An extension of a property from mixtilinear incircle

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An extension of a property from mixtilinear incircle

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#1 h Dec 27, 2014, 6:40 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate points. $AP$ cuts $(O)$ again at $D$ . $DE$ is diameter of $(O)$ . $\text{[math]}$ cuts $(O)$ again at $F$ . Prove that $BC, DF$ and the line passing through $P$ and perpendicular to $AQ$ are concurrent.

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#2 h Dec 27, 2014, 10:41 PM • 6 Y

Y by Arab, buratinogigle, Adventure10, Mango247, MS_asdfgzxcvb, and 1 other user

Let $AQ$ cut $(O)$ again at $D'$ and let $X \equiv DF \cap BC.$ If we fix $AD,AD',$ the relation between $P$ and the intersection of $BC$ with the perpendicular from $P$ to $AD'$ is obviously a perspectivity and $EP \ \overline{\wedge} \ EQ \equiv EF \ \overline{\wedge} \ DF \equiv DX.$ So it suffices to show that the desired concurrency holds for at least 3 positions of $P.$

When $P \equiv AD \cap BC,$ then $X \equiv P$ and $Q \equiv A \equiv F$ $\Longrightarrow$ the concurrency trivially holds. When $P$ is at infinity, then $Q \equiv F \equiv D'$ $\Longrightarrow$ $DF \parallel BC$ $\Longrightarrow$ $X$ is at infinity $\Longrightarrow$ the concurrency holds. Finally, when $P \equiv A,$ then $Q \equiv AD' \cap BC$ and we have $\angle FAD'=\angle FDD'=\angle FXQ$ $\Longrightarrow$ $AXFQ$ is cyclic $\Longrightarrow$ $\angle XAQ=\angle DFQ=90^{\circ},$ i.e. $AX \perp AQ$ $\Longrightarrow$ the concurrency holds.

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#3 h Dec 28, 2014, 1:54 AM • 6 Y

Y by buratinogigle, Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users

My solution:

Let $X$ be the projection of $P$ on $AQ$ .
Let $Y$ be the projection of $Q$ on $BC$ .
Let $Z=AQ \cap \odot (ABC), T=YZ \cap \odot (ABC), U=AP \cap BC, V=DF \cap BC$ .

It's suffices to prove $V \in PX$

Easy to see $F, Q, V, Y$ lie on a circle with diameter $VQ$ .

From Reim theorem ( for $T-Y-Z$ and $A-U-D$ ) we get $A, T, U, Y$ are concyclic .
From my post at Concurrent with PQ line (lemma 3) we get $A, T, P, X$ all lie on $\odot (AP)$ .
From Reim theorem ( for $F-D-V$ and $T-Z-Y$ ) we get $T \in \odot (VQ)$ . ... $(\star)$

Since $\angle AXT=90^{\circ}-\angle TAP=90^{\circ}-\angle TYU=\angle QYT$ ,
so we get $T, Q, X, Y$ are concyclic ,
hence combine with $(\star)$ we get $F, T, Q, V, X, Y$ all lie on $\odot (VQ)$ ,
so we get $\angle AXV=90^{\circ} =\angle AXP$ . ie. $X, P, V$ are collinear

Q.E.D

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#4 h Feb 4, 2015, 8:18 AM • 3 Y

Y by shinichiman, Adventure10, Mango247

Lemma. Let $ABC$ be a triangle inscribed circle $(O)$ . $P,Q$ are isogonal conjugate points. $AP$ cuts $(O)$ agian at $M$ . $QM$ cuts $BC$ at $E$ then $PE\parallel AQ$ .

Proof by my pupil Phan Anh Quan. $AQ$ cuts $(O)$ again at $N$ and cut $BC$ at $H$ . From $P,Q$ are isogonal conjugate we have $\triangle CHN\sim\triangle ACM$ and $\triangle CPM\sim\triangle QCN$ (g.g) deduce $HN.AM=CM.CN=QN.PM$ , therefore $\dfrac{MP}{MA}=\dfrac{NH}{NQ}=\dfrac{ME}{MQ}$ , thus $PE\parallel AQ$ . We are done.

Proof of the problem by my pupil Trinh Huy Vu. Let $FD$ cut $BC$ at $S$ , we will prove that $PS\perp AQ$ , ineed. Let $QD$ cuts $BC$ at $R$ . $T$ is projection of $R$ on $FD$ . From the lemma, $PR\parallel AQ$ . From this, $\dfrac{DT}{DF}=\dfrac{DR}{DQ}=\dfrac{DP}{DA}$ deduce $PT\parallel AF$ . We have angle charsing,

$\angle TPR=\angle FAQ=\angle BAQ-\angle BAF=\angle PAC-\angle BDF=\angle DBC-\angle BDF=\angle TSR.$

Therfore, $STRP$ is cyclic or $SP\perp PR\parallel AQ$ . We are done.

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#5 h Feb 9, 2015, 3:44 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

buratinogigle wrote:

Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate points. $AP$ cuts $(O)$ again at $D$ . $DE$ is diameter of $(O)$ . $\text{[math]}$ cuts $(O)$ again at $F$ . Prove that $BC, DF$ and the line passing through $P$ and perpendicular to $AQ$ are concurrent.

$AQ$ cuts $(O)$ again at $M$ , $DF$ cuts $BC$ at $T$ . Since $P,Q$ are isogonal conjugates wrt $\triangle ABC$ so $\triangle DBP \sim \triangle MQB \; ( \text{A.A})$ . Hence $MQ \cdot DP=MB \cdot DB$ . We can also prove that $\triangle MFB \sim \triangle DBT \; ( \text{A.A})$ so $MB \cdot DB= MF \cdot DT$ . Thus, $MF \cdot DT=MQ \cdot DP$ or $\frac{MQ}{MF}= \frac{DT}{DP}$ . From here we get $\triangle MQF \sim \triangle DTP \; ( \text{S.A.S})$ . Therefore $\angle DPT= \angle MFQ= \angle MAE=90^{\circ}- \angle PAQ$ . This follows $TP\perp AQ$ .

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#6 h May 10, 2015, 3:40 PM • 3 Y

Y by shinichiman, Adventure10, Mango247

A better extension

Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate point. $AP$ cuts $(O)$ again at $D$ . $E$ is a point on $(O)$ . $\text{[math]}$ cuts $(O)$ again at $F$ . $DF$ cuts $BC$ at $M$ . $AQ$ cuts $(FMQ)$ again at $N$ . Prove that $M,N,P$ are colliear.

The solution is the same as in #4 or in #5 of shinichiman.

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#7 h May 10, 2015, 10:18 PM • 2 Y

Y by buratinogigle, Adventure10

We can also solve the extension with the same method I gave in my previous post.

Fix $D$ and $E$ and animate $P.$ If we redefine $N \in AQ,$ such that $\angle (NA,NP)=\angle (FE,FD)=\text{const},$ then all lines $NP$ are parallel to each other $\Longrightarrow$ series $P,N$ are similar, but clearly $P \ \overline{\wedge} \ Q \ \overline{\wedge} \ F \ \overline{\wedge} \ M.$ So it suffices to prove that $M,N,P$ are collinear for at least 3 positions of $P.$

When $P$ is at infinity and $P \equiv D$ the collinearity trivially holds. When $P \equiv A,$ then $Q \in BC$ and we have $\angle AQB=\angle ACD=\angle AFD$ making $AQFM$ cyclic $\Longrightarrow$ $\angle (AQ,AM)=\angle (FE,FD)$ $\Longrightarrow$ the collinearity holds. Thus $M,N,P$ are collinear for any $P.$

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