Hard inequality

by JK1603JK, Apr 29, 2025, 4:24 AM

Let $a,b,c>0$ and $a^2+b^2+c^2=2(a+b+c).$ Find the minimum $P=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

Interesting number theory

by giangtruong13, Apr 28, 2025, 4:15 PM

Let $a,b$ be integer numbers $\geq 3$ satisfy that: $a^2=b^3+ab$ . Prove that:
a) $a,b$ are even
b) $4b+1$ is a perfect square number
c) $a$ can’t be any power $\geq 1$ of a positive integer number

number theory

Hard Inequality Problem

by Omerking, Apr 28, 2025, 3:51 PM

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ is given where $a,b,c$ are positive reals. Prove that:
$\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}} \le \frac{3}{\sqrt{2}}$

Coincide

by giangtruong13, Apr 27, 2025, 4:05 PM

Let $ABCD$ be a trapezoid inscribed in circle $(O)$ , $AD||BC, AD < BC$ . Let $P$ is the symmetric point of $A$ across $BC$ , $AP$ intersects $BC$ at $K$ . Let $M$ is midpoint of $BC$ and $H$ is orthocenter of triangle $ABC$ . On $BD$ take a point $F$ so that $AF||HM$ . Prove that: $FK,AC,PD$ coincide

geometry

trapezoid

Arbitrary point on BC and its relation with orthocenter

by falantrng, Apr 27, 2025, 11:47 AM

In an acute-angled triangle $ABC$ , $H$ be the orthocenter of it and $D$ be any point on the side $BC$ . The points $E, F$ are on the segments $AB, AC$ , respectively, such that the points $A, B, D, F$ and $A, C, D, E$ are cyclic. The segments $BF$ and $CE$ intersect at $P.$ $L$ is a point on $HA$ such that $LC$ is tangent to the circumcircle of triangle $PBC$ at $C.$ $BH$ and $CP$ intersect at $X$ . Prove that the points $D, X,$ and $L$ lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus

This post has been edited 1 time. Last edited by falantrng, Sunday at 4:38 PM

geometry

BMO

Find points with sames integer distances as given

by nAalniaOMliO, Jul 17, 2024, 9:44 PM

Points $A_1, \ldots A_n$ with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points $B_1, \ldots ,B_n$ with integer coordinates such that $A_iA_j=B_iB_j$ for every pair $1 \leq i \leq j \leq n$
N. Sheshko, D. Zmiaikou

This post has been edited 1 time. Last edited by nAalniaOMliO, Oct 31, 2024, 10:12 AM

combinatorics

number theory

d | \overline{aabbcc} iff d | \overline{abc} where d is two digit number

by parmenides51, Mar 14, 2020, 2:23 PM

Determine the largest two-digit number $d$ with the following property:
for any six-digit number $\overline{aabbcc}$ number $d$ is a divisor of the number $\overline{aabbcc}$ if and only if the number $d$ is a divisor of the corresponding three-digit number $\overline{abc}$ .

Note The numbers $a \ne 0, b$ and $c$ need not be different.

A Weird but Interesting Distribution pt II

by 3ch03s, Mar 27, 2016, 12:47 AM

In the first part I define an interesting distribution. Let $X_n,n\in \mathbb{N}$ a bernoulli iid sequence with parameter $p \in ]0,1[$ , and let
$R=\sum_{n \in \mathbb{N}} \frac{X_n}{2^n}$

We called it $X\sim rm(p)$ .
We also showed some interesting properties of his probability distribution (cumulated) function, and we give some clues about the behavior of the distribution when p=1/2.

I claim when p=1/2, $R$ has the uniform distribution (which is in fact, the lebesgue measure in [0,1]). Also we give this property of cdf:

$F(r)=(1-p)F(2r)+pF(2r-1)$

Then we can compute any value of this distribution using dyadic aproximations of any number $r$ in [0,1]. Is pretty clear the sequence of aproximations has the form $x_n=\sum_{i=1}^n a_i/2^i$ . this implies that for any number of the form $r=k/2^n$ , with $k<2^n$ natural, F(r) has a "closed" form, because we are dealing with a finite number of bernoulli variables in that case.

we know every number $x$ in [0,1] has a unique (in the sense of unique sequence of 0's and 1's without getting finite 0's) binary representation, with this, we can compute $F(r)$ if $r=k/2^n$ , which using the recurssion formula leads to:

$F(k/2^n)=\sum_{\sum x_i/2^i \leq k/2^n} p^{\sum x_i}(1-p)^{n-\sum x_k}$

Another aproach to the formula is thinking in the secuence of 0 and 1 for the finite dyadic representation.
So for every aproximation $r_n$ we can compute using this recurssion:

$F(x_n)=F(x_{n-1})+a_i(p^{\sum x_i})(1-p)^{n-\sum x_i}$

¿What happens when $p=1/2$ ?

the formula for dyadic numbers just reduces to:

$F(k/2^n)=\sum_{\sum x_i/2^i \leq k/2^n} 1/2^n=k/2^n$ , hence, any dyadic number $r$ has the propierty $F(r)=r$ , then, the recursion for any dyadic approximation gives $F(x_n)=x_n$ , So in the limit, by density arguments that i will justificate soon, we have $F(x)=x$ for any $x \in [0,1]$ , so, for p=1/2, the distribution is in fact, the uniform distribution and hence has a pdf.

But... what about the image of X? are [0,1]?... in fact, because the representation, the uniform distribution does not go trough the whole interval, just, a select set of numbers in [0,1] with the property: $x \in [0,1], x=\sum x_n/2^n|\frac 1n \sum x_i \to 1/2$ .

So we define:

$A_p=\{ x \in {0,1}, x=\sum x_n/2_n, x_n \in \{0,1\}, \textrm{such that } \frac 1n\sum x_n \to p\}$

and hence $[0,1]=\bigcup_{p \in [0,1]} A_p$ .

if $X \sim rm(p)$ then the image of X is $A_p$ .
its clear that $A_p$ and $A_q$ are disjoint if $p \neq q$ . We will use this property using the fact if p=1/2, $X \sim U[0,1]$ , and hence the probability measure is in fact, the Lebesgue measure in [0,1], so, exists a pdf by radon-nykodym theorem. Let $\lambda$ the Lebesgue measure in [0,1], write:
$[0,1]=A_{1/2} \cup \bigcup_{p \neq 1/2} A_p$
Clearly $A_{1/2}$ and the rest of the union are disjoint.
so:
$\lambda([0,1])=1=\lambda (A_{1/2})+\lambda (\bigcup_{p \neq 1/2} A_p)$

Functional Equation

by JSGandora, Mar 17, 2013, 5:47 PM

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying
$f(x+f(y))=x+f(f(y))$
for all real numbers $x$ and $y$ , with the additional constraint $f(2004)=2005$ .

function

algebra

IMO ShortList 2008, Number Theory problem 3

by April, Jul 9, 2009, 10:27 PM

Let $a_0$ , $a_1$ , $a_2$ , $\ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\gcd (a_i, a_{i + 1}) > a_{i - 1}$ . Prove that $a_n\ge 2^n$ for all $n\ge 0$ .

Proposed by Morteza Saghafian, Iran

greatest common divisor

number theory

Sequence

IMO Shortlist

Impossible divisibility

by pohoatza, Jun 7, 2008, 5:21 PM

Let $m,\ n \geq 3$ be positive odd integers. Prove that $2^{m}-1$ doesn't divide $3^{n}-1$ .

modular arithmetic

number theory

Quadratic Residues

quadratic reciprocity

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hola. Mi nombre es claudio, soy chileno (y la ctm) Ingeniero civil Matematico con menci??n en estadistica aceptado en el doctorado de Estadpistica de la puc. Mucho Floyd Buki TB-303 y Lacie Baskerville y disco house.
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9 shouts

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