Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
function
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
n x n square and strawberries
pohoatza   19
N 20 minutes ago by atdaotlohbh
Source: IMO Shortlist 2006, Combinatorics 4, AIMO 2007, TST 4, P2
A cake has the form of an $ n$ x $ n$ square composed of $ n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$.

Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $ \mathcal{A}$ by performing a number of switches, defined as follows:

A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.
19 replies
pohoatza
Jun 28, 2007
atdaotlohbh
20 minutes ago
J
H
Consecutive squares are floors
ICE_CNME_4   7
N 30 minutes ago by ICE_CNME_4

Determine how many positive integers \( n \) have the property that both
\[ \left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor \]are consecutive perfect squares.
7 replies
ICE_CNME_4
6 hours ago
ICE_CNME_4
30 minutes ago
J
H
Hard geo finale with the cursed line
hakN   12
N an hour ago by ihategeo_1969
Source: 2024 Turkey TST P9
In a scalene triangle $ABC,$ $I$ is the incenter and $O$ is the circumcenter. The line $IO$ intersects the lines $BC,CA,AB$ at points $D,E,F$ respectively. Let $A_1$ be the intersection of $BE$ and $CF$. The points $B_1$ and $C_1$ are defined similarly. The incircle of $ABC$ is tangent to sides $BC,CA,AB$ at points $X,Y,Z$ respectively. Let the lines $XA_1, YB_1$ and $ZC_1$ intersect $IO$ at points $A_2,B_2,C_2$ respectively. Prove that the circles with diameters $AA_2,BB_2$ and $CC_2$ have a common point.
12 replies
hakN
Mar 18, 2024
ihategeo_1969
an hour ago
J
H
two lines passsing through the midpoint
miiirz30   1
N an hour ago by optimusprime154
Source: 2025 Euler Olympiad, Round 2
Points $A$, $B$, $C$, and $D$ lie on a line in that order, and points $E$ and $F$ are located outside the line such that $EA=EB$, $FC=FD$ and $EF \parallel AD$. Let the circumcircles of triangles $ABF$ and $CDE$ intersect at points $P$ and $Q$, and the circumcircles of triangles $ACF$ and $BDE$ intersect at points $M$ and $N$. Prove that the lines $PQ$ and $MN$ pass through the midpoint of segment $EF$.

Proposed by Giorgi Arabidze, Georgia
1 reply
miiirz30
Today at 10:23 AM
optimusprime154
an hour ago
J
H
Upper bound on products in sequence
tapir1729   11
N an hour ago by HamstPan38825
Source: TSTST 2024, problem 7
An infinite sequence $a_1$, $a_2$, $a_3$, $\ldots$ of real numbers satisfies
\[ a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3} \]for every positive integer $n$. Prove that there exists a real number $C$ such that $a_{n} a_{n+1} < C$ for every positive integer $n$.

Merlijn Staps
11 replies
tapir1729
Jun 24, 2024
HamstPan38825
an hour ago
J
H
Long and wacky inequality
Royal_mhyasd   4
N an hour ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
4 replies
Royal_mhyasd
May 12, 2025
Royal_mhyasd
an hour ago
J
H
perpendicular diagonals criterion for a cyclic quadrilateral
parmenides51   3
N an hour ago by PEKKA
Source: Sharygin 2005 Finals 9.1
The quadrangle $ABCD$ is inscribed in a circle whose center $O$ lies inside it.
Prove that if $\angle BAO = \angle DAC$, then the diagonals of the quadrilateral are perpendicular.
3 replies
parmenides51
Aug 26, 2019
PEKKA
an hour ago
J
H
functional inequality with equality
miiirz30   3
N 2 hours ago by genius_007
Source: 2025 Euler Olympiad, Round 2
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that the following two conditions hold:

1. For all real numbers $a$ and $b$ satisfying $a^2 + b^2 = 1$, We have $f(x) + f(y) \geq f(ax + by)$ for all real numbers $x, y$.

2. For all real numbers $x$ and $y$, there exist real numbers $a$ and $b$, such that $a^2 + b^2 = 1$ and $f(x) + f(y) = f(ax + by)$.

Proposed by Zaza Melikidze, Georgia
3 replies
miiirz30
Today at 10:32 AM
genius_007
2 hours ago
J
H
JBMO Shortlist 2023 N6
Orestis_Lignos   4
N 2 hours ago by MR.1
Source: JBMO Shortlist 2023, N6
Version 1. Find all primes $p$ satisfying the following conditions:

(i) $\frac{p+1}{2}$ is a prime number.
(ii) There are at least three distinct positive integers $n$ for which $\frac{p^2+n}{p+n^2}$ is an integer.

Version 2. Let $p \neq 5$ be a prime number such that $\frac{p+1}{2}$ is also a prime. Suppose there exist positive integers $a <b$ such that $\frac{p^2+a}{p+a^2}$ and $\frac{p^2+b}{p+b^2}$ are integers. Show that $b=(a-1)^2+1$.
4 replies
Orestis_Lignos
Jun 28, 2024
MR.1
2 hours ago
J
H
functional equation with exponentials
produit   7
N 2 hours ago by GreekIdiot
Find all solutions of the real valued functional equation:
f(\sqrt{x^2+y^2})=f(x)f(y).
Here we do not assume f is continuous
7 replies
produit
Today at 12:46 PM
GreekIdiot
2 hours ago
J
H
J
H
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
J
Equivalent definition for C^1 functions
G H J
Reveal topic tags
calculus
derivative
real analysis
function
L
G H BBookmark kLocked kLocked NReply
Source: Romania NMO 2025 11.3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ciobi_
28 posts
Ciobi_
#1 h  Apr 2, 2025, 1:54 PM
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KAME06
160 posts
KAME06
#2 h  Apr 3, 2025, 2:17 AM
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.
Z K Y
N Quick Reply
G
H
=
a