parmenides51 wrote: Given an acute triangle $ABC$, $(c)$ its circumcircle with center $O$ and $H$ the orthocenter of the triangle $ABC$. The line $AO$ intersects $(c)$ at the point $D$. Let $D_1, D_2$ and $H_2, H_3$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $AB, AC$ respectively. Let $(c_1)$ be the circumcircle of the triangle $AD_1D_2$. Suppose that the line $AH$ intersects again $(c_1)$ at the point $U$, the line $H_2H_3$ intersects the segment $D_1D_2$ at the point $K_1$ and the line $DH_3$ intersects the segment $UD_2$ at the point $L_1$. Prove that one of the intersection points of the circumcircles of the triangles $D_1K_1H_2$ and $UDL_1$ lies on the line $K_1L_1$. I guess the last $UDL_1$ should be $UAL_1$

No, I believe it is correct

Attachments:

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -15.140045138345492, xmax = 29.40535315233497, ymin = -34.564260807240935, ymax = 24.382502382486873; /* image dimensions */ /* draw figures */ draw(circle((13.38,-2.82), 7.29068756236778)); draw((10.86931228617467,4.024747791972381)--(6.649058619743866,-5.6215269528633645)); draw((6.649058619743866,-5.6215269528633645)--(20.002523916274797,-5.868983783235425)); draw((20.002523916274797,-5.868983783235425)--(10.86931228617467,4.024747791972381)); draw(circle((10.532700329421585,-14.139778425151565), 18.167644874942166)); draw((-2.5925704743376006,-1.578306113754345)--(24.114360118724264,-2.073219774498469)); draw((10.196088372668504,-32.30430464227551)--(24.114360118724264,-2.073219774498469)); draw((2.172194100187394,-1.6666033094932933)--(16.70981185441136,3.6658675708254997)); draw((16.70981185441136,3.6658675708254997)--(15.162386384574592,-21.517291756199914)); draw(circle((0.025657050987431918,11.104416464592848), 12.95015704050292), red); draw(circle((-16.133292858711275,-13.645623387000333), 32.27046176014981), blue); draw((-10.81799818419979,18.184085137213344)--(15.162386384574592,-21.517291756199914), linetype("2 2")); draw((10.86931228617467,4.024747791972381)--(10.196088372668504,-32.30430464227551)); draw((20.002523916274797,-5.868983783235425)--(6.646119467274403,-0.025545115224891373)); draw((6.649058619743866,-5.6215269528633645)--(16.70981185441136,3.6658675708254997)); draw((-2.5925704743376006,-1.578306113754345)--(15.89068771382533,-9.664747791972381)); draw((15.89068771382533,-9.664747791972381)--(24.114360118724264,-2.073219774498469)); draw((10.86931228617467,4.024747791972381)--(15.89068771382533,-9.664747791972381)); draw((10.86931228617467,4.024747791972381)--(24.114360118724264,-2.073219774498469)); draw((10.86931228617467,4.024747791972381)--(-2.5925704743376006,-1.578306113754345)); draw((-2.5925704743376006,-1.578306113754345)--(10.196088372668504,-32.30430464227551)); draw((-10.81799818419979,18.184085137213344)--(10.86931228617467,4.024747791972381)); draw((10.86931228617467,4.024747791972381)--(16.70981185441136,3.6658675708254997)); draw((6.646119467274403,-0.025545115224891373)--(15.89068771382533,-9.664747791972381)); draw((10.86931228617467,4.024747791972381)--(6.646119467274403,-0.025545115224891373)); /* dots and labels */ dot((10.86931228617467,4.024747791972381),linewidth(3.pt) + dotstyle); label("$A$", (11.968406436331684,4.968897739163345), NE * labelscalefactor); dot((6.649058619743866,-5.6215269528633645),linewidth(3.pt) + dotstyle); label("$B$", (4.555939208880894,-4.2084426376805055), NE * labelscalefactor); dot((20.002523916274797,-5.868983783235425),linewidth(3.pt) + dotstyle); label("$C$", (19.52206351573392,-8.09116356634521), NE * labelscalefactor); dot((10.76089482219333,-1.8257629441264067),linewidth(3.pt) + dotstyle); label("$H$", (11.474241954501633,-4.279037563656227), NE * labelscalefactor); dot((15.89068771382533,-9.664747791972381),linewidth(3.pt) + dotstyle); label("$D$", (16.698266476705047,-11.197340309276976), NE * labelscalefactor); dot((6.646119467274403,-0.025545115224891373),linewidth(3.pt) + dotstyle); label("$H_2$", (4.273559504978007,0.1684427728142539), NE * labelscalefactor); dot((-2.5925704743376006,-1.578306113754345),linewidth(3.pt) + dotstyle); label("$D_1$", (-3.562477278327114,-0.25512678304007763), NE * labelscalefactor); dot((24.114360118724264,-2.073219774498469),linewidth(3.pt) + dotstyle); label("$D_2$", (24.393113408058724,-1.667025302554516), NE * labelscalefactor); dot((16.70981185441136,3.6658675708254997),linewidth(3.pt) + dotstyle); label("$H_3$", (17.26302588451082,4.333543405381847), NE * labelscalefactor); dot((10.196088372668504,-32.30430464227551),linewidth(3.pt) + dotstyle); label("$U$", (10.98007747267158,-33.928906473459435), NE * labelscalefactor); dot((2.172194100187394,-1.6666033094932933),linewidth(3.pt) + dotstyle); label("$K_1$", (2.085116799730631,-1.1022658947487407), NE * labelscalefactor); dot((15.162386384574592,-21.517291756199914),linewidth(3.pt) + dotstyle); label("$L_1$", (16.06291214292355,-22.139553835513873), NE * labelscalefactor); dot((-10.81799818419979,18.184085137213344),linewidth(3.pt) + dotstyle); label("$P$", (-12.245653173340896,19.08788293430773), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] It is well known that $H_2,H_3\in (ABC)$ We have $BC\parallel D_1D_2$ so $AH\bot D_1D_2$ and by Steiner's line we have $H\in \overline {D_1D_2}$ By angle chasing $\angle AD_1H=\angle ABH=\angle ACH=\angle AD_2H\implies H\quad\text{is the midpoint of}\ \overline{D_1D_2}$ and hence we have $AD_1=AD_2$ and $UD_1=UD_2$. $\textbf{Claim}: \boxed{ADL_1U, AH_2K_1D \quad\text{concyclic}}$ $\textit{Proof}$ Note that since $AD_1UD_2$ concyclic $$\angle ADH_3=\angle ABH=\angle AD_1H=\angle AUD_2=\angle AUL_1\implies ADL_1U\quad\text{concyclic}$$And $\angle AH_2H_3=\angle ABH=\angle AD_1H=\angle AD_1K_1\implies AH_2K_1D\quad\text{concyclic}$. $\quad\blacksquare$ Assume $(UDL_1)\cap(D_1K_1H_2)$ at $A$ and $P$ with $P\not \equiv A$ $$\implies \angle APK_1=\angle AD_1K_1=\angle AD_1H=\angle ABH_3=\angle ADH_3=\angle AUL_1=\angle APL_1$$$$\implies \overline {P,K_1,L_1}\implies \text{the proof is complete}$$

It's well known that $H$ is midpoint of $D_1D_2$. Claim1 : $AHBD_1$ and $AHCD_2$ are cyclic. Proof : $\angle ABD_1 = \angle ABD = \angle 90 = \angle AHD_1$ so $AHBD_1$ is cyclic and we'll prove other one with same approach. Claim2 : $D_1K_1H_2A$ and $UL_1DA$ are cyclic. Proof : $\angle AD_1K_1 = \angle AD_1H = \angle ABH = \angle ABH_3 = \angle AH_2H_3$ so $D_1K_1H_2A$ is cyclic and we'll prove other one with same approach. Let $S$ be other intersection point of the circumcircles of the triangles $D_1K_1H_2$ and $UDL_1$. Claim3 : $L_1,K_1,S$ are collinear. Proof : $\angle ASK_1 = \angle AD_1K_1 = \angle AUD_2 = \angle AUL_1 = \angle ASL_1$. we're Done.

Good problem but it's way too easy for a G8 level. Anyway, here is my solution. Let $(D_1K_1H_2)$ and $(UDL_1)$ intersect at point $T$. It suffices to prove that $T$ lies on the line $K_1L_1$. One can easily prove that $H$ lies on the Steiner's line of $\triangle{ABC}$. But there's another way to prove this. Notice that $D$ is the antipode of $A$ w.r.t $(ABC)$, so it must be $\angle{ABD}=\angle{ACD}=90^{\circ}$. Let $M$ be the midpoint of $BC$, so it is well-known that $D,M,H$ are collinear and $DM=MH$ (because that $HBDC$ is a parallelogram). Note that $D,B,D_1$ and $D,C,D_2$ are collinear since that $AB\perp BD$ and $AC\perp CD$. Now since that $DB=BD_1$ and $DC=CD_2$, there exists a homothety $h$ centered at $D$ and with scale 2 that maps $\triangle{DBC}\rightarrow\triangle{DD_1D_2}$. Since $DM=MH$ and $M$ lies on $BC$, then $H$ must be the midpoint of $D_1D_2$. Now I claim that $A$ lies on both $(D_1K_1H_2)$ and $(UDL_1)$. Let $\angle{AD_1K_1}=\alpha$ and $\angle{HD_1B}=\beta$. Observe that $\angle{HBC}=\angle{FBC}=90^{\circ}-\angle{FCB}=90^{\circ}-\angle{ACB}=90^{\circ}-\angle{ADB}=90^{\circ}-\angle{AD_1B}=90^{\circ}-(\angle{AD_1H}+\angle{HD_1B})=90^{\circ}-\alpha-\beta$. Also, observe that since $H_2C\parallel D_1D$ ($HBDC$ is a parallelogram), then $\angle{HCB}=\angle{H_2CB}=\angle{HD_1B}=\beta$. Moreover, $\angle{AH_2H_3}=\angle{ABH_3}=\angle{ABF}=90^{\circ}-\angle{BAF}=90^{\circ}-\angle{GAF}=90^{\circ}-(180^{\circ}-\angle{GHF})=90^{\circ}-(180^{\circ}-\angle{BHC})=90^{\circ}-(\angle{HBC}+\angle{HCB})=90^{\circ}-(90^{\circ}-\alpha-\beta+\beta)=\alpha$. Finally, observe that $\angle{AD_1K_1}+\angle{AH_2K_1}=\angle{AD_1K_1}+(180^{\circ}-\angle{AH_2K_1})=\alpha+(180^{\circ}-\alpha)=180^{\circ}$ $\Longrightarrow$ $AD_1K_1H_2$ is cyclic. Similarly, we can obtain that $AUL_1D$ is also cyclic. Since that $H$ is the midpoint of arc $D_1D_2$ at $(c_1)$ and $U=AH\cap(c_1)$, hence $AU$ is the diameter of that circle. Consider a circle with diameter $UD_2$ which is $(D_2HU)$. Since that $UD_2\perp AD_2$, thus $AD_2$ is tangent to $(D_2HU)$ at $D_2$. By Alternate Segment Theorem, then $\angle{ATL_1}=\angle{AUL_1}=\angle{HUD_2}=\angle{AD_2H}=\angle{AD_1H}$. For the final act, let's do the angle-chasing $\angle{ATK_1}=\angle{AD_1K_1}=\angle{AD_1H}=\angle{AUL_1}=\angle{ATL_1}$ $\Longrightarrow$ $T,K_1,L_1$ are collinear. We're done. $\blacksquare$

Attachments:

Wow uhh this was trivial… [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + blue; defaultpen(dps); /* default pen style */ pen dotstyle = blue; /* point style */ real xmin = -60.65525524978904, xmax = 57.729484665231695, ymin = -36.753288519162304, ymax = 28.35831843409934; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((8.960122710498915,-4.440292897186127)--(-3.65474201105606,9.308214696311097)--(-8.878057334132327,-4.602184043675263)--cycle, linewidth(0.3) + ccqqqq); /* draw figures */ draw(circle((0,0), 10), linewidth(0.3) + qqwuqq); draw((8.960122710498915,-4.440292897186127)--(-3.65474201105606,9.308214696311097), linewidth(0.3) + ccqqqq); draw((-3.65474201105606,9.308214696311097)--(-8.878057334132327,-4.602184043675263), linewidth(0.3) + ccqqqq); draw((-8.878057334132327,-4.602184043675263)--(8.960122710498915,-4.440292897186127), linewidth(0.3) + ccqqqq); draw((-21.410856679320716,0.10384660896057518)--(14.265503409941768,0.4276289019388452), linewidth(0.3) + ffvvqq); draw(circle((-3.4540272226256925,-12.80779727399918), 22.11692274927057), linewidth(0.3) + qqwuqq); draw(circle((-18.745638175409322,16.69119656928972), 16.800106201450717), linewidth(0.3) + qqwuqq); draw(circle((-33.310116556472835,-13.07875745714821), 37.15666510370267), linewidth(0.3) + qqwuqq); draw((-33.8365343397626,24.074178442268305)--(2.276924132194524,-23.764275537795932), linewidth(0.3) + ffvvqq); draw((-15.779805103784033,0.15495145223619022)--(5.347311807413569,8.450222271295315), linewidth(0.3) + ffvvqq); draw((-9.668199114698707,2.5545891799933695)--(-3.65474201105606,9.308214696311097), linewidth(0.3) + ffvvqq); draw((-21.410856679320716,0.10384660896057518)--(-3.65474201105606,9.308214696311097), linewidth(0.3) + ffvvqq); draw((-21.410856679320716,0.10384660896057518)--(3.65474201105606,-9.308214696311097), linewidth(0.3) + ffvvqq); draw((-33.8365343397626,24.074178442268305)--(3.65474201105606,-9.308214696311097), linewidth(0.3) + ffvvqq); draw((-33.8365343397626,24.074178442268305)--(-3.65474201105606,9.308214696311097), linewidth(0.3) + ffvvqq); draw((5.347311807413569,8.450222271295315)--(2.276924132194524,-23.764275537795932), linewidth(0.3) + ffvvqq); draw((-3.2533124341953132,-34.92380924430945)--(14.265503409941768,0.4276289019388452), linewidth(0.3) + ffvvqq); draw((-33.8365343397626,24.074178442268305)--(-3.2533124341953132,-34.92380924430945), linewidth(0.3) + ffvvqq); draw((-3.65474201105606,9.308214696311097)--(2.276924132194524,-23.764275537795932), linewidth(0.3) + ffvvqq); draw((-3.65474201105606,9.308214696311097)--(3.65474201105606,-9.308214696311097), linewidth(0.3) + ffvvqq); draw((-3.65474201105606,9.308214696311097)--(-3.2533124341953132,-34.92380924430945), linewidth(0.3) + ffvvqq); draw((3.65474201105606,-9.308214696311097)--(-3.2533124341953132,-34.92380924430945), linewidth(0.3) + ffvvqq); draw((5.347311807413569,8.450222271295315)--(14.265503409941768,0.4276289019388452), linewidth(0.3) + ffvvqq); draw((-3.65474201105606,9.308214696311097)--(5.347311807413569,8.450222271295315), linewidth(0.3) + ffvvqq); /* dots and labels */ dot((8.960122710498915,-4.440292897186127),dotstyle); label("$C$", (9.358219840578688,-3.5500684961212112), NE * labelscalefactor); dot((-3.65474201105606,9.308214696311097),dotstyle); label("$A$", (-3.312646853450874,10.230655134611721), NE * labelscalefactor); dot((-8.878057334132327,-4.602184043675263),dotstyle); label("$B$", (-8.491979224733031,-3.642556574179822), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.38687626889352345,0.8893592506920823), NE * labelscalefactor); dot((-3.57267663468947,0.26573775544971073),dotstyle); label("$H$", (-3.2201587753922642,1.166823484867913), NE * labelscalefactor); dot((3.65474201105606,-9.308214696311097),dotstyle); label("$D$", (3.993911313179311,-8.359448555168946), NE * labelscalefactor); dot((-21.410856679320716,0.10384660896057518),dotstyle); label("$D_1$", (-21.07035784070398,1.074335406809303), NE * labelscalefactor); dot((14.265503409941768,0.4276289019388452),dotstyle); label("$D_2$", (14.630040289919453,1.3517996409851338), NE * labelscalefactor); dot((-9.668199114698707,2.5545891799933695),dotstyle); label("$H_2$", (-9.32437192726052,3.47902543633317), NE * labelscalefactor); dot((5.347311807413569,8.450222271295315),dotstyle); label("$H_3$", (5.7511847962928995,9.398262432084229), NE * labelscalefactor); dot((-3.2533124341953132,-34.92380924430945),dotstyle); label("$U$", (-2.8502064631578246,-33.978646177403995), NE * labelscalefactor); dot((-15.779805103784033,0.15495145223619022),dotstyle); label("$K$", (-15.428585079128776,1.074335406809303), NE * labelscalefactor); dot((2.276924132194524,-23.764275537795932),dotstyle); label("$L$", (2.606590142300162,-22.88007681037076), NE * labelscalefactor); dot((-33.8365343397626,24.074178442268305),dotstyle); label("$P$", (-33.463760300557716,25.028747623989368), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Drop subscripts on $K$ and $L$ (I don’t see the point for retaining them), and let $(D_1KH_2)$ and $(UDL)$ intersect at points $P$ and $Q$; I claim one of these points is $A$ and the other one lies on $KL$. 0: $D_1-H-D_2$ Recall that $D_1D_2$ is the Steiner line of $D$ w.r.t. $\triangle ABC$. However notice that since $B$, $C$ are the midpoints of $AD_1$, $AD_2$ ($D$ point diametrically opposite $A$) then $D_1D_2$ is parallel to $BC$, and so since it is well known the line through the reflection of $H$ over $BC$ and the midpoint of $BC$ (which is just $D$) is parallel to $BC$, the reflection of $D$ over $BC$ and $H$ both lie on $D_1D_2$. 1: $A\in(D_1KH_2),(UDL)$ Relatively straightforward; $\measuredangle AH_2K=\measuredangle AH_2H_3=\measuredangle H_2H_3A=\measuredangle H_2DA=\measuredangle AD_1H=\measuredangle AD_1K$. Similarly with $A\in(UDL)$. 2: $P-K-L$ $\measuredangle APK=\measuredangle AD_1D_2=\measuredangle AUD_2=\measuredangle AUL=\measuredangle APL$, done.