Let $O$ be the center of the equilateral triangle $ABC$. Pick two points $P_1$ and $P_2$ other than $B$, $O$, $C$ on the circle $\odot(BOC)$ so that on this circle $B$, $P_1$, $P_2$, $O$, $C$ are placed in this order. Extensions of $BP_1$ and $CP_1$ intersects respectively with side $CA$ and $AB$ at points $R$ and $S$. Line $AP_1$ and $RS$ intersects at point $Q_1$. Analogously point $Q_2$ is defined. Let $\odot(OP_1Q_1)$ and $\odot(OP_2Q_2)$ meet again at point $U$ other than $O$. Prove that $2\,\angle Q_2UQ_1 + \angle Q_2OQ_1 = 360^\circ$. Remark. $\odot(XYZ)$ denotes the circumcircle of triangle $XYZ$.
Problem
Source: 2020 Taiwan TST
Tags: geometry proposed, geometry
TelvCohl
30.03.2020 18:10
Clearly, $ P_1 $ lies on $ \odot (ARS) $ and $ \triangle ABR \stackrel{+}{\cong} \triangle BCS, \triangle ACS \stackrel{+}{\cong} \triangle CBR, $ so $ AR = BS, AS = CR $ and therefore $ O $ is the midpoint of arc $ RS $ in $ \odot (ARS). $ Let $ T $ be the intersection of $ BC, RS, $ then $$ \frac{Q_1R}{Q_1S} = \frac{TR}{TS} = \frac{CR}{BS} = \frac{AS}{AR}\ , $$so note that $ AO $ is the bisector of $ \angle RAS $ we get $ \measuredangle (AO, RS) = \measuredangle (RS, OQ_1) $ and $ RS $ is tangent to $ \odot (OP_1Q_1), $ hence $$ 2\angle OUQ_1 = \angle (OQ_1, RS) + \angle (RS, AO) = 180^{\circ} - \angle AOQ_1. \qquad \blacksquare $$
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fcomoreira
02.04.2020 02:29
Where can I find the full Taiwan TST?
k12byda5h
03.10.2020 17:24
Obviously, $A,P_1,R,S$ lie on a circle $\omega_1$ with center $O_1$. $\triangle BSC \overset{+}{\cong} \triangle ARB$. Hence, $AR=BS$ and $AS=CR$. Let $\odot(ASO)$ intersect $AC$ again at $R'$. By Ptolemy theorem, $R=R'$. Hence, $O$ lies on $\omega_1$. Since $Q_1$ is the pole of $BC$ wrt. $\omega_1$, $O_1Q_1 \bot BC$ and $R(\omega_1)^2/O_1Q_1=dist(Q,BC)=AO_1$ (Since $O_1$ lies on perpendicular bisector of $AO$). Hence, $\triangle AO_1O \overset{+}{\cong} \triangle O_1Q_1O$. Let $\measuredangle OAP_1 = \alpha$. So, $\measuredangle AP_1O = 90^{\circ}-\alpha$ and $\measuredangle Q_1OA = 2\alpha$. Use angle chasing to complete the proof. [asy][asy] size(300); pair A,B,C,O,R,S,O_1,Q_1,P_0,Q; A=dir(90); B=dir(210); C=dir(330); O=circumcenter(A,B,C); P_0=0.7*B+0.3*C; path ooo = circumcircle(B,O,C); pair P = IP(circumcircle(B,O,C), A--P_0); dot("$P_1$",P,dir(270)); draw(unitcircle,blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(350)); dot("$O$", O, dir(320)); pair T = circumcenter(A,O,P); R = extension(B,P,A,C); S = extension(C,P,A,B); Q=foot(T,B,C); Q_1=extension(A,P,T,Q); dot("$R$",R,dir(30)); dot("$S$",S,dir(200)); draw(circumcircle(A,O,P),heavycyan); dot("$O_1$",T,dir(230)); draw(A--B--C--cycle,darkcyan); draw(C--S,heavymagenta); draw(B--R,heavymagenta); dot("$Q_1$",Q_1,dir(180)); draw(A--T--O,red); draw(T--Q_1,orange); draw(Q_1--O,orange); draw(A--P,deepblue); draw(A--O,fuchsia+dashed); draw(Q--Q_1,fuchsia+dashed); pair M_0=dir(270); draw(arc(M_0,length(C-M_0),20,160),mediumblue); [/asy][/asy]
Aryan-23
30.11.2020 02:40
Beautiful and hard! My solution is quite similar to Telv Cohl's so I'll hide it.
Drop the subscripts.
We start by proving that $A,R,O,S,P$ lie on a circle. Note that $\angle RPO= \angle (PB,OP)=\angle OCB=30^{\circ}=\angle OAR$. Hence $R\in \odot (AOP)$. Similarly $S\in \odot (AOP)$. Note that $AO$ bisects $\angle RAS$. Let $AO$ hit $RS$ at $L$.
Next we prove that $AS=CR$ and $AR=BS$ and $SC=RB$.
Note that if $\angle ABP=\theta$, then we have : $\angle PBC = 60-\theta \implies \angle PCB = 180-(60-\theta+120)=
\theta$. This gives us $ARB \stackrel {+}{\cong} BSC$, which gives $BS=AR$. Similarly $CR=AS$.
Now we claim that $RS$ is tangent to $\odot(POQ)$ at $Q$. Define $T=RS\cap BC$.
Since $(TQ;RS)=-1$ we have $\tfrac {QS}{QR}=\tfrac {TS}{TR}$
By Menelaus on $\triangle ASR $ with transversal $\overline {TBC}$ we get : $ \tfrac {AB}{BS}\cdot \tfrac {TR}{TS} \cdot \tfrac {RC}{AC}=1 \implies \tfrac {TS}{TR}= \tfrac {BS}{CR}= \tfrac {AR}{AS}$
Hence we have $\tfrac {AR}{AS}=\tfrac {QS}{QR}$. This means that if $Q$ reflected over the midpoint of $RS$ is $L'$, then $AL'$ bisects $\angle SAR$. Hence $Q$ is the reflection of $L$ over the perpendicular bisector of $RS$.
We have $\measuredangle {RQO}= \measuredangle {SLO}=\measuredangle ASO =\measuredangle {APO}$. This proves the claim.
Finally, for any point $U$ on arc $\widehat {PO}$ of $\odot (POQ)$, we have :
$$2\angle QUO = 2 \angle OLS = 180 - \angle APO $$
Doing the same for $P=P_2$ and summing, we are done. $\square$
guptaamitu1
09.08.2021 19:45
By simple angle chasing our problem is equivalent to proving $2 \angle AP_1O + \angle AOQ_1 = 2 \angle AP_2O + \angle AOQ_2$. We will in fact prove $2 \angle AP_10 + \angle AOQ_1 = 180^\circ$ (then that would analogously imply $2 \angle AP_2O + \angle AOQ_2 = 180^\circ$ and we would be done). Claim 1: points $A,S,P_1,O,R$ lie on some circle $\omega_1$ (say). proof: Note that $\angle SAO = 30^\circ = \angle OBC = \angle OP_1C = 180^\circ - \angle OP_1S$ so $S \in \odot(AOP_1)$. Similarly, $R \in \odot(AOP_1)$. This proves our claim. $\square$ [asy][asy] size(200); pair A=dir(90),B=dir(-150),C=dir(-30),O=(0,0),P1=waypoint(circumcircle(B,O,C),0.3),R=extension(B,P1,A,C),S=extension(C,P1,A,B),Q1=extension(R,S,A,P1),L=extension(A,O,R,S),T=2*foot(A,circumcenter(A,R,S),(0,0))-A; dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(-70)); dot("$P_1$",P1,dir(-90)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); dot("$Q_1$",Q1,dir(160)); dot("$L$",L,dir(70)); dot("$T$",T,dir(T)); draw(CP(circumcenter(B,O,C),B,15,165)^^unitcircle,red); draw(circumcircle(A,R,S),green); draw(A--B--C--A^^B--R^^C--S^^O--A--P1^^R--S^^O--T,magenta); [/asy][/asy] Claim 2: $AR = BS$. proof: $BP_1/CP_1 = (B,C;P_1,O) \stackrel{B}{=} (A,C;R,\text{midpoint}(\overline{AC})) = AR/CR$. Similarly, $CP_1/BP_1 = AS/BS$. So $AR/CR = BS/AS$ which implies our claim (as $AB = AC$). $\square$ Claim 3: (Key Claim) $L = \overline{AO} \cap \overline{RS}$ and $Q_1$ are isotomic conjugate wrt segment $RS$ (i.e. midpoint of segments $LQ_1,RS$ coincide). proof: Let $k = AR/CR = BS/CR$. As $\overline{AL}$ is the internal angle bisector of $\angle SAR$, so $RL/SL = k$. So it suffices to show $SQ_1 / RQ_1 = k$. Using Ceva's Theorem in $\triangle ARS$ wrt points $Q_1,B,C$ we obtain $$\frac{SQ_1}{RQ_1} = \frac{BS}{BA} \cdot \frac{CA}{CR} = \frac{BS}{CR} = k$$This proves our claim. $\square$[asy][asy] size(200); pair A=dir(90),B=dir(-150),C=dir(-30),O=(0,0),P1=waypoint(circumcircle(B,O,C),0.3),R=extension(B,P1,A,C),S=extension(C,P1,A,B),Q1=extension(R,S,A,P1),L=extension(A,O,R,S),T=2*foot(A,circumcenter(A,R,S),(0,0))-A; dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(-70)); dot("$P_1$",P1,dir(-90)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); dot("$Q_1$",Q1,dir(160)); dot("$L$",L,dir(70)); dot("$T$",T,dir(T)); draw(CP(circumcenter(B,O,C),B,15,165)^^unitcircle,red); draw(circumcircle(A,R,S),green); draw(A--B--C--A^^B--R^^C--S^^O--A--P1^^R--S^^O--T,magenta); [/asy][/asy] Now we are ready to finish the proof. As $\overline{AO}$ is the internal angle bisector of $\angle SAR$, so $O$ is the midpoint of arc $\widehat{RS}$ of $\omega_1$ not containing $A$. So if we let $T = \overline{OQ_1} \omega_1 \ne O$ then by Claim 3 we obtain $T$ is the reflection of $A$ perpendicular bisector of segment $\overline{BC}$. Hence, \begin{align*} 2 \angle AP_1O + \angle AOQ_1 & = 2(\angle AP_1R+ \angle RP_1O) + \angle AOT = 2 \angle AP_1R + 2 \angle RP_1O + (\angle AOS - \angle TOS) \\ &= 2 \angle ASR + 2 \angle RAO + (\angle ARS - \angle ASR) = \angle ASR + \angle SAR + \angle ARS \\ & = 180^\circ \end{align*}This completes the proof of the problem. $\blacksquare$