Geos are definitely cool

So, I  did G1s. Here are solutions to a few of them.

I think they were really INMO level. Not much theory was needed. It's true that almost everything was angle chase-able. Although $2017$ G$1$ was an exception, as it had pappus. But the rest were nice. The hardest would also be $2017$ G$1$. It's not hard-hard, but hard in terms of theoretically.

The second hardest would probably be $2018$ G$1.$  Anand had proof with no words. I will add the image here. One should definitely check! 

Then, the third hardest would probably be $2009$ G$1.$ The right construction was hard for me to guess. Then $2005$ G1. And then $2007$G$1,$ since the right use of similarities were required.  

Also I, fortunately(?) didn't use any ggb, so I would recommend making your own diagrams. 

And before starting, I would like to thanks every one of you who comes here almost daily and views this blog! I hope this blog is helpful to you in some way. Yeah, we crossed 6k views! 💖

So here are 5+1 problems, which I would say are perfect for INMO!

2019 G1: Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

Proof: 

•  Redefine $FF\cap (ADFGE)=T.$
•  Then we get $\angle ABF=\angle DFT= 180-\angle TAD\implies AT||BC$
•  Then we show that $TG$ is tangent to $(GEC),$ then $\angle TGE=\angle TAE=\angle ACG.$

2015 G1: Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.

Proof: 

• Draw ||gm $IAHA'$
•  To show that $IJ=IA'$
• Note that $CHBM$ cyclic and $\angle IGC=\angle IJC=90-C.$
• But $\angle HAA'= \angle AAI'=90-C.$
•  So $IJ=IA'=AH.$

2016 G1: Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

Proof:

•  Redefine $M$ as a point such that $CDEM$ is ||gm. We let $\angle MCD=\theta$
•   So we get $CM=DF=EA$ and $EM=CD=DA.$ Hence $DEMA$ is isosceles trapezoid. Hence $DM=EA.$ 
•    So we get $\angle DME=180-\angle CMD-\angle EMA=\theta.$ 
•     Now, we have $$\angle EMF=\angle EMA=\theta =\angle MAB \implies ME||BA$$
•    So $$\angle \theta =\angle EBA=\angle BEM.$$
•      So if we define $Z=BE\cap MA$ then $\angle BZA=180-2\theta \implies Z=F\implies B-F-E$
•     Hence $MF=FE$ and $\Delta MEF$ congruent to $\delta MED.$ 
•     Hence $CM=DE=MF$Hence $M$ is the midpoint of $CF.$

2012 G1: Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proof: 

• Note that $AP^2=AF\cdot AB=AH\cdot AD=AE\cdot AC=AR^2.$ Note that $\angle BQD=C-X=\angle PAF\implies (AQPF).$
•     We use inversion at $A$ with radius $AH\cdot AD$ so $P\rightarrow P,~~R\rightarrow R,~~F\rightarrow B,~~H\rightarrow D,~~E\rightarrow C.$
•     Define $PB\cap FD\Leftrightarrow (PFA)\cap (ABH).$
•      So let $Q'= (PFA)\cap (ABH)$
•   Note that $C=180-\angle AHB=\angle AQ'B\implies \angle AQ'B=C.$
•     note that $\angle ARF=\angle APF=\angle FQ'A=X$ So we have $FQ'B=C-X$
•      Now note that $C-X=\angle PAF=\angle PQ'F.$
•      So $Q'-P-B\implies Q'=Q.$
•   So $AQ=AR=AP.$

2001 G1: In the plane we are given two circles intersecting at $X$ and $Y$. Prove that there exist four points with the following property:

(P) For every circle touching the two given circles at $A$ and $B$, and meeting the line $XY$ at $C$ and $D$, each of the lines $AC$, $AD$, $BC$, $BD$ passes through one of these points.

Proof:  

•  Define $AD\cap (O_1)=G,~~BD\cap (O_2)=H,~~AC\cap (O_1)=E,~~BC\cap (O_2)=F$
•    Note that $\angle GAE=\angle GAD,~~\angle EO_1A=\angle AO_3C\implies \angle EGA=\angle ADC\implies GE||XY$
•     Similarly we get HF||XY, so $HF||XY||GE.$
•      by pop we have $ABGH,~~ABEF$ cyclic.
•      Also we have now $GAH\sim DAC$ and so.
•      Now we angle chase, we get $$\angle EFB=\angle BAC=\angle BDC=\angle BHD\implies EF\text{~tangent to~~}(O_2).$$
•     Similar for others too, so we get $EF,GH$ as the common tangent. Hence fixed.

And the 2018 G1.

2018 G1: Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Anand is really the best

So that's it for this week's post, hope you have a great week! And see you all in next week's post!

Also, do comment if you want a non-geo or geo post. I would probably prefer a non-geo, number-theory post. If geo is requested then I will post the solutions to another 5 of the ISL G1s.

On a side note, I have opened " Anand is the best Fanclub" :P.

Hope you enjoyed it. See you all soon 

Sunaina 💜