Similarity through arc midpoint in right triangle

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Similarity through arc midpoint in right triangle

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Source: Iranian Geometry Olympiad 2016 Medium 4

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510 posts

#1 h May 26, 2017, 11:15 AM • 5 Y

Y by itslumi, tony88, Adventure10, Rounak_iitr, ItsBesi

Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ( $\angle A = 90^{\circ}$ ). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$ . Suppose that $M$ is the midpoint of the minor arc $AB$ , and $PM$ intersects $\omega$ for the second time in $Q$ . The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$ . Prove that $\angle PKC = 90^{\circ}$ .

Proposed by Davood Vakili

This post has been edited 1 time. Last edited by cjquines0, May 26, 2017, 11:15 AM

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600 posts

#2 h May 26, 2017, 11:44 AM • 2 Y

Y by Adventure10, Mango247

Here's the full problems and solution link:
http://igo-official.ir/wp-content/uploads/2016/12/igo_2016_-_problems__solutions_-_english.pdf

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101 posts

#3 h May 26, 2017, 11:45 AM • 5 Y

Y by Pluto1708, GeoKing, Adventure10, Mango247, redred123

We will use the well-known fact that the tangent to the circumcircle of the triangle $ABC$ divides the side $BC$ in ratio $-\frac {AB^2}{AC^2}$ . Before applying it, let note a Lemma.
Lemma: $\frac{AQ}{CQ}=\frac{AB}{AC}$ .
Proof: This can be proven considering the inversion with center $P$ and radius $PA$ , and as the circle of inversion is orthogonal to $\omega$ , $(M, Q)$ and $(B,C)$ are excanging places. So, by applying the inverison distance formula, we get $\frac{AQ}{CQ}=\frac{AM\times BP}{BM\times AP}=\frac{BP}{AP}=\frac{AB}{AC}$ .
Further, we easily compute the ratios $\frac{AK}{KC}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{AC^2}=\frac{BP}{CP}$ , and by Thales, we have the conclusion.

This post has been edited 2 times. Last edited by aleksam, May 26, 2017, 11:47 AM

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#4 h Sep 15, 2019, 8:48 AM • 2 Y

Y by tony88, Adventure10

I too have inversion solution but it is different.
Let $K'$ be on $AC$ such that $\angle PK'C = 90^{\circ}$ . Let $N$ be midpoint of $BC$ . We shall prove that $K'Q$ is tangent to $\omega$ .
Perform an inversion with centre $P$ and radius $PA$ .
Then $\omega$ gets mapped to $\omega$ itself. $K'$ goes to $PK' \cap AN$ say $L$ and $Q$ goes to $M$ . So line $K'Q$ gets mapped to $\odot (PLM)$ .
Now note that $MN$ is perpendicular bisector of $AB$ and hence $MN$ is perpendicular bisector of $PL$ .
However circumcentre of $\odot (PLM)$ lies on perpendicular bisector of $PL$ .
Also $N$ is circumcentre of $\omega$ .
$\therefore$ The centres of $\omega , \odot (PLM)$ and their intersection pts. are collinear.
So $\omega$ and $\odot (PLM)$ are tangent to each other. $\therefore$ before inversion $K'Q$ and $\omega$ must have been tangent to each other as desired.

This post has been edited 4 times. Last edited by Euler365, Sep 15, 2019, 9:29 AM

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#5 h Aug 4, 2020, 5:24 PM • 5 Y

Y by Jafarly8097, itslumi, shalomrav, popdit, Professor33

I am not sure if this works for $AB>AC$ but I hope it does.

Let $AQ$ and $MC$ meet at $R$ . Pascal on $AQQMCA$ shows that $R$ , $K$ and $P$ are collinear. Pascal on $ABCMMQ$ shows that $PR$ is parallel to $AB$ . Thus $PK$ is parallel to $AB$ and $\angle PKC=90^{\circ}$ .

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JustKeepRunning

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JustKeepRunning

#6 h Jun 12, 2021, 9:49 PM

Y by

@above, how can you use pascal if the lines $AB$ and $MM$ do not even intersect? I have not learned about this use of pascal before, so could you explain?

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Mahdi_Mashayekhi

697 posts

Mahdi_Mashayekhi

#7 h Dec 27, 2021, 7:41 AM

Y by

we have to prove PK || AB or PB/PC = KA/KC.
first note that PMA and PAQ are similar. PMB and PCQ are similar. PBA and PAC are similar.

PB/PC = (AB/AC)^2 and KA/KC = (QA/QC)^2
so we have to prove AB/AC = QA/QC.
AB/AC = PB/PA
PA/PQ = AM/AQ and PQ/PB = QC/BM so PB/PA = QA/QC.
we're Done.

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5 posts

tony88

#8 h Apr 26, 2022, 8:47 AM

Y by

Euler365 wrote:

I too have inversion solution but it is different.

Thank you for your very good inversion solution!!

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lian_the_noob12

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lian_the_noob12

#9 h Jul 22, 2023, 10:13 PM • 2 Y

Y by Rounak_iitr, LeYohan

Finally Solved With $\textbf{Pascal's Theorem}$

$\color{red} \boxed{\textbf{SOLUTION}}$

Let, $AQ \cap MC \equiv X, AB \cap AC \equiv Y$

By $\textbf{Pascal's Theorem}$ on $Q,Q,A,A,C,M$ We get $QQ \cap AC \equiv K, QA \cap CM \equiv X, AA \cap MQ \equiv P$
So, $P,X,K$ are collinear.

We need to show $PK \equiv PX \parallel AB$
As $M$ is the midpoint of minor arc $AB, CM$ is the angle bisector of $\angle ACB$
Let, $\angle ACM=\angle BCM=\alpha \implies \angle ABC=90-2\alpha$
Now,
$\angle PQX=\angle AQM=\angle ACM=\angle BCM=\angle PCX$ $\implies PQCX$ cyclic
Therefore,
$\angle CXK=\angle PQC=\angle PQX + \angle XQC = \alpha + \angle ABC=\alpha + 90- 2\alpha=90-\alpha$ So,
$\angle AYC=90- \alpha= \angle CXK=\angle YXK \implies AY \parallel PX \implies AB \parallel PX \implies AB \parallel PK \implies \angle PKC=\angle BAC=90 \blacksquare$

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327 posts

#11 h Dec 7, 2023, 4:46 PM

Y by

It suffices to show that $PK\parallel AB$ . This can be done with 3 applications of Pascal's Theorem. Let $T=MA\cap QB$ and $X=QA\cap CM$ .

Pascal's on $QQMACB$ yields $K, P, T$ are collinear.

Pascal's on $ABCMMQ$ yields $PX\parallel AB$ (since $MM$ intersects $AB$ at the point at infinity along that line)

Pascal's on $QQAACM$ yields $K,X,P$ are collinear.

Therefore, $K,P,T,X$ are collinear and these lie on a line parallel to $AB$ , so we are done.

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#12 h Oct 22, 2024, 4:52 AM

Y by

It's just angle chasing

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147 posts

#13 h Apr 28, 2025, 8:38 PM • 1 Y

Y by sami1618

Nice problem!
From Pascal Theorem on $(Q,Q,M,C,A,A)$ we get that the intersections of these lines are collinear:
$QQ \cap AC=\{K\}$ $QM \cap AA=\{P\}$ $MC \cap AQ=\{X\}$
Hence: Points $\overline{K-P-X}$ are collinear

Claim Points $P$ , $C$ , $Q$ and $X$ are concyclic
Claim
$\angle CXQ \equiv \angle CXA \stackrel{\triangle CXA}{=} 180-\angle XAC-\angle XCA=\angle CAQ-\angle MCA \stackrel{\omega}{=} \angle CMQ-\angle MCA=\angle CMQ-\angle BCM=$
$=180-\angle PMQ-\angle BCM \equiv 180-\angle PMQ-\angle PCM \stackrel{\triangle PMC}{=} \angle MPC \equiv \angle QPC \implies \angle CXQ=\angle CPQ \implies$
Points $P$ , $C$ , $Q$ and $X$ are concyclic $\square$

Claim $\angle PKC=90^{\circ}$
Claim
$\angle PKC \equiv \angle XKA \stackrel{\triangle XKA}{=} 180-\angle KXA-\angle KAX=\angle PXA-\angle QAC \equiv \angle PXQ-\angle QAC \stackrel{\odot(PCQX)}{=} 180-\angle PCQ-\angle QAC$
$\equiv 180-\angle BCQ-\angle QAC \stackrel{\omega}{=} \angle QAB-\angle QAC=\angle BAC=90^{\circ} \implies$
$\angle PKC=90^{\circ} \blacksquare$
Nice problem!
From Pascal Theorem on $(Q,Q,M,C,A,A)$ we get that the intersections of these lines are collinear:
$QQ \cap AC=\{K\}$ $QM \cap AA=\{P\}$ $MC \cap AQ=\{X\}$
Hence: Points $\overline{K-P-X}$ are collinear

Claim Points $P$ , $C$ , $Q$ and $X$ are concyclic
Claim
$\angle CXQ \equiv \angle CXA \stackrel{\triangle CXA}{=} 180-\angle XAC-\angle XCA=\angle CAQ-\angle MCA \stackrel{\omega}{=} \angle CMQ-\angle MCA=\angle CMQ-\angle BCM=$
$=180-\angle PMQ-\angle BCM \equiv 180-\angle PMQ-\angle PCM \stackrel{\triangle PMC}{=} \angle MPC \equiv \angle QPC \implies \angle CXQ=\angle CPQ \implies$
Points $P$ , $C$ , $Q$ and $X$ are concyclic $\square$

Claim $\angle PKC=90^{\circ}$
Claim
$\angle PKC \equiv \angle XKA \stackrel{\triangle XKA}{=} 180-\angle KXA-\angle KAX=\angle PXA-\angle QAC \equiv \angle PXQ-\angle QAC \stackrel{\odot(PCQX)}{=} 180-\angle PCQ-\angle QAC$
$\equiv 180-\angle BCQ-\angle QAC \stackrel{\omega}{=} \angle QAB-\angle QAC=\angle BAC=90^{\circ} \implies$
$\angle PKC=90^{\circ} \blacksquare$

Edit: I just relaized that you can finish this way easier.
After you prove that points $P$ , $C$ , $Q$ and $X$ are concyclic, we also have points $B$ , $A$ , $Q$ and $C$ are concyclic combining with $\overline{Q-A-X}$ and $\overline{C-B-P}$ are collinear by Reim's Theorem you get that: $BA \parallel PX \implies BA \parallel PK \implies \angle PKC=\angle BAC=90 \implies \angle PKC=90^{\circ}$

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