I need some help regarding the following questions; I have a solution to each of these but there's something missing to each of them:

1) If X1xX2 is a topological space and t is in X1, then show that (t)xX2 considered with the subspace topology induced from the product topology on X1xX2 is homeomorphic to X2:

The subspace topology makes the mapping i:{t}×X_2→X_1×X_2 continuous, and the product topology is the smallest topology that makes the mapping π:X_1×X_2→X_2 continuous, so their composition mapping π∙i:{t}×X_2→X_2 is also continuous. I think I just need to prove their inverse is also continuous, which I'm trying to figure out how to.

2) Prove X is connected iff the only continuous functions f:X→Z are constant.

I can prove the 'only if' bit. With the 'if,' I tried contradiction by supposing  X is not connected and so can be split into 2 non-trivial closed and open disjoint sets, but I just can't prove they cannot be non-trivial in the end.

Can someone please help me with these questions and tell me if I'm going down the right direction?
Thanks

1. let $ f : X\to X\times Y$  st $ f(x) \equal{} (x,y_{0})$  then $ f$ is continuous
then clearly $ f(X) \equal{} X\times y_{0}$, $ f$ is 1-1 and onto $ X\times y_{0}$,
to prove that $ g \equal{} f^{ \minus{} 1}$ is continuous, take $ O$ to be open in $ X$ then $ g^{ \minus{} 1}(O) \equal{} O\times y_{0} \equal{} (O\times Y)\cap (X\times y_{0})$ which is open in $ X\times y_{0}$

2. $ f : X\to \mathbb{Z}$ is continuous then $ f(X)$ is connected in $ \mathbb{Z}$, and it is known that connected components of $ \mathbb{Z}$ is just one-point sets , so $ f$ must be constant .

now consider the other direction, $ X \equal{} A\cup B$ where $ A, B$ are non-empty [b]disjoint[/b] subset of $ X$
define $ f : X \equal{} A\cup B \to \mathbb{Z}$ , with $ f(A) \equal{} \{0\}, f(B) \equal{} \{1\}$ which are open in $ \mathbb{Z}$  then clearly $ f$ is continuous, not constant.


Does anyone know how to make an image that looks nice (isn't pixilated) for a wide range of zoom levels? (Like 100%, 125%, 150%.)

Whenever I use an image in one of my PDF documents, it looks great at 100%. However, it becomes ugly as soon as you get past 115%.

If you're not sure what I'm talking about, try zooming in a lot on this document. The images still look really nice.
[url=http://www.usamts.org/Tests/USAMTSProblems_18_1.pdf]USAMTS Round 1 Problems[/url]

But this document was created using LaTeX, so images are .eps, that's why they looks nice... This is a type of images called [b]Vectorial Images[/b].

what is the best program (but not really complicated and involved) to use to make image files and geometry renderings to include in latex files?

is that program MetaPost useful or is it too complicated to learn just for the sake of making a few good drawings?

Hmmm...Just a word of warning...if you build files with pictures in JPG, it seems the pictures get progressively worse. I used paint on this one picture...and saved it in JPG (actually...maybe it's JPEG...uh...no sure what's right...) and after building my latex document 4 times, it was like someone sprinkled pepper on the image...I guess that's the best way I can describe it. Has anyone else had this problem?

I use a french software called [url=http://perso.orange.fr/Fradin.Patrick/TeXgraph/Telecharger.html]TeXgraph[/url], but there is other soft like gnplot
http://www.mathlinks.ro/Forum/viewtopic.php?t=111694

[quote="DPatrick"]Correct -- the images in the USAMTS problem document that you cite are EPS (Encapsulated PostScript) files created using [url=http://www.tug.org/metapost.html]MetaPost[/url].  So they will look sharp at any resolution.[/quote]
How can I obtain MetaPost?
I've visited the link you've given and downloaded the file, but my computer can't seem to open a .tar file. Is there any program I can obtain that can do this, or is there another way I can obtain the program?
(Also, I have Windows XP, in case you were wondering.)
Thanks!

[url=http://www.winzip.com/]WinZip[/url], [url=http://www.7-Zip.org/]7-zip[/url], [url=http://www.zipgenius.it/eng/index.php]ZipGenius[/url] and other compression programs should be able to extract from a tar file. They are also a lot faster at zipping & unzipping than XP!


search
inequalities proposed
inequalities

Prove that the following ineq holds for all positive reals $ x,y,z$

$ \frac{x}{\sqrt{x\plus{}y}}\plus{}\frac{y}{\sqrt{y\plus{}z}}\plus{}\frac{z}{\sqrt{z\plus{}x}}\leq \frac{5}{4} \sqrt{x\plus{}y\plus{}z}$

Already posted. See [url]http://www.mathlinks.ro/viewtopic.php?t=152848[/url].

[quote="FOURRIER"]Prove that the following ineq holds for all positive reals $ x,y,z$

$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq \frac {5}{4} \sqrt {x \plus{} y \plus{} z}$[/quote]
See at http://www.mathlinks.ro/viewtopic.php?search_id=1688548402&t=1459


linear algebra
matrix
inequalities
combinatorics proposed
combinatorics

One tornament have n players. Each player play with other exactly 1 time.
For each 2 players A,B exists exactly t players that loses to A and B.
Prove that n = 4t + 3.

Please, give me a solution for this problem.  :)

[quote="grobber"][quote="Lopes"]
For each 2 players A,B exists exactly t times that loses to A and B.
[/quote]

Do you mean $t$ players instead of $t$ times? :?[/quote]

sorry,sorry..
see again and help me, please.

This problem is really hard?  :(
Please, help me  :(

Hmm...I don't think that to hurry people is the best way to receive an answer...

Pierre.

I think we should be able to prove, that every team won the same number of games and therefore lost the same number of games, so we would have, every team won $(n-1)/2$ games. So counting the number of trios $(a,b,c)$ such that $a$ and $c$ defeated $b$, we would have $\binom{n}{2}t=n\binom{\frac{n-1}{2}}{2}$ from where we would obtain $n=4t+3$. However, i cant prove anything about the regularity of the degrees, so i guess I will have to wait until someone post a Linear Algebra solutions, becuase it is the only that seems to work, and it seems because it can be traslated into nice matrix condition I guess becuase i dont now LA at all.

ok..a thanks  :)
when you solve post your solution, please  :)

ok here is one inequality:
 count the number of triples $(a,b,c)$ such that $a \rightarrow c,b \rightarrow c$.choosing $c$ first the number of such triples equals $\displaystyle \sum_{v \in V} (d^-(v))^2$(where $d^-(v)$ denotes the indegree).on the other hand, we can choose $a$ and $b$ first.IF $a=b$ then the number of such triples is $\displaystyle \sum_{v \in V} d^-(v)=n(n-1)/2$.if $a \neq b$, the number of triples is $n(n-1)t$,so we have
 $\displaystyle \sum_{v \in V} (d^-(v))^2=n(n-1)t+n(n-1)/2$. now
 $\displaystyle \frac{\sum_{v \in V} (d^-(v))^2}{n}\geq (\frac{\sum_{v \in V} d^-(v)}{n})^2$,so we have $(n-1)t+(n-1)/2 \geq (\frac{n-1}{2})^2 \Rightarrow t+1/2\geq (n-1)/4 \\ \Rightarrow 4t+3 \geq n$.
Now if there is equality attained then there must be equality in the inequality above.which comes to saying what Pascual2005 said.

Maybe I am missing something but if the equality attaines then we are done  :?

This solution is not complete..
please, give me a complete solution.

I have got a weak lower bound :(
Indeed, $\frac{n(n-1)(n-2)}{6}\geq \frac{n(n-1)}{2}\cdot t$, so $n\geq 3t+2$.

ok here is the remaining part of the solution:
  firstly we already have $n \leq 4t+3$.now we'll prove that for every vertex $v,d^+(v) \geq 2t+1$;then summing over all the vertices we get $n(n-1)/2 \geq n(2t+1) $ and that is the reverse inequality.
 to see this consider any vertex $v$ and the set $N^+(v)=\{w|v\rightarrow w\}$;for each fixed $w \in N^+(v)$ we have that there are exactly $t$ vertices $u$ such that $v\rightarrow u,w\rightarrow u$. but then since $N^+(v)$ includes all the vertices such that $v\rightarrow u$ it follows that in the induced subgraph on $N^+(v)$, every vertex has outdegree atleast $t$.then again summing over all $w \in N^+(v)$ we have $\displaystyle \sum_{w \in N^+(v)} |N^+(w)\cap N^+(v)| \geq d^+(v)t \Rightarrow d^+(v)(d^+(v)-1)/2 \geq d^+(v)t \\ \Rightarrow d^+(v) \geq 2t+1$
and we are through.


$1^{\circ}.\ \triangle ABC\Longrightarrow \underline {\overline {\left| A=90^{\circ}\Longleftrightarrow \ b+c=2\cdot (R+r)\Longleftrightarrow 1+cosB+cosC=\frac{r_a}{R}\right| }}. $
I ask for a geometrical interpretation.

$2^{\circ}$. Ascertain the form of the triangle ABC for which exists the relation $cs(s-c)=S(a+b-R).$

I believe the only one you can get is by actually constructing triangle s etc....

When I will return here (15.o3.o6, U.S.A.) I will offer the geometrical construct.
Is it O.K. ? Good luck !


[quote="mat12"]does anyone know how to compute cyclic generators?
for example
how many cyclic generators  of (Z/227Z)* are there?[/quote]

I can't understand your notation can you explain better?
Thx  ;)

TCM

I think he wants to know how many generators the multiplicative group of invertible elements of $\mathbb Z_{227}$ has. For a prime $p$, there are $\varphi(p-1)$ generators for its multiplicative group of non-zero residues (in general, there are $\varphi(d)$ elements of order $d$ for all divisors $d$ of $p-1$ in that group).

P.S. $227$ is a prime, right? :)

mat12: DON'T post same topic in five different subforums !!  :mad:


search
number theory proposed
number theory

Let $ k$ and $ d$ be integers such that $ k>1$ and $ 0 \le d<9$.
Prove that there exists a positive integer $ n$ such that the decimal representation of the number $ 2^n$ has the digit $ d$ in the position $ k$(from the right) :)

A more general result here (post #3) :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1604429076&t=23417

Pierre.

Do anyone has a simple solution?
(I don't understand in #2 solution,It's too hard for me!)

Pierre, the post you are referring to doesnt seem to tell much about the above problem!


LaTeX
calculus
real analysis
trigonometry
algebra
function
domain

I'm teaching an undergraduate Fourier Analysis course this spring, and I'm using my own notes. If any of you are interested, I have put the notes on line at the following address: http://www.csulb.edu/~kmerry/

When you get there, click on "Notes for Fourier Analysis." There's a separate file for each chapter, and the whole thing is about 150 pages (of fairly kludgy $ \text{\LaTeX},$ from a file originally written in MS Word.) It's posted in .pdf form.

Some comments:

1. The notes are heavily influenced by and borrow somewhat from a textbook by David Kammler.

2. The official prerequisites for the course are just a differential equations course, although a student should have a little more than that to get something from the course. The audience usually includes a physics major or two - they're not all math majors. In particular, I am not insisting on a real analysis or advanced calculus course as prerequisite, and I'm certainly not expecting students to know the Lebesgue integral. (Which doesn't stop me from using norms: $ \|f\|_1,\|f\|_2,\|f\|_\infty.$)

3. The notation depends on complex exponentials, not sines and cosines. But I include a section up front introducing that.

4. I would characterize my attitude towards serious analysis in these notes as "Eulerian": calculate first, justify later, if at all.

5. The notes attempt to do Fourier analysis on four different domains simultaneously, as one subject. The four domains are $ \mathbb{R},\mathbb{T}\ ( = \mathbb{R}/(P\mathbb{Z})),\mathbb{Z},$ and $ \mathbb{Z}/(N\mathbb{Z}).$

6. I include a section on distributions.

If you find the places in which I've responded on this site to a question about Fourier series or the Fourier transform, you'll see that I've generally used language and attitudes that come from these notes.

I fixed a few typos and posted the cleaned-up versions in the same place.

Perhaps this is a lot to ask, but would it be possible to provide solutions for the problems, or at least sketches of solutions?


Prove that: $ \left ( C_n^1\right)^2\plus{}2\left ( C_n^2\right)^2\plus{}...\plus{}n\left ( C_n^n\right)^2\equal{}\frac{(2n\minus{}1)!}{\left [ (n\minus{}1)!\right]^2}$

$ \sum_{i\equal{}1}^{n} i \binom{n}{i}^2 \equal{} n\sum_{i\equal{}1}^{n} \binom{n\minus{}1}{i\minus{}1}\binom{n}{n\minus{}i}$
And this is just $ n$ multiplied by the coefficient of $ x^{(i\minus{}1)\plus{}(n\minus{}i)}\equal{}x^{n\minus{}1}$ in the product $ (1\plus{}x)^{n\minus{}1}(1\plus{}x)^{n}\equal{}(1\plus{}x)^{2n\minus{}1}$
Which is $ \binom{2n\minus{}1}{n\minus{}1}$.
So we have $ n \cdot \binom{2n\minus{}1}{n\minus{}1} \equal{} n\frac{(2n\minus{}1)!}{(n\minus{}1)!n!} \equal{} \frac{(2n\minus{}1)!}{\left [ (n\minus{}1)!\right]^2}$.

(after the first manipulation it becomes just a special case of Vandermonde's identity)


[quote="Ulanbek_Kyzylorda KTL"]In an acute triangle ABC bisector AD is equal to AC and AD is perpendicular to OH, where O is circumcenter of triangle ABC and H is orthocenter. Find angles of triangle.[/quote]

Let's $ \angle A\equal{}4\alpha$
we have $ \angle CAD\equal{}\angle BAD\equal{}2\alpha$
$ \angle CAH\equal{}\angle DAH\equal{}\alpha$
$ \angle ACB\equal{} 90^o\minus{}\alpha$
$ \angle ABC\equal{}90^o\minus{}3\alpha$
It's known that $ \angle DAH\equal{}\angle DAO\equal{}\alpha$(becouse points $ H$ and $ O$ are isogonal conjugate with respect to $ \triangle ABC$) and we have $ AD \perp OH$ therefore $ AH\equal{}AO\equal{}R$
also it's known that $ AH\equal{}2R \cdot cos(4\alpha)$
So $ R\equal{}2R\cdot cos(4\alpha)$
or $ cos(4\alpha)\equal{}\frac{1}{2}$
$ 4\alpha\equal{}60^o$
$ \alpha\equal{}15^o$
$ \angle BAC\equal{}60^o$, $ \angle ACB\equal{}75^o$, $ \angle ABC\equal{}45^o$

AC=AD  :arrow: AN is height <CAN=<NAD. We know that  O is circumcenter, so OA=OB=r and OC=OA=OB. Denote intersection point of AD and OH K and let <KAO=b(beta) and <CAH=a(alpha).OA=OC :arrow: <ACO=<CAO hence 2a+b=4a-b  :arrow: b=a. From here we notice that triangle AHO is isosceles, so AH=OA=OC=OB
In triangle ACH   AC/sin(3a+90)=AH/sin(90-4a)   AH=AC*cos4a/cos3a
In triangle AOC   <AOC=180-6a  :arrow: AC/sin(180-6a)=OC/sin3a  :arrow: AC/sin6a=OC/sin3a  :arrow: AC/ 2*sin3a*cos3a= OC/sin3a   AC/2*cos3a=OC=AH  :arrow: AC/2*cos3a = AC*cos4a/cos3a  :arrow: cos4a=1/2 (a<90) a=15 so the answer is <A=60 <C=75 <B=45


linear algebra
matrix
vector
linear algebra unsolved

Show that there does not exist an orthogonal matrix with rational entries that reduces
xy to diagonal form.

That would imply orthogonal diagonalisatoin of
0  1/2
1/2  0

Compute the eigenvalues and then the eigenvectors
you will find that while the eigenvalues are rational you cannot find scalar multipels of them in the rationals that are unit vectors


inequalities
geometry
circumcircle
geometry proposed

Show that in any triangle $ ABC$ the following inequality holds: $ \frac{b^2\plus{}c^2}{m_a}\plus{}\frac{c^2\plus{}a^2}{m_b}\plus{}\frac{a^2\plus{}b^2}{m_c}\le 12R$ .

Let $ M$ denote the midpoint of $ BC$ and let the median $ AM$ cut the circumcircle $ (O)$ of $ \triangle ABC$ at $ P.$ From the power of $ M$ to $ (O)$ we get $ AM \cdot MP \equal{} BM \cdot MC \equal{} \frac {_1}{^4}a^2$

$\Longrightarrow  AP \equal{} AM \plus{} MP \equal{} \frac {AM^2 \plus{} \frac {1}{4}a^2}{AM} \equal{} \frac {\frac {1}{2}(b^2 \plus{} c^2) \minus{} \frac {1}{4}a^2 \plus{} \frac {1}{4}a^2}{m_a} \equal{} \frac {b^2 \plus{} c^2}{2m_a}$

Since a diameter  is greater or equal than the chord $ AP,$ we have $ 2R \ge \frac {b^2 \plus{} c^2}{2m_a}$

By cyclic exchange we have: $ 2R \ge \frac {a^2 \plus{} c^2}{2m_b} \ , \ 2R \ge \frac {a^2 \plus{} b^2}{2m_c}$

$ \Longrightarrow  12R \ge \frac {b^2 \plus{} c^2}{m_a} \plus{} \frac {a^2 \plus{} c^2}{m_b} \plus{} \frac {a^2 \plus{} b^2}{m_c}$


This is an old bonus question from my math class. Kinda easy but still fun.

I have eight supposedly gold coins.  One is a forgery, however, and is lighter than the others.  The only way to tell which one is to weigh them.

The science department has a great balance that will let me do that, but since they’re envious of my vast wealth (seven legitimate gold coins, after all!) they’re going to charge me $\$$ 100 for each use of the balance.  It’s the kind where you put objects on both sides and see if they balance

I get an anonymous e-mail telling me I can determine the bogus coin with only two weighings, but it doesn’t say how.  Help!  Tell me how!

[hide] Put 3 coins on each side of the scale. If they balance out, you know the fake is one of the two you didn't use. Weigh them against eachother to find the lighter one. If the original 6 didn't balance out, the fake is in one of the 3 on the lighter side. Weigh any two of those against eachother. If they balance, the one left out is the fake. If they don't, the lighter one is the fake. [/hide]


quadratics
algebra
function
domain
calculus
calculus computations

Draw the domain of $ (a,\ b)$ for which the quadratic equation with real coefficients : $ x^2\plus{}ax\plus{}b\equal{}0$ has more than one solution in $ \minus{}1<x<1$.

[quote="kunny"]Draw the domain of $ (a,\ b)$ for which the quadratic equation with real coefficients : $ x^2 \plus{} ax \plus{} b \equal{} 0$ has more than one solution in $ \minus{} 1 < x < 1$.[/quote]

The domain is V :

$ V \equal{} \{  M \left( x \; ; \; y \right) \in \Re ^2 \; : \; \left(  \minus{}1 < x < 1 \right) \wedge  \left( y < \frac{x^2}{4} \right) \wedge \left( y > x \minus{} 1 \right)  \wedge  \left( y > \minus{}x \plus{} 1 \right)  \}$

sorry ! I don't know how to paste the image on forum :D

[quote="shyong"][quote="brahman"][quote="kunny"]Draw the domain of $ (a,\ b)$ for which the quadratic equation with real coefficients : $ x^2 \plus{} ax \plus{} b \equal{} 0$ has more than one solution in $ \minus{} 1 < x < 1$.[/quote]

The domain is V :

$ V \equal{} \{ M \left( x \; ; \; y \right) \in \Re ^2 \; : \; \left( \minus{} 1 < x < 1 \right) \wedge \left( y < \frac {x^2}{4} \right) \wedge \left( y > x \minus{} 1 \right) \wedge \left( y > \minus{} x \plus{} 1 \right) \}$

sorry ! I don't know how to paste the image on forum :D[/quote]

I think is $ V \equal{} \{ M \left( x \; ; \; y \right) \in \Re ^2 \; : \; \left( \minus{} 2 < x < 2 \right) \wedge \left( y < \frac {x^2}{4} \right) \wedge \left( y > x \minus{} 1 \right) \wedge \left( y > \minus{} x \plus{} 1 \right) \}$[/quote]

uhm .... may be ... you're right ! :D

$ \minus{}1 < \minus{} \frac{a}{2} < 1 \;\; \Leftrightarrow \;\; \minus{}2 < a < 2 \;\; \left( it's \; not\; \Leftrightarrow \;\; \minus{}1 < a < 1 \right)$

It means "and."  For example, $ \{(x, y) \mid (x > 0) \wedge (y > 0)\}$ is the first quadrant of the Cartesian plane.

Thank you. Regrettably I haven't received the correct answer. First I thought this problem should be posted in Intermidate Forum, but this problem doesn't seem to be easy, so I have posted in the forum.

$ x^2 \plus{} ax \plus{} b \equal{} 0 \Rightarrow x \equal{} \frac { \minus{} a \pm \sqrt {a^2 \minus{} 4b}}{2}$
In order for $ \minus{} 1 < x < 1$ for both zeros:

[hide]$ \minus{} 1 < \frac { \minus{} a \pm \sqrt {a^2 \minus{} 4b}}{2} < 1$
$ \minus{} 2 < \minus{} a \pm \sqrt {a^2 \minus{} 4b} < 2$
$ a \minus{} 2 < \pm \sqrt {a^2 \minus{} 4b} < a \plus{} 2$

$ \Leftrightarrow a \minus{} 2 < \minus{} \sqrt {a^2 \minus{} 4b} \leq 0$ and $ 0 \leq \sqrt {a^2 \minus{} 4b} < a \plus{} 2$
$ \Leftrightarrow a^2 \minus{} 4a \plus{} 4 > a^2 \minus{} 4b \geq 0$ and $ 0 \leq a^2 \minus{} 4b < a^2 \plus{} 4a \plus{} 4$ and $ \minus{}2 < a < 2$
$ \Leftrightarrow a \minus{} 1 < b \leq \frac {a^2}{4}$ and $ \minus{} a \minus{} 1 < b \leq \frac {a^2}{4}$ with $ \minus{}2 < a < 2$

So the domain, (a,b), is the set of points such that: $ b \leq \frac {a^2}{4}$, $ b > \minus{} a \minus{} 1$, $ b > a \minus{} 1$, and $ \minus{}2 < a < 2$.[/hide]

Sorry, now your solution seems to be true, when I saw your solution, I couldn't see last line. So I tried to delete my post, but I couldn't that, that is inconvinient.
Anyway, you should consider the case for which the qudratic equation has only one solution.


find  all  positive integers  n such  that \[d_{5}^{2}+d_{6}^{2}-1 = n\] ,where \[\begin{array}{l}1 = d_{1}< d_{2}< ... < d_{k}= n \\ \\ \end{array}\] are  all  the  positive divisors  of  n.


Let S be the set of natural numbers with no prime divisors greater than 3 (i.e. the smallest elements of S are 1, 2, 3, 4, 6, 8, 9, 12). What is the sum of the reciprocals of S?

Let the sum be A(for answer, since S is already taken).

A=(1+1/2+1/4+1/8+...)(1+1/3+1/9+1/27+...)=2*(3/2)=3.

Euler used this method for all primes to prove in a very interesting way that there are an infinite number of primes. Being on the topic of there being an infinite number of primes, I love this proof:

:Sigma: n=1 to :inf: 1/n^2 = Product <we need a Pi like the :Sigma:> p prime (1-1/p^2)^-1=:pi::^2:/6. But that's irrational! Thus there are infinitely many primes in the product, which is all the primes. Yay!

Interesting. This problem appeared in a problem-solving section of a calculus textbook that an instructor posed. The solution was found in about 3 minutes, to his surprise.
Darn you ComplexZeta :)

In one of our ARML practices, we had a problem that was just like this. I think it was from the TRML.


Let $ a$, $ b$, $ c$ be three positive real numbers such that $ a\plus{}b\plus{}c\equal{}3$.
Prove that \[ \frac{a}{ab\plus{}1}\plus{}\frac{b}{bc\plus{}1}\plus{}\frac{c}{ca\plus{}1}\ge\frac{3}{2}\]

[quote="April"]Let $ a$, $ b$, $ c$ be three positive real numbers such that $ a\plus{}b\plus{}c \equal{} 3$.
Prove that
\[ \frac{a}{ab\plus{}1}\plus{}\frac{b}{bc\plus{}1}\plus{}\frac{c}{ca\plus{}1}\ge\frac{3}{2}\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689


every time thales or fambior posts, god kills a kitten

this is way too easy
way way way too easy

Whenever someone abuses the priveledge to massload bots, God kills a kitten.

Whenever I can't think of a legitimate forum game idea, God kills a kitten.

[quote="randomdragoon"]Whenever __________________________________, God kills a kitten.

Fill the blank!

Whenever you say "I lost" more times than necessary to make as many people lose the game as possible, God kills a kitten.[/quote]
omg you evil
I lost

[quote="Sly Si"]Whenever you write the sentence, "Whenever you write the sentence, "Whenever you write the sentence, "Whenever you write the sentence ...", God kills a kitten", God kills a kitten", God kills a kitten", God kills a kitten", God kills a kitten[/quote]

That's not a well defined sentence ... recall Russel's paradox! :roll:

Everytim god is bored, god kills a kitten.

Whenever someone fails a math test, God kills a kitten.

[quote="now a ranger"] :mad:  :mad:  :mad: [/quote]

we were trying to consciously avoid the obvious answer

good job sir

Whenever you get a math problem wrong, god kills a kitten.  :o

Whenever you fight MISSINGNO, God kills a kitten.

whenever you PULL, HARDER, STRINGS, MARTYR, god kills a kitten

[quote="vulgarfraction"]every time someone pronounces 'euler' wrong god kills a kitten :P[/quote]
ew, ler

Everytime someone steals my library card, God kills a kitten.

when thames (the guy who screws up GFF) posts. God kills a kitten

WHY KITTENS.

Everytime you kill a kitten,  :mad:  :mad:  :mad:

When the kitten asks for it, god kills a kitten... :D

[quote="vulgarfraction"]every time someone pronounces 'euler' wrong god kills a kitten :P[/quote]
Then I killed a couple of kittens in eighth grade.
 :mad:  :mad:  :mad:

What?  Just keeping up the running gag.

When God kills a kitten, God kills a kitten.

Whenever Treething makes a "treething's favorite number problem", God kills a kitten

Each time you breathe, a ninja tries to do a backflip. If the ninja fails, God kills a kitten.

Whenever my friend Tommy yells at me, God kills a kitten.

Kittens must be dying at an alrming rate.  :huh:  :maybe:

especiially with that pet food recall in USA because of poisoning...

hurt junggi heal klebian

wait, this isn't the hurt-heal.

Whenever I become constipated, God kills a kitten.

when kitten convert to atheism, god kills a kitten

When the kitty proves that pi is rational, God kills that kitty


The diagonals of the trapezoid $ ABCD$ with bases $ AB$ and $ CD$ are mutually perpendicular and intersect at point $ O$.Let $ M$ and $ B$ be points on the half lines $ OA$ and $ OB$,respectively,such that the angles $ \angle{ANC}$ and $ \angle{BMD}$ are right angles.Let $ E$ be the midpoint of the line segment $ MN$.Prove that
the line $ OE$ is perpendicular to the line $ AB$.

$ AB \parallel DC \Longrightarrow$ $ \frac{OC}{OA} \equal{} \frac{OD}{OB} \Longrightarrow$ the $ \triangle ANC \sim \triangle BMD$ with $ \angle ANC \equal{} \angle BMD$ and altitudes NO, MO are (oppositely) similar, $ \angle MCN \equal{} \angle MDN \Longrightarrow$ CDMN is cyclic with perpendicular diagonals $ CM \perp DN,$ its diagonal intersection is identical with its anticenter, the intersection of maltitudes, and $ OE \perp (DC \parallel AB)$ is one of its maltitudes.

[quote="yetti"]$ AB \parallel DC \Longrightarrow$ $ \frac {OC}{OA} \equal{} \frac {OD}{OB} \Longrightarrow$ the $ \triangle ANC \sim \triangle BMD$ with $ \angle ANC \equal{} \angle BMD$ and altitudes NO, MO are (oppositely) similar, $ \angle MCN \equal{} \angle MDN \Longrightarrow$ CDMN is cyclic with perpendicular diagonals $ CM \perp DN,$ its diagonal intersection is identical with its anticenter, the intersection of maltitudes, and $ OE \perp (DC \parallel AB)$ is one of its maltitudes.[/quote]
Why $ \frac{OC}{OA}\equal{}\frac{OD}{OB} \rightarrow \triangle ANC \sim \triangle BMD$?

[quote="apollo"][quote="yetti"]$ AB \parallel DC \Longrightarrow$ $ \frac {OC}{OA} \equal{} \frac {OD}{OB} \Longrightarrow$ the $ \triangle ANC \sim \triangle BMD$ with $ \angle ANC \equal{} \angle BMD$ and altitudes NO, MO are (oppositely) similar, $ \angle MCN \equal{} \angle MDN \Longrightarrow$ CDMN is cyclic with perpendicular diagonals $ CM \perp DN,$ its diagonal intersection is identical with its anticenter, the intersection of maltitudes, and $ OE \perp (DC \parallel AB)$ is one of its maltitudes.[/quote]
Why $ \frac {OC}{OA} \equal{} \frac {OD}{OB} \rightarrow \triangle ANC \sim \triangle BMD$?[/quote]
Sorry,I saw it.


Round 3 of the USAMTS is now available at [url]http://www.usamts.org[/url]


Let $\omega_1,\omega_2, . . . ,\omega_k$ be distinct real numbers with a nonzero sum. Prove that there exist integers $n_1, n_2, . . . , n_k$ such that $\sum_{i=1}^k n_i\omega_i>0$, and for any non-identical permutation $\pi$ of $\{1, 2,\dots, k\}$ we have
\[\sum_{i=1}^k n_i\omega_{\pi(i)}<0.\]

Without loss of generality, let $\omega_1 > \omega_2 > \ldots > \omega_k$, and let $S = \omega_1 + \omega_2 + \ldots + \omega_k$. It is clear that of all sums of the form $2 \omega_{\pi(1)} + \omega_{\pi(2)} + \ldots + \omega_{\pi(n)}$, $2\omega_1 + \omega_2 + \ldots + \omega_n$ is greatest and $\omega_1 + 2\omega_2 + \omega_3 + \ldots + \omega_n$ is the second greatest. Because the rationals are dense over the reals, we can find some rational number $\epsilon = \frac{a}{b}, b>0$ such that
 \[ \omega_1 + 2\omega_2 + \ldots + \omega_n + S \epsilon < 0 < 2\omega_1 + \omega_2 + \ldots + \omega_n + S \epsilon\]
Then
\begin{align*}
(2+\epsilon)\omega_1 + (1+\epsilon)\omega_2 + \ldots + \omega_n
&> 0 > (1 + \epsilon)\omega_1 + (2+\epsilon)\omega_2 + (1+\epsilon)\omega_3 + \ldots + (1+\epsilon)\omega_n \\
&\geq (2+\epsilon)\omega_{\pi(1)} + (1+\epsilon)\omega_{\pi(2)} + \ldots + (1+\epsilon)\omega_{\pi(n)} \end{align*}
for any permutation $\pi$ distinct from the identity. Multiplying both sides by $b$ shows that the sequence $n_1 = 2b + a = b(2 + \epsilon)$ and $n_i = b + a = b(1 + \epsilon)$ for $2 \leq i \leq k$ suffices.


Prove that
$sec^{2}\theta+sin2\theta+2tan\theta\sqrt{1+sin2\theta}> 9sin^{\frac{4}{3}}\theta$
for all $\theta$.

If $\theta=\frac{2\pi}{3}$ then $\sec^{2}\theta+\sin \theta+2 \tan \theta \sqrt{1+\sin 2\theta}<9 \sin^{2}\theta$
Because,
$\sec^{2}\theta+\sin \theta+2 \tan \theta \sqrt{1+\sin 2\theta}= 4-\frac{\sqrt{3}}{2}-\sqrt{12-6\sqrt{3}}<4$
$9 \sin^{2}\theta = \frac{27}{4}= 6.75$

You misread the question.  It's $sin2\theta$, not $sin\theta$.

I'm sorry.
I have done a typo.

$\sec^{2}\theta+\sin 2\theta+2 \tan \theta \sqrt{1+\sin 2\theta}= 4-\frac{\sqrt{3}}{2}-\sqrt{12-6\sqrt{3}}<4$

Blue line is $\sec^{2}\theta+\sin 2\theta+2 \tan \theta \sqrt{1+\sin 2\theta}$
Red line is $9\sin^{2}\theta$

Oops, the question had another mistype.  It should say "for $0 < \theta < \frac{\pi}{2}$."  It's too late to edit it.

[hide="Solution"]Simplify the left side.
$1+tan^{2}\theta+2sin\theta cos\theta+2tan\theta \sqrt{cos^{2}\theta+sin^{2}\theta+2sin\theta cos\theta}$
$cos^{2}\theta+sin^{2}\theta+2sin\theta cos\theta+2tan\theta(sin\theta+cos\theta)+tan^{2}\theta$
$(cos\theta+sin\theta)^{2}+2(tan\theta)(sin\theta+cos\theta)+tan^{2}\theta$
So we have
$(cos\theta+sin\theta+tan\theta)^{2}> 9sin^{\frac{4}{3}}\theta$
$cos\theta+sin\theta+tan\theta > 3sin^{\frac{2}{3}}\theta$
$\frac{cos\theta+sin\theta+tan\theta}{3}> \sqrt[3]{sin^{2}\theta}$
$\frac{cos\theta+sin\theta+tan\theta}{3}> \sqrt[3]{sin\theta cos\theta tan\theta}$
which is true by AM-GM if $sin\theta$, $cos\theta$, and $tan\theta$ are all positive.  The restrictions said for $0 < \theta < \pi$, for which all trigonometric functions are positive.  However, AM-GM states $AM \ge GM$, not $AM > GM$.  For our problem to work, we simply need to show that the equality case ($sin\theta = cos\theta = tan\theta$) is impossible.  It is easy to show that $cos\theta$ and $sin\theta$ are only equal when the are $\frac{\sqrt{2}}{2}$.  In this case $tan\theta = 1$ and thus the equality case is impossible.
[/hide]


Find the lowest $ n\in\mathbb{N}$ such that  $ 2008|\underbrace{888...888}_{\text{2008n}}$

[quote="zai"]How to proof if $ n < 25$ then $ 2008\not|\underbrace{888...888}_{\text{2008n}}$  :([/quote]

Show $ 2008\not|\underbrace{8888}$
and    $ 2008\not|\underbrace{88888}$

What is a convenient way to find the smallest $ n$?

I tried by hand and cant find a solution for $ n< 10^{13}$

minimal T, suth that $ 251|10^{T}\minus{}1$ is 50, therefore $ 50|2008n$ or $ 25|n$.

[quote="Rust"]$ 251|10^{2008n}\minus{}1$ or $ 50|2008n$ or $ 25|n$.[/quote]

May I ask, how did you make that connection?

And does that mean...
[hide]From $ 25|n$, that implies $ min(n)\equal{}25$?[/hide]


The motto is:

"Christmas Spirit is about giving,
Friendship is about sharing,
Relationship about receiving"

Um, what are we supposed to post on this topic?

People, stop reviving old topics! (Especially you, armalite46!)


x+y+z=1, x,y,z are positive reals
Find the minimum of

$\frac{{\sqrt x }}{{1 - x}} + \frac{{\sqrt y }}{{1 - y}} + \frac{{\sqrt z }}{{1 - z}}$

[quote="nttu"]I think this is an easy and well-known problem .
 We will prove $ \frac{\sqrt{x}}{1-x} \geq \frac{3\sqrt3}{2}x $
 It's equivalent to $ \frac{2}{3\sqrt3} \geq \sqrt{x}(1-x) $
 It is followed by using AM-GM with : 2x , 1-x , 1-x
  So min = $ 3\sqrt{3}/2 $[/quote]

this is exactly my solution, while i am not sure the ans is 3sqrt 3/2 actually...
thanks for verifying it for me. :D

[quote="nttu"]
 It is followed by using AM-GM with : 2x , 1-x , 1-x [/quote]

but how ? I get only
$\frac{2x+(1-x)+(1-x)}{3}=\frac{2}{3} \geq \sqrt[3]{2x(1-x)^2}$
where's my mistake ?

Megus, your expression $A \ge B$ is equal to Nttu expression.

You must do $A^{3/2}  \geq B^{3/2}$

thanks - I must've been blind  :blush:  :blush:  :blush:

Alternately, if one doesn't see the AM-GM right away, we have that:

$\displaystyle \frac {\sqrt x}{1-x} \geq \frac {3\sqrt 3}{2} x \Rightarrow 0\geq (3x-1)^2(3x-4)x$

which is clearly true for $x,y,z \leq 1$


superior algebra
superior algebra solved

I got this as homework:

If we have x3=x for all x in a ring A then the ring is commutative.

[i]unknown source[/i]

My teacher likes to discourage us, so he said: " this is awfully hard" :D I think I might have seen it before, but i don't know where.

More generally, a theorem of Jacobson states that :
If for each x in a ring A there exists an integer n > 1, which may depend on x, such that x n  = x, then the ring is commutative.

You will find a solution of your problem and references on Jacobson's theorem at :
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/herstein

Pierre.

nice theorem :) but I guess it's not that easy to prove. we have had in Romania in contests the cases n=2,3,4,5,6,8 (this are the ones I've seen) :)

[quote="pbornsztein"]Romanians don't like number 7??? :D

Pierre.[/quote]

7 is my favorite number. Even my phone number has lots of 7 in it.
My birthday has a 7 in it. And I'm pretty sure I'm romanian. I'm not proud of it, but that is what I am  :D (a big smile).

Thanks for the solution, Mihai!

I did find a solution myself:

Lemma 1: If we have a ring s.t x2 =0 =>x=0 then every idempotent element is in Z(A):

Assume x is idempotent, i.e. x2 =x. We have
(xyx-xy)2 =0 and (xyx-yx)2 =0, so xyx-xy=xyx-yx, so xy=yx, for any y in A.

Lemma 2: If in a ring xn=x then xn-1 is in Z(A).

We just apply Lemma 1 to the ring.

Now to our problem. From Lemma 2 we get x*y2=y2*x  for all x and y in A.

We have (x+1)3=x+1 for all x, so 3x2+3x=0 for all x. (*)

(-x2-x)2 = x4 + x2 +2x3 = 2x2+2x=
-x2-x from (*). This means that -x2-x is idempotent, so it's in Z(A) for all x in A (Lemma 1), and since x2 is also in Z(A), it means that x is in Z(A) for all x, so we're done.

[quote="grobber"]Thanks for the solution, Mihai!
[/quote]

I would have liked to solve this problem, but I didn't. I think your thanks are for Moubinool.
I think this problem was given in a Titeica contest, and I think I had it too in the form:p prime and  x p=x for all x, than A is commutative, but I'm not sure.
I had this or a similar problem in my 11th grade at the Titeica teams contest. But the team had a 12th grade member, 11&12 grade problems, and I couldn't touch the problem. It was too hot for me back then:D. It is a 12th grade problem and I've expected the 12th grade member to solve it. If I remember it correctly, no team solved that problem. However, my teammate solved the problem for p=2 and 3!!! and the team got some precious points that allowed us to win the contest.

Sorry Moubinool! I didn't take a good look to see who solved the problem! :D


geometry
geometric transformation
reflection

So I've been thinking about this for far too long...I'd like to know how to trisect a line segment with a straight-edge and non-collapsing compass in 6 steps (or less).  Trisection means finding BOTH trisection points - though, of course, if you find one, you're one step away from getting the other one too.  (A step is drawing a new line or a new circle.  Picking a random point and extending an existing line do not count as steps.)  There's a book that tells me this is possible.  I can do it 1 way in 7 steps, about 6 ways in 8 steps, and several more ways in even more steps.  No one in university seems to know anything about purely Euclidean geometry, so I'm referring the question to this forum.

What do you mean by 'non-collapsing compass'  :?:

If you mean that it is of fixed length, I can't solve it in 6 steps
If it is a usual compass, then it's ok

No, a nice compass.  As in, you can draw a circle with any radius that's a distance between two points you already have (not only a distance from the center of the circle to another point).  So if you already have points A, B, and C, you can draw a circle centered at A with radius BC.  Sorry if "non-collapsing compass" was confusing...that was way more explanation than it was worth.  So how do you do it in 6 steps?

ok then..

Let $ AM$ be the given segment. We'll construct a triangle $ ABC$ with $ M,N$ being the midpoints of $ BC$ and $ CA$ respectively

Pick an arbitrary point $ N$ out of the segment $ AM$ (not a step)

[b]Step 1: [/b] Draw the ray $ AN$
[b]Step 2: [/b] Reflect $ A$ through $ N$ to get the point $ C$ (requires a circle centered at $ N$, passing $ A$)
[b]Step 3: [/b] Draw the ray $ CM$
[b]Step 4: [/b] Reflect $ C$ through $ M$ to get the point $ B$
[b]Step 5: [/b] Draw the line $ BN$, This intersects $ AM$ at the centroid $ G$
[b]Step 6: [/b] Reflect $ M$ through $ G$ to get the point $ G'$

Then of course $ MG=GG'=G'A$

Ahhh...ok, thanks!  :lol:  I was beating around the bush as usual.

I just saw this yesterday.

To trisect segment $ AB$,
1) Construct circle $ \omega_{1}$ with centre $ A$ and radius $ AB$,
2) Construct circle $ \omega_{2}$ with centre $ B$ and radius $ AB$,
Let $ C$ be an intersection of $ \omega_{1}$ and $ \omega_{2}$,
Extend AB to meet $ \omega_{2}$ at E,
3) Extend CA to meet $ \omega_{1}$ at D,
4) Let $ F$ lie on $ \omega_{2}$ and $ DE$,
5) Let $ G$ lie on $ CF$ and $ AB$,
6) Let $ G'$ lie on $ AB$ and the circle with centre $ G$ and radius $ AG$ such that $ G'$ $ \neq$ $ A$,

Then $ AG$ = $ GG'$ = $ G'B$.

I am wondering if anyone has an elegant proof that this is in fact a trisection. I have a trigonometric proof which is not so bad but it hides the elegance of the construction.

given line AB
1 build square ABCD
2 bisect CD at E
3 draw diagonal AC
4 draw line BE intersects AC at F
5 project F to AB then we have BF = 1/3 AB

chichi, how do you build a square in one step?

carpo, here's a proof:

Let $ \angle DCF=x$ and let $ AB=r$, the radius of $ w_{1}$ and $ w_{2}$.  We know $ \triangle ABC$ is equilateral, so $ \angle BAC=60^\circ$.  Then in $ \triangle AGC$, $ \angle AGC=120^\circ-x$.  Now $ \angle BAD=120^\circ$, and $ AB=AD$, so $ \triangle ABD$ is isosceles and $ \angle ABD=\angle ADB=30^\circ$.  Let $ CF$ meet $ DB$ at $ I$.  Then in $ \triangle GBI$, $ \angle BGI=120^\circ-x$, and $ \angle GBI=30^\circ$, so $ \angle GIB=30^\circ+x$.  Now $ CD$ is a diameter of $ w_{1}$, so $ \angle CBI=\angle CBD=90^\circ$.  $ AE$ is a diameter of $ w_{2}$, so $ \angle DCE=\angle ACE=90^\circ$.  $ \angle CEB=\frac{1}{2}\angle ABC=30^\circ$, as $ \angle CBE$ and $ \angle ABC$ are the subtended and sector angles of chord $ AC$ in $ w_{2}$.  $ \angle AEF=\angle ACF=x$, since both are subtended by chord $ AF$ in $ w_{2}$.  So $ \angle CED=30^\circ+x$, and $ \triangle DCE \sim \triangle CBI$ (AA).  From $ \triangle ACE$, $ CE=\sqrt{3}r$.  So $ \frac{BI}{CE}=\frac{BC}{CD}=\frac{r}{2r}=\frac{1}{2}$.  So $ BI=\frac{\sqrt{3}}{2}r$.  But from $ \triangle CBD$, $ BD=\sqrt{3}$.  So $ I$ is the midpoint of $ BD$.  So in $ \triangle CBD$, $ CI$ and $ AB$ are medians, so $ AG=2BG$ as desired.

The construction with the centroid is so cool! I love it.


hi,
Prove, if $ a,b,c \in \mathbb{R}$, $ a > 0$ and
\[ T(x) \equal{} ax^{2} \plus{} bcx \plus{} b^{3} \plus{} c^{3} \minus{} 4abc,\]
has complex roots, than one of the polynoms $ T_{1}(x) \equal{} ax^{2} \plus{} bx \plus{} c,$ $ T_{2}(x) \equal{} ax^{2} \plus{} cx \plus{} b,$ has only strict positive values.

[quote="doicu"]hi,
Prove, if $ a,b,c \in \mathbb{R}$, $ a > 0$ and
\[ T(x) \equal{} ax^{2} \plus{} bcx \plus{} b^{3} \plus{} c^{3} \minus{} 4abc,
\]
has complex roots, than one of the polynoms $ T_{1}(x) \equal{} ax^{2} \plus{} bx \plus{} c,$ $ T_{2}(x) \equal{} ax^{2} \plus{} cx \plus{} b,$ has only strict positive values.[/quote]
$ T(x) \equal{} ax^{2} \plus{} bcx \plus{} b^{3} \plus{} c^{3} \minus{} 4abc$ has complex roots,so
$ b^2c^2\minus{}4a(b^3\plus{}c^3\minus{}4abc)<0$
<=>$ (b^2\minus{}4ac)(c^2\minus{}4ab)<0$
=>$ b^2\minus{}4ac<0$ or $ c^2\minus{}4ab<0$
Note that $ a>0$ so
If $ b^2\minus{}4ac<0$ then $ T_{1}(x) \equal{} ax^{2} \plus{} bx \plus{} c$ has only strict positive values
If $ c^2\minus{}4ab<0$ then $ T_{2}(x) \equal{} ax^{2} \plus{} cx \plus{} b$ has only strict positive values
Done!

But every polynomial has complex roots!!

Abusedly so, here the proposer meant by [b]complex[/b] roots to say that the polynomial has [b]no real [/b]roots.


Rules of the game:
1) You must say three words, and only three words.
2) The three words must make sense when connected to the three words above you.
3) No sick/perverted things.

Ok I'll start

Once upon a...

When you say make sense, do you mean just grammatically correct or not weird.

Like, am I allowed to say

He lick desks

Anyway....

of very tasty

of a pi

(The message is too small. Please make the message longer before submitting.)

He saw a...
And I mean it has to make sense with the three words above you.

who hate pianogirl

EDIT: Two households, both with yongyi781 in naughty positions with cranberry juice, went to the fair. At the fair, this bump is containing a baby. Yongyi meets tinytim over some food. Tinytim then challenges Yongyi to a views contest. Tinytim then decides that his avatar is really cool and shames Yongyi because he did not make a blueberry pie for the hobos. Yongyi takes out (^_^)'s blue shoes, and burns them. Then, laughing evilly, they started crying over the hills and tripped on gauss1181's head. Fortunately, pacman2812 was there playing soccer with myyellowducky82 and xoangieexo. xoangieexo kicks the ball into the pit next to the emo club that ends the scenario to receive payment for work at the emomelon where emo people dance hokey pokeys and make IMO. IMO was bloody because a bomb-exam with spikes exploded and no one was alive after. And then they, the emo-zombies, bit dannyhamtx and killed Me, so I, the very greatest zombie killed Lawrence Johnson because he did not gave really bad grammar book to his parents, therefore he haves an F in math exam. Meanwhile, the world is being attacked by mutant aliens who hate pianogirl

um...

why did you two revive a topic from almost a year ago when there's already a different 3 word game going on?

Two households, both with yongyi781 in naughty positions with cranberry juice, went to the fair. At the fair, this bump is containing a baby. Yongyi meets tinytim over some food. Tinytim then challenges Yongyi to a views contest. Tinytim then decides that his avatar is really cool and shames Yongyi because he did not make a blueberry pie for the hobos. Yongyi takes out (^_^)'s blue shoes, and burns them. Then, laughing evilly, they started crying over the hills and tripped on gauss1181's head. Fortunately, pacman2812 was there playing soccer with myyellowducky82 and xoangieexo. xoangieexo kicks the ball into the pit next to the emo club that ends the scenario to receive payment for work at the emomelon where emo people dance hokey pokeys and make IMO. IMO was bloody because a bomb-exam with spikes exploded and no one was alive after. And then they, the emo-zombies, bit dannyhamtx and killed Me, so I, the very greatest zombie killed Lawrence Johnson because he did not gave really bad grammar book to his parents, therefore he haves an F in math exam. Meanwhile, the world is being attacked by mutant aliens who hate pianogirl and destroy splatyango because he/she wants to die. But the aliens were like wussup and promptly ate


Find $ lim \sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2\plus{}...\plus{}\sqrt{2\plus{}\sqrt{2008}}}}}$

If it equals x, then $ x \equal{} \sqrt {2 \plus{} x}$ hence x=2 :wink:

[quote="thanhnam2902"]Find $ lim \sqrt {2 + \sqrt {2 + \sqrt {2 + ... + \sqrt {2 + \sqrt {2008}}}}}$[/quote]
I question what this problem means.  Are you saying that the "..." implies an infinite continuation?  If so, then I don't think the problem is possible.  You're saying that in an infinite chain of $ \sqrt{2+\sqrt{2+...}}$, the chain ends with a $ \sqrt{2008}$.  Clearly, an infinite chain does not end, so this can't be true.
I understand that a limit is being sought, but you're asking for an infinite limit of something finite and infinite at the same time (which is impossible).
A more appropriate problem would be to find $ \lim\sqrt{2+\sqrt{2+\sqrt{2+...}}}$, with the "..." implying an infinite continuation.  Either way, a $ \sqrt{2008}$ would be meaningless.

How about the following:
Let $ f(x)\equal{}\sqrt{2\plus{}x}$. Find $ \lim_{n\to\infty} f^n(\sqrt{2008})$.


[quote="Peter"]The sequence $ \{a_{n}\}_{n \ge 1}$ is defined by
\[ a_{1} \equal{} 1, \; a_{2} \equal{} 2, \; a_{3} \equal{} 24, \; a_{n} \equal{} \frac { 6a_{n \minus{} 1}^{2}a_{n \minus{} 3} \minus{} 8a_{n \minus{} 1}a_{n \minus{} 2}^{2}}{a_{n \minus{} 2}a_{n \minus{} 3}}\ \ \ \ (n\ge4).
\]
Show that $ a_{n}$ is an integer for all $ n$, and show that $ n|a_{n}$ for every $ n\in\mathbb{N}$.[/quote]
$ v_n \equal{} \frac {a_n}{a_{n \minus{} 1}}$
$ v_1 \equal{} 2,v_2 \equal{} 12$
$ v_n \equal{} 6v_{n \minus{} 1} \minus{} 8v_{n \minus{} 2}$
Then we has $ v_n \equal{} 4^n \minus{} 2^n$
$ a_n \equal{} (4^n \minus{} 2^n)a_{n \minus{} 1}$
So $ a_n \equal{} 2^{\frac {n(n \plus{} 1)}{2}}\prod _{k \equal{} 1}^n(2^k \minus{} 1)$
Result $ n|a_n$ follow from Euler theorem.


How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

Post problems using the "Ask on Forum" feature, not before you've attempted it :wink:

We start by thinking of the ways we "create" zeros. This is achieved simply by noting that 2*5=10. Since clearly in the given product there are fewer fives than twos, we are counting the number of fives in the product.

In 100!, there are 100/5=20 + 4 (because each multiple of 25 produces 2 fives) = $ 24$ fives produced.

In 200!, there are 200/5=40 + 8 +1 (because $ 125\equal{}5^3$ produces a 3rd five) = $ 49$ fives produced.

Following a similar method, we find 300! = $ 74$ fives produced.

Therefore, our final answer is simply $ 24\plus{}49\plus{}74\equal{}\boxed{147}$.

Hope this made sense!


1.  To crease means to just fold.  For example, when you mail a letter you crease the letter twice so it will fit in the envelope.  To wrinkle is more like crumpling that letter.  Creasing is more gentle, deliberate, and neat.  Wrinkling is more rugged and disorganized.

2. Well 'puckered brows' means that his eyebrows were kinda cringing/wrinkled.  Generally, I think of 'pucker' (or 'puckered') in terms of a person eating a lemon.

[quote="Silverfalcon"]I was reading this short story and came up with two things that I don't understand. Please help (I used dictionary to help me but didn't get much out of it).

1. What is the difference between creased and wrinkled? The sentence had something like "He was creased but not wrinkled."

2. What does this sentence mean? -> John considered with puckered brows.[/quote]

I think that in these instances, it is actually the connotation of the word(s) that is important, not the denotation.

I'm not sure of the context, but I would guess that "He was creased but not wrinkled" means that he has some age/experience under his belt, but he isn't old. The word "creased" suggests that he's probably experienced some defining event at one point, leaving a permanent mark on him. On the other hand, "wrinkled" would suggest that he's old and worn. (It's just a guess; there's no way for me to know without the text in front of me.)

"Puckered" in itself means that it has become "gathered, contracted." In this case, his "puckered brows" probably suggest some sort of perplexity. (Another wild guess.)

If you give me more context, I can probably give you a better answer.


induction
USAMTS
modular arithmetic
LaTeX

What did you score?

I got 5 [b]5[/b] 4 5 5.

I missed number 3 because it wasn't in a closed form.

I got 22 (5 2 5 5 5).......
no commended solution.. :(

24 55455 no commended. I got a point taken off of 3 'cause I subtracted the original vertices for some unknown reason.

I got a [b]5[/b] 3 5 [b]5 5[/b], and 23 composite

I got three commended solutions: #1, #4, and #5

i wish i got a 24 though so I could be on the leaderboard though  :(

I got 5 5 5 5 5.
I'm really happy I got a 25 but not so happy I didn't get any comended solutions.

5 5 [b]5[/b] 5 5

Apparently, my solution for #4 was one the grader hadn't seen before.  The comments for number three were "Nice job on the induction!"
The rest were just "Good job", etc.

5 4 5 5 [b]5[/b]
I'm so glad they only deducted 1 point for number 2.

25, no commended.  I'm happy, especially since this is my first time doing the USAMTS.  Now that I'm more  comfortable with the problems, I'm aiming for commended solutions in later rounds.

5 4 5 1 -

Actually better than I expected! I thought that I would get an extra point or two off on number 2.

5 3 5 [b]5[/b] 5

23 total, Better than I expected, thought I'd get a 1 on number 2. Has USAMTS put logged Entry Forms yet? They still haven't loggged mine.[/b]

Got a 25. Guess I better appeal...=s

5 [b][color=darkred]5[/color][/b] [b][color=darkred]5[/color][/b] 5 5

#3 I had a nice solution to that one, I kind of expected it.
#2 was a surprise though - I thought that was one of my less-nice solutions (I mean, come on, I used casework!)

5+5+[color=red]5[/color]+[color=red]5[/color]+2 = 22
I'm not complainging since my solution to #5 was BS.

4 [b]5[/b] 5 5 5

Wow, I got commended on the one I thought I lost points on and lost a point for the one I didn't think I had to think about!  Oh well, I stlil have a chance for gold!  :D

[b]5 5 5 5 -[/b]

So yeah, I turned in only 4 problems, but all of them got commended.

Hooah.

5 [b]5[/b] 5 [b]5 5[/b]

I guess I really don't have anything to complain about. Someday I'll get 5 commendations in one round though  :P

[quote="Ciaccona"]
Perhaps you need an unusal approach in addition to a well-written solution? I got a commended, but I used a different method than the MathJam and everyone I know.[/quote]

woah....your solution owns     :10:

i think i have a pretty unique #1... (aka no mods)

Let the number we are looking for be $S$. We have that $S=d_{n}d_{n-1}d_{n-2}...d_{2}4$. Where each $d_{k}$ represents the digit of the decimal expansion of $S$. Let us construct a rational repeating decimal $N$ such that $N=0.d_{n}d_{n-1}d_{n-2}...d_{2}4d_{n}d_{n-1}...$. For the condition of the problem, we have that $4N$ must equal $0.4d_{n}d_{n-1}d_{n-2}...d_{3}d_{2}...$.

Thus, we have the following equations:

$(1)$ $N=0.d_{n}d_{n-1}d_{n-2}...d_{2}4d_{n}d_{n-1}...$.

$(2)$ $4N=0.4d_{n}d_{n-1}d_{n-2}...d_{3}d_{2}...$.

Let $(3)$ be $10 \cdot 4N=40N$. Thus, we now have:

$(1)$ $N=0.d_{n}d_{n-1}d_{n-2}...d_{2}4d_{n}d_{n-1}...$.

$(3)$ $40N=4.d_{n}d_{n-1}d_{n-2}...d_{2}4...$

$(3)-(1)$ gives us $39N=4$ and $N=.102564102564102564...$.

The repeating part of $N$ is the smallest such number that we desire. Thus, $S=102564$.

Checking, we see that $4 \cdot 102564=410256$ and indeed satisfies our conditions.

55555, no commended but im happy.
this is the first time ive gotten a perfect.

[quote="mustafa"]Yeah, I'm planning on getting a 25 with 5 commended solutions, so watch out....... :rotfl:[/quote]

Yeah we're going to bounce back from our mediocre scores and end up with 15 commended solutions at the end of the year...

Sigh, I was hoping for at least one point on number 4, though I didn't anticipate getting 3 on number 2...

[quote="Iversonfan2007"]i think i have a pretty unique #1... (aka no mods)[/quote]

Wow...that's a really, really nice solution.

[quote="me@home"][quote="perfectnumber628"]i got 25, none commended.  wow it seems like massive amounts of people got 25.[/quote]
As predicted... right after we turned in solutions, people started a poll about how they thought they did and something like 18 people predicted 25's... it was like 40% of the vote... and everyone said theres no way those people are right
I think most of us were, though[/quote]
That someone=me.
I guess I was wrong. There have been 10 25s so far...

5,3,3,2,2

total=15  :blush:

That was better than what I expected!!

[quote="anirudh"]I didn't even turn in a paper. I didn't have time to latex it up.  :([/quote]

You should have just written it, that's what I did. And they gave you like 6 weeks. Anyway, what do you think you would have gotten? I'm curious because you're in middle school and want to know what middle schoolers got.


for all real $x,y,z$ prove or disprove that

$x^4+y^4+z^4+3(x^2y^2+x^2z^2+y^2z^2)\geq 2(x^3(y+z)+y^3(x+z)+z^3(x+y))$

It is a particular case of the more general inequality

$\sum x^4+r(r+2)\sum y^2z^2 +(1-r^2)xyz\sum x \geq (r+1)\sum yz(y^2+z^2)$,

where $r$ is a real number.

obviously, it's ture by muirhead inequality.

We may write also the inequality

$\sum x^4+r(r+2)\sum y^2z^2 +(1-r^2)xyz\sum x \geq (r+1)\sum yz(y^2+z^2)$,

as

$\frac{1}{2}\sum (y-z)^2(x+rx-y-z)^2 \geq 0$.


Find all triples $\text{(x, y, z)}$ of integers that satisfy the following system of equations: \[x^{3}-4x^{2}-16x+60 = y\]  \[y^{3}-4y^{2}-16y+60 = z\]  \[z^{3}-4z^{2}-16z+60 = x\]

Abridged solution

[hide]

Add four to each side, and factor, the first equation factors as , and the rest similarly. Multiply all together, and you get  the solutions follow quickly from this if you remember that they must all be integers.

[/hide]


Να λυθεί η τριγωνομετρική εξίσωση
$ sin^{2008}(x)\plus{}cos^{2009}(x)\plus{}cos^{2}(x)\equal{}2$

λυση σε hidden
[hide]η εξισωση γραφεται
$ sin^{2008}{x} \minus{} sin^2{x} \equal{} 1 \minus{} cos^{2009} {x}$

$ \leftrightarrow sin^2{x}( sin^{2006}{x} \minus{} 1 ) \equal{} 1 \minus{} cos^{2009}{x}$

ομως,

$ sin^2{x}( sin^{2006}{x} \minus{} 1 ) \leq0$

$ 1 \minus{} cos^{2009}{x} \geq 0$

που σημαινει οτι εμεις αναζητουμε τα x τετοια ωστε
$ sin^2{x}( sin^{2006}{x} \minus{} 1 ) \equal{} 0$
$ 1 \minus{} cos^{2009}{x} \equal{} 0$

συστημα με λυση
$ cosx \equal{} 1 \equal{} > x \equal{} 2kp$ κ ακεραιος, $ p \equal{} 3.141...$[/hide]


Find all $n\in \mathbb{N}$ such that:
\[\frac{3^n-1}{2^n-1}\in \mathbb{N}\]
Maybe it is classical! ;)

$7=6\cdot 1+1|6\cdot 2+2=14$.

(it's meant to be a counterexample, jpsteenstra :))

here is a proof using Jacobi Symbols and Jacobi Reciprocity:

for notational purposes, [tex]\left(\frac{A}{B}\right)[/tex] is a Jacobi Symbol.

Now for n = 1, it is obvious, so consider n > 1 and assume for the sake of contradiction, that for some n > 1, [tex]2^n-1|3^n-1[/tex]

then n must be odd, for if n were even, then 2^n-1 must be divisible by 3, while 3^n-1 is not.

then we have [tex]3^n-1\equiv 0\mod (2^n-1)[/tex] which implies [tex] 3^n\equiv 1\mod (2^n-1)[/tex] and so 3 must be a square [tex]\mod (2^n-1) [/tex] since n is odd, and

[tex]3^\left(\frac{n+1}{2}\right)^2[/tex][tex] \equiv 3 \mod (2^n-1)[/tex]

therefore [tex]\left(\frac{3}{2^n-1}\right) = 1[/tex] and since n is odd, we also have that [tex]\left(\frac{2^n-1}{3}\right)=1[/tex]

and by Jacobi Reciprocity:

[tex]1=\left(\frac{3}{2^n-1}\right)*\left(\frac{2^n-1}{3}\right) = (-1)^{\frac{2^n-2}{2}*\frac{3-1}{2}} = -1[/tex] contradiction

I think another proof can be given using Newton's Formula for $(2+1)^n$ and $2^n$ but the results won't be probably so nice in appearence


The sequence $ a_1, a_2, a_3, ...$ satisfies $ a_{4n\plus{}1} \equal{} 1, a_{4n\plus{}3} \equal{} 0, a_{2n} \equal{} a_n$. Show that it is not periodic.

[quote="tdl"]The sequence $ a_1, a_2, a_3, ...$ satisfies $ a_{4n \plus{} 1} \equal{} 1, a_{4n \plus{} 3} \equal{} 0, a_{2n} \equal{} a_n$. Show that it is not periodic.[/quote]

Assume $ a_{n\plus{}p}\equal{}a_n$ $ \forall n$ and for some period $ p>0$.

Let then $ p\equal{}2^aq$ with odd $ q$.
Let then $ n\equal{}2^{a\plus{}2}(q\plus{}2)$

Using periodicity, we have $ a_{n\plus{}p}\equal{}a_n\equal{}a_{q\plus{}2}$
But $ a_{n\plus{}p}\equal{}a_{2^a(q\plus{}4(q\plus{}2))}\equal{}a_{q\plus{}4(q\plus{}2)}$

But $ q\plus{}2$ and $ q\plus{}4(q\plus{}2)$ are two odd numbers different modulus 4. So $ a_{q\plus{}2}\neq a_{q\plus{}4(q\plus{}2)}$ (one is 0, the other is 1).

And so no such $ p>0$ exists and the sequence is not periodic.


Find all $f: Q\to R$ satisfying $f(xy)=f(x)f(y)$ and $f(x+y)\le\max(f(x),f(y))$


This problem isn't to hard because the absolute value is always positive
[hide]

$  \mid q + 7 \mid - 4 = 2  $

add 4 on each side:

$  \mid q + 7 \mid = 6 $

So:

$ q = -1  $[/hide]


Let T be a linear transformation of an n dimensional vectorspace into itself such that T has n+1 eigen vectors, (every n of them are linearly independent). Prove that T is a multiple of the identity.

I don't really understand what you meant.

First, $ T$ has $ n\plus{}1$ eigenvectors and each $ n$ of them are linearly independent is equivalent to the fact that the matrix representing $ T$ is diagonalizable and one of the eigenvector is a multiple of one of the others.

So, I can take any diagonalizable matrix (just choose any diagonal matrix $ D$ and let $ T \equal{} P^{\minus{}1} D P$) and it's done: $ T$ satisfies your condition and is not a multiple of identity.

If one of the eigenvectors is a multiple of the others, then it's not true that any $ n$ of them are linearly independent.

It is true that $ T$ is diagonalizable - what can you deduce about its eigenvalues?


Ring Theory
superior algebra
superior algebra unsolved

[color=olive]Let f : R ! S be a homomorphism. Let T = { f(r) : r 2 R}.
T is called the image of f. Show that T is a subring of S.
Hint: In this problem, you can use the following facts: f(0R) = 0S, f(−x) = −f(x)
for x 2 R, and f(x − y) = f(x) − f(y) for x, y 2 R. These facts were stated in
class for isomorphisms only but the same properties (with the same proofs) hold
for homomorphisms.[/color]

Thanks very much

Why does this belong to the olympiad section?


Consider the function $ f(x) \equal{} 2x \minus{} sinx$. Prove that there exist a real number $ \large b \in R$ and functions $ g,h$ such that :
i) $ \large f(g(x)) \equal{} x \forall x \in R$.
ii) $ h$ is a periodic function.
iii) $ g(x) \equal{} b.x \plus{} h(x) \forall x \in R$

I have a nice solution, and I will become happy if someone give more solution, thank!!! (my English is so poor!) :blush:

Because f -bijection, g is bijection too. Therefore work $ b\equal{}0,h\equal{}g$.

Im' sorry.
I've just edit that $ h$ is a periodic function.

Obviosly work $ b\equal{}\frac 12, h\equal{}g\minus{}\frac 12 x$.

Yes, that right!
But I want a clear solution  :blush: (my solution is so clearly!.. :blush: )

Dtrung, please give your post a relevant subject title.  The title "..." does not say anything about your post.


function
limit
real analysis
real analysis unsolved

Let $ a\geqslant 0$, $ x_1 \equal{} 1 \plus{} a$ and $ x_{n\plus{}1} \equal{} x_n \plus{} \frac{n}{x_n}$.
1. Show that $ \lim_{n\to \infty} n(x_n \minus{} n)$ exists and is non negative. Denote that limit $ b$.
2. Denote by $ f : \mathbb{R}_\plus{} \to \mathbb{R}_\plus{}$ the function defined by $ b \equal{} f(a)$. Is $ f$ continuous?

This is question Q650 in RMS Mars 2009 page 100
the author of this question is Alain Tissier

If any one of you have a solution you can send in PDF or TEX at

al.tissier@laposte.net

for publication in RMS


Show there are infinitely many non-primes in $ f(\mathbb{N})$ for any $ f \in \mathbb{Z}[x]$ of degree $ \geq 1$.

What if $ f \in \mathbb{Z}[x_1,...,x_n]$

Let $ f$ is non constandt and $ A\equal{}|f(x_1,x_2,...,x_n)|>1$, then exist prime p, suth that $ p|A$.
If  $ deg_{x_i}f>0$, then for all k $ p|f(x_1,..,x_i\plus{}kp<x_{i\plus{}1},..x_n)$ and for infinetely many k $ |f(x_1,..,x_i\plus{}kp,..,x_n)|>p$.

$ f(x) | f(kf(x) \plus{} x)$ for $ k \in \mathbb{Z}$.


Knowing that p + q = 2, proves that:

$ \ 3p^qq^p + p^pq^q \le 4$

I believe that this is impossible ( $ \ p,q\in \Re$)

Not true. For $ p \equal{} \minus{} 2$ and $ q \equal{} 4$ you get the value of 67.. However, it is true for $ p,q \geq 0$..


group theory
abstract algebra
function
number theory
prime numbers
superior algebra
superior algebra unsolved

Let $p_{1},p_{2},...,p_{n}$ be prime numbers and $G$ a group such that $p_{i}|card(G), \forall i=1,2,3,...,n$ such that for every $p_{i}$ there exist $x_{i},y_{i}\in G$ where $y_{i}$ is not a power of $x_{i}$ such that $ord(x_{i})=ord(y_{i})=p_{i}$.
Prove that:  $|G|\geq 1+\prod_{i=1}^{n}(p_{i}^{2}-1)$

For all $i=1,\ldots,n$ let $H_{i}=\langle x_{i}\rangle$ and $K_{i}=\langle y_{i}\rangle$.
We have $|H_{i}|=|K_{i}|=p_{i}$ and since $y_{i}$ is not a power of $x_{i}$ and the primes $p_{i}$ are distinct we also have

$i)$ $H_{i}\cap K_{i}=(1)$ and

$ii)$ $H_{i}\cap H_{j}=K_{i}\cap K_{j}=(1)$ for all $i\neq j$.

Hence, there exists a one--to-one map from

\[\prod_{i=1}^{n}(H_{i}\times K_{i})\setminus\{(1,1)\}\longrightarrow G\]

Since $|H_{i}\times K_{i}|=p_{i}^{2}$ we obtain $|G|\geq 1+\prod_{i=1}^{n}(p_{i}^{2}-1)$.

[quote="anenciu"]
Hence, there exists a one--to-one map from
\[\prod_{i=1}^{n}(H_{i}\times K_{i})\setminus\{(1,1)\}\longrightarrow G \]
.[/quote]
Can you kind show the one-to-one map explicitly? Thank you

I think I made it harder and unclearer than I wanted. The idea is that $A_{i}=H_{i}K_{i}$ is a subgroup of $G$ of order $p_{i}^{2}$ for all $i=1,\ldots,n$ (here we use the fact that $y_{i}$ is not a power of $x_{i}$). Furthermore, for all $i\neq j$ we have $A_{i}\cap A_{j}=(1)$. So $G$ cotains subgroups $A_{1},\ldots, A_{n}$ each of order $p_{i}^{2}$ and any two have in common just the identity. So, if we let $B_{i}=A_{i}\setminus (1)$ then the $B_{i}$'s have no elemnts in comon and they have order $p_{i}^{2}-1$.Since $G$ is a group it will contain elements of the form $\prod_{i=1}^{n}b_{i}$ where $b_{i}\in B_{i}$ and 1 which is not of the above form (product of elements from B_i's). Hence, the conclusion follows now.

[quote="anenciu"][...] The idea is that $A_{i}=H_{i}K_{i}$ is a subgroup of $G$ [...][/quote]

That is wrong in general.

Here is another approach: Sylows theorem tells us, that the number of subgroups of order $p^{k}$ is congruent 1 mod p (provided $p^{k}$ is a divisor of $|G|$).
Especially we have at least $p_{i}+1$ subgroups of order $p_{i}$ (if there was only one, $y_{i}$ would be power of $x_{i}$). They contain together $(p_{i}-1)(p_{i}+1)$ elements of order $p_{i}$. So G has at least $\prod_{i=1}^{n}{(p_{i}^{2}-1)}$ elements of prime order. Add the neutral element and we're done.

Actually I think that Gockel's proof is very much the same like anenciu's one from the point he has shown that $G$ has $p_{i}^{2}-1$ elements of order $p_{i}$, we should find a injective function $\prod A_{i}\to G$ where $A_{i}=\{ x\in G | ord(x)=p_{i}\}$

Maybe I'm blind :blush: ,But can somebody kindly show me it is an injective?

[quote="zhaobin"][quote]Here is another approach: Sylows theorem tells us, that the number of subgroups of order $p^{k}$ is congruent 1 mod p (provided $p^{k}$ is a divisor of $|G|$).
Especially we have at least $p_{i}+1$ subgroups of order $p_{i}$ (if there was only one, $y_{i}$ would be power of $x_{i}$). They contain together $(p_{i}-1)(p_{i}+1)$ elements of order $p_{i}$. So G has at least $\prod_{i=1}^{n}{(p_{i}^{2}-1)}$ elements of prime order. Add the neutral element and we're done.[/quote]
But from what your wrote the elements is at least  $1+\sum_{i=1}^{n}{(p_{i}^{2}-1)}$...[/quote]

f***...
You're right, I proved that G contains at least the sum, not the product...

[quote="Gockel"][quote="anenciu"][...] The idea is that $A_{i}=H_{i}K_{i}$ is a subgroup of $G$ [...][/quote][/quote]

That is wrong in general.

You are right. I made a mistake thinking that $x_{i}$ and $y_{i}$ commmute and we DO NOT KNOW this for this problem. Sorry! I will have to think of another proof.

It's too late now, but there's no point wasting your time when the smallest possible counterexample works ($ A_4$ has 3 elements of order 2 and 8 of order 3, in a group of order 12).

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172678]here[/url].


modular arithmetic
algebra
binomial theorem

[quote="mdk"]Find the last two digits of N=11^10-9[/quote]
[hide="Not Sure"]$ 11^{1}=11$
$ 11^{2}=121$
$ 11^{3}=1331$
$ 11^{4}=14641$ You see the pattern? Every time its $ 11^{n}$, the last two digits are going to be n, and 1. So, since its 10, its going to be 01. Subtracting that from nine, taking one from the hundreds place, we have the two last digits are 92. [/hide]

[hide]the last digit is always going to be 1

suppose the tens digit of $ 11^{n}$ is $ a$.

...a 1
X 1 1
-----
(a+1) 1 mod 100

so it just adds one to the tens digit every time[/hide]

[hide="Approach requiring the least effort and the most theory"] $ gcd(n, 100) = 1 \implies n^{20}\equiv 1 \bmod 100$ by Carmichael's Theorem.  Hence we know that $ 11^{10}\equiv 01, 99 \bmod 100$.  Because we also know that $ 11^{10}\equiv 1 \bmod 10$, it follows that

$ 11^{10}\equiv 01 \bmod 100 \implies$
$ 11^{10}-9 \equiv \boxed{92}\bmod 100$ [/hide]


induction
number theory proposed
number theory

$ a)$ Prove that there isn't any sequence $ a_1<a_2<...<a_n<...$ of integers such that any difference $ a_i\minus{}a_j$ divides the sum $ a_i\plus{}a_j.$
$ b)$ Prove that for any positive integer $ p$, $ p \ge 2$ there is a finite sequence of $ p$ terms so that the property from $ a)$ takes place.

[hide]a) Fix $ j$ and take arbitrary $ i \equal{} j \plus{} k$ with $ k \geq 1$. Then $ a_i \minus{} a_j \mid a_i \plus{} a_j \equal{} (a_i \minus{} a_j) \plus{} 2a_j$, hence $ a_i \minus{} a_j \mid 2a_j$. But since the sequence is strictly increasing, $ a_i \minus{} a_j \geq i \minus{} j \equal{} k$. As $ k$ is arbitrary large, the only possibility is $ a_j \equal{} 0$, absurd, since $ j$ can be chosen at will.

b) We will proceed by induction on $ p$. For $ p \equal{} 2$ one can take $ a_1 \equal{} 1$ and $ a_2 \equal{} 2$. Assume a sequence built up to $ p$  terms, with $ a_1 > 0$. Take now $ a'_i \equal{} \frac {a_i} {a_i \plus{} 1} \prod_{k \equal{} 1}^p (a_k \plus{} 1)$ for all $ 1 \leq i \leq p$, and $ a'_{p \plus{} 1} \equal{} \prod_{k \equal{} 1}^p (a_k \plus{} 1)$. We still have $ 0 < a'_1 < a'_2 < \cdots < a'_p < a'_{p \plus{} 1}$. Then, for $ 1 \leq j < i \leq p$ we have $ a'_i \minus{} a'_j \equal{} (\frac {a_i} {a_i \plus{} 1} \minus{} \frac {a_j} {a_j \plus{} 1})\prod_{k \equal{} 1}^p (a_k \plus{} 1) \equal{} \frac {a_i \minus{} a_j} {(a_i \plus{} 1)(a_j \plus{} 1)}\prod_{k \equal{} 1}^p (a_k \plus{} 1)$ and also $ a'_i \plus{} a'_j \equal{} (\frac {a_i} {a_i \plus{} 1} \plus{} \frac {a_j} {a_j \plus{} 1})\prod_{k \equal{} 1}^p (a_k \plus{} 1) \equal{} \frac {a_i \plus{} a_j \plus{} 2a_ia_j} {(a_i \plus{} 1)(a_j \plus{} 1)}\prod_{k \equal{} 1}^p (a_k \plus{} 1)$. But $ a_i \plus{} a_j \plus{} 2a_ia_j \equal{} (a_i \plus{} a_j)(1 \plus{} a_j) \plus{} (a_i \minus{} a_j)a_j$, divisible by $ a_i \minus{} a_j$ by the induction hypothesis, therefore $ a'_i \minus{} a'_j \mid a'_i \plus{} a'_j$.
Now, for $ 1 \leq i \leq p$, we have $ a'_{p \plus{} 1} \minus{} a'_i \equal{} (1 \minus{} \frac {a_i} {a_i \plus{} 1})\prod_{k \equal{} 1}^p (a_k \plus{} 1) \equal{} \frac {1} {a_i \plus{} 1}\prod_{k \equal{} 1}^p (a_k \plus{} 1)$, while $ a'_{p \plus{} 1} \plus{} a'_i \equal{} (1 \plus{} \frac {a_i} {a_i \plus{} 1})\prod_{k \equal{} 1}^p (a_k \plus{} 1) \equal{} \frac {2a_i \plus{} 1} {a_i \plus{} 1}\prod_{k \equal{} 1}^p (a_k \plus{} 1)$, so clearly $ a'_{p \plus{} 1} \minus{} a'_i \mid a'_{p \plus{} 1} \plus{} a'_i$. Induction is complete. Since the sequence changes altogether when increasing the number of terms, there is no contradiction with the result at point a).
[/hide]


geometry
geometric transformation
reflection
circumcircle
trigonometry
homothety
parallelogram

[quote="Rushil"]In an acute angled triangle $ABC$, $\angle A = 30 ^{\circ}$, $H$ is the orthocenter, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2 BC$.[/quote]

Since HM = MT, the point T is the reflection of the point H in the point M. On the other hand, since M is the midpoint of the segment BC, the point C is the reflection of the point B in the point M. Since reflection in a point maps any line to a parallel line, we thus have CT || BH. But since H is the orthocenter of triangle ABC, we have $BH\perp CA$, and thus $CT\perp CA$. Thus, < ACT = 90º. Similarly, < ABT = 90º. Hence, the points C and B lie on the circle with diameter AT. In other words, the circle with diameter AT passes through the points A, B, C; thus, this circle is the circumcircle of triangle ABC. Hence, the segment AT is a diameter of the circumcircle of triangle ABC, so that we have AT = 2R, where R is the radius of the circumcircle of triangle ABC. But on the other hand, by the extended law of sines, $BC=a=2R\sin A=2R\sin 30^{\circ}=2R\cdot\frac12=R$, so that $AT=2R=2\cdot BC$. That's all.

  darij

Just for the end of the solution : instead of use of the law of sines, we have $\angle BOC=2\angle BAC$ (angle inscribed !) and then $\angle BOC=60^{\circ}$. Then triangle $BOC$ is equilateral, so $BC=BO=\frac{AT}{2}$
($O$ is the center of the cercle circumscribed to $ABC$.)

[quote="Rushil"]In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$, $H$ is the orthocenter, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2 BC$.[/quote]
Well, the homothety $H(H,2)$ take the nine point circle to the circumcircle, since the midpoints of $AH,BH,CH$ lie on the nine point circle. Also, it is well known that if $P$ is the midpoint of $AH$, then $PM$ is a diameter of the nine point circle. Therefore, $M$ goes to the point diametrically opposite to $A$. Thus, $AT = 2R$ and since $BC = R$ by easy law of sines, we're done. :)

My solution:
ATC is right triangle;
TC=BH...(TBCH is a parallelogram)
BH=2OE (O is circumcenter;E is midpoint of AC)
OAE is a right traingle
OE^2=R^2-1/4*b^2
AT^2=AC^2+CT^2
        =b^2+4(OE^2)
        =b^2+4(R^2-b^2/4)
        =4R^2
AT=2R

Let the circumcentre of $\Delta ABC$be at the origin.If $z_1,z_2,z_3$ be the complex numbers representing $A,B,C$ respectively.Then, $H=z_1+z_2+z_3$ and $M=\frac{z_2+z_3}{2}$.If $x$ represents the point $T$,then $\frac{x+z_1+z_2+z_3}{2}=\frac{z_2+z_3}{2}$
$\Rightarrow x=-z_1$.Therefore $AT=2R$
Since $BC=2RSinA=R$.
$AT=2BC$

$BTCH$ is a parallelogram so $\angle TBM=\angle MCH$,but $\angle MCH =\angle ABH$,so $BT$ is a tangent of $(ABD)${$D=AH \cap BC$},so $AB$ is perpendicular to $BT$,so $AT$ is the diameter of $ABTC$, ($ABTC$ cyclic because $\angle BTC=\angle BHC$),also $BOC$ is regular($O$ is the circumcenter),so $BO=BC$,hence done.

BM=MC and HM=MT.
so,BHCT is a parallelogram.
let X be a point on AC passes through the point H.
angle BXC=90 degree.
so, angle TCA is 90 degree.
And Y is a point on AB also passes through point H.
angle TYB=90 degree.
so,angle TBA=90 degree
Now we know that ABTC is cyclic.
AT is the diameter of that circle.
Let O be the circum center of that circle.so, angle BOC=2*angle BAC=60 degree.
OB=OC. and angle BOC=60 degree.
Then angle OBC=angle OCB=60 degree.
Now,OBC is a equilateral triangle.
AT=2*AO
AO=BO=CO.,
BC=OB=OC.
so,AT=2*BC.(proved)

Take a point F to be the midpoint of AH.Then AT = 2FM. So the problem reduces to show FM = BC.Now consider the Circumcentre O of the Triangle.
Then, 2OM = AH => OM = FH. Also angle chasing shows angle AMH = 90. Therefore, FM = FH = OM = BC (BM = MC = OM/2).So we are done.

Given, $HM=MT,\quad BM=MC$.
Thus, if we join $HC,CT,TB,BH,$ the diagonals of quadrilateral $CTBH$ bisect each other $\implies HCTB$ is a parallelogram.
In $\Delta ABE, \angle AEB= 90^{\circ}$ (as $BE$ is altitude). By angle sum property, $\angle ABE= 90^{\circ}-A$.
Similarly in $\Delta CEB$, $\angle CBE=90^{\circ}-C$. In $\Delta FCB, \angle FCB=90^{\circ}-B$.

$BH \parallel CT$ (parallelogram), $HC\parallel BT$
$\implies \angle HCB=\angle CBT=90^{\circ}-B$ (alternate angles)
$\implies \angle HBC=\angle TCB=90^{\circ}-C$ (alternate angles)
Therefore, $\angle ACT=270^{\circ}-(A+B+C)=90^{\circ}$
$\angle ABT= 90^{\circ} \implies ABTC$ is cyclic.
As $\angle ACT=90^{\circ}$, diameter= $AT=2R$. $T$ lies on the circumcircle of $\Delta ABC$.
$\angle A=30^{\circ}$
By sine rule, $\frac{a}{\sin A}=2R \implies \frac{BC}{\sin 30^{\circ}}=AT$

$\implies BC=AT/2 \implies AT=2BC$. Hence, proved!

Observe that from the Orthocentre Reflection Lemma in EGMO Ch 1, $AT$ is nothing but the circumdiameter of $\Delta ABC$. Now letting $R$ be the circumradius, and $a$ to be the length $BC$, we get $$2BC = 2a = 2 \cdot (2R \sin A) = 2 \cdot \left(2R \sin \frac{\pi}{6} \right) = 2R = AT,$$ so we are done. $\square$


Solve for x: $ 2^{x} \equal{} x2^{x\plus{}3}$

if you divide both sides by $ 2^x$, then wouldn't the problem reduce to
$ 1\equal{}x2^x\minus{}3\minus{}x$ because of division by exponents

this would reduce to $ 1\equal{}x2^3$
so $ 1\equal{}x8$
so $ x\equal{}1/8$

im gonna guess this is dead wrong, but im gonna go with it for now...

Triple post? Really?

I have no idea what you did, but you're right.  :huh:

$ 1\equal{}x\cdot 2^{x\minus{}x\plus{}3}\equal{}8x\implies \boxed{x\equal{}\frac{1}{8}}$.

holy crap...sry!
how did i do that? :huh:

Perhaps you did triple click on "Submit" button.


Just with compass (no ruler) find the center of a given circle.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=24888 .

  darij

thanks mr.darij for answering this much fast(wow)


linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

prove that for any $n\in \mathbb{N}$ and prime number p,there exist a group with order $p^n$ such that its center contains $p$ or $p^2$ or $p^3$ members (i'm not sure that i translate center correctly).

Is the center the set $\{x: xy=yx$ $\forall y\}$? If so, you translated correctly.

Here is a construction:
Let $G_m$ be the set of $m\times m$ matrices $A$ over the field of $p$ elements satisfying $A=I+N$ with $N$ strictly upper-triangular. This is a group under matrix multiplication, and its center has order $p$. Also, $G_m$ has $p^{\frac{m(m-1)}{2}}$ elements. For completeness, let $G_1$ be the trivial group.
(The center of $G_m$ is the subset of matrices which are zero except on the diagonal and in the upper right corner.)

By a well-known result, we can express $n$ as a sum of at most three triangular numbers; $n=\frac{a(a-1)}{2}+\frac{b(b-1)}{2}+\frac{c(c-1)}{2}$ (with the possibility that some of $a,b,c$ are $1$). Let $H_n=G_a \times G_b \times G_c$; then the center of $H_n$ is a product of one, two, or three cyclic groups each of order $p$.

Here's an improved result: for all $n>4$, there is a group with $p^n$ elements and $p$ elements in its center.

Again, consider $m \times m$ matrices over the field of $p$ elements of the form $I+N$, where $N$ is strictly upper-triangular. We restrict this further; the potentially nonzero elements $a_{ij}$ of $N$ satisfy $f(i) \le j$, with $f(2)=1$, $f(m)=m-1$, $f(j) < j$, and $f(j) \le f(j+1)$. By adjusting $f$, we can give this group $p^n$ elements if $2m-1 \le n \le \frac{m(m-1)}{2}$. In all cases, its center is the same group of $p$ elements.
The only unresolved case here is order $p^4$; groups of order $p^2$ are abelian so their center cannot have only $p$ elements, but that is not guaranteed for $p^4$.


Any idea what the cutoff will be this year for qualifying to AIME?

Just take the AMC....it is [i]so[/i] much easier to qualify.

[quote="chess64"]I hope it's lower than 68... if it is, then I have to get perfect scores on rounds 2 and 3 :([/quote]

Why not Round 4?

The deadline for the fourth round is March 13, and AIME and AIME II take place on March 7 and March 22, respectively.

The cutoff is determined by the combined score of the first 3 rounds.


[quote="Yimin Ge"]For positive reals $ a,b,c$ with $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$ prove:
\[ a\plus{}b\plus{}c\plus{}\frac{1}{abc}\geq 4\sqrt{3}\]
[/quote]
We have $ abc\leq\frac{1}{3\sqrt{3}}$ =>$ \frac{1}{abc}\geq 3\sqrt{3}$
$ a\plus{}b\plus{}c\plus{}\frac{1}{9abc}\geq\frac{4}{\sqrt{3}}$
Done :P !

[quote=Yimin Ge]For positive reals $a,b,c$ with $a^2+b^2+c^2=1$ prove:
\[ a+b+c+\frac{1}{abc}\geq 4\sqrt{3} \][/quote]
By AM-GM, we have
$a+b+c+\frac{1}{abc}\geq12\sqrt[12]{abc(\frac{1}{9abc})^9}=\frac{4\sqrt{3}}{3\sqrt[3]{a^2b^2c^2}}\geq\frac{4\sqrt{3}}{a^2+b^2+c^2}=4\sqrt{3}.$
[url=http://www.artofproblemsolving.com/community/c6h611053p3632912]Strengthen of Macedonia 1999[/url]


function
integration
real analysis
real analysis unsolved

Is there a family of continuous functions $f_n:[0,1]->[0,1]$ such that $\int_{0}^{1}{f_n(x)dx}$ goes to 0 and the sequence $ (f_n(x))$ diverges for all $x$?

Of course, but it's much easier to draw than to explain. I'll modify this post if I decide to do it :).


prove that the sum of the reciprocals of the lengths of the altitudes of a triangle equals the reciprocal of the length of the inradius.

I don't know if you want complete geometric reasoning, but this is my proof in any case:

[hide]$ \frac{1}{h_a}\plus{}\frac{1}{h_b}\plus{}\frac{1}{h_c}\equal{}\frac{1}{2S}\left(AB\plus{}BC\plus{}CA\right)$

$ \equal{}\frac{2s}{2S}\equal{}\frac{1}{r}$[/hide]


Let $ \lambda\geq 1$ be a natural number. Find the limit

$ \lim\limits_{n\rightarrow \infty}\left(\frac{1}{n}\right)^{n\lambda}\plus{}\left(\frac{2}{n}\right)^{n\lambda}\plus{}\cdots\plus{}\left(\frac{n\minus{}1}{n}\right)^{n\lambda}.$

For $ \lambda >0$ it equal  $ \frac{1}{e^{\lambda}\plus{}1}.$

No, answer is $ \frac {1}{e^{\lambda} \minus{} 1 }$
 :roll:
[hide="hint"]    approx ...              http://www.mathlinks.ro/viewtopic.php?t=200869                                        [/hide]

Yes. \[ \sum_{k\ge 1}e^{\minus{}k\lambda}\equal{}\frac{1}{e^{\lambda}\minus{}1}.\]

[quote="Rust"]Yes.
\[ \sum_{k\ge 1}e^{ \minus{} k\lambda} \equal{} \frac {1}{e^{\lambda} \minus{} 1}.
\]
[/quote]
and how can we use it to solve this problem ? :huh:


did anyone get this resolved??  Can you post the details as to how to get admitted in writing the SATs on the reserved day (and where?)

Try to talk to your guidance counselor; he should be able to help you if he wants, especially since it sounds like a lot of people from your school will be going to ARML.  I didn't really get the situation resolved.  I'm just taking it with another nearby team.

Yeah, I was planning on taking SATII Math on ARML day... I didn't exactly get it resolved, I'm just going to take this SAT in the fall.  I don't know what grade you're in, but I guess this isn't really an option for you... try to see if it is, though (I mean, check other SAT dates...though you've probably done this already...)

unfortunately, tis a burden for all of us ARMLers...I was planning on taking chem this year :-(

[quote="cats..."]Caroline, did you, or others from your school, try to talk to your guidance counselors?  I'm pretty surprised AAST won't arrange something.[/quote]

Well as far as I know, it's not the school that organizes it, it's the testing board thingy... They mandate the dates the tests have to be administered and stuff.  I'm not even sure which tests my school administers, as I'll probably be taking the regular SAT at my town high school which I live right next to.  ;)

[quote]Well as far as I know, it's not the school that organizes it, it's the testing board thingy...[/quote]

Actually, the school coordinates the rescheduling with Collegeboard, so I would just ask your school about it (and contact Collegeboard, informing them of the conflict).

[quote="**********"]the college board does not reschedule SATs for conflicts unless they're religious or something...tough luck...u gotta take them next fall i guess...

im taking mine this may (i.e. in 3 days.)[/quote]

Not true. Just ask any of the seasoned ARML takers around here.

[quote="**********"]the college board does not reschedule SATs for conflicts unless they're religious or something...tough luck...u gotta take them next fall i guess...

im taking mine this may (i.e. in 3 days.)[/quote]

Yep, Fiery is correct -- your school can reschedule for sanctioned academic conflicts as well, such as taking ARML.  I happen to know this because I've taken a rescheduled SAT twice due to ARML, as have many other people who have been on the NYC team.

Ooh thanks, I wasn't aware of this - I'll talk to guidance tomorrow and see if they can do anything :)

hmmm...i still think ur wrong...

MAYBE it's because i've already talked to both my school and Collegeboard and they specifically said there weren't any alt. testing dates (except like THREE MONTHS from now)... but i dunno, maybe both my college counselor and the collegeboard rep i talked to are utterly incompetent, which could very well be the case

anyways, tell me how it goes for you...

How can people who have taken the test on an alternate date because of ARML be wrong about the possibility of getting it changed?

[quote="gauss202"]Maybe the rule has changed since then, or maybe it was just a special exception made for them.[/quote]

They're doing it again this year, unless I am sorely mistaken.  I think they passed out forms about it when I was at NYSML.  Special exception seems possible but rather unlikely, to be honest.  Unfortunately, I don't have any knowledge of how to go about finding out about how NYC does it.

i was just told that no team can take it anymore.  I was going to take it with the Eastern MA team, but the coach was just called by ETS today, saying that they can't give an alternate date anymore.


inequalities
LaTeX
inequalities unsolved

Let $ a,b,c$ be non-negative numbers, such that $ a\plus{}b\plus{}c\plus{}abc\equal{}4$

Prove that:
\[ 3abc \le ab \plus{} bc \plus{} ca \le abc \plus{} 4\]

The first part is easy:
By AM-GM, we know $ \frac {ab \plus{} bc \plus{} ca}{3}\geqslant\sqrt [3]{a^2b^2c^2}$ and it suffices to prove that $ abc\leqslant 1$. This is clear from the condition because if $ abc > 1$, then also $ a \plus{} b \plus{} c > 3$ (from AM-GM), and, adding these two, $ a \plus{} b \plus{} c \plus{} abc > 4$. Thus, $ abc > 1$ is impossible and the first part is proved. We have equality iff $ a\equal{}b\equal{}c\equal{}1$.

My solution:

First Part:

By AM-GM we have:
        $ \frac{a\plus{}b\plus{}c\plus{}abc}{4}\geq\sqrt[4]{(abc)^2}\Rightarrow{1\geq abc}$ (1)

In the inequality we have:

   $ 3\leq\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}$ or  $ \frac{3}{\frac{1}{a} \plus{}  \frac{1}{b} \plus{}  \frac{1}{c}} \leq \sqrt[3]{abc}\leq 1$

By HM-GM and (1)

Solution to this problem
[b]
First Part[/b]

By using AM - GM inequality, we have

$ ab \plus{} bc \plus{} ca \ge 3\sqrt [3]{a^2b^2c^2}$

So it suffices to prove that

$ \sqrt [3]{a^2b^2c^2} \ge abc$ or $ latex abc \le 1$

By using AM - GM inequality we have

$ 1 \equal{} \dfrac{a \plus{} b \plus{} c \plus{} abc}{4} \ge \sqrt [4]{a^2b^2c^2} \Leftrightarrow abc \le 1$

The proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$

[b]Second Part[/b]

Without loss of generality, we may assume that $ c \equal{} max${$ a,b,c$}. From the condition, it is clear that $ c \ge 1$ because if $ c < 1$ then also $ a, b < 1$, and $ a \plus{} b \plus{} c \plus{} abc < 4$ (which is a contradiction)

By using AM - GM inequality, we have

$ 4 \equal{} a \plus{} b \plus{} c(1 \plus{} ab) \ge 2\sqrt {c(a \plus{} b)(1 \plus{} ab)}$

$ \Leftrightarrow c(a \plus{} b)(1 \plus{} ab) \equal{} bc \plus{} ca \plus{} abc(a \plus{} b) \le 4$

$ \Leftrightarrow ab \plus{} bc \plus{} ca \le 4 \plus{} ab \minus{} abc(a \plus{} b)$

So it suffices to prove that

$ ab \minus{} abc(a \plus{} b) \le abc$

$ \Leftrightarrow ab[c(a \plus{} b \plus{} 1) \minus{} 1] \ge 0$ (which is true)

The proof is completed. Equality holds if $ (a,b,c) \equal{} (0,2,2)$ or any cyclic permutation.

[quote="leviethai"]Let $ a,b,c$ be non-negative numbers, such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4$

Prove that:
\[ 3abc \le ab \plus{} bc \plus{} ca \le abc \plus{} 4\]
[/quote]
The second part can be strengthened. Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}abc\equal{}4,$ then
$ ab\plus{}bc\plus{}ca \le a\plus{}b\plus{}c.$

This inequality is equivalent to Vietnamese TST 1996 problem.

[quote="can_hang2007"][quote="leviethai"]Let $ a,b,c$ be non-negative numbers, such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4$

Prove that:
\[ 3abc \le ab \plus{} bc \plus{} ca \le abc \plus{} 4\]
[/quote]
The second part can be strengthened. Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4,$ then
$ ab \plus{} bc \plus{} ca \le a \plus{} b \plus{} c.$

This inequality is equivalent to Vietnamese TST 1996 problem.[/quote]

There is simple solution for this problem.

Assume for the sake of contradiction that $ ab \plus{} bc \plus{} ca>a\plus{}b\plus{}c$. Using Schur:
\[ \frac {9abc}{a \plus{} b \plus{} c}\ge 4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2 > abc(a \plus{} b \plus{} c)\]
Hence $ a \plus{} b \plus{} c < 3$, but using again the condition you get $ a \plus{} b \plus{} c\ge 3$.

[quote="can_hang2007"][quote="leviethai"]Let $ a,b,c$ be non-negative numbers, such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4$

Prove that:
\[ 3abc \le ab \plus{} bc \plus{} ca \le abc \plus{} 4\]
[/quote]
The second part can be strengthened. Actually, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \plus{} abc \equal{} 4,$ then
$ ab \plus{} bc \plus{} ca \le a \plus{} b \plus{} c.$

This inequality is equivalent to Vietnamese TST 1996 problem.[/quote]

Thanks to Can hang2007 to reveal my  mistake   :P   :lol:

[quote="Abdek"]
On the other hand we have the identity

$ \frac {(x \plus{} y \minus{} z)(y \plus{} z \minus{} x)}{xz} \plus{} \frac {(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)}{xy} \plus{} \frac {(z \plus{} x \minus{} y)(x \plus{} y \minus{} z)}{yz} \plus{} \frac {(x \plus{} y \minus{} z)(y \plus{} z \minus{} x)(z \plus{} x \minus{} y)}{xyz} \equal{} 4$[/quote]
I think you have some mistakes. This identity is related to $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4,$ while the original condition is $ a \plus{} b \plus{} c \plus{} abc \equal{} 4.$

[quote="Pain rinnegan"]Assume for the sake of contradiction that $ a \plus{} b \plus{} c > ab \plus{} bc \plus{} ca$. [/quote]
There is a little typo in your writing, Pain. It must be $ ab \plus{} bc \plus{} ca > a \plus{} b \plus{} c.$


Let $ N \equal{}$ {1,2,...}. Find all $ n \in N$ such that :
If $ (x,n) \equal{} 1$ then $ x^2 \equiv 1 (mod n)$

If prime $ p|n$, then for any $ 0<x<p$ $ x\equal{}1$ or $ x\equal{}p\minus{}1$. Therefore $ p\equal{}2$ or $ p\equal{}3$.
Therefore $ n\equal{}2^k3^m$. $ x\equal{}5$ give $ n|5^2\minus{}1\equal{}24$ and any divisor of 24 satisfyed these condition.

[quote="Rust"]If prime $ p|n$, then for any $ 0 < x < p$ $ x \equal{} 1$ or $ x \equal{} p \minus{} 1$. Therefore $ p \equal{} 2$ or $ p \equal{} 3$.
Therefore $ n \equal{} 2^k3^m$. $ x \equal{} 5$ give $ n|5^2 \minus{} 1 \equal{} 24$ and any divisor of 24 satisfyed these condition.[/quote]
I think your solution is wrong because "if prime $ p|n$ then for any $ 0 < x < p$ $ x \equal{} 1$ or $ x \equal{} p \minus{} 1$" but $ (x,n)$ can different 1.
For example: $ n \equal{} 30$, $ p \equal{} 5$ then $ x \equal{} p \minus{} 1 \equal{} 4$ but $ (x,n) \equal{} (4,30) \equal{} 2 \neq 1$

You forget $ (x,n)\equal{}1,n|x^2\minus{}1$. If $ p|n$ then $ (x,p)\equal{}1, p|x^2\minus{}1\equal{}(x\minus{}1)(x\plus{}1\to p\equal{}2 \ or \ p\equal{}3$.


(n^4)+((n-1)^4)...+1^4
What's the formula for this?

I know (n(n-1)(2n-1))/6 is for squares, and (n+(n+1)...+1)^2 is for cubes, but I'm not sure what/if the forumla for ^4 is. Anyone know? I tried to come up with it myself but failed miserably.

Dang, that'll be hard to remember. Thanks though :-P


The number 88 is what fractional part of 888?

 :o


Leila is preparing boxed lunches for a group of 150 people. The lunches are the same except for the sandwiches. The four sandwich options are ham, turkey, roast beef or vegetarian. From past experience, she knows she will need 30$ \%$ turkey, $ \frac{1}{3}$ as many vegetarian as turkey and equal quantities of ham and roast beef. How many roast beef sandwiches will she make?

Because she has four options, they make up 100% of all of the sandwiches she will make. One-third of 30% is clearly 10%, so 10% of the sandwiches will be vegetarian. Subtraction the amounts of turkey and vegetarian sandwiches from the total 100%, (100-(30+10))%=(100-40)%=60%. This 60% is to be evenly divided up into ham and roast beef sandwiches, so there will be 30% of each. The number of roast beef sandwiches to be made is therefore 30% of 150, which is $ \boxed{45}$.


geometry AMC AIME MATHCOUNTS LaTeX trigonometry ARML

hey guys... i m in need of solved problems.... if neone has compiled a word document or a .pdf file for number theory or geometry... pls do send it to me... i think someone has made a .pdf file on Problem Solving Marathons... if neone has .. pls do send me...

raul_vrules@yahoo.com

Please do not use so much chatspeak in your posts.  You do not have to use perfect grammar, but it is much nicer and easier to read when you do.   :)

Thanks,
[color=red]b-flat[/color]

Try
[code]Hey guys, I'm in need of solved problems. If anyone has compiled a word document or a .pdf file for number theory or geometry, please do send it to me. I think someone has made a .pdf file on Problem Solving Marathons. If anyone has, please do send them to me.[/code]

Try the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Top.php]AoPS resources section[/url]

If you look at many of the mock contests that have taken place, most of authors have posted solutions to their contests.

hey dude thanks a lot for dat...

act. it wud be very nice if neone can give a compiled .pdf file on marathons... . pls guys.... need ur help

haha yeah i wish some kind soul had made a pdf for the marathons. but hey we r all here asking for a pdf cause we couldnt be bothered to make one. i guess no one did :P haha

I don't see the point in anyone making one since it's always right here and you can access it whenever.

[quote="akimatsu"]but i prefer to print it out, sit down somewhere with pencil and paper to do the problems. not next to my computer where there are wires and everything[/quote]

Wires?  :huh: I have plenty of room in front of my moniter to put notebooks, calculators, and other mathematical instruments.

[quote="akimatsu"]but i prefer to print it out, sit down somewhere with pencil and paper to do the problems. not next to my computer where there are wires and everything[/quote]
Yeah i like to go sit on the couch and work problems. Desk = eww.

Regardless of your likes, dislikes, and printer problems, the main point is that the person asking the question would like a pdf of the marathons, and unfortunately there is none yet.
Considering the fact that I have nothing to do for 4-5 hours each day, I might try to make one... though that's not a guarantee...

gr8 work dude... many ppl will be very gr8ful to u for dat effort of urs.... thanx a millions dude.. in advance

alright.. actually i was told to write like dis

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=marathon&t=46376]Getting Started Forum Marathon[/url]

[url=http://www.mathideas.org/public/math/marathon.pdf]Intermediate Forum AIME Marathon[/url]

I once made a mathcounts one as well. If you are interested, PM me...

There are no answers to the AIME marathon?

I don't know if you guys still want these, but here is another.

GS2 is the one I made
Klebian made GS3
Playtoxin made GS1

Sorry, but I somehow cannot view these documents.

Ok. Ive done the MOEMs problem of the day...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109826

[quote="raul_v20"][b] guys wer r d solns for dis marathons. has neone already done it???????. is neone plannin to do dem..???? it wud be gr8 guys if we get d solns as well[/b][/quote]
Yeah, ok, as soon as you stop using chatspeak and as soon as I have some of my life to myself.

[quote="ragnarok23"]Ok, doing the MathCounts Marathon would be a lot easier IF IT WEREN'T LOCKED.[/quote]
PM a mod and ask if it can be unlocked temporarily. Make sure they post something threatening when it is unlocked.

[quote="akimatsu"]is anyone doing the solutions for these? or has this been sorta abandoned.

anyway thanks for the questions now i can print them out  :lol:[/quote]

I'm sorry, I kinda stopped mine, and I doubt I'll ever get around to finishing it.  :(

http://web.mit.edu/hmmt/www/datafiles/problems/

check that site out. u can print problem straight from it in various topics

I'll work on doing the mathcounts forum marathon...

Wait...can you give me a copy once it is done, or should i also make one

When you make something like this, it is always great to post the TeX file along with the PDF, so that other people can fix/add stuff.  Also, I made a marathon one time, now to find it...

EDIT: Here are the solutions, now to find the problems... :roll:

SC ARML Marathon:
Note that solutions are not complete, so if you edit it, please repost the new PDF as well as the new TeX.  Thanks!

[quote="Totally Zealous"]Well I have bad eyesight so when i work at the computer i have to stare really hard and my eyesight gets worse.
My printer's broken. :(  :([/quote]

I really understand how that feels. The Mac screen stopped it.

raul_v20... This is not a chatting website. people are actually trying to read your posts. Even though some people use chatspeak, not everyone does.

Friendly reminder: Please use regular typing.

P.S. I'm not good at making pdf's.  :noo:


For $a,b,c$ reals positive such that $a+b+c=3$ prove by Shur that

$\displaystyle (\sum {ab})^2-3abc \sum {a} \ge abc(3+3abc-2\sum{ab})$

Sorry for this post, but I thought it could be interesting as Vasc said he had solved it by Shur, but probably he was wrong as it can be easily solved by AM-GM........................or am I wrong? :?

Please can anyone check it?

[quote]After  homogeneization you obtain

$\sum a^2b^2(a^2+b^2) +2\sum a^3b^3 +2abc\sum ab(a+b) \geq 18a^2b^2c^2 +2abc \sum a^3$

and this can be solved by AM-GM[/quote]

Thank you very much.

[quote="manlio"]Sorry for this post, but I thought it could be interesting as Vasc said he had solved it by Shur, but probably he was wrong as it can be easily solved by AM-GM........................or am I wrong? :?

Please can anyone check it?

$\sum a^2b^2(a^2+b^2) +2\sum a^3b^3 +2abc\sum ab(a+b) \geq 18a^2b^2c^2 +2abc \sum a^3$

and this can be solved by AM-GM
Thank you very much.[/quote]

Manlio is right ,this ineq is easy to prove bay AM-GM:
Because:
$\sum{a^4(b^2+c^2)} \ge 2\sum{a^4bc}=2abc\sum{a^3}$

Manlio, you are right, I thought Cauchy, but I wrote Schur.
But I verified now that we can also use Schur's ineq. to prove that

$\di\displaystyle \displaystyle (\sum {ab})^2+2abc \sum {ab} \ge 3abc(4+abc)$.


geometry
combinatorics unsolved
combinatorics

A convex polygon is such that it cannot cover any triangle of area 1/4. Prove that it can be covered by some triangle of area 1.

Let $ P$ be the convex polygon.
It is not difficult to prove that among all the triangles which can be covered by $ P$ there is one, say $ T$, with maximal area and that its vertices are vertices of $ P$.
Let $ T \equal{} A_1A_2A_3$ (the vertices are not supposed to be consecutive on $ P$).
Let $ \Delta_1$ be the line through $ A_1$ which is parallel to the line $ A_2A_3$.
Similarly, we define $ \Delta_2$ and $ \Delta_3$.
Let $ C_1$ be the intersection of $ \Delta_2$ and $ \Delta_3$. Define $ C_2$ ad $ C_3$ similarly.

Now, it is easy to prove that $ C_1C_2C_3$ covers $ P$ and that $ A_1,A_2,A_3$ are the respective midpoints of segments $ C_2C_3,C_3C_1,C_1C_2$ .
Since $ T$ is supposed to have area not greater than $ \frac14$, the conclusion follows.

Pierre.


Hi! I have a question about Mafia.
I would've posted it in the Mafia board, but I wasn't sure this would get read.
So.......I joined AoPS recently and am interested in playing Mafia. Am I supposed to start out with a "noob game?" I haven't seen any recently. Should I go ahead and join the next game?
Thanks for any advice you may offer.

Go ahead and join next (don't miss the signups, they are really quick!). And you couldn't post a thread there, only mods can.

Well, I see that there are mixed feelings here, so I guess I will join whichever game comes first.  :lol:
By the way, you are still welcome to voice opinions.
I'd like to hear form many points of view.

Eh hem...
If you join a noob game:
Read the rules, the games and the roles, attempt to outsmart the experienced player :P
If you join a regular game:
Again, Read the rules, the games and the roles... over and over again. try your best and attempt to start discussion, dead game = bad game. Avoid any noob moves, (bandwagon, noob card, etc...).