Reflection of D moves on a line

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Reflection of D moves on a line

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Source: KöMaL A. 705

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#1 h Nov 11, 2017, 3:45 PM • 2 Y

Y by Adventure10, Mango247

Triangle $ABC$ has orthocenter $H$ . Let $D$ be a point distinct from the vertices on the circumcircle of $ABC$ . Suppose that circle $BHD$ meets $AB$ at $P\ne B$ , and circle $CHD$ meets $AC$ at $Q\ne C$ . Prove that as $D$ moves on the circumcircle, the reflection of $D$ across line $PQ$ also moves on a fixed circle.

Michael Ren

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RC.

#3 h Nov 16, 2017, 6:41 AM • 3 Y

Y by AlastorMoody, Adventure10, Funcshun840

Here's the outline I will complete it later.

Denote by $O$ the circumcenter of $\Delta ABC.$

$P, H, Q$ are collinear.
Redefine the problem:- Triangle $ABC$ has orthocenter $H$ . Let $D$ be a point distinct from the vertices on the circumcircle of $ABC$ . Suppose that circle $BHD$ meets $AB$ at $P\ne B.$ Prove that as $D$ moves on the circumcircle, the reflection of $D$ across line $HP$ (i.e $D'$ ) also moves on a fixed circle.
Note that:- $\Delta DHD' \sim \Delta DOA$ $\therefore \Delta DAD' \sim \Delta DOH$
Rest is .... Lemon squeeze, although lemon is out of season in India.

This post has been edited 2 times. Last edited by RC., Nov 16, 2017, 8:34 AM

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#5 h Jul 5, 2021, 8:41 PM • 1 Y

Y by Funcshun840

Ankoganit wrote:

Triangle $ABC$ has orthocenter $H$ . Let $D$ be a point distinct from the vertices on the circumcircle of $ABC$ . Suppose that circle $BHD$ meets $AB$ at $P\ne B$ , and circle $CHD$ meets $AC$ at $Q\ne C$ . Prove that as $D$ moves on the circumcircle, the reflection of $D$ across line $PQ$ also moves on a fixed circle.

Michael Ren

I prove a more general result.

Now we prove that the following result.

$\textbf{Lemma}:$ $T$ is a random point inside triangle $\triangle ABC$ and the circumcircle of $\triangle BTC$ intersects line $AB$ at a second point $D$ . $C'$ is the reflection of $C$ wrt line $DT$ . now as $C$ traverses the circumcircle of $\triangle ABC$ then $C'$ is moving on a circle.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5.533356307592783cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ real xmin = -2.5967225715269686, xmax = 2.9366337360658137, ymin = -1.4500901983026113, ymax = 1.497749970650213; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pair A = (-0.38182256002963183,0.9242356477935798), B = (-0.8712073830836633,-0.49091516136753727), T = (-0.4884619389510197,0.0716836776184096), C = (0.8243241283679402,-0.5661181249442873), D = (-0.7910498768987462,-0.25912423857179945), G = (0.1367007089328688,0.9906123945203044), O = (0.,0.); filldraw((-1.2404832561853285,1.3225483131102684)--A--C--cycle, qqffff + opacity(0.10000000149011612), linewidth(0.) + qqffff); filldraw(C--O--T--cycle, qqffff + opacity(0.10000000149011612), linewidth(0.) + qqffff); /* draw figures */ draw(circle(O, 1.), linewidth(0.4)); draw(B--T, linewidth(0.4)); draw(A--B, linewidth(0.4)); draw(circle((-0.02893771651931373,-0.6524318709577974), 0.8576163703494034), linewidth(0.4)); draw(A--C, linewidth(0.4)); draw(B--C, linewidth(0.4)); draw(T--D, linewidth(0.4)); draw(C--T, linewidth(0.4)); draw(T--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(T--G, linewidth(0.4)); draw(D--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(A--G, linewidth(0.4)); draw(C--D, linewidth(0.4)); draw(C--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(A--O, linewidth(0.4)); draw(C--O, linewidth(0.4)); draw(A--(-1.2404832561853285,1.3225483131102684), linewidth(0.4)); draw(O--T, linewidth(0.4)); draw(C--G, linewidth(0.4)); /* dots and labels */ dot(A,linewidth(4.pt)); label("$A$", (-0.37001275416763546,1.0052707061473545), NE * labelscalefactor); dot(B,linewidth(4.pt)); label("$B$", (-1.0454128883429783,-0.6023223215512621), NE * labelscalefactor); dot(T,linewidth(4.pt)); label("$T$", (-0.4438846438430636,-0.13094931124138326), NE * labelscalefactor); dot(C,linewidth(4.pt)); label("$C$", (0.9033979154754588,-0.6374994118728948), NE * labelscalefactor); dot(D,linewidth(4.pt)); label("$D$", (-0.9258107812494281,-0.2611045454314244), NE * labelscalefactor); dot((-1.2404832561853285,1.3225483131102684),linewidth(4.pt)); label("$C'$", (-1.3690421193019968,1.2128155390449877), NE * labelscalefactor); dot(G,linewidth(4.pt)); label("$G$", (0.19282069097848353,1.0650717596941301), NE * labelscalefactor); dot(O,linewidth(4.pt)); label("$O$", (0.013417530338158087,0.02734759520596411), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

$\textbf{proof:}$ note that $\triangle COT \sim \triangle CAC'$ ,on the other hand we have:
$\frac{OT}{R}=\frac{AC'}{AC}$ now since the ratio between $AC'$ and $AC$ is constant then $C'$ acts like $C$ and moves on a circle. $\blacksquare$

back to the problem,
Consider that $H$ is a random point in the problem. redefine the problem resp.
note that $P,H,Q$ are collinear now guess what! we throw away the point $C$ and $Q$ . now apply the lemma on triangle $\triangle ADC$ and we are done.

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#6 h Oct 9, 2024, 4:31 AM

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Very funny problem, I really liked it ngl.
Let $Q_A$ the A-queue point, $AH \cap (ABC)=H_A$ , let $AA'$ be diameter in $(ABC)$ , $O$ center of $(ABC)$ . Now let $H'$ relfection of $A$ over $H$ and let $DH \cap (ABC)=D_2$ and $D'H \cap D_2A'=D_1$ .
Claim 1: $D$ is miquel point of $D'AOH$
Proof: From reims we have $PH \parallel AD_2 \parallel QH$ so $P,H,Q$ are colinear. Also note that triangles $DHD'$ and $DOA$ are isosceles but also $\angle DHD'=2\angle PHD=2\angle ACD=\angle AOD$ therefore we have the desired similar triangles which imply the claim.
The Finish: Trivially $Q_A, H, A'$ are colinear, now note that from invert at $H$ with radius $-\sqrt{AH \cdot HH_A}$ we have that $Q_A \to A', A \to H_A, D \to D_2$ and now since $\angle DD'H=\angle DAO=\angle DQ_AH$ meaning that $D'Q_AHD$ is cyclic so we have that $D' \to D_1$ . From easy case check we can guess that we want to prove that $(DQ_AA)$ is fixed, which inverting means we want $(H_AD_1A')$ to be fixed, now note that $D_2$ is the reflection of $D_1$ over $PQ$ because $D,D'$ are reflections over $PQ$ , but now by midbase (since $AD_2 \parallel PQ$ ) we have that $H'D_1 \parallel AD_2$ as well so by Reim's we have $H_AHD_1A'$ cyclic which is now fixed because we have 3 fixed points on the circle thus we are done :cool:

:cool:

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#7 h Apr 29, 2025, 5:20 PM

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Let $D'$ be the reflection of $D$ over $PQ$ . Let $A',A_1,R,U,V$ be the antipode of $A$ , reflection of $A$ over $OH$ , $A-$ queue point, reflection of $A$ over $AA'$ , reflection of $D$ over $OH$ . Let $DH\cap (ABC)=F$ . Since $H$ is the circumcenter of $(DD'V)$ ,
$\measuredangle DVD'=\measuredangle DHP=180-\measuredangle ABD=\measuredangle DVA$ Hence $A,V,D'$ are collinear. Also $\measuredangle A'FH+\measuredangle FHQ=\measuredangle A'AD+\measuredangle DCA=90$ and since the inversion at $H$ with radius $\sqrt{-HA.HR}$ swaps $D,R,D'$ with $F,A',D'^*$ where $HF=HD^*$ , and $HD,HD'$ are reflections over $HQ$ we see that $F,D'^*,A'$ are collinear which implies $H,D,D',R$ are concyclic. Thus, $(A,A';U,V)=-1=(RA,RH;RD',RD)=(A,A';RD'\cap (ABC),V)$ hence $D',R,U$ are collinear.
$\measuredangle AD'R=\measuredangle ARU-\measuredangle RAD'=\measuredangle DA'A-\measuredangle RA'V=\measuredangle DA'R-\measuredangle AA'V=\measuredangle DA'R-\measuredangle DA'A_1=\measuredangle A_1A'R$ Thus, $\measuredangle AD'R$ is constant while $D$ moves on the circumcircle. Since $A,R$ are fixed, we see that $D'$ lies on a fixed circle as desired. $\blacksquare$

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