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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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

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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
Yesterday at 11:16 PM
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Line AT passes through either S_1 or S_2
v_Enhance   88
N 8 minutes ago by bjump
Source: USA December TST for 57th IMO 2016, Problem 2
Let $ABC$ be a scalene triangle with circumcircle $\Omega$, and suppose the incircle of $ABC$ touches $BC$ at $D$. The angle bisector of $\angle A$ meets $BC$ and $\Omega$ at $E$ and $F$. The circumcircle of $\triangle DEF$ intersects the $A$-excircle at $S_1$, $S_2$, and $\Omega$ at $T \neq F$. Prove that line $AT$ passes through either $S_1$ or $S_2$.

Proposed by Evan Chen
88 replies
v_Enhance
Dec 21, 2015
bjump
8 minutes ago
Inequality with a,b,c
GeoMorocco   4
N 13 minutes ago by Natrium
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
4 replies
GeoMorocco
Apr 11, 2025
Natrium
13 minutes ago
China Northern MO 2009 p4 CNMO
parkjungmin   1
N 40 minutes ago by WallyWalrus
Source: China Northern MO 2009 p4 CNMO P4
The problem is too difficult.
1 reply
parkjungmin
Apr 30, 2025
WallyWalrus
40 minutes ago
Polynomial Squares
zacchro   26
N 41 minutes ago by Mathandski
Source: USA December TST for IMO 2017, Problem 3, by Alison Miller
Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P + \lambda Q$ is the square of a polynomial.

Alison Miller
26 replies
zacchro
Dec 11, 2016
Mathandski
41 minutes ago
Mmo 9-10 graders P5
Bet667   8
N 43 minutes ago by User141208
Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$
8 replies
Bet667
Apr 3, 2025
User141208
43 minutes ago
Tangent to two circles
Mamadi   1
N 43 minutes ago by ricarlos
Source: Own
Two circles \( w_1 \) and \( w_2 \) intersect each other at \( M \) and \( N \). The common tangent to two circles nearer to \( M \) touch \( w_1 \) and \( w_2 \) at \( A \) and \( B \) respectively. Let \( C \) and \( D \) be the reflection of \( A \) and \( B \) respectively with respect to \( M \). The circumcircle of the triangle \( DCM \) intersect circles \( w_1 \) and \( w_2 \) respectively at points \( E \) and \( F \) (both distinct from \( M \)). Show that the line \( EF \) is the second tangent to \( w_1 \) and \( w_2 \).
1 reply
Mamadi
Today at 7:01 AM
ricarlos
43 minutes ago
China Northern MO 2009 p4 CNMO
parkjungmin   2
N an hour ago by WallyWalrus
Source: China Northern MO 2009 p4 CNMO
China Northern MO 2009 p4 CNMO

The problem is too difficult.
Is there anyone who can help me?
2 replies
parkjungmin
Apr 30, 2025
WallyWalrus
an hour ago
Problem 4
codyj   86
N an hour ago by Mathgloggers
Source: IMO 2015 #4
Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Greece
86 replies
codyj
Jul 11, 2015
Mathgloggers
an hour ago
Israeli Mathematical Olympiad 1995
YanYau   24
N an hour ago by bjump
Source: Israeli Mathematical Olympiad 1995
Let $PQ$ be the diameter of semicircle $H$. Circle $O$ is internally tangent to $H$ and tangent to $PQ$ at $C$. Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB\perp PQ$ and is tangent to $O$. Prove that $AC$ bisects $\angle PAB$
24 replies
YanYau
Apr 8, 2016
bjump
an hour ago
P(x), integer, integer roots, P(0) =-1,P(3) = 128
parmenides51   3
N an hour ago by Rohit-2006
Source: Nordic Mathematical Contest 1989 #1
Find a polynomial $P$ of lowest possible degree such that
(a) $P$ has integer coefficients,
(b) all roots of $P$ are integers,
(c) $P(0) = -1$,
(d) $P(3) = 128$.
3 replies
parmenides51
Oct 5, 2017
Rohit-2006
an hour ago
2017 CGMO P1
smy2012   9
N an hour ago by Bardia7003
Source: 2017 CGMO P1
(1) Find all positive integer $n$ such that for any odd integer $a$, we have $4\mid a^n-1$
(2) Find all positive integer $n$ such that for any odd integer $a$, we have $2^{2017}\mid a^n-1$
9 replies
smy2012
Aug 13, 2017
Bardia7003
an hour ago
Euler's function
luutrongphuc   1
N 2 hours ago by luutrongphuc
Find all real numbers \(\alpha\) such that for every positive real \(c\), there exists an integer \(n>1\) satisfying
\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]
1 reply
luutrongphuc
2 hours ago
luutrongphuc
2 hours ago
Square problem
Jackson0423   1
N 2 hours ago by maromex
Construct a square such that the distances from an interior point to the vertices (in clockwise order) are
1,2,3,4, respectively.
1 reply
Jackson0423
2 hours ago
maromex
2 hours ago
Sequence with infinite primes which we see again and again and again
Assassino9931   4
N 2 hours ago by SimplisticFormulas
Source: Balkan MO Shortlist 2024 N6
Let $c$ be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence $a_1 = c$, $a_{n+1} = a_n^3 + c$.
4 replies
Assassino9931
Apr 27, 2025
SimplisticFormulas
2 hours ago
Another config geo with concurrent lines
a_507_bc   15
N Apr 5, 2025 by E50
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.
15 replies
a_507_bc
May 3, 2024
E50
Apr 5, 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
677 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
716 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
1389 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
19 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
59 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)); 
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[/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
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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
104 posts
#13
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Draq
11 posts
#14
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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
20 posts
#15
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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
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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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