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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 3:18 PM
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
Yesterday at 3:18 PM
0 replies
(Original version) Same number of divisors
MNJ2357   2
N 3 minutes ago by john0512
Source: 2024 Korea Summer Program Practice Test P8 (original version)
For a positive integer \( n \), let \( \tau(n) \) denote the number of positive divisors of \( n \). Determine whether there exists a positive integer triple \( a, b, c \) such that there are exactly $1012$ positive integers \( K \) not greater than $2024$ that satisfies the following: the equation
\[ \tau(x) = \tau(y) = \tau(z) = \tau(ax + by + cz) = K \]holds for some positive integers $x,y,z$.
2 replies
MNJ2357
Aug 12, 2024
john0512
3 minutes ago
An inequality problem
Arithmetic_fighter   0
23 minutes ago
Given $a,b,c \in \mathbb R$ such that $a^2+b^2+c^2=3$. Prove that
$$\frac{a b}{c^2+a^2+1}+\frac{b c}{a^2+b^2+1}+\frac{c a}{b^2+c^2+1} \leq 1$$
0 replies
Arithmetic_fighter
23 minutes ago
0 replies
Geometry :3c
popop614   3
N 23 minutes ago by ItzsleepyXD
Source: MINE :<
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
3 replies
popop614
4 hours ago
ItzsleepyXD
23 minutes ago
Game About Passing Pencils
WilliamSChen   0
30 minutes ago
A group of $n$ children sit in a circle facing inward with $n > 2$, and each child starts with an arbitrary even number of pencils. Each minute, each child simultaneously passes exactly half of all of their pencils to the child to their right. Then, all children that have an odd number of pencils receive one more pencil.
Prove that after a finite amount of time, the children will all have the same number of pencils.

I do not know the source.
0 replies
WilliamSChen
30 minutes ago
0 replies
Easy matrix equation involving invertibility
Ciobi_   1
N Yesterday at 8:08 PM by loup blanc
Source: Romania NMO 2025 11.2
Let $n$ be a positive integer, and $a,b$ be two complex numbers such that $a \neq 1$ and $b^k \neq 1$, for any $k \in \{1,2,\dots ,n\}$. The matrices $A,B \in \mathcal{M}_n(\mathbb{C})$ satisfy the relation $BA=a I_n + bAB$. Prove that $A$ and $B$ are invertible.
1 reply
Ciobi_
Yesterday at 1:46 PM
loup blanc
Yesterday at 8:08 PM
linear algebra
ay19bme   5
N Yesterday at 6:13 PM by loup blanc
Does the matrix equation $X^3=mI_2$ is solvable over $M_{2}(\mathbb{Z})$ for every $m\in \mathbb{Z}$. Here $X\in M_{2}(\mathbb{Z})$, $I_2=\begin{pmatrix} 1& 0\\0 & 1\end{pmatrix}$.
5 replies
ay19bme
Yesterday at 7:06 AM
loup blanc
Yesterday at 6:13 PM
Polynomial meets geometry
chirita.andrei   0
Yesterday at 5:42 PM
Source: Own. Proposed for Romanian National Olympiad 2025.
(a) Let $A,B,C$ be collinear points (in order) and $D$ a point in plane. Consider the disc $\mathcal{D}$ of center $D$ and radius $kBD$, for some $k\in(0,1)$. Prove that $\mathcal{D}\cap [AC]$ is either the empty set or a segment of length at most $2kAC$.
(b) Let $n$ be a positive integer and $P(X)\in\mathbb{C}[X]$ be a polynomial of degree $n$. Prove that \[\sup_{x\in[0,1]}|P(x)|\le(2n+1)^{n+1}\int\limits_{0}^{1}|P(x)|\mathrm{d}x.\]
0 replies
chirita.andrei
Yesterday at 5:42 PM
0 replies
Integral inequality with differentiable function
Ciobi_   1
N Yesterday at 3:35 PM by MS_asdfgzxcvb
Source: Romania NMO 2025 12.2
Let $f \colon [0,1] \to \mathbb{R} $ be a differentiable function such that its derivative is an integrable function on $[0,1]$, and $f(1)=0$. Prove that \[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]
1 reply
Ciobi_
Yesterday at 2:29 PM
MS_asdfgzxcvb
Yesterday at 3:35 PM
On coefficients of a polynomial over a finite field
Ciobi_   0
Yesterday at 2:59 PM
Source: Romania NMO 2025 12.4
Let $p$ be an odd prime number, and $k$ be an odd number not divisible by $p$. Consider a field $K$ be a field with $kp+1$ elements, and $A = \{x_1,x_2, \dots, x_t\}$ be the set of elements of $K^*$, whose order is not $k$ in the multiplicative group $(K^*,\cdot)$. Prove that the polynomial $P(X)=(X+x_1)(X+x_2)\dots(X+x_t)$ has at least $p$ coefficients equal to $1$.
0 replies
Ciobi_
Yesterday at 2:59 PM
0 replies
On non-negativeness of continuous and polynomial functions
Ciobi_   0
Yesterday at 2:51 PM
Source: Romania NMO 2025 12.3
a) Let $a\in \mathbb{R}$ and $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous function for which there exists an antiderivative $F \colon \mathbb{R} \to \mathbb{R} $, such that $F(x)+a\cdot f(x) \geq 0$, for any $x \in \mathbb{R}$, and$ \lim_{|x| \to \infty} \frac{F(x)}{e^{|\alpha \cdot x|}}=0$ holds for any $\alpha \in \mathbb{R}^*$. Prove that $F(x) \geq 0$ for all $x \in \mathbb{R}$.
b) Let $n\geq 2$ be a positive integer, $g \in \mathbb{R}[X]$, $g = X^n + a_1X^{n-1}+ \dots + a_{n-1}X+a_n$ be a polynomial with all of its roots being real, and $f \colon \mathbb{R} \to \mathbb{R}$ a polynomial function such that $f(x)+a_1\cdot f'(x)+a_2\cdot f^{(2)}(x)+\dots+a_n\cdot f^{(n)}(x) \geq 0$ for any $x \in \mathbb{R}$. Prove that $f(x) \geq 0$ for all $x \in \mathbb{R}$.
0 replies
Ciobi_
Yesterday at 2:51 PM
0 replies
Proving AB-BA is singular from given conditions
Ciobi_   0
Yesterday at 2:04 PM
Source: Romania NMO 2025 11.4
Let $A,B \in \mathcal{M}_n(\mathbb{C})$ be two matrices such that $A+B=AB+BA$. Prove that:
a) if $n$ is odd, then $\det(AB-BA)=0$;
b) if $\text{tr}(A)\neq \text{tr}(B)$, then $\det(AB-BA)=0$.
0 replies
Ciobi_
Yesterday at 2:04 PM
0 replies
RREF of some matrices
tommy2007   3
N Yesterday at 1:51 PM by tommy2007
for $\forall n \in \mathbb{N},$
what is the maximum integer that appears in one of the Reduced Row Echelon Forms of $n \times n$ matrices which has only $-1$ and $1$ for their entries?
3 replies
tommy2007
Yesterday at 6:57 AM
tommy2007
Yesterday at 1:51 PM
Finding pairs of functions of class C^2 with a certain property
Ciobi_   0
Yesterday at 1:31 PM
Source: Romania NMO 2025 11.1
Find all pairs of twice differentiable functions $f,g \colon \mathbb{R} \to \mathbb{R}$, with their second derivative being continuous, such that the following holds for all $x,y \in \mathbb{R}$: \[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]
0 replies
Ciobi_
Yesterday at 1:31 PM
0 replies
Inverse of absolute value function
MetaphysicalWukong   2
N Yesterday at 12:32 PM by paxtonw
how does the function have an inverse for k= 101, 203, 305, 509, 611 and 713?

how do we deduce this without graphing software?
2 replies
MetaphysicalWukong
Yesterday at 7:21 AM
paxtonw
Yesterday at 12:32 PM
A colouring game on a grid
Tintarn   2
N Mar 30, 2025 by math-olympiad-clown
Source: Baltic Way 2024, Problem 8
Let $a$, $b$, $n$ be positive integers such that $a + b \leq n^2$. Alice and Bob play a game on an (initially uncoloured) $n\times n$ grid as follows:
- First, Alice paints $a$ cells green.
- Then, Bob paints $b$ other (i.e.uncoloured) cells blue.
Alice wins if she can find a path of non-blue cells starting with the bottom left cell and ending with the top right cell (where a path is a sequence of cells such that any two consecutive ones have a common side), otherwise Bob wins. Determine, in terms of $a$, $b$ and $n$, who has a winning strategy.
2 replies
Tintarn
Nov 16, 2024
math-olympiad-clown
Mar 30, 2025
A colouring game on a grid
G H J
Source: Baltic Way 2024, Problem 8
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Tintarn
9029 posts
#1
Y by
Let $a$, $b$, $n$ be positive integers such that $a + b \leq n^2$. Alice and Bob play a game on an (initially uncoloured) $n\times n$ grid as follows:
- First, Alice paints $a$ cells green.
- Then, Bob paints $b$ other (i.e.uncoloured) cells blue.
Alice wins if she can find a path of non-blue cells starting with the bottom left cell and ending with the top right cell (where a path is a sequence of cells such that any two consecutive ones have a common side), otherwise Bob wins. Determine, in terms of $a$, $b$ and $n$, who has a winning strategy.
Z K Y
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NiuNiuBaba
3 posts
#2
Y by
Denote each cell as (i, j) for row i, column j, so bottom left is (1,1), top right is (n, n)
We say all a(i, j) build a wall when i+j=k+1, k is integer and 1<=k<=2n-1. There are 2n-1 walls.
It's obvious that a path from (1,1) to (n, n) will across all walls: considering the sum of i, j of each cell on a "path", every step you move on the path, the sum +1 or -1, and the start number is 2, the end number is 2n, so each of 2~2n mush appears on the path("discrete version of IVT", not sure if it's the name in English)
1.obviously, if a>=2n-1, Alice can paint a green path from (1,1) to (n, n). Alice Win
2. if a<2n-1, there must be some "walls" whose all cells are not colored by Alice(as there are 2n-1 walls), we call them "non-green" walls. The max of their minimum length is [a/2]+1. So,
2.1. if b>= [a/2]+1, Bob can pick the non-green wall whose length is minimum, and color all of its cells to blue, and Bob Win.
2.2. if b<[a/2]+1, Alice can color the bottom row cells from(1, 1) to (1, [a/2]), and the right column cells form (n, n) to (n, [a/2]+1). Before Bob paints any cell, there are at least [a/2]+1 uncolored paths to connect these 2 green tiles, and these paths are not crossing each others. (I just don't know how to insert a pic here, so hopefully below ASCII chart (for n=5, a=6) can show the idea, I use '.' and '|' to show the paths, 'x' is placeholder to make the chart looks ok):
.........G
| .......G
| | .....G
| | |xxx|
GGG..|
To cut off all of these paths, Bob needs paint at least [a/2]+1 cells, one cell for each path. it means if b<[a/2]+1, Alice will Win.
In summary, Bob will win only if (a<2n-1)&&(b>= [a/2]+1) is true, otherwise, Alice will win.
Z K Y
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math-olympiad-clown
17 posts
#3
Y by
is this ok? :wacko:

we separate the question into two cases
Case1. a+b=n^2 , in this case it's obvious to see that a must be greater than 2n-2 due to connectivity.

Case2. a+b<n^2
2-1 : if Alice has more than 2n-1 green blocks then obviously Alice wins.

2-2: if a is smaller than 2n-1 and b is greater than n-1 then Bob wins
(because Bob can "cut" any diagonal in any squares in the diagram and a is smaller than 2n-1 implies a is not absolutely connected by green bolcks)

2-3 : if a is smaller than 2n-1 and b=k which is smaller than n ,then we deduce that a must at least be 2k.

We call the absolutely connected green blocks from the bottom left corner "left dragon" and define the "right dragon" analogously.
now we define another noun called "HP" which is the number of non-coloring blocks surrounded by the dragon.
because the dragon must be absolutely connected , the "HP" of a dragon will be less than the number of its length+1 . we go back to the situation, b=k and if a is smaller than 2k, that means there must be a dragon which its length is at most k-1 so its "HP" will be at most k . then when Bob started to play , he can use all the blue blocks to suffocate the dragon , hence Bob wins.
and its easy to construct the example for a=2k and b=k and no dragon got suffocated.
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