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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Mar 2, 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
USA Canada math camp
Bread10   15
N 6 minutes ago by akliu
How difficult is it to get into USA Canada math camp? What should be expected from an accepted applicant in terms of the qualifying quiz, essays and other awards or math context?
15 replies
1 viewing
Bread10
Mar 2, 2025
akliu
6 minutes ago
questions from a first-time applicant to math camps
akliu   18
N 25 minutes ago by akliu
hey!! im a first time applicant for a lot of math camps (namely: usa-canada mathcamp, PROMYS, Ross, MathILY, HCSSiM), and I was just wondering:

1. how much of an effect would being a first-time applicant have on making these math camps individually?
2. I spent a huge amount of effort (like 50 or something hours) on the USA-Canada Mathcamp application quiz in particular, but I'm pretty worried because supposedly almost no first-time applicants get into the camp. Are there any first-time applicants that you know of, and what did their applications (as in, qualifying quiz solutions) look like?
3. Additionally, a lot of people give off the impression that not doing the full problem set will screw your application over, except in rare cases. How much do you think a fakesolve would impact my PROMYS application chances?

thanks in advance!
18 replies
akliu
Mar 12, 2025
akliu
25 minutes ago
conics ew
math31415926535   31
N 37 minutes ago by Magnetoninja
Source: 2022 AIME II Problem 12
Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that $$\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.$$Find the least possible value of $a+b.$
31 replies
math31415926535
Feb 17, 2022
Magnetoninja
37 minutes ago
high tech FE as J1?!
imagien_bad   45
N an hour ago by gladIasked
Source: USAJMO 2025/1
Let $\mathbb Z$ be the set of integers, and let $f\colon \mathbb Z \to \mathbb Z$ be a function. Prove that there are infinitely many integers $c$ such that the function $g\colon \mathbb Z \to \mathbb Z$ defined by $g(x) = f(x) + cx$ is not bijective.
Note: A function $g\colon \mathbb Z \to \mathbb Z$ is bijective if for every integer $b$, there exists exactly one integer $a$ such that $g(a) = b$.
45 replies
+1 w
imagien_bad
Today at 12:00 PM
gladIasked
an hour ago
Numbers from 1 to 15 with rare properties
hectorleo123   1
N an hour ago by EmersonSoriano
Source: 2015 Peru Cono Sur TST P2
Let $a, b, c$ and $d$ be elements of the set $\{ 1, 2, 3,\ldots , 2014, 2015 \}$ such that $a < b < c < d$, $a + b$ is a divisor of $c + d$, and $a + c$ is a divisor of $b + d$. Determine the largest value that $a$ can take.
1 reply
hectorleo123
Jul 10, 2023
EmersonSoriano
an hour ago
Number Theory
MuradSafarli   3
N 2 hours ago by MuradSafarli
find all natural numbers \( (a, b) \) such that the following equation holds:

\[
7^a + 1 = 2b^2
\]
3 replies
MuradSafarli
3 hours ago
MuradSafarli
2 hours ago
Abelkonkurransen 2025 2a
Lil_flip38   1
N 3 hours ago by RANDOM__USER
Source: Abelkonkurransen
A teacher asks each of eleven pupils to write a positive integer with at most four digits, each on a separate yellow sticky note. Show that if all the numbers are different, the teacher can always submit two or more of the eleven stickers so that the average of the numbers on the selected notes are not an integer.
1 reply
Lil_flip38
Today at 11:10 AM
RANDOM__USER
3 hours ago
Abelkonkurransen 2025 2b
Lil_flip38   3
N 3 hours ago by alexanderhamilton124
Source: abelkonkurransen
Which positive integers $a$ have the property that \(n!-a\) is a perfect square for infinitely many positive integers \(n\)?
3 replies
Lil_flip38
Today at 11:12 AM
alexanderhamilton124
3 hours ago
Oi! These lines concur
Rg230403   18
N 4 hours ago by HoRI_DA_GRe8
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
18 replies
Rg230403
May 10, 2021
HoRI_DA_GRe8
4 hours ago
Incircle
PDHT   0
4 hours ago
Source: Nguyen Minh Ha
Given a triangle \(ABC\) that is not isosceles at \(A\), let \((I)\) be its incircle, which is tangent to \(BC, CA, AB\) at \(D, E, F\), respectively. The lines \(DE\) and \(DF\) intersect the line passing through \(A\) and parallel to \(BC\) at \(M\) and \(N\), respectively. The lines passing through \(M, N\) and perpendicular to \(MN\) intersect \(IF\) and \(IE\) at \(Q\) and \(P\), respectively.

Prove that \(D, P, Q\) are collinear and that \(PF, QE, DI\) are concurrent.
0 replies
PDHT
4 hours ago
0 replies
Surprisingly low answer to the question what is the maximum
mshtand1   2
N 5 hours ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 8.6, 10.5
Given $2025$ positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining $2024$ numbers that are not hidden. What is the maximum number of the $2025$ written numbers that can be perfect squares?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
sarjinius
5 hours ago
Functional Equation
AnhQuang_67   5
N 5 hours ago by megarnie
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$f(xf(y)+2y)=f(f(y))+f(xy)+xf(y), \forall x, y \in \mathbb{R}$$
5 replies
AnhQuang_67
Today at 3:57 PM
megarnie
5 hours ago
hard ............ (2)
Noname23   1
N 5 hours ago by Amkan2022
problem
1 reply
Noname23
6 hours ago
Amkan2022
5 hours ago
Fun issue about Euler’s function
luutrongphuc   2
N 6 hours ago by vi144
Let $p$ is a prime number and $n,\alpha$ are positive integers. Prove that there exist infinitely $a$ such that $\phi(a),\phi(a+1),…,\phi(a+n)$ are divisible by $p^{\alpha}$
2 replies
luutrongphuc
Today at 11:27 AM
vi144
6 hours ago
Erecting Rectangles
franchester   101
N Mar 15, 2025 by Ilikeminecraft
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that \[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.
101 replies
franchester
Apr 15, 2021
Ilikeminecraft
Mar 15, 2025
Erecting Rectangles
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
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lian_the_noob12
173 posts
#101
Y by
$\color{magenta} \boxed{\textbf{SOLUTION P1}}$
$\color{red} \textbf{Geo Marabot Solve 8}$

The angle condition implies the circumcircles of the three rectangles intersect at some point $P$
We have, $\angle B_2PC+\angle A_1PC=90+90=180$ and similar for other three we are done that the lines concur at $P\blacksquare$
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peace09
5411 posts
#102
Y by
Claim. The rectangles' circumcircles concur.
Proof. Let $(BCC_1B_2)$ and $(CAA_1C_2)$ meet again at $O$. Note $\sum_{\text{cyc}}\angle BOC=360^\circ$ and \[(\angle BOC,\angle COA)=(180^\circ-\angle BC_1C,180^\circ-\angle CA_1A),\]whence the angle condition gives $\angle AOB=180^\circ-\angle AB_1B$. Then $O\in(ABB_1A_2)$ as claimed.

The rest is easy: $\angle APB_1=\angle APC_2=90^\circ$ and so $O\in B_1C_2$ and so on. In all, $O$ is the desired point of concurrency. $\square$
This post has been edited 1 time. Last edited by peace09, Feb 25, 2025, 8:43 PM
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HamstPan38825
8854 posts
#103
Y by
Let $P = (ABB_1A_2) \cap (AC_1CA)$; the given angle condition then yields that $P$ lies on $(BCB_2C_1)$ too. It follows that $\angle BPC_1 +\angle BPA_2 = 90^\circ + 90^\circ = 180^\circ$, so $P$ lies on $\overline{A_2C_1}$ and cyclic permutations.
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setts
3 posts
#105
Y by
This problem is trivial when you notice that

<ABD = 180 - <AB_1 B
<BDC = 360 - <ADB - <ADC.
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alsk
28 posts
#106
Y by
Slightly different finish.

If we let $A \neq P = (ABB_1A_2) \cap (AC_1C_2A_1)$, the given angle condition gives us that $P \in (BCC_1B_2)$, so all three circumcircles of the rectangles concur.

Furthermore, notice that $BA_2, BC_1$ are diameters of their respective circles, and both circles go through $B, P$. Thus, $A_2C_1$ goes through $P$ (can be shown by similar triangles with circumcenters), and we can use the same logic to find that $P$ lies on all three lines.
This post has been edited 1 time. Last edited by alsk, Mar 12, 2024, 2:09 AM
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Mr.Sharkman
487 posts
#107
Y by
Solution
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xHypotenuse
739 posts
#108
Y by
redacted
This post has been edited 1 time. Last edited by xHypotenuse, Apr 23, 2024, 6:01 PM
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Markas
105 posts
#109
Y by
Let $\angle BC_1C = \alpha$, $\angle CA_1A = \beta$ and $\angle AB_1B = 180 - \alpha - \beta$. Every rectangle is a cyclic quadrilateral. Lets draw the circumcircles around the rectangles $BCC_1B_2$, $CAA_1C_2$, $ABB_1A_2$ and name them $\omega$, $\omega_1$ and $\omega_2$, respectively.

Let $\omega$ $\cap$ $\omega_1$ = P, where P is a point other than C.

Claim: P $\in$ $\omega_2$

Proof: $\angle APB = 360 - \angle APC - \angle BPC$. Since $A_1APC$ is cyclic, $\angle APC = 180 - \angle AA_1C = 180 - \beta$. Since $BC_1CP$ is cyclic, $\angle BPC = 180 - \angle BC_1C = 180 - \alpha$ $\Rightarrow$ $\angle APB = 360 - (180 - \beta) - (180 - \alpha) = \alpha + \beta$. Now $\angle APB + \angle AB_1B = \alpha + \beta + 180 - \alpha - \beta = 180$ $\Rightarrow$ $AB_1BP$ is cyclic $\Rightarrow$ P $\in$ $\omega_2$.

Now we want to show that P lies on $B_1C_2$, $A_2C_1$ and $A_1B_2$. We have $\angle A_1PC + \angle CPB_2 = 90^\circ + 90^\circ = 180^\circ$, since $A_1C$ is a diameter in $\omega$ and $B_2C$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_1B_2$. Also $\angle B_1PA + \angle APC_2 =  90^\circ + 90^\circ = 180^\circ$, since $AB_1$ is a diameter in $\omega_2$ and $AC_2$ is a diameter in $\omega$ $\Rightarrow$ P lies on $B_1C_2$. Finally $\angle A_2PB + \angle BPC_1 = 90^\circ + 90^\circ = 180^\circ$, since $A_2B$ is a diameter in $\omega_2$ and $BC_1$ is a diameter in $\omega_1$ $\Rightarrow$ P lies on $A_2C_1$. This means that $B_1C_2$, $A_2C_1$ and $A_1B_2$ are concurrent at point P $\Rightarrow$ we are ready.
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shendrew7
792 posts
#110
Y by
First notice the circumcircles of the rectangles are forced to meet at a single point, say $K$, from the angle condition. Then we have
\[\angle B_1KX_2 = \angle B_1KA + \angle AKC_2 = \angle AA_2B_1 + \angle AA_1C_2 = 90 + 90 = 180,\]
so $X$ lies on $B_1C_2$, and analogously the other two lines. $\blacksquare$
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eg4334
605 posts
#111
Y by
Immediately constructed circumcircles for potential radical axis, turned out to be a lot different.

Let $(AA_2B_1B) \cap (BB_2C_1C) = D$. $$\angle ADC = 360 - \angle ADB - \angle CDB  = \angle AB_1B + \angle CC_1B = 180 - \angle AA_1C$$so $A_1ADC$ is cyclic and the three circumcircles of the rectangles concur. Also, $B_1DC_2$ are collinear and cyclic variations hold by angle chasing: $$\angle B_1DC_2 = \angle B_1DA + \angle ADC_2 = 180$$so done.
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trolled767
3 posts
#112
Y by
Let the circumcircles of rectangles intersect at a point O.
Proof:
let circles ABB1A2 and AA1CC2 intersect at point O,
angle BOC is 360 - 180 - 180 + angleCA1A + angle AB1B
which = angle BC1C (mod 180)
Now trivially angle BOA2 = 90; and so is angle BOC1, cyclically all those lines pass through O.
hence proved
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Saucepan_man02
1299 posts
#113
Y by
(Easy P1 tho):

Let $P$ denote the intersection of $(ABB_1A_2), (BCC_1B_2)$. We will prove that $P$ lies on $(AA_1C_2C)$.
Notice that: $\angle BPC = 180^\circ - \angle B C_1 C$ and $\angle APB = 180^\circ - \angle A B_1 B$ which forces $\angle APC = 180^\circ - \angle A A_1 C$ and therefore $P$ lies on $(AA_1C_2C)$.

Notice that: $CP \perp A_1P$ and $CP \perp B_2P$. Thus, $A_1 B_2$ passes through point $P$. Similarly, the other two lines pass through $P$ and we are done.
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Mathandski
715 posts
#114
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Note: This was transcribed by ChatGPT so there may be typoes

Let \( BCC_1B_2 \) and \( CAA_1C_2 \) meet at \( P \). Let the intersections of \( \overline{A_2 C} \cap \overline{CA_1} = O_B \) and \( \overline{B C_1} \cap \overline{C B_2} = O_A \).

Since \( C A A_1 C_2 \), \( A B B_1 A_2 \) are rectangles, we have \( O_B, O_C \) are the centers of the respective circumcircles.

We use directed angles to show \( P \in \odot(A B B_1 A_2) \).

Since \( \angle B C_1 C \), \( \angle C_1 A A \), and \( \angle A B_1 B \) are in the same direction as each other as the rectangles are all erected outside \( \triangle A B C \),

\[
\angle B C_1 C + \angle C A_1 A + \angle A B_1 B = 180^\circ
\]
\[
\Rightarrow \measuredangle B C_1 C + \measuredangle C A_1 A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P C + \angle C P A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P A + \measuredangle A B_1 B = 0
\]
\[
\Rightarrow \measuredangle B P A = \measuredangle B B_1 A \Rightarrow (A P B B_1) \text{ cyclic.}
\]
It now suffices to show \( P \in \overline(A_1 B_2) \) as the other two follow analogously.

Indeed, the midpoint \( M \) of \( \overline{C P} \) satisfies \( M \in \overline{O_A O_B} \) since \( \overline{C P} \) is the shared chord of \( \odot(B C C_1 B_2) \) and \( \odot(C A A_1 C_2) \).

A homothety at \( C \) of scale \( 2 \) suffices.
This post has been edited 2 times. Last edited by Mathandski, Feb 28, 2025, 12:19 AM
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bjump
973 posts
#116 • 2 Y
Y by KenWuMath, imagien_bad
Consider drawing the circles $(ABB_1A_2)$, $(BCC_1B_2)$, $(CAA_1C_2)$. Let $(ABB_1A_2) \cap  (BCC_1B_2) = X_{\text{onk}}$. Now we have
$$\angle AX_{\text{onk}}B = 360^{\circ}- \angle BX_{\text{onk}}C - \angle CX_{\text{onk}}A = \angle AB_1B + \angle BC_1C = 180^\circ - \angle AA_1C  $$Therefore the $3$ drawn circles concur at $X_{\text{onk}}$.

Now the line between the centers of $(AA_2B_1B)$, and $(AA_1C_2C)$, intersects $AX_{\text{onk}}$ at the midpoint of $AX_{\text{onk}}$ since $B_1$, and $C_2$, are the reflections of $A$ over the centers of the circles respectively. By homothety $X_{\text{onk}} \in B_1C_2$. By a symmetry all the desired lines concur at $X_{\text{onk}}$.
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Ilikeminecraft
300 posts
#117
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pretty easy???

Let $P = (ABB_1A_2)\cap (AA_1C_2C).$ We have that $\angle BPC = 360 - \angle BPA - \angle APC = \angle AB_1B + \angle CA_1A = 180 - \angle BC_1C,$ so all 3 circles concur at $P$.
Thus, $\angle APC = \angle APB + \angle BPC_1 = 180$
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