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Source: Iranian third round 2015 geometry problem 3

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824 posts

andria

#1 h Sep 10, 2015, 11:52 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle. consider an arbitrary point $P$ on the plain of $\triangle ABC$ . Let $R,Q$ be the reflections of $P$ wrt $AB,AC$ respectively. Let $RQ\cap BC=T$ . Prove that $\angle APB=\angle APC$ if and if only $\angle APT=90^{\circ}$ .

This post has been edited 1 time. Last edited by andria, Sep 11, 2015, 9:12 AM

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#2 h Sep 10, 2015, 9:44 PM • 3 Y

Y by techniqq, Adventure10, Mango247

Let the perpendicular to $PA$ at $P$ cut $AC,AB,BC$ at $Y,Z,T^*,$ resp and let $X \equiv RZ \cap QY.$ Since $AC,AB$ are perpendicular bisectors of $PQ,PR,$ then $A$ is the center of $\odot(PQR)$ $\Longrightarrow$ $YZ,$ $QY,$ $RZ$ are tangents at $P,Q,R$ $\Longrightarrow$ $XP$ is the polar of $T^*$ WRT $\odot(PQR)$ $\Longrightarrow$ $X(Z,Y,P,T^*)=-1$ $\Longrightarrow$ $A(B,C,P,T^*)=-1$ or $P(B,C,A,T^*)=-1.$ As a result, $\angle APT=90^{\circ}$ $\Longleftrightarrow$ $T \equiv T^*$ $\Longleftrightarrow$ $PA$ bisect $\angle BPC$ $\Longleftrightarrow$ $\angle APB=\angle APC.$

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4147 posts

#3 h Sep 11, 2015, 3:43 PM • 2 Y

Y by Adventure10, Mango247

Sorry there is a flaw in the previous resolution. $T^*$ is the pole of $XP$ WRT $\odot(PQR)$ only when $T \equiv T^*,$ though this can be easily fixed keeping the same notations.

If $\angle APT=90^{\circ}$ $\Longrightarrow$ $T$ is the pole of $XP$ WRT $\odot(PQR)$ $\Longrightarrow$ $X(Z,Y,P,T)=-1$ $\Longrightarrow$ $P(B,C,A,T)=-1$ $\Longrightarrow$ $AP$ bisects $\angle BPC.$ Conversely if $AP$ bisects $\angle BPC$ $\Longrightarrow$ $P(B,C,A,T^*)=-1$ or $A(B,C,P,T^*)=-1$ $\Longrightarrow$ perpendiculars from $P$ to $AB,AC,AP,AT^*$ form a harmonic pencil as well. If $D \in \odot(PQR)$ is the reflection of $P$ on $AT^*,$ then $P(Q,R,D,T^*)=-1$ $\Longrightarrow$ $PQDR$ is harmonic $\Longrightarrow$ $Q,R,T^*$ are collinear $\Longrightarrow$ $T \equiv T^*$ $\Longrightarrow$ $\angle APT=90^{\circ}.$

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#4 h Sep 11, 2015, 6:34 PM • 4 Y

Y by Crystal., enhanced, Adventure10, Mango247

After performing the Inversion with center $A$ we get the following equivalent problem :

Given a $\triangle ABC$ and an arbitrary point $P$ . Let $Q,$ $R$ be the reflection of $P$ in $CA,$ $AB$ , respectively. Let $T$ be the second intersection of $\odot (ABC)$ and $\odot (AP)$ . Prove that $A,$ $Q,$ $R,$ $T$ are concyclic if and only if $\angle PBA$ $=$ $\angle PCA$ .
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Proof :

Let $Y,$ $Z$ be the projection of $P$ on $CA,$ $AB$ , respectively. Since $A$ lies on the perpendicular bisector $CA,$ $AB$ of $PQ,$ $PR$ , so $A$ is the circumcenter of $\triangle PQR$ $\Longrightarrow$ $\angle AQR$ $=$ $\angle ARQ$ $=$ $90^{\circ}$ $-$ $\angle BAC$ , hence $T$ lies on $\odot (AQR)$ iff $\angle QTP$ $=$ $\angle RTP$ $=$ $\angle BAC$ . On the other hand, since $T$ is the Miquel point of the complete quadrilateral $\{ BC, CA, AB, YZ \}$ , so $\triangle TYC$ $\sim$ $\triangle TZB$ $\Longrightarrow$ $\angle QTP$ $=$ $\angle RTP$ $=$ $\angle BAC$ iff $\triangle TYC$ $\cup$ $(P,Q)$ $\sim$ $\triangle TZB$ $\cup$ $(R,P)$ (notice $PQ$ $\perp$ $CY,$ $RP$ $\perp$ $BZ$ and $Y,$ $Z$ is the midpoint of $PQ,$ $RP$ , respectively.) iff $\angle PBA$ $=$ $\angle PBZ$ $=$ $\angle RBZ$ $=$ $\angle QCY$ $=$ $\angle PCY$ $=$ $\angle PCA$ .

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824 posts

andria

#5 h Sep 11, 2015, 7:21 PM • 2 Y

Y by Adventure10, Mango247

Different solution by inversion:
Consider an inversion $\Psi$ with center $P$ . After performing $\Psi$ we get the following problem:
Problem:
Let $PBC$ be a triangle. Consider an arbitrary point $A$ . Let $R,Q$ be the circumcenters of $\triangle PAB,\triangle PAC$ respectively. Let $T$ be a point on circumcircle of $\triangle ABC$ such that $PT\perp PA$ then $RPTQ$ is cyclic if and if only $\angle APB=\angle APC$ .
Proof:
Let $O$ be the circumcenter of $\triangle PBC$ then $RO,OQ$ are perpendicular bisectors of $PB,PC$ respectively. since $RQ$ is perpendicular bisector of $PA$ so $PT\parallel RQ$ . also $OP=OT$ . Now if:

1) $RPTQ$ is cyclic then it is isosceles trapezoid so $PR=QT$ and $\angle OPR=\angle OTQ$ hence $\triangle OTQ=\triangle OPR\Longrightarrow OR=OQ\Longrightarrow \angle APC=\angle APB$ .

2) $\angle APC=\angle APB$ then $\angle OQR=\angle ORQ\Longrightarrow OR=OQ$ Also since $T$ is midpoint of arc $BPC$ of $\odot(PBC)$ we get $\angle TOQ=\angle POR=\angle C$ hence $\triangle OTQ=\triangle OPR\Longrightarrow TQ=PR\Longrightarrow RPTQ$ is isosceles trapezoid so it is cyclic.
DONE

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290 posts

jred

#6 h Oct 15, 2017, 2:55 PM • 2 Y

Y by Adventure10, Mango247

Dose anyone notice that the claim is not true when $P$ lies on the extension of segment $BC$ ? I wonder if the translation is correct.

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328 posts

#7 h Aug 5, 2018, 9:58 AM • 2 Y

Y by Adventure10, Mango247

a bit similar to Luis Gonzalez's solution perhaps:

lemma: for two points $P,Q$ on a circle $\omega$ and $A$ another point on $\omega$ then if $(AA,PP)=S$ , $(AA,QQ)=T$ , $(PQ,AA)=B$ $\implies (STAB)=-1$

Back to problem: let $A'=(AP,BC)$ we easily see that $A \equiv O_{PQR}$ and if then let $T'$ be a point on $BC$ such as that $PT' \perp AP$ so its tangent to $\odot PQR$ and let $PT'$ meet $AB, AC$ respectively at $Z, Y$ then if $PA$ bisects $\angle BPC$ then easy to see $(BCA'T')=(ZYPT')=-1$ and according to the lemma $(ZYPU)=-1$ in which $U$ is the cross point of $ZY$ and $QR \implies U \equiv T' \equiv T$ and the same for the converse !

This post has been edited 3 times. Last edited by H.HAFEZI2000, Aug 5, 2018, 10:00 AM

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Arefe

#8 h Aug 1, 2020, 8:00 AM • 2 Y

Y by r_ef, hakN

Let $K$ be a point on $RQ$ such that $\angle{KPA}=90$ and let $M$ midpoint of $PK$ and $L$ foot of perpendicular of $P$ to $AK$ and $E,F$ be the midpoint of $PR , PQ$ .
$ALFPE$ is cyclic ( $\omega$ ) and $PK$ is tangent to $\omega$ .Because $MP=ML$ and $MP$ is tangent to $\omega$ , $ML$ is tangent , too . So $(FE,LP)=-1$ , so $A(BC,KP)=-1$
so if we get $X$ and $D$ intersection of $AK , AP$ with $BC$ , we can proof the else easily

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