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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 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
Transform the sequence
steven_zhang123   0
17 minutes ago
Given a sequence of \( n \) real numbers \( a_1, a_2, \ldots, a_n \), we can select a real number \( \alpha \) and transform the sequence into \( |a_1 - \alpha|, |a_2 - \alpha|, \ldots, |a_n - \alpha| \). This transformation can be performed multiple times, with each chosen real number \( \alpha \) potentially being different
(i) Prove that it is possible to transform the sequence into all zeros after a finite number of such transformations.
(ii) To ensure that the above result can be achieved for any given initial sequence, what is the minimum number of transformations required?
0 replies
+1 w
steven_zhang123
17 minutes ago
0 replies
prove that any quadrilateral satisfying this inequality is a trapezoid
mqoi_KOLA   3
N 17 minutes ago by mqoi_KOLA
Prove that any Trapezoid/trapzium satisfies the given inequality$$
|r - p| < q + s < r + p
$$where $p,r$ are lengths of parallel sides and $q,s$ are other two sides.
3 replies
mqoi_KOLA
Yesterday at 3:48 AM
mqoi_KOLA
17 minutes ago
Number Theory Chain!
JetFire008   41
N 20 minutes ago by pooh123
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
41 replies
JetFire008
Apr 7, 2025
pooh123
20 minutes ago
Find the angle
Alfombraking   0
42 minutes ago
Inside a right triangle ABC at , point Q is located, which belongs to the bisector of angle C. On the extension of BQ, point P is located from which PM⊥CQ(M en CQ) is drawn, such that BP=2(MC). If AQ=BC, then the measure of angle BAQ is.
0 replies
Alfombraking
42 minutes ago
0 replies
IMO Problem 4
iandrei   105
N an hour ago by cj13609517288
Source: IMO ShortList 2003, geometry problem 1
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
105 replies
iandrei
Jul 14, 2003
cj13609517288
an hour ago
Inspired by old results
sqing   2
N an hour ago by sqing
Source: Own
Let $ a,b  $ be reals such that $  a^2+b^2=2$ . Prove that
$$ (a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2 $$$$ (a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)\leq 2 $$
2 replies
sqing
2 hours ago
sqing
an hour ago
NEPAL TST 2025 DAY 2
Tony_stark0094   5
N an hour ago by iStud
Consider an acute triangle $\Delta ABC$. Let $D$ and $E$ be the feet of the altitudes from $A$ to $BC$ and from $B$ to $AC$ respectively.

Define $D_1$ and $D_2$ as the reflections of $D$ across lines $AB$ and $AC$, respectively. Let $\Gamma$ be the circumcircle of $\Delta AD_1D_2$. Denote by $P$ the second intersection of line $D_1B$ with $\Gamma$, and by $Q$ the intersection of ray $EB$ with $\Gamma$.

If $O$ is the circumcenter of $\Delta ABC$, prove that $O$, $D$, and $Q$ are collinear if and only if quadrilateral $BCQP$ can be inscribed within a circle.

$\textbf{Proposed by Kritesh Dhakal, Nepal.}$
5 replies
Tony_stark0094
Yesterday at 8:40 AM
iStud
an hour ago
Inspired by KHOMNYO2
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 $ and $ a^2+b^2=\frac{5}{2}. $ Prove that $$ 2a + 2b + \frac{1}{a} + \frac{1}{b}  +\frac{ab}{\sqrt 2}\geq 5\sqrt 2$$$$ a +  b +\frac{2}{a} + \frac{2}{b}  + ab\geq \frac{5}{4} + \frac{13}{\sqrt 5} $$$$ a +  b +\frac{2}{a} + \frac{2}{b}  +  \frac{ab}{\sqrt 2}\geq \frac{5}{4\sqrt 2} + \frac{13}{\sqrt 5} $$
2 replies
sqing
Mar 28, 2025
sqing
2 hours ago
USAMO 2003 Problem 4
MithsApprentice   71
N 2 hours ago by LeYohan
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.
71 replies
MithsApprentice
Sep 27, 2005
LeYohan
2 hours ago
Inspired by giangtruong13
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d\geq 0 ,a-b+d=21 $ and $ a+3b+4c=101 $. Prove that
$$ 61\leq  a+b+2c+d\leq \frac{265}{3}$$$$- \frac{2121}{2}\leq   ab+bc-2cd+da\leq \frac{14045}{12}$$$$\frac{519506-7471\sqrt{7471}}{27}\leq   ab+bc-2cd+3da\leq 33620$$
6 replies
sqing
Apr 11, 2025
sqing
2 hours ago
Inspired by Ruji2018252
sqing   4
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ a^2+b^2+c^2-2a-4b-4c=7. $ Prove that
$$ -4\leq 2a+b+2c\leq 20$$$$5-4\sqrt 3\leq a+b+c\leq 5+4\sqrt 3$$$$ 11-4\sqrt {14}\leq a+2b+3c\leq 11+4\sqrt {14}$$
4 replies
sqing
Apr 10, 2025
sqing
2 hours ago
Inspired by Abelkonkurransen 2025
sqing   0
2 hours ago
Source: Own
Let $ a,b,c $ be real numbers such that $ \frac{a}{bc}+\frac{4b}{ca}+\frac{c}{ab}=24. $ Prove that
$$\frac{1}{a}+\frac{1}{2b}+\frac{1}{ c}\geq -6$$$$\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}\geq 1-\sqrt{73} $$$$\frac{1}{a}+\frac{1}{4b}+\frac{1}{ c}\geq \frac{3}{2}(1-\sqrt{33} )$$
0 replies
sqing
2 hours ago
0 replies
Problem 16
Nguyenhuyen AG   37
N 2 hours ago by flower417477
Let $a,b,c >0 $ such that $abc\ge1 $. Prove that
$\frac{1}{a^4+b^3+c^2}+\frac{1}{b^4+c^3+a^2}+\frac{1}{c^4+a^3+b^2}\le 1$
37 replies
Nguyenhuyen AG
May 3, 2010
flower417477
2 hours ago
Sets With a Given Property
oVlad   3
N 3 hours ago by flower417477
Source: Romania TST 2025 Day 1 P4
Determine the sets $S{}$ of positive integers satisfying the following two conditions:
[list=a]
[*]For any positive integers $a, b, c{}$, if $ab + bc + ca{}$ is in $S$, then so are $a + b + c{}$ and $abc$; and
[*]The set $S{}$ contains an integer $N \geqslant 160$ such that $N-2$ is not divisible by $4$.
[/list]
Bogdan Blaga, United Kingdom
3 replies
oVlad
Apr 9, 2025
flower417477
3 hours ago
P_1X bisects <Q_1P_1Q_2, start with concentric circles
parmenides51   6
N Oct 20, 2024 by Fermat_Fanatic108
Source: Canadian Junior Mathematical Olympiad - CJMO 2021 p1
Let $C_1$ and $C_2$ be two concentric circles with $C_1$ inside $C_2$. Let $P_1$ and $P_2$ be two points on $C_1$ that are not diametrically opposite. Extend the segment $P_1P_2$ past $P_2$ until it meets the circle $C_2$ in $Q_2$. The tangent to $C_2$ at $Q_2$ and the tangent to $C_1$ at $P_1$ meet in a point $X$. Draw from X the second tangent to $C_2$ which meets $C_2$ at the point $Q_1$. Show that $P_1X$ bisects angle $Q_1P_1Q_2$.
6 replies
parmenides51
May 29, 2021
Fermat_Fanatic108
Oct 20, 2024
P_1X bisects <Q_1P_1Q_2, start with concentric circles
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G H BBookmark kLocked kLocked NReply
Source: Canadian Junior Mathematical Olympiad - CJMO 2021 p1
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parmenides51
30630 posts
#1 • 3 Y
Y by centslordm, HWenslawski, fastandfuriuos
Let $C_1$ and $C_2$ be two concentric circles with $C_1$ inside $C_2$. Let $P_1$ and $P_2$ be two points on $C_1$ that are not diametrically opposite. Extend the segment $P_1P_2$ past $P_2$ until it meets the circle $C_2$ in $Q_2$. The tangent to $C_2$ at $Q_2$ and the tangent to $C_1$ at $P_1$ meet in a point $X$. Draw from X the second tangent to $C_2$ which meets $C_2$ at the point $Q_1$. Show that $P_1X$ bisects angle $Q_1P_1Q_2$.
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MrOreoJuice
594 posts
#2 • 3 Y
Y by centslordm, HWenslawski, Mango247
Canada has junior Olympiads too?
Anyways, let $O$ be the center of the concentric circles.
$$\angle OQ_1X = \angle OQ_2X = 90^\circ$$$$\angle OP_1X = \angle OQ_2X = 90^\circ$$Thus all the $5$ points $O, P_1 , Q_1 , X , Q_2$ lie on a circle with diameter $OX$.
$$\angle X  P_1 Q_1  = \angle XQ_2 Q_1 = \angle X Q_1 Q_2 = \angle X P_1 Q_2.$$
This post has been edited 1 time. Last edited by MrOreoJuice, May 29, 2021, 3:41 PM
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parmenides51
30630 posts
#3 • 1 Y
Y by centslordm
It started last year, this one is the second one, and it has common problems with Canadian MO. See here for the rest problems.
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Bluesoul
892 posts
#4 • 1 Y
Y by paintingredflagsgreen3761
Notice $X,Q_1,P_1,O,Q_2$ are concyclic then angle chase with equal arcs($\widehat{XQ_2}=\widehat{XQ_1}$)
This post has been edited 1 time. Last edited by Bluesoul, Mar 2, 2022, 12:52 AM
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Mogmog8
1080 posts
#5 • 1 Y
Y by centslordm
[asy]
size(7cm);
import geometry;

point O=(-3,1);
point P_1=(3,1);
point Q_2=(-8,7);
circle C_1=circle(O,length(segment(O,P_1)));
circle C_2=circle(O,length(segment(O,Q_2)));
point P_2=intersectionpoints(C_1,P_1--Q_2)[1];
circle omega=circle(O,Q_2,P_1);
point Q_1=intersectionpoints(C_2,omega)[0];
point X=intersectionpoints(bisector(Q_1,Q_2),omega)[1];

draw(C_1); 
draw(C_2); 
draw(omega,dotted); 
draw(P_1--Q_2); 
draw(Q_1--X--Q_2); 
draw(X--O); 
draw(O--P_1--Q_1); 
draw(O--Q_2);

dot(O^^P_1^^Q_2^^P_2^^Q_1^^X);

label("$O$",O,SW); 
label("$P_1$",P_1,SW); 
label("$Q_2$",Q_2,NW); 
label("$P_2$",P_2,N); 
label("$Q_1$",Q_1,SE); 
label("$X$",X,NE);
label("$\mathcal{C}_1$",(-7,-3),N);
label("$\mathcal{C}_2$",(-6.5,-6),N);
[/asy]
Let $O$ be the center of circles $\mathcal{C}_1$ and $\mathcal{C}_2.$ Since $\angle OQ_{\{1,2\}}X=\angle OP_1X=90,$ we know $OP_1Q_1XQ_2$ is cyclic. Hence, $$\angle Q_1P_1X=\tfrac{1}{2}\widehat{XQ_1}=\tfrac{1}{2}\widehat{XQ_2}=\angle XP_1Q_2.$$$\square$
This post has been edited 2 times. Last edited by Mogmog8, Mar 5, 2022, 6:13 PM
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parmenides51
30630 posts
#6
Y by
same as others
This post has been edited 4 times. Last edited by parmenides51, May 10, 2024, 7:31 PM
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Fermat_Fanatic108
48 posts
#7
Y by
[ See attachment ]

It is clear that,

$ \angle O Q_2 X = \angle O P_1 X = \angle O Q_1 X = 90^\circ      $

Thus , we find that $ X Q_1 P_1 O Q_2$ lie on a circle

Since, $ Q_1 P_1 Q_2 X $ is a cyclic quad $ \implies \angle X Q_1 Q_2 = \angle X P_1 Q_2 $
$\hspace{48mm}     $ $ \implies \angle Q_1 Q_2 X  = \angle Q_1 P_1 X$

Since, $ X Q_1 = X Q_2 \implies \angle X Q_1 Q_2 = \angle X Q_2 Q_1 $

This tells us, $ \angle X P_1 Q_2 = \angle X P_1 Q_1 $

$ \blacksquare $
Attachments:
This post has been edited 1 time. Last edited by Fermat_Fanatic108, Oct 20, 2024, 5:21 PM
Reason: .
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