The incircle of a triangle $ABC$ touches $BC$ at point $D$. Let $M$ be the midpoint of arc $\widehat{BAC}$ of the circumcircle, and $P$, $Q$ be the projections of $M$ to the external bisectors of angles $B$ and $C$ respectively. Prove that the line $PQ$ bisects $AD$.
Problem
Source: Sharygin Finals 2023 10.5
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
MathLuis
03.08.2023 00:55
Sketch: Draw $M_B, M_C$ sre the midpoints of minor arcs $AC, AB$ in $(BAC)$ respecitivily, let $B', C'$ the B, C antipodes in $(ABC)$, since its known that $MM_BIM_C$ is a paralelogram (use I-E lemma to prove this) we have $M, M_C, P$ colinear and $M, M_B, Q$ colinear, let $AB' \cap MM_C=K$ and $AC' \cap MM_B=L$, now by single angle chase u can prove that $\angle AKM_C=90-\angle ABI=\angle AIM_C$ so $AKIM_C$ is cyclic so by Reim's we have $KI \parallel B'C \perp BC$ so $K,I,D$ are colinear, in a similary way u prove $L, I, D$ colinear so from both results we get that $AKDBP$ and $ALQCD$ are cyclic with diameters $BK, CL$ respecitvily, and now its easy to prove $AP=PD$ using arcs and angle chase (simlar for $AQ=QD$) all the angles can be computed, hence we are done .
SQTHUSH
03.08.2023 08:34
Let $PB\cap QC=J$,$CI\cap \odot(ABC)=K,PQ\cap CK=T$ Consider that $\angle MPJ=\angle MAJ=\angle MQJ=90^{\circ}$ Hence $A,P,J,Q,M$ are cyclic Note that $\angle CBJ=90^{\circ}-\dfrac{1}{2}\angle ABC,\angle BCJ=90^{\circ}-\dfrac{1}{2}\angle ACB$ Hence $J$ is the circumcenter of $A-$excircle Suppose $JM\cap \odot(ABC)=E,AJ\cap \odot(ABC)=R$ $\mathrm{Lemma.} \angle DAR=\angle RAE$ $\mathrm{Proof.}$ Let $\angle RD\cap \odot(ABC) = X,AR\cap BC=Y$ Note that $RD\cdot RX=RY\cdot RA=RI^{2}$ So $A,X,D,Y$ are cyclic Hence $\angle AXI=\angle AYC-\angle DIY=90^{\circ}$ Note that $\angle XEJ=90^{\circ}+\angle XAR=\angle XIJ$ Hence $X,I,E,J$ are cyclic Note that $I,B,J,C$ are cyclic By root axis theorem we can know that $XE,BC,IR$ are concurrent Hence $\angle RAE=\angle DXY=\angle DAR$ Hence $\angle PAD+\angle APQ=\angle PMR+\angle IBC=\angle KAR+\angle IBC=90^{\circ}$ So $AD\bot PQ$ Consider that $\angle DAQ=90^{\circ}-\angle AQP=90^{\circ}-\angle BJI=90^{\circ}-\angle ICB=\angle BCJ=180^{\circ}-\angle BCQ$ Hence $A,D,C,Q$ are cyclic Consider that $\angle ACT=\angle ICB=\angle BJI=\angle AQT$ Hence $A,T,D,C,Q$ are cyclic Which means that $\angle TQD=\angle TCD=\angle ACT=\angle AQT \Rightarrow PQ$ bisects $AD$
VicKmath7
03.08.2023 19:44
Add the excenters $I_b, I_c$ and notice that $P, Q$ are midpoints of $BI_c, CI_b$. We shall show that $PQ$ is the perpendicular bisector of $AD$, so we can rephrase the problem to the following orthocenter configuration:
Let $ABC$ be a triangle with orthic triangle $DEF$ and midpoints $M, N$ of $BF, CE$. If $R$ is the foot of the perpendicular from $H$ to $EF$, then show that $MN$ is the perpendicular bisector of $DR$.
Let $EF \cap BC=S$, $Q$ be the midpoint of $SH$, $AH \cap (ABC)=A'$ and $T=(AH) \cap (ABC)$. We have that $Q, M, N$ lie on the Newton-Gauss line for the quadrilateral $BEFC$ and $Q$ is the center of the circle $(SDHR)$, so $QR=QD$. We are left to show that $MN \perp RD$, which will finish the problem.
It is easy to see that the spiral similarity at $T$ sending $EF$ to $BC$ sends $H$ to $A'$ and $R$ to $D$, so $\angle (RD, AB)=\angle BTD$. Moreover, the spiral similarity at $T$ sending $BF$ to $CE$ sends $M$ to $N$, so $\angle (MN, BC)=\angle BTM$, so $\angle (MN, AB)=\angle AMN=\angle ABC-\angle BTM$, so we should show that $\angle ABC-\angle BTM=90^{o}-\angle BTD$, which rewrites as $\angle MTD=\angle BTD-\angle BTM=90^{o}-\angle ABC=\angle MAD$. But this is trivial as if $L$ is the midpoint of $BC$, then $ATMDN$ is cyclic with diameter $AL$, done.
melowmolly
06.08.2023 12:38
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Let $J$ be the midpoint of arc ${BC}, MP\cap CI \equiv E, MQ \cap BI \equiv F, DI \cap MP \equiv G, DI \equiv MQ \equiv H$
We have $E, F \in (ABC)$ by easy angle chasing
Since $\angle EGI=\angle EMJ=\angle EAJ\Rightarrow A,G,I,E$ are concyclic
$\Rightarrow \angle BAG=\angle BAJ+\angle GAJ=\angle BCJ+\angle MEC=\angle MBC+\angle BCJ=90\Rightarrow A,G,D,B,P$ are concyclic, similarly $A,H,Q,C,D$ are concyclic
So $\angle PAD=\angle PDA\Rightarrow PA=PD, \angle QAD=\angle QDA\Rightarrow QA=QD\Rightarrow PQ$ is the bisector of $AD$.
trinhquockhanh
06.08.2023 16:15
$\text{solved by using a property of touchpoint between the mixtilinear excircle and the circumcircle}$
paccongnong19hy
06.08.2023 18:51
Let $I$ be the incenter and $I_A, I_B, I_C$ be the $A$-excenter, $B$-excenter, $C$-excenter of $\triangle ABC$.
By angle chasing, we get $\triangle IBC \cup \{D\} \sim \triangle II_CI_B \cup \{A\}$, so $\dfrac{\overline{DC}}{\overline{DB}}=\dfrac{\overline{AI_B}}{\overline{AI_C}}$ .
It's well-known that $M$ is the midpoint of $I_BI_C$, from which $P,Q$ are the midpoints of $I_CB, I_BC$.
Let $\{R\}=PQ\cap AD$, by ERIQ we have $\dfrac{\overline{RD}}{\overline{RA}}=\dfrac{\overline{QC}}{\overline{QI_B}}=-1$, this means that $R$ is the midpoint of $AD$.
P2nisic
14.04.2024 00:21
kamatadu wrote: The incircle of a triangle $ABC$ touches $BC$ at point $D$. Let $M$ be the midpoint of arc $\widehat{BAC}$ of the circumcircle, and $P$, $Q$ be the projections of $M$ to the external bisectors of angles $B$ and $C$ respectively. Prove that the line $PQ$ bisects $AD$. Interesting! All the problem is to prove that $\angle AI_AM=90-\frac{\angle A}{2}+\angle C\Leftrightarrow \angle AMI_A=\frac{\angle A}{2}+\angle C$ which is not so well know as $\angle AMI=\angle ALI$ where $L$ is the midpoint of $BC$ with the same method we can prove that $\angle AMI_A=\angle I_ALC= \angle ADB=frac{\angle A}{2}+\angle C$ (We use that $AD//I_AL$). So leting $E=(AMQI_A)\cap AD$ we get that $E,L,M$ are collinear and $LMQC$ lie on the same circle by the perpediculars. Now $\angle AQC=\angle AQI_A=90-\angle AI_AM=\angle ADB$ so $A,Q,C,I_A$ lie on the same circle and $CQ$ is external bisectors of angles $B$ so we will get that $QD=QA$ similary we have $PD=PA$ which gives the result.
Attachments:
axolotlx7
08.06.2024 19:05
Let $I$ be the incenter and $\Delta I_AI_BI_C$ be the excentral triangle. Let $BC \cap I_BI_C = Z$. Note that $A,M,P,Q,I_A$ concyclic and $I,A,Z,D$ concyclic, and it is well-known that $I$ is the orthocenter of $\Delta I_AMZ$, so \[ \measuredangle BPA = \measuredangle I_APA = \measuredangle I_AMA = \measuredangle ZIA = \measuredangle ZDA = \measuredangle BDA \]implies $A,D,B,P$ concyclic. As $BP$ is the external angle bisector of $\angle ABD$, we have $PA = PD$. Similarly $QA = QD$ so $PQ$ is the perpendicular bisector of $AD$ and the conclusion follows.