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k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
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
Incenters on an inscribed quadrilateral
AlperenINAN   2
N 28 minutes ago by EmersonSoriano
Source: 2023 Turkey Junior National Olympiad P2
Let $ABCD$ be an inscribed quadrilateral. Let the incenters of $BAD$ and $CAD$ be $I$ and $J$ respectively. Let the intersection point of the line that passes through $I$ and perpendicular to $BD$ and the line that passes through $J$ and perpendicular to $AC$ be $K$. Prove that $KI=KJ$
2 replies
AlperenINAN
Dec 22, 2023
EmersonSoriano
28 minutes ago
Really classical inequatily from canada
shobber   78
N 29 minutes ago by Tony_stark0094
Source: Canada 2002
Prove that for all positive real numbers $a$, $b$, and $c$,
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c \]
and determine when equality occurs.
78 replies
shobber
Mar 5, 2006
Tony_stark0094
29 minutes ago
They copied their problem!
pokmui9909   9
N 44 minutes ago by Mapism
Source: FKMO 2025 P1
Sequence $a_1, a_2, a_3, \cdots$ satisfies the following condition.

(Condition) For all positive integer $n$, $\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1$ holds.

For a positive integer $m = 1001 \cdot 2^{2025}$, compute $a_m$.
9 replies
pokmui9909
Mar 29, 2025
Mapism
44 minutes ago
Normal but good inequality
giangtruong13   0
44 minutes ago
Source: From a province
Let $a,b,c> 0$ satisfy that $a+b+c=3abc$. Prove that: $$\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5} $$
0 replies
giangtruong13
44 minutes ago
0 replies
Count the distinct values in 2025 fractions
Stuttgarden   1
N an hour ago by RagvaloD
Source: Spain MO 2025 P1
Determine the number of distinct values which appear in the sequence \[\left\lfloor\frac{2025}{1}\right\rfloor,\left\lfloor\frac{2025}{2}\right\rfloor,\left\lfloor\frac{2025}{3}\right\rfloor,\dots,\left\lfloor\frac{2025}{2024}\right\rfloor,\left\lfloor\frac{2025}{2025}\right\rfloor.\]
1 reply
Stuttgarden
4 hours ago
RagvaloD
an hour ago
Geometry Ratio
steven_zhang123   0
an hour ago
Source: 0
In triangle \( \triangle PQR \), \( PQ = PR \), and \( \angle P = 120^\circ \). Points \( M \) and \( N \) are located on \( PQ \) and \( PR \) respectively, such that \( PQ = 2 \cdot PM \) and \( \angle PMN = \angle NQR \). Find the ratio of \( PN \) to \( NR \).
0 replies
steven_zhang123
an hour ago
0 replies
IMO Shortlist 2013, Geometry #2
lyukhson   78
N an hour ago by numbertheory97
Source: IMO Shortlist 2013, Geometry #2
Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.
78 replies
lyukhson
Jul 9, 2014
numbertheory97
an hour ago
Romania TST 2021 Day 1 P4
oVlad   21
N an hour ago by ravengsd
Determine all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following relationship for all real numbers $x$ and $y$\[f(xf(y)-f(x))=2f(x)+xy.\]
21 replies
oVlad
May 15, 2021
ravengsd
an hour ago
geometry
Duc15_g-yh   2
N an hour ago by Duc15_g-yh
Source: Original
Title: Geometry Problem – Equal Angles and Concurrency

Post Content:

Hi everyone,

I need help solving this geometry problem:

Given a triangle ABC, let (C_1) be the excircle touching BC, CA, AB at X, P, Q respectively. Similarly, let (C_2) be the excircle touching CA, AB, BC at Y, M, N respectively.
1. Prove that \angle YMN = \angle XQP.
2. Let S be the intersection of MN and PQ. Prove that MY, PX, and SC are concurrent.

Any hints or full solutions would be greatly appreciated!

Thanks in advance!
2 replies
Duc15_g-yh
2 hours ago
Duc15_g-yh
an hour ago
Functional equation over nonzero reals
Stuttgarden   1
N an hour ago by pco
Source: Spain MO 2025 P6
Let $\mathbb{R}_{\neq 0}$ be the set of nonzero real numbers. Find all functions $f:\mathbb{R}_{\neq 0}\rightarrow\mathbb{R}_{\neq 0}$ such that, for all $x,y\in\mathbb{R}_{\neq 0}$, \[(x-y)f(y^2)+f\left(xy\,f\left(\frac{x^2}{y}\right)\right)=f(y^2f(y)).\]
1 reply
1 viewing
Stuttgarden
4 hours ago
pco
an hour ago
A geometry problem
Lttgeometry   1
N 2 hours ago by Lttgeometry
Given a non-isosceles triangle $ABC$ that is inscribed in $(O)$ . The incircle $(I)$ is tangent to $BC,CA,AB$ at $D,E,F$ respectively. A line through $A$ parallel to $BC$ intersects $(O)$ at $T$, and $TD$ intersects $(O)$ again at $J$. Let $N$ is the midpoint of $BC$. $P,Q$ be the second intersection of $JE,JF$ with $(O)$. $AI$ intersects $(O)$ again at $M$. Prove that the line passing through $A$ perpendicular to $PQ$ bisects $MN$.
1 reply
Lttgeometry
Yesterday at 4:02 PM
Lttgeometry
2 hours ago
thanks u!
Ruji2018252   1
N 2 hours ago by bluefingreen
Find $x$
\[3\sqrt[3]{(2x^2-x-1)^2}-12\sqrt[3]{2x^2-x-1} +12-12x=x^3+2x^2-4x+1\]
1 reply
Ruji2018252
4 hours ago
bluefingreen
2 hours ago
Easy geo
kooooo   2
N 2 hours ago by alexmay
Source: own
In triangle $ABC$, let $O$ and $H$ be the circumcenter and orthocenter, respectively. Let $M$ and $N$ be the midpoints of $AC$ and $AB$, respectively, and let $D$ and $E$ be the feet of the perpendiculars from $B$ and $C$ to the opposite sides, respectively. Show that if $X$ is the intersection of $MN$ and $DE$, then $AX$ is perpendicular to $OH$.
2 replies
kooooo
Jul 31, 2024
alexmay
2 hours ago
k VERY HARD MATH PROBLEM!
slimshadyyy.3.60   42
N 2 hours ago by giangtruong13
Let a ≥b ≥c ≥0 be real numbers such that a^2 +b^2 +c^2 +abc = 4. Prove that
a+b+c+(√a−√c)^2 ≥3.
42 replies
+1 w
slimshadyyy.3.60
Saturday at 10:49 PM
giangtruong13
2 hours ago
Balkan MO 2014 shortlist-G1
dizzy   14
N Oct 4, 2024 by onyqz
Source: Balkan MO 2014 shortlist-G1
Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$
14 replies
dizzy
Jun 10, 2015
onyqz
Oct 4, 2024
Balkan MO 2014 shortlist-G1
G H J
G H BBookmark kLocked kLocked NReply
Source: Balkan MO 2014 shortlist-G1
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dizzy
181 posts
#1 • 4 Y
Y by Adventure10, Mango247, lian_the_noob12, Rounak_iitr
Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$
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huynguyen
535 posts
#2 • 3 Y
Y by mamavuabo, Adventure10, Mango247
i don't think it's too hard, so please check my solution carefully:
In fact, we just have to prove that $\frac{AB}{AM}=\frac{BP}{PM}$.
By the law of sine, $\frac{AB}{AM}=\frac{sin\widehat{AMB}}{sin\widehat{ABM}}$ and $\frac{BP}{PM}=\frac{sin\widehat{BMP}}{sin\widehat{PBM}}$
Since $\widehat{PBM}+\widehat{ABM}=180$ so $sinPBM=sinABM$
Since the two triangles $ABM$ and $NCM$ are similar, we get $\widehat{BMP}=\widehat{CMN}=\widehat{AMB}$ and as a result, $sin\widehat{BMP}=sin\widehat{AMB}$
So we get the desired property. (Q.E.D)
This post has been edited 1 time. Last edited by huynguyen, Jun 10, 2015, 12:21 PM
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tranquanghuy7198
253 posts
#3 • 3 Y
Y by huynguyen, Adventure10, Mango247
My solution:
Notice that: $\triangle{CNM} \sim \triangle{BAM}$ (1)
We have:
$\frac{AB}{BP}.\frac{PM}{MN}.\frac{NC}{CA} = 1$ (Menelaus)
$\Rightarrow \frac{BP}{PM} = \frac{CN}{NM}$
$\Rightarrow \frac{BP}{PM} = \frac{BA}{AM}$ (because of (1))
$\Rightarrow$ Bisectors of $\angle{BAM}, \angle{BPM}$ intersect on $BC$
Q.E.D
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samithayohan
41 posts
#4 • 2 Y
Y by sydneymark, Adventure10
I think that it is unnecessary to use Menelaus Theorem. Just observe that $\triangle$$PBM$$\sim$$\triangle$$ACM$
Hence $\implies$ $PB$$/$$PM$$=$$AM$$/$$AC$ But since $AB$$=$$AC$ from the converse of the angle bisector theorem we are done.
It's too easy :-D
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Gryphos
1699 posts
#5 • 2 Y
Y by Adventure10, Mango247
Well, it´s G1, that´s usually the easiest geometry problem in the shortlist.
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TelvCohl
2312 posts
#6 • 4 Y
Y by AlastorMoody, enhanced, Adventure10, Mango247
(See attachment)
Internal bisector of $ \angle BAM $, external bisector of $ \angle MPA $, external bisector $ BC $ of $ \angle AMP $ are concurrent at A-excenter of $ \triangle APM $ .
Attachments:
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AdithyaBhaskar
652 posts
#7 • 2 Y
Y by Adventure10, Mango247
Hey, where can we get Balkan Mo Shortlists???
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dizzy
181 posts
#8 • 2 Y
Y by Adventure10, Mango247
I will post them
Z K Y
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AlastorMoody
2125 posts
#9 • 3 Y
Y by amar_04, Adventure10, Mango247
Nice & Beautiful
Balkan MO SL 2014 G1 wrote:
Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$
Solution: The condition $\angle BAM=\angle MNC$ $\implies$ $\angle AMB=\angle NMC$. Hence, $BM$ bisects $\angle AMN$ $\implies$ bisectors of $\angle BAM$, $\angle BPM$ concur at the $P-$excenter WRT $\Delta APM$
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kvs
620 posts
#10 • 1 Y
Y by Adventure10
It suffices to show that line $BC$ bisects $\angle{AMP}$, since this implies the incenter of $\triangle{APM}$ lies on $BC$ as desired. Since $\angle{MNC}=\angle{ABM}$ and $\angle{ABM}=\angle{NCM}$ (since $\triangle{ABC}$ is isosceles), $\angle{NMC}=\angle{AMB}=\angle{BMP}$, as desired.
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Aimingformygoal
126 posts
#11
Y by
Solution
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hectorleo123
339 posts
#12
Y by
dizzy wrote:
Let $ABC$ be an isosceles triangle, in which $AB=AC$ , and let $M$ and $N$ be two points on the sides $BC$ and $AC$, respectively such that $\angle BAM = \angle MNC$. Suppose that the lines $MN$ and $AB$ intersects at $P$. Prove that the bisectors of the angles $\angle BAM$ and $\angle BPM$ intersects at a point lying on the line $BC$
Let $T=$ bisector of $\angle BAC \cap BC$
The problem is the same as proving that $TP$ is the bisector of $\angle BPM$
Let $NC=a, MN=c$ and $MC=b$
Since $\triangle BAM \cong \triangle CNM$
$\Rightarrow AB=ak, AM=ck$
Since $AT$ is a bisector of $\angle BAM$
$\Rightarrow \frac{BT}{TM} = \frac {a}{c}$
$\Rightarrow PT$ is bisector of $\angle BPM \Leftrightarrow \frac{BP}{PM} = \frac{a}{c}$
Let $2 \alpha = \angle MNC$ and $\beta=\angle MCN$
$\Rightarrow \angle PBM = \beta$ and $\angle PMB=2 \alpha + \beta$
In $\triangle MNC:$
$\frac{sin \beta}{sin( 2 \alpha + \beta )}=\frac{c}{a}$
In $\triangle BPM:$
$\frac{BP}{PM}=\frac{sin( 2 \alpha + \beta )}{sin \beta} = \frac{a}{c} _\blacksquare$
This post has been edited 3 times. Last edited by hectorleo123, May 20, 2023, 12:57 PM
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GeorgeRP
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#13
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Let $\angle MBA = \beta$, $\angle BAM = \alpha$. Then, $\angle BMA = 180 - \alpha - \beta$ and $\angle BMN = \alpha + \beta \Rightarrow \frac{BA}{AM} = \frac{\sin(180 - \alpha - \beta)}{\sin(\beta)} = \frac{\sin(\alpha + \beta)}{\sin(\beta)} = \frac{BP}{PM}$. Therefore, the two bisectors meet on $BC$.
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mahmudlusenan
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#14 • 1 Y
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let $\angle MAC=2\beta$ , $\angle BAM = 2\alpha$ so $\angle MNC= 2\alpha$ we can see that $\triangle ABM \sim\triangle NCM$
so $\angle AMB =\angle NMC =90-\alpha+\beta$
Then it is easy to see that $BM$ is external angle bisector of $\angle AMP$.
angle bisector of $\angle BAM$ is also external angle bisector of $\angle MAP$
Let $X$ is intersection of $BM$ and angle bisector of $\angle BAM$ so X is excenter of $\triangle AMP$
Then $PT$ is angle bisector of $\angle APM$.
This post has been edited 5 times. Last edited by mahmudlusenan, Nov 14, 2024, 5:28 PM
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onyqz
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#15
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solution
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