I don't think so. AoPS occasionally offers free Math Jams but those are for the AMC and stuff.

If my lessons you mean information, then yes. Look at the transcripts for math jams and frequent the forums. These are both free and you will learn quite a bit.

[quote="Nick31"]are there websites that offer free lessons ? I'm coming from a musical background and there's tons of free info on the internet. I'm wondering if there are any good maths or science lessons on the web ?[/quote]
sometimes i google the stuff i want to learn.

btw for information about a single topic, wikipedia is surprisingly good. on undergrad-level topics, they usually give a full exposition of the subject at hand, while for more obscure grad/postgrad research topics, it seems that more often than not they'll still have some information

[quote="myyellowducky82"]but is it true that tutor .com gives free lessons in all of the subjects they offer? (i havent tried it; i've just read about it)[/quote]

http://www.tutor.com/pricing/default.aspx

i didnt read the fine print (im assuming that u wont either), but the shiny silver piggy banks did catch my eye

[quote="Bictor717"]If by lessons you mean class materials and videotaped lectures, MIT has its [url=http://ocw.mit.edu/OcwWeb/web/home/home/index.htm]OpenCourseWare[/url]. There are quite a few videos on their Youtube account if you'd rather just view them without downloading anything.[/quote]

interesting, I just downloaded a bunch of their videos

You should check out http://www.contestmath.com/index.html.  It's maintained primarily by [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=6133]mysmartmouth[/url] I believe.

Encyclopaedias have [i]never[/i] been acceptable sources for research.  People who try to rely on it for research are typically just being too lazy to conduct real research.  (People who rely on it for a quick background read or links to possible sources are using it appropriately.)  In fact, in the age of Wikipedia and Google, "filtering out nonsense" is an even more useful skill than it was in the past.  To quote a comment I've seen posted on [url=http://www.realclimate.org/]Realclimate[/url], "Google doesn't have a 'wisdom' button."

What you think this has to do with the subject at hand, though, is beyond me.

[quote="dragon96"]Why do teachers hate wikipedia so much? It usually provides the most information when I'm doing research (at least in non-school related research, that is).[/quote]
Anyone can edit it, basically. But in some subjects Wikipedia can be accurate!

Wikipedia is actually proving to be more and more of a reliable source. Some artcicles were people would like to make fun of other people (such as George W. Bush) are locked.

Also if you try to write something vulgar, the wikipedia staff edit these phrases out very quickly.

Each article also has a list of links and sources. Wikipedia articles are edited almost daily

Wikipedia is completely surpassing sources like Britannica because it is giving more information from just as reliable sources.

The fact that anyone can edit Wikipedia is true. But there are not that many people who would
want to mess with wikipedia information.
If you are doing a research project, Wikipedia will have a list of links at the bottom of the links that are usually credible. And if teachers do not want you to use Wikipedia than you can just click on those links and use those.
If an article does not list any sources or does not "follow Wikipedia's quality standards" then Wikipedia will warn you at the top of the page.

[quote="guineapiggies"][quote="dragon96"]Why do teachers hate wikipedia so much? It usually provides the most information when I'm doing research (at least in non-school related research, that is).[/quote]

lol we have this argument at least 10 times a month. so the thing is. you CAN trust wikipedia. BUT always check with another website. i've seen some pretty random stuff on wiki. :maybe:[/quote]
I always check with 3 other websites.


i have tests in 1994,1995,1996,1997,1998,2003,2004,2005. i am going to upload all these files(1994 and 1996 have already been uploaded). if the ones having tests in 1993,1999,2000,2001,2002 will upload these files, i will be very pleased. Or if you think this,editing and translating, is hard for u, i can if u send them via mail xeroxia@gmail.com
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112316]1994[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112715]1995[/url]

neden türkçe konuşmuyorsun.
ben aslında bütün aralıkları yazacaktım yazın ama okuldan almayı unuttum.haftasonları fırsat bulduğum sürece dedigin yılları yazmaya çalışacagım.


Let a,b,c be positive reals. Determine min value of
$A=\frac{x^3y^3}{z^4(x^2+y^2)}+\frac{x^3z^3}{y^4(x^2+z^2)}+\frac{z^3y^3}{x^4(z^2+y^2)}$

use cauchy,we should only need to prove:
$2(x^2y^2+y^2z^2+z^2x^2)^2 \ge 3 \sum_{sym} x^4y^3z$
$\iff \sum x^4 (y^2+yz+z^2)(y-z)^2 \ge 2\sum x^4 yz (y-z)^2$
$\iff \sum x^4 (y^2-yz+z^2)(y-z)^2 \ge 0$
obvious
hope I am right. :)

[quote="Lovasz"][quote="zhaobin"]use cauchy,we should only need to prove:
$2(x^2y^2+y^2z^2+z^2x^2)^2 \ge 3 \sum_{sym} x^4y^3z$
[/quote]
How did you obtain this one?[/quote]
ok,sorry.the minimum is 3/2.
then we will prove $lhs \ge 3/2$
use cauchy: we have:$(\sum_{sym} x^4y^3z)(\frac{x^3y^3}{z^4(x^2+y^2)}+\frac{x^3z^3}{y^4(x^2+z^2)}+\frac{z^3y^3}{x^4(z^2+y^2)} )=$
$(\sum_{sym} x^4y^3z)(\frac{x^4y^4}{z^4xy(x^2+y^2)}+\frac{x^4z^4}{y^4zx(x^2+z^2)}+\frac{z^4y^4}{x^4yz(z^2+y^2)} ) \ge (x^2y^2+y^2z^2+z^2x^2)^2$
so I get that one. :)

Ok, that right! This is my solution
Let $X=\frac{1}{x},Y=\frac{1}{y},Z=\frac{1}{z}$, it becomes $A=\sum_{cyc}\frac{Z^4}{XY(X^2+Y^2)}$.
Since $XY(X^2+Y^2) \leq X^4+Y^4$ then it becomes very easy!


2 diagonals of a regular nonagon (a 9-sided polygon) are chosen.  What is the probability that their intersection lies inside the nonagon?

In every group of four points, there is one unique pair of intersecting diagonals, which there are $ \dbinom94$ of.

There are $ \frac{9\cdot6}{2}\equal{}27$ diagonals, so there are $ \dbinom{27}2$ pairs of diagonals total.

Finally, our probability is $ \frac{\dbinom94}{\dbinom{27}2}\equal{}\boxed{\frac{14}{39}}$.

There are $\frac{9\cdot8}{2}$ diagonals.

[quote=r0518][quote=look4pilot][quote=cardesigner06]Another one revived  :([/quote]

yes[/quote]

yeah[/quote]

Please just please don't respond to this

every point on the nonagon has $56$ which is $(6+10+12)\cdot2$, so the total will be $\frac{56\cdot9}{2\cdot2}=126$(This is to prevent the over counting of the $2$ pairs of diagonals). The number diagonals is $\frac{(9-3)\cdot9}{2}=27$, so there is $\binom{27}{2}$ ways to choose $2$ diagonals, which is $351$. The final answer is $\frac{126}{351}=\boxed{\frac{14}{39}}$.

[quote=cardesigner06]Howcome people go back to old problems already solved[/quote]

This isn't "going back". This is an alcumus problem so the people who respond just got the problem from alcumus. We have no control over which problem we get for a certain section.

[quote=daveyoshi][quote=cardesigner06]Howcome people go back to old problems already solved[/quote]

This is "going back". This is an alcumus problem so the robots who respond just got the problem from lol alcumus. We have a lot - tons -  of control over which problem we get for a certain section.[/quote]

[quote=cardesigner06]Another one revived  :([/quote]

anorher wat

[quote=IAmTheHazard]There is some vagueness in this question. Do intersections on the nonagon count as "inside"? For example:
[img]https://i.imgur.com/UqNUV2Y.png[/img]
Diagonals $BI$ and $BE$ intersect at $B$. Does this count as "inside"?[/quote]

can someone answer this i had the same question

I don't think so. In my mind, B should be "on" the nonagon since AB is a segment that does not count and B is on AB.

[hide=Solution]Note any 4 vertices form 2 intersecting diagonals. Hence, there are 9C4=126 successes. For the total, there are 9C2-9=27 diagonals, so 27C2=351 pairs. Answer is 126/351=14/39.[/hide]

[quote=peelybonehead]For a general $n$-sided convex polygon, it is just $$\frac{\binom{n}{4}}{\binom{\frac{n(n+3)}{2}}{2}}.$$[/quote]

Should be $$\frac{\binom{n}{4}}{\binom{\frac{n(n-3)}{2}}{2}}.$$

I did case work with 3 cases and divided them by 2 because of overcounting since every line has 2 points. Then, I divided everything by the total number of pairs and got the exact answer.


function
limit
real analysis
real analysis unsolved

Assume $ f(x)$ is a real-valued function defined on $ \mathbb{R}$, satisfied:
$ \lim_{h\rightarrow 0}\left[f(x \plus{} h) \minus{} f(x \minus{} h)\right] \equal{} 0$, for $ \forall x \in \mathbb{R}$, can we come to the conclusion that: $ f(x) \in C(\mathbb{R}) ?$

No.  Any even function discontinuous at $ 0$ is a counterexample, e.g. $ f(x) \equal{} \begin{cases} x^{\minus{}2} & x\neq 0 \\
0 & x \equal{} 0\end{cases}$.

This also never detects any removable discontinuity.

However, it IS true that $ \lim_{h\to 0}\left|f(x\plus{}h)\minus{}f(x)\right|\equal{}0$ for $ x\in (a,b)$ implies $ \lim_{h\rightarrow 0}\left[f(x\plus{}h)\minus{}f(x\minus{}h)\right] \equal{} 0$ (not too hard to prove).


vector
quadratics
algebra proposed
algebra

Let the non -parallel vectors $\vec{a},\ \vec{b}$ . Solve the equation

$\frac{\left|\vec{a}\right|^{2}+x\vec{a}\ \vec{b}}{\left|\vec{a}\right|\left|\vec{a}+x\vec{b}\right|}=\frac{\left|\vec{b}\right|^{2}+\vec{a}\ \vec{b}}{\left|\vec{b}\right|\left|\vec{a}+\vec{b}\right|}$

This is not only the solution  :wink:  . There are other ones .Rust you didn't take all the cases .

Did you consider it as a quadratic ?
The roots I found are if $x>0$ then $x=\frac{\left|\vec{a}\right|^{2}}{\left|\vec{b}\right|^{2}}$ and

if $x<0$ then $x=\frac{-\left|\vec{a}\right|^{2}}{\left|\vec{b}\right|^{2}+2\vec{a}\ \vec{b}}$

My mistake is:
[quote="Rust"]$cos(\vec{b},\vec{b}+\vec{a})=cos(\vec{a},\vec{a}+\vec{b})$, therefore x=1.[/quote]
Yes. x is root quadratic equation.


function
quadratics
number theory proposed
number theory

It is given a prime number $ p$.
Find all functions $ f: \mathbb{Z}\to \mathbb{R}$ which observe the conditions:
- $ f(x \plus{} p) \equal{} f(x)$;
- $ f(xy) \equal{} f(x)f(y)$.

From $ f(x\plus{}p)\equal{}f(x)$ , we know $ f(x)$ is periodic $ mod \;\; p$.
Substitute $ y\equal{}0$ into the second equation we get $ f(x)\equal{}1$ for all $ x$(a trivial solution) or $ f(0)\equal{}0$,
Substitute $ y\equal{}1$ into the second equation we get $ f(x)\equal{}0$ for all $ x$(another trivial solution) or $ f(1)\equal{}1$
We also know $ |f(x)| \le 1$ for all $ x$ $ \dots \dots \dots 3$ or $ f(x^n)\equal{}f(x)*f(x)*f(x)*\dots$ which is not bounded as $ n \to \infty$

Now choose $ x$ in the second equation such that $ p \mid x$, then there must be a $ y$, where $ y \neq 0$ but may $ \equal{}x(mod \;\; p)$, such that $ xy\equal{}1(mod \;\; p)$ so for every such $ x$ we can find $ y$ such that $ f(x)f(y)\equal{}f(xy)\equal{}f(1)\equal{}1$ , so the value of $ f(x)$ can only be $ 1$ or $ \minus{}1$ from $ 3$.

Obviously values of $ f(x)$ cannot be all $ \minus{}1$.So $ f(x)\equal{}0$, some of the $ f(x)$ ,say for $ x\equal{}\{a_n\}$are $ 1$ and the other $ f(x)$ , say $ x\equal{}\{b_n\}$ are $ \minus{}1$,
Of course all quadratic residues $ \{c_n\}$ $ \neq 0 (mod \;\; p)$ must be in $ \{a_n\}$ by setting $ x\equal{}y$ in the $ 2nd$ equation.No. of quadratic residues = no. of quadratic residues also.

Now assume that $ \{a_n\}$ consists of not only $ \{c_n\}$ , say $ k$ also, then we have all $ {a_n}$ times $ k$ gives all non-residues (by Euler's criterion), so $ f(x)\equal{}1$ for all non-residues (which is the trivial solution given before) also , contradiction.
Thus $ \begin{cases} f(0)\equal{}0 \\
f(x)\equal{}1 \;\; for \;\; x \;\; being \;\; a \;\; quadratic \;\; residue \\
f(x)\equal{}\minus{}1 \;\; for \;\; x \;\; being \;\; a \;\; quadratic \;\; non\minus{}residue \end{cases}$

[quote="BG Yoda"]It is given a prime number $ p$.
Find all functions $ f: \mathbb{Z}\to \mathbb{R}$ which observe the conditions:
- $ f(x + p) = f(x)$;
- $ f(xy) = f(x)f(y)$.[/quote]
The first condition is $ x \equiv y \bmod p \Rightarrow f(x) = f(y)$.
First see $ f(1)^2 = f(1)$ and thus $ f(1) = 0$ or $ f(1) = 1$.
$ f(1) = 0$: Let $ a$ be a residue $ \bmod p$ not equal to $ 0$. Then $ f(a)^{p - 1} = f(a)f(a)^{p - 2} = f(a^2)f(a)^{p - 3} = f(a^{p - 1})$ But $ a^{p - 1} \equiv 1 \bmod p$ so $ f(a)^{p - 1} = f(1)$. Thus $ f(a) = 0$. So we have two solutions:
$ f(n) =  \{\begin{array}{cc} n \equiv 0 & 0 or 1 \\
\text{else} & 0 \end{array}$
$ f(1) = 1$: Then in the same manner as before: $ f(a) = \pm 1$ for all $ a \equiv 2,3,...,p - 1 \bmod p$. Let $ a$ be a generator $ \bmod p$. If $ f(a) = 1$ then $ f(1) = f(2) = ... = f(p - 1) = 1$
$ f(n) =  \{\begin{array}{cc} n \equiv 0 & 1 \\
\text{else} & 1 \end{array}$
If $ f(a) = - 1$. Then all quadratic residues is $ 1$ and all non quadratic residues is $ - 1$. And we must have $ f(0) = 0$ of course.

So the last solution is: $ f(a) =  ( \frac {a}{p}  )$.

Edit: Hmm.. Beaten :P

Exactly  :)
Do you know where is this problem from, by the way?


The sum of sixteen positive numbers is 100 and the sum of their squares is 1000. Prove that none of the sixteen numbers is greater then 25.


let $ ABC$ a triangle and  $ H$  is the orthocenter and $ G$ is the center of gravity.
prove that $ tan(A).tan(B)\equal{}3$ $ \equal{}>$ $ (HG)$ and $ (AB)$ are parallel.

call K the projection of C on AB, then we have: $ \frac{\overline{AK}}{\overline{HK}}\equal{}3 \ \Longleftrightarrow \ \frac{\overline{KB} \tan B}{\overline{KB} \cot A} \equal{} 3 \ \Longleftrightarrow \ \tan A \cdot \tan B \equal{} 3$


Given triangle $ABC$ and $P$ is an arbitrary point inside it, $B',B''$ are midpoint of $CA,PB$ similarly $C',C''$ are midpoint of $AB,PC$ line $MA\cap BC=\{A'\}$ prove that $S_{A'B'B''}+S_{A'C'C''}-\frac{S_{PBC}}{4}=const$ forall $P$ inside triangle!

[quote="gemath"]  line $MA\cap BC=\{A'\}$ [/quote]
what is $M$

Sorry it is $P$ , I rewrite it here
Given triangle $ABC$ and $P$ is an arbitrary point inside it, $B',B''$ are midpoint of $CA,PB$ similarly $C',C''$ are midpoint of $AB,PC$ line $PA\cap BC=\{A'\}$ prove that $S_{A'B'B''}+S_{A'C'C''}-\frac{S_{PBC}}{4}=const$ forall $P$ inside triangle!


What smallest quantity of corner ( 4 square blocks which cut one out) can be cut out of the 8x8 chessboard so that it is no more possible to cut any such corner out of the remaining part of square ?

Example

o
oo

oo
o

oo
  o

  o
oo

[quote="JBL"]This is now the third time you've posted this problem -- multiple posting is quite rude, so please stop it.[/quote]

sorry but Any solutions please


floor function
continued fraction
number theory theorems
number theory

Can we describe all solutions of the Pell's $x^2-D y^{2}=1$ equation?

See : http://mathworld.wolfram.com/PellEquation.html . :)

Yes we can:
Let $a , b > 0$ be the smallest positive solution (ordered by the size of $a$ or $b$ or $a+b$ or whatever) to $a^2-Db^2=1$.
Then consider the number $a+\sqrt{D} b$: all solutions are of type $x+\sqrt{D} y = \pm (a+\sqrt{D} b)^n$ for integers $n$ then (as long as $D$ isn't a square, we have $x+\sqrt{D} y = w+\sqrt{D} t$ iff $x=w$ and $y+t$ for all integers $x,y,w,t$).
So the question is: how to find (it exists everytime) this smallest solution¿

For that use continued fractions:
Find the representation $\sqrt{D} = [a_0 ; \overline{ a_1 , a_2 , ... , a_n}]$ of $\sqrt{D}$ as continued fraction with minimal period length $n$.
This means:
$\sqrt{D} = a_0 + \frac{1}{a_1+\frac{1}{a_2+ \frac{1}{...}}}$, where $a_{n+k}=a_k$ for all $k>0$.
To find this, just set $x_0 = \sqrt{D}$ and compute:
$a_k : = \left\lfloor x_k \right\rfloor$ (with $\left\lfloor x \right\rfloor$ being the greatest integer $\leq x$).
$x_{k+1} : = \frac{1}{x_k-a_k}$
This will give the representation above, and will get cyclic (lets avoid the proof, it's ugly).

Now we compute the $n$-th approximate fraction $\frac{p_n}{q_n}$ of $\sqrt{D}$:
$p_{-1}=0 , p_0=1$
$q_{-1}=1 , q_0=0$
$p_{k+1} = a_k p_k + p_{k-1}$
$q_{k+1} = a_k p_k + q_{k-1}$
Then the smallest solution to $x^2-Dy^2 = \pm 1$ will be $x=p_n , y=q_n$, and when $x^2-Dy^2=-1$, then the smallest solution to $a^2-Db^2=1$ will be $a=x^2+Dy^2 , b= 2xy$.

Note that this is generalized by Dirichlet's Theorem on units in number fields.

[b]the information you want can be found here:

http://www.matf.bg.ac.yu/~matic/competitions/

there is an article talking about the full form solutions of the Pell equations you need to know about.

 :D [/b]


geometry
trigonometry
trapezoid
perpendicular bisector
geometry unsolved

[hide="PROBLEM 0 "]

$ ABCD$ convex quarilateral, $ AC$ and $ BD$ are diagonals. If $ \angle BAC \equal{} 10^o , \angle BDA \equal{} 50^o , AB \equal{} BD$ and $ AC \equal{} AD$ then $ \angle BCA \equal{} ?$[/hide]

Let $ O\in AC\cup BD$. Do some angle chasing:

$ \angle ADO\equal{}50^{\circ}, \angle DOA\equal{}90^{\circ},\angle OAD\equal{}40^{\circ},\angle COD\equal{}90^{\circ},\angle ODC\equal{}20^{\circ},\angle DCO\equal{}70^{\circ},\angle BOA\equal{}90^{\circ},\angle OAB\equal{}10^{\circ},\angle ABO\equal{}80^{\circ}$

Let $ AD\equal{}1$. We have $ AO\equal{}\cos 40^{\circ},DO\equal{}\cos 50^{\circ}$. Thus $ CO\equal{}\tan20^{\circ}\cos50^{\circ}$ and $ BO\equal{}\tan10^{\circ}\cos 40^{\circ}$. It is not difficult to prove that $ \tan 30^{\circ}\equal{}\frac{\tan10^{\circ}\cos 40^{\circ}}{\tan20^{\circ}\cos50^{\circ}}$. So $ \angle ACB\equal{}30^{\circ}$.

[quote="Johan Gunardi"]Let $ O\in AC\cup BD$. Do some angle chasing:

$ \angle ADO \equal{} 50^{\circ}, \angle DOA \equal{} 90^{\circ},\angle OAD \equal{} 40^{\circ},\angle COD \equal{} 90^{\circ},\angle ODC \equal{} 20^{\circ},\angle DCO \equal{} 70^{\circ},\angle BOA \equal{} 90^{\circ},\angle OAB \equal{} 10^{\circ},\angle ABO \equal{} 80^{\circ}$

Let $ AD \equal{} 1$. We have $ AO \equal{} \cos 40^{\circ},DO \equal{} \cos 50^{\circ}$. Thus $ CO \equal{} \tan20^{\circ}\cos50^{\circ}$ and $ BO \equal{} \tan10^{\circ}\cos 40^{\circ}$. It is not difficult to prove that $ \tan 30^{\circ} \equal{} \frac {\tan10^{\circ}\cos 40^{\circ}}{\tan20^{\circ}\cos50^{\circ}}$. So $ \angle ACB \equal{} 30^{\circ}$.[/quote]
It looks difficult for me

Take E the symmetric of B w.r.t. perpendicular bisector of CD; the triangle ABE is equilateral, the triangle BDE is isosceles 20-80-80, BCDE is isosceles trapezoid, hence <CBE = 80 degs, <DBE = 20 degs, hence <CBD = 60 degs and, with AC and BD perpendicular, <ACB = 30 degs.

Note: This is my present to those celebrating the Easter Day on 19th April.

Best regards,
sunken rock

[quote=Virgil Nicula][quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ ABCD$ be a convex quarilateral for which $ BA \equal{} BD$ , $ AC \equal{} AD$ ,

$ m(\angle BAC) < 30^{\circ}$ , and  $ m(\angle BAC) \plus{} m(\angle BDA) \equal{} 60^{\circ}$ . Ascertain $ m(\angle BCA)$ .[/color]

[hide="Proof."]
[color=darkblue]Denote $ x \equal{} m(\angle BAC)$ . Observe that $ \underline {BA \equal{} BD}$ and $ \underline {m(\angle BAC) \plus{} m(\angle BDA) \equal{} 60^{\circ}}$ $ \implies$

$ m(\angle BAD) \equal{} m(\angle BDA) \equal{} 60^{\circ} \minus{} x$ , $ m(\angle CAD) \equal{} 60 \minus{} 2x$ and $ m(\angle DBA) \equal{} 60 \plus{} 2x$ .  Construct

the equilateral $ \triangle ABE$ so that $ AD$ separates $ B$ , $ E$ . Observe that $ m(\angle DBE) \equal{} 2x$ , $ m(\angle DAE) \equal{} x$ ,

the triangle $ DBE$ is $ B$-isosceles, i.e. $ m(\angle BED) \equal{} m(\angle BDE) \equal{} 90^{\circ} \minus{} x$ and $ \boxed {\ m(\angle ADE) \equal{} 30^{\circ}\ }$ .

Observe that  $ ABC\equiv AED$ because $ AB \equal{} AE$ , $ \underline{AC \equal{} AD}$ and $ m(\angle BAC) \equal{} m(\angle EAD) \equal{} x$ .

Hence $ BC \equal{} ED$ and $ m(\angle BCA) \equal{} m(\angle EDA) \equal{} 30^{\circ}$ , i.e. $ m(\angle BCA) \equal{} 30^{\circ}$ .

[b][u]Particular case[/u].[/b] For $ m(\angle BAC) \equal{} 10^{\circ}$ we"ll obtain the proposed problem.[/color][/hide][/quote][/quote]

Construct circle of diameter $AD$, take a random point on it, closer to $D$; reflection of $D$ about $M$ is $C$. Now construct the equilateral triangle $ADN$, $N$ and $C$ on same side of $AD$. The angle bisector of $\angle CAN$ intersects $ON$ at $B$. As constructed $\angle BAC=\angle BAN=\angle BDN$ ( symmetry ), $\angle BDN+\angle BDA=\angle ADN=60^\circ$. Now by symmetry $\angle ACB=\angle ANB=30^\circ$, done.

Best regards,
sunken rock


limit
inequalities
ratio
calculus
geometric series

prove .9 repeating (.99999999999999...)=1 (without saying that by definition it has to be)

Well...

0.9(repeating)/9 = 0.1(repeating) = 1/9 times 9 = 1

That's probably wrong, but...

Or, you could solve it using the formula for an infinite geometric series (I think it's a/(1-r), but I'm not 100% sure)

EDIT:

0.999999999999999999 divided by 9 = 0.1111111111111111111111

That equals 1/10 + 1/(10)^2 + 1/(10)^3 +... 1/(10)^n

Now, plugging those values in to the infinite geometric series formula you get 1/9 as the value of that.

Don't forget to multiply it by 9 to get the original number.

1/9 times 9 = 1

I hate to say this but this appeared before.

http://www.artofproblemsolving.com/Forum/topic-19502.html

This topic has been asked many times. ;)

[hide]I've seen it done a couple diff ways before, but Potter's proof has always been my favorite approach.  Much like treething's approach you can say
.9bar/3 is .3bar which is 1/3 times 3 is one[/hide]

sorry I'm new here and I didn't see that previous post, I apologize

That's okay. :)

This post was stopped sometimes ago so it's obvious. I'm just doing a job of a moderator.

Just a tip:

If you click on that link, you'll see many beautiful solutions. :)

The proof is simple. By definition, $.999\ldots=\sum^{\infty}_{k=1}\frac{9}{10^k}=\lim_{n\to\infty}\sum^{n}_{k=1}\frac{9}{10^k}$. The rest is basic calculus.

In the other thread the argument was repeated many times that $1=3\cdot\frac{1}{3}=3\cdot.333\ldots=.999\ldots$ or some variation of that. But of course that "proof" is based on the assumption that $\frac{1}{3}=.333\ldots$, which is pretty much what you're trying to prove...  :(

[quote="shinwoo"]let's suppose that
1 didn't equal to 0.99999999........

then 1 - 0.999..... > 0
right?[/quote]Right, but now you'll need some sort of definition for .99999... which I gave in my post.[quote]let x = 1 - 0.9999....... > 0
then
x > 0.000000..001, that is not 0 at some n th digit[/quote]How does "x > 0.000000..001" follow from your definition of x? Also, 0.000...001 is not actually a defined number..[quote]
now 1 = 0.999999...... + x > 0.99999.... + 0.000000..001
                                                ^
                                                 |
                                              = 1.00000.....000099999 .... > 1
                                             which proves that
0.9...... equals to 1[/quote]Sorry but I have no idea what you mean by "1.00000.....000099999 .... > 1" or how it follows from the previous line or how it proves anything.

[quote="shinwoo"]hm...
How am I supposed to explain this...
okay

since 1 doesn't equal to .999999999999..... , there must be a number, that exists between that infinte .9's and 1
when I said something about 0.000000000000....001 , i meant like a real tiny number that has a limit...[/quote]I know what you meant by that. The objection I raised with it is that your number is not actually defined. There were other parts that I did not understand.

Then think simply this way

to prove something, u have to think about a situation that's opposite of the statement..

to prove that .99999999..... repeating = 1

then is there any specific number that exisits between the repeating number and an integer 1?

AntonioMainenti : I don't really understand what u were saying.. so could you approach the objection in a little easier way to understand?

[quote="shinwoo"]AntonioMainenti : I don't really understand what u were saying.. so could you approach the objection in a little easier way to understand?[/quote]Gladly.

First, the key to understanding .9999... = 1, I think, is understanding what .999..... actually means, or how it is defined in mathematics.

Second, you introduce the inequality x > .0000...0001, where the 1 follows an infinite string of 0's, but that infinitely small "number" is not defined; it has no meaning.

Last, I just have no idea what this says: 1.00000.....000099999 .... > 1.

The easiest way is the infinite sum as mentioned before or multiplying it by 10.

AntonioMainenti : I think u objected too good.. I can't really think of how to define any numbers that exsist....( I don't really know how to define thing )

wow...

this is harder than I thought since, every numbers I state are supposed to be defined.. I would like to admire AntonioMainenti in alot of ways...

DPopov : interesting...where did u learnt that method? or how did u figure it out?? It's very advanced math..

I asked Joe about this topic....

and he said it was always debated and also almost impossible to "win" this argument..

some people believe .999.... as just a number that goes forever but not exactly nor equals to 1.

some people say, they are the same things because there is not specific number that exsists between .999.... and 1

So, I would like to say, it's impossible to prove that .999... equals to 1 since not all the mathmaticians agree with it..

[quote="shinwoo"]DPopov : interesting...where did u learnt that method? or how did u figure it out?? It's very advanced math..[/quote]I also showed that method but not to completion. It's just basic calculus.

Yes, I just finished the proof from what Antonio had. It's a simple geometric series with ratio of $\frac{1}{10}$ between consecutive terms.

AntonioMainenti and DPopov: you guys are both amazing... I never really learned how to use sigmas.... I only know how to read it :blush:
so.. You guys are both amazing and highly advanced in math.....

:lol:

sorry for laughing but it's cool that you admire someone who can write sigmas.

i absolutely agree with antonio and dpopov. how can you prove anything about 0.999... if you don't know how it's defined?

and mathematicians fully agree that 0.999... = 1 since once we defined it, it's just very basic calculus to prove the equality, as dpopov did it.

however, in the other thread i wrote that there's another way to define some other numbers(which are not the real numbers we work with): non-standard analysis. in non-standard analysis, we actually have infinitesimal small numbers. but working with them is quite difficult since e.g. you have to accept the existence of an "ultrafilter" which needs the axiom of choice(at least i think so).

btw, non-standard analysis shows as well that your proofs 0.999... = 9* 0.111.... = 9* 1/9 = 1 don't work: in non-standard analysis, the equality 0.111... = 1/9 is wrong as well. so, it's as dpopov said: proving 0.111... = 1/9 is exactly what we want to do.

Just a further note about what Peter just said: an infinitesimal number would be a number,  :varepsilon:, such that  0 < :varepsilon: < 1/n for all positive integers n.  There is no such number in our conventional system -- in fact, it's a standard type of proof in analysis to show that something is 0 by showing that it's less than all terms of a sequence like 1, 1/2, 1/3, 1/4, ...  So, for example, the "number" that some of you mentioned, 0.000...0001, which isn't well-defined in our standard system, was an attempt at introducing this idea of infinitesimals (even if you didn't realize it).

hm...

I thought it was still debated, somehow.....

well, I'm not that good at math

I didn't think that we had to define a number before we give the proof..

oh well :blush:

It is still debated in the sense that there are still many people that do not believe that $.\bar{9}=1$.  But mathematicians have come to the agreement that $.\bar{9}$ does equal $1$.  If you ever try convincing someone that .9999=1 then you will probably find out how stubborn people can be.


Does anyone know really good sites that explain all the definitons and basic principles of circular non-euclidean geometry :?:


[quote="roah"]Triangle ABC, AQ - is a bisector, AR - altitude.
We know that BQ is equal to b and CQ is equal to c.
What is the maximum altitude AR?[/quote]

$ AQ$ is the bisector,then: $ \frac{AB}{AC}\equal{}\frac{BQ}{QC} \Longleftrightarrow \frac{c}{b}\equal{} \frac {b}{c} \Longleftrightarrow b^2\equal{}c^2 \Longleftrightarrow b\equal{}c$
anyway as u said we have $ AB\plus{}AC\equal{}b\plus{}c\equal{}BQ\plus{}QC\equal{}BC$,and it means$ A,B,C$ are colinear!
correct ure problem!

Sorry,

i meant only that lengths of BQ and CQ are known.

b and c should be replaced by x and y.


i was thinking about buying the introductory geometry and counting books here, but was wondering how useful they would be for me. i found AOPS vol.1 to be a great book that was challenging for me. i only understood parts of AOPS vol. 2, but, then i got ambitious and bought Art and Craft of Problem Solving.(what a mistake that was :blush: ) do the books cover material found in AOPS volumes 1 and 2, or different material?

I don't have the books but I'm pretty sure they provide a more indepth coverage of geometry and counting, respectively, compared to AoPS V1/V2.  AoPS V1 and V2 only have a couple chapters each on counting and geometry whereas the subject books are entirely on that subject.  There will be a little bit of overlap, I assume, but there should be so much more stuff in the subject books that it won't even matter.  Additionally, from what I can tell, they should be right at your problem solving level (i.e. they will challenge you but won't discourage you that much).

As for AaCoPS, don't kick yourself for buying that book.  It's a totally awesome book and I am sure that you will love it once you get to that level.  Put it aside for now, but hopefully in the not so distant future you will get a lot of use out of it.

thanks for your help. :) i'll buy the books now.


probability
advanced fields
advanced fields unsolved

As we know $ \pi=3.141592653589793\ldots$. Now, using $ \pi$'s decimal expansion we construct a sequence:
\[ a_{1}=0.1,\ a_{2}=0.41,\ a_{3}=0.592,\ a_{4}=0.6535\]
and so on. Find all limit points of $ a_{n}$.

[quote="Lewap"]As we know $ \pi=3.141592653589793\ldots$. Now, using $ \pi$'s decimal expansion we construct a sequence:
\[ a_{1}=0.1,\ a_{2}=0.41,\ a_{3}=0.592,\ a_{4}=0.6535 \]
and so on. Find all limit points of $ a_{n}$.[/quote]

How do we even know that it converges? If it converges then for any $ \epsilon >0$ we can choose $ N>0$ so that $ n,m>N$ imply $ |s_{n}-s_{m}| < \epsilon$. Which [i]at least[/i] means for any $ \epsilon > 0$ we can fnd $ M\in \mathbb{N}$ so that $ |s_{M+1}-s_{M}|<\epsilon$. That means that those two adjancent elements must be arbitrary close. How can you possibly show that?

Well, truth is that I have no idea for the solution but I know that it is possible to show that is sequence has got at least one limit point. I'm not sure that we agree with the notion  'limit point'. By limit point I understand any number which is a limit of some subsequence of a given sequence. So our $ a_{n}$ doesn't have to be convergent. It just has to have a convergent subsequence. But I don't think that this makes things any easier...

That sequence doesn't converge, and it's almost certainly dense in $ [0,1]$. If we built a similar sequence for a number chosen uniformly at random from some interval, it would be dense in $ [0,1]$ with probability 1. There's no reason to expect $ \pi$ to be an exception, but actually proving this is absurdly hard.

uhm..
how can you be so sure that it doesn't converge, jmerry?
ok, i know it's almost certain ( :p ) that it does, but... i don't think a proof for this fact would be so much easier than showing that its accumulation points are all the interval $ [0,1]$.


From http://my.opera.com/dotuananh/blog/show.dml/1237716 :P
Cái này dành cho các em nhỏ nha, các anh già rồi thì chịu. :P

Theo yêu cầu của nhiều bạn quan tâm đến việc học tiếng Anh, du học và xin học bổng tại Hoa Kỳ, Viện Giáo Dục Quốc Tế IIE Việt Nam đã lập danh sách và tóm tắt những thông tin cơ bản về một số các website hữu ích dành riêng cho bậc học Graduate để các bạn tham khảo. IIE khuyến khích việc phổ biến rộng rãi những thông tin này. Tuy nhiên, khi đăng tải bản danh sách này, xin ghi rõ nguồn cung cấp:Viện Giáo Dục Quốc Tế, IIE Việt Nam, http://www.iievn.org.

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Một lời cảm ơn chân thành cho những đóng góp của anh  :D

[quote="N.T.TUAN"]Hãy cảm ơn anh bằng những link hay.  :lol:[/quote]
Em đóng góp thêm 1 trang nữa: http://smartweb.vn/
Tiện đây anh Tuân cho em hỏi muốn cài latex cho forum invisionfree thì phải làm thế nào ạ?

[quote="NguyenDungTN"]
Tiện đây anh Tuân cho em hỏi muốn cài latex cho forum invisionfree thì phải làm thế nào ạ?[/quote]
Anh không biết em ạ.

[quote="zaizai-hoang"]thêm cái code latex trong ACP là có ngay  :D Dễ như ăn hạt mễ  :D ngày hôm nay chán như con gà rán :D[/quote]
Cu Hoàng có thể post code lên được ko? Tiện thể hướng dẫn cụ thể luôn!


calculus
integration
function
logarithms
algebra
functional equation
topology

Find an increasing continous function $f: [0,+\infty) \to [0,1]$ satisfying:

1. $f(x)=\int_{0}^{x}f\left(\frac{x}{1-y}\right) dy$, $x \in [0,1)$
2. $f(x)=1$, $x \in [1,+\infty)$

We may assume that $f$ is infinitely differentiable increasing function and we can extend it such that $f: R \to [0,1]$ satisfying:

1. $f(x)=0$, $x \in (-\infty, 0]$
2. $f(x)=\int_{0}^{x}f(\frac{x}{1-y}) dy$, $x \in (0,1)$
3. $f(x)=1$, $x \in [1,+\infty)$

I think that exists exactly one such function but it is almost impossible to express it...
No ideas?

[quote="fedja"]Something is weird here. We have $f\le 1$ everywhere, whence $f(x)\le x$ on $[0,1]$ due to the functional equation, whence
$f(1-)\le\int_{0}^{1}\min\left\{\frac y{1-y},1\right\}<1$, so $f$ must be discontinuous at $1$. Perhaps, the condition of continuity on the entire $[0,+\infty)$ should be dropped?[/quote]

I am really sorry but I've made a terrible typo mistake:  :oops:

Instead of:
$\int_{0}^{x}f(\frac{y}{1-y}) dy$
it should be read:
$\int_{0}^{x}f(\frac{x}{1-y}) dy$

Then you are right when saying that there is exactly one such continuous function: just consider the metric space of all continuous functions on $[0,+\infty)$ that are $0$ at $0$ and $1$ beyond $1$ with the distance $d(f,g)=\max|f-g|$ and show that the right hand side of your functional equation defines a contraction on this space with the coefficient $\frac{1}{2}$. Most likely, you are also right when saying that there is no reasonable expression for this function.

One more thing: quoting the previous post entirely never makes sense: unless otherwise specified, your post is considered to be an answer to the previous post in the thread by default. :)

Well the original problem is:

When positive integer $n$ tends to $+\infty$, what percentage from all possible $n!$ permutations (of $n$ objects) has the length of the longest cycle at most a fraction $x$ from $n$ (as function of $x$) ?

This leads to:

Determine $f$ - increasing continous function $f: R \to [0,1]$ such that:

1. $f(x)=0$, $x \in (-\infty, 0]$
2. $f(x)=x \int_{x}^{\frac{x}{1-x}}\frac{f(t)}{t^{2}}dt$, $x \in (0,1)$
3. $f(x)=1$, $x \in [1,+\infty)$

So far I've managed to show that for $x \in (0,1]$, $f$ is a multiform on intervals like $I_{k}=[\frac{1}{k+1}, \frac{1}{k}]$ as $f_{k}$, where $k$ - positive integer:

For $x \in I_{1}=[\frac{1}{2},1]$:

$f_{1}(x)=x \int_{x}^{\frac{x}{1-x}}\frac{f(t)}{t^{2}}dt$, $x \in [\frac{1}{2},1]$
${f_{1}(x)=(2x-1)+x \int_{x}^{1}}\frac{f_{1}(t)}{t^{2}}dt$, $x \in [\frac{1}{2},1]$

Then $f_{1}(x)=1+lnx$.

Generally, for $x \in I_{k}=[\frac{1}{k+1}, \frac{1}{k}]$:
${f_{k}(x)=x \int_{x}^{\frac{1}{k}}}\frac{f_{k}(t)}{t^{2}}dt+x \int_{\frac{1}{k}}^{\frac{x}{1-x}}\frac{f_{k-1}(t)}{t^{2}}dt$, $x \in [\frac{1}{k+1}, \frac{1}{k}]$

But it's almost a pain in the... to determine even $f_{2}(x)$.

I think this one is a very beautifull problem!

Consider the original problem.

We may easy solve it for the case $x>1/2$. Indeed, let's count permutations which have a cycle of length greater than $nx$. But there can be only one such cycle. Suppose $k>nx$. Then the number of permutations which have a cycle of length $k$ is $C_{n}^{k}\cdot (k-1)!\cdot (n-k)!=\frac{n!}{k}$. So we have a sum $\sum_{k=[nx]+1}^{n}\frac{1}{k}$ and it tends to $-\ln x$. Hence the answer is $f(x)=1+\ln x$ for $x>1/2$.

Right?


probability geometry conics ellipse rectangle ratio calculus

[color=cyan]Two real numbers x and y are chosen such that |x|<2 and |y|<3.  Compute the probability that (x/2)^2 + (y/3)^2 < 1.[/color]

[hide] The probability is equivalent to the area of an ellipse x^2/4+ y^2/9 = 1 over the area of the rectangle with vertices at (2,0) (-2,0) (0,3) (0,-3). I believe it comes out to (pi)/4.[/hide]

[color=cyan]Why is that your belief?  (Not that it's wrong, just how did you come to that conclusion.  Please don't say you used calculus.)[/color]

Spoiler:

[hide]That is because the condition "|x|<2 and |y|<3" is met by the rectangle and the condition "(x/2)^2 + (y/3)^2 < 1" is met by the ellipse. Since the ellipse is completely inside the rectangle, we can simply take the ratio of the areas.[/hide]

What I meant was, where did she get [hide]the area of the ellipse from?  It's fairly non-trivial.[/hide]

The area of the ellipse? Isn't that ab(pi)?

[color=blue]Yes.  Can you give me an argument as to why?[/color]

Hmm...
The ellipse is a stretched circle. In the case of this problem the ellipse is a circle with radius 2, but elongated in one dimension... assume that the original circle was filled by an infinite number of little rectangles... the sum of the area of those rectangles would be equal to the area of the circle. Because of the elongation, all those rectangles would increase in length and retain the same width. The increase in length would be by a proportion of a/r, or half the major axis/ the radius. So the new area would be a/r* r* r* pi, or (the original area of the circle) times (the factor by which the ellipse was elongated), or ar(pi). The original radius of the circle is half the length of the minor axis in the ellipse, or b. So the area is given by ab (pi).

I'm not sure how accurate the above proof is. Do I need the argument with the rectangles?

[color=cyan]No, that was really all I was looking for.  You can state it in terms of the rectangles that one uses when taking an integral, of course.  That concept of "stretching" is known as Cavalieri's Principle, or at least is very closely related.

A neat problem using the principle:  given two cylinders of equal radii r, intersecting so that their axes are perpendicular.  Compute the volume of the solid of intersection.[/color]

One way you can think of it is to consider an elliptical cross section of a cylinder. Let the semiminor axis be the radius of the cylinder, and then you can make some right triangles to form the semimajor axis and thus the ellipse you want. Then you can use some ratios to show that the area of an ellipse is :pi: ab. Alternatively, you can do a trig substitution with an integral to find the area.


Let f(x) =  \sqrt (2+x) +  \sqrt (2-x), for x in [-1,1].

We prove that  \sqrt 3   \leq  f(x)  \leq  2 \sqrt 2, and that these constants are best possible.

The upper bound comes directly from the concavity of g(x) =  \sqrt x, and equality occurs iff x = 0.
The lower bound comes from the fact that f is even, so that wlog we may assume that x  \geq 0. But, it is easy to verify that f is increasing on [0,1], so that f(x)  \geq f(1) =  \sqrt 3, as claimed. Note that equality occurs iff x = -1 or x = 1.

Pierre.

I have only one question: Where do these inequalities come from? Don't you think they are a little bit too easy?

Well these inequalities  are nice and they will be in an increasing order of difficulty!!!

cheers! :D :D

and where are they coming from? are they your own creation?


geometry
incenter
inradius
trigonometry
analytic geometry
geometry unsolved

Let the incircle of an equilateral triangle $ABC$ touch the sides $AB$, $AC$, $BC$ respectively at $C'$, $B'$ and $A'$. Let $M$ be any point on the minor arc $B'C'$, and $H, K,L$ the orthogonal projections of $M$ onto the sides $BC$,$AC$ and $AB$, respectively. Prove that:

$\boxed{\sqrt{MH}= \sqrt{MK}+\sqrt{ML}}$

Let I be the incenter, r the inradius, and denote $2\theta = \angle C'IM,\ 0 \le 2\theta \le 120^\circ.$ Then $\angle B'IM = 120^\circ-2\theta,$ $\angle A'IM = 120^\circ+2\theta,$ and

$\sqrt{MH}= \sqrt{r[1-\cos(120^\circ+2\theta)]}= \sqrt{2r}\sin(60^\circ+\theta) > 0$

$\sqrt{MK}= \sqrt{r[1-\cos(120^\circ-2\theta)]}= \sqrt{2r}\sin(60^\circ-\theta) > 0$

$\sqrt{ML}= \sqrt{r[1-\cos 2\theta]}= \sqrt{2r}\sin \theta > 0$

Since $\sin(60^\circ-\theta)+\sin \theta = \sin(60^\circ+\theta),$ the conclusion follows.

Since triangle $ABC$ is an equilateral triangle, so B' and C' are the midpoints of AC and AB respectively. Hence Triangle AB'C' is also an equilateral triangle and $B'C' \parallel BC$. Let MH intersects B'C' at point D, and the altitude of side B'C' of triangle AB'C' be $h'$, then $MD \cdot B'C'+ML \cdot AC'+MK \cdot AB' = 2S_{\triangle{AB'C'}}= B'C' \cdot h'$. Because AB'C' is equilateral, so we have $MD+ML+MK=h'$.

Let the altitude of side BC of triangle ABC be $h$, then $h=2h'$. Since $MH+MK+ML=h=2h'=2MD+2ML+2MK$. Therefore $MH=2MD+ML+MK$.

Connect MC' and MB', then, since LC' is tangent to the circle, so $\angle{LC'M}= \angle{DB'M}$. Because $\angle{MLC'}=\angle{MDB'}=90^{o}$, so there is $\triangle{MLC'}\sim \triangle{MDB'}$. Therefore $\frac{ML}{MC'}=\frac{MD}{MB'}$. Which gives $\frac{ML}{MD}=\frac{MC'}{MB'}$. Similarly, the similarity of triangle MKB' and triangle MDC' gives $\frac{MD}{MK}=\frac{MC'}{MB'}$. Hence we have $MD^{2}= ML \cdot MK$. So $MD= \sqrt{ML \cdot MK}$.

Now use the identity we have proved above, we have $MH=ML+2\sqrt{ML \cdot MK}+MK = \left( \sqrt{ML}+\sqrt{MK}\right)^{2}$, take off the square sign we finally get $\sqrt{MH}= \sqrt{ML}+\sqrt{MK}$.

[color=darkblue][b]Lemma.[/b] If the positive numbers $x$, $y$, $z$ verify the relations $x\ge y$, $x\ge z$, $x+y+z=1$ and $xy+yz+zx=\frac{1}{4}$,  then $\boxed{\ \sqrt x=\sqrt y+\sqrt z\ }\ .$[/color]

[color=darkblue][b]Lemma 1.[/b] If the positive numbers $x$, $y$, $z$ verify the relations $x\ge y$, $x\ge z$, $x+y+z=1$ and $xy+yz+zx=\frac{1}{4}$,  then $\boxed{\ \sqrt x=\sqrt y+\sqrt z\ }\ .$

[b]Lemma 2.[/b] The point $P$ has the barycentrical coordinates $(x,y,z)$ w.r.t. the equilateral triangle $ABC$ for which $AB=1\ .$.
Denote the incircle $w$ of the given triangle and the projections $H$, $K$, $L$ of the point $P$ to the sides $[BC]$, $[CA]$, $[AB]$ respectively. Then :
$1.\blacktriangleright\ \frac{x}{PH}=\frac{y}{PK}=\frac{z}{PL}\ .$
$2.\blacktriangleright\ P\in w\Longleftrightarrow xy+yz+zx=\frac{1}{4}\ .$

[b]Proof of the proposed problem.[/b] Let $P(x,y,z)$ be a point in the plane of the equilateral triangle $ABC$ for which $AB=1\ .$ Thus, $\boxed{\ x+y+z=1\ }$ and $P\in w\ ,\ PH\ge PK\ ,\ PH\ge PL\Longrightarrow x\ge y\ ,\ x\ge z$ and $\boxed{\ xy+yz+zx=\frac{1}{4}\ }\ .$ Therefore, $\sqrt x=\sqrt y+\sqrt z$, i.e. $\sqrt{PH}=\sqrt{PK}+\sqrt{PL}\ .$[/color]


You are given 4 points pairwise connected by segments (a $K_4$). We place then in a 1 X 1 square. What is the probabilty that none of the segments cross?

The segments will not cross if and only if one of the points lies within the triangle created by the other three.  The probability of the point's not being inside the triangle is the area of the triangle.  So we can reduce the problem to a question of the average area of the triangle created by 3 randomly chosen points in a 1 x 1 square.


[quote="m1357-1357"]Prove that there is a number $ n^{5000}$ which starts with 5000 ones.[/quote]

Let $ k\equal{}1111....11$ ($ 5000$ ones ) . Then we only need to prove that we can always find a suitable positive integer $ x$ so that there is a $ 5000$ th power between $ 10^xk$ and $ 10^x(k\plus{}1)$.

And this is equivalent to prove that we can always find $ x$ such that

$ \lfloor \sqrt[5000]{10^x(k\plus{}1)} \rfloor > \sqrt[5000]{10^xk}$

To prove this , we only need to show $ \sqrt[5000]{10^x(k\plus{}1)} > \sqrt[5000]{10^xk}\plus{}1$ is possible for some $ x$ , which is equivalent to

$ 10^x(k\plus{}1) > 10^xk\plus{} 5000 (10^xk)^{\frac{4999}{5000}}\plus{}...\plus{}5000 (10^xk)^{\frac{1}{5000}}\plus{}1$,

which clearly holds for any large enough $ x$, done.


x and y are distinct positive integers, $ x < y$ and $ x^{2}\plus{}y^{3}$ is divisible by $ x^{3}\plus{}y^{2}$.
Are there infinity such pair (x, y) or not?  :maybe:

I just noticed that there are one such pair (2, 6)  :idea:

If all we're interested in is infinitely many (not necessarily "all"), then we can make some simplifying assumptions.  For example, you could suppose that $ y \equal{} cx$ for some integer $ c$.  This leads to a system that eventually lets you choose a value for $ x$ in terms of $ c$ to get infinitely many solutions.

And is it possible to find the second or the third pair of solutions?

The computer likes (7, 63), (13, 39), (15, 60), (17, 68), (19, 95) and (27, 54).  (Note that not [i]every[/i] solution is of the form $ y \equal{} cx$ -- the first counterexample my computer found is (64, 224).  There are 21 solutions with both $ x, y$ less than 1000.


inequalities
trigonometry
geometry unsolved
geometry

[quote="shobber"]In acute triangle $ ABC$, show that:

$ \sin^3{A}\cos^2{(B \minus{} C)} \plus{} \sin^3{B}\cos^2{(C \minus{} A)} \plus{} \sin^3{C}\cos^2{(A \minus{} B)} \leq 3\sin{A} \sin{B} \sin{C}$

and find out when the equality holds.[/quote]
We only need to prove the equality:
$ \sum_{cyc}sin^3Acos(B \minus{} C) \equal{} 3sinAsinBsinC$
$ 4sin^3Acos(B \minus{} C) \equal{} 4sin^2(B \plus{} C)sin(B \minus{} C) \equal{} 2sin^2A(sin2B \plus{} sin2C) \equal{} (1 \minus{} cosA)(sin2B \plus{} sin2C) \equal{} sin2B \plus{} sin2C \minus{} sin2Bcos2A \minus{} sin2Ccos2A$
It follows that:
$ \sum_{cyc}4sin^3Acos(B \minus{} C) \equal{} \sum_{cyc}(sin2B \plus{} sin2C) \minus{} \sum_{cyc}sin2Bcos2A \minus{} \sum_{cyc}sin2Ccos2A$
$ \equal{} 2\sum_{cyc}sin2A \minus{} \sum_{cyc}(sin2Acos2B \plus{} cos2Asin2B)$
$ \equal{} 2\sum_{cyc}sin2A \plus{} \sum_{cyc}sin2C$
$ \equal{} 3(sin2A \plus{} sin2B \plus{} sin2C)$
$ \equal{} 12sinAsinBsinC$
The equality occurs iff $ cos(A \minus{} B) \equal{} cos(B \minus{} C) \equal{} cos(C \minus{} A) \equal{} 1$<=>ABC is an equilateral triangle

We will prove the miraculous identity
$$\sum_{cyc} \sin^3 A \cos (B-C) = 3 \sin A \sin B \sin C.\qquad (1)$$
This would clearly imply the desired inequality.
First of all, applying the Cosine Subtraction Identity and dividing both sides by $\sin A \sin B \sin C$, we see that $(1)$ is equivalent to:
$$\sum_{cyc} \sin^2 A (\cot B \cot C + 1) = 3,$$
which in turn is equivalent to
$$\sum_{cyc} \sin^2 A \cot B \cot C = \sum_{cyc} \cos^2 A$$
after a subtraction.
After multiplying by $\tan A \tan B \tan C$, we need only to show that:
$$\tan A \tan B \tan C \sum_{cyc} \cos^2 A = \sum_{cyc} \tan^3 A \cos^2 A. \qquad (2)$$
Now, set $x, y, z$ to be $\tan A, \tan B, \tan C$ respectively. Using the identity $\cos^2 \theta = \frac{1}{1+ \tan^2 \theta}$ for all angles $\theta$, we can rewrite $(2)$ as:
$$xyz \sum_{cyc} \frac{1}{1+x^2} = \sum_{cyc} \frac{x^3}{1+x^2}. \qquad(3)$$
Now, we cite the well-known identity that $x+y+z = xyz$. With this in mind, we know that:
$$\frac{1}{1+x^2} = \frac{x+y+z}{xyz + x^2(x+y+z)} = \frac{x+y+z}{x(x+y)(x+z)},$$
with similar relations for $y, z.$ Plugging this relations into $(3)$, we now want:
$$xyz \sum_{cyc} \frac{x+y+z}{x(x+y)(x+z)} = \sum_{cyc} \frac{x^3(x+y+z)}{x(x+y)(x+z)}. \qquad(4).$$
After multiplying through by $(x+y)(y+z)(z+x),$ it's easy to find that $(4)$ is simply equivalent to:
$$\sum_{cyc} (x+y+z)yz(y+z) = \sum_{cyc} (x+y+z) x^2(y+z),$$
which is easily verified to be true. We are now done.

$\square$


probability
integration
geometry
calculus
inequalities
analytic geometry

When two real numbers are selected between $5$ and $8$, what is the probability that they differ by less than one? Generalize.

[quote="vishalarul"][hide="Inclusive or exclusive?"]If it is inclusive,

If you pick $5$ with probability $\frac{1}{4}$, you must pick a $6$, with a probability $\frac{1}{4}$. The total probability is $\frac{1}{16}$

If you pick $6$, pick $7$ or $5$. The total probability is $\frac{1}{8}$

If you pick $7$, pick $6$ or $8$. The total probability is $\frac{1}{8}$

If you pick $8$, pick $7$. The total probability is $\frac{1}{16}$

Summing it up, I get that the answer is $\frac{3}{8}$

If it is exclusive, you can easily check that the answer is $\frac{1}{2}$[/hide][/quote]

It doesn't matter if it is inclusive or exclusive because the probability of picking 5 or 8 is 0.  Note that the numbers don't have to be integers.  Also note that the numbers $x$ and $x+1$ differ by $1\not<1$.

[hide="hint"]Can you express this geometrically?[/hide]

[quote="vishalarul"][quote="Quevvy"][quote="vishalarul"][hide="Inclusive or exclusive?"]If it is inclusive,

If you pick $5$ with probability $\frac{1}{4}$, you must pick a $6$, with a probability $\frac{1}{4}$. The total probability is $\frac{1}{16}$

If you pick $6$, pick $7$ or $5$. The total probability is $\frac{1}{8}$

If you pick $7$, pick $6$ or $8$. The total probability is $\frac{1}{8}$

If you pick $8$, pick $7$. The total probability is $\frac{1}{16}$

Summing it up, I get that the answer is $\frac{3}{8}$

If it is exclusive, you can easily check that the answer is $\frac{1}{2}$[/hide][/quote]
The original problem said real numbers, not integers.[/quote]

Oops.

This problem is the same if the two numbers were $0$ and $3$, so let's try that.

$3$ cases:

$0<n\leq1$
$1<n<2$
$2\leq n<3$

In the first case, $1<n+1\leq 2$, so the probability of picking such a number is $\frac{1}{3}$. We can't do $n-1$

In the second case, $2<n+1<3$, so the probability is $\frac{1}{3}$. $0<n-1<1$, so the probability is $\frac{1}{3}$. Together, we get $\frac{2}{3}$

Similarly, in the third case, the probability is $\frac{1}{3}$.

Is the answer $\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}= \frac{2}{27}$[/quote]

thats almost right. youre not supposed to multiply them

I think the answer should be

[hide]
$\frac{5}{9}$

I got this by graphing $y<x+1$ and $y>x-1$ and findinig the area of it within the region $5<x<8$ and $5<y<8$. I got that to be $5$, and then I divided it by the total area of that region, $9$.

[/hide]

[quote="Altheman"]for $y\in[0,1]$, we have that $x\in[0,y+1]$

for $y\in[1,2]$, we have $x\in[y-1,y+1]$

for $y\in[2,3]$, we have $x\in[y-1,3]$

so, we have $\frac{1}{3}\int_{0}^{1}\frac{y+1}{3}\, dy+\frac{1}{3}\int_{1}^{2}\frac{2}{3}\, dy+\frac{1}{3}\int_{2}^{3}\frac{4-y}{3}\, dy=\frac{5}{9}$

uugh...counting the complement would have been nicer...oh well...also the generalization is pretty trivial using this method, which is pretty much the same as teh geometric method, but once you get into more variables, the only practical way is to use integrals...

EDIT: uugh...i miscalculated the last integral...[/quote]

i would say that was calc, but its just a fancy way of finding the area under the curve so i wont delet it.  btw $\int_{0}^{1}\frac{x+1}{3}\, dx$ is calculated be finding the area under the line $\frac{x+1}{3}$ from $x=0$ to $x=1$.

Draw a picture. Kinda squarish. doesn't matter. label one side from 5 through 8 and the other side from 5 through 8. find the area of the whole thing (9). find the area of it where they differ by less than 1: $|x-y|<1$ bound that area. divide.

as you guys may or may not have deduced, the integral that i used is the same thing as finding the area satisfied by certain inequalities on a cartesian plane, however, the reason that i chose to use integrals is because it easily generalizes...what if we chose 4 variables? the geometric solution becomes much more difficult then...however the integral is readily applied


complex numbers
complex analysis
complex analysis unsolved

Prove that if $ a,b,c$ are complex numbers and $ |a|\equal{}|b|\equal{}|c|\equal{}r\ne 0$ then $ \left|\frac{ab\plus{}bc\plus{}ac}{a\plus{}b\plus{}c}\right|\equal{}r$.

[hide]$ \left \vert \frac {ab \plus{} bc \plus{} ac}{a \plus{} b \plus{} c}\right\vert \equal{} r\left \vert \frac {e^{i(\theta _1 \plus{} \theta _2)} \plus{} e^{i(\theta _1 \plus{} \theta _3)} \plus{} e^{i(\theta _2 \plus{} \theta _3)}}{e^{i\theta _1} \plus{} e^{i\theta _2} \plus{} e^{i\theta _3}}\right \vert \\
 \\
 \\
\equal{} r\sqrt {\frac {\left(e^{i(\theta _1 \plus{} \theta _2)} \plus{} e^{i(\theta _1 \plus{} \theta _3)} \plus{} e^{i(\theta _2 \plus{} \theta _3)}\right)\left(e^{ \minus{} i(\theta _1 \plus{} \theta _2)} \plus{} e^{ \minus{} i(\theta _1 \plus{} \theta _3)} \plus{} e^{ \minus{} i(\theta _2 \plus{} \theta _3)}\right)}{\left(e^{i\theta _1} \plus{} e^{i\theta _2} \plus{} e^{i\theta _3}\right)\left(e^{ \minus{} i\theta _1} \plus{} e^{ \minus{} i\theta _2} \plus{} e^{ \minus{} i\theta _3}\right)}} \\
 \\
 \\
\equal{} r\sqrt {\frac {\sum_{j \equal{} 1}^3\sum_{k \equal{} 1}^3 e^{i\theta_j \minus{} i\theta_k}}{\sum_{j \equal{} 1}^3\sum_{k \equal{} 1}^3 e^{i\theta_j \minus{} i\theta_k}}} \equal{} r$
Does this have some nice geometric interpretation?[/hide]


superior algebra
superior algebra solved

1.: I can't open the link, possibly some error :(
2.: Whatever it is, you
  a) should write it down here in the forum
  b) not post it in the solved problems section


One of the following four-digit numbers is not divisible by 4: 3544, 3554, 3564, 3572, 3576.  What is the product of the units digit and the tens digit of that number?

[hide]3554 is not divisible by 4 so 4*5=20[/hide]

[hide]
3544, 3554, 3564, 3572, 3576
divisible by 4 if the tens and ones digit creates a number that is divisivble by 4, so 3554 is it

$5\cdot4=20$[/hide]

[hide]3,554 isn't divisible by 4 since 54 isn't.
5*4=20
the answer is [b]20[/b].[/hide]

A number is is divisible by 4 if the last 2 digits is divisible by 4.

3544, 3554, 3564, 3572, 3576

3544--> 44 is divisible by 4. hence 3544 is divisible by 4.
3554--> 54 is not divisible by 4. hence 3554 is not divisible by 4.
72 n 76 are divisible by 4.

answer:

4*5 = 20.

risinglegend and venkat please hide your answers!  :)


Problem: Let $ X$ be an infinite set and $ \tau$ a topology on $ X$. If every infinite subset of $ X$ is in $ \tau$, prove that $ \tau$ is the discrete topology.

The following is my idea. First we know $ X$ is a discrete topology if and only if $ \{x\}\in\tau$ for all $ x\in X$. Thus, for any $ x\in X$ if we can choose two infinite subset $ A$ and $ B$ of $ X$ such that $ A\cap B\equal{}\{x\}$, then we are done.

However, this method requires the axiom of choice. Actually, can we prove it without using the axiom of choice and why?

you shouldn't need it. $ X$ is infinite iff it contains a copy of $ \mathbb{N}$ and iff $ X\setminus\{x\}$ is infinite.
now consider a copy of the naturals inside $ X\setminus x$ and the two sets of "odd" elements and "even" elements, and add $ x$ to both. they're infinite and their intersection is $ x$.

"$ X$ is infinite $ \iff X$ contains a countable subset" requires a form of the AC too. At least the countable axiom of choice.

So, how're we defining "infinite", then? Usually: in bijection with a proper subset of itself... but
we're not even allowed to say "let $ f$ be such a bijection" without using the axiom of choice.

Not a lot we can even work with here.

Also, FWIW, most of topology requires axiom of choice-ish things, so I don't think
you're going to be able to avoid it...

An infinite set has at least one element.
An infinite set, minus one element, is still infinite.
These two should be enough to finish your problem.
Choice not needed, but definition of "infinite" is needed.

[quote="Nukular"]So, how're we defining "infinite", then? Usually: in bijection with a proper subset of itself... but
we're not even allowed to say "let $ f$ be such a bijection" without using the axiom of choice.[/quote]

yes, we're allowed to! for every [i]finite[/i] collection of nonempty sets we can choose an element from each of them, without worrying of axiom of choice! it's induction + choosing from one nonempty set an element (and this is allowed by definition).


How much is a prime numbers $ p$, where $ p\equal{}21*k\plus{}17$?


1.
A tsunami destroyed 4/5 of forest. if 10,800 trees aere lost, how many trees survied.

2.
Find three consecutive even integers for which 3 times the largest is twise the sum of the two smallest intehers.

3.
Solve:

(for Y) M-3N=Y+N

4.
2x+8(x-3)=4-(3x+2)

5.

Evaluate: c-bc if b= -1 and c=-2

[hide="Answer to 1"]If $\frac45$ is 10,800 trees, then $\frac15$ of the trees is 2700.  There were $1-\frac45=\frac15$, or $\boxed{2700}$ trees that survived.[/hide]

[hide="Answer to 2"]We know that

$3\times(n+2)=2\times(n+(n-2))$
$3n+6=4n-4$
$-n=-10$
$n=10$

The numbers are therefore:
$(10-2), 10, (10+2)\Rightarrow8,10,12$[/hide]

[hide="Answer to 3"]$M-3N=Y+N$
$M-4N=Y$
$\boxed{Y=M-4N}$[/hide]

[hide="Answer to 4"]$2x+8(x-3)=4-(3x+2)$
$2x+8x-24=4-3x-2$
$10x-24=2-3x$
$13x=26$
$\boxed{x=2}$[/hide]

[hide="Answer to 5"]$c-bc$
We know that $b=-1$ and $c=-2$.
Substitute into the equation to get:
$-2-(-1)\times-2$
$-2-2=\boxed{-4}$[/hide]

[quote=MCrawford said]Please limit posts to one problem per post.[/quote]

[hide="1"]2700[/hide]
[hide="2"]8,10,12[/hide]
[hide="3"]M-4N[/hide]
[hide="4"]2[/hide]
[hide="5"]-4[/hide]

Wolfie, for further reference, its uno dos tres cuatro cinco. :)

[quote="Wolfie"]:blush: I was going to do something original for each one. Then I forgot.

And I think you should check your Uno.[/quote]

Uno is now edited. Thanks!


geometry ratio articles calculus probability Support search

the only carreers is accounting and teaching
research is to hard
accounting is getting replaced by bots

um...have you looked at the other threads of this forum?  traders, quants, and actuaries are all very good careers in math.

I don't think research is "too hard" except insofar as it doesn't mesh with some people's personalities.

[quote="jimli"]the only carreers is accounting and teaching[/quote]I know mathematicians that works in NASA, NSA, Bell Labs, Intel and other goverment organizations and private companies. Honestly I don't think that  a mathematician on NSA or Bell Labs works doing accounting and teaching.

[quote="jimli"]research is to hard[/quote]There is nothing that works for everybody, but if you like math and you enjoy research, it shouldn't be that hard.

[quote="jimli"]accounting is getting replaced by bots[/quote]bots are tools, not accounters.

[b]BTW[/b]: This is my post 1234  :D

I am an accounting major, taking my first accounting class. There is practically no math involved.

If nothing else, mathy people can always move into a sciency field and apply their math skills.  I'm sure most scientific research involves a certain degree of mathematical problem solving ability and background.

If lawyers dont relate to logics or math in any ways, how does being a good problem solver help being an actuary, per se?[/code]

[quote="xxreddevilzxx"]how does being a good problem solver help being an actuary, per se?[/quote]An actuary's main task is to predict the likelihood of future events by examining past events.

Insurance companies rely on actuaries to establish insurance rates.  If the premiums are set too high, customers will buy from competitors.  If the premiums are too low, the company will not be able to pay future claims, perhaps even going out of business.

For auto insurance, for example, actuaries decide on rates by looking at variables that predict the likelihood of an accident.  Variables include the driver's age, gender, neighborhood, car model, experience, accident history, etc.  Actuaries make decisions about which variables are important and which are not by examining historical data.  They then determine how much weight to assign each variable.  They have to take into account the fact that customers hate to be asked an endless stream of questions about their driving habits and that customers can be unreliable in their answers.

Suppose you're an actuary assigned to set insurance rates for the new World Trade Center complex.  The owner, of course, wants to insure against future acts of terror.  How do you decide what a reasonable annual premium is?  How do you define "act of terror"?  How do you distinguish an "act of terror" from an "act of war", which is usually excluded from insurance coverage?  You would need to examine the motives of the 9/11 attackers.  You would need to predict whether terrorists would choose the new Trade Center as a future target.  You would need to look at the probability of various types of attack: bombings, biochemical attacks, airplanes crashing into buildings, etc.  You would need to analyze the ability of the buildings to withstand such attacks.  How much damage would be sustained?  How many people would be injured or killed?

The type of problem solving actuaries do is very different from that of pure mathematicians.  It involves heavy use of probability and statistics as well as real world knowledge of the industries they're insuring.

[quote="djimenez"][quote="qvc122"]Usually, people who are very good with math/ physics plus good language skills, often make to top 5% in their medical school and law school classes. Logic is not only behind math and science.[/quote]But those are not MATH carriers, and actually are not quite related.[/quote]
True. Real math positions (research positions, those I truly consider as mathematicians) are very limited. I believe there are about one thousand students graduate each year with PhD in math and there are about 200 academic job openings in universities in US. In other words, the surply is more than the demand. One thing that academic job is attractive for people with math PhD is the flexible schedule. An academic mathematician cannot really provide a family well if he/ she is the lone income source in a family. That is just reality.
When you are a student, and people consider you as super smart, it sure makes you happy and feel so good about yourself. After you start you job and can't provide you family well, or you are one of those who did not end up in a "real" 4-year university but instead in a 2-year community college, you will not have good feelings any more. My previous note is just a suggestion that when you are facing a career choice, don't be narrow minded. I feel very sorry to see smart math people struggle in real life because of their poor ability to adjust. I don't mean to offend anyone.

For those interested in an academic career in math, the [i]AMS 2004 Annual Survey of the Mathematical Sciences[/i] reports that there were 1041 new doctoral recipients in 2003-04.  739 found employment in the U.S.: 614 in academic positions, 99 in business, 26 in government.  The unemployment rate among new PhDs was 5.7%.

In 2004 the median faculty salary for all tenured or tenure-track math faculty was:

[code]Math Department        Asst   / Assoc  / Full
Group I Public Univ    63,060 / 69,730 / 100,910
Group I Private Univ   61,070 / 75,920 / 113,900
Group II Univ          56,860 / 64,070 /  86,350
Group III Univ         53,630 / 59,530 /  76,940
Master's Institution   48,690 / 58,730 /  75,760
Bachelor's Institution 47,310 / 55,990 /  70,640

Statistics Dept        65,100 / 71,200 / 103,070
Biostatistics Dept     66,710 / 77,500 / 110,630
Applied Math Dept      52,500 / 58,130 /  90,830
[/code]([i]Note: Math departments are ranked by "scholarly quality of program faculty".  Group I: 48 depts ranked 3.00-5.00.  Group II: 56 depts ranked 2.00-2.99.  Group III: 73 depts ranked 0.00-1.99.[/i])

I have the greatest respect for those who wish to become math professors but it's not the most lucrative of careers.  Starting salaries at four-year colleges can fall below $\$50,000.  These days universities seem to be moving toward fewer tenured positions and more temporary instructor/visitor positions which pay even less than the figures above.

[quote="qvc122"]I feel very sorry to see smart math people struggle in real life because of their poor ability to adjust.[/quote]

And the stereotyping of math majors continues...

[quote="qvc122"]
True. Real math positions (research positions, those I truly consider as mathematicians) are very limited. I believe there are about one thousand students graduate each year with PhD in math and there are about 200 academic job openings in universities in US.... An academic mathematician cannot really provide a family well if he/ she is the lone income source in a family. That is just reality.
[/quote]

(1) I consider anyone who has done original mathematics research to be mathematicians.  (That's not a necessary condition for me, but it is a sufficient one.)  So there are plenty of undergraduate mathematicians out there, in my book.
(2) By your definition, very few professors of mathematics are mathematicians, because there are almost no research positions in mathematics departments.  Most positions are some combination of research and teaching.
(3) There are lots and lots of mathematicians (my definition) at colleges---that is, institutions without graduate programs---who are productive researchers.
(4) I know several colleagues who are the sole income source for their families, and they seem to provide for their children just fine.

My apology if my opinion upsets you. I do sense all that pride here. I personally started in math and physics and got quite far. I also married an academic mathematician, who publishes actively and does review as a referee for a few journals all year round. You can see where my stereotype came from, not entirely unfounded. I am glad that the job market opened up quite a bit, much better than 4 or 5 years ago in university positions.  Everyone is entitled to what they like to do/ to choose for their career. I can only wish you all best.

[quote="qvc122"]Btw, djimenez, what do you plan to do after finishing your graduate school? Any preference regarding academia vs. industry? How many more years would it take you to finish your degree? Are you working toward a PhD in math? :?:[/quote]Well, let's go step by step. What do I plan to do after finishing my gradschool, well, recently in a conference, someone working in Standford told me that if I continue doing what I am doing, he may considered to work a postdoc there for me (it sounds great! even when it was told very informally for a professor), and there are a few research labs I would like to apply (Bell Labs just as an example). About Mathematical Finance... well, I think I could, but I don't know much about it, I would have to learn a lot (what is not bad at all) and I don't know if someone would give me such kind of job without much (at least theoretical) knowledge about finance. Then, the answer is, I really don't know, even when academia is, in my personal opinion, a good option. But with academia, I would not like to end up in a place where the emphasis is teaching and you have basically no time to do your research (I can name a few places where I know it is that way).

Second, I hope that I would finish either in the spring or fall of 2007. May be even before, but we'll see. I am working toward a Ph.D. in math, I started in the fall of 2003.

Best,

[quote="djimenez"] Then, the answer is, I really don't know, even when academia is, in my personal opinion, a good option. But with academia, I would not like to end up in a place where the emphasis is teaching and you have basically no time to do your research (I can name a few places where I know it is that way).
[/quote]

So, here's my guideline for you:  Don't apply to tenure-track positions which have greater than a 3/2 teaching load, unless you have graders for all of your courses.  (Then 3/3 would be okay, but still difficult.)  There are lots of places where the emphasis is on teaching, but there's still plenty of support for doing research.

Djimenez,

My husband works for his second academic job now as an academic mathematician. He worked for a 4-year state collage for 13 years until  two years ago. He was a full professor when he left that university. He did not make 100k then (much less than that) since he was newly promoted full professor. You really have to be in full professor position for another 10 year to make that much. Statistics sometimes does not reflect the real situation especially when you combine state universities and private universities. Some full professors with titles of chairman or vice chair would make more so they skew the statistics too. If the statistics result is done by survey, some people who do not make a lot would not participate in the survey. Those are all the factors which may potentially make the real averages higher than they really are. My husband made a big sacrifice to move to where we were (me and two daughters) for my career two years ago. When you are a tenured professor and you want to move to a different university in a specific city, you will have a major difficulty as a math professor. To make the story shorter, he gave up the full professor and his tenure to start an assistant professor position again at the current state university, taking about 30% salary cut, since that is the only university job available. His starting salary two years ago was 49k for nine months (you may potentially get two more months income by doing summer teaching or having grants). The good thing is that he is still doing exactly same job in terms of teaching and research.

Regarding teaching load, it is not bad at all as an academic mathematician if you end up in a regular 4-year collage. Usually it is 8 credit hours per week (roughly 8 hours teaching). Math is something if you know your material well, you don’t really need to prepare a lot for class, not like teaching biology, which you really have to spend lots of time to prepare. Community collages may be different, which you may teach 20 hours per week week. My husband always has time to do research and always publishes in good journals. If you want to have time to do research, I do believe you would like to get into a 4-year collage with graduate program. If you can do a posdoc with a well know name, it is very helpful for an academic position, especially when you start to apply for grant.

University system is the main job buffer for PhD mathematicians. They do absorb most of the newly PhD graduates. Doing finance is great. However, it may not suit everyone. Financial market can only absorb small portion of PhD mathematicians and it is less stable. My husband likes the flexibility of his job. He actually took one accounting class when he was with the original univ. and got top score. But I didn’t think he liked the nature of the practice as an accountant. Some people are born to do more creative /pioneering work than routine work. If you wife has a good job, you don’t need to make too much. You don’t have to be a bread winner at home, honestly.  :)

Thank you for reading my story, wild octagon.

The purpose of the story is not to discourage Djimenez, but to serve as guidance in certain aspects. Djimenez's situation is very similar to when my husband and I initially came to United States when we were in our mid 20s. I am sure Djimenez and my husband would both get PhD from Princeton if they were educated in the US from their childhood. But there is no reason to dwell on ifs. We can always hope that our children are capable of fulfilling their own dreams.

To wild octagon, regarding what is enough in one of the previous message. My kids do take private piano lessons and the older one plays violin too, probably just like your children.

Yes, my kids play cello and violin.  Music and math often go together, don't they?

I think it's interesting that money is a bit of a taboo topic in our capitalist country.  Parents often don't discuss finance with their children.  Students often graduate from high school and college without taking a single course in economics.

When I was attending college, most students believed that a career in business or finance was "bad" while a career in a non-profit organization was "good".  I've worked at several non-profits and I've been surprised at the amount of energy devoted to begging -- I mean asking -- for donations from wealthy patrons.

Recently I've begun to think that the best way to "do good" is to follow the path of my college classmate who entered the world of finance and became a successful money manager.  He now makes million-dollar contributions to worthy causes.  By earning lots of money and donating it, he's probably making more of a difference than if he had chosen a career with a non-profit.  Ironic, isn't it?

[quote="wild octagon"]Recently I've begun to think that the best way to "do good" is to follow the path of my college classmate who entered the world of finance and became a successful money manager.  He now makes million-dollar contributions to worthy causes.  By earning lots of money and donating it, he's probably making more of a difference than if he had chosen a career with a non-profit.  Ironic, isn't it?[/quote]

But.. after someone donates the money, there have to be people to accept it and implement it wisely. I would say that both roles are necessary, and both are equally important.

[quote="gauss202"]But.. after someone donates the money, there have to be people to accept it and implement it wisely. I would say that both roles are necessary, and both are equally important.[/quote]Yes, of course.  It's just that with a large endowment you have the means to hire many such people.

[quote="gauss202"]But.. after someone donates the money, there have to be people to accept it and implement it wisely. I would say that both roles are necessary, and both are equally important.[/quote]

I don't see why either role is necessary.  There is no reason that both activities can't take place within the same organization.  I think that such a reorganization of responsibilities/labor is a natural progression from the status quo.

there are many subjects that tie into math implicitly.
music, as mentioned before, ties into math because it requires a quick mental mind to calculate fractions and positions of notes.
applications in biology include chromosomes and genetics, as well as population curves.
applications in chemistry include using calculus to derive rate-time formulas for kinetics.
there are many many applications for physics, but i dont know of many, because i am only a high school sophomore, and i didnt take physics yet.
math, reading, and writing are the 3 most important subjects that we use everyday. we could probably have school and do without all the other subjects and be successful (although that would make school infinitely boring). there aren't many interesting careers that explicitly use mathematics, but that should never prevent encouragement from continuing math courses

[quote="bobthebuilder"]there aren't many interesting careers that explicitly use mathematics,[/quote]

I couldn't disagree more.  Here are a few: engineering (or which ther are many varieties), artificial intelligence, programming (of many varieties), architecture, mathematicians, physics, genetics, statistics (there are an abundance of varieties of stats careers), quantitative analysis in any financial field, and many others.

I hope very few students have this perception of mathematics.

[quote="MCrawford"]
I hope very few students have this perception of mathematics.[/quote]

Unfortunately, most do.

Opinion post alert!!! :P  Read at your own risk!

 I firmly believe that if you choose your career solely based on how much money you can get, you will be most likely to be extremely miserable at some stage of your life.  You really should consider choosing a career, by asking yourself one question: [i]If for what ever reason the job suddenly isn't lucrative money-wise, and I was forced to work even after retirement, would I be prepared to work in this career to my dying day?[/i]  If you can answer [i]yes[/i] to this, then it should be a career worth pursuing.  If not, find a career that you are more likely to say yes to, last thing you want to do is end up with a low pay job, that you hate.

 What about if you have to support a family?  Having a degree in mathematics won't limit your choices, and you can still make a career choice outside mathematics, even if it is only to keep your afloat for a difficult period of time.

Take this argument for example,

 Even [b]if[/b] by comparison to another field, mathematics doesn't provide as many choices - involving mathematics.  The studying of mathematics and getting a degree in Mathematics will open more doors to you than most other fields can.  This is simply because anyone who has completed a degree in mathematics, are [b]probably[/b] more likely to be able to solve any other kind of problem, due to way they are used to solving mathematical problem.  If I were a manager, I'd pick someone with a degree in Mathematics - and train him where necessary, simply because he is [b]likely[/b] to have problem solving skills.

 I love mathematics, and will do it till my very last day.

BTW, Actuary is probably the most well paid job in the mathematical field, and involves a lot of mathematics and statistics :)

[quote="Ixeus"]
 I firmly believe that if you choose your career solely based on how much money you can get, you will be most likely to be extremely miserable at some stage of your life. [/quote]

I agree entirely and would like to add something to this thought.  Successful people in life tend to enjoy what they do.  That's a good reason to take a look at many subjects while you're in school, including math (though AoPSers already do).  I feel that more and more people find something enjoyable to do with math when it involves math and some other field.

Add math + any subject and you usually get a career.

Add math + computing + any subject and you're probably in extremely high demand.

Math is a subject, not a career (for most people).  If you enjoy it at all, be sure to take a look around and see how you might enjoy using it.


This problem is from one of the workout ones from a MATHCOUNTS handbook.

Number Eight:

The Rotokas of Papua New Guinea, have eleven letters in their alphabet. The letters are: A, B, E, G, I, K, O, P, R, T, U. Suppose license plates of five letters utilize only letters in the Rotoka alphabet. How many license plates of five letters are possible that begin with either G or K, end with T, and have no letters that repeat?

I read the MATHCOUNTS solution given, but I didn't understand it...also, I know that the MATHCOUNTS solutions are usually slow processes...are there any shortcuts to this problem?

the mathcounts solution isn't hard. It's a good way.

The first letter can be either G or K, so that's 2.
The second letter must be different from the first, and can't be T (reserved for last spot), so that's only 9 possibilities.
The third letter must be different from the first, the second, and it can't be T, so it leaves 8 possibilities.
Same idea with the fourth letter, with 7 possibilities.
The last letter must be T or 1 possibility.

2*9*8*7*1 = 1008.

If you aren't familiar with these concepts, nice explanations can be found in the AoPS Volume 1, A probability book, or Algebra 2 textbook.  :lol:

[hide]

Since there is either a G or a K, the first choice is 2. Then for the middle 3; 9*8*7.Since the last has to be a T.

$2 * 9 * 8 * 7 * 1 = \framebox{1008}$[/hide]


topology
function
real analysis
real analysis unsolved

Let $ X$ be a connect topology space such that the $ T_1,T_4$ axiom with more than two elements,Prove that $ X$ is uncountable.

Thank you.

Hmm...seems to me that you only need $ T_{3\frac {1}{2}}$.

Let $ x,y\in X$ with $ x\neq y$. Then $ \{x\}$ is closed and $ y\notin \{x\}$, so there exists a continuous function $ f: X\to [0,1]$ such that $ f(x) \equal{} 0$ and $ f(y) \equal{} 1$. Suppose $ X$ is only countable. Then the image of $ f$ has only a countable number of elements in $ [0,1]$. Let $ r\in [0,1]$ be one of the points the image misses.

Check that $ f^{ \minus{} 1}([0,r))$ and $ f^{ \minus{} 1}((r,1])$ form a disconnection of $ X$, a contradiction. Thus $ X$ must be uncountable.

Thank you,That's the point :lol:
For $ T_4$ we should use the Urysohn Lemma.


hello, you want to solve the system for $ x,y,z$? And $ \alpha,\beta,\gamma$ are real constants?
Sonnhard.

[quote="anorithdialga"]solve the system for reals
\[ \begin{array}{l} yz \equal{} \alpha (y \plus{} z) \plus{} \alpha \\
zx \equal{} \alpha (z \plus{} x) \plus{} \beta \\
xy \equal{} \alpha (x \plus{} y) \plus{} \gamma \\
\end{array}
\]
[/quote]
Is this what you meant?  It kind of ruins the symmetry of the equation.  It gives a nice solution, though:
[hide="Solution"]
Allow $ \alpha \equal{} a$, $ \beta \equal{} b$, $ \gamma \equal{} c$.  Then we get that $ zx \minus{} a(x \plus{} z) \plus{} a^2 \equal{} a^2 \plus{} b$, $ yz \minus{} a(y \plus{} z)\plus{}a^2 \equal{} a^2 \plus{} a$, $ xy \minus{} a(x \plus{} y) \plus{} a^2 \equal{} a^2 \plus{} c$.  This becomes $ (z \minus{} a)(x \minus{} a) \equal{} a^2 \plus{} b$, $ (y \minus{} a)(z \minus{} a) \equal{} a^2 \plus{} a$, and $ (x \minus{} a)(y \minus{} a) \equal{} a^2 \plus{} c$.  Letting $ x \minus{} a \equal{} l$, $ y \minus{} a \equal{} m$, and $ z \minus{} a \equal{} n$, this becomes $ nl \equal{} a^2 \plus{} b$, $ mn \equal{} a^2 \plus{} a$, and $ lm \equal{} a^2 \plus{} c$.  Multiplying together and taking the squareroot give sthat $ mnl \equal{} \sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}$.  Dividing to find the values of $ m$, $ n$, and $ n$ give that $ m \equal{} \frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} b}$, $ l \equal{} \frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} a}$, and $ n \equal{} \frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} c}$.  We conclude that\[ x \equal{} \frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} a} \plus{} a\]\[ y \equal{} \frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} b} \plus{} a\]\[ z\equal{}\frac {\sqrt {(a^2 \plus{} b)(a^2 \plus{} a)(a^2 \plus{} c)}}{a^2 \plus{} c} \plus{} a\][/hide]


Jenny has a spinner with 3 equally likely outcomes numbered 1 through 3.  If Jenny spins the spinner 6 times, what is the probability that she gets more 1's than 3's?

The probability that he gets more ones than threes is the same as the probability that she gets more threes than ones. Let the probability that she gets more ones than threes be $ x$ and let the probability that she gets the same number of ones as threes by $ y$. Since these are mutually exclusive, $ 2x\plus{}y\equal{}1$.

Now we do a bit of casework to find $ y$:

All 2's (0 ones and threes): $ \frac{1}{3^6}$

4 2's, one 1 and one 3: $ \frac{30}{3^6}$ (we pick the spin that we spin one (6 possibilities) and then pick the spin we get 3 (5 possibilities))

two 2's, two 1's, and two 3's: $ \frac{90}{3^6}$

zero 2's, 3 ones, and 3 threes: $ \frac{20}{3^6}$

$ y\equal{}\frac{141}{3^6}\equal{}\frac{141}{729}$

from that, we can solve for $ x$.


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sincop   :P

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Consider two functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ and $ g: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$ f(x\minus{}1)\equal{}2x\minus{}3\plus{}g(1)\minus{}f(1)$ and $ g(x\minus{}1)\equal{}4x\plus{}5\minus{}g(1)\minus{}f(1)$, for all $ x \in \mathbb{R}$.

$ a)$ Determine $ f$ and $ g$

$ b)$ Find, if there exists, the intersection point between the graphs of these functions


i got selected for the interview!!! (so did my friend d1994)
congrats to all the others who got thru
can some seniors help me with some tips for the interview

hey i have been selected for the kvpy interview this year. i would be grateful if the seniors could give me some tips regarding the interview.

[quote="asish2000"]where r u from ..

btw nuclear .. do the same team of board members go about conducting the interviews??[/quote]
 :?:  :|

KVPY 2009 interview schedules out
visit http://www.iisc.ernet.in/kvpy/interview%2009%20schedule.htm

Best of Luck  :thumbup: to all selected.

@ashish
they keep changing
not all of them but some of them

@Div
there are only a few tips that I can give you but you'll already have heard most of them
Be cool and confident but not a stud.Be polite and try to look as innocent/smart as you can.
As for the questions-they are very basic and do not require much class XI knowledge (except maybe a little bit of organic)
Also they don't expect you to know the answer to each question but you should still try as it may be in the right direction and then the interviewees will themselves offer help.

thanx a lot
btw i am from Delhi and have just joined this forum.

I was talking about the board I faced.

They were mostly from our state only..

Well,
The interviewers will ask those subjects (mostly) that you have mentioned as ur favourite subject.. Even if u get stuck at a question, they'll help u to reach the answer.. Dont think that being unable to answer lowers yuor chances.. It makes them look at ur thinking process.. Be prepared to use a whiteboard.. They might ask u to draw graphs, circuits ... or reactions (in org).. I remember I had written phy as my fav.. and one of them asked me quite a lot on mechs and heat and td.. Dont panic.. If you are at the beginning in the day's schedule.. be prepared for the worst (they tend to have longest interviews in the morning abt 1 hr) then on they reduce it to 40 mins.. (I was the last in the days schedule and it got over in 35 mins) the first guy was there for abt 55 mins..

@ashish
Hi im in 11th,
Did they ask u to write anything on the board??? :(
Thats scary! :o  What did they ask in maths?Is it only organic in chem :( .I dont know anything in organic,yet to start)

Board is easier I reckon coz you don't have the old foogies staring at your face.
Also, I'm a living specimen to prove that the KVPY interview are not designed to test your knowledge alone. If your bloody hell at science, good for you, if not, don't worry, you just need to reach the cutoff through a combination of your wit, communication and rocketmanship :P

Well yeah board is easier.. u dont have to look at them directly.. In maths they told me to draw the graphs of y=sinx, sin^(-1)x, lsinxl (where in physics have you seen this graph ?), sinlxl.. They also asked me to draw the circuits of full wave rectifier and half wave rectifier

Other questions that I vaguely remember,

What is the blue color when iodine reacts with starch? how is it formed?
No. of water of crystallization in Potash Alum (its formula)
(luckily this much only)

Bio--they asked one two questions from Xth like what is the digestive enzyme in saliva

Phy - he shaked a bottle and asked if we can increase the temp of the water by heating? How is it possible.

next was to draw the free body diag of the bottle resting on the table.

Is the forces mg and R action-reaction pairs...?

I guess this much only..

@abhitayal
Thanx for writing everything down,now iam a lot clearer on how to proceed,i too dread chemistry and hope they dont ask me organic or something.Thanx again abhitayal :D

@abhitayal : thanx a ton man...reading that gave me some confidence
btw were u selected?
And how many guys are selected after the interview?

Last yr (SA stream) from 526 selected from interview 196 were selected for scholarship.. dunno abt the other streams though

@ abhitayal
thanx for the interview

can anyone be so kind to do the same :help:

[quote="abhitayal"]Yu're welcome!!!
You can read nucleur's interview :P
visit http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251255[/quote]
Damn! I STILL haven't completed it  :rotfl:
I'll try to do that on Jan 1.

just had my kvpy interview...wasnt very long ..20 mins approx.
went off fine i guess...wasnt able to answer a maths question thats it
they asked me questions only from maths and organic as i had mentioned them as my favourites...more on it later...all u folks also tell me bout your interview..

you are in which stream?I also had m,ine well?questions were easy ?

Hiiii everyone....at present im in MAMC New Delhi doing MBBS. I want to apply for the kvpy this year. So i need some suggestions regarding topic selection and how to peoceed with the research project......if anybody can help

Something really strange hapenned yesterday. I got a call from Bangalore that ive been selected for KVPY Interviews on 21st March In Bangalore. I gave the aptitude test on 1st November like everyone else. But I dont understand this ... Are they considering more students for interviews or what???

Note : Im in Std. XI SA Stream.

Anyone has any idea about this?

there was some problem
issues were raised saying aptitude test was not corrected properly
so they uploaded the paper and their solutions online and asked for any corrections
so u might have made it to the interview after the corrections
this is also the reson for delay in final result of KVPY this year
and ur date of the interview says that results might not be out till April

BTW
congrats and best of luck for the interview

Congrats!
All the best.
BTW
Where can I find the aptitude paper and the solutions?

Two people from my class(11th grade) have been called for KVPY interview aswell on March 19th at IISc ...SA stream.
 From the letter they have gotten it appears another 112 have been called for the interview in addition to the 518 that have been interviewed already.
It is not very clear what is going on with KVPY this year.

@Swanand
visit iisc.ernet.in/kvpy/
and see the links 'Stream SA 2009' and 'Stream SB 2009' for the papers
and go to 'Answer Key - 2009' for key

Thank You friends  :) .

I guess Abhi is right , the website shows ~ 500 people selected for interviews , while the letter i received indicates ~ 600.
Good luck to all others.  :!:

[quote="abhitayal"]If anybody interested in reading my interview here it is..[/quote]
were u selected

@atuleshjha: [quote="abhitayal"]@Div
You're welcome
and yes! I was selected[/quote]Post no. 17 in this topic. :D  :D  :D  :P

hey guys..
i hav my interview in 4 days..for SA
any final tips..?

Stay cool, don't tension much. It will be all right.  :thumbup:  :thumbup:


modular arithmetic
algebra
binomial theorem

Find the remainder when $99999999^{100000000}$ is divided by 1000.[/hide]

I'm assuming this is based on Mock AMC E #24?


You can use $\sin ^2 \theta=\frac{1-\cos 2\theta}{2},\cos ^ 2 \theta=\frac{1+\cos 2 \theta}{2}$


How many elements has the group $SL_{2}(Z/nZ)$?


number theory
greatest common divisor
number theory unsolved

Let $ x,y$ be integers
such that
$ \frac {x^{3} \plus{} 1}{y \plus{} 1} \plus{} \frac {y^{3} \plus{} 1}{x \plus{} 1}$ is a integer
Prove that
$ \frac {x^{2007} \plus{} 1}{y \plus{} 1}$ is an integer.

let $ t\equal{}\gcd(x,y)$, then the property $ \frac{x^3\plus{}1}{y\plus{}1}\plus{}\frac{y^3\plus{}1}{x\plus{}1}\in\mathbb{Z}$ is equivalent to

$ \frac{x\plus{}1}{t}\frac{y\plus{}1}{t}\,\left|\,(x^2\minus{}x\plus{}1)\left(\frac{x\plus{}1}{t}\right)^2\plus{}(y^2\minus{}y\plus{}1)\left(\frac{y\plus{}1}{t}\right)^2\right.$ which yields $ \left.\frac{y\plus{}1}{t}\,\right|\,x^2\minus{}x\plus{}1$, i.e. $ y\plus{}1\,\left|\,x^3\plus{}1\right.$, but then $ y\plus{}1\,\left|\,(x^3\plus{}1)\sum_{n\equal{}0}^{668}(\minus{}x^3)^n\equal{} x^{2007}\plus{}1\right.$.

[quote="hjbrasch"]
$ \frac {x \plus{} 1}{t}\frac {y \plus{} 1}{t}\,\left|\,(x^2 \minus{} x \plus{} 1)\left(\frac {x \plus{} 1}{t}\right)^2 \plus{} (y^2 \minus{} y \plus{} 1)\left(\frac {y \plus{} 1}{t}\right)^2\right.$ which yields $ \left.\frac {y \plus{} 1}{t}\,\right|\,x^2 \minus{} x \plus{} 1$, i.e. $ y \plus{} 1\,\left|\,x^3 \plus{} 1\right.$[/quote]

Did you proved that:
$ \frac{y\plus{}1}{t}$ does not devide $ (\frac{x\plus{}1}{t})^2)$?
Or equivalently:
$ y\plus{}1$ does not devide $ \frac{(x\plus{}1)^2}{t}$ ?

Thank you very very much for the link TUAN  :lol:

answer to FOURRIER's question, yes it is bovious:

$ t\equal{}\gcd(x\plus{}1,y\plus{}1)\Longrightarrow 1\equal{}gcd\left(\frac{x\plus{}1}{t},\frac{y\plus{}1}{t}\right)\Longrightarrow 1\equal{}gcd\left(\frac{x\plus{}1}{t},(\left(\frac{y\plus{}1}{t}\right)^2\right)$


Prove that for all a,b,c,x>0 we have the inequality:
(ax^2+bx+c)(a+bx+cx^2)>=(a+b+c)^2x^2

[i]Titu Andreescu[/i]

Cauchy would turn in his grave if you don't see this one :D

well Cauchy works for the triplets: (sqrt(a);sqrt(bx);sqrt(c)x) and (sqrt(a)x;sqrt(bx);sqrt(c));

that seems the only simple solution.. I was looking for another approach.. but this is the easiest..

cheers! 10x! :D

Just for joke:

LHS - RHS = (x-1)^2*(x^2*c*a+c*b*x+b*x*a+2*x*c*a+c*a) >=0

 :D

Namdung