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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.

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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
Problem 3
EthanWYX2009   5
N 16 minutes ago by parkjungmin
Source: 2023 China Second Round P3
Find the smallest positive integer ${k}$ with the following properties $:{}{}{}{}{}$If each positive integer is arbitrarily colored red or blue${}{}{},$
there may be ${}{}{}{}9$ distinct red positive integers $x_1,x_2,\cdots ,x_9,$ satisfying
$$x_1+x_2+\cdots +x_8<x_9,$$or there are $10{}{}{}{}{}{}$ distinct blue positive integers $y_1,y_2,\cdots ,y_{10}$ satisfiying
$${y_1+y_2+\cdots +y_9<y_{10}}.$$
5 replies
EthanWYX2009
Sep 10, 2023
parkjungmin
16 minutes ago
Inspired by old results
sqing   1
N 32 minutes ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
1 reply
+1 w
sqing
an hour ago
sqing
32 minutes ago
Made from a well-known result
m4thbl3nd3r   0
an hour ago
1. Let $a,b,c>0$ such that $$\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}=3+a+b+c.$$Prove that $$\sqrt{\frac{a+b}{2}}+\sqrt{\frac{b+c}{2}}+\sqrt{\frac{c+a}{2}}\ge ab+bc+ca.$$2. Let $x,y,z$ be sidelengths of a triangle such that $$x^2+y^2+z^2+6=2(xy+yz+zx).$$Prove that $$2\sqrt{2x}+2\sqrt{2y}+2\sqrt{2z}+(x-y)^2+(y-z)^2+(z-x)^2\ge x^2+y^2+z^2.$$
0 replies
+1 w
m4thbl3nd3r
an hour ago
0 replies
Interesting inequalities
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
5 replies
sqing
Yesterday at 12:44 PM
sqing
an hour ago
Long and wacky inequality
Royal_mhyasd   5
N 2 hours ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
5 replies
1 viewing
Royal_mhyasd
May 12, 2025
Royal_mhyasd
2 hours ago
Polynomials algebra
Foxellar   2
N 2 hours ago by elizhang101412
\textbf{9.} The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
4 hours ago
elizhang101412
2 hours ago
Collinear points
tenplusten   2
N 3 hours ago by Blackbeam999
Let $A,B,C$ be three collinear points and $D,E,F$ three other collinear points. Let $G,H,I$ be the intersection of the lines $BE,CF$ $AD,CF$ and $AD,CE$,respectively. If $AI=HD$ and $CH=GF$.Prove that $BI=GE$



I hope you will use Pappus theorem in your solutions.
2 replies
tenplusten
Jun 20, 2016
Blackbeam999
3 hours ago
Some number theory
EeEeRUT   4
N 3 hours ago by juckter
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 $p$.
4 replies
EeEeRUT
May 14, 2025
juckter
3 hours ago
Simple Geometry
AbdulWaheed   0
3 hours ago
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 $ABC$.
0 replies
AbdulWaheed
3 hours ago
0 replies
Graph Theory
ABCD1728   0
3 hours ago
Can anyone provide the PDF version of "Graphs: an introduction" by Radu Bumbacea (XYZ press), thanks!
0 replies
ABCD1728
3 hours ago
0 replies
Number theory for people who love theory
Assassino9931   3
N 4 hours ago by NamelyOrange
Source: Bulgaria RMM TST 2019
Prove that there is no positive integer $n$ such that $2^n + 1$ divides $5^n-1$.
3 replies
Assassino9931
Jul 31, 2024
NamelyOrange
4 hours ago
Interesting inequalities
sqing   0
4 hours ago
Source: Own
Let $ a,b> 0 ,   a+b+a^2+b^2=2.$ Prove that
$$ab+ \frac{k}{a+b+ab} \geq \frac{3-k+(k-1)\sqrt{5}}{2}$$Where $ k\geq 2. $
$$ab+ \frac{2}{a+b+ab} \geq \frac{1+\sqrt{5}}{2}$$$$ab+ \frac{3}{a+b+ab} \geq  \sqrt{5} $$
0 replies
sqing
4 hours ago
0 replies
Stability of Additive Cauchy Equation
doanquangdang   1
N 5 hours ago by jasperE3
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
|f(x+y)-f(x)-f(y)-x y| \leq \varepsilon\left(|x|^p+|y|^p\right)
$$for some $\varepsilon>0,$ $p \in[0,1)$ and for all $x, y \in \mathbb{R}$, then there exists a unique solution $a: \mathbb{R} \rightarrow \mathbb{R}$ of the functional equation $a(x+y)=$ $a(x)+a(y)$ for all $x, y \in \mathbb{R}$ such that
$$
\left|f(x)-a(x)-\frac{1}{2} x^2\right| \leq \frac{2}{2-2^p} \varepsilon|x|^p
$$for all $x \in \mathbb{R}$.
1 reply
doanquangdang
Aug 16, 2024
jasperE3
5 hours ago
Polynomials with common roots and coefficients
VicKmath7   10
N 6 hours ago by math-olympiad-clown
Source: Balkan MO SL 2020 A3
Let $P(x), Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$.

Demetres Christofides, Cyprus
10 replies
VicKmath7
Sep 9, 2021
math-olympiad-clown
6 hours ago
IMO 2008, Question 1
orl   156
N May 11, 2025 by Siddharthmaybe
Source: IMO Shortlist 2008, G1
Let $ H$ be the orthocenter of an acute-angled triangle $ ABC$. The circle $ \Gamma_{A}$ centered at the midpoint of $ BC$ and passing through $ H$ intersects the sideline $ BC$ at points $ A_{1}$ and $ A_{2}$. Similarly, define the points $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$.

Prove that the six points $ A_{1}$, $ A_{2}$, $ B_{1}$, $ B_{2}$, $ C_{1}$ and $ C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia
156 replies
orl
Jul 16, 2008
Siddharthmaybe
May 11, 2025
IMO 2008, Question 1
G H J
Source: IMO Shortlist 2008, G1
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G0d_0f_D34th_h3r3
22 posts
#150 • 1 Y
Y by cubres
Let's prove this nice problem :).

If we prove that $B_1$, $B_2$, $C_1$ and $C_2$ are concyclic then rest of the cases follow similarly.
Let $\omega_2$ and $\omega_3$ to be $(HB_1B_2)$ and $HC_1C_2$ respectively.


Lemma: $A$ lies on the radical axis of $\omega_2$ and $\omega_3$.
Proof We know that $\omega_2$ and $\omega_3$ have centers $M_2$ and $M_3$ respectively.

Since, $M_1$ and $M_2$ are midpoints of $AC$ and $AB$, so
$$M_2M_3 \parallel BC$$Also since line $AH \perp BC$, so
$$AH \perp M_2M_3$$Hence, $A$ lies on the radical axis of $\omega_2$ and $\omega_3$.

Since, $B_1$, $B_2$ and $C_1$, $C_2$ lie on $\omega_2$ and $\omega_3$ and $B_1B_2$ and $C_1C_2$ intersect on their radical axis.
Therefore, $B_1$, $B_2$, $C_1$ and $C_2$ are concyclic.
Attachments:
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dudade
139 posts
#151 • 1 Y
Y by cubres
Let $M$ and $N$ be the midpoints of $BC$ and $CA$, respectively. Since $NM \perp CH$, then the second intersection of $\left(A_1HA_2\right)$ and $\left(B_1HB_2\right)$, say $I$, lies on $CH$.

By Power of Point, on $\left(A_1HA_2\right)$ and $\left(B_1HB_2\right)$ yields:
\[
\begin{cases}
A_1HA_2 &: A_1C \cdot CA_2 = CI \cdot CH \\
B_1HB_2 &: B_1C \cdot CB_2 = CI \cdot CH.
\end{cases}
\]Equating yields $A_1C \cdot CA_2 = B_1C \cdot CB_2$ which, by Power of Point, implies $A_1A_2B_1B_2$ is cyclic. Similarly, $B_1B_2C_1C_2$ and $C_1C_2A_1A_2$ are both cyclic, thus $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, and $C_2$ are concyclic, as desired.
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Jndd
1417 posts
#152 • 1 Y
Y by cubres
okayy here's my hilariously bad writeup originally done on mathdash:

If $A_1, A_2, B_1, B_2, C_1, C_2$ were concyclic, then their center would have to be the circumcenter of $ABC$, which we will call $O$, since the perpendicular bisector of $A_1A_2$ is the same as the perpendicular bisector of $BC$, and likewise for all the other sides of $ABC$.

Let $D$, $E$, and $F$ be the midpoints of sides $BC$, $AC$, and $AB$. In order to show that the distance from $O$ to all of $A_1, A_2, B_1, B_2, C_1, C_2$ are the same, we will show that $OA_1^2 = DA_1^2+DO^2=DH^2+DO^2$ and the same with the other points are equal. Let $r$ be the radius of the circumcircle of $ABC$, so by Power of Point, we have $(r+DO)(r-DO)=BD^2$, giving $DO^2=r^2-BD^2$, so $DO^2+DH^2=r^2-(BD^2-DH^2)$.

Let $X_D$ be the point on $\Gamma_A$ inside $ABC$ such that $BX_D$ is tangent to $\Gamma_A$ and let $X_F$ be the point on $\Gamma_C$ inside $ABC$ such that $BX_F$ is tangent to $\Gamma_C$. Since $BD^2-DH^2=BX_D^2$ and $BF^2-FH^2=BX_F^2$, we want to show $BX_D=BX_F$. This is simply true because $B$ lies on the radical axis of $\Gamma_A$ and $\Gamma_C$ since $H$ definitely lies on the radical axis, and since that radical axis is perpendicular to the line connecting their centers, $B$ also lies on it since $BH\perp DF$ because $BH\perp AC$.

Now that we know $BX_D=BX_F$, we get $OA_1^2=DO^2+DH^2=r^2-(BD^2-DH^2) = r^2 - (BF^2-FH^2)=FO^2+FH^2=OC_2$, meaning $OA_1=OC_2$. Therefore, using symmetry, we now have that all of $A_1, A_2, B_1, B_2, C_1, C_2$ have the same distance from $O$, as desired.
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RedFireTruck
4243 posts
#153 • 1 Y
Y by cubres
https://mathdash.s3.us-east-2.amazonaws.com/user-uploads/65c15bae768c20d4b06de6f0/1722320783919-11

clearly the center of such a circle would have to be at the circumcenter (intersection of green lines) because the circles are centered at midpoints so the perp. bisector of $A_1A_2$ would be the same as the perp. bisector of $BC$, for example

now it just suffices to show that the sum of squares of the purple lengths in the picture doesn't depend on which midpoint they are on (in the example they are on the midpoint of $BC$)

let the circumcenter be $0$ and the vertices be $a,b,c$ s.t. $|a|=|b|=|c|=1$ and the orthocenter is $a+b+c$

then the sum of the squares of the purple lengths in the picture is $$|\frac{b+c}{2}|^2+|(a+b+c)-\frac{b+c}{2}|^2=|\frac{b+c}{2}|^2+|a+\frac{b+c}{2}|^2$$$$=\frac{b+c}{2}\frac{\frac1b+\frac1c}{2}+(a+\frac{b+c}{2})(\frac{1}{a}+\frac{\frac1b+\frac1c}{2})$$$$=2+\frac{\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}}{2}$$which is symmetric in $a,b,c$, as desired.
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ezpotd
1286 posts
#154 • 1 Y
Y by cubres
Claim: $A_2A_1B_1B_2$ and cyclic variants are all cyclic.

Proof: Let $X_C$ be the intersection of $(HA_1A_2), (HB_1B_2)$, clearly $X_C$ is the reflection of $H$ over $M_AM_B$, so $X_CH$ is just the $C$ altitude, and by radax we are done.

Now assume these three circles $A_1A_2B_1B_2$ and cyclic variants are all distinct, then their radaxes should concur but they happen to precisely be the sides of the triangle, so the three circles are the same and all six points are cyclic.
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MagicalToaster53
159 posts
#155 • 1 Y
Y by cubres
Suppose $BH$ and $CH$ meet $\overline{CA}$ and $\overline{AB}$ at $E$ and $F$, respectively. Then observe $\Gamma_A \cap \Gamma_B = X$ lies on $CH$. Indeed, $XH$ is the radical axis of $\Gamma_A$ and $\Gamma_B$, we find $XH \perp M_AM_B \implies XH \perp AB$. Hence $C$ has equal power from both $\Gamma_A$ and $\Gamma_B$ so that $CA_2 \times CA_1 = CB_1 \times CB_2 \implies A_2, A_1, B_1, B_2$ are concyclic. Similarly $B_1, B_2, C_1, C_2$ and $C_1, C_2, A_1, A_2$ are separately concyclic, which follows from symmetry. Hence $A_1, A_2, B_1, B_2, C_1, C_2$ all lie on a circle (and further the circumcenter of such a circle is the circumcenter of $ABC$), as desired. $\blacksquare$
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megahertz13
3184 posts
#156 • 1 Y
Y by cubres
Let $X$, $Y$, and $Z$ be the midpoints of the sides opposite $A$, $B$, and $C$, respectively. Also, let $\omega_1, \omega_2$ be the circles centered at $Y$ and $Z$, respectively. Suppose that $\omega_1$ and $\omega_2$ intersect again at $P$.

Since $PH\perp YZ$ and $AH\perp BC\implies AH\perp YZ$, we have $P, A, H$ collinear.

Now, $A$ lies on the radical axis of $\omega_1,\omega_2$, so $B_1, B_2, C_1,C_2$ are concyclic. Similarly, we can find that $A_1, A_2, B_1, B_2$ are concyclic, as are $C_1, C_2, A_1, A_2$. Notice that the center of those circles is the circumcenter $O$ of $ABC$. Since those three circles have the same radius of $$OA_1=OA_2=OB_1=OB_2=OC_1=OC_2,$$they are the same circle and we are done.
This post has been edited 1 time. Last edited by megahertz13, Nov 26, 2024, 6:16 PM
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gladIasked
648 posts
#157 • 1 Y
Y by cubres
Let $M$ be the midpoint of $AC$ and $N$ be the midpoint of $AB$. Note that $AH\perp BC$ and $BC\parallel MN$ by similar triangles. Thus, $AH\perp MN$. Consider the circles $(C_1HC_2)$ and $(B_1HB_2)$; note that the centers of these two circles are $N$ and $M$, respectively. However, because $H$ lies on both circles and $AH\perp MN$, we know that $AH$ is the radical axis of the two circles. Therefore, by power of a point, $B_1B_2C_1C_2$ is cyclic. We can similarly conclude that $A_1A_2B_1B_2$ and $C_1C_2A_1A_2$ are cyclic.

Assume for the sake of contradiction that $(A_1A_2B_1B_2)\ne (B_1B_2C_1C_2)$. Clearly, $AC$ is the radical axis of these two circles. However, the power of $B$ with respect to $(B_1B_2C_1C_2)$ is $BC_2\cdot BC_1 = BA_1\cdot BA_2$ by power of a point on circle $(C_1C_2A_1A_2)$. Clearly, the power of $B$ with respect to $(A_1A_2B_1B_2)$ is $BA_1\cdot BA_2$, so in fact $B$ must lie on the radical axis of $(A_1A_2B_1B_2)$ and $(B_1B_2C_1C_2)$, or just $AC$. This is a contradiction, so $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic. $\blacksquare$
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Vedoral
89 posts
#158 • 1 Y
Y by cubres
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Maximilian113
575 posts
#159 • 1 Y
Y by cubres
Let $D, E, F$ be the midpoints opposite $A, B, C.$ We begin by showing that $C_2, C_1, B_2, B_1$ are concyclic. Let $P$ be the second intersection of the circles, then clearly $PH \perp EF$ but since $EF \parallel BC$ by the Midpoint Theorem, it follows that $PH \perp BC,$ so $A$ lies on $PH.$ Therefore, $A$ lies on the radical axis of $\Gamma_B, \Gamma_C$ and by Power of a Point our claim is proven.

Similarly, $B_1, B_2, A_1, A_2$ are concyclic, and $C_1, C_2, A_1, A_2$ are too. But their centers are clearly $O,$ the circumcenter of $\triangle ABC,$ so we are done. QED
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clarkculus
244 posts
#160 • 2 Y
Y by centslordm, cubres
Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Because $AH\perp BC$, $AH\perp MN$, and because $H$ lies on the radical axis of $\Gamma_B$ and $\Gamma_C$, $A$ also lies on the radical axis of $\Gamma_B$ and $\Gamma_C$, implying $(AC_1)(AC_2)=(AB_1)(AB_2)$. By the Converse of Power of a Point, this implies $C_1,C_2,B_1,B_2$ are concyclic. Similarly, $A_1,A_2,C_1,C_2$ are concyclic, which finishes.
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QueenArwen
110 posts
#161 • 1 Y
Y by cubres
Let $M_1$ be the midpoint of $AB$ and $M_2$ be the midpoint of $AC$. Since $M_1M_2$ is parallel to $BC$, $AH$ is perpendicular to it, hence $A$ lies on the radical axis of $\Gamma_B$ and $\Gamma_C$ (since $H$ lies on both these circles). Since $C_1C_2$ and $B_1B_2$ intersect on the radical axis, $B_1B_2C_1C_2$ is cyclic by a well-known theorem. Similarly $A_1A_2B_1B_2$ and $A_1A_2C_1C_2$ are also cyclic. Since the perpendicular bisectors of $C_1C_2$ and $B_1B_2$ are the same as those of $AB$ and $AC$ respectively, the circumcenter of cyclic quadrilateral $B_1B_2C_1C_2$ and $\triangle{ABC}$ is the same, which we call $O$. Similarly $O$ is also the circumcenter of $A_1A_2B_1B_2$ and $A_1A_2C_1C_2$, so $OA_1 = OA_2 = OB_1=OB_2=OC_1=OC_2$, hence $A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic.
This post has been edited 2 times. Last edited by QueenArwen, Apr 3, 2025, 6:03 AM
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eg4334
637 posts
#162 • 1 Y
Y by cubres
Clearly we need to only prove it true for two circles, lets say the $A$ and $B$ ones. Then, it suffices for $C$ to have the same power wrt to both, or $C, H, K$ to be collinear where $K$ is their second intersection. But $HK$ (because perpendicular to line joining centers) and $HC$ (orthocenter) are both perpendicular to $AB$ and the result follows.
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Ilikeminecraft
658 posts
#163 • 1 Y
Y by cubres
Let the circles through the midpoints of $AB, BC, AC$ be $\omega_C, \omega_A, \omega_B,$ respectively.

I will prove that $A_1, A_2, B_1, B_2$ is cyclic. Notice that the radical axis of $\omega_A, \omega_B$ must be perpendicular to $EF,$ and it also runs through $H.$ Thus, $C$ also lies on this radical axis. However, because $CH, AC, BC,$ are concurrent, we have that there must exist a circle running through $A_1, A_2, B_1, B_2.$

Similarly, there is a circle going through $A_1, A_2, C_1, C_2,$ and another one going through $B_1, B_2, C_1, C_2.$ However, in all circles, the center is clearly $O,$ the circumcenter. Thus, we are done.
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Siddharthmaybe
117 posts
#164 • 1 Y
Y by alexanderhamilton124
Let $M_A, M_B, M_C$ be the midpoints of $BC,CA,AB$, then by midpoint theorem, $M_AM_B \perp HC$, also $\Gamma_A \cap\Gamma_B = H$ (and some other point). Now these conditions are enough to deduce that $HC$ is the radax of $\Gamma_A, \Gamma_B$. So evidently, $CA_1 \times CA_2 = CB_1 \times CB_2$ or $A_1A_2B_1B_2$ is cyclic and similarly for $A_1A_2C_1C_2, B_1B_2C_1C_2$. Now assume FTSOC that they have distinct circumcircles, then the radax of the first two circles is the common chord $\overleftrightarrow{A_1A_2} \equiv \overleftrightarrow{BC}$, similarly the lines $AB, CA$ are also the radical axes of the other two pairs, now by radical center they must be concurrent which means that $ABC$ is degenerate $\implies\impliedby$.
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