Configuration with three similar triangles

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Configuration with three similar triangles

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Source: IGO 2022 Intermediate P3

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#1 h Dec 13, 2022, 6:39 PM

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Let $O$ be the circumcenter of triangle $ABC$ . Arbitrary points $M$ and $N$ lie on the sides $AC$ and $BC$ , respectively. Points $P$ and $Q$ lie in the same half-plane as point $C$ with respect to the line $MN$ , and satisfy $\triangle CMN \sim \triangle PAN \sim \triangle QMB$ (in this exact order). Prove that $OP=OQ$ .

Proposed by Medeubek Kungozhin, Kazakhstan

This post has been edited 3 times. Last edited by a_507_bc, Dec 23, 2022, 8:48 AM

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#2 h Dec 13, 2022, 7:19 PM

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Here is a sketch.
Obviously $ANCP$ and $BMCQ$ are cyclic. Extend $MN$ to meet $(ACP)$ and $(BCQ)$ at $X, Y$ . The angle conditions imply that $ACPX$ and $BCQY$ are isosceles trapezoids. If we prove that $XPQY$ is cyclic, we will be done, as $O$ would be its center ( $O$ lies on the perpendicular bisectors of $XP$ and $YQ$ using the isosceles trapezoids). Notice that $P-C-Q$ by an easy angle chase. To finish, note that $\angle YXP= \angle NCQ= 180- \angle YQP$ , so we are done.

This post has been edited 2 times. Last edited by VicKmath7, Dec 13, 2022, 7:21 PM

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#3 h Dec 21, 2022, 1:15 PM • 1 Y

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Spent 1.5 hours on this, then realized its an easy complex bash (only 1 page, with big handwriting

)

This post has been edited 1 time. Last edited by LLL2019, Jan 4, 2023, 1:23 AM

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AlanLG

#6 h Jan 4, 2023, 4:03 AM

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Note that $N$ and $M$ are the center of the spiral similarity that sends $MC$ to $AP$ and $NC$ to $BQ$ , respectively, therefore we have
$\triangle NMC\sim \triangle NAP\hspace{1cm} \triangle NMA\sim \triangle NCP \hspace{1cm}P\in\left(ACN\right)$ $\triangle MNC\sim \triangle MBQ\hspace{1cm} \triangle MNB\sim \triangle MCQ \hspace{1cm}Q\in\left(BCM\right)$
Lemma 1 $P, C, Q$ are collinear
Proof:

By $APCN$ cyclic and $\triangle NMC\sim \triangle NAP$
$\angle PCA=\angle PNA=CNM$ By $BMCQ$ cyclic and $\triangle MNB\sim \triangle MCQ$
$\angle BMQ=\angle MCQ=\angle MNB$
Therefore $\angle MCQ+\angle PCA=\angle MNB+\angle CNM=180^\circ$
$\hspace{20cm} \blacksquare$

Lemma 2 If $R=PQ\cap \left(ABC\right)$ , then $A$ is the center of the spiral similarity that sends $PR$ to $NB$
Proof:

By $ARCB$ and $APCN$ cyclic,
$\angle PRA=\angle NBA\hspace{1cm}\angle APR=\angle ANB$ $\hspace{20cm} \blacksquare$

Lemma 3 $CQ=PR$
Proof:

By $\triangle MNB\sim \triangle MCQ$
$\frac{CQ}{MC}=\frac{NB}{MN}$
And by Lemma 2 we have $\triangle APR\sim \triangle ANB$ and $\triangle ARB\sim \triangle APN\sim MCN$
$\frac{PR}{BN}=\frac{AR}{AB}=\frac{AP}{AN}$ Therefore,
$\frac{PR}{CQ}=\frac{MN}{MC}\cdot \frac{AP}{AN}=1$ $\hspace{20cm} \blacksquare$

Finally by PoP from $P$ and $Q$ to $(ABC)$

$OP^2=PR\cdot PC=PR\cdot (PR+RC)=QC\cdot QR=OQ^2$ $\hspace{20cm}\blacksquare$

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#7 h Oct 15, 2023, 2:56 AM

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Very nice problem

Let, $\angle MCN= \alpha, \angle MNC=\phi, \angle NMC= \beta$
So $\alpha + \beta + \phi = 180^{\circ}$
Claim 1: A,N,C,P are concyclic and B,M,C,Q are concyclic.
Proof:
Since $\triangle MNC\sim \triangle ANP$ , so $\angle MCN=\angle APN = \alpha$
$\rightarrow$ A,N,C,P are concyclic. $\square$

Since $\triangle MNC\sim \triangle MBQ$ , so $\angle MCN=\angle MQB = \alpha$
$\rightarrow$ B,M,C,Q are concyclic. $\square$

Claim 2: Q,C,P are collinear
Proof: Let's see what

By Claim 1 $\angle ANP = \angle ACP = \phi$
By Claim 1 $\angle BCQ = \angle BMQ = \beta$

In $C$ , $\angle ACP + \angle MCN + \angle BCQ = \alpha + \beta + \phi = 180^{\circ}$ , so Q,C,P are collinear $\square$ .

Let $R: PQ \cap \odot ABC$

So $\angle BAR = \angle BCR = \beta, \angle ACB = \angle MCN = \angle ARB = \alpha$
So $\triangle ABR \sim \triangle MNC$

Let's see what $\triangle BMN \sim \triangle QMC$ , and $\triangle ABN \sim \triangle ARP$
So $\triangle MCN \sim \triangle ARB$

Claim 3: $CQ = PR$
Proof:
By $\triangle BMN \sim \triangle QMC$
$\frac{CQ}{MC}= \frac{BN}{MN}$ By $\triangle ABN \sim \triangle ARP$
$\frac{RP}{AR}=\frac{BN}{AB}$ Now
$\frac{CQ}{RP}=\frac{BN \cdot MC \cdot AB}{MN \cdot BN \cdot AR}=\frac{MC \cdot AB}{MN\cdot AR}=1$
So $CQ=RP$ $\square$

Finishing:
Since $O$ belongs to the bisector of $RC$ , but since $CQ= RP$ then $CR$ and $PQ$ have the same midpoint, then since $O$ is equidistant from $R$ and $C$ then $O$ also equidistant from $P$ and $Q$ , finally $OP=OQ$ , as desired.

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#8 h Aug 23, 2024, 6:57 AM

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Let's present a different approach.
As proven above we have that ANCP and BMCQ are cyclic. And P, C, and Q are collinear. Also by similarities given we have that $\bigtriangleup MCQ\sim\bigtriangleup MNB$ and $\bigtriangleup NCP\sim\bigtriangleup NMA$ .
By the Law of Cosines in $\bigtriangleup OCQ$ and $\bigtriangleup OCP$ it suffices that $OP^2=OQ^2 \iff CP^2-CQ^2=-2R\cdot CP\cos\angle OCP +2R\cdot CQ\cos\angle OCQ\iff |CP^2-CQ^2|=|(CP+CQ)(CP-CQ)|=2R|\cos\angle OCQ|(CP+CQ) \iff |CP-CQ|=2R|\cos\angle OCQ|$ where $R$ denotes the circumradius of $(ABC)$ . Note that $\angle (MN,AB)=|90^{\circ}-\angle OCQ|$ , hence $|\cos \angle OCQ|=\sin\angle (MN,AB)$ . So it suffices to show that $|CP-CQ|=2R\sin\angle (MN,AB)$ . From the similarities of triangles above, we get that $CP=\frac{NC}{NM}AM=\frac{\sin\angle CMN}{\sin\angle C}AM$ and similarly $CQ=\frac{\sin\angle CNM}{\sin C}BN$ . So now it suffices to show that $|\frac{\sin\angle CMN}{\sin\angle (MN,AB)}AM - \frac{\sin\angle CNM}{\sin\angle (MN,AB)}BN|=2R\sin\angle C$ . But note that $2R\sin\angle C=AB$ . Let $J$ be the intersection of $MN$ and $AB$ , then $|\frac{\sin\angle CMN}{\sin\angle (MN,AB)}AM - \frac{\sin\angle CNM}{\sin\angle (MN,AB)}BN| =| \frac{JA}{AM}AM-\frac{JB}{BN}BN|=|JA-JB|=|AB|=AB$ , and we are done. I did not consider the $MN\parallel BC$ , but it's not hard.

This post has been edited 1 time. Last edited by Circumcircle, Aug 23, 2024, 7:20 AM
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