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

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

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

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

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Gives typical russian combinatorics vibes
Sadigly   4
N 4 minutes ago by lbd4203
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
4 replies
Sadigly
May 8, 2025
lbd4203
4 minutes ago
Product of Sum
shobber   4
N 5 minutes ago by alexanderchew
Source: CGMO 2006
Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]
4 replies
shobber
Aug 9, 2006
alexanderchew
5 minutes ago
Prove that two different boards can be obtained
hectorleo123   1
N 16 minutes ago by Joalro178
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
1 reply
hectorleo123
Sep 15, 2023
Joalro178
16 minutes ago
Italian WinterCamps test07 Problem4
mattilgale   90
N 28 minutes ago by mathwiz_1207
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
90 replies
mattilgale
Jan 29, 2007
mathwiz_1207
28 minutes ago
Iran TST P8
TheBarioBario   8
N an hour ago by Mysteriouxxx
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
8 replies
TheBarioBario
Apr 2, 2022
Mysteriouxxx
an hour ago
IMO 2010 Problem 6
mavropnevma   42
N an hour ago by awesomeming327.
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
42 replies
1 viewing
mavropnevma
Jul 8, 2010
awesomeming327.
an hour ago
PJ // AC iff BC^2 = AC· QC
parmenides51   1
N 2 hours ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1998 OMM P5
The tangents at points $B$ and $C$ on a given circle meet at point $A$. Let $Q$ be a point on segment $AC$ and let $BQ$ meet the circle again at $P$. The line through $Q $ parallel to $AB$ intersects $BC$ at $J$. Prove that $PJ$ is parallel to $AC$ if and only if $BC^2 = AC\cdot QC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
2 hours ago
Self-evident inequality trick
Lukaluce   10
N 3 hours ago by ytChen
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
10 replies
Lukaluce
Sunday at 3:34 PM
ytChen
3 hours ago
Power Of Factorials
Kassuno   181
N 3 hours ago by SomeonecoolLovesMaths
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
181 replies
Kassuno
Jul 17, 2019
SomeonecoolLovesMaths
3 hours ago
Gergonne point Harmonic quadrilateral
niwobin   4
N 4 hours ago by on_gale
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
4 replies
niwobin
May 17, 2025
on_gale
4 hours ago
NCG Returns!
blacksheep2003   64
N 4 hours ago by SomeonecoolLovesMaths
Source: USEMO 2020 Problem 1
Which positive integers can be written in the form \[\frac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}\]for positive integers $x$, $y$, $z$?
64 replies
blacksheep2003
Oct 24, 2020
SomeonecoolLovesMaths
4 hours ago
Binomial stuff
Arne   2
N 4 hours ago by Speedysolver1
Source: Belgian IMO preparation
Let $p$ be prime, let $n$ be a positive integer, show that \[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]
2 replies
Arne
Apr 4, 2006
Speedysolver1
4 hours ago
Geometry hard problem
noneofyou34   1
N 4 hours ago by Lil_flip38
Let ABC be a triangle with incircle Γ. The tangency points of Γ with sides BC, CA, AB are A1, B1, C1 respectively. Line B1C1 intersects line BC at point A2. Similarly, points B2 and C2 are constructed. Prove that the perpendicular lines from A2, B2, C2 to lines AA1, BB1, CC1 respectively are concurret.
1 reply
noneofyou34
Yesterday at 3:13 PM
Lil_flip38
4 hours ago
segment of projections is half as sidelength, right triangle inscribed in right
parmenides51   3
N 4 hours ago by NumberzAndStuff
Source: 2020 Austrian Federal Competition For Advanced Students, Part 1, p2
Let $ABC$ be a right triangle with a right angle in $C$ and a circumcenter $U$. On the sides $AC$ and $BC$, the points $D$ and $E$ lie in such a way that $\angle EUD = 90 ^o$. Let $F$ and $G$ be the projection of $D$ and $E$ on $AB$, respectively. Prove that $FG$ is half as long as $AB$.

(Walther Janous)
3 replies
parmenides51
Nov 22, 2020
NumberzAndStuff
4 hours ago
Another config geo with concurrent lines
a_507_bc   17
N May 12, 2025 by Rayvhs
Source: BMO SL 2023 G5
Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
17 replies
a_507_bc
May 3, 2024
Rayvhs
May 12, 2025
Another config geo with concurrent lines
G H J
G H BBookmark kLocked kLocked NReply
Source: BMO SL 2023 G5
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a_507_bc
678 posts
#1 • 2 Y
Y by lian_the_noob12, Rounak_iitr
Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$. Let $Y$ be the point where the external bisector of $\angle BXC$ intersects with $AC$. Let $K$ be the projection of $X$ onto $BY$. Prove that the lines $AK, XO, BC$ have a common point.
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bin_sherlo
733 posts
#2 • 3 Y
Y by Om245, farhad.fritl, At777
$XO\cap BC=S$. Let $A'$ be the antipode of $A$ in $(ABC)$. $X,A',C$ are collinear. Let the tangents at $B,C$ to $(ABC)$ intersect at $T$.

Claim $1$: $X\in (BOC)$.
Proof: $\angle OXC=\angle SXB=\angle XSB-\angle XCS=\angle B-(90-\angle C)=90-\angle A=\angle OBC$ which gives that $X\in (BOC)$.$\square$

$\angle BXO=90-\angle A=\angle OXC$ thus $OX$ is the interior angle bisector of $\angle BXC\implies OX\perp XY$

Claim $2$: $K\in (ABC)$.
Proof: $\angle BKC=180-\angle CKY=180-\angle CXY=90+\angle OXC=180-\angle A$ Hence $ABCK$ is cyclic.$\square$

Claim $2$ gives that $X,K,B'$ are collinear where $B'$ is the antipode of $B$ on $(ABC)$.

Finishing Claim: $A,S,K$ are collinear.
Proof: Pascal at $A'AKB'BC$ gives that $O,AK\cap BC,X$ are collinear $\iff AK\cap BC=S$ as desired.$\blacksquare$
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VicKmath7
1390 posts
#3
Y by
Solution
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Orestis_Lignos
558 posts
#4 • 1 Y
Y by parmenides51
Proposed by Jason Prodromidis and Minos Margaritis, Greece :)
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squarc_rs3v2m
46 posts
#5 • 2 Y
Y by Scilyse, ohiorizzler1434
All of the above solutions are incorrect, because the problem is false.

The official solution assumes that $XO$ internally bisects $\angle BXC$, but this does not hold in general (see attached image below) and while $BOCX$ is necessarily cyclic, $O$ is an arc midpoint but not necessarily the midpoint of the arc $BC$ not containing $X$.

This failure can also occur when $\angle B$, $\angle C$ are acute and $\angle A$ is obtuse. However, the problem is fine when $\triangle ABC$ is constrained to be acute overall.

https://cdn.discordapp.com/attachments/799475903569592340/1236120831638372483/image.png?ex=6636dabc&is=6635893c&hm=51afb14d812cad898d8a0f488476759f38c60ee3a953d0ae583cdf57c69c1f1d&
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Tellocan
35 posts
#6
Y by
squarc_rs3v2m wrote:
All of the above solutions are incorrect, because the problem is false.

The official solution assumes that $XO$ internally bisects $\angle BXC$, but this does not hold in general (see attached image below) and while $BOCX$ is necessarily cyclic, $O$ is an arc midpoint but not necessarily the midpoint of the arc $BC$ not containing $X$.

This failure can also occur when $\angle B$, $\angle C$ are acute and $\angle A$ is obtuse. However, the problem is fine when $\triangle ABC$ is constrained to be acute overall.

https://cdn.discordapp.com/attachments/799475903569592340/1236120831638372483/image.png?ex=6636dabc&is=6635893c&hm=51afb14d812cad898d8a0f488476759f38c60ee3a953d0ae583cdf57c69c1f1d&

İ agree, the problem statement must include the triangle being acute.
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matematica007
17 posts
#7
Y by
How $\angle XKY=\angle XCY=90^{\circ}$ we have that $XKCY$ is cyclic.
Point $X$ is the intersection of the parallel line from $O$ to $AB$ with the perpendicular line to $AC$ from $C$ so $\angle OXC=90^{\circ}-\angle A=\angle OBC=\angle OCB$ so $OBXC$ is cyclic. This implies that $\angle BXC=180^{\circ}-2\angle A$ so $\angle YXC=\angle YKC=\angle A$ so $\angle BKC= 180^{\circ}-\angle A$ so $ABKC$ is cyclic.

We will prove that $AOKX$ is cyclic. After some easy angle chasing we have $\angle XKA=90^{\circ}+\angle C= \angle XOA $ so $AOKX$ is cyclic.

New we have that $AK$ is radical axis of circles $(ABKC)$ and $(XKOA)$ , $XO$ is radical axis of circles $(XBOC)$ and $(XKOA)$ and $BC$ is radical axis of circles $(ABKC)$ and $(XBOC)$.So the lines $AK, XO, BC$ have a common point.
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motannoir
171 posts
#8 • 1 Y
Y by Assassino9931
Sketch: Let $S=XO\cap BC$ ;prove by angles that $BOCX$ is cyclic ,then it is trvial that $XKCY$ is cyclic,use this to prove $K\in (ABC)$ and now prove(again by angles) that $AOKX$ is cyclic.Now use radical axes on $AOKX,ABKC,BOCX$ to get that $AK,OX,BC$ are concurrent
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Z4ADies
64 posts
#9
Y by
It is very trivial problem for G5
$\angle CAO=\angle OCA= x$
$\angle OBC=\angle OCB=y$
$\angle OAB=\angle OBA=90-x-y$
From parallelity of $OX,AB$ $\implies$ $\angle ABO=\angle BOX=90-x-y$
Similarly from perpendicularity of $AC,AX$ $\implies$ $\angle BCX=90-x-y$
Thus, $BOCX$ is cyclic that will be our hint to prove concurrency with radical axes are concurr at some point.
We have to deduce $ABKC$ and $AOKX$ cyclic.
$(.....1)$ $KXYC$ is cyclic because $XK$ is perpendicular to $BY$ and $AY$ is perpendicular to $XC$
From $(.....1)$ $\angle CXY=\angle CKY=90-y$ and we have $\angle BAC=90-y$ thus $ABKC$ cyclic.
Let $\angle OXK= z$ $\implies$ $\angle KXC=y-z$ from $(.....1)$ $\angle KYA=y-z$ , $\angle AKY=180-x-y$ , and $\angle YAO=x$ it gives us $\angle KAO=z$ which means $AOXK$ cyclic.
We are done.
This post has been edited 3 times. Last edited by Z4ADies, May 10, 2024, 12:25 PM
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NuMBeRaToRiC
22 posts
#10
Y by
By easy angle chasing we get that $XBOC$ cyclic. Let $XO\cap BC=D$, $AK\cap XD=M$ and . By BMO 2022 G2 we get that $M$ midpoint of $XD$. Incenter $I$ and excenter $I'$ of triangle $BXC$ lies on $ABC$. From $(X,D;I,I')=-1$ so $MX^2=MD^2=MI\cdot MI'$ so $\angle XCD=\angle FKD$. By easy angle chasing we get that $\angle FCB=90$.So $\angle FKD=\angle XCD=\angle FCA=\angle FKA$, from this we get that $K, D, A$ collinear.We are done.
This post has been edited 2 times. Last edited by NuMBeRaToRiC, Jun 26, 2024, 7:22 PM
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Ihatecombin
64 posts
#11
Y by
Nice problem, had a fun time doing it.
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
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 = -5.974725338137207, xmax = 30.033727935533033, ymin = -11.465676723098179, ymax = 12.099250581265741;  /* image dimensions */

 /* draw figures */
draw(circle((5,5), 5), linewidth(0.4)); 
draw((3.322736366511161,9.71028520408009)--(0.6972183465910291,2.4532235977441674), linewidth(0.4)); 
draw((0.6972183465910291,2.4532235977441674)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); 
draw((9.302781653408971,2.4532235977441674)--(3.322736366511161,9.71028520408009), linewidth(0.4)); 
draw((0.6972183465910291,2.4532235977441674)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); 
draw((0.6972183465910291,2.4532235977441674)--(19.665169827022172,-10.122013838908856), linewidth(0.4)); 
draw((1.859615688652539,-3.6801774895883588)--(4.329421138984925,0.04517164866836985), linewidth(0.4)); 
draw((0.6972183465910291,2.4532235977441674)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); 
draw((4.078606783358132,2.4532235977441683)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); 
draw((4.078606783358132,2.4532235977441683)--(4.329421138984925,0.04517164866836985), linewidth(0.4)); 
draw((4.329421138984925,0.04517164866836985)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); 
draw((4.329421138984925,0.04517164866836985)--(3.6872409900399328,1.3714691968320367), linewidth(0.4)); 
draw((3.322736366511161,9.71028520408009)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); 
draw((0.6972183465910251,7.5467764022558255)--(3.6872409900399328,1.3714691968320367), linewidth(0.4)); 
draw((0.6972183465910251,7.5467764022558255)--(6.677263633488835,0.2897147959199078), linewidth(0.4)); 
draw((0.6972183465910251,7.5467764022558255)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); 
draw((5,5)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); 
draw((1.859615688652539,-3.6801774895883588)--(9.302781653408971,2.4532235977441674), linewidth(0.4)); 
draw((19.665169827022172,-10.122013838908856)--(1.859615688652539,-3.6801774895883588), linewidth(0.4)); 
draw((9.302781653408971,2.4532235977441674)--(19.665169827022172,-10.122013838908856), linewidth(0.4)); 
draw(circle((1.2784170176217802,-0.6134769459221007), 3.1212888555420335), linewidth(0.6) + green); 
draw(circle((5.377935208423484,1.3714691968320383), 1.6906942183835514), linewidth(0.6) + red); 
 /* dots and labels */
dot((5,5),linewidth(3pt) + dotstyle); 
label("$O$", (5.094827972055157,5.151615248405972), NE * labelscalefactor); 
dot((3.322736366511161,9.71028520408009),linewidth(3pt) + dotstyle); 
label("$A$", (3.4356911761481044,9.869785511765965), NE * labelscalefactor); 
dot((0.6972183465910291,2.4532235977441674),linewidth(3pt) + dotstyle); 
label("$B$", (0.7914419076712397,2.6110620296736675), NE * labelscalefactor); 
dot((9.302781653408971,2.4532235977441674),linewidth(3pt) + dotstyle); 
label("$C$", (9.398214036439073,2.6110620296736675), NE * labelscalefactor); 
dot((1.859615688652539,-3.6801774895883588),linewidth(3pt) + dotstyle); 
label("$X$", (1.9580224672933857,-3.5329289176687406), NE * labelscalefactor); 
dot((19.665169827022172,-10.122013838908856),linewidth(3pt) + dotstyle); 
label("$Y$", (19.76781901085815,-9.962084001807632), NE * labelscalefactor); 
dot((4.329421138984925,0.04517164866836985),linewidth(3pt) + dotstyle); 
label("$K$", (4.420803648717916,0.20012887312158348), NE * labelscalefactor); 
dot((4.078606783358132,2.4532235977441683),linewidth(3pt) + dotstyle); 
label("$D$", (4.187487536793487,2.6110620296736675), NE * labelscalefactor); 
dot((6.677263633488835,0.2897147959199078),linewidth(3pt) + dotstyle); 
label("$A'$", (6.779888780398257,0.4334449850459787), NE * labelscalefactor); 
dot((3.6872409900399328,1.3714691968320367),linewidth(3pt) + dotstyle); 
label("$M$", (3.798627350252772,1.5222535073598231), NE * labelscalefactor); 
dot((0.6972183465910251,7.5467764022558255),linewidth(3pt) + dotstyle); 
label("$C'$", (0.7914419076712397,7.692168467138275), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
Sketch of solution
We now begin with the proof. All points not found in the problem statement are defined in the sketch of the solution. We begin with a small claim.
Claim 1: $XC \cap (ABC) = A'$
Proof
Claim 2: $OX$ bisects $\angle BDA'$ and $\angle BXA'$
Proof
Claim 3: $BMKX$ is cyclic
Proof
Claim 3: $MKA'D$ is cyclic
Proof
Now we are done, just notice that by claim $4$ we have $\angle DKA' = \angle DMA' = 90$, however we also have $\angle AKA' = 90$, thus it must be true that $A-D-K$.
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Nari_Tom
117 posts
#12
Y by
Similar sketch.

Let $OX\cap BC=X'$. Let $\omega$ be the center with the diameter $OX'$. Let $Z', Y'$ be the intersection of $(AOB),(AOC)$ with $\omega$. Let $Y,Z$ be the intersections of $OY', OZ'$ with $AB,AC$, respectively. We will get that $X$ is the foot of the altitude of $\triangle OYZ$. Let $ZC \cap BY=K$. Then $K$ lies on the $(ABC)$.

Let $BC \cap ZY=D$ and $M$ be the midpoint of $BC$. Then we have that $(AD,AK;AB,AC)=-1$. We want to show that $A-X'-K$. So it's suffices to prove $(B,C;D,X')=-1$.

Since $ZY$ is the radical axis of $(ABC),\omega$ and $D$ lies on it we will get that $DB*DC=DX'*DM$ which only happens when $(B,C;D,X')=-1$, and we are done.
This post has been edited 1 time. Last edited by Nari_Tom, Jan 30, 2025, 5:40 AM
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SimplisticFormulas
118 posts
#13
Y by
Click to reveal hidden text
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Draq
13 posts
#14
Y by
Let $\angle CAB =x \angle CBA=y \angle ACB=z$ $\because AB $is parallel to $DX$ $\therefore \angle ABD= \angle BDX $
which is equal to y and $\because AC$ is perpendicular to $ XC$ $\angle  ACB = 90-  \angle BCX $ $ \therefore$ $ \angle BCX=90-z$
İn triangle $ DCX$ $\angle BDX=\angle DXC+ \angle  DCX$ $\therefore$ $\angle DXC = 90-x=\angle OXC= \angle OBC $ $OBXC$ is cyclic
$\angle XCY =\angle XKY =90$ also $\angle OXB=\angle OCB=90-x$ $XD$ angle bisector also $XY$ external angle bisector so $XD$ is perpendicular to $XY$ so $\angle CXY =x$ then $\angle CKY=x$ which means $ABKC $ is cyclic then $\angle AKB=\angle ACB=z$ so $\angle AKX=90+z$ and $ \angle OAK=90-z$ because of parallel $\angle AOX=90+z$ so AOKX cyclic and by radical axis theorem AK BC XO are concurrent at point $D$
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optimusprime154
23 posts
#15
Y by
claim 1: \(BXOC\) concyclic:
proof: we know that \(\angle XOB = \angle OBA = 90 - \angle C\) and also \(\angle XCB = 90 - \angle C\) finishing the claim.
from now on, let \(S\) = \(OX \cap BC\) and \(K\) = \(XY \cap AB\) , \(R\) = \(CX \cap BK'\) and \(T\) = \(KC \cap BY\) first notice that \(OX \perp KY\) since \(OX\) is the bisector of \(\angle BXC\) , we know \(BXAY\) cyclic because \(\angle K'XB = \angle A\) we also know \(K'XCA\) cyclic for obvious reasons.
claim 2: \(BTXK'\) cyclic.
this just follows from the fact that \(\angle XBT = \angle XBY = \angle XAY = \angle XAC = \angle XK'C = \angle XK'T\)
this means \(\angle K'TB = \angle K'XB = \angle A\) meaning \(T\) lies on the circle \(ABC\) . we Also got \(\angle XTB = \angle XK'B = 90\) so \(T\) is the foot from \(X\) to \(BY\)
claim 3: \(A, S, T\) collinear.
we first prove that \(XTOA\) is cyclic. which just follows from \(\angle XTA = \angle XTB + \angle BTA = 90 + \angle C = \angle XOA\)
now we let \(S'\) denote \(AT \cap BC\) we know that \(X, S', O\) must be collinear by PoP on \(ABTC\), \(XBOC\) , \(XTOA\) , since we know \(T\) is the foot from \(X\) to \(BY\), And \(AT\) \(BC\) \(OK\) concur At \(S\) so we're done!
This post has been edited 1 time. Last edited by optimusprime154, Mar 10, 2025, 3:29 PM
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E50
8 posts
#16
Y by
Claim 1 : $O,B,X,C$ concyclic

Proof : Let $Z=OX \cap BC$. Then $\angle OXC = \angle ZXC  =90^\circ -\angle XZC = 90^\circ - \angle BAC =  \frac{180^\circ -\angle BOC}{2}=\angle OBC \Longrightarrow O,B,X,C $ concyclic.

Claim 2 : $A,B,K,C$ concyclic.

Proof : Since $\angle XKY=90^\circ = \angle XCY \Longrightarrow X,K,C,Y$ cincyclic. Since $OB =OC$ and $O,B,X,C$ concyclic ; $\angle BXO = \angle CXO \Longrightarrow \angle CKY = \angle CXY =90^\circ - \angle OXC = \angle BAC \Longrightarrow A,B,K,C$ concyclic.

Claim 3 : $A,O,K,X$ concyclic.

Proof : Let $\angle ACB = \alpha$. Then $\angle AKX  = 90^\circ + \alpha$, $\angle BOX = \angle BCX = 90^\circ - \alpha$, $\angle AOB = 2\angle ACB = 2\alpha \Longrightarrow \angle AOX = 90^\circ+\alpha \Longrightarrow \angle AKX = \angle AOX \Longrightarrow A,O,K,X  $ concyclic.

And by Radical Axis Concurrence Theorem on $(AOKX),(ABKC),(OBXC)$ we obtain $AK,XO,BC$ have a common point as desired.
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Tkn
33 posts
#17
Y by
Solved with EeEeRUT.
[asy]
import geometry;
import graph;

real markscalefactor=0.025;

size(10cm);
defaultpen(fontsize(10pt));

path rightanglemark(pair A, pair B, pair C, real s=8) {
    pair P,Q,R;
    P=s*markscalefactor*unit(A-B)+B;
    R=s*markscalefactor*unit(C-B)+B;
    Q=P+R-B;
    return P--Q--R;
}

pair A = (-0.75,2);
pair B = (-1,0);
pair C = (2,0);

path circ1 = circumcircle(A,B,C);

pair O = extension((B+C)/2,rotate(90,B)*C-B+(B+C)/2,(C+A)/2,rotate(90,C)*A-C+(A+C)/2);
pair X1 = extension(O+2*(B-A),O,C,rotate(90,C)*A);
pair Y1 = extension(A,C,rotate(90,X1)*O,X1);
pair K = intersectionpoints(circ1, B--Y1)[1];

path circ2 = circumcircle(X1,K,C);

pair T = extension(A,B,X1,Y1);
pair P = extension(A,K,B,C);

path circ3 = circumcircle(B,T,X1);
path circ4 = circumcircle(A,C,T);

draw(A--B--C--cycle, black);
draw(O--X1, red+dashed);
draw(B--X1--C, black);
draw(X1--Y1--C, black);
draw(B--Y1, blue);
draw(A--K, red+dashed);
draw(B--T--X1, black);
draw(T--C, royalblue+dashed);
draw(X1--K, black);

draw(circ1);
draw(circ2, orange);
draw(circ3, heavymagenta);
draw(circ4, heavycyan);

draw(rightanglemark(B,T,Y1));

dot(A);
dot(B);
dot(C);
dot(O);
dot(X1);
dot(Y1);
dot(K);
dot(T);
dot(P);

label("$A$", A, N, black);
label("$B$", B, 1.5W, black);
label("$C$", C, 2S, black);
label("$X$", X1, SE, black);
label("$Y$", Y1, SW, black);
label("$K$", K, 2N+0.4E, black);
label("$T$", T, SW, black);
label("$P$", P, N+2W, black);
label("$O$", O, SE, black);
[/asy]
Before we begin with the proof, we define the following points:
  • $P$ is the intersection of $\overline{BC}$ and $\overline{AK}$.
  • $T$ is the intersection of the line $AB$ and $XY$.

Claim 1. $\overline{OX}$ bisects $\angle{BXC}$.
Proof. Notice that $\angle{OXC}=90^{\circ}-\angle{BAC}=\angle{OBC}=\angle{OCB}$. So, $OCXB$ is a cyclic quadrilateral. Which also implies $\angle{OXB}=\angle{OCB}=\angle{OXC}$. The first claim is complete.
Claim 2. $K\in (ABC)$.
Proof. It is easy to see that $K,X,Y,C$ are concylic because $\angle{XKY}=\angle{XCY}=90^{\circ}$.
From the first claim, $\overline{OX}$ is an angle bisector of $\angle{BXC}$. So, we can redefine the external angle bisector of $\angle{BXC}$ to be a line perpendicular to $\overline{KX}$. Which means ${TY}\perp AB$. The angle chasing from $\angle{A}$ gives:
$$\angle{BAC}=90^{\circ}-AYT=\angle{YKC}.$$Since, $B,K$ and $Y$ are collinear. Therefore, $K\in(ABC)$.
Claim 3. $T,K$ and $C$ are collinear.
Proof. Observe that $BKXT$ and $ACXT$ are cyclic quadrilateral. Perform an angle chasing:
$$\angle{TKX}=\angle{TBX}=\angle{BXO}=90^{\circ}-\angle{BAC}\implies \angle{BKC}=90^{\circ}+TKX$$Therefore, $T,K$ and $C$ are collinear.
Since $X\in (BKT),(KCY)$, and $T=KC\cap AB, Y=BK\cap AC$, $X$ is a miquel point of $ABKC$. It is well-known that $O,P$ and $X$ are collinear. In conclusion, $AK,XO$ and $BC$ are concurrent.
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Rayvhs
20 posts
#18
Y by
This was a fun problem.
First, note that
\[\angle BOX = \angle ABO = \angle BAO = 90^\circ - \angle ACB = \angle BCX,\]so $BOCX$ is cyclic.
Next, note that $\angle XKY = 90^\circ = \angle XCY$, so $XKCY$ is cyclic.
Also, observe that
\[\angle BXC = 180^\circ - \angle BOC = 180^\circ - 2\angle A,\]and since $XY$ is an external angle bisector, $\angle CXY = \angle A$.
Because $XYCK$ is cyclic, we have $\angle CKY = \angle A$, so $ABKC$ is cyclic.
Now, observe
\[\angle AOX = \angle AOB + \angle BOX = 2\angle ACB + \angle BCX = \angle ACB + 90^\circ,\]\[\angle AKX = \angle AKB + \angle BKX = \angle ACB + 90^\circ,\]thus $AOKX$ is cyclic.

Finally, by radical axes theorem on $\odot(BOCX)$, $\odot(ABKC)$, and $\odot(AOKX)$,
we conclude that $BC$, $AK$, and $OX$ are concurrent.
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