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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inspired by old results
sqing   0
2 minutes ago
Source: Own
Let $ a,b> 0, a^2+b^2+ab=3 .$ Prove that
$$ \frac {8} {3} \geq(a+b)^2(\frac {a} {b^2+a+1}+\frac {b} {a^2+b+1}) \geq  \frac {3(3-\sqrt 3)} {2} $$
0 replies
1 viewing
sqing
2 minutes ago
0 replies
Interesting geometrical configuration
SeboS   1
N 4 minutes ago by Lil_flip38
Let $ABC$ be a triangle with orthocentre $H$, and altitudes $D,E,F$ on sides $(BC), (AC), (AB)$
If $$ (HBC) \cap (HEF)=X $$$$(AHC) \cap (DHF)=Y $$$$(ABH) \cap (DEH)=Z $$Prove that the points $ H, X, Y, Z $ are cyclic
1 reply
SeboS
12 minutes ago
Lil_flip38
4 minutes ago
4 var inequality
ehuseyinyigit   2
N 8 minutes ago by ehuseyinyigit
Source: Own
Let $a,b,c,d$ be positive real numbers. Prove
$$\sum{a^2}+3\sum{a^2b^2c^2}+\sum{ab^2c}+12abcd$$$$\geq 2abcd[(a+c)(b+d)-ac-bd]+3\sum{a^2bc}+\sum{abc^2}$$
2 replies
1 viewing
ehuseyinyigit
Yesterday at 2:53 PM
ehuseyinyigit
8 minutes ago
Interesting functional equation
TheUltimate123   14
N 9 minutes ago by math-olympiad-clown
Source: ELMO Shortlist 2023 A2
Let \(\mathbb R_{>0}\) denote the set of positive real numbers. Find all functions \(f:\mathbb R_{>0}\to\mathbb R_{>0}\) such that for all positive real numbers \(x\) and \(y\), \[f(xy+1)=f(x)f\left(\frac1x+f\left(\frac1y\right)\right).\]
Proposed by Luke Robitaille
14 replies
TheUltimate123
Jun 29, 2023
math-olympiad-clown
9 minutes ago
A specific case of my previous conjecture
Rhapsodies_pro   1
N 24 minutes ago by nexu
Source: n=4
Prove that \(3\) is the largest value of the constant \(k\) such that \[{ab+ac+ad+bc+bd+cd-6}\leqslant{k{\left(a+b+c+d-1\right)}{\left(a+b+c+d-4\right)}}\]holds for any nonnegative real numbers \(a, b, c, d\) satisfying \({a^2+b^2+c^2+d^2+5abcd}\geqslant9\).
1 reply
Rhapsodies_pro
Wednesday at 4:38 PM
nexu
24 minutes ago
3 var inequality
ehuseyinyigit   9
N 26 minutes ago by nexu
Source: Own
Let $x,y,z$ be positive real numbers. Prove that

$$\dfrac{x^3+72xy^2}{z^3+x^2y}+\dfrac{y^3+72yz^2}{x^3+y^2z}+\dfrac{z^3+72zx^2}{y^3+z^2x}\geq \dfrac{15}{2}+\dfrac{102xyz(x+y+z)}{x^3y+y^3z+z^3x}$$
9 replies
ehuseyinyigit
Jul 21, 2025
nexu
26 minutes ago
Four variables
Nguyenhuyen_AG   3
N 32 minutes ago by nexu
Let $a,\,b,\,c,\,d$ non-negative real numbers. Prove that
\[\frac{abc}{(a+b+c)^3}+\frac{bcd}{(b+c+d)^3}+\frac{cda}{(c+d+a)^3}+\frac{dab}{(d+a+b)^3} \leqslant \frac{(a+b+c+d)^2}{27(a^2+b^2+c^2+d^2)}.\]
3 replies
Nguyenhuyen_AG
Jul 23, 2025
nexu
32 minutes ago
Inequality
SunnyEvan   5
N 33 minutes ago by nexu
Source: Own
Let $ a,b,c >0$, such that: $ a+b+c=3 .$ Prove that:
$$ 2025 \geq (628-96(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}))(ab+bc+ca)+864\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$$When does the equality hold ?
5 replies
SunnyEvan
Jul 18, 2025
nexu
33 minutes ago
The refinement of GMA 567
mihaig   6
N 37 minutes ago by mihaig
Source: Own
Let $a_1,\ldots, a_{n}\geq0~~(n\geq4)$ be real numbers such that
$$\sum_{i=1}^{n}{a_i^2}+(n^2-3n+1)\prod_{i=1}^{n}{a_i}\geq(n-1)^2.$$Prove
$$\left(\sum_{i=1}^{n}{a_i}\right)^2+\frac{2n-1}{(n-1)^3}\cdot\sum_{1\leq i<j\leq n}{\left(a_i-a_j\right)^2}\geq n^2.$$
6 replies
mihaig
Yesterday at 11:22 AM
mihaig
37 minutes ago
IMO ShortList 1999, number theory problem 1
orl   65
N 42 minutes ago by SSS_123
Source: IMO ShortList 1999, number theory problem 1
Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p-1}$ is a divisor of $ (p-1)^{x}+1$.
65 replies
orl
Nov 13, 2004
SSS_123
42 minutes ago
Tough with weird constraint
mihaig   1
N 42 minutes ago by nexu
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\left(a+b+c+d\right)^2+\frac7{27}\cdot\sum_{\text{sym}}{\left(a-b\right)^2}=16.$$Prove
$$a^2+b^2+c^2+d^2+5abcd\leq9.$$
1 reply
mihaig
Today at 4:08 AM
nexu
42 minutes ago
Four variables (3)
Nguyenhuyen_AG   2
N 44 minutes ago by nexu
Let $a,\,b,\,c,\,d$ be non-negative real numbers. Prove that
\[\frac{24a^3 + 49bcd}{(b + c + d)^3} + \frac{24b^3+49cda }{(c + d + a)^3} + \frac{24c^3+49dab}{(d + a + b)^3} + \frac{24d^3+49abc}{(a + b + c)^3} \geqslant \frac{292}{27}.\]
2 replies
Nguyenhuyen_AG
Yesterday at 9:25 AM
nexu
44 minutes ago
5-var inequality
sqing   3
N an hour ago by nexu
Source: Zhaobin
Let $ a,b,c,d,e>0 . $ Prove that$$\frac {a^2}{b} +\frac {b^2}{c} +\frac {c^2}{d} +\frac {d^2}{e} +\frac {e^2}{a}  \geq \sqrt{5(a^2+b^2+c^2+d^2+e^2)}$$
3 replies
sqing
Jun 18, 2025
nexu
an hour ago
Integer-Valued FE comes again
lminsl   217
N an hour ago by TigerOnion
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
217 replies
lminsl
Jul 16, 2019
TigerOnion
an hour 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
679 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.
Z K Y
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bin_sherlo
755 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
1392 posts
#3
Y by
Solution
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Orestis_Lignos
560 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
19 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
175 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
65 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
27 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
77 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
161 posts
#13
Y by
Click to reveal hidden text
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Draq
14 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
33 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
49 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
69 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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