{1 + d^2} \in \mathbb{R}$ since $ \frac {d}{1 + d^2} = \frac {\frac {1}{d}}{\frac {1}{d^2} + 1} = \overline{\frac {d}{1 + d^2}}$ so $ I \in AC$, which concludes: $ z_I = d^2 - d$, since the point is unique. It is easy to show that $ E$ and $ F$ are constructed by rotating $ A$ $ 60 ^\circ$ around $ O$ positivily and negatively respectively. Let $ \omega = e^{i\frac {\pi}{6}}$, then $ e = \omega^2d^2 = u^2, u = \omega d$ and $ f = \overline{\omega}^2d^2 = v^2, v = \omega^5 d$. Since $ c = w^2, w = - i$, and $ u,v,w$ the incenter of $ \triangle CEF$ is $ - uv - vw - wu$. (Wellknown. See the training materials from imomath.com) And $ - uv - vw - wu = - \omega^6d^2 + i(\omega d + \omega^5 d) = d^2 + i^2d = d^2 - d = z_I$. So $ I$ is the incenter of $ \triangle CEF$.

enough to show $ AD=AF=AI$ and observe just B D A I C O, easily find $ OA=AI$ by elementary angle chasing. so we are done

grobber wrote: Triangles $ AOE,AOF$ are equilateral because $ AE = OE = OA,\ AF = OF = OA$. The condition $ \angle AOB < 120^{\circ}\iff \angle AOC > 60^{\circ}$ simply means that $ F$ (assume $ F$ is on the same side of $ OA$ as $ C$) lies on the arc $ AC$ which doesn't contain $ E$, so $ CA$ is the internal bisector of $ \angle ECF$ because $ A$ is the midpoint of the arc $ EF$ (without the condition, $ AC$ would be the external bisector). This means that $ J$ lies on the bisector of $ \angle ECF$. My solution was the same up to that point (though I think the $ J$ should have been $ I$ throughout the proof .) Here's how I finished it: We have that $ m \angle OAC = m \angle ACO = \frac{m \angle BOA}{2} = m \angle DOA$, so $ DO || AI$. Hence, $ DOIA$ is a parallelogram, so $ DO = AI$. On the other hand, $ \triangle AOF$ is equilateral, so we have that $ AO = AI = AF$, that is, $ AI = AF$, so $ \triangle AIF$ is isosceles. Let $ m \angle AFE = \alpha$ and let $ m \angle EFI = \beta$. $ m \angle AIF = \alpha + \beta$ since $ \triangle AIF$ is isosceles, so $ m \angle FIC = 180 - \alpha - \beta$. Also, $ \alpha = m \angle EFA = m \angle AEF = m\angle ACF$, so $ m \angle IFE = 180 - (180 - \alpha - \beta + \alpha) = \beta$. Since $ m \angle EFI = m \angle IFC$, $ I$ lies on the bisector of $ \angle EFC$. But $ I$ also lies on the bisector of $ \angle ECF$, so $ I$ is the incenter of $ \triangle ECF$.

My solution is slightly different... unfortunately I missed the fact that $DO\|AI$... WLOG let $E$ be on minor arc $BD$. Let $OI$ hit the circle at $X$ and $Y$ so that $Y$ is on minor arc $FC$. First we note that $AE=AF$, so $I$ is on the angle bisector of $\angle{ECF}$. Since $AO=AE=AF$, it suffices to show that $AO=AI$, which would imply that $A$, the midpoint of minor arc $EF$, is the circumcenter of $\triangle{EIF}$ and $I$ is thus the incenter of $\triangle{ECF}$ (since the incenter is on this circle, and we know that $I$ is on the angle bisector of $\angle{ECF}$). Using the facts that $AD=AB$ and $AD$ is parallel to $XY$, we have (interpret all arcs as those not containing any point in the interior of minor arc $CX$) \begin{align*} 2\angle{AOI}=2\widehat{AY}=\widehat{AY}+\widehat{DX}&=\widehat{AY}+\widehat{DB}+\widehat{BX}\\ &=\widehat{AY}+\widehat{DA}+\widehat{CY}=\widehat{DY}+\widehat{CY}=\widehat{AX}+\widehat{CY}=2\angle{AIO}, \end{align*}as desired.

Hmm.. I did exactly as Zhero did up until his angle-chase. Note that $AF=AO=AI=AE$ so $FEOI$ is cyclic with centre $A$. Then $2\angle FEI=\angle FAI=\angle FAC=\angle FEC$ so $IE$ bisects $\angle FEC$. This was much easier than G1!?

Since $DO\perp AB$, we conclude $DO\parallel AC$ and, with $AD\parallel OI$, $ADOI$ is a parallelogram, i.e. $AI=OD$, so $AE=AI=AF$ and $I$ is, indeed, the incenter of $\triangle CEF$. Best regards, sunken rock

My solution : Let's show that $I$ is the incenter of the triangle $\triangle{CEF}$: First, it's easy to check that $OA=OE=AE=OF=AF$ and hence the quarilateral $AEOF$ is a rhombus, and we have $A$ is the middle of the arc $EF$ and since $A$ lies in the perpendicular bisector of $EF$, it follow from a well-known fact that $CI$ is the bisector of $\angle{ECF}$. Otherwise, from the condition $OI\parallel AD$, we have : $\angle{AIO}=\pi-\angle{CIO}=\pi-\angle{CAD}=\angle{DBC}$. Now let $A'$ be the symetric of $A$ wrt $O$, then $\angle{AOI}=\angle{DAO}=\angle{DAA'}=\frac{\pi}{2}-\angle{ACD}=\frac{\pi}{2}-\angle{DCB}=\angle{DBC}=\angle{AIO}$ since $CD$ is the bisector of $\angle{ACB}$. Hence wde get : $AO=AI$ which means that points $E$,$I$,$O$ and $F$ are concylic ( In particular in circle with center $A$ ), then : $\angle{FEI}=\frac{1}{2}\angle{FAI}=\frac{1}{2}\angle{FEC}$. And so $EI$ is the bisector of $\angle{FEC}$ and so we are done !

Very easy solution: Since <BOD = <BCA then DO||AC and therefore AIOD is a parallelogram. Therefore AI = OD = OA = OE = OF = AF = AO = AE and therefore A is the centre of the circumcircle of quadrilateral FOIE. Therefore <FIE = 180º - <FAE / 2 = 120º since < FCE = 60º and < FAE = 120º (since FAEO is a 60º-rhombus) then if I' is the incentre of CFE, then C, I, I' are collinear since < FCI = arc AF = arc AE = < ACE and FIEI' are concyclic since < FIE = 120º = 90 + <FCE / 2 = < FI'E then I = I'. Done.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.46cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.300000000000004, xmax = 11.16000000000001, ymin = 1.480000000000002, ymax = 6.300000000000006; /* image dimensions */ /* draw figures */ draw(circle((2.180000000000002,1.960000000000002), 4.201190307520005)); draw((-2.020000000000002,1.860000000000002)--(6.380000000000007,2.060000000000002)); draw((1.057644715918773,6.008495846150148)--(2.180000000000002,1.960000000000002)); draw((-1.887277891922414,3.012259734989030)--(5.124922607841190,4.956236111161120)); draw((1.057644715918773,6.008495846150148)--(-1.194073337803743,4.463123870510994)); draw((1.057644715918773,6.008495846150148)--(6.380000000000007,2.060000000000002)); draw((-1.887277891922414,3.012259734989030)--(6.380000000000007,2.060000000000002)); draw((5.124922607841190,4.956236111161120)--(6.380000000000007,2.060000000000002)); draw((2.180000000000002,1.960000000000002)--(4.431718053722520,3.505371975639156)); draw((-1.194073337803743,4.463123870510994)--(2.180000000000002,1.960000000000002)); draw((1.057644715918773,6.008495846150148)--(-1.887277891922414,3.012259734989030)); draw((1.057644715918773,6.008495846150148)--(5.124922607841190,4.956236111161120)); draw((5.124922607841190,4.956236111161120)--(2.180000000000002,1.960000000000002)); draw((2.180000000000002,1.960000000000002)--(-1.887277891922414,3.012259734989030)); draw((4.431718053722520,3.505371975639156)--(5.124922607841190,4.956236111161120)); draw((4.431718053722520,3.505371975639156)--(-1.887277891922414,3.012259734989030)); /* dots and labels */ dot((2.180000000000002,1.960000000000002),dotstyle); label("$O$", (2.260000000000003,2.080000000000003), NE * labelscalefactor); dot((-2.020000000000002,1.860000000000002),dotstyle); label("$B$", (-1.940000000000002,1.980000000000003), NE * labelscalefactor); dot((6.380000000000007,2.060000000000002),dotstyle); label("$C$", (6.460000000000007,2.180000000000003), NE * labelscalefactor); dot((1.057644715918773,6.008495846150148),dotstyle); label("$A$", (1.140000000000002,6.060000000000006), NE * labelscalefactor); dot((-1.194073337803743,4.463123870510994),dotstyle); label("$D$", (-1.120000000000001,4.580000000000005), NE * labelscalefactor); dot((-1.887277891922414,3.012259734989030),dotstyle); label("$E$", (-1.800000000000002,3.140000000000004), NE * labelscalefactor); dot((5.124922607841190,4.956236111161120),dotstyle); label("$F$", (5.200000000000006,5.080000000000004), NE * labelscalefactor); dot((4.431718053722520,3.505371975639156),dotstyle); label("$I$", (4.520000000000006,3.620000000000004), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\angle DOB = \theta$. Notice that $\angle AIO = \angle IOC + \angle ICO = \theta + 90 - 1.5 \theta = \angle AOI$, so $AO = AI = AF = AE$. Therefore, $EFIO$ is cyclic, giving us $\angle EIF = \angle EOF = 120^{\circ}$. Notice that $\angle ECF = 60^{\circ}$, so $I$ must lie on the circumcircle formed by the incenter of $CEF$ and $E, F$. But $\angle ACE = \angle ACF = 30^{\circ}$, so $I$ lies on the angle bisector as well. Therefore, $I$ is the incenter of $CEF$

This is easily done with complex numbers as well. WLOG $ S $ is the unit circle. WLOG let $ A, B, C, D, I, E, F $ have complex coordinates $ a^2, -1, 1, d, x, e, f $ respectively. Clearly $ d = -ia $. Since $ OD \perp AB $ by definition and since $ AI \perp AB $ since $ BC $ is a diameter of $ S $ we have that quadrilateral $ ADOI $ is a parallelogram and so we have that $ x = a^2 + ai $. Now, note that $ E, F \in S $ and that the projections of either point onto $ AO $ has complex coordinate $ \frac{a^2}{2} $. This implies that $ e, f $ are the roots of the equation $ z^2 - a^2z + a^4 = 0 $. Therefore by Vieta's formulas $ e + f = a^2 $ and $ ef = a^4 $. Now since the circumcircle of $ CEF $ is $ S $ it suffices to show that $ x = -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) $. Choosing the appropriate sign for the square roots given that $ \angle{AOB} < 120 $ we have that $ \sqrt{ef} = -a^2 $ and since $ (\sqrt{e} + \sqrt{f})^2 = e + f + 2\sqrt{ef} = -a^2 $ we have that $ \sqrt{c}(\sqrt{e} + \sqrt{f}) = -ai $ so $ -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) = a^2 + ai $ as desired. This is literally the worst solution possible but I felt that I had to post it for its ridiculousness. In fact, you can easily prove that $ ADOI $ is a parallelogram with solely complex numbers (it requires solving a simple linear system of two equations).

A bit too easy for an IMO #2? By http://www.mit.edu/~evanchen/handouts/Fact5/Fact5.pdf as usual it suffices to show that $AJ = AE = AF$, but obviously $AE = AF = AO$ so we simply wish to check that $\angle AOJ$ is isosceles which is angle chasing.

Oops I can't see perpendicular lines... We claim that $DO\parallel AI$. Invert about $A$ with radius $AE=AF=AO$. The image of $D$ is $AD \cap EF \equiv D'$, and we want to show that $AC$ is tangent to $(AD'O)$. But $\angle CAO=90^{\circ}-\angle ABC=\angle ACB=2\angle ACD=\angle AOD=\angle AD'O$ as desired. Now $ADOI$ is a parallelogram, so that $AI=DO=AO$. Hence, by a converse of the incenter-excenter lemma, we are done$.\:\blacksquare\:$

Dear Mathlinkers, I have also this reference http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=550362 Sincerely Jean-Louis

We do not use complex numbers. By the Perpendicularity Lemma, $AE^2+FO^2=AF^2+EO^2 \implies AE = AF \implies \angle EOA = \angle FOA$. Because $\angle AOB < 120^{\circ}$, $A$ is the midpoint of $\overarc{EF}$ that does not contain $C$, and $CA$ is the angle bisector of $\angle ECF$. Now, I claim $AE = AF = AI$. Let $AO \cap EF = X$. Observe that $\triangle AFX \equiv \triangle OFX$, so $AF=OF$. Also, $ADOI$ is a parallelogram, since $\angle ACB = \frac{\overarc{AB}}{2} = \overarc{BD} = \angle DOB \implies DO \parallel AI$. Hence $AI = DO = FO$. So we have that $AE = AF = AI$. The Incenter-Excenter Lemma, also known as Fact 5, completes the proof.

Let $F$ be on minor arc $AC$. Note that $CI$ is the bisector of $\angle ECF$ because $A$ is the midpoint of arc $EF$. Line $OD$ bisects segment $AB$, so $DO||AC$ and $ADOI$ is a parallelogram. Since $AO=OD=AI$, $FIOE$ is cyclic with center $A$. Then $2\angle EFI=\angle EAI=\angle EAC=\angle EFC$, so $FI$ is the bisectof of $\angle EFC$. Hence, $I$ is the incenter of $EFC$.

Here is my solution using only angle-chasing (I hope true ) Let's do some substitions $\angle FEI=x$,$\angle IEC=y$, $\angle DAE =\beta$ $\angle BOE =\alpha$ Easy to prove that triangles $OAE$,$OAF$ are equilateral. It is clear that $OA \cap EF$ is midpoint of $EF$.So $\angle AEF=\angle AFE =30^{\circ}$ which means $CI$ is the angle-bisector of $\angle ECF$. Now I claim that $2\beta =x+y$. $\frac{\angle BOE }{2}=\angle EAB=\angle DAB-\angle DAE =\angle DBA-\angle DAE =15^{\circ}+\frac{\alpha}{4}-\beta =\frac{\alpha}{2}$ $\Longrightarrow$ $4\beta+\alpha =60^{\circ}$ ($1$) If we look at the sum of angles of $IOC$ we ll get $2x+2y+\alpha =60^{\circ}$ ($2$) By 1 and 2 we get $2\beta =x+y$ $\angle AIO = 60^{\circ}+x+y-\beta =60+\beta =\angle AOI$ So $AI=AO=AF$ Thus $AIF$ is isosceles. $\angle IAF=x+y$ $\angle EFI=90^{\circ}-\frac{x+y}{2}-30^{\circ}=60^{\circ}-\frac{x+y}{2}$ and since $\angle EFC =120^{\circ}-x-y$ so $FI$ is the angle-bisector of $\angle EFC$ in triangle $EFC$.Thus $I$ is the incenter of triangle $EFC$.

[asy][asy] size(12 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.051483785518323, xmax = 7.212262187154099, ymin = -5.334205171826316, ymax = 13.717611539018502; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); pen zzttqq = rgb(0.6,0.2,0); draw(arc((-0.2645619625868887,3.470484547086684),0.4978697746736456,-174.72106310504066,-94.39638699435766)--(-0.2645619625868887,3.470484547086684)--cycle, linewidth(2) + blue); draw(arc((-1.9151046479455704,3.3179803058821387),0.4978697746736456,-75.04573921572357,5.278936894959368)--(-1.9151046479455704,3.3179803058821387)--cycle, linewidth(2) + blue); draw(arc((-0.6425802971003163,-1.446355450299897),0.4978697746736456,5.278936894959391,85.60361300564237)--(-0.6425802971003163,-1.446355450299897)--cycle, linewidth(2) + blue); draw(arc((2.5147312459492235,-0.6030572388919284),0.4978697746736456,124.3049085629105,204.62958467359348)--(2.5147312459492235,-0.6030572388919284)--cycle, linewidth(2) + blue); draw((-0.10255990359711326,0.9850779718092986)--(-0.07557332701301833,1.3360891980557879)--(-0.42658455325950767,1.3630757746398827)--(-0.4535711298436026,1.0120645483933934)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw(circle((-0.2645619625868887,3.470484547086684), 4.931350060696237), linewidth(2)); draw((-0.6425802971003163,-1.446355450299897)--(-0.2645619625868887,3.470484547086684), linewidth(2) + ccqqqq); draw((-4.711679473923795,1.3394380291783055)--(3.804537214236589,0.6846910676084812), linewidth(2)); draw((-3.043855171123001,7.544026333065296)--(-4.711679473923795,1.3394380291783055), linewidth(2)); draw((-3.043855171123001,7.544026333065296)--(3.804537214236589,0.6846910676084812), linewidth(2)); draw((-0.6425802971003163,-1.446355450299897)--(1.007962388258365,-1.2938512090953513), linewidth(2)); draw((1.007962388258365,-1.2938512090953513)--(2.5147312459492235,-0.6030572388919284), linewidth(2)); draw((-4.711679473923795,1.3394380291783055)--(-0.2645619625868887,3.470484547086684), linewidth(2) + ccqqqq); draw((-0.2645619625868887,3.470484547086684)--(3.804537214236589,0.6846910676084812), linewidth(2)); draw((-4.711679473923795,1.3394380291783055)--(-0.6425802971003163,-1.446355450299897), linewidth(2) + ccqqqq); draw((-0.6425802971003163,-1.446355450299897)--(3.804537214236589,0.6846910676084812), linewidth(2) + ccqqqq); draw((-0.2645619625868887,3.470484547086684)--(-3.043855171123001,7.544026333065296), linewidth(2)); draw((-0.2645619625868887,3.470484547086684)--(2.5147312459492235,-0.6030572388919284), linewidth(2)); draw((-1.9151046479455704,3.3179803058821387)--(-0.6425802971003163,-1.446355450299897), linewidth(2) + ccqqqq); draw((-1.9151046479455704,3.3179803058821387)--(-3.043855171123001,7.544026333065296), linewidth(2)); draw((-1.9151046479455704,3.3179803058821387)--(-0.2645619625868887,3.470484547086684), linewidth(2)); draw(circle((-0.6425802971003189,-1.4463554502998919), 4.931350060696231), linewidth(2) + linetype("4 4") + zzttff); /* dots and labels */ dot((-3.043855171123001,7.544026333065296),dotstyle); label("$C$", (-2.977472534499847,7.709982924623185), NE * labelscalefactor); dot((-0.2645619625868887,3.470484547086684),linewidth(4pt) + dotstyle); label("$O$", (-0.2059974554832199,3.6108551131435074), NE * labelscalefactor); dot((2.5147312459492235,-0.6030572388919284),linewidth(4pt) + dotstyle); label("$B$", (2.5820732826891954,-0.47167703918038223), NE * labelscalefactor); dot((-0.6425802971003163,-1.446355450299897),dotstyle); label("$A$", (-0.57110195691056,-1.2848643378140026), NE * labelscalefactor); dot((1.007962388258365,-1.2938512090953513),linewidth(4pt) + dotstyle); label("$D$", (1.0718682995124704,-1.1686947237234853), NE * labelscalefactor); dot((3.804537214236589,0.6846910676084812),linewidth(4pt) + dotstyle); label("$E$", (3.876534696840674,0.822784374971095), NE * labelscalefactor); dot((-4.711679473923795,1.3394380291783055),linewidth(4pt) + dotstyle); label("$F$", (-5.616182340270169,0.5738494876342725), NE * labelscalefactor); dot((-1.9151046479455704,3.3179803058821387),linewidth(4pt) + dotstyle); label("$I$", (-1.8489677119062504,3.4448985215856256), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

orl wrote: The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$ First, note that AOE and AOF are equilateral (perp. bisectors from vertice). Trivially CI is an angle bisector since they both are inscribed angles of 60 degrees, and <ECF=60 degrees. Lemma: ADOI is a parallelogram. Proof: <DOB=2<DCB=<ACB, so by corresponding angles DO//AC, two pairs of parallel lines make a parallelogram. Now, we see that EF perp. AO, EF a chord, so EF is bisected by AO, but so is DI (diagonals are bisected in parallelogram), all concurrent at the midpoint of AO=midpoint of EF=midpoint of DI, so EDFI is a parallelogram. Now, we have 120=<EDF=<EIF, but there exists the unique point incenter iff in a triangle ABC incenter I we have <BIC=90+1/2<BAC, which is true, since 120=90+30, so I is the incenter of CEF, as desired. $\blacksquare$ Additional facts we could've derived: OI is the interior angle bisector of <FOG, since we know <EOF=<EIF, EOIF is cyclic => 1/2<FOG=<FEG=<FEI=<FOI=<FOH, as desired. Edit: Because angle AOB never exceeds 120 degrees, it can be seen that E and F lie on opposite sides of AO

lol this is so stupid Note that $ADOI$ is a parallelogram since \[\angle BOD = \frac{\angle BOA}{2} = \angle BCA\]and $AFOE$ is a rhombus, so \[AE = AF = FO = OD = AI\]and Fact 5 finishes. $\blacksquare$

Claim: $A$ is the circumcenter of $\triangle EIF$. Let $P$ be the intersection of $AD$ and $BC$. Let $\angle ACD=\angle BCD=\alpha$. Then, $$\angle BAD=\alpha,$$so $$\angle APB=90-3\alpha.$$Thus, $\angle IOC=90-3\alpha,$ and since $\angle AOB=4\alpha$ we have $\angle AOI=90-\alpha$. However, $\angle AIO=90-3\alpha+2\alpha=90-\alpha$ as well, so $AO=AI$. Clearly, $AO=AE=AF$ since $\triangle AEO$ and $\triangle AFO$ are equilateral, so $AE=AF=AI$, thus $A$ is the circumcenter of $\triangle EFI$. Consider $\triangle CEF$. $I$ lies on $CA$ (since $A$ is the arc midpoint), as well as the circle centered at $A$ passing through $E$ and $F$. Thus, $I$ is either the incenter or the $C$-excenter. Claim: $I$ is inside the circle $S$. Note that $ADOI$ is a parallelogram since $\angle DOB=\angle ICO=2\alpha$ and we already know the other sides are parallel. Thus, $I$ is the reflection of $D$ over the midpoint of $AO$. Clearly, $\angle DAO<60$, thus $I$ lies inside the circle, hence done.

Note that triangles $AFO$ and $AEO$ are equilateral triangles so we get some easy angles. In particular, $\angle FCA=\angle ECA$. Claim. $DAJO$ is a parallelogram. Proof. $DA\parallel OJ$ by definition and \[\angle BOD=2\angle BCD=\angle BCA \Longrightarrow DO\parallel AJ.\] Let the midpoint of $AO$ be $M$. Then a reflection about point $M$ takes $F$ to $E$, $A$ to $O$, and $J$ to $D$. Therefore, \[\angle FJC=180^{\circ}-\angle FJA=180^{\circ}-\angle EDO =180^{\circ}-\left(90^{\circ}-\frac12\angle EOD\right) =90^{\circ}+\frac12\angle EOD.\]However, \[\angle FEC=\angle FAC=60^{\circ}+\angle OAC=60^{\circ}+\angle ACB=60^{\circ}+\frac12\angle AOB=60^{\circ}+\angle DOA=\angle DOE\;\blacksquare\]

how did this make the shortlist Note that $AEOF$ is a rhombus so $$\angle ACE = \angle AFE = \angle FEA = \angle FCA,$$and $I$ lies on the internal bisector of $\angle ECF$. But because $$\angle DAC = 180^{\circ} - \angle DBC = 180^{\circ} - \angle ADO,$$we find that $ADOI$ is a parallelogram, and thus $AI = DO = FO = AF = AE$. So, $A$ is the center of $(EFI)$, and since $A$ is also the midpoint of arc $EF$ not containing $C$, the conclusion follows by fact 5.

hello, $AI=DO=AO=AE=AF$ so by Fact 5 done.

$\color{magenta}\boxed{\textbf{SOLUTION G3}}$ $\color{green} \textbf{Claim 1:}$ $CA$ is the angle bisector of $\angle ECF$ $\textbf{proof :}$ As $EF$ is the perpendicular bisector of $OA$, $A$ is the midpoint of arc $EF$ Hence done $\square$ $\color{green} \textbf{Claim 2:}$ $\triangle AOF$ is equilateral $\textbf{proof :}$ Let $OA \cap EF \equiv X$ $\triangle AXF$ and $\triangle OXF$ are congruent. Hence $AF=OF=OA \square$ $\color{green} \textbf{Claim 3:}$ $ADOI$ is a parallelogram $\textbf{proof :}$ $\angle DAC = 180- \angle DBC = 180 - \angle ADO$ and we have $AD \parallel OI \square$ So, $$AI = DO = FO = AF= AE$$And by Incenter-Excenter Lemma we are done $\blacksquare$

It is easy to prove $CI$ is angle bisector. Just observe $AI=DO=AE$ and we are done.

Prove that: $AI=AF=AE$

We can see that $EO = EA$ since this is the perpendicular bisector. Also, we can see that $EO = OA$. Thus, $\triangle OEA$ is equilateral. Similarly, $\triangle OAF$ is equilateral. Now notice that $J$ lies on the angle bisector of $\angle ECF$. This is because $A$ is the midpoint of arc $EF$. Now since $OD$ hits $AB$ at a $90$ degree angle and same with $CA$, we can see that $OD \parallel CA$. Thus $J$ is the reflection of $D$ over the midpoint of $OA$. This means that $\angle EJF = \angle EDF = 120$. Thus we can conclude. $\blacksquare$

Claim: $CI$ is the angle bisector of $\angle ECF$. Proof. In cyclic quadrilateral $EAFC$, we have $\angle ACF = \angle AEF = \angle AFE = \angle ACE$, and the result follows. $\blacksquare$ Claim: $ADOI$ is a parallelogram. Proof. $D$ lies on the perpendicular bisector of $AB$, so we have $OD \perp AB$. However, $AC \perp AB$ since $BC$ is the diameter of $S$. Thus, $OD \parallel AC$, or $OD \parallel AI$, and combining with the given $OI \parallel AD$, the claim is shown. $\blacksquare$ Now, since $AO$ and $EF$ bisect one another, $AEOF$ is a rhombus, which implies that $AF = AO = OD = AI$. Let $I'$ be the incenter of $\angle CEF$. By the Incenter-Excenter Lemma in $\Delta CEF$, $I'$ is the unique point on the angle bisector of $\angle ECF$ such that $AI' = AE = AF$, but $AI = AE = AF$ and $CI$ is the angle bisector of $\angle ECF$, so $I' = I$, as desired. $\blacksquare$

Claim: $ADOI$ is a parallelogram. Proof: We already know by definition that $OI \parallel DA;$ it suffices to show that $DO \parallel AI.$ But $$\measuredangle DOA = 2\measuredangle DCA = \measuredangle BCA = \measuredangle OCA = \measuredangle CAO,$$which implies the claim. Since $DO = AO,$ we get that $AO = AI.$ But it is clear (say by properties of regular hexagons) that $AE = AF = AO.$ Thus $AE = AF = AI.$ Combining this with the fact that $CI$ bisects $\angle ECF$ clearly, we conclude by the Incenter-Excenter Lemma that $I$ is the incenter of $CEF.$

Simple angle chasing reveals that triangles $ODA$ and $AIO$ are congruent, and thus $AI = AO = r$, where $r$ is the radius of the circle. Furthermore, $AF = FO = AO = AE$, so $F,I,E$ lie on a circle with center $A$. Lastly, notice that $A$ is the arc midpoint of arc $FE$, so the incenter-excenter lemma applies.

It is clear that $\triangle AEO, \triangle AFO$ are equilateral. Therefore, $$AE=AF \implies \angle ECA = \angle FCA, \text{ and } OE=OA=OF,$$so the by Incenter-Excenter Lemma the desired conclusion follows. QED