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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
2025 IMO P4
sarjinius   10
N a few seconds ago by Maximilian113
Source: 2025 IMO P4
A proper divisor of a positive integer $N$ is a positive divisor of $N$ other than $N$ itself.

The infinite sequence $a_1,a_2,\dots$ consists of positive integers, each of which has at least three proper divisors. For each $n\ge1$, the integer $a_{n+1}$ is the sum of the three largest proper divisors of $a_n$.

Determine all possible values of $a_1$.
10 replies
+27 w
sarjinius
an hour ago
Maximilian113
a few seconds ago
IMO 2025 P2
sarjinius   52
N 2 minutes ago by Sleepy_Head
Source: 2025 IMO P2
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.
52 replies
sarjinius
Yesterday at 3:38 AM
Sleepy_Head
2 minutes ago
i can't think of a good title for this
Scilyse   6
N 13 minutes ago by pingupignu
Source: 2024 IMOSL G5
Let $ABC$ be a triangle with incentre $I$, and let $\Omega$ be the circumcircle of triangle $BIC$. Let $K$ be a point in the interior of segment $BC$ such that $\angle BAK < \angle KAC$. Suppose that the angle bisector of $\angle BKA$ intersects $\Omega$ at points $W$ and $X$ such that $A$ and $W$ lie on the same side of $BC$, and that the angle bisector of $\angle CKA$ intersects $\Omega$ at points $Y$ and $Z$ such that $A$ and $Y$ lie on the same side of $BC$.

Prove that $\angle WAY = \angle ZAX$.
6 replies
Scilyse
2 hours ago
pingupignu
13 minutes ago
orang NT
KevinYang2.71   11
N 17 minutes ago by InterLoop
Source: ISL 2024 N1
Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have $d+1\mid n$ or $d+1$ is prime.
11 replies
+1 w
KevinYang2.71
2 hours ago
InterLoop
17 minutes ago
f(D(x)), inf f(x) and sup f(x)
aktyw19   1
N Jul 9, 2025 by Mathzeus1024
find f(D(x)), inf f(x) and sup f(x)
$ f(x) = \{\begin{array}{cc}\sqrt {x}\cdot\arctan\frac {1}{x}\mbox { dla } x > 0 \\
0\mbox{ dla } x = 0 \end{array}$
where D(x)-domain f
1 reply
aktyw19
Mar 23, 2010
Mathzeus1024
Jul 9, 2025
Today's calculation of Integral 599
Kunihiko_Chikaya   6
N Jul 9, 2025 by aaravdodhia
Evaluate $\int_0^{\frac{\pi}{6}} \frac{e^x(\sin x+\cos x+\cos 3x)}{\cos^ 2 {2x}}\ dx$.

created by kunny
6 replies
Kunihiko_Chikaya
Apr 27, 2010
aaravdodhia
Jul 9, 2025
Functions
mclolikoi   3
N Jun 2, 2025 by Mathzeus1024
Let us consider f as the following function : $ f(x)= \frac {x - \sqrt{2}}{ [x  \sqrt{2} ] } $

1- Find the definition domain $ D_f $

2-Prove that $ f $ is continous on $ \sqrt{2} $

3-Study the continuity of $ f $ on $ \frac {3 \sqrt{2} }{2} $

4-Then draw the geometric representation of $ f $ on $ ] \frac {1}{ \sqrt{2} } ; 2 \sqrt{2} [ $
3 replies
mclolikoi
Sep 23, 2012
Mathzeus1024
Jun 2, 2025
functional equation
pratyush   4
N May 20, 2025 by Mathzeus1024
For the functional equation $f(x-y)=\frac{f(x)}{f(y)}$, if f ' (0)=p and f ' (5)=q, then prove f ' (-5) = q
4 replies
pratyush
Apr 4, 2014
Mathzeus1024
May 20, 2025
B.Math 2008-Integration .
mynamearzo   14
N May 14, 2025 by Levieee
Source: 10+2
Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function . Suppose
\[f(x)=\frac{1}{t} \int^t_0 (f(x+y)-f(y))\,dy\]
$\forall x\in \mathbb{R}$ and all $t>0$ . Then show that there exists a constant $c$ such that $f(x)=cx\ \forall x$
14 replies
mynamearzo
Apr 16, 2012
Levieee
May 14, 2025
Investigating functions
mikejoe   1
N May 14, 2025 by Mathzeus1024
Source: Edwards and Penney
Investigate the function $f(x) = (x-2) \sqrt{x+1}$
Also determine its domain and range.
1 reply
mikejoe
Nov 2, 2012
Mathzeus1024
May 14, 2025
Find the marginal profit..
ArmiAldi   1
N May 11, 2025 by Juno_34
Source: can someone help me
The total profit selling x units of books is P(x) = (6x - 7)(9x - 8) .
Find the marginal average profit function?
1 reply
ArmiAldi
Mar 2, 2008
Juno_34
May 11, 2025
Time required for draining out water
Kunihiko_Chikaya   1
N Apr 29, 2025 by Mathzeus1024
Source: Kyoto University 2nd-stage entry exam 2006 /Science, Problem
Let $H>0,\ R>0$ and $O$ be the origin, $P\ (R,\ 0\ ,H).$ In space given the vessel formed by the revolution of the segment $OP$ about $z$-axis. Denote the hight from $O$ to the surface of water by $h.$ When the vessel is filled with water, drain the water out such that the displacement per time is $\sqrt{h}$ at the time of $h,$ that is to say, if the total volume of the water drained out till the time $t$ is $V(t),$ then $\frac{dV}{dt}=\sqrt{h}$ holds. Find the time required for draining out water.
1 reply
Kunihiko_Chikaya
Mar 14, 2006
Mathzeus1024
Apr 29, 2025
2 Curves with a common tangent line
Kunihiko_Chikaya   1
N Apr 25, 2025 by Mathzeus1024
Source: 2010 Waseda University entrance exam/Science and Technology
Let $ a,\ b$ be real numbers. Consider the following curves on $ xy$ plane.

\[ y=e^{|x|}\ \ \ \ \ \ \ \ \ \ [1]\]

\[ \ \ \ \ \ \ y=ax+b\ \ \ \ \ \ \ [2]\]

(1) Suppose those curves have a common tangent line, express $ b$ interms of $ a$, sketch the graph on $ ab$ plane.

(2) Denote $ b=f(a)$ the graph of (1). For a constant $ p$, find the maximum value of $ pa+f(a)$ and the value of $ a$ giving the maximum value.
1 reply
Kunihiko_Chikaya
Feb 17, 2010
Mathzeus1024
Apr 25, 2025
I.S.I. B.Math.(Hons.) Admission test : 2010 Problem 5
mynamearzo   17
N Apr 24, 2025 by P162008
Let $a_1>a_2>.....>a_r$ be positive real numbers .
Compute $\lim_{n\to \infty} (a_1^n+a_2^n+.....+a_r^n)^{\frac{1}{n}}$
17 replies
mynamearzo
Apr 10, 2012
P162008
Apr 24, 2025
Concurrency
Dadgarnia   36
N Jun 28, 2025 by HamstPan38825
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
36 replies
Dadgarnia
Mar 12, 2020
HamstPan38825
Jun 28, 2025
Concurrency
G H J
Source: Iranian TST 2020, second exam day 2, problem 4
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Dadgarnia
164 posts
#1 • 10 Y
Y by Purple_Planet, itslumi, jhu08, mathematicsy, Mango247, Mango247, ItsBesi, cubres, Rounak_iitr, Kurin
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 • 9 Y
Y by mueller.25, amar_04, Mathasocean, Purple_Planet, agwwtl03, IFA, jhu08, sabkx, cubres
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]
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Mathematicsislovely
245 posts
#7 • 4 Y
Y by Purple_Planet, jhu08, PRMOisTheHardestExam, cubres
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 • 8 Y
Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam, cubres
Solution
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Aryan-23
558 posts
#9 • 6 Y
Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam, cubres
Solution
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AwesomeYRY
579 posts
#10 • 3 Y
Y by jhu08, PRMOisTheHardestExam, cubres
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
464 posts
#11 • 3 Y
Y by jhu08, PRMOisTheHardestExam, cubres
[asy]
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[/asy]
Remarks
Attachments:
This post has been edited 1 time. Last edited by SatisfiedMagma, Aug 13, 2021, 3:06 PM
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MrOreoJuice
597 posts
#12 • 5 Y
Y by jhu08, SatisfiedMagma, PRMOisTheHardestExam, vrondoS, cubres
$\angle QIA = \angle QCI = \angle ICB = 90^\circ - \angle (IC , \text{angle bisector}) \implies \angle QIC = 90^\circ \implies N$ 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 • 3 Y
Y by PRMOisTheHardestExam, jhu08, cubres
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 • 3 Y
Y by PRMOisTheHardestExam, jhu08, cubres
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 • 2 Y
Y by jhu08, cubres
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
7414 posts
#16 • 1 Y
Y by cubres
Solution
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guptaamitu1
668 posts
#17 • 3 Y
Y by PRMOisTheHardestExam, Lantien.C, cubres
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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[/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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BarisKoyuncu
577 posts
#18 • 2 Y
Y by teomihai, cubres
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 • 1 Y
Y by cubres
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 • 2 Y
Y by MrOreoJuice, cubres
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 • 1 Y
Y by cubres
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
668 posts
#22 • 1 Y
Y by cubres
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
5006 posts
#23 • 2 Y
Y by Lantien.C, cubres
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 • 1 Y
Y by cubres
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
Attachments:
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ihatemath123
3490 posts
#25 • 2 Y
Y by happypi31415, cubres
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 • 1 Y
Y by cubres
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
7584 posts
#27 • 1 Y
Y by cubres
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
1070 posts
#28 • 1 Y
Y by cubres
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 • 1 Y
Y by cubres
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
816 posts
#30 • 1 Y
Y by cubres
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
2535 posts
#31 • 1 Y
Y by cubres
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
91 posts
#32 • 1 Y
Y by cubres
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
777 posts
#33 • 1 Y
Y by cubres
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
379 posts
#36 • 1 Y
Y by cubres
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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Mathandski
776 posts
#37 • 1 Y
Y by cubres
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]

To prove $(AJBC)$ harmonic let $J'$ be midpoint of minor arc $BC$ and $J'I = J'C$ by 1.18. $JI^2 = JH \cdot JD$ by Power of a Point but $J'C^2 = J'H \cdot J'D$ by Shooting lemma
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Blackbeam999
58 posts
#38 • 1 Y
Y by cubres
AwesomeYRY wrote:
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 $\blacksquare$

Why does AI perpendicular to NI?
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deraxenrovalo
9 posts
#39 • 1 Y
Y by cubres
This one is entertaining! :coolspeak:

Let $AD$ cut $BC$ at $Z$ and $J$ be the midpoint of the $\overset{\frown}{BC}$ opposite to $A$ on $(ABC)$.
Let $P$ be the reflection of $B$ across $MJ$ and $PB$ cut $IC$ at $Q$.

Under the inversion $\mathcal{I}_J^{-\mathcal{P}_{J/(BIC)}}$ :
$$A\longleftrightarrow A'\quad;\quad B\longleftrightarrow B\quad;\quad C\longleftrightarrow C\quad;\quad D\longleftrightarrow D'\quad;\quad I\longleftrightarrow I$$$$M\longleftrightarrow M'\quad;\quad N\longleftrightarrow N'\quad;\quad Z\longleftrightarrow Z'$$$$AD\longleftrightarrow (JA'D')\quad;\quad BC\longleftrightarrow (ABC)\quad;\quad MN\longleftrightarrow (JM'N')$$It is clear that $M'$, $Q$, $N'$, $J$ are concyclic since $\mathcal{P}_{J/(AB)}=\mathcal{P}_{J/\omega}=-\mathcal{P}_{J/(BIC)}$.
Therefore it suffices to show that $Q$, $D'$, $Z'$ are collinear.
Let $BZ'$ and $CZ'$ cut $(BIC)$ again at $E$ and $F$, respectively.
Noticed that :
$$\widehat{PID'}=\widehat{PIC}\;+\;\widehat{CID'}=\widehat{PBC}\;+\;\widehat{CDD'}=\widehat{PBC}\;+\;\widehat{CBJ}=\widehat{PBJ}=\widehat{PA'D'}$$This implies $I$, $P$, $D'$, $A'$ are concyclic.
Moreover, we have :
$${AC}^2=\overline{AA'}\cdot\overline{AJ}\Longleftrightarrow\mathcal{P}_{A/(BIC)}=\mathcal{P}_{A/(JA'D')}\quad;\quad\overline{ZB}\cdot\overline{ZC}=\overline{ZD'}\cdot\overline{ZA'}\Longleftrightarrow\mathcal{P}_{Z/(BIC)}=\mathcal{P}_{Z/(JA'D')}$$Which means $AZ$ is radical axis of $(BIC)$ and $(JA'D')$, thus $Z$ is the radical center of $(BIC)$, $(JA'D')$ and $(IA'D')$.
So $I$, $P$, $Z$ are collinear and note that $EF$ is reflected to $BC$ over $ZJ$, we have $IP$, $BC$ and $EF$ are concurrent at $Z$.
Hence :
$$B(A,Q,Z',C)\overset{B}{=}(B,P,E,C)\overset{Z}{=}(C,I,F,B)\overset{C}{=}C(A,Q,Z',B)$$Therefore $Q$, $D'$, $Z'$ are collinear as desired.$\qquad\blacksquare$
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Buh_-1235
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#40 • 1 Y
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Let $J$ be a point beyond C on line $AC$ and $K$ be the intersection of lines $BC$ & $AD$.It is really easy to see that $N$ is the center of $(CIQD)$.Now: \[\angle ABC =\angle CDK =\angle ACB =\angle KCJ \]Hence $\angle CDK=\angle KCJ $ which implies $NC$ is tangent to $(CDK)$. As $NC=ND$,$ND$ is tangent to that circle as well.So $KN$ is the $K$-symmedian of $\triangle CDK$. Now as $CD$ is antiparallel to $AB$,$KN$ passes through the midpoint of $AB$ or the point $M$.So we are done!!!
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sadat465
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#41
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Pretty easy using harmonic bundle.
First let (CID) interest BC at F. And AD interest BC at E. MN interest BC at E'.
By ceva you can prove AF,BE,MC concurrent.
Implies B,F,C,E' harmonic.
Again,
It is easy to prove B,F,C,E also HM.
(FDC=90, FD bisect BDC)
Implies E=E".
Solved
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sadat465
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#42
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Blackbeam999 wrote:
AwesomeYRY wrote:
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 $\blacksquare$

Why does AI perpendicular to NI?

AI is Tangent. And N is the Center OF (CID)
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HamstPan38825
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#43
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Let $E = \overline{MN} \cap \overline{BC}$.

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

Proof: Because $\angle INC = 2\angle IQC = 2\angle PIC = 180^\circ - \angle C$. $\blacksquare$

Let $R$ be the second intersection of $\omega$ with $\overline{BC}$. Since triangle $NCR$ is isosceles, $\overline{NR} \parallel \overline{AB}$, so $E$ is the exsimilicenter of $(AB)$ and $\omega$.

Now let $F$ be the reflection of $R$ over $N$. Then by homothety, $E$ lies on $\overline{AF}$. Let $D'$ be the intersection of $\overline{AF}$ and $\omega$; by Reim's theorem, $\overline{AB}$ and $\overline{CD'}$ are antiparallel, so $D' = D$. This completes the proof.
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