An Extension of Turkey TST 2015

by AlastorMoody, Feb 6, 2019, 7:21 PM

Original Problem:
Turkey TST 2015 Day 2 P1 wrote:
Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$.

Lemma 1: In $\Delta ABC$, $AB=AC$ and $AD,AE$ are produced to intersect $BC$ at $F,T$, then, Points $F,D,E,T$ are concyclic
Proof: Let $\measuredangle DAB= \alpha$ and $\measuredangle ABC=x$, then, $\measuredangle AFB=x -\alpha$ and $\measuredangle DET=\measuredangle DEF +\measuredangle FET =180^{\circ}-x+\alpha$ $\implies$ implies the conclusion $\qquad \blacksquare$

Solution: Using (Lemma 1) $\implies$ $G$ lies on $BC$, Now Perform Inversion $\Phi$ around $\omega $ be a circle centered at $A$ with radius $AB$, then, $\Phi$ swaps $F \longleftrightarrow D$, $G \longleftrightarrow E$ and $B,C$ remain invariant under $\Phi$, hence, $E$ lies on the polar of $G$ WRT $\omega$, hence, $AC^2=AE \cdot AG$ $\implies$ $AC$ is tangent to $\odot (CGE)$ $\qquad \blacksquare$

Remark: Similarly, Trivial to see, $AB$ tangent to $\odot (FDB)$ $\implies$ $AD \cdot AF =AB^2=AC^2=AE \cdot AG$

Corollary 1: $AC$ is tangent $\odot (FDC)$ and $AB$ is tangent to $\odot (GBE)$
Corollary 2: $\odot (FBD) , \odot (GBE) $ and $\odot (FDC) , \odot (GCE)$ are tangents to each other

Combining,
Property: $A$ is the radical center WRT $\odot (FBD), \odot (FDC) , \odot (GBE)$ and $\odot (GCE)$

Lemma 2: In $\Delta ABC$, $AB=AC$ and $AD,AE$ are produced to intersect $BC$ at $F,T$, Let $FE \cap \odot (ABC) =P$ and $DT \cap \odot (ABC) =Q$, then, $PQ || BC$
Proof: $\angle DTF=\angle DEF=\angle DEP=\angle DQP \Longrightarrow PQ||BC \qquad \blacksquare $

Which in-turn helps us show, $PB=QC$

Somewhat similar type of Problem is as follows:
Costa Rica Finals /CentroAmerican 2003 P2 wrote:
Let $AB$ be a diameter of circle $\omega$. $\ell$ is the tangent line to $\omega$ at $B$. Take two points $C$, $D$ on $\ell$ such that $B$ is between $C$ and $D$. $E$, $F$ are the intersections of $\omega$ and $AC$, $AD$, respectively, and $G$, $H$ are the intersections of $\omega$ and $CF$, $DE$, respectively. Prove that $AH=AG$.
Solution:$\angle ACD=\angle ABE=\angle AFE=180^{\circ}-\angle DFE$ $\implies$ Points $C,D,E,F$ are concyclic $\implies$ $\angle EDC=\angle EFC$ $=$ $\angle EHG$ $\implies$ $HG||CD$ $\implies$ $AB \perp HG$ $\implies $ $AH=HG$ $\qquad \blacksquare$

We can find an extension of the above problem by extending $AG,AH$ to intersect $CD$ at $M,N$, Four Cyclic Quadrilaterals pop-up in the diagram:
Remark 1: Points $M,D,A,E$ and $N,F,A,C$ are concyclic
Proof: $MN||GH$ $\implies$ $\angle DMA=\angle HGA=\angle DEA \Longrightarrow $ $MDAE$ is cyclic, the other one follows similarly! $\qquad \blacksquare$
Remark 2: Points $N,H,F,D$ and $M,E,G,C$ are concyclic
Proof: $\angle ACD=\angle EFA=\angle AGE=180^{\circ}-\angle MGE$ $\implies$ $MCEG$ is cyclic, the other one follows similarly! $\qquad \blacksquare$
This post has been edited 27 times. Last edited by AlastorMoody, Feb 12, 2019, 8:39 PM

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    by bookstuffthanks, Jul 31, 2024, 12:05 PM

  • hello fellow moody!!

    by crazyeyemoody907, Oct 31, 2023, 1:55 AM

  • @below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study :( . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read

    by kamatadu, Jan 3, 2023, 1:25 PM

  • Lots of good stuffs here.

    by amar_04, Dec 30, 2022, 2:31 PM

  • But even if he went to jee he could continue with this.

    Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows

    by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM

  • Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
    Btw @below finally everyone falls to the monopoly of JEE :) Coz IIT's are the best in India.

    by BVKRB-, Feb 1, 2022, 12:57 PM

  • Kukuku first shout of 2022,why did this guy left this and went for trashy JEE

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