A crowded geometry

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Source: Iran TST1 Day1 P2

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#1 h Feb 22, 2020, 5:12 PM • 2 Y

Y by k12byda5h, Adventure10

Let $O$ be the circumcenter of the triangle $ABC$ . Points $D,E$ are on sides $AC,AB$ and points $P,Q,R,S$ are given in plane such that $P,C$ and $R,C$ are on different sides of $AB$ and pints $Q,B$ and $S,B$ are on different sides of $AC$ such that $R,S$ lie on circumcircle of $DAP,EAQ$ and $\triangle BCE \sim \triangle ADQ , \triangle CBD \sim \triangle AEP$ (In that order), $\angle ARE=\angle ASD=\angle BAC$ , If $RS\| PQ$ prove that $RE ,DS$ are concurrent on $AO$ .

Proposed by Alireza Dadgarnia

This post has been edited 2 times. Last edited by matinyousefi, Mar 12, 2020, 5:39 PM

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lminsl

#2 h Feb 23, 2020, 5:51 AM

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Simple angle chasing gives that $\overleftrightarrow{APQ}$ is the tangent line of $A$ WRT $\odot(ABC)$ . Now let $K$ be the center of sprial similarity sending $\triangle BCD$ to $\triangle EAP$ . Then

$\measuredangle KBC=\measuredangle KEA$ , hence $\odot(KBE)$ is tangent to $BC$ ;
$\measuredangle KCD=\measuredangle KAP$ , hence $K$ lies on $\odot(ABC)$ ;
$\measuredangle KDB=\measuredangle KPE$ , and since $\angle BDC=\angle APE \implies X=PE \cap BD \in \odot(APD)$ , $K$ lies on $\odot(APD)$ .

Thus $\angle AKE=\angle CKB=\angle BAC$ . Since there is a unique point $R$ lying on $\odot(APD)$ such that $\angle ARE=\angle BAC$ and $R, C$ are on different sides of $AB$ , we have $K \equiv R$ .
Similarly if $L$ is the center of sprial similarity sending $\triangle CBD$ to $\triangle AEP$ , then $L \equiv S$ .

Construct points $X,Y \in \odot(ABC)$ such that $BX \parallel AC$ and $CY \parallel AB$ . Let $J=KY \cap LX$ . From $\angle BKY=\angle ABC$ , $(K, E, Y)$ are collinear; so $J=RE \cap DS$ . Now note that $AX=BC=AY \implies AO \perp XY$ , so $X, Y$ are symmetric WRT $\overleftrightarrow{AO}$ . Also $K$ and $L$ are symmetric WRT $AO$ since $KL \parallel PQ \perp AO$ . Thus, by symmetry, $J \equiv KY \cap LX$ lies on $AO$ . We're done.

This post has been edited 3 times. Last edited by lminsl, Feb 24, 2020, 2:35 AM

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#4 h Apr 5, 2020, 12:55 PM • 2 Y

Y by YS_PARKGONG, JH_K2IMO

IMO 2000, BB'//AC, CC'//AB, B',C' are on circiumcircle of ABC, B'C'//A tangent line.
If, R,S are on circiumcircle, REB', SDC' colinear and RS//PQ//B'C'-> RE,DS,AO meet one point!
And I prove (R is on circiumcircle) use center of spiral similarity!
I think this problem is so beautiful!

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#5 h Jul 29, 2020, 8:17 AM

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Let $l$ be tangent line of $A$ WRT $\odot(ABC)$ . Let $Z\in l$ and $T\in l$ such that $\triangle ZAD \sim \triangle ABC$ and $\triangle TEA \sim \triangle ABC$ (In that order).
According to the problem $P, Q\in l$ .

Claim: $\odot(ZAD),\odot(ABC),\odot(AQE)$ are concurrent in $S$ .
Proof: Note that $\odot(ZAD) \cap \odot(AQE) \equiv S$ ; and $S$ is unique. Let $\odot(ABC) \cap \odot(AZD)\equiv S^*$ .
$\angle ZAS^*=\angle ABS^*$ , so $S^*$ is a corresponding point on $\odot(ZAD)$ and $\odot(ABC)$ , where $\triangle ZAD \sim \triangle ABC$ . $(\star)$
$\triangle ABC \sim \triangle ZAD , \triangle BEC \sim \triangle AQD , (\star) \implies \triangle QZS^* \sim \triangle EAS^* \implies \angle ZQS^*=\angle AES^* \implies AQS^*E$ is cyclic. So $S^* \equiv S$ . And we’re done with the claim.
And similarly $\odot(TEA),\odot(ABC),\odot(APD)$ are concurrent in $R$ .

$RS \parallel PQ \implies RS \parallel l \implies AS=AR \implies AO$ is perpendicular bisector of $RS$ . $\implies \angle ARS =\angle ASR$ .
So;
$\angle ERS=\angle ERA-\angle SRA= \angle DSA-\angle RSA=\angle DSR$ .
So $(SD \cap RE) \in AO$ (cause $AO$ is perpendicular bisector of $RS$ ).
And we’re done. $\square$

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#6 h Jul 1, 2021, 1:14 PM

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Can someone draw a diagram, please?

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