[b]Problem 1[/b]
Let $ABC$ be a triangle with $\widehat{ABC}+\widehat{ACB}=60^\circ$. Suppose the internal bisector of angle $\widehat{BAC}$ meets the circumcircle of triangle $ABC$ at $P$. Prove that $AP> BC$.

[b]Problem 2[/b]
Triangle $ABC$ with area $\mathcal{A}$ has an obtuse angle. Let $d$ denote the length of its longest vertice. Show that $\sqrt{\mathcal{A}}\leq\frac{d}{2}$. When does equality occur?

For #2 I am assuming you meant longest edge instead of longest vertex

[hide="2"]

Let A be the obtuse angle. The side oppose it is the longest side.

I need to show that $A\le\frac{d^{2}}{4}$

The obtuse angle is below the line cutting the square in half. (consider a semi-circle with diameter BC; if A is inside it, it is obtuse; on it it, it is right, outside it, it is acute)

if the height of the triangle is h, then $h<\frac{d}{2}$ and then

$A=\frac{dh}{2}<\frac{d^{2}}{4}$

(equality occurs if A is on the line bisecting the square, which can only happen if A is acute or right. So equality cannot occur.)

[asy]size(100); pair A,B,C,D,E; A=(60,35); B=(0,0); C=(100,0); D=(100,100); E=(0,100); draw(B--C--D--E--cycle); draw(B--A--C); draw((0,50)--(100,50)); label("A",A,N); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("d",(50,0),S);[/asy]

[/hide]

I assumed that the angle was not strictly obtuse, so equality occurs iff $ABC$ is an isoceles right triangle.

There is by the way another way to prove problem 2, which i personnally find quite beautiful

[hide="Question 2"]
On the one hand, it is known that, in a triangle with an obtuse angle, we have $a^{2}+b^{2}\leq c^{2}$, where $c$ is the longest side. On the other hand, we have $a^{2}+b^{2}\geq 2ab\geq 2ab\cdot\sin{\alpha}=\mathcal{A}$ where $\alpha$ is the obtuse   angle. Indeed, we have $\sin{\alpha}\leq 1$. Then, by combining these tow identities, we are done.
[/hide]

[hide="1"]
The condition is simply $\angle BAC=120^\circ$, so $\angle BAP=\angle CAP=60^\circ$. But we're also given quadrilateral $ABPC$ cyclic, so $\angle BCP=\angle BAP$ and $\angle CBP=\angle CAP$. This means that $\triangle BCP$ is equilateral!

Now we have $BC=BP$, so the problem is reduced to $AP>BP$, which follows from
\[\angle ABP=\angle ABC+\angle BCP>\angle BCP=\angle BAP\]
[/hide]

For problem 1, it is a more or less well-known theorem (by the way easily established using Ptolemy) that for any point (here $A$) on the circumcircle of an equilateral triangle (here $PBC$), we have $AP = AB+AC > BC$ by triangular inequality.


limit
inequalities
factorial
real analysis
real analysis unsolved

Let $ K_{k}\equal{}\sqrt{k}\frac{1\cdot{3}\cdot{5}\cdots{(2k\minus{}3)}\cdot{(2k\minus{}1)}}{2\cdot{4}\cdot{6}\cdots{(2k\minus{}2)}\cdot{(2k)}}$. $ K_{k}>{0}$ is trivial for all natural number $ k$.

$ K_{k}^2\equal{}k\cdot\frac{1\cdot1\cdot3\cdot3\cdot5\cdot5\cdots{(2k\minus{}3)}\cdot{(2k\minus{}3)}\cdot{(2k\minus{}1)}\cdot{(2k\minus{}1)}}{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots{(2k\minus{}2)}\cdot{(2k\minus{}2)}\cdot{(2k)}\cdot{(2k)}}$
$ <k\cdot\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots{(2k\minus{}3)}\cdot{(2k\minus{}2)}\cdot{(2k\minus{}1)}\cdot{(2k)}}{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdots{(2k\minus{}2)}\cdot{(2k\minus{}1)}\cdot{(2k)}\cdot{(2k\plus{}1)}}$.       We used the inequality $ \frac{n}{m}<\frac{n\plus{}1}{m\plus{}1}\quad (when\quad{m,n>0, \frac{n}{m}<1})$.

By telescoping $ 0<K_{k}^2<\frac{k}{2k\plus{}1}$. So, we get $ 0<\lim_{k\rightarrow\infty}K_{k}<\frac{1}{\sqrt{2}}<\infty$.

[quote="daniel1024"]$ 0 < \lim_{k\rightarrow\infty}\sqrt {k}\frac {2k \minus{} 1}{2k}\frac {2k \minus{} 3}{2k \minus{} 2}\frac {2k \minus{} 5}{2k \minus{} 4}...\frac {1}{2} < \infty$[/quote]

Use factorial and Stirling for an equivalent of $ n!$


The edge lengths of a right rectangular prism are in the ratio of 2:4:5. The total surface area of the solid is 1900 square cm. What is the volume of the solid?

this question i don't think i have gotten right on FTW yet, so idk if this solution is correct

let's say that the edge lengths of the prism are those numbers, and call this our pretend prism. therefore, the total surface area would be...
$ 2 \times 4 \times 2 \plus{} 2 \times 5 \times 2 \plus{} 4 \times 5 \times 2 \equal{} 76$

since the actual surface area of the prism is 1900 square sm, then that means that the ratio of our pretend prism to the actual prism is $ \frac {1} {25}$, so we have to increase each of the edges in our pretend prism by 25
since the edges of our pretend prism are 2, 4, and 5, we have to multiply those by 25, making the side lengths of the actual prism 50, 100, and 125
from there, we multiply those numbers together to get the volume
$ 50 \times 100 \times 125 \equal{} 625000$

as i said, i haven't gotten this right on FTW yet, so would someone plz confirm

No, because it's area, the factor is squared , so you only multiply by 5

oh
thanks mewto
that means the ratio of the sides is $ \frac {1} {5}$
so we multiply the sides by 5

so the sides of the actual prism is 10, 20, and 25, which means the volume is
$ 10 \times 20 \times 25 \equal{} 5000$

thanks mewto!


[quote="arqady"][quote="ali666"]let \[x,y,z\] be positive number such that \[{0}\leq{x+y+z}\leq{\frac{1}{3}}\] show that: \[\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq{\frac{\sqrt{2}}{2}}\] [/quote]
It is not true. Try $x\rightarrow0^{+},$ $y\rightarrow\frac{1}{4}^{-}$ and $z=\frac{1}{12}.$ :wink:[/quote]
sorry :oops:


number theory
least common multiple
superior algebra
superior algebra solved

This a short, nice and quite easy (if you get the right idea) problem:

Let A be a finite ring. Prove that there exist natural numbers m>p>=1 s.t. a m = a p for all a in A.

Don't misunderstand this: the same m and p have to work for all a in A :D

[i]Romanian NMO/final round/Targu Mures 2001/problem 2/12'th grade/proposed by Ion Savu[/i]

Normally when we have a finite ring we consider x->ax or a^j, in the
problem we see exponent so consider the second one.

Let A={a_1,a_2,...,a_s}
{j\in N,  (a_1)^j} is  in A finite there exist m_1, p_1 \in N s.t
(a_1)^m_1=(a_1)^p_1
......................................................................................

{j\in N,  (a_s)^j} is  in A finite there exist m_s, p_s \in N s.t
(a_s)^m_s=(a_s)^p_s

m,p is  depend of (m_1,...,m_s,p_1,...,p_s)

for all a \in A, a^m=a^p
But I have no idea how to built m,n .
lcm ?

I'll give you 2 solns. Here's mine (the other one, from the booklet where I found the problem, a bit later):

[color=white]

We have some m_i>p_i>=1 s.t. a_i m_i = a_i p_i for all i from 1 to n,. where a_i are the elmts of the ring. We may assume m_i>=2p_i, because if m_i<2p_i we just leave p_i the same and choose 2m_i-p_i as the new m_i. We repeat this to make m_i as large as we want, so we can eventually make it >=2p_i. Thus, we may assume m_i>=2p_i.

Take t_i=m_i-2p_i>=0. We have a_i k_i = a_i 2k_i , where k_i=p_i+t_i. this holds for all i from 1 to n.

We just take m= 2* \pi k_i and p= \pi k_i.

[/color]

Here's the other solution:

[color=white]

For all a_i in A there exist s_i, t_i s.t. a_i s_i = a_i s_i+t_i, so a_i k = a_i k+t_i*t for all k>=s_i and t in N. We take p>=max{s_1, s_2, .., s_n} and q= lcm{t_1, t_2, .., t_n} and we have a p = a p+q for all a in A, thus we may take m=p+q.

[/color]

This one's shorter.


Les paso este problema, trae 2 dificultades, pero en escencia es el mismo. No vean la versión fácil a menos que no les salga la versión difícil.
Versión difícil
Sea ABC un triángulo y P un punto en su interior. Se construye el punto M fuera del triángulo de manera que $<ABC=<CBM$ y que $<ACP=<BCM$ Supóngase que el área del triángulo ABC es la misma que del cuadrilátero PCMB. Demostrar que BMC es semejante a ACP

Versión Fácil:
[hide] Sea ABC un triángulo y P un punto en su interior. La paralela a BC por P intersecta a AB en E y al circuncírculo de APC en D. Supóngase que el área del triángulo ABC es la misma que la suma de las áreas de AED y PBC. Demostrar que EAD es semejante a ACP[/hide]

Como $ \angle ABC \equal{} \angle CBM,$ la simetría axial respecto a $ BC$ lleva $ M$ a un punto $ Q$ sobre $ AB$  tal que $ CQ$ es isogonal de $ CP$ respecto a $ \angle C,$ ya que $ \angle ACP \equal{} \angle BCM \equal{} \angle QCB.$  Entonces $ [PCMB] \equal{} [ABC]$ implica:

$ [PBC] \plus{} [BQC] \equal{} [APB] \plus{} [APC] \plus{} [PBC]$

$  [BQC] \equal{} [APB] \plus{} [APC] \equal{} [ABC] \minus{} [PBC] \equal{} [BQC] \plus{} [AQC] \minus{} [BPC]$

$[AQC] \equal{} [BPC] \Longrightarrow AC \cdot QC \sin \angle ACQ \equal{} PC \cdot BC \sin \angle PCB$

Pero como $ CP$ y $ CQ$ son rayos isogonales con respecto al ángulo $ C,$ entonces $ \angle ACQ \equal{} \angle PCB.$ Así

$ \frac {QC}{PC} \equal{} \frac {BC}{AC}  \Longrightarrow \ \triangle CMB \cong \triangle CQB \sim \triangle CPA.$

Mi solución es muy parecida sólo que uso la idea clave hacia afuera como se podrá observar en la figura adjunta:
Dado que $ \Delta ABC\cong\Delta A'BC$, lo que hace que $ [ABC] \equal{} [A'BC]$  luego $ [BPCM] \equal{} [A'BC]$ por lo que $ [BCP] \equal{} [MA'C]$  luego como ambos triángulos poseen un ángulo igual se usa la propiedad de áreas
$ \frac {[BCP]}{[MA'C]} \equal{} \frac {PCxBC}{A'CxMC} \equal{} (\frac {PC}{AC})x(\frac {BC}{MC}) \equal{} 1 \Longrightarrow\ \Delta PCA\sim\Delta MCB$

Nota:
La propiedad mencionada puede ser usada para áreas de triángulos con ángulos iguales o suplementarios, más aún este teorema se puede generalizar para volumenes de piramides cuyo ángulo solido sea el mismo.

_____________________________________________________
[color=green][b]Centro Juvenil de Entrenamiento Olímpico[/b][/color]
http://www.cjeo-peru.com

[b][i]Enunciado:[/i][/b]
[b]Sea ABC un triángulo y P un punto en su interior. La paralela a BC por P intersecta a AB en E y al circuncírculo de APC en D. Supóngase que el área del triángulo ABC es la misma que la suma de las áreas de AED y PBC. Demostrar que EAD es semejante a ACP[/b]
[b][i]Solución[/i][/b]
Como ED//BC  entonces $ [BEC] \equal{} [BPC]$ y siendo que $ [ABC] \equal{} [AED] \plus{} [PBC]$
Tenemos que $ [BEC] \plus{} [AEQ] \plus{} [EQC] \equal{} [AEQ] \plus{} [QAD] \plus{} [BEC]$
por lo que: $ [EQC] \equal{} [AQD]$
Así de esta forma EA//CD por tanto $ \angle AED \equal{} \angle EDC$ y puesto que APCD es conciclico concluimos que $ \Delta EAD \sim \Delta APC$

Nota:
A propio criterio personal me ha gustado más este ejercicio que el de la versión dificil, más aún me parece muy educativo.

_______________________________________________________
[color=green][b]Centro Juvenil de Entrenamiento Olímpico[/b][/color]
http://www.cjeo-peru.com


[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point inwardly to it. The line segments $ AP,\ BP,\ CP,$ intersect its circumcircle $ (O),$ at points $ A',\ B',\ C'$ respectively and let $ D,\ E,\ F$ be, the midpoints of the side-segments $ BC,\ AC,\ AB,$ respectively. We denote the points $ D'\equiv B'C'\cap DP,$ and $ E'\equiv A'C'\cap EP,$ and $ F'\equiv A'B'\cap FP.$ Prove that the line segments $ A'D',\ B'E',\ C'F',$ are concurrent at one point so be it $ Q,$ lies on the line segment $ GP,$ where $ G$ is the centroid of $ \bigtriangleup ABC.$[/color][/size][/b]

Kostas Vittas.

Let ABC a triangle and let P and Q two points on the same plane of ABC. Let D, E, F the intersection of AP, BP, CP with BC, CA, AB and let D', E', C' the intersection of AQ, BQ, CQ with the circumcircle of ABC. Let R, S, T the intersection of QD, QE e QF with E'F', F'D' e D'E'.

1) Prove that D'R, E'S, F'T concur.
2) Call K the point of concurrence, so prove that K, P, Q are collinear.

[url=http://img519.imageshack.us/img519/9977/generalizationofproblemgo9.png]figure of the problem[/url]

[quote="¬[ƒ(Gabriel)³²¹º]¼"]Let ABC a triangle and let P and Q two points on the same plane of ABC. Let D, E, F the intersection of AP, BP, CP with BC, CA, AB and let A', B', C' the intersection of AQ, BQ, CQ with the circumcircle of ABC. Let R, S, T the intersection of QD, QE e QF with E'F', F'D' e D'E'.

1) Prove that A'R, B'S, C'T concur.
2) Call K the point of concurrence, so prove that K, P, Q are collinear.[/quote]
Thank you dear [b]Gabriel[/b] for your interest.

Although I know that it is true (by the drawing) the generalization of the problem you mentioned, however I have not in mind any proof about it.

Best regards, Kostas Vittas.

Hello!

The Gabriel's generalization of the Vittasko's problem:
Let $ ABC$ a triangle and let $ P$ and $ Q$ two points on the same plane of $ ABC$.
Let $ D,E,F$ the intersection of $ AP, BP, CP$ with $ BC, CA, AB$ and let $ A', B', C'$
the intersection of $ AQ, BQ, CQ$ with the circumcircle of $ ABC$. Let $ D', E', F'$
the intersection of $ QD, QE, QF$ with $ B'C', C'A',  A'B'$.
1. Prove that $ A'D', B'E', C'F'$ concur.
2. Call $ P'$ the point of concurrence, so prove that $ P, Q, P'$ are collinear.

Solution:
1. $ \frac{D'B'}{D'C'}\equal{}\frac{QB'\cdot sin\left(\angle D'QB'\right)}{QC'\cdot sin\angle\left(D'QC'\right)}\equal{}\frac{QC\cdot sin\left(\angle DQB\right)}{QB\cdot sin\angle\left(DQC\right)}\equal{}$
$ \equal{}\frac{QC}{QB}\cdot\frac{\frac{DB}{QB}}{\frac{DC}{QC}}\equal{}\frac{QC^{2}}{QB^{2}}\cdot\frac{DB}{DC}\Rightarrow\frac{D'B'}{D'C'}\equal{}\frac{QC^{2}}{QB^{2}}\cdot\frac{DB}{DC}$   (1)
Similarly $ \Rightarrow\frac{E'C'}{E'A'}\equal{}\frac{QA^{2}}{QC^{2}}\cdot\frac{EC}{EA}$   (2) and $ \frac{F'A'}{F'B'}\equal{}\frac{QB^{2}}{QA^{2}}\cdot\frac{FA}{FB}$   (3)

(1), (2), (3) $ \Rightarrow\frac{D'B'}{D'C'}\cdot\frac{E'C'}{E'A'}\cdot\frac{F'A'}{F'B'}\equal{}\frac{QC^{2}}{QB^{2}}\cdot\frac{DB}{DC}\cdot\frac{QA^{2}}{QC^{2}}\cdot\frac{EC}{EA}\cdot\frac{QB^{2}}{QA^{2}}\cdot\frac{FA}{FB}\equal{}$
$ \equal{}\frac{DB}{DC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}\equal{}1\Rightarrow A'D', B'E', C'F'$ are concurrent.

2.Let $ P'$ be the point of concurrency of $ A'D', B'E', C'F'$ and $ \left\{P"\right\}\equal{}PQ\cap A'D'$. Let's prove that $ P'\equal{}P"$.
$ \frac{P'A'}{P'D'}\equal{}\frac{A'B'\cdot sin\left(\angle A'B'E'\right)}{D'B'\cdot sin\left(\angle C'B'E'\right)}\equal{}\frac{A'B'\cdot\frac{E'A'}{A'B'}}{D'B'\cdot\frac{E'C'}{B'C'}}\equal{}$
$ \equal{}\frac{E'A'}{E'C'}\cdot\frac{B'C'}{D'B'}\equal{}\frac{QC^{2}}{QA^{2}}\cdot\frac{EA}{EC}\cdot\frac{B'C'}{D'B'}\Rightarrow$
$ \Rightarrow\frac{P'A'}{P'D'}\equal{}\frac{QC^{2}}{QA^{2}}\cdot\frac{EA}{EC}\cdot\frac{B'C'}{D'B'}$   (4)

$ \frac{P"A'}{P"D'}\equal{}\frac{QA'\cdot sin\left(\angle A'QP"\right)}{QD'\cdot sin\left(\angle D'QP"\right)}\equal{}\frac{QA'}{QD'}\cdot\frac{sin\left(\angle AQP\right)}{sin\left(\angle DQP\right)}\equal{}$
$ \equal{}\frac{QA'}{QD'}\cdot\frac{\frac{PA}{QA}}{\frac{PD}{QD}}\equal{}\frac{QD}{QD'}\cdot\frac{PA}{PD}\cdot\frac{QA'}{QA}\Rightarrow$
$ \Rightarrow\frac{P"A'}{P"D'}\equal{}\frac{QD}{QD'}\cdot\frac{PA}{PD}\cdot\frac{QA'}{QA}$   (5)

$ \frac{QD}{QD'}\equal{}\frac{DB\cdot sin\left(\angle QBC\right)}{D'B'\cdot sin\left(\angle QB'C'\right)}\equal{}\frac{DB}{D'B'}\cdot\frac{sin\left(\angle QBC\right)}{sin\left(\angle QCB\right)}\Rightarrow$
$ \Rightarrow\frac{QD}{QD'}\equal{}\frac{DB}{D'B'}\cdot\frac{QC}{QB}$   (6)

$ \frac{PA}{PD}\equal{}\frac{AB\cdot sin\left(\angle ABE\right)}{DB\cdot sin\left(\angle CBE\right)}\equal{}\frac{AB}{DB}\cdot\frac{\frac{EA}{AB}}{\frac{EC}{BC}}\equal{}\frac{BC}{DB}\cdot\frac{EA}{EC}\Rightarrow$
$ \Rightarrow\frac{PA}{PD}\equal{}\frac{BC}{DB}\cdot\frac{EA}{EC}$   (7)

$ \frac{QA'}{QA}\equal{}\frac{QA'\cdot QA}{QA^{2}}\equal{}\frac{QC'\cdot QC}{QA^{2}}\equal{}\frac{\frac{QB\cdot B'C'}{BC}\cdot QC}{QA^{2}}\Rightarrow$
$ \Rightarrow\frac{QA'}{QA}\equal{}\frac{QB\cdot QC\cdot B'C'}{BC\cdot QA^{2}}$   (8)

(5), (6), (7), (8) $ \Rightarrow\frac{P"A'}{P"D'}\equal{}\frac{DB}{D'B'}\cdot\frac{QC}{QB}\cdot\frac{BC}{DB}\cdot\frac{EA}{EC}\cdot\frac{QB\cdot QC\cdot B'C'}{BC\cdot QA^{2}}\Rightarrow$
$ \Rightarrow\frac{P"A'}{P"D'}\equal{}\frac{QC^{2}}{QA^{2}}\cdot\frac{EA}{EC}\cdot\frac{B'C'}{D'B'}$   (9)

(4), (9) $ \Rightarrow\frac{P'A'}{P'D'}\equal{}\frac{P"A'}{P"D'}\Rightarrow P'\equal{}P"$. So $ P'\in PQ$.

Best regards, Petrisor Neagoe :)


algebra
polynomial
geometry
inradius
circumcircle
national olympiad

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=140292
Let $a_{1}, a_{2}, \ldots, a_{n}\in [0;1]$. If $S=a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}$ then prove that
\[\frac{a_{1}}{2n+1+S-a_{1}^{3}}+\frac{a_{2}}{2n+1+S-a_{2}^{3}}+\ldots+\frac{a_{n}}{2n+1+S-a_{n}^{3}}\leq \frac{1}{3}. \]
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=140291
Find all polynomials $f\in \mathbb{Z}[X]$ such that if $p$ is prime then $f(p)$ is also prime.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=140293
Consider a triangle $ABC$, with corresponding sides $a,b,c$, inradius $r$ and circumradius $R$. If $r_{A}, r_{B}, r_{C}$ are the radii of the respective excircles of the triangle, show that
\[a^{2}\left(\frac 2{r_{A}}-\frac{r}{r_{B}r_{C}}\right)+b^{2}\left(\frac 2{r_{B}}-\frac{r}{r_{A}r_{C}}\right)+c^{2}\left(\frac 2{r_{C}}-\frac{r}{r_{A}r_{B}}\right)=4(R+3r). \]
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=140290
We are given $n$ distinct points in the plane. Consider the number $\tau(n)$ of segments of length $1$ joining pairs of these points. Show that $\tau(n)\leq \frac{n^{2}}3$.

Thanks to iura  :wink:


Rewrite this sum as $\sum_{j=2}^\infty\sum_{k=2}^\infty\frac{j!\cdot k!}{(j+k)!}$.
Now, treating $j$ as a constant, the inner sum telescopes:
$\sum_{k=2}^\infty\frac{j!}{(k+1)(k+2)\cdots(k+j)}= \sum_{k=2}^\infty j!\cdot\frac1{j-1}\cdot\left(\frac1{k+1}-\frac1{k+j}\right) \cdot\frac1{(k+2)(k+3)\cdots(k+j-1)}$
$=\frac{j!}{j-1}\left(\frac1{3\cdot4\cdots(j+1)}\right)=\frac2{(j-1)(j+1)}$
$\sum_{j=2}^\infty\frac2{(j-1)(j+1)}=\sum_{j=2}^\infty\frac1{j-1}-\frac1{j+1}=1+\frac12.$


floor function
function
ceiling function
algebra

I see a lot of thing say the fractional part of $ x$ and it represented like this $ \left\{ x \right\}$ and I see also the integer part and it represented like this $ \left\lfloor x \right\rfloor$
So what this should mean and give example if possible  :)
Thanks in advance  :)

You got the two mixed up.

Integer part: $ \lfloor x\rfloor$ or $ [x]$ are used.
Fractional part: $ \{x\}$ is used.

So we always have $ 0\le\{x\}<1$ and $ \lfloor x\rfloor\in\mathbb Z$ such that $ x\equal{}\lfloor x\rfloor\plus{}\{x\}$.

Like $ 5.6\equal{}\lfloor5.6\rfloor\plus{}\{5.6\}\equal{}5\plus{}0.6\equal{}5.6$, and $ \minus{}5.6\equal{}\lfloor\minus{}5.6\rfloor\plus{}\{\minus{}5.6\}\equal{}\minus{}6\plus{}0.4\equal{}\minus{}5.6$.

The integer part of a number is defined as the greatest integer that is less than or equal to a number. (This is called the greatest integer function or floor function.)

It's represented as either $ [x]$ or $ \lfloor x\rfloor$.

$ \lfloor 1.1\rfloor \equal{} 1$
$ \lfloor \minus{}1.1\rfloor \equal{} \minus{}2$
$ \lfloor 1\rfloor \equal{} 1$
$ \lfloor \pi\rfloor \equal{} 3$

There is also the least integer function, which refers to the least integer that is greater than or equal to a number.

$ \lceil 1.1\rceil \equal{} 2$
$ \lceil \minus{}1.1\rceil \equal{} \minus{}1$
$ \lceil 1\rceil \equal{} 1$
$ \lceil \pi\rceil \equal{} 4$

Finally, the fractional part is defined as follows: $ \{x\} \equal{} x \minus{} \lfloor x\rfloor$.

$ \{1.1\} \equal{} 0.1$
$ \{\minus{}1.1\} \equal{} 0.9$
$ \{1\} \equal{} 0$

Hope that helps.

Thanks a lot "math154" and "sunehra" you explanations was wonderful    :D
And for "math154" I change the mistake in the my first message thanks for knowing it  :)

One more thing can it be $ \left\lfloor x \right\rfloor  \cdot \left\{ x \right\}\equal{}?$ LOL


I am starting (or planning :P ) a rating system for mathematically talented young minds (a.k.a. most of you) in USA (will later be extended).

There will be provisional ratings which you will start with, and they may be adjusted:

choose one of the following (one that gets you the highest rating):

(i) 1200 +(chapter MC score)+ (state MC score) *2 + (nats MC score)*3
(ii)1200 +(AMC score; -20 if AMC 10) +(AIME score)*10 + (USAMO score)*10
(iii) 1200 + 12*(ARML score) + 8* (highest USAMTS 1 round score)

My rating would be 1368, through option (ii). My other ratings are 1330 and 1284

Please post your ratings and I will pair you up accordingly.

I will get 16 people, and I will provide problems (MC level to hard AIME level)

3 problems per round and I hope all of you join!

My highest is 1200. I have only done AMC 8 and I don't know my scores, although it would not have mattered much.

I misunderstood at first danny. I thought this was a ftw tourney. Since it is not ftw as I thought as I was talking to you today, I figure I have time for this. My rating is 1349.5.

Hunter: I thought you said you only made a 2 on aime this year......... Or did you take the amc's/aime last year too?

MY RATING IS 1337!!!! (not really) its not that great. 1353, which is ~150 less than my goal.

i think MC should be worth more b/c if you look at my goals for this year, im going to get 1500+ using AMC. lets say i get 46 on all the mc tests, then its 1476, so my actual MC rating wont even be close to that.

I wonder if you could revise the rating format a bit?  maybe letting us choose which couple of events we would like to use towards our rating, and changing the multiplying factors accordingly?  in the current system, for eg, my arml score is unusable b/c I didn't do USAMTS.

using the current system, my rankings r:

1. 1398
2. 1360
3. 1248

btw, I mite participate in the tourney, idk, I posted here cuz i like the rating idea  :D

I'll join, I guess.

1443, 1458.5, 1448

Apparently, they are surprisingly (and coincidentally) close.

I didn't do AMC last year, so my best rating through option (i) is...

1297.  :noo:

[quote="hunter34"]hurdler, I am counting both AMC 10 scores (132,150) and the AIME score.[/quote]

You only choose one. So your rating would be 1352.

EDIT:

Current signups:

1. hunter34
2. Alysero
3. hurdler
4. abacadaea
5. Quickster94
6. andersonw
7. pythag011
8. lotsofmath

Halfway!

um, sry to say, I will not be able to participate, I've got way too much other math stuff to do, sry  :(

(i) 1200 (didn't do MC yet)

(ii) 1364

(iii) 1248

So option 2 I GUESS.

Hmm, bogtro, I thought you didn't make AIME?

Wait. Are these due TOMMOROW!?!?!??!

Or is it a wekk from tomorrow.

OK yay more time.

I think I shall not procrastinate til next weekend and do these tonight.

How come no one got eliminated during the last round? I was pretty sure to be beaten. And now I'm facing pythag011...oh well let's hope he makes stupid mistakes  :P

there is a competition between 3rd place and 4th place so you are still in

and pythag011 better be triv :/

AGHGHGHGHH!?!?!?

20 WAS ACTUALLY ATTAINABLE!?!?!?!?!?

I've never seen a problem where an incredibly easy to get maximum is attainable....

ARGARGARGARGARGARGARGARGARGARGARG

[quote="pythag011"]AGHGHGHGHH!?!?!?

20 WAS ACTUALLY ATTAINABLE!?!?!?!?!?
[/quote]

Lol I was like that too. I spent so much time trying to get 20, but always failed around 16, so I put that as my answer.

Please extend deadline.

*cough*iTest*cough*

itest was pretty fun. I shall hopefully not be lazy now and just go ahead and type up what I have for this round

It's ok. I haven't even really started yet. These don't look as hard as last round's, but they definitely will take me a while.

yeah time

you can have more time until September 27th 12:00 A.M EDT.

have fun (oh, btw, 3 is noted to be incredibly hard; i haven't solved it yet D:)

BTW, also use this problem as a bonus:

if the prime factorization of a number m is $ p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$

Define f(m) as $ e_1^{p_1}e_2^{p_2}\ldots e_k^{p_k}$

This problem should be pretty hard.

of course this is a hard question. No one in the world could solve it - :) Pythag, you never asked the question - you just gave some info

You're right. The question is to find all n such that f(n) = n.

Haven't started yet  :mad:

Too much math going on...


Provide a synthesis of cyclohexanecarboxilic acid from benzene.

my pitiful guess:

1. Gatterman-Koch Formylation. Benzene -> benzaldehyde.
2. Catalytic Hydrogenation of Benzaldehyde. -> hydroxymethylcyclohexane.
3. Bichromic acid. -> cyclohexanecarboxylic acid.

Yes, that's correct, but the purpose was to use benzene as the source of all carbons.

this mite sound stupid but what do you mean. the extra carbon ought to come frm benzene? :maybe: coz that doesnt seem like a good synthetic approach.(no offense)

Yes, that's right: all carbons must come from benzene. Of course this synthesis would never be done in the laboratory: it's purpose (and of many organic synthesis problems) is simply to make students think of the several options available when confronted with real, hard synthetic problems.

Should I post my proposal?

Is the extra carbon generated by coupling benzene then oxidising with acidic KMnO4?

with itself(benzene) using diazotisation

Oxidise the biphenyl with acidic KMnO4 => benzoic acid is got

I don't think the oxidation of biphenyl with KMnO4 will yield cleanly benzoic acid. At least, harsh conditions must be employed, and in that case a mixture of products is to be expected. There is a much better way.


Let $ a,b,c$ be distinct nonnegative real numbers. Prove that

$ (a) \ \ \ \  (\frac {b\plus{}c\minus{}2a}{b\minus{}c})^2\plus{}(\frac {c\plus{}a\minus{}2b}{c\minus{}a})^2\plus{}(\frac {a\plus{}b\minus{}2c}{a\minus{}b})^2 \ge 18$;

$ (b) \ \ \ \  (\frac {b\plus{}c\minus{}a}{b\minus{}c})^2\plus{}(\frac {c\plus{}a\minus{}b}{c\minus{}a})^2\plus{}(\frac {a\plus{}b\minus{}c}{a\minus{}b})^2 \ge 8$.

[quote="Vasc"]Let $ a,b,c$ be distinct nonnegative real numbers. Prove that

$ (a) \ \ \ \ (\frac {b \plus{} c \minus{} 2a}{b \minus{} c})^2 \plus{} (\frac {c \plus{} a \minus{} 2b}{c \minus{} a})^2 \plus{} (\frac {a \plus{} b \minus{} 2c}{a \minus{} b})^2 \ge 18$;[/quote]
$ \Leftrightarrow (\frac {2(b \minus{} c)}{a \minus{} b} \plus{} \frac {2(c \minus{} a)}{b \minus{} c} \plus{} \frac {2(a \minus{} b)}{c \minus{} a} \plus{} 3)^2\ge 0$, It is obvious true.

[quote="Vasc"]$ (b) \ \ \ \ (\frac {b\plus{}c\minus{}a}{b\minus{}c})^2\plus{}(\frac {c\plus{}a\minus{}b}{c\minus{}a})^2\plus{}(\frac {a\plus{}b\minus{}c}{a\minus{}b})^2 \ge 8$[/quote]

$ \Leftrightarrow (\frac{b\plus{}c\minus{}a}{b\minus{}c}\plus{}\frac{c\plus{}a\minus{}b}{c\minus{}a}\plus{}\frac{a\plus{}b\minus{}c}{a\minus{}b})^2\ge0$, It is obvious, too.

I have found a general problem (of your first ineq) from tranquoc's one. See it here: (#8)  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=256766
@Vasc: [hide]In our new book: "MIC Shotlist and topics of ineqs" , we have a topic named "The strange ineqs" contain the ineq like that. And in this topic, I have about 15 strange ineqs. After the book finish, we very want send to you some chapters of book for your comments. Thank you before, my Bro  :) [/hide]

[quote="Vasc"]
If $ a,b,c$ are real numbers, then

$ \frac {(a \minus{} b)^2}{(b \minus{} c)^2} \plus{} \frac {(b \minus{} c)^2}{(c \minus{} a)^2} \plus{} \frac {(c \minus{} a)^2}{(a \minus{} b)^2} \ge 5$.
[/quote]
I have seems it already. It is obvious because it is equivalent with
$ (\frac{a\minus{}b}{b\minus{}c}\plus{}\frac{b\minus{}c}{c\minus{}a}\plus{}\frac{c\minus{}a}{a\minus{}b}\plus{}1)^2\ge0$.
But I don't know how I can solve the inspired from Can_hang's one..

And nguoivn, very nice generation! :lol:

[quote="Vasc"]Yes, NextPeace. :lol:
I tested my theorem for this strong inequalities and I found that it solves them. :wink:

The following is a little strange:

If $ a,b,c$ are real numbers, then

$ \frac {(a \minus{} b)^2}{(b \minus{} c)^2} \plus{} \frac {(b \minus{} c)^2}{(c \minus{} a)^2} \plus{} \frac {(c \minus{} a)^2}{(a \minus{} b)^2} \ge 5$.

On the other hand, I think the following inequality (inspired from Can_hang's one) is less strong but more hard:

If $ a,b,c$ are the side-lengths of a non-isosceles triangle, then

$ \sum \frac {4abc \minus{} a^3}{(b \minus{} c)^2} \ge 3(a \plus{} b \plus{} c)$.[/quote]
The inequality is equivalent to
\[ \sum \frac{a(b\plus{}c\minus{}a)}{(b\minus{}c)^2} \ge 4.\]
Setting $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y$ where $ x,y,z \ge 0$. THis inquality becomes
\[ \frac{x(y\plus{}z)}{(y\minus{}z)^2} \plus{}\frac{y(z\plus{}x)}{(z\minus{}x)^2} \plus{}\frac{z(x\plus{}y)}{(x\minus{}y)^2} \ge 2,\]
which is trivial, since if we assume $ z\equal{}\min\{x,y,z\}$, then
\[ \frac{x(y\plus{}z)}{(y\minus{}z)^2} \plus{}\frac{y(z\plus{}x)}{(z\minus{}x)^2} \ge \frac{x(y\plus{}z)}{y^2}\plus{}\frac{y(x\plus{}z)}{x^2} \ge \frac{xy}{y^2}\plus{}\frac{xy}{x^2} \ge 2.\]
Very nice inequality, VasC. ;) :)

[quote="vasc"]
I think the following inequality (inspired from Can_hang's one) is less strong but more hard:
If $ a,b,c$ are the side-lengths of a non-isosceles triangle, then
$ \sum \frac {4abc \minus{} a^3}{(b \minus{} c)^2} \ge 3(a \plus{} b \plus{} c)$.
[quote="can_hang2007"] The inequality is equivalent to
\[ \sum \frac {a(b \plus{} c \minus{} a)}{(b \minus{} c)^2} \ge 4.
\]
Setting $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y$ where $ x,y,z \ge 0$. THis inquality becomes
\[ \frac {x(y \plus{} z)}{(y \minus{} z)^2} \plus{} \frac {y(z \plus{} x)}{(z \minus{} x)^2} \plus{} \frac {z(x \plus{} y)}{(x \minus{} y)^2} \ge 2,
\]
which is trivial, since if we assume $ z \equal{} \min\{x,y,z\}$, then
\[ \frac {x(y \plus{} z)}{(y \minus{} z)^2} \plus{} \frac {y(z \plus{} x)}{(z \minus{} x)^2} \ge \frac {x(y \plus{} z)}{y^2} \plus{} \frac {y(x \plus{} z)}{x^2} \ge \frac {xy}{y^2} \plus{} \frac {xy}{x^2} \ge 2.
\]
Very nice inequality, VasC. ;) :) [/quote][/quote]
Nice solution, Can_hang.
I think the following generalization holds:

Let $ a,b,c$ be the side-lengths of a non-isosceles triangle. Prove that:
(a) If $ k\ge 4$ then

$ \sum \frac {kabc \minus{} a^3}{(b \minus{} c)^2} \ge [k\minus{}1\minus{}\sqrt{k(k\minus{}4)}](a \plus{} b \plus{} c)$;

(b) If $ k\ge 0$, then

$ \sum \frac {kabc \plus{} a^3}{(b \minus{} c)^2} \ge [k\plus{}1\plus{}\sqrt{k(k\plus{}4)}](a \plus{} b \plus{} c)$.

What about finding minimum value of :

$ \frac{(a^2\minus{}bc)^2}{(b^2\minus{}ac)^2}\plus{}\frac{(b^2\minus{}ac)^2}{(c^2\minus{}ab)^2}\plus{}\frac{(c^2\minus{}ab)^2}{(a^2\minus{}bc)^2}$

Do you think it is 3?  :maybe:

[quote="shaaam"]What about finding minimum value of :

$ \frac {(a^2 \minus{} bc)^2}{(b^2 \minus{} ac)^2} \plus{} \frac {(b^2 \minus{} ac)^2}{(c^2 \minus{} ab)^2} \plus{} \frac {(c^2 \minus{} ab)^2}{(a^2 \minus{} bc)^2}$

Do you think it is 3?  :maybe:[/quote]
Not, shaam. It still be $ 5$. We just need to put $ a\equal{}x/y,b\equal{}y/z,c\equal{}z/x$. ;) :)

[quote="shaaam"]What about finding minimum value of :

$ \frac {(a^2 \minus{} bc)^2}{(b^2 \minus{} ac)^2} \plus{} \frac {(b^2 \minus{} ac)^2}{(c^2 \minus{} ab)^2} \plus{} \frac {(c^2 \minus{} ab)^2}{(a^2 \minus{} bc)^2}$

Do you think it is 3?  :maybe:[/quote]

Indeed, for $ a\equal{}0$ and $ b\equal{}c\equal{}1$ we find 3.

Similarly prove or disprove for $ k,\ell \in\mathbb R$

\[ \frac {(a^2 \minus{}kbc)^2}{(b^2 \minus{}\ell ac)^2} \plus{} \frac {(b^2 \minus{}kac)^2}{(c^2 \minus{}\ell ab)^2} \plus{} \frac {(c^2 \minus{}kab)^2}{(a^2 \minus{}\ell bc)^2}\ge k^2\plus{}1\plus{}\frac1{\ell ^2}\] :)

[quote="Vasc"][quote="shaaam"]What about finding minimum value of :

$ \frac {(a^2 \minus{} bc)^2}{(b^2 \minus{} ac)^2} \plus{} \frac {(b^2 \minus{} ac)^2}{(c^2 \minus{} ab)^2} \plus{} \frac {(c^2 \minus{} ab)^2}{(a^2 \minus{} bc)^2}$

Do you think it is 3?  :maybe:[/quote]

Indeed, for $ a \equal{} 0$ and $ b \equal{} c \equal{} 1$ we find 3.[/quote]
Sorry, I have got a mistake  :(  :blush:


[quote="SAPOSTO"]I have a question.
Find the minimal number of colours in which could be colored natural nummbers from 1 to 2004 so that there aren't three different numbers a, b and c coloured in one and the same colour such that a divides b and b divides c.
My question is if 6 is the answer of the problem. If you write solution please first write the if 6 is the number and than hide your solution.[/quote]

Yes, answer is 6(i have solution...i will post soon)

[quote="SAPOSTO"]Yes! I managed to find the solution.
Since b must be equal to ak (for natural k) and c must be bm (for natural m) c=akm, so $ c\ge 4a$. Now we can easily construct the sets $ \{2004,\ldots,502\}$, $ \{501,\ldots,126\}$, $ \{125,\ldots,32\}$, $ \{31,\ldots,9\}$, $ \{8,\ldots,3\}$, $ \{1,2\}$. If we colour each set in different colour we get that 6 is possible number of colours. If we assume we can take less than 6 colours we shall consider the numbers $ 1,2,2^{2},\ldots,2^{10}$. These numbers are 11 and since we have less than 6 colours there must be at least three numbers coloured in one and the same colour. The smallest of these numbers divides the bigger than it, and the biggest is divisible by the smaller than it - impossible. So the answer of the problem is 6.[/quote]

=) I have same solution :lol: , but my example for $ 6$ is different.


Let $ A, B$ be measure zero sets on the real line. Is $ A\plus{}B$ always a measure zero set?

No. An example: If $ C$ is the standard middle-third Cantor set, $ C\plus{}C$ is the entire interval $ [0,2]$.


function
real analysis
real analysis unsolved

[quote="eugene"]Show that all continuous functions of two variables may be expressed as the composition of functions of $ 1$ variable together with addition.

* This is supposed to be solved using Functional Analysis.[/quote]

More general result is known as Kolmogorov's superposition theorem, namely that each $ f \in C([0,1]^{N})$ can be expressed in the following form: $  f(x_{1},x_{2},...,x_{N}) = \sum_{i=1}^{2N+1}g( \sum_{j=1}^{N}\lambda_{j}\phi_{i}(x_{j}))$ where $ g$ and $ \phi_{i}$ are continuous functions.

[quote="eugene"]Do you know the proof of this result or any reference.[/quote]

I dont know a proof :(, but in this article is some reference http://www.math.technion.ac.il/~pinkus/papers/acta.ps (page 185).


complex analysis
function
calculus
derivative
integration
complex analysis unsolved

Hi!

Please, give counterexample to the following theorem:

$ \textbf{Theorem.}$ If $ \{f_n\}_{n \equal{} 1}^{\infty}$ is a sequence of holomorphic functions that converges uniformly to a function $ f$ in every compact subset of $ \Omega$, then $ f$ is holomorphic in $ \Omega$ and the sequence of $ k^{th}$ derivatives $ \{f^{(k)}_n\}_{n \equal{} 1}^{\infty}$ converges uniformly to $ f^{(k)}$ on every compact subset of $ \Omega$.

Why are you asking for a counterexample to a true statement? Prove it with the Cauchy integral formula.

Hi, I'm sorry, I mean that the conditions in the theorem are necessary, if don't consider any statement in the condition then the proposition is not true. I would like counterexample in such case. :roll:

If the conditions fail, then the result obviously fails for $ k\equal{}0$. You need to be a little more specific about what "weaker" version you mean in order to ask a nontrivial question.


I am new to this forum and have a question on how to post problems that I write in latex.  Once I know how to do that, I will post a question from a local math competition about logarithms.

I don't know if I'm posting this in the right place.  If I'm not, I appologize and will post in the right place in the future.

Any help would be appreciated!

     -- Pawel

You probably should have posted this question in the LaTeX forum.  But since you are new we forgive you.  About $\LaTeX$, there is a guide on this site which you can find [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url].

Basically what you need to know if all you want to do is use it on the forum is to put dollar signs around the equations that you want in LaTeX.  Then if need to do stuff like fractions, matrices, etc. you can look them up in the LaTeX guide.

So, for example, say you wanted to write the following equation in LaTeX:

sqrt(x)/3 = sin (45 degrees)

Your code would look like this:

[code]
$\$ \frac{\sqrt{x}}{3} = \sin 45^\circ. $\$[/code]

check out valentin's $\LaTeX$ tutorial. i learned it really quickly. You will start to get a feel for common symbols that you use, then if you forget one, you can check on the symbols page. $\LaTeX$ can actually get kinda hairy when you do really advanced things though.

That is this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]tutorial[/url] ;)


$2007^{3}+3\cdot2007^{2}\cdot2993+3\cdot2007\cdot2993^{2}+2993^{3}$

No, I do not want you to type this into your calculator.

Is there a way you can simplify this to help solve the problem? I do greatly hope it's not brute force :D .

-jorian

I think you can use simon's favorite factoring trick.

[hide]this was quite simple, since the powers are written out.

it is (2007+2993)^3=5000^3=125000000000[/hide]

[quote="jhredsox"]but it's adding 3*2007^2...

jorian[/quote]

$(x+y)^{3}\neq x^{3}+x^{2}y+xy^{2}+y^{3}$, rather
$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$. As a generalization,

$(x+y)^{n}=\sum_{k=0}^{n}{n\choose k}x^{n}y^{n-k}$. Proven easily by combinatorial argument. As a hint, considr how each term is generated from $(x+y)(x+y)\cdots(x+y)$.

[quote="13375P34K43V312"]$(x+y)^{3}\neq x^{3}+x^{2}y+xy^{2}+y^{3}$, rather
$(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$. As a generalization,

$(x+y)^{n}=\sum_{k=0}^{n}{n\choose k}x^{n}y^{n-k}$. Proven easily by combinatorial argument. As a hint, considr how each term is generated from $(x+y)(x+y)\cdots(x+y)$.[/quote]
My math teacher taught us the binomial theorem in 5th grade.

Previous discussions on this Forum have concluded that discussions concerning the general Binomial Theorem belong in the High School forums.  I think you are more likely to find a question involving the Binomial Theorem on the AIME than on MATHCOUNTS.  Back to MATHCOUNTS...  For State or Nationals, it's probably a good idea to know a few special forms of binomial expansion.

$(x+y)^{1}= x+y$

$(x+y)^{2}= x^{2}+2xy+y^{2}$

$(x+y)^{3}= x^{3}+3x^{2}y+3xy^{2}+y^{3},$ and

$(x+1)^{3}= x^{3}+3x^{2}+3x+1$


floor function
number theory unsolved
number theory

But, if $n$ is prime, your number isn't an integer :(

I think what grodig meant was $\lfloor \frac{(n-1)!}{n(n+1)} \rfloor$, the greatest integer function...

we love partial fractions (= at least, the way i did it...

i did it using some ugly method with partial fractions.


The roots of $ Ax^2 \plus{} Bx \plus{} C \equal{} 0$ are $ r$ and $ s$. For the roots of
\[ x^2 \plus{} px \plus{} q \equal{} 0
\]
to be $ r^2$ and $ s^2$, $ p$ must equal:

$ \textbf{(A)}\ \frac{B^2 \minus{} 4AC}{A^2}\qquad
\textbf{(B)}\ \frac{B^2 \minus{} 2AC}{A^2}\qquad
\textbf{(C)}\ \frac{2AC \minus{} B^2}{A^2}\qquad \\
\textbf{(D)}\ B^2 \minus{} 2C\qquad
\textbf{(E)}\ 2C \minus{} B^2$

Using Vieta's Formulas, we get the following equations:

$ r\plus{}s\equal{}\minus{}\frac{B}{A}$
$ rs\equal{}\frac{C}{A}$
$ r^2s^2\equal{}q$
$ r^2\plus{}s^2\equal{}\minus{}p$

Squaring the first equation, we get

$ r^2\plus{}2rs\plus{}s^2\equal{}\frac{B^2}{A^2}$
$ r^2\plus{}s^2\equal{}\minus{}p\equal{}\frac{B^2}{A^2}\minus{}\frac{2C}{A}$
$ p\equal{}\frac{2C}{A}\minus{}\frac{B^2}{A^2}\equal{}\boxed{\frac{2AC\minus{}B^2}{A^2}}$

Answer: C


E facuta de mine inegalitatea, nu e grea. Bineinteles toti ceilalti sunteti bine-veniti :wink:

$a,b,c\geq 0, a^{2}+b^{2}+c^{2}=3$

Demonstrati ca $a^{3}b^{2}+b^{3}c^{2}+c^{3}a^{2}+3 \geq 16abc(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1})$

O mai puteam complica umpicut, dar am lasat-o asa pana la urma

Thanks  :wink: .

Dar nu cred ca e corecta.Mai uita-te o data peste ea.De ex daca $a=b=c=1$ este verificata conditia dar inegalitatea nu are loc
$6\geq 16\cdot \frac{3}{2}$ :?:

Voi demonstra ca: $\sum \frac{a}{a+1}\leq \frac{3}{2}$ $\Leftrightarrow$ $\sum \left( 1-\frac{1}{a+1}\right)\leq \frac{3}{2}$ $\Leftrightarrow$
$\sum \frac{1}{1+a}\geq \frac{3}{2}$
Dar din CBS avem $\sum \frac{1}{1+a}\geq \frac{9}{\sum a+3}\geq \frac{3}{2}$ $\Leftrightarrow$ $3\geq a+b+c$ $\Leftrightarrow$ $3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}$ ,evident.
Asadar este suficient sa aratam ca
$\sum a^{3}b^{2}+3\geq 6abc$
Dar deoarece $3\geq a+b+c$ $\Rightarrow$
$\sum a^{3}b^{2}+3\geq \sum a^{3}b^{2}+\sum a$
Dar din AM-GM $\Rightarrow$ $\sum a^{3}b^{2}+\sum a\geq 6\sqrt[6]{a^{6}b^{6}c^{6}}=6abc$.qed.

PS:Cu cat incerci sa "complici" o inegalitate cu atat ea devine mai "slaba" si poate fi atacata in multe moduri.
Ai cumva alta solutie ?

da, e o solutie. eu pornisem de la $\sum_{cyc}b^{2}(a^{2}-1)(a-1) \geq 0$.Destul de slabuta, nu e suficient de "stronga" :D o sa pregatesc si altele


function
real analysis
algebra proposed
algebra

find all functions $f: R\rightarrow R$ such that for all reals like$x,y$,we have $f(x+y)=f(x)+f(y)$.
(supose $f$ is not continuous).

This is too famous to be discussed really

[b]but  consider $f$ does not need  to be continuous .and i am interested of a desired function that is not continuous.
if  you know find [/b]it

Take any Hamel basis $\mathcal{B}$ for $\mathbb{R}$ over $\mathbb{Q}$, define $f$ arbitrarily on $\mathcal{B}$, and extend it to all of $\mathbb{R}$ by linearity.  Of course, since you can't explicitly describe a Hamel basis, as far as I know, this is as good as you're going to get.

It's possible to prove that if $f$ is continuous at at least one point, bounded on a non trivial interval, monotonic or even Lebesgue-mesurable then $f$ is linear.

Non linear solutions of Cauchy equation are really, really, ill-mannered   :P

How do you get continuity just from $f(0)=0$¿

[quote="leepakhin"]But since $f(0)=0$, $f$ is continuous at one point $0$, and by a certain theorem (whose name I can't remember; I think Soarer knows that), it is continuous everywhere.[/quote]
:?:  :?:
The weird solutions constructed with Hamel basis also have $f(0)=0$ but there are not continuous anywhere

[quote="leepakhin"]But since $f(0)=0$, $f$ is continuous at one point $0$[/quote]
 :?:


Stump your parents!  Stump your math teachers!  The following "Foxtrot" cartoon was published on October 11, 2005.  What is the next number in the "Hut!" sequence given in the final panel?  For extra credit, what is the proper response to the final panel?
[img]http://www.research.att.com/~njas/sequences/a001608.gif[/img]

[quote="Ars"]The next one is 5 and it is the Perrin sequence.[/quote]
Yep.  For the Fibonacci sequence, the first two terms are given, by definition.  Then the sum of any two consecutive terms is the next subsequent term of the series ($0+1=1$, $1+1=2$, $1+2=3$, $2+3=5$, $3+5=8$, etc.).

With the unknown sequence in the fourth panel, the first three terms are given.  Then the sum of any two consecutive terms is the second subsequent term of the series ($3+0=3$, $0+2=2$, $2+3=5$, $3+2=5$, $2+5=7$, etc.).  As Ars said, this is called the Perrin Sequence.

Marcus must be a pretty smart guy!

[quote="Treething"]Hut 83[/quote]

Uhh.... where did that come from  :huh:


hello

please help me with this question thanx

Determine the number of paths of length n between any two adjacent vertices in K3,3 , for all n ≥ 0.

I assume $K_{3, 3}$ means a complete bipartite graph with two groups of $3$ vertices.

[hide] On each step, you clearly must go from one group to the other one, so you can only get to a vertex in the same group if you have taken an even number of steps and the opposite for vertices in the other group.

So for vertices in the same group, $n$ even case: Each step, you have three possible moves except the last one (you have to make it to the target vertex), so this is just $3^{n-1}$.

For the opposite group, $n$ odd case has the exact same argument, so it is also $3^{n-1}$.

If you are trying to get to a vertex in the same group with a path with odd length or a vertex in the opposite group with a path of even length, this is impossible.[/hide]


http://results.gov.in/result2announced.aspx  :D

chennai's result coming tomorrow,without any impulse... :rotfl:

But I found this on the net...which says that results out on friday.??

http://www.ptinews.com/pti%5Cptisite.nsf/0/27BCC9746C513DD5652575BB00437312?OpenDocument

We will see who is right tomorrow  :)

EDIT : Oh this must be for the delhi region

hey do u know when will the class 10th results be out??????

i mean do u have any info on net regarding it..... if yes

then plz give the link.....

thankx..........

 :)  :)  :)

:blush: super!!!

@others??highest..... :P

I think V Prashant of PS School,Chennai may be the topper
with 490/500 . He is also getting about 400 [hide="..."]( marks not rank )[/hide] in IIT JEE

Highest at my school - (488/500)

hey when will the class 10th board results be out

no one replied to my this question in my previous post..... :mad:

plz reply.....

thankx in advance.........

Probably on 27th or 29th
No official word though....ALL the Best  :)

lol.....i hope i pass.....
i wonder when people will realise dat board toppers just get "jhooti shan"

I think the top mark from chennai is 490.
Someone from SBOA scored 490 as well.

@nora:brilliant marks :omighty:

to others,best of luck to u :thumbup:

hey wen did nora post her marks???

@frank..
well-u r ofc right ,but see V Prashant

i remember last year Aops was full of studs from PS school
wats the full form??problem solving school?? :rotfl:

hey ritu

plz tell me when will the tenth results be out......

i am feeling like hell.......

waise bhi pass to hona hi nahi hai

compartment aayegi........... :(  :(  :o

par fir bhi....

bata do plz bhaiya
result kab aa raha hai

ya koi bhi bata do jisko pata hai.......

thankx....

rite now,even CBSE is not sure abt the date...how can i tell u?? :rotfl:

i was telling abt % marks :oops:

@jas:be tension free,the usual trend is:CBSE(XII)>AIEEE~IITJEE>then 10th
best of luck :)

[quote="skand"]@nora:brilliant marks :omighty:

.. blah [/quote]

[i]But when did i mention my marks ??[/i]  :o

@ _MM_ : [b]CONGRATS !! [/b]  :coolspeak:

[b]That's a really good score !![/b]

hi jaswinder,i'm shibin george.. me too waiting anxiously for the result.....
please inform your %.... :blush:

I GOT A NICE HUNDRED IN pHY AND MATHS :D

flunked in bio and eng

total 475 :(

congrats ritu

any news abt class 10th results???

plz reply fast

........

thankx

[b]Congrats Rituraj !!!! [/b]   :thumbup:

My school topper got 485

and no idea about 10th results

well

i m asking abt PCM highest..

i got 99% :D  :P

Congrats ritu.... Looks like we have the same marks(475)
Your PCM marks are impressive......Mine( 95% each  :oops: )

Our school has around 25 students getting more than 95% and almost everyone above 90%

oops,sorry 4 d typo  :oops:

i as talking abt _MM_

@nora:brilliant marks  :thumbup:  now

@ skand : I still din tell my marks, they're not so good compared to these guyz  :blush:

^ That was my school topper's score

[quote="Nora.91"]even [b]this[/b] girl got 99 %[/quote]

i was referring to dis  :P

and u r a girl,right??/

hey skand is there any official word abt class 10th results as of now??????? :mad:  :mad:  :mad:

plz reply fast

My MTNL net is finally back.

 I got 89.4% agg. PCM 92.67% . Actually more than I what I really deserve in CBSE.

well good gracious..

i think 100 in 2 subjs is good

my teachers were telling me to get my bio paper reevaluated (i had expected atleast 99)

I [hide="replied"]
well who bothers abt board marks
anything above 75pc is fine :P
doesnt matter weather i get 95 or 96.6 :rotfl: [/hide]

Ritu buddy stop trying to act cool, we all know what kind of material you are.  :P

i m not made of any metal alchemist's alloys( :P ),thats the same material as u guys :rotfl:
V prashant may b mettalic :rotfl:

nora..do post ur marks..


As you proceed from figure to figure, each shaded square is divided into four congruent squares and the upper left square of the four is painted white. By continuing the pattern, what fractional part of the tenth figure will be shaded? Express your answer as a common fraction in which the numerator and denominator are expressed in prime factored form using exponents.

[asy]path g = (0,0)--(4,0)--(4,-4)--(0,-4)--cycle; real k,B; pair A,C; unitsize(.5cm);
for(int i = 0; i < 4; ++i){
A = (6*i,0);
fill(shift(A)*g,grey);
k = 4*(1/2)^i;
for(int p = 1; p < i+1; ++p){
B = 1/2^p;
    for(int n = 0; n < 2^(p-1); ++n){
        for(int c = 0; c < 2^(p-1); ++c){
C=(n*8*B,-c*8*B);
fill(shift(A+C)*(scale(B)*g),white);
        }
    }
    }
for(int m = 0; m < 2^i + 1; ++m){
draw(A+(k*m,0)--A+(k*m,-4),black);
draw(A+(0,-k*m)--A+(4,-k*m),black);
}
}
label("1",(2,-4),S); label("2",(8,-4),S); label("3",(14,-4),S); label("4",(20,-4),S);[/asy]

Examine the first few stages: 1, 3/4, 9/16, 27/64 is shaded. This is a geometric sequence of ratio 3/4. So, the tenth figure will be shaded 3^9/4^9, which in prime factorized form, is $ \frac{3^9}{2^{18}}$.


Anyone know how to save a file as a jpeg or gif or pdf or something?...gsd's realli dont do too much outside the sketchpad...

I couldn't find the command but I think I've heard it's possible...

[edit] I mean I don't think sketchpad can make PDF's...but I have acrobat so that should do it...I just need it in jpeg or gif...

I often "save as" .wmf files, which can be inluded in Word as vector images.

I just copy the relevant parts of diagrams into Paint where I can save them in a variety of forms.  If you want to put it into a word processor (either Word or WordPerfect, for sure -- maybe others, too) you can just copy and paste into the document.  Dunno about making .pdfs or anything.

You can use Go2PDF or CutePDF or PDFFactory to make PDFs.  There's a link to Go2PDF in the downloads section of the LaTeX pages.


(a) Solve in $\mathbb C$ the system $|z| = \left| \frac{a}{z}\right| = |z-a|$, where $a \in \mathbb R, \, a>0$.
(b) Solve the equation $\log_{x}\left( x+10.25 \right) = \lg 10.5$.

${\{\alpha ,\beta \}\in \mathbb{R}}$
${z=\alpha+i \beta }$

${|z|=\sqrt{\alpha^{2}+\beta^{2}}}$
${|z-a|=\sqrt{(\alpha-a)^{2}+\beta^{2}}}$
${\left|\frac{a}{z}\right|=\frac{a}{\sqrt{\alpha^{2}+\beta^{2}}}}$

${\sqrt{\alpha^{2}+\beta^{2}}=\sqrt{(\alpha-a)^{2}+\beta^{2}}\text{ }\Longleftrightarrow \text{ }\alpha =\frac{a}{2}}$

${\sqrt{\alpha^{2}+\beta^{2}}=\frac{a}{\sqrt{\alpha^{2}+\beta^{2}}}\text{ }\Longleftrightarrow \text{ }\beta^{2}+\alpha^{2}=a\text{ }\Longleftrightarrow \text{ }\beta =\pm \frac{1}{2}\sqrt{4 a-a^{2}}}$

and for the second problem

x is approximately $3$

i dont think there is a full-form for the solution

:)

[quote="perfect_radio"]Solve in $\mathbb C$ the system $|z|=\frac{a}{|z|}= |z-a|$, where $a \in \mathbb R^{*}_{+}$.[/quote]
[color=darkblue][b]Proof.[/b] $\{\begin{array}{c}z\overline z=a\\\ (z-a)(\overline z-a)=z\overline z\end{array}\|$ $\Longleftrightarrow$ $\{\begin{array}{c}z\overline z=a\\\ z+\overline z=a\end{array}\|$ $\Longleftrightarrow$ $z^{2}-az+a=0$.

Thus appear three cases : $\{\begin{array}{ccc}a\in (0,4) & \Longrightarrow & z\in\{\frac{a\pm i\sqrt{4a-a^{2}}}{2}\}\\\ a=4 & \Longrightarrow & z=2\\\ a\in (4,\infty ) & \Longrightarrow & z\in\{\frac{a\pm\sqrt{a^{2}-4a}}{2}\}\end{array}$
[/color]

Is that a base-$10$ logarithm on the RHS of part (b)?

Yes, $\lg x = \log_{10}x$. I probably should have mentioned it, since I haven't seen myself this notation for a long time.

if the basis of lg is 10, then, the solution is simple 100

i thoug the basis was e. the most usual thing...=


$ a_i$ are perfect squares and
$ \sum_{i \equal{} 1}^{17}a_i \equal{} 1917$  then find  $ \sum_{i \equal{} 1}^{17}\sqrt {a_i}$ :roll:

im not sure that only one a sequence can be determined by this summation but i could easily find a counterexample.
if we look at 1917=0(mod 9)
i had an idea that all ai can be divisible by 3
so if we set bi as 9.ai

b1+b2+...+bn=213

and thats easy by giving b1=196 b2=16 b3=1    rest is 0.

so a1=196.9 a2=16.9 b3=9

and sum of square roots become

14.3+4.3+3+0+0+0+...+0=57


When is equating coefficients valid? I was doing a problem involving logarithms and I equated their parameters (kinda the same thing as equating coefficients) and another member said that it is not necessarily true. Thanks in advance.

Equating coefficients is really useful for things like partial fraction decomposition.  If you have several variables, or variables rasised to different powers equal to some other expression with the same variables, it is useful.


abstract algebra
vector
group theory
superior algebra
superior algebra unsolved

Can we characterize those groups $ G$ for which $ AutG$ acts transitively on the set of non-identity elements of $ G$? It's easy for finite groups, for infinite ones I would be interested to see any possible approaches, thanks in advance.

I don't think there is going to a be a complete answer here.  If the group is periodic abelian, then it is of course a vector space over some $ \mathbb{Z}/p\mathbb{Z}$.  But even if it is periodic, it is not currently known if it is abelian or not (same as asking if the center is trivial or not).  At least, that's what I surmised from this article (which *is* 20 years old, however):
http://www.jstor.org/stable/2274766

When the group is torsion-free, I think it is even more difficult.  For example, *any* vector space over *any* field with char=0 would work.

Perhaps someone out there can shed more light on this than I did.

Steve

Are any examples known that are not vector spaces over a field?

Such a group is characteristically simple, and there are nonabelian characteristically simple groups of exponent p, but I don't know about the action of its automorphism group.  I think McLain's characteristically simple group is not an example.

[quote="EBz89"]Sorry for being a pain, but could you give some reference for McLain's group? When I started to think of this problem I tried for some time to find what I now know to be called a characteristically simple (non-abelian) group and of course I failed. (I'm on holiday at them moment so don't have access to the usual resources so I can't get on Springer etc...)[/quote]

Sure, but as I said, McLain's characteristically simple group is not an example.  The original papers are:

McLain, D. H. "A characteristically-simple group." Proc. Cambridge Philos. Soc. 50, (1954). 641-642.
http://www.ams.org/mathscinet-getitem?mr=64045
http://dx.doi.org/10.1017/S0305004100029819

Roseblade, James E. "The automorphism group of McLain's characteristically simple group." Math. Z. 82, (1963). 267-282.
http://www.ams.org/mathscinet-getitem?mr=159889
http://dx.doi.org/10.1007/BF01111395

It is described in textbook form in Robinson's Course in the Theory of Groups.  It has a short discussion of characteristically simple groups on pages 84–85 of Chapter 3.3.  McLain's characteristically simple group is discussed on pages 347–349 of Chapter 12.1

I was thinking recently about a post Jack Schmidt made on the Ask-An-Algebraist forum a little bit ago
(link: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1959.0001).  From a result in Robinson's "Course on the Theory of Groups", we can take any torsion-free group and embed it into an infinite simple group, all of whose non-trivial elements are conjugate; that is, a group $ G$ such that the *inner* automorphisms act transitively on $ G\minus{}\{1\}$.  So it would seem the torsion-free part of the OP is quite complex.

Also, by choosing to embed a non-abelian group, we see there are examples which are not vector spaces.

Steve

Good call.  That technique also suffices to produce non-abelian p-groups all of whose elements are conjugate in some larger group, but that larger group may not contain the original as a normal subgroup. This is contained in Corollary 1 of the original paper:

Higman, Graham; Neumann, B. H.; Neumann, Hanna. "Embedding theorems for groups." J. London Math. Soc. 24, (1949). 247–254.
http://www.ams.org/mathscinet-getitem?mr=32641
http://dx.doi.org/10.1112/jlms/s1-24.4.247

Actually, reading Osin's preprint I see that there cannot exist a periodic group with exactly two conjugacy classes other than Z/2Z:  Such a periodic group is of exponent p for some prime p.  If p=2, then the group is abelian, so is Z/2Z.  If p>2, then there are elements a,b in the group such that a^b=a*a, but then a = a^(b^p) = a^(2^p), so 1 = 2^p mod p, contradicting 2 = 2^p mod p.

So that means one has to do something else in the periodic case: inner automorphisms cannot suffice.


I know by computer program that $<AEB = <ACD$. but I can't prove .  :(
$BE$ and $GC$ are tangent line.
Help me please. !!!
(Err... I can show that $<ABE = <ADC$  ;) )

$\measuredangle ADC = \measuredangle BAD + \measuredangle ABD$

(sum of 2 angles of the $\triangle ABD$ equals to its external angle at the remaining vertex $D$)

$\measuredangle BAD = \measuredangle DBE$

(angle spanned by the arc $BD$ equals to the angle between the chord $BD$ and the tangent to the circle $(F)$ at the point $B$)

$\measuredangle ADC = \measuredangle DBE + \measuredangle ABD = \measuredangle ABE$

You know this.

$\measuredangle ACD = \measuredangle DAE$

(angle spanned by the arc $AD$ equals to the angle between the chord $AD$ and the tangent to the circle $(G)$ at the point $A$)

$\measuredangle ADC = 180^o - (\measuredangle ACD + \measuredangle DAC) = 180^o - (\measuredangle ACD + \measuredangle DCE) =  180^o - \measuredangle ACE$

Consequently,

$\measuredangle ABE + \measuredangle ACE = 180^o$

which means that the quadrilateral $ABEC$ is cyclic. The angles $\measuredangle AEB = \measuredangle ACB = \measuredangle ACE$ are equal, because they span the common arc $AB$ of the quadrilateral circumcircle.

Q.E.D.

More stuff:

This circumcircle is also the circumcircle of the right angle triangles $\triangle EGC$, $\measuredangle EGA$, i.e., the center $G$ of the circle $(G)$ is on this circumcircle and the circumcenter is the midpoint of the segment $EG$. This implies that the angle $\measuredangle GBE$ is right and the angle $\measuredangle GBD = \measuredangle ABE - 90^o = \measuredangle ADC - 90^o$.

The center line $GF$ is perpendicular to the common chord $AD$ of the circles $(G), (F)$, hence, it forms the angle $\measuredangle ADC - 90^o$ with the line $BD$, just like the line $GB$. Therefore, the parallel lines $GF, GB$, both passing through the point $G$, are identical, i.e., the point $B$ is on the center line $GF$. Since the line $GB \equiv GF$ is perpendicular to both the lines $BE$ and $AD$, these 2 lines are parallel.

Errr... $yetti, \overline{EA}$ is not tangent line.
So $\angle ACD \neq \angle DAE$
However I find proof from your explanation. Thanks.  :D

Well, the line $AE$ looks like a tangent to the circle $(G)$ in your drawing, just like the line $BE$ looks like a tangent to the circle $(F)$. But if $AE$ is not a tangent to $(G)$, then

$\measuredangle AGC = 360^o - 2 \measuredangle ADC$

because the angle $\measuredangle ADC$ spans the opposite (shorter) arc $AC$ of the circle $(G)$ than the one corresponding to the central angle $\measuredangle AGC$. The triangle $\triangle ACG$ is isosceles, so

$\measuredangle ACG = \frac 1 2 (180^o - \measuredangle AGC) = \frac 1 2 [180^o - (360^o - 2\measuredangle ADC)] =$
$= \measuredangle ADC - 90^o = \measuredangle ABE - 90^o = \measuredangle ABF$

$\measuredangle ABE + \measuredangle ACE = (90^o + \measuredangle ABF) + (90^o - \measuredangle ACG) = 180^o$

The quadrilateral $ABEC$ is still cyclic and the angles $\measuredangle AEB = \measuredangle ACB$ are equal, because they span the common arc $AB$. Of course, the center $G$ of the circle $(G)$ is no longer on the circumcircle, the center line $GF$ does not pass through the point $B$ and the tangent $BE$ is not parallel to the common chord $AD$ of the circles $(F), (G)$.

[quote="DPopov"]Both angles intersect the same arc, $AB$, this should be enough to prove it.[/quote]

I don't understand. Can you give me more explanation. ?  :blush:

[quote="DPopov"]There is a theorem that two angles which intersect the same arc are congruent.[/quote]
Only if you first have that the points are on the circle.  Thus, you need to show that the 4 points are concyclic before you can apply that fact.


search parameterization Stanford college Putnam

how do u register for an IQ test?
i never took one
 :o  :o  :o

I don't know how you can register for an official IQ test. If you want to see how is an IQ test you should simply search one on the internet. There are a lot of IQ tests. I think that the result of that test is important only if you want to go to a university or you need it somewhere else. In my opinion, the inteligence can not be scored, because it is a quality, not a result.

IQ tests are not important while applying to a university. In fact, many American universities will find it a turn-off if you include an IQ score because it conflicts with meritocratic values. [url=http://www.tickle.com]Tickle.com[/url] has a reasonably "accurate" online IQ test, but most online IQ tests are garbage.

Keep in mind that intelligence is multifaceted, and any measure of it will vary depending on the index used. The value of an IQ test depends on one's confidence that the test meaningfully isolates [i]g[/i] (and belief that the universal [i]g[/i] exists and is meaningful in the first place).

Yeah I find that most online IQ tests are very inaccurate. Mostly by the standard deviation one gets by taking a variety of them... Not sure how to register for a real one though.

Yeah
Yeah, they are right
I strongly agree with 'TZF' and others who assert that the so-called IQ tests are 'garbage', because I personally feel that these tests can't accurately gauge a person's innate abilities, let alone intelligence; Intelligence is a complex concept, and it can't be understood as easily through an IQ test.

I also personally feel that if a person scores highly on an IQ test, it doesn't necessarily mean that he is 'more intelligent' than those who have scored less than him on that particular test, and vice-versa.

But I do feel IQ tests can predict a certain 'type' of intelligence (to some extent) that one may possess.

I am also piqued by the fact--- What is a true test of a person's intelligence? Is is: Good/Great Academic achievements, scores on standardized tests like GRE/SAT/GMAT or something like winning a gold/silver/bronze medal at IMO/IPHO or something else?

I hope people can resolve the above query of mine, but I can assert that there exists a correlation (not necessarily a strong one) between academic achievements and intelligence though academic achievements depend upon several other parameters like: determination, hard work.

[quote="Gen8"]I am also piqued by the fact--- What is a true test of a person's intelligence? Is is: Good/Great Academic achievements, scores on standardized tests like GRE/SAT/GMAT or something like winning a gold/silver/bronze medal at IMO/IPHO or something else?[/quote]
There isn't a true test of someone's intelligence. You simply can't quantify intelligence, just like you can't quantify love. But where almost everyone agrees that love can't be measured, many people believe that intelligence is accurately represented by a number.

I honestly feel like Binet is rolling in his grave by our modern misuse of intelligence testing...

While I'm not a proponent of intelligence tests, I also don't think it's fair to dismiss them as useless. Though intelligence is multifaceted (and though we can't easily define what it is), there is a subtlety here.

Intelligence tests will aim to measure what is known as the g-factor (or just [i]g[/i]). The idea behind this is an interesting, though risky. Empirical evidence suggests that among a population, intelligence across different dimensions tend to correlate. For example, suppose a test measured verbal and quantitative reasoning abilities. If the verbal scores are plotted against their corresponding quantitative scores, we expect a positive correlation.

An overview of the [url=http://en.wikipedia.org/wiki/G_(factor)]general intelligence factor, 'g'[/url]
A specified model dividing Spearman's [i]g[/i] into [url=http://en.wikipedia.org/wiki/Fluid_and_crystallized_intelligence]fluid and crystallized intelligence, 'gF' and 'gC'[/url].

[b]@ max_tm:[/b] Actually, I think Binet would applaud the amazing progress that has been made since his original test. Whatever IQ tests measure, (hopefully [i]g[/i] or [i]gF[/i]), they seem to be doing reasonably well. Their results are certainly much more reliable and more scientifically sound than Binet's pioneering work. I myself am impressed by how scientists can make a test that accurately measures something, and keep improving these tests, even though they don't know what that something is.

Binet was a strong critic of the way they were used in immigration and such back in the day, but he would probably be more complaisant about how they are being used now. In particular, people realize that IQ doesn't get you too far, especially in a meritocratic society. People who blame their failure or on, or attribute others' success to, primarily IQ, are seen as whiners.

We also now have some pretty reasonable standardized tests to measure what Binet originally wanted to measure with his IQ tests. At least, the SAT math section scores seem to correlate pretty well with mathematical ability, and verbal section scores, with verbal ability. The problem, of course, is that we now have public school systems that try to teach toward the frameworks of these tests, and teachers who are unable to accomplish this without teaching very superficially. That, however, is a problem with the teachers and not with standardized testing. But this is an issue to be discussed in another thread.

I guess the 'Putnam' examination should be should be administered as an intelligence determining test,  :lol: .

[quote="Gen8"]I guess the 'Putnam' examination should be should be administered as an intelligence determining test.[/quote]

But the median score on that exam is 0! What does that mean?  :lol:

I think that intelligence tests are useful when they are interpreted by the correct person. A random person can't say that "I have x IQ" they can say that they got x score on an IQ test though. But a psychologist would be able to look an intelligence tests and say "I think this person may be mentally challenged, we should do more tests to get help." There are certain uses for IQ tests but ranking yourself above others with your score is not one of them.

Well, maybe IQ tests can't really be trusted because there is so much more to success in life and "true" intelligence than what's on an IQ test. While IQ tests are useful for certain purposes, it shouldn't be treated as a Holy Grail of intelligence, and even of academic skills, because in order to truly attain knowledge, one must be able to apply it in any and all relevant situations.
On the other hand, IQ tests can't be bashed for being pointless, because they do have their uses. People just think they are stupid because they see IQ tests as having been touted (in the past) as the only indicator of success, and then not living up to that expectation.
While there are ways to test the extraneous variables that make up human intelligence as a whole, a true IQ test probably focuses more on the retention of knowledge.

TZF, you make a good point. You're right that I was alluding to the Stanford-Binet version and it's role in trying to strengthen eugenics, but most people's opinion on the subject has changed since then  :).

[b]@ All:[/b] I've attached an interesting discussion on IQ. It takes a few pages to warm up. You can skip right to page 7 if you are impatient, but it does not hurt to skim the first few pages. Please do not reproduce it, it is copyrighted material.

[b]@ Gen8:[/b] The Putnam would not make a good IQ test even if the grading scheme were adjusted. It is overly dependent on dedicated knowledge (ie, advanced mathematics). Also, math competitions rely on a ridiculous level of [i]gC[/i] -- perhaps even more than [i]gF[/i].

[b]@ evilhamster:[/b] [i]Re: your first comment --[/i] An ideal IQ test will measure [i]g[/i] without requiring any contextually-specific dedicated processes or knowledge. Why does the exclusion of these processes/knowledge mean the IQ test cannot be trusted? It would still measure what it was made to measure.

[i]Re: your last comment --[/i] A true IQ test is supposed to be a measure of raw reasoning ability, [i]gF[/i], which is also a reasonable measure of [i]g[/i]. Consider Raven's Progressive Matrices.

[i]General:[/i]  You are also mixing up intelligence and knowledge. Intelligence is only dependent on knowledge in the sense that [i]gC[/i] can usually be applied more readily by a person with a larger span of knowledge.


Ας βάλω και γω ενα θεματάκι...
Ορίζω ως $ a_n$ το πρώτο μη μηδενικό ψηφίο του $ n!$, έτσι ώστε $ a_1\equal{}1, a_2\equal{}2, a_3\equal{}6, a_4\equal{}4, a_5\equal{}2 ....$.
Είναι η ακολουθία $ a_n$ περιοδική?

Γειά σας. Έστω ότι είναι περιοδική, δηλαδή ότι υπάρχει Τ φυσικός ώστε το τελευταίο ψηφίο του (n+T)! να είναι ίδιο με εκείνο του n! για κάθε φυσικό n. Δείχνουμε ότι αυτό είναι άτοπο για την περίπτωση n=1, δηλαδή ότι δεν θα προκύψει ξανά άσσος ως πρώτο μη μηδενικό ψηφίο. Αρκεί να σκεφτούμε ότι το πρώτο από δεξιά μη μηδενικό ψηφίο του n! προκύπτει πολλαπλασιάζοντας το αντίστοιχο μη μηδενικό ψηφίο του n με το μη μηδενικό του (n-1)!. Άρα από n=2 και έπειτα το πρώτο μη μηδενικό ψηφίο θα είναι άρτιος, δηλαδή όχι άσσος. Δεν είναι λοιπόν περιοδική.

[quote="xp"][b]kosta GT [/b]καλως ηρθες στην παρεα μας
Το $ 17$ παραγοντικο ειναι $ 355687428100000$ και το τελευταιο μη μηδενικο του ψηφιο ειναι $ 1$.......
Ξαναδες το
[b]Βιlly[/b] καλο διαβασμα[/quote]

To 17! δεν μπορεί να έχει τόσα μηδενικά στο τελος... Έχει συνολικά 3 πεντάρια στο γινόμενο και άρα το πρώτο μη μηδενικό ψηφίο είναι το 4 από το τέλος.. Κάπου έχεις κι εσύ λάθος στον υπολογισμό σου :P

Όσο για τη λύση και εμένα δεν με εμπνέει αρκετά.

[b]Εdit:[/b] Όντως το 17! ισούται με 355687428096000 και άρα το επιχείρημά σου δεν ισχύει.

Επιπλέον μπορούμε να εκφράσουμε το γενικό όρο της ακολουθίας στη μορφή:

$ a_n \equiv \frac {n!} {10^k} (mod 10)$, όπου:

$ k \equal{} \lfloor \frac {n}{5} \rfloor \plus{} \lfloor \frac {n}{5^2} \rfloor \plus{} \lfloor \frac {n}{5^3} \rfloor \plus{} ...$

Αν δεν το χει χρησιμοποιήσει κάποιος ως αύριο για να αποδείξει αν είναι ή όχι περιοδική θα το παλέψω τότε...

Η πρόταση του Κώστα μου φαίνεται σωστή αλλά το επιχείρημα μου φαίνεται λάθος. Για παράδειγμα, αν ν! λήγει σε 02 και πολλαπλασιάσουμε με το ν+1 που είναι πολλαπλάσιο του 5, τότε το πρώτο μη μηδενικό ψηφίο θα είναι περιττό.

Επειδή όμως πάντα υπάρχουν περισσότερα δυαρια από πεντάρια, μπορούμε να κάνουμε τον πολλαπλασιασμο με διαφορετική σειρά και να κρατήσουμε ένα δύο για να πολλαπλασιάσουμε στο τέλος. Άρα το τελευταίο μη μηδενικό ψηφίο θα είναι πάντα άρτιος (για ν>1)

Νομίζω όμως πως η άσκηση θα ήταν πιο ενδιαφέρουσα αν ρωτούσε αν είναι eventually periodic, δηληδή αν υπάρχει Τ και Ν ώστε $ a_{n\plus{}T} \equal{} a_n$ για καθε $ n \geqslant N$. (Για παράδειγμα αν είναι της μορφής 1,2,6,4,2,6,4,...)

Απεδειξα οτι το κοπμπιουτερακι μου κανει λαθος
Οντως μου βγαζει το νουμερο που εγραψα
Εγινε και αυτο..

[quote="Demetres"]Νομίζω όμως πως η άσκηση θα ήταν πιο ενδιαφέρουσα αν ρωτούσε αν είναι eventually periodic, δηληδή αν υπάρχει Τ και Ν ώστε $ a_{n \plus{} T} \equal{} a_n$ για καθε $ n \geqslant N$. (Για παράδειγμα αν είναι της μορφής 1,2,6,4,2,6,4,...)[/quote]

[hide]Fact για Trivial Pursuit: αν $ a_n$ το τελευταίο μη μηδενικό ψηφίο του $ n!$, ο αριθμός $ 0.a_1a_2a_3\ldots$ είναι υπερβατικός :!:[/hide]

Cheerio,

Durandal 1707

ναι, υπερβατικοί είναι  transcendental

[quote="wallace"]Παναγιώτη υπερβατικοί είναι οι  [b]transcendental[/b]?   ή μπερδεύω την ορολογία?[/quote]

Ναι... α, και για να μην παρεξηγηθώ, εννοείται πως δεν το βάζω για άσκηση! Θέλει ένα πολύ πρόσφατο (και αν θυμάμαι καλά δυσκολούτσικο) θεώρημα για να αποδειχθεί αυτό.

Η πρόταση του Dem πάντως δεν είναι unfeasible... είναι λίγο μπελάς στα logistics αλλά μπορεί να δειχθεί στοιχειωδώς ότι το τελευταίο μη μηδενικό ψηφίο του $ n!$ δεν επαναλαμβάνεται περιοδικά.

Cheerio,

Durandal 1707

Για το ότι είναι άρρητος δεν είναι τελικά τόσο δύσκολο όσο φοβόμουν:

Ας είναι Τ περίοδος της ακολουθίας. Δεν παίρνω το Τ να είναι η μικρότερη περίοδος, αλλά το διαλέγω με τέτοιο τρόπο ώστε το τελευταίο μη μεδενικό ψηφίο του Τ να είναι $ t \neq 0 \bmod 5$. (Μπορεί νε επιτευχθεί πολλαπλασιάζοντας την μικρότερη περίοδο με αρκετά μεγάλη δύναμη του 2.)

Παίρνω Ν αρκετά μεγάλο ώστε
1) 10^Ν έχει περισσότερα μηδενικά από το Τ
2) $ a_n \equal{} a_{n\plus{}T}$ για κάθε $ n \geq 10^N\minus{}1$.

Έχουμε
3) $ a_{10^N \minus{} 1} \equal{} a_{10^N \minus{} 1 \plus{} T} \equal{} a_{10^N \minus{} 1 \plus{} 2T}$
4) $ a_{10^N} \equal{} a_{10^N \plus{} T} \equal{} a_{10^N \plus{} 2T}$

και

5) $ a_{10^N \minus{} 1} \equal{} a_{10^N }$
6) $ a_{10^N \plus{} T} \equal{} ta_{10^N \minus{} 1 \plus{} T} \bmod 5$, άρα $ t \equal{} 1 \bmod 5$
7) $ a_{10^N \plus{} 2T} \equal{} 2ta_{10^N \minus{} 1 \plus{} 2T} \bmod 5$ άρα $ t \equal{} 2 \bmod 5$

Άτοπο

Παναγιώτη, έχεις καμια παραπομπή για την υπερβατικότητα;

Έτσι ακριβώς Dem, well done :)

Το θεώρημα που χρησιμοποιείς είναι ένα αρκετά καινούριο αποτέλεσμα των Adamczewski-Bugeaud-Luca. Μετά από λίγο googling, ένα (εντελώς readable) reference είναι [url=http://home.wlu.edu/~dresdeng/papers/lnzd2.pdf]το paper του Dresden[/url].

Cheerio,

Durandal 1707

Παναγιώτη, ευχαριστώ.


click [url]http://olympiads.hbcse.tifr.res.in/uploads/inmo-2009[/url] :)
i am one of the oldies out here
i have a very dubious record as far as inmo's are concerned.
i appeared at it for 4 times in my classes 8th,9th ,10th and 11th
but didnt qualify in all those 4 chances.
so i didnt give rmo in my 12th :(
i am pardesi's statemate and batchmate

CONGRATS TO ALL THE INMO AWARDEES 2009

[quote="anitsahu"]click [url]http://olympiads.hbcse.tifr.res.in/uploads/inmo-2009[/url]
i have a very dubious record as far as inmo's are concerned.
i appeared at it for 4 times in my classes 8th,9th ,10th and 11th
[/quote]
You cleared RMO in class 8!!!! You must be a genius lol

nopes
until last year orissa used to have its own qs  paper which used to be very easy
so its something i would call "luck by chance" :D

I think you were one of the best ppl here and one who had achieved brilliant ranks,etc(in FIITJEE AITS probably)

and you proved yourself by clearing RMO 4 times also  :omighty:

Hey all,
         Congrats to those who cleared INMO. After a long time, Ill not be part of IMOTC, but still hope all of you have great fun in the camp. Do well, and at the same time, chill out. It will definitely be an experience of a lifetime.
P.S: Chill Anit, you rock!!
Ever urs,
Chappli.. :)

hey guys...congrats to everyone who cleared the inmo...

i've also made it thru....thot that i should get to know a few ppl i'll be meetin there..

btw i'm vinit from pune, class XII :)


algebra
function
domain
partial fractions
complex analysis
complex analysis unsolved

1)find the power series expansion for

$ f(z) = \frac {z-1}_{z^{2}-2z+2}$ around z=1.

What is the radius of convergence of this power series?

2) Using rouche's theorem, find the number of zeros of:

$ z^{4}-4z^{3}+6z^{2}-4z+3$ in the domain $ 0<z<2$

There's a general methodology to these if you can't find a "quick fix":  Express it in terms of partial fractions:

$ f(z)\equal{}\frac{A_0}{z\minus{}z_0}\plus{}\frac{A_1}{z\minus{}z_1}\plus{}...\plus{}\frac{A_n}{z\minus{}z_n}$

now, in order to expand it about the point $ c$:

$ \equal{}\frac{A_0}{z\minus{}c\plus{}c\minus{}z_0}\plus{}\frac{A_1}{z\minus{}c\plus{}c\minus{}z_1}\plus{}...\plus{}\frac{A_n}{z\minus{}c\plus{}c\minus{}z_n}$

or:

$ \frac{A_0}{c\minus{}z_0}\left[\frac{1}{1\minus{}(\minus{}\frac{z\minus{}c}{c\minus{}z_0})}\right]\plus{}\text{rest of um'}$

alright then, as long as:

$ |z\minus{}c|<|c\minus{}z_0|\;\text{and}\;|z\minus{}c|<|c\minus{}z_1|\;
\text{and rest of them}$ which is essentially the distance of c to the nearest singularity, can expand each as a geometric series.  For the first one I'd get:

$ \frac{A_0}{c\minus{}z_0}\sum_{n\equal{}0}^{\infty}(\minus{}1)^n(\frac{z\minus{}c}{c\minus{}z_0})^n$

When I let $ c\equal{}1$ and do that for your problem I get:

$ f(z)\equal{}\frac{i}{2}\sum_{n\equal{}0}^{\infty}\left[(\minus{}i)^n\minus{}i^n\right](z\minus{}1)^n$

which if you study it for a while, when n is even, the i's cancel and when n is odd the sign of the powers alternate and so we obtain:

$ f(z)\equal{}(z\minus{}1)\minus{}(z\minus{}1)^3\plus{}(z\minus{}1)^5\plus{}...;\quad |z\minus{}1|<1$

Oh great . . . just write it as:

$ \frac{z\minus{}1}{1\plus{}(z\minus{}1)^2}\equal{}\frac{z\minus{}1}{1\minus{}(\minus{}(z\minus{}1)^2)}\equal{}
(z\minus{}1)\frac{1}{1\minus{}(\minus{}(z\minus{}1)^2))}$

so for $ |z\minus{}1|<1$ it's:

$ \sum_{n\equal{}0}^{\infty}(\minus{}1)^n(z\minus{}1)^{2n\plus{}1}\equal{}(z\minus{}1)\minus{}(z\minus{}1)^3\plus{}(z\minus{}1)^5\plus{}...$

For the domain of the second problem do you mean $ 0<|z|<2$ ?

If so...

If we let $ f(z)\equal{}z^4\minus{}4z^3\plus{}6z^2\minus{}4z\plus{}3$, and $ \gamma\equal{}\{w\in\mathbb{C} : |w|\equal{}2\}$. Then we should be able to find a nice $ g(z)$ such that $ |g(z)\plus{}f(z)|<|g(z)|$ for all $ z\in\gamma$. Then by Rouche's Theorem the number of zeros inside $ \gamma$ of $ g(z)$ is the same as the number of zeros of $ f(z)$ inside $ \gamma$.

Let $ g(z)\equal{}\minus{}z^4$, then $ |g(z)\plus{}f(z)|\equal{}|\minus{}13|\equal{}13<16\equal{}|g(z)|$ on $ \gamma$, so since $ g(z)$ has four zeros inside the curve, we have that $ f(z)$ also has four zeros inside the curve. Thus by Rouche's Theorem $ z^4\minus{}4z^3\plus{}6z^2\minus{}4z\plus{}3$ has four zeros in $ 0<|z|<2$ (note that we know none of the four are $ 0$ itself by plugging that in).

[quote="HilbertThm90"]

Let $ g(z) \equal{} \minus{} z^4$, then $ |g(z) \plus{} f(z)| \equal{} | \minus{} 13|$ on $ \gamma$.[/quote]

I don't see how you can say that.  Looks to me the most that can be said about $ |g\plus{}f|$ on $ |z|\equal{}2$ is:

$ |g\plus{}f|\leq 67$

Also, I think there are only 2 zeros inside $ |z|\equal{}2$

Notice that the zeros of $ f$ are precisely $ 1\plus{}\epsilon\sqrt[4]2$ for $ \epsilon^4\equal{}\minus{}1$. owk


Given $xyz =1$, find the sum:

\[ \frac{1}{1+x+xy} + \frac{1}{1+y+yz} + \frac{1}{1+z+xz}  \]

[hide]Plugging in $x=y=z=1$ (since what $x$, $y$, and $z$ are shouldn't matter if the problem is valid) yields $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\boxed{1}$.[/hide]

ehh maybe a more "rigorous" solution please haha

or fine $\textbf{prove}$ that our expression always equals $1$

:P

[hide="Proof"]We'll make a multiple use of the fact $xyz=1$

\begin{eqnarray*}\frac{1}{1+y+yz} &=& \frac{1}{xyz+y+yz}\\ &=& \frac{1}{y(1+z+zx)}\\ &=& \frac{1}{y(xyz+z+zx)}\\ &=& \frac{1}{yz(1+x+xy)}\\ &=& \frac{xyz}{yz(1+x+xy)}\\ &=& \frac{x}{1+x+xy}\end{eqnarray*}

\begin{eqnarray*}\frac{1}{1+z+zx} &=& \frac{1}{xyz+z+zx}\\ &=& \frac{1}{z(1+x+xy)}\\ &=& \frac{xyz}{z(1+x+xy)}\\ &=& \frac{xy}{1+x+xy}\end{eqnarray*}

Substituting these into the initial expression, we get

$S=\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy}=1$[/hide]

[hide="Alternative"]Substitute $z=\frac{1}{xy}$ and simplify[/hide]

[hide]Let $S=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}$.

Multiplying the top and bottom of each fraction by the variable that produces an $xyz$ term, $S=\frac{z}{1+z+xz}+\frac{x}{1+x+xy}+\frac{y}{1+y+yz}$.

Multiplying again, $S=\frac{yz}{1+y+yz}+\frac{xz}{1+z+xz}+\frac{xy}{1+x+xy}$.

Adding all three forms of $S$ and grouping terms, $3S=\frac{1+x+xy}{1+x+xy}+\frac{1+y+yz}{1+y+yz}+\frac{1+z+xz}{1+z+xz}=3\Rightarrow S=\boxed{1}$.[/hide]


Three congruent circles $G_{1}$, $G_{2}$, $G_{3}$ have a common point $P$.
Further, define $G_{2}\cap G_{3}=\left\{A,\ P\right\}$, $G_{3}\cap G_{1}=\left\{B,\ P\right\}$, $G_{1}\cap G_{2}=\left\{C,\ P\right\}$.
[b](1)[/b] Prove that the point $P$ is the orthocenter of triangle $ABC$.
[b](2)[/b] Prove that the circumcircle of triangle $ABC$ is congruent to the given circles $G_{1}$, $G_{2}$, $G_{3}$.

Hehehe... In Romanian books this is called Titeica's problem (Gheorghe Titeica was a really cool Romanian mathematician). I say 'in Romanian' books because there are some people who think he's not the first one to have discovered this.

There are tons of solns, but, since I know you (Arne) are starting to enjoy inversion, I'll give a solution using this beautiful method :D

Consider any inversion of pole P. It turns circles Gi into 3 lines di, which are equidistant from P. A turns into A' = d2  \cap d3 and so on (we get b' and C' in the same way). In the triangle A'B'C'  P is the incenter. We want to prove that AP perpend. to BC, but, since inversion conserves the angle of 2 curves, it's equivalent to proving that the image of PA, ehich is the same line PA=PA' (as lines, not segments) is "perpend." to the image of BC, which is circle PB'C'.

The perpendicularity between a line (PA') and a circle (PB'C') translates to the fact that the line passes through the center of the circle, but this can be derived pretty quickly by an angle-chase in triangle A'B'C', where P is the incenter. This proves (a).

(b) follows quickly if we use (a): the radius of circle PBC is BC/2sinBPC = BC/2sinBAC (because BAC=Pi-BPC) = radius of the circle ABC.

This problem is also known in Romanian as 'the problem of the 5 Lei coin' (nasty English translation  :D ). Leu = Romanian money (it means 'lion'). This is because, apparently, Titeica used such a coin during a contest to draw the circles and to find this theorem.

It was the logo of IMO 1999 in Bucharest :)

I read about Johnson theorem and johnson circles today.
Is it possible to solve this problem in reverse as follows:

In a given ABC triangle, find point P so that three circles could be drawn all passing through given point of P and have the same radius.

For a given triangle, are these circles unique?


ειναι το νουμερο 12 της εξωγηινης ολυμπιαδας του Ραλλη.  καθε θετικος ακεραιος ν μπορει να γραφει ως α^3+β^3/γ^3+δ^3, α,β,γ,δ  φυσικοι  :)

Αν δεν κάνω λάθος το πρόβλημα λέει :
Να δείξετε  ότι κάθε [u]ρητός [/u]  αριθμός μπορεί να γραφεί στη μορφή (α^3+β^3)/(c^3+d^3)

εγω λεω το θεμα οπως το λεει ο Ραλλης. φυσικα και γενικευεται στους ρητους οπως ακριβος στην IMOshortlist 1999. ας το λυσουμε για καθε ρητο λοιπον

Δείτε εδώ για τη λύση . Το πόστ του zabelman
http://www.mathlinks.ro/Forum/viewtopic.php?p=131812#p131812


geometry
parallelogram
cyclic quadrilateral
similar triangles
geometry solved

Take the points X and Y on the sides AB and BC of a cyclic quadrilateral ABCD, such that XBYD is a paralelogram. If M and N are the midpoints of AC and BD (respectively) and AC and XY meet at L prove that M,N,L,D lie on a circle.

This problem is problem 4 of the LMO 2002, Elimination Round, 10 grade.

Working with directed angles modulo 180, we have < DCA = < DBA, since the quadrilateral ABCD is cyclic. On the other hand, the parallelogram XBYD yields XB || YD; hence, < DBA = < BDY, and thus < DCA = < BDY = - < YDB. Also, < DAC = < DBC (again from the cyclic quadrilateral ABCD), and < DBC = - < YBD (trivial), so that < DAC = - < YBD. The angle equations < DCA = - < YDB and < DAC = - < YBD together yield that the triangles DCA and YDB are inversely similar. The points M and N are corresponding points in these two triangles, being the midpoints of their respective sides AC and BD. Since corresponding points in inversely similar triangles form oppositely equal angles, we thus have < DMC = - < YNB. In other words, < DML = < DNL. Hence, the points M, N, L, D lie on one circle.

This is probably the shortest solution I have ever posted ;-)

  Darij


If a 4-digit number is inserted between the two zeroes of 2008, an 8-digit is formed. If the 8-digit number is divisible by the 4-digit number, find their quotient.

If $ A$ is 4 digit number, 8 digit number is $ 20000008\plus{}100A\equal{}2^3*7*19*18797\plus{}100F$ and quotent is $ \frac{20000008}{A}\plus{}100$.
20000008 had unique 4 digit divisor $ 2^3*7*19\equal{}A$, therefore quotent is $ 18797\plus{}100\equal{}18897$.

I thought there were solutions other than 18897, so I tried looking for factors of 18797. But I can see form your factorization that 18797 is prime. That's a relief.


3. There are six distinct ordered pairs of nonzero digits (b,x) such that the sum of the digits of [i]b^x[/i] is equal to [i]b + x[/i]. What is the sum of all the [i]b^x[/i]  possibilities?

why did you post 4 problems in 2 different forums?
anyway, i think mods are key to solving this problem

Brute Force. (I guess there might be an algebraic approach, like "if b^x is the number mn [=10m + n]...")

[quote="fishythefish"](5,2) (2,2) (-1,2) (5,3) (6,3) (8,2)[/quote]Brute-forcing it, I find (7,3) and all the fishy pairs, except (-1,2) that is not a "pair of  .. digits" as the problem asked. It's easy enough to add up all six $ b^x$ results. However, even after noticing interesting details about the sum, I don't see an elegant solution.

Oops! I was just thinking about single-digit nonzero integers. Yeah, if you take out (-1,2) and add (7,3), you should have the correct answer.

P.S. "...the fishy pairs..." I like that.  :lol:


given 30 numbers all distinct arranged randomly.find the probabilty that the 20th n.o is the largest

the correct answer is 1/30

but how i thought about it was dat on comparing the 20th with each remaining number either it is larger or smaller  and  hence the prob of it being larger dan all of them is (1/2)^29

where am i going wrong?

Look at it this way:

There is 1 largest number. It can be placed in slot 1, slot 2, slot 3, slot 4, ...., or slot 30. The probability that it is in slot 20 is 1/30.

[quote="abhishek.nanda"]the problem is not justifying the solution but i want to know y i am wrong.[/quote]

Because your solution requires more complexity. If we have the numbers 1-30, There's not a 1/2 chance that the 20th number will be greater than 1. It's 29/30. You can't apply what's known as the principle of indifference and say that if there's two choices, both have a 1/2 probability.

i somewat get ur point can u explain exactly why i cant apply the 'priciple of indifference'.
please explain the term also(principle of.....)

If you compare the 20th term and the first term, there's a $\frac{1}{2}$ chance that the 20th term is greater.

Now let's compare the 20th term and the second term.  Let $t_{n}$ be the $n$th term.  If we arrange $t_{1}$, $t_{2}$, and $t_{20}$ in increasing order, there are three possibilities:  (Remember that $t_{20}$ is greater than $t_{1}$, so $t_{20}$ has to appear later in the sequence.)

$t_{1},t_{2},t_{20}$
$t_{1},t_{20},t_{2}$
$t_{2},t_{1},t_{20}$

Therefore, the probability that $t_{20}$ is greater than $t_{2}$ is $\frac{2}{3}$.  It is NOT $\frac{1}{2}$.  Basically, this is because we already know that the 20th term is greater than the 1st term.  This makes it more likely that the 20th term is greater than the 2nd term.

In fact, if we continued using this highly impractical method of comparing the 20th term to each other term one by one, we would find that the probability is $\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{29}{30}=\frac{1}{30}$.  :)

[quote="abhishek.nanda"]the problem is not justifying the solution but i want to know y i am wrong.[/quote]

You're assuming that each comparison is independent of the others. In fact, they are not, as shown in matt276eagles' solution; therefore, you cannot just multiply the probability of $\frac{1}{2}$ by itself a bunch of times.

Another solution:  The total number of ways to arrange these $\ 30$ numbers is $\ 30!$.  If we claim the one in the twentieth position is the highest then there are $\ 29!$ ways to arrange the rest of the numbers, so $\ \frac{29!}{30!} = \frac{1}{30}$.  Also, using this method its easy to generalize this result for $\ n$ numbers.


function
algebra
domain
calculus
derivative
calculus computations

[quote="Anurag Agrawal"]If $f(x)=-1$ for $x<0$ and $f(x)=1$ for $x\geq 0$
$g(x)=0$ for all $x$,
is $gof(x)$ continuous at $x=0 ?$[/quote]
Yes, because $g_{0}f\equiv 0$.

This was what I thought. :) But according to the definition of continuity of $gof$ $f(x)$ should also be continuous at $x=0$..
 so what's your say on that??? :maybe:

Whatever this thing is that you think is the definition, it's wrong.  The continuity of $g \circ f$ does not imply either the continuity of $g$ or the continuity of $f$.  Also, you can make the symbols nicer with \circ, as in previous sentence.

I think you have a point since every definition should be if and only if (may be I am wrong here). But this seems not to be the case here. So I guess we can say that a mapping composition (be it continuous or not) of two functions does not tell us much about those functions themselves, that is, it is not true that if a composition of f and g is continuous, then both f and g need to be continuous.

This brings to mind something that has always intrigued me.

The composition g(f(x)) of two functions may be differentiable at a point (or on a domain) even though one (or both) of them is not. Does anyone know any examples of this? (Aside from trivial definitions of g)

What I'm trying to say is g(f(x)) may have a derivative that isn't computable by chain rule.

What if you take $f(x)$ to be as in your original post and let $g(x) = |x|$. Then $g(f(x))$ is differentiable at $x = 0$. Yet neither $g(x)$ nor $f(x)$ is at $x =0$.

Well, both of your functions have to be differentiable in the first place in order to apply the chain rule.

There are, of course, lots and lots of such examples.  Let $g$ be [i]any[/i] bijection between $\bf R$ and $\bf R$; indeed, let it be some such bijection which is [i]nowhere[/i] continuous.  Even then the map $g \circ g^{-1}$ is a $C^\infty$ function (has infinitely many derivatives).

Frequently, properties of a composition arise from properties of the things being composed.  It is much, much rarer that one can say anything like the converse, exactly because of examples like the above.  For example, given any sort of simple property and a function $f$ which has that property, I can write $f = (f\circ g) \circ g^{-1}$, where the $g$ is one of the bijections of the previous paragraph.  It's pretty clear that for a great many properties you care to think of, I can choose $g$ so that neither $f \circ g$ nor $g^{-1}$ have that property.

Really sorry for the late response;
but I must say that the above discussions have aroused many doubts in my mind which I despirately wan't to be cleared. :(

Let $f$ be a mapping from a set $D$ to $R$ where $f(x)=1/x$ for all x belonging to $D$ and $g(x)$ from $R$ to $R$ defined as $g(x)=0$ for all x belonging to $R$.
Now $g\circ f(x)=g(f(x))$ is a mapping from $D$ to $R$.Hence by extending $D$ even to a maximum length of $R-\{0\}$ we are not able to define $g\circ f$ at one point i.e. $x=0$ which shows that the composite function depends on the history of the functions used.
So how could you conclude anything about the continuity and the differentiability of $g\circ f$ without having any idea of these properties of the functions involved in the composition?

$g\circ f$ is equal to zero everywhere except for $x=0$, where it is not defined. Therefore, $g\circ f$ is continuous everywhere except for $0$.


inequalities
geometry
geometric transformation
reflection
inequalities proposed

Yes, this problem is very hard. I am waiting for a solution without expanding.

Why do everybody still? Nobody ? :wink:  :arrow: Let kill it  :D

[quote="zaizai-hoang"]$\bigstar$ [Hungkhtn's problem] Let $a,b,c$ are real number. Prove that:
$\frac{ab}{4a^{2}+b^{2}+4c^{2}}+\frac{bc}{4b^{2}+c^{2}+4a^{2}}+\frac{ca}{4c^{2}+a^{2}+4b^{2}}\le \frac{1}{3}$

It is interesting problem. I think it not easy and solution for it is not simple. What do you think?[/quote]

I had a very complicated solution to this one but for the following slighter (and even more beautiful) can be proved very very nice

\[\sqrt{\frac{ab}{4a^{2}+b^{2}+4c^{2}}}+\sqrt{\frac{bc}{4b^{2}+c^{2}+4a^{2}}}+\sqrt{\frac{ca}{4c^{2}+a^{2}+4b^{2}}}\le 1.\]

[quote="toanhocmuonmau"][quote="hungkhtn"][quote="zaizai-hoang"]$\bigstar$ [Hungkhtn's problem] Let $a,b,c$ are real number. Prove that:
$\frac{ab}{4a^{2}+b^{2}+4c^{2}}+\frac{bc}{4b^{2}+c^{2}+4a^{2}}+\frac{ca}{4c^{2}+a^{2}+4b^{2}}\le \frac{1}{3}$

It is interesting problem. I think it not easy and solution for it is not simple. What do you think?[/quote]

I had a very complicated solution to this one but for the following slighter (and even more beautiful) can be proved very very nice
\[\sqrt{\frac{ab}{4a^{2}+b^{2}+4c^{2}}}+\sqrt{\frac{bc}{4b^{2}+c^{2}+4a^{2}}}+\sqrt{\frac{ca}{4c^{2}+a^{2}+4b^{2}}}\le 1. \]
[/quote]
[hide="hint"]Just using Cauchy Schwarz inequality,
\[\left(\sum\sqrt{\frac{ab}{4a^{2}+b^{2}+4c^{2}}}\right)^{2}\le \left(\sum (a^{2}+4b^{2}+4c^{2})\right)\left(\sum\frac{ab}{(4a^{2}+b^{2}+4c^{2})(a^{2}+4b^{2}+4c^{2})}\right)=\frac{9(a^{2}+b^{2}+c^{2})\sum ab(4a^{2}+4b^{2}+c^{2})}{(4a^{2}+4b^{2}+c^{2})(4b^{2}+4c^{2}+a^{2})(4c^{2}+4a^{2}+b^{2})}\]
It suffices to prove
\[\frac{9(a^{2}+b^{2}+c^{2})\sum ab(4a^{2}+4b^{2}+c^{2})}{(4a^{2}+4b^{2}+c^{2})(4b^{2}+4c^{2}+a^{2})(4c^{2}+4a^{2}+b^{2})}\le 1 \]
which is easy.
[/hide]
For the orginal problem, see the solution in the note I have sent to you, Zaizai. ;)[/quote]

Yes, it is also my solution. But I use Jensen (instead of Cauchy-Schwarz). The left part is nice, too, not so easy. Who can complete it?  :)

[quote="zaizai-hoang"]$\bigstar$ [Hungkhtn's problem] Let $a,b,c$ are real number. Prove that:
$\frac{ab}{4a^{2}+b^{2}+4c^{2}}+\frac{bc}{4b^{2}+c^{2}+4a^{2}}+\frac{ca}{4c^{2}+a^{2}+4b^{2}}\le \frac{1}{3}$
[/quote]
This is my 245-th post
Use two lemmas here
\[\bigstar \]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=147258[/url]
And
\[\bigstar \]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=123742[/url]
The conclusion follows!

Nice, The Soul!
But my second inequality is unsolved until now on Mathlinks Forum. The identity I posted without proof is not a proper solution.
But there is a very nice proper solution. For this reason I proposed this inequality in Mathematical Reflections Journal.

The first inequality the Soul post belongs to me. I think he gets it from toanhocmuonmau's article who I send this problem to. Both of this inequality and my initial posts belongs to my article sent to mathematics and youth Magazines (vietnamese) a long time ago, which will be issued in the near time,

Well, why sometime you say that "this inequality belongs to me" but i can't? I created it independent, not take in toanhocmuonmau's articles! I'm not a thief! I'm sure that!

[quote="The soul of rock"]I'm sorry for spam but I've checked in toanhocmuonmau's article in Mathematical Reflections magazine but I found nothing like mine![/quote]

H...m I have something to say to you, the Soul. I did say that I send this problem to Toanhocmuonmau (and his article is not in Math reflection, it is only shared by someone), and I also did send this problem (in an article to Math and Youth Magazines) a long time ago (unpublished now, but in the future) because it is the fact and moreover, in the past, some of the pupils posted my problem in my Vietnamese book without mentioned its author. So I must say. I can believe that you also create it independently because I understand you, too.  :lol: We are Vietnamese friend, so keep peace in Mathlinks? It is better? Itsn't it?

Now return to the problem, we can prove the left by SOS. I think is is very nice. Manlio, you can prove it easily by yourself. If not, please tell me tomorrow.

[quote="The soul of rock"]Yes, I don't want to spam more and dont' want the other thinks some bad about Vietnamese! And toanhocmuonmau hadn't sent me any thing because I'm not his friend or brother![/quote]

No, I never think that Mathlinkers think badly about Vietnamese,   :roll: So why you think that...

To make this post not becoming a spam  :lol:, I will give a hint for the rest (following THMM and Manlio's posts)

WLOG, assume that $a^{2}+b^{2}+c^{2}=3$. It remains to prove that

\[3\sum ab(4-c^{2}) \le \prod (4-a^{2})\]
or
\[36\sum a^{3}(b+c)+9\sum a^{2}bc+9a^{2}b^{2}c^{2}\le 16\sum a^{4}+68\sum a^{2}b^{2}.\]

By AM-GM, we have
\[9a^{2}b^{2}c^{2}\le abc(a+b+c)(a^{2}+b^{2}+c^{2})=3abc(a+b+c)\]
So it remains to prove that

\[9\sum a^{3}(b+c)+3\sum a^{2}bc \le 4\sum a^{4}+17\sum a^{2}b^{2}.\]

This follow SOS method.

I think you are making mistake in writing to SOS form, manlio.  :lol:

[quote="manlio"]I have obtained

$\sum (a-b)^{2}(4a^{2}+4b^{2}+3c^{2}-10ab) \ge 0$

and SOS doesn't work, sorry. :(

I tried to use

If $S_{a}+S_{b}+S_{c}\ge 0$ and $S_{a}S_{b}+S_{b}S_{c}+S_{c}S_{a}\ge 0$ then $S \ge 0$[/quote]

Ok. I should finish the last step. It should be
\[\succ \lt(2a^{2}+2b^{2}+\frac{3c^{2}}{2}-5ab\rt)(a-b)^{2}\ge 0.\]
This is easy by SOS, Manlio.

Thank you. Now I see how to solve it. :)


geometry
3D geometry
function
number theory proposed
number theory

Consider the cubical lattice of all the points of the form (i, j, k), where each of 1 ≤ i, j, k ≤ N  is an integer.

Part 1. How many cubes are there whose vertices are lattice points and whose edges are parallel to the edges of the big cube? (as function of N)

Part 2. In addition if the condition that the edges of the cubes are parallel to the edges of the big cube is dropped - how many cubes are there whose vertices are lattice points?

It does seem interesting.

I just have the first part (which was easy to begin with :)): there are $(n-1)^3$ cubes of side $1$, $(n-2)^3$ cubes of side $2$, and so on. In total, there are $1^3+2^3+\ldots+(n-1)^3=\frac{(n-1)^2n^2}4$ cubes.


function
limit
real analysis
real analysis unsolved

I have a function of the form \[ f(x)\equal{}\frac{1}{a\plus{}i x}\minus{}\frac{1}{a\minus{}i x}\] Now with $ \lim_{a\rightarrow 0}$ we get $ f(x)\equal{}\frac{2}{i x}$. If we expand $ f(x)$ using power series we get \[ f(x)\equal{}\frac{\minus{}2}{a}\left[\frac{j x}{a}\plus{}\left(\frac{i x}{a}\right)^{3}\plus{}\cdots\right]\] Now how to prove that \[ \lim_{a\rightarrow 0}\frac{\minus{}2}{a}\left[\frac{j x}{a}\plus{}\left(\frac{i x}{a}\right)^{3}\plus{}\cdots\right]\equal{}\frac{2}{i x}\]

regards.
bhupala

[quote="bhupala"]I have a function of the form
\[ f(x)\equal{}\frac{1}{a\plus{}i x}\minus{}\frac{1}{a\minus{}i x}\]
Now with $ \lim_{a\rightarrow 0}$ we get $ f(x) \equal{}\frac{2}{i x}$. If we expand $ f(x)$ using power series we get
\[ f(x)\equal{}\frac{\minus{}2}{a}\left[\frac{j x}{a}\plus{}\left(\frac{i x}{a}\right)^{3}\plus{}\cdots\right]\]
[/quote]
The series converges only if $ |x| < |a|$. But then you can't take the  $ \lim_{a\rightarrow 0}$. So the statement cannot be proved.


select the random variable that is [b]not[/b] discrete
a)numbers are assigned according to the length of time it will take to get served at a bank
b)numbers are assigned to the number of days before a share price stops increasing
please help
Thanks

(a) would be the one thats not discrete.