2 lines and a circumcircle concurrent

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Source: BMO Shortlist 2017 G5

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30628 posts

#1 h Jul 14, 2019, 9:09 AM • 7 Y

Y by jhu08, tiendung2006, itslumi, ImSh95, Adventure10, Mango247, GeoKing

Let $ABC$ be an acute angled triangle with orthocenter $H$ . centroid $G$ and circumcircle $\omega$ . Let $D$ and $M$ respectively be the intersection of lines $AH$ and $AG$ with side $BC$ . Rays $MH$ and $DG$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $PD$ and $QM$ intersect on $\omega$ .

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#2 h Jul 14, 2019, 10:14 AM • 6 Y

Y by jhu08, ImSh95, Adventure10, Mango247, GeoKing, ehuseyinyigit

It is well-known that's $PA \parallel BC$ . And if the second intersection of $HM$ and $\omega$ is $T$ then $TM=MH$ . By easy angle chasing (or because $HD = DK$ implies that $K$ lies on $\omega$ and $\angle AKT = \angle HDM = 90^o$ ) we have that $AT$ is a diameter of $\omega$ . So $\angle MQT = \angle DAM$ (since $AD$ and $QT$ are simmetric about the perpendicular bisector of $BC$ ) and $\angle APM = 90^o = \angle ADM$ implies $APDM$ cyclic and $\angle DPM = \angle DAM$ so we are done.

This post has been edited 2 times. Last edited by Daniil02, Jul 14, 2019, 10:15 AM

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#3 h Jul 14, 2019, 12:45 PM • 4 Y

Y by jhu08, ImSh95, Adventure10, GeoKing

Let $S$ $\equiv$ $QM$ $\cap$ $\omega$ ; $I$ $\equiv$ $AP$ $\cap$ $BC$
We have: $\dfrac{BS}{CS} = \dfrac{BM}{CM} . \dfrac{QC}{QB} = \dfrac{AB}{AC}$
Then: $\dfrac{BP}{CP} = \dfrac{IB}{IC} . \dfrac{AC}{AB} = \dfrac{BD}{CD} . \dfrac{CS}{BS}$ or $D$ , $P$ , $S$ are collinear
So: intersection of $DP$ and $QM$ lies on $\omega$

This post has been edited 2 times. Last edited by khanhnx, Jul 14, 2019, 2:29 PM

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1915 posts

#4 h Jan 2, 2020, 9:42 PM • 6 Y

Y by GeoMetrix, strawberry_circle, jhu08, ImSh95, Adventure10, GeoKing

BMO Shortlist 2017 G5 wrote:

Let $ABC$ be an acute angled triangle with orthocenter $H$ . centroid $G$ and circumcircle $\omega$ . Let $D$ and $M$ respectively be the intersection of lines $AH$ and $AG$ with side $BC$ . Rays $MH$ and $DG$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $PD$ and $QM$ intersect on $\omega$ .

Notice that a Homothety $\mathcal H$ at $G$ with a scale factor of $-\frac{1}{2}$ sends $\odot(X_5)$ to $\odot(ABC)$ , so $\mathcal H:D\leftrightarrow Q$ . Hence, we get that $QA\|BC$ .

Now introduce the orthic triangle $DEF$ of $\triangle ABC$ . Note that $P$ is the $A-\text{Queue Point}$ of $\triangle ABC$ , hence, $\angle APH=90^\circ$ . Now Radical Axis Theorem on $\odot(ABC),\odot(BC)$ and $\odot(AH)$ we get that $AP,EF,BC$ are concurrent at the $A-\text{Expoint} (X_A)$ of $\triangle ABC$ . Let $PD\cap\odot(ABC)=T$ and $QM\cap\odot(ABC)=T'$ . Now notice that $-1=(X_AD;BC)\overset{P}{=}(AT;BC)\implies ABTC\text{ is a Harmonic Quadrilateral.}$ Similarly we get that $-1=(BC;M\infty_{BC})\overset{Q}{=}(BC;T'A)\implies ABT'C\text{ is a Harmonic Quadrilateral.}$ So from this we can conclude that $\boxed{T'\equiv T}$ . Hence, $PD\cap QM\in\odot(ABC). \blacksquare$

This post has been edited 4 times. Last edited by amar_04, Jan 2, 2020, 9:54 PM

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256 posts

#5 h Aug 28, 2020, 3:06 PM • 3 Y

Y by mkomisarova, jhu08, ImSh95

It is well - known that $P$ is $A$ Queue point and that $AQ \parallel BC$ . Assume that $F$ intersects $\odot(ABC)$ at point $F$ .
Claim: $F$ is intersection of $A$ symmediaa and $\odot(ABC)$
Proof: Consider $X = PA \cap BC$ . It is well - known that $X$ is $A$ Ex point. Therefore:
$-1 = (X,D;B,C) \overset{P}{=} (A,F;B,C)$ This proves our claim:
Now assume that $QM \cap \odot(ABC) = F'$ . Note that:
$-1 = (\infty_{BC},M;B,C) \overset {Q}{=} (A,F';B,C)$ This proves that $F = F'$ . As a result that lines $PD \cap QM$ lies on $\odot(ABC)$ as desired.

This post has been edited 1 time. Last edited by Kimchiks926, Aug 28, 2020, 3:07 PM
Reason: typo

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380 posts

#6 h Aug 28, 2020, 3:26 PM • 3 Y

Y by Aryan-23, jhu08, ImSh95

We use the well known facts that $AQ \parallel BC$ and $HM$ passes through the antipode of $A$ on $\omega$ , which we denote by $A'$ .

From here, we use complex numbers with $\omega$ as the unit circle. As usual, let lowercase letters denote the complex coordindates of their respective uppercase letter. Since $AQ \parallel BC$ , we have $q = \frac{a}{bc}$ . From the midpoint formula we have $m = \frac{b+c}{2}$ . Since $D$ is the foot from $A$ to $BC$ ,
$d = \frac{1}{2} \left(a + b + c - \frac{bc}{a} \right)$ Because $M$ lies on chord $PA'$ , we have that
$p = \frac{m + a}{1 + a \overline{m}} = \frac{ bc(2a + b +c)} {2bc + a(b+c)}$
Let $X$ be the second intersection of $QM$ with $\omega$ . Then $X$ is on chord $QM$ , so
$x = \frac{m + q}{1 + q \overline{m}} = \frac{a(b+c) + 2bc}{a - b - c} .$
We want to check that $D$ is on chord $PX$ . This is equivalent to
$p + x = d + px \overline{d}$ Checking this is a straightforward computation.

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primesarespecial

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primesarespecial

#7 h Aug 4, 2021, 12:28 PM • 2 Y

Y by jhu08, ImSh95

We consider a projective transformation fixing $\omega$ and sending $M$ to the center of $\omega$ .
Now the problem is much simplified.
Let $\omega$ be a circle with diameter $BC$ and $A$ be a point on it.Let $D$ be the feet of $A$ on $BC$ and $G$ be the centroid of $A,B,C$ .
Let $Q$ be the intersection of $DG$ and $\omega$ .Prove that $AD$ and $QM$ intersect on $\omega$ .

Easily, $AQBC$ is an isosceles trap.
Now, $AD \cap \omega$ be $E$ ,thus $\angle EAQ=90$ and so $\text{[math]}$ is a diameter and we are done.
(Just wanted to add that the points P and Q are well known and a complex bash would finish it but this is just shorter.)

This post has been edited 1 time. Last edited by primesarespecial, Aug 4, 2021, 3:51 PM

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1162 posts

#8 h Dec 15, 2021, 1:39 AM • 1 Y

Y by ImSh95

Let $PD$ meet $\omega$ again at $K$ . It's well-known, i.e. by 2011 G4, that $AQ \parallel BC$ . Thus, symmetry and isogonality implies $QM$ and the $A$ -Symmedian meet on $\omega$ . In addition, the Orthocenter Reflection Lemma yields $P$ as the $A$ -Queue point.

Let $BH$ meet $CA$ at $E$ and $CH$ meet $AB$ at $F$ . By Radical Axes, we know $AP, BC, EF$ are concurrent at some point $T$ , so Ceva-Menelaus gives $-1 = (T, D; B, C) \overset{P}{=} (A, K; B, C)$ which clearly finishes. $\blacksquare$

Remark: APMO 2012/4 directly implies $-1 = (A, K; B, C)$ , but I included a proof for the sake of completeness.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#9 h Feb 22, 2022, 1:03 PM • 1 Y

Y by ImSh95

Claim1 : $AQ || BC$ .
Proof : Let $K$ be a point on $\omega$ such that $AK || BC$ . we have $AK = 2DM$ , $AG = 2GM$ and $\angle AKG = \angle MDG$ so $AGK$ and $MDG$ are similar so $\angle AGK = \angle MGD$ so $K,G,D$ are collinear so $K$ is $Q$ .

Claim2 : $APDM$ is cyclic.
Proof : It's well-known that reflection of $H$ across $M$ and $D$ lies on $\omega$ so If $S$ is reflection of $H$ across $M$ then $AS$ is diameter of $\omega$ so $\angle APM = \angle APS = \angle 90 = \angle ADM$ .

we will prove $\angle MQA + \angle DPA = \angle 180$ . $\angle MQA = \angle MAQ = \angle AMB$ and $\angle DPA = \angle AMC$ so we're Done.

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322 posts

BVKRB-

#10 h Mar 25, 2022, 6:16 PM • 1 Y

Y by ImSh95

Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy)

It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$
Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$ , this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$
Came here from The problem proposer’s YT channel!

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465 posts

#11 h Mar 26, 2022, 11:25 AM • 2 Y

Y by ImSh95, HoripodoKrishno

Is it just me or does it seem hella familiar with some problem, maybe i did this one before o wot?

Let $Q' = AA \cap \odot ABC$ , $X = A-$ symmedian $\cap \odot ABC$ . It is well known that $AQ \parallel BC$ , that $AMXQ'$ is cyclic by the $A-$ Apollonius circle wrt $BC$ and that $P$ lies on $\odot AH$ .

$\measuredangle APM = \measuredangle APH = 90^{\circ} = \measuredangle ADM \implies APDM$ is cyclic.
Now perform an Inversion with center $A$ and with radius $\sqrt{AB \cdot AC}$ followed by a reflection wrt the Angle Bisector of $\angle BAC$ .

$\begin{align*} Q \xleftrightarrow{} Q'\\ M \xleftrightarrow{} X\\ \end{align*}$ Thus $\overline{X - M - Q}$ are collinear.

Now, $\measuredangle APD = \measuredangle AMD = \measuredangle AMC = \measuredangle ABX = \measuredangle APX \implies \overline{P - D - X}\text{ are collinear.}$ Thus $X$ is our desired concurrency point.

This post has been edited 1 time. Last edited by kamatadu, Mar 27, 2022, 8:18 AM
Reason: forgor :skull:

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173 posts

#12 h Apr 29, 2022, 9:04 PM • 1 Y

Y by ImSh95

BVKRB- wrote:

Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy)

It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$
Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$ , this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$
Came here from The problem proposer’s YT channel!

Sorry but can you tell me which lemma is that? I tried finding in EGMO, but didn't get it

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322 posts

BVKRB-

#13 h Jun 7, 2022, 5:04 PM

Y by

MathsinMaths wrote:

BVKRB- wrote:

Wow! I am surprised to not see a really short and simple solution involving the butterfly theorem (I got to know this method by seeing a part of the solution to ISL 2016 G2, after that, this was pretty easy)

It is well known (EGMO Lemma in chapter 3) that $AQ \parallel BC$ and let $HM \cap \omega = X$ and let $XQ \omega = Y$
Notice that by orthocenter reflections lemma and reflection symmetry we get $XQ \perp BC$ , this means $MD = DY$ now, the butterfly theorem finishes $\blacksquare$
Came here from The problem proposer’s YT channel!

Sorry but can you tell me which lemma is that? I tried finding in EGMO, but didn't get it

Sorry for 1 month late reply, I didn't see the thread

It is EGMO Problem 3.24.

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597 posts

eibc

#14 h Sep 25, 2023, 7:58 PM

Y by

Note that

$P$ is the $A$ -Queue point, so it's well known that if $T = \overline{AP} \cap \overline{BC}$ then $(T, D; B, C) = -1$ (as $T$ is the $A$ -Ex point)
$G$ is the insimilicenter of the nine-point circle of $\triangle ABC$ and $\omega$ , so $\overline{AQ} \parallel \overline{BC}$

Thus, we have
$\begin{align*} -1 &= (T, D; B, C) \overset{P}{=} (A, \overline{PD} \cap \omega; B, C), \\ -1 &= (M, \infty_{BC}; B, C) \overset{Q}{=} (\overline{QM} \cap \omega, A; B, C), \end{align*}$ which finishes.

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#15 h Jun 18, 2024, 7:56 PM

Y by

Posting for storage.
We know that $M-H-Q_a=P$ and $AQ \parallel BC$ . Let $A-symmedian \cap (ABC)$ at $K$ and take pencil from $P$ . So, $ABKC$ is harmonic.Also, $(B,C;M,W^{\infty} )=-1$ . Take pencil from $Q$ , $(QM \cap (ABC), A;B, C) =-1$
Thus, $QM \cap (ABC) \equiv K$
No need for diagram just lemmas

This post has been edited 3 times. Last edited by Z4ADies, Jun 18, 2024, 8:00 PM

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#16 h Jun 22, 2024, 10:04 AM

Y by

Recall that $P$ is the $A$ -Queue point and that $Q$ is the reflection of $A$ across the perpendicular bisector of $BC$ .
Let $K$ be the point on $\omega$ such that $(B,C;A,K)=-1$ . We claim that $P-D-K$ and $Q-M-K$ .
Let $X$ be the $A$ -Ex point. Then $-1=(B,C;X,D) \stackrel{P}{=} (B,C;A,PD \cap \omega)$ , so $P-D-K$ .
$AK$ is the $A$ -symmedian, so $\angle BQK = \angle BAK = \angle CAM = \angle BQM$ , which implies that $Q-M-K$ .
Hence proved.

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Eka01

#17 h Aug 9, 2024, 1:06 PM • 1 Y

Y by Sammy27

It is well known that $Q$ is the point on $(ABC)$ such that $AQ|| BC$ so $(B,C;M,P_\infty)=-1$ . Projecting this through $Q$ onto $(ABC)$ gives us that $QM$ meets the circle at a point $X$ such that $ABXC$ is harmonic.

Let $AP \cap BC= T$ , which is well known to be $A$ expoint. Then the fact that $(B,C;D,T)=-1$ is also well known. Projecting this bundle onto the circumcircle, if $PD \cap (ABC) =X'$ , then $ABX'C$ is also harmonic.
Hence, $X \equiv X'$ which is what we needed to show.

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#18 h Aug 9, 2024, 2:48 PM

Y by

Let $S$ be second intersection of $PD$ with $(ABC)$ and $J \equiv AP \cap BC$ . We have $\dfrac{PB}{PC} \cdot \dfrac{SB}{SC} = \dfrac{DB}{DC} = \dfrac{JB}{JC} = \dfrac{PB}{PC} \cdot \dfrac{AB}{AC}$ . Then $\dfrac{AB}{AC} = \dfrac{SB}{SC}$ . But $G$ is center of homothety that transforms NPC of $\triangle ABC$ to $(ABC)$ so it's easy to see that $AQ \parallel BC$ . Hence $\dfrac{AB}{AC} = \dfrac{QC}{QB},$ so $\dfrac{SB}{SC} \cdot \dfrac{QB}{QC} = 1 = \dfrac{MB}{MC}$ or $Q, M, S$ are collinear

This post has been edited 2 times. Last edited by khanhnx, Aug 9, 2024, 2:48 PM

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