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Source: Centroamerican Olympiad 2016, Problem 6

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241 posts

#1 h Jun 20, 2016, 7:09 PM • 1 Y

Y by Adventure10

Let $\triangle ABC$ be triangle with incenter $I$ and circumcircle $\Gamma$ . Let $M=BI\cap \Gamma$ and $N=CI\cap \Gamma$ , the line parallel to $MN$ through $I$ cuts $AB$ , $AC$ in $P$ and $Q$ . Prove that the circumradius of $\odot (BNP)$ and $\odot (CMQ)$ are equal.

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#2 h Jun 20, 2016, 8:03 PM • 2 Y

Y by Adventure10, Mango247

Let $R_1$ be the radius of $\triangle MQC$ . Let $R_2$ be the radius of $\triangle NPB$ .

Using the fact that $NB=NI=NA$ , $MC=MI=MA$ , and since $NM \parallel PQ$ , we get:

$\angle PIB= \angle MIQ = \angle IMN= \angle NCA= \angle NAB= \angle NBA$

$\angle QIC= \angle NIP= \angle INM= \angle MBA= \angle MAC= \angle MCA$

Therefore, by trig ceva in $\triangle NBI$ and $\triangle MIC$ , we get:

$\frac{ \sin \angle BNP}{ \sin \angle PNI} = \frac{ \sin \angle IMQ}{ \sin \angle QMC}$

By Two Equal Angles Lemma we get $\angle BNP= \angle IMQ$

We now easily see that $\angle ANP+ \angle AMQ= 180$ . And $AP= AQ$ .

By law of sines in $\triangle ANP$ and $\triangle AMQ$ , we get:

$\frac{NP}{ \sin \angle NAP} = \frac{AP}{ \sin \angle ANP} = \frac{AQ}{ \sin \angle AMQ} = \frac{MQ}{ \sin \angle MAQ}$ .

By law of sines in $\triangle MQC$ and $\triangle NPB$ we get:

$2R_1= \frac{MQ}{ \sin \angle MCQ} = \frac{NP}{ \sin \angle NBP} = 2R_2$

$\Longrightarrow R_1=R_2$

Done!

This post has been edited 1 time. Last edited by Math_CYCR, Sep 25, 2016, 2:30 PM

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#3 h Jun 22, 2016, 12:51 AM • 2 Y

Y by Adventure10, Mango247

Let $T = AI\cap\odot(ABC)$ . It's well known that $AT\perp MN$ , so $AI\perp PQ$ .

Recall that by Fact 5 $BN = NI$ and $CM = MI$ ; combined with the fact that $\angle BNC = \angle BMC$ we get that $\triangle BNI\sim\triangle ICM$ . Furthermore, remark that $\angle API = 90^\circ-\tfrac12A = 180^\circ - \left(90^\circ+\dfrac12\angle A\right) = 180^\circ - \angle BIC = \angle NIB.$ Thus $\angle PBI = \angle NIP$ , so $\angle NBP=\angle PIB$ . Similar reasoning works for the other triangle $\triangle CMI$ , so in fact $P$ and $Q$ are corresponding points in $\triangle BNI$ and $\triangle IMC$ respectively.

As such, $\dfrac{R_{\odot(MQC)}}{R_{\odot(BPN)}} = \left(\dfrac{R_{\odot(MQC)}}{R_{\odot(MQI)}}\right)\left(\dfrac{CM}{CN}\right) = \left(\dfrac{\sin\angle MCA}{\sin\angle ABN}\right)\left(\dfrac{MA}{AN}\right) = 1,$ where the last equality is from Extended Law of Sines. Done. $\blacksquare$

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#4 h Jun 22, 2016, 2:23 AM • 2 Y

Y by edfearay123, Adventure10

1) $\angle NBP=\angle NAP$ And since $\triangle BNP$ shares $NP$ with $\triangle NPA$ , they have both the same circumradius. Analogously $\triangle CQM$ and $\triangle AQM$ have the same circumradius.
So its enough to prove $\triangle AQM$ and $\triangle NPA$ have the same circumradius.

2) It´s known that $NI=NA$ and $MI=MA \Rightarrow MN\perp AI\Rightarrow PQ\perp AI$
Since $AI$ its the angle bisector of $\angle BAC\Rightarrow AP=AQ$ So its enough to prove that $\angle ANP=180-\angle AMQ \Leftrightarrow NP\cap MQ$ is on $\Gamma$

3) Consider $NP\cap \Gamma=D$ and $MD\cap AC=Q'$
From Pascal we get $AB\cap ND=P$ , $BM\cap CN=I$ and $AC\cap MD=Q'$ are collinear but $P, I, Q$ are collinear $\Rightarrow Q'=Q$
$Q.E.D$

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doxuanlong15052000

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doxuanlong15052000

#5 h Jun 22, 2016, 3:44 AM • 5 Y

Y by phantranhuongth, Ghost_rider, Adventure10, Mango247, AlexCenteno2007

My solution:
Let $NP$ cut $MQ$ at $R$ . From Pascal's theorem, we have $R$ lies on $(O)$ . We have $\angle RNB=\angle RMB$ and $\angle NBP=\angle NMI=\angle MIQ$ $\Longrightarrow$ $\triangle NPB\sim \triangle MQI$ $\Longrightarrow$ $\frac {NP}{MQ}=\frac {NB}{MI}=\frac {NI}{MI}=\frac {sin\frac {\angle C}{2}}{sin\frac {\angle B}{2}}$ $\Longrightarrow$ the circumradius of $\odot (BNP)$ and $\odot (CMQ)$ are equal

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PROF65

#6 h Jun 22, 2016, 2:43 PM • 2 Y

Y by Adventure10, Mango247

@doxuanlong1505 apart from some typos that you can rectify, nice solution.

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#7 h Jun 25, 2016, 4:19 AM • 2 Y

Y by Adventure10, Mango247

Let $\odot (CMQ)$ $\cap$ $NC$ = $D$ . $\angle NBP$ = $\angle$ $ICA$ . It´s easy to see: $AP$ = $AQ$ $\longrightarrow$ $PI$ = $IQ$ . If prove $NP$ = $QD$ the circumradius of $\odot (BNP)$ and $\odot (CMQ)$ are equal. Let $NP$ $\cap$ $(O)$ = $E$ . It´s easy to see $IPBE$ is cyclic. By angle chasing $IQCE$ is cyclic $\longrightarrow$ $E$ , $Q$ and $M$ are collinear. Then: $\angle$ $PNI$ = $\angle$ $IDQ$ . $\triangle$ $NPI$ $\cong$ $\triangle$ $IDQ$ $\longrightarrow$ $NP$ = $QD$ , done.

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Ilove_mathematics

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Ilove_mathematics

#8 h Nov 4, 2018, 4:45 AM • 2 Y

Y by Adventure10, Mango247

Math_CYCR wrote:

Let $R_1$ be the radius of $\triangle MQC$ . Let $R_2$ be the radius of $\triangle NPB$ .

Using the fact that $NB=NI=NA$ , $MC=MI=MA$ , and since $NM \parallel PQ$ , we get:

$\angle PIB= \angle MIQ = \angle IMN= \angle NCA= \angle NAB= \angle NBA$

$\angle QIC= \angle NIP= \angle INM= \angle MBA= \angle MAC= \angle MCA$

Therefore, by trig ceva in $\triangle NBI$ and $\triangle MIC$ , we get:

$\frac{ \sin \angle BNP}{ \sin \angle PNI} = \frac{ \sin \angle IMQ}{ \sin \angle QMC}$

By Two Equal Angles Lemma we get $\angle BNP= \angle IMQ$

We now easily see that $\angle ANP+ \angle AMQ= 180$ . And $AP= AQ$ .

By law of sines in $\triangle ANP$ and $\triangle AMQ$ , we get:

$\frac{NP}{ \sin \angle NAP} = \frac{AP}{ \sin \angle ANP} = \frac{AQ}{ \sin \angle AMQ} = \frac{MQ}{ \sin \angle MAQ}$ .

By law of sines in $\triangle MQC$ and $\triangle NPB$ we get:

$2R_1= \frac{MQ}{ \sin \angle MCQ} = \frac{NP}{ \sin \angle NBP} = 2R_2$

$\Longrightarrow R_1=R_2$

Done!

OMG MY FRIEND, I THOUGHT SAME THING! My solution is identical, so it would be useless post my solution...

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jbaca

#9 h Nov 18, 2018, 4:53 PM • 2 Y

Y by Adventure10, AlexCenteno2007

It's easy to infer that $AI$ bisects $\overline{PQ}$ . Let $E$ be the second intersection point of $BI$ and $(CQM$ ) and $D$ the second intersection of $CI$ and $(CQM)$ . Note that $\angle IEQ=\angle MCA=\angle IBP\therefore PB\parallel EQ$ so $BI=IE$ and hence $BPEQ$ is a parallelogram, thus $EQ=PB$ . Observe that $\angle IED=\angle MCI=\angle NBI\therefore NB\parallel ED$ therefore $\frac{IN}{ID}=\frac{BI}{IE}=1$ which implies $NI=ID$ , then $NB=ED$ . Finally, since $DN$ and $PQ$ bisect each other at $I$ , so $NQDP$ is a parallelogram with $NP=QD$ , then $\bigtriangleup NPB\cong \bigtriangleup DQE$ , which gives the required result.

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#10 h Mar 30, 2025, 8:42 PM

Y by

Obviously $\overline{AI} \perp \overline{PQ}$ . Let $T=\overline{MQ} \cap \overline{NP} \cap (ABC)$ be the $A$ -Mixtilinear intouch point by the Shooting Lemma. Now since $\measuredangle NBP=\measuredangle NBA=\measuredangle BAN=\measuredangle PAN$ , the corresponding $\widehat{PN}$ arcs are equal in measure so $R(NBP)=R(NAP)$ . Similarly $R(CQM)=R(AQM)$ . Let $K=(AQM) \cap (NAP)$ and $\measuredangle AKQ=\measuredangle AMT=\measuredangle ANT=\measuredangle AKP$ and so $K \in \overline{PQ}$ . To finish just see that $\measuredangle APK=\measuredangle KQA$ and again the corresponding $\widehat{AK}$ arcs are equal in measure so $R(ANP)=R(AMQ)$ and done.

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