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Source: Iranian TST 2020, second exam day 2, problem 4

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164 posts

#1 h Mar 12, 2020, 10:54 AM • 7 Y

Y by Purple_Planet, itslumi, jhu08, mathematicsy, Mango247, Mango247, ItsBesi

Let $ABC$ be an isosceles triangle ( $AB=AC$ ) with incenter $I$ . Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$ . $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$ , respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$ . Prove that $AD$ , $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia

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#5 h Mar 12, 2020, 12:28 PM • 8 Y

Y by mueller.25, amar_04, Mathasocean, Purple_Planet, agwwtl03, IFA, jhu08, sabkx

Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$ . Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$ ). Now since $\angle NCH=\angle NHC=\angle ABC$ $\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$ . But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$ . Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$ . But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$ . Done $\blacksquare$ .

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Mathematicsislovely

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Mathematicsislovely

#7 h Mar 26, 2020, 9:56 PM • 3 Y

Y by Purple_Planet, jhu08, PRMOisTheHardestExam

CLAIM: $N$ is the centre of $\omega=\odot(CIQ)$ .
Proof: If $N'$ is the centre of $\omega$ then $N'C=N'I$ and $\angle N'IA=90^\circ$ implies $N'I$ and $BC$ are parallel .Let $IN'$ cut $AC$ at $N''$ .Then $\angle N''IC=\angle ICB=\angle ICN''$ .It means $N''C=N''I$ .So $N''\equiv N'$ .Now the circle centre at $N''$ and with radius $N''C=N''I$ is the unique circle tangent to $AI$ at $I$ .So $N''=N$ ..
$\square$

Now let $\omega=\odot(CIQ)$ cut $BC$ at $X$ then $NX=NC$ implies $\angle NXH=\angle NXC=\angle ABC$ so $NX$ and $AB$ are parallel.So the line joining $C$ and the intersection of $BN$ and $AX$ passes through midpoint of $AB$ .In other words $AX$ , $BN$ , $CM$ are concurrent at a point.So let $F$ is the intersection of $MN$ and $BC$ .So $(B,C;X,F)=-1$ and let M be the midpoint of $BC$ .Then $FC.FB=FX.FM$ [since $F$ is the inverse image of $X$ w.r.t circle with diameter $BC$ ].

Now assume $Y= XN \cap AD$ .We claim that $XYDC$ is cyclic.Indeed, as $XN$ and $AB$ are parallels so $\angle XYD=\angle BAD=180^\circ-\angle XCD$ .
Now as $XY$ passes through center $N$ of $omega=\odot(CIQ)$ so $\angle ADX=\angle XDY=90^\circ$ and together with $\angle AMXY=90^\circ$ We get $AMXD$ is cyclic.
Let $F'= BC\cap AD$ .So $F'D.F'A=F'C.F'B$ and $F'D.F'A=F'X.F'M$ . So we have $F'C.F'B=F'X.F'M$ .But we have previously proved that $FC.FB=FX.FM$ so $F \equiv F'$ . So $BC,AD,MN$ concur at a point $F$ .

This post has been edited 1 time. Last edited by Mathematicsislovely, Mar 26, 2020, 9:58 PM

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2125 posts

#8 h Apr 12, 2020, 8:48 AM • 7 Y

Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam

Solution

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558 posts

#9 h Jun 30, 2020, 12:24 PM • 5 Y

Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam

Solution

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579 posts

#10 h May 17, 2021, 8:37 PM • 2 Y

Y by jhu08, PRMOisTheHardestExam

Claim: N is the circumcenter of (CIQ)
Proof:
Note that $IN\perp, AI, BC\perp AI \Longrightarrow NI\parallel BC$ . Thus, $\angle NIC = \angle BCI = \angle ICA = \angle ICN$ Thus, $NI=NC$ , so the circle with center $N$ and radius $NI$ is both tangent to $AI$ at $I$ and passes through $C$ , so we have sufficiently redefined $\omega$ . $\square$

Now, define $X=\omega \cap BC$ and $Y=\omega \cap AD$

Claim 1: $NX\parallel AB$
This clearly follows from
$\angle CNX = \angle NCX = \angle ACB=\angle ABC$ $\square$

Claim 2: $XY\parallel AB$
We angle chase (basically a rederivation of Reim's)
$\angle DAB = \angle DCB = \angle DCX = \angle DYX$ $\square$

Combining these, we have that $N,X,Y$ are collinear. Since, $X,Y\in \omega$ and $N$ is the center of $\omega$ , we have that $N$ is the midpoint of $XY$ . Since $M$ is the midpoint of $AB$ , combined with $YX\parallel AB$ , we have that $AY,MN,BX$ are concurrent which finishes $\blacksquare$

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#11 h Aug 13, 2021, 3:04 PM • 2 Y

Y by jhu08, PRMOisTheHardestExam

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Remarks

Attachments:

This post has been edited 1 time. Last edited by SatisfiedMagma, Aug 13, 2021, 3:06 PM

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#12 h Aug 13, 2021, 4:35 PM • 4 Y

Y by jhu08, SatisfiedMagma, PRMOisTheHardestExam, vrondoS

$\angle QIA = \angle QCI = \angle ICB = 90^\circ - \angle (IC , \text{angle bisector}) \implies \angle QIC = 90^\circ \implies N$ is the circumcenter of $\omega$ .
Let $E=BC \cap \omega$ and $F=AD \cap \omega$ .
$\angle FEC = 180^\circ - \angle FDC = \angle ABC = \angle NCE = \angle NEC$ Thus $\overline{E-N-F}$ are collinear also $EF \parallel AB$ so by homothety we are done.

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2811 posts

#13 h Sep 25, 2021, 5:05 PM • 2 Y

Y by PRMOisTheHardestExam, jhu08

To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$ . We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$ . By Reim's Theorem, we get that $EF \parallel AB$ . To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

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RM1729

#14 h Nov 9, 2021, 6:13 PM • 2 Y

Y by PRMOisTheHardestExam, jhu08

Consider $P$ as $AI \cap BC$ .

Claim
$N$ is the centre of the circle $\omega$

Proof
We angle chase.

Note that since $\angle IQC = \angle PIC = 90^{\circ} - \angle ICP = 90^{\circ} - \angle C/2$ and $\angle QCI = \angle C/2$

We have that $\angle QIC = 90^{\circ}$ and thus $QC$ is a diameter

But since $N$ is the midpoint of $QC$ it must be the centre of the circle

We now define $X$ as $AD\cap \omega (\neq D)$ and $Y$ as $BC\cap \omega (\neq C)$

Note that $\angle NXC = \angle NCX$ but $\angle NCX = \angle ABC$ since the triangle is isosceles. Thus $\angle NXC = \angle ABC$ and so $NX||AB$

However we also have that $XY||AB$ by Reim's Theorem. Combining these we have that $ABXY$ is a trapezium and $M$ and $N$ are the midpoints of its parallel sides. A simple homothety argument proves that $AY, BX,MN$ that is $AD,BC,MN$ are concurrent.

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#15 h Dec 6, 2021, 5:30 AM • 1 Y

Y by jhu08

srisainandan6 wrote:

To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$ . We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$ . By Reim's Theorem, we get that $EF \parallel AB$ . To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?

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DottedCaculator

7356 posts

DottedCaculator

#16 h Dec 23, 2021, 3:34 PM

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Solution

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#17 h Feb 14, 2022, 4:19 PM • 2 Y

Y by PRMOisTheHardestExam, Lantien.C

Here's a solution with radical axes ( $\overline{MN}$ will become radical axes of two circles)

Let $\Omega = \odot(ABC)$ and $A',C'$ be antipode of $A,C$ wrt $\Omega$ . $\angle QIC = \angle AIC - \angle ACI = 90^\circ$ . Thus $\overline{QC}$ is a diameter of $\omega$ , consequently $N$ is its center. So $\overline{CO},\overline{QD}$ intersect at $C'$ . Let $T = \overline{AD} \cap \overline{BC}$ , $\Gamma = \odot(BIC)$ and $X$ be a point of $\overline{AC}$ such that $\overline{QD}$ bisects $\angle ADX$ , let $\gamma = \odot(AXD)$ . hand drawn figure
$[asy] size(250); pair A=dir(90),B=dir(-130),C=dir(-50),O=(0,0),I=incenter(A,B,C),N=extension(A,C,I,I+C-B),Q=2*N-C,Cp=-C,D=foot(C,Cp,Q),T=extension(A,D,B,C),M=1/2*(A+B),Ap=-A,X=foot(I,A,C); draw(unitcircle^^circumcircle(C,I,Q),cyan); draw(CP(circumcenter(A,X,D),X,-210,-60 ),purple); draw(CP(Ap,I,-10,190 ),purple); dot("$A$",A,dir(110)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$O$",O,dir(0)); dot("$I$",I,dir(130)); dot("$N$",N,dir(-120)); dot("$Q$",Q,dir(50)); dot("$C'$",Cp,dir(Cp)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$T$",T,dir(T)); dot("$A'$",Ap,dir(Ap)); dot("$X$",X,dir(50)); draw(C--A--B--T^^A--Ap,red); draw(M--T,purple); draw(T--A^^Cp--D--X,green); draw(C--Cp,green); [/asy]$
Claim: Points $M,N,T$ lie on radical axes $\ell$ of $\Gamma,\gamma$ .
Proof: $T \in \ell$ is direct. For $M$ : Note $\Gamma$ is tangent to $\overline{BC}$ as it has center $A'$ . Note $\angle OAC = \angle OCA = \angle C'CA = \angle C'DA = \angle QDA$ As $\angle BAC = 2 \angle OAC = 2 \angle QDA = \angle XDA$ , so $\gamma$ is tangent to $\overline{AB}$ . Hence power of $M$ wrt both $\Gamma,\gamma$ equals $MB^2 = MA^2$ . For $N$ : We will mostly focus of $\triangle ADX$ . Observe
$\angle NDX = \angle NDQ - \angle XDQ = \angle NQD - \angle ADQ = \angle QAD = \angle NAD$ So $\overline{ND}$ is tangent to $\gamma$ (experts may also directly note that $N$ is the center of $D$ -appolonius circle wrt $\triangle ADX$ ). Hence $NX \cdot NA = ND^2 = NC^2$ . Laslty, recall $\overline{NC}$ is tangent to $\Gamma$ (as $A'$ is its center). This proves our Claim. $\square$

It follows points $M, N,T$ are collinear, solving our problem. $\blacksquare$

Motivation

This post has been edited 1 time. Last edited by guptaamitu1, Feb 14, 2022, 4:19 PM

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#18 h Mar 7, 2022, 8:24 AM • 1 Y

Y by teomihai

Simple angle chasing gives us that $N$ is the center of $(IDC)$ .
Let $AD\cap BC=S$ . We have
$\angle ASC=\angle ACB-\angle CAD=\angle ABC-\angle CAD=\angle ACD$ Hence, $NC$ is tangent to $(DCS)$ . Since $|NC|=|ND|$ , we know that $ND$ is tangent to $(DCS)$ as well. Then, $SN$ is symmedian in $DSC$ . Since $DC$ and $AB$ are antiparallels wrt $DSC$ , we find that $SN$ bisects $AB$ , done.

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#19 h Mar 7, 2022, 8:36 AM

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A kind of generalization

This post has been edited 1 time. Last edited by BarisKoyuncu, Mar 7, 2022, 8:36 AM

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#20 h Mar 7, 2022, 10:27 AM • 1 Y

Y by MrOreoJuice

BarisKoyuncu wrote:

A kind of generalization

Proof. All angles are oriented. From $\angle ADB=\angle ACB=\angle CBA=\angle SBA$ we obtain $\angle CDN=\angle NCD=\angle ABD=\angle CBS.$ Therefore $SN$ is a symmedian in $CDS$ and so bisects $AB.$

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dusicheng20080513

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dusicheng20080513

#21 h Mar 7, 2022, 11:09 AM

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GeoMetrix wrote:

Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$ . Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$ ). Now since $\angle NCH=\angle NHC=\angle ABC$ $\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$ . But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$ . Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$ . But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$ . Done $\blacksquare$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.006881897432205, xmax = 25.634412731501875, ymin = -8.577795838543958, ymax = 14.83226710342647; /* image dimensions */ pen ffqqff = rgb(1,0,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.5534973808986277,9.160392608982256)--(-3.5857910956748746,-1.8868970241407366), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((-3.5857910956748746,-1.8868970241407366)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw(circle((5.807241390273251,2.031588419358404), 3.8827498197705226), linewidth(0.4) + wvvxds); draw(circle((1.9067865071150452,2.276975553541132), 6.892477317034895), linewidth(0.4) + dtsfsf); draw(circle((3.5081332893469326,4.891548128823499), 4.6950649335634775), linewidth(0.4) + wvvxds); draw((8.19899262334367,5.090227365365952)--(5.807241390273251,2.031588419358404), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(8.19899262334367,5.090227365365952), linewidth(0.4) + qqffff); draw((8.19899262334367,5.090227365365952)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + wrwrwr); draw((3.8176841460738786,5.365866172483034)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw((1.9295955015644004,1.832569513956508)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((5.807241390273251,2.031588419358404)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((1.9295955015644004,1.832569513956508)--(2.2600756333314593,-4.606441501899994), linewidth(0.4) + ffqqff); draw((2.2600756333314593,-4.606441501899994)--(8.19899262334367,5.090227365365952), linewidth(0.4) + green); draw((-1.0161468573881236,3.63674779242076)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.5534973808986277,9.160392608982256)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.9295955015644004,1.832569513956508)--(5.807241390273251,2.031588419358404), linewidth(0.4) + wrwrwr); draw((7.796798634472622,-1.3026893337662244)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); /* dots and labels */ dot((1.5534973808986277,9.160392608982256),dotstyle); label("$A$", (1.6396828828887742,9.383171533517269), NE * labelscalefactor); dot((-3.5857910956748746,-1.8868970241407366),dotstyle); label("$B$", (-3.4819992193005715,-1.6553396638953617), NE * labelscalefactor); dot((7.796798634472622,-1.3026893337662244),dotstyle); label("$C$", (7.88392544583195,-1.0706727572527435), NE * labelscalefactor); dot((1.9295955015644004,1.832569513956508),linewidth(4pt) + dotstyle); label("$I$", (2.0138697031400508,2.01636850982028), NE * labelscalefactor); dot((5.807241390273251,2.031588419358404),linewidth(4pt) + dotstyle); label("$N$", (5.896057963247044,2.2268485962116227), NE * labelscalefactor); dot((3.8176841460738786,5.365866172483034),linewidth(4pt) + dotstyle); label("$Q$", (3.9081904806621375,5.547756625941694), NE * labelscalefactor); dot((8.19899262334367,5.090227365365952),linewidth(4pt) + dotstyle); label("D", (8.28149894234893,5.267116510753237), NE * labelscalefactor); dot((-1.0161468573881236,3.63674779242076),linewidth(4pt) + dotstyle); label("$M$", (-0.9328515063387511,3.8171425822795437), NE * labelscalefactor); dot((17.798762343332783,-0.7893418218123583),linewidth(4pt) + dotstyle); label("Q", (17.893422887553594,-0.602939231938649), NE * labelscalefactor); dot((4.675148007685625,3.928851637608016),linewidth(4pt) + dotstyle); label("$F$", (4.773497502493214,4.121169373733705), NE * labelscalefactor); dot((4.169500681742824,-1.488859213336944),linewidth(4pt) + dotstyle); label("$H$", (4.258990624647709,-1.3045395199097907), NE * labelscalefactor); dot((2.2600756333314593,-4.606441501899994),linewidth(4pt) + dotstyle); label("$J$", (2.3646698471256222,-4.414967463248519), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

When you use Geogebra-Asymptote conversion, how did you doit. Why did mine have an error?

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656 posts

#22 h Mar 7, 2022, 11:31 AM

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Firstly you have convert it from Geogebra Classic and not Geogebra Geometry. If you find easier to draw in Geogebra Geomtery (as I also do), then save it and reopen it in Geogebra Classic. After the conversion, you have to do some slight edits. Like changing the size and defaultpen.

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#23 h Apr 14, 2022, 4:02 PM • 1 Y

Y by Lantien.C

We begin with the following key claim.

Claim: $N$ is the center of $\omega$ .
Proof: Let $N'$ be the intersection of the perpendicular to $\overline{AI}$ at $I$ with $\overline{AC}$ . Then we have
$\angle N'IC=\angle ICB=\angle ICA=\angle ICN',$ so $N'I=N'C$ and thus $N'$ is the center of $\omega$ . Then $Q$ is the reflection of $C$ over $N'$ , hence $N'$ is the midpoint of $\overline{CQ}$ and we have $N=N'$ , which implies the desired result.

Now let $P \neq D$ be the intersection of $\omega$ with $\overline{AD}$ , $R \neq C$ be the intersection of $\omega$ with $\overline{BC}$ , and $X=\overline{AD} \cap \overline{BC}$ . By Reim's, we have $\overline{AB} \parallel \overline{PR}$ , so
$\angle QPR=\angle QCR=\angle ABC=\angle PRC,$ so we have $\overline{PQ} \parallel \overline{CR}$ as well. Then we have $90^\circ=\angle QPC=\angle PCR$ , so $CPQR$ is a rectangle, and $N$ is the midpoint of $\overline{PR}$ . But triangles $\triangle XAB$ and $\triangle XPR$ are homothetic, which implies that $\overline{MN}$ passes through $X$ as well, hence $\overline{AD}$ , $\overline{MN}$ , and $\overline{BC}$ concur. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Apr 14, 2022, 4:02 PM

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#24 h Jun 5, 2023, 3:37 PM

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mohamad021 wrote:

srisainandan6 wrote:

To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$ . We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$ . By Reim's Theorem, we get that $EF \parallel AB$ . To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?

JI//FH，as the picture says

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3448 posts

#25 h Sep 8, 2023, 3:03 AM • 1 Y

Y by happypi31415

By angle chasing, $\angle QIC = 90^{\circ}$ , hence $\angle QDC = 90^{\circ}$ , hence $\angle ADQ = 90 - \angle B$ , hence $(ADQ)$ is tangent to $\overline{AI}$ .

If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$ , then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$ , hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$ , and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$ , hence they all concur by the radical axis theorem.

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614 posts

trk08

#26 h Oct 16, 2023, 6:45 PM

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Claim:
$N$ is the center of $(CIQ)$
Proof:
By angle-chasing, we can see that:
$\angle CQI=\angle CIF=90^{\circ}-\angle ICF=90^{\circ}-\angle ICQ.$ Therefore, $\angle QIC=90^{\circ}$ , or $N$ is the center $\blacksquare$

Claim:
The line parallel to $AB$ that passes through $N$ goes through $E$ and $F$ .
Proof:

Denote $E=(CQD)\cap BC$ and $F=(CQD)\cap AD$ . By Reim's theorem, $EF$ is parallel to $AB$ .

Also, $\triangle{NEC}$ is isosceles, so it is similar to $\triangle{ABC}$ . As a result, $NE\parallel AB$ . Therefore, $N,E,F$ are all collinear and parallel to $AB$ $\blacksquare$

As a result, we can take a homothety at $T=AD\cap BC$ , which sends $N$ to $M$ , implying the desired result $\square$

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7585 posts

#27 h Oct 29, 2023, 1:54 AM

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Note that $N$ is the center of $\omega$ . Let $E$ be the other intersection of $\omega$ with $BC$ and let $E'$ be its antipode wrt $\omega$ . Let $F$ be the midpoint of $BC$ . It suffices to show that $AFED$ is cyclic as this implies $A,E',D$ collinear.

Here's the interesting part.

Claim: Let $ABCD$ be a cyclic quadrilateral. Let $F$ and $E$ be points where $B,F,E,C$ are on $BC$ in that order. If $\angle BAF=\angle CDE$ then $AFED$ is cyclic.
Proof: Angle chase.

Now we're done; just apply the claim here.

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1030 posts

bjump

#28 h Nov 24, 2023, 2:32 PM

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Claim: $N$ is the center of $\omega$ .
Proof Let $O$ denote the center of $\omega$ . $\angle ICQ= \angle ICB = \tfrac{1}{2} \angle QOI$ . Since $OI \perp AI$ , then $OI \parallel BC$ . Since $\angle QOI = \angle ICQ + \angle ICB= \angle BCA$ , $O$ lies on $AC$ . Since $OQ=OC$ , then $O$ is the midpoint of $CQ$ . So $O= N$ . $\square$

By Reims on $\omega$ and $(ABC)$ with lines $AD$ and $BC$ we get that the line through $D$ parallel to $AB$ intersects $BC$ on $\omega$ at a point we will call $P$ then let $DP\cap MN= Q$ then by homothety $BP$ , $AD$ , $MQ$ are concurrent so $MN$ , $BC$ , and $AD$ are concurrent.

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LuciferMichelson

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LuciferMichelson

#29 h Nov 24, 2023, 5:01 PM

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Define $R$ as midpoint of $BC$
Easy to see $N$ is center of $DQIC$ .
After that define $S=(DIQ) \cap BC$
Angle chasing shows $SN//AB$ so $AS,BN,CM$ concurrent.
Now we should show that $(B,C;S,AD \cap BC)=-1$
Let $AD \cap BC= K'$ so $(B,C;S,AD \cap BC)=-1$ is equal to $KD.KA=KS.KR$ and from angle chasing it is easy that show $ADSR$ is cyclic.

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799 posts

#30 h Dec 23, 2023, 9:18 PM

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If $N'$ is the intersection of $AC$ with the line through $I$ parallel $BC$ , we know $N'I \perp AI$ and
$\angle ACI = \angle ICB = \angle NIC,$
so $N'$ is the center of $\omega$ . Therefore $N'$ is the midpoint of $CQ$ , so $N' = N$ .

Next we define the intersections of $AC$ and $BC$ with $\omega$ as $K$ and $L$ . Looking at $CD$ , Reim's tells us that $AB \parallel KL$ , which then gives
$\angle LKQ = \angle LCQ = \angle ABC = \angle KLC,$
so $KQ \parallel BC$ as well. As a result, $\angle KCL = \angle QKC = \angle QLC = 90$ , so $KL$ is a diameter of $\omega$ , and hence passes through $M$ . We finish by noting the homothety which maps $KL$ to $AB$ also maps corresponding midpoints $N$ to $M$ .

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#31 h Mar 15, 2025, 12:38 AM

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We begin with a simple, but important claim.

Claim: $N$ is the center of $\omega$ .

Proof: Let $\angle IAC = \alpha$ . Note that $\angle ICA = 90-\tfrac{\alpha}{2}$ , and we also have $\angle ICA = \angle AIQ$ from the tangency condition. Therefore,

$\angle CQI = \angle IAQ+\angle AIQ = \alpha + \left(90-\frac{\alpha}{2} \right) = 90+\frac{\alpha}{2},$
which means $\angle CQI + \angle ICQ = 90^\circ$ , or $\angle CIQ = 90^\circ$ . This means that $\overline{CQ}$ is a diameter of $\omega$ , and the midpoint of $\overline{CQ}$ is the center of $\omega$ , which is $N$ . $\square$

Let $C' \neq C = \overline{AC} \cap \omega$ and $D' \neq D = \overline{AD} \cap \omega$ . Note that lines $\overline{AB}$ and $\overline{CD}$ are antiparallel and $\overline{CD}$ and $\overline{C'D'}$ are also antiparallel. By Reim's Theorem, we have $\overline{AB} \parallel \overline{C'D'}$ .

It suffices to show that $N$ is the midpoint of $\overline{C'D'}$ ; if so, the three lines will concur at the center of the homothety that maps $\overline{AB}$ to $\overline{D'C'}$ . However, we simply angle chase to find

$\angle NC'C = \angle NCC' = \angle ACB = \angle ABC,$
so $\overline{NC'} \parallel \overline{AB}$ . This means $N$ lies on $\overline{C'D'}$ , and $N$ is the center of $\omega$ , which implies $N$ bisects $\overline{C'D'}$ . $\blacksquare$

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zuat.e

#32 h Apr 23, 2025, 3:22 PM

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First, let $N'$ be the center of $(CQDI)$ . Note that $IN'\parallel BC$ , therefore $\measuredangle N'CI=\measuredangle CIN'=\measuredangle ICB$ , hence $N'$ lies on $AC$ and consequently $N'\equiv N$ .

Let $R = AD\cap BC$ and let $E,F = BC, AD\cap (CQDI)$ . We will prove that $R-N-M$ are collinear. The main claim is the following:
Claim: $EF$ is a diameter of $(CQID)$ and parallel to $AB$
Proof: $\measuredangle CEN=\measuredangle NCE=\measuredangle ACB=\measuredangle CBA$ and if we define $F'=EN\cap AD$ , $\measuredangle EFD=\measuredangle BAR=\measuredangle RCD$ , from which it follows $F=F'$ and $EF$ is a diameter of $(CQID)$ parallel to AB.

Now consider the homothety centered at $R$ sending $\triangle RFE$ to $\triangle RAB$ . As both $RM$ and $RN$ are the respective medians of $\triangle RFE$ and $\triangle RAB$ ,
$X_R: N\mapsto M$ hence $R-N-M$ are collinear, as desired.

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753 posts

#33 h May 3, 2025, 2:30 AM

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ihatemath123 wrote:

By angle chasing, $\angle QIC = 90^{\circ}$ , hence $\angle QDC = 90^{\circ}$ , hence $\angle ADQ = 90 - \angle B$ , hence $(ADQ)$ is tangent to $\overline{AI}$ .

If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$ , then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$ , hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$ , and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$ , hence they all concur by the radical axis theorem.

This solution is amazing!

I was wondering if a radical axis solution was possible

This post has been edited 1 time. Last edited by happypi31415, May 3, 2025, 2:31 AM

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#36 h May 3, 2025, 11:34 PM

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Let $AI \cap BC = T$ , clearly
$\angle QIC = 180^\circ - \angle AIQ - \angle CIT = 180^\circ - \angle ACI - \angle CIT = 90^\circ$ so $N$ is the center of $(CIQ)$ . Now define $X = AD \cap BC$ , we require $N \in XM$ to finish. Since $XM$ is the $X$ -median in $\triangle XAB$ it must be the $X$ -symmedian of $\triangle XCD$ . But
$\begin{align*} \angle CXD &= \angle ACB - \angle CAX \\ &= \angle ACB - (180^\circ - \angle ADC - \angle ACD) \\ &= \angle ACB - \angle ABC + \angle ACD \\ &= \angle NCD = \angle NDC. \end{align*}$ Hence, $NC$ and $ND$ are tangents to $(XCD)$ , so $N$ lies on the $X$ -symmedian of $\triangle XCD$ . Therefore, $N \in XM$ and we're finished.

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