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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Inspired by 2025 Beijing
sqing   6
N 10 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
10 minutes ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 25 minutes ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
25 minutes ago
Non-linear Recursive Sequence
amogususususus   2
N 29 minutes ago by wh0nix
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
2 replies
amogususususus
Jan 24, 2025
wh0nix
29 minutes ago
Nice "if and only if" function problem
ICE_CNME_4   14
N 33 minutes ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
14 replies
ICE_CNME_4
Friday at 7:23 PM
wh0nix
33 minutes ago
2-var inequality
sqing   1
N 42 minutes ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
1 reply
sqing
an hour ago
sqing
42 minutes ago
Balkan Mathematical Olympiad
ABCD1728   1
N an hour ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
Yesterday at 11:27 PM
ABCD1728
an hour ago
Functional equation
shactal   0
an hour ago
Let $f:\mathbb R\to \mathbb R$ a function satifying $$f(x+2xy) = f(x) + 2f(xy)$$for all $x,y\in \mathbb R$.
If $f(1991)=a$, then what is $f(1992)$, the answer is in terms of $a$.
0 replies
shactal
an hour ago
0 replies
area of quadrilateral
AlanLG   1
N an hour ago by Altronrren
Source: 3rd National Women´s Contest of Mexican Mathematics Olympiad 2024 , level 1+2 p5
Consider the acute-angled triangle \(ABC\). The segment \(BC\) measures 40 units. Let \(H\) be the orthocenter of triangle \(ABC\) and \(O\) its circumcenter. Let \(D\) be the foot of the altitude from \(A\) and \(E\) the foot of the altitude from \(B\). Additionally, point \(D\) divides the segment \(BC\) such that \(\frac{BD}{DC} = \frac{3}{5}\). If the perpendicular bisector of segment \(AC\) passes through point \(D\), calculate the area of quadrilateral \(DHEO\).
1 reply
AlanLG
Jun 14, 2024
Altronrren
an hour ago
Inspired by 2025 SXTB
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b  $ be real number such that $ a^2+b^2=\frac12. $ Prove that
$$-\frac{\sqrt{9+6\sqrt 3}}{2}\leq(a+1)^2- (b-1)^2\leq\frac{\sqrt{9+6\sqrt 3}}{2}$$Let $ x $ be real number . Prove that
$$-\frac{2\sqrt 2}{3}\leq \frac{x}{x^2+1}+ \frac{ 2x}{x^2+4} \leq\frac{2\sqrt 2}{3}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
IMO Shortlist 2014 G2
hajimbrak   14
N 2 hours ago by ezpotd
Let $ABC$ be a triangle. The points $K, L,$ and $M$ lie on the segments $BC, CA,$ and $AB,$ respectively, such that the lines $AK, BL,$ and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM, BMK,$ and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$.

Proposed by Estonia
14 replies
hajimbrak
Jul 11, 2015
ezpotd
2 hours ago
Divisiblity...
TUAN2k8   0
2 hours ago
Source: Own
Let $m$ and $n$ be two positive integer numbers such that $m \le n$.Prove that $\binom{n}{m}$ divides $lcm(1,2,...,n)$
0 replies
TUAN2k8
2 hours ago
0 replies
interesting diophantiic fe in natural numbers
skellyrah   4
N 2 hours ago by aidan0626
Find all functions \( f : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \),
\[
mn + f(n!) = f(f(n))! + n \cdot \gcd(f(m), m!).
\]
4 replies
skellyrah
Yesterday at 8:01 AM
aidan0626
2 hours ago
IMO 2010 Problem 4
mavropnevma   128
N 2 hours ago by ezpotd
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Proposed by Marcin E. Kuczma, Poland
128 replies
mavropnevma
Jul 8, 2010
ezpotd
2 hours ago
Simple Geometry
AbdulWaheed   5
N 2 hours ago by Adywastaken
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 $ABC$.
5 replies
AbdulWaheed
May 23, 2025
Adywastaken
2 hours ago
Similar triangles formed by angular condition
Mahdi_Mashayekhi   5
N Apr 21, 2025 by sami1618
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
sami1618
Apr 21, 2025
Similar triangles formed by angular condition
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran 2025 second round P3
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Mahdi_Mashayekhi
696 posts
#1 • 2 Y
Y by Rounak_iitr, Parsia--
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
Z K Y
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gghx
1072 posts
#2
Y by
Let $S=BP\cap CI$ and $T=BI\cap CP$. Note that $$\angle BPC=\angle ABP+\angle ACP=\angle ACI+\angle ABI=\angle BIC,$$hence $BPIC$ is cyclic.

We now prove that $S$ is the circumcenter of $\triangle B'IX$. This is true because $$\angle SB'I=\angle BAI + \angle ABP=\frac{1}{2}(\angle A + \angle C)=\angle IAC+\angle ACI = \angle SIB',$$hence $SI=SB'$. Furthermore, $$\angle ISB'=180^\circ-\angle A - \angle C=\angle B=2\angle ABI=2\angle B'XI,$$hence $S$ is the circumcenter of $\triangle B'IX$ as desired.

Now, $\angle SXI=\angle SIX=\angle SIT=\angle SPT$, hence $SXTP$ is cyclic. Similarly, $SYTP$ is cyclic, so $SXTPY$ is cyclic.

We are now ready to finish the question. We have $$\angle YPX=\angle XSI=180^\circ-2\angle SIX=\angle A$$and $$\angle YXP=180^\circ-\angle PSI=\angle ABP+\angle ACI=\angle C,$$so triangles $PXY$ and $ABC$ are similar.
Z K Y
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ItzsleepyXD
149 posts
#3
Y by
quite similar (or same IDK) to @above

since $\triangle ABB' \sim \triangle ACI$ and $\triangle ACC' \sim \triangle ABI$
implies that $\triangle BB'I \sim \triangle CIC'$

Let $E = BB' \cap IY , F= CC' \cap IX$ .
So $EB'=EI , FC'=FI$
Known that $\angle B'EI = \angle ABC = 2 \angle B'XI$ implies that $E =$ center of $(B'IX)$.
so $ \angle EXI = \angle EIX = \angle EPF$ so $E,X,F,P$ concyclic.
same as $E,Y,F,P$ concyclic .
so $E,Y,X,F,P$ concyclic.
thus $\angle PXY = \angle PEY = \angle ABC$ and $ \angle PYX = 180^{\circ} - \angle PFX = \angle ACB$
Conclude that $\triangle PXY \sim \triangle ABC$ . done $\square$
Z K Y
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mathuz
1525 posts
#4
Y by
Consider the intersection $C'Y\cap B'X = O(.)$. Then $O$ is the circumcenter of $PB'C'$, and it suffices to show that $O$ lies on the circumcircle of $PXY$.
Z K Y
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bin_sherlo
733 posts
#5
Y by
Let $BP\cap CI=U,PC\cap BI=V,AI\cap BC=D$. Note that $\measuredangle CPB=90+\frac{\measuredangle A}{2}=\measuredangle CIB$ hence $B,I,C,P$ are concyclic.
By Menelaus at $B'UBDCI$ we get $\frac{B'B}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}$. Hence $BX=BB'.\frac{\sin \frac{\measuredangle C}{2}}{\sin \frac{\measuredangle B}{2}}=\frac{BB'.IB}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}.IB$ Also $\frac{BV}{BP}=\frac{\cos \frac{\measuredangle A}{2}}{\sin \measuredangle C}$ thus,
\[\frac{BX.BV}{BP}=\frac{BI}{\sin \measuredangle B}.\cos \frac{\measuredangle A}{2}=BU\]which implies $X\in (PUV)$. Similarily $Y\in (PUV)$. Hence $\measuredangle XYP=\measuredangle XVC=\measuredangle C$ and $\measuredangle PXY=\measuredangle PUY=\measuredangle B$ as desired.$\blacksquare$
Z K Y
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sami1618
913 posts
#6
Y by
Here is a different approach rephrasing the problem in terms of reference triangle $PB'C'$. When I first drew the diagram the points $P$, $X$, and $Y$ were all very close together so this solution was motivated by drawing a diagram consisting of the points other than $A$, $B$, and $C$. :)
It is not hard to show that $\angle B'PC'=90^{\circ}-\tfrac{1}{2}\angle A$, $\angle PB'C'=90^{\circ}-\tfrac{1}{2}\angle B$, and $\angle PC'B'=90^{\circ}-\tfrac{1}{2}\angle C$. Since $\angle BPC=\angle BIC$, it must be that $I$ lies on the interior of segment $B'C'$. Because of this, it is also not hard to see that $X$, $Y$, and $P$ all lie on the same side of segment $AI$.
[asy]

import geometry;

size(10cm);
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair Ep=B+C-D;
pair J=I+Ep-D;
pair P=isogonalconjugate(triangle(A,B,C),J);
pair Bp=intersectionpoint(line(B,P),line(A,I));
pair Cp=intersectionpoint(line(C,P),line(A,I));
pair X=intersectionpoint(line(Bp,B+Bp-A), line(B,I));
pair Y=intersectionpoint(line(Cp,C+Cp-A), line(C,I));
pair O=intersectionpoint(line(Cp,Y), line(Bp,X));
point[] Ap=intersectionpoints(circle(Bp,O,Cp),line(O,I));
pair Ap=Ap[1];
pair Op=circumcenter(Ap,Bp,Cp);
pair M_a=2*Op-O;
point[] N_c=intersectionpoints(circle(Bp,O,Cp),line(P,Cp));
pair N_c=N_c[0];
point[] N_b=intersectionpoints(circle(Bp,O,Cp),line(P,Bp));
pair N_b=N_b[0];


draw(A--B--C--cycle, black);
draw(B--Bp,black); draw(C--P,black); draw(A--Cp,black); draw(B--I,black); 
draw(C--Y,black); draw(Bp--X); draw(Cp--Y);  draw(P--Bp--Cp--cycle, black);


dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(50));

dot("$P$", P, dir(140));
dot("$B'$", Bp, dir(40));
dot("$C'$", Cp, dir(290));
dot("$X$", X, dir(310));
dot("$Y$", Y, dir(110));
[/asy]
Now notice that $\triangle B'IX\sim\triangle AIB$ and $\triangle C'IY\sim \triangle AIC$. We will now focus on the acute reference triangle $PB'C'$. Let $O$ be the circumcenter of $PB'C'$. Let $X'$ be the second intersection of line $B'O$ with the circumcircle of triangle $POC'$ and let $Y'$ be the second intersection of line $C'O$ with the circumcircle of triangle $POB'$. It is easy to show that $\triangle PX'C'\sim$ $\triangle PB'Y'\sim$ $\triangle ABC$. Now notice that $\triangle B'C'X'\sim$ $\triangle AIB\sim$ $\triangle B'IX$. Thus $X$ is the point on segment $B'X'$ with $IX\parallel C'X'$. Similarly, $Y$ is the point along segment $C'Y'$ with $IY\parallel B'Y'$.
[asy]
import geometry;

size(10cm);

pair P=dir(100);
pair Bp=dir(270+55);
pair Cp=dir(270-55);
pair O=(0,0);
pair I=intersectionpoint(line(O,P),line(Bp,Cp));
pair Xp[]=intersectionpoints(line(Bp,O),circle(P,O,Cp));
pair Xp=Xp[1];
pair Yp[]=intersectionpoints(line(Cp,O),circle(P,O,Bp));
pair Yp=Yp[1];
pair X=intersectionpoint(line(I,I+Xp-Cp), line(Bp,Xp));
pair Y=intersectionpoint(line(I,I+Yp-Bp), line(Cp,Yp));

fill(P--Xp--Cp--cycle,palered+white);
fill(P--Yp--Bp--cycle,palered+white);
fill(P--X--Y--cycle, palered);

draw(P--X--Y--cycle);
draw(P--Bp--Cp--cycle);
draw(circle(P,O,Cp));
draw(circle(P,O,Bp));
draw(Bp--Xp); draw(Cp--Yp);
draw(X--I--Y);
draw(Cp--Xp--P--Yp--Bp);

dot("P",P,2*dir(P));
dot("B'",Bp,dir(270));
dot("C'",Cp,dir(270));

dot("O",O,dir(270));
dot("I",I,dir(270));
dot("X'",Xp,dir(150));
dot("Y'",Yp,dir(30));
dot("X",X,dir(220));
dot("Y",Y,dir(-40));
[/asy]
Then we have that $$\frac{X'X}{XB'}=\frac{C'I}{IB'}=\frac{C'Y}{YY'}.$$Thus triangle $PXY$ is a linear combination of triangle $PX'C'$ and $PB'Y'$. Since these triangles are both similar to $ABC$, it is a well-known result that $PXY$ must be similar to $ABC$ as well.
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