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Italian WinterCamps test07 Problem4
mattilgale   90
N an hour ago by mathwiz_1207
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
90 replies
mattilgale
Jan 29, 2007
mathwiz_1207
an hour ago
Iran TST P8
TheBarioBario   8
N 2 hours ago by Mysteriouxxx
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
8 replies
TheBarioBario
Apr 2, 2022
Mysteriouxxx
2 hours ago
IMO 2010 Problem 6
mavropnevma   42
N 2 hours ago by awesomeming327.
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
42 replies
mavropnevma
Jul 8, 2010
awesomeming327.
2 hours ago
PJ // AC iff BC^2 = AC· QC
parmenides51   1
N 3 hours ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1998 OMM P5
The tangents at points $B$ and $C$ on a given circle meet at point $A$. Let $Q$ be a point on segment $AC$ and let $BQ$ meet the circle again at $P$. The line through $Q $ parallel to $AB$ intersects $BC$ at $J$. Prove that $PJ$ is parallel to $AC$ if and only if $BC^2 = AC\cdot QC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
3 hours ago
Self-evident inequality trick
Lukaluce   10
N 3 hours ago by ytChen
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
10 replies
Lukaluce
Sunday at 3:34 PM
ytChen
3 hours ago
Power Of Factorials
Kassuno   181
N 3 hours ago by SomeonecoolLovesMaths
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
181 replies
Kassuno
Jul 17, 2019
SomeonecoolLovesMaths
3 hours ago
Gergonne point Harmonic quadrilateral
niwobin   4
N 4 hours ago by on_gale
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
4 replies
niwobin
May 17, 2025
on_gale
4 hours ago
NCG Returns!
blacksheep2003   64
N 4 hours ago by SomeonecoolLovesMaths
Source: USEMO 2020 Problem 1
Which positive integers can be written in the form \[\frac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}\]for positive integers $x$, $y$, $z$?
64 replies
blacksheep2003
Oct 24, 2020
SomeonecoolLovesMaths
4 hours ago
Binomial stuff
Arne   2
N 4 hours ago by Speedysolver1
Source: Belgian IMO preparation
Let $p$ be prime, let $n$ be a positive integer, show that \[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]
2 replies
Arne
Apr 4, 2006
Speedysolver1
4 hours ago
Geometry hard problem
noneofyou34   1
N 5 hours ago by Lil_flip38
Let ABC be a triangle with incircle Γ. The tangency points of Γ with sides BC, CA, AB are A1, B1, C1 respectively. Line B1C1 intersects line BC at point A2. Similarly, points B2 and C2 are constructed. Prove that the perpendicular lines from A2, B2, C2 to lines AA1, BB1, CC1 respectively are concurret.
1 reply
noneofyou34
Yesterday at 3:13 PM
Lil_flip38
5 hours ago
Concurrency
Dadgarnia   29
N May 3, 2025 by blueprimes
Source: Iranian TST 2020, second exam day 2, problem 4
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
29 replies
Dadgarnia
Mar 12, 2020
blueprimes
May 3, 2025
Concurrency
G H J
Source: Iranian TST 2020, second exam day 2, problem 4
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Dadgarnia
164 posts
#1 • 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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GeoMetrix
924 posts
#5 • 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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[/asy]
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Mathematicsislovely
245 posts
#7 • 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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AlastorMoody
2125 posts
#8 • 7 Y
Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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Aryan-23
558 posts
#9 • 5 Y
Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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AwesomeYRY
579 posts
#10 • 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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SatisfiedMagma
461 posts
#11 • 2 Y
Y by jhu08, PRMOisTheHardestExam
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dot((6.591226634147988,-2.219627553277049),linewidth(4.pt) + dotstyle); 
label("$X$", (6.261,-2.5097147419492565), NE * labelscalefactor); 
dot((5.081025846916564,0.8007740211858014),linewidth(4.pt) + dotstyle); 
label("$M$", (4.693,0.8978983016140666), NE * labelscalefactor); 
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label("$E$", (9.77,1.618206749846964), NE * labelscalefactor); 
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Remarks
Attachments:
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MrOreoJuice
594 posts
#12 • 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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srisainandan6
2811 posts
#13 • 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
63 posts
#14 • 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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mohamad021
1 post
#15 • 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
#16
Y by
Solution
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guptaamitu1
656 posts
#17 • 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]
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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
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BarisKoyuncu
577 posts
#18 • 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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BarisKoyuncu
577 posts
#19
Y by
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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JAnatolGT_00
559 posts
#20 • 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
36 posts
#21
Y by
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]
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[/asy]

When you use Geogebra-Asymptote conversion, how did you doit. Why did mine have an error?
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guptaamitu1
656 posts
#22
Y by
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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IAmTheHazard
5003 posts
#23 • 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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Lantien.C
7 posts
#24
Y by
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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ihatemath123
3448 posts
#25 • 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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trk08
614 posts
#26
Y by
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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asdf334
7585 posts
#27
Y by
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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bjump
1030 posts
#28
Y by
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
18 posts
#29
Y by
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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shendrew7
799 posts
#30
Y by
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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joshualiu315
2534 posts
#31
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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
65 posts
#32
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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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happypi31415
753 posts
#33
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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! :D 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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blueprimes
356 posts
#36
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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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