2018 Brazilian Olympic Revenge Problem 4

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2018 Brazilian Olympic Revenge Problem 4

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Source: 2018 Brazilian Olympic Revenge Problem 4

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rsa365

#1 h Jul 15, 2018, 12:45 PM • 3 Y

Y by tiendung2006, Adventure10, Mango247

Let $ABC$ be a triangle with acute angles with incircle $\omega$ and incenter $I$ . Let $T_A,T_B$ and $T_C$ be the points where $\omega$ touches the sides of $ABC$ . Let $l_A$ be a line through $A$ parallel to $BC$ , and let $l_B$ and $l_C$ be analogously defined. Let $L_A$ be the second intersection of $AI$ to the circumcircle of $ABC$ , and define $L_B$ and $L_C$ analogously. Let $P_A = T_BT_C \cap l_A$ and $P_B, P_C$ be defined analogously. Let $S_A = P_BT_B \cap P_CT_C$ and define $S_B, S_C$ analogously. Prove that $S_AL_A, S_BL_B$ and $S_CL_C$ meet at a point.

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#2 h Jul 15, 2018, 1:45 PM • 2 Y

Y by Adventure10, Mango247

I remember having solved this a few months back, on AoPS.

Edit : It was here.

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PROF65

#4 h Jul 15, 2018, 6:25 PM • 3 Y

Y by dimdo, Adventure10, Mango247

rsa365 wrote:

Let $ABC$ be a triangle with acute angles with incircle $\omega$ and incenter $I$ . Let $T_A,T_B$ and $T_C$ be the points where $\omega$ touches the sides of $ABC$ . Let $l_A$ be a line through $A$ parallel to $BC$ , and let $l_B$ and $l_C$ be analogously defined. Let $L_A$ be the second intersection of $AI$ to the circumcircle of $ABC$ , and define $L_B$ and $L_C$ analogously. Let $P_A = T_BT_C \cap l_A$ and $P_B, P_C$ be defined analogously. Let $S_A = P_BT_B \cap P_CT_C$ and define $S_B, S_C$ analogously. Prove that $S_AL_A, S_BL_B$ and $S_CL_C$ meet at a point.

use pole and polar remark that $K_A$ , the intersection of the $A$ -median and $T_BT_C$ is on the intersection of $IT_A$ then the pole of $\ell_A$ is $K_A$ so the polar of $P_A$ is the median say $AM_a$ thus the polar of $M_a$ is $P_AT_A$ similarly we get that $M_b$ is $P_BT_B$ , $M_c$ is $P_CT_C$ whence the polar of $S_A$ is $M_bM_c$ in the other hand the pole of $M_bM_c$ is $H_a$ the orthocenter of $IBC$ indeed $H_a$ is the symmetric of the excenter $I_a$ in $M_a$ then its polar pass through the pole of $M_aI_a$ and is parallel to $BC$ but the polar of $I_a$ is the $T_A$ -midline of $T_AT_BT_C$ thus it cuts $T_AP_A$ at its midpoint which is a point of the $A$ -midline of $ABC$ hence the polar of $H_a$ is the $A$ -midline $M_bM_c$ of $ABC$ thus $S_A=H_a$ which means that $S_AL_A, S_BL_B$ and $S_CL_C$ are the euler lines of $IBC,ICA,IAB$ which are concurrent
RH HAS

This post has been edited 2 times. Last edited by PROF65, Jul 16, 2018, 11:11 AM

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dimdo

#6 h Jan 29, 2019, 2:42 PM • 1 Y

Y by Adventure10

PROF65 wrote:

rsa365 wrote:

Let $ABC$ be a triangle with acute angles with incircle $\omega$ and incenter $I$ . Let $T_A,T_B$ and $T_C$ be the points where $\omega$ touches the sides of $ABC$ . Let $l_A$ be a line through $A$ parallel to $BC$ , and let $l_B$ and $l_C$ be analogously defined. Let $L_A$ be the second intersection of $AI$ to the circumcircle of $ABC$ , and define $L_B$ and $L_C$ analogously. Let $P_A = T_BT_C \cap l_A$ and $P_B, P_C$ be defined analogously. Let $S_A = P_BT_B \cap P_CT_C$ and define $S_B, S_C$ analogously. Prove that $S_AL_A, S_BL_B$ and $S_CL_C$ meet at a point.

use pole and polar remark that $K_A$ , the intersection of the $A$ -median and $T_BT_C$ is on the intersection of $IT_A$ then the pole of $\ell_A$ is $K_A$ so the polar of $P_A$ is the median say $AM_a$ thus the polar of $M_a$ is $P_AT_A$ similarly we get that $M_b$ is $P_BT_B$ , $M_c$ is $P_CT_C$ whence the polar of $S_A$ is $M_bM_c$ in the other hand the pole of $M_bM_c$ is $H_a$ the orthocenter of $IBC$ indeed $H_a$ is the symmetric of the excenter $I_a$ in $M_a$ then its polar pass through the pole of $M_aI_a$ and is parallel to $BC$ but the polar of $I_a$ is the $T_A$ -midline of $T_AT_BT_C$ thus it cuts $T_AP_A$ at its midpoint which is a point of the $A$ -midline of $ABC$ hence the polar of $H_a$ is the $A$ -midline $M_bM_c$ of $ABC$ thus $S_A=H_a$ which means that $S_AL_A, S_BL_B$ and $S_CL_C$ are the euler lines of $IBC,ICA,IAB$ which are concurrent
RH HAS

Why the euler lines of IAB,IBC,ICA meet at a point?

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#7 h Jan 29, 2019, 2:51 PM • 1 Y

Y by Adventure10

dimdo wrote:

PROF65 wrote:

rsa365 wrote:

Let $ABC$ be a triangle with acute angles with incircle $\omega$ and incenter $I$ . Let $T_A,T_B$ and $T_C$ be the points where $\omega$ touches the sides of $ABC$ . Let $l_A$ be a line through $A$ parallel to $BC$ , and let $l_B$ and $l_C$ be analogously defined. Let $L_A$ be the second intersection of $AI$ to the circumcircle of $ABC$ , and define $L_B$ and $L_C$ analogously. Let $P_A = T_BT_C \cap l_A$ and $P_B, P_C$ be defined analogously. Let $S_A = P_BT_B \cap P_CT_C$ and define $S_B, S_C$ analogously. Prove that $S_AL_A, S_BL_B$ and $S_CL_C$ meet at a point.

use pole and polar remark that $K_A$ , the intersection of the $A$ -median and $T_BT_C$ is on the intersection of $IT_A$ then the pole of $\ell_A$ is $K_A$ so the polar of $P_A$ is the median say $AM_a$ thus the polar of $M_a$ is $P_AT_A$ similarly we get that $M_b$ is $P_BT_B$ , $M_c$ is $P_CT_C$ whence the polar of $S_A$ is $M_bM_c$ in the other hand the pole of $M_bM_c$ is $H_a$ the orthocenter of $IBC$ indeed $H_a$ is the symmetric of the excenter $I_a$ in $M_a$ then its polar pass through the pole of $M_aI_a$ and is parallel to $BC$ but the polar of $I_a$ is the $T_A$ -midline of $T_AT_BT_C$ thus it cuts $T_AP_A$ at its midpoint which is a point of the $A$ -midline of $ABC$ hence the polar of $H_a$ is the $A$ -midline $M_bM_c$ of $ABC$ thus $S_A=H_a$ which means that $S_AL_A, S_BL_B$ and $S_CL_C$ are the euler lines of $IBC,ICA,IAB$ which are concurrent
RH HAS

Why the euler lines of IAB,IBC,ICA meet at a point?

It is well known fact and the intersection point is called Schiffler point. Also it is easily proved by bary bash.

If i am not mistaken the coordinates of Schiffler point must be $(\frac{b+c-a}{b+c}, \frac{c+a-b}{c+a}, \frac{a+b-c}{a+b})$ .

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#8 h Feb 23, 2019, 10:10 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
Firstly, we have a lemma:
Lemma: Given $\triangle$ $ABC$ with incircle ( $I$ ) touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ . $N$ is tangency point of $A$ - excircle with $BC$ . $T$ is intersection which near $N$ of $AN$ and ( $I$ ), $K$ is intersection of $EF$ and $DT$ . Prove that: $AK$ $\parallel$ $BC$
About the proof for this problem, you can see here
Back to main problem: Let $M_A$ , $M_B$ , $M_C$ be midpoint of $BC$ , $CA$ , $AB$
From the above lemma, we have: $T_BP_B$ , $T_CP_C$ are polars of $M_B$ , $M_C$ with respect to ( $I$ ), respectively
So: $M_BM_C$ is polar of $S_A$ WRT ( $I$ ) or $S_A$ is orthocenter of $\triangle$ $IBC$
Similarly: $S_B$ is orthocenter of $\triangle$ $ICA$ , $S_C$ is orthocenter of $\triangle$ $IAB$
But: $L_A$ , $L_B$ , $L_C$ are circumcenters of $\triangle$ $IBC$ , $\triangle$ $ICA$ , $\triangle$ $IAB$ so $S_AL_A$ , $S_BL_B$ , $S_CL_C$ are Euler line of $\triangle$ $IBC$ , $\triangle$ $ICA$ , $\triangle$ $IAB$
Hence: $S_AL_A$ , $S_BL_B$ , $S_CL_C$ concurrent at Schiffler point of $\triangle$ $ABC$

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