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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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0 replies
jwelsh
Jul 1, 2025
0 replies
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sumdigits
teomihai   6
N 14 minutes ago by teomihai
Find sum digits all number 1 to 2001.
6 replies
teomihai
Yesterday at 11:27 AM
teomihai
14 minutes ago
J
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Sequences...
TheDarkPrince   16
N 26 minutes ago by mqoi_KOLA
Source: RMO 2018 P6
Define a sequence $\{a_n\}_{n\geq 1}$ of real numbers by \[a_1=2,\qquad a_{n+1} = \frac{a_n^2+1}{2}, \text{ for } n\geq 1.\]Prove that \[\sum_{j=1}^{N} \frac{1}{a_j + 1} < 1\]for every natural number $N$.
16 replies
TheDarkPrince
Oct 28, 2018
mqoi_KOLA
26 minutes ago
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Some of my less-seen proposals
navid   4
N 28 minutes ago by shaboon
Dear friends,

Since 2003, I have had several nice days in AOPS-- i.e., Mathlinks; as some of you may remember. I decided to share you some of my less-seen proposals. Some of them may be considered as some early ethudes; several of them already appeared on some competitions or journals. I hope you like them and this be a good starting point for working on them. Please take a look at the following link.

https://drive.google.com/file/d/1bntcjZAHZ-WN1lfGbNbz0uyFvhBTMEhz/view?usp=sharing

Best regards,
Navid.
4 replies
3 viewing
navid
Yesterday at 7:08 PM
shaboon
28 minutes ago
J
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Interesting inequality
sqing   2
N 33 minutes ago by pooh123
Source: Own
Let $ a,b,c>0,(a+b+1)\left(\frac{1}{a} + \frac{1}{b} +1\right)= 10. $ Prove that
$$ \frac{4\sqrt{2}}{5} \geq\frac{ \sqrt{a} + \sqrt{b} }{a+b+ab}\geq \frac{1 }{2\sqrt{2}}$$$$ \frac{4(1+\sqrt{2})}{5} \geq \frac{ \sqrt{a} + \sqrt{b} +1}{a+b+ab}\geq \frac{1+2\sqrt{2} }{8}$$
2 replies
sqing
Jul 29, 2025
pooh123
33 minutes ago
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2024 NYMA = New Years Mock AIME #10 complex ineq
parmenides51   3
N an hour ago by douqile
Two complex numbers $z_1$ and $z_2$ are chosen uniformly and at random from the disk $0\le |z| \le 1$. Suppose $z_1 = a + bi$ and $z_2 = c + di$, where $a, b, c, d$ are all real numbers. The probability that
$$\frac{a^2 + b^2} {2abcd + a^2c^2 + b^2d^2} \le \frac{a^2 + b^2}{ - 2abcd + a^2d^2 + b^2c^2} \le 4$$can be written as $\frac{p+q\pi -r\sqrt{s}}{ i \pi}$ , where $p, q, r, s, t$ are positive integers, gcd $(p, q, r, t) =1$, and $s$ is not divisible by the square of any prime. Compute $pqrst$.
3 replies
parmenides51
Feb 13, 2024
douqile
an hour ago
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AIMO 2025 olympiad
Unknown0108   1
N an hour ago by Unknown0108
If you are participating in Asia International olympiad 2025 can we share our knowledge cuz I am participating too
1 reply
Unknown0108
Jul 29, 2025
Unknown0108
an hour ago
J
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Perfect square
Ecrin_eren   3
N 2 hours ago by Solar Plexsus


Find all integer values of n such that for every pair of integers (a, b) satisfying:

n·a² + a = (n + 1)·b² + b

the number |a − b| is always a perfect square.




3 replies
Ecrin_eren
Jul 28, 2025
Solar Plexsus
2 hours ago
J
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1998 St. Petersburg
radioactiverascal90210   0
2 hours ago
In the plane are given several squares with parallel sides, such that among any $n$ of them, there exist four having a common point. Prove that the squares can be divided into at most $n$ − $3$ groups, such that
all of the squares in a group have a common point.
0 replies
radioactiverascal90210
2 hours ago
0 replies
J
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Inequality proof
radioactiverascal90210   0
2 hours ago
There are $n$ points $P_1$, $P_2$ . . . , $P_n$ in the interval $[-1,1]$ where the
points are consecutively arranged from left to right. Let $a_i$ be the product of all
distances from $P_i$ to the other $n$ − $1$ points. Prove that
$\sum_{i=1}^n$ $\frac{1}{a_i}$ $\ge$ $2^{n-1}$
0 replies
radioactiverascal90210
2 hours ago
0 replies
J
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If OAB and OAC share equal angles and sides, why aren't they congruent?
Merkane   0
Today at 4:33 AM

Problem 1.39 (CGMO 2012/5). Let ABC be a triangle. The incircle of ABC is tangent
to AB and AC at D and E respectively. Let O denote the circumcenter of BCI .
Prove that ∠ODB = ∠OEC. Hints: 643 89 Sol: p.241

While I have solved the problem, I encountered a step that seems logically sound but leads to a contradiction, and I would like help identifying the flaw.

Here is the reasoning I followed:

The quadrilateral ABOC is cyclic.

OB = OC.

∠OAB = ∠OCB.
Similarly, ∠OAC = ∠OBC.

From symmetry and the above, it seems that ∠OAB = ∠OAC.

Since OA is a shared side, I concluded that triangle OAB ≅ triangle OAC.


But clearly, OAB and OAC are not congruent.
Where exactly is the logical error in this argument?
0 replies
Merkane
Today at 4:33 AM
0 replies
J
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Geometry Problem
Rice_Farmer   0
Today at 3:09 AM
Let $w_1$ ad $w_2$ be two circles intersecting at $P$ and $Q.$ The tangent like closer to $Q$ touches $w_1$ and $w_2$ at $M$ and $N$ respectively. If $PQ=3,NQ=2,$ and $MN=PN,$ find $QM.$
0 replies
1 viewing
Rice_Farmer
Today at 3:09 AM
0 replies
J
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A writing game
Ecrin_eren   2
N Today at 1:51 AM by Ecrin_eren


There is an integer greater than 1 written on the board in A’s house. Every morning when A wakes up, he erases the number n on the board and does the following:

If there is a positive integer m such that m^3= n, then he writes m on the board.

Otherwise, he writes 2n+1 on the board.


Since A repeats this process infinitely many times, prove that among all the numbers A has written and will write on the board, there are infinitely many greater than 10^100.





2 replies
Ecrin_eren
Jul 28, 2025
Ecrin_eren
Today at 1:51 AM
J
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Inequalities
sqing   11
N Today at 1:47 AM by sqing
Let $ a,b,c\geq 0, \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{3}{2}.$ Prove that
$$ \left(a+b+c-\frac{17}{6}\right)^2+9abc \geq\frac{325}{36}$$$$ \left(a+b+c-\frac{5}{2}\right)^2+12abc \geq\frac{49}{4}$$$$\left(a+b+c-\frac{14}{5}\right)^2+\frac{49}{5}abc \geq\frac{49}{5}$$
11 replies
sqing
Jun 30, 2025
sqing
Today at 1:47 AM
J
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Inequalities
sqing   8
N Today at 1:22 AM by sqing
Let $ a,b> 0, a^2+b^2+ab=3 .$ Prove that
$$ (a+b+1)^2(\frac {a} {b^2+1}+\frac {b} {a^2+1})\geq 9$$Let $ a,b> 0, a+b+ab=3 .$ Prove that
$$(a+b+1)^2(\frac {a+1} {b^2+1}+\frac {b+1} {a^2+1})\geq 18$$Let $ a,b> 0, a+b+2ab=4.$ Prove that
$$(a+b+1)^2(\frac {a} {b^2+1}+\frac {b} {a^2+1})\geq 9$$$$ (a+b+1)^2(\frac {a+1} {b^2+1}+\frac {b+1} {a^2+1}) \geq 18$$
8 replies
sqing
Jul 25, 2025
sqing
Today at 1:22 AM
J
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J
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An algorithm for discovering prime numbers?
Lukaluce   4
N May 30, 2025 by alexanderhamilton124
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
4 replies
Lukaluce
May 18, 2025
alexanderhamilton124
May 30, 2025
J
An algorithm for discovering prime numbers?
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Reveal topic tags
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Source: 2025 Junior Macedonian Mathematical Olympiad P3
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Lukaluce
286 posts
Lukaluce
#1 h  May 18, 2025, 3:31 PM
Y by
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
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grupyorum
1466 posts
grupyorum
#2 h  May 18, 2025, 6:31 PM
Y by
We first show that there is an $n_0$ and $\epsilon\in\{-1,1\}$ such that for every $n\ge n_0$, $p_{n+1} = 2p_n+\epsilon$.

To see this, suppose $p_1>3$. If $p_1\equiv 1\pmod{3}$ then $p_{n+1}=2p_n-1$ must hold necessarily (otherwise $3\mid 2p_n+1$ but $p_n>3$). Likewise if $p_1\equiv -1\pmod{3}$ then $p_{n+1}\equiv 2p_n+1$ must hold. If $p_1\le 3$, then $p_j>3$ for some $j>1$, so the same argument carries through. Shifting if necessary, we will analyze the sequence $p_{n+1} =2p_n-1$ and $p_{n+1}=2p_n+1$ for $p_1>3$.

Case 1. Let $p_{n+1} = 2p_n-1$ for $n\ge 1$. Set $b_n:=p_n-1$ to obtain $b_{n+1} = 2b_n$. Iterating, we find $b_n = 2^{n-1}b_1$. Consequently, $p_n = 2^{n-1}(p_1-1)+1$. Taking $n=k(p_1-1)+1$ for suitably large $k$, Fermat's theorem asserts $2^{n-1}\equiv 1\pmod{p_1}$. So, $p_1\mid p_n$ but $p_n>p_1$, hence $p_n$ cannot be a prime.

Case 2. Let $p_{n+1}=2p_n+1$ for $n\ge 1$. Set $b_n:=p_n+1$ to obtain $p_n = 2^{n-1}(p_1+1)-1$. The same choice of $n$ ensures $p_1\mid p_n$, a contradiction.

So, no such infinite sequence exists.

Remark. This is an old Bulgarian problem (between 2003-2010 I think), though I don't remember the exact year.
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Assassino9931
1550 posts
Assassino9931
#3 h  May 29, 2025, 4:17 PM
Y by
@above Hm, haven't seen this in Bulgaria, but it is popular from Baltic Way 2004.
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TopGbulliedU
24 posts
TopGbulliedU
#4 h  May 29, 2025, 5:15 PM • 1 Y
Y by alexanderhamilton124
hahaha I was in the comp,after i got out I told everyone that nobody could solve this after the results came it was only me :-D
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alexanderhamilton124
407 posts
alexanderhamilton124
#5 h  May 30, 2025, 2:47 PM
Y by
TopGbulliedU wrote:
hahaha I was in the comp,after i got out I told everyone that nobody could solve this after the results came it was only me :-D

orz gj man
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