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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
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   2
N 9 minutes ago by korncrazy
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
korncrazy
9 minutes ago
A sharp one with 4 var
mihaig   0
an hour ago
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\left(a+b+c+d-1\right)^2+7\leq\frac83\cdot\left(ab+bc+ca+ad+bd+cd\right).$$Prove
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}\leq2.$$
0 replies
mihaig
an hour ago
0 replies
A geometry problem
Lttgeometry   0
an hour ago
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 $AQ$.
0 replies
Lttgeometry
an hour ago
0 replies
A sharp one with 3 var
mihaig   9
N an hour ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
9 replies
mihaig
May 13, 2025
mihaig
an hour ago
Van der Corput Inequality
EthanWYX2009   0
an hour ago
Source: en.wikipedia.org/wiki/Van_der_Corput_inequality
Let $V$ be a real or complex inner product space. Suppose that ${\displaystyle v,u_{1},\dots ,u_{n}\in V} $ and that ${\displaystyle \|v\|=1}.$ Then$${\displaystyle \displaystyle \left(\sum _{i=1}^{n}|\langle v,u_{i}\rangle |\right)^{2}\leq \sum _{i,j=1}^{n}|\langle u_{i},u_{j}\rangle |.}$$
0 replies
EthanWYX2009
an hour ago
0 replies
Hard Functional Equation in the Complex Numbers
yaybanana   6
N 2 hours ago by jasperE3
Source: Own
Find all functions $f:\mathbb {C}\rightarrow \mathbb {C}$, s.t :

$f(xf(y)) + f(x^2+y) = f(x+y)x + f(f(y))$

for all $x,y \in \mathbb{C}$
6 replies
yaybanana
Apr 9, 2025
jasperE3
2 hours ago
FE with conditions on $x,y$
Adywastaken   3
N 2 hours ago by jasperE3
Source: OAO
Find all functions $f:\mathbb{R_{+}}\rightarrow \mathbb{R_{+}}$ such that $\forall y>x>0$,
\[
f(x^2+f(y))=f(xf(x))+y
\]
3 replies
Adywastaken
Friday at 6:18 PM
jasperE3
2 hours ago
Point satisfies triple property
62861   36
N 2 hours ago by cursed_tangent1434
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
36 replies
62861
Jan 22, 2018
cursed_tangent1434
2 hours ago
Inspired by 2025 Xinjiang
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 12+8\sqrt 2 $$
1 reply
sqing
Yesterday at 5:32 PM
sqing
2 hours ago
Inspired by 2025 Beijing
sqing   4
N 2 hours ago by pooh123
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
4 replies
sqing
Yesterday at 4:56 PM
pooh123
2 hours ago
Find the minimum
sqing   27
N 3 hours ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
27 replies
sqing
Sep 4, 2018
sqing
3 hours ago
Sharygin 2025 CR P15
Gengar_in_Galar   7
N 3 hours ago by Giant_PT
Source: Sharygin 2025
A point $C$ lies on the bisector of an acute angle with vertex $S$. Let $P$, $Q$ be the projections of $C$ to the sidelines of the angle. The circle centered at $C$ with radius $PQ$ meets the sidelines at points $A$ and $B$ such that $SA\ne SB$. Prove that the circle with center $A$ touching $SB$ and the circle with center $B$ touching $SA$ are tangent.
Proposed by: A.Zaslavsky
7 replies
Gengar_in_Galar
Mar 10, 2025
Giant_PT
3 hours ago
Inequality olympiad algebra
Foxellar   1
N 3 hours ago by sqing
Given that \( a, b, c \) are nonzero real numbers such that
\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]let \( M \) be the maximum value of the expression
\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]Determine the sum of the numerator and denominator of the simplified fraction representing \( M \).
1 reply
Foxellar
Yesterday at 10:01 PM
sqing
3 hours ago
Inspired by RMO 2006
sqing   2
N 3 hours ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
2 replies
sqing
Yesterday at 3:24 PM
sqing
3 hours ago
Two circles, a tangent line and a parallel
Valentin Vornicu   105
N May 15, 2025 by Fly_into_the_sky
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
105 replies
Valentin Vornicu
Oct 24, 2005
Fly_into_the_sky
May 15, 2025
Two circles, a tangent line and a parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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Valentin Vornicu
7301 posts
#1 • 19 Y
Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only., ItsBesi, cubres
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
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arqady
30258 posts
#2 • 29 Y
Y by jam10307, shiningsunnyday, claserken, shawnee03, futurestar, Richangles, myh2910, OlympusHero, User582032, starchan, sotpidot, FaThEr-SqUiRrEl, centslordm, Adventure10, megarnie, asdf334, Numbertheorydog, HWenslawski, rayfish, Kanimet0, ImSh95, Mango247, CoC_Ali, Rounak_iitr, Math_.only., ehuseyinyigit, cubres, and 2 other users
let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$
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yetti
2643 posts
#3 • 10 Y
Y by JasperL, FaThEr-SqUiRrEl, centslordm, Adventure10, Designerd, jhu08, ImSh95, Rounak_iitr, cubres, and 1 other user
The radical axis MN of the circles $(G_1), (G_2)$ cuts the tangent length segment AB at its midpoint C. Since $PQ \parallel AB$, the triangles $\triangle ABN \sim \triangle PQN$ are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii $G_1A, G_2B$ of the circles $(G_1), (G_2)$ are perpendicular to the tangent AB, i.e, also to the chords AM, DM. Hence they cut these chords at their midpoints, which means that the triangles $\triangle ACM, \triangle BDM$ are isosceles and the angles $\angle ACM = \angle AMC, \angle BDM = \angle BMD$ are equal. Since the lines $AB \parallel CD$ are parallel and $M \in CD$, the angles $\angle MAB = \angle AMC = \angle ACM = \angle EAB$ are equal and similarly, the angles $\angle MBA = \angle BMD = \angle BDM = \angle EBA$ are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals $AB \perp EM$ are perpendicular to each other. Consequently, the lines $EM \perp CD \equiv PQ$ are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle $\triangle EPQ$ is isosceles with EP = EQ.
Attachments:
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Arne
3660 posts
#4 • 8 Y
Y by mihajlon, OlympusHero, FaThEr-SqUiRrEl, Adventure10, ImSh95, cubres, and 2 other users
An "extra" question:

Prove that $EN$ bisects $\angle{CND}$.
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Zhero
2043 posts
#5 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
Since $ AB$ is the common tangent of $ G_1$ and $ G_2$ and $ MN$ is the radical axis of $ G_1$ and $ G_2$, $ MN$ bisects $ AB$. A homothety centered at $ N$ that maps $ A$ to $ P$ and $ B$ to $ Q$ therefore maps the midpoint of $ AB$ to $ PQ$, so we have that $ M$ is the midpoint of $ PQ$.

It is sufficient to show that $ EM \perp CD$. Let $ R$ and $ S$ be the projections of $ A$ and $ B$ onto $ CD$, respectively; since $ AB || CD$, $ AR = BS$. We first note that $ AC = AM$, since $ m \angle AMC = m \angle EAB = m \angle ACM$ (as $ AB$ is tangent to $ G_1$). We can similarly deduce that $ BM = BD$.

It also follows from this that $ CM = 2CR$ and $ DM = 2DS$. Let $ E'$ be the point such that $ E'M \perp CD$ and $ E'M = 2AR = 2BS$. A homothety centered at $ C$ that maps $ R$ to $ M$ must therefore map $ A$ to $ E'$, and a homothety centered at $ D$ must therefore map $ B$ to $ E'$. But this means that $ E'$ lies on both $ AC$ and $ BD$, implying that $ E' = E$. It follows that $ EM \perp CD$, so we are done.
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Pikachu!!!
58 posts
#6 • 6 Y
Y by S117, FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
Arne wrote:
An "extra" question:

Prove that $ EN$ bisects $ \angle{CND}$.

This extra question is also beautiful one.

First, from what we know Let $ \angle{ANM}=\angle{BAM}=\angle{BAE}=x$ and $ \angle{BNM}=\angle{ABM}=\angle{ABE}=y$

Therefore, $ \angle{AEB}+\angle{ANB}= (\pi-x-y)+(x+y)=\pi$ $ \longrightarrow$ $ A,N,B,E$ is concyclic.

It follows that $ \angle{BNE}=\angle{BAE}=x$ and it is known that $ \angle{BND}=\angle{BNM}=y$
We get $ \angle{DNE}=x+y$

Similarly, we also have $ \angle{CNE}=x+y$ :lol: :lol: :lol:
This is what we want to prove.
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arshakus
769 posts
#7 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
$K$ is the midpint of $AB$. $NM$ is radical axis for $G_1,G_2$ circles. also we know that radical axis passes through the midpoints of comment tangents. Thus $N,M,K$ lies on the same line. $PQ||AB=>M$ is the midpoint of $PQ$.
Also it is easy to note this number of equations.
$\angle{PCA}=\angle{ANM}=\angle{BAE}=\angle{MAB}=\angle{PMA}=\angle{CNA}=\angle{ENB}$[*] the last equation follows from the fact that $AENB$ is cyclic.
Similarly we can get in the opposite side. Thus $\triangle {AMB}=\triangle {AEB}=>EM\perp PQ=>EP=EQ$
@Arne
from [*] it is easy to note it.
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arshakus
769 posts
#8 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
One more "extra" question!)
Prove that $2AB=CD$
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siavosh
29 posts
#9 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres
arshakus wrote:
One more "extra" question!)
Prove that $2AB=CD$
In your solution we prove that $ \triangle{AMB}=\triangle{AEB}=> EB=BM=BD$
and $ CD\parallel AB =>  \frac{EB}{ED}=\frac{AB}{CD}=2 $
:)
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JSGandora
4216 posts
#10 • 6 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, Rounak_iitr, cubres
Let $\angle ACM=\alpha$, then since $AB||CD$, we have $\angle EAB=\angle ACM=\alpha$ and since $AB$ is tangent to circle $G_1$, we have $\angle BAM=\angle ACM=\alpha$. Thus $\angle EAB=\angle BAM=\alpha$. Similarly, if we let $\angle BDM=\beta$, we find $\angle EBA=\angle ABM=\beta$.

Now let the intersection of $EM$ and $AB$ be $F$, then $\angle AFE=90^\circ$ and $\angle CME=\angle AFE=90^\circ$ and therefore proving $EP=EQ$ is equivalent to proving $PM=QM$ since $EM\perp PQ$.

Let the intersection of ray $NM$ and $AB$ be point $G$, then $GA=GB$ because $G$ is on the radical axis of the two circles and thus $GA^2=GB^2\implies GA=GB$. Additionally, since $AB||CD$, we have $\triangle AGN~\triangle PMN$ and $\triangle BGN~\triangle QMN$ and therefore $PM=PQ$ as desired. $\blacksquare$
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BBAI
563 posts
#11 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
As the tangent is parallel to the chords So it follows that $ \triangle ACM$ and $ \triangle BDM$ are isosceles and $EA=AM=AC$ so $EM$ is perpendicular to $PQ$. Define $Z=MN \cap AB$.$Z$ lies on radical axis ,so $Z$ is the midpoint of $AB$. And hence $M$ is the midpoint of $PQ$. So done.
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subham1729
1479 posts
#12 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
OMG!! got a proof without bashing :O - if $MN$ meets $AB$ at $K$ then $AK^2=BK^2\implies AK=KB\implies PM=MQ\implies \angle{CMN}=\frac {\pi}{2}\implies EP=EQ$
This post has been edited 1 time. Last edited by subham1729, May 3, 2013, 12:04 PM
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BBAI
563 posts
#13 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, cubres
I think $E,M,N$ are not at all collinear.
I have drawn a ggb diagram. and checked it.
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sunken rock
4401 posts
#14 • 6 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, cubres, and 1 other user
Yet additional requirement:

the quadrilateral $ANBE$ is harmonic!


Best regards,
sunken rock
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exmath89
2572 posts
#15 • 7 Y
Y by Math-Star., FaThEr-SqUiRrEl, ImSh95, Adventure10, Rounak_iitr, cubres, and 1 other user
Solution
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G
H
=
a