Let $ABC$ be a triangle and $N$ be a point that differs from $A,B,C$. Let $A_b$ be the reflection of $A$ through $NB$, and $B_a$ be the reflection of $B$ through $NA$. Similarly, we define $B_c, C_b, A_c, C_a$. Let $m_a$ be the line through $N$ and perpendicular to $B_cC_b$. Define similarly $m_b, m_c$. a) Assume that $N$ is the orthocenter of $\triangle ABC$, show that the respective reflection of $m_a, m_b, m_c$ through the bisector of angles $\angle BNC, \angle CNA, \angle ANB$ are the same line. b) Assume that $N$ is the nine-point center of $\triangle ABC$, show that the respective reflection of $m_a, m_b, m_c$ through $BC, CA, AB$ concur.
Problem
Source: Vietnam TST 2021 P3
Tags: geometry, geometric transformation, reflection
01.04.2021 15:58
sorry,i don't understand part a,what does "are the same" mean?
01.04.2021 21:46
They are the same line. I'll make that clear.
02.04.2021 13:46
Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$
04.04.2021 23:13
20.05.2021 12:00
Groupsolved with L567, For part a) We claim that the desired line is the Euler line of triangle ABC. Let $E,F$ be the feet of the perpendicular from $B,C$ respectively. Proof. Let $l_1,l_2$ be the lines through $A$ parallel to the Euler line and the perpendicular from $A$ to $B_1C_1$. Now, if we can show that $l_1,l_2$ are isogonal in $\angle BAC$, then the desired result will follow as then we will have $\angle(l_2,BH)=\angle (l_2, AC)+90\angle (l_1,AB)+90=\angle (l_1, CH)$ and the isogonality follows. Now, to prove that these two lines are isogonal, we just need that $l_1$ passes through the circumcenter of $(AB_1C_1)$. Let this point be $O_A$. We claim that $O_AAOH$ is a parallelogram! Let $O'$ be the midpoint of $AH$. So, we would just need that the midpoint of $O_A$ is the midpoint of $AH$. But, consider the spiral similarity that takes $BF$ to $CE$. Let this center of spiral similarlity be $S$.Then, $SO_AO\sim SB_1B$ and $SO_AO'\sim SB_1F$ and $SO'O\sim SFB$. Thus, $O'$ is the midpoint of $OO_A$ and the result follows. Now, for part $b)$, my angle chase is very similar to khanhnx's above angle chase so I will omit that.
30.03.2022 18:17
khanhnx wrote: Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$ Can you explain me how $180^{\circ} - \angle{BAC}=\angle{O'_bNO'_c}?$
04.07.2022 16:40
Jack125 wrote: khanhnx wrote: Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$ Can you explain me how $180^{\circ} - \angle{BAC}=\angle{O'_bNO'_c}?$ I think you can see here