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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by Philippine 2025
sqing   0
3 minutes ago
Source: Own
Let $ a,b,c,d $ be real numbers . Prove that
$$\frac{(a-1)(b-3)(c-3)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{7+4\sqrt 3}{144}$$$$\frac{(a-1)(b-2)(c-2)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{11+4\sqrt 7}{432}$$


0 replies
+1 w
sqing
3 minutes ago
0 replies
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such : $(f
guramuta   1
N 15 minutes ago by jasperE3
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such :
$(f(x-y))^2= (f(x))^2 - 2f(xy) + (f(y))^2$
1 reply
1 viewing
guramuta
an hour ago
jasperE3
15 minutes ago
Quadrangle, nine-point conic, Steiner line
kosmonauten3114   0
15 minutes ago
Source: My own
Let $P_1P_2P_3P_4$ be a general quadrangle which does not form an orthocentric system. Let $P$, $I$, $M$, $T$ be the Euler-Poncelet point ($\text{QA-P2}$), isogonal center ($\text{QA-P4}$), midray homothetic center ($\text{QA-P8}$), inscribed square axes crosspoint ($\text{QA-P23}$) of $P_1P_2P_3P_4$, respectively.
Let $H_1$ be the orthocenter of $\triangle{P_2P_3P_4}$, and define $H_2$, $H_3$, $H_4$ cyclically.
Let $A_{ij}=P_iP_j \cap H_iH_j$ ($\{i, j\} \in \{1, 2, 3, 4\}, i<j$).
Let $B_{ij}=P_iP_j \cap H_kH_l$ ($\{i, j, k, l\} \in \{1, 2, 3, 4\}, i<j$).
Then, the 12 points $A_{12}$, $A_{13}$, $A_{14}$, $A_{23}$, $A_{24}$, $A_{34}$, $B_{12}$, $B_{13}$, $B_{14}$, $B_{23}$, $B_{24}$, $B_{34}$ lie on the same conic, here denoted by $\mathcal{C}_1$.
Let $\mathcal{C}_2$ be the nine-point conic of $P_1P_2P_3P_4$.
Suppose that $\mathcal{C}_1$ and $\mathcal{C}_2$ have 4 distinct intersection points, and let $U$, $V$, $W$ be the intersections, other than $P$, of $\mathcal{C}_1$ and $\mathcal{C}_2$.

Prove that the Steiner line of $P$ with respect to $\triangle{UVW}$ passes through $I$ and $M$, and show that the center of $\mathcal{C}_1$ and the orthocenter of $\triangle{UVW}$ coincide with $T$.
0 replies
kosmonauten3114
15 minutes ago
0 replies
Inspired by Philippine 2025
sqing   1
N 30 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+1)(c^2+1)} \ge -\frac{7+5\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+2)(c^2+2)} \ge -\frac{5+3\sqrt 3}{32}$$
1 reply
1 viewing
sqing
38 minutes ago
sqing
30 minutes ago
Metric space
wiseman   3
N May 21, 2025 by alinazarboland
Source: IMS 2014 - Day1 - Problem4
Let $(X,d)$ be a metric space and $f:X \to X$ be a function such that $\forall x,y\in X : d(f(x),f(y))=d(x,y)$.
$\text{a})$ Prove that for all $x \in X$, $\lim_{n \rightarrow +\infty} \frac{d(x,f^n(x))}{n}$ exists, where $f^n(x)$ is $\underbrace{f(f(\cdots f(x)}_{n \text{times}} \cdots ))$.
$\text{b})$ Prove that the amount of the limit does not depend on choosing $x$.
3 replies
wiseman
Oct 2, 2014
alinazarboland
May 21, 2025
Double integration
Tricky123   2
N May 21, 2025 by Mathzeus1024
Q)
\[\iint_{R} \sin(xy) \,dx\,dy, \quad R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]\]
How to solve the problem like this I am using the substitution method but its seems like very complicated in the last
Please help me
2 replies
Tricky123
May 18, 2025
Mathzeus1024
May 21, 2025
Find solution of IVP
neerajbhauryal   2
N May 15, 2025 by Mathzeus1024
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
2 replies
neerajbhauryal
Sep 23, 2014
Mathzeus1024
May 15, 2025
fourier series?
keroro902   2
N May 15, 2025 by Mathzeus1024
f(x)=$\sum _{n=0}^{\infty } \text{cos}(nx)/2^{n}$
f(x) = ?
thanks
2 replies
keroro902
May 14, 2010
Mathzeus1024
May 15, 2025
uniformly continuous of multivariable function
keroro902   1
N May 14, 2025 by Mathzeus1024
How can I determine which of the following functions are uniformly continuous on the given domain A?

$f \left( x, y \right) = \frac{x^3 + y^2}{x^2 + y}$ , $A = \left\{ \left( x, y
\right) \in \mathbb m{R}^2 \left|  \right.  \left| y \right| \leq \frac{x^2}{2}
%Error. "nocomma" is a bad command.
, x^2 + y^2 < 1 \right\}$

$g \left( x, y \right) = \frac{y^2 + 4 x^2}{y^2 - 4 x^2 - 1}$, $A = \left\{
\left( x, y \right) \in \mathbb m{R}^2 \left| 0 \leq x^2 - y^2 \leqslant 1
\right\} \right.$
1 reply
keroro902
Nov 2, 2012
Mathzeus1024
May 14, 2025
Miklos Schweitzer 1971_5
ehsan2004   2
N May 9, 2025 by pi_quadrat_sechstel
Let $ \lambda_1 \leq \lambda_2 \leq...$ be a positive sequence and let $ K$ be a constant such that \[  \sum_{k=1}^{n-1} \lambda^2_k < K \lambda^2_n \;(n=1,2,...).\] Prove that there exists a constant $ K'$ such that \[  \sum_{k=1}^{n-1} \lambda_k < K' \lambda_n \;(n=1,2,...).\]

L. Leindler
2 replies
ehsan2004
Oct 29, 2008
pi_quadrat_sechstel
May 9, 2025
Miklos Schweitzer 1968_9
ehsan2004   1
N May 8, 2025 by pi_quadrat_sechstel
Let $ f(x)$ be a real function such that
\[ \lim_{x \rightarrow +\infty} \frac{f(x)}{e^x}=1\]
and $ |f''(x)|\leq c|f'(x)|$ for all sufficiently large $ x$. Prove that \[ \lim_{x \rightarrow +\infty} \frac{f'(x)}{e^x}=1.\]

P. Erdos
1 reply
ehsan2004
Oct 8, 2008
pi_quadrat_sechstel
May 8, 2025
Range of 2 parameters and Convergency of Improper Integral
Kunihiko_Chikaya   3
N May 1, 2025 by Mathzeus1024
Source: 2012 Kyoto University Master Course in Mathematics
Let $\alpha,\ \beta$ be real numbers. Find the ranges of $\alpha,\ \beta$ such that the improper integral $\int_1^{\infty} \frac{x^{\alpha}\ln x}{(1+x)^{\beta}}$ converges.
3 replies
Kunihiko_Chikaya
Aug 21, 2012
Mathzeus1024
May 1, 2025
f'(1)>1 implies f has a fixed point in (0,1)
Sayan   13
N Apr 27, 2025 by Apple_maths60
Source: ISI(BS) 2010 #4
A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.
13 replies
Sayan
May 17, 2012
Apple_maths60
Apr 27, 2025
Maximum value of ∫_0^1 e^x logf(x) dx when ∫_0^1 f(x) dx =1
tom-nowy   2
N Apr 26, 2025 by Dattier
Source: 1982 Niigata University entrance exam
Let $\mathcal{F}$ be the set of continuous functions $f: [0,1] \to (0, \infty )$ such that $ \int_0^1 f(x) \, \mathrm dx =1 $. For $f \in \mathcal{F}$, let $$I(f)=\int_0^1 e^x \log f(x) \, \mathrm dx .$$Determine $\max_{f \in \mathcal{F}}I(f)$.
2 replies
tom-nowy
Jul 7, 2013
Dattier
Apr 26, 2025
Really easy "divide the cube in cubes" problem
cyshine   4
N Nov 8, 2009 by davicoelhoamorim
Source: Brazilian Math Olympiad, Problem 4
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
4 replies
cyshine
Oct 22, 2009
davicoelhoamorim
Nov 8, 2009
Really easy "divide the cube in cubes" problem
G H J
Source: Brazilian Math Olympiad, Problem 4
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cyshine
236 posts
#1 • 3 Y
Y by Davi-8191, Adventure10, Mango247
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
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Mathias_DK
1312 posts
#2 • 2 Y
Y by Adventure10, Mango247
cyshine wrote:
Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.
We can partion any cube into $ 8$ smaller cubes or $ 27$ smaller cubes easily.($ 2^3=8,3^3=27$)

So if we can split a cube into $ m$ smaller cubes we can also split it into $ m+7$ and $ m+26$ smaller cubes. Since we can "split" it into $ 1$ cube we can split it into $ 1+7x+26y$ for any $ x,y \in \mathbb{N}_0$. $ 1+7x+26y = n \iff x = \frac{n-1-26y}{7}$. Let $ 0 \le y \le 6$ be such that $ 26y \equiv n-1 \bmod 7 \iff y \equiv 3n-3 \bmod 7$. Let $ n_0 = 1 + 26 \cdot 6$, then $ x \ge 0$ and we can split it into $ n$ smaller cubes whenever $ n \ge n_0$. QED
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mszew
1046 posts
#3 • 1 Y
Y by Adventure10
check http://www.research.att.com/~njas/sequences/A014544

for minimum $ n_0=48$[/hide]
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jmerry
12096 posts
#4 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Mathias' argument gives us that $ n_0=1+(7-1)(26-1)+1=152$ works; $ 150$ is the largest positive integer that can't be written as a sum of sevens and 26s. Obviously, we can do better.

First, that $ 26$ could be $ 19$; rather than replacing a $ 1\times 1\times 1$ cube with $ 27$ $ \frac13\times\frac13\times\frac13$ cubes, we replace it with one $ \frac23\times\frac23\times\frac23$ and $ 19$ $ \frac13\times\frac13\times\frac13$ cubes. Since adding $ 26$ is possible by doing the above and then splitting the large cube into eight pieces, we don't need to include it separately. From sevens and nineteens, we get a new estimate of $ n_0\ge 6\cdot 18+2=110$.
For the next improvement, consider breaking a $ 1\times 1\times 1$ into a $ \frac34\times\frac34\times\frac34$ and $ 37$ $ \frac14\times\frac14\times\frac14$ cubes; this allows us to add $ 37$. This is very useful, since $ 19\equiv -2\mod 7$ and $ 37\equiv 2\mod 7$. The multiples $ 0,19,38,57,76,37,74$ form a complete system of residues mod $ 7$, and $ 69$ is the last number we can't get as a sum of copies of $ 7,19,37$. The new estimate is $ n_0\ge 71$.

Finally, there are ways to dissect a cube into $ 49$, $ 51$, or $ 54$ smaller cubes, which effectively replace my dissections into $ 77,$ $ 58,$ and $ 75$ cubes respectively. You can look them up at mszew's link, or try to find them yourself if you're feeling clever. That gets $ n_0$ down to $ 48$.
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davicoelhoamorim
2 posts
#5 • 1 Y
Y by Adventure10
We can partion any cube into 8 and 27 cubes. So, we can partion any cube into 8+7k, 8k+26, 27+7K and 27+7k.
8+26k and 27+26k form a complete set of residue classes modulo 7. So, we form all numbers greater than 152 (n0=152).. If we can partion any cube into x cubes, we can partion into x+7k
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