Let there be an acute triangle $ABC$ with orthocenter $H$. Let $BH, CH$ hit the circumcircle of $\triangle ABC$ at $D, E$. Let $P$ be a point on $AB$, between $B$ and the foot of the perpendicular from $C$ to $AB$. Let $PH \cap AC = Q$. Now $\triangle AEP$'s circumcircle hits $CH$ at $S$, $\triangle ADQ$'s circumcircle hits $BH$ at $R$, and $\triangle AEP$'s circumcircle hits $\triangle ADQ$'s circumcircle at $J (\not=A)$. Prove that $RS$ is the perpendicular bisector of $HJ$.
Problem
Source: 2016 Korea Winter Camp 2nd Test #5
Tags: geometry, perpendicular bisector
25.01.2016 15:31
Hello. [asy][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.212314724489112, xmax = 14.025087753507485, ymin = -3.2629713717026645, ymax = 8.187137925058535; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((5.296151743635104,5.344495266665992)--(0.,0.)--(6.54,0.)--cycle, aqaqaq); /* draw figures */ draw((5.296151743635104,5.344495266665992)--(0.,0.), sqsqsq); draw((0.,0.)--(6.54,0.), sqsqsq); draw((6.54,0.)--(5.296151743635104,5.344495266665992), sqsqsq); draw(circle((3.27,2.0559491071791953), 3.8626191543188577)); draw((0.,0.)--(7.111769103613479,1.6551537905501157)); draw((1.184423276980099,5.3071329629013695)--(6.54,0.)); draw((1.9608817798530844,1.978780801269739)--(6.305697782460652,1.006736220596542)); draw(circle((3.25201928702729,4.034729908448934), 2.4277485029843913)); draw(circle((5.31569188201593,3.062685327775736), 2.281893602740122)); draw((5.296151743635104,5.344495266665992)--(3.5439829136539505,1.624601274025045)); draw((5.296151743635104,5.344495266665992)--(7.111769103613479,1.6551537905501157)); draw((6.305697782460652,1.006736220596542)--(7.111769103613479,1.6551537905501157)); draw((1.9608817798530844,1.978780801269739)--(1.184423276980099,5.3071329629013695)); draw((1.184423276980099,5.3071329629013695)--(5.296151743635104,5.344495266665992)); draw((3.5439829136539505,1.624601274025045)--(1.184423276980099,5.3071329629013695)); draw((4.325685981571209,1.006736220596541)--(3.5439829136539505,1.624601274025045)); draw((5.296151743635104,5.344495266665992)--(5.296151743635105,1.2325970523076015)); draw((5.296151743635104,5.344495266665992)--(4.543156794201497,1.978780801269738), linewidth(1.6) + linetype("2 2") + red); draw((4.543156794201497,1.978780801269738)--(4.325685981571209,1.006736220596541), linewidth(1.6) + linetype("2 2") + qqwuqq); /* dots and labels */ dot((5.296151743635104,5.344495266665992),linewidth(3.pt) + dotstyle); label("$A$", (5.313875032715717,5.467261464199978), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.2970589249995056,-0.16269260862612614), NE * labelscalefactor); dot((6.54,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.64528309047865,-0.08661214818253013), NE * labelscalefactor); dot((5.296151743635105,1.2325970523076015),linewidth(3.pt) + dotstyle); label("$H$", (5.389955493159313,1.3589166002457937), NE * labelscalefactor); dot((7.111769103613479,1.6551537905501157),linewidth(3.pt) + dotstyle); label("$D$", (7.234906658916521,1.4730372909111877), NE * labelscalefactor); dot((1.184423276980099,5.3071329629013695),linewidth(3.pt) + dotstyle); label("$E$", (0.9202284420980341,5.315100543312785), NE * labelscalefactor); dot((1.9608817798530844,1.978780801269739),linewidth(3.pt) + dotstyle); label("$P$", (1.871234197642987,1.6632384420201778), NE * labelscalefactor); dot((6.305697782460652,1.006736220596542),linewidth(3.pt) + dotstyle); label("$Q$", (6.398021594036963,0.7883131469188238), NE * labelscalefactor); dot((4.543156794201497,1.978780801269738),linewidth(3.pt) + dotstyle); label("$S$", (4.686211234056048,1.8534395931291678), NE * labelscalefactor); dot((4.325685981571209,1.006736220596541),linewidth(3.pt) + dotstyle); label("$R$", (4.172668126061773,0.6551723411425309), NE * labelscalefactor); dot((3.5439829136539505,1.624601274025045),linewidth(3.pt) + dotstyle); label("$J$", (3.3548031762931134,1.2638160246912988), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Step 1:We will show that $J\in PQ$. Suppose that the circumcircle of $\triangle{AEP}$ intersects $PQ$ at $J'$.It suffices to show that $ADQJ'$ is cyclic. We have $\angle{AJ'Q}=\angle{AEP}$ hence,it suffices to show that $\angle{ADQ}+\angle{AEP}=180^{\circ}$. We know that $D,E$ are the symmetric of $H$ with respect to $AC,AB$ respectively,whence we get $\angle{ADQ}=\angle{AHQ}$ and $\angle{AEP}=\angle{AHP}$ yielding $\angle{AEP}+\angle{ADQ}=\angle{AHQ}+\angle{AHP}=180^{\circ}$ and we are done. Step 2: We will show that $AS\perp PQ$ and $AR\perp PQ$. $H,E$ are symmetric with respect to $AB$,hence $\angle{HEP}=\angle{EHP}\Leftrightarrow \angle{SAP}=\angle{EHP}$. However,$HE\perp AB\Rightarrow \angle{EHP}=90^{\circ}-\angle{APH}\Rightarrow \angle{SAP}=90^{\circ}-\angle{APH}\Rightarrow AS\perp PQ$. We can show that $AR\perp PQ$ in a similar way. Step 3:From the previous steps we obtain that $A,S,R$ are collinear and $\overline{ASR}\perp HJ$. In order to end the proof,we shall show that $AJ=AH$. This follows from the fact that $\angle{HJA}=\angle{AEP}$ (from the cyclic quadrilateral $AEPJ$) and $\angle{AHJ}=\angle{AEP}$ (from the symmetry of $H,E$ with respect to $AB$).
10.04.2016 19:48
1) $H$ is the orthocenter of $ASP$ and $QRA$ by REIM theorem we get that $(PS)//(BC)$ therefore $(AH)\perp(PS)$ and since $(SH)\perp(AP)$ we get that $H$ is the orthocenter of $APS$ the same way for $QRA$. 2) $Q$,$J$ and $P$ collinear we have $\angle{QJA}=\angle{QRA}=180-\angle{QHA}=\angle{AHP}=\angle{AEP}=180-\angle{AJP}$ thus $Q$,$J$ and $P$ are collinear 3)$A$,$R$ and $S$ are collinear we have $(QP)\perp(AR)$ and $(PQ)\perp(AS)$ thus $A$,$R$ and $S$ are collinear 4)$H$,$E$ and $D$ belong to the circle with center $A$ clear that $AH=AD=AE$ 5)$D$,$J$,$H$ and $E$ concyclic we have $\angle{DHE}=\angle{DHA}+\angle{AHE}=90-\angle{QAH}+90-\angle{PAH}=180-\angle{DAQ}-\angle{PAE}=180-\angle{DJQ}-\angle{PJE}=\angle{DJE}$ conclusion $AJ=AH$ and $(AS)\perp(JH)$ therefore $RS$ is the perpendicular bisector of $JH$
12.04.2016 14:02
[asy][asy] //Asymptote is really cool! import graph; size(9cm); import olympiad; pair A=(3,8),B=(0,0),C=(10,0),R,S,J; pair O=circumcenter(A,B,C); path circ=circumcircle(A,B,C); pair G=(A+B+C)/3; pair H=3*G-2*O; pair E=reflect(A,B)*H; pair D=reflect(A,C)*H; pair foot=extension(A,B,C,H); pair P=(0.7)*(foot)+(0.3)*(B); pair Q=extension(P,H,A,C); path p=circumcircle(A,E,P); path q=circumcircle(A,D,Q); pair Afoot=extension(A,H,B,C); pair x[]=intersectionpoints(q,B--D); R=x[1]; pair y[]=intersectionpoints(p,C--E); S=y[1]; pair z[]=intersectionpoints(p,q); if (z[0]==A) { J=z[0]; } else { J=z[1]; } draw(A--Afoot,cyan);draw(A--B--C--cycle,heavymagenta);draw(circ,cyan);draw(B--D,blue);draw(C--E,blue);draw(p,orange);draw(q,orange);draw(P--Q,heavygreen);draw(A--R--S,brown+dashed); draw(A--P--S--R,blue);draw(A--J,magenta);draw(A--D--Q,magenta);draw(A--E--P,magenta); dot(A);dot(B);dot(C);dot(D);dot(E);dot(H);dot(P);dot(Q);dot(R);dot(S);dot(J); label("$A$",A,N);label("$B$",B,SW);label("$C$",C,SE);label("$H$",H,NW);label("$D$",D,NE);label("$E$",E,W);label("$R$",R,NNE);label("$S$",S,SW);label("$P$",P,SW);label("$Q$",Q,ESE);label("$J$",J,SSE); [/asy][/asy] We shall use the well-known and useful lemma: Lemma:The reflection of the orthocenter in any of the sides of a triangle lies on its circumcircle. Or its paraphrase: Lemma': If the line joining one of the vertices (say $A$) and the orthcenter $H$ of a triangle $ABC$ meets its circumcenter at $X$, then $X$ is the reflection of $H$ in line $BC$. Applied to the given configuration, this implies, for example, that $E$ is the reflection of $H$ in $AB$. We can also infer that the triangle $APE$ is essentially the mirror image of $APH$, $AB$ being the mirror. Let's see how this and several other consequences of our Lemma help us. But first make a few angle chases: We have \begin{align*} \angle AJQ+\angle AJP &=\pi-\angle ADQ+\pi=\angle AEP\\ &=2\pi-\angle AHQ-\angle AHP\quad\text{[By our ``Mirror" observation]}\\ &=2\pi-\pi=\pi\end{align*}Hence, $P,J,Q$ are collinear. $\square$ Also observe that \begin{align*} \angle RAQ+\angle PAS &=\angle RDQ+\angle SEP\\ &=\angle DHQ+\angle EHP\quad\text{[Again by ``Mirror"]}\\ &=\angle DHQ+\angle QHC\\ &=\angle DHC=\angle BAC\end{align*}This means nothing but that $A,R,S$ are collinear. $\square$ Now we also have \begin{align*} \angle QPA+\angle SAP &=\angle QPA+\angle SEP\\ & =\angle HPA+\angle EHP\quad\text{[``Mirror" once more!]}\\ &=\frac{\pi}{2}\end{align*}Therefore, $PQ\perp AS$. $\square$ Let's make another observation:$\angle APS=\angle AES=\angle AEC=\angle ABC$, so $PS||BC$. Also, since $AH\perp BC$, we must have $AH\perp PS$. $\square$ Now let's consider the triangle $ASP$ and apply the things we have got so far. We have $AH\perp PS$ and $PH\perp AS$, therefore $H$ is the orthcenter of $ASP$. But the line $PH$ meets the circumcircle of this triangle at $J$, so by our Lemma, $J$ is the reflection of $H$ in line $AS$, or the line $RS$. But this means that $RS$ is the perpendicular bisector of $HJ$ ! So we are done. $\blacksquare$