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Source: Iran MO 3rd round 2017 finals - Geometry P3

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164 posts

#1 h Sep 3, 2017, 5:36 PM • 2 Y

Y by Abdollahpour, Adventure10

In triangle $ABC$ points $P$ and $Q$ lies on the external bisector of $\angle A$ such that $B$ and $P$ lies on the same side of $AC$ . Perpendicular from $P$ to $AB$ and $Q$ to $AC$ intersect at $X$ . Points $P'$ and $Q'$ lies on $PB$ and $QC$ such that $PX=P'X$ and $QX=Q'X$ . Point $T$ is the midpoint of arc $BC$ (does not contain $A$ ) of the circumcircle of $ABC$ . Prove that $P',Q'$ and $T$ are collinear if and only if $\angle PBA+\angle QCA=90^{\circ}$ .

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#2 h Sep 4, 2017, 3:05 PM • 2 Y

Y by Adventure10, Mango247

Any solutions?

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andria

#3 h Sep 6, 2017, 3:30 PM • 4 Y

Y by buratinogigle, Akatsuki1010, Adventure10, Mango247

Solution for part b:

Lemma: Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ . $AC\cap BD=P,AB\cap CD=E,AD\cap BC=F$ and $Q$ is the center of spiral similarity that sends $AD$ to $CB$ .
(a) $E,Q,O$ are collinear.
(b) $Q$ is second intersection of $\odot(AOB),\odot(DOC)$ .

Proof:
Let $M=OP\cap EF$ be the miquel point of $ABCD$ . Since $FP$ is polar of $E$ WRT $(ABCD)\Longrightarrow EO\perp FP$ . Let $EO\cap FP=Q'$ then $P$ is orthocenter of $\triangle EFO\Longrightarrow FP.FQ=FM.FE=FD.FA=FC.FB\Longrightarrow Q'$ lies on $\odot(PAD),\odot(PBC)\Longrightarrow Q'=Q$ . Also note that $ED.EC=EA.AB=EM.EF=EQ.EO\Longrightarrow Q\in\{\odot(OAB)\cap \odot(OCD)\}$ . $\blacksquare$
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Back to the problem:
Assume that $\angle PBA+\angle QCA=90^{\circ}$ . Since $\angle XPQ=\angle XQP=\frac{\angle A}{2}\Longrightarrow XP=XQ=XP'=XQ'\Longrightarrow PQQ'P'$ is cyclic quadrilateral with center $X$ . Note that :
$\angle PXP'+\angle QXQ'=(180-2\angle XPB)+(180-2\angle XQC)=180^\circ\Longrightarrow PQ'\perp QP'$ (assume that $PQ'\cap QP'=V$ ). The tangents from $B,C$ to $\odot(ABC)$ intersect at $W$ . Let $\omega(W,WB)$ cut $AB,AC$ at $R,L$ respectively. Since $\angle A+\angle PSQ=\angle PBA+\angle QCA=90^\circ\Longrightarrow \angle S=90^\circ-\angle A\Longrightarrow S\in \omega$
Let $U$ be the second intersection of $\odot(XPP'),\odot(XQQ')$ .
From the lemma $X,U,S$ are collinear and $U$ is spiral similarity sending $P'Q'$ to $QP$ so from $PQ'\perp QP'\Longrightarrow \angle PUQ=\angle P'UQ'=90^\circ$ . since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear. similarly $L,U,P$ are collinear. Since $\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is cyclic. similarly $QNUC$ is cyclic.
Let $T'$ be the second intersection of $\odot(BUP')$ and $\odot(CUQ')$ then since $\angle UTP'+\angle UTQ'=\angle UCS+\angle UBS=180^\circ\Longrightarrow P',T',Q'$ are collinear.
$\angle T'BU=\angle UP'Q'=\angle UQP=\angle NCU=90^\circ-\frac{\angle A}{2}-\angle UCB\Longrightarrow \angle T'BC=90^\circ-\frac{\angle A}{2}-\angle UCB-\angle UBC=90^\circ-\frac{\angle A}{2}-(90^\circ-\angle A)=\frac{\angle A}{2}$ Similarly $\angle T'CB=\frac{\angle A}{2}$ . Hence $T'=T$ . $\blacksquare$

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571 posts

#4 h Sep 7, 2017, 6:30 AM • 2 Y

Y by Adventure10, Mango247

$\textbf{Proof :}$

I think it's enough to prove that if $PBA + QCA = 90$ then $P'Q'T$ are collinear.

$PQ$ is angle bisector of $\angle A$ , so $XP =XQ = XP' = XQ'$ and $X$ is center of $PQP'Q'$ . Let $A', B', C'$ be the second intersection points of $(ABC)$ with $PQ, PP', QQ'$ respectively. Easy to see that $A'C'\parallel PQ'$ and $A'B'\parallel QP'$ . Also from simple angle computation from $PBA + QCA = 90$ we get that $QP'\perp Q'P$ . Let $Y$ be on $P'Q'$ , such that $P'Y/YQ' = QA/AP$ . So we see that $YC'\parallel P'Q\parallel B'A'$ and $YB'\parallel Q'P\parallel C'A'$ . So $A'B'C'Y$ is a rectangle. Easy to see that $A'$ is midpoint of bigger arc $BC$ of $(ABC)$ and $Y$ is symmetric to $A'$ wrt center of $(ABC)$ , so $Y$ is midpoint of smaller arc $BC$ of $(ABC)$ and $Y\in P'Q'$ . Done

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290 posts

jred

#5 h Nov 17, 2017, 3:38 AM • 1 Y

Y by Adventure10

Mosquitall wrote:

$\textbf{Proof :}$

I think it's enough to prove that if $PBA + QCA = 90$ then $P'Q'T$ are collinear.

$PQ$ is angle bisector of $\angle A$ , so $XP =XQ = XP' = XQ'$ and $X$ is center of $PQP'Q'$ . Let $A', B', C'$ be the second intersection points of $(ABC)$ with $PQ, PP', QQ'$ respectively. Easy to see that $A'C'\parallel PQ'$ and $A'B'\parallel QP'$ . Also from simple angle computation from $PBA + QCA = 90$ we get that $QP'\perp Q'P$ . Let $Y$ be on $P'Q'$ , such that $P'Y/YQ' = QA/AP$ . So we see that $YC'\parallel P'Q\parallel B'A'$ and $YB'\parallel Q'P\parallel C'A'$ . So $A'B'C'Y$ is a rectangle. Easy to see that $A'$ is midpoint of bigger arc $BC$ of $(ABC)$ and $Y$ is symmetric to $A'$ wrt center of $(ABC)$ , so $Y$ is midpoint of smaller arc $BC$ of $(ABC)$ and $Y\in P'Q'$ . Done

Why it's enough to show the "if" part? On the contrary, I found the real difficult part of this problem is the "only if" part. Still hoping someone post a solution for that ...

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28 posts

#6 h Dec 16, 2017, 5:45 PM • 1 Y

Y by Adventure10

Let $D$ be so that $DT$ is a diameter.
For the only if direction, we'll prove that for a fixed $Q$ , there can be at most one $P$ on $AD$ such that $P', Q', T$ are collinear. Then, by part (a), we'd be done.

Vary $P$ along the line $AD$ .
Claim. As $P$ varies, the line $P'Q'$ passes through a fixed point $Y$ . If this were true, then $Q'$ would be uniquely defined as the intersection of $YT$ and $QC$ . From here, $X$ is uniquely defined, and therefore so is $P$ .

Therefore it suffices to prove our claim.

Proof of Claim.

Let $P_1, Q_1$ be the intersections of $PB, QC$ respectively with the circumcircle of $ABC$ .

By the above posts, $DP_1 \parallel P'Q$ , and $DQ_1 \parallel PQ'$ . Thus, as $AP_1BD$ is cyclic, $\angle PAB = \angle PP_1D = \angle PP'Q$ , and so $ABP'Q$ is cyclic, say on circle $\Omega$ .. Let $Y'$ be the second intersection of $P'Q'$ with $\Omega$ . Note that $\Omega$ , $\angle AQC$ are both fixed since $Q$ is fixed. Therefore, as $\angle BQY$ = $180 - \angle BP'Y$ = $180 - \angle PP'Q'$ = $\angle APC$ . Therefore, $Y$ is fixed since $B, Q, \Omega$ , and $\angle BQY$ are all fixed.

This post has been edited 2 times. Last edited by ashologe, Dec 17, 2017, 1:46 PM

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370 posts

#7 h Jan 20, 2018, 5:31 PM • 3 Y

Y by gemcl, Adventure10, Mango247

andria wrote:

since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear.

I can't see the links between them.

andria wrote:

$\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is concylic.

Neither does this one make sense.
Write it carefully or don't write down your solution please.

This post has been edited 1 time. Last edited by liekkas, Jan 20, 2018, 5:41 PM

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293 posts

#9 h May 18, 2019, 8:07 PM • 2 Y

Y by Adventure10, Mango247

Mosquitall wrote:

$P'Y/YQ' = QA'/AP'$ .

I think it should be like this. By the way beautiful solution.

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55 posts

yds

#10 h Sep 22, 2020, 12:36 AM

Y by

liekkas wrote:

andria wrote:

since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear.

I can't see the links between them.

andria wrote:

$\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is concylic.

Neither does this one make sense.
Write it carefully or don't write down your solution please.

Can you give a new solution?

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#11 h Jul 7, 2022, 3:04 PM

Y by

Part $1 : \angle PBA+\angle QCA=90^{\circ}$ .
Let $PQ'$ and $P'Q$ meet at $K$ and Let $ABC$ meet $PQ$ and $PP'$ and $QQ'$ at $T'$ and $R$ and $S$ and Let $BS$ meet $AT$ at $Y$ .
Claim $: \angle PKQ = \angle 90$ .
Proof $:$ Note that $\angle PKQ = \angle QP'P + \angle Q'PP' = \frac{\angle PXQ + \angle P'XQ'}{2}= \angle 90$ .
Claim $: \angle PBS = \angle 90 = \angle QCR$ .
Proof $:$ Note that $\angle ACR = \angle ABR = \angle ABP = \angle 90 - \angle QCA$ . we prove the other one with same approach.
Note that $\angle TAT' = \angle 90$ so $PBYA$ is cyclic so $\angle BPY = \angle BAY = \angle BAT = \angle 90 - \angle PAB = \angle XPA$ .
Note that $PACQ'$ is cyclic so $\angle APQ' = \angle QCA = \angle ABS = \angle ABY = \angle APY$ so $P,Y,Q'$ are collinear.
Claim $: T'S || PQ'$ and $T'R || P'Q$ .
Proof $: \angle APY = \angle ABY = \angle ABS = \angle ST'Q$ and we prove the other one with same approach.
Note that $\angle T'ST = \angle 90 = \angle T'RT$ so $T'R = TS$ so $\frac{ST}{QP'} = \frac{RT'}{QP'} = \frac{PT'}{PQ} = \frac{Q'S}{Q'Q}$ which implies $Q',T,P'$ are collinear.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jul 8, 2022, 8:44 AM

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