Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inequality em981
oldbeginner   16
N a few seconds ago by mihaig
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
16 replies
oldbeginner
Sep 22, 2016
mihaig
a few seconds ago
Nice FE over R+
doanquangdang   4
N a minute ago by jasperE3
Source: collect
Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that
\[x+f(yf(x)+1)=xf(x+y)+yf(yf(x))\]for all $x,y>0.$
4 replies
doanquangdang
Jul 19, 2022
jasperE3
a minute ago
right triangle, midpoints, two circles, find angle
star-1ord   0
4 minutes ago
Source: Estonia Final Round 2025 8-3
In the right triangle $ABC$, $M$ is the midpoint of the hypotenuse $AB$. Point $D$ is chosen on the leg $BC$ so that the line segment $DM$ meets $(ACD)$ again at $K$ ($K\neq D$). Let $L$ be the reflection of $K$ in $M$. The circles $(ACD)$ and $(BCL)$ meet again at $N$ ($N\neq C$). Find the measure of $\angle KNL$.
0 replies
star-1ord
4 minutes ago
0 replies
interesting functional equation
tabel   3
N 6 minutes ago by waterbottle432
Source: random romanian contest
Determine all functions \( f : (0, \infty) \to (0, \infty) \) that satisfy the functional equation:
\[
f(f(x)(1 + y)) = f(x) + f(xy), \quad \forall x, y > 0.
\]
3 replies
tabel
an hour ago
waterbottle432
6 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   4
N 22 minutes ago by GreenTea2593
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
4 replies
OgnjenTesic
May 22, 2025
GreenTea2593
22 minutes ago
pairs (m, n) such that a fractional expression is an integer
cielblue   2
N 39 minutes ago by cielblue
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
2 replies
cielblue
Yesterday at 8:38 PM
cielblue
39 minutes ago
Sociable set of people
jgnr   23
N an hour ago by quantam13
Source: RMM 2012 day 1 problem 1
Given a finite number of boys and girls, a sociable set of boys is a set of boys such that every girl knows at least one boy in that set; and a sociable set of girls is a set of girls such that every boy knows at least one girl in that set. Prove that the number of sociable sets of boys and the number of sociable sets of girls have the same parity. (Acquaintance is assumed to be mutual.)

(Poland) Marek Cygan
23 replies
jgnr
Mar 3, 2012
quantam13
an hour ago
diophantine equation
m4thbl3nd3r   0
an hour ago
Find all positive integers $n,k$ such that $$5^{2n+1}-5^n+1=k^2$$
0 replies
m4thbl3nd3r
an hour ago
0 replies
A geometry problem
Lttgeometry   1
N an hour ago by Funcshun840
Triangle $ABC$ has two isogonal conjugate points $P$ and $Q$. The circle $(BPC)$ intersects circle $(AP)$ at $R \neq P$, and the circle $(BQC)$ intersects circle $(AQ)$ at $S\neq Q$. Prove that $R$ and $S$ are isogonal conjugates in triangle $ABC$.
Note: Circle $(AP)$ is the circle with diameter $AP$, Circle $(AQ)$ is the circle with diameter $AQ$.
1 reply
Lttgeometry
Today at 4:03 AM
Funcshun840
an hour ago
Functional equation
shobber   19
N 3 hours ago by Unique_solver
Source: Canada 2002
Let $\mathbb N = \{0,1,2,\ldots\}$. Determine all functions $f: \mathbb N \to \mathbb N$ such that
\[ xf(y) + yf(x) = (x+y) f(x^2+y^2)  \]
for all $x$ and $y$ in $\mathbb N$.
19 replies
shobber
Mar 5, 2006
Unique_solver
3 hours ago
Prove the inequality
Butterfly   0
3 hours ago
Let $a,b,c$ be real numbers such that $a+b+c=3$. Prove $$a^3b+b^3c+c^3a\le \frac{9}{32}(63+5\sqrt{105}).$$
0 replies
Butterfly
3 hours ago
0 replies
Functional equation
shactal   1
N 3 hours ago by ariopro1387
Let $f:\mathbb R\to \mathbb R$ a function satifying $$f(x+2xy) = f(x) + 2f(xy)$$for all $x,y\in \mathbb R$.
If $f(1991)=a$, then what is $f(1992)$, the answer is in terms of $a$.
1 reply
shactal
5 hours ago
ariopro1387
3 hours ago
interesting diophantiic fe in natural numbers
skellyrah   5
N 3 hours ago by skellyrah
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
5 replies
skellyrah
Yesterday at 8:01 AM
skellyrah
3 hours ago
Non-linear Recursive Sequence
amogususususus   3
N 3 hours ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
3 hours ago
Constant Angle Sum
i3435   6
N Apr 16, 2025 by bin_sherlo
Source: AMASCIWLOFRIAA1PD (mock oly geo contest) P3
Let $ABC$ be a triangle with circumcircle $\Omega$, $A$-angle bisector $l_A$, and $A$-median $m_A$. Suppose that $l_A$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$. A line $l$ parallel to $\overline{BC}$ meets $l_A$, $m_A$ at $G$, $N$ respectively, so that $G$ is between $A$ and $D$. The circle with diameter $\overline{AG}$ meets $\Omega$ again at $J$.

As $l$ varies, show that $\angle AMN + \angle DJG$ is constant.

MP8148
6 replies
i3435
May 11, 2021
bin_sherlo
Apr 16, 2025
Constant Angle Sum
G H J
G H BBookmark kLocked kLocked NReply
Source: AMASCIWLOFRIAA1PD (mock oly geo contest) P3
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i3435
1350 posts
#1 • 4 Y
Y by amar_04, sotpidot, centslordm, MS_asdfgzxcvb
Let $ABC$ be a triangle with circumcircle $\Omega$, $A$-angle bisector $l_A$, and $A$-median $m_A$. Suppose that $l_A$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$. A line $l$ parallel to $\overline{BC}$ meets $l_A$, $m_A$ at $G$, $N$ respectively, so that $G$ is between $A$ and $D$. The circle with diameter $\overline{AG}$ meets $\Omega$ again at $J$.

As $l$ varies, show that $\angle AMN + \angle DJG$ is constant.

MP8148
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amar_04
1916 posts
#2 • 4 Y
Y by Mathematicsislovely, Bumblebee60, centslordm, MS_asdfgzxcvb
[asy]
defaultpen(fontsize(9pt));
size(5cm);

pair A,B,C,I,X,D,M,O,A1,M1,V,J,P1,Q1,N;

A=dir(135);
B=dir(210);
C=dir(330);
I=incenter(A,B,C);
X=(B+C)/2;
O=circumcenter(A,B,C);
D=extension(A,I,B,C);
M=-A+2*foot(O,A,I);
A1=-A+2O;
M1=-M+2O;
V=-M1+2*foot(O,M1,D);
J=-A1+2*foot(O,A1,I);
P1=(D+X)/2;
Q1=extension(A,P1,I,X);
N=extension(D,Q1,A,X);

draw(A--B--C--A);
draw(circumcircle(A,B,C));
fill(A--V--D--cycle,0.8*black+0.5*white);
fill(A--X--M--cycle,0.8*black+0.5*white);
draw(A--M);
draw(A--A1--J);
draw(M--M1--V);
draw(A--X);
draw(I--N);
draw(arc(circumcenter(J,D,V),circumradius(J,D,V),-40,100));
draw(circumcircle(A,D,X));
draw(A--V--D--A);
draw(A--X--M--A);
draw(V--I);
draw(M--N);

dot("$A$" , A , dir(A));
dot("$B$" , B , dir(B));
dot("$C$" , C , dir(C));
dot("$G$" , I , dir(250));
dot("$X$" , X , dir(310));
dot("$M$" , M , dir(M));
dot("$D$" , D , dir(220));
dot("$A^*$" , A1 , dir(A1));
dot("$M_A$" , M1 , dir(M1));
dot("$V$" , V , dir(270));
dot("$J$" , J , dir(J));
dot("$O$" , O , dir(0));
dot("$N$" , N , dir(80));

[/asy]
Claim: As $G$ varies over $\ell_A$, $\measuredangle NMA+\measuredangle DJG=\angle\left(\frac{B-C}{2}\right)$ which is constant as $\Delta ABC$ is fixed.

Proof: Let $A^*$ be the $A-$ antipode of $\odot(ABC)$ and $X$ be the midpoint of $BC$. Clearly $A^*,G,J$ are collinear. Let $M_A$ be the midpoint of the major arc $\widehat{BAC}$ and let $\overline{M_AD}\cap\odot(ABC)=V$. Now notice that $OM_A=OM$ and $AO=A^*O$, this $AM_A^*M$ is a rectangle $\implies \overline{AD}\parallel\overline{A^*M_A}$, thus by the converse of Reims we have $J,G,D,V$ are concyclic and $-1=(AB,AC;AD,AM_A)\overset{A}{=}(BC;MM_A)\overset{D}{=}(AV;BC)\implies\overline{AV}$ is the $A-$ Symmedian of $\Delta ABC$. Now, notice that $\measuredangle MAX=\measuredangle VAD$ and $\measuredangle DVA=\measuredangle XMA$ and as $GN\parallel\overline{BC}$, $\frac{GA}{GD}=\frac{NA}{NX}\implies\Delta AVD\cup G\stackrel{+}{\sim}\Delta AMX\cup N$. Thus, $\measuredangle NMA+\measuredangle DJG=\measuredangle NMA+\measuredangle DVG=\measuredangle NMA+\measuredangle XMN=\measuredangle M_AMA=\measuredangle M_ABA=\angle\left(\frac{B-C}{2}\right)$ which is constant as $\Delta ABC$ is fixed. $\blacksquare$
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MP8148
888 posts
#3 • 2 Y
Y by centslordm, Mango247
Here is an extension for this problem. I wasn't able to prove it, but geogebra tells me that it's true.

Let $l$ meet $\overline{AC}$, $\overline{AB}$ at $E$, $F$, and let $\overline{JD}$ meet $\Omega$ again at $K$. Show that $\overline{AK}$ passes through the $M$-Dumpty point in $\triangle MEF$.
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sotpidot
290 posts
#4 • 1 Y
Y by centslordm
Epic problem.

Let $m_A$ meet $\Omega$ again at $Y$, and let $X$ be the reflection of $Y$ over $OM$. Let $m_A$ meet the midpoint of $BC$ at $E$ and let $AB < AC$ WLOG. Here, $\overarc{AB}$ will denote the angle subtending any minor arc $AB$, and $\overbrace{AB}$ will denote the angle subtending any major arc $AB$.

Lemma 1: $DEMX$ is cyclic.

Proof: Notice $\angle EYM = \angle AYM = \overarc{AM}$, and that $\angle EDM = \overarc{AB} + \overarc{CM}$. Since $M$ is the midpoint of $\overarc{BC}$, $\overarc{BM} = \overarc{CM}$ and thus $\angle EDM = \overarc{AM} = \angle EYM$. Then, since $X$ is the reflection of $Y$ over $OM$, we have $\angle EXM = \angle EDM$ and thus $DEMX$ is cyclic. $\square$

Lemma 2: $XDGJ$ is cyclic.

Proof: Since $DEMX$ is cyclic and $\angle DEM = 90^\circ$, $\angle DXM = 90^\circ$. Then, we have $\angle AMX + \angle MDX + 90^\circ = 180^\circ$. We see that $\angle AMX = \overarc{AX}$, and thus $\angle MDX + 90^\circ = \overbrace{AX} = \angle AJX$. Since $AG$ is the diameter of the circumcircle of $\triangle GAJ$, $\angle AJG = 90^\circ$, and thus $\angle GJX = \angle MDX$. It then follows that $\angle GDX + \angle GJX = 180^\circ$. $\square$

Lemma 3: Let $MN$ meet $\Omega$ again at $R$. Then, $R$, $G$, $X$ are collinear.

Proof: We first claim that $RNGA$ is cyclic. Let the two points at which $GN$ meets $\Omega$ be $P$ and $Q$. Then $M$ is the midpoint of $\overarc{PMQ}$, since $PQ \parallel BC$. Then there exists an inversion about $M$ w.r.t some arbitrary circle with radius $r$ that transforms $\Omega$ to the line $PQ$. This implies that $MN \cdot MR = MG \cdot MA = r^2$, and thus $RNGA$ is cyclic. Then let $RG$ meet $\Omega$ again at a point $X'$. Since $\angle NRG = \angle NAG$, $\angle MRX' = \angle MAY$, and thus $X' = X$. $\square$

Now, we consider $\triangle XRM$, where we see trivially that $X$ and $M$ are fixed as $\ell$ varies. Since $XDGJ$ is cyclic, we have $\angle DJG = \angle DXG$, and we also see that $\angle AMN = \angle GMN$. Then, we see that: $$\angle AMN + \angle DJG = \angle GMN + \angle DXG = 180^\circ - \angle XRM - \angle DXM - \angle DMX.$$Since $D$, $X$, $M$ are fixed, and $\angle XRM$ is constant since $R$ lies on $\Omega$, we can conclude that $\angle AMN + \angle DJG$ is constant as $\ell$ varies. $\blacksquare$
This post has been edited 1 time. Last edited by sotpidot, May 12, 2021, 9:00 PM
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SerdarBozdag
892 posts
#5 • 3 Y
Y by Mango247, Mango247, Mango247
Let $E$ be the midpoint of $BC$. Let $L$ and $O$ be points on $AB$ and $AC$ respectively such that $AOGLJ$ is cyclic. We will prove $\angle GJD= \angle NME$ and $\angle AMN=\angle MJD$. This finishes the solution because $\angle AMN + \angle DJG=\angle AME$ is constant.

Let the foot of perpendicular from $A$ to $HF$ be $I$. $I$ is on $AOGLJ$. Let $A'$ be the antipode of $A$. Observe that $J-G-A'$ is collinear. We have $\angle IJG=\angle IAG=\angle A'AM=\angle A'JM=\angle MJG \implies \textbf{J-I-M}$ are collinear.

$\angle EMA=\angle MAI=\angle GJM$. Proving $\frac{sin(\angle MJD)}{sin(\angle DJG)}=\frac{sin(\angle AMN)}{sin(\angle NME)}$ finishes the solution because $\frac{sin(x)}{sin(k-x)}$ is injective where k is constant.($0<x<k<90$).

Observe that $J$ is a spiral similarity center which takes triangle $LGO$ to triangle $BMC$. We have $MD\cdot MA=MB^2$, $sin(A)=OL/AG$, $ME\cdot BC=sin(A)\cdot MB^2$. The last two comes from the area of $BMC$ and $ALGO$.


$$\frac{sin(\angle MJD)}{sin(\angle DJG)} =\frac{DM\cdot JG }{DG\cdot JM} = \frac{DM\cdot OL }{DG\cdot BC} =\frac{\frac{MB^2}{MA}\cdot AG \cdot ME  }{DG\cdot MB^2}=\frac{AG\cdot ME }{DG\cdot MA}=\frac{AN\cdot ME }{DG\cdot AM}=\frac{sin(\angle AMN)}{sin(\angle NME)}$$$\square$
This post has been edited 5 times. Last edited by SerdarBozdag, Jun 3, 2021, 9:11 AM
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Vitriol
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#6 • 2 Y
Y by GeoKing, MS_asdfgzxcvb
Let $A'$ be the $A$-antipode on $\Omega$. Let $JD$ hit $\Omega$ again at $T_1$, let $T_1'$ be the reflection of $T_1$ over $AA'$, and let $T_1''$ be the $T_1'$-antipode on $\Omega$. Let $MN$ hit $\Omega$ again at $T_2$. Note that $\measuredangle GJA = \pi /2 = \measuredangle A'JA$, so $A'-G-J$. Furthermore,
\[ \measuredangle DJG = \measuredangle T_1JA' = -\measuredangle T_1'JA' = -\measuredangle T_1''JA = \measuredangle AMT_1'',\]while
\[ \measuredangle NMA = \measuredangle T_2MA,\]so it suffices to show that $\measuredangle T_2MT_1''$ is constant.

Animate $G$ on $l_A$. We have that $G \stackrel{\infty_{BC}}\mapsto N \mapsto MN = MT_2$ is projective. In addition, $G \stackrel{A'}\mapsto J \stackrel{D}\mapsto T_1 \stackrel{\infty_{AA}}\mapsto T_1' \stackrel{O}\mapsto T_1'' \mapsto MT_1''$ is projective. Thus, $\deg MT_2 = \deg MT_1'' = 1$. The following lemma finishes the problem:

Lemma: Let $k{}$ be a constant, and let $A{}$, $B{}$, $C{}$, and $D{}$ be moving points on a line $\ell$. Then, the statement ``$(A, B; C, D) = k$'' has degree at most $\deg A + \deg B + \deg C + \deg D$.
Proof: This is just writing it out. Take a projective transformation onto $\mathbb P^1$. Then, the statement ``$\deg A = d$'' is equivalent to $A = n_A(t) / d_A(t)$ for polynomials $n_A, d_A$ with $\max \{\deg n_A, \deg d_A\} = d$. Similarly, we can write $B = n_B(t) / d_B(t)$, $C = n_C(t) / d_C(t)$, and $D = n_D(t) / d_D(t)$. Then,
\[ (A, B; C, D) = \frac{A-C}{B-C} \div \frac{A-D}{B-D} =  \frac{n_A(t) d_C(t) - n_C(t) d_A(t)}{n_B(t) d_C(t) - n_C(t) d_B(t)} \div \frac{n_A(t) d_D(t) - n_D(t) d_A(t)}{n_B(t) d_D(t) - n_D(t) d_B(t)},\]so ``$(A, B; C, D) = k$'' is just,
\[ (n_A(t) d_C(t) - n_C(t) d_A(t))(n_B(t) d_D(t) - n_D(t) d_B(t)) - k(n_B(t) d_C(t) - n_C(t) d_B(t))(n_A(t) d_D(t) - n_D(t) d_A(t)) = 0,\]which has at most the desired degree. $\square$

Observe that this also implies that if $(A, B; C, D)$ is the same for $\deg A + \deg B + \deg C + \deg D + 1$ values of $t{}$, then it is always constant (alternatively, you can just pick that one value, let's say $k{}$, and then the above lemma will tell you that $(A, B; C, D) = k$ is always true since it is true for more than $\deg A + \deg B + \deg C + \deg D$ values of $t{}$). Finally, observe that the dual will also be true.

Now for the problem, let $I$ and $J$ be the circular points at infinity. By Laguerre's formula,
\[ \measuredangle T_2 MT_1'' = \frac1{2i} \log (MT_2, MT_1''; MI, MJ),\]so it suffices to show that $\measuredangle T_2 MT_1''$ is constant for $3{}$ values of $t{}$.
  • if $G = A$, then $N = T_2 = A$, and $J = A$ as well. Thus, $T_1 = M$, so $T_1'$ is the reflection of $M$ over $AA'$, and $T_1''$ is the point such that $AA'T_1T_1''$ is a cyclic isosceles trapezoid. Thus, $\measuredangle T_2MT_1'' = \measuredangle MAA'$.
  • if $G = D$, then $N$ is the midpoint of $BC$, so $MT_2$ is the perpendicular bisector of $BC$. In addition, $J-D-A'$, so $T_1=A'$, and $T_1' = A'$, so $T_1'' = A$. Thus, $\measuredangle T_2MT_1'' = \measuredangle OMA = \measuredangle MAO = \measuredangle MAA'$.
  • if $G = M$, then $N = MM \cap m_A$, so $MT_2$ is the tangent to $\Omega$ at $M$. In addition, $J = G$, so $T_1 = A$, and $T_1'=A$, so $T_1''=A'$. Thus, $\measuredangle T_2MT_1'' = \measuredangle MAA'$ (by tangency).
We are done. $\blacksquare$
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bin_sherlo
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#7
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Let $S$ be the intersection of $A-$symmedian and $(ABC)$, $K$ be the midpoint of $BC$, $L$ be the midpoint of arc $BAC$, $A'$ be the antipode of $A$, $AK\cap (ABC)=W,SG\cap (ABC)=P,JD\cap (ABC)=Q$.
Pascal at $SPMMAW$ gives $G,PM\cap AW,BC_{\infty}$ are collinear thus, $P,M,N$ are collinear. Pascal at $PQJA'LS$ implies $PQ\cap A'L,D,G$ are collinear and since $A'L\parallel DG$ we get $PQ\parallel A'L$. Hence $\measuredangle GJD+\measuredangle AMN=\measuredangle A'JQ+\measuredangle AMP=\measuredangle PML+\measuredangle AMP=\frac{\measuredangle B-\measuredangle C}{2}$ as desired.$\blacksquare$
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