[hide=" click to reveal hidden content"]

By Pythagoras' theorem,

$ (x\minus{}y)^{2} \plus{} x^{2} \equal{} (x\plus{}y)^{2}$

$ x^{2} \minus{}2xy \plus{} y^{2} \plus{} x^{2} \equal{} x^{2} \plus{} 2xy \plus{} y^{2}$

$ x^{2} \equal{} 4xy$

$ x\equal{}4y$

$ \frac{x}{y} \equal{} 4$.

$ x: y\equal{}4: 1$

[/hide]

[quote="sri340"]The sides of a right triangle have lengths x – y, x, x + y, where x>y>0. Find the ratio of x : y.[/quote]
[hide]Let the right triangle be a right triangle with sides 3, 4, and 5.  Therefore, $ x \equal{} 4$ and
$ y \equal{} 1.$  $ x \minus{} y \equal{} 3$, $ x \equal{} 4$, $ x \plus{} 1 \equal{} 5$.  Therefore, the ratio of $ x: y$ is $ 4: 1$.[/hide]

[hide="Click here to reveal hidden content"]

All primitive Pythagorean Triples have side lengths of a=p^2-q^2,b=2pq, and c=p^2+q^2, where a^2+b^2=c^2. x=p^2, and y=q^2. We are trying to find what p^2/q^2 is. Since 2pq=x, because x=b, x=p^2, so 2pq=p^2. Dividing by p gives 2q=p. Squaring gives 4q^2=p^2, thus p^2/q^2=4/1, so the ratio is [b]4:1[/b].[/hide]


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

If E is a finite galois extension of F and there exists only one intermediate field between E and F, then Gal(E/F) is cyclic.  Can someone help me with this problem and tell me why it is important to have only one intermediate field between E and F.  Thanks

Work in the Galois group only:
Let $\sigma \in G\backslash H$ with $G$ the Galois group and $H$ the subgroup corresponding to that single intermediate field.
Then the subgroup generated by $\sigma$ is surely not $\{ e \}$ or $H$, thus it is $G$.

I am sorry, I am not following your argument.  I am very new at this galois theory.  Can you explain this little more.

The intermediate fields between (or equal to) $F$ and $E$ correspond bijectively to the subgroups of $G=\mathrm{Gal}(E|F)$.
Thus we are given a group $G$ with three subgroups:
- $G$ corresponding to $F$
- $\{e\}$ corresponding to $E$
- some subgroup $H$ corresponding to the single intermediate field

Now as $H \neq G$, we get that there is $\sigma \in G \backslash H$.
This $\sigma$ generates a subgroup $S$ of $G$ (by it's powers), and we want to show that it generates $G$ (thus $S=G$) itself (we are done then as we want to show that $G$ is cyclic, in other words generated by one element).
But $S$ can't be $\{e\}$ or $H$ as $\sigma \in S$ but $\sigma \not\in \{e\}, H$.


ABC is an acute-angled triangle with circumcenter O. The circumcircle of ABO intersects AC and BC at M and N. Show that the circumradii of ABO and MNC are the same.

It's easy to see that $\angle ACB=\angle NBC=\angle MAC$, so $NO,MO$ are the perpendicular bisectors of $BC,AC$ respectively. In particular, this means that $MO\perp AC,\ NO\perp BC$, so $O$ is the orthocenter of $CMN$. Since the diameter of $(ABO)$ equals $\frac{MN}{\sin \angle MON}$, the diameter of $(CMN)$ equals $\frac{MN}{\sin\angle MAN}$, and $\angle MAN=\pi-\angle MON$, we get the required result.


The Middle School Basics forum is no more.  Because the level of content was all either appropriate for the MOEMS or MATHCOUNTS forums (mostly MATHCOUNTS), we moved all posts to the MATHCOUNTS forum and deleted the Basics forum.

I was wondering how the Mathcounts Forum had several thousand posts added in a very short amount of time  :D

I think that was a good decision, but what will happen to the moderators of the Basic forum?

Those moderators can PM me if they are interested in reassignment.

This was a very good decision, giving the mediocre activity in The Basics forum lately.

[hide="off topic"]MCrawford' signature reminds me of one of Hawking's books. I think it argued a similiar view.  :D [/hide]

why mediocre???? it is for young childs, and there are not a big amount of children surfing the internet, and less in a math community... they're more interested in games and music, like everyone when we were child

I think Knuth is talking about the quality and quantity of the posts due to the fact that most experienced AoPSers post the same levels of problems in the MATHCOUNTS forum and spend more time there discussing problems.

There is still gonna be the classroom forum for younger students which is to stay (right?)

[quote="Treething"]Heh. I thought that I was banned from another forum...

Are you considering me for moderatorship yet :roll: ?[/quote]

Well, maybe it would be better to PM him, because that's more private there than chatting and asking here


Are negative probabilities (and probabilities greater than 1) inherently self-contradictory or are they merely frowned upon because of a lack of a real-world counterpart?

Probability is defined in terms of [url=http://en.wikipedia.org/wiki/Probability_distribution]probability distributions[/url], which are non-negative by definition.  The problem isn't the lack of a "real-world" counterpart; there is simply no way to interpret a negative probability within the framework of probability theory.

The same goes for probabilities greater than 1. A probability of 1 means it happen every time, so a probability greater than one would imply the outcome happens more times than the event.

I guess what I'm wondering is, if you removed the requirement that all probabilities be between 0 and 1, would the rest of the laws of probability still be consistant?

Be specific.  Some of the "laws" of probability are actually [url=http://en.wikipedia.org/wiki/Probability_axioms]axioms[/url] (including the non-negative rule).


probability AMC USA(J)MO USAMO MATHCOUNTS

On a given circle, six points $A$, $B$, $C$, $D$, $E$, and $F$ are chosen at random, independently and uniformly with respect to arc length Determine the probability that the two triangles $ABC$ and $DEF$ are disjoin, i.e., have no common points.

[hide]
Given any six points, there are $\binom{6}{3} = 20$ ways of making two triangles that don't share points. It's easy to see that the vertices of each triangle must be consecutive vertices among the six, so there are $6$ ways such that they are disjoint. Since any set of six points is equally likely to be chosen, the total probability is $\frac{6}{20} = \frac{3}{10}$.[/hide]

Yeah, I did it exactly the same way as paladin. This problem seemed astoundingly easy for the USAMO.

how was that a USAMO qustion????? thats like mathcounts maybe. lol


topology
real analysis
real analysis unsolved

In a metric space where the closure of any open ball is the closed ball , show that any open ball is connected

If you show path connected, then you have connected. There should be a clear path connecting any two points of an open ball. If $ X$ is your metric space and your ball has radius $ \varepsilon$ about $ z$, then let $ u,v\in B_{\varepsilon}(z)$, the path $ f: [0,1]\times X\to X$ by $ f(t,x)\equal{}tv\plus{}(1\minus{}t)u$ should work.


induction
number theory unsolved
number theory

Suppose that $a_1,a_2,...,a_n\in \mathbb{R}$. Prove that there exist $x\in \mathbb{R}$ s.t. all of the number $x+a_1,x+a_2,...,x+a_n$ are the irrational.

I think we can easily proceed by induction (I hope I'm not doing now any mistake but let's give it a try  :P ):

For $n=1$ it is trivial so we can assume that we can find such a number for $n$ numbers and we'll prove it for $n+1$. If there is no rational numbers among given $n+1$ numbers we choose $x$ to be zero. So we can suppose that $a_{n+1}$ is rational. Using induction hypothesis there is $x$ such that:

$x+a_1,...,x+a_n$ are all irrational and we have that $x+a_{n+1}$ rational. But now I think it is clear that we can find number $b$ irrational such that $x+b+a_1,...,x+b+a_n$ are still irrational (this can be proved by induction as well). Hence the result (because $x+b+a_{n+1}$ is irrational).

Here's the strongest version of this problem you can get :): if $c$ is the continuum cardinality, and $A\subset\mathbb R$ with $|A|<c$, then there is a real $x$ s.t. $x+A\subset\mathbb R\setminus\mathbb Q$.

Consider $\mathbb{R/Q}-a_i$ for each $i$.  Say that the intersection of these sets is empty.  Taking complements, we get that $\mathbb{R}=\cup_i \overline{\mathbb{R/Q}-a_i}$, which shows that an uncountable set is the countable union of countable sets, a contradiction.

(By $\mathbb{R/Q}-a_i$, I mean of course $\{x-a_i:x\in\mathbb{R/Q}\}$)


algebra
polynomial
Gauss
function
complex analysis
complex analysis unsolved

Find all integer $ n > 1$ such that the nonzero roots of $ (z \plus{} 1)^n \equal{} z^n \plus{} 1$ all lie on the unit circle.

Although my solution does not use complex analysis, the problem was given on a complex analysis final and it seems like a solution along these lines should exist (argument principle and so forth?) -- if mods think otherwise, of course, please do move it.

Hmm.  I don't have much experience with complex analysis, so please let me know if this argument is wrong:

If the nonzero roots of $ f(z) \equal{} (z \plus{} 1)^n \minus{} z^n \minus{} 1$ lie on the unit circle then $ f'(z) \equal{} n(z \plus{} 1)^{n\minus{}1} \minus{} nz^{n\minus{}1}$ should have roots within a convex polygon inscribed in the unit circle by the [url=http://en.wikipedia.org/wiki/Gauss-Lucas_theorem]Gauss-Lucas theorem[/url].  These roots $ r$ have the property that $ \frac{r \plus{} 1}{r}$ is an $ (n\minus{}1)^{th}$ root of unity.  On the other hand, the inverse function $ z \mapsto \frac{1}{z \minus{} 1}$ takes the unit circle to the line $ x \equal{} \minus{} \frac{1}{2}$, so in order for every $ (n\minus{}1)^{th}$ root of unity to map into the unit circle it is necessary that $ n \minus{} 1 \le 6$.  I think casework suffices from here.


I recently downloaded the complete MikTeX 2.7 distribution and when opening TeXnicCenter it asks to locate full path of the directory. I thought it was MikTeX 2.7 - MiKTeX - bin, but it says "please enter the directory where the executable files (tex.exe, latex.exe, etc.) are located, as your distribution's directory."

Then what's its location?

You should find the exectuable files in C:\Program Files\MiKTeX 2.7\miktex\bin if you installed MiKTeX there. Go there in Windows Explorer and check that is the right path and contains latex.exe etc. If not then you may have installed MiKTeX to somewhere else in whch case you need <[i]where you installed MiKTeX[/i]>\miktex\bin


Guys: [url]http://www.artofproblemsolving.com/Forum/index.php?f=249[/url]

The SOUTH CAROLINA flag listing is the NORTH CAROLINA flag.

This is very much not cool. we dunn like being associated with them northies.

This is an SC flag that should pretty easily be resized for that forum.

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/6/69/Flag_of_South_Carolina.svg/125px-Flag_of_South_Carolina.svg.png[/img]

Oh, haha that's because it used to be called North & South Carolina, then we convinced somebody to split NC off into their own forum, and I guess the flag never got changed.


If B=n*n*n*n + n*n + 1 is a prime nr. , find n

[hide="Solution"]Using t0rajir0u's hint, we have:
$n^{4}+n^{2}+1 = (n^{2}+1)^{2}-n^{2}= (n^{2}+n+1)(n^{2}-n+1)$

So $B$ will be prime if and only if one of those factors are $1$:

$n^{2}+n+1=1$
$n^{2}-n+1=1$

So $n^{2}\pm n=0$:
$n^{2}\pm n=0$
$n(n\pm 1)=0$

So $n=-1,0,1$. So $B$ would be:$3,1,3$. Since $1$ is not prime, we find that the answers are $n=\pm 1$.[/hide]


The AoPSWiki will be up soon, so I wanted to take a moment and discuss wiki contributions.

We have not yet decided on a documentation [url=http://en.wikipedia.org/wiki/License]license[/url].  However, AoPSers should note that when you contribute to the AoPSWiki, you relinquish copyright over the material you submit.  This is necessary for a properly functioning wiki.  Expect other contributers to alter your words and the organization of articles you create.

Keep in mind also that since this is a publicly accessible education project, the time you spend contributing to the AoPSWiki is public service -- and it's the kind of public service that AoPSers will be good at.  We plan to come up with a system, if necessary, to validate (to the degree possible) service hours for students in honor societies.

Before getting started contributing to the AoPSWiki, click through a few random pages using the navigation window.  See what kind of content is already started and which links are still red (article not yet written).

I would also like to outline a few projects that AoPSers can contribute to immediately when the AoPSWiki launches.

[b]Competitions[/b]

I took all the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]math competitions[/url] from the AoPS resources area and created an organized set of links for them.  Just search for "mathematics competitions" and dig through the internal links.  If you know of any contests nationally, internationally, or locally that aren't listed, please list them and add a link when you can find it.

The same goes for chemistry competitions, physics competitions, technology competitions, biology competitions, engineering competitions, and other academic competitions.  I populated these lists to a degree, but they are very far from being comprehensive.

There are also scores of links to competition pages that have not yet been built or are currently just article "stubs".  Please help build these pages in your own words (don't just copy from Wikipedia, which is against the rules anyway).

[b]Scholarships[/b]

Getting other people to pay for your education is cool!  I've seeded several pages with math, science, and technology scholarships and helping to expand these lists is a great service to the AoPS community.

[b]Content[/b]

Out of sheer time constraints, I have not yet added any real math, problem solving, or otherwise educational content to the AoPSWiki.  Pick your favorite topic and write an article.  However, try to follow these guidelines:
* Do your best to delineate sections for Introductory, Intermediate, Olympiad students, and professionals (if the content gets to that level).
* Don't write articles about your unpublished theories or theorems.  Save it for the journals.  Controversial content of that kind does not serve the AoPS student community.
* Keep articles well-organized.
* Don't be too ambitious.  It takes a long time to compose a good math article.  Don't just write slop, but don't expect to write a book in the first go.  Do what you can, and when your motivation wanes, leave the article for somebody else to pick up and expand.

Additionally, I've also created some pages for summer programs, though I'm sure they could be made more thorough.  This goes for other pages that aren't coming to mind as well -- just explore and see where you can add knowledge.

Probably the biggest need right now is for those of you with the most wiki experience to help start building the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki_tutorial]AoPSWiki tutorial[/url] page.


Show that $ \binom{n}{k} < \left(\frac{en}{k}\right)^k$, where $ e$ is the natural number.

My method - induction.
For $ k\equal{}1$: $ \binom{n}{1}\equal{}n < (\frac{en}{1})^1\equal{}en$ (it follows that we can divide the RHS by $ e$)
Let's assume this is true for some $ k$ and all $ n \ge k$. Then it is also true for $ k\plus{}1$ and all $ n \ge k\plus{}1$:

$ \binom{n\plus{}1}{k\plus{}1}\equal{}\binom{n}{k}\frac{n\plus{}1}{k\plus{}1} <\left(\frac{en}{k}\right)^{k}\frac{n\plus{}1}{k\plus{}1}$
The last expression is $ \le \left(\frac{e(n\plus{}1)}{k\plus{}1}\right)^{k\plus{}1}$ if and only if $ ( \frac{(1\plus{}\frac{1}{k})}{(1\plus{}\frac{1}{n})})^k < e$ (just rearranging). The numerator ($ (1\plus{}\frac{1}{k})^k$) is less then $ e$ (a known 'fact', this sequence tends to $ e$, it is only left to show that it is increasing), and the denominator is trivially $ \ge 1$. So indeed $ ( \frac{(1\plus{}\frac{1}{k})}{(1\plus{}\frac{1}{n})})^k < e$ and the last expression is bounded by what is needed.


Maybe this has been posted and solved before.
"On a remote Norwegian mountain top, there is a huge checkerboard, 1000 squares wide and 1000 squares long, surrounded by steep cliffs to the north, south, east, and west. Each square is marked with an arrow pointing in one of the eight compass directions, so (with some possible exception of some squares on the edges) each square has an arrow pointing to one of its eight nearest neighbors. The arrows on squares sharing an edge differ by at most 45°. A lemming is placed randomly on one of the squares, and it jumps from square to square following the arrows. Prove that the poor creature will eventually plunge from a cliff to its death."

Is it a trick question? LEmmings usually commit suicide because they have very bad eyesight and can't see the edge of a cliff. So how would they be able to see the arrows.

This is a hard one ... But I think it all boils down to proving that there can't be a closed circuit of arrows and that at least one edge arrow has to point away from the board.

[quote="SplashD"]can't it be a closed circuit?

[u]__|E |SE|__
NE|__|__|S
N |__|__|SW
__|NW|W|__[/u][/quote]
But what about the squares in the middle???

The squares in the middle DO matter, since we have to make sure that each square only differs by 45 degrees.

They do matter because you can't fill them in:

The top left has to be NE or E, and the top right has to be SE or S. This forces the top left to be E and the top right to be SE. (This all follows from the rule that each pair of boxes that shares an edge differs by at most 45 degrees.)

Similarly, the bottom left square has to be NW and the bottom right square has to be W.

This is a contradiction because the squares in the middle differ by more than 45 degrees. Therefore, this arrangement cannot be done.


$ A$ is a moving point on the curve $ y\equal{}x^2\plus{}2$ ,  $ B$ is a moving point  on the straight line $ y\equal{}x$ , and point $ P(2,3)$ .

Find the least value of $ PA\plus{}PB$ ?

I take it that their rates will affect their movement and the distance between them?

hello
$ PA\plus{}PB$ will be least when $ P,A,B$  will be collinear and line $ PAB$ will be normal to both the curves.
any such line could be $ y\plus{}x\equal{}C$
after putting $ P$ we will get $ y\plus{}x\equal{}5$
$ A\equal{}(\frac{\sqrt{13}\minus{}1}{2},\frac{11\minus{}\sqrt{13}}{2}),B\equal{}(\frac{5}{2},\frac{5}{2})$

so least value of $ PA\plus{}PB \equal{}\sqrt{19\minus{}5\sqrt{3}} \plus{}\frac{1}{\sqrt2}$
hey vinskman i may commit mistake in calculation so kindly check that and tell me.
thank u

Jai ho!!! AR :first:


What integer can be added to both the numerator and denominator of $ \frac34$ to obtain a fraction equal to $ \frac78$?

We set up an equation: $ \frac{3\plus{}x}{4\plus{}x}\equal{}\frac{7}{8}$.  Cross multiplying (multiplying each side by $ 4\plus{}x$ and by $ 8$), we get $ 24\plus{}8x\equal{}28\plus{}7x$.  Subtracting $ 24$ and $ 7x$ from both sides, we get $ x\equal{}\boxed{4}$.  Or you can see that both numerators are one less than their respective denominator, and that $ 3$ and $ 4$ are each $ \boxed{4}$ less than $ 7$ and $ 8$, respectively.

this can be done way more easier just [color=#FF0000]LOOK[/color] for what [color=#4040BF]EQUALS[/color] 7/8 and adding [color=#800080]FOUR[/color]  gives 7/8 instead of [color=#40BF00]WORKING[/color] out all the stuff the other perosn did previosly so the answer is [hide][color=#000000][size=200]40[hide][size=200][color=#4040FF]PLEASE RATE MY POST IF YOU FULLY UNDERSTAND MY LOGIC AND PLEASE RATE IT GOOD !!!!THANK YOU SO MUCH!!!!!![/color][/size][/hide][/size][/color][/hide]

[size=200] :starwars:  :starwars:  :starwars:  :starwars: [/size]   THE POWER OF MATH   : ) THROUGH STARWARS !!!!!!!!

Hello, of course that is just a coincidence regarding the problem.  If you had another problem that could not be solved by such guess-and-check you would probably not solve it...

And there is no need to put in big black letters "Please rate my post etc. etc." because we already know that :)

[hide=solution]We notice 4-3=1 and 8-7=1 since we add the same number to each side 7-3=$\boxed{4}$[/hide]


Find dy/dx if xy^(1/2) + y^2=x

Need help please.

$\frac{d}{dx} x\cdot \sqrt{y} + x\cdot \frac{d}{dx} \left(\sqrt{y}\right) + \frac{d}{dx} \left(2y\right)=\frac{d}{dx} x$

$\sqrt{y}+\frac{x}{2\sqrt{y}}\cdot \frac{dy}{dx}+2y\cdot \frac{dy}{dx}=1\Rightarrow \frac{dy}{dx} = \frac{1-\sqrt{y}}{\frac{x}{2\sqrt{y}}+2y} = \frac{2(\sqrt{y}-y)}{x+4y^{3/2}}$

[hide="The Multivariate Calc way"]$f(x,y) = x\sqrt{y}+y^2-x=0$

By the chain rule we have $\frac{dy}{dx} = -\frac{f_x}{f_y}$

$f_x = \sqrt{y} -1\text{   and   }f_y=\frac{x}{2\sqrt{y}}+2y\Rightarrow \frac{dy}{dx} = \frac{1-\sqrt{y}}{\frac{x}{2\sqrt{y}}+2y}$[/hide]

[quote="DJIntel"]Appreciate it thanks, but is it "wrong" to have the 2(y-rooty)=2y^(2/2)-2y^(1/2) to equal 2y^(1/2)?[/quote]
Yes, it's wrong. You've just invented another of the False Laws of Algebra.

To emphasize the point, try letting $y=9$.

$2(y-\sqrt{y})=2(9-3)=2\cdot 6=12.$

$2y^{1/2}=2\cdot3=6.$ Not the same, and there was never a good reason to think it would be the same.


[quote="ajai"]The third term of an arithmatic sequence is -1, and the ninth term is 20. What is the thirteenth term?[/quote]
[hide]We have that the common difference between each term is $\frac{20-(-1)}{9-3}=\frac{7}{2}$, so the thirteenth term is $20+4\times \frac{7}{2}=\boxed{34}$, the answer.[/hide]

[hide]There is a difference of 6 between 3 and 9 and a difference of 4 between 9 and 13.  So, we know the number that must be added to the 9th term to get the 13th term must be $\frac{2}{3}$ of the difference between the 3rd term and 9th term.  Since the difference is 21, $\frac{2}{3}$ of that is 14, so we add 14 to 20 to get $34$[/hide]

[hide]The common difference is $\frac{20+1}{9-3}=\frac{7}{2}$, SO THE ANSWER is just $20+(13-9)*\frac{7}{2}=34$.

$34$[/hide]


a(n)=\sum_{k=2 to n}ln(1- 1/k^2) for n \geq 2
a) Study the monotony and boundaries of a(n)
b) Show that a(n) < 1/n -ln2 for n \geq 2
c)Compute lim a(n)
----------------------------------------------------------
a)b)c) same time
a(n)=\sum_{k=2 to n}(ln(k+1)-lnk + ln(k-1)-lnk) telescoping sum
we have an explicit expression of
a(n)=ln(n+1)-ln2+ln1-ln n= ln(1+ 1/n) -ln2
a(n) is decreasing like 1/n , a(n)>-ln2 so a(n) is convergent to -ln2
From ln(x+1) <x for x>0 we have b)

Remark: we have an asymptotic expansion of
a(n) = -ln2 + 1/n -1/(2n^2) +o(1/n^2) when n-->+oo
Further question give an equivalent of  a(n)^n
Mr.Omarjee

For the a_n^n consider the product representation of a_n.


Let $ f: \mathbb{R} \to \mathbb{R}$ be a function such that for all $ x,y \in \mathbb{R}$, $ f(x^3\plus{}y^3)\equal{}(x\plus{}y)(f(x)^2\minus{}f(x)f(y)\plus{}f(y)^2)$

Prove that for all $ x \in \mathbb{R}$, $ f(1996x)\equal{}1996f(x)$

Here is some indication for those who need help:

[hide]These are the steps from the official solution:

Let $ S$ be the set $ S\equal{}\{ a>0: f(ax)\equal{}af(x), \forall x \in \mathbb{R} \}$

$ a \in S \Rightarrow a^{\frac{1}{3}}\in S$ and $ a,b \in S  \Rightarrow a\plus{}b \in S$[/hide]


linear algebra
matrix
quadratics
linear algebra unsolved

Let $M$ be a real symetric matrix.
$d_1 \leq d_2 \leq .. \leq d_n$ where $\{d_1,..,d_n\}=\{M_{1,1},..,M_{n,n}\}$ are the entries of the diagonal of $M$.
${\lambda}_1 \leq ... \leq {\lambda}_n$ the eigenvalues of $M$.
Prove that
${\lambda}_1 \leq d_1$
${\lambda}_1+{\lambda}_2 \leq d_1+d_2$
...
${\lambda}_1+..+{\lambda}_{n-1} \leq d_1+..+d_{n-1}$

Since the matrix is orthogonally diagonalizable, it all comes down to this (after writing it as $A\cdot diag(\lambda_i)\cdot A^{-1}$ and performing the computations):

Given $A=(a_{ij})$, an orthogonal real matrix, and the reals $\lambda_1\le\lambda_2\le\ldots\le \lambda_n$, show that $S=\sum_{i=1}^k\sum_{j=1}^n\lambda_j a_{ij}^2$ is at least $\lambda_1+\ldots+\lambda_k$, for every $1\le k\le n-1$.

If we fix $k$, we may assume that $\lambda_i=\lambda_k,\ \forall i\ge k$. With this convention, it looks like (I haven't really checked my computations) we have $S-\sum_{i=1}^k \lambda_i$ is equal to $\sum_{j=1}^{k-1}(\lambda_k-\lambda_j)(1-S_j)$, where $S_j=\sum_{i=1}^ka_{ij}^2$, and that expression is clearly $\ge 0$.
Like I said, it's not been checked properly, so sorry if it's wrong :).

A different approach:
Let $E$ an euclidian espace of dim $n$,$q$ a quadratic forme associed to the auto adjoint endomorphism $u$.Let $\lambda_1,..,\lambda_n$ the eigenvalues in their increasing order.For every $p$ we define $F_p$as being the set of subspaces of dim $p$ of $E$.These are only the definitions which make the solution easier to be written.
Ok,now here are some few steps leading to the solution:
1) Show that $\lambda_p=inf_{F\in F_p}(sup_{x\in F}\frac{q(x)}{||x||^2})$.

2) Let  $m\in [1,..,n-1],E'$ a subspace of dim $n-m$,$q'$ the restriction of $q$ to $E'$,$u'....$ .
 Let $\lambda_1',..,\lambda_{n-m}'$ the eigenvalues of $u'$.Show:   $\forall p \in [1,n-m],\lambda_p\leq \lambda_p' \leq \lambda_{p+m}$.
(Hint:You can consider $F'_p$ and see some obvious inclusions in their coresponding sets defined at the begining)

3) Apply these results to the real symmetric matrix $M$ considering the matrice $M'$ obtained by deleting the $i^{th}$ column and line.And so on(i mean...so on with the process of deleting the $i^{th}$ column and line...and ordering the eigenvalues of the corresponding matrices).

For more details,just post your request...


logarithms
number theory
arithmetic sequence
prime numbers
number theory theorems

Hi guys ;),I want asking you about [b]Analitic and Algebraic number theory[/b],I want learn  it for using in elementary number theory(in IMO and other olimpiads :blush: ),but I dont know any solution with using Analitic and Algebraic number theory for olimpiad problems :? ,can you give me any example?(if ther are :lol: )
[b]Thank you[/b].

[quote="ZetaX"]Well, what do you know about algebra and complex analysis¿ (really using them for killing problems requires some knowledge)

Some easy (not needing that much) examples:

- http://www.mathlinks.ro/Forum/viewtopic.php?t=60555
- solve $x^3+y^3=z^3$ or $x^3+y^3=2z^3$
- http://www.mathlinks.ro/Forum/viewtopic.php?t=54066
- some problems needing one of Dirichlet's theorems (primes in arithmetic progression; units in number fields and similar ones), like http://www.mathlinks.ro/Forum/viewtopic.php?t=64322 or solving equations that are norms of number fields[/quote]
Thank you $ZetaX$ for replay :) .
I saw your examples,but I know,Gues's integer numbers($a+bi$), Dirichlet's theorem and  Lagrange's simvol($(\frac{a}{p})$) :blush: ,I  was means more hard examples,like prove of Dirichlet's theorem(about prime numbers in seqanses),realy I don't know it,but I know that is wery difficult :(  :)(but the theorem is most elementar :D ),so can you give me that kind of examples :oops:  :lol: .
Thank you ;) .

[quote="{x}"]When it comes to analytic number theory:
Do you mean stuff like this:
\[ \left( \ln \ln x \right) - 1 \leqslant \sum_{p \leq n} \frac{1}{p} \leqslant \ln \ln x + c  \]
where $c$ is an absolute constant and $p$ runs through primes.
If yes, then take a look at this great book: http://www.princeton.edu/~ppollack/notes/notes.pdf
First chapter is elementary (like the exemple above) analytic number theory.
Then it goes into more "obscure" stuff like Dirichlets theorem, or the prime number theorem.

When it comes to algebraic number theory, you need a reasonable background in algebra to get started.[/quote]
Thanks a lot  {x} ;)  :)


i was just wondering if people knew the significance of the awards given for usnco (like honors, high honors, etc.)

I would guess that being high honors (Top 50) is worth a lot less than making the study camp because it's sort of a similar difference between USAMO and MOP. Both the chemistry and math study camps would be considered "hooks", in my opinion, but getting high honors or being a USAMO qualifier is not.
Unfortunately for me, I did not make the study camp this year, although I felt I was really close. Had I not misread and messed up on some simple things on open ended questions, I may have had a better shot. Anyways, I was rather disappointed to not get a phone call on May 1st  :(

ah me too. but it's still a better feeling to be in high honors than honors. then i know it's sort of possible to make it next year :)

in response to the original question, high honors is top 50 (so all study camp participants actually get this), and honors is 100-51.

i just talked to my chem teacher, and i believe he said high honors was top 150. so actually 51-150. (top sixth).


superior algebra
superior algebra unsolved

Let $ R$ be a ring with 1. Every element in $ R$ can be written as product of idempotent ($ u^n\equal{}u$ for some $ n$) elements. Prove that $ R$ is commutative

[quote="Omid Hatami"]

Let $ R$ be a ring with 1. Every element in $ R$ can be written as product of idempotent ($ u^n \equal{} u$ for some $ n$) elements. Prove that $ R$ is commutative[/quote]

The problem, as you wrote it, is false I'm afraid! :yup: A counter-example is the ring of $ 2 \times 2$ upper triangular matrices with entries in $ \mathbb{Z}/2\mathbb{Z}.$ The idempotent here means what it

usually means, i.e. $ u^2 \equal{} u.$ In this case the problem is not hard to solve: obviously it suffices to show that every idempotent is central. Suppose first that $ ab \equal{} 1,$ for some

$ a,b \in R.$ We claim that $ a \equal{} b \equal{} 1.$ Suppose the claim is false. Then $ a \equal{} e_1e_2 \cdots e_k,$ where $ e_j$ are idempotents and $ e_1 \neq 1.$ Let $ e \equal{} e_2 \cdots e_kb.$ We have $ e_1e \equal{} 1$ and hence

$ 1 \minus{} e_1 \equal{} (1 \minus{} e_1)e_1e \equal{} (e_1 \minus{} e_1^2)e \equal{} 0.$ Thus $ e_1 \equal{} 1.$ Contradiction! Now suppose $ x^2 \equal{} 0,$ for some $ x \in R.$ Then $ (1 \minus{} x)(1 \plus{} x) \equal{} 1$ and therefore $ x \equal{} 0,$ by what we just

proved. Finally let $ e$ be an idempotent. Then for any $ z \in R: (ez \minus{} eze)^2 \equal{} (ze \minus{} eze)^2 \equal{} 0$ and so $ ez \equal{} ze.$


In any triangle  ABC  prove  that  :
4[ABC]*sqrt(ab+bc+ca)<=a^3+b^3+c^3<= [R*(ab+bc+ca)^2]/(4[ABC])
THANKS

For the first part:

 \sum a 3 \geq  (\sum a 2 )( \sum a)/3 \geq (\sum a 2 ) [\sqrt (\sum ab)]/ \sqrt 3 \geq 4[ABC] \sqrt (\sum ab), where we used the known inequality  \sum a 2 \geq 4 \sqrt 3 [ABC].


Salam bacheha ... Kasi az emtehanaye payan dore khabar nadare ? fek konam bayad tahala tamoom shode bashe ... ye ensane kheirkhahi peyda she soala ro bezare :)

$ 1\minus{}$ farz konid $ p$ addad aval bashad . tedad adad haye na hamnehesht be hang $ p$ ke ba binahayat az azaye donbale zir hamnehesht hastand ra biyabid :  $ 1, 11, 111, 1111, ...$

$ 2\minus{}$ Farz konid $ G$ yek gorooh bashad va $ a_1, a_2, .. , a_m$ tamam azaye $ G$ bashand ke daraye martabe $ n$ hastand sabet konid $ \phi (n) | m$

$ 3\minus{}$ Farz konid $ m, n \in N$ va $ (m,n) \equal{}1$ sabet konid moadele zir binahayat javab dar $ N$ darad :
$ x^m t^n \plus{} y^m s^n \equal{} v^m k^n$

$ 4\minus{}$ Farz konid $ p$ addadi aval bashad va $ n<p$ addadi tabie bashad. baraye har $ a_1, a_2, ..., a_n, c \in Z_p$ ke hame anha sefr nistand safhe dar $ Z^n _p$ ingoone tarif mishavad :  ( $ Z^n _p \equal{} Z_p * Z_p * Z_p * ... * Z_p$ (n bar) )
$ S\equal{} [ (x_1, x_2, .., x_n) \in  Z^n _p  |  a_1 x_1 \plus{} a_2 x_2 \plus{}...\plus{} a_n x_n \equiv c (mod p) ]$

Farz konid $ S_1, S_2, ..., S_k$ safhe haiy dar $ (Z^n)_p$ bashand ke $ \bigcup _i  S_i \equal{} Z^n _p$ ( $ i$ az $ 1$ ta $ k$)
Hadeaghal meghdar $ k$ ra biyabid

$ 5\minus{}$ kochektarin addad Sahih va namanfi $ n$ ra biyabid ke  baraye an , Tabe ghir sabet $ f : Z_\rightarrow[0,\infty)$ ba sharayet zir  vojood dashte bashad :
$ a) f(xy)\equal{} f(x) f(y)$
$ b) 2 f(x^2\plus{}y^2)\minus{}f(x)\minus{}f(y) \in [0, 1, ..., n]   x,y \in Z$
Va baraye in $ n$ tamam tavabe ba in khasiat ra biyabid

hi darse vizheye emsal chi boode?? midooni??

khaheshan tor khoda soalat marhale 3 dars haye  dige ra begzarid . :(


function
geometry
advanced fields
advanced fields solved

Let $V \in k^{n}$ be a hypersurface, that is not a hyperplane.Suppose that $p \in V$ is smooth.Suppose further that $I(V) = (f)$.Show that the variety $V \cap T_{p}(V)$ has a singular point at $p$ or the restriction of $f$ to $T_{p}(V)$ has an irreducible factor of multiplicity $\geq 2$

*$T_{p}(V)$ is a tangent space of $V$ at $p$.

Let $g=f|_{T_{p}V}$. Assume that $g$ has no irreducible multiple factors. This means that $g$ is reduced, so it gives the equation of $V\cap T_{p}V$.
It is easy to see that all the partials of $g$ vanish in $p$ just by writing them in terms of the partials of $f$ keeping an eye for the equation of $T_{p}V$. This means that $p$ is singular for $V\cap T_{p}V$.

Thanks for the answer.But could you please explain a bit more, just in terms of definitions, cause i'm a newbie an algebraic geometry.

[quote="amfulger"]Let $g=f|_{T_{p}V}$. Assume that $g$ has no irreducible multiple factors. This means that $g$ is reduced, so it gives the equation of $V\cap T_{p}V$.[/quote]

What exactly do you mean here, what are these equations?

[quote="amfulger"]
It is easy to see that all the partials of $g$ vanish in $p$ just by writing them in terms of the partials of $f$ keeping an eye for the equation of $T_{p}V$.
[/quote]
So $T_{p}(V) = V ( \sum \frac{\partial{f}}{\partial{x_{i}}}(p)(x_{i}-p_{i}) )$. The partial of $g$ is just the partial of $f$ on $T_{p}(V)$ and why all $\frac{\partial{f}}{\partial{x_{i}}}(p)$ must be zero ?

[quote="amfulger"]
This means that $p$ is singular for $V\cap T_{p}V$.[/quote]

How does it follow ?

The problem is that for me it's quite difficult to go beyond the definitions at the moment, so the definition of singular point i'm using:

A point $p$ is called singular if $dim T_{p}(V) \neq dim_{p}(V)$ - local dimension at $p$ (the maximal irreducible component of $V$,containing $p$).

Thanks in advance and sorry for stupid questions.

I think I should ask this first:
Is $k$ algebraically closed? I am not sure, but I think it would be easier for me to explain the first question if $k$ is algebraically closed.
Basically $V\cap T_{P}V$ is a hypersurface in $T_{P}V$ given by one equation (i.e. I(V)=(h) and $h$ is the equation) and this equation has to be $g$ if $g$ is reduced by the Nullenstellensatz. But this only works for algebraically closed $k$.

If you can have a variety $V$ in $k^{n}$ whose defining ideal, $I(V)=(f_{1},\ldots,f_{k})$, then $T_{P}V$ is the kernell of $J(f_{1},\ldots,f_{k})(p)$.
In case $I(V)=(f)$ you can explain this easier by saying that
$T_{P}V$ is given by the equation $\sum_{i}=1^{n}\frac{\partial f}{\partial x_{i}}(p)(X_{i}-p_{i})=0$. This is just what you said.

Let's look at $g$. You should look it as defined on $T_{P}V$.
The easiest way is to assume that $T_{P}V$ is given by the equation $x_{1}=0$ and $P=0$.
Then you can think $g$ as a function of the variables $x_{2},\ldots, x_{n}$ and
$g(x_{2},\ldots,x_{n})=f(0,x_{2},\ldots,x_{n})$. There is no partial $\frac{\partial g}{\partial x_{1}}$. It makes no sense. All the partials of $g$ in $P$ are indeed the partials of $f$ (wrt $x_{2},\ldots,x_{n}$), but these are $0$ because they are the corresponding coefficients in the equation of $T_{P}V$ which are $0$.

If $V$ is a hypersurface given by the equation $f=0$, then a point $p$ in $V$ is singular if and only if all the partials of $f$ at $p$ are $0$. I leave this as an exercise.

I hope it is a bit more clear. I'll look some more into the first stament so see if I can explain it without the Nullenstellensatz

Thanks a lot, but i still have some questions.
Yes $k$ is supposed to be algebraically closed.

[quote="amfulger"]
Basically $V\cap T_{P}V$ is a hypersurface in $T_{P}V$ given by one equation (i.e. I(V)=(h) and $h$ is the equation) and this equation has to be $g$ if $g$ is reduced by the Nullenstellensatz. But this only works for algebraically closed $k$.
[/quote]

I don't quite understand this, and how do we use here that $g$ is actually reduced ?

[quote="amfulger"]
Let's look at $g$. You should look it as defined on $T_{P}V$.
The easiest way is to assume that $T_{P}V$ is given by the equation $x_{1}=0$ and $P=0$.
[/quote]

Sorry for my stupid questions, but why can we just assume that $T_{p}V$ is just defined by $x_{1}= 0$ ?

[quote="amfulger"]
All the partials of $g$ in $P$ are indeed the partials of $f$ (wrt $x_{2},\ldots,x_{n}$), but these are $0$ because they are the corresponding coefficients in the equation of $T_{P}V$ which are $0$.
[/quote]

Why are the partials of $f$ zero? We know that on $T_{p}V$ their combination  $\sum \frac{\partial{f}}{\partial{x_{i}}}(p)(x_{i}-p) = 0$ for all $x=(x_{1},x_{2}...,x_{n}) \in T_{p}V$ , but how does it imply that they are all zero? ($\frac{\partial{f}}{\partial{x_{i}}}(p) = 0$ ?)

Thanks in advance and again sorry for my stupid questions.

1. Hypersurfaces have dimension $n-1$, so $T_{P}V$ is isomorphic to $k^{n-1}$ because $P$ is smooth.

If you accept that the ideal of $V\cap T_{P}V$ in $T_{P}V$ is of the form $(h)$ and that $V(g)=V\cap T_{P}V$, then by the Nullenstellensatz, $\sqrt{(g)}=(h)$ which implies that $g$ and $h$ have the same prime factors. But $g$ is reduced, hence $g|h$ from which it is easy to conclude that $g$ and $h$ differ by an invertible element, so $(g)=(h)$ which means that $g$ is an equation for $V\cap T_{P}V$.

2. We can assume that $T_{P}V$ is $x_{1}=0$ because there exist linear (affine) automorphisms of $k^{n}$ that move $T_{P}V$ onto $x_{1}=0$. These affine isomorphisms don't really change anything in the geometry the varieties in question.

3. In this case, $\frac{\partial f}{\partial x_{i}}(p)$ are all zero for all $i=\overline{2,n}$ because I assumed that the equation of $T_{P}V$ is $x_{1}=0$ and $P=0$.
In this case on the one hand you have
$1\cdot x_{1}+0\cdot x_{2}+\ldots+0\cdot x_{n}=0$
on the other hand you have
$\frac{\partial f}{\partial x_{1}}(0)(x_{1}-0)+\frac{\partial f}{\partial x_{2}}(0)(x_{2}-0)+\ldots+\frac{\partial f}{\partial x_{n}}(0)(x_{n}-0)=0$.
Keeping in mind that the above are identities on $T_{P}V$ we conclude that all the partials wrt $x_{\geq 2}$ are $0$. Which means that ALL the partials of $g$ in $0$ are $0$, so $0$ is not smooth for the variety $V\cap T_{P}V$ regarded as a subvariety of $T_{P}V$.
--------------------------------------------------------------------------------
I don't want to bring a formal proof without assuming $T_{P}V=\{x_{1}=0\}$.
Instead, I leave you with this problem.
Let $f: k^{3}\to k$, $f(x,y,z)=(x+y+z+1)^{2}-1$.
Define $g: k^{2}\to k$ by $g(x,y)=f(x,y,-x-y)$.
Compute the partials of $g$.
Find a connection to the problem above.


inequalities
function
inequalities proposed

Let $a,b,c,d$ be positive real numbers such that: $a\le1$, $a+b\le5$, $a+b+c\le14$ and $a+b+c+d\le30$. Then:
                                                 $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le10.$

see here also
http://www.mathlinks.ro/Forum/viewtopic.php?t=146592

[quote="N.T.TUAN"]Use Abel's lemma, we can prove
If $a_{1}+a_{2}+...+a_{k}\leq b_{1}+b_{2}+...+b_{k}\forall 1\leq k\leq n, b_{1}\leq b_{2}\leq...\leq b_{n}$ and $a_{i}>0,b_{i}>0$ then $\sum_{i=1}^{n}\sqrt{a_{i}}\leq \sum_{i=1}^{n}\sqrt{b_{i}}$.[/quote]
A similar thing:
The hypothesis tells us that $(a;b;c;d)\succ(1;4;9;16)$, so by applying Karamata for the concave function $f(x)=\sqrt{x}$, we obtain $f(a)+f(b)+f(c)+f(d)\leq f(1)+f(4)+f(9)+f(16)=10$


conics
parabola
analytic geometry
graphing lines
slope
geometry
perpendicular bisector

Find the equation of the circle through $ (1,0)$ and touching parabola $ y\equal{}x^{2}$ at $ (2,4)$.

The 'general' equation of a circle is: [b] x^2 +y^2 +2gx +2fy +c=0 [/b] where 'g' and 'f' are the co-ordinates of the center.

Since the circle passes through the point [b](1,0) [/b], and touches the parabola at [b] (2,4) [/b] -- The point (1,0) lies on the circle, so the equation of the circle passing through (1,0) is: [b] 2g+c=-1 [/b]

The point (2,4) also lies on the circle -- It also lies on the parabola. Substitute (2,4) in the 'general' equation of a circle, we get: [b] 2g+8f=-19 [/b]

Finally,After some 'working'-- I got the equation of the circle as given below:
[b] x^2+y^2+5x-6y-6=0 [/b]

 I think the above equation is correct since it's obvious that the points [b] (1,0) and (2,4) [/b] lie on the circle. These two points 'satisfy' the equation.

I think I might have committed an inadvertent error somewhere, if it's so, do correct me.

hello, i have the following worksheet
ansatz $ (x\minus{}x_M)^2\plus{}(y\minus{}y_M)^2\equal{}r^2$ with $ y'(2)\equal{}4$ as the slope of the parabola in the point $ P(2;4)$, for the circle we get $ y'(x)\equal{}\pm\frac{1}{2}(r^2\minus{}(x\minus{}x_M)^2)^{(\minus{}1/2)}\cdot(\minus{}2(x\minus{}x_M))$
solving this i have got for the radius $ \frac{17}{16}\sqrt{17}$.
Sonnhard.

Well, I would like to know whether my 'reasoning' was correct or was it fallible?

It's not correct because it's not tangent to the parabola at $ (2,4)$. Without the tangent condition there are infinitely many circles that go through $ (1,0)$ and $ (2,4)$.
One can use [url=http://www.cut-the-knot.org/Curriculum/Geometry/ParabolaFocal.shtml]this property of parabolas[/url] to [hide="solve it without calculus."]Using the labels in the link (right below the diagram, select all of the boxes under "Show"), the [url=http://en.wikipedia.org/wiki/Parabola#Derivation_of_the_focus]focus[/url] $ F$ is $ \left(0,\frac14\right)$ and the point $ A'$ on the directrix is $ \left(2,\minus{}\frac14\right)$. Also, $ A$ is $ (2,4)$, and $ T$ is $ (1,0)$ (the midpoint of $ F$ and $ A'$).

Because the bisector of angle $ FAA'$ is $ AT$, the tangent line to the parabola at $ (2,4)$ (and thus to the circle at $ (2,4)$) is $ y\minus{}0\equal{}4(x\minus{}1)\implies y\equal{}4x\minus{}4$.

The equation of a circle is $ (x\minus{}a)^2\plus{}(y\minus{}b)^2\equal{}r^2$ for center $ (a,b)$ and radius $ r$. Since the circle passes through $ (1,0)$ and $ (2,4)$, it satisfies $ (1\minus{}a)^2\plus{}b^2\equal{}r^2$ and $ (2\minus{}a)^2\plus{}(4\minus{}b)^2\equal{}r^2$. As it is tangent to the line $ y\equal{}4x\minus{}4$ at $ (2,4)$, its center lies on the perpendicular to $ y\equal{}4x\minus{}4$ at $ (2,4)$. Noting that $ (a,b)$ thus satisfies the perpendicular line's equation, we can solve for $ a,b,r$.[/hide]

Let the slope of the tangent will be $ m$ so the equation of tangent will be $ y \minus{} 4 \equal{} m(x \minus{} 2)$ now we take a point circle on the line whose cordinates are $ (2,4)$ and radius is zero.There will be infinite circles through this point, we use the equation of family of circles
$ (x \minus{} 2)^{2} \plus{} (y \minus{} 4)^{2} \plus{} \beta(y \minus{} 4 \minus{} m(x \minus{} 2)) \equal{} 0$,we can easily find the slope $ m$ also note that $ (1,0)$ put these points in the equation to get the value of $ \beta$ and that will be the final equation of the circle..
Done :showoff:

[b]A reasonable approach:[/b] You know that the tangent line to the parabola at (2,4) is going to be the same as the tangent to the circle, which is in turn perpendicular to the radius of the circle. So, the normal line to the parabola at (2,4) contains the center of the circle.

You also know that the center of the circle is equidistant from (1,0) and (2,4), so the perpendicular bisector of these two points contains the center of the circle.

The intersection of these two lines is the center of the circle.

[b]Problem we run into:[/b] Note that both of the two lines have the same slope: -1/4. However, one contains (2,4) and the other does not. So, the lines are parallel and do not intersect, and there is no point which can be a candidate for being the center of the circle; i.e., there is no such circle.

The line connecting (1,0) and (2,4) is in fact the tangent line to the parabola at (2,4). Thus, the circle would have to have two points in common with its tangent line: (1,0) and (2,4). This is impossible, so there is no such circle.

Can u plz tell me of which two lines are u talking about

In the attached image, the red curve is the parabola (or whatever curve you want your circle to be tangent to). We want the circle to be tangent at point T on this parabola.

The green line is the tangent line to the parabola at T. It must also be the tangent line to the circle at point T. So, the center of the circle must lie on the orange line, which crosses the green line at a right angle at point T. That was the first of the two lines I was talking about.

However, we know point P is also on the circle. If the circle is to have center C, it must be true that CT = CP, and thus, C must be on the perpendicular bisector of segment TP. I've made segment TP cyan, with it's median marked as point M, and drawn the perpendicular bisector in blue. This is the second line.

Since C must be on the first (orange) line and the second (blue), line, it must be at their intersection -- thus, we've located the center of the circle.

[b]That is,[/b] if such an intersection exists. In this case, however, the point P we're given is on the tangent line (the green one) -- for example, perhaps where point P* is. Imagine dragging point P over to where P* is. The slope of the blue line would decrease until the blue line is parallel to the orange line when P=P*. Now, the blue and orange lines don't intersect.

It might help if you visualize what's happening to the circle. As P moves to P*, the slope of the blue line is decreasing, so the intersection of the blue and orange lines (aka the center C of the circle) moves further and further to the right along the orange line -- but since the circle must go through T, the radius of the circle increases without bound.

The part of the circle we can see without "zooming out" looks more and more like the tangent line. In the limit, it "is" the tangent line -- but we are looking for a circle with finite radius, and we certainly can't make one that goes through P*.

[b]Note 1:[/b] [i]If you have Geometer's Sketchpad, draw two perpendicular lines intersecting at T, and an independent point P. Then, create segment TP, then point M, and then the perpendicular bisector of TP through M, in that order. Define C to be the intersection of the appropriate lines, and draw circle CP. Move P around, and see what happens to C and the circle![/i]

[b]Note:[/b] [i]I made P and P* on the x-axis, but they certainly don't have to be on the x-axis -- P can be anywhere (except on the green line) and P* can be anywhere on the green line.[/i]


geometry
circumcircle
projective geometry
perpendicular bisector

Let $\triangle ABC$ with $D\in BC$, $E\in CA$, and $F\in AB$ such that $FD=FE$. Let $P\in AD\cap BE$ Prove the equivalence: $PECD$ cyclic $\iff$ $FE$, $FD$ are tangents to the circumcircle of $\triangle CDE$.

This problem is nice, congratulations.

Here is my solution:
Call $Z$ the intersection point of $PC$ and $ED$.

1) If $PECD$ is cyclic, then $AB$ is the polar line of $Z$ wrt the circumcircle of $PECD$.
$F$ lies on both the polar line of $Z$ and the perpendicular bisector of $[ED]$. Therefore $FE,FD$ have to be the tangents.

2) If the two segments are tangents: Call $P'$ the intersection of $AD$ and the circumcircle of $\Delta CDE$ and $Z'=P'C \cap ED$
Now clearly $AF$ is the polar of $Z'$ wrt the circumcircle. Therefore, the intersection of $AF$ and $CD$ must lie on $EP'$.
This implies that $P=P'$. Done.

For $\rightarrow$, apply pascal on $EEPDDC$, then $EE\cap DD=F'\in AB$. But since $F$ and $F'$ both lie on the perpendicular bisector of $DE$ and line $AB$, they are the same point.

For $\leftarrow$, let $AD$ intersect the circumcircle of $\triangle CDE$ at $P'$. Let $EP'\cap DC=B'$. Pascal on $EEP'DC$ shows that $AFB'$ are collinear. But $B$ and $B'$ are both the point that lies on $AF$ and $CD$, so they are the same. Likewise, $P$ and $P'$ are both the point that lies on $BE$ and $AD$, so they are the same. Since $P'DCE$ is cyclic, $PDCE$ is cyclic.

Our proofs are completely similar. You call it Pascal's theorem, I call it poles and polars  :P
It seems different but in this case, the two methods are essentially the same I think.

Hmm...it also occured to me that our solutions were quite similar. I am not sure how to articulate that. One thing to note, I think pascal's theorem is stronger than your direct use of pole-polar locus stuff. I considered a degenerate case, when two pairs of points were the same. This is very interesting.


The asteroid belt between Mars and Jupiter consists of many fragments (which some space scientists think come from a planet that once orbited the Sun but was destroyed).

(a) If the center of mass of the asteroid belt (where the planet would have been) is about three times farther from the Sun than the Earth is, how long would it have taken this hypothetical planet to orbit the Sun?

(b) Can we use these data to deduce the mass of this planet? If so, how?

(a) is  Kepler's Laws question: the orbital period is proportional to the 3/2 power of the major axis. (Or you could just assume circular motion, start with $ \frac{v^2}r$ being proportional to $ \frac1{r^2},$ and working from there to get the same result.) The orbital period of the hypothetical planet - and it would still be some kind of average of the orbital periods of the asteroids - would be about $ 3\sqrt{3}$ years, or about 1900 days.

As for (b), I don't see any way to get at the mass from this information. As long as the mass is much smaller than the mass of the sun, the orbital period depends only on the radius, and not on the mass of the orbiting body. And if you want to invoke any modern theories of planetary formation as to what the mass of a planet in that location should be, you're going to run into the fact that those same theories of planetary formation will tell you that the asteroids probably never were consolidated into a larger planetary body - they just never coalesced in the first place (with perturbations from Jupiter's gravity being an important consideration.)


To get the indention like they have it use \begin{enumerate} and every time you want to start a new problem type \item. It will automatically number it for you.  If you want the indention like that but want to change the number to say a letter you can use \begin{description} and for each item put \item[description] where you put the letter in where it says description.  Also note that when using the description environment the stuff in brackets after \item is optional.

Oh yeah, don't forget the \end{enumerater} or \end{description} at the end.

He asked how to justify the text not enumerate it automatically.
Sorry I really don't have any clues, I thought that it is automatically justified on some classes.


algebra
polynomial
inequalities
algebra unsolved

[quote="easternlatincup"]If the polynomial $ f(n) \equal{} x^n \plus{} a_1x^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 1}x \plus{} 1$ has nonnegative coefficients and $ n$ real roots, prove that $ f(2) > 3^n$[/quote]
[hide="for what it's worth..."]
Since the coefficients are nonnegative, the roots are all non-positive...so write the polynomial as $ (x \plus{} r_1)(x \plus{} r_2)\ldots (x \plus{} r_n)$, $ r_i\le 0$. We need to show $ (2 \plus{} r_1)(2 \plus{} r_2)\ldots (2 \plus{} r_n) > 3^n$, or $ \left(\frac {2 \plus{} r_1}{3}\right)\left(\frac {2 \plus{} r_2}{3}\right)\ldots\left(\frac {2 \plus{} r_n}{3}\right) > 1$

Start from $ r_1 \equal{} r_2 \equal{} \ldots \equal{} r_n \equal{} 1$, the $ f(2) > 3^n$ doesn't hold for $ f(x) \equal{} x \plus{} 1$...if we have $ f(x) \equal{} (x \plus{} 1)^n$ and treat the $ \minus{} 1$ as 'distinct' roots then we get $ f(2) \equal{} 3^n$ so we are just making $ f(2) > 3^n$ stronger if we consider roots with multiplicity as distinct roots...
If $ n$ is even, $ r_1r_2\ldots r_n \equal{} 1$ and we can show that
\[ \left(\frac {2 \plus{} (r_1 \plus{} \epsilon)}{3}\right)\left(\frac {2 \plus{} \frac {1}{r_1 \plus{} \epsilon}}{3}\right)\ldots\left(\frac {2 \plus{} r_n}{3}\right)\]
\[ > \left(\frac {2 \plus{} r_1}{3}\right)\left(\frac {2 \plus{} r_2}{3}\right)\ldots\left(\frac {2 \plus{} r_n}{3}\right)\]
\[ > 1\]
with our strongest case at $ r_1 \equal{} r_2 \equal{} \ldots \equal{} r_n \equal{} 1$...we can do something similar for odd $ n$...[/hide]

Well, first thing I'm sorry about the problem statement. Of course the $ >$ has to be changed to the $ \geq$

Second thing, I don't think your solution is complete. You just showed for a particular case...

[quote="t0rajir0u"]Expand $ \prod (2 \plus{} r_i)$ as a polynomial in $ 2$ and use AM-GM on every coefficient.[/quote]
It is the Holder inequalities :wink:

[quote="easternlatincup"]If the polynomial $ f(n) \equal{} x^n \plus{} a_1x^{n \minus{} 1} \plus{} ... \plus{} a_{n \minus{} 1}x \plus{} 1$ has nonnegative coefficients and $ n$ real roots, prove that $ f(2) \geq 3^n$[/quote]
In general if $ x\geq 0$ then   $ f(x) \geq (x\plus{}1)^n$
  :cool:


According to a study, the duration of a match in the World Cup is
approximate normally distributed with the mean 111 minutes and standard
deviation 5 minutes ( including the break between the halves ).

Q1. Estimate the probability that more than 30 matches will have the
duration between 110 and 118 minutes ( assume there are 64 matches ).

Q2. A researcher divides all the matches into 3 categories:
    S ( those matches less than 105 minutes )
    M ( those matches between 105 and 115 minutes )
    L ( those matches more than 115 minutes )
What is the probability that there will be 10S, 44M and 10L in the World
Cup finals tournament?

[hide="hint1"]
you probably want a solution that doesn't use the calculator so try finding the z-scores for the bounds 110 and 118.
remember that zscores are equal to (x-mean)/std
find the probability that a score is inbetween those two...use a chart or calc (check DISTR on ti83)
then do what you usually do when you have a probability of a single event and want the probability of ATLEAST a certain number of successes. [/hide]
[hide="hint2"]
Q2 is nearly the same approach. this time your success variable will change from S to M to L[/hide]

Dont really understand for the hint 1..can u please explain alittle bitmore?

thank in advance!

Ha, our last unit last year besides conics.  Basic stats...  Hated it.

Let me try to help explain hint 1
[hide]
Ok for question 1 you have 2 x-values you use to derive the z-score.  These are 110 and 118.  You then subtract each of these values from the mean value or 111 minutes.  Afterwards you divide by the SD or 5 minutes.  This gives you the Z-score which you can transform into a percentage by looking at a Z-table.  Remember the Z-table gives you the % from the mean on a bellcurve.
[/hide]

how about question 2..can u give more hint on that question please?..thanks in advance!


trigonometry
function
analytic geometry
geometry
trig identities
Law of Sines
similar triangles

[color=cyan]This is actually a proof of the converse of what the problem asks:  that is, given that angles FDH and HDE are equal, show that BE and CF are concurrent with the altitude from A.  It is followed by what I hope is a valid explanation of how the converse implies the statement to be proven.

Proof:
Let the measure of BCA be c and the measure of ABC be b, and let the measure of FDH and HDE equal x.  Then
BDF = 90 - x = EDC
CAD = 90 - c
DAB = 90 - b
DEA = 90 + c - x
AFD = 90 + b - x
CED = 90 - c + x
DFB = 90 - b + x

Applying the Law of Sines to triangles ADE and AFD, we get
EA/sin(x) = AD/sin(90 + c - x)  and  AF/sin(x) = DA/sin(90 + b - x), respectively.  Dividing the first equation by the second,
EA/AF = sin(90 + b - x)/sin(90 + c - x)

Applying the Law of Sines to triangles CED and FBD, we get
CE/sin(90 - x) = DC/sin(90 - c + x)  and
FB/sin(90 - x) = BD/sin(90 - b + x), respectively.  Dividing the second equation by the first,
FB/ED = (BD*sin(90 - c + x)) / (DC*sin(90 - b + x))

(FB/AF)*(DC/BD)*(EA/CE) = (EA/AF)*(DC/BD)*(FB/CE), which by substitution equals
(sin(90 + b - x)/sin(90 + c - x)) * (DC/BD) *
(BD*sin(90 - c + x))/(DC*sin(90 - b + x)).
The sine function is symmetric about 90 degrees.  Therefore, this expression is equal to 1.  By Ceva's theorem, the three lines AD, BE, and CF are concurrent, Q.E.D.

Now, that was the converse of the original statement.  However, I believe I can turn it around to prove the original statement as follows: as x approaches 0, H approaches A.  As x approaches 90, H approaches D.  H must vary continuously with x -- it can't simply skip around on the altitude.  (That is not something I can rigorously prove, but it is true.)  Thus, every point of the altitude is included in the above proof.  Therefore, if we take a point H on the altitude and draw lines BH and CH, we already know that the angles are equal.  So the theorem is true by its converse.

I know how to prove the theorem directly in the case where BE and CF are also altitudes, but not generally.[/color]

Its not too hard to prove this using coordinates - of course there is always a nicer solution but here is this one anyway:

Let D be (0,0), A is (0,b), B is (-1,0), C is (a,0), and H is (0,m).
(We can scale the triangle so B is (-1,0)).
Then line BH is y=(-m/a)(x-a) and line AC is y=b(x+1), so they intersect where x(b + m/a) = m-b, ie x=(m-b)/(b+m/a), or x=a(m-b)/(ab+m). y=b(x+1) which equals bm(a+1)/(ab+m).

Similarly, CH is y=m(x+1) and AB is y=(-b/a)(x-a) and these intersect when x(m + b/a) = b-m, ie x=a(b-m)/(am+b). Then y=mb(a+1)/(ma+b).

Now the gradient from F to D (the origin) is bm(a+1)/(ab+m) divided by a(m-b)/(ab+m) which simplifies to bm(a+1)/(a(m-b)), and the gradient from E to D is mb(a+1)/(ma+b) divided by a(b-m)/(am+b), which is mb(a+1)/(a(b-m)). This are negatives of each other, thus angle FDH equals angle HDE.

You can also solve this using similar triangles - drop altitudes from E and F down onto BC and you get a bunch of them. Give lots of side lengths a letter, do a bit of algebra and you can prove that the two angles you want are equal as well.

Let  [tex]l[/tex] be the line parallel to BC and passing through the vertex  A.

Let X be the intersection of the line [tex]l[/tex] with the line DF.

Let Y be the intersection of the line [tex]l[/tex] with the line DE.

Then [tex]\displaystyle \frac{AX}{BD}=\frac{AF}{FB}[/tex] since  :tri: AFX :sim:  :tri: BFD. Similarly, [tex]\displaystyle \frac{DC}{AY}=\frac{CE}{EA}[/tex].

So [tex]\displaystyle \frac{AX}{AY}=\frac{AX}{BD}\cdot\frac{BD}{DC}\cdot\frac{DC}{AY}=\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1[/tex] by Ceva's Theorem.

Hence AX=AY. So  :tri: XAD :cong:  :tri: YAD. Therefore :ang: FDH =  :ang: XDA =  :ang: YDA =  :ang: EDH

Q.E.D.


[hide]ACB
ACFB
ACDFB
ACDEFB

ADCB
ADCFE
ADEFB
ADEFCB

8?[/hide]

no, not 8. The answer is ...
[hide=":?:"]10[/hide]

You can list systamatically

You can find how many ways go through C.  Then list how many ways go through D but not C.

Or how many ways you can go if the first is D.  Then how many ways you can go if the first is C.

I hate these counting problems: "how many ways can you go without revisting..." "how many rectangles"  "how many triangles"  "what is the probability that a randomly selected square from a NbyN grid of dots..."

Try using a similar method to those ones with the rectangles (how many ways are there to go from a to b if you can move only up and to the right, etc., where you add as you go), and dvide the problem into cases.

If you first go to C, then there is 1 way to go to D that doesn't end up revisting, then 1 way to get to E, 3 ways to get to F (from C, D, and E), and 1+3=4 ways to get to B.

If you go first to D and then to C, there is one way to get to F and then 2 ways to get to B.

If you go first to D and then somewhere besides C, there is one way to get to E, two ways to get to F, then 2 ways to get to C, and 2+2=4 ways to get to B.

Therefore there are 4+2+4=10 total ways.

PS: This makes a lot more sense if you look at the picture.


superior algebra
superior algebra solved

Let $(G,\cdot)$ be a group and $f,g:G\to G$, $f(x)=x^n$, $g(x)=x^{2n}$ morphisms, $n\in\mathbb{N}^*$, $f$ surjective (onto map).
a) Prove that $G$ is abelian.
b) Give such and example of group $G$ s.t. for any $n\in\mathbb{N}^*$ the conditions from the hypothesis are satisfied.

a) since $f$ is surjective, for all $a,b \in G$ there exists $x,y \in G , x^n=a , y^n=b$
This gives $a^2 b^2=g(x)g(y)=g(xy)=(ab)^2=abab$ leading to $ab=ba$, so its abelian.
b) simple example: $G= e $  ;)
infinite example: $G=\mathbb{R}^+$


In parallelogram $ ABCD$, $ AB \equal{} 38$ cm, $ BC \equal{} 3y^3$ cm, $ CD \equal{} 2x \plus{}4$ cm, and $ AD \equal{} 24$ cm. What is the product of $ x$ and $ y$?

Just for clarification, parallel sides of a parallelogram are obviously $  \boxed{=}  $  .

Basic geometry.


limit
complex numbers
complex analysis
complex analysis unsolved

Let $ L(z) \equal{} \frac {z\minus{}a}{1\minus{}a^*z}$ where $ a \in D(0,1)$ and $ a^*$ denotes the complex conjugate of $ a$.

Define a sequence of maps by $ L_{j\plus{}1}(z) \equal{} L(L_j(z))$ where $ L_1(z) \equal{} L(z)$

Prove that $ \lim L_j(z)$ exists, uniformly on compact subsets of $ D(0,1)$ and find the limit function.

It is possible to compute $ L_j(z)$ in closed form by computing $ \left[ \begin{array}{cc} 1 & \minus{}a \\ \minus{}a^* & 1 \end{array} \right]^j$.  What are its eigenvalues?


1. Albatross says, "Find a power of ten that is a close approximate of 3^30"

2. Albatross says, "Find a power of ten that is a close approximate of 2^60"

[hide="Hint 1"]3^2  :approx: 10[/hide]

[hide="Hint 2"]2^10 :approx: 1000[/hide]

Helpful on MATHCOUNTS.

[quote="A+MATH"]Helpful on MATHCOUNTS.[/quote]
A+:  The second problem and hint are helpful on MATHCOUNTS and other competitions.  As far as determining "the number of digits" in any tenth power of 2, this hint works on numbers [i]almost[/i] up to 300 digits long.  [Let's let the readers experiment.]

The first problem and hint are helpful in a similar way.  As far as determining "the number of digits" in an even power of 3, this hint works on numbers up to about 20 digits long.

But you now have received two different answers for Problem 1.  Are they both correct?  What did you mean by a "close approximate"?

[hide]10^15 should be the good one[/hide]

[hide="To A Plus Math"][quote="A+MATH"]10^15 should be the good one[/quote]
$\hbox{But }10^{14}<205891132094649<10^{15}\hbox{.  So, }10^{14}\hbox{ is the closer approximation.}$[/hide]


Suppose we have a quadrilateral $ ABCD$ inscribed in a circle. The tangents to this circle at the points $ A$ and $ C$ intersect at a point $ P$, such that this point $ P$ does not lie on the line $ BD$, and such that $ PA^{2}=PB\cdot PD$. Prove that the line $ BD$ passes through the midpoint of the segment $ AC$.

Let $O$ be the center of the circle.
Suppose that $PD$ cuts the circle at $D'$ and $PB$ cuts the circle at $B'$.
Then $DD'BB'$ is a trapezoid. Since $AC$ is the polar of $P$ w.r.t the circle it follows that it passes through the point $X=BD\cap{B'D'}$. But the points $O,X,P$ are collinear, thus $X$ is the midpoint of $AC$.


algebra
polynomial
algebra proposed
range

Here is a problem with a completely crazy idea. Sorry if you find the solution in 2 minutes:
    Is there a polynomial $f$ with possibly many variables such that the set of positive values that the polynomial takes when its variables are restricted to integer values is exactly the set of the numbers $ ((a^n+b^n)/2)^2)$ where $a=1+\sqrt{2}, b=1-\sqrt{2}$?

I heard about something like this and about polynomials that produce only primes and negative numbers, but probably the proofs for these results are quite difficult. The solution of this problem has 3 lines!

I saw the polynomial which produces primes. It fits on half a page :). Of course, there's the matter of proving that it produces only primes :).

Come on, guys, the proof is so simple.  :D

[quote="Ravi B"]That result was proved by Matiyasevich (spelling?) as a lemma in his proof of Hilbert's 10th problem.[/quote]
Yes, at least in Russian your spelling is correct.

It is not easy. You shouldn't say that. There are many people who didn't solved it.

I know, I was just joking, Myth. Too bad I posted, it would have been a great problem for a contest: nice idea, short solution.

[quote="Ravi B"]A related result: There is a polynomial in several variables that, when restricted to integer inputs and positive outputs, has range the set of Fibonacci numbers.  That result was proved by Matiyasevich (spelling?) as a lemma in his proof of Hilbert's 10th problem.[/quote]

Polynomial expression of "$n$ is a Fibonacci number" has (in substance) appeared in competitions and was surely known long before Matijasevich.

[hide="Hint"] $F_{n+1}F_{n-1} - F_n F_n = \pm 1$ [/hide]

Matijasevich's amazing new result was that the relation $n = F_m$ is expressible as the solvability (in integers) of a polynomial $F(n,m,a,b,c,...) = 0$.

Yes, you're right.  My memory was faulty.

[quote="Leva1980"]The answer is $ p(x,y) \equal{} (2 \minus{} (x^{2} \minus{} 2y^{2})^{2}) x^{2}$, isn't it?
 :P[/quote]
[quote="harazi"]Yup, I told you it's easy. But isn't it nice? :D[/quote]
it's very nice :) here you're right

but if it is so simple, could you tell how to prove it and what is this formula come from (how to discover it) :?:


[color=blue][i]Από τα γερμανικά (Januar 2007) αποδίδω στη γλώσσα μας ένα πρόβλημα στη διαιρετότητα :[/i][/color]

 Να βρεθούν οι θετικοί ακέραιοι $ a , b , c$ ,ώστε $ (a\plus{}1) /  (b\plus{}c) , (b\plus{}1) / (c\plus{}a) , (c\plus{}1) / (a\plus{}b)$, όπου το $ /$ δηλώνει  το '' διαιρεί ''.

 Μπάμπης


algebra
polynomial
geometry
3D geometry
superior algebra
superior algebra solved

Sorry, this is not going to be a nice problem but rather a request to do an ugly computation, since I have no idea about Mathematica, Maple etc.:

Consider the ring $\mathbb{Z}_{2}\left[X,Y\right]$, where $\mathbb{Z}_{2}$ means $\mathbb{Z}\slash 2\mathbb{Z}$ (not $2$-adic integers). I would like to know whether the polynomial $X^{2}$ lies in the ideal $\left(X^{2}-Y^{3},X^{3},Y^{4}\right)$ of this ring.

Background: I had a conjecture that in a finite ring, every element which is a square and a cube simultaneously must be a $6$-th power. (For the rings $\mathbb{Z}_{n}$, this is equivalent to [url=http://www.problem-solving.be/pen/viewtopic.php?t=464&search_id=1229747076]an old ISL problem[/url].) I suspect that this conjecture is wrong, but I would like to have a counterexample, and if the polynomial $X^{2}$ does [b]not[/b] lie in the ideal $\left(X^{2}-Y^{3},X^{3},Y^{4}\right)$ (what I suspect), then $\mathbb{Z}_{2}\left[X,Y\right]\slash\left(X^{2}-Y^{3},X^{3},Y^{4}\right)$ would be such a counterexample.

Thanks in advance for any help (Mathematica source code would be great, too, since I intend to learn that one day).

  Darij

I would suggest downloading Macaulay2, it's a freeware computational algebraic geometry package.

Then you could do the following:

i1: R = (ZZ/2)[X,Y]

i2: I = ideal(X^2-Y^3,X^3,Y^4)

i3: K = gb I

i4: (X^2)%K

This script is pretty easy to follow, i1 establishes what ring we are working over, i2 sets the ideal, i3 lets K be a groebner basis for I, and i4 finds the remainder when dividing X^2 by K.  The remainder is X^2 btw, so X^2 does not belong to that ideal.

Thank you!!
So the conjecture should be disproven.

  Darij, installing cygwin and watching free disk space fade away ;)

glad I can help.  I'm not sure if you were downloading cygwin for the purpose of running Macaulay2, but I'm pretty sure you don't need to go through all that.

Try following the directions at

[url]http://www.commalg.org/m2win.shtml[/url]

it's been a while, but I think that's where I learned to install it.

You might have done it by now but if it is of any help I used Singular and I got that $X^{2}$ is not in that ideal.


How do you solve problems like:

How many zeroes are in the base $ 8$ representation of $ 120!$?

I am wondering which is correct approach below or whether both of the are incorrect:

1) $ \displaystyle \left\lfloor\frac{120}{8}\right\rfloor\plus{}\left\lfloor\frac{15}{8}\right\rfloor\equal{}\boxed{16}$

2) $ \displaystyle \left\lfloor\frac{120}{2}\right\rfloor\plus{}\left\lfloor\frac{60}{2}\right\rfloor\plus{}\left\lfloor\frac{30}{2}\right\rfloor\plus{}\left\lfloor\frac{15}{2}\right\rfloor\plus{}\left\lfloor\frac{7}{2}\right\rfloor\plus{}\left\lfloor\frac{3}{2}\right\rfloor\equal{}116$.

Now, we are looking for how many groups of $ 3$ $ 2$s there are.

$ \displaystyle \therefore \left\lfloor\frac{116}{3}\right\rfloor\equal{}\boxed{38}$.

Thank you  :)

The second one is fine.  The first one is not; that method only works for primes.

As a matter of fact, I would hazard a guess that there are more than 38 zeroes in 120! (in base 8). It just so happens that there are at most 38 at the end.

[quote="pytheagle"]Well, I'm just curious, but is there a way to find the TOTAL number of zeroes in a factorial without, obviously, multiplying it out?[/quote]

There is a way to find out how many zeros at the END of 120!
Your third method is correct.

However, I don't think there is an easy way to find how many zeros in the WHOLE 120!  (Read Mewto's comment above).

Can someone explain to me how he got the second line? Step by step?

Basically what he did is since 8 is $ 2^3$ he divided to see how many 2's there are [b][u]without[/u][/b] taking remainders:
$ \frac{120}{2} \plus{} \frac{120}{2^2} \plus{} \frac{120}{2^3} ... \equal{} 116$
But since you only want $ 2^3$ you divide by 3 = $ \frac{116}{3} \equal{} \boxed{38}$


geometry
3D geometry
search
number theory
combinatorics unsolved
combinatorics

Prob::::: Consider $n$ consecutive integers none of which are Perfect CUBES. Find the smallest value of $n$ such that there always exist three integers $a,b,c$ among them such that $abc$ is a [b]perfect cube[/b].

This came up in connection with a Number theory problem! please help me with this! I dont know whether this is even solvable or not!

I have seen someone do this problem before, I think.  Try and search for it.  I clearly remember the clever substitution

$a = x^2 y$
$b = y^2 z$
$c = z^2 x$

What search keyword do you suggest? I cant think of one that will give a narrow search!

EDIT::: I've done some basic searching --- havent found anything yet!!

Hye people, please help me with this! I cant locate the solved problem on Mathlinks! Please help me with the solution!!!

Please help !!!!

Hello Rushil, I think that there is no minimum value of $n$ because:

Imagine you have such an $n$. You take $n$ primes over $n$. You can find $n$ consecutive integers so that only the first is divisible by the first prime but not by the 3rd power of this prime, only the second number is divisible by the second prime but not by the 3rd power of this prime and so on. (because of the chinese rest-theorem or what its name is in English)
If you take any 3 of these numbers you will never get an perfect cube (if I understand this word in the right way) because any of these prime factors over $n$ isn't in the 3rd, 6th or another 3k-th power.
If I didn't do an error, there is no minimum $n$ for which you can find $abc=k^{3}$.

Naphthalin


salam doostan kasi rahi be gheir az enekeas baraye masaleye zir dare?
noghteye $b$ motealegh be pare khate $ac$ ast.se nimdayreye be ghotrhaye $ab$&$ac$&$bc$ ra dar nazar begirim ke har se dar yek tarafe parekhate $ac$hastand
sabet konidke shoahaye do dayere ke har kodam  ham bar do nim dayere vaham bar khate amood bar$ac$ dar noghteye $b$momas ast ba yekdigar barabarand.

ghasieye fisaghoores baraye kodoom mosalas?

in masale ba enekas vahshatnake....

yani mahshar hall mishe...

adam mitooneh haftehha bahsh erza beshe ;)  :rotfl:

[quote="amir2"]ba ghazyeye Fisaghoores ham mitooni esbat koni.[/quote]
i got it. ;)


geometry
parallelogram
trapezoid
geometry unsolved

In a trapezoid $ ABCD$ with $ AD \parallel{} BC$, $ O_1,O_2,O_3,O_4$ denote the circles with diameters $ AB,BC,CD,DA$, respectively. Show that there exists a circle with center inside the trapezoid which is tangent to all the four circles $ O_1,... ,O_4$ if and only if $ ABCD$ is a parallelogram.


Let be given f : [0, Pi/2]→R is continuous, and f(0) > 0, ∫(0..Pi/2) f(t) dt ≤ 1. Prove that for some z in (0, Pi/2) we have f(z) = sin z.

<i>Source VUMC 2003 </i>

Do I miss something or this problem is very easy?
  Let g(x)=f(x)-sinx. Then the integral of g from 0 to pi/2 is at most 0. Since the integral is g(c)*pi/2 for a certain c in (0,pi/2), we can find c such that g(c )<=0. Since g(0)>0 and g is continuos, the conclusion follows.

Nice solution Harazi

Here another one suppose for any x in (0; \pi /2) f(x)=/=sinx
sin0 = 0, f(0)>0  the curve of f is strictely above of the curve of
sinx on (0; \pi /2)
 \int _(0; \pi /2)sinxdx <  \int _(0; \pi /2)f(x)dx  \leq 1
1 <   \int _(0; \pi /2)f(x)dx  \leq 1 wich is a contradiction


Can someone give me a proof for the following problem.  I think the solution incorporates the pigeonhole principle:

Given any n real numbers arranged in any order left to right, we can either find [tex]\lfloor{\sqrt{n}}\rfloor+1[/tex] increasing or [tex]\lfloor{\sqrt{n}}\rfloor+1[/tex] decreasing.

The following theorem and the proof are well-known.

Theorem: Given a sequence of distinct real numbers, if the sequence has $n^2+1$ members, then it contains a subsequence of length $n+1$ that is either strictly increasing or strictly decreasing.

For each element $a_k$ in the given sequence, define a pair $(i_k, d_k)$ where $i_k$ is the length of the longest strictly increasing sequence starting with $a_k$ and $d_k$ is the length of the longest strictly decreasing sequence starting with $a_k$.

Suppose the theorem is false.  Then $i_k<n+1$ and $d_k<n+1$ for all $k$.

Thus, there are most $n$ distinct $i_k$ and $n$ distinct $d_k$, so there are at most $n^2$ distinct pairs.  But there are $n^2+1$ pairs, so by the pigeonhole principle, two of the pairs are identical.  That is, $(i_k, d_k) = (i_m, d_m)$ for some $k$ and $m$.

If $a_k$ < $a_m$, then there is a strictly increasing sequence of length $i_m+1$ starting with $a_k$, (namely the sequence for $a_m$ which has length $i_m$ plus $a_k$ on the front),  but the $(i_k, d_k)$ pair claims that the maximum length sequence is $i_k = i_m$.  Contradiction!

Similarly, if $a_k > a_m$, then the $d_k$ leads to a contradiction.

So our assumption was false, and the theorem is proved.

I have a feeling that generalises to ab+1 members -> a decreasing or b increasing in a row. I can't be totally sure of that.


calculus
integration
limit
real analysis
real analysis unsolved

find
$ \lim_{n\rightarrow \infty}\sqrt[n]{\int_{0}^{1}x^{\frac{n(n\plus{}1)}{2}}(1\minus{}x)(1\minus{}x^{2})\cdots (1\minus{}x^{n})dx}$

try to comprise this product with $ (1\minus{}x)^{n}$ and $ (1\minus{}x^{n})^{n}$.
you will get beta function. I'm not sure but I think it will works.

[quote="Extremal"]try to comprise this product with $ (1 \minus{} x)^{n}$ and $ (1 \minus{} x^{n})^{n}$.
you will get beta function. I'm not sure but I think it will works.[/quote]

$ \sqrt[n]{\frac{\Gamma(n\plus{}1)\Gamma(\frac{n(n\plus{}1)}{2}\plus{}1)}{\Gamma(n\plus{}2\plus{}\frac{n(n\plus{}1)}{2})}}\leq A(n)\leq \sqrt[n]{\frac{\Gamma(n\plus{}1)\Gamma(\frac{(n\plus{}1)}{2}\plus{}\frac{1}{n})}{n\Gamma(\frac{1}{n}\plus{}\frac{3(n\plus{}1)}{2})}}$

when we set $ n\to \infty$:

$ 0\leq A(\infty)\leq \frac{2}{9}\sqrt{3}$ :(


trigonometry
inequalities unsolved
inequalities

Find the minimum and the maxiumum of E;
where E=(sin(x)-sin(y)/(1-sin(x)*sin(y)) for x,y in [0,pi/2] and x<>y.

[quote="Lagrangia"]Find the minimum and the maxiumum of E;
where $E=\frac{\sin(x)-\sin(y)}{1-\sin(x)\sin(y)}$ for $x,y \in [0,\frac{\pi}2]$ and $x\not=y$.[/quote]Sorry for my stupid remark but... Lagrangia, could it be there is a typo? Since I think this is not trig, as we can put $a=\sin(x), b=\sin(y)$ and have:

Find the minimum and maximum of $E=\frac{a-b}{1-ab}$, with $a,b \in [0,1]$, which is trivially $-1$ and $1$?


Prove that the difference between any two odd perfect squares is divisible by 8

[hide]We have two odd integers. Let them be $2n+1$ and $2n+3$,where n is an integer and n isn't -1.
$(2n+1)^2=4n^2+4n+1$
$(2n+3)^2=4n^2+12n+9$
Take the difference of the two to get:
$8n+8$, which is divisible by 8 for all integers n.

Obviously, if the two odd integers were -1 and 1 it wouldn't work, but I'm not sure what "perfect" means, so it might be included in there.[/hide]

[hide]let our two odd integers be (2x+1) and (2x-1).
so the difference of the squares of these numbers:
((2x+1)-(2x-1))((2x+1)+(2x-1)) = (2)(4x) = 8x...
regardless of the modulo 8 residue of x, 8x cong. to 0 (mod 8).
.: the difference of the squares of two consecutive odd integers is divisible by 8[/hide]
[quote]Obviously, if the two odd integers were -1 and 1 it wouldn't work, but I'm not sure what "perfect" means, so it might be included in there.[/quote]
Of course it would work! 0 is divisible by 8. and perfect square means: the square root of a perfect square is an integer

Hm, somehow I thought 1-1=2. That's what tripped me up.

[quote]1-1=2[/quote]
interesting...so I would assume 1+1 = 0 and we are dealing with an imaginary function where: a+b really is a-b and a-b is really a+b.

[quote="h_s_potter2002"]I tried another way.

[hide]We can call one number $2x+2y+1$, and another $2x+1$, where $x$ and $y$ are integers.  The difference of their squares is $4y^2+8xy+4y$, which factors out to $4y(y+2x+1)$.  Now, we can consider two cases, if $y$ is odd, or if $y$ is even.

If $y$ is odd, then $y+2x+1$ has to be even, and $4$ times an even number is always divisible by $8$.

If $y$ is even, then $4y$ is divisible by $8$ becuase of the reason stated above.[/hide][/quote]

Hm, I like your way better. At first, I also thought that the odd numbers that were going to be squared had to be consecutive. Luckily, it all worked out for the best.

Said in simple terms: [b]Square of any odd number is 1 modulo 8.[/b]

Why because: (2n+1)^2 = 4n^2+4n+1 = 4*n*(n+1) +1, Now either n or n+1 is divsible by 2, so 4n(n+1) is divisible by 8.

Hence the difference is always multiple of 8.


geometry
geometric transformation
reflection
search
articles
algorithm
algebra

Hey everyone.  We've now spent a week as the National Online Math League, and one thing is quite clear:

The new name really sucks.

National Online Math League is too long, too stuffy, and too serious.  So the question is this:  what new, snappy, possibly humorous name should we give the league?

Here are the rules:

1. No obscenity
2. Should reflect the cleverness and creativity of our players
3. Should be shorter than National Online Math League
4. Should be funny and/or catchy (preferably both)
5. Should have a mathy flair to it

All right!  I want to hear your suggestions.  Go to work!

I forgot to mention, I don't want it to have my name in it.  I want this to be a project that other people can get in on in the future, and I don't think it will attract people as much if it's named after me.

I don't see the problem with NOML.  For example, Art of Problem Solving is a good name for a book, but a terrible name for an online math community.  However, the abbreviation AoPS, the most common usage, works well.  "National Online Math League" is unwieldy, but NOML (rhymes with pommel when spoken as an acronym) does an admirable job.

I think I agree with Thoobik -- I mean, AJHSME wasn't a nice name for a contest, either.  I vote stick with it but only use the acronym.  Sort of like the SAT, for example.

But IMB (pronounced like limb) sounds nowhere as cool as NOML (pronounced like pommel)...

Besides, if we wanted Internet something or another, there's the grand AHSIMC

Okay, so based on what I'm hearing, "I actually liked NOML" should be an option.

Obviously if I hated it I would never have chosen it.  I'm not deeply attached to it but I think it's okay.  But the people who aren't in favor of it have been much louder than those who are.

I vociferously clamour in favour of NOML!

Ehh.  NOML doesn't do it for me.  How about we call it "The Big MAC (MAthematics Competition)."  Or anything but NOML.

I commend you for making this type of evaluation, as this is something that we recently went through with our competition, the AHSIMC, in branding it the iTest.

As our lawyers have described, purely descriptive names are very difficult to defend on an infringement basis, so you're also doing yourself a favor to come up with something catchy in this regard.  (I think a lawyer would have had a heart attack at seeing your previous ESPN web design...I would definitely recommend staying away from that in the future, for the sake of all the work you're doing.)

While I can't offer you a suggestion for a new name, all I'll suggest is to spend some time thinking about the image you want your new name to conjure up in your current and future participants.  You seem to do a good job fostering a sense of community among your league members - perhaps that could be a starting point.

I also have to mention that I noticed your reference to your league's lack of "lavish gaming consoles" to award to participants, an obvious reference to my competition.  There's nothing lavish about a $150 prize.  At all.  It just underscores the absolutely abysmal state of prizes that are awarded for most math competitions, a topic I could talk about at length.

A prize isn't very important for your league, as like I said earlier, its all about the camraderie among your participants and having pride in beating these other people you know who are in the league, as well as representing your area.  But one of the best prizes I've ever seen is the Mandlebrot's decks of playing cards they give out...inexpensive and clever.  Let me suggest to you that something like that makes a lot of sense, and is MUCH better than a certificate or ribbon or whatever at the top end of your competition.

Hope this helps.  Good luck.

he probably does. even I can think of an alternative (like MERP) and I'm not even trying

Come now, we want a name which keeps some self-respect.  MERP and its like are right out.

I already suggested "The Big MAC (MAth Competition)" but I suppose that's bad because of legal issues (as if McDonald's would sue us for defaming their name or something).

Maybe iMac (Internet MAth Competition).  But again some lawyers might hunt us down and kill us or something.

Perhaps even T-MAC (The MAth Competition).  If you don't know who Tracy McGrady is, you fail at life.

MAPS (Math And Problem Solving)

PSI (Problem Solving on the Internet).  By the way, my zealot owns your 4 zerglings and/or 3 marines.

I know there exists [u]some[/u] creativity out there...

i dont find anything wrong with NOML but i just want to add that not evry1 in it lives in america :D

Leg enhancements+Level 3 shields/ground weapons/ground armor own all.  And then you break out the goons.

if we do iMac we'll have apple breathing down our backs (anyways, iMac has been taken by Apple already)
how abou iMath...
or The Bowl (i know stupid... but with capital letters, it sounds impressive)
or something that will become engrained into our memories... like ARML or AIME, etc, etc, etc.

is there any way I can sign up for this thing? perhaps next year?

Send Cody an email and you'll get sent to the draft.

NOML sounds alright... although it would look better as NoML(i just like it when acronyms have smaller cases in between).

how about BIG 6PAC OF DOOMMMMMMMMMMMMMMMMMMMMM

...noML looks allright.... emphasizing ML... but it also looks bad because it looks like the name says:

no ML!!!!

whoa I just went to a really funny website


conics
parabola
USAMTS
pigeonhole principle
induction
calculus
geometry

Before I start the discussion, I must ask a question:
When can we start discussing?

[quote="perfect628"]Last year he posted the problems online the day after, and then discussion was allowed.[/quote] I posted the problems last year because I was the question writer last year.  There is a different question writer each year.  Feel free to discuss the contest!

so, what did you guys get for the last one? (number eight)
I think there were two cases - that it's either an intersection of two lines or a single intersection of two parabolas, and it was sort of easy to miss one of them...
so, the right answer was 8pi, i think...

Thoughts/Outlines on each problem:

[hide="1"]1a was one of the three good problems on the test, albeit with a well-known idea. Basically you know that Q is the midpoint of arc BC by inscribed angles so OQ is the perpendicular bisector of BC, and also <OQT=90, so the result follows. 1b is pretty boring; just connect the centers and get some right angles. Computation.[/hide]

[hide="2"]So I've totally never seen these problems before. a is 499 and b is 8040.[/hide]

[hide="3"]3a was just a silly exercise, 1000/1019. 3b was another one of the three good problems; you can take f(1)=a+b+c and f(0)=c and if c>0 by the intermediate value theorem you have a root between 0 and 1.[/hide]

[hide="4"]4a was trivial, you have -1 and 1 mod 6, and 4b was the last of the good problems. Basically you get that you need P1+P2/2 to be prime, which is between P1 and P2, contradiction. 4c/d were just stupid casework; the answers are 14 and 38 respectively.[/hide]

[hide="5"]This is just an annoying exercise that's error-prone. Again, not an original idea. The answer is like $ \dfrac{3^{d}\plus{}1}{3^{d}\minus{}d}$ where $ d$ is the number of dimensions.[/hide]

[hide="6"]Another unoriginal one! USAMTS 18.1.4! It's a decent problem, but too many people have seen this idea, even if it wasn't from USAMTS. By Pigeonhole some two colors in each column are the same and by Pigeonhole again you get that some two columns have the same two places and colors for the two same colors.[/hide]

[hide="7"]You guessed it...unoriginal. I'd be surprised if less than half the competitors had seen this problem before. Just induction on the left side and for the right side strenghten the bound to 2sqrtn-1/sqrtn and induct. Or you can just use calculus, which also kills it.[/hide]

[hide="8"]Oh darn you can have two lines? I did not know that. Well apart from realizing that (which is just kind of silly because all it's testing is whether or not you get caught by some trick, evidently I did), this problem is just pure computation.[/hide]

So I got all of them except that second case on #8. But the test was way too easy; I finished in less than half the time and even re-copied all of my solutions. Apart from being easy, the problems were pretty uninteresting and most of them were well-known ideas (or well-known problems themselves).

This competition was way too easy.  Other than the trick on #8, there's really no other hard problem.  I'm predicting that there's going to be a ton of ties in top 20 based on whether you saw the trick on #8 or not.

[hide="And I did something a bit different for #7"]
$ \frac{1}{2\sqrt{a}}>\frac{1}{\sqrt{a}\plus{}\sqrt{a\plus{}1}}\equal{}\sqrt{a\plus{}1}\minus{}\sqrt{a}$

and

$ \frac{1}{2\sqrt{a}}<\frac{1}{\sqrt{a}\plus{}\sqrt{a\minus{}1}}\equal{}\sqrt{a}\minus{}\sqrt{a\minus{}1}$

Both sums telescope.[/hide]

Darn on the #8 thing...Also on #8: I did not see it until about 6 minutes were left in the test...I solved the two parabola case, but probably wasn't very clear, and missed the two lines. Otherwise I finished 1-7 in very little time...I'd seen essentially all the ideas before.

Wait...
On number 8, the two $ x^2$ coefficients were $ sin^2$ and $ cos^2$, right?
They can't both be 0...

On the other hand, they can be equal, in which case the parabolas aren't tangent but still only intersect once...

I qualified with a 128, but forgot about round two and missed it . :|
What did I miss? Were there prizes?


It is equivalent to
(x-y) 2 (11x 2  + 18xy + 11y 2 )/80  \geq  0 which clearly holds since x 2  + y 2   \geq  -2xy, so that the second factor is at least 2(x 2  + y 2 ).

Pierre.


[hide="Answer"]I'm not entirely sure how to use latex, so I'll just post the answer: a = 2b/3.[/hide][/hide]


geometry rectangle MATHCOUNTS ratio rotation AMC USA(J)MO

(Remember, MathCounts problems are for middle schoolers.)

The length of a rectangle is $\ell$ and its width is $w$.  Find the ratio of the volume of the solid generated when the rectangular region is rotated around its length to the volume of the solid generated when the rectangular region is rotated around its width.

[hide] using the formula for the volume of a cylinder, it's width/length.[/hide]

Yep.  By the way, are you participating in MathCounts this year?  Because you have qualified for USAMO, I would guess that you would make a strong contender for MathCounts.

Lol.  USAMO does not equal MathCounts.

I finished 7th place at state mathcounts.

But none of those silly goose heads who beat me got into the USAMO, so what now!!!!!!!

Mr. Crawford reassured me that my USAMO qualification is far more significant than making it into national mathcounts, so now really what now!!!!!  :)

Yes, USAMO is more impressive.  Congrats!

(But are you sure those other students are silly goose heads? ;) )


function
inequalities unsolved
inequalities

Find the minimum of

$ \frac{c}{b\plus{}a} \plus{}\frac{b}{c\plus{}d}$

where $ a\plus{}d\equal{}b\plus{}c$, $ c,b>0$, $ a,d \ge 0$

[quote="malcolm"]Find the minimum of

$ \frac {c}{b \plus{} a} \plus{} \frac {b}{c \plus{} d}$

where $ a \plus{} d \equal{} b \plus{} c$, $ c,b > 0$, $ a,d \ge 0$[/quote]

without loss of generality, we can assume that $ b \le c$.

(1), If $ b \equal{} c$, then we have

$ \frac {c}{b \plus{} a} \plus{} \frac {b}{c \plus{} d} \equal{} b\left(\frac {1}{b \plus{} a} \plus{} \frac {1}{b \plus{} d}\right) \ge \frac {4b}{2b \plus{} a \plus{} d} \equal{} 1$,

when $ a \equal{} b \equal{} c \equal{} d$ get "=";

(2), If $ b < c$, then $ \frac {c}{b \plus{} a} \plus{} \frac {b}{c \plus{} d} \equal{} \frac {c}{b \plus{} a} \plus{} \frac {b}{b \plus{} 2c \minus{} a}$,

let $ f(a) \equal{} \frac {c}{b \plus{} a} \plus{} \frac {b}{b \plus{} 2c \minus{} a}$, $ 0 \le a \le b \plus{} c$, we have

$ f'(a) \equal{} \frac {(b \minus{} c)a^2 \plus{} 2(b^2 \plus{} bc \plus{} 2c^2)a \plus{} b^3 \minus{} b^2c \minus{} 4bc^2 \minus{} 4c^3}{(a \plus{} b)^2 (a \minus{} b \minus{} 2 c)^2}$,

setting the roots of $ f'(a) \equal{} 0$ are $ a_1,a_2$ and assume that $ a_1\le a_2$, then we have

$ a_1 \equal{} \frac {{b^2 \plus{} bc \plus{} 2c^2 \minus{} 2(b \plus{} c)\sqrt {bc} }}{{c \minus{} b}} \ge \frac {{b^2 \plus{} bc \plus{} 2c^2 \minus{} (b \plus{} c)^2 }}{{c \minus{} b}} \equal{} b > 0$

$ a_2 \equal{} \frac {{b^2 \plus{} bc \plus{} 2c^2 \plus{} 2(b \plus{} c)\sqrt {bc} }}{{c \minus{} b}} > \frac {{c^2 \minus{} b^2 }}{{c \minus{} b}} \equal{} b \plus{} c$

thus,

(i), if $ b \plus{} c \ge a_1 \iff 4b^3 \plus{} 4b^2c \plus{} bc^2 \ge c^3$, then we have $ f(a) \ge f(a_1) \equal{} \frac {1}{2} \plus{} \frac {\sqrt {bc}}{b \plus{} c}$;

(ii), if $ b \plus{} c < a_1 \iff 4b^3 \plus{} 4b^2c \plus{} bc^2 < c^3$, then we have $ f(a) \ge f(b \plus{} c) \equal{} \frac {b}{c} \plus{} \frac {c}{2b \plus{} c}$.

let $ c \equal{} kb,(k > 1)$, i.e.

(i), when $ 4 \plus{} 4k \plus{} k^2 \ge k^3$, we have

$ \frac {c}{b \plus{} a} \plus{} \frac {b}{c \plus{} d} \ge \frac {1}{2} \plus{} \frac {\sqrt {k}}{1 \plus{} k}$;

(ii), when $ 4 \plus{} 4k \plus{} k^2 < k^3$, we have

$ \frac {c}{b \plus{} a} \plus{} \frac {b}{c \plus{} d} \ge \frac {1}{k} \plus{} \frac {k}{2 \plus{} k}$.

so we just need to find the minimum of this function:
\[ g\left( k \right) \equal{} \left\{ {\begin{array}{*{20}c} {\frac {1}{2} \plus{} \frac {{\sqrt k }}{{1 \plus{} k}},} & {4 \plus{} 4k \plus{} k^2 \ge k^3 \wedge k > 1} \\
{\frac {1}{k} \plus{} \frac {k}{{2 \plus{} k}},} & {4 \plus{} 4k \plus{} k^2 < k^3 \wedge k > 1} \\
\end{array}} \right.\]
Actually,

$ 4 \plus{} 4k \plus{} k^2 \ge k^3 \wedge k > 1 \iff k \in (1,k_0]$ and $ 4 \plus{} 4k \plus{} k^2 < k^3 \wedge k > 1 \iff k \in (k_0, \plus{} \infty)$, where $ k_0 \approx 2.8751$ is a root of $ 4 \plus{} 4k \plus{} k^2 \equal{} k^3$,

easy to prove $ g(k)$ is continuous function, and $ {\frac {1}{2} \plus{} \frac {{\sqrt k }}{{1 \plus{} k}}}$ is decreasing in $ [1,k_0)$,

and $ \left( {\frac {1}{k} \plus{} \frac {k}{{2 \plus{} k}}} \right)^\prime \equal{} \frac {{k^2 \minus{} 4k \minus{} 4}}{{k^2 (k \plus{} 2)^2 }} \equal{} 0 \Rightarrow k \equal{} 2 \plus{} 2\sqrt 2 > k_0$ we get $ {\frac {1}{k} \plus{} \frac {k}{{2 \plus{} k}}}$ is decreasing in $ (k_0 \plus{} 2 \plus{} 2\sqrt 2)$ and increasing in $ (2 \plus{} 2\sqrt 2, \plus{} \infty)$,

therfore, we get that $ g(k)$ is decreasing in $ (1 \plus{} 2 \plus{} 2\sqrt 2)$ and increasing in $ (2 \plus{} 2\sqrt 2, \plus{} \infty)$, so we have

$ g(k)_{\min} \equal{} g(2 \plus{} 2\sqrt 2) \equal{} \sqrt {2} \minus{} \frac {1}{2}.$

when $ a \equal{} 3 \plus{} 2\sqrt 2, b \equal{} 1, c \equal{} 2 \plus{} 2\sqrt {2},d \equal{} 0$ get the minimum.

done.

awesomeness, a little more bashy than the official solution, but the official one doesn't cover all the cases. tyvm!


Let p be a prime. Find all solutions in positive integers of the equation :

2/a + 3/b = 5/p

$ \frac{2}{a}\plus{}\frac{3}{b}\equal{}\frac{5}{p}$
$ \Leftrightarrow (5b\minus{}3p)(5a\minus{}2p)\equal{}6p^2$
$ \Rightarrow 5b\minus{}3p|6p^2$
Consider $ p\geq 5$
[u]Case 1[/u] $ 5b\minus{}3p\equal{}1,5a\minus{}2p\equal{}6p^2$
Imply that $ b\equal{}\frac{3p\plus{}1}{5},a\equal{}\frac{6p^2\plus{}2p}{5}$
So $ 5|3p\plus{}1$
[u]Case 2[/u] $ 5b\minus{}3p\equal{}2,5a\minus{}2p\equal{}3p^2$
..
Solve with other case $ \tau(6p^2)\equal{}12$ case
The case $ p\in\{2,3\}$ is easy.


which of the following is a factor of  12 a^2 + 13 ab -35b^2

a) 4a +5b  b) 3a-7b  c) 5a-4b  d) 7a +3b  e) none of these

Factoring the expression, we have $ (4a\minus{}5b)(3a\plus{}7b)$. Since none of the answer choices are factors, the answer is E.

One little advice:
Why don't you post in the Middle School Contests (MATHCOUNTS, AMC 8, MOEMS, Ages 10-13) forum?Besides,your problems are somewhat of the same type.Why don't you actually try to look up in a text-book instead of posting away madly.I mean,you come to the forum and post 5 problems at a time.I am sure that would be helpful to you.No use having the highest no of posts here.If you have a problem you have people to help you.


Νίκο, ευχαριστούμε.

  α) Που πήγε το 4ο θέμα ; Δε φαίνεται.

  β) Μήπως έχεις να στείλεις και τα θέματα των μεγάλων ;

  γ) καλή μελέτη και [b]καλή επιτυχία[/b]!

   Από γεωμετρία να διαβάσετε καλά << Ομοιότητα - μετρικές σχέχεις - δύναμη σημείου και εγγράψιμα>>.Τα προτιμούν οι Ρουμάνοι και μπαίνουν συχνά.

  Μπάμπης

κ. Στεργίου

 Έτσι ακριβώς δόθηκε το χαρτί απο την ΕΜΕ .
 Μάλλον έγινε μπέρδεμα με τα νούμερα .Τα θέματα είναι τέσσερα.
 Tα θέματα των μεγάλων τα έχει βάλει σε τόπικ ο Σιλουανός .


inequalities
real analysis
real analysis theorems

Suppose f :[a,b]->R satisfies:
for every e>0, there is d>0 : Sum (k=1_n) (b_k-a_k) <d => Sum (k=1_n) |f(b_k)-f(a_k)|<e

(ie disjointness is removed in the definition of absolute continuity)

Prove f is of Lipschitz type.

Let $\epsilon >0$. Then $\exists d>0$ satisfying the hypothesis.
Now assume the result to be false; this implies in particular: $\exists x,y \in [a,b], |f(x)-f(y)|>\frac{2 \epsilon}{d}(x-y)$ (we can take $x>y$).

Let $n$ be a positive integer such that $\frac{x-y}{n} < \frac{d}{2}$.

Then partitioning $[x,y]$ into $n$ intervals with the same length, at least one of them (say $[j,k]$) is such that $|f(j)-f(k)| > \frac{2 \epsilon}{dn}(x-y)=\frac{4 \epsilon}{d}(j-k)$.

Let $N$ be the greatest positive integer satisfying $N(j-k) \leq d$. We then have $d-N(j-k)<\frac{d}{2}$, i.e $\frac{N(j-k)}{d}>\frac{1}{2}$, and by hypothesis:

$N|f(j)-f(k)|<\epsilon$, but also $N|f(j)-f(k)|>N\frac{2 \epsilon}{d}(j-k) > \epsilon$. Combining the two inequalities yields $\epsilon>\epsilon$, which is absurd.


yes obviously it is 0...
but the proof is necessary :)


probability
floor function
continued fraction
advanced fields
advanced fields open

Hello,

I think this should be a known resoult for someone, but it does not seem clear to me. Suppose that you have $X$ an absolutely continuous random variable, with probability density $g$, and suppose that $\{\bullet\}$ is the fractional part (i.e, $x=\lfloor x\rfloor+\{x\}$). Then, the question is, given $\alpha>0$, are the random variables $\{X\}$ and $\{\alpha X\}$ independent. The answer, if $\alpha\in\mathbb{Q}$ is no. My advisor says that if $\alpha$ is irrational, then the answer is still no, but I am not sure about it. I have tried (and I haven't succed yet) to prove either way. I am sure that the fact that $\{n\alpha\}_{n\in\mathbb{N}}$ is dense in $[0,1]$ forces the rvs to be independent, but   :huh:  :huh:  :huh: :?  :?  :?  :what?:  :what?:  :what?:  :wacko:

Consider some interval $[a,b]$ such that $\lfloor a\rfloor=\lfloor b\rfloor$, $\lfloor \alpha a\rfloor=\lfloor \alpha b\rfloor$ and $X$ has a positive probability c of being in $[a,b].$
Then, in the joint distribution $(\{X\},\{\alpha X\})$, the diagonal segment from $(\{a\},\{\alpha a\})$ to $(\{b\},\{\alpha b\})$ has zero measure and positive probability. Since $(\{X\}$ and $\{\alpha X\})$ are absolutely continuous, the joint distribution would be absolutely continuous if they were independent. The contradiction here proves that they cannot be independent for any choice of $X$ and $\alpha$.

JMerry,

Thank you very much. I feel somehow embarased, because your solution, even when it is very clever, it is also very simple, and I had been struggling two days with Weyl's distribution theorem and continued fraction approximations, and that kind of stuff  :blush: , but it is solved.

Thanks. Best regards,


Find the sum of the absolute value of all solutions to the equation $ (\log_{10}  x^2)^2 \equal{} \log_{10} x^4$

$ (2\log x)^2\equal{}4\log x$

$ 4(\log x)^2 \equal{} 4\log x$

This is satisfied when $ \log x \equal{}0,1$, so $ x\equal{}10,1$. Also, we need to include the negative values, since in the first equation we have a $ x^2$ and $ x^4$ term. So, the sum of the absolute valuse is $ 22$