Tilted Students Thoroughly Splash Tiger part 2

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Source: ELMO 2024/5

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DottedCaculator

7355 posts

DottedCaculator

#1 h Jun 21, 2024, 4:17 PM • 9 Y

Y by NO_SQUARES, centslordm, crazyeyemoody907, bjump, Rounak_iitr, nmoon_nya, ehuseyinyigit, chirita.andrei, MS_asdfgzxcvb

In triangle $ABC$ with $AB<AC$ and $AB+AC=2BC$ , let $M$ be the midpoint of $\overline{BC}$ . Choose point $P$ on the extension of $\overline{BA}$ past $A$ and point $Q$ on segment $\overline{AC}$ such that $M$ lies on $\overline{PQ}$ . Let $X$ be on the opposite side of $\overline{AB}$ from $C$ such that $\overline{AX} \parallel \overline{BC}$ and $AX=AP=AQ$ . Let $\overline{BX}$ intersect the circumcircle of $BMQ$ again at $Y \neq B$ , and let $\overline{CX}$ intersect the circumcircle of $CMP$ again at $Z \neq C$ . Prove that $A$ , $Y$ , and $Z$ are collinear.

Tiger Zhang

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#2 h Jun 21, 2024, 4:17 PM • 4 Y

Y by centslordm, crazyeyemoody907, Rounak_iitr, MS_asdfgzxcvb

Quite shrimple?

Claim: $BC = CQ$ , and similarly $BP = BC$ .
Proof. By directed lengths, we get that $BQ = b \cdot \frac{\frac{a}{2}}{a \cdot \frac{b}{b+c}} = \frac{b+c}{2}$ . $\blacksquare$

Claim: $BQ, PC, AX$ concur on $(XPQ)$ .
Proof. Note that $\measuredangle (AX, QB) = \measuredangle CBQ = \measuredangle BQC = \measuredangle (BQ, QA)$ . This implies that $W = BQ \cap AX$ lies on $(XPQ)$ . By symmetry, we similarly get $W$ lies on $PC$ . $\blacksquare$
As such, it follows that $\measuredangle QBM + \measuredangle BCP = \measuredangle QWP$ which implies that the circles $(BMQ), (CMP), (PXQ)$ concur at some point $D$ . Now, by triangle Miquel on $BXC$ we get that $(XYZD)$ is cyclic.

Claim: $AX, AD$ are tangents to $(XYZD)$ .
Proof. Since $\measuredangle MDQ = \measuredangle MBW = \measuredangle XWQ = \measuredangle XDQ$ we get that $M$ lies on $XD$ .
Note that since $\measuredangle AXD = \measuredangle AXM = \measuredangle DMB = \measuredangle DYB = \measuredangle DYX$ , it follows that $AX$ is a tangent.
By orthogonality it follows that $AD$ must also be a tangent. $\blacksquare$
As such, since $(BC;M\infty) \overset{X}= (YZ;DX) = -1$ this implies $A$ lies on $YZ$ .

This post has been edited 1 time. Last edited by YaoAOPS, Jun 21, 2024, 4:19 PM

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CyclicISLscelesTrapezoid

372 posts

CyclicISLscelesTrapezoid

#3 h Jun 21, 2024, 4:20 PM • 11 Y

Y by centslordm, CT17, khina, crazyeyemoody907, iamnotgentle, v4913, Rounak_iitr, cosdealfa, Yiyj1, Sedro, mrtheory

My problem!

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Let $\measuredangle$ denote directed angles modulo $180^\circ$ .

By Menelaus, we have
$\frac{BP}{PA} \cdot \frac{AQ}{QC} \cdot \frac{CM}{MB}=1,$ and since $BM=CM$ and $AP=AQ$ , this gives $BP=CQ$ . If we let $BP=CQ=x$ and $AP=AQ=y$ , then we have $x-y=AB$ and $x+y=AC$ , so $x=\tfrac{AB+AC}{2}=BC$ .

Let circle centered at $A$ through $X$ , $P$ , and $Q$ intersect $\overline{XM}$ again at $R$ .

Claim: $R$ lies on the circumcircles of $BMQ$ and $CMP$ .
Proof: We have
$\angle MRQ=180^\circ-\angle XRQ=\frac{\angle XAQ}{2}=\frac{180^\circ-\angle ACB}{2}=\angle MBQ,$ so $BMQR$ is cyclic. Analogously, $CMRP$ is cyclic. $\square$

Claim: $XYRZ$ is cyclic and its circumcircle is tangent to $\overline{AX}$ .
Proof: We have
$\measuredangle XYR=\measuredangle BYR=\measuredangle BMR=\measuredangle AXR,$ so the circumcircle of $XRY$ is tangent to $\overline{AX}$ . Analogously, the circumcircle of $XRZ$ is tangent to $\overline{AX}$ , so the circle through $X$ and $R$ tangent to $\overline{AX}$ passes through $Y$ and $Z$ , as desired. $\square$

Let $\infty_{BC}$ be the point at infinity on $\overline{BC}$ . We have $(X,R;Y,Z)\overset{X}{=}(\infty_{BC},M;B,C)=-1$ , so $XRYZ$ is harmonic. Since $\overline{AX}$ is tangent to its circumcircle, we know that $\overline{AR}$ is tangent as well, so $A$ , $Y$ , and $Z$ are collinear by the symmedian configuration. $\blacksquare$

This post has been edited 7 times. Last edited by CyclicISLscelesTrapezoid, Sep 2, 2024, 10:04 PM

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DottedCaculator

7355 posts

DottedCaculator

#4 h Jun 21, 2024, 4:23 PM • 3 Y

Y by centslordm, parola, mrtheory

By lengths, $AP=AQ=\frac{b-c}2$ so $BP=CQ=a$ , which gives $P=(2a:c-b:0)$ and $Q=(2a:0:b-c)$ . We get $X=(2a:b-c:c-b)$ . The circumcircle of $BMQ$ is $-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{(2b-a)(b-c)}4x+\frac12a^2z\right)=0$ so line $BX$ intersects this at $(2a:y:c-b)$ , implying
$y(-a^2(c-b)-2ac^2)-2ab^2(c-b)+\frac14(2a+c-b+y)((2b-a)(b-c)2a+2a^2(c-b))=0$ or
$y=\frac{(b-c)(c-2b)}{(b-2c)}.$
Similarly, $CX$ intersects this at $(2a:b-c:z)$ where $z=\frac{(c-b)(b-2c)}{c-2b}$ , so $A$ , $Y$ , and $Z$ collinear is equivalent to $\frac y{c-b}=\frac{b-c}z$ , or $yz=-(b-c)^2$ , which is true.

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#5 h Jun 21, 2024, 4:25 PM • 1 Y

Y by centslordm

Nice problem!
First we will prove that $BP=BC=CQ$ . Note that by Menelaus's theorem we have $1=\frac{BM}{MC} \cdot \frac{CQ}{QA} \cdot \frac{AP}{PB} = \frac{CQ}{PB},$ so $CQ=PB$ . Also $CQ+PB=(CQ+QA)+(PB-PA)=AC+AB=2BC \Rightarrow CQ=PB=BC.$

Let $D=(BMQ) \cap (CMP)$ . Now we will prove that $AD=AP=AQ$ . Note that this condition is equivalent to condition that $A$ is center of circle $(PQD)$ . Since $AP=AQ$ , it's enough to prove that $\angle PAQ=2 \angle PDQ$ . Note that $D$ is center of spiral similarity of segments $PQ$ and $CB$ , so $\angle PQD=\angle BCD$ and $\angle PCD=\angle QBD$ , so $\angle BDC=180^\circ-\angle DBC-\angle DCB=180^\circ - (\angle CBQ + \angle BCP)=180^\circ-(90^\circ - \frac{1}{2}\angle BCA + 90^\circ - \frac{1}{2}\angle CBA)=\frac{1}{2} (\angle ABC + \angle ACB)=\frac{1}{2}\angle QAP.$ So, really $\angle PAQ=2 \angle PDQ$ and $AP=AQ=AD$ .

Note that if $BQ \cap AX=E$ , then $\angle AEQ=\angle CBQ=\angle CQB=\angle AQE$ and so $AQ=AE \Rightarrow E \in (PQD).$ Now note that since $AD=AE=AP=AQ=AX$ , $PEQDX$ is cyclic and then $\angle AXD = \angle DQB = \angle DMB$ and so $M,D,X$ are collinear. By PoP we get $XY \cdot XB = XD\cdot XM = XC \cdot XZ$ , so $BYZC$ is cyclic.

We have $\angle DYX = \angle BMD = \angle CZD$ , so $XYDZ$ is cyclic. Now $\angle AXZ = \angle XCB = \angle ZYX$ , so $AX$ is tangent to circle $(XYDZ)$ . Since $AD=AX$ , line $AD$ is also tangent to circle $(XYDZ)$ . By this reason it is enough to prove that $XYDZ$ is harmonic quadrilateral.

Note that $\Delta XZD \sim \Delta XMC$ and $\Delta XDY \sim \Delta XBM$ , so $\frac{XZ}{ZD}=\frac{XM}{MC}=\frac{XM}{MB}=\frac{XY}{YD} \Rightarrow XZ \cdot DY = XY \cdot ZD$ and so we are done!

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#6 h Jun 21, 2024, 4:27 PM • 4 Y

Y by centslordm, Number_theory060222, GeoKing, Muaaz.SY

First, notice that $P$ and $Q$ both lie on line through $M$ parallel to the angle bisector of $\angle BAC$ . Thus, if $P_1 = 2P-B$ and $Q_1=2Q-C$ , then $AP_1=AC$ and $AQ_1=AB$ . Thus, $BP = CQ = \tfrac{AB+AC}2 = BC$

Now, let $XM$ intersect $\odot(XPQ)$ at $T$ . We have
$\angle QTM = 180^\circ - \angle QTX = \frac{\angle QAX}2 = 90^\circ - \frac{\angle C}2 = \angle QBM,$ so $T\in\odot(BMQ)$ . Similarly, $T\in\odot(CMP)$ . By power of point from $X$ , we get $BCYZ$ concyclic. Moreover, by Miquel's theorem, we get $XYZT$ concyclic.

Finally, notice that since $XM$ is median of $\triangle XBC$ , we get that $XT$ is symmedian of $\triangle XYZ$ . Thus, $XYZT$ is harmonic quadriltaral. Since we have $XA$ tangent to $\odot(XYZ)$ and $AX=AT$ , it follows that $A\in YZ$ , done.

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#7 h Jun 21, 2024, 4:38 PM • 4 Y

Y by centslordm, Number_theory060222, Rounak_iitr, ehuseyinyigit

Nice Problem! Looks like I over-complicated lol.
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dot((-2.8164648333091575,4.609786710663822),linewidth(4pt) + dotstyle); label("$F$", (-2.679478315490082,4.8589936808966225), NE * labelscalefactor); dot((-4.2792524285896185,4.728524257125601),linewidth(4pt) + dotstyle); label("$M_{BC}$", (-4.155040697525012,4.98730345324749), NE * labelscalefactor); dot((-3.7007739538427322,-6.2844624165595775),linewidth(4pt) + dotstyle); label("$M_A$", (-3.577646721946126,-6.015259525839423), NE * labelscalefactor); dot((-1.9476903294155312,-1.5312146350493545),linewidth(4pt) + dotstyle); label("$I$", (-1.8133873521217538,-1.267797948857315), NE * labelscalefactor); dot((-4.482095530124239,8.59022088296326),linewidth(4pt) + dotstyle); label("$L$", (-4.347505356051307,8.836596623773524), NE * labelscalefactor); dot((-5.453857578269934,-11.037710198069801),linewidth(4pt) + dotstyle); label("$I_A$", (-5.309828648682783,-10.794798545909249), NE * labelscalefactor); dot((-4.189099939018675,3.0122146456421968),linewidth(4pt) + dotstyle); label("$S$", (-4.058808368261864,3.255121526510775), NE * labelscalefactor); dot((-1.1932300134959146,3.169578521256201),linewidth(4pt) + dotstyle); label("$T$", (-1.075606161104289,3.4155087419493597), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
If $I_A$ is the excenter of $\triangle ABC$ and $W$ is reflection of $I_A$ in $M$ . Let $BW \cap AC = Q'$ and $CW \cap AB = P'$

$\textbf{Claim 1:}$ $P' \equiv P$ and $Q' \equiv Q$ .

$\textbf{Proof:}$ By reflection $\measuredangle Q'CB = 90 - \frac{C}{2}$ so $BC = CQ' = BP'$ ; $AP = BC-AB = AC-BC = AQ$ and by converse of menelaus : $\frac{BM}{MC} \cdot \frac{BP}{AP} \cdot \frac{AQ}{CQ} = 1$ , we get $M,P,Q$ are collinear which implies the claim as there exists an unique choice of $P,Q$ .

$\textbf{Claim 2:}$ $AW \parallel BC$ , $AW = AP=AQ$

$\textbf{Proof:}$ let $M_A$ denote the arc midpoint of arc $BC$ , by ptolemy $M_AB(AB+AC) = AM_A(BC)$ so $2M_AB=AM_A$ which implies $M_AI_A=M_AI=AI$ by incenter-excenter lemma.Let $AI \cap BC = D$ , then it is well known that $-1=(BA,BD;BI,BI_A) = (AD;II_A) = \frac{IA}{ID} \cdot \frac{I_AD}{I_AA} = \frac{3 \cdot ID}{I_AD}$ , therefore $AD = I_AD$ , now reflect $I_A$ about $B,C$ to get $I_A'$ and $I_A''$ then $\overline{AI_A'I_A''}$ is collinear and parallel to $BC$ as $I_AD=AD$ so we conclude $AW \parallel BC$ , now for $AW=AP$ consider homothety of half at $I_A$ to get $AW = 2MD = BC-AC=AP=AQ$

Let $L = CW \cap MM_A$ , let $D \in (ABC)$ such that $AD \parallel BC$ , let $T = PQ \cap AD$ , let $S = AD \cap MM_A$ .

$\textbf{Claim 3:}$ $L \in (CMPY)$ and $\overline{LPX}$ are collinear.

$\textbf{Proof:}$ $\measuredangle CLM = \measuredangle C/2 = \measuredangle (90- A/2 - B/2) = \measuredangle BPC - \measuredangle APQ = \measuredangle CPM$ . $90^{\circ}= \measuredangle CPL = \measuredangle (90-B/2) + \measuredangle (B/2) = \measuredangle CPB + \measuredangle BPX = \measuredangle CPX$
$\textbf{Claim 4:}$ $APQSM_{BC}$ is cyclic where $M_{BC}$ is arc midpoint of $BC$ containing $A$ .

$\textbf{Proof:}$ $APQM_{BC}$ is cyclic as $M_{BC}QC \equiv M_{BC}PB$ , notice that $\overline{MPQ}$ is simson line of $M_{BC}$ as $\measuredangle AQM_{BC} = 180-\measuredangle (90-A/2)-\measuredangle A/2 = 90$ so $(APQ)$ is circle with diameter $M_{BC}A$ which implies $S \in (APQ)$ as well.

$\textbf{Claim 5:}$ $PXYT$ is cyclic.

$\textbf{Proof:}$ $\measuredangle TPY = \measuredangle MPY = \measuredangle MCY = \measuredangle MCX = \measuredangle TXY$
$\textbf{Claim 6:}$ $STYZM$ is cyclic.

$\textbf{Proof:}$ It suffices to prove that $STYM$ is cyclic and the conclusion follows by symmetry.
$\measuredangle MYT = \measuredangle MYP - \measuredangle PYT = 180- \measuredangle MCP$ $- \measuredangle PXA = 180-\measuredangle (90-B/2) - \measuredangle (B/2) = 90^{\circ} = \measuredangle MST$
$\textbf{Claim 7:}$ $BCYZ$ is cyclic.

$\textbf{Proof:}$ $\measuredangle YSW = \measuredangle YST = \measuredangle YMP = \measuredangle YCW$ so $CSWY$ is cyclic, and similarly $SQZB$ is cyclic, so $XC \cdot XY = XS \cdot XW =XB \cdot XZ$

$\textbf{Claim 8:}$ $AX$ is tangent to $(XYZ)$

$\textbf{Proof:}$ $\measuredangle AXY = \measuredangle YCB = \measuredangle XZY$
Now we note that $S,T$ are inverses WRT $(APQ)$ because $A,S,T$ are collinear and $S \in (APQ)$ so $AS \cdot AT = AX^2$ so $A$ lies on radical axes of $(STYZM)$ and $(XYZ)$ , that is $YZ$ , hence we're done. $\blacksquare$

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#8 h Jun 21, 2024, 5:02 PM • 1 Y

Y by Rounak_iitr

Solution

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OronSH

#9 h Jun 21, 2024, 6:43 PM • 2 Y

Y by ihatemath123, Jack_w

Very nice, just add the reflection $T$ of $X$ over $A,$ the reflection of $A$ over $T,$ the midpoint of $AT,$ the reflection of $B$ over $C,$ the intersections of the perpendicular bisector of $BC$ with lines $BQ$ and $CP,$ the foot from $A$ to $BC,$ the intouch point, the extouch point, the incenter, the orthocenter, the centroid, the circumcenter and the nagel point and now it is trivial (what I did in contest)

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ddami

#10 h Jun 21, 2024, 9:55 PM

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Similarly to others we can prove that lines BQ and CP meet at a point D at the circumcircle of XPQ. Here is another way to finish

Let T be the intersection of PQ with AX.
Let A' and X' be the reflections of A and X through D, respectively
Let W be the reflection of X through A'.

After a few computations we can get A'M perpendicular to BC. Thus DX'CB is a trapezoid.
Note that XYQT and XZTP are cyclic
Then angles CZT, XPT, XDQ and the supplementary of DX'C are equal. Hence TZCX' is cyclic
Then XZ * XC = XT * XX' = XA * XW and thus angles XAZ and WCX are equal
Likewise we obtain angles XAY and WBX are equal.
Since XBCW is a trapezoid we get that angles WCX and WBX are equal, thus angles XAZ and XAY are equal. Conclusion follows

Note: The computations I did to get A'M perpendicular to BC involve the hypothesis AC + AB = 2BC. Thus if the tangency point of BC with the incircle of ABC divides BC in two segments of lenghts 2b < 2c, then AB = 3b + c, AC = b + 3c, AA' = 2c - 2b, and from there we may compute AM and the altitude from A to BC

This post has been edited 1 time. Last edited by ddami, Jun 21, 2024, 9:58 PM
Reason: misspelled

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Ywgh1

#11 h Jun 22, 2024, 9:54 AM • 1 Y

Y by Sedro

Any explanation to the title ?

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#12 h Jun 23, 2024, 3:08 AM

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#13 h Jun 23, 2024, 1:14 PM • 1 Y

Y by Rounak_iitr

Let me also add one sol. without details)
One can easily observe that $PB=BC=CQ$ (by Menelaus th). Let $BQ \cap PC=T$ , by cheva we have $X - A - T$ and simply we have $(XPQT)$ is cyclic. Next step is proving $(BYZC)$ is cyclic. Take the miquel point of $BMCTPQ$ and call that point $R$ . By angle chasing we have $\angle XRT=\angle TRM=90$ , so $X - R - M$ . Then by PoP $(BYZC)$ is cyclic, then $AX$ is tangent to $(XYZ)$ . Since $AX=AQ$ , it is enough to prove that $\angle AQZ=\angle ZYQ$ (because by proving this, we will prove that $A$ lies on radical axes of circles $(XYZ)$ and $(YZQ)$ , which means $A - Y - Z$ ). Let $PM \cap XT=F$ and $PM \cap CX=G$ , then $(PFZX)$ and $(YZQG)$ are cyclic, and rest follows from angle chasing, so we are done:).

This post has been edited 1 time. Last edited by X.Allaberdiyev, Jun 23, 2024, 1:16 PM
Reason: Typo

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#14 h Jul 5, 2024, 8:03 PM • 2 Y

Y by ehuseyinyigit, Rounak_iitr

CyclicISLscelesTrapezoid wrote:

My problem!

Let $\measuredangle$ denote directed angles modulo $180^\circ$ .

By Menelaus, we have
$\frac{BP}{PA} \cdot \frac{AQ}{QC} \cdot \frac{CM}{MB}=1,$ and since $BM=CM$ and $AP=AQ$ , this gives $BP=CQ$ . If we let $BP=CQ=x$ and $AP=AQ=y$ , then we have $x-y=AB$ and $x+y=AC$ , so $x=\tfrac{AB+AC}{2}=BC$ .

Let $T$ be the reflection of $X$ over $A$ . Notice that $APT \sim BPC$ and $AQT \sim CQB$ by SAS, so $\overline{BQ}$ and $\overline{CP}$ intersect at $T$ . Let the circumcircles of $BMQ$ and $CMP$ intersect again at $R$ .

Claim: $R$ lies on the circle centered at $A$ through $P$ , $Q$ , $X$ , and $T$ .
Proof: We have
$\measuredangle PRQ=\measuredangle PRM+\measuredangle MRQ=\measuredangle PCM+\measuredangle MBQ=\measuredangle CTB=\measuredangle PTQ,$ so $PTQR$ is cyclic, as desired.

Since $\measuredangle BMR=\measuredangle BQR=\measuredangle TQR=\measuredangle TXR$ , we obtain that $X$ , $R$ , and $M$ are collinear.

Claim: $XYRZ$ is cyclic and its circumcircle is tangent to $\overline{AX}$ .
Proof: We have
$\measuredangle XYR=\measuredangle BYR=\measuredangle BMR=\measuredangle AXR,$ so the circumcircle of $XRY$ is tangent to $\overline{AX}$ . Analogously, the circumcircle of $XRZ$ is tangent to $\overline{AX}$ , so the circle through $X$ and $R$ tangent to $\overline{AX}$ passes through $Y$ and $Z$ , as desired.

Let $\infty_{BC}$ be the point at infinity on $\overline{BC}$ . We have $(X,R;Y,Z)\overset{X}{=}(\infty_{BC},M;B,C)=-1$ , so $XRYZ$ is harmonic. Since $\overline{AX}$ is tangent to its circumcircle, we know that $\overline{AR}$ is tangent as well, so $A$ , $Y$ , and $Z$ are collinear by the symmedian configuration.

ADMITS
*throws water balloon*
i didnt see anyone throw water balloons at MOP this year, can someone confirm that tiger was hit with a water balloon?

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#15 h Sep 2, 2024, 10:37 PM

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Define $K$ and $L$ such that

$LB=LC$ and $KB=KC$ ,
$A,K,L$ lie on the same side of $BC$ ,
$\angle BLC=\angle ABC$ and $\angle BKC=\angle ACB$ .

Claim: $CQ=BC=BP$ .
Proof. By menelaus on $ABC$ and $P,Q,M$ , $1=\frac{BM}{MC}\cdot \frac{CQ}{QA}\cdot \frac{AP}{PB}=\frac{CQ}{PB}$ so $CQ=PB=y$ since $BM=MC$ , $AP=AQ=x$ . Now, $AB=BP-AP=y-x$ and $AC=y+x$ so $AB+AC=2y=2BC$ so $y=BC$ . $\blacksquare$

Claim: $K$ lies on $XP, BQ$ , and $(PMC)$ .
Proof. Since $\angle QBA=90^{\circ} - \angle \frac{\angle C}{2}$ from $CQ=CB$ , $K$ lies on $BQ$ as $\angle KBC=90^{\circ} - \frac{\angle C}{2}$ . Now, $\angle MQC=\frac{\angle A}{2}$ as $AP=AQ$ . Thus $\angle CMP = \angle CMQ = 180^{\circ} - \frac{\angle A}{2}-\angle C$ . As $\angle PCM=90^{\circ} - \frac{\angle B}{2}$ , $\angle MPC=180^{\circ} - \angle CMP - \angle PCM = \frac{\angle C}{2}$ But since $\angle MKC = \frac{\angle C}{2}$ , $K$ lies on $(CMP)$ .
Now, as $A$ is the circumcenter of $XPQ$ and $\angle XAQ=180^{\circ} - \angle C$ , we get $\angle XPQ=90^{\circ} - \frac{\angle C}{2}$ . But $\angle MPK=180^{\circ} - \angle KMC = 180 ^{\circ} - \left(90^{\circ} - \frac{\angle C}{2}\right) = 180 ^{\circ} - \angle XPQ,$ so $K$ lies on $XP$ . $\blacksquare$

Similar results hold for $L$ , so since $\measuredangle QLK=\measuredangle QLM=\measuredangle QBM=\measuredangle KBC=\measuredangle MCK=\measuredangle MPK=\measuredangle QPK,$ $PKQL$ is cyclic, so by radical axis, $X$ lies on the radical axis of $(PCM)$ , $(BMQ)$ , $(PKQL)$ .

Now, invert at $X$ with radius $\sqrt{\operatorname{pow}_{(PKQL)}X}$ which fixes $(PCM)$ , $(BMQ)$ , $(PKQL)$ . It sends $Y$ to $B$ , $Z$ to $C$ , and $(XPQ)$ to $KL$ , the perpendicular bisector of $BC$ . As $A$ is the circumcenter of $XPQ$ , $A$ goes to $X'$ , the reflection of $X$ over $KL$ . Thus $AYZ$ inverts to $XBCX'$ , an isosceles trapezoid, so inverting back we conclude.

This post has been edited 1 time. Last edited by GrantStar, Sep 2, 2024, 10:38 PM

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#16 h Nov 9, 2024, 4:41 PM

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Let $E=\overline{XQ}\cap (BMQ)$ , $F=\overline{XP}\cap (CMP)$
note that $\overline{PQ}\parallel \overline{AD}$ where $D$ is the intersection of $\angle BAC$ bisector and $BC$
from $\frac{CQ}{CA}=\frac{CM}{CD}$ we can get $CQ=BC=BP$
some angle chase shows that $M$ , $E$ , $F$ are collinear, $BYZC$ is cyclic and $\overline{BEF} \perp \overline{BC}$
Now let $R$ be the reflection of $X$ over $\overline {ME}$ , abviously $XBCR$ is cyclic.
$\angle QER=\angle XER=2\angle XEF=2\angle QBM=180-\angle QAX$
so $XA.XR=XY.XB$
To finish note that he inversion centered at $X$ with radius $\sqrt{XY.XB}$ sends $(XBCR)$ to a line $\overline{YZA}$ as needed.

This post has been edited 1 time. Last edited by Muaaz.SY, Nov 9, 2024, 4:44 PM

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awesomeming327.

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awesomeming327.

#18 h Jan 27, 2025, 5:28 AM

Y by

Claim 1: $BP=BC=QC$ .

By Menelaus, we have
$\frac{AP}{BP}\cdot \frac{BM}{CM}\cdot \frac{CQ}{AQ}=1$ But since $AP=AQ$ and $BM=CM$ , we must have $BP=CQ$ . Then note that
$BP+CQ=BA+AP+CA-CQ=BA+CA=2BC$ so our claim is proved.

Let $D$ be the reflection of $X$ across $A$ , which is on $(XPQ)$ . Then $\triangle PAD$ and $\triangle PBC$ are similar and $AD\parallel BC$ so $D$ lies on $CP$ . Similarly, it lies on $BQ$ . Then, by Miquel Point on complete quadrilateral $PDCMBQ$ : $(BQM)$ , $(CMP)$ , $(DPQ)$ , $(DBC)$ concur at a point, call it $S$ . We have
$\angle SYX=\angle SMB=\angle SZC$ so $(XYZ)$ also passes through $S$ .

Claim 2: $BCZY$ cyclic.

Note that since
$\angle DXS=\angle CPS=\angle BMS$ we have $S$ lies on $XM$ . Thus by radical axis we are done.

Claim 3: $AX$ and $AS$ are tangent to $(XYZ)$ .

We have $\angle AXZ=\angle XCB=\angle XYZ$ , so $AX$ is tangent. Since $AX=AS$ , $AS$ is also tangent.

Claim 4: $XYSZ$ is harmonic.

Note that since $\tfrac{XY}{XZ}=\tfrac{XC}{XB}$ it suffices to show that $SZ\cdot CX=SY\cdot BX$ . This is clear:
$SZ\cdot CX=[SXC]\csc(\angle SZX)=[SXB]\csc(\angle SYX)=SY\cdot BX$ because $XS$ is the median and because $\angle SZX$ and $\angle SYX$ are supplements.

Since $XYSZ$ is harmonic, $YZ$ passes through $A$ . We are done.

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#20 h Apr 10, 2025, 3:00 PM

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It's not all over , we are not rusted to dust :wallbash_red:

:wallbash_red:

ELMO 2024 P5 wrote:

In triangle $ABC$ with $AB<AC$ and $AB+AC=2BC$ , let $M$ be the midpoint of $\overline{BC}$ . Choose point $P$ on the extension of $\overline{BA}$ past $A$ and point $Q$ on segment $\overline{AC}$ such that $M$ lies on $\overline{PQ}$ . Let $X$ be on the opposite side of $\overline{AB}$ from $C$ such that $\overline{AX} \parallel \overline{BC}$ and $AX=AP=AQ$ . Let $\overline{BX}$ intersect the circumcircle of $BMQ$ again at $Y \neq B$ , and let $\overline{CX}$ intersect the circumcircle of $CMP$ again at $Z \neq C$ . Prove that $A$ , $Y$ , and $Z$ are collinear.

Tiger Zhang

Note that if $I$ is the Incentre ,then $PQ \parallel AI$ , using angle bisector theorem and parallel ratios we get $AP=AQ=AX=AX'=c-b$ , where $A$ is the midpoint of $XX'$ .This also gives $BP=BC=CQ$

Claim : $\overline{B-X'-Q}$ and $\overline{C-X'-P}$
Proof : Note that,
$\angle BQC=\frac{180^{\circ}-\angle ACB}{2}=90^{\circ}-\frac{\angle QCB}{2}=90^{\circ}-\frac{\angle QAX'}{2}=\angle AX'Q \implies B-X'-Q$ Similarly we also have $C-X'-P$ and hence our claim is proved $\square$

Now using miquel on complete quadrilateral $QX'CM$ we have that $\odot(\triangle BMQ),\odot(\triangle CMP),\odot(PX'QX)$ meet at a certain point.Call it $D$ .

Now we observe that ,
$\angle XDP=\angle XX'P=\angle MCP=180^{\circ}-\angle PDM \implies X \in MD$ From here using power of point gives, $XZ\cdot XC=XD \cdot XM=XY \cdot XB \implies BYZC$ is cyclic.

Claim : $XYDZ$ is cyclic and $AX$ is tangent to the circle.
Proof : We can angle chase both the parts;
$\angle XYD=\angle BMD=\angle CZD=180^{\circ}-\angle XZD \implies \text{ Part 1 }$ $\angle AXD=\angle XMB=\angle DMB=\angle XYD \implies \text{ Part 2 } \square$ Note that since $AX=AD$ ( $A$ is the centre of $(PX'QDX)$ ) , we get that $AD$ is tangent to $(XYDZ)$ . So it suffices to prove that the quadrilateral $XYDZ$ is harmonic. That directly follows from ,
$(Y,Z;X,D) \stackrel{X}=(B,C;M, \infty_{BC})=-1 \text{ } \blacksquare$

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#21 h May 1, 2025, 7:05 PM

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Let $H$ the orthocenter of $\triangle BIC$ where $I$ is the incenter of $\triangle ABC$ , let $HI \cap BC=D$ and let $D'$ reflection of $D$ over $I$ , redefine $IM \cap AH=X$ , also let altitude from $A$ to $BC$ hit $IM, BC$ at $G, L$ while letting $AI \cap BC=K$ and also let $AD' \cap BC=T$ , and let $I_A$ the A-excenter.
Claim 1: $AH \parallel BC$ .
Proof: First notice that from homothety $I_AT \perp BC$ so $-1=(A, K; I, I_A) \overset{\infty_{BC}}{=} (L, K; D, T)$ but we also have that $\frac{LD}{DK}=\frac{AI}{IK}=2$ so from the cross ratio we have $LK=KT$ thus from homothety also trivially notice $IM \parallel AT$ so $AGID'$ is a parallelogram where $AK$ bisects $GD'$ too which means that $GD' \parallel BC$ and thus from parallelograms we have $LD=GD'=TM$ and thus $DK=KM$ which gives that $MD' \parallel AI$ also remember it is well known that $ID, EF, AM$ are concurren at $J$ (you can drop parallel from $J$ to $BC$ and angle chase to win) and in addition since we now have $AD'MI$ is a parallelogram it happens that $J$ is the intersection of its diagonals and thus $AJ=JM$ and $IJ=JD$ however this means that $J$ lies on the A-midbase which from Iran Lemma spam is well-known to be the polar of $H$ w.r.t. incircle and we are done as by La'Hire it means $AH$ is polar of $J$ w.r.t. incircle and as seen to lie on $ID$ you can conclude $AH \perp ID$ so $AH \parallel BC$ as desired.
Claim 2: $BH \cap AC=Q$ and $CH \cap AB=P$ .
Proof: Define those points as seen above, then it is clear that $BP=BC=CQ$ by reflections and however from thales we can also have $AP=AQ=AH$ and now to finish you simply throw menelaus to see that $P,Q,M$ are colinear and in fact $D'$ lies on this line too as seen above.
The finish: Now just note that since $DK=KM$ from thales we have $XA=AH$ so $X$ is also the one from the problem statement and now if you let $N$ to be the H-queue point of $\triangle BHC$ then as $\angle XNH=90$ we have $AN=AH=AX$ too but also notice that $N$ is miquelpoint of $BPQC$ as a result and thus it lies on both $(CMP), (BMQ)$ and also on $IM$ too.
So now just notice from Miquelpoint theorem and PoP that $BYNM, MNZC, BYZC, XYNZ$ are all cyclic so we also have that $AX$ is tangent to $(XYNZ)$ but also $-1=(B, C: M, \infty_{BC}) \overset{X}{=} (Y, Z; N, X)$ which shows that not only $AN$ is tangent to $(XYNZ)$ it now also with this shows that $Y,Z,A$ are colinear as desired thus we are done :cool:

:cool:

.

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#22 h Yesterday at 2:12 PM

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Nice problem :love:

:love:

By Ceva's theorem, we have $PB = BC = CQ$ (see 2022 AIME I P14).

Let $R$ be the intersection of lines $BQ$ and $CP$ —equivalently, $R$ is the orthocenter of $\triangle BIC$ . By angle chasing, $R$ lies on $(PQX)$ . Let $W$ be the second intersection of $(MBQ)$ and $(MCP)$ , so $W$ is the Miquel point of $PQBC$ . In particular, $W$ lies on $(PQX)$ and $(BRC)$ .

Claim: $W$ is the $R$ -queue point in $\triangle RBC$ .
Proof: Let $Q'$ and $P'$ be the midpoint of $\overline{BQ}$ and $\overline{CP}$ , respectively. Then, since $W$ is the Miquel point of quadrilateral $PQBC$ , it is also the Miquel point of quadrilateral $P'Q'BC$ . But $P'$ and $Q'$ are the feet from $B$ and $C$ to opposite sides in $\triangle RBC$ , so by definition $W$ is the $R$ -queue point.

Claim: Points $W$ , $X$ and $M$ are collinear.
Proof: From the above claim, it follows that $W$ , $I$ and $M$ are collinear. So, it suffices to show that $M$ , $I$ and $X$ are collinear. Let $N$ be the circumcenter of $\triangle BIC$ . Point $R$ lies on line $AX$ because $IR = 2MN$ and then some dumb $AB + AC = 2BC$ stuff. So, $R$ and $X$ are reflections about $A$ . Some dumb mass point stuff probably finishes.

Claim: We have that $(XYZ)$ is tangent to $\overline{AX}$ .
Proof: We have $\measuredangle XYZ = \measuredangle BCX = \measuredangle AXZ$ .

Point $W$ lies on $(XYZ)$ by Miquel's theorem on $\triangle XBC$ . Projecting $BMC \infty$ through $X$ onto $(XYZ)$ gives us that $XYWZ$ is harmonic. Since $\overline{XA}$ is tangent to $(XYZ)$ and $AX = AW$ , it follows that $\overline{WA}$ is also tangent to $(XYZ)$ . The collinearity follows.

remark

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