Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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)!

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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
Difficult combinatorics problem
shactal   5
N 6 minutes ago by shactal
Can someone help me with this problem? Let $n\in \mathbb N^*$. We call a distribution the act of distributing the integers from $1$
to $n^2$ represented by tokens to players $A_1$ to $A_n$ so that they all have the same number of tokens in their urns.
We say that $A_i$ beats $A_j$ when, when $A_i$ and $A_j$ each draw a token from their urn, $A_i$ has a strictly greater chance of drawing a larger number than $A_j$. We then denote $A_i>A_j$. A distribution is said to be chicken-fox-viper when $A_1>A_2>\ldots>A_n>A_1$ What is $R(n)$
, the number of chicken-fox-viper distributions?
5 replies
shactal
Yesterday at 10:40 AM
shactal
6 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N 9 minutes ago by Ash_the_Bash07
Let $\triangle ABC$ satisfy $AB<AC$. The circumcircle $(O)$ and the incircle $(I)$ of $\triangle ABC$ are tangent to the sides $AC,AB$ at $E,F$, respectively. The line $BI$ meets $EF$ at $M$ and intersects $AC$ at $P$, while the line $BO$ meets $CM$ at $Q$. Construct the common external tangent $\ell$ (different from $BC$) to the incircles of the triangles $PBC$ and $QBC$. Show that $\ell$ is parallel to the line $PQ$.
1 reply
ZeltaQN2008
12 minutes ago
Ash_the_Bash07
9 minutes ago
Point inside parallelogram
BigSams   21
N 16 minutes ago by Want-to-study-in-NTU-MATH
Source: Canadian Mathematical Olympiad - 1997 - Problem 4.
The point $O$ is situated inside the parallelogram $ABCD$ such that $\angle AOB+\angle COD=180^{\circ}$. Prove that $\angle OBC=\angle ODC$.
21 replies
BigSams
May 7, 2011
Want-to-study-in-NTU-MATH
16 minutes ago
Geometry
MathsII-enjoy   1
N 18 minutes ago by MathsII-enjoy
Given triangle $ABC$ inscribed in $(O)$ with $M$ being the midpoint of $BC$. The tangents at $B, C$ of $(O)$ intersect at $D$. Let $N$ be the projection of $O$ onto $AD$. On the perpendicular bisector of $BC$, take a point $K$ that is not on $(O)$ and different from M. Circle $(KBC)$ intersects $AK$ at $F$. Lines $NF$ and $AM$ intersect at $E$. Prove that $AEF$ is an isosceles triangle.
1 reply
MathsII-enjoy
May 15, 2025
MathsII-enjoy
18 minutes ago
Probably a good lemma
Zavyk09   5
N 26 minutes ago by Orzify
Source: found when solving exercises
Let $ABC$ be a triangle with circumcircle $\omega$. Arbitrary points $E, F$ on $AC, AB$ respectively. Circumcircle $\Omega$ of triangle $AEF$ intersects $\omega$ at $P \ne A$. $BE$ intersects $CF$ at $I$. $PI$ cuts $\Omega$ and $\omega$ at $K, L$ respectively. Construct parallelogram $KFRE$. Prove that $A, R, L$ are collinear.
5 replies
Zavyk09
Yesterday at 12:50 PM
Orzify
26 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
an hour ago
Two perpendiculars
jayme   2
N an hour ago by jayme
Source: Own?
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. D the pole of BC wrt 0
4. B', C' the symmetrics of B, C wrt AC, AB
5. 1b, 1c the circumcircles of the triangles BB'D, CC'D
6. J the center of 1b
7. V the second point of intersection of DJ and 1c.

Prove : CV is perpendicular to BC.

Sincerely
Jean-Louis
2 replies
jayme
5 hours ago
jayme
an hour ago
IMO 2009, Problem 3
orl   52
N an hour ago by N3bula
Source: IMO 2009, Problem 3
Suppose that $ s_1,s_2,s_3, \ldots$ is a strictly increasing sequence of positive integers such that the sub-sequences \[s_{s_1},\, s_{s_2},\, s_{s_3},\, \ldots\qquad\text{and}\qquad s_{s_1+1},\, s_{s_2+1},\, s_{s_3+1},\, \ldots\] are both arithmetic progressions. Prove that the sequence $ s_1, s_2, s_3, \ldots$ is itself an arithmetic progression.

Proposed by Gabriel Carroll, USA
52 replies
orl
Jul 15, 2009
N3bula
an hour ago
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such : $f(
guramuta   2
N an hour ago by GreekIdiot
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such :
$f(x+yf(x)) + f(xf(y)-y) = f(x) - f(y) + 2xy$
2 replies
guramuta
Yesterday at 2:18 PM
GreekIdiot
an hour ago
Another triangle
Rushil   14
N 2 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 1991 Problem 1
Let $P$ be an interior point of a triangle $ABC$ and $AP,BP,CP$ meet the sides $BC,CA,AB$ in $D,E,F$ respectively. Show that \[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.  \]
Remark
14 replies
Rushil
Oct 15, 2005
SomeonecoolLovesMaths
2 hours ago
13rd ibmo - rep. dominicana 1998/q2.
carlosbr   6
N 2 hours ago by fearsum_fyz
Source: Spanish Communities
The circumference inscribed on the triangle $ABC$ is tangent to the sides $BC$, $CA$ and $AB$ on the points $D$, $E$ and $F$, respectively. $AD$ intersect the circumference on the point $Q$. Show that the line $EQ$ meet the segment $AF$ at its midpoint if and only if $AC=BC$.
6 replies
carlosbr
Apr 16, 2006
fearsum_fyz
2 hours ago
Inspired by old results
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a, b\geq 0, a+b=2. $ Prove that
$$\frac {24}{25} < \frac {1}{a^3 + 1 + ab}+\frac {1}{b^3 +1 + ab} +\frac {1}{a^3 + b^3 + ab} \leq \frac {89}{72}$$Let $ a, b\geq 0,  a+b+ab=3. $ Prove that
$$\frac {4}{5} < \frac {1}{a^3 + 1 + ab}+\frac {1}{b^3 +1 + ab} +\frac {1}{a^3 + b^3 + ab} \leq \frac {811}{756}$$Let $ a, b\geq 0,  a+b+ab=2. $ Prove that
$$\frac {23}{20} < \frac {1}{a^3 + 1 + ab}+\frac {1}{b^3 +1 + ab} +\frac {1}{a^3 + b^3 + ab} \leq \frac {404+291\sqrt{3}}{506}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
RMM 2013 Problem 6
dr_Civot   15
N 3 hours ago by N3bula
A token is placed at each vertex of a regular $2n$-gon. A move consists in choosing an edge of the $2n$-gon and swapping the two tokens placed at the endpoints of that edge. After a finite number of moves have been performed, it turns out that every two tokens have been swapped exactly once. Prove that some edge has never been chosen.
15 replies
dr_Civot
Mar 3, 2013
N3bula
3 hours ago
3-variable inequality with min(ab,bc,ca)>=1
mathwizard888   72
N 3 hours ago by math-olympiad-clown
Source: 2016 IMO Shortlist A1
Let $a$, $b$, $c$ be positive real numbers such that $\min(ab,bc,ca) \ge 1$. Prove that $$\sqrt[3]{(a^2+1)(b^2+1)(c^2+1)} \le \left(\frac{a+b+c}{3}\right)^2 + 1.$$
Proposed by Tigran Margaryan, Armenia
72 replies
mathwizard888
Jul 19, 2017
math-olympiad-clown
3 hours ago
Concurrency
Dadgarnia   29
N May 3, 2025 by blueprimes
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
29 replies
Dadgarnia
Mar 12, 2020
blueprimes
May 3, 2025
Concurrency
G H J
Source: Iranian TST 2020, second exam day 2, problem 4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(14cm); 
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 = -10.006881897432205, xmax = 25.634412731501875, ymin = -8.577795838543958, ymax = 14.83226710342647;  /* image dimensions */
pen ffqqff = rgb(1,0,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); 
 /* draw figures */
draw((1.5534973808986277,9.160392608982256)--(-3.5857910956748746,-1.8868970241407366), linewidth(0.8)); 
draw((1.5534973808986277,9.160392608982256)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); 
draw((-3.5857910956748746,-1.8868970241407366)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); 
draw((1.5534973808986277,9.160392608982256)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); 
draw(circle((5.807241390273251,2.031588419358404), 3.8827498197705226), linewidth(0.4) + wvvxds); 
draw(circle((1.9067865071150452,2.276975553541132), 6.892477317034895), linewidth(0.4) + dtsfsf); 
draw(circle((3.5081332893469326,4.891548128823499), 4.6950649335634775), linewidth(0.4) + wvvxds); 
draw((8.19899262334367,5.090227365365952)--(5.807241390273251,2.031588419358404), linewidth(0.4) + qqffff); 
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); 
 /* dots and labels */
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]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AlastorMoody
2125 posts
#8 • 7 Y
Y by GeoMetrix, Purple_Planet, SenatorPauline, Aryan-23, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aryan-23
558 posts
#9 • 5 Y
Y by AlastorMoody, Siddharth03, Muaaz.SY, jhu08, PRMOisTheHardestExam
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SatisfiedMagma
461 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 */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = 1.7630505196437911, xmax = 17.69294889402503, ymin = -4.7304, ymax = 4.139286318662105;  /* image dimensions */
pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffqqff = rgb(1.,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); 

filldraw((6.775903779376068,3.721548042371603)--(3.38614791445706,-2.12)--(9.796305353838916,-2.3192551065540985)--cycle, invisible, linewidth(1.) + blue); 
filldraw((6.944386092416243,-0.3160792577621652)--(6.9535157076039695,-0.022374008615092755)--(6.659810458456897,-0.013244393427366385)--(6.65068084326917,-0.30694964257443885)--cycle, invisible, linewidth(1) + qqwuqq); 
filldraw((6.884931883295061,-2.2287571684647753)--(6.894061498482787,-1.935051919317703)--(6.600356249335714,-1.9259223041299767)--(6.591226634147988,-2.219627553277049)--cycle, invisible, linewidth(0.4) + qqwuqq); 
filldraw((9.62150852293415,1.5152303300388892)--(9.612378907746429,1.2215250808918166)--(9.9060841568935,1.2123954657040958)--(9.915213772081222,1.5061007148511685)--cycle, invisible, linewidth(0.4) + qqwuqq); 
filldraw((8.026350584687354,-2.264237343034512)--(8.03548019987508,-1.9705320938874396)--(7.741774950728007,-1.9614024786997133)--(7.732645335540281,-2.2551077278467857)--cycle, invisible, linewidth(0.4) + qqwuqq); 
 /* draw figures */
draw(circle((6.656692229708716,-0.11355970054063641), 3.8369600979576926), linewidth(0.4) + ffxfqq); 
draw((6.775903779376068,3.721548042371603)--(3.38614791445706,-2.12), linewidth(0.4) + blue); 
draw((3.38614791445706,-2.12)--(9.796305353838916,-2.3192551065540985), linewidth(0.4) + blue); 
draw((9.796305353838916,-2.3192551065540985)--(6.775903779376068,3.721548042371603), linewidth(0.4) + blue); 
draw((6.65068084326917,-0.30694964257443885)--(9.796305353838916,-2.3192551065540985), linewidth(0.4)); 
draw((6.775903779376068,3.721548042371603)--(6.591226634147988,-2.219627553277049), linewidth(0.4)); 
draw(circle((8.823929553810784,-0.37450350649783165), 2.1742983885386487), linewidth(0.4) + red); 
draw((8.823929553810784,-0.37450350649783165)--(6.65068084326917,-0.30694964257443885), linewidth(0.4)); 
draw((7.85155375378265,1.5702480935584389)--(7.732645335540281,-2.2551077278467857), linewidth(0.4) + green); 
draw((10.230134161477556,1.2838577580577875)--(9.796305353838916,-2.3192551065540985), linewidth(0.4)); 
draw((7.732645335540281,-2.2551077278467857)--(10.230134161477556,1.2838577580577875), linewidth(0.4)); 
draw((7.85155375378265,1.5702480935584389)--(10.230134161477556,1.2838577580577875), linewidth(0.4)); 
draw((6.65068084326917,-0.30694964257443885)--(10.230134161477556,1.2838577580577875), linewidth(0.4)); 
draw((6.65068084326917,-0.30694964257443885)--(7.85155375378265,1.5702480935584389), linewidth(0.4) + xfqqff); 
draw((6.65068084326917,-0.30694964257443885)--(7.732645335540281,-2.2551077278467857), linewidth(0.4) + xfqqff); 
draw((6.537480680041361,-3.9486674434528757)--(7.732645335540281,-2.2551077278467857), linewidth(0.4)); 
draw((6.537480680041361,-3.9486674434528757)--(6.591226634147988,-2.219627553277049), linewidth(0.4)); 
draw((9.915213772081222,1.5061007148511685)--(7.85155375378265,1.5702480935584389), linewidth(0.4) + blue); 
draw((9.915213772081222,1.5061007148511685)--(9.796305353838916,-2.3192551065540985), linewidth(0.4) + green); 
draw((7.732645335540281,-2.2551077278467857)--(9.915213772081222,1.5061007148511685), linewidth(0.4) + blue); 
draw(circle((7.254274557458178,0.7332201572624102), 3.026374423339594), linewidth(0.4) + ffqqff); 
draw((5.081025846916564,0.8007740211858014)--(15.591017496944685,-2.4993795413193407), linewidth(0.4) + linetype("4 4") + blue); 
draw((6.775903779376068,3.721548042371603)--(15.591017496944685,-2.4993795413193407), linewidth(0.4) + blue); 
draw((9.796305353838916,-2.3192551065540985)--(15.591017496944685,-2.4993795413193407), linewidth(0.4) + blue); 
 /* dots and labels */
dot((6.775903779376068,3.721548042371603),dotstyle); 
label("$A$", (6.832913828359907,3.862244607803299), NE * labelscalefactor); 
dot((3.38614791445706,-2.12),dotstyle); 
label("$B$", (2.98,-2.3434897154339724), NE * labelscalefactor); 
dot((9.796305353838916,-2.3192551065540985),dotstyle); 
label("$C$", (9.880372647806754,-2.620531426292779), NE * labelscalefactor); 
dot((6.65068084326917,-0.30694964257443885),linewidth(4.pt) + dotstyle); 
label("$I$", (6.445055433157582,-0.3349373117076234), NE * labelscalefactor); 
dot((6.591226634147988,-2.219627553277049),linewidth(4.pt) + dotstyle); 
label("$X$", (6.261,-2.5097147419492565), NE * labelscalefactor); 
dot((5.081025846916564,0.8007740211858014),linewidth(4.pt) + dotstyle); 
label("$M$", (4.693,0.8978983016140666), NE * labelscalefactor); 
dot((8.823929553810784,-0.37450350649783165),linewidth(4.pt) + dotstyle); 
label("$N$", (8.744501633285656,-0.08559977193469738), NE * labelscalefactor); 
dot((10.230134161477556,1.2838577580577875),linewidth(4.pt) + dotstyle); 
label("$D$", (10.282083128552019,1.3965733811599186), NE * labelscalefactor); 
dot((7.85155375378265,1.5702480935584389),linewidth(4.pt) + dotstyle); 
label("$Q$", (7.8441160729945425,1.7567276052763674), NE * labelscalefactor); 
dot((7.732645335540281,-2.2551077278467857),linewidth(4.pt) + dotstyle); 
label("$K$", (7.705595217565141,-2.5651230841210175), NE * labelscalefactor); 
dot((9.915213772081222,1.5061007148511685),linewidth(4.pt) + dotstyle); 
label("$E$", (9.77,1.618206749846964), NE * labelscalefactor); 
dot((6.537480680041361,-3.9486674434528757),linewidth(4.pt) + dotstyle); 
label("$F$", (6.33423874881406,-4.28), NE * labelscalefactor); 
dot((15.591017496944685,-2.4993795413193407),linewidth(4.pt) + dotstyle); 
label("$G$", (15.64284023366988,-2.3850459720627932), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Remarks
Attachments:
This post has been edited 1 time. Last edited by SatisfiedMagma, Aug 13, 2021, 3:06 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
DottedCaculator
7356 posts
#16
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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);
draw(unitcircle^^circumcircle(C,I,Q),cyan);
draw(CP(circumcenter(A,X,D),X,-210,-60 ),purple);
draw(CP(Ap,I,-10,190 ),purple);
dot("$A$",A,dir(110));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$O$",O,dir(0));
dot("$I$",I,dir(130));
dot("$N$",N,dir(-120));
dot("$Q$",Q,dir(50));
dot("$C'$",Cp,dir(Cp));
dot("$D$",D,dir(D));
dot("$M$",M,dir(M));
dot("$T$",T,dir(T));
dot("$A'$",Ap,dir(Ap));
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
This post has been edited 1 time. Last edited by guptaamitu1, Feb 14, 2022, 4:19 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $AB.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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); 
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 = -10.006881897432205, xmax = 25.634412731501875, ymin = -8.577795838543958, ymax = 14.83226710342647;  /* image dimensions */
pen ffqqff = rgb(1,0,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); 
 /* draw figures */
draw((1.5534973808986277,9.160392608982256)--(-3.5857910956748746,-1.8868970241407366), linewidth(0.8)); 
draw((1.5534973808986277,9.160392608982256)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); 
draw((-3.5857910956748746,-1.8868970241407366)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); 
draw((1.5534973808986277,9.160392608982256)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); 
draw(circle((5.807241390273251,2.031588419358404), 3.8827498197705226), linewidth(0.4) + wvvxds); 
draw(circle((1.9067865071150452,2.276975553541132), 6.892477317034895), linewidth(0.4) + dtsfsf); 
draw(circle((3.5081332893469326,4.891548128823499), 4.6950649335634775), linewidth(0.4) + wvvxds); 
draw((8.19899262334367,5.090227365365952)--(5.807241390273251,2.031588419358404), linewidth(0.4) + qqffff); 
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); 
 /* dots and labels */
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?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5003 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3448 posts
#25 • 1 Y
Y by happypi31415
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bjump
1030 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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
799 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 $M$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joshualiu315
2534 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 $\overline{C'D'}$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zuat.e
65 posts
#32
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
happypi31415
753 posts
#33
Y by
ihatemath123 wrote:
By angle chasing, $\angle QIC = 90^{\circ}$, hence $\angle QDC = 90^{\circ}$, hence $\angle ADQ = 90 - \angle B$, hence $(ADQ)$ is tangent to $\overline{AI}$.

If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$, then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$, hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$, and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$, hence they all concur by the radical axis theorem.

This solution is amazing! :D I was wondering if a radical axis solution was possible
This post has been edited 1 time. Last edited by happypi31415, May 3, 2025, 2:31 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueprimes
356 posts
#36
Y by
Let $AI \cap BC = T$, clearly
\[ \angle QIC = 180^\circ - \angle AIQ - \angle CIT = 180^\circ - \angle ACI - \angle CIT = 90^\circ \]so $N$ is the center of $(CIQ)$. Now define $X = AD \cap BC$, we require $N \in XM$ to finish. Since $XM$ is the $X$-median in $\triangle XAB$ it must be the $X$-symmedian of $\triangle XCD$. But
\begin{align*}
\angle CXD &= \angle ACB - \angle CAX \\
&= \angle ACB - (180^\circ - \angle ADC - \angle ACD) \\
&= \angle ACB - \angle ABC + \angle ACD \\
&= \angle NCD = \angle NDC.
\end{align*}Hence, $NC$ and $ND$ are tangents to $(XCD)$, so $N$ lies on the $X$-symmedian of $\triangle XCD$. Therefore, $N \in XM$ and we're finished.
Z K Y
N Quick Reply
G
H
=
a