The altitudes $AH_1,BH_2,CH_3$ of an acute-angled triangle $ABC$ meet at point $H$. Points $P$ and $Q$ are the reflections of $H_2$ and $H_3$ with respect to $H$. The circumcircle of triangle $PH_1Q$ meets for the second time $BH_2$ and $CH_3$ at points $R$ and $S$. Prove that $RS$ is a medial line of triangle $ABC$.
Problem
Source: Sharygin 2018
Tags: geometry
Vrangr
04.04.2018 22:45
I complex bashed this one too, define $R'$ and $S'$ as the intersection point of the medial line and the altitudes, then prove $P, Q, R', S', H_1$ to be concyclic.
Tumon2001
04.04.2018 22:58
Solution: Let's start with a nice lemma. Lemma: Let $\triangle ABC$ be a triangle with incenter $I $ and $A-\text{excenter} $ $I_a $. Let $B'$ and $C'$ be the reflections of $B $ and $C $ in $I $ respectively. Let the circumcircle of $\triangle AB'C'$ intersect $BI $ at $B_1$ and $CI $ at $C_1$. Then, $B_1C_1$ is the $\perp $ bisector of $AI_a $. Proof of the lemma: Let $AC'$ intersect $BC $ at $M $ and $I_aC $ at $U $. Note that $BCB'C'$ is a parallelogram. Angle chasing shows that $ABB_1M$ is cyclic. Claim: $CC'\parallel I_aM $. Proof of the claim: Note that $\odot (CII_a) $ is the $C $-Apollonius circle of $\triangle ACX $, where $X\equiv AI\cap BC $. So, $(-1)=(A,X;I,I _a)\stackrel {C}\doublebarwedge (A,M;C',U)\stackrel{I_a}\doublebarwedge (I,Y;C',C) $, where $Y\equiv CC'\cap MI_a $. As $I $ is the midpoint of $CC'$, it follows that $Y\in\infty\implies CC'\parallel I_aM $. Back to the proof of the lemma: Since $ABB_1M $ is cyclic, so, angle chasing gives $B_1A=B_1M $. Also, $2\angle AI_aM=2\angle AIC'=\angle AB_1M $. Thus, $B_1$ is the circumcenter of $\triangle AMI_a $. Analogously, $C_1$ is the circumcenter of $\triangle ANI_a $, where $N\equiv AB'\cap BC $. Hence, $B_1$ and $C_1$ lie on the $\perp $ bisector of $AI_a $ and the lemma follows. Main problem: It's well-known that $H $ and $A $ are the incenter and the $H_1-\text {excenter} $ respectively of $\triangle H_1H_2H_3$. The above lemma implies that $RS $ is the $\perp $ bisector of $AH_1$. The result now follows immediately.
anantmudgal09
04.04.2018 23:02
TheDarkPrince wrote: The altitudes $AH_1,BH_2,CH_3$ of an acute-angled triangle $ABC$ meet at point $H$. Points $P$ and $Q$ are the reflections of $H_2$ and $H_3$ with respect to $H$. The circumcircle of triangle $PH_1Q$ meets for the second time $BH_2$ and $CH_3$ at points $R$ and $S$. Prove that $RS$ is a medial line of triangle $ABC$. Let $M, N$ be the midpoints of $AB, AC$ respectively. Instead define $R,S$ to be the meeting points of $BH_2, CH_3$ with the $A$-medial line. Now by $PQ \parallel H_2H_3$ and $RS \parallel BC$ we conclude $PQRS$ is cyclic. Observe that $\angle MH_1H=\angle MCB=\angle MSH$ (directed angles) proving $MSHH_1$ is cyclic. Now we go on to prove that $H_1$ indeed lies on $\odot(PQRS)$. To that end, we require $\angle H_PH_2=\angle H_1SR=180^{\circ}-\angle MSH_1=180^{\circ}-\angle MHH_1=\angle AHM$. Now we prove the crucial claim: Claim: $\triangle MHA \sim \triangle H_1PH_2$ (Proof) Observe $\angle HAM=\angle H_1H_2P$ and $$\frac{AH}{AM}=\frac{2AH}{AB}=\frac{2HH_2}{H_1H_2}=\frac{PH_2}{H_1H_2}$$proving the claim. $\blacksquare$ Thus, the problem falls, just like Winterfell.
TheDarkPrince
05.04.2018 09:51
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As $HQ=HH_3$ and $HP=HH_2$, $PQH_2H_3$ is a parallelogram. So, \[\angle HCH_1=\angle H_3AH=\angle HH_2H_3=\angle HPQ=\angle RPQ=\angle RSQ = \angle RSC.\]So, $RS||BC$.
Let $RS$ meet $AC$ and $AB$ at $E$ and $F$ respectively. Let $EC=kAC$ and $BF=kAB$. Also let circle $PH_1Q$ meet $BC$ again at $D$ (if $BC$ is tangent to $(PH_1Q)$, $D\equiv H_1$).
From power of the point, $BP\cdot BR=BD\cdot BH_1$ and $CQ\cdot CS=CD\cdot CH_1$. So, $BC=BD+DC=\frac{BP\cdot BR}{BH_1}+\frac{CQ\cdot CS}{CH_1}$.
As $RE||BC$ and $SF||BC$, $BR=EC\tan C=kAC\tan C$ and $CS=kAB\tan B$. We can find all the lengths $BH_1,CH_1,BH_2,CH_2,HH_2,HH_3$.
We have \[BC=\frac{BP\cdot kAC\tan C}{AB\cos B}+\frac{CQ\cdot kAB\tan B}{AC\cos C}.\]
So,
\begin{align*}
BC&=\frac{BP\cdot kAC\tan C}{AB\cos B}+\frac{CQ\cdot kAB\tan B}{AC\cos C}\\
&= \frac{kAC^2\sin C(BC\sin C-2AB\cos A\cot C)+kAB^2\sin B(BC\sin B-2AC\cos A\cot B)}{AB\cdot AC\cos B\cos C}\\
&= \frac{k(AC^2\cdot BC\sin^2 C-2AB\cdot AC^2\cos A\cos C+AB^2\cdot BC\sin ^2 B-2AC\cdot AB^2\cos A \cos B)}{AB\cos B\cdot AC\cos C}
\end{align*}
We need $k=\frac{1}{2}$. So we need \[2BC\cdot AB\cdot AC\cos B\cos C=AC^2\cdot BC\sin ^2C+AB^2\cdot BC\sin^2 B - 2AB\cdot AC(AB\cos A\cos C+AC\cos A\cos B).\]
So we need \[\sin A\sin B\sin C\cos B\cos C=\sin A\sin^2 B\sin^2 C -(\sin^2 B\sin C\cos A\cos C+\sin^2 C\sin B\cos A\cos B\]or $\sin(A+B+C)=\sin A\cos B\cos C+\sin B\cos C\cos A+\sin C\cos A\cos B-\sin A\sin B\sin C=0$ which is clearly true.
So, $k=\frac{1}{2}$ which gives $AE=EC$ and $AF=FB$. So, $RS$ is the medial line.
Proved.
RC.
07.04.2018 05:49
The only proof I formalised, maybe the same as any above. OP wrote: The altitudes $AH_1,BH_2,CH_3$ of an acute-angled triangle $ABC$ meet at point $H$. Points $P$ and $Q$ are the reflections of $H_2$ and $H_3$ with respect to $H$. The circumcircle of triangle $PH_1Q$ meets for the second time $BH_2$ and $CH_3$ at points $R$ and $S$. Prove that $RS$ is a medial line of triangle $ABC$.
\(RS || BC\)
Proof:- By Reim's:
\(PQ || H_2H_3\) and \(PQRS\) is cyclic. \(\Rightarrow H_2H_3RS\) is cyclic.
Now, \(H_2H_3RS\) is cyclic also, \(BCH_2H_3\) is cyclic \(\Rightarrow RS || H_2H_3\)
\(M \in \overleftrightarrow{RS}\) where \(M=\) midpoint of \(AC.\)
Proof:- Let \(H_2H_3 \cap AH = D, PQ \cap AH = E, H_1M \cap CH = F.\)
We have:- \(\angle PH_1S = \angle PRS = \angle CBR = \angle 90^{\circ} - \angle C =\angle AH_1M\)
\(\Rightarrow \angle PH_1S + \angle HH_1S = \angle HH_1E + \angle AH_1M\)
\(\Rightarrow \angle PH_1H = \angle SH_1M\)
Also, \(\angle H_1PE = \angle H_1SF\)
\(\therefore \Delta PH_1E \sim \Delta SH_1F - (i)\)
Now, \(\Delta H_1H_3D \sim \Delta CMF\)
\(\Rightarrow \dfrac{H_1H_3}{H_3D}= \dfrac{CM}{MF}\)
\(\Rightarrow \dfrac{HH_1}{HD}= \dfrac{H_1M}{MF}\)
\(\Rightarrow \dfrac{HH_1}{HE}= \dfrac{H_1M}{MF} - (ii)\)
\(Eq^{n} [i, ii] \Longrightarrow \Delta PH_1H \sim \Delta SH_1M\)
\(\Rightarrow \angle H_1PH = \angle H_1SM\)
\(\Rightarrow \angle H_1SR = \angle H_1SM\)
\(\Longrightarrow M \in \overleftrightarrow{RS}\)
sharp0407
13.04.2018 19:20
TO RC why 90-C=AH1M ???
yzhiyu
03.07.2018 14:02
Tumon2001 wrote: Solution: Let's start with a nice lemma. Lemma: Let $\triangle ABC$ be a triangle with incenter $I $ and $A-\text{excenter} $ $I_a $. Let $B'$ and $C'$ be the reflections of $B $ and $C $ in $I $ respectively. Let the circumcircle of $\triangle AB'C'$ intersect $BI $ at $B_1$ and $CI $ at $C_1$. Then, $B_1C_1$ is the $\perp $ bisector of $AI_a $. Another proof for your lemma: Denote by $M_{B}$ and $M_{C}$ the midpoints of arcs $ABC$ and $ACB$ respectively. It's well-known that $M_{B}M_{C}$ is the $\perp$ bisector of $AI_a$. Suppose that it intersects $BI$ and $CI$ at $B_2$ and $C_2$, respectively, we will prove that $A, B', B_2, C', C_2$ are concyclic and thus $B_2=B_1, C_2=C_2$. Suppose $AI_a$ intersects the arc $BC$ at $P$ and thus $P$ is the midpoint of the arc $BC$. It's easy to show that $B'C'C_2B_2$ is cyclic, and we proceed to prove that $AC'B_2C_2$ is cyclic. Let $G$ lie on the line $M_BM_C$ such that $A, C, G, C_2$ are concyclic. Recall that $M_BM_C$ is the $\perp$ bisector of $AI_a$. Angle chasing shows that $\triangle AIC\sim \triangle AM_CG$ and that $BI\parallel PM_C \parallel I_aG$. We know that $IP=PI_a$ and thus $M_C$ is the midpoint of $B_2 G$. So $\triangle AC'I \sim \triangle AB_2M_C$ and we get what we wanted.