calculus
integration
function
induction
search
Functional Analysis
real analysis

$a \in [0;1]$ an arbitrary real.
find all continuous functions $f$ that satisfie,
$f(x) = \int_0^{ax} f(t) dt$

Using power serie one obtains the solution
\[f(x)=\sum \frac{1}{n!}a^{i(i-1)/2}x^i,\]
which converges for every $x$ if $a\in[0,1]$.

Of course $f$ times a constant is also a solution. I don't know if $f$ has a better representation in a closed form.

Actually the answer is $f=0$ because, assuming that $|f|\leqslant M$ on $[-R,R]$, we get by induction
\[|f(x)|\leqslant \frac{M}{i!}a^{i(i+1)/2}|x|^i,\quad x\in[-R,R] \]
and the right hand side tends to $0$ if $a\in[0,1]$.

Xamog's function (with $(i-1)$ replaced by $(i+1)$ and $n$ replaced by $i$) is the unique solution of the equation $f(x)=1+\int_0^{ax}f(t)\,dt$.

An interesting question is whether there are non-trivial solutions for $a>1$.

or we can observe that $f -> \int_0^{ax} f(t) dt$ is a $a$-contraction in the set of continuous function with the uniform norm (banach space) and conclude with a fixed point theorem that the only solution is $f=0$.
(for the case $0<a<1$)

[quote="alekk"]or we can observe that $f -> \int_0^{ax} f(t) dt$ is a $a$-contraction in the set of continuous function with the uniform norm [/quote]
Unfortunately, this is literally true only if we consider a sufficiently short interval around $0$.  :)

If $\|f\|_{C[-R,R]}=M$, then the norm of the image in $C[-R,R]$ can be as large as $aRM$ (the integral does not exceed the maximum of the function [b]times the length of the interval of integration[/b]), so the mapping is a contraction in the uniform norm only if $aR<1$. I admit that this can be easily repaired by considering a slightly different norm or by some other tricks. I just meant that some extra care should be taken to make the mapping a contraction on large intervals. And, of course, when trying to reduce the problem to the theorem about contractive mappings, one has to work with finite intervals because $C(\mathbb R)$ is not a normed space.

:blush:  I forgot to say that $f \in C([0;1], \mathbb{R})$ ..
but it is interesting to search all $f \in C(\mathbb{R}, \mathbb{R})$ that satifies this condition.


Let  $a,b$ positive integers such that  $\frac{a}{b}<\sqrt{5}$

Prove that  $\sqrt{5}-\frac{a}{b}>\frac{1}{4ab}$

[hide="Proof"]If ${a\over b}\leq\frac{\sqrt{5}}{3}$, then $\sqrt{5}-\frac{a}{b}\geq\frac{2\sqrt{5}}{3}>\frac{1}{4}\geq\frac{1}{4ab}$

If ${a\over b}>\frac{\sqrt{5}}{3}$, then $b\sqrt{5}<3a\implies b\sqrt{5}+a<4a.$ On the other hand, $|5b^2-a^2|\geq 1$ for all natural $a,b$, since $5x^2-y^2=0$ can't have natural solution, on the reason of $\sqrt{5}$ being irrational.

Hence $b\sqrt{5}-a=\frac{5b^2-a^2}{b\sqrt{5}+a}>\frac{1}{4a}\implies \sqrt{5}-{a\over b}>\frac{1}{4ab}.$[/hide]


could we please have a geometry math jam? :shock: because I am really really really really bad at geometry.

In other words, I have no idea how to attack a problem unless it is a blatantly obvious use of a formula.

erm. by geometry, I mean AIME/USAMO level geometry.

I'll try to get some geometry in the upcoming USAMTS Math Jam, and we'll try to find some for the next 6+ difficulty Jam we have.

Can u post geometric figures in that classroom?

We provide links - you click on them and a diagram appears in a browser.


If the equation x^4 - x^3 + x + 1 = 0 has roots a, b, c, d, show that 1/a + 1/b is a root of x^6 + 3x^5 + 3x^4 + x^3 - 5x^2 - 5x - 2 = 0.


Prove that $2^n$ divides $(n+1)(n+2)....(2n)$.
It is obvious by induction, but theres a cute combinatorial interpretation.

[hide]The fraction is $\frac{(2n)!}{n!}$.

Now just use the formula for power of a prime dividing a factorial; for the denominator it is $\sum_{k=1}^\infty \left\lfloor \frac{n}{2^k} \right\rfloor$, and for the numerator it is $\sum_{k=1}^\infty \left\lfloor \frac{2n}{2^k} \right\rfloor$.

The first term in the summation for the numerator is $n$. The rest of the terms are the same as those in the summation for the denominator.

It follows that the product is divisible by $2^n$ (and not $2^{n+1}$).[/hide]

[quote="scorpius119"][hide]
The fraction is $\frac{(2n)!}{n!}$.
[/hide][/quote]

Alternatively this is equal to $n!{{2n}\choose{n}}$


Hi,

the most important tutorials you need to have are provided onto the net at:

- http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf (general tutorial)

- http://ctan.cict.fr/info/companion-rev/ch8.pdf (math formulae)

Please browse the CTAN server it is really a valuable source. These links, standard resources on LaTex, can be recommended too:

- http://www.latex-project.org/intro.html
- http://www.tug.org.in/tutorials.html

BTW which packagges are installed ?  :)

The standard packages used are (unless Valentin has added more):

\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}


superior algebra
superior algebra unsolved

assume that for each $x$ in a ring $R$ there exists an integer $n\geq 2$such that $x=x^2+\dots +x^n$.must $R$ be commutative?.


[img]http://img132.imageshack.us/img132/8680/sumpj3.jpg[/img]

What can I do with the upper bound of the summation?

[hide="Solution"]
We can expand the summation and write it as:
$ (2^{n\minus{}1}\plus{}\cdots\plus{}2^2\plus{}2^1\plus{}2^0)\plus{}(2^{n\minus{}2}\plus{}\cdots\plus{}2^0)\plus{}\cdots\plus{}(2^1\plus{}2^0)\plus{}(2^0).$

Then we can rewrite the summation as:
\[ \sum_{k\equal{}0}^{n\minus{}1} (n\minus{}k)2^k\equal{}\sum_{k\equal{}0}^{n\minus{}1} \left(\sum_{i\equal{}0}^k 2^i\right)\equal{}\sum_{k\equal{}0}^{n\minus{}1} \left(2^{k\plus{}1}\minus{}1\right)\equal{}2\times \left(\sum_{k\equal{}0}^{n\minus{}1} 2^k\right)\minus{}n\equal{}2\left(2^n\minus{}1\right)\minus{}n.\]
[/hide]


Let $a,b,c$ be reals with $0 \le a \le 1$ and $ab+(1-a)c \ge 0$. The equation

$ae^{-bx}+(1-a)e^{-cx}= 1$

has a positive real solution for $x$. Let $x(a,b,c)$ be the solution for some $a,b,c$. For a part of my research project, I want to find a way of determining the lesser of the two values $x(a,b,c)$ and $x(a,b',c')$ for any $a, b, c$ and $a, b', c'$ satisfying the conditions above.

Anyone have any ideas?

Trial-error says
a=1, b=0, c=any number
a=0, b=any number, c=0

Of course, this is not a real way for solution.

[quote="Salviati"]Trial-error says
a=1, b=0, c=any number that is $\geq 0$
a=0, b=any number, c=0

Of course, this is not a real way for solution.[/quote]
So u r saying u tried all the $\infty$ different cases?

[quote="rem"]So u r saying u tried all the $\infty$ different cases?[/quote]
No, and I don't [i]have to[/i].

Let me clarify. $a, b, c$ are given constants; I'm solving the equation for $x$. There is also a mistake in the problem statement. I want to compare

$x(a,b,c)$ and $x(a,b',c')$.

So, for example, which is bigger:

$x\left(\frac{1}{3},-3, 2\right)$ or $x\left(\frac{1}{3},-4, 3\right)$?


vector
geometry
circumcircle
geometric transformation
reflection
parallelogram
trapezoid

Let $ ABCD$ be a cyclic quadrilateral with $ AB>CD$ and $ BC>AD$. Take points $ X$ and $ Y$ on the sides $ AB$ and $ BC$, respectively, so that $ AX\equal{}CD$ and $ AD\equal{}CY$. Let $ M$ be the midpoint of  $ XY$. Prove that $ AMC$ is a right angle.

Phew.  I've been working on this one for while and still can't get anywhere.   :(

Vectors!

Take the origin to be the circumcenter of $ ABCD$.

$ X=(1-\alpha) A + \alpha B$ and \[ |C-D|=|X-A|=\alpha|B-A|\]
So, \[ X = A + \frac{(B-A)}{|B-A|}|C-D|\]
Likewise,
\[ Y = C + \frac{(B-C)}{|B-C|}|A-D|\]

First: as $ A,B,C,D$ are concyclic, $ \angle(AD,CD)=180-\angle(AB,CB)$
Thus, \[ \frac{(B-A)\cdot(B-C)}{|B-A\parallel{}B-C|}=-\frac{(D-A)\cdot(D-A)}{|D-A\parallel{}D-C|}\qquad (1)\]

We wish to prove that $ A,\frac{X+Y}{2},\text{ and }C$ form a right triangle - that is,
\[ \left(\frac{X+Y}{2}-A\right)\cdot\left(\frac{X+Y}{2}-C\right)=0\]

This is clearly equivalent to:\[ \left(\frac{(B-A)}{|B-A|}|C-D|+\frac{(B-C)}{|B-C|}|A-D|\right)^2=(A-C)^2\]

Let's expand the left side and appeal to $ (1)$.
\[ \begin{eqnarray*}\left(\frac{(B-A)}{|B-A|}|C-D|+\frac{(B-C)}{|B-C|}|A-D|\right)^2 &=& |C-D|^2+|A-D|^2+ \ 2\left[\frac{(B-A)\cdot(B-C)}{|B-A\parallel{}B-C|}\right]|C-D\parallel{}A-D|\\ &=& |C-D|^2+|A-D|^2 - 2(C-D)\cdot(A-D)\qquad\text{from }(1) \\ &=& \left( (C-D) - (A-D) \right)^2 = (A-C)^2\]

which therefore establishes that $ AM$ is perpendicular to $ AC$, i.e. the goal.

Let $ A'$ be the reflection of $ A$ with respect to $ M$. In this case, the quadrilateral $ AXA'Y$ is a parallelogram and therefore $ YA'\equal{}AX\equal{}CD$, $ CY \equal{} AD$ and $ \angle{A'YC}\equal{}180^{\circ} \minus{} \angle{A'YB} \equal{} 180^{\circ} \minus{}\angle{CBA} \equal{} \angle{CDA}$. Thus, the triangles $ ACD$ and $ A'CY$ are congruent, hence $ AC \equal{} A'C$, and since $ M$ is the midpoint of $ AA'$, it follows that $ CM \perp AM$.

I'm posting my solution, which I've sent to bboypa yesterday, just to show how two solutions of two different people can be similar, even in wording:

Let $ A'$ be the reflection of the point $ A$ through the point $ M$. Then quadrilateral $ AXA'Y$ is a parallelogram since its diagonals bisects each other. Therefore $ YA' \equal{} AX \equal{} CD$ and $ YA' \parallel AX$. It follows that $ \angle CYA' \equal{} 180^{\circ} \minus{} \angle ABC \equal{} \angle ADC$ since quadrilateral $ ABCD$ is cyclic. The given condition $ CY \equal{} AD$ implies that triangles $ CYA'$ and $ ADC$ are congruent (Side - Angle - Side). It means that $ CA' \equal{} CA$ and the triangle $ ACA'$ is isosceles. Since $ AM\equal{}MA'$ it follows that $ CM \perp AM$ and the proof is finished.

[quote="TomciO"]I'm posting my solution, which I've sent to bboypa yesterday, just to show how two solutions of two different people can be similar, even in wording:

Let $ A'$ be the reflection of the point $ A$ through the point $ M$. Then quadrilateral $ AXA'Y$ is a parallelogram since its diagonals bisects each other. Therefore $ YA' \equal{} AX \equal{} CD$ and $ YA' \parallel AX$. It follows that $ \angle CYA' \equal{} 180^{\circ} \minus{} \angle ABC \equal{} \angle ADC$ since quadrilateral $ ABCD$ is cyclic. The given condition $ CY \equal{} AD$ implies that triangles $ CYA'$ and $ ADC$ are congruent (Side - Angle - Side). It means that $ CA' \equal{} CA$ and the triangle $ ACA'$ is isosceles. Since $ AM \equal{} MA'$ it follows that $ CM \perp AM$ and the proof is finished.[/quote]

Excellent solution Tomci0 & pohoatza!

@pohoatza..i'll attend also you next round if you have enough time  :wink:

I proved that pythagorean theorem is true in the triangle $ \triangle AMC$ by using AL-KACHI and median theorems.

I don't know reflections, can someone give me a link? -- Wikipedia isn't very clear, i found.

I've just noticed that you are talking about what i call "central symetrie". ^^"

Seeing Pohoatza's, one could think: "What else could be added?"

However, here's my approach:
Name K and L the intersection of AB and CD, respectively BC and AD. Also name N and P the middle of AD and CX respectively.
I shall use the following properties:
1) A quadrilateral having 2 opposite sides equal, has the bisector of the angle made by them parallel to the line joining the midpoints of the other 2 sides
2) In a cyclic quadrilateral KABCLD, the bisectors of the angles BKC and BLA are perpendicular. (I do not see necessary to prove the above statements, they are well known)
According to (1), NP is parallel to the inner angle bisector of <BKC, so we intend to prove AM||NP, or AMPN an isosceles trapezoid. It has AN=MP, both being half of AD, and MP||CY. If the angle bisector of <BKC cuts AD and BC at N' and P', then the angles AKN' and BKP' are supplementary (add 180 deg), the triangles KAD and KCB being similar, therefore <ANP = <NPM and AMPN is isosceles trapezoid, consequently AM ||NP||KP'.
Same way CM is parallel to the other angle bisector.

Best regards,
sunken rock


LaTeX
trigonometry
real analysis
real analysis unsolved

Show that the following sum is convergent :

[img]http://xcalibermath.files.wordpress.com/2008/11/picture-13.png[/img]

This forum has [[Latex]] enabled (see that link and also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]), which will make your formulas easier to read.  For this problem, it's enough to use $ 0 \leq \sin x \leq x$ for $ x \in [0, 1]$ together with things you probably know about series of the form $ \sum_{n \equal{} 1}^\infty \frac{1}{n^p}$.

[quote="JBL"]This forum has [[Latex]] enabled (see that link and also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]), which will make your formulas easier to read.  For this problem, it's enough to use $ 0 \leq \sin x \leq x$ for $ x \in [0, 1]$ together with things you probably know about series of the form $ \sum_{n \equal{} 1}^\infty \frac {1}{n^p}$.[/quote]
Let me write the solution,to help everybody understand:
As our friend said we have $ 0 \leq \sin(\frac{1}{n}) \leq \frac{1}{n}$ so if we put $ \frac{1}{n}$ instead of $ sin(\frac{1}{n})$,we have to prove that $ \sum_{n\minus{}1}^\infty \frac{\frac{1}{n}}{\sqrt{n}}$ is convergence.it doesn't matter that our sum begins from $ 1$ or $ n\minus{}1$ becuase the difference is finte sentenses and it doesn't do anything with convergence!So we prove that $ \sum_{1}^\infty \frac{1}{n^{\frac{3}{2}}}$ is convergence.The series in the form of $ \sum_{1}^\infty \frac{1}{n^{p}}$ are convergence if $ p>1$ and here we have $ \frac{3}{2}>1$ so our seire is convergence.


Find all right triangles with integer leg lenghths and hypothenuse of length $ \sqrt{2006}$.

[quote="modularmarc101"]Find all right triangles with integer leg lenghths and hypothenuse of length $ \sqrt {2006}$.[/quote]

Such a triangle doesn't exist.

Let $ x,y$ be the length of two legs.

Then $ x^2\plus{}y^2\equal{}2006\equal{}2 \times 17 \times 59$

As $ 59\equal{}3(mod 4)$, $ 59$ must divide both $ x$ and $ y$

$ \Rightarrow 59^2 |2006$, which is wrong , so no such triangle exists.


limit
function
calculus
derivative
real analysis
real analysis unsolved

Let $N$ be a norm on a finite dimensional vectorial space. Is it true than it has a "Gateux differential" at any point and for any direction? ( "Gateaux differential" means here that we restrict do right derivatives).

So we're supposed to find out whether for any choice of $x$ and $u$, the limit $\lim_{t\searrow 0}\frac{\|x+tu\|-\|x\|}t$ exists and is finite? (that $\searrow$ means that $t$ decreases to $0$). If so, I think it's true.

Denote that expression after $\lim$ by $f(t)$. Let $t_1,t_2>0$, and $\lambda\in(0,1)$. We have $f(\lambda t_1+(1-\lambda)t_2)\le\frac{\lambda(\|x+t_1u\|-\|x\|)+(1-\lambda)(\|x+t_2u\|-\|x\|)}{\lambda t_1+(1-\lambda)t_2}$ (because of the convexity of the norm). This last expression is not larger than $\max(f(t_1),f(t_2))$. We have thus found that the concinuous function $f: (0,\infty)\to\mathbb R$ reaches its maximum on any interval $[t_1,t_2]\subset(0,\infty)$ in one of the endpoints of the interval. There are thus two cases: either this function is montonic on the whole $(0,\infty)$, or it reaches a global minimum in $t_0>0$, and it's decreasing on $(0,t_0)$ and increasing on $(t_0,\infty)$. In both cases, it has a limit in $0$. The only trouble we could run into is this limit being $\pm\infty$, but this is impossible, since $|f(t)|\le\|u\|$, as can easily be checked.

I only used that $tf(t)$ is convex and vanishes in 0. Since any convex function has right derivatives everywhere, the conclusion follows.

Silly me! :) We don't even need to mention convexity and all that. Just notice that $f$ is increasing on $(0,\infty)$ (and this can be shown by a simple application of the triangle inequality).


On each of three days the probability of rain is $ 50\%$. Assuming independence of events, express as a decimal percent the probability that it will rain exactly two of the three days.


vector
geometry
3D geometry
tetrahedron
Pythagorean Theorem
linear algebra

19. If [i]a[/i], [i]b[/i], and [i]c[/i] are the x-intercept, the y-intercept, and the z-intercept of a plane, respectively, and [i]d[/i] is the distance from the origin to the plane, show that:

$ \frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$

---

Hello there,

I am showing my work below. Could you please tell me where I've erred because my final step is different from the equation above?

Thank you.

---

I let:

a = (x, 0, 0)
b = (0, y, 0)
c = (0, 0, z)

Now:

First direction vector = [0, y, 0] - [x, 0, 0] = [-x, y, 0]

Second " " = [0, 0, z] - [0, y, 0] = [0, -y, z]

Therefore:

$ \vec{\ell} = [x, 0, 0] + s[-x, y, 0] + t[0, -y, z]$

Conversion of $ \vec{\ell}$ into scalar form:

$ [-x, y, 0] \times [0, -y, z] = [yz, xz, xy]$.

$ xyz + xyz + xyz + D = 0$

Subsitution of $ (x, 0, 0)$ to find D:

$ x^2yz + D = 0 \Rightarrow D = -x^2yz$

Since [i]d[/i] = $ \frac{| Ax + By + C + D |}{\sqrt{A^2 + B^2 + C^2}}$, and using (x, 0, 0):

[i]d[/i] = $ \frac{| xyz - x^2yz |}{\sqrt{y^2z^2 + x^2y^2 + x^2z^2}}$

[i]d[/i] = $ \frac{\sqrt{(xyz)^2 - (x^2yz)^2}}{\sqrt{y^2z^2 + x^2y^2 + x^2z^2}}$

$ d^2 = \frac{x^2y^2z^2 - (x^2yz)^2}{y^2z^2 + x^2y^2 + x^2z^2}$

$ \frac{1}{d^2} = \frac{y^2z^2 + x^2y^2 + x^2z^2}{x^2y^2z^2 - x^4y^2z^2}$

If I let a = x, b = y, and c = z, then the equation given by the question is:

$ \frac{1}{d^2} = \frac{y^2z^2 + x^2y^2 + x^2z^2}{x^2y^2z^2}$

However, my equation has an additional $ -x^4y^2z^2$.

[quote="catalan"]
a = (x, 0, 0)
b = (0, y, 0)
c = (0, 0, z)

[/quote]

Why do you set like that? You can directly write the equation of the plane as $ \frac{x}{a}\plus{}\frac{y}{b}\plus{}\frac{z}{c}\equal{}1.$

We also can use  pure geometry to prove this nice relation:

Let $ A,B,C$ be the traces on the axes $x,y,z$. $\overline{OA} \equal{} a$ , $\overline{OB} \equal{} b$ , $\overline{OC} \equal{} c$

Volume of the tetrahedron $OABC$ is given by $ [OABC] \equal{} \frac {_1}{^3}d \cdot [ABC]$

But, since trihedron $O (ABC)$ is orthogonal we have $ [OABC] \equal{} \frac {_1}{^6}abc$

$\Longrightarrow \ 2d \cdot [ABC] \equal{} abc \ \Longrightarrow \ \frac {1}{d^2} \equal{} \frac {4[ABC]^2}{a^2b^2c^2}$

Plane $ \alpha$ passing though $OA$ orthogonal to the plane $ OBC$ cuts $ BC$ at the foot $ H$ of the altitude $ AH \equal{} h$ from $ A$ to the edge $ BC$. By Pythagorean theorem we get then

$ [ABC]^2 \equal{} \frac {_1}{^4}h^2(b^2 \plus{} c^2) , \ h^2 \equal{} a^2 \plus{} OX^2 \equal{} a^2 \plus{} \frac {b^2c^2}{ b^2 \plus{} c^2}$

$\Longrightarrow \ [ABC]^2 \equal{} \frac {_1}{^4}(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)$

$\Longrightarrow \  \frac {1}{d^2} \equal{} \frac { a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2}{a^2b^2c^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$

For $ A(a,\ 0,\ 0),\ B(0,\ b,\ 0),\ C(0,\ 0,\ c)\ (abc\neq 0)$, Decarte Gua's Theorem holds:

$ \boxed{(\triangle{ABC})^2\equal{}(\triangle{OAB})^2\plus{}(\triangle{OBC})^2\plus{}(\triangle{OCA})^2}$


number theory
relatively prime
number theory proposed

Show that the equation $ x^2\plus{}y^5\equal{}z^3$ has infinitely many solutions in nonzero integers $ x,y,z$.

[quote="moldovan"]Show that the equation $ x^2 \plus{} y^5 \equal{} z^3$ has infinitely many solutions in nonzero integers $ x,y,z$.[/quote]

$ x\equal{}2^{15a\plus{}10}$, $ y\equal{}2^{6a\plus{}4}$, $ z\equal{}2^{10a\plus{}7}$, where $ a$ is a positive integer. Then $ x^2\plus{}y^5\equal{}2^{30a\plus{}20}\plus{}2^{30a\plus{}20}\equal{}2^{30a\plus{}21}\equal{}z^3$.

@MTM :  this  is   a  soln  by  substitution.

but can u provide  a more general solution ?

The most general solution is quite hard to find because, if you find a solution $ (x_0,y_0,z_0)$ to $ n^4x^2\plus{}y^5\equal{}n^4z^3$, for any positive integer $ n$, then $ (n^3x_0,n^2y_0,n^2z_0)$ is a solution to $ x^2\plus{}y^5\equal{}z^3$, so it seems difficult (if it is at all possible) to claim that any two numbers out of $ (x,y,z)$ are relatively prime, or that if a certain prime divides both, then which is the multiplicity with which it does, or any other similar statement.  Simplifying the equation this way also seems difficult.  Finally, I have played around a little bit with remainders to see if anything seemed to come out of it, and at the first glance it does not seem to give much additional information.  I think that the problem is stated as it is (ie, prove the infinitude of the solutions, rather than finding a general solution) for some reason...  It is of course always desirable to find the general solution, but I am afraid that in this case it is quite hard if possible at all.

equations of the form $ x^a\plus{}y^b\equal{}z^c$ tend to have infinitely many [url=http://www.mathlinks.ro/viewtopic.php?t=150638]solutions[/url] ... sometimes :D


inequalities
algebra
polynomial
inequalities proposed

[quote="toanhocmuonmau"]Let $a,b,c$ be positive real numbers. Prove that
\[\underset{cyc}{\sum}\frac{a^{3}}{2a^{2}-ab+2b^{2}}\ge \frac{a+b+c}{3}\]
:)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271

Your solution is very nice and strange, Aqardy. But, Can you solve it by using SOS? I used SOS to solve it! :) In this problem, the theorem of Hungkhtn is failed. :D But, i have a way to overcome it. How about you? :)

I think argady's solution used to SOS method  :D
[quote]In this problem, the theorem of Hungkhtn is failed[/quote]
Really? I will try... :o

[quote="toanhocmuonmau"]Let $ x,y,z$ be positive real numbers. Prove that

$ \frac {x^{3}}{2x^{2} \minus{} xy \plus{} 2y^{2}}\plus{}\frac {y^{3}}{2y^{2} \minus{} yz \plus{} 2z^{2}}\plus{}\frac {z^{3}}{2z^{2} \minus{} zx \plus{} 2x^{2}}\ge \frac {x \plus{} y \plus{} z}{3}.$[/quote]$ \sum_{cyc}\frac{x^3}{2x^2\minus{}xy\plus{}2y^2}\minus{}\frac{x\plus{}y\plus{}z}{3}\equiv\frac{8F(x,y,z)}{3(2x^2\minus{}xy\plus{}2y^2)(2y^2\minus{}yz\plus{}2z^2)(2z^2\minus{}zx\plus{}2x^2)}$.

Without loss of generality, we assume $ x\equal{}\min\{x,y,z\}$, then

$ F(x,y,z)\equal{}F(x,x\plus{}s,x\plus{}t)$

$ \equal{}18(s^2\minus{}st\plus{}t^2)x^5\plus{}9(3s^3\plus{}5s^2t\minus{}4st^2\plus{}3t^3)x^4$

$ \plus{}3(7s^4\plus{}22s^3t\plus{}9s^2t^2\minus{}14st^3\plus{}7t^4)x^3$

$ \plus{}3(2s^5\plus{}13s^4t\plus{}17s^3t^2\minus{}7s^2t^3\minus{}2st^4\plus{}2t^5)x^2$

$ \plus{}3st(2s^4\plus{}12s^3t\minus{}2s^2t^2\minus{}3st^3\plus{}2t^4)x\plus{}2s^2t^2(2s^3\plus{}4s^2t\minus{}5st^2\plus{}2t^3)\geq 0$,

which is true for

$ 2s^5\plus{}13s^4t\plus{}17s^3t^2\minus{}7s^2t^3\minus{}2st^4\plus{}2t^5$

$ \equal{}s^3t^2\plus{}(2s\plus{}t)(s^4\plus{}6s^3t\plus{}5s^2t^2\minus{}6st^3\plus{}2t^4)\geq 0$,

$ 2s^4\plus{}12s^3t\minus{}2s^2t^2\minus{}3st^3\plus{}2t^4\equal{}s^4\plus{}8s^3t\plus{}(s\plus{}t)^2(s\minus{}t)^2\plus{}t(s\plus{}t)(2s\minus{}t)^2\geq 0$.

By the way, for all positive real numbers $ x,y,z$, the following inequality

$ \frac {x^{3}}{x^{2} \plus{}kxy \plus{} y^{2}}\plus{}\frac {y^{3}}{y^{2} \plus{} kyz \plus{} z^{2}}\plus{}\frac {z^{3}}{z^{2} \plus{}kzx \plus{} x^{2}}\ge \frac {x \plus{} y \plus{} z}{2\plus{}k}$

holds if and only if $ k\geq k_{0}\equal{} \minus{}0.662559\cdots$, which is a root of the irreducible polynomial

$ 729k^{17}\minus{}2889k^{16}\minus{}5771k^{15}\plus{}34066k^{14}\plus{}16813k^{13}\minus{}192131k^{12}\minus{}25716k^{11}$

$ \plus{}678459k^{10}\plus{}137803k^{9}\minus{}1568946k^{8}\minus{}888013k^{7}\plus{}2069899k^{6}\plus{}2790923k^{5}$

$ \plus{}86614k^{4}\minus{}2956448k^{3}\minus{}3464400k^{2}\minus{}1778800k\minus{}351200$;

with euqality if $ k\equal{}k_{0}$, $ x\equal{}1$, $ y\equal{}2.98559\cdots$, which is a root of the irreducible polynomial

$ 1458y^{51}\plus{}2862y^{50}\minus{}287725y^{49}\minus{}2292447y^{48}\minus{}3389726y^{47}\plus{}35029716y^{46}$

$ \plus{}191856227y^{45}\plus{}241984111y^{44}\minus{}1125621085y^{43}\minus{}5505453737y^{42}$

$ \minus{}8532450271y^{41}\plus{}8454375689y^{40}\plus{}65602099289y^{39}\plus{}130519390177y^{38}$

$ \plus{}81117260652y^{37}\minus{}185757926602y^{36}\minus{}541590214013y^{35}\minus{}705812878707y^{34}$

$ \minus{}878525333672y^{33}\minus{}1991344532544y^{32}\minus{}4090380074568y^{31}\minus{}3669795432380y^{30}$

$ \plus{}5394798247218y^{29}\plus{}24717623845842y^{28}\plus{}45281990356958y^{27}$

$ \plus{}52184214131982y^{26}\plus{}37885327761978y^{25}\plus{}9754039178242y^{24}$

$ \minus{}15465116379282y^{23}\minus{}24110693356474y^{22}\minus{}16873207892348y^{21}$

$ \minus{}4529464269040y^{20}\plus{}4026019761432y^{19}\plus{}6084313906464y^{18}$

$ \plus{}1802241070859y^{17}\minus{}1366432312511y^{16}\minus{}329760432402y^{15}\plus{}89611807936y^{14}$

$ \minus{}54720300541y^{13}\plus{}20403168935y^{12}\minus{}4333850545y^{11}\plus{}1491466779y^{10}$

$ \plus{}2591357253y^{9}\minus{}2171718139y^{8}\plus{}591451185y^{7}\plus{}31309185y^{6}\minus{}74219784y^{5}$

$ \plus{}25778002y^{4}\minus{}4623261y^{3}\plus{}353061y^{2}\plus{}5994y\minus{}1458$,

$ z\equal{}5.66003\cdots$, which is a root of the irreducible polynomial

$ 1458z^{51}\minus{}5994z^{50}\minus{}353061z^{49}\plus{}4623261z^{48}\minus{}25778002z^{47}\plus{}74219784z^{46}$

$ \minus{}31309185z^{45}\minus{}591451185z^{44}\plus{}2171718139z^{43}\minus{}2591357253z^{42}\minus{}1491466779z^{41}$

$ \plus{}4333850545z^{40}\minus{}20403168935z^{39}\plus{}54720300541z^{38}\minus{}89611807936z^{37}$

$ \plus{}329760432402z^{36}\plus{}1366432312511z^{35}\minus{}1802241070859z^{34}\minus{}6084313906464z^{33}$

$ \minus{}4026019761432z^{32}\plus{}4529464269040z^{31}\plus{}16873207892348z^{30}$

$ \plus{}24110693356474z^{29}\plus{}15465116379282z^{28}\minus{}9754039178242z^{27}$

$ \minus{}37885327761978z^{26}\minus{}52184214131982z^{25}\minus{}45281990356958z^{24}$

$ \minus{}24717623845842z^{23}\minus{}5394798247218z^{22}\plus{}3669795432380z^{21}$

$ \plus{}4090380074568z^{20}\plus{}1991344532544z^{19}\plus{}878525333672z^{18}\plus{}705812878707z^{17}$

$ \plus{}541590214013z^{16}\plus{}185757926602z^{15}\minus{}81117260652z^{14}\minus{}130519390177z^{13}$

$ \minus{}65602099289z^{12}\minus{}8454375689z^{11}\plus{}8532450271z^{10}\plus{}5505453737z^{9}$

$ \plus{}1125621085z^{8}\minus{}241984111z^{7}\minus{}191856227z^{6}\minus{}35029716z^{5}\plus{}3389726z^{4}$

$ \plus{}2292447z^{3}\plus{}287725z^{2}\minus{}2862z\minus{}1458$.


geometry
incenter
inradius
rectangle
geometry unsolved

In triangle ABC, angle B is a right angle, E is the incenter, r is the inradius, EF is perpendicular to AE, FG is perpendicular to BC, ED is perpendicular to CE, and DH is perpendicular to AB. Prove that H, E, and G are collinear.

Hello everyone! :)
Just a small clarification, I guess this should be:
Let $\triangle ABC$ have $\angle B = 90^{o}$ and incentre $E$. Let $F$ and $D$ be on $AC$ such that $\angle AEF = \angle CED = 90^{o}$, and $G$ and $H$ be the feet of the perpendiculars from $F$ onto $BC$ and $D$ onto $AB$ respectively. Prove that $H, E, G$ are collinear. But please correct me if I'm wrong.
Anyway, I solved it like this:
Let $EF$ meet $AB$ at $R$, $P$ be the foot of the perpendicular from F to $AB$, and $AC$ be tangent to the incircle at $Y$. Since $\angle AER = \angle AEF = 90^{o}$ and $AE$ bisects $\angle A$, we get $\triangle AER \equiv \triangle AEF$ and $RE=EF$. Also, $\angle PRF = 90^{o}-\frac{\angle A}{2}= \angle EFY$ as well as $\angle EYF = 90^{o}= \angle FPR$ shows that $\triangle EFY \approx \triangle FRP$, so that $\frac{PF}{RF}=\frac{EY}{EF}$, and $RF=2EF$ and $EY=r$ (the radius of the incircle) show that $PF = 2r$. The right angles at $P, B, G$ show that $PBGF$ is a rectangle, so $BG = 2r$. Similarly, $BH = 2r$. Thus $G$ and $H$ are opposite vertices of a square with all sides tangent to the incircle, so $GH$ passes through $E$.
Thanks!


I'm not quite sure how this concept can be applied.  For instance

M&IQ 1991

$ f(1 \minus{} x) \equal{} f(x) \plus{} 1 \minus{} 2x$

I did $ a\rightarrow 1 \minus{} x$, $ b\rightarrow x$ to get $ f(a) \minus{} f(b) \equal{} a \minus{} b$.
$ f(a) \plus{} a \equal{} f(b) \plus{} b$
If $ g(x)\equal{}f(x)\plus{}x$, then $ g(1\minus{}a)\equal{}g(a)$.
If $ h(x)\equal{}g(x\plus{}\frac{1}{2})$, then $ h(\frac{1}{2}\minus{}x)\equal{}h(\frac{1}{2}\plus{}x)$.
$ j(x)\equal{}h(x\minus{}\frac{1}{2})$, then $ j(x)\equal{}j(\minus{}x)$.

So would my answer be any even function translated $ \frac{1}{2}$ units right?

[quote="azn6021023"] $ f(a) \minus{} f(b) \equal{} a \minus{} b$.
$ f(a) \plus{} a \equal{} f(b) \plus{} b$
[/quote]

This isn't equivalent.

My bad.  It should be $ f(a)\minus{}a\equal{}f(b)\minus{}b$.


Prove that $ \sum_{n \equal{} 4}\frac {1}{2^{n \minus{} 2} \plus{} ( \minus{} 1)^{n \minus{} 1}} < 7/12$

I don't konw how to type ....

[quote="boxedexe"]Is your sum supposed to range from some number to $ n\geq 4$?  It seems that your $ n$ is given, and so cannot vary.[/quote]

Sorry,I didn't express well at first....Now I have editted it ....Many thanks.

Anyone have a solution,or give me some tips......I couldn't work it out.... :(


trigonometry function Gauss inequalities algebra polynomial calculus

This problem is taken from University of South Carolina High School Math contest. This may be easy to you guys but it's tough to me because I haven't dealt with trigonometry yet. I only know some basic rules.

   x+(1/x) = 2 cos 12 :^o:
   What is the value of x :^5: + (1/x :^5: )?

Thanks in advance.

The first step[hide] is to figure out how to express (x :^5: + (1/x :^5:) in terms of (x + (1/x)). [/hide] Do you know how to do that?

You could just [hide]use the identity x^(n+1)+x^(-(n+1))= (x^n + x^-n)(x+1/x) - (x^(n-1) + x^(-(n-1))), and compute x^5 + 1/x^5 recursively.[/hide]

Right, that's the usual strategy, and it works.  But in this special case, with the problem already involving a trig function, I think it will be easier to [hide]solve the first equation for x.  It comes out to a nice solution.  Then you can evaluate its fifth power pretty quickly.[/hide]

Right. This problem is actually easy. I did try Ravi B's method but when i caculate 2 cos 12 :^o: , the calculator came up with an error so I needed to ask for help. What I did wrong was I forgot to change from the radian mode to the degree mode.

Thanks, guys.

There are several ways to solve this one, but the quickest way is to use DeMoivre's Theorem: eit = cos (t) + i*sin (t).

Let x = e12i = cos (12) + i*sin (12).
Then 1/x = e-12i = cos (12) - i*sin (12),
and so x + 1/x = 2cos (12).

This x (or 1/x) is the solution to the first equation that's given (notice that it is complex, which is why your calculator might have given you an error).

Now use DeMoivre's Theorem again to say:
x5 = cos (5*12) + i*sin (5*12),
and 1/x5 = cos (5*12) - i*sin (5*12).
So x5 + 1/x5 = 2*cos (60) = 1.

[Note: Everything is done in degrees here.]

Huh, cute.  Your calculator (if scientific) won't be able to do it, though.

gauss, nice solution, i haven't seen that one before!

At the risk of being overkill, I'll mention that this topic is related to something called Chebyshev polynomials: see [url]http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html[/url].  The nth Chebyshev polynomial expresses cos(nt) as a polynomial in cos(t).  Equivalently, as gauss showed, it expresses (xn + 1/xn)/2 as a polynomial in (x + 1/x)/2.

i finanly get how the e^(i*x) (euler's formula) is proved since i'm leraning taylor polynomials at school

summation((i*x)^n/n!,0,infinity)

also you can take the derivative of the expression a=(cos(x)+i*sin(x))

-sin(x) +i*cos(x) dx = i *(cos(x)+i*(sin(x))) dx = i*a dx

Int(da\a)=Int(i*dx). so ln(a)=i*x a=e^(i*x)


geometry
geometric transformation
reflection
angle bisector
geometry solved

[quote="behzad"]Let $ABC$ be a triangle and $M$ be midpoint of $BC$ and $X$ be where incircle touches $BC$.$A'$ is where the symmetric of $AX$ with respect to angle bisector of $\angle A$ , intersects the incircle($A'$ is closer to $A$ than second intersection).$l_a$ is the line which passes through $A',M$. $l_b,l_c$ are defined such as $l_a$.Prove that $l_a,l_b,l_c$ are concurrent.[/quote]

The answer is that simple: The lines $l_a$, $l_b$, $l_c$ are not concurrent.
What is true (but I guess you know that) is that they are concurrent if A' is the point of intersection of the reflection of AX in the angle bisector of angle CAB with the incircle such that A' is [b]farther from A[/b] than the second intersection (and similarly, of course, B' and C'). But in fact, in this case, A' is simply the reflection of the point X in the angle bisector of angle CAB, and we arrive at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1307]IMO 1982 problem 2[/url].

  darij


geometry
geometric transformation
rotation
analytic geometry
reflection
3D geometry
frustum

Quadrilateral $ABCD$ is constructed where $A(0,0)$, $B(3,-1)$, $C(6,2)$, and $D(2,3)$ are the points on a Cartesian coordinate plane. What is the volume of the locus of points $ABCD$ rotated about the x-axis?

[quote="Altheman"]the volume is: the distance from the centroid to the axis of rotation * the area of the quadrilateral[/quote]
Could you please explain this a little more? I want to make sure I know why everything works.
Sorry for my stupidity. :(

[quote="Farenhajt"]I'd always recommend sketching a picture. See the attached diagram.

First you see that the part under the x-axis won't affect the total volume, which is proven if you reflect $B$ over to $B'$ and see that $AB'E$ falls completely within $AECD$. Therefore you disregard it, and the rest is all cones :)

Basically, you have: Cone $AD_{1}D$ plus cone $DD_{1}F$ minus cone $CC_{1}F$ minus cone $CC_{1}E$.

Or, if you're fluent in frustums: Cone $AD_{1}D$ plus frustum $DD_{1}C_{1}C$ minus cone $CC_{1}E$.

Of course, you need to find points $E$ and $F$ first, but I'll leave that to you :)[/quote]
This is exactly what I had, but I didn't get the right answer. Anyways, I'll just check over my work knowing that I was doing the right thing. Thanks.

[url]http://mathworld.wolfram.com/PappussCentroidTheorem.html[/url]

this is where my formula comes from

There IS a possible catch though, depending on what's the convention.

The natural approach is to disregard $ABE$, but some authors consider areas lying opposite the rotation axis to produce NEGATIVE volume, hence we would actually be having a pentagon $AB'ECD$ rotating. If that's the case, you would still have to subtract two cones which have $BB'$ for diameter and $A, E$ as apices. The results are ${86\over 3}\pi$ and ${82\over 3}\pi$ respectively.

Farenhajt

How did you get that image into .GIF format? I can't attach anything I draw using Geometer's Sketchpad because .gsp is not allowed.

First use Save As in Geometer's Sketchpad to save your pic in .wmf or .emf format. Then use any graphic/browsing software (such as ACDSee or similar) to open .wmf/.emf, then go to Save As and choose .gif in File Format/Save As Type.

Using Altheman's formula (a.k.a. Pappus's Centroid Theorem), I got the volume to be $\frac{45\pi}{2}$.  Did I do that right?

Doesn't seem so. Please describe what centroid have you calculated and how.

[quote="delta"][quote="Altheman"]the volume is: the distance from the centroid to the axis of rotation * the area of the quadrilateral[/quote]
This is not quite true as the axis has to be [b]external[/b][/quote]

Or, which is the same, the generatrix of the body must be reduced to $AECD$ (see the diagram)

Yeah, I noticed that part way through and did it for $AECD$.  Then I calculated the centroid by averaging the coordinates of the vertices, and found the area using the Shoelace Theorem.  I got the area of $AECD$ to be 9, and the centroid to be $\left(3,\frac{5}{4}\right)$.  According to the centroid theorem, that would make the volume $V=2\pi(9)\left(\frac{5}{4}\right)=\frac{45\pi}{2}$

The catch is following: The geometric centroid (the "average" of all the points) and the physical center of mass (material point which can replace the whole figure) need not coincide, especially in asymmetric figures. Those two always coincide in triangles, but in general case the latter must be found using calculus.

[hide="An illustration involving calculus, can be skipped"]
For a trapezoid defined by $y=px+q, x=a, x=b$ we have

$C_{x}=\frac{\int_{a}^{b}xydx}{\int_{a}^{b}ydx}=\frac{p{b^{3}-a^{3}\over 3}+q{b^{2}-a^{2}\over 2}}{p{b^{2}-a^{2}\over 2}+q(b-a)}\neq{a+b\over 2}$[/hide]

Ah, that would explain it...

In that case, it's not really worth it to use that method.  The addition and subtraction of cones is simpler...


A positive integer $ n$ is [i]powerful[/i] if for every prime $ p$ dividing $ n$, we have that $ p^2$ divides $ n$. Show that for any $ k>1$ there exist $ k$ consecutive integers, none of which is powerful.

Even easier than http://www.mathlinks.ro/Forum/viewtopic.php?t=171179 .
And posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=105016 .


give $ a,b,c \in R$ $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ , $ k \in R$
Find min: $ A \equal{} ab \plus{} bc \plus{} kca$ :roll:

$ a,b,c\geq0$ , $ a^2\plus{}b^2\plus{}c^2\equal{}1 , k,l>0$
Find max , min $ F\equal{}ab\plus{}kbc\plus{}lca$


Whats your opinion on this? Sometimes its sad that we are forced to be hypocrites, but I really get pissed when people do it on purpose. I really can't go upto my maths teacher and say "You really don't know any math"..... :(

I have more than a dozen classmates *ahem* who come before every exam and say that they are so scared and that they haven't studied. And you can obviously guess that they will be the toppers....

And ofcourse day before, day before yesterday 8-9PM was supposed to be earth hour. All news channels were ranting about all day how imperative for us it was to switch off our lights to avoid CO2 emission.. But then their channel itself will be running from 8-9, how nice... with all lights on ofc.
And radio mirchi is another hypocrite, their campaign was "Lights off, mirchi on"... and won't the mirchi station use any light from 8-9??

And why leave out our politicans, who say that they understand the pulse of our nation's poorest? or our govt employees, who hardly do any work (referring to the majority) but still want a salary on par with IT. Let them goto a pvt org and see how tough life is...

Why forget the nice people, who say "I love physics... I hope many people get into research into physics/maths" and they themselves end up getting a plum job in some IT company after B.Tech..

Sorry for the rant but what says you?

That's why they say in tamil "Mathavalukku ella advice, namakkunu varum pothu athellam thooki kuppaila podanum".

Well thats the way things are, esp abt the not finished studying thing that i hear from my own fraud friends every time except for maybe chappli ;) You know, you might have posted this somewhere else ;) :D

what u said applies in a greater meaning to u madness!!!!!

think abt it.(don post your reply here,call me over the phone)

i will post here bcos u chose not to say that over the phone :P
Well citations and examples might only save you for all you know, otw ud be a piece of dead meat the next time i see you, and fortunately or unfortunately, that is before the jee :D

poda mudichinga pannu!!!!!!!!!!!!!

PS why are u hidden?????

Wow. Shre, emotions overload coming from you for the past so many days. Though I will not let that take away from this topic, I somewhat agree. Earth Hour, well. I did switch off all the lights, no laptop too :D

As for everything else, its always for the other, never for one self. I am not making any sense. ;)

[quote="shruthibaskaran"]
As for everything else, its always for the other, never for one self. I am not making any sense. ;)[/quote]

Actually you are. Our behaviour is just to impact others and is not we may exactly want to do.

And that I believe is the meaning of the word 'hypocrisy'.

[quote="Cyber-Shreyas"][quote="shruthibaskaran"]
As for everything else, its always for the other, never for one self. I am not making any sense. ;)[/quote]

Our behaviour is just to impact others and is not we may exactly want to do.
[/quote]
or made so to impact others ;) :D

Oh very well. Glad someone made sense of what I said.

we r real hypocrites--just we want to proove we aren't :D

i agree with shre's comments on toppers!!

i mean, the way they say that they have studied nothing(!), they are getting 0(!) and all, it makes ur blood boil!

abt the physics lover BTECH getter category, i would like to comment that i do love physics, but i wouldn't want to make a living out of it, and would rather do a B-Tech, because i love applied physics rather than pure physics, and mind you, its not that B-Tech has no physics!!

P.S. i too have noticed a lot of hypocrisy in this world, and there is no need to throw politicians into this category. bcoz apart from being hypocrites, they are liars, thieves, pompous fools and corrupt! (not all of them though)

I refrain from commenting on such a sensitive issue as hypocritism....
It might affect my votebank in the coming elections....

go on......
u will win the elections :rotfl:  :rotfl:

[quote="rituraj007"]go on......
u will win the elections[/quote]

This is hypocrisy.  :rotfl:  :rotfl:

[quote]even pardesi is becoming a nerd of the nth order[/quote]

Find the eror in the error in the order of order of magnitude of the probability of the event ...

[quote="VARUNARASU"]
and rite again abt hypocrites in chennai, the biggest one being madness. :P  he is one the biggest nerds mankind has ever seen. i bet he has a book by his side even when he's online. no wonder he's always at loggerheads wid anit sahu. he cannot stand the thought of another supernerd :rotfl:[/quote]

WTH!! Dont make me change some syllables man! Im swearing n hope u can here me wherever the - you are :mad: :cursing:

[quote="hash_include"]
testified by shridhar, good friend of madness/rohit/sandeep ramesh [/quote]
You wont live after today's chem class da, im sure of that. OMG! ME a supernerd? Have they changed that definition too? :wallbash_red: :bomb:

As to what varun said about Anit, we'll he told me in his first pm that he'll get single digit marks or sth. He even gave me his address and asked me to deliver the pizza and seemed d a m n confident that he wont get thro at all. Maybe he was lying all the while.

Well i deleted that pm in a fit of anger!

See this:
[quote="anitsahu"] so have u conceded defeat yes or no
i knew u would
as u dont stand a chance
anyways my address is
ADDRESS
Send me the pizza by the evening bcos i have exam that day
[/quote]

Here the defeat he's referring to was the bet thingy we had, i was d a m n confident that he being a supernerd would surely get thro, but seems that fortune wasnt his way that day

EDITED: Okay but i wanted to show that im saying the truth thats all

@Soumya: Answer is Zero. Pardesi has successfully transmorfigured into a super-nerd.

As far as anit goes. He doesn't behave like an egoistic fellow here. Infact he behaves too hypocritical. God knows why.

[quote="Varun"]i bet he has a book by his side even when he's online.[/quote]
Bingo. Bang on Target. Are you psychic? Madness is also a nerd of the nth order but $ \lim_{n \to \infty}$.

@madness: Please delete the address. Its not good to post personal details in a public forum. I'm serious. [/quote]

And check this http://123iitjee.net/iitjee/course/view.php?id=17. 90% of the challenges there, have anits name as the cracker.

Look who's talking thats all i can say :D

PS: To all in chennai, dont forget that you are all gonna meet me once tomorrow before the jee so if you wanna live to write the exam, shut up to say the least :D :mad:

pardesi is a meganerd now......

i heard anit has exam papers pasted at his bathrooms's doors & walls

ah..well i don't aggree toppers r hypocrites :( --hypocrites means assuming u r virtuous when u r not so :)
since toppers being virtuous try not to show it (unless exaggerated),they can't b called hypocrites in ne way

 :lol:

[quote="rituraj007"]

i heard anit has exam papers pasted at his bathrooms's doors & walls

[/quote]

 :rotfl:  :rotfl:  :rotfl:

and i don't kno much about rohit...................... :rotfl:  :heli: :chief:

Lol :D (Rohit, I wont be there tomorrow :D :D Lets see what you can manage)

[quote="madness"]
PS: To all in chennai, dont forget that you are all gonna meet me once tomorrow before the jee so if you wanna live to write the exam, shut up to say the least :D :mad:[/quote]
i am sure rohit has scared the s.h.i.t out of everyone with his threats, and they are too scared to even come out of their room leave alone meeting him tomorrow.  :rotfl:

i know that i will be branded hypo but fing hell my keyboard isnt working. godogod dunno was wrong. i amposting using on screen keyboard utilty which is dead slow... god im unable to evenlogin. using guest account.

varun we are dead scared of madness es uge ahd has a ferocious moustache and beard........ :rotfl:  :rotfl:  :rotfl:

[quote="Chronoz"]varun we are dead scared of madness es uge ahd has a ferocious moustache and beard........ :rotfl:  :rotfl:  :rotfl:[/quote]

Absolutely right :D And ofc, all know that i hold the secret key that can release the attache and make him attach on others so they are all scared to death :rotfl: :D

yes,yes,...... :whistling: :whistling: :whistling:


Out of all positive real numbers x and y, what is the smallest value of x+y for which
x + y <=xy?

Isn't there just a simple [hide="theorem?"]
By AM-GM, $ \frac{x\plus{}y}2\ge\sqrt{xy} \longrightarrow (x\plus{}y)^2\ge4xy$. Now someone extend that for me while I say the answer is just 4.
[/hide]


inequalities
geometry
perimeter
trigonometry
area of a triangle
Heron\u0027s formula
trig identities

Let $ x,y,z\ge 0$ and $ xy\plus{}yz\plus{}zx\equal{}1$.
Prove that:
                          $ \sqrt{x^2\plus{}xy\plus{}y^2}\plus{}\sqrt{y^2\plus{}yz\plus{}z^2}\plus{}\sqrt{z^2\plus{}zx\plus{}x^2} \ge 3$

I wonder we can use Minkowski Inequality.

Sorry, kunny, I don't know what is Minkowski inequality.
I searched it but I'm dull and I can't understand it.
Can you give me Minkowski inequality with three variables and prove it.
Thanks a lot

another way to do it is to start out with the trivial inequality on the individual terms $ \sqrt{x^2\plus{}xy\plus{}y^2}$, etc, then use the cauchy-schwarz inequality.  :)

[quote="TZF"][quote="Pain rinnegan"]
$ LHS\ge 3\sqrt [6]{(x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)}\ge 3\sqrt {xy \plus{} yz \plus{} zx} \equal{} 3$
[/quote]

You probably meant to have a sixth root over the second-to-last term instead of a squareroot, so that the first inequality is by AM-GM and the second is by the trivial inequality.[/quote]

I meant exactly what i wrote . Firstly i applied AM-GM:

\[ LHS\ge 3\sqrt [6]{(x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)}\]

And then Holder:

\[ (xy\plus{}y^2\plus{}x^2)(y^2\plus{}yz\plus{}z^2)(x^2\plus{}z^2\plus{}zx)\ge (xy\plus{}yz\plus{}zx)^3\]

which is equivalent to:

\[ 3\sqrt[6]{(x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)}\ge 3\sqrt {xy \plus{} yz \plus{} zx}\]

therefore i didn't made any typo.

It is equivalent to saying a triangle with area $ \frac{\sqrt{3}}{4}$ must have perimeter at least $ 3$. Try to figure it out!

[quote="theSA"]it was coming out almost,
$ \sum {\sqrt {x^2 \plus{} y^2 \plus{} xy} } \ge \sum {\sqrt {3xy} } \ge \sqrt {\left( {\sum {3xy} } \right) \plus{} 6} \equal{} \sqrt 9 \equal{} 3$.

but I can`t finish to prove  $ \sqrt {xyz} (\sqrt x \plus{} \sqrt y \plus{} \sqrt z ) \ge 1$.

Well if someone prove this for me then we have simple solution,if not,then my solution is wrong. :oops:[/quote]

It's wrong.

Yes it is wrong.Using Cauchy Schwarz:
$ \left( {\sum {\sqrt {xy} \sqrt {xz} } } \right)^2  \le \left( {\sum {xy} } \right)^2  \equal{} 1$, which contradicts to my previous post.

[quote="dgreenb801"]It is equivalent to saying a triangle with area $ \frac {\sqrt {3}}{4}$ must have perimeter at least $ 3$. Try to figure it out![/quote]
Okay, I will post the full solution.
Take three segments of length $ x$,$ y$, and $ z$ that are joined at a common endpoint and separated at $ 120$ degrees from each other, and connect the other endpoints to form a triangle. Then, by the law of cosines, the perimeter of the triangle is the left side of the given inequality. Also, the area of the triangle is the sum of the areas of the three inner triangles. Using the formula $ area \equal{} \frac {1}{2}absinC$, we find the area of the triangle is $ \frac {\sqrt {3}}{4}(xy \plus{} yz \plus{} zx) \equal{} \frac {\sqrt {3}}{4}$.
So we have to show a triangle with area $ \frac {\sqrt {3}}{4}$ must have perimeter at least $ 3$
I will show that for a fixed perimeter, the area of a triangle is maximized when it is equilateral. When the perimeter is fixed, so is the semiperimeter$ \equal{} s$, and by AM-GM we have $ (s \minus{} a)(s \minus{} b)(s \minus{} c) \le (\frac {(s \minus{} a) \plus{} (s \minus{} b) \plus{} (s \minus{} c)}{3})^3 \equal{} (\frac {s}{3})^3$, which is fixed, and equality holds when $ a \equal{} b \equal{} c$. But by Heron's formula, the area of the triangle is $ \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$, so the area is maximized when $ a \equal{} b \equal{} c$. Thus, I have shown that for a fixed perimeter, the area of a triangle is maximized when it is equilateral, and it is easy to see from this that a triangle of a fixed area has minimal perimeter when it's equilateral. Thus, a triangle with area $ \frac {\sqrt {3}}{4}$ must have perimeter at least $ 3$.

[quote="Unidranix"]Let $ x,y,z\ge 0$ and $ xy \plus{} yz \plus{} zx \equal{} 1$.
Prove that:
                          $ \sqrt {x^2 \plus{} xy \plus{} y^2} \plus{} \sqrt {y^2 \plus{} yz \plus{} z^2} \plus{} \sqrt {z^2 \plus{} zx \plus{} x^2} \ge 3$[/quote]
Isn't this from "Old and New Inequalities"..??

[hide="My Solution....very elementary..."]

Using the identity
$ a^2\plus{}ab\plus{}b^2\equal{}\frac{3}{4}(a\plus{}b)^2\plus{}\frac{1}{4}(a\minus{}b)^2 \ge \frac{3}{4}(a\plus{}b)^2$

We have
\[ RHS \ge \sqrt{3}(a\plus{}b\plus{}c) ....(*)\]
Now, $ (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ca) \equal{} 3 \Rightarrow a\plus{}b\plus{}c \ge \sqrt{3}$
Plugging this into $ (*)$...we get our result...
$ Q.E.D.$
 :lol:  :D
[/hide]

[quote="Unidranix"]I can't imagine that my problem got a lot of attention.
Thank all of you alot especially Rijul saini,drgreenb801 and pain rinnegan.[/quote]
Actually this is one of the starting problems of the book "Old and New Inequalities" bu Titu Andreescu...
And You're quite welcome BTW  :wink:

[quote="not_trig"][quote="Rijul saini"]Hey...wait if $ a,b,c$ are [b]non-negative[/b] then you can't use AM-GM or HOLDER[/quote]

yes you certainly can...[/quote]
Well, doesn't these inequalities clearly state that the numbers should be [b]POSITIVE REALS[/b]??

[quote="Poincare"][quote="dgreenb801"]It is equivalent to saying a triangle with area $ \frac {\sqrt {3}}{4}$ must have perimeter at least $ 3$. Try to figure it out![/quote]

How'd you come up with that?![/quote]
Please poincare...this is pure spam....please read all the solutions
You will find that he has all ready posted his answer

[quote="Rijul saini"]Hey...wait if $ a,b,c$ are [b]non-negative[/b] then you can't use AM-GM or HOLDER[/quote]

I think here you're spamming , because the case when one or two or all the varibles are 0 is trivial. So usually when people solve inequalities there is no big difference between the words "positive" and "non-negative".


Three-time post:
http://www.mathlinks.ro/viewtopic.php?t=204492
http://www.mathlinks.ro/viewtopic.php?t=209773
If you want a solution which only needs basic geometric properties, you can consult my construction in the following.
[hide="My construction"]Draw the line d though A and H. Construct the symmetric line $ d'$ of $ d$ wrt the line AI. Let T be the midpoint of AH and K be the orthogonal projection of I on $ d$. The line through point I and parallel to $ d'$ meets $ d$ at Q. Let K' on the line $ d$ such that KK'=QT (K between A and K'). The circle of diameter AK' cuts the line IK at R. The line through R and parallel to $ d$ meets the circle of diameter AT at S. Let S' be the orthogonal of S on $ d$. The circle $ (I;AS')$ is just the incircle of triangle ABC. Maybe everything becomes easy now.  :) [/hide]
Tell me if you need a proof for this construction.  :wink:
Best regards.

"The line through R and parallel to d meets the circle of diameter AT at S"   ?

 :(

Oh, yes. Maybe there are two cases in the figure.
[quote="mr.danh"]
Draw the line d though A and H. Construct the symmetric line $ d'$ of $ d$ wrt the line AI. Let T be the midpoint of AH and K be the orthogonal projection of I on $ d$. The line through point I and parallel to $ d'$ meets $ d$ at Q. Let K' on the line $ d$ such that KK'=QT (K between A and K'). The circle of diameter AK' cuts the line IK at R.[/quote]
+) First case (Figure 1): Q between A and T
This case relative to my first post has been showed:
The line through R and parallel to $ d$ meets the circle of diameter AT at S. Let S' be the orthogonal of S on $ d$. The circle $ (I;AS')$ is just the incircle of triangle ABC.
+) Second case (Figure 2): T between A and Q
This case responds jrrbc's figure.
The perp bisector of AT meets the line through R and parallel to $ d$ at L. The circle (L;LA) meets the line LR at S. Let S' be the orthogonal of S on $ d$. The circle $ (I;AS')$ is just the incircle of triangle ABC.
Is it clearly?  :)

how can i construct the triangle after incircle ?
with perp to the 2 sides (tangents to incircle through A) through H ? but i haven't got the correct triangle.

 :(

[quote="mr.danh"]
The perp bisector of AT meets the line through R and parallel to $ d$ at L. The circle (L;LA) meets the line LR at S. Let S' be the orthogonal of S on $ d$. The circle $ (I;AS')$ is just the incircle of triangle ABC.[/quote]
You perhaps made a mistake in your figure. It must be the perp bisector of AT, not the perp line to AT at T. OK?


[quote="nguyenvuthanhha"][i] Find all nonnegative integers m such that

  $ A \ \equal{} \ 4^{2m \plus{} 1} \ \plus{} \ 1$ has less then or equal $ 2$ divisors which are primes[/i][/quote]
$ A \equal{} 4*(2^m)^4 \plus{} 1 \equal{} (2*(2^m)^2\plus{}1)^2 \minus{} (2*(2^m))^2 \equal{}$ $ (2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1)(2^{2m\plus{}1}\plus{}2^{m\plus{}1}\plus{}1)$.
And $ (2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1, 2^{2m\plus{}1}\plus{}2^{m\plus{}1}\plus{}1) \equal{}$ $ (2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1, 2^{m\plus{}2}) \equal{} (\minus{}2^{m\plus{}1}\plus{}1, 2^{m\plus{}2}) \equal{} (\minus{}2^{m\plus{}1}\plus{}1, 2) \equal{} 1$.
$ m\equal{}0$: gives $ A \equal{} 4^1\plus{}1 \equal{} 5$ which satifies the condition.
For $ m > 0$ we get $ 2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1 > 0$. So we have $ 2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1 \equal{} p^x$ and $ 2^{2m\plus{}1}\plus{}2^{m\plus{}1}\plus{}1 \equal{} q^y$ for some primes $ p,q$. But $ 5 \mid A$ so one of the primes have to be $ 5$.

$ p \equal{} 5$: $ 2^{2m\plus{}1}\minus{}2^{m\plus{}1}\plus{}1 \equal{} 5^x$ Then $ 2^{m\plus{}1}(2^m\minus{}1) \equal{} 5^x \minus{} 1$ and thus $ 2^{m\plus{}1} \mid 5^x \minus{} 1$. Let $ x \equal{} (2k\plus{}1)*2^n$. Set $ a \equal{} 5^{2k\plus{}1}$. Then $ 5^x \minus{} 1 \equal{} (a^{2^{n\minus{}1}}\plus{}1)(a^{2^{n\minus{}2}}\plus{}1) \cdots (a\plus{}1)(a\minus{}1)$. Obviously $ a\minus{}1 \equiv 4 \bmod 8$ and $ a^{2k}\plus{}1 \equiv 2 \bmod 4$ for all $ k$. So $ 2^{n\plus{}2} \mid 5^{(2k\plus{}1)2^n}\minus{}1$ and $ 2^{n\plus{}3} \nmid 5^{(2k\plus{}1)2^n}\minus{}1$. But $ 2^{2m\plus{}2} > 2^{2m\plus{}1}\minus{}2^{m\plus{}1} \equal{} 5^x \minus{} 1$. On the other hand $ m\plus{} \le k\plus{}2$, when $ x \equal{} (2j\plus{}1)2^k$ So $ 2^{2(k\plus{}2)} > 5^x \minus{} 1 \ge 5^{2^k} \minus{} 1$. So $ 4^{k\plus{}2} \plus{} 1 > 5^{2^k}$. Giving $ k \le 1$. So $ m\plus{}1 \le k\plus{}2\equal{}3$ so $ m \le 2$. Both $ m\equal{}1$ and $ m\equal{}2$ are solutions.

$ q\equal{}5$ is handled the same way, and we get the only solutions: $ m \in \{0,1,2\}$

Btw. Are you sure this is from "Hồ Chí Minh city"? It was in Baltic way about five years ago.


[hide]Taking $2$ to the power of each side and using change-of-base, we get
$\log_8 x = 2^{\log_8 (log_2 x)} = (\log_2 x)^3$
$x=8^{(\log_2 x)^{(\log_2 x)(\log_2 x)}} = (x^3)^{(\log_2 x)(\log_2 x)}$
$x=x^{3(\log_2 x)^2}$, therefore $3(\log_2 x)^2 = 1$ and $(\log_2 x)^2 = \boxed{\frac{1}{3}}$[/hide]

Erm... I thought that AIME answers were all integer values from 000 - 999? I disagree... but then again, my answer could be wrong ;)

[hide]We have:

$\log_2 (\log_8 x) = \log_2 ((\log_2 x)^{1/3})$

Equating, we have:

$\log_8 x = (\log_2 x)^{1/3}$

Then, we have:

$\log_2 x^{1/3} = (\log_2 x)^{1/3}$

Letting $\log_2 x = X$, we have:

$\frac{1}{3} X = X^{1/3}$

$\frac{1}{27}X^3 - X = 0$

Thus, $X = \sqrt{27}$. We then have:

$\log_2 x = \sqrt{27}$, which yields as $(\log_2 x)^2 = \boxed{27}$[/hide]

Note how useful the power rule is... specifically:

$\log_{a^n} b^n = \log_a b$

EDIT: the above formula was written incorrectly  :D

Yeah I got 27, where did cincodemayo mess up?

[quote=cincodemayo5590]Taking $2$ to the power of each side and using change-of-base, we get
$\log_8 x = 2^{\log_8 (log_2 x)} = (\log_2 x)^3$
[/quote]

Step 2 should instead be
$\log_8 x = 2^{\log_8 (log_2 x)} = (\log_2 x)^{1/3}$

p2: Let $\log_8(x) = y$, then $\log_2(x) = 3y$. Then $\log_2{y} = \log_8{3y}$, or $y = \sqrt[3]{3y}$, equivalent to $y^3 = 3y \implies y^2 = 3$. The answer is $(3y)^2 = 9y^2 = \boxed{27}$.


The first six terms of a sequence are $0,1,2,3,4,5$. Each subsequent term is the last digit of the sum of the six previous terms. In other words, the seventh term is $5$, the eighth term is $0$, etc. Can the subsequence $1,3,5,7,9$ occur anywhere?

Nice question.

Hint:
[hide]think even and oneven[/hide]

[hide="I agree with Meghrabi"]By induction, it can be proved that the sequence $\mod 2$ is periodic:
$0,1,0,1,0,1,1,$
$0,1,0,1,0,1,1,$
$0,1,0,1,0,1,1,\dots$
Notice that there won't be more than two consecutive odd numbers.[/hide]

Consider the invariant $I=2(x_{1}+2x_{2}+3x_{3}+4x_{4}+6x_{6})\pmod{10}$ for a sextuple $(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})$.  Then seek contradiction about the sequence $\hdots, 1,3,5,7,9,\hdots$

Boxeded, that reminds me strongly of a problem in Engel :) However, they are not exactly the same; this time we only know $5$ of the terms. Therefore, it is a little more complicated.

Not really.  Find a few of numbers, in terms of variables, on the left side of the sequence $1,3,5,7,9$, and then you get $2x\equiv 5\pmod{10}$, contradiction.


[quote="yif man12"]angrier than a drill seargent[/quote]
er dont similes use like or as?

[quote="Rep123max"]
Metaphor:
1 using the word excitement, 1 using the word embarrasment, 1 using the word loudness, 1 using the word ignorance, and 1 using the word wealth.

Any help would be appreciated![/quote]
the excitement of______________(whatever your favorite thing to do is).
Ear destablizing Loudness
Ignorance of a rock.
Wealth fit for nations.

No similies don't always use like or as. yif's is a good one.

the _________ was as shy as a student at a new school.
The _________ had a lot of motherly understanding.
Probably not that good. I thought you just said anything that describes an inanimate object with human qualities, like the rock is as coarse as hair or something like that.

you know waht all of these things are, right?
Personification is when someone makes an animal talk or someting like that.  giving an object a personality of a human or somethin like that.

he was as angry as a bull who had just caught sight of a bright red flag waving in the distance

oooooo nice one. you could change my first one to "as angry as" if you want.

thank you...and yes, i'm pretty sure similes have to have 'like' or 'as' in them to be classified as a simile


Fie functia $f:[0,1] \to [0,1]$ continua ,astfel incat :\[
\forall x \in [0,1],f(x) \in \{ \sin f(x),f(\sin x)\}
\]Sa se demonstreze ca f este constanta.

Suppose $f\not\equiv 0$ and $f(x)\neq 0$ for some $x\in[0,1]$. Then $f(x)\neq \sin f(x)$, since $t>\sin t$ on $(0,1]$. It follows that $f(x)=f(\sin x)\neq 0$. In its turn $f(\sin x)=f(\sin\sin x)$ and so on. But sequence $\sin^{(n)}x$ converges to 0. It follows that $f(x)=f(0)$. Thus if $f(x)\neq 0$ then $f(x)=0$. It implies that $f(x)\equiv const$.


induction
geometry
3D geometry
number theory

prove that $ n^3 \minus{} n \plus{} 3$ is divisible by 3 for any natural number n.

Clearly, If $ n\equal{}1$, $ n^3\minus{}n\plus{}3\equal{}3$, wihch is divisible by $ 3$.

Assume that this statement is true for some natural number $ k$.

We will show that if it works for $ k$, it works for $ k\plus{}1$.

Plugging in $ k\plus{}1$ gives $ k^3\plus{}3k^2\plus{}3k\plus{}1\minus{}k\minus{}1\plus{}3\equal{}k^3\plus{}3k^2\plus{}2k\plus{}3$.

From our inductive hypothesis, $ k^3\minus{}k\plus{}3$ is divisible by $ 3$.

Therefore, since $ k^3\plus{}3k^2\plus{}2k\plus{}3\equal{}(k^3\minus{}k\plus{}3)\plus{}3k\plus{}3k^2$, then all of the

terms are divisible by $ 3$, and hence the sum is divisible by $ 3$.

Having achieved this result, the induction is complete.

QED.

Another way:

We wish that $ n^3\minus{}n\plus{}3 \equiv 0 (mod 3) \implies n^3 \equiv n (mod 3)$

If $ n$ is not a multiple of $ 3$, then we can divide by $ n$ to get

$ n^2 \equiv 1 (mod 3)$.

If $ n \equiv 1 (mod 3)$, then $ n\equal{}3k\plus{}1$ for some integer $ k$.

Plugging this in gives $ 9k^2\plus{}6k\plus{}1 \equiv 1 (mod 3)$, which is clearly true.

If $ n \equiv 2 (mod 3)$, then $ n\equal{}3k\plus{}2$ for some integer $ k$.

Plugging this in gives $ 9k^2\plus{}12k\plus{}4 \equiv 1 (mod 3)$, which is again clearly true.

Clearly, if $ n$ is a multiple of $ 3$, then $ n^3 \equiv n (mod 3)$, since both sides

will be multiples of $ 3$.

The proof is complete, since we have exhausted all cases.

QED.

The modular argument is a bit too long.  All positive integers are equivalent to $ \pm 1, 0\mod {3}$.  Notice how the cubes of the numbers are equal to the numbers.  This implies that for any $ n$, $ n^3 \minus{} n \equiv 0\mod {3}$.  And the constant is 3, which is divisible by 3.

[hide="One line"] $ \frac {n^3 \minus{} n \plus{} 3}{3} \equal{} 2{n \plus{} 1 \choose 3} \plus{} 1$ [/hide]
[hide="Generalization"] This is [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url] with $ p \equal{} 3$. [/hide]

[hide]
$ n^3 \minus{} n \equal{} n (n^2 \minus{} 1) \equal{} n (n \minus{} 1) (n \plus{} 1)$ The product of 3 consecutive terms is always divisible by 3. Hence the result.
.If you know Fermat's Little Theorem you can directly state that $ n^3 \minus{} n$ is divisible by 3 as 3 is a prime[/hide]

My solution : We wish to prove $ n^3 \minus{} n \plus{} 3 \equal{} 0 (mod 3)$ Since 3 is divisible by 3, this is identical to proving that $ n^3 \minus{} n \equal{} 0 (mod p)$ Which is true by Fermat's Little Theorem.
Q.E.D.

btw this is Fermats Little Theorem: for a prime p $ a^p \equal{} a (mod p)$

let us consder ,
                  n^3 - n + 3
                 n(n^2 -1) +3
                 n(n + 1)(n _ 1) + 3
as (n - 1), n , (n = 1) are three consecutive intezers
 we know that product of any thre consecutive intezers is divisible bye 3
  there fore {n(n + 1)(n _ 1)} is divisible by three
   therefore,
   let x be any intezer such that {n(n + 1)(n _ 1)} = 3X
then,
                  n(n + 1)(n _ 1) + 3 = 3X + 3
                  n(n + 1)(n _ 1) + 3 = 3(X+1)
  from above equation we can understand n(n + 1)(n _ 1) + 3 is divisible by 3
     hence {n^3 - n +3} is divisible by three


conics
parabola
geometry
circumcircle
geometry proposed

Assume points $P,Q$ lie outside a parabola s.t. the focus of the parabola lies on $PQ$. From $P,Q$ we draw four tangents to the parabola. These four lines intersect at four points (apart from $P,Q$). Show that the four points lie on a circle.

In fact, let A, B, C, D be the four points where the four tangents to the parabola from the points P and Q intersect each other (apart from the points P and Q). Hereby, let the lines BC and DA be the two tangents from P, and let the lines AB and CD be the two tangents from Q. We have to prove that the points A, B, C, D lie on one circle.

Let F be the focus of our parabola. There is a well-known theorem (mentioned in [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=4200]another MathLinks thread[/url]) stating that if three tangents to a parabola form a triangle, then the circumcircle of this triangle passes through the focus of the parabola. Hence, for instance, the circumcircle of the triangle BCQ, whose sidelines are all tangents to our parabola, passes through the focus F of the parabola. In other words, the points B, C, Q, F lie on one circle, and thus < BFQ = < BCQ, where angles are considered to be directed angles modulo 180. Similarly, < BFP = < BAP. But we have assumed that the line PQ passes through the focus F of the parabola; hence, < BFQ = < BFP. Therefore, < BCQ = < BAP. In other words, < BCD = < BAD. This yields that the points A, B, C, D lie on one circle, and the problem is solved.

  Darij


[color=blue] [size=150]$8^{th}$  Iberoamerican Olympiad[/size]
Ciudad de Mexico, MEXICO. [1993]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1[/b]. Un número es capicúa, si cuando éste es escrito en notación decimal, se puede leer igual de izquierda a derecha como viceversa. Sea $x_1<x_2<\cdots<x_i<x_{i+1},\cdots$ la secuencia de todos los números capicúas. Para cada $i,$ definamos $y_i=x_{i+1}-x_i.$ Cuántos primos diferentes contienen el conjunto $\{y_1,y_2, \ldots\} \ ?.$[/quote]
$0 < 1 < \cdot\cdot\cdot < 9 < 11 < \cdot\cdot\cdot < 99 < \cdot\cdot\cdot n$ $\Longrightarrow$ Los posibles valores de $y$ son $1, 2, 11.$ Si $x_{i}$ tiene $n$ digitos $(n \geq 3),$ entonces $x_{i+1}$ termina con los mismos dígitos que $x_{i},$ i.e éste es  divisible por $10,$ ó no. En tal caso, los dos capicúas consecutivos pueden ser de la forma $x_{i} = a9 \cdots 9a,$ $x_{i+1} = b0 \cdots 0b$ con $b = a + 1$ y $y_{i} = 11.$ Si $x_{i+1}$ tiene un dígito mas que $x_{i},$ entonces éstos tienen la forma $x_{i} = 9a \cdots a9,$ $x_{i+1} = 1b \cdots b1$ con $y_{i} = 2.$ Por ende, el conjunto contiene dos primos diferentes.

[quote][b]Problema 2.[/b] Probar que para cada polígono convexo con área menor o igual a 1, existe un paralelogramo con área 2 que contiene el polígono.[/quote]
Probaremos que existe un paralelogramo que contiene al polígono y cuya área es a lo mas dos véces el área de éste. Consideremos un polígono con 4 o mas lados, puesto que el caso del triángulo es trivial. Sean $A,B,C,D$ los vértices del polígono que forman el cuadrilátero con mayor área. Sea $PQRS$ el paralelogramo formado por las paralelas de $B,D$ a $AC$ y las paralelas de $A,C$ a $BD.$ El área de $PQRS$ es en efecto dos veces el área de $ABCD$ y contiene al polígono, ya que por lo menos si hay otro vértice $E$ fuera de $PQRS$ de modo que $A$ y $E$ yacen en el mismo lado de $BD,$ entonces $|\triangle EBD|>|\triangle ABD|$ $\Longrightarrow$ $|EBCD|>|ABCD|,$ que contradice la hipotesis inicial.

[quote][b]Problema 3.[/b] Sea $\mathbb{N}^*=\{1,2,\ldots\}.$ Hallar todas las funciones $f: \mathbb{N}^*\rightarrow \mathbb{N}^*,$ tales que si $x<y,$ entonces $f(x)<f(y).$ Además $f\left(yf(x)\right)=x^2f(xy)$ para todo  $x,y \in\mathbb{N}^*.$[/quote]
Note que la primera condición, implica que $f$ es inyectiva. Si $y=1,$ entonces $f(f(x))=x^2f(x).$ Si $y=f(z),$ entonces $f(f(x)f(z))=x^2f(xf(z))=x^2z^2f(xz)=f(f(xz)).$ Pero como $f$ es inyectiva, entonces $f(xz)=f(x)f(z).$ Supongamos que $f(k) >k^2,$ para algún $k.$ Entonces, $f( f(k) ) > f(k^2) = f(k \cdot k) = f(k)^2.$ Pero $f(f(k)) = k^2 f(k),$ así $k^2 > f(k),$ que es contradicción. Analogamente, supongamos que $f(k)<k^2.$ Entonces $k^2 f(k) = f(f(k)) < f(k^2) = f(k)^2,$ así $k^2 < f(k),$ que es contradicción. Por tanto, se debe tener $f(k)=k^2.$

[quote="carlosbr"][b]Problema 4.[/b] Sea $\triangle ABC$ un triangulo equilatero, $\Gamma$ es su incírculo y sean $D$ y $E$ dos puntos en $AB$ y $AC$ tales que $DE$ es tangente a $\Gamma.$ Probar que $\frac {_{AD}}{^{DB}} \plus{} \frac {_{AE}}{^{EC}} \equal{} 1.$[/quote]
Sea $P \equiv BE \cap CD , Q \equiv AP \cap BC$  y $M,N$ los puntos de tangencia de $\Gamma$ con $AB,AC.$ Como el cuadrilátero $DECB$ es tangencial, entonces por teorema de Newton se tiene $P \equiv MN \cap BE \cap CD$ y en vista que $\triangle ABC$ es equilátero, $MN$ es la paralela media de $BC$ $\Longrightarrow$ $P$ es punto medio de $AQ.$ Luego aplicando el teorema de Van Aubel al triángulo $\triangle DEQ$ ceviano del punto $P$ resulta $\frac {\overline{AD}}{\overline{DB}} \plus{} \frac {\overline{AE}}{\overline{EC}} \equal{} \frac {\overline{AP}}{\overline{PQ}} \equal{} 1.$

[quote][b]Problema 5.[/b] Sean $P$ y $Q$ dos puntos distintos en el plano. Denotemos por $m(PQ)$ la mediatriz de $\overline{PQ}.$ Sea $S$ un subconjunto finito del plano, con mas de un elemento, el cual satisface que: Si $P$ y $Q$ están en $S,$ entonces $m(PQ)$ intersecta a $S.$ Si $\overline{P_1Q_1},$ $ \overline{P_2Q_2},$ $\overline{P_3Q_3}$ son tres segmentos diferentes, tal que sus extremos están en $S,$ entonces, si existe un punto en $m(P_1Q_1),$ $m(P_2Q_2)$ y $m(P_3Q_3),$ éste no pertencece a $S.$ Halle el número de puntos que $S$ puede tener. [/quote]
Note que hay $\frac{_1}{^2}n(n-1)$ pares de puntos y cada uno de éstos tiene un punto de $S$ sobre su mediatriz. Pero cada punto de $S$ está a lo mas sobre dos mediatrices, así $2n \ge \frac{_1}{^2}n(n-1)$ $\Longrightarrow$ $n \le 5.$ Es facil ver que el triángulo equilátero y el pentágono regular muestran que los casos $n=3,5$ son posibles. El caso $n=2,$ queda evidentemente descartado. Ahora, si asumimos que $n=4,$ hay 6 pares de puntos, entonces al menos un punto de $S$ debe estar sobre dos mediatrices. Sin perdida de generalidad, pongamos que $A$ está sobre las mediatrices de $\overline{BC}$ y $\overline{BD},$ pero también está sobre la mediatriz de $\overline{CD}$ (A es circuncentro de BCD), lo cual es contradicción.


The numbers on a standard six-faced die are arranged such that
numbers on opposite faces always add to seven. The product of
the numbers appearing on the four lateral faces of a rolled die is
calculated (ignoring the numbers on the top and bottom). What
is the maximum possible value of this product?

You want to have the maximum product, so we want all of the numbers to be as close as possible to each other (the sum is going to be the same no matter what).  Therefore, we decide that the extremes, $ 1$ and $ 6$, will be on the top and bottom.  Therefore, our answer is just $ 2\cdot3\cdot4\cdot5\equal{}\boxed{120}$


topology
real analysis
real analysis unsolved

prove that every convergent sequence of real numbers has a limit.

As "a sequence is convergent if it has a limit"?  :huh:

{a_n} is convergent if for all e > 0, there exists M>0 such that for all n,m>M abs(a_m-a_n) < e

Every real Cauchy sequence converges? i.e. $ \mathbb{R}$ is a complete metric space? This is what you mean I'm assuming?

Okay. Most people would call that a [url=http://en.wikipedia.org/wiki/Cauchy_sequence]Cauchy sequence[/url]. You want to prove that every Cauchy sequence of real numbers has a limit. (This property is expressed by saying that "$ \mathbb R$ is [url=http://en.wikipedia.org/wiki/Complete_metric_space]complete[/url]"). The proof will depend on your [url=http://en.wikipedia.org/wiki/Construction_of_real_numbers]construction of real numbers[/url]. One way to define real numbers is: a real number is an equivalence class of Cauchy sequences of rational numbers. This approach makes the statement you want to prove a tautology.

[quote="mlok"]One way to define real numbers is: a real number is an equivalence class of Cauchy sequences of rational numbers. This approach makes the statement you want to prove a tautology.[/quote]

It is not just a tautology, but rather somewhat complicated problem. In this case, what we have to show is that a Cauchy sequence of real numbers (each of which is an equivalence class of Cauchy sequence of rational numbers) has a limit.

Agreed. :)
What I should have said is: the proof of completeness is easier with some definitions of real numbers than with others.


ratio
continued fraction
real analysis
real analysis unsolved

ok, i was doing this problem in a textbook, which i am doing on myself since i will be a grad student soon and i need to prepare myself.  i looked at a proof given in the text, but i couldn't understand it; it's too complicated, and it didn't even seem to use dirichlet approx theorem.

does anyone know a elementary proof of this theorem, one that just uses contradiction, uses well ordering principle and done method?

the theorem is stated here.

http://en.wikipedia.org/wiki/Hurwitz%27s_theorem

Getting all the way to $ \sqrt{5}$ is the hard part. With a constant of $ 1$, every convergent of the continued fraction works. With a constant of $ 2$, all rational approximations that work are convergents, and some might not be good enough. At $ \sqrt{5}$, you have the golden ratio as the "equality"case; it's just a little bigger half the time, and just a little smaller the other half.

Incidentally, the constant in the denominator is improved from $ 1$ to $ 2$ by considering [b]pairs[/b] of consecutive convergents, and then improved to $ \sqrt{5}$ by considering [b]triples[/b] of consecutive convergents. One would expect to get an even better estimate by choosing one of [b]four[/b] convergents, but that just doesn't happen.

The proof (with $ \sqrt{5}$) in Khinchine's book on continued fractions is too long to be reproduced here, and is probably similar to the proof you read. You can derive the estimate with $ 2$ easily from the basic properties of [url=http://en.wikipedia.org/wiki/Continued_fraction]continued fractions[/url].


what is the mechanism for aldol using acid catalysts??
[hide="My opinion"]
it should be similar to the halogenation of carbonyl compound.see my post [url=http://www.mathlinks.ro/viewtopic.php?t=185538]here[/url].except that the formed carbanion attacks the carbonyl group...
but agian it is sykes thats creating the prblem.he says it is the enol form of the compound that attacks the enolate ion of that compound to form a carbocation attached to $ \ce{OH}$ whcich loses $ \ce{H}$ to form a aldol[/hide]

I think in aldol which is acid catalysed we need to consider the enol form attacking the protonated aldehyde/ketone.

ok...it seems fine but odd to me at least :)

yes this was tere in sykes....thanks...but still it feels a bit 'unobvious' to me

Why not obvious? Don't forget that the enol form is a minor component in an acidified solution of carbonyl.

so that gets protaonated and and there is resonance stabilazation for the carboation??

the $ \ce{O}$ gets protonated and there is resonace between that $ \ce{O}$ and the carbon it is attached to but the charge mainly resides on $ \ce{O}$

Very good, that's why the carbonyl carbon is even more electrophilic when the carbonyl oxygen is protonated, and we have a good nucleophile at hand: an enol.

Great. Now you are ready to explain what aromatic product can be formed when acetone is heated in sulphuric acid.

carcul the answer is not a big deal for me since i know that it is sym tri methyl benzene....
what i care is my mechanism...

i beleive first the aldol product oh one of the $ \alpha$ hydrohen is formed then again the hydrogen from oth group reacts now we have a intramolecular michael addition where the last carbon atom becomes a carbanion due to the available base($ \gamma$ hydrogen)  thus generating a 6 membered ring namely $ 1\minus{}hydroxy 3,5,5 trimethyl cyclohex 1,3 diene$ which is resonance stabilized then there is protonation again to the keto hydroxy group which is lost...then wagner rearrangement occurs to give the reqd. product...


Okay, how do I put boxes in my answer? I tried $ \Box$ but no boxes appeared. (This is for question 1).

Okay...i guess you can't see what I typed
I typed
\Box
surrounded by two money marks. But nothing shows up in my PDF file.
In case the \Box shows up as a box, I typed
\.B.o.x. remove the .'s

Yes. You will find \phantom very useful here. Use it to make the space as big as whatever you want to put in there.
Suppose for example you want a box that is wide and tall enough to take the formula $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$ then you type \boxed{\phantom{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}} which gives $ \boxed{\phantom{x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}}}$
On the other hand \boxed{\phantom{x=1}} gives $ \boxed{\phantom{x \equal{} 1}}$
\phantom makes room for the text but doesn't actually show the text.


Okay, I'm new here, so I may be wrong, but I believe my question belongs in this section.

If 100% is 10^10, 50% is 10^(10/3), 25% is 10^(10/27), 12.5% is 10^(10/81) and so forth, what is 18.75%? I've noticed that each time the percentage is cut in half, the power factor is reduced by three. But, I have no idea how to get the values in between the numbers. If you know an equation that works for this, please help!

I have no idea what you mean by $ 100\% \equal{} 10^{10}$ and so forth. However, $ 18.75 \equal{} 12.5 \plus{} 6.25 \equal{} 3\cdot6.25$. I have no idea what to do from here because the percentages are totally confusing me.

I'm not sure what you mean but there seems to be a $ 10^{\frac{10}{9}}$ missing. If so, your pattern then suggests that $ 12.5 \plus{} 6.25$ gives $ 10^{\frac{10}{81}} \plus{} 10^{\frac{10}{243}}$.

This problem is extremely poorly posed, and the reconstruction the other posters have gone for doesn't make any sense.  (If we were meant to get 18.75 as 12.5 + 6.25, then we should surely also get 12.5 as 6.25 + 6.25, but this doesn't match the proposed pattern (correcting it for the omission that AndrewTom noted by replacing 10/27 and 10/81 with 10/9 and 10/27, respectively).)  The most plausible associated problem is the following:

Suppose that $ f \colon \textbf{R}_{> 0} \to \textbf{R}$ satisfies $ f(x^{1/3}) \equal{} \frac {f(x)}{2}$ for all $ x$ and $ f(10^{10}) \equal{} a$.  Find $ f(x)$.

Unfortunately, this problem is also underdetermined: we'll need some additional property like continuity at some point in order to make the proper conclusion for $ x > 1$ (or $ x \geq 1$, or whatever).  On the other hand, it's not incredibly hard to come up with the unique function continuous on $ \textbf{R}_{\geq 1}$ that satisfies this condition.

[quote="Ink70"]I'm sorry guys, there was something missing. I had to re-read the problem to see if I missed something, and I did. But, it is still just as confusing.

100% of a number is 10^10, 50% of a number is 10^(10/3), 25% of a number is 10^(10/9), 12.5% is 10^(10/27), 6.25% of a number is 10^(10/81). What is 18.75% of a number?

Thanks for all of your time![/quote]
um. Calculate $ 0.1875 * 10^{10}$ which is $ 1875000000$. I probably am wrong.

I'm pretty sure this problem has somehow redefined the definition of percentages.

Every time you multiply the percentage by 2, you multiply the exponent by 3. So if you multiply the percentage by 3, then you multiply the exponent by 4.5, or 9/2? Therefore, that would make $ 18.75 \equal{} 3\cdot6.25 \equal{} 10^{\frac {10}{81}\cdot\frac{9}{2}} \equal{} 10^{\frac {5}{9}}$.

When you multiply the percentage by $ x$, you multiply the exponent by $ x\log_23$.\[ 18.75\equal{}6.25\cdot3\]\[ 10^{\frac{10}{81}3\log_23}\approx\boxed{3.86}\]


limit
logarithms
Euler
calculus
integration
function
calculus computations

\[ \lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{k\cdot ln(k)}= ?\]

By the Prime Number theorem we know that $ p_{n}$~$ n \ln n$, or if you are unfamiliar with that notation,
$ \lim_{n\to \infty}\frac{p_{n}}{n \ln n}= 1$.

Hence the series in question is comparible to the series $ \sum_{n=1}^{\infty}\frac{1}{p_{n}}$ which Euler famously proved that it diverges.

Gib Z, I think you're using a much stronger result, because the Prime Number theorem is a lot harder to prove that the problem itself.
Just do the integral test.

Integral test or condensation test. Handling this particular series is a strong point for both of those tests.

Another method for proving that this series diverges is the following one:

$ \frac{1}{(k+1)\ln(k+1)}\leq \ln\ln(k+1)-\ln\ln k\leq \frac{1}{k\ln k}$ for $ k\geq 2$ and hence

$ \ln \ln(n+1)-\ln\ln 2\leq \sum_{k=2}^{n}\frac{1}{k\ln k}$, and the result follows.

That method is essentially the integral test.

The integral test is really nice when you have something like $ \frac{1}{n\ln n}$ or $ \frac{1}{n\ln (n)\cdot\ln(\ln n)}$ where you can make a $ u$ substitution to get $ \frac{1}{u}$. In fact, my prof. used to talk about how creepy it was that you could keep tacking on these $ \ln\ln\ldots\ln n$ terms as many times as you'd like and all of them go to infinity, yet no matter how many there are it still diverges by the integral test.

Although, as I noted recently in another thread, you also have to begin your summation later and later, which (for me at least) somewhat spoils the effect.  For example, the question given at the beginning of this thread is malphrased: the first term of the series is $ \frac{1}{0}$.  This can be rectified in general by replacing the function $ \ln(x)$ everywhere with the function $ \ln(1+x)$, and you can start all of your series at 1.

To be different, we could also use the Ermakov Test, with $ f(x) = \frac{1}{x \ln x}$:

$ \lim_{x\to\infty}\frac{e^{x}\cdot \frac{1}{e^{x}\ln(e^{x})}}{\frac{1}{x \ln x}}= \lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{x \ln x}}= \lim_{x\to\infty}\ln x =+\infty$

and so the series diverges.


Find a formula for the largest number of regions space can be divided into by n planes.

I've seen a similar problem to this, but it was a 2-D problem, not 3-D...

[hide="2D problem and solution"]
Problem: What is the maximum number of non-overlapping regions created by $n$ distinct lines in a plane?

Solution: When there are no lines on the plane, there is only 1 region. If we add 1 line to the plane, that gives us 1 more region. If we add the 2nd line to the plane, that gives us 2 more regions, and the 3rd line gives us 3 more, and so on. So, the maximum number of non-overlapping regions created by $n$ lines is $1+(1+2+...+n)=1+\frac{n(n+1)}{2}$.[/hide]
Does this method work in this problem? :?

I believe the problem becomes almost exactly the same just increasing the dimension of each element by one (1 to 2-D and 2-D to 3-D).

If this is correct then here is a nice problem:

What is the maximum number of regions the plane can be divided into by $n$ Vees where a Vee is defined as two rays which meet at a point. The angle in between them can be any positive number.


If $ a$ and $ b$ are positive and $ a\not\equal{} 1,\,b\not\equal{} 1$, then the value of $ b^{\log_b{a}}$ is:

$ \textbf{(A)}\ \text{dependent upon }{b} \qquad
\textbf{(B)}\ \text{dependent upon }{a}\qquad
\textbf{(C)}\ \text{dependent upon }{a}\text{ and }{b}\qquad
\textbf{(D)}\ \text{zero}\qquad
\textbf{(E)}\ \text{one}$

the above is a logarithmic idenity and therefore, the answer would be the value of $ a$, making the answer $ B$


For each $p \in \mathbb{R}^{2}$, consider a "off-center disc" $D$ whose center is at $q$, whose radius is $r > 0$, and $r = \frac{|p-q|}{10}$.

Let $E$ be a measurable subset of $\mathbb{R}^{2}$. Prove that $p$ is a density point of $E$ [b]if and only if[/b] the concentration of E in each off-center disc $D$ associated to $p$ tends to 1 as $r \rightarrow 0$.

(Here the definition of density point is: a point $x$ is a density point of $E$ if $lim_{B\downarrow x}\frac{\mu(B\cap E)}{\mu(B)}= 1$, for any ball $B$ [b]contained[/b] $x$).

One direction is easy: if $p$ is a density point for $E$, then $\frac{\mu(D(q,r)\setminus E)}{\mu(D(q,r))}\le \frac{\mu(B(x,11r)\setminus E)}{\mu(D(q,r))}=11^{2}\frac{\mu(B(x,11r)\setminus E)}{\mu(B(x,11r))}\to 0$ as $r\to 0$.

For the opposite direction, we need to cover $B(x,R)$ with off-center disks $D_{j}$ in an efficient way, so that the estimate $\mu(B(x,R)\setminus E)\le \sum_{j}\mu(D_{j}\setminus E)$ can be used. Any suggestions?

You do not need to cover the entire $B(x,R)$, just most of it ;).

For the converse: suppose that the concentration of $E$ in each off-center disc $D$ associated to $p$ tends to $1$ as $r\to 0$. Let $x_{1},\dots,x_{200}$ be evenly distributed points on the circle of radius $r$ centered at $p$. The off-center disks $D(x_{i})$ cover the annulus $A=\left\{x\colon \frac{19}{20}r\le |x-p|\le \frac{21}{20}r\right\}$ because any point $x\in A$ is at the distance at most $\frac{r}{20}+\frac{2\pi r}{200}<\frac{r}{10}$ from the center of some $D(x_{i})$. Rescaling the whole configuration by the factor of $\left(\frac{19}{20}\right)^{n}$ and taking a union over $n\ge 0$, we cover $B(p,r)\setminus\{p\}$ by off-center disks $D_{n,i}$. Therefore, $\mu(B(x,r)\setminus E)\le \sum_{n,i}\mu(D_{n,i}\setminus E)$. For any $\epsilon>0$ there exists $r>0$ such that $\sum_{n,i}\mu(D_{n,i}\setminus E)<\epsilon \sum_{n,i}\mu(D_{n,i})=200\epsilon \pi r^{2}\sum_{n\ge 0}\left(\frac{19}{20}\right)^{2n}=\frac{80000}{39}\epsilon \pi r^{2}$.  :lol:


topology
group theory
abstract algebra
advanced fields
advanced fields open

Hello, I'd like you can help me to elucidate if the following statement is true or not:

[i]Let $ G$ and $ H$ be two topological groups.   Suppose $ p: G \rightarrow H$ is an epimorphism and also a covering map.    Then, $ G/ker(p)$ and $ H$ are homeomorphic.[/i]

Of course, if it is true I'll appreciate some hints and if not some counterexample.

Thank you guys

I believe I have worked out a solution to this problem although I am not sure it is correct (I believe that in fact, that the two spaces are homeomorphic). In my proposed solution, I am assuming that topological groups are Hausdorff (I know that some authors use this convention):

[hide]

Let  be a covering map which is also an epimorphism of groups. Because  is a covering map, we know it is in particular, an open map. Let  be the normal subgroup of G defined to be  where  is the identity element of . Define a left  action on  by composition of elements. Then the projection map  is also an open map. Now define a map  by the following: . This map is an isomorphism: First, it is clear that this map is a homomorphism and it is also surjective. Now, we want to show it is injective: Let  Then we get that  This means that  but N is just the fiber of the identity element so we get  as desired. Now, we must demonstrate that  is a homeomorphism. Let  be an open set in , by continuity,  is open in G. Because  is an open map, we know  is open in G/N so  is indeed continuous. Now to show  is open, repeat the same argument switching the places where  and  appear.

[/hide]

the homomorphism theorem for groups shows that $ p : G \to H$ induces an isomorphism of groups $ G / ker(p) \to H$, and by the definition of the quotient topology, it's continuous since $ p$ is continuous. every open subset in $ G/ker(p)$ is the image of some open subset in $ G$, so that the openness of $ p$ yields the openness of $ G/ker(p) \to H$. thus $ G/ker(p) \cong H$ as topological groups.


inequalities
function
inequalities proposed

[hide]Suppose $ n\ge2$. Let $ a_2, ..., a_n$ be fixed.
The left side becomes a function in $ a_1$. Note that it is an increasing function.

Indeed, it suffices with verifying that if $ y\ge x\ge 0$ and $ b\ge 1\ge a, c\ge 0$ the inequality $ \dfrac{y}{ay\plus{}b}\plus{}\dfrac{c}{cy\plus{}b}\minus{}\dfrac{x}{ax\plus{}b}\minus{}\dfrac{c}{cx\plus{}b}\ge 0$.
This is true because
$ \dfrac{y}{ay\plus{}b}\plus{}\dfrac{c}{cy\plus{}b}\minus{}\dfrac{x}{ax\plus{}b}\minus{}\dfrac{c}{cx\plus{}b}\equal{}\dfrac{(y\minus{}x)(xyc^2(b\minus{}a^2)\plus{}(xbc\plus{}ybc)(b\minus{}ca)\plus{}b^3\minus{}b^2c^2)}{(ay\plus{}b)(ax\plus{}b)(cy\plus{}b)(cx\plus{}b)}$

Taking $ a\equal{}a_2$, $ b\equal{}n\minus{}1$ and $ c\equal{}a_n$ proves what we wanted (increasing function).

Analogously we prove that the left side of the inequality is an increasing function in each variable so the maximun value of the expression is taken when all variables are maximized. In this case the expression equals one. So we are done.[/hide]

:)


function
inequalities
probability and stats

I tried (without succes ) to prove or disapprove , that for an increasing functions $ f$ and  $ g$ and a gaussien random variable $ X$ we have $ Correlation(f(X),g(X)) \geq 0$  !

for every $ x,y$:
\[ (f(x) \minus{} f(y))\cdot (g(x) \minus{} g(y)) \equal{} f(x)g(x) \plus{} f(y)g(y) \minus{} f(x)g(y) \minus{} f(y)g(x) \geq 0\]
so that if you choose $ X,Y$ two independent Gaussian random variables, this gives you:
\[ E[f(X)g(X)] \geq E[f(X)] \cdot E[g(X)]\]
Taking $ \hat{f}(x) \equal{} f(x) \minus{} E[f(X)]$ and $ \hat{g}(x) \equal{} g(x) \minus{} E[g(X)]$, it gives you the result.

ps: $ X$ does not need to be Gaussian

Yes, thank to you alekk, that's the inequality I used but I couldn't separe the expectations, I should made the independance hypothesis, i think that the inequality is a commun result in variance reduction !


vector
linear algebra
matrix
linear algebra unsolved

V is a finite dimensional subspace with dimension greater than 1. Show that the set of noninvertible operators on V is not a subspace of the vector space consisting of all operators on V.

Illustration in $F^2$ (for any field $F$):

Let $A=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\,,\,B=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$

$A$ and $B$ are both singular; $A+B$ is the identity operator, which is invertible.

The example is really quite general; all you need is a nonzero idempotent $P$ which is not the identity. Then $P$ is singular and $I-P$ is also a nontrivial idempotent which is also singular. And their sum is the identity.


I noticed that the introductory algebra book is coming out pretty soon.  I have not taken a rigorous algebra class, but I think that I might know a lot of what is in the book.  How hard will the book be exactly?

I think it depends on your level of mastery.  What grade are you in?  Have you taken Algebra I?  Algebra II?  Do you have AoPS Volumes 1 and 2?  If not, I would recommend getting those first.  If so, do you understand all of Chapter 6 in Volume 1?

Answer these questions, and we can probably help you...more effectively.

- Sean

I am in 7th grade taking the intro to geometry class.  I have not taken any algebra class, but I understand the chapters on linear and quadratic equations in volume 1 (I assume chapter six is one of those but I don't have the book with me at the moment).
I have AoPS volume 1 and 2.  I have worked through volume 1 (other than the geometry- Im doing geometry in the intro to geometry book).  I am on the first chapter of volume 2, and it looks like a lot of it will be very hard for me.  However, I might work through it with my tutor.

What other books do you plan on publishing or selling within the next half-year? Do you have an estimate of what month each of them will come out?

These are guesses:

Intro Algebra: Feb/Mar 2007
Interm Counting: Apr/May 2007
Interm Algebra: May/June 2007

[quote="bpms"][quote="13375P34K43V312"]I heard somewhere that Interm Trig/Complex and Interm Algebra would be put in the same book...[/quote]
I have heard stuff contradicting that...[/quote]

We thought about it for a while, but have since decided against it.  There's just too much!

[quote="rrusczyk"]If you can cover everything in AoPS Volume 1 very easily, then much of the Intro book will be review for you.  The Intermediate book should be out by next summer, though :)[/quote]
what would be in the intermediate algebra book?  would some of that be in volume two?

Some will be in AoPS2, but there will be much more in the Interm Algebra book than what is in AoPS2.

Can you tell us what topics(like the titles of the chapters) will be included in each of the intermediate books?

[quote="daermon"]so what you're saying is that I should skip the intro algebra and get intermediate algebra?[/quote]

What I've always thought is that if you can work through the Geometry book with no huge problems, you know enough algebra to be able to go through interm algebra. Of course, I can't say.


function
vector
trigonometry
linear algebra

Show that the set of functions consisting of all linear combinations of sin(x) and cos(x) is a vector space.

The set of functions consisting of all linear combinations of [i]any[/i] collection of functions is a vector space, so this is a strangely specific question.  Go through the vector space axioms; which ones are you having trouble with?

Just so I'm understanding here, is this the same as showing that

$ \{k_1 \sin x \plus{} k_2 \cos x \ \ | \ \ k_1, k_2, x \in \mathbb{R}\}$

is a vector space? And to do this, I show that every possible linear combination obeys the axioms of a vector space?

You don't show that every possible linear combination obeys the axioms of vector space but the whole set of linear combinations.

Aha well yeah, that's what I meant. I'd be here for a while if I had to show it for every possible linear combination.

EDIT: right i've got it now, thanks for your help!

[quote="rorylando45"]I'd be here for a while if I had to show it for every possible linear combination.[/quote]
Then pick one, how would you show that it obeys the axioms of vector space?

I basically just showed that the set is closed under both addition and scalar multiplication. The additional axioms (commutativity, associativity etc.) are more or less trivial so I just referenced those.

(asinx + bcosx) + (csinx + dcosx) = (a+c)sinx + (b+d)cosx

k(asinx + bcosx) = (ka)sinx + (kd)cosx


$ ABCD$ is a convex quadrangle.diagonal $ BD$ bisects neither the angle $ <ABC$   nor the angle $ <CDA$ .$ P$ is a point in $ ABCD$ such that $ < PBC$ = $ < DBA$ and $ < PDC$ = $ < BDA$. prove that $ ABCD$ is cyclic if and only if $ AP$ = $ CP$

[quote]BD diameter doesn t half <ABC and <CDA[/quote]

Sorry, didn't undertand.

Posted [url=http://www.mathlinks.ro/viewtopic.php?p=99759#99759]before[/url]  :wink:

I am sorry for my bad english

it means the diagonal $ BD$ bisects neither the angle $ <ABC$ nor the angle $ <CDA$

[quote="behdad.math.math"]$ ABCD$ is a convex quadrangle.diagonal $ BD$ bisects neither the angle $ < ABC$   nor the angle $ < CDA$ .$ P$ is a point in $ ABCD$ such that $ < PBC$ = $ < DBA$ and $ < PDC$ = $ < BDA$. prove that $ ABCD$ is cyclic if and only if $ AP$ = $ CP$[/quote]

first, if ABCD is cyclic:
let $ DP$ and $ BP$ intersect the circunference at $ X$and $ Y$ ,respectively.The triangle $ XYC$ is
congruent to $ ABD$.Notice also that triangles $ BPD$ and $ PXY$ are also congruent,since $ XY\equal{}BD$
and they have equal angles.Therefore  triangles $ BPA$ and $ XPY$ are congruent, and the result follows.

now , if $ AP\equal{}PC$  consider the segment $ XY$ with $ X$ belonging to the prolongation of $ DP$ and $ Y$
belonging to the prolongation os $ BP$, in such way that triangles $ XYP$ and $ BPD$ are congruent.then the quadrilateral $ BXDY$ is cyclic. if $ A$ and $ C$ belong to this circunference, the problem is clearly done.But I couldn´t prove it yet... :|


A sequence $ (x_{n})$ is said to be of bounded variation if there exists a constant $ M>0$ such that

$ |x_{2}\minus{}x_{1}|\plus{}|x_{3}\minus{}x_{2}|\plus{}\cdots\plus{}|x_{n}\minus{}x_{n\minus{}1}|<M$, $ n\equal{}2,3,\cdots$.

Prove or disprove:

a) If a  sequence in $ R$ is of bounded variation then it converges.

b) If a sequence in $ R$ converges then it is of bounded variation.

a)Right,Use Cauchy's rule.
If not there exist $ a>0$,for every $ N>0$,there exist $ n,m$,sucht that $ |x_n\minus{}x_m|>a$
You can get the contradiction.
b)Wrong,consider the sequence,$ a_n\equal{}\frac{1}{n}$,n odd.$ a_n\equal{}0$,n even

Let $ y_n\equal{}x_{n\plus{}1}\minus{}x_n,$ thus turning the sequence $ (x_n)$ into the series $ \sum y_n.$

a) is precisely the true statement that an absolutely convergent series converges.

b) is false, because it is possible to have a conditionally convergent series - convergent but not absolutely convergent.


Let $ \alpha$ and $ \beta$ be the roots of

$ z \plus{} \frac{1}{z} \equal{} 2 ( cos \phi \plus{} i sin \phi ) \ \ 0 < \phi < \pi$

(a) Show that  $ \alpha \plus{} i$ and $ \beta \plus{} i$ have the same argument and $ | \alpha \minus{} i | \equal{} | \beta \minus{} i |$ .

(b) Find the locus of the roots $ \alpha$ and $ \beta$ in [b]C[/b] as $ \phi$ varies from 0 to 2 $ \pi$ .

( where [b]C[/b] is the set of complex nos. )

Thank You  :)

[b]I'm sorry i think i posted this in the wrong forum ![/b]  :o

[color=green]Mod: Looks appropriate to me. Even correctly placed topics can fail to get attention for whatever reason, though.[/color]


superior algebra
superior algebra unsolved

Let $ P$ be a 2-group and $ Z(P)$ the centre of $ P$. If $ g\in P\minus{}Z(P)$, then wether or not every element of the coset $ gZ(P)$ has same order?

Not; in the modular group of order 16, $ P\equal{}\langle s,t\bigm| s^8 \equal{} t^2 \equal{} 1, st \equal{} ts^5\rangle$, we have $ s^2\in Z(P), t\not\in Z(P)$, but $ t$ has order 2 and $ s^2 t$ has order 4.


Let s(n) be the sum of the positive divisors of n. For example, s(6) = 1 + 2 + 3 + 6 = 12. Show that there are

infinitely many n such that s(n)/n > #####)/m for all m < n.

Did you copy again or where does the ##### come from and what does it mean¿
Please read at least what you post before clicking on submit.


Galois Theory
superior algebra
superior algebra unsolved

Supose that $(K(a): K)<\infty$ ($K$-field). Prove that there exist only finitely many pairwise distinct fields $M$ such that $K \subset M \subset K(a)$.

[url]http://planetmath.org/encyclopedia/ProofOfPrimitiveElementTheorem2.html[/url]
Primitive element theorem: A finite extension $K/F$ is simple iff there are only finitely many fields $L$ with $F\subset L\subset K$.

I know, but I tried not to use fundamental theorem of Galois theory since we haven't had it yet, but thanks anyway.

@Captain: The solution posted in the link is not exactly the solution to the problem I've posted. If $K(a)$ is not a separable extension of $K$ then there is no Galois extension $K'$ of $K$ such that $K(a) \subset K'$ since $K'$ is a separable extension of $K$.
Edit: I've noticed that separability is not necessary, excuse me.