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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.

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[*]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
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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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IMO Genre Predictions
ohiorizzler1434   30
N 2 minutes ago by Iwowowl253
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
30 replies
ohiorizzler1434
Saturday at 6:51 AM
Iwowowl253
2 minutes ago
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2015 solutions for quotient function!
raxu   49
N 8 minutes ago by blueprimes
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
49 replies
raxu
Jun 26, 2015
blueprimes
8 minutes ago
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something...
SunnyEvan   0
21 minutes ago
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
0 replies
SunnyEvan
21 minutes ago
0 replies
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Cosine of polynomial is polynomial of cosine
yofro   1
N 40 minutes ago by yofro
Source: 2025 HMIC #2
Find all polynomials $P$ with real coefficients for which there exists a polynomial $Q$ with real coefficients such that for all real $t$, $$\cos(P(t))=Q(\cos t).$$
1 reply
yofro
an hour ago
yofro
40 minutes ago
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Problem 1
randomusername   72
N an hour ago by blueprimes
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
72 replies
+1 w
randomusername
Jul 10, 2015
blueprimes
an hour ago
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Permutation with no two prefix sums dividing each other
Assassino9931   2
N 2 hours ago by Assassino9931
Source: Bulgaria Team Contest, March 2025, oVlad
Does there exist an infinite sequence of positive integers $a_1, a_2 \ldots$, such that every positive integer appears exactly once as a member of the sequence and $a_1 + a_2 + \cdots + a_i$ divides $a_1 + a_2 + \cdots + a_j$ if and only if $i=j$?
2 replies
Assassino9931
2 hours ago
Assassino9931
2 hours ago
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Sum of Good Indices
raxu   25
N 4 hours ago by cj13609517288
Source: TSTST 2015 Problem 1
Let $a_1, a_2, \dots, a_n$ be a sequence of real numbers, and let $m$ be a fixed positive integer less than $n$. We say an index $k$ with $1\le k\le n$ is good if there exists some $\ell$ with $1\le \ell \le m$ such that $a_k+a_{k+1}+...+a_{k+\ell-1}\ge0$, where the indices are taken modulo $n$. Let $T$ be the set of all good indices. Prove that $\sum\limits_{k \in T}a_k \ge 0$.

Proposed by Mark Sellke
25 replies
raxu
Jun 26, 2015
cj13609517288
4 hours ago
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f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)
dangerousliri   7
N 4 hours ago by jasperE3
Source: FEOO, Shortlist A4
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}$ such that for any positive real numbers $x, y$ and $z$,
$$f(x^2+y^2+z^2)=f(xy)+f(yz)+f(zx)$$
Proposed by ThE-dArK-lOrD, and Papon Tan Lapate, Thailand
7 replies
dangerousliri
May 31, 2020
jasperE3
4 hours ago
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foldina a rectangle paper 3 times
parmenides51   1
N 5 hours ago by TheBaiano
Source: 2023 May Olympiad L2 p4
Matías has a rectangular sheet of paper $ABCD$, with $AB<AD$.Initially, he folds the sheet along a straight line $AE$, where $E$ is a point on the side $DC$ , so that vertex $D$ is located on side $BC$, as shown in the figure. Then folds the sheet again along a straight line $AF$, where $F$ is a point on side $BC$, so that vertex $B$ lies on the line $AE$; and finally folds the sheet along the line $EF$. Matías observed that the vertices $B$ and $C$ were located on the same point of segment $AE$ after making the folds. Calculate the measure of the angle $\angle DAE$.
IMAGE
1 reply
parmenides51
Mar 24, 2024
TheBaiano
5 hours ago
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Divisibility NT
reni_wee   0
5 hours ago
Source: Japan 1996, ONTCP
Let $m,n$ be relatively prime positive integers. Calculate $gcd(5^m+7^m, 5^n+7^n).$
0 replies
reni_wee
5 hours ago
0 replies
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Modular arithmetic at mod n
electrovector   3
N 5 hours ago by Primeniyazidayi
Source: 2021 Turkey JBMO TST P6
Integers $a_1, a_2, \dots a_n$ are different at $\text{mod n}$. If $a_1, a_2-a_1, a_3-a_2, \dots a_n-a_{n-1}$ are also different at $\text{mod n}$, we call the ordered $n$-tuple $(a_1, a_2, \dots a_n)$ lucky. For which positive integers $n$, one can find a lucky $n$-tuple?
3 replies
electrovector
May 24, 2021
Primeniyazidayi
5 hours ago
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Sequences problem
BBNoDollar   3
N Yesterday at 6:45 PM by BBNoDollar
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
3 replies
BBNoDollar
Saturday at 5:53 PM
BBNoDollar
Yesterday at 6:45 PM
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Arbitrary point on BC and its relation with orthocenter
falantrng   33
N Yesterday at 6:44 PM by Thapakazi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
33 replies
falantrng
Apr 27, 2025
Thapakazi
Yesterday at 6:44 PM
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Inequality
lgx57   4
N Yesterday at 6:31 PM by GeoMorocco
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
4 replies
lgx57
Saturday at 3:14 PM
GeoMorocco
Yesterday at 6:31 PM
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Irrational equation
giangtruong13   5
N Apr 25, 2025 by pooh123
Solve the equation : $$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$
5 replies
giangtruong13
Apr 24, 2025
pooh123
Apr 25, 2025
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Irrational equation
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Reveal topic tags
equation
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giangtruong13
141 posts
giangtruong13
#1 h  Apr 24, 2025, 1:44 PM
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Solve the equation : $$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$
This post has been edited 1 time. Last edited by giangtruong13, Apr 24, 2025, 1:44 PM
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Tuvshuu
11 posts
Tuvshuu
#2 h  Apr 24, 2025, 2:20 PM
Y by
$x = 4$.
I think it's only solution
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Tuvshuu
11 posts
Tuvshuu
#3 h  Apr 24, 2025, 2:37 PM
Y by
Let $f(x) = (\sqrt{x} + 1)[2 - (x - 6) \sqrt{x - 3}]$.
We can easily check $f''(x) \leq 0$ then $f$ is concave function.
$x + 8$ is the $f$'s tangent on the $x = 4$ then only solution is $x = 4$ ($f'(4) = 1$ and $f(4) = 12$).
Attachments:
This post has been edited 7 times. Last edited by Tuvshuu, Apr 24, 2025, 2:43 PM
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navier3072
112 posts
navier3072
#4 h  Apr 24, 2025, 2:58 PM
Y by
How did you show $f''(x)\leq 0$? It seems like a very complicated function
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giangtruong13
141 posts
giangtruong13
#5 h  Apr 25, 2025, 8:19 AM
Y by
boom chalaka
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pooh123
38 posts
pooh123
#6 h  Apr 25, 2025, 1:31 PM
Y by
giangtruong13 wrote:
Solve the equation : $$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$

The domain of the equation is \( x \geq 3 \). We have:

\[ (\sqrt{x} + 1)\left[2 - (x-6)\sqrt{x-3}\right] = x + 8 \]
\[ \Leftrightarrow x + 8 + (\sqrt{x} + 1)\left[(x-6)\sqrt{x-3} - 2\right] = 0 \]
\[ \Leftrightarrow x + 8 - 4\sqrt{x} - 4 + (\sqrt{x} + 1)\left[(x-6)\sqrt{x-3} + 2\right] = 0 \]
\[ \Leftrightarrow (\sqrt{x} - 2)^2 + (\sqrt{x} + 1)\left[(x-3)\sqrt{x-3} - 3\sqrt{x-3} + 2\right] = 0 \]
\[ \Leftrightarrow (\sqrt{x} - 2)^2 + (\sqrt{x} + 1)(\sqrt{x-3}+2)(\sqrt{x-3}-1)^2 = 0 \]
Since each term on the left-hand side is non-negative, the left-hand side must be non-negative.
Therefore, \(\sqrt{x} - 2 = 0\) and \(\sqrt{x-3} - 1 = 0\), so \(x = 4\), which is in the domain.

Hence the only solution to the equation is \(x = 4\).
This post has been edited 2 times. Last edited by pooh123, Apr 25, 2025, 1:42 PM
Reason: typo
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