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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
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 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
(Easy version) Same number of divisors
MNJ2357   4
N 15 minutes ago by thdwlgh1229
Source: 2024 Korea Summer Program Practice Test Junior P8
For a positive integer \( n \), let \( \tau(n) \) denote the number of positive divisors of \( n \). Determine all positive integers \( K \) such that the equation
\[ \tau(x) = \tau(y) = \tau(z) = \tau(2x + 3y + 3z) = K \]holds for some positive integers $x,y,z$.
4 replies
MNJ2357
Aug 12, 2024
thdwlgh1229
15 minutes ago
Circumcenters
rkm0959   14
N 15 minutes ago by MathLuis
Source: 2015 Final Korean Mathematical Olympiad Day 1 Problem 2
In a triangle $\triangle ABC$ with incenter $I$, the incircle meets lines $BC, CA, AB$ at $D, E, F$ respectively.
Define the circumcenter of $\triangle IAB$ and $\triangle IAC$ $O_1$ and $O_2$ respectively.
Let the two intersections of the circumcircle of $\triangle ABC$ and line $EF$ be $P, Q$.
Prove that the circumcenter of $\triangle DPQ$ lies on the line $O_1O_2$.
14 replies
rkm0959
Mar 21, 2015
MathLuis
15 minutes ago
Infinite sequence of polynomials
OronSH   13
N 19 minutes ago by monval
Source: 2024 USEMO/4
Find all sequences $a_1$, $a_2$, $\dots$ of nonnegative integers such that for all positive integers $n$, the polynomial \[1+x^{a_1}+x^{a_2}+\dots+x^{a_n}\]has at least one integer root. (Here $x^0=1$.)

Kornpholkrit Weraarchakul
13 replies
OronSH
Oct 27, 2024
monval
19 minutes ago
Bounded function satisfying averaging condition
62861   42
N 25 minutes ago by KevinYang2.71
Source: USA Winter Team Selection Test #1 for IMO 2018, Problem 2
Find all functions $f\colon \mathbb{Z}^2 \to [0, 1]$ such that for any integers $x$ and $y$,
\[f(x, y) = \frac{f(x - 1, y) + f(x, y - 1)}{2}.\]
Proposed by Yang Liu and Michael Kural
42 replies
62861
Dec 11, 2017
KevinYang2.71
25 minutes ago
IMO Shortlist 2017 A1
math90   84
N 26 minutes ago by ray66
Source: IMO Shortlist 2017
Let $a_1,a_2,\ldots a_n,k$, and $M$ be positive integers such that
$$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=k\quad\text{and}\quad a_1a_2\cdots a_n=M.$$If $M>1$, prove that the polynomial
$$P(x)=M(x+1)^k-(x+a_1)(x+a_2)\cdots (x+a_n)$$has no positive roots.
84 replies
+1 w
math90
Jul 10, 2018
ray66
26 minutes ago
Generalization P2 IMO 2025
hn111009   1
N an hour ago by keglesnit
Source: Le Viet An
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 $B$. Point $C\neq A$ lies on $\Omega$ and point $D\neq A$ lies on $\Gamma$, so that $\overrightarrow{MC}\uparrow \downarrow \overrightarrow{ND}$. 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 $BEF$.
1 reply
hn111009
Yesterday at 5:06 PM
keglesnit
an hour ago
Polynomial - odd or even coefficients
perfect_radio   7
N an hour ago by OutKast
Source: Romanian TST 2 2007, Problem 1
Let
\[f = X^{n}+a_{n-1}X^{n-1}+\ldots+a_{1}X+a_{0}\]
be an integer polynomial of degree $n \geq 3$ such that $a_{k}+a_{n-k}$ is even for all $k \in \overline{1,n-1}$ and $a_{0}$ is even.
Suppose that $f = gh$, where $g,h$ are integer polynomials and $\deg g \leq \deg h$ and all the coefficients of $h$ are odd.
Prove that $f$ has an integer root.
7 replies
perfect_radio
Apr 15, 2007
OutKast
an hour ago
Gcd Functional equation
EeEeRUT   13
N 2 hours ago by monval
Source: ISL 2024 N7
Let $\mathbb{Z}_{>0}$ denote the set of positive integers. Let $f : \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ be a function satisfying the following property: for $m,n \in \mathbb{Z}_{>0}$, the equation
\[
f(mn)^2 = f(m^2)f(f(n))f(mf(n))
\]holds if and only if $m$ and $n$ are coprime.

For each positive integer $n$, determine all the possible values of $f(n)$.
13 replies
EeEeRUT
Jul 16, 2025
monval
2 hours ago
Functional equation
shactal   4
N 2 hours ago by shactal
Source: Own
Hello, I found this functional equation that I can't solve, and I haven't got any hints. Could someone try and find the solution, it's actually quite difficult:
Find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that, for all $x, y \in \mathbb{R} $,
$$
f(x + f(y)) + f(y + f(x)) = f(x \, f(y) + y \, f(x)) + f(x + y)$$Thank you.
4 replies
shactal
Jul 22, 2025
shactal
2 hours ago
Classic Functional Equation
rkm0959   16
N 2 hours ago by MathLuis
Source: 2015 Final Korean Mathematical Olympiad Day 1 Problem 1
Find all functions $f: R \rightarrow R$ such that
$f(x^{2015} + (f(y))^{2015}) = (f(x))^{2015} + y^{2015}$ holds for all reals $x, y$
16 replies
rkm0959
Mar 21, 2015
MathLuis
2 hours ago
Number theory game
SYBARUPEMULA   1
N 2 hours ago by SYBARUPEMULA
Given number $N$ in the whiteboard. Anna and Bob play by turns to divide the number in the board by a prime divisor of $N$ and replace $N$ with the new number. The new number must have as many divisors of the form $4k + 1$ as divisors of the form $4k + 3$. The first player that can't make a move loses. Anna plays first, determine which player has the winning strategy if given $N = 9999^{99}$.

(Note: integer $X$ is the form of $4k + 1$ if $X \equiv 1 \mod 4$.)
1 reply
SYBARUPEMULA
Yesterday at 6:25 AM
SYBARUPEMULA
2 hours ago
The circumcenter
Fermat -Euler   5
N 2 hours ago by YaoAOPS
Source: IMO Shortlist 1994, G3
A circle $ C$ has two parallel tangents $ L'$ and$ L"$. A circle $ C'$ touches $ L'$ at $ A$ and $ C$ at $ X$. A circle $ C"$ touches $ L"$ at $ B$, $ C$ at $ Y$ and $ C'$ at $ Z$. The lines $ AY$ and $ BX$ meet at $ Q$. Show that $ Q$ is the circumcenter of $ XYZ$
5 replies
Fermat -Euler
Oct 22, 2005
YaoAOPS
2 hours ago
An inequality
huytran08   7
N 2 hours ago by Victoria_Discalceata1
Given $a,b,c>0$. Prove that:
$$(a^4+b^4 + c^4)^2 \geq 3abc(a^5+b^5+c^5)$$
7 replies
huytran08
Yesterday at 8:33 AM
Victoria_Discalceata1
2 hours ago
Functional equation
socrates   32
N 2 hours ago by Fly_into_the_sky
Source: Baltic Way 2014, Problem 4
Find all functions $f$ defined on all real numbers and taking real values such that \[f(f(y)) + f(x - y) = f(xf(y) - x),\] for all real numbers $x, y.$
32 replies
socrates
Nov 11, 2014
Fly_into_the_sky
2 hours ago
Random Points = Problem
kingu   5
N May 31, 2025 by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
kingu
Apr 27, 2024
happypi31415
May 31, 2025
Random Points = Problem
G H J
G H BBookmark kLocked kLocked NReply
Source: Chinese Geometry Handout
The post below has been deleted. Click to close.
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kingu
220 posts
#1 • 4 Y
Y by Rounak_iitr, buddyram, GeoKing, ItsBesi
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
Z K Y
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cursed_tangent1434
737 posts
#2 • 3 Y
Y by Shreyasharma, Rounak_iitr, GeoKing
Solved with kingu. Very nice problem. We were really embarrassed by how long it evaded us once we found the solution.

We first show the following important claim.

Claim : The quadrilaterals $AGEN$ and $MAEF$ are cyclic.
Proof : For this first, we simply note that
\[\measuredangle NAE = \measuredangle ABE = \measuredangle DBE  = \measuredangle NGE  \]and for the second note that
\[\measuredangle EFM = \measuredangle EFD = \measuredangle NGE = \measuredangle  NAE\]where the last equality is from the previously established concyclicity. Thus, both parts are proved and we have our claim.

Now, its just a matter of angle chasing. We note that
\[\measuredangle  NMD = \measuredangle AMF = \measuredangle AEF = \measuredangle AED + \measuredangle DEF = \measuredangle AED + \measuredangle DGF = \measuredangle AED + \measuredangle NGA   = \measuredangle AED + \measuredangle NEA = \measuredangle NED\]which implies that the points $N$ , $E$ , $D$ and $M$ are concyclic.

We are left to deal with $L$ which we will do as follows. Simply note that,
\[\measuredangle NML = \measuredangle AMC = \measuredangle NAC + \measuredangle ACL = \measuredangle ABC + \measuredangle AEL = \measuredangle DBF + \measuredangle AEL = \measuredangle NGA + \measuredangle AEL = \measuredangle NEA + \measuredangle AEL = \measuredangle NEL \]which then implies that $L$ lies on $(NEM)$. Thus, we are done.
Z K Y
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Shreyasharma
688 posts
#3 • 2 Y
Y by cursed_tangent1434, Rounak_iitr
[asy] 
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(8cm); 
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 = -6.279959829984758, xmax = 4.683599076677082, ymin = -3.3545640553934764, ymax = 5.472559889950314;  /* image dimensions */
pen ccqqqq = rgb(0.8,0,0); pen qqwwzz = rgb(0,0.4,0.6); 
 /* draw figures */
draw((-2.7289948738723533,-0.9208382624102394)--(-2.1523401062787375,2.8310990234356757), linewidth(0.7) + ccqqqq); 
draw((-2.1523401062787375,2.8310990234356757)--(-0.9175572083736604,-0.46589860995516075), linewidth(0.7) + ccqqqq); 
draw(circle((-1.806980067781941,1.6607746427676848), 1.220218305130859), linewidth(0.7) + qqwwzz); 
draw(circle((0.3909202925190619,0.6042038741711437), 1.6903350841995062), linewidth(0.7) + qqwwzz); 
draw(circle((-3.4088051058555418,1.7859687898225678), 2.7908683898611724), linewidth(0.7) + qqwwzz); 
draw((-3.696508040412313,4.561968319897044)--(-0.9175572083736604,-0.46589860995516075), linewidth(0.7) + ccqqqq); 
draw((-3.696508040412313,4.561968319897044)--(-2.7289948738723533,-0.9208382624102394), linewidth(0.7) + ccqqqq); 
draw((-2.7289948738723533,-0.9208382624102394)--(-0.623155624053535,1.9565657829452703), linewidth(0.7) + ccqqqq); 
draw((-0.623155624053535,1.9565657829452703)--(2.049742849421233,0.2793340208632457), linewidth(0.7) + ccqqqq); 
draw(circle((-0.29671037190742056,-0.49162974573266305), 2.4698639333549055), linewidth(0.7) + qqwwzz); 
draw((-3.696508040412313,4.561968319897044)--(-0.623155624053535,1.9565657829452703), linewidth(0.7) + ccqqqq); 
draw((-2.7289948738723533,-0.9208382624102394)--(2.049742849421233,0.2793340208632457), linewidth(0.7) + ccqqqq); 
draw((-2.4878448906463215,0.6481761018037537)--(2.049742849421233,0.2793340208632457), linewidth(0.7) + ccqqqq); 
draw((-2.7289948738723533,-0.9208382624102394)--(-0.683486394306744,1.184649997493327), linewidth(0.7) + ccqqqq);
 /* dots and labels */
dot((-2.7289948738723533,-0.9208382624102394),dotstyle); 
label("$A$", (-3.045036968642753,-1.3119238035783847), NE * labelscalefactor); 
dot((-2.1523401062787375,2.8310990234356757),dotstyle); 
label("$E$", (-2.5926903697329838,2.940102843057113), NE * labelscalefactor); 
dot((-0.9175572083736604,-0.46589860995516075),dotstyle); 
label("$D$", (-1.1961914215460196,-0.8995772048732832), NE * labelscalefactor); 
dot((-2.4878448906463215,0.6481761018037537),dotstyle); 
label("$B$", (-2.989977361342428,0.4639703052313937), NE * labelscalefactor); 
dot((-1.298593158231549,0.5515066578450925),dotstyle); 
label("$C$", (-1.2254673682632824,0.8202948142988752), NE * labelscalefactor); 
dot((2.049742849421233,0.2793340208632457),linewidth(4pt) + dotstyle); 
label("$F$", (2.265780002204128,0.27216455497209485), NE * labelscalefactor); 
dot((-0.683486394306744,1.184649997493327),linewidth(4pt) + dotstyle); 
label("$T$", (-0.5376395281114941,1.001678930620342), NE * labelscalefactor); 
dot((-0.623155624053535,1.9565657829452703),linewidth(4pt) + dotstyle); 
label("$M$", (-0.5689044299365753,2.252275003160194), NE * labelscalefactor); 
dot((-3.696508040412313,4.561968319897044),linewidth(4pt) + dotstyle); 
label("$X$", (-3.899610951861642,4.732623880364234), NE * labelscalefactor); 
dot((-1.2112892518629763,0.06554138915213582),linewidth(4pt) + dotstyle); 
label("$Y$", (-1.4588404397968325,-0.48018204373300654), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Diagram with the point naming of the rewritten problem.
Claim: $E$ is the Miquel point of $ANDF$.
Proof. Note that this is equivalent to showing the following rephrasing:
Rephrased problem wrote:
Consider quadrilateral $ABCD$, and let $\overline{AB} \cap \overline{CD}$ and $\overline{AD} \cap \overline{BC}$ be the points $E$ and $F$. Let $M$ be the Miquel point of complete quadrilateral $AFCE$. If $T = \overline{AC} \cap (BECM)$, then show that $(ATM)$ is tangent to $\overline{AD}$.
This is true as,
\begin{align*}
\angle TMA &= 180 - (\angle MTC + \angle MAC) \\
&= 180 -  (180 - \angle MED) - \angle MAC\\
&= \angle MBF - \angle MAC\\
& = \angle MAF - \angle MAC = \angle TAF
\end{align*}proving the claim. $\square$

Continuing in our rephrased wording we may define the points,
  • $X = \overline{ET} \cap (MTA)$
  • $Y = \overline{XD} \cap (MTA)$.
The problem is then finished if we can show:

Claim: $Y$ lies on $(MCDF)$.
Proof. Note that,
\begin{align*}
\angle MYD = 180 - \angle MYX = 180 - \angle MAX
\end{align*}and hence it suffices to show $\angle MAX = \angle MFD$. We will show this by demonstrating $\triangle XMA \sim \triangle AMF$. To do this first observe from $\overline{AD}$ tangent to $(AMX)$,
\begin{align*}
\angle MXA =  \angle MAF
\end{align*}We also have,
\begin{align*}
\angle XMA = \angle XTA = \angle ETC = \angle EMC = \angle AMF
\end{align*}and hence the claim follows and we're done. $\square$
Remarks: The problem seemed to be written unaturally, so it made sense to look for a change of reference. This basically seemed to work because $C$ seemed unimportant, whereas the Miquel point $E$ seemed to be the main focus.
This post has been edited 4 times. Last edited by Shreyasharma, Apr 27, 2024, 3:41 AM
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ItsBesi
159 posts
#4
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Let $(ABC)=\Omega$
Claim: Points $A$,$M$,$E$ and $F$ are concyclic.
Proof:
$\angle AMF \equiv \angle AMD \stackrel{\triangle AMD}{=} 180-\angle MAD-\angle MDA \equiv 180-\angle MAB-\angle MDA  \stackrel{\Omega}{=} 180- \angle ACB-\angle MDA=(180-\angle ACB)-\angle BDF$
$\stackrel{\Omega}{=} \angle AEB-\angle BDF  \stackrel{\omega}{=} \angle AEB-\angle BEF=(\angle AEF+\angle BEF)-\angle BEF=\angle AEF \implies$ Points $A$,$M$,$E$ and $F$ are concyclic $\square$.
Let $\odot (AMEF)=\Gamma_1$

Claim: Points $A$,$G$,$E$ and $N$ are concyclic.
Proof:
$\angle NGE \equiv \angle DGE \stackrel{\omega}{=} \angle DFE \equiv \angle MFE \stackrel{\Gamma_1}{=} \angle MAE \equiv \angle NAE \implies \angle NGE=\angle NAE \implies$ Points $A$,$G$,$E$ and $N$ are concyclic. $\square$
Let $\odot(AGEN)=\Gamma_2$

Claim: Points $M$,$N$,$E$ and $D$ are concyclic.
Proof:
$\angle MND \equiv \angle ANG  \stackrel{\Gamma_2}{=}\angle AEG=\angle AED+\angle DEG  \stackrel{\omega}{=} \angle AED+\angle DFG \equiv \angle AED+\angle MFA  \stackrel{\Gamma_1}{=} \angle AED+\angle MEA=\angle MED \implies$
$ \angle MND=\angle MED \implies$ Points $M$,$N$,$E$ and $D$ are concyclic. $\square$ Let $\odot(MNED)=\sigma$

Claim: $L \in \sigma$
Proof:
$\angle MLE=180-\angle ELC  \stackrel{\Omega}{=} \angle EBC \equiv \angle EBF \stackrel{\omega}{=} 180-\angle EGF  \stackrel{\Gamma_2}{=} 180-\angle ENA \equiv 180-\angle ENM \implies \angle MLE=180-\angle ENM \implies \angle MLE+\angle MNE=180 \implies$ Points $M$,$N$,$E$ and $L$ are cocyclic $\equiv L \in \odot(MNE) \iff L \in \sigma$ $\square$

Hence Points $M$ , $L$ , $D$ , $E$ and $N$ are concyclic. $\blacksquare$

Remark: Very nice problem, I really enjoyed solving it. I was looking for a Spiral Sim solution but couldn't find it.
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zuat.e
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$\measuredangle NAE=\measuredangle ACE=\measuredangle ABE=\measuredangle DGE=\measuredangle NGE$, so $AGEN$ is cyclic, implying $\measuredangle ENA=\measuredangle EGA=\measuredangle EDF$, hence $NEDM$ is cyclic.

Finally, $\measuredangle EDF=\measuredangle EBF=\measuredangle EAC=\measuredangle ELC$, hence $MLDE$ is cyclic, yielding the concyclicity of $MLDEN$, as desired.
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happypi31415
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First, notice that $EGAN$ is cyclic because $\angle NAE = \angle ACE = \angle EPA = \angle EGD$. Then, $NMDE$ is cyclic because $\angle MNE = \angle EGF = \angle EDF$ and we have $MNEL$ cyclic because $\angle MNE = \angle ELC = 180-\angle EPF = \angle EDF$, done.
This post has been edited 1 time. Last edited by happypi31415, May 31, 2025, 5:38 PM
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