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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.

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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
Quadrangle, nine-point conic, Steiner line
kosmonauten3114   0
a few seconds ago
Source: My own
Let $P_1P_2P_3P_4$ be a general quadrangle which does not form an orthocentric system. Let $P$, $I$, $M$, $T$ be the Euler-Poncelet point ($\text{QA-P2}$), isogonal center ($\text{QA-P4}$), midray homothetic center ($\text{QA-P8}$), inscribed square axes crosspoint ($\text{QA-P23}$) of $P_1P_2P_3P_4$, respectively.
Let $H_1$ be the orthocenter of $\triangle{P_2P_3P_4}$, and define $H_2$, $H_3$, $H_4$ cyclically.
Let $A_{ij}=P_iP_j \cap H_iH_j$ ($\{i, j\} \in \{1, 2, 3, 4\}, i<j$).
Let $B_{ij}=P_iP_j \cap H_kH_l$ ($\{i, j, k, l\} \in \{1, 2, 3, 4\}, i<j$).
Then, the 12 points $A_{12}$, $A_{13}$, $A_{14}$, $A_{23}$, $A_{24}$, $A_{34}$, $B_{12}$, $B_{13}$, $B_{14}$, $B_{23}$, $B_{24}$, $B_{34}$ lie on the same conic, here denoted by $\mathcal{C}_1$.
Let $\mathcal{C}_2$ be the nine-point conic of $P_1P_2P_3P_4$.
Suppose that $\mathcal{C}_1$ and $\mathcal{C}_2$ have 4 distinct intersection points, and let $U$, $V$, $W$ be the intersections, other than $P$, of $\mathcal{C}_1$ and $\mathcal{C}_2$.

Prove that the Steiner line of $P$ with respect to $\triangle{UVW}$ passes through $I$ and $M$, and show that the center of $\mathcal{C}_1$ and the orthocenter of $\triangle{UVW}$ coincide with $T$.
0 replies
kosmonauten3114
a few seconds ago
0 replies
Inspired by Philippine 2025
sqing   1
N 15 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+1)(c^2+1)} \ge -\frac{7+5\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+2)(c^2+2)} \ge -\frac{5+3\sqrt 3}{32}$$
1 reply
+1 w
sqing
23 minutes ago
sqing
15 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N 26 minutes ago by Funcshun840
Source: IDK
Let triangle \(ABC\) be inscribed in circle \((O)\). Let \((I_a)\) be the \(A\)-excircle of triangle \(ABC\), which is tangent to \(BC\), the extension of \(AB\), and the extension of \(AC\). Let \(BE\) and \(CF\) be the angle bisectors of triangle \(ABC\). Let \(EF\) intersect \((O)\) at two points \(S\) and \(T\).

a) Prove that circle \((O)\) bisects the segments \(I_aT\) and \(I_aS\).
b) Prove that \(S\) and \(T\) are the points of tangency of the common external tangents of circles \((O)\) and \((I_a)\) .

1 reply
ZeltaQN2008
Yesterday at 3:07 PM
Funcshun840
26 minutes ago
Random walk
EthanWYX2009   2
N 41 minutes ago by mathematical-forest
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
2 replies
EthanWYX2009
3 hours ago
mathematical-forest
41 minutes ago
2025 KMO Inequality
Jackson0423   2
N an hour ago by sqing
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
2 replies
Jackson0423
Yesterday at 4:32 PM
sqing
an hour ago
Geometry Concurrence
KHOMNYO2   1
N an hour ago by Funcshun840
Given triangle $XYZ$ such that $XY \neq XZ$. Let excircle-$X$ be tangent with $YZ, ZX, XY$ at points $U, V, W$ respectively. Let $R$ and $S$ be points on the segment $XZ, XY$ respectively such that $RS$ is parallel to $YZ$. Lastly, let $\gamma$ be the circle that is externally tangent with the excircle-$X$ on point $T$. Prove that $VW, UT$, and $RS$ concur at a point.
1 reply
KHOMNYO2
3 hours ago
Funcshun840
an hour ago
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such : $(f
guramuta   0
an hour ago
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such :
$(f(x-y))^2= (f(x))^2 - 2f(xy) + (f(y))^2$
0 replies
guramuta
an hour ago
0 replies
Nice one
Blacklord   9
N an hour ago by Stear14
Source: ....
Find all integers numbers (a,b,c) such that
$a/b + b/c + c/a =3$
9 replies
Blacklord
Jan 26, 2017
Stear14
an hour ago
Cauchy FE
InftyByond   2
N an hour ago by jasperE3
For the Cauchy FE, is x^nf(x)=f(x^(n+1)) enouh to go from rationals to reals? thats what wikipedia says but idk really
2 replies
InftyByond
Mar 1, 2025
jasperE3
an hour ago
Lemma on tangency involving a parallelogram with orthocenter
Gimbrint   1
N 2 hours ago by Captainscrubz
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$ and orthocenter $H$. Let $M$ be the midpoint of $AC$. Line $BH$ intersects $\omega$ again at $L\neq B$, and line $ML$ intersects $\omega$ again at $P\neq L$. Points $D$ and $E$ lie on $AB$ and $BC$ respectively, such that $BEHD$ is a parallelogram.

Prove that $BP$ is tangent to the circumcircle of triangle $BDE$.
1 reply
Gimbrint
4 hours ago
Captainscrubz
2 hours ago
Simple Geometry
AbdulWaheed   2
N 2 hours ago by Sadigly
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
2 replies
AbdulWaheed
Today at 5:15 AM
Sadigly
2 hours ago
inequality thing
BinariouslyRandom   2
N 2 hours ago by BinariouslyRandom
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.

answer
2 replies
+1 w
BinariouslyRandom
4 hours ago
BinariouslyRandom
2 hours ago
inequality
mathematical-forest   1
N 3 hours ago by lbh_qys
Positive real numbers $x_{1} ,x_{2} \cdots ,x_{n}$,satisfied $\sum_{i=1}^{n}x_{i} =1$
Proof:$$\sum_{i=1}^{n} \frac{\min  \left \{  x_{i-1},x_{i}\right \}\max \left \{  x_{i},x_{i+1}\right \}  }{x_{i}} \le 1$$
1 reply
mathematical-forest
3 hours ago
lbh_qys
3 hours ago
Some number theory
EeEeRUT   5
N 3 hours ago by shafikbara48593762
Source: Thailand MO 2025 P9
Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.
5 replies
EeEeRUT
May 14, 2025
shafikbara48593762
3 hours ago
Circles with same radical axis
Jalil_Huseynov   9
N Apr 17, 2025 by Nari_Tom
Source: DGO 2021, Individual stage, Day2 P3
Let $O$ be the circumcenter of triangle $ABC$. The altitudes from $A, B, C$ of triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at $A_1, B_1, C_1$ respectively. $AO, BO, CO$ meets $BC, CA, AB$ at $A_2, B_2, C_2$ respectively. Prove that the circumcircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$ share two common points.

Proporsed by wassupevery1
9 replies
Jalil_Huseynov
Dec 26, 2021
Nari_Tom
Apr 17, 2025
Circles with same radical axis
G H J
G H BBookmark kLocked kLocked NReply
Source: DGO 2021, Individual stage, Day2 P3
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Jalil_Huseynov
439 posts
#1
Y by
Let $O$ be the circumcenter of triangle $ABC$. The altitudes from $A, B, C$ of triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at $A_1, B_1, C_1$ respectively. $AO, BO, CO$ meets $BC, CA, AB$ at $A_2, B_2, C_2$ respectively. Prove that the circumcircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$ share two common points.

Proporsed by wassupevery1
This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 28, 2021, 12:33 PM
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Davsch
381 posts
#2 • 1 Y
Y by PRMOisTheHardestExam
We use complex numbers with $(ABC)$ as the unit circle. Then $a_1=-\frac{bc}a,a_2=\frac{a^2(b+c)}{a^2+bc}$. The condition $Z\in (AA_1A_2)$ is equivalent to $\frac{(z-a)(a^3b+a^3c+a^2bc+b^2c^2)}{a(az+bc)}=\frac{(a\bar z-1)(a^3+bc(a+b+c))}{bc\bar z+a}$, or, after diving by $a^2+bc$, \[z\bar z(b^2c^2-a^4)+za(a+b)(a+c)-\bar zabc(a+b)(a+c)-a(b+c)(a^2-bc)=0.\]Analogous equations hold for $Z\in (BB_1B_2),(CC_1C_2)$. We will now show that these equations are linearly dependent. We compute that (to evaluate the $3\times 3$-determinants, we always subtract row $1$ from rows $2,3$)
\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)\\b(b+a)(b+c)&-abc(b+a)(b+c)\end{pmatrix}=-abc(a+b)^2(b+c)(c+a)(a-b)\neq 0,\]\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&b^2c^2-a^4\\b(b+a)(b+c)&-abc(b+a)(b+c)&c^2a^2-b^4\\c(c+a)(c+b)&-abc(c+a)(c+b)&a^2b^2-c^4\end{pmatrix}\]\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(a-b)(a+b)(a^2+b^2+c^2)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(a-c)(a+b)(a^2+b^2+c^2)\end{pmatrix}=0,\]\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&a(b+c)(a^2-bc)\\b(b+a)(b+c)&-abc(b+a)(b+c)&b(c+a)(b^2-ca)\\c(c+a)(c+b)&-abc(c+a)(c+b)&c(a+b)(c^2-ab)\end{pmatrix}\]\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(b-a)(b+a)(ab+bc+ca)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(c-a)(c+a)(ab+bc+ca)\end{pmatrix}=0,\]as desired.
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BelieverofMaths
263 posts
#3 • 2 Y
Y by Noob_at_math_69_level, vuanhnshn
is there any geomerical proof of this one ?
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bin_sherlo
733 posts
#4 • 1 Y
Y by ehuseyinyigit
Nice problem!
Replace $A_2,B_2,C_2$ with $D,E,F$. Let $H$ be the orthocenter of $\triangle ABC$. Set $(BB_1E)\cap (CC_1F)=P,Q$.
Claim: $H$ lies on $PQ$.
Proof:
\[Pow(H,(BB_1EPQ))=HB.HB_1=HC.HC_1=Pow(H,(CC_1FPQ))\]Thus, $H$ lies on the radical axis of $(BB_1E)$ and $(CC_1F)$ which is $PQ$.$\square$
Claim: $A,A_1,P,Q$ are concyclic.
Proof:
\[HA.HA_1=HB.HB_1=HP.HQ\]Which gives the desired result.$\square$
Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\angle BAC$.
New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. $BC$ intersects the reflection of $AO$ with respect to $AB,AC$ at $B_1,C_1$ respectively. Parallel lines to $AB_1,AC_1$ through $C,B$ intersect $AB,AC$ at $E,F$ respectively. $D$ is on $(ABC)$ which holds $AD\perp BC$. Prove that $D$ lies on the radical axis of $(B_1EC)$ and $(C_1FB)$.
Lemma: $ABC$ is a triangle whose circumcenter is $O$. $AO\cap (BOC)=G,BO\cap AC=E$. $C_1$ is the intersection of the altitude from $C$ to $AB$ and $(ABC)$. Then, $C_1,E,C,G$ are concyclic.
Proof: Let $(GCE)\cap AG=S$.
\[\measuredangle ESG=180-\measuredangle GCA=\measuredangle C=90-\measuredangle GAB\]Hence $AS\perp AB\perp CC_1$. Also $\measuredangle OES=90-\measuredangle ABS=\measuredangle C=\measuredangle ESO$ which implies $OE=OS$. Combining this with $OC_1=OC,$ we conclude that $CESC_1$ is an isosceles trapezoid. Thus, $G,C,E,C_1,S$ are concyclic.$\square$
Let $B_1F\cap C_1E=T$. Let $(B_1EC)\cap EC_1=L,(C_1BF)\cap B_1F=K$. By inverting the configuration of the lemma with $\sqrt{bc}$, we get that $H,B,C_1,E$ are concyclic.
\[\measuredangle ECB_1=90-\measuredangle A=\measuredangle EC_1H=\measuredangle EC_1B+\measuredangle BC_1H=\measuredangle EC_1B+\measuredangle DC_1B_1=\measuredangle EC_1B+\measuredangle C_1B_1D=\measuredangle(EC_1,B_1D)\]Hence $B_1,D,L$ are collinear. Similarily, $C_1,D,K$ are collinear.
Claim: $TB_1=TC_1$.
Proof:
\[\frac{BB_1}{BF}=\frac{BC}{BF}.\frac{AB_1}{CE}=\frac{BC}{CE}.\frac{AB_1}{BF}=\frac{CC_1}{CE}\]And $\measuredangle FBB_1=90+\measuredangle A=\measuredangle C_1CE$ subsequently $FBB_1\sim ECC_1$. So $\measuredangle BB_1F=\measuredangle EC_1C$ which yields $TB_1=TC_1$.$\square$
Claim: $TK=TL$.
Proof: Let $M=(B_1EC)\cap B_1F,N=(C_1FB)\cap C_1E$.
\[\measuredangle EMB_1=\measuredangle ECB_1=90-\measuredangle A=\measuredangle C_1BF=\measuredangle C_1NF\]Thus, $M,N,E,F$ are concyclic. By using this,
\[TM.TF.TB_1.TK=(TK.TF)(TM.TB_1)=(TN.TC_1)(TE.TL)=(TN.TE).TC_1.TL=TM.TF.TB_1.TL\]Hence $TK=TL$.$\square$
Since $TK=TL$ and $TB_1=TC_1,$ we see that $B_1C_1LK$ is an isosceles trapezoid whose diagonals intersect at $D$.
\[Pow(D,(B_1EC))=DB_1.DL=DC_1.DK=Pow(D,(C_1FB))\]Which completes our proof as desired.$\blacksquare$
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soryn
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#5
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Very nice problem and very nices solutions!
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breloje17fr
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To continue, prove that the same is true when A1, B1 and C1 are the feet of the altitudes and A2, B2 and C2 are the antipodes of A, B and C on the circumcircle, and that the radical axis then is the Euler's line.
On the figure below, the red circles and the green circles corespond to the initial problem and the continuation, respectively.
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Nari_Tom
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Two beautiful proof for this beautiful problem. Here i provides two lemmas that technically solves the problem. How to prove them is your choice, it can be shorter may be. But i have nicer proofs here.

Lemma 1: (just for the sake of proving lemma 2). $O$ is the circumcircle of $\triangle ABC$. $AO \cap (BC)=A_2$, and $S$ be the intersection of $A$ altitude with $(ABC)$. Let's assume $B_2$ and $C_2$ defined similarly as $A_2$. Let $B'$ and $C'$ be the antipodes of $B$ and $C$ respectively. Let $D=SA_2 \cap (ABC)$, $D'=DO \cap 
(ABC)$. And let $B'C'$ and tangent of $(ABC)$ at $A$ intersect at $F$. Then $F$ lies on $B_2C_2$.


Lemma 2: Let's assume all the points defined same as Lemma 1. Let $X$ be the projection of $A_2$ to line $B_2C_2$. Then $X$ lies on $(AA_1A_2)$, which is actually follows from $F$ lies on D'S.

Proof for the lemma 1:
For the convenience let's assume that $B_2C_2 \ AA=F$, then let's prove that $F$ lies on $B'C'$. By the converse of Menelaus theorem it suffices to prove that $\frac{FC_2}{FB_2} \cdot \frac{C'O}{C'C_2} \cdot \frac{B'B_2}{B'O}=\frac{FC_2}{FB_2} \cdot \frac{B'B_2}{C'C_2}=1$, which can turn to mess if you don't use trigonometry instantly. In $\triangle AC_2B_2$:
by extended sines theorem, $\frac{FC_2}{FB_2}=\frac{AC_2}{AB_2} \cdot \frac{sin \gamma}{sin \beta}$. Now we can express everything in terms of $\triangle ABC$, so it's not hard for you.

Proof for the lemma 2:
I wont use any analytic technique here. Once you realize $FAA_2S$ is cyclic with diameter $A_2F$, we just need to prove that $F-D'-S$ are collinear. Let's forget about $B_2, C_2$ and use lemma 1. Then new definition of $F$ will be $F=C'B' \cap AA$. Since there is too many antipodes let's use inversion which swaps them. (which is the combination of inversion with $(ABC)$, and reflection with point $O$).

Let $O$ be the circumcenter of $\triangle ABC$. Let $S$ be the intersection of $(ABC)$ and $A$ altitude. Let $A'$ and $S'$ be the antipodes of $A$ and $S$. Let $A_2=AO \cap BC$ and $D=SA_2 \cap (ABC)$. Let $F'$ be the intersection of $(OBC)$ and circle with diameter $A'O$. Prove that $OS'DF'$ are concyclic.

Let $F=OF' \cap BC$. By regular inversion at $(ABC)$ (we're not using that, just motivational thing), we know that $FA'$ is tangent to $(ABC)$. By some angle chase $DA_2A'F$ is cyclic, which proves the collinearity $F-D-S'$. And we're done.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:12 AM
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Nari_Tom
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#8
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Actually they should mention that $\triangle ABC$ is not isosceles, at least not equilateral for this one.

Let $H$ and $H_2$ be the orthocenters of $\triangle ABC, \triangle A_2B_2C_2$, respectively.
Since $Pow(H, (AA_1A_2))=Pow(H, (BB_1B_2))=Pow(H, (CC_1C_2))=HA \cdot HA_1=HB \cdot HB_1= HC \cdot HC_1$, thus $H$ is radical center of these circles. By using the lemma 2 and, easy power of a point we can deduce that $H_2$ is also a radical center of these circles. Which means circles$ (AA_1A_2)$, $(BB_1B_2)$, $(CC_1C_2)$ have two different radical centers $\implies$ these circles have common radical axis.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:58 AM
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Nari_Tom
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#9
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Another solution using very nice lemma.
Lemma: Let $ABCD$ be a convex quadrilateral. Let $E=AB \cap CD$, $F=AD \cap BC$. let $H_1, H_2, H_3, H_4$ be the four orthocenters of triangles formed by lines $AB,AD, CB,CD$. Circles with diameter $AC, BD, EF$ have common radical axis, which passes through these orthocenters.
(Proof consists some power of a power argument, which is not hard. So you will do it yourselves.)

Let's solve the main problem. Let $X, Y, Z$ be the antipodes of $A_2, B_2, C_2$ in circles $(AA_1A_2), (BB_1B_2), (CC_1C_2)$, respectively. We proved $X, Y, Z$ lies on the lines $B_2C_2, A_2C_2, A_2B_2$, respectively. So let's just prove that $X-Y-Z$ are collinear. While proving the lemma 1, we've proved that $\frac{XC_2}{XB_2}=\frac{AC_2}{AB_2} \cdot \frac{AB}{AC}$. So just by the Menelaus theorem on the $\triangle A_2B_2C_2$ and points $X, Y, Z$, we're done.
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Nari_Tom
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#10
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For the second solution some should prove these circles should intersect (not tangent and have intersections with each other), For the first solution some should prove $H$ and $H_2$ are different. May be i will comeback for these later
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