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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)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
Irritative methods
ILOVEMYFAMILY   0
an hour ago
Let $A \in \mathbb{R}^{n \times n}$. Prove that:
1) If $A$ is a diagonally dominant matrix, then both the Jacobi and Gauss-Seidel methods converge, and the Gauss-Seidel method converges faster in the sense that $\rho(T_{GS}) < \rho(T_J)$, where $T_{GS}$ and $T_J$ are the iteration matrices defined by $x^{(k+1)}=Tx^k+b$ for each method
2) If $A$ is a symmetric positive definite matrix, then both the Jacobi and Gauss-Seidel methods converge.
0 replies
ILOVEMYFAMILY
an hour ago
0 replies
Spectral radius
ILOVEMYFAMILY   0
an hour ago
Let $A \in \mathbb{R}^{n \times n}$. The spectral radius of $A$, denoted by $\rho(A)$, is defined as
\[
\rho(A) = \max_i |\lambda_i|
\]where $\lambda_i$ are all the eigenvalues of the matrix $A$.
Let $A \in \mathbb{R}^{n \times n}$. There exists a norm $\|\cdot\|$ such that $\|A\| < 1$ if and only if the spectral radius of $A$ satisfies the condition $\rho(A) < 1$.
0 replies
ILOVEMYFAMILY
an hour ago
0 replies
irritative method
ILOVEMYFAMILY   2
N an hour ago by ILOVEMYFAMILY
If $\|T\| < 1$ for any natural matrix norm and $c$ is a given vector, then the sequence $\{x^{(k)}\}_{k=0}^{\infty}$ defined by
\[
x^{(k)} = T x^{(k-1)} + c
\]converges, for any $x^{(0)} \in \mathbb{R}^n$, to a vector $x \in \mathbb{R}^n$, with $x = Tx + c$, and the following error bounds hold:
a) $\|x - x^{(k)}\| \leq \|T\|^k \|x^{(0)} - x\|$
b) $\|x - x^{(k)}\| \leq \frac{\|T\|^k}{1 - \|T\|} \|x^{(1)} - x^{(0)}\|$
2 replies
ILOVEMYFAMILY
5 hours ago
ILOVEMYFAMILY
an hour ago
Number.theory problem
menseggerofgod   1
N 3 hours ago by lpieleanu
Let n a number positive integer and let
1=d_{1} < d_{2} < ...... < d_{k-2} < d_{k-1} < d_{k} = n
Are the divisors of the number n . Find all the possible values of n if d_{k-2} =21d_{2}
1 reply
menseggerofgod
Today at 2:43 AM
lpieleanu
3 hours ago
gcd and lcm of factorials
Razorrizelim   1
N 3 hours ago by KolunG
Given that \( \gcd(a,b) = 22! \) and \( \text{lcm}(a,b) = 33! \), how many ordered pairs \( (a,b) \) of positive integers satisfy this condition?\[ \textbf{(A)}\ 498 \quad \textbf{(B)}\ 504 \quad \textbf{(C)}\ 512 \quad \textbf{(D)}\ 524 \quad \textbf{(E)}\ 532 \]
1 reply
Razorrizelim
Today at 8:36 AM
KolunG
3 hours ago
Trigonometry equation practice
ehz2701   19
N 4 hours ago by vanstraelen
There is a lack of trigonometric bash practice, and I want to see techniques to do these problems. So here are 10 kinds of problems that are usually out in the wild. How do you tackle these problems? (I had ChatGPT write a code for this.). Please give me some general techniques to solve these kinds of problems, especially set 2b. I’ll add more later.

Leaderboard and Solved Problems

problem set 1a (1-10)

problem set 2a (1-20)

problem set 2b (1-20)
answers 2b

General techniques so far:

Trick 1: one thing to keep in mind is

[center] $\frac{1}{2} = \cos 36 - \sin 18$. [/center]

Many of these seem to be reducible to this. The half can be written as $\cos 60 = \sin 30$, and $\cos 36 = \sin 54$, $\sin 18 = \cos 72$. This is proven in solution 1a-1. We will refer to this as Trick 1.
19 replies
ehz2701
Jul 12, 2025
vanstraelen
4 hours ago
Basic Inequalities Doubt
JetFire008   1
N Today at 8:30 AM by LayZee
If $x+y=1$, find the maximum value of $x^2+y^2=1$.
I saw a solution to this question where Titu's lemma was applied and the answer was $\frac{1}{2}$. But my doubt is can't we apply other inequality to get the maximum result? or did they us titu's lemma because the given information can fit only in this lemma?
1 reply
JetFire008
Today at 7:52 AM
LayZee
Today at 8:30 AM
Minimal of xy subjected to a constraint!
persamaankuadrat   3
N Today at 7:53 AM by LayZee
Let $x,y$ be positive real numbers such that

$$x+y^{2}+x^{3} = 1481$$
Find the minimal value of $xy$
3 replies
persamaankuadrat
Yesterday at 1:44 PM
LayZee
Today at 7:53 AM
a + b + c + \fracs
SYBARUPEMULA   3
N Today at 7:45 AM by SYBARUPEMULA
Given $a, b, c > 0$ such that $9a + 5b + 9c = 218$,
find the smallest value of

$$a + b + c + \frac{40}{a + 6} + \frac{72}{b + 8} + \frac{10}{c + 2}$$
3 replies
SYBARUPEMULA
Wednesday at 4:16 PM
SYBARUPEMULA
Today at 7:45 AM
Inequalities
sqing   5
N Today at 5:58 AM by sqing
Let $ a,b,c> 0,a+b+c+ab+bc+ca=6 .$ Prove that
$$(a^2+b^2+c^2)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$Let $ a,b,c> 0,ab+bc+ca=3 .$ Prove that
$$(a^2+b^2+c^2)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$Let $ a,b,c> 0,a+b+c=3 .$ Prove that
$$( ab+bc+ca)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$
5 replies
sqing
Today at 1:52 AM
sqing
Today at 5:58 AM
A strange number theory problem
Minty_SH   0
Today at 5:50 AM
Let $a,b,c,d$ be positive integers such that $a>b>c>d$, and that $(a+b-c+d) \mid (ab+cd)$
Prove or disprove: for any positive integer $m$ and positive odd number $n$, $a^nb^m+c^md^n$ is never a prime number.
0 replies
Minty_SH
Today at 5:50 AM
0 replies
Qualms about Inequality
amplreneo   5
N Today at 3:36 AM by mudkip42
Is this inequality true?

$$\frac{x}{2x+y+z} + \frac{y}{x+2y+z} + \frac{z}{x+y+2z} \leq \frac{3}{4}$$
5 replies
amplreneo
Aug 28, 2015
mudkip42
Today at 3:36 AM
A rare number theory's problem
menseggerofgod   0
Today at 3:14 AM
We say that a positive integer "a" is derived from the positive integer b if the number "a" is equal to a number obtained from a permutation of the digits of the number b . For example, 4202 and 2420 are both derived from 2024 , instead 4402 is not derived from 2024 . We say that a positive integer n is peculiar if there exist positive integers a and b such that n = a - b where b is a derivative of a from the following list:
1, 2, 3,........,2024
How many numbers are peculiar?
0 replies
menseggerofgod
Today at 3:14 AM
0 replies
analysis
We2592   1
N Today at 12:56 AM by alexheinis
Q) find the value of the integration $I=\int_{a}^{b} \frac{e^{-|x|}}{1+(sinhx)^2}$
1 reply
We2592
Yesterday at 12:13 PM
alexheinis
Today at 12:56 AM
Equivalent definition for C^1 functions
Ciobi_   1
N Apr 3, 2025 by KAME06
Source: Romania NMO 2025 11.3
Prove that, for a function $f \colon \mathbb{R} \to \mathbb{R}$, the following $2$ statements are equivalent:
a) $f$ is differentiable, with continuous first derivative.
b) For any $a\in\mathbb{R}$ and for any two sequences $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$, convergent to $a$, such that $x_n \neq y_n$ for any positive integer $n$, the sequence $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
1 reply
Ciobi_
Apr 2, 2025
KAME06
Apr 3, 2025
Equivalent definition for C^1 functions
G H J
G H BBookmark kLocked kLocked NReply
Source: Romania NMO 2025 11.3
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Ciobi_
33 posts
#1
Y by
Prove that, for a function $f \colon \mathbb{R} \to \mathbb{R}$, the following $2$ statements are equivalent:
a) $f$ is differentiable, with continuous first derivative.
b) For any $a\in\mathbb{R}$ and for any two sequences $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$, convergent to $a$, such that $x_n \neq y_n$ for any positive integer $n$, the sequence $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
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KAME06
163 posts
#2
Y by
Let $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$ sequences that converge to $a$ and $x_n \neq y_n$ for any $n$.
First, note that $(x_n)$ or $(y_n)$ mustn't have an $n$ such that $x_n=a$ or $y_n=a$ (because $x_n \neq y_n$ and they are convergent to $a$), so WLOG consider $y_n \neq a$ for any $n$.
(a)$\rightarrow$(b):
If $f$ is differentiable, with continuous first derivative, then the function $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.
That implies that: $\lim_{x \rightarrow a} \frac{f(a)-f(b)}{a-b}$=$\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)$.
As $(x_n)$ and $(y_n)$ converge, for large enough $n$, $|x_n-a|<\frac{\epsilon}{2}$, $|y_n-a|<\frac{\epsilon}{2}$, so $|x_n-y_n|<\epsilon$.
That implies (and the fact that that $y_n \neq x_n$ for any $n$) if $n \rightarrow \infty$, then $y_n \rightarrow x_n$, so:
$\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$. The limit exists.
We conclude that $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
(b) $\rightarrow$ (a):
As above: $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}\right)$.
$\lim_{n \rightarrow \infty}f(x_n)$ must be defined because the limit exist. If $f(x_n)$ doesn't converge to $f(a)$, notice that $b \rightarrow a$ and $a-b$ converges to $0$ and it can be as little as we want, so $\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}$ is $\infty$ or $-\infty$, but $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$ converges. Contradiction.
Then $f(x_n)$ converges to $f(a)$ and $b \rightarrow a$, so $\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{b \rightarrow a} \frac{f(a)-f(b)}{a-b}$.
We conclude that $f$ is differentiable and that $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.
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