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#1 h Aug 5, 2015, 6:19 AM • 2 Y

Y by Adventure10, Mango247

Let $\triangle ABC$ with incenter $I$ . $AI, BI, CI$ cuts $BC, CA, AB$ at $D, E, F$ , res. Let $X$ is orthocenter of $\triangle DEF$ . Prove that $IX$ is parallel to Euler line of $\triangle ABC$

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A-B-C

#2 h Aug 5, 2015, 12:27 PM • 1 Y

Y by Adventure10

It has been posted here http://www.artofproblemsolving.com/community/c6h171583
You can see the proof of Jean-Louis Ayme in his website.

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#3 h Aug 5, 2015, 3:21 PM • 9 Y

Y by mineiraojose, A-B-C, enhanced, bobaboby1, noway, AlastorMoody, tarzanjunior, enzoP14, Adventure10

My solution :

Lemma 1 :

Let $I$ be the incenter of $\triangle ABC$ and $\triangle XYZ$ be the incentral triangle of $\triangle ABC$ .
Let $D, E, F$ be the reflection of $I$ in $BC, CA, AB$ , resp and $P \equiv AD \cap BE \cap CF$ .

Then $IP$ passes through the orthocenter $H$ of $\triangle XYZ$

Proof :

Let $Q \equiv YH \cap ZX, R \equiv XH \cap YZ, S \equiv AX \cap YZ$ .
Let $T \equiv YZ \cap BC, U \equiv AD \cap BC, V \equiv BE \cap CA$ .

From $U(A,I;S,X)=-1$ and $\angle IUX=\angle DUX \Longrightarrow US \perp UX \equiv BC$ ,
so combine $AT \perp AI \Longrightarrow A, S, T, U$ are concyclic $\Longrightarrow \angle IUS=\angle AUS=\angle ATS$ ,
hence notice $A, R, T, X$ are concyclic $\Longrightarrow \angle IUS=\angle ATS=\angle AXR=\angle RUS \Longrightarrow I \in UR$ ,
so from Desargue theorem (for $\triangle ABU$ and $\triangle XYR$ ) $\Longrightarrow XR \cap AP$ lie on the antiorthic axis of $\triangle ABC$ .

Similarly, we can prove $I \in VQ$ and $YQ \cap BP$ lie on the the antiorthic axis of $\triangle ABC$ ,
so from Desargue theorem (for $\triangle ABP$ and $\triangle XYH$ ) $\Longrightarrow AX \cap BY \in PH$ . i.e. $P, I, H$ are collinear
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Lemma 2 :

Let $I$ be the incenter of $\triangle ABC$ .
Let $D, E, F$ be the reflection of $I$ in $BC, CA, AB$ , respectively .

Then $IP$ is parallel to the Euler line $\mathcal{E}$ of $\triangle ABC$ where $P \equiv AD \cap BE \cap CF$

Proof :

Let $Y \equiv FD \cap CA, Z \equiv DE \cap AB$ .
Let $\triangle A_1B_1C_1, \triangle A_2B_2C_2$ be the intouch triangle, orthic triangle of $\triangle ABC$ , respectively .
Let $T$ be the midpoint of $AI$ and $\overline{A_3B_3C_3}$ be the orthic axis of $\triangle ABC$ ( $A_3\in BC, B_3\in CA,C_3\in AB$ ) .

Since $\odot (Y, YI), \odot (Z, ZI)$ is the image of $BE, CF$ , resp under the inversion $\mathbf{I}(\odot (DEF))$ ,
so $IP$ is the radical axis of $\odot (Y, YI), \odot (Z, ZI) \Longrightarrow IP \perp YZ$ (also follows from Sondat theorem) . (1)

From $TC_1 \parallel AF, A_1C_1 \parallel DF \Longrightarrow \angle AFD=\angle TC_1A_1$ ,
so notice $T$ is the center of $\odot (AI) \Longrightarrow \angle AFD=90^{\circ}-\tfrac{1}{2} \angle A+\tfrac{1}{2} \angle B$ ,

hence by sine rule we get $AY=AF \cdot \frac{\sin \angle AFD}{\sin \angle AYD}=AI \cdot \frac{\cos \tfrac{1}{2} (\angle A-\angle B)}{\sin \tfrac{1}{2} (\angle A-\angle C)}$ .

Similarly, $AZ=AI \cdot \frac{\cos \tfrac{1}{2} (\angle A-\angle C)}{\sin \tfrac{1}{2} (\angle A-\angle B)} \Longrightarrow \frac{AY}{AZ}=\frac{ \sin \tfrac{1}{2} (\angle A-\angle B) \cdot \cos \tfrac{1}{2} (\angle A-\angle B)}{\sin \tfrac{1}{2} (\angle A-\angle C) \cdot \cos \tfrac{1}{2} (\angle A-\angle C)}=\frac{\sin (\angle A-\angle B)}{\sin (\angle A-\angle C)}$ . (2)

Since $AB_3=AC_2 \cdot \frac{\sin \angle AC_2B_3}{\sin \angle AB_3C_2}= \frac{AC_2 \cdot \sin \angle C}{\sin (\angle A-\angle C) } , AC_3=AB_2 \cdot \frac{\sin \angle AB_2C_3}{\sin \angle AC_3B_2}=\frac{AB_2 \cdot \sin \angle B}{\sin (\angle A-\angle B) }$ ,

so we get $\frac{AB_3}{AC_3}=\frac{AC_2 \cdot \sin \angle C}{AB_2 \cdot \sin \angle B} \cdot \frac{\sin (\angle A-\angle B)}{\sin (\angle A-\angle C)}=\frac{AC_2 \cdot AB}{AB_2 \cdot AC} \cdot \frac{\sin (\angle A-\angle B)}{\sin (\angle A-\angle C)}=\frac{\sin (\angle A-\angle B)}{\sin (\angle A-\angle C)}$ ,

hence combine (2) $\Longrightarrow B_3C_3 \parallel YZ \Longrightarrow YZ \perp \mathcal{E}$ ( $\because B_3C_3$ is perpendicular to $\mathcal{E}$ (well-known)) $\Longrightarrow IP \parallel \mathcal{E}$ (from (1)) .
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Back to the main problem :

Let $I_a, I_b, I_c$ be the reflection of $I$ in $BC, CA, AB$ , resp and $P \equiv AI_a \cap BI_b \cap CI_c$ .

From lemma 1 $\Longrightarrow P \in IX$ , so from lemma 2 $\Longrightarrow IX \equiv IP$ is parallel to the Euler line of $\triangle ABC$ .

Q.E.D
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P.S. The fact $AI_a, BI_b, CI_c$ are concurrent simplified follows from Jacobi theorem

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#4 h Aug 5, 2015, 3:25 PM • 3 Y

Y by bobaboby1, Adventure10, Mango247

My solution is based on some properties that appearred on ETC.
$X=X_{500}$ , $I=X_1$
From http://faculty.evansville.edu/ck6/encyclopedia/glossary.html and http://faculty.evansville.edu/ck6/encyclopedia/ETC.html(see $X_{35},X_{79},X_{500}$ , we have $X_{500}$ is crosssum of $X_{1},X_{79}$ so $X_{500},X_1,X_{79}$ are collinear.
So we have to show that $X_1X_{79}$ is parallel to Euler line of $\triangle ABC$ .
Also from here http://faculty.evansville.edu/ck6/encyclopedia/ETC.html, if $A',B',C'$ are reflections of $I$ in $BC,CA,AB$ then $AA',BB',CC'$ are concurrent at $X_{79}$ .
From properties of Neuberg cubic,http://bernard.gibert.pagesperso-orange.fr/Exemples/k001.html, $P$ lies on Neuberg cubic, $P_a,P_b,P_c$ are reflections of $P$ in $BC,CA,AB$ then $AP_a,BP_b,CP_c$ are concurrent at a point on $PP*$ where $P*$ is isogonal conjugate of $P$ WRT $\triangle ABC$ , furthermore, $PP*$ is parallel to Euler line of $\triangle ABC$ so $X_{1}X_{79}\parallel X_{3}X_{4}$ .

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#5 h Aug 5, 2015, 3:43 PM • 4 Y

Y by mineiraojose, bobaboby1, Adventure10, Mango247

This proof is wrong.
______
A synthetic proof for lemma 2 in Telv's post:
$H_a$ is orthocenter of $\triangle IBC$ , $AI$ intersects $\odot (ABC)$ at $O_a\ne A$ then $O_a$ is circumcenter of $\triangle IBC$
$O_b,O_c$ is determined similarly.
Since $BH_a,CH_a\perp IC,IB$ , and $IO_{c}=BO_c$ , $IO_{b}=CO_b$ then $H_a$ is circumcenter of $\triangle IO_bO_c$
Similarly $\Rightarrow$ $O_aH_a$ , $O_bH_b$ , $O_cH_c$ are concurrent at Kosnita point $S$ of $\triangle O_aO_bO_c$ (also Schiffler point and Schiffler point lie on Euler line)
$A,B,C$ are circumcenters of $\triangle IEF,\triangle IFD,\triangle IDE$ so $AD,BE,CF$ are concurrent at Kosnita point $P$ of $\triangle DEF$
$\triangle DEF$ and $\triangle O_aO_bO_c$ are homothetic, $I$ , $O$ are their circumcenters
$\Longrightarrow$ $IP\parallel OS$ , or $IP$ is parallel to Euler line of $\triangle ABC$

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This post has been edited 3 times. Last edited by A-B-C, Dec 6, 2021, 4:45 AM
Reason: wrong

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#6 h Sep 13, 2015, 10:48 PM • 6 Y

Y by mineiraojose, bobaboby1, enzoP14, enhanced, Adventure10, Mango247

TelvCohl wrote:

Lemma 2 :

Let $I$ be the incenter of $\triangle ABC$ .
Let $D, E, F$ be the reflection of $I$ in $BC, CA, AB$ , respectively .

Then $IP$ is parallel to the Euler line $\mathcal{E}$ of $\triangle ABC$ where $P \equiv AD \cap BE \cap CF$

Another proof to the Lemma 2 at post #3 :

Lemma :

Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ and $Y$ $\equiv$ $CA$ $\cap$ $FD,$ $Z$ $\equiv$ $AB$ $\cap$ $DE$ . Let $T$ be the intersection of the bisector of $\angle DYC$ and $\angle DZB$ . Let $I_a$ be the reflection of the Incenter $I$ of $\triangle ABC$ in $BC$ . Then $A,$ $T,$ $I_a$ are collinear.

Proof :

Let $K$ $\equiv$ $AI_a$ $\cap$ $BC,$ $U$ $\equiv$ $BI$ $\cap$ $CA,$ $V$ $\equiv$ $CI$ $\cap$ $AB$ . Let $P,$ $Q$ be the projection of $U$ on $AB,$ $BC$ , respectively and $R,$ $S$ be the projection of $V$ on $BC,$ $CA$ , respectively. From $AP$ $/$ $FP$ $=$ $AU$ $/$ $CU$ $=$ $AB$ $/$ $CB$ $=$ $AD$ $/$ $CF$ $=$ $AY$ $/$ $FY$ we get $YP$ is the bisector of $\angle DYC$ , so $Y,$ $T,$ $P$ are collinear. Similarly, we can prove $Q$ $\in$ $YT,$ $R$ $\in$ $ZT$ and $S$ $\in$ $ZT$ .

Let $T_1$ $\equiv$ $PQ$ $\cap$ $AI_a,$ $T_2$ $\equiv$ $RS$ $\cap$ $AI_a$ . From Menelaus theorem (for $\triangle ABK$ and $\overline{QT_1P}$ ) and notice $BP$ $=$ $BQ$ we get $AT_1$ $/$ $KT_1$ $=$ $AP$ $/$ $KQ$ ... (1). Similarly, we can prove $AT_2$ $/$ $KT_2$ $=$ $AS$ $/$ $KR$ ... (2). On the other hand, if $G$ $\equiv$ $AI$ $\cap$ $UV$ then from $(KA, KI; KG, BC)=-1$ and $BC$ bisect $\angle AKI$ we get $GK$ $\perp$ $BC$ , so $KQ$ $/$ $KR$ $=$ $GU$ $/$ $GV$ $=$ $AU$ $/$ $AV$ $=$ $AP$ $/$ $AS$ , hence combine with (1) and (2) we conclude that $T_1$ $\equiv$ $T_2$ $\equiv$ $T$ . i.e. $A,$ $T,$ $I_a$ are collinear
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Back to the main proof :

Let $\triangle XYZ$ be the orthic triangle of $\triangle ABC$ . Let $X_1$ be the intersection of the bisector of $\angle (CA, ZX)$ and the bisector of $\angle (AB, XY)$ (define $Y_1,$ $Z_1$ , similarly). From the Lemma $\Longrightarrow$ $X_1$ $\in$ $AD,$ $Y_1$ $\in$ $BE$ and $Z_1$ $\in$ $CF$ , so the perspectrix of any two triangles of $\{ \triangle ABC, \triangle DEF, \triangle X_1Y_1Z_1 \}$ are concurrent, hence the perspectrix $\tau$ of $\triangle ABC,$ $\triangle DEF$ is parallel to the Orthic axis of $\triangle ABC$ (notice $\triangle DEF$ and $\triangle X_1Y_1Z_1$ are homothetic).

From Sondat's theorem we get $\tau$ is perependicular to the line connecting the orthology center $I$ of $\{ \triangle ABC, \triangle DEF \}$ and the perspector $P$ of $\triangle ABC,$ $\triangle DEF$ , so we conclude that $IP$ is parallel to the Euler line of $\triangle ABC$ .

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#7 h Mar 1, 2018, 4:45 PM • 3 Y

Y by bobaboby1, Adventure10, Mango247

A-B-C wrote:

$BI=BO_c,CI=CO_b$

Thanks for your reply, but you simply went wrong here. Please check it carefully:)

This post has been edited 1 time. Last edited by liekkas, Mar 1, 2018, 4:45 PM

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