Tangent at Anti Steiner point

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ABCCBA

#1 h Apr 10, 2019, 4:15 PM • 3 Y

Y by tiendung2006, GeoKing, Adventure10

$\triangle ABC$ inscibes $(O),$ a point $P \in$ arc $BC$ of $(O)$ not contain $A.$ The Steiner line of $P$ wrt $\triangle ABC$ cuts $AB, AC$ at $E, F,$ resp. The line through $E \perp OB$ and through $F \perp OC$ cut each other at $R.$ Prove that $PR$ is tangent to $(O)$

This post has been edited 2 times. Last edited by ABCCBA, Apr 20, 2019, 5:40 AM
Reason: P lies on arc BC, not lies on (O)

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Paramizo_Dicrominique

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Paramizo_Dicrominique

#2 h Jul 14, 2023, 9:41 AM • 1 Y

Y by GeoKing

$\textbf{Solution:}$
Let $PE,PF$ cut $(ABC)$ again at $C',B'$ . $H$ is the orthocenter of $\triangle{ABC}$ .
Since $P$ is the Anti-Steiner point of $EF$ wrt. $\triangle{ABC}$ so it is well known that $\overline{C,H,C'},\overline{B,H,B'}$ .
$X,Y$ are the intersections of $RE,RF$ with $OB,OC$ . $K,L$ are the intersections of $PE,PF$ with $OB,OC$ .
Since $\angle{KPL} = \angle{C'PB'} = 180^{\circ} - \angle{C'AB'} = 180^{\circ} - 2\angle{BAC} = 180^{\circ} - \angle{BOC} \implies PKOL$ is cyclic.
We have $\angle{PEX} =90^{\circ} - \angle{EKX} = 90^{\circ} - \angle{FLY} = \angle{PFY} \implies \angle{PEX} = \angle{PFY}$ . $(\star)$
We will prove $\frac{EP}{EX} = \frac{FP}{FY}$ which is equivalent to $\frac{\frac{EB \cdot EA}{EC'}}{EX} = \frac{\frac{FC \cdot FA}{FB'}}{FY}$ $\Longleftrightarrow \frac{EX \cdot EC'}{EB \cdot EA} = \frac{FY \cdot FB'}{FC \cdot FA}$ $\Longleftrightarrow \frac{sin(ABO)}{\frac{sin(PBA)}{sin(BAC')}} = \frac{sin(ACO)}{\frac{sin(PCA)}{sin(CAB')}}$ $\Longleftrightarrow sin(ABO) \cdot sin(BAC') = sin(ACO) \cdot sin(CAB') \Longleftrightarrow sin(ABO) \cdot sin(BAH)$ $= sin(ACO) \cdot sin(CAH)$ which is true. $(\star \star)$
From $(\star),(\star \star) \implies$ $\triangle{PXE} \overset{+}{\sim} \triangle{PYF}$ so by spiral homothety $P$ lie on $(RXY)$ , also it is easy to see $O$ lie on $(RXY)$ so $P,R,Y,O,X$ lie on a circle with diameter $OR$ $\implies PR \perp PO \implies$ $PR$ is tangent to $(ABC)$ .
$Q.E.D.$

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#3 h Jul 14, 2023, 12:10 PM • 2 Y

Y by Paramizo_Dicrominique, Vnmboyacgn

A generalization for this nice problem.

Let $ABC$ be a triangle and points $E$ and $F$ lie on the lines $CA$ and $AB$ , respectively. Let $P$ be a point lying on the line $EF$ . Let $Q$ be an isogonal conjugate of $P$ with respect to triangle $ABC$ . Perpendicular lines from $E$ and $F$ to $QC$ and $QB$ meet at $R$ . It is well-known that the incenter $S$ of the triangle is bound by reflections of the line $EF$ in the lines $BC$ , $CA$ , and $AB$ lying on circumcircle $(O)$ of $ABC$ . $RS$ meets $(O)$ again at $T$ . Prove that $QT\perp RS$ .

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Paramizo_Dicrominique

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Paramizo_Dicrominique

#4 h Jul 19, 2023, 6:43 AM

Y by

buratinogigle wrote:

A generalization for this nice problem.

Let $ABC$ be a triangle and points $E$ and $F$ lie on the lines $CA$ and $AB$ , respectively. Let $P$ be a point lying on the line $EF$ . Let $Q$ be an isogonal conjugate of $P$ with respect to triangle $ABC$ . Perpendicular lines from $E$ and $F$ to $QC$ and $QB$ meet at $R$ . It is well-known that the incenter $S$ of the triangle is bound by reflections of the line $EF$ in the lines $BC$ , $CA$ , and $AB$ lying on circumcircle $(O)$ of $ABC$ . $RS$ meets $(O)$ again at $T$ . Prove that $QT\perp RS$ .

The idea of this problem is same to the previous problem but we have to prove some lemmas, I will post my solution soon.

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Paramizo_Dicrominique

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Paramizo_Dicrominique

#5 h Jul 19, 2023, 7:13 AM • 1 Y

Y by buratinogigle

Paramizo_Dicrominique wrote:

buratinogigle wrote:

A generalization for this nice problem.

Let $ABC$ be a triangle and points $E$ and $F$ lie on the lines $CA$ and $AB$ , respectively. Let $P$ be a point lying on the line $EF$ . Let $Q$ be an isogonal conjugate of $P$ with respect to triangle $ABC$ . Perpendicular lines from $E$ and $F$ to $QC$ and $QB$ meet at $R$ . It is well-known that the incenter $S$ of the triangle is bound by reflections of the line $EF$ in the lines $BC$ , $CA$ , and $AB$ lying on circumcircle $(O)$ of $ABC$ . $RS$ meets $(O)$ again at $T$ . Prove that $QT\perp RS$ .

The idea of this problem is same to the previous problem but we have to prove some lemmas, I will post my solution soon.

$\textbf{Solution:}$
From Iran MO 1995 Round 2 https://artofproblemsolving.com/community/c6h379391p2097984 we get $S$ lie on $(ABC)$ .
Let $EF$ cut $BC$ at $D$ . Let the tangent from $A$ to $(AEF)$ intersects $(ABC)$ again at $W$ .
$\textbf{Lemma 1:}$ Let $\triangle{ABC}$ be a triangle with $D,E,F$ lie on $BC,CA,AB$ and they line on a line $\alpha$ . The reflection of $\alpha$ over $AC,AB$ intersects at $X$ . Prove that: $AX$ is perpendicular to the tangent from $A$ to $(AEF)$ .
$\textbf{Proof:}$ Let $(AEF)$ cut $AX$ again at $I$ . It is easy to show that $I$ is the incenter of $\triangle{XEF}$ and $A$ is the $X-$ excenter of $\triangle{XEF}$ so $\angle{AFI} = 90^{\circ} \implies AW \perp AX$ and we are done.
Let $\mathcal{C}$ be circumconic of $\triangle{ABC}$ and isogonal conjugate to $\overline{D,E,F}$ .By simple angle chasing it is not hard to show $W$ lie on $\mathcal{C}$ .
We will redefine $T$ .
Let $BP,CP$ cut $(ABC)$ again at $L,K$ and $LE,KF$ intersects at $T$ . By conversed Pascal then $T$ lie on $(ABC)$ .
$\textbf{Claim:}$ $\overline{W,Q,T}$
$\textbf{Proof:}$ Let $S,R$ be the isogonal conjugate pair and $R$ is the intersection of $AQ$ and $EF$ . Then $S$ lie on $\mathcal{C}$ . Let $\mathcal{C'}$ be the circumconic of $\triangle{ABC}$ passes through $P,R$ .
From Lemma 2.2 of https://artofproblemsolving.com/community/c6h366218p28082252 we get $WQ,YR$ intersects on $(ABC)$ at point $T'$ .
Again from Lemma 3 of https://artofproblemsolving.com/community/c6h366218p28082252 we get $SQ,ZT',BC$ are concurrent with $Z$ is the intersection of $AP$ and $(ABC)$ . Appyling Lemma 3 of the mentioned link again we get $PR,BC,ZT'$ are concurrent. Hence we concluded that $PR,ZT',BC,SQ$ are concurrent at $D$ .
Applying Pascal theorem for $\binom{Z,B,K}{C,T,A}$ then it is not hard to show $T \equiv T'$ .
(Let $DT$ cut $(ABC)$ again at $Z'$ then by pascal $Z \equiv Z'$ so $T \equiv T'$ )
WIth the exactly same proof of the previous problem we can prove $T,G,H,Q,R$ lie on a circle with $G,H$ being the intersections of $QB,QC$ with $RF,RE$ . Hence $TQ \perp TR$ .
By $\textbf{Lemma 1}$ so $\angle{WAS} = 90^{\circ}$ $\implies$ $TW \perp TS$ hence $\overline{T,R,S}$ and we also had $QT \perp RS$ .
$Q.E.D.$

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This post has been edited 2 times. Last edited by Paramizo_Dicrominique, Jul 19, 2023, 7:19 AM

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