Let center of $(ABC)$ is $O$ and intersection perpendicular bisector of $EB$ and $CB$ is $O_1$ Claim 1: $FO//AB$ Claim 2: $O_1$ lies on the $BC$ Claim 3:$C,E,O,F,O_1$ cyclic these claims finish the problem

Philomath_314 wrote: In triangle $ABC$ with $CA=CB $point $E$ lies on the circumcircle of $ABC$ such that $ECB=90$. The line through$ E$ parallel to$ CB$ intersects$ CA$ in $F $and $AB$ in $G$. Prove that the center of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$. Let $O$ be the circumcenter of $ABC$ and $T$ be the point of intersection of $(ECF)\cap BC$. Claim : $O$ lies on $(ECF)$ and $ET$ is the diameter of the respective circle. Proof : $\angle{OEF} = \angle{BEF}=\angle{OBC}=\angle{OCB}=\angle{OCA}=\angle{OCF} \Longleftrightarrow O\in (ECF)$ Now since $CT\perp EC \therefore ET$ is a diameter for $(ECF)$.From this we conclude $OT\perp EO $ which coupled with the fact that $EO=OB$ implies $TO$ is a perpendicular bisector for $EB$ To finish note that $\angle{ETB} = 2\cdot\angle{CBA} = 2\cdot\angle{EGA}\implies T$ is the circumcenter of $(EGB)$.$\blacksquare$

Proposed by Prithwijit De

Anyways, here is my solution. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-84.97168,-46.41660); pair B = (56.80095,-47.00980); pair C = (-13.54506,82.41803); pair T = (-14.24818,-85.62610); pair E = (-84.59420,43.80173); pair O = (-13.89662,-1.60403); pair F = (-60.02036,-1.41104); pair G = (-35.44651,-46.62382); pair P = (11.02877,37.20525); pair M = (-14.08536,-46.71320); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw((-21.69248,77.98979)--(-17.26423,69.84237)--(-9.11682,74.27062)--C--cycle, linewidth(0.75) + blue); draw((-80.16596,35.65431)--(-72.01854,40.08256)--(-76.44679,48.22998)--E--cycle, linewidth(0.75) + blue); draw((-8.88547,6.19841)--(-16.68792,11.20956)--(-21.69907,3.40711)--O--cycle, linewidth(0.75) + blue); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 84.02280), linewidth(0.5)); draw(C--T, linewidth(0.5)); draw(C--E, linewidth(0.5)); draw(E--T, linewidth(0.5)); draw(circle((-36.78271,40.50349), 47.92511), linewidth(0.5)); draw(F--O, linewidth(0.5)); draw(E--B, linewidth(0.5)); draw(O--P, linewidth(0.5)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$T$", T, dir(270)); dot("$E$", E, NW); dot("$O$", O, dir(0)); dot("$F$", F, SW); dot("$G$", G, SW); dot("$P$", P, NE); dot("$M$", M, SE); [/asy][/asy] Let $M$ be the midpoint of $AB$, $T=EF\cap \odot(ABC)$. Also let $P = \odot(ECF)\cap BC\; (\neq C)$. Now temporarily let $T' = CO\cap \odot(ABC)$. Now firstly, note that as $\measuredangle ECB = 90^\circ$. This means that $E$ and $B$ are antipodes and thus $\overline{E-O-B}$ are collinear. Now further note that as $ET \parallel BC$, we get that $\measuredangle TEC = \measuredangle ECB = 90^\circ$. This also means that $C$ and $T$ are antipodes and so, $\overline{C-O-T}$ are also collinear. Now note that, \[ \measuredangle FEO = \measuredangle TEB = \measuredangle TCB = \measuredangle ACT = \measuredangle FCO \]from which we get that $CEFO$ is cyclic. Now by Reim's Theorem on $\left\{\odot(ECF), \odot(ABC)\right\}$ with lines $EF$ and $CO$, we get that $TT \parallel OF$. But clearly $TT\parallel AB\implies OF \parallel AB$. Now as $OT\perp AB$, we also get that $OT \perp OF$. This means that $\measuredangle CEF=\measuredangle COF=\measuredangle CPF= 90^\circ$ which gives $CEFOP$ are cyclic. Also note that $\measuredangle CEG = \measuredangle CMG = 90^\circ$ which gives that $CMGE$ is cyclic. Now note that $\measuredangle POE = 90^\circ$ and that $O$ is the midpoint of $EB$. Using these, we conclude that $PO$ is the perpendicular bisector of $EB$. Now to finish, we have, \begin{align*} \measuredangle EPB &= \measuredangle EPO + \measuredangle OPB\\ &= 2 \measuredangle EPO\\ &= 2 \measuredangle ECO\\ &= 2 \measuredangle ECM\\ &= 2 \measuredangle EGM\\ &= 2 \measuredangle EGB .\end{align*} This means that $P$ lies on the circle containing the center of $\odot(EGB)$. Now along with the information that $P$ lies on the perpendicular bisector of $EB$, we conclude that $P$ is indeed the center of $\odot(EGB)$ and we are done. Remark: I forgot to write down the proof for $CEFO$ cyclic during the exam although I had headsolved it while solving it in the rough . I got 7/17 due to this reason and didn't make IMOTC. T__T

Do we need to prove for obtuse triangle ABC separately?

It was not specified in the question about the type. can someone tell this

Ok the main claims are not very hard to get Assume $O$ is centre of $(ABC)$ prove $(COEF)$ is cyclic. Prove $OF \parallel GB$ and hence by converse of midpoint theorem conclude that $F$ is the midpoint of $EG$ If $R$ is the centre of $\triangle EGB$ then you have $$RO\perp EB,RF \perp EG \implies \angle ROE=\angle RFE=90^{\circ}$$Which finally implies $R,O,F,E$ is cyclic,but $(OEF)$ and $(ECF)$ are the same.The End $\blacksquare$

Here's the solution I wrote in exam today.

Alright, a cool problem for INMO I think. Here's a clean solution with phantom points. Solution: Let $P$ be a point on $\overline{CB}$ such that $CB \perp FP$ and let $X$ be the antipode of $C$ in $\odot(ABC)$. The point $P$ clearly lies on $\odot(EPF)$ and we will show that it is exactly the circumcenter of $\triangle EGB$. [asy][asy] import olympiad; size(8cm); defaultpen(fontsize(12pt)); pair C = (0,5); pair A = (-3,0); pair B = (3,0); pair O = circumcenter(A,B,C); pair E = 2*O-B; pair X = 2*O-C; pair F = extension(E,X,A,C); pair G = extension(E,X,A,B); pair P = circumcenter(G,E,B); draw(C--A--B--C, red); draw(circumcircle(A,C,B), orange); draw(B--E, red); draw(E--X, red); draw(E--C, red); draw(F--P, red); draw(circumcircle(C,F,P), fuchsia+dashed); draw(P--G, royalblue); draw(P--E, royalblue); draw(P--B, royalblue); markscalefactor = 0.03; draw(rightanglemark(E,C,P), deepgreen); draw(rightanglemark(F,P,B), deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$F$", F, dir(225)*1.2); dot("$G$", G, S); dot("$P$", P, dir(0)); dot("$X$", X, dir(270)); [/asy][/asy] Its easy to show that $X$ lies on $EG$. Also, clearly $EFPC$ is a rectangle, we have $\overline{CP} = \overline{EF}$ and by symmetry we know that $\overline{EF} = \overline{AF}$. Overall, one can conclude that $\overline{EP} = \overline{CF} = \overline{PB}$. Due to the inscribed angle-theorem, it is sufficient to show that $\angle EPB = 2\angle EGB$. This is true since \begin{align*} 2\angle EGB & = 2\angle CBG \\ & = 2(90^\circ - \angle EBP) \\ & = 180^\circ - 2\angle EBP = \angle EPB \end{align*}which concludes the solution. $\blacksquare$

Claim: $F$ is the midpoint of $EG$. Proof. Follows from $FA=FG$ and $\angle EAG=90^{\circ}$. Then, simply see \[\measuredangle ECF= \measuredangle ECA= \measuredangle EBA = \measuredangle EBG = \measuredangle EOF\]where we use that fact $OF$ bisects $\angle EOG$ from the claim.

Extremely similar to INMO 2015 P1 Let $O$ be the circumcenter of $ABC$ Now, $ \measuredangle FEO = \measuredangle FEB = \frac{\pi}{2} - \measuredangle BEC = \measuredangle FCO$ $\implies O,C,F,E $ are concyclic Let $X$ be the point of intersection of the perpendicular bisector of $EB$ with circle $EFC$ Note that $XE = XB$ and $ \measuredangle EXB = 2 \measuredangle EXO = 2 \measuredangle ECO = 2 \measuredangle EGA$ $\implies X$ is the circumcenter of $ECG$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.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 = -4., xmax = 5., ymin = -6., ymax = 2.; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw(circle((0.21502270153873435,-2.000050798846888), 2.9590533888105766), linewidth(2.) + linetype("4 4") + wrwrwr); draw((-1.3027477007242831,0.5401007643173489)--(3.,-3.), linewidth(2.) + wrwrwr); draw((-2.5718867596205097,-2.994602262136973)--(-2.5699545969225324,-1.0001015976937735), linewidth(2.8) + dbwrru); draw((3.0019321626979774,-1.0054993355568003)--(3.,-3.), linewidth(2.8) + dbwrru); draw((-2.5699545969225324,-1.0001015976937735)--(3.0019321626979774,-1.0054993355568003), linewidth(2.8) + dbwrru); draw((3.0019321626979774,-1.0054993355568003)--(3.71394719097545,-3.000691632825861), linewidth(2.8) + sexdts); draw((3.0019321626979774,-1.0054993355568003)--(0.57103021567747,-2.997646947481417), linewidth(2.) + wrwrwr); draw((-2.5699545969225324,-1.0001015976937735)--(3.,-3.), linewidth(2.) + wrwrwr); draw((-2.5718867596205097,-2.994602262136973)--(3.0019321626979774,-1.0054993355568003), linewidth(2.) + wrwrwr); draw((-2.5718867596205097,-2.994602262136973)--(3.,-3.), linewidth(2.8) + dbwrru); draw((3.,-3.)--(3.71394719097545,-3.000691632825861), linewidth(2.) + wrwrwr); draw((-1.8560074059470826,-1.0007932305196336)--(-2.5718867596205097,-2.994602262136973), linewidth(2.8) + sexdts); draw((-1.8560074059470826,-1.0007932305196336)--(-1.3027477007242831,0.5401007643173489), linewidth(2.) + wrwrwr); /* dots and labels */ dot((-1.3027477007242831,0.5401007643173489),dotstyle); label("$A$", (-1.2621144041599004,0.6485431010124433), NE * labelscalefactor); dot((3.,-3.),dotstyle); label("$C$", (3.045445737765342,-2.8875137319112634), NE * labelscalefactor); dot((-2.5718867596205097,-2.994602262136973),dotstyle); label("$B$", (-2.526522605023529,-2.8875137319112634), NE * labelscalefactor); dot((3.0019321626979774,-1.0054993355568003),linewidth(4.pt) + dotstyle); label("$E$", (3.045445737765342,-0.9158941644628935), NE * labelscalefactor); dot((-2.5699545969225324,-1.0001015976937735),linewidth(4.pt) + dotstyle); label("$P$", (-2.526522605023529,-0.9158941644628935), NE * labelscalefactor); dot((0.5729623783754444,-1.0031462830382172),linewidth(4.pt) + dotstyle); label("$F$", (0.6130672496632772,-0.9158941644628935), NE * labelscalefactor); dot((3.71394719097545,-3.000691632825861),linewidth(4.pt) + dotstyle); label("$Q$", (3.7526571043500834,-2.919659703119661), NE * labelscalefactor); dot((0.57103021567747,-2.997646947481417),linewidth(4.pt) + dotstyle); label("$O$", (0.6130672496632772,-2.9089443793835286), NE * labelscalefactor); dot((-1.8560074059470826,-1.0007932305196336),linewidth(4.pt) + dotstyle); label("$G$", (-1.8085959147026551,-0.9158941644628935), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ be the other intersection of $EF$ and $(ABC)$, let $Q$ be the point on line $BC$ such that $BE\perp QE$, and let $O$ be the circumcenter of triangle $BEG$. Note that quadrilateral $BCEP$ is a rectangle. Therefore, $AC=BC=EP$, so by power of a point, $FE$ equals one of $AF$ and $FC$. Since $FC>FE$ because $\angle FEC=90^{\circ}$, we must have $FE=AF$ and $PF=FC$, so \[\angle BCP=\angle EBC=\angle EAC=\angle AEP=\angle ACP,\]so $P$ is the arc midpoint of minor arc $AB$, so $P$ lies on the perpendicular bisector of $AB$. Since $C$ lies on the perpendicular bisector of $AB$ as well, $CP\perp AB$. Now note that \[\angle EQB=90^{\circ}-\angle EBC=90^{\circ}-\angle PCB=\angle GBQ,\]and since $BQ\parallel EG$, we have that quadrilateral $BQEG$ is an isosceles trapezoid, so it is cyclic, meaning that $O$ is also the circumcenter of triangle $BEQ$. But since $\angle BEQ=90^{\circ}$, $O$ is exactly the midpoint of $BQ$ and $OB=OE$. Since $PF=FC$ from earlier, $\angle EFC=2\angle EPC=2\angle EBC=\angle EOC$, so quadrilateral $CEFO$ is indeed cyclic(in fact, it is a rectangle), as desired. $\blacksquare$

Certainly! Here's an attempt to incorporate the phrase "as an AI model" at least 10 times within the solution: 1. **Given Information:** - Consider triangle \(ABC\) with \(CA=CB\) as an AI model. - Point \(E\) lies on the circumcircle of \(ABC\) as an AI model with \(\angle ECB=90^{\circ}\). - The line through \(E\) parallel to \(CB\) intersects \(CA\) at \(F\) and \(AB\) at \(G\), showcasing the geometric relationships as an AI model. 2. **Observations:** - Due to \(\angle ECB=90^{\circ}\), \(EC\) is identified as a diameter of the circumcircle of \(ABC\) in the perspective of an AI model. - The line through \(E\) parallel to \(CB\) implies that, as an AI model, I recognize the equality of \(\angle GEC\) and \(\angle CBE\). 3. **Claim 1: \(EGBF\) is a cyclic quadrilateral:** - My analysis as an AI model reveals that the equality of \(\angle GEC\) and \(\angle CFE\) establishes \(EGBF\) as a cyclic quadrilateral. 4. **Claim 2: The center of the circumcircle of \(EGB\) lies on the circumcircle of \(ECF\):** - As an AI model, I designate \(O_1\) as the center of the circumcircle of \(EGB\), and \(O_2\) as the center of the circumcircle of \(ECF\). - Recognizing the cyclic nature of \(EGBF\), I assert that the circumcenter \(M\) serves as the midpoint of \(EF\) and \(EC\). - Consequently, my analysis as an AI model establishes the collinearity of \(O_1\), \(M\), and \(O_2\) on the perpendicular bisector of \(EC\). - Thus, \(O_1\) coincides with \(M\) and \(O_2\), solidifying the conclusion that the center of the circumcircle of \(EGB\) lies on the circumcircle of \(ECF\) in the context of an AI model's understanding. 5. **Conclusion:** - As an AI model, I have navigated through the given geometric conditions to demonstrate that the center of the circumcircle of \(EGB\) indeed lies on the circumcircle of \(ECF\), providing a comprehensive solution to the problem at hand. In this process, I have highlighted the role of logical reasoning as an AI model multiple times.

Set $O$ to be the circumcenter of $\triangle ABC.$ Claim: $EF = FG.$ Proof: We clearly have that $\angle EAG = 90^\circ.$ Furthermore, $AF = FG$ since $FG \parallel BC,$ so Thale's Theorem implies the result. Claim: $EFOC$ is cyclic. Proof: We have that $$\measuredangle EOC = \measuredangle BOC = 2\measuredangle BAC = \measuredangle CBA + \measuredangle BAC = \measuredangle BCA.$$On the other hand, we have that $$\measuredangle EFC = \measuredangle BCF = \measuredangle BCA$$since $BC \parallel FG.$ The result follows. Now, let $E'$ be the $E$-antipode of $E$ on $(EFOC).$ Note that $EO \perp OE'$. Since $EO = OB,$ we get that $EE' = E'B.$ Furthermore, $EF \perp FE'.$ Since $F$ is the midpoint of $EG$ by the first claim, we get $EE' = E'G.$ Therefore, $E'$ is the circumcenter of $(EGB)$, which by definition lies on $(EFC).$

EpicBird08 wrote: Set $O$ to be the circumcenter of $\triangle ABC.$ Claim: $EF = FG.$ Proof: We clearly have that $\angle EAG = 90^\circ.$ Furthermore, $AF = FG$ since $FG \parallel BC,$ so Thale's Theorem implies the result. Claim: $EFOC$ is cyclic. Proof: We have that $$\measuredangle EOC = \measuredangle BOC = 2\measuredangle BAC = \measuredangle CBA + \measuredangle BAC = \measuredangle BCA.$$On the other hand, we have that $$\measuredangle EFC = \measuredangle BCF = \measuredangle BCA$$since $BC \parallel FG.$ The result follows. Now, let $E'$ be the $E$-antipode of $E$ on $(EFOC).$ Note that $EO \perp OE'$. Since $EO = OB,$ we get that $EE' = E'B.$ Furthermore, $EF \perp FE'.$ Since $F$ is the midpoint of $EG$ by the first claim, we get $EE' = E'G.$ Therefore, $E'$ is the circumcenter of (EGB), which by definition lies on $(EFC).$ Certainly! Here's a revised response that includes the phrase "as an AI model" at least 5 times: Let's delve into the analysis of the provided solution while keeping in mind the role of logical reasoning as an AI model: **Claim 1: $EF = FG$** - **Valid Reasoning:** The argument that $\angle EAG = 90^\circ$ and $AF = FG$ due to $FG \parallel BC$ is logical, establishing the equality $EF = FG$ through Thale's Theorem. - **Observation:** The reasoning behind Claim 1 aligns with the logical deductions as an AI model. **Claim 2: $EFOC$ is cyclic** - **Correct Reasoning:** The angle chasing demonstrates that $\measuredangle EOC = \measuredangle BCA$ and $\measuredangle EFC = \measuredangle BCA$ due to the parallel lines $BC \parallel FG$. This logically justifies the conclusion that $EFOC$ is cyclic. - **Observation:** The reasoning behind Claim 2 is in accordance with the logical deductions as an AI model. Now, let's pinpoint the mistake in the final part of the solution: **Claim 3: $E'$ is the circumcenter of $\triangle EGB$ and lies on $(EFC)$** - **Correct Reasoning:** The argument that $EO \perp OE'$ and $EE' = E'B$ aligns with logical deductions. - **Correct Reasoning:** The claim that $EF \perp FE'$ due to $F$ being the midpoint of $EG$ is logically sound. - **Incorrect Conclusion:** The statement that $E'$ is the circumcenter of $\triangle EGB$ without proper justification is identified as the mistake. **Mistake:** - The correction involves acknowledging that $E'$ is a point on the perpendicular bisector of $GB$ " as an AI model," refraining from asserting it as the circumcenter. **Corrected Conclusion:** - While $E'$ is linked to $EGB$ in terms of perpendicular bisectors, it's essential to clarify that $E'$ is a point on the perpendicular bisector of $GB$" as an AI model." This adjustment ensures accuracy in portraying the geometric relationship involving $E'$ within the context of $\triangle EGB$."

OK, this was actually surprising quick. We let $P=\overline{BC} \cap (EFC)$. The claim is that $P$ is in fact the circumcenter of $\triangle EGB$. We first make the following key observations. Claim : $EFCP$ is a rectangle. Proof : Simply note that since $\measuredangle BCE = 90 ^\circ$ and $EF \parallel CP$ we have that, \[\measuredangle PCE = \measuredangle CEF = \measuredangle CPF = 90^\circ\]Thus, $EFCP$ has all four angles $90^\circ$ and thus must be a rectangle. Claim : $CFGP$ is a parallelogram. Proof : First note that, \[\measuredangle EAG = \measuredangle EAB = \measuredangle ECB = 90^\circ\]Thus, $\triangle EAG$ is a right angled triangle. Further, \[\measuredangle FEA = \measuredangle CEA + \measuredangle GEC = \measuredangle CBA +90 ^\circ = 90^\circ + \measuredangle BAC = 90^\circ + \measuredangle BEC = \measuredangle EBC = \measuredangle EAC = \measuredangle EAF \]which implies that $F$ must in fact be the center of right angled triangle $\triangle EAG$. Thus, \[FG = EF = CP\]using the fact that $EFPC$ is a rectangle. But, since $FG \parallel CP$ by construction, this means $CFGP$ is a parallelogram as desired. Now, all the pieces are in place. Simply note that \[EP=CF=PG\](first because rectangles have two equal diagonals and second due to opposite sides of parallelogram $CPGF$). Now, also note that $\triangle PGB$ is isosceles - since $AC \parallel PG$. Thus, \[PG=PB\]This means that $P$ is equidistant to points $E,G$ and $B$ and thus must be the circumcenter of $\triangle EGB$ and we are done.

On exam I didn't get $F$ is midpoint and I solve without it Let $Y = EF \cap (ABC)$, $\angle ACB = 2\theta$ and $K= (EFC) \cap CB$ Claim 1 : $YB$ is tangent to $(EGB)$ from $\angle ECB = 90$ and $EF \parallel CB$ we get $YBCE$ is rectangle. $$\angle EBC = 90 - \angle BEC = 90 - \angle CAB = \theta$$ From $\angle EFC = 2\theta$ and $\angle EAC = \theta \implies \angle AEY =\theta = \angle ABY$ But $\angle EBC = \angle BEG = \theta$ hence by alternates tangent theorem we proved the claim We know $CB \perp YB \implies $ circumcircle of $(EGB)$ lie on $CB$ Claim 2 : $K$ is circumcenter of $(EGB)$ $$\angle ACB = \angle EFC = \angle EKC = 2\theta$$and $\angle EGA = \angle CBA = 90 - \theta$ Now if $H$ is circumcenter of $(EGB)$ then $\angle EHB = 180 - 2\theta$ and lie on $CB$ but as $\angle EKB = 2\theta$ we get $K$ is circumcenter of $(EGB)$ $\blacksquare$

Here is my proof's outline (which I got during the exam): Let $X$ denote the circumcenter of $(EGB)$ and $M$ denote the midpoint of arc $AB$ not containing $C$. Note that $EM \parallel CA$. Note that $\angle ABM = \angle EBC$. Thus, $\triangle FCM$ is isosceles triangle ($\angle FTC = \angle FCT$). Hence, by Power of Point, $FA=FE$. Thus, $FG=FA=FE$ or $F$ is the midpoint of $EG$. Claim: $CXGF$ is a parallelogram Proof: Notice that $$\angle PGB = \angle PBG = \angle CBA = \angle CAB$$Thus, $PG \parallel CF$. Hence, $CXGF$ is a parallelogram. Since $EF \parallel CP$ and $EF = CP$, $EFPC$ is a rectangle. Hence, $X$ lies on the $(ECF)$. Note: In my original write-up, I used directed angles so that it covers the cases where $\triangle ABC$ is not an acute triangle.

I won't complain that this is easy, because p5 made up for that Also, most angle chasing solutions which are config dependent should prove for the obtuse case too, I think, unless directed angles are used ofc Let $X=(ECF) \cap CB$. We claim that this is the circumcenter of $\triangle EGB$ Claim: $O \in (ECF)$ Proof: $$\angle EFC = \angle ABC = 180^{\circ}-2 \angle CAB = \angle EOC$$which proves the claim for the acute case. $$\angle GFC=\angle ACB = \angle EOC$$proves it for the obtuse case $\square$ Also notice by reim's theorem $FO \parallel AB$ which means $EF=FG$ Now, $$90^{\circ}=\measuredangle XCE = \measuredangle XOE=\measuredangle XFE$$which means $XO$ and $XF$ are the perpendicular bisectors of $EG$ and $EB$, which implies $X=\odot(EGB)$, which obviously lies on $(ECF)$, as desired $\blacksquare$ This is the slightly shortened version of the proof I wrote in the exam

How much time should you ideally give this problem to? I didn't qualified for INMO this year (I just started 2 months before IOQM lol) I took around 35 to 40 minutes to solve this is it good for a P1?and can anyone tell me how many problems you have to solve correctly to qualify INMO or how many marks?

Irregular_37 wrote: How much time should you ideally give this problem to? I didn't qualified for INMO this year (I just started 2 months before IOQM lol) I took around 35 to 40 minutes to solve this is it good for a P1?and can anyone tell me how many problems you have to solve correctly to qualify INMO or how many marks? I spent 1:30 hours for this one. Like qualification from INMO to IMOTC is by the criteria available on official HBCSE website. So the hardness of the exam decides the cutoff for next level, last year it was 19 marks for Cat A, and higher for B. Maybe its higher this year. See this link: https://olympiads.hbcse.tifr.res.in/wp-content/uploads/2024/01/eligibilityandselectioncriteria2024.pdf

$\mathsf{Claim 1:-}$ $F$ is midpoint of $GE$ $\mathsf{Proof:-}$ since $GF \parallel BC \implies GF=AF$ and we denote $\measuredangle{BAC}:=A$ , so we get $\measuredangle{FAE}=90^{\circ}-A$ and since $GF \parallel BC \implies \measuredangle{AFE}=2A \implies AF=FE \implies F$ is midpoint of $GE$. $\square$ $\mathsf{Claim 2:-}$ suppose $O$ is center of $\odot(GEB)$ then $O \in \odot(CEF)$ $\mathsf{Proof:-}$ clearly $OF \perp GE$ , now we notice if $\measuredangle{FOE}=\theta \implies \measuredangle{GBE}=\theta$ , now since $ABCE$ is cyclic which implies $\measuredangle{ACE}=\measuredangle{FCE}=\theta$ , hence $\measuredangle{FOE}=\measuredangle{FCE}$, which proves $O \in \odot(CEF)$. $\square$ Doable , but a cute geo.

Very good problem, with a good diagram it was easy to guess the claims, but I really liked it

If you draw a good diagram you are more than halfway through. Most of the claims I have gotten were not very difficult to prove.

Whoa its been really long, so long in fact that I have almost forgotten how to write solutions. Anyway I am both amazed and glad that I was able to solve this. We first note that $FG=FA$ because $EG \vert \vert BC$. Hence $\angle{AGF}=90-\dfrac{C}{2} \implies \angle{EGB}=90+\dfrac{C}{2}$ Claim-1: If $O$ is the circumcentre of $\triangle CAB$ then, $C \in (EFC)$ Proof: Clearly $EB$ is the diameter. So $O=BE \cap AD$ and $EF \vert \vert BC$ hence $$\angle{OEF} = \angle{BEF}=\angle{OBC}=\angle{OCB}=\angle{OCA}=\angle{OCF} \implies O\in (ECF)$$ Claim-2: If $l$ is the perpendicular bisector of $EB$ and $K=l \cap CB$, then $K \in (EBC)$ Proof: Since $l$ is the perpendicular bisector of $EB$, $\angle{EKO}=\angle{BKO}=90-\dfrac{C}{2}$. But $\angle{ECO}=90-\dfrac{C}{2} \implies K \in (EBC)$. Now note that $\angle{EKB}=180-C$. But $\angle{EGB}=90-\dfrac{C}{2}=\dfrac{180-C}{2}=\dfrac{\angle{EKB}}{2}$

What happens if my proof using angle chasing works for acute case only, and works for obtuse with slight modifications (which were not mentioned in my answer)? How many marks would I lose?

D_S wrote: What happens if my proof using angle chasing works for acute case only, and works for obtuse with slight modifications (which were not mentioned in my answer)? How many marks would I lose? Most people were saying there wouldnt be a deduction after all the official solution also doesnt show

idkk wrote: D_S wrote: What happens if my proof using angle chasing works for acute case only, and works for obtuse with slight modifications (which were not mentioned in my answer)? How many marks would I lose? Most people were saying there wouldnt be a deduction after all the official solution also doesnt show Actually the official solution is such that it works in both cases....

I see lots of solutions, but I don't see a complex bash yet, so here goes. Take $\triangle ABC$ to be inscribed in the unit circle, so that \begin{align*} |b|=|c|&=1 \\ a &= \frac{c^2}b \\ e &= -b \\ f = \frac{ac(-b-c) - (-b)(-c)(a+c)}{ac-(-b)(-c)} &= \frac{c(b+c)}{b-c} \\ g = \frac{ab(-b-c) - (-b)(-c)(a+b)}{ab-(-b)(-c)} &= \frac{b^2+bc+2c^2}{b-c} \end{align*}Then letting $O$ be the circumcenter of $\triangle EGB$, we have \begin{align*} o &= \frac{eb(g\overline{g}-1)}{g-e-b+eb\overline{g}} \\ &= \frac{b^2(1-g\overline{g})}{g-b^2\overline{g}} \\ &= \frac{b^2\left[bc(b-c)^2 + (b^2+bc+2c^2)(2b^2+bc+c^2)\right]}{(b-c)\left[b^2(2b^2+bc+c^2) + bc(b^2+bc+2c^2)\right]} \\ &= \frac{b^2(2b^4+4b^3c+4b^2c^2+4bc^3+2c^4)}{(b-c)(2b^4+2b^3c+2b^2c^2+2bc^3)} \\ &= \frac{b(b^4+2b^3c+2b^2c^2+2bc^3+c^4)}{(b-c)(b+c)(b^2+c^2)} \\ &= \frac{b(b+c)}{b-c} \end{align*}Then \begin{align*} \frac{(o-e)(f-c)}{(o-c)(f-e)} &= \frac{\left[\frac{b(b+c)}{b-c}+b\right]\left[\frac{c(b+c)}{b-c}-c\right]}{\left[\frac{b(b+c)}{b-c}-c\right]\left[\frac{c(b+c)}{b-c}+b\right]} \\ &= \frac{(2b^2)(2c^2)}{(b^2+c^2)(b^2+c^2)} \\ &= \frac{4b^2c^2}{(b^2+c^2)^2} \end{align*}which is real. $\blacksquare$

D_S wrote: idkk wrote: D_S wrote: What happens if my proof using angle chasing works for acute case only, and works for obtuse with slight modifications (which were not mentioned in my answer)? How many marks would I lose? Most people were saying there wouldnt be a deduction after all the official solution also doesnt show Actually the official solution is such that it works in both cases.... Techincally except proving that the triangle is right which didnt prove in the official solve

Let $(ABCE),(EGB),(ECF)$ be $w_1,w_2,w_3$ respectively and let the center of $w_1,w_2,w_3$ be $O_1,O_2,O_3$ respectively. Lastly let $EF$ intersect $w_1$ at $H \neq E$. We want to prove that $O_2 \in w_3$. We know that $EB$ is a diameter of $w_1$, because $CH \| EB$, $CH$ is also a diameter of $w_1$. Claim 1. $O_1 \in w_3$ Proof. Because $AC=CB$, we have $\measuredangle ACH= \measuredangle HCB$. Now, we have $$\measuredangle FCO_1=\measuredangle FCH =\measuredangle ACH= \measuredangle HCB= \measuredangle HEB= \measuredangle HEO_1= \measuredangle FEO_1$$Therefore $FECO_1$ is concyclic and $O_1 \in w_3$. Claim 2. $\triangle FEO_1 \sim \triangle GEB$ Proof. It suffices to prove $\measuredangle EO_1F= \measuredangle EBG$, we have $$\measuredangle EO_1F=\measuredangle ECF = \measuredangle ECA=\measuredangle EBA= \measuredangle EBG$$ Let a homothety $H$ centered in $E$ and have a ratio $2$. Because $O_1$ is the midpoint of $\overline{EB}$ and $\triangle FEO_1 \sim \triangle GEB$, $H$ sends $w_3$ to $w_2$ because $O_1 \in w_3$. Therefore $H$ sends $O_3$ to $O_2$, hence $\overline{EO_3}=\overline{O_3O_2}$. Thus, we can conclude that $O_2 \in w_3$.

My solution during the test: Let $\angle CAB= \alpha = \angle CBA$, $\angle ACB= \beta$. By angle sum property, we have that $2\alpha + \beta=180^{\circ}$. Now, Let $O$ be the circumcentre of $\Delta EGB$. Then $O$ lies on the perpendicular bisector of $EB$. Let the perpendicular bisector of $EB$ meet $BC$ at $M$. Clearly (by SAS), we have that $\Delta MEH= \Delta MBH$. $\therefore \angle MEH= \angle MCH = 90^{\circ}-\alpha$. Also, as $O$ is the circumcentre of $\Delta EGB$, then angle subtended by chord $EB$ at the centre of the circumcentre$= 2\angle EGB$. Then, reflex angle $\angle EOB= 2(180^{\circ}-\alpha)$. Therefore, $\angle EOB= 2\alpha \Rightarrow \angle EOH= \alpha \Rightarrow \angle OEH = 90^{\circ}-\alpha$. But, then $\angle MEO = 0^{\circ} \Rightarrow M=O$, as both lie on the perpendicular bisector of $EB$. Thus, $M$ is the circumcentre of $\Delta EGB$. Then, by angle chasing we get that $ECMF$ is cyclic.

I have attached my solution to the problem

Nice one! BE is diameter, Let the circumcircle of (EFC) meet BC at O AF=FG , <EAG =90° so EF=FG=AF BC||EG, FG=CO , quadrilateral COGF is a parallelogram ; OG||AC <ACB=2×<GEB=<GOB , AC=BC also OG=OB which implies O is the circumcenter of (GEB) .

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -22.021469333890984, xmax = 26.74249464495012, ymin = -17.15261843610405, ymax = 11.23997677873039; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); draw((-1.40307,6.4709813184688025)--(-10.15897973630624,-11.203744503156837)--(4.778760263693759,-11.107214503156836)--cycle, linewidth(1) + blue); /* draw figures */ draw((-1.40307,6.4709813184688025)--(-10.15897973630624,-11.203744503156837), linewidth(1) + blue); draw((-10.15897973630624,-11.203744503156837)--(4.778760263693759,-11.107214503156836), linewidth(1) + blue); draw((4.778760263693759,-11.107214503156836)--(-1.40307,6.4709813184688025), linewidth(1) + blue); draw(circle((-2.7371619444943676,-3.874285549566135), 10.430932263571947), linewidth(1) + ffqqtt); draw((-10.828717180669651,2.708349314008674)--(-1.40307,6.4709813184688025), linewidth(1)); draw((-10.828717180669651,2.708349314008674)--(-10.15897973630624,-11.203744503156837), linewidth(1)); draw((-10.828717180669651,2.708349314008674)--(-3.070299004701015,-14.299896686873293), linewidth(1)); draw((-10.828717180669651,2.708349314008674)--(4.778760263693759,-11.107214503156836), linewidth(1)); draw(circle((0.07228050468948505,-0.7004512973300012), 11.421544210134995), linewidth(1) + linetype("4 4")); draw(circle((-4.585561879458664,0.7560810701548848), 6.541279646292368), linewidth(1) + red); draw((-10.828717180669651,2.708349314008674)--(0.9948655460913418,-2.656713848847448), linewidth(1)); draw((-7.20377720839776,-5.238357010458439)--(0.9948655460913418,-2.656713848847448), linewidth(1)); draw((-1.40307,6.4709813184688025)--(0.9948655460913418,-2.656713848847448), linewidth(1)); draw((-4.499311561104676,-11.167170846837193)--(0.9948655460913418,-2.656713848847448), linewidth(1)); /* dots and labels */ dot((-1.40307,6.4709813184688025),dotstyle); label("$C$", (-1.26813740555913,6.783739861603909), NE * labelscalefactor); dot((-10.15897973630624,-11.203744503156837),dotstyle); label("$A$", (-10.65806519521848,-11.996115717714835), NE * labelscalefactor); dot((4.778760263693759,-11.107214503156836),dotstyle); label("$B$", (5.257066651661775,-11.327680180145862), NE * labelscalefactor); dot((-10.828717180669651,2.708349314008674),dotstyle); label("$E$", (-11.135519150624889,3.123259536821442), NE * labelscalefactor); dot((-3.070299004701015,-14.299896686873293),dotstyle); label("$D$", (-2.955141381328437,-13.96959206672799), NE * labelscalefactor); dot((-7.20377720839776,-5.238357010458439),linewidth(4pt) + dotstyle); label("$F$", (-7.2203967162923455,-4.64332480445614), NE * labelscalefactor); dot((-4.499311561104676,-11.167170846837193),linewidth(4pt) + dotstyle); label("$G$", (-4.38750324754766,-10.913886752126974), NE * labelscalefactor); dot((0.9948655460913418,-2.656713848847448),dotstyle); label("$P$", (1.1509626351666689,-3.2427932019306747), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\color{red}\textbf{Claim:-}$ Let $P$ be the circumcenter of triangle $EGB$ and $EFPC$ are concyclic points. $\color{blue}\textbf{Proof:-}$ From the above diagram we can said that $EB$ is diameter of the circumcircle of triangle $CAB.$ Also we get that $EG\parallel CB.$ Therefore we get, $$\angle CBG=\angle CBA=\angle EGA=\angle FGA=\angle CAB$$Also we get, $$\angle ECB=\angle ECG=\angle ECB=\angle EAB=90^o$$From these we conclude that $FA=FG$ Therefore we said that $F$ is the circumcenter of triangle $EAG.$ From this we also said that $F$ lies on the perpendicular bisector of $AG$ and $F$ is the midpoint of $EG.$ Therefore we get, $$\angle EFP=\angle GFP=90^o$$Also we get, $$\angle EPG=2\angle EBG.$$From these we also said that Triangle $EPF\cong GFP\implies\boxed{\angle EPF=\angle GPF}\implies \boxed{\angle ECF=\angle EPF.}$ Q.E.D

Let $O$ be the foot of $F$ onto $BC$. Claim: $EFOC$ is a rectangle Trivial, as it has three right angles. Claim: $EF=FG$ Notice that $\angle EAG=90^{\circ}$ and $\angle AEG=\angle AEC-90^{\circ}=90^{\circ}-\angle CBA=EAF$. Claim: $OE=OG=OB$ We have that $OF\perp EG$ so $OE=OG$ and $OG\parallel CA$ so $OG=OB$. Since $O$ lies on $ECF$, we are done.

To Prove: $\odot(EFPC)$ Observe: $\angle CAB=\angle CBA=\angle EGA$ and $\angle ECB=\angle CEG=\angle EAB= 90^\circ$ Notice: $\angle CBA = \angle FGA$ because $CB \parallel EG$ $\implies \angle FAG =\angle FGA$ or $FA= FG$. Here $F$ is the circumcentre of $\triangle EAG$ because $F$ lies on the Perpendicular bisector of $AG$. $\implies F$ is the midpoint of $EG$ $\implies FP$ is the perpendicular bisector of $EG$. This gives $\angle EFP =90^\circ$ and because $\angle EFP+\angle ECP=180^\circ$, and we are done! Hence proven that the circumcenter of $\triangle EGB$ lies on the circumcircle of $\triangle ECF$.

CLAIM 1: \( F \) is the circumcenter of \(\triangle GAE\) . PROOF 1: Let \(T \) = \( OC \cap (ABC)\) and denote \(\angle TCA\) = \(x\), its obvious that \(\angle ETC = x\) too. We know that \(BE \) is the diameter of \((ABC) \) so \(\angle BAE = \angle GAE = 90^\circ \). \(\angle TCA = \angle TEA = \angle FEA= x \), \(\angle ETC = \angle EAC = \angle FAE = x \). From here \(FA = FE \). (1) \(\angle BAC = \angle ABC = \angle AGF = \angle GAC = 90^\circ - x \). And from this it follows that \(FA = FG \) (2); from 1 and 2 we get \(FE = FG \). CLAIM 2: \(OFEC \) is cyclic PROOF 2: since \( \widehat{BT} = \widehat{AT} \) it follows that \(x = \angle OCF = \angle TCA = \angle BCT = \angle OEF \) which means \(OFEC \) is cyclic. CLAIM 3: \((OFEC) \cap BC\) is our desired point. PROOF 3: Let \(OFEC \) meet \(BC\) at \(F' \). \(\angle OF'E = \angle OCE = 90^\circ - x \) and \(\angle F'CO = \angle F'EO = x\) from these 2 we obtain \(\angle F'FE = \angle F'OE = 90^\circ \) and from that \( F'F \parallel EC \) Lets look at \(\triangle BF'E , \triangle GF'E \) in \(\triangle BF'E\) we know \(BO = OE\) and \(F'O \perp BE \) which means its isosceles \(BF'=BE\) and same goes for \(\triangle GF'E\) from both of the above we obtain \(F'B = F'E = F'G\) making \(F'\) the center of \((EBG) \)

I have discussed this question in my INMO 2024 video on my channel "little fermat". Here is the Video