Prove that the circumcentre appears on a circumcircle

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Prove that the circumcentre appears on a circumcircle

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Source: Lusophon Mathematical Olympiad 2024 Day 1 Problem 3

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#1 h Jul 26, 2024, 3:58 PM

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Let $ABC$ be a triangle with incentre $I$ . A line $r$ that passes through $I$ intersects the circumcircles of triangles $AIB$ and $AIC$ at points $P$ and $Q$ , respectively. Prove that the circumcentre of triangle $APQ$ is on the circumcircle of $ABC$ .

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#3 h Jul 26, 2024, 4:44 PM

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Oops, this was for p6.

This post has been edited 1 time. Last edited by weaving2, Jul 26, 2024, 4:45 PM
Reason: asd

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#4 h Jul 26, 2024, 4:57 PM • 1 Y

Y by VicKmath7

Let $\angle CIQ = \varphi$ and $\angle BIP=\theta$ . Clearly, we have that $\theta+ \varphi = 90^{\circ}-\frac{\alpha}{2}.$ Let $PB\cap(ABC)=T.$

Claim 1: $T\in CQ.$

Proof: It suffices to show that $\angle TCI = \angle IAQ.$ Now angle chasing $\angle TBI = \angle PAI = \theta+\frac{\alpha}{2}$ and since $\angle BTC = 180-\alpha,$ we get that $\angle TCI = 360^{\circ}-\angle BTC - \angle TBI - \angle BIC = 90^{\circ}-\theta.$ However, $\angle IAQ = \frac{\alpha}{2}+\varphi = 90^{\circ}-\theta,$ so we are done.

Claim 2: $T$ is the circumcenter of $\triangle APQ.$

Proof: Firstly, note that $\angle CQI = \angle CAI = \angle BAI = \angle BPI$ . Hence, $\angle TPQ = \angle TQP$ , i.e. $TP=TQ$ . Moreover, we have that $\angle ATC = \angle ABC = \beta$ and since $\angle CQA = \frac{\alpha+\gamma}{2}=90-\frac{\beta}{2},$ we get that $\triangle TAQ$ is also isosceles. Thus, $TA=TQ=TP,$ finishing the problem.

This post has been edited 2 times. Last edited by rstenetbg, Jul 26, 2024, 4:58 PM

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#5 h Jul 26, 2024, 5:36 PM • 1 Y

Y by iMqther_

Let $D=BP\cap CQ$ . Then $\measuredangle DPQ=\measuredangle BPI=\measuredangle BAI=\measuredangle IAC=\measuredangle IQC=\measuredangle PQD,$ so $DP=DQ$ and $\measuredangle PDQ=\measuredangle PAQ$ , which means that $D\in (ABC)$ . Then $\measuredangle DPA=\measuredangle BPA=\measuredangle BIA=90^\circ -\measuredangle ICB$ and $\measuredangle ADP=\measuredangle ADB=\measuredangle ACB=2\measuredangle ICB,$ so $\measuredangle PAD=90^\circ -\measuredangle ICB=\measuredangle DPA$ . Hence $DA=DP=DQ$ , so $D$ is the circumcentre of $APQ$ , and it lies on the circumcircle of $ABC$ . Done.

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#6 h Jul 28, 2024, 1:39 PM • 1 Y

Y by dinisteixeiraa

Cool problem.

Let $M=CI\cap(ABC)$ and $N=BI\cap(ABC)$ . It is well known that $A,B,C,M,N$ are concyclic. Also $M$ and $N$ are the circumcenters of $\triangle AIB$ and $\triangle AIC$ respectively.
Now one needs to suspect that the circumcentre of $\triangle APQ$ can be found by the intersections of the lines $CQ$ and $PB$ and the rest is angle chasing.

Let $S=PB\cap(ABC), \angle ABC=2b, \angle ACB=2c, \angle BAC=2a$ . My method to show that $S,C,Q$ are collinear is the same as @rstenetbg(The first comment).
Now we have $\angle BAI=\angle BPI=\angle CAI= \angle CQI=a$ which means $\angle SPQ=\angle SQP=a$ $\implies SQ=SP$ .
Let $\angle CIQ=f$ then $\angle CIQ=\angle MIP=\angle MPI=f$ but since $\angle IAP=\angle IBA=b$ we will have $\angle MAP=\angle MPA=b-f$ and also $\angle MAB=\angle MBA=\angle MCB=c$ .
Now lets denote $\angle BAS=m$ , since $ABSC$ is cyclic then $\angle SCB=m$ .
$\triangle SAP$ is isosceles $\iff$ $\angle SAP=\angle SPA \iff a+b=b-f+c+m \iff f+a=c+m$ which is true because $\angle MCS=\angle QIC+ \angle CIQ$ which is the same as $\angle SCB+ \angle MCB= a+f$ so $f+a=m+c$ therefore $SA=SP$ $\implies SA=SQ=SP$ $Q.E.D \square$

Attachments:

This post has been edited 3 times. Last edited by iMqther_, Aug 9, 2024, 8:22 PM
Reason: typo

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#7 h Jul 29, 2024, 7:43 PM • 1 Y

Y by iMqther_

Solution during contest:
Let $M$ be the midpoint of the arc $\overarc{AB}$ not containing $C$ and $O$ be the center of $(APQ)$ . By the Incenter-Excenter Lemma we conclude that $M$ is the center of $(AIBP)$ . Hence $M$ belongs to the perpendicular
bissector of $AP$ . $O$ also does because it is center of $(APQ)$ .
Thus $OM$ is the perpendicular bissector of $AP\rightarrow\angle{AOM}=\angle{POM}=\frac{\angle{AOP}}{2}$ .
Because $O$ is center, we also have $\angle{AQP}=\frac{\angle{AOP}}{2}\rightarrow \angle{AOM}=\angle{AQP}$ . But $\angle{AQP}=\angle{AQI}=\angle{ACI}$ , by the inscribed angle.
Finally, $\angle{ACI}=\angle{ACM}$ , because $C, I, M$ are collinear. Hence $\angle{AOM}=\angle{ACM}\rightarrow O\in(ACM)=(ABC)$ , as we wanted.

This post has been edited 3 times. Last edited by Thelink_20, Aug 12, 2024, 10:05 PM

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