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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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0 replies
jlacosta
Apr 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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Combinatoric
spiderman0   3
N 22 minutes ago by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
3 replies
1 viewing
spiderman0
Apr 22, 2025
MathBot101101
22 minutes ago
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set of sum of three or fewer powers of 2, 2024 TMC AIME Mock #13
parmenides51   4
N 24 minutes ago by maromex
Let $S$ denote the set of all positive integers that can be expressed as a sum of three or fewer powers of $2$. Let $N$ be the smallest positive integer that cannot be expressed in the form $a-b$, where $a, b \in S$. Find the remainder when $N$ is divided by $1000$.
4 replies
parmenides51
Yesterday at 8:16 PM
maromex
24 minutes ago
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easy functional
B1t   7
N 35 minutes ago by MathLuis
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[ f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)). \]
7 replies
B1t
Yesterday at 6:45 AM
MathLuis
35 minutes ago
J
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Dou Fang Geometry in Taiwan TST
Li4   6
N 36 minutes ago by CrazyInMath
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
6 replies
+1 w
Li4
Yesterday at 5:03 AM
CrazyInMath
36 minutes ago
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Sum of Series
P162008   0
41 minutes ago
Let $S = \sum_{0 \le i < j \le n} \sum P(i)P(j)$

Now, Consider $a_{ij} = P(i)P(j)$ to be an element of a $(n + 1) \times (n + 1)$ sqaure matrix.

Also, we know that in the principal diagonal $i = j$ and $a_{ij} = a_{ji}$ since the matrix is symmetrical in nature.

Due to the symmetrical nature of the matrix, sum of all the elements of the lower triangle will be equal to sum of all the elements of the upper triangle.

Then, $\sum_{i=0}^{n} \sum_{j=0}^{n} P(i)P(j) = \sum_{i=0}^{n} P(i)^2 + 2\left(\sum_{0 \le i < j \le n} \sum P(i)P(j)\right)$

$\therefore \boxed{\sum_{0 \le i < j \le n} \sum P(i)P(j) = \frac{1}{2}\left[\sum_{i=0}^{n} \sum_{j=0}^{n} P(i)P(j) - \sum_{i=0}^{n} P(i)^2\right]}$
0 replies
P162008
41 minutes ago
0 replies
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inquality
Doanh   1
N 43 minutes ago by sqing
Given that \( x, y, z \) are positive real numbers satisfying the condition \( xy + yz + zx = 1 \),
find the maximum value of the expression:
\[ P = \frac{1}{1+x^2} + \frac{1}{1+y^2} + \frac{z}{1+z^2} \]
1 reply
Doanh
2 hours ago
sqing
43 minutes ago
J
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Weird ninja points collinearity
americancheeseburger4281   1
N an hour ago by MathLuis
Source: Someone I know
For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle $ABC$, define:
[list]
[*]$O$ as its circumcentre, $H$ as its orthocentre and $N_9$ as its nine-point centre.
[*]$M_a$, $M_b$ and $M_c$ to be the midpoint of the smaller arcs.
[*]$G$ as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
[*]$S$ as the ninja point of $\Delta M_aM_bM_c$
[*]$K$ as the ninja point of the contact triangle
[/list]
Prove that:
$(a)$ Points $K$, $N_9$ and $I$ are collinear, that is $K$ is the Feuerbach point.
$(b)$ Points $H$, $G$ and $S$ are collinear
1 reply
americancheeseburger4281
6 hours ago
MathLuis
an hour ago
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To Mixtilinear or not to mixtilinear
ihategeo_1969   2
N an hour ago by Ilikeminecraft
Source: Evan Chen's Stream Twitch Solves ISL Episode 6
Let $ABC$ be a triangle and let $T$ be the contact point of the $A$-mixtilinear incircle with the circumcircle, and let $T'$ be the reflection of $T$ over $BC$. Prove that the nine-point circle of $T'BC$ is tangent to the incircle.
2 replies
ihategeo_1969
Jan 8, 2025
Ilikeminecraft
an hour ago
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Continued fraction
tapir1729   10
N an hour ago by EpicBird08
Source: TSTST 2024, problem 2
Let $p$ be an odd prime number. Suppose $P$ and $Q$ are polynomials with integer coefficients such that $P(0)=Q(0)=1$, there is no nonconstant polynomial dividing both $P$ and $Q$, and
\[ 1 + \cfrac{x}{1 + \cfrac{2x}{1 + \cfrac{\ddots}{1 + (p-1)x}}}=\frac{P(x)}{Q(x)}. \]Show that all coefficients of $P$ except for the constant coefficient are divisible by $p$, and all coefficients of $Q$ are not divisible by $p$.

Andrew Gu
10 replies
tapir1729
Jun 24, 2024
EpicBird08
an hour ago
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Inspired by old results
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b\geq 0 $ and $ a+b=2. $ Prove that
$$a^3b^2( \frac 32a+ b ) \leq 80\sqrt{5}-176$$$$a^3b^2( \frac 38a+ b ) \leq \frac{64}{125}(5- \sqrt{5})$$$$a^3b^2( \frac 34a+ b ) \leq 1088-768\sqrt{2}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
J
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2025 Caucasus MO Juniors P7
BR1F1SZ   3
N 2 hours ago by Bergo1305
Source: Caucasus MO
It is known that from segments of lengths $a$, $b$ and $c$, a triangle can be formed. Could it happen that from segments of lengths $$\sqrt{a^2 + \frac{2}{3} bc},\quad \sqrt{b^2 + \frac{2}{3} ca}\quad \text{and} \quad \sqrt{c^2 + \frac{2}{3} ab},$$a right-angled triangle can be formed?
3 replies
BR1F1SZ
Mar 26, 2025
Bergo1305
2 hours ago
J
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Inequalities
Scientist10   5
N 2 hours ago by Bergo1305
If $x, y, z \in \mathbb{R}$, then prove that the following inequality holds:
\[ \sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x \]
5 replies
Scientist10
Apr 23, 2025
Bergo1305
2 hours ago
J
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On a conditon for Hamilton Graphs
flower417477   0
2 hours ago
Prove that: for a graph $G$,if for any vertex $u,v,w$ for which $dist(u,v)=2$ and $uw,vw\in E(G)$ there's $d(u)+d(v)\geq|N(u)\cup N(v)\cup N(w)|$,then $G$ is a Hamiltonian graph
0 replies
flower417477
2 hours ago
0 replies
J
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Inspired by old results
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ a+b+c=3. $ Prove that
$$ \frac{2}{a}+\frac {2}{ab}+\frac{1}{abc}\geq 4$$$$ \frac{1}{a}+\frac {1}{ab}+\frac{2}{abc}\geq 2+\sqrt 3$$$$ \frac{3}{a}+\frac {3}{ab}+\frac{1}{abc}\geq\frac {7+\sqrt {13}}{2}$$$$ \frac{1}{a}+\frac {1}{ab}+\frac{3}{abc}\geq\frac {5+\sqrt {21}}{2}$$$$ \frac{1}{a}+\frac {1}{ab}+\frac{4}{abc}\geq 3+2\sqrt 2$$
3 replies
sqing
Yesterday at 12:30 PM
sqing
2 hours ago
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Congruence
Ecrin_eren   3
N Apr 7, 2025 by lbh_qys
Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.

3 replies
Ecrin_eren
Apr 3, 2025
lbh_qys
Apr 7, 2025
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Congruence
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Reveal topic tags
number theory
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Ecrin_eren
56 posts
Ecrin_eren
#1 h  Apr 3, 2025, 10:34 AM
Y by
Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.
Z K Y
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Ecrin_eren
56 posts
Ecrin_eren
#2 h  Apr 4, 2025, 1:39 PM
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Bump bump
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Ecrin_eren
56 posts
Ecrin_eren
#3 h  Apr 6, 2025, 8:42 AM
Y by
Any ideas?
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lbh_qys
552 posts
lbh_qys
#4 h  Apr 7, 2025, 3:39 AM
Y by
If $y\neq0,1$, then the original equation asserts that

\[ \left( 3x(1-y) y^{-1} \right)^2 = 3(1-y)(y+3) \]
The number of solutions satisfying this equation is given by

\[ \left(\frac{3 (1-y)(y+3)}{p}\right) + 1, \]
where $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol.

Moreover, since

\[ \sum_{y=0}^{p-1} \left(\frac{(y-1)(y+3)}{p}\right) = -1, \]
it follows that the summation for this part is

\[ \sum_{y=2}^{p-1}\left( \left(\frac{3 (1-y)(y+3)}{p}\right) + 1 \right) = p-2+\left(\frac{-3}{p}\right)\left(\sum_{y=0}^{p-1} \left(\frac{(y-1)(y+3)}{p}\right) - \left(\frac{-3}{p}\right) - \left(\frac{0}{p}\right) \right) = p - 2 - \left(\frac{-3}{p}\right) - 1. \]
When $y = 0$, it is evident that $x = 0$ is the unique solution; and when $y = 1$, there is evidently no solution. Therefore, the total number of solutions is

\[ p - 2 - \left(\frac{-3}{p}\right) - 1 + 1 = p - 2 - \left(\frac{-3}{p}\right) = p - 2 - \left(\frac{p}{3}\right). \]
Since $p = 41 \equiv 2 \pmod{3}$, it follows that $\left(\frac{p}{3}\right) = -1$. Consequently, the number of solutions is

\[ p - 2 - (-1) = p - 1 = 40. \]
This post has been edited 1 time. Last edited by lbh_qys, Apr 7, 2025, 3:41 AM
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