$\triangle ABC$ is an acute triangle and its orthocenter is $H$. The circumcircle of $\triangle ABH$ intersects line $BC$ at $D$. Lines $DH$ and $AC$ meets at $P$, and the circumcenter of $\triangle ADP$ is $Q$. Prove that the circumcenter of $\triangle ABH$ lies on the circumcircle of $\triangle BDQ$.
Problem
Source: 2015 Final Korean Mathematical Olympiad Day 2 Problem 4
Tags: geometry, circumcircle
22.03.2015 07:45
My solution: Let $ T $ be the center of $ \mathcal{W} $ . From $ \angle HDC=\angle HAB=\angle HCD \Longrightarrow D $ is the reflection of $ C $ in $ AH $ . Since $ \angle APD=\angle ACB+\angle CDP=\angle ACB+90^{\circ}-\angle CBA=90^{\circ}-\angle DAB $ , so $ AB $ pass through the center $ Q $ of $ \odot (ADP) \Longrightarrow AB $ is the diameter of $ \odot (ADP) $ , hence from $ \angle DTB=2\angle DAB=\angle DQB $ we get $ T $ lie on $ \odot (BDQ) $ . Q.E.D
22.03.2015 10:50
Dear Mathlinkers, the main idea of this problem is based on the Morley's circle... See: http://jl.ayme.pagesperso-orange.fr/Docs/Les%20cercles%20de%20Morley,%20Euler,%20Mannheim%20et%20Miquel.pdf Sincerely Jean-Louis
23.03.2015 11:26
my solution = let $O$ be circumcentre of triangle $ABD$ than clearly $AD$ is radical axis of of circle of $ABD,ADP$ and hence $AD$ is perepndicular to $OQ$ and thus $\angle OQD=\angle APD$ now angle chasing gives $\angle DAP=180-2C$ and $\angle HBA=\angle ADH=90-A$ and hence $\angle APD=\angle OQD = C-B+90$ now $\angle BOD=2\angle BAD=2(A+C-180)$ and hence $\angle OBD=90-\angle BOD=90+B-C$ and thus $\angle OBD+\angle OQD=(90+B-C) +(90+C-B) = 180$ and hence $BDQO$ is cyclic quad. we are done
10.09.2015 23:16
Let $O'$ be the circumcenter of $\triangle ADP$. Note that $O'Q$ is the perpendicular bisector of $AD$ and let it meet $BC$ at $T$. Now by power of a point, it shall suffice to showing $TD.TB=TO'.TQ$ Now just trig bash!!!
17.09.2015 10:58
Dear Mathlinkers, I use http://www.artofproblemsolving.com/community/c6h1142448_a_circumcenter_on_bc as a lemma and conclude with the Morley cicle mantioned above. Sincerely Jean-Louis
13.03.2017 00:09
Let $BH\cap CA=E$. $\angle DAB=\angle DHB$(A,H,B,D concyclic) $=\angle PHE=90^{\circ}-\angle APD.$ Also, $Q$ being circumcentre of $\triangle ADP$, we have $\angle DAQ=90^{\circ}-\angle APD=\angle DAB.$ So $Q$ lies on $AB$, giving $\angle BQD=2\angle DAB=\angle DOB$, where $O$ is the circumcentre of $\triangle ADB$. Hence proved.
20.03.2018 17:57
Three years ago, I cartesian bashed this problem. My re-solve is the same as above
13.04.2019 13:00
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 16... Sincerely Jean-Louis
16.04.2019 22:10
WLOG, Assume $D$ lies outside $\Delta ABC$ (Though this also works for $D$ inside $\Delta ABC$). Let the perpendicular to $AC$ at $P$ intersect $AB$ at $A'$, then, $\angle ADP$ $=$ $\angle ABH$ $=$ $\angle AA'P$ $\implies$ $D$ $\in$ $\odot (AA'P)$. Hence, $AA'$ is the diameter of $\odot (ADP)$ $\implies$ $Q$ $\in$ $AB$. Let $O'$ be the center of $\odot (ABH)$ $\implies$ $\angle DO'B$ $=$ $\angle DAB$ $=$ $\angle DQB$ $\implies$ $O'$ $\in$ $\odot (BDQ)$
19.05.2020 08:59
Claim: $\angle DHF=\angle B$. Proof: $\angle DHF=180-\angle DHA=180-\angle DBA=\angle B$. $\square$ Claim: $Q\in AB$. Proof: Now, \begin{align*} \angle DAB&=\angle DHB=\angle DHF-\angle BHF=\angle B-(90-\angle HBC)=\angle B-\angle C \\ \angle DAQ&= \tfrac12(180-\angle DQA) = 90-\angle DPA = 90-(180-\angle AHP-\angle PAH) \\ &= 90-(180-\angle DHF - (90-\angle C)) \\ &= 90-(180-\angle B-(90-\angle C)) \\ &= \angle B-\angle C. \end{align*}Hence $\angle DAB=\angle DAQ$, so $Q\in AB$. $\square$ We want to show $O'BDQ$ is cyclic, where $O'$ is the circumcenter of $(ABH)$. We have \begin{align*} \angle DO'B&=2\angle DAB \\ \angle DQB&=180-\angle DQA=2\angle DAQ=2\angle DAB. \end{align*}The second equality follows since $QA=QD$ and $Q\in AB$. We are done.
07.10.2020 09:50
[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -24.96660579925842, xmax = 15.48599776224397, ymin = -15.768276060431432, ymax = 18.594688255253395; /* image dimensions */ /* draw figures */draw(circle((-15.103283787706305,2.7400144498824512), 8.255243083650193), linewidth(0.8)); draw(circle((-15.130415285106306,-1.818077113317177), 4.55817231098856), linewidth(0.8)); draw((-8.375963639670925,7.524594054820753)--(-11.645519949032499,-4.756178424244665), linewidth(0.8)); draw((-18.650039841722155,-4.714484853454846)--(1.7516863917661438,-4.835923700082753), linewidth(0.8)); draw((1.7516863917661438,-4.835923700082753)--(-8.375963639670925,7.524594054820753), linewidth(0.8)); draw((-18.650039841722155,-4.714484853454846)--(-1.83976272949642,-0.45265902452608153), linewidth(0.8)); /* dots and labels */dot((-8.375963639670925,7.524594054820753),dotstyle); label("$A$", (-7.712627936108655,7.937819575072659), NE * labelscalefactor); dot((-11.645519949032499,-4.756178424244665),dotstyle); label("$B$", (-11.156173938207873,-5.908860138631566), NE * labelscalefactor); dot((1.7516863917661438,-4.835923700082753),dotstyle); label("$C$", (2.7267536281500266,-4.712681001060258), NE * labelscalefactor); dot((-18.650039841722155,-4.714484853454846),dotstyle); label("$D$", (-19.49318004855335,-5.655125170055833), NE * labelscalefactor); dot((-8.433397594943035,-2.1243104308939427),linewidth(4pt) + dotstyle); label("$H$", (-8.11135431529909,-3.1177754842985146), NE * labelscalefactor); label("$c$", (-19.239445079977617,9.061503007336613), NE * labelscalefactor); dot((-1.83976272949642,-0.45265902452608153),linewidth(4pt) + dotstyle); label("$P$", (-1.188014458446979,0.18077910718599924), NE * labelscalefactor); dot((-10.645948139979769,-1.0016891902417233),linewidth(4pt) + dotstyle); label("$Q$", (-9.706259832060834,-0.9066564724242802), NE * labelscalefactor); dot((-15.103283787706305,2.7400144498824512),linewidth(4pt) + dotstyle); label("$O$", (-15.614659814610018,3.3343422880558093), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM. $Q,A,B$ are collinear. Proof. Angle Chasing. $\angle QAD=90^{\circ}-\angle APD=90^{\circ}-(\angle AHD)+\angle HAC=B-C=\angle BAD$ $\blacksquare$ Now $\angle QOD=\angle ABC=\angle QBC$ as desired.
10.06.2022 00:28
Let $F=AB\cap (PAD)$. Notice $\angle DHF=\angle DAF=\angle DAB=\angle DHB$, which means $BH\parallel FP$. Since $BH\perp AC$ therefore $FP\perp AC$. Meaning $Q$ lies on $AF\equiv AB$. Let $E$ be the circumcenter of $(ABH)$. Now $\angle BED=2\angle BAD=\angle BQD$.
10.01.2024 01:56
We ignore configuration issues. I think this is for $\angle C>\angle B$ and everything acute. $\angle CDP=\angle HAB=90^\circ-\angle B$, so $\angle CPD=90^\circ+\angle B-\angle C$. Thus $\angle AQD=180^\circ+2\angle B-2\angle C$, and $\angle DAQ=\angle C-\angle B$. On the other hand, $\angle DAB=\angle DHB=180^\circ-(90^\circ-\angle C)-\angle BDH=90^\circ+\angle C-180^\circ+(90^\circ-\angle B)=\angle C-\angle B$, so $Q$ lies on $\overline{AB}$. Thus $\angle BQD=2\angle C-2\angle B$. It thus suffices to show that $\angle BAD=\angle C-\angle B$, or $\angle BDA=180^\circ-\angle C$, but this follows since $\angle BDA=\angle BHA$. $\blacksquare$