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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.

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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:
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0 replies
jlacosta
Mar 2, 2025
0 replies
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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
J
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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
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Maximizing score of permutations
navi_09220114   4
N 6 minutes ago by Manteca
Source: Malaysian IMO TST 2023 P2
Let $a_1, a_2, \cdots, a_n$ be a sequence of real numbers with $a_1+a_2+\cdots+a_n=0$. Define the score $S(\sigma)$ of a permutation $\sigma=(b_1, \cdots b_n)$ of $(a_1, \cdots a_n)$ to be the minima of the sum $$(x_1-b_1)^2+\cdots+(x_n-b_n)^2$$over all real numbers $x_1\le \cdots \le x_n$.

Prove that $S(\sigma)$ attains the maxima over all permutations $\sigma$, if and only if for all $1\le k\le n$, $$b_1+b_2+\cdots+b_k\ge 0.$$
Proposed by Anzo Teh Zhao Yang
4 replies
navi_09220114
Apr 29, 2023
Manteca
6 minutes ago
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not all sufficiently large integers are clean
ABCDE   26
N 24 minutes ago by mathfun07
Source: 2015 IMO Shortlist C6, Original 2015 IMO #6
Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.
26 replies
ABCDE
Jul 7, 2016
mathfun07
24 minutes ago
J
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Nice problem
hanzo.ei   5
N 40 minutes ago by RandomMathGuy500
Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that
\[ f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1, \quad \forall x, y \in \mathbb{R}. \]
5 replies
+1 w
hanzo.ei
Today at 4:31 PM
RandomMathGuy500
40 minutes ago
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Maximizing
steven_zhang123   1
N 43 minutes ago by RagvaloD
Source: China TST 2001 Quiz 5 P2
Find the largest positive real number \( c \) such that for any positive integer \( n \), satisfies \(\{ \sqrt{7n} \} \geq \frac{c}{\sqrt{7n}}\).
1 reply
steven_zhang123
Today at 12:56 AM
RagvaloD
43 minutes ago
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number theory
karimeow   1
N an hour ago by RagvaloD
Prove that there exist infinitely many positive integers m such that the equation (xz+1)(yz+1) = mz^3 + 1 has infinitely many positive integer solutions.
1 reply
karimeow
Today at 8:14 AM
RagvaloD
an hour ago
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Poland 2017 P1
j___d   18
N an hour ago by Avron
Points $P$ and $Q$ lie respectively on sides $AB$ and $AC$ of a triangle $ABC$ and $BP=CQ$. Segments $BQ$ and $CP$ cross at $R$. Circumscribed circles of triangles $BPR$ and $CQR$ cross again at point $S$ different from $R$. Prove that point $S$ lies on the bisector of angle $BAC$.
18 replies
j___d
Apr 4, 2017
Avron
an hour ago
J
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Nice problem
hanzo.ei   2
N an hour ago by socrates
Given two positive integers \( m, n \) satisfying \( m > n \) and their sum is an even number, consider the quadratic polynomial:

\[ P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1). \]
Prove that all roots of \( P(x) \) are positive integers but are not perfect squares.
2 replies
hanzo.ei
Today at 2:58 PM
socrates
an hour ago
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Kvant 898 NT
Anto0110   4
N 2 hours ago by Hertz
Source: Kvant 898
Find all odd integers \(0 < a < b < c < d\) such that
\[ ad = bc, \quad a + d = 2^k, \quad b + c = 2^m \]for some positive integers \(k\) and \(m\).
4 replies
Anto0110
Jul 27, 2024
Hertz
2 hours ago
J
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prove that a chord is tangent to the incircle
ihategeo_1969   1
N 2 hours ago by ihategeo_1969
Source: SORY 2019 P6
Let $ABC$ be a triangle with incenter $I$ and intouch triangle $DEF$. Let $P$ be the foot of the perpendicular from $D$ onto $EF$. Assume that $BP$, $CP$ intersect the sides $AC$, $AB$ in $Y,Z$ respectively. Finally, let the rays $IP$, $YZ$ meet the circumcircle of $\triangle ABC$ in $R$, $X$ respectively. Prove that the tangent from $X$ to the incircle and the line $RD$ meet on the circumcircle of $\triangle ABC$.

Proposed by Aditya Khurmi
1 reply
ihategeo_1969
2 hours ago
ihategeo_1969
2 hours ago
J
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PQ parallel to BC
keyree10   16
N 2 hours ago by Mathworld314
Source: INMO 2010 Problem 1
Let $ ABC$ be a triangle with circum-circle $ \Gamma$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \Gamma$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.
16 replies
keyree10
Jan 18, 2010
Mathworld314
2 hours ago
J
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2x+1 is a perfect square but the following x+1 integers are not.
Sumgato   9
N 2 hours ago by lksb
Source: Spain Mathematical Olympiad 2018 P1
Find all positive integers $x$ such that $2x+1$ is a perfect square but none of the integers $2x+2, 2x+3, \ldots, 3x+2$ are perfect squares.
9 replies
Sumgato
Mar 17, 2018
lksb
2 hours ago
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ortho conf DEF, radius MD, intersect ME,MF, collinear H,K,L
star-1ord   1
N 2 hours ago by Davut1102
Source: Estonia Final Round 2025 12-3
Let $ABC$ be an acute-angled triangle with $|AB|<|AC|$. The altitudes $AD,BE$ and $CF$ intersect at $H$. Let $M$ be the midpoint of $BC$. Point $K$ is chosen on the extension of $EM$ beyond $M$ and point $L$ is chosen on the segment $FM$ such that $|MK|=|ML|=|MD|$. Prove that points $K, L$ and $H$ are collinear.

a little harder version
1 reply
star-1ord
5 hours ago
Davut1102
2 hours ago
J
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Right angles on incircle
DynamoBlaze   38
N 2 hours ago by ehuseyinyigit
Source: RMO 2018 P6
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that
$\text{(i)}$ $B,C,Y,X$ are concyclic.
$\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.
38 replies
DynamoBlaze
Oct 7, 2018
ehuseyinyigit
2 hours ago
J
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inequality
ehuseyinyigit   3
N 3 hours ago by ehuseyinyigit
Source: Nice
For all positive real numbers $a,b$ and $c$, prove
$$\sum_{cyc}{\dfrac{1}{b\left(a^4+a^3c+b^2c^2\right)}}\geq \dfrac{27}{(a+b+c)(a^2+b^2+c^2)^2}$$
3 replies
ehuseyinyigit
Feb 3, 2025
ehuseyinyigit
3 hours ago
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J
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With $a,b,c\ge 0$
khanhsy   12
N Sep 24, 2017 by pym123
If $a,b,c$ are nonnegative real numbers, then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ab}{(a+b)^2}\ge 2$$
12 replies
khanhsy
Sep 24, 2017
pym123
Sep 24, 2017
J
With $a,b,c\ge 0$
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Reveal topic tags
algebra
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khanhsy
126 posts
khanhsy
#1 h  Sep 24, 2017, 12:52 PM • 2 Y
Y by Adventure10, Mango247
If $a,b,c$ are nonnegative real numbers, then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ab}{(a+b)^2}\ge 2$$
This post has been edited 1 time. Last edited by khanhsy, Sep 24, 2017, 2:02 PM
Reason: edit
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dipdas
955 posts
dipdas
#2 h  Sep 24, 2017, 1:03 PM • 2 Y
Y by Adventure10, Mango247
$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$.......(i)
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dipdas
955 posts
dipdas
#3 h  Sep 24, 2017, 1:15 PM • 2 Y
Y by Adventure10, Mango247
khanhsy wrote:
If $a,b,c$ are nonnegative real numbers, then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{27abc}{(a+b+c)^3}\ge 2$$

is the inequality correct?
take a=1,b=c=0
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pym123
10 posts
pym123
#4 h  Sep 24, 2017, 1:17 PM • 2 Y
Y by Adventure10, Mango247
How could you divide a number with 0?
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dipdas
955 posts
dipdas
#5 h  Sep 24, 2017, 1:18 PM • 2 Y
Y by Adventure10, Mango247
pym123 wrote:
How could you divide a number with 0?

so your inequality is not correct since the numbers are non negative
This post has been edited 1 time. Last edited by dipdas, Sep 24, 2017, 1:19 PM
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knm2608
468 posts
knm2608
#6 h  Sep 24, 2017, 1:21 PM • 1 Y
Y by Adventure10
I guess the OP means that no two of $a,b,c$ are zero.

By the way
dipdas wrote:
$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$.......(i)
this approach doesn't help.
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dipdas
955 posts
dipdas
#7 h  Sep 24, 2017, 1:22 PM • 1 Y
Y by Adventure10
knm2608 wrote:
By the way
dipdas wrote:
$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$.......(i)
this approach doesn't help.
i know
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pym123
10 posts
pym123
#9 h  Sep 24, 2017, 1:26 PM • 2 Y
Y by Adventure10, Mango247
The inequality is equivalent to
$\sum a^5 + 3\sum a^4b +3\sum ab^4 + 9\sum a^2b^2c \ge 4\sum a^3b^2 + 4\sum a^2b^3 + 12\sum a^3bc$
This post has been edited 1 time. Last edited by pym123, Sep 24, 2017, 1:29 PM
Reason: typo
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knm2608
468 posts
knm2608
#10 h  Sep 24, 2017, 1:30 PM • 2 Y
Y by Adventure10, Mango247
By $uvw$ we only have to check two cases.
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khanhsy
126 posts
khanhsy
#11 h  Sep 24, 2017, 1:50 PM • 2 Y
Y by Adventure10, Mango247
Sorry. That is no correct? :)
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dipdas
955 posts
dipdas
#12 h  Sep 24, 2017, 2:06 PM • 3 Y
Y by khanhsy, Adventure10, Mango247
khanhsy wrote:
Sorry. That is no correct? :)

i already told you :mad:
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khanhsy
126 posts
khanhsy
#13 h  Sep 24, 2017, 2:44 PM • 2 Y
Y by Adventure10, Mango247
dipdas wrote:
khanhsy wrote:
Sorry. That is no correct? :)

i already told you :mad:

Thank you, Dipdas! I have fixed my post.
This post has been edited 1 time. Last edited by khanhsy, Sep 24, 2017, 2:45 PM
Reason: .
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pym123
10 posts
pym123
#14 h  Sep 24, 2017, 3:35 PM • 2 Y
Y by khanhsy, Adventure10
This inequality is equivalent to
$a^4 + 2a^2b^2 + b^4 + a^2c^2 + 2abc^2 +b^2c^2 \ge 2a^3c + 2abc^2 + 2b^3c + 2ab^2c$
And it is equivalent to $(a^2 + b^2)^2 + c^2(a + b)^2 \ge 2c(a^2 + b^2)(a + b)$
Which is correct by AM - GM.
This post has been edited 1 time. Last edited by pym123, Sep 24, 2017, 3:35 PM
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