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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]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
Yesterday at 11:16 PM
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
A sequence containing every natural number exactly once
Pomegranat   3
N 4 minutes ago by YaoAOPS
Source: Own
Does there exist a sequence \( \{a_n\}_{n=1}^{\infty} \), which is a permutation of the natural numbers (that is, each natural number appears exactly once), such that for every \( n \in \mathbb{N} \), the sum of the first \( n \) terms is divisible by \( n \)?
3 replies
Pomegranat
2 hours ago
YaoAOPS
4 minutes ago
An inequality with n of Maximum with Minimum
Qing-Cloud   2
N 5 minutes ago by Qing-Cloud
Let $ n $ be a positive integer divisible by 3, and let $ x_1, x_2, \dots, x_n $ be non-negative real numbers satisfying $ \sum_{i=1}^n x_i = 1 $. Define
\[
M_n = \min_{1 \le i \le n} \{x_i x_{3i}\},  
\]where the indices are taken modulo $ n $. Determine the maximum possible value of $ M_n $.
2 replies
1 viewing
Qing-Cloud
Yesterday at 2:18 PM
Qing-Cloud
5 minutes ago
Iranian Second Round P2
AlirezaOpmc   10
N 31 minutes ago by Lemmas
Source: Iran 2nd-round MO 2018 - P2
Let $n$ be odd natural number and $x_1,x_2,\cdots,x_n$ be pairwise distinct numbers. Prove that someone can divide the difference of these number into two sets with equal sum.
( $X=\{\mid x_i-x_j \mid | i<j\}$ )
10 replies
AlirezaOpmc
Apr 26, 2018
Lemmas
31 minutes ago
Inequality with 4 variables
bel.jad5   3
N an hour ago by sqing
Source: Own
Let $a$,$b$,$c$ $d$ positive real numbers. Prove that:
\[ \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \geq 4+\frac{8(a-d)^2}{(a+b+c+d)^2}\]
3 replies
bel.jad5
Sep 5, 2018
sqing
an hour ago
Cyclic symmetric ineq.with the range of variables (13_09_01)
Lastnightstar   5
N an hour ago by sqing
(1)If $a,b,c\in{[2-\sqrt{3},2+\sqrt{3}]},$then \[2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq \frac{b}{a}+\frac{c}{b}+\frac{a}{c}+3\]

(2)If $a,b,c,d\in{[\frac{1}{2},2]},$then \[2(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})\geq \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}+4\]
5 replies
Lastnightstar
Sep 1, 2013
sqing
an hour ago
Function equation
LeDuonggg   2
N an hour ago by jasperE3
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
2 replies
LeDuonggg
Yesterday at 2:59 PM
jasperE3
an hour ago
Trigonometry article for geometry
xytunghoanh   2
N an hour ago by appuk
Does anyone have any articles on using trigonometry to prove geometry problems (Law of Sines, Ceva's Theorem in trigonometric form,..) that they can share with me?
Thanks!
2 replies
1 viewing
xytunghoanh
3 hours ago
appuk
an hour ago
Queue geo
vincentwant   4
N 2 hours ago by vincentwant
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
4 replies
vincentwant
Wednesday at 3:54 PM
vincentwant
2 hours ago
Do not try to bash on beautiful geometry
ItzsleepyXD   9
N 2 hours ago by Captainscrubz
Source: Own , Mock Thailand Mathematic Olympiad P9
Let $ABC$be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$
9 replies
ItzsleepyXD
Wednesday at 9:30 AM
Captainscrubz
2 hours ago
centroid lies outside of triangle (not clickbait)
Scilyse   1
N 3 hours ago by LoloChen
Source: 数之谜 January (CHN TST Mock) Problem 5
Let $P$ be a convex polygon with centroid $G$, and let $\mathcal P$ be the set of vertices of $P$. Let $\mathcal X$ be the set of triangles with vertices all in $\mathcal P$. We sort the elements $\triangle ABC$ of $\mathcal X$ into the following three types:
[list]
[*] (Type 1) $G$ lies in the strict interior of $\triangle ABC$; let $\mathcal A$ be the set of triangles of this type.
[*] (Type 2) $G$ lies in the strict exterior of $\triangle ABC$; let $\mathcal B$ be the set of triangles of this type.
[*] (Type 3) $G$ lies on the boundary of $\triangle ABC$.
[/list]
For any triangle $T$, denote by $S_T$ the area of $T$. Prove that \[\sum_{T \in \mathcal A} S_T \geq \sum_{T \in \mathcal B} S_T.\]
1 reply
Scilyse
Jan 26, 2025
LoloChen
3 hours ago
4 lines concurrent
Zavyk09   6
N 3 hours ago by hectorleo123
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
6 replies
Zavyk09
Apr 9, 2025
hectorleo123
3 hours ago
No More than √㏑x㏑㏑x Digits
EthanWYX2009   4
N 3 hours ago by tom-nowy
Source: 2024 April 谜之竞赛-3
Let $f(x)\in\mathbb Z[x]$ have positive integer leading coefficient. Show that there exists infinte positive integer $x,$ such that the number of digit that doesn'r equal to $9$ is no more than $\mathcal O(\sqrt{\ln x\ln\ln x}).$

Created by Chunji Wang, Zhenyu Dong
4 replies
EthanWYX2009
Mar 24, 2025
tom-nowy
3 hours ago
Old hard problem
ItzsleepyXD   1
N 3 hours ago by ItzsleepyXD
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
1 reply
ItzsleepyXD
Apr 25, 2025
ItzsleepyXD
3 hours ago
Existence of a solution of a diophantine equation
syk0526   5
N 3 hours ago by cursed_tangent1434
Source: North Korea Team Selection Test 2013 #6
Show that $ x^3 + x+ a^2 = y^2 $ has at least one pair of positive integer solution $ (x,y) $ for each positive integer $ a $.
5 replies
syk0526
May 17, 2014
cursed_tangent1434
3 hours ago
Concurrency
Dadgarnia   27
N Apr 23, 2025 by zuat.e
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
27 replies
Dadgarnia
Mar 12, 2020
zuat.e
Apr 23, 2025
Concurrency
G H J
Source: Iranian TST 2020, second exam day 2, problem 4
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Dadgarnia
164 posts
#1 • 7 Y
Y by Purple_Planet, itslumi, jhu08, mathematicsy, Mango247, Mango247, ItsBesi
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
Z K Y
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GeoMetrix
924 posts
#5 • 8 Y
Y by mueller.25, amar_04, Mathasocean, Purple_Planet, agwwtl03, IFA, jhu08, sabkx
Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$.

[asy]
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
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Mathematicsislovely
245 posts
#7 • 3 Y
Y by Purple_Planet, jhu08, PRMOisTheHardestExam
CLAIM: $N$ is the centre of $\omega=\odot(CIQ)$.
Proof: If $N'$ is the centre of $\omega$ then $N'C=N'I$ and $\angle N'IA=90^\circ$ implies $N'I$ and $BC$ are parallel .Let $IN'$ cut $AC$ at $N''$.Then $$\angle N''IC=\angle ICB=\angle ICN''$$.It means $N''C=N''I$.So $N''\equiv N'$.Now the circle centre at $N''$ and with radius $N''C=N''I$ is the unique circle tangent to $AI$ at $I$.So $N''=N$..
$\square$

Now let $\omega=\odot(CIQ)$ cut $BC$ at $X$ then $NX=NC$ implies $\angle NXH=\angle NXC=\angle ABC$ so $NX$ and $AB$ are parallel.So the line joining $C$ and the intersection of $BN$ and $AX$ passes through midpoint of $AB$.In other words $AX$,$BN$,$CM$ are concurrent at a point.So let $F$ is the intersection of $MN$ and $BC$.So $(B,C;X,F)=-1$ and let M be the midpoint of $BC$.Then $FC.FB=FX.FM$ [since $F$ is the inverse image of $X$ w.r.t circle with diameter $BC$].

Now assume $Y= XN \cap AD$.We claim that $XYDC$ is cyclic.Indeed, as $XN$ and $AB$ are parallels so $\angle XYD=\angle BAD=180^\circ-\angle XCD$.
Now as $XY$ passes through center $N$ of $omega=\odot(CIQ)$ so $\angle ADX=\angle XDY=90^\circ$ and together with $\angle AMXY=90^\circ$ We get $AMXD$ is cyclic.
Let $F'= BC\cap AD$.So $F'D.F'A=F'C.F'B$ and $F'D.F'A=F'X.F'M$. So we have $F'C.F'B=F'X.F'M$.But we have previously proved that $FC.FB=FX.FM$ so $F \equiv F'$. So $BC,AD,MN$ concur at a point $F$.
This post has been edited 1 time. Last edited by Mathematicsislovely, Mar 26, 2020, 9:58 PM
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AlastorMoody
2125 posts
#8 • 7 Y
Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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Aryan-23
558 posts
#9 • 5 Y
Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
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AwesomeYRY
579 posts
#10 • 2 Y
Y by jhu08, PRMOisTheHardestExam
Claim: N is the circumcenter of (CIQ)
Proof:
Note that $IN\perp, AI, BC\perp AI \Longrightarrow NI\parallel BC$. Thus, \[\angle NIC = \angle BCI = \angle ICA = \angle ICN\]Thus, $NI=NC$, so the circle with center $N$ and radius $NI$ is both tangent to $AI$ at $I$ and passes through $C$, so we have sufficiently redefined $\omega$. $\square$

Now, define $X=\omega \cap BC$ and $Y=\omega \cap AD$

Claim 1: $NX\parallel AB$
This clearly follows from
\[\angle CNX = \angle NCX = \angle ACB=\angle ABC\]$\square$

Claim 2: $XY\parallel AB$
We angle chase (basically a rederivation of Reim's)
\[\angle DAB = \angle DCB = \angle DCX = \angle DYX\]$\square$

Combining these, we have that $N,X,Y$ are collinear. Since, $X,Y\in \omega$ and $N$ is the center of $\omega$, we have that $N$ is the midpoint of $XY$. Since $M$ is the midpoint of $AB$, combined with $YX\parallel AB$, we have that $AY,MN,BX$ are concurrent which finishes $\blacksquare$
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SatisfiedMagma
458 posts
#11 • 2 Y
Y by jhu08, PRMOisTheHardestExam
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15.929898374381239cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
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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
594 posts
#12 • 4 Y
Y by jhu08, SatisfiedMagma, PRMOisTheHardestExam, vrondoS
$\angle QIA = \angle QCI = \angle ICB = 90^\circ - \angle (IC , \text{angle bisector}) \implies \angle QIC = 90^\circ \implies N$ is the circumcenter of $\omega$.
Let $E=BC \cap \omega$ and $F=AD \cap \omega$.
$$\angle FEC = 180^\circ - \angle FDC = \angle ABC = \angle NCE = \angle NEC$$Thus $\overline{E-N-F}$ are collinear also $EF \parallel AB$ so by homothety we are done.
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srisainandan6
2811 posts
#13 • 2 Y
Y by PRMOisTheHardestExam, jhu08
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$
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RM1729
63 posts
#14 • 2 Y
Y by PRMOisTheHardestExam, jhu08
Consider $P$ as $AI \cap BC$.

Claim
$N$ is the centre of the circle $\omega$


Proof
We angle chase.

Note that since $ \angle IQC = \angle PIC = 90^{\circ} - \angle ICP = 90^{\circ} - \angle C/2 $ and $\angle QCI = \angle C/2$

We have that $\angle QIC = 90^{\circ}$ and thus $QC$ is a diameter

But since $N$ is the midpoint of $QC$ it must be the centre of the circle



We now define $X$ as $AD\cap \omega (\neq D)$ and $Y$ as $BC\cap \omega (\neq C)$

Note that $\angle NXC = \angle NCX$ but $\angle NCX = \angle ABC$ since the triangle is isosceles. Thus $\angle NXC = \angle ABC$ and so $NX||AB$

However we also have that $XY||AB$ by Reim's Theorem. Combining these we have that $ABXY$ is a trapezium and $M$ and $N$ are the midpoints of its parallel sides. A simple homothety argument proves that $AY, BX,MN$ that is $AD,BC,MN$ are concurrent.
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mohamad021
1 post
#15 • 1 Y
Y by jhu08
srisainandan6 wrote:
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?
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DottedCaculator
7345 posts
#16
Y by
Solution
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guptaamitu1
656 posts
#17 • 2 Y
Y by PRMOisTheHardestExam, Lantien.C
Here's a solution with radical axes ($\overline{MN}$ will become radical axes of two circles)
Let $\Omega = \odot(ABC)$ and $A',C'$ be antipode of $A,C$ wrt $\Omega$. $\angle QIC = \angle AIC - \angle ACI = 90^\circ$. Thus $\overline{QC}$ is a diameter of $\omega$, consequently $N$ is its center. So $\overline{CO},\overline{QD}$ intersect at $C'$. Let $T = \overline{AD} \cap \overline{BC}$, $\Gamma = \odot(BIC)$ and $X$ be a point of $\overline{AC}$ such that $\overline{QD}$ bisects $\angle ADX$, let $\gamma = \odot(AXD)$. hand drawn figure
[asy]
size(250);
pair A=dir(90),B=dir(-130),C=dir(-50),O=(0,0),I=incenter(A,B,C),N=extension(A,C,I,I+C-B),Q=2*N-C,Cp=-C,D=foot(C,Cp,Q),T=extension(A,D,B,C),M=1/2*(A+B),Ap=-A,X=foot(I,A,C);
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dot("$N$",N,dir(-120));
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dot("$C'$",Cp,dir(Cp));
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dot("$T$",T,dir(T));
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dot("$X$",X,dir(50));
draw(C--A--B--T^^A--Ap,red);
draw(M--T,purple);
draw(T--A^^Cp--D--X,green);
draw(C--Cp,green);
[/asy]
Claim: Points $M,N,T$ lie on radical axes $\ell$ of $\Gamma,\gamma$.
Proof: $T \in \ell$ is direct. For $M$: Note $\Gamma$ is tangent to $\overline{BC}$ as it has center $A'$. Note $$\angle OAC = \angle OCA = \angle C'CA = \angle C'DA = \angle QDA$$As $\angle BAC = 2 \angle OAC = 2 \angle QDA = \angle XDA$, so $\gamma$ is tangent to $\overline{AB}$. Hence power of $M$ wrt both $\Gamma,\gamma$ equals $MB^2 = MA^2$. For $N$: We will mostly focus of $\triangle ADX$. Observe
$$ \angle NDX = \angle NDQ - \angle XDQ = \angle NQD - \angle ADQ = \angle QAD = \angle NAD $$So $\overline{ND}$ is tangent to $\gamma$ (experts may also directly note that $N$ is the center of $D$-appolonius circle wrt $\triangle ADX$). Hence $NX \cdot NA = ND^2 = NC^2$. Laslty, recall $\overline{NC}$ is tangent to $\Gamma$ (as $A'$ is its center). This proves our Claim. $\square$

It follows points $M, N,T$ are collinear, solving our problem. $\blacksquare$

Motivation
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BarisKoyuncu
577 posts
#18 • 1 Y
Y by teomihai
Simple angle chasing gives us that $N$ is the center of $(IDC)$.
Let $AD\cap BC=S$. We have
$$\angle ASC=\angle ACB-\angle CAD=\angle ABC-\angle CAD=\angle ACD$$Hence, $NC$ is tangent to $(DCS)$. Since $|NC|=|ND|$, we know that $ND$ is tangent to $(DCS)$ as well. Then, $SN$ is symmedian in $DSC$. Since $DC$ and $AB$ are antiparallels wrt $DSC$, we find that $SN$ bisects $AB$, done.
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BarisKoyuncu
577 posts
#19
Y by
A kind of generalization
This post has been edited 1 time. Last edited by BarisKoyuncu, Mar 7, 2022, 8:36 AM
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JAnatolGT_00
559 posts
#20 • 1 Y
Y by MrOreoJuice
BarisKoyuncu wrote:
A kind of generalization

Proof. All angles are oriented. From $\angle ADB=\angle ACB=\angle CBA=\angle SBA$ we obtain $$\angle CDN=\angle NCD=\angle ABD=\angle CBS.$$Therefore $SN$ is a symmedian in $CDS$ and so bisects $AB.$
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dusicheng20080513
36 posts
#21
Y by
GeoMetrix wrote:
Nice and easy.
Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(14cm); 
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draw((3.8176841460738786,5.365866172483034)--(8.19899262334367,5.090227365365952), linewidth(0.4) + qqffff); 
draw((8.19899262334367,5.090227365365952)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + wrwrwr); 
draw((3.8176841460738786,5.365866172483034)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); 
draw((1.9295955015644004,1.832569513956508)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + qqffff); 
draw((3.8176841460738786,5.365866172483034)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); 
draw((5.807241390273251,2.031588419358404)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); 
draw((1.9295955015644004,1.832569513956508)--(2.2600756333314593,-4.606441501899994), linewidth(0.4) + ffqqff); 
draw((2.2600756333314593,-4.606441501899994)--(8.19899262334367,5.090227365365952), linewidth(0.4) + green); 
draw((-1.0161468573881236,3.63674779242076)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); 
draw((1.5534973808986277,9.160392608982256)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); 
draw((1.9295955015644004,1.832569513956508)--(5.807241390273251,2.031588419358404), linewidth(0.4) + wrwrwr); 
draw((7.796798634472622,-1.3026893337662244)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); 
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dot((1.5534973808986277,9.160392608982256),dotstyle); 
label("$A$", (1.6396828828887742,9.383171533517269), NE * labelscalefactor); 
dot((-3.5857910956748746,-1.8868970241407366),dotstyle); 
label("$B$", (-3.4819992193005715,-1.6553396638953617), NE * labelscalefactor); 
dot((7.796798634472622,-1.3026893337662244),dotstyle); 
label("$C$", (7.88392544583195,-1.0706727572527435), NE * labelscalefactor); 
dot((1.9295955015644004,1.832569513956508),linewidth(4pt) + dotstyle); 
label("$I$", (2.0138697031400508,2.01636850982028), NE * labelscalefactor); 
dot((5.807241390273251,2.031588419358404),linewidth(4pt) + dotstyle); 
label("$N$", (5.896057963247044,2.2268485962116227), NE * labelscalefactor); 
dot((3.8176841460738786,5.365866172483034),linewidth(4pt) + dotstyle); 
label("$Q$", (3.9081904806621375,5.547756625941694), NE * labelscalefactor); 
dot((8.19899262334367,5.090227365365952),linewidth(4pt) + dotstyle); 
label("D", (8.28149894234893,5.267116510753237), NE * labelscalefactor); 
dot((-1.0161468573881236,3.63674779242076),linewidth(4pt) + dotstyle); 
label("$M$", (-0.9328515063387511,3.8171425822795437), NE * labelscalefactor); 
dot((17.798762343332783,-0.7893418218123583),linewidth(4pt) + dotstyle); 
label("Q", (17.893422887553594,-0.602939231938649), NE * labelscalefactor); 
dot((4.675148007685625,3.928851637608016),linewidth(4pt) + dotstyle); 
label("$F$", (4.773497502493214,4.121169373733705), NE * labelscalefactor); 
dot((4.169500681742824,-1.488859213336944),linewidth(4pt) + dotstyle); 
label("$H$", (4.258990624647709,-1.3045395199097907), NE * labelscalefactor); 
dot((2.2600756333314593,-4.606441501899994),linewidth(4pt) + dotstyle); 
label("$J$", (2.3646698471256222,-4.414967463248519), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

When you use Geogebra-Asymptote conversion, how did you doit. Why did mine have an error?
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guptaamitu1
656 posts
#22
Y by
Firstly you have convert it from Geogebra Classic and not Geogebra Geometry. If you find easier to draw in Geogebra Geomtery (as I also do), then save it and reopen it in Geogebra Classic. After the conversion, you have to do some slight edits. Like changing the size and defaultpen.
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IAmTheHazard
5001 posts
#23 • 1 Y
Y by Lantien.C
We begin with the following key claim.

Claim: $N$ is the center of $\omega$.
Proof: Let $N'$ be the intersection of the perpendicular to $\overline{AI}$ at $I$ with $\overline{AC}$. Then we have
$$\angle N'IC=\angle ICB=\angle ICA=\angle ICN',$$so $N'I=N'C$ and thus $N'$ is the center of $\omega$. Then $Q$ is the reflection of $C$ over $N'$, hence $N'$ is the midpoint of $\overline{CQ}$ and we have $N=N'$, which implies the desired result.

Now let $P \neq D$ be the intersection of $\omega$ with $\overline{AD}$, $R \neq C$ be the intersection of $\omega$ with $\overline{BC}$, and $X=\overline{AD} \cap \overline{BC}$. By Reim's, we have $\overline{AB} \parallel \overline{PR}$, so
$$\angle QPR=\angle QCR=\angle ABC=\angle PRC,$$so we have $\overline{PQ} \parallel \overline{CR}$ as well. Then we have $90^\circ=\angle QPC=\angle PCR$, so $CPQR$ is a rectangle, and $N$ is the midpoint of $\overline{PR}$. But triangles $\triangle XAB$ and $\triangle XPR$ are homothetic, which implies that $\overline{MN}$ passes through $X$ as well, hence $\overline{AD}$, $\overline{MN}$, and $\overline{BC}$ concur. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Apr 14, 2022, 4:02 PM
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Lantien.C
7 posts
#24
Y by
mohamad021 wrote:
srisainandan6 wrote:
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles.

Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$

Can you explain Reim's theorem please?
JI//FH,as the picture says
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ihatemath123
3446 posts
#25
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By angle chasing, $\angle QIC = 90^{\circ}$, hence $\angle QDC = 90^{\circ}$, hence $\angle ADQ = 90 - \angle B$, hence $(ADQ)$ is tangent to $\overline{AI}$.

If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$, then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$, hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$, and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$, hence they all concur by the radical axis theorem.
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trk08
614 posts
#26
Y by
Claim:
$N$ is the center of $(CIQ)$
Proof:
By angle-chasing, we can see that:
\[\angle CQI=\angle CIF=90^{\circ}-\angle ICF=90^{\circ}-\angle ICQ.\]Therefore, $\angle QIC=90^{\circ}$, or $N$ is the center $\blacksquare$

Claim:
The line parallel to $AB$ that passes through $N$ goes through $E$ and $F$.
Proof:

Denote $E=(CQD)\cap BC$ and $F=(CQD)\cap AD$. By Reim's theorem, $EF$ is parallel to $AB$.

Also, $\triangle{NEC}$ is isosceles, so it is similar to $\triangle{ABC}$. As a result, $NE\parallel AB$. Therefore, $N,E,F$ are all collinear and parallel to $AB$ $\blacksquare$

As a result, we can take a homothety at $T=AD\cap BC$, which sends $N$ to $M$, implying the desired result $\square$
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asdf334
7585 posts
#27
Y by
Note that $N$ is the center of $\omega$. Let $E$ be the other intersection of $\omega$ with $BC$ and let $E'$ be its antipode wrt $\omega$. Let $F$ be the midpoint of $BC$. It suffices to show that $AFED$ is cyclic as this implies $A,E',D$ collinear.

Here's the interesting part.

Claim: Let $ABCD$ be a cyclic quadrilateral. Let $F$ and $E$ be points where $B,F,E,C$ are on $BC$ in that order. If $\angle BAF=\angle CDE$ then $AFED$ is cyclic.
Proof: Angle chase.

Now we're done; just apply the claim here.
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bjump
1010 posts
#28
Y by
Claim:$N$ is the center of $\omega$.
Proof Let $O$ denote the center of $\omega$. $\angle ICQ= \angle ICB = \tfrac{1}{2} \angle QOI$. Since $OI \perp AI$, then $OI \parallel BC$. Since $\angle QOI = \angle  ICQ + \angle ICB= \angle BCA$, $O$ lies on $AC$. Since $OQ=OC$, then $O$ is the midpoint of $CQ$. So $O= N$. $\square$

By Reims on $\omega$ and $(ABC)$ with lines $AD$ and $BC$ we get that the line through $D$ parallel to $AB$ intersects $BC$ on $\omega$ at a point we will call $P$ then let $DP\cap MN= Q$ then by homothety $BP$, $AD$, $MQ$ are concurrent so $MN$, $BC$, and $AD$ are concurrent.
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LuciferMichelson
18 posts
#29
Y by
Define $R$ as midpoint of $BC$
Easy to see $N$ is center of $DQIC$.
After that define $S=(DIQ) \cap BC$
Angle chasing shows $SN//AB$ so $AS,BN,CM$ concurrent.
Now we should show that $(B,C;S,AD \cap BC)=-1$
Let $AD \cap BC= K'$ so $(B,C;S,AD \cap BC)=-1$ is equal to $KD.KA=KS.KR$ and from angle chasing it is easy that show $ADSR$ is cyclic.
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shendrew7
794 posts
#30
Y by
If $N'$ is the intersection of $AC$ with the line through $I$ parallel $BC$, we know $N'I \perp AI$ and
\[\angle ACI = \angle ICB = \angle NIC,\]
so $N'$ is the center of $\omega$. Therefore $N'$ is the midpoint of $CQ$, so $N' = N$.

Next we define the intersections of $AC$ and $BC$ with $\omega$ as $K$ and $L$. Looking at $CD$, Reim's tells us that $AB \parallel KL$, which then gives
\[\angle LKQ = \angle LCQ = \angle ABC = \angle KLC,\]
so $KQ \parallel BC$ as well. As a result, $\angle KCL = \angle QKC = \angle QLC = 90$, so $KL$ is a diameter of $\omega$, and hence passes through $M$. We finish by noting the homothety which maps $KL$ to $AB$ also maps corresponding midpoints $N$ to $M$.
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joshualiu315
2533 posts
#31
Y by
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
55 posts
#32
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First, let $N'$ be the center of $(CQDI)$. Note that $IN'\parallel BC$, therefore $\measuredangle N'CI=\measuredangle CIN'=\measuredangle ICB$, hence $N'$ lies on $AC$ and consequently $N'\equiv N$.

Let $R = AD\cap BC$ and let $E,F = BC, AD\cap (CQDI)$. We will prove that $R-N-M$ are collinear. The main claim is the following:
Claim: $EF$ is a diameter of $(CQID)$ and parallel to $AB$
Proof: $\measuredangle CEN=\measuredangle NCE=\measuredangle ACB=\measuredangle CBA$ and if we define $F'=EN\cap AD$, $\measuredangle EFD=\measuredangle BAR=\measuredangle RCD$, from which it follows $F=F'$ and $EF$ is a diameter of $(CQID)$ parallel to AB.

Now consider the homothety centered at $R$ sending $\triangle RFE$ to $\triangle RAB$. As both $RM$ and $RN$ are the respective medians of $\triangle RFE$ and $\triangle RAB$,
\[X_R: N\mapsto M\]hence $R-N-M$ are collinear, as desired.
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