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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
Funcional equation problem
khongphaiwminh   0
5 minutes ago
Determine all functions $f \colon \mathbb R^+ \to \mathbb R^+$ that satisfy the equation
$$f(x+f(y))=f(x+y)+f(y)$$for any $x, y \in \mathbb R^+$. Note that $\mathbb R^+ \stackrel{\text{def}}{=} \{x \in \mathbb R \mid x > 0\}$.
0 replies
khongphaiwminh
5 minutes ago
0 replies
4 Variables Cyclic Ineq
nataliaonline75   0
5 minutes ago
Prove that for every $x,y,z,w$ non-negative real numbers, then we have:
$\frac{x-y}{xy+2y+1}+\frac{y-z}{yz+2z+1} + \frac{z-w}{zw+2w+1} + \frac{w-x}{wx+2x+1} \geq 0$
0 replies
nataliaonline75
5 minutes ago
0 replies
Number theory
MathsII-enjoy   2
N 24 minutes ago by SimplisticFormulas
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
2 replies
MathsII-enjoy
Yesterday at 3:22 PM
SimplisticFormulas
24 minutes ago
National diophantine equation
KAME06   2
N 26 minutes ago by damyan
Source: OMEC Ecuador National Olympiad Final Round 2024 N3 P5 day 2
Find all triples of non-negative integer numbers $(E, C, U)$ such that $EC \ge 1$ and:
$$2^{3^E}+3^{2^C}=593 \cdot 5^U$$
2 replies
KAME06
Feb 28, 2025
damyan
26 minutes ago
IMO ShortList 1998, number theory problem 1
orl   55
N 36 minutes ago by reni_wee
Source: IMO ShortList 1998, number theory problem 1
Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
55 replies
orl
Oct 22, 2004
reni_wee
36 minutes ago
IMO Genre Predictions
ohiorizzler1434   55
N 41 minutes ago by NoSignOfTheta
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
55 replies
ohiorizzler1434
May 3, 2025
NoSignOfTheta
41 minutes ago
Number of sets S
Jackson0423   2
N an hour ago by Jackson0423
Let \( S \) be a set consisting of non-negative integers such that:

1. \( 0 \in S \),
2. For any \( k \in S \), both \( k + 9 \in S \) and \( k + 10 \in S \).

Find the number of such sets \( S \).
2 replies
Jackson0423
an hour ago
Jackson0423
an hour ago
F.E....can you solve it?
Jackson0423   16
N an hour ago by jasperE3
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that
\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]for all real numbers \( x \) satisfying \( f(x) \neq 1 \).
16 replies
Jackson0423
Yesterday at 1:27 PM
jasperE3
an hour ago
Find all positive a,b
shobber   14
N an hour ago by reni_wee
Source: APMO 2002
Find all positive integers $a$ and $b$ such that
\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]
are both integers.
14 replies
shobber
Apr 8, 2006
reni_wee
an hour ago
Geo metry
TUAN2k8   2
N an hour ago by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
2 replies
TUAN2k8
Today at 10:33 AM
TUAN2k8
an hour ago
(not so) small set of residues generates all of F_p upon applying Q many times
62861   14
N an hour ago by john0512
Source: RMM 2019 Problem 6
Find all pairs of integers $(c, d)$, both greater than 1, such that the following holds:

For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\big(\tfrac{2c-1}{2c+1}\big)p$ integers, such that
\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]contains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).
14 replies
62861
Feb 24, 2019
john0512
an hour ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   13
N an hour ago by SimplisticFormulas
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
13 replies
parmenides51
Jul 25, 2018
SimplisticFormulas
an hour ago
Functional equation of nonzero reals
proglote   5
N an hour ago by TheHimMan
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
5 replies
proglote
Oct 24, 2013
TheHimMan
an hour ago
5-th powers is a no-go - JBMO Shortlist
WakeUp   8
N an hour ago by sansgankrsngupta
Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

Note
8 replies
1 viewing
WakeUp
Oct 30, 2010
sansgankrsngupta
an hour ago
MN bisects segment CH
parmenides51   6
N Jan 25, 2025 by raffigm
Source: JBMO Shortlist 2018 G1
Let $H$ be the orthocentre of an acute triangle $ABC$ with $BC > AC$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $CB$ intersects $\Gamma$ at the point $D$, which is on the arc $AB$ not containing $C$. The circle with centre $C$ and radius $CA$ intersects the segment $CD$ at the point $K$. The line parallel to $BD$ through $K$, intersects $AB$ at point $L$. If $M$ is the midpoint of $AB$ and $N$ is the foot of the perpendicular from $H$ to $CL$, prove that the line $MN$ bisects the segment $CH$.
6 replies
parmenides51
Jul 22, 2019
raffigm
Jan 25, 2025
MN bisects segment CH
G H J
G H BBookmark kLocked kLocked NReply
Source: JBMO Shortlist 2018 G1
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parmenides51
30651 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $H$ be the orthocentre of an acute triangle $ABC$ with $BC > AC$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $CB$ intersects $\Gamma$ at the point $D$, which is on the arc $AB$ not containing $C$. The circle with centre $C$ and radius $CA$ intersects the segment $CD$ at the point $K$. The line parallel to $BD$ through $K$, intersects $AB$ at point $L$. If $M$ is the midpoint of $AB$ and $N$ is the foot of the perpendicular from $H$ to $CL$, prove that the line $MN$ bisects the segment $CH$.
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Steve12345
619 posts
#2 • 2 Y
Y by Adventure10, Mango247
Related topics:
- Bosnia and Herzegovina Regional Olympiad 2018 : https://artofproblemsolving.com/community/c6h1709471p11016808
- European Mathematical Cup 2018 , Junior P3 : https://artofproblemsolving.com/community/c6h1758655p11490969
- Danube Mathematical Competition 2014 , Junior P3 : https://artofproblemsolving.com/community/c6h1703980p10958876
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microsoft_office_word
66 posts
#5 • 5 Y
Y by Euiseu, fastlikearabbit, Grumpah, augustin_p, Adventure10
Let $P$ is the midpoint of $CH$. We need to prove that $P-N-M$ are colliniar. We perform homotety from $H$ with ratio $2$. $P$ goes to $C$. $M$ goes to antipode of $C$ wrt $	\circ ABC$, and let $N$ goes to $Q$. We need now to prove that $Q$ lies on diameter from $C$. This can be easily proved with angle chasing and some similarities.
Proof with angle chasing and some similarities
[asy]
import graph; size(8cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -12.009090909090915, xmax = 10.172727272727274, ymin = -5.835454545454546, ymax = 7.346363636363635;  /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); 
 /* draw figures */
draw(circle((-2.2200116993272885,0.740298332845862), 5.706489632560346), linewidth(0.5) + wrwrwr); 
draw((-6.3,4.73)--(-7.22,-2.01), linewidth(0.5) + wrwrwr); 
draw((-7.22,-2.01)--(2.98,-1.61), linewidth(0.5) + wrwrwr); 
draw((2.98,-1.61)--(-6.3,4.73), linewidth(0.5) + wrwrwr); 
draw((-6.3,4.73)--(0.2460429784135627,-4.405826252986779), linewidth(0.5) + wrwrwr); 
draw((2.98,-1.61)--(0.2460429784135627,-4.405826252986779), linewidth(0.5) + wrwrwr); 
draw((-2.3379346965024017,-0.7995604283541823)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); 
draw((-6.3,4.73)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); 
draw((-6.3,4.73)--(-6.099976601345422,-0.370596665691723), linewidth(0.5) + wrwrwr); 
draw((-6.099976601345422,-0.370596665691723)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); 
draw((-6.099976601345422,-0.370596665691723)--(-4.37519129392871,0.3952858493872016), linewidth(0.5) + wrwrwr); 
draw((-6.199988300672711,2.1797016671541387)--(-2.12,-1.81), linewidth(0.5) + linetype("2 2") + wrwrwr); 
draw((-3.374106427962564,-1.859180644233826)--(-0.6831768080801006,0.8926445003478276), linewidth(0.5) + wrwrwr); 
draw((-7.22,-2.01)--(-0.6831768080801006,0.8926445003478276), linewidth(0.5) + wrwrwr); 
draw((-6.099976601345422,-0.370596665691723)--(-2.650405986511998,1.1611683644661264), linewidth(0.5) + wrwrwr); 
draw((-2.650405986511998,1.1611683644661264)--(-6.3,4.73), linewidth(0.5) + wrwrwr); 
 /* dots and labels */
dot((-6.3,4.73),dotstyle); 
label("$C$", (-6.736363636363641,4.819090909090908), NE * labelscalefactor); 
dot((-7.22,-2.01),dotstyle); 
label("$A$", (-7.918181818181823,-2.1809090909090916), NE * labelscalefactor); 
dot((2.98,-1.61),dotstyle); 
label("$B$", (3.1545454545454534,-1.8354545454545461), NE * labelscalefactor); 
dot((0.2460429784135627,-4.405826252986779),linewidth(4pt) + dotstyle); 
label("$D$", (0.0454545454545434,-5.180909090909091), NE * labelscalefactor); 
dot((-2.3379346965024017,-0.7995604283541823),linewidth(4pt) + dotstyle); 
label("$K$", (-2.0818181818181847,-0.8718181818181826), NE * labelscalefactor); 
dot((-3.374106427962564,-1.859180644233826),linewidth(4pt) + dotstyle); 
label("$L$", (-3.4090909090909123,-2.308181818181819), NE * labelscalefactor); 
dot((-6.099976601345422,-0.370596665691723),linewidth(4pt) + dotstyle); 
label("$H$", (-6.5,-0.3990909090909099), NE * labelscalefactor); 
dot((-4.37519129392871,0.3952858493872016),linewidth(4pt) + dotstyle); 
label("$N$", (-4.3,0.5463636363636355), NE * labelscalefactor); 
dot((-2.12,-1.81),linewidth(4pt) + dotstyle); 
label("$M$", (-2.0818181818181847,-2.1990909090909097), NE * labelscalefactor); 
dot((-6.199988300672711,2.1797016671541387),linewidth(4pt) + dotstyle); 
label("$P$", (-6.118181818181823,2.3281818181818172), NE * labelscalefactor); 
dot((-0.6831768080801006,0.8926445003478276),linewidth(4pt) + dotstyle); 
label("$W$", (-0.8272727272727296,1.0736363636363628), NE * labelscalefactor); 
dot((-2.650405986511998,1.1611683644661264),linewidth(4pt) + dotstyle); 
label("$Q$", (-2.572727272727276,1.31), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
This post has been edited 3 times. Last edited by microsoft_office_word, Feb 14, 2020, 11:01 PM
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Nymoldin
57 posts
#7
Y by
What was the morality of this act?
This post has been edited 2 times. Last edited by Nymoldin, Jan 8, 2022, 10:18 PM
Reason: Cmon,who cares this post?
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celilcelil
164 posts
#8
Y by
Let $O$ be circumcenter of $\triangle ABC$ and $T$ midpoint of $CH$. We know that $TM\parallel CO$. After writing $\angle CBA=\alpha$ and $\angle ABD=2\beta$ and doing some angle chasing we can find that $\angle HTN=\angle HCO$. It means that $TN\parallel CO$ $\implies$ $T-N-M$ are collinear.
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StefanSebez
53 posts
#9
Y by
$\angle LKD=\angle KDB=\angle CDB=\angle CAB=\angle CAL$
$\implies ALKC$ cyclic
let $\angle BAC=x$ and $\angle ABC=y$
$\angle ACD=\angle ACB-\angle DCB=180-x-y-(180-2x)=x-y$
$\angle CAK=\angle CKA=90-\frac{\angle ACD}{2}=90+\frac{y}{2}-\frac{x}{2}$
$\angle ACL=180-\angle LAC-\angle CLA=180-x-\angle CKA=180-x-90+\frac{x}{2}-\frac{y}{2}=90-\frac{x}{2}-\frac{y}{2}=\frac{\angle ACB}{2}$
let $A_1$ and $B_1$ be foot of perpendiculars from $A$ and $B$ to $BC$ and $AC$ respectively
let $E$ be midpoint of $A_1B_1$
let $O$ be midpoint of $HC$
$\angle HB_1C=90=\angle HNC=\angle HA_1C$
$\implies HNA_1CB_1$ cyclic with center $O$
$\angle NB_1A_1=\angle NCA_1=\angle NCB_1=\angle NA_1B_1$
$\implies NB_1=NA_1$
$M, E, O$ collinear because they lie on Gauss line of $B_1HA_1C$
$\angle AB_1B=90=\angle AA_1B$
$\implies AB_1A_1B$ cyclic with center $M$
$A_1B_1$ is radical axis of circles $(B_1HNA_1C)$ and $(AB_1A_1B)$
so $OM$ is perpendicular to $A_1B_1$, but because it passes through $E$, midpoint of $A_1B_1$, it is perpendicular bisector of $A_1B_1$
but $N$ passes through perpendicular bisector of $A_1B_1$ because $NB_1=NA_1$
$\implies M, N, O$ collinear
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raffigm
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let $X,Y,$ and $Z$ be the foot of perpendiculars from $C,A,$ and $B$ to $AB, BC,$ and $AC$ respectively. Also, let $P$ and $E$ be the midpoint of $CH$ and $BC$ respectively, which results in $P$ and $E$ are the circumcenters of $(CHN)$ and $(CXB)$ respectively, along with $PE||BZ$ and $EM||AC$

Claim 1: $ALKC$ is cyclic
since $LK||BD$, we have that $\angle CAB = \angle CDB = \angle LKD = 180 -\angle LKC$, as desired. Furthermore, since $AC=CK$, then $\angle ALC = \angle CKA = \angle CAK = \angle CLK$, making $\angle CLA = \frac{\angle KLA}{2}$

Claim 2: $PXME$ is cyclic
From $\angle PEM = 180 - \angle MEB - \angle PEC = 180 - \angle ACB -\angle ZBC = 90$, we have that $\angle PXM = \angle PEM = 90$, therefore $PXME$ is cyclic

For finishing up, just notice that $\angle XPN = 2\angle HCN = 2(90 - \angle HLC) = 180 - \angle ALK = \angle KLM = \angle ABD$, and $\angle XPM =\angle XEM = \angle XEB - \angle MEB = 2\angle XCB - \angle ACB = 2(90 -\angle XBC) - (180 - \angle CAB - \angle CBA) = 180 -2(\angle CBD - \angle ABD) - (180 - \angle CDB - (\angle (CBD - \angle (ABD)) = 180 -2\angle CBD +2\angle ABD - 180 + \angle CBD + \angle CBD -\angle ABD = \angle ABD$, which reveals that $\angle XPM = \angle XPN$, so $M,N,P$ are colinear, thus done :-D
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