linear algebra
matrix
LaTeX
algebra
polynomial
quadratics
linear algebra unsolved

Hi, is there a general method for finding out whether two matrices A and B are similar,and if yes, then what is the similarity transform P such that B=PAP^(-1)?
Thanx in advance..

Two matrices are similar iff their rank is the same........
I know 1 method: look to the first matrix as the matrix of quadratik form and find the transformation Xto its normal from, the find the transformation matrix Y that lead the 2nd to the normal(may be you say diagonal) from.
then$P=XY^{-1}$ :?:

Thanx, your answer is perfect.But actually I wanted something particular,so I should rephrase my question:
Can any 3X3 matrix be converted by a similarity transform to a matrix of the form:
 a    1     0
 b    0     1
 c     0     0
(Sorry I dont know how to write in LaTex here)
Thanx a lot again..
[/list][/code]

Not any, e.g. the identity matrix is not similar to such a matrix. A 3x3 matrix is similar to a matrix of such form iff its minimal polynomial's degree is equal to 3.

[quote="rustam"]Two matrices are similar iff their rank is the same........[/quote]
It is not true in general. The identity matrix is not similar to the diagonal matrix with $1,2,3\ldots$ on the diagonal.

Thanx lofar!!Can you give a proof of the claim that it is possible if and only if the matrix has a minimal polynomial of degree 3?Also, what would be the similarity transformation that would take a 3X3 matrix with a minimal polynomial of degree 3 to a matrix of such form?

Your problem is connected with the [url=http://en.wikipedia.org/wiki/Frobenius_normal_form]Frobenius normal form (rational canonical form)[/url] of a matrix. This theory is very similar to the theory of Jordan normal form. Matrix in Frobenius normal form has a block-diagonal structure. Each block is a [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url] of some polynomial. This theory is not very difficult, but it is rather extensive, so it is impossible to give a complete answer right in this forum.

[quote="lofar"]
[quote="rustam"]Two matrices are similar iff their rank is the same........[/quote]
It is not true in general. The identity matrix is not similar to the diagonal matrix with $1,2,3\ldots$ on the diagonal.[/quote]

I read in the book Gantmacher "Matrix Theory"(or the theory of matrix, Nauka, 2nd edition, 1967) the defenition:
We call two matrices A and B similar if there exists matrix P such that $A=PBP^{-1}$.
Then there follows theorem: [b]Any two quadratic matrices nxn are  similar iff their rank is the same [/b]

Or do I miss somethink? :?

Please, could you post the proof of your conjecture. May be i have misunderstood you...

It is wrong that two matices are similar iff they have the same rank.
Take for example two invertible matrices whose determinants are different.

But the right result is that two matrices are equivalent iff they have the same ranK (A et B are equivalent iff there exist P et Q two invertible matrices such that A=PBQ).


In how many ways can 8 students be divided into 2 groups, if each group must consist of at least 3 students?

$x_{1}+x_{2}=8$ if $3\leq x_{i}|i\in\{1,2\}$

change the variables

say $x_{1}= y_{1}+3$
$x_{2}= y_{2}+3$

$y_{1}+3+y_{2}+3 = 8$
$y_{1}+y_{2}= 2$

the answer to that Q is $D (2,2) = \binom{2-1+2}{2-1}= \binom{3}{2}= \frac{3*2}{2!}= 3$

of course you have to choose which students will go into which group
(that is important because every student is different. If they were objects, and were all the same, then the answer would be just 3)

the first 3 students to group $x_{1}$ can be chosen in $\binom{8}{3}$ ways

the other 3 students to group $x_{2}$ can be chosen in $\binom{5}{3}$ ways
and you multiply that by the number of ways you can arrange the other 2 students (which is 3)

so the answer would be $\binom{8}{3}* \binom{5}{3}* 3 = \frac{8*7*6}{3!}* \frac{5*4*3}{3!}*3 = 56*10*3=1680$

[quote="Lena"][quote="aviateurpilot"]it's $\sum_{k=3}^{5}\binom{8}{k}$[/quote]

how does $\binom{8}{3}+\binom{8}{4}+\binom{8}{5}$ fit in the question ?[/quote]

A:first group; B:second group

A | B
3 | 5 => $\binom{8}{5}$
4 | 4 => $\binom{8}{4}$
5 | 3 => $\binom{8}{3}$


inequalities
function
logarithms
number theory
prime numbers
number theory proposed

Is it strictly true that $ \vartheta(x) \leq x$ for all x > 0?

Dusart stated that this was true up to $ 10^{11}$, but didn't say whether it was true afterwards.

Remember that $ \vartheta(x)$ is the natural logarithm of the product of all prime numbers not exceeding x.

UPDATE: I found on Wikipedia that Erhard Schmidt proved that for any real positive K,

$ \psi(x) \minus{} x > K\sqrt{x}$ infinitely many times.

So if we set $ \psi(x) \minus{} x > 2\sqrt{x}$ then that would definitely imply $ \vartheta(x) > x$, right?


AMC AIME USA(J)MO USAMO HMMT function geometry

Is there a lot concepts in AIME connected to precal?
Do you think Aops2 is enough for AIME preparation?

There are some concepts in the AIME that are 'covered' in precalculus, but by and large, most of your training for the AIME will have to come outside the typical classroom.  While the standard curriculum may give you the basic tools needed for some of the problems, it won't show you how to use them to tackle harder problems.  (Also, precalculus is in no way *required* for success on the AIME.  Every year, many, many students qualify for the USAMO without ever taking a precal class.)

Volume 2 will be helpful, but don't just study from the book - do lots of practice problems.  Past AIME tests are essential practice, and the old HMMT tests are very good, too.

Yeah...you shouldn't really think about the topics covered on the AIME in terms of conventional high school classes, as much of what you need to know is never actually covered in school.

Well, if I'd say you had to take a "course" before taking the AIME, I'd probably say Art of Problem Solving I and II  :D *cough* get the books if you don't have them *cough*

It depends when you say enough... the book is enough for you to get at least a 12 if you have mastered all of the concepts extremely well. But to get questions 13 through 15, get Engel or something, because those questions are usually Olympiad-level.

No, one does not have to go through "precal" before AIME.

As you can see from prior years' AIME contests, a few questions with
(1) complex numbers
(2) trigonometric functions, graphs and identities
(3) sequences and series
(4) function notation
occur on the AIME.  A student who wishes to solve these problems or understand the published solutions needs to be familiar with these mathematical concepts.  These topics are typically, but not always or exclusively, covered in a "precal" course.  Of course, it does not matter where or when or how the student learns the concepts.   You should find your best learning plan and follow that.

Remember too, AMC 10, AMC 12 and AIME questions often cross course and curricular boundaries, combining ideas that occur in several courses.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="Kent Merryfield"]I remember a certain student, a 9th grader taking geometry at the time, correctly answering an AIME question that involved complex roots of unity and trigonometric identities (half-angle formulas), all tied together in a probability question. I came to the conclusion that anyone who could answer that particular question didn't really need a precalculus course and could safely skip over it.[/quote]

But there could [i]possibly[/i] be some "precalculus" topics that they have not been exposed to -- like Inverse Functions, Logarithms, Polar Functions, Matrices, Induction, etc. Granted, if they'd learned that much on their own already its likely they would learn all of those topics on their own in due time also. But someone should at least give them some guidance to make sure those topics get covered before they get to Calculus.

I think logarithms and inverse functions show up in Alg 2, and if you know that well enough there's no reason why you can solve an AIME problem about it. Generally, I find that the logs questions are among the easiest on the AIME.

At my school an 8th grader taking Geometry made USAMO (aops: redcomet46).

I think this question's easier when reworded.

Can student cover all the concepts, including the ones in precalc, by Art of Problem Solving books?

The answer is YES. Some chapters overlap in two volumes so if you haven't exposed yourself to complex numbers or such topics, I suggest get both volumes and learn the materials.

Moreover, there is NO question that tests limit, continuity, derivative, or integral or those calculus materials (I'm saying these because precalc book does briefly cover this) so yes like people above said, outside classroom work is way more important than classroom curriculum. I mean, you learn good deal of concepts in the school but AMC never acts nice and puts them in format as "ok, I'm going to give you logarithrm problem so simplify first and use one of the properties of logarithrm."  :rotfl:


geometry
geometric transformation
rotation
pigeonhole principle

A magician's assistant holds a standard deck of 52 cards. An audience member selects five cards from the deck and gives them to the assistant. The assistant reads off four of the cards (for example, five of clubs, seven of diamonds, jack of clubs, nine of hearts). Find a way that the assistant and the magician can collaborate so that the magician will always be able to announce the fifth card correctly.

Ah! This is a fun question. It's a lot like the game "Broom Over". :)

Would you mind describing this game? I haven't heard of it before.

Well, its easy to tell which suit its going to be. Because we have 5 cards, we are going to have at least 2 cards that are the same suit. So all the assistant has to do is pick one of the cards with a shared suit (like hearts) and leave that one behind. Then, the first card he announces will be a hearts one. Ill work on the specific card one as im eating.

there are 4!=24 way of encoding a number.  but 52-5=47 cards.  then the assistant can cheat a little and perhaps asks a question with even/odd number of words that that will consittute 2*24=48 numbers encoded.

Well, what exactly can the assistant do?  I mean, it's theoretically possible to design a simple code for every card, based on the number of rotations he applies to each card or something.  Or some modular or digit-based forumla (digits mod 2?)  Actually, that's probably it -- use a base 2 system for naming each of the numbers of the cards 1 to 13, by some method (number of rotations, or face-up vs. face-down, for example) combined with Gauss's idea for suit.

I'm not exactly sure what constitutes "cheating" and what constitutes acceptable behavior.

The assistant only has flexibility in which cards he reads (4 out of 5) and which order he reads them.

I'm sorry, I didn't really read the question very closely ... I'll think about this some more.

Well, I don't know if this is the simplest solution, but I think it works.

Spoiler: [hide]As jelyman said, by pigeonhole, some suit has at least two cards, call them x and y.  Think of the 13 ranks as being laid out in a circle.  The two cards x and y must be within distance 6 of each other on this circle.  Without loss of generality, suppose that going from x to y clockwise is at most 6 units.  Then the assistant will read x first, and make y the 5th card.  There are 6 choices for y.  But with the other 3 cards, there are 6 relative orderings we can choose.  (Put an arbitrary total ordering on all of the cards.)  So map each of the 6 possible orderings to the numbers 1..6.  Choose the ordering that gives the distance from x to y.

I don't know if what I said makes sense, but I'm feeling lazy.[/hide]

spoiler:[hide]

I was thinking along the same lines as Ravi, you choose the card of the same suit that is a distance less than 6 from the other card, and then you can use lexicographic permutations of the other 3 cards to tell what number to add to that card mod 13, so you would arrange it by number first, and then by suit, and get the 6 permutations.[/hide]

Those solutions look good Ravi B and zscool. I haven't found any other way.

Wait -- but how does he indicate clockwise vs. counterclockwise?  You said that he could go WLOG clockwise -- but what if it isn't clockwise?  He needs a whole new set of directions, doesn't he?

[hide]They can decide that it will always go clockwise (for example) and then decide which of the two in that suit the magician must guess. I didn't word that very well, but do you understand what I mean?[/hide]

Oh, I see.  That was rather obvious, yes.  Since he has the choice, it is WLOG.

Now..., for a real puzzle let us change the rules slightly:

Find a way that the assistant and the magician can collaborate so that the magician will always be able to announce the fifth card correctly IF an
audience member  is also allowed to select the fifth card (instead of the assistant).

Not true, see my reply on this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=15841


A rectangular picture frame is made from one-inch-wide pieces of wood.  The area of just the frame is $ 18$ inches, and one of the outer edges of the frame $ 5$ inches long.  What is the sum of the lengths of the four interior edges of the frame?

[asy]size(5cm,5cm);

draw((0,0)--(5,0)--(5,7)--(0,7)--(0,0));

draw((1,1)--(4,1)--(4,6)--(1,6)--(1,1));

fill(((1,1)--(4,1)--(4,6)--(1,6)--cycle),darkblue);

draw (shift(0, 0.5)*((0,7)--(5,7)), Bars);

label("$5''$",(2.5,7.5),N);

draw (shift(0, -0.5)*((4,0)--(5,0)), Bars);

label("$1''$",(4.5,-0.5),S);

draw (shift(0.5,0)*((5,0)--(5,1)), Bars);

label("$1''$",(5.5,0.5),E);[/asy]

One side of the interior picture is 5-1-1=3 Let the other side of the picture be x.
Subtracting the area of the picture from the area of the entire frame+picture is equal to 18.
So, 5(x+2)-3x=18
Solving, x=4
So, the lengths of the interior sides of the frame are 3, 3, 4, 4.
Adding these up gives us [b]14 inches [/b]as the answer.


[quote="hanschinaman"]$0<a,b,c\leq 1$ such that $a^{2}+b^{2}+c^{2}=2$, prove
$\frac{1-a^{2}}{b}+\frac{1-b^{2}}{c}+\frac{1-c^{2}}{a}\leq 1.25$[/quote]
Let $x=b^{2}+c^{2}-a^{2},$ $y=a^{2}+c^{2}-b^{2}$ and $z=a^{2}+b^{2}-c^{2}.$
Hence, $x\geq0,$ $y\geq0,$ $z\geq0$ and $x+y+z=2$ and $\frac{1-a^{2}}{b}+\frac{1-b^{2}}{c}+\frac{1-c^{2}}{a}\leq 1.25\Leftrightarrow$
$\Leftrightarrow\frac{x}{\sqrt{x+z}}+\frac{y}{\sqrt{y+x}}+\frac{z}{\sqrt{z+y}}\leq\frac{5}{4}\cdot\sqrt{x+y+z},$ which obviously true. :D
( It's  known Garfunkel's inequality )

Sorry, but I haven't heard of the Garfunkel's inequality, can you tell me where I can find a proof of it, thanks!


search
abstract algebra
group theory
superior algebra
superior algebra unsolved

Hi,
It is easy to find the list of groups of order (i.e. < 30), for instance http://opensourcemath.org/gap/small_groups.html
I'd like to either find the list for some bigger groups
To start with, what are the groups of order
32
50
54

or to find out a way to determine them with GAP.
Thank you.

The problem is that there are a whole lot of those groups, and many of them don't have standard names. For example, [url=http://www.math.uga.edu/~lvalero/cohohtml/groups_32_index.htm]here[/url] are the 51 groups of order $ 32$.

How did I find it? Google search, first result.

I would have thought groups of order 50 were fairly easy to deal with. The Abelian groups $ C_{50}$ and $ C_{10}\times C_5$ and the dihedral group $ D_{25}$ are the immediate candidates. There’s also the group $ D_5\times C_5$ as well as the cyclic extension of $ C_5\times C_5$ by $ C_2$ induced by the automorphism $ x\mapsto x^{ \minus{} 1}$ of $ C_5\times C_5.$ The last two groups are clearly not isomorphic to each other, and as they don’t have any cyclic subgroup of order 25, they aren’t isomorphic to $ D_{25}$ either.

I’m sure these are the only groups of order 50. :D

In general, powers of 2 are by far the worst. 54 is pretty easy, too: there's a normal subgroup of order 27 (five possibilities), and then an element of order two which acts by conjugation as some automorphism.

[quote="Fleury"]Thanks jmerry and ValentineA,
I take it that there is no list of groups of order <100 for instance, nor a "magic GAP command" to list them.[/quote]

gap> List(AllSmallGroups([50]),StructureDescription);
[ "D50", "C50", "C5 x D10", "(C5 x C5) : C2", "C10 x C5" ]

There are 1032 groups of order <100, so probably you don't want a list of them all.

gap> Sum([1..99],NrSmallGroups);
1032

If you skip orders 48, 64, 80, and 96, then there are only 379 such groups, which might be a little easier to digest.

Perfect, thanks a lot JackSchnidt, I'll install GAP right now.


A tanker carries $ 10,000$ gallons of gasoline, Gasoline weights $ 5.99$ pounds per gallon at $ 60$ degrees Fahrenheit and $ 6.25$ pounds per gallon at zero degrees Fahrenheit. How many pounds heavier is the tanker when it carries a full load at the colder temperature than at the warmer temperature.

It's 0.26 pounds heavier for one gallon, multiply by 10000 to get 2600 pounds.


vector
inequalities proposed
inequalities

[quote="thanthuong"]show that : x, y, z are real numbers.
                   X+Y+Z = 0
                   X^2+Y^2+Z^2=6
Find the max , min :  X^2.Y+Y^2. Z +Z^2.X[/quote]

Let $f(x,y,z) = x^{2}y+y^{2}z+z^{2}x, \, g(x,y,z)=x+y+z=0$, and $h(x,y,z)=x^{2}+y^{2}+z^{2}=6$. According to the method of Lagrange multipliers, the extrema of $f(x,y,z)$ occur when:

\[\vec{\nabla f}(x,y,z)=\lambda\vec{\nabla g}(x,y,z)+\mu\vec{\nabla h}(x,y,z), \, g(x,y,z)=0, \mbox{ and }h(x,y,z)=6\]

the vector equation is

\[< 2xy+z^{2},2yz+x^{2},2xz+x^{2}\geq \lambda< 1,1,1>+\mu< 2x,2y,2z>\]

component-wise these are

\[2xy+z^{2}=\lambda+2\mu x\qquad\qquad (1) \]
\[2yz+x^{2}=\lambda+2\mu y\qquad\qquad (2) \]
\[2xz+x^{2}=\lambda+2\mu z\qquad\qquad (3) \]

summing (1), (2), and (3) gives

\[2(xy+yz+xz)+x^{2}+y^{2}+z^{2}=3\lambda+2\mu (x+y+z)\]

but $x^{2}+y^{2}+z^{2}=6$ and $x+y+z=0$ so the prior equation becomes

\[2(xy+yz+xz)+6=3\lambda\qquad\qquad (4) \]

...break... may post more later, or perhaps one of you mathlinkers will finish it?


LaTeX quadratics inequalities logarithms

[quote="Magnara"]Find the smallest value with $x>0$ of $5^x+5^{\frac{1}{x}}$.[/quote]

 AM-GM like this:$5^x+5^{\frac{1}{x}}\geq2\sqrt{5^(x+\frac1x)}$ but $x+\frac{1}{x}\geq2$ the conclusion follows.We have equality when $x=1$ ;)

Here's an interresting problem:

   Solve the equation: $4^x+4^(\frac{1}{x}) =2(2^x +2 ^(\frac{1}{x})) $ ;)

Sorry, I didn't use AM-GM, I got $x=1$ simply because it was fairly obvious because a fractional answer would simply make $5^{\frac{1}{x}}$ larger.

[quote="cezar lupu"]Here's an interresting problem:

   Solve the equation: $4^x+4^(\frac{1}{x}) =2(2^x +2 ^(\frac{1}{x})) $ ;)[/quote]

$x=1$ That should be the only answer. Because:

$4^x+\frac{4}{x}=2^{x+1}+\frac{4}{x}$ Thus the $\frac{4}{x}$ cancels and you have this left: $4^x=2^{x+1}$, thus $2^{2x}=2^{x+1}$, so $x=1$.

You avoid my first question,and about the other equation I would love to see a proof.What do you think? ;)  ;)  ;)

[quote="cezar lupu"]can u elaborate? ;)[/quote]

What do you mean? :| I just thought it was obvious, I mean, if $x$ was a fraction, when it was divided by $1$, it would become the reciprocal of that, thus the power of $5$ would get larger.

[quote="cezar lupu"]Ok,have you solved the second equation? ;)[/quote]

The one you made, yes...I answered it right on top...$x=1$.

[quote="cezar lupu"]Ohhh,it's my fault.My bad Latex again. It's not $4. 1/x$ it is $4^ {1/x}$ and for $2$ the same thing.Can you post it
 correctly,please? ;)[/quote]

OKay:

$4^x+4^{\frac{1}{x}}=2(2^x+2^{\frac{1}{x}})$

[quote="riddler"]Is the answer $x=1 and -1/2$?[/quote]

$\frac{-1}{2}$ would not work because it would not be equal to $0$ when you plug it in riddler ;) And if your talking about the minimum problem proposed by magnara, remember that $x>0$.

[quote="riddler"][quote="Iversonfan2005"][quote="cezar lupu"]Yeah,with $x$ real number.Thanks a lot and good luck in solving it. ;)[/quote]
Answer

[hide]$x=1$ because:

$4^x+4^{\frac{1}{x}}=2^{x+1}+2^{\frac{x+1}{x}}$
$2^{2x}+2^{\frac{2}{x}}=2^{x+1}+2^{\frac{x+1}{x}}$
$2x^2+2=x^2+2x+1$
$x^2-2x+1=0$
$x=1$[/hide][/quote]
How dyou just equate the equation?????
That is not possible......... or is it?[/quote]

I believe it is, worked out didn't it? :lol:

[quote="cezar lupu"]Hey,IversonFan2005 you are wrong.Riddler is right. You just can't do that. :o  :o  :coolspeak:[/quote]

 :o But $x=1$ does work out? That should be the only solution right? :o

[quote="riddler"][quote="Iversonfan2005"][quote="cezar lupu"]Yes.In fact $x=1$ is the only solution of the equation,but I want to see a proof if you don't mind. ;) .[/quote]

I thought my solution was fine? Why can't you just equate it like I did? :?[/quote]
You cant just equate things like that because .... there are exceptions  where this doesnt work at all. But I cant think of them.[/quote]

Oh well, that's how I always solved to find the answers to problems like these, however, if you want another solution, lemme work on it.

[quote="cezar lupu"]Well,I'd like to see those "lemme". ;)[/quote]

Well, okay, I don't really have another solution, guess I just got lucky :blush: HOw would you do it, Cezar?

I got to the $a^2+b^2=2a+2b$ bit then I got stuck but is there a simpler way to do it....  without Cauchy thingy.

I guess by calculations you got $(a-b)^2\leq0$.Am I wrong? :what?:

No I guessed it after that and got a quadratic equation which gave $x=1 and -1/2$. :blush:

[quote="riddler"]No I guessed it after that and got a quadratic equation which gave $x=1 and 2$. :blush:[/quote]

Yes, but $2$ obviously wouldn't work, so it would be $1$ as the only solution.

[quote="cezar lupu"]Here it is: we denote: $2^x=a>0$ and $ 2^{\frac{1}{x}} =b $. So, the equation becomes $a^2+b^2=2(a+b) $.
Let's assume that $x>0$ because if $x<0$ we obtain a contradiction.By Cauchy-Schwarz,we have $2(a^2+b^2)\geq(a+b)^2 $ and we deduce $a+b\leq2$ and from here is the same idea on applying AM-GM. ;)[/quote]
i think i grab your idea until Cauchy. i don't know Cauchy but i think you got it from $a^2+b^2=2ab$ to get $2(a^2+b^2)\geq(a+b)^2$ and so $a+b\leq 4$ no? you put wrong number. by AM-GM, $a+b=2^x+2^{\frac1{x}}\geq 2\sqrt{2^{x+\frac1{x}}}\geq 4$. Hence $a+b=4$ and only value for $x$ is $x=1$

now im trying to solve your second equation...

[quote="cezar lupu"]In connection to minimum here is an interresting problem:

   Solve the following equation: $3a^{2x}=4a^{3x}-a^{6x}$ where $a>0$ and $a\neq1$

  Cheers, :D[/quote]
[hide="hiding is fun"]since $a>0$, divide through by $a^{2x}$ to get $3=4a^x-a^{4x}$. let $t=a^x$.
then $t^4-4t+3=0\Rightarrow (t-1)(t^3+t^2+t-3)=(t-1)^2(t^2+2t+3)=0$. now $t^2+2t+3=(t+1)^2+2>0$ for any $t$, $t=1$ gives $x=0$
im not sure what inequality you want us to use here thou...  :?[/hide]

$3a^{2x} = 4a^{3x} - a^{6x}$

$\Rightarrow \frac{3a^{2x} + a^{6x}}{4} = a^{3x}$

By "weighted" AM-GM we get:

$\Rightarrow \frac{3a^{2x} + a^{6x}}{4} \geq \sqrt[4]{(a^{2x})^3 \cdot (a^{6x})^1} = a^{3x}$

With equality $\text{iff } a^{2x} = a^{6x}$. Since $a>0$, we can divide both sides by $a^{2x}$ to get $a^{4x} = 1$. Taking the natural log of both sides we get $\ln(a^{4x}) = 4x \ln a= \ln 1 = 0$. $a \neq 1 \Rightarrow \ln a \neq 0$, which means we can divide both sides by $\ln a$ without changing anything. Doing so, we get $4x = \frac{0}{\ln a} = 0 \Rightarrow x = 0$.

***I know this is intuitively true, but I used natural logs to make it a little more rigorous because sometimes intuition is not good for you.

[quote="Pork_Chop8"]$3a^{2x} = 4a^{6x} - a^{3x}$

$\Rightarrow \frac{3a^{2x} + a^{3x}}{4} = a^{6x}$

By "weighted" AM-GM we get:

$\Rightarrow \frac{3a^{2x} + a^{3x}}{4} \geq \sqrt[4]{(a^{2x})^3 \cdot (a^{3x})^1} =\textdollar$[/quote]
nope, thats the wrong equation you are using, 6x 3x mix-up and there seems to have a little problem with the $\LaTeX$

Cezar, you triple posted and double posted in the same thread!   :bruce:

I'm gonna edit them.  :police:

P.S. Could you edit your exponents so they show up in $\LaTeX$ right?

I propose some dinky little excerise, decided to check on the thread, and there are 37 replies.  I was expecting 1.

Your thread transformed in another one  :w00tb: This is stange...


In $ \triangle ABC$, $ AB \equal{}5$ cm, $ BC\equal{}10$ cm, and the altitude drawn to $ \overline{AB}$ is 8 cm. What is the number of centimeters in the length of the altitude to $ \overline{BC}$?

The area is:

$ \frac {8\times 5}{2}\equal{}20$

If the altitude to BC is x, then $ \frac{10x}{2}\equal{}20$, so x=4.


calculus
integration
logarithms
calculus computations

can anybody help with those integrals?? :)

a)int(1/(exp{x}+1))dx

b)int(sin4xcos5x)dx

c)int(cosx/(((sinx)^3)+sinx)))[/url][/quote][/u]

$ \int \frac{dx}{e^x\plus{}1} \equal{} \int \frac{e^{\minus{}x} dx}{1\plus{}e^{\minus{}x}} \equal{} \minus{}\ln|1\plus{}e^{\minus{}x}|\plus{}C \equal{} x \minus{} \ln|1\plus{}e^x|\plus{}C$

[quote="kcn2rivers"]$ \int \frac {dx}{e^x \plus{} 1} \equal{} \int \frac {e^{ \minus{} x} dx}{1 \plus{} e^{ \minus{} x}} \equal{} \minus{} \ln|1 \plus{} e^{ \minus{} x}| \plus{} C \equal{} x \minus{} \ln|1 \plus{} e^x| \plus{} C$[/quote]

can you please give me some details about that?

$ \frac {1}{e^x \plus{} 1} \equal{} \frac {1}{e^x \plus{} 1} \cdot \frac {e^{ \minus{} x}}{e^{ \minus{} x}} \equal{} \frac {e^{ \minus{} x}}{e^x e^{ \minus{} x} \plus{} e^{ \minus{} x}} \equal{} \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}}$.

Alternatively,

$ \frac {1}{e^x \plus{} 1} \equal{} \frac {e^x \plus{} 1 \minus{} e^x}{e^x \plus{} 1} \equal{} \frac {e^x \plus{} 1}{e^x \plus{} 1} \minus{} \frac {e^x}{e^x \plus{} 1} \equal{} 1 \minus{} \frac {e^x}{e^x \plus{} 1}$,

from which it follows that

$ \int \frac {dx}{e^x \plus{} 1} \equal{} \int \left( 1 \minus{} \frac {e^x}{e^x \plus{} 1}\right) dx \equal{} x \minus{} \log (e^x \plus{} 1) \plus{} C$,

with the substitution $ u \equal{} e^x \plus{} 1$, $ du \equal{} e^x \, dx$.

$ \int sin(4x)cos(5x) dx\equal{}\frac{1}{2}\int(sin(9x)\minus{}sin(x))dx\equal{}\frac{1}{18}(9cos(x)\minus{}cos(9x)$

let $ t\equal{}sin(x)$ so $ dt\equal{}cos(x)dx$

$ \int \frac{cos(x)}{sin^3(x)\plus{}sin(x)}dx\equal{}\int \frac{dt}{t(1\plus{}t^2)}\equal{}ln(t)\minus{}\frac{1}{2}ln(1\plus{}t^2)\equal{}ln(sin(x))\minus{}\frac{1}{2}ln(1\plus{}sin^2(x))$

[quote="mehdi cherif"]let $ t \equal{} sin(x)$ so $ dt \equal{} cos(x)dx$

$ \int \frac {cos(x)}{sin^3(x) \plus{} sin(x)}dx \equal{} \int \frac {dt}{t(1 \plus{} t^2)} \equal{} ln(t) \minus{} \frac {1}{2}ln(1 \plus{} t^2) \equal{} ln(sin(x)) \minus{} \frac {1}{2}ln(1 \plus{} sin^2(x))$[/quote]

thanks a lot!! :)  but can you explain this:
$ ln(t)\minus{}\frac{1}{2}ln(1\plus{}t^2)$[/quote]
how did you get there?pleaseee.. :!:

hello, you must observe that
$ \frac {1}{t^3 \plus{} t} \equal{} \frac {1}{t} \minus{} \frac {t}{1 \plus{} t^2}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you muwst observe that
$ \frac {1}{t^3 \plus{} t} \equal{} \frac {1}{t} \minus{} \frac {t}{1 \plus{} t^2}$
Sonnhard.[/quote]

ok..thnk you vry much! :lol:


A homogeneous thin rod of length $ l$ (see figure) is tied with an inextensible massless string to a wall. The point at which the string is tied to the  rod is at a distance of $ \frac {l}{3}$ from the end of rod touching the wall. The angle made by the string with the wall is $ \alpha$ and that made with the rod is $ (\pi - \beta)$.

Given that the rod is in equilibrium, find all possible values of the coefficient of friction $ \mu$.

[hide="My attempt:"]
First of all, I get only one value of $ \mu$ (against the proposed "[b]all values[/b]") and that too a wrong one (although most of the terms in my answer are the same as those in the solution, some terms are mysteriously missing!)

I proceeded as follows, (refer figure):

Equilibrium along $ X$ and $ Y$ axes,

\begin{eqnarray}
mg = \mu N + T\cos \alpha
\\
N = T\sin \alpha
\end{eqnarray}

Rotational equilibrium:

$ \frac {l}{6}\sin \beta (T \cos \alpha) + \frac {l}{2} \cos \beta (\mu N) + frac{l}{6} \cos \beta (T\sin \alpha) = \frac {l}{2} (N \cos \beta)$

Simplifying further,

$ \mu = 1 - \frac {\sin (\alpha + \beta)}{3\sin \alpha \cos \beta}$
[/hide]

Could you fix the Latex in your "Rotational equilibrium" equation? Seems strange...
Maybe posting (hiding) the solution you've mentioned would help...

[quote="I"]
Rotational equilibrium:

$ \frac {l}{6}\sin \beta (T \cos \alpha) \plus{} \frac {l}{2} \cos \beta (\mu N) \plus{} \color{red}frac{l}{6} \cos \beta (T\sin \alpha)$ $ \equal{} \frac {l}{2} (N \cos \beta)$

[/quote]

Sorry, it is:

Rotational equilibrium:

$ \frac {l}{6}\sin \beta (T \cos \alpha) \plus{} \frac {l}{2} \color{red}\sin \beta (\mu N) \plus{} \color{red}{\frac {l}{6} \cos \beta (T\sin \alpha)}$ $ \equal{} \frac {l}{2} (N \cos \beta)$

[quote="I"]
The angle made by the string with the wall is $ \alpha$ and that made with the rod is $ (\pi \minus{} \beta)$.

[/quote]

Actually, I must say:

[b]The angle made by the string with the wall is $ \alpha$ and that made with the rod is $ \color{red}(\pi \minus{} (\alpha \plus{} \beta))$.[/b]

Simply put, the angle between the rod and the wall is $ \beta$

I get

$ \frac{l}6 \sin{\beta}(T \cos \alpha)\plus{} \frac{l}6 \cos{\beta}(T \sin{\alpha}) \plus{} \frac{l}2 \color{red}\sin{\beta}\color{black}(\mu N) \equal{} \frac{l}2 (N \cos{\beta})$

By the way, in your drawing, isn't the angle between the string and the rod $ (\pi \minus{} \alpha \minus{} \beta)$ rather than $ (\pi \minus{} \beta)$?

[quote="ffao"]I get

$ \frac {l}6 \sin{\beta}(T \cos \alpha) \plus{} \frac {l}6 \cos{\beta}(T \sin{\alpha}) \plus{} \frac {l}2 \color{red}\sin{\beta}\color{black}(\mu N) \equal{} \frac {l}2 (N \cos{\beta})$

[/quote]

Yes you are right!

I made a mistake typing it onto the screen out of my rough notebook.

[quote="ffao"]
By the way, in your drawing, isn't the angle between the string and the rod $ (\pi \minus{} \alpha \minus{} \beta)$ rather than $ (\pi \minus{} \beta)$?

[/quote]

Yes I wanted to say angle between rod and wall is $ \beta$ but since I wrote it in a hurry, I made another mistake. My apologies for that.

Are we allowed to use $ m$ and $ g$ as part of the solution? Just wondering...

My simplifications of the equations led to a different solution as well...

[hide]
$ \mu \equal{} \frac1{3} \plus{} \frac{\sin (\alpha \minus{} \beta)}{3 \sin{\alpha}\sin{\beta}}$
[/hide]

EDIT: Good point about $ \mu$ being dimensionless...

[quote="ffao"]Are we allowed to use $ m$ and $ g$ as part of the solution? Just wondering...

[/quote]

If you mean as a part of your answer, I dont think that can be dimensionally correct.

[quote="ffao"]
$ \mu \equal{} \frac {2 \cos \beta \sin \alpha \minus{} \cos \alpha \sin \beta}{3 \sin \alpha \sin \beta}$

[/quote]

Just as I said, some terms are similar to the solution, some are missing. There are supposed to be three answers, as far as I remember! I'll post the answers soon.

Till then keep trying...keep thinking.

The moment  becomes easier if we take it in relation to the left extreme of the rod.

[hide] $ \mu \ge \frac {2 \cos \beta \sin \alpha \minus{} \cos \alpha \sin \beta}{3 \sin \alpha \sin \beta}\;\;\;$  if   $ \;\;\; 2 \tan \alpha > \tan \beta$

$ \mu \ge \frac {\cos \alpha \sin \beta \minus{} 2 \cos \beta \sin \alpha}{3 \sin \alpha \sin \beta}\;\;\;$  if   $ \;\;\; 2 \tan \alpha < \tan \beta$

$ \mu \equal{} 0 \;\;\;$  if  $ \;\;\; 2 \tan \alpha \equal{} \tan \beta$ [/hide]

Well, my teacher says that you first assume the friction to take a random direction -- if you're wrong, the worst that will happen is a negative value for $ \mu$. I got $ \mu \equal{} \frac {2 \cos \beta \sin \alpha \minus{} \cos \alpha \sin \beta}{3 \sin \alpha \sin \beta}$ -- I should have assured it was positive. Immanuel's answer is simply $ \mu \equal{} \left|\frac {2 \cos \beta \sin \alpha \minus{} \cos \alpha \sin \beta}{3 \sin \alpha \sin \beta}\right|$.


I was doing the 2007 state sprint rounds, and I couldn't do a problem.

I didn't understand the solution either. Could someone explain it to me?

http://www.mathcounts.org/webarticles/articlefiles/510-07_S_Sprint.pdf

number 26.

Wait, I have a solution, I'm just drawing the diagrams.

EDIT:

Done, took me 20 minutes in word to type everything up.


When I go to read any profile, I get  this:

Could not obtain ranks information

DEBUG MODE

SQL Error : 1054 Unknown column 'c.cat_user_id' in 'where clause'

SELECT count(*) AS pic_total FROM phpbb_album_cat AS c, phpbb_album AS p WHERE c.cat_user_id = 2 AND p.pic_cat_id = c.cat_id

Line : 86
File : usercp_viewprofile.php

Just figured I should report this.

Yeah it's normal. I am upgrading some of the forums functionalities. It will be a while until you will be able to access someone's profile.


Let $\ ABCDEF$ be a hexagon with parallel opposite sides, and let  $\ FC\cap AB=X_{1}, FC\cap ED=X_{2}, AD\cap EF=Y_{1},$
 $\ AD\cap BC=Y_{2}, BE\cap CD=Z_{1}, BE\cap AF=Z_{2}$.
Prove that if $\ X_{1}, Y_{1}, Z_{1}$ are collinear then $\ X_{2}, Y_{2}, Z_{2}$ are also collinear and in this case the lines $\ X_{1}Y_{1}Z_{1}$ and $\ X_{2}Y_{2}Z_{2}$ are parallel.

let $CD\cap EF=P$ and $AF\cap BC=Q$ if $X_{1},Y_{1},Z_{1}$ are collinear then by menelaous theorem in triangle $CPF$ we have :$\frac{CX_{1}}{FX_{1}}.\frac{FY_{1}}{PY_{1}}.\frac{PZ_{1}}{CZ_{1}}=1$
by angle chasing we have:

$\Delta BCX_{1}\sim \Delta EFX_{2}\Longrightarrow FX_{2}=\frac{EF}{BC}.CX_{1}$

$\Delta AFX_{1}\sim \Delta CDX_{2}\Longrightarrow CX_{2}=\frac{CD}{AF}.FX_{1}$

$\Delta CDY_{2}\sim \Delta AFY_{1}\Longrightarrow CY_{2}=\frac{CD}{AF}.FY_{1}$

$\Delta AQY_{2}\sim \Delta PDY_{1}\Longrightarrow QY_{2}=\frac{AQ}{PD}.PY_{1}$

$\Delta QBZ_{2}\sim \Delta EPZ_{1}\Longrightarrow QZ_{2}=\frac{BQ}{EP}.PZ_{1}$

$\Delta EFZ_{2}\sim \Delta BCz_{1}\Longrightarrow FZ_{2}=\frac{EF}{BC}.CZ_{1}$

hence:

$\frac{FX_{2}}{CX_{2}}.\frac{CY_{2}}{QY_{2}}.\frac{QZ_{2}}{FZ_{2}}=(\frac{CX_{1}}{FX_{1}}.\frac{FY_{1}}{PY_{1}}.\frac{PZ_{1}}{CZ_{1}}).(\frac{EF.AF}{BC.CD}.\frac{CD.PD}{AF.AQ}.\frac{BQ.BC}{EP.EF})=\frac{PD.BQ}{AQ.EP}$

and because $\Delta PDE \sim \Delta ABQ \Longrightarrow \frac{PD}{AQ}=\frac{PE}{BQ}\Longrightarrow \frac{PD.BQ}{PE.AQ}=1$

hence $X_{2},Y_{2},Z_{2}$ are collinear.
now notice that :$\frac{PZ_{1}}{QZ_{2}}=\frac{EP}{BQ}=\frac{PD}{AQ}=\frac{PY_{1}}{QY_{2}}\Longrightarrow Y_{1}Z_{1}\parallel Y_{2}Z_{2}$

Let $ U_{\infty},V_{\infty},T_{\infty}$ be the infinity points of $ FA,AB,BC$  and $ \tau_{\infty} \equiv U_{\infty}V_{\infty}T_{\infty}$ is the line at infinite. By the converse of Pascal theorem $ A,B,C,D,E,F$ lie on a same conic. By  Pascal theorem for hexagon $ BADCFE,$ the intersections $ X_1 \equiv AB \cap FC ,$ $Y_1 \equiv AD \cap FE$ and $ Z_1\equiv EB \cap DC$ are collinear. Analogously, in the hexagon $ EDAFCB,$ the intersections $ X_2 \equiv ED \cap CF,$ $ Y_2 \equiv DA \cap CB$ and $ Z_2 \equiv AF \cap BE$ are collinear. In the hexagon $ AFEBCD,$ the intersections $ P_1 \equiv FA \cap CB,$ $ P_2 \equiv EF \cap DC$ and $ P_3 \equiv EB \cap DA$ are collinear $ \Longrightarrow$  degenerate $ \triangle Y_1Y_2T_{\infty}$ and $ \triangle Z_1Z_2U_{\infty}$ are homologic $ \Longrightarrow$ $ P \equiv Y_1Z_1 \cap Y_2Z_2 \cap \tau_{\infty}$ $ \Longrightarrow$ $ P$ is at infinity $ \Longrightarrow$ $ X_1Y_1Z_1 \parallel X_2Y_2Z_2.$


Construct a triangle given the lengths of the three altitudes.

Yes, classical. Construct a triangle, whose sides are proportional to $\frac{1}{h_i}$. This triangle is similar to the required one. The rest is clear.

Since we know $\frac{h_a}{h_b}$, we know $\frac ab=\frac{h_b}{h_a}$. In the same manner, we find $\frac bc,\frac ca$, and it's pretty easy from here.


calculus trigonometry logarithms algebra polynomial geometry AMC

[quote="anirudh"]I have a few questions related to the curriculum of AMC contests.

For AMC 10's, do you need to know any 'real math' above what level?

For AMC 12's, does it include pre-calculus?

thanks.[/quote]

The curriculum of AMC 10 includes anything before Algebra 2.

The curriculum of AMC 12 includes all pre-calculus mathematics.

Some people like to say that the curriculum of the AMC tests is far beyond the high school curriculum either way because of the difficulty of the problems, although that's more that you need better problem-solving ability, not more advanced concepts/technical stuff.

[quote="anirudh"]Oh. Okay. So almost all of the stuff that is labeled as AMC 10 mock is actually amc 12 level (including mine)[/quote]None of your problems really involved Pre-Calc material.  :maybe:  The largest difference between AMC 12 and 10 is more advanced trigonometry, logarithms (??), polynomials, and more involved problem-solving. The 2 problems with logs in your Mock were trivial, with the right formulas, and would never appear in an AMC 12 (or AMC 10, since they don't include logs).

[quote="junggi"][quote="anirudh"]Hello but mine had trig in it. And it had logs which aren't part of amc 10 curriculum, accordig to calc ruz.

Also, I avoided PreCalc because I thought it might be too hard for an AMC 10.[/quote]
You're contradicting yourself completely in that post.
If trig and logs are precalc, then you did not avoid precalc.
And I'm sure they introduce you to basic trig,logs, and complex numbers in algebraII? heck, they taught us trig in geometry class.
And your trig,log,complex number "problems" were solvable if you had the very basic knowledge(like definitions) of those subjects.(who can say that they know logs if they can't do #9,#20, and who can say that they know trig/complex numbers if they can't do #10,11?)
Lastly, what on earth are you trying to get out of this? you want your mock AMC10 to be considered a 12? what benefit do you get from that?[/quote]

yeah, in a way that is a paradox

[quote="anirudh"]Algebra II should be enough to solve the problems in my mock. And I never wanted mine to become an AMC 12. :wink:[/quote]
from your posts here, you clearly wanted yours to be "AMC12 level," whatever that is.

(example:"Hello but mine had trig in it. And it had logs which aren't part of amc 10 curriculum, accordig to calc ruz. ")

Yes. I know. But I didn't want to make it an amc 12.

[quote="anirudh"]Algebra II should be enough to solve the problems in my mock. And I never wanted mine to become an AMC 12. :wink:[/quote]

If you read my post carefully, you'll notice that AMC 10 does not include Algebra II.

[quote="sapphyre571"]is the polar coordinate system fair game for the AMC series?[/quote]
I think it is for the AIME, not 100% sure about that though, but if so then it is for the AMC 12.


Hi Albanian Eagle. Here is my probkem present for you

[u][b]Problem.[/b][/u] Prove that for all non-negative real numbers $a,b,c$ such that $a^2+b^2+c^2=3$ then the following inequality is always true
\[ \frac{a^2\sqrt{b+c}}{\sqrt{a^2+bc}}+\frac{b^2\sqrt{c+a}}{\sqrt{c^2+ab}}+\frac{c^2\sqrt{a+b}}{\sqrt{c^2+ab}}\le 3.  \]

Hope you and other mathlinks members will like it.

hi hunghtn! Thank you for this nice problem.
I found this solution
apply cauchy and get
$\sum \frac{a^2 \sqrt{b+c}}{\sqrt{a^2+bc}}\le \sqrt{(a^2+b^2+c^2)(\sum \frac{a^2(b+c)}{a^2+bc})}$
so it is enough to show that $\sum \frac{a^2(b+c)}{a^2+bc}\le 3$ by homogenising we see that it is equivalent to showing for any positive real numbers
$(\sum \frac{a^2(b+c)}{a^2+bc})^2\le 3(a^2+b^2+c^2)$
We can show something stronger ;
$\sum \frac{a^2(b+c)}{a^2+bc}\le a+b+c$
this one is true because after expansion it reduces to $\sum a^5 bc \geq \sum a^3 b^3 c$

[quote="Lovasz"][quote="hungkhtn"]Notice that we have equality when $a=b=c$ or $a=b,c=0$.[/quote]
So that SOS worked.[/quote]

Yes... But a detailed solution is better, isn't it?

[quote="hungkhtn"]OK. The stronger one is that
\[ \frac{a^3(b+c)}{a^2+bc}+\frac{b^3(c+a)}{b^2+ac}+\frac{c^3(a+b)}{c^2+ab}\le a^2+b^2+c^2.  \]
Notice that we have equality when $a=b=c$ or $a=b,c=0$.[/quote]

I use Schur Extension  to solve it

WLOG let $a\geq b\geq c$ then the inequality is equivalent to

$\sum \left(a^2-\frac{a^3(b+c)}{a^2+bc}\right)\geq 0\Longleftrightarrow\sum \frac{a^2}{a^2+bc}(a-b)(a-c)\geq 0$

if we let $x=\frac{a^2}{a^2+bc}$ , $y=\frac{b^2}{b^2+ac}$ , $z=\frac{c^2}{c^2+ab}$ (clearly $x,y,z\geq 0)$

then $x-y=\frac{c(a^3-b^3)}{(a^2+bc)(b^2+ac)}\geq 0$ so $x\geq y$

So
$x(a-b)(a-c)+y(b-a)(b-c)+z(c-a)(c-b)=(x-y)(a-b)(a-c)+y(a-b)^2+z(c-a)(c-b)\geq 0$
which is clearly true .


Just out of curiosity, is anyone else going to be taking the SAT Biology Subject test anytime soon?

ive taken both e and m and theyre not very difficult tests. if you prepared well for the ap exam, you should be ready for both of the biology sat 2's. i got an 800 on both of them and the curve is pretty nice-- i think you can miss 2 or 3 and get an 800.

wow, you're a hardcore bio person.  I felt well-prepared for the AP bio but I actually heard bio had one of hte hardest curves.  If you look at the percentiles, 800 bio has one of the highest...99+ I believe.  By contrast, physics is really easy and chem is pretty easy as well.

bio E 800 is 99+%, bio M is 98%. the curves are okay, the percentiles are high because few people do really well on them.

The curve is actually more generous I think, allowing up to five wrong sometimes.  It varies from year to year.  I just took it last weekend and I probably got either a 790 or an 800.  I made a few stupid mistakes.   :P

I took it last weekend as well, but doubt I did quite that well.  I am not in AP Bio as my school doesn't let you take it as a freshman and the regular (freshmen) biology classes are at best a joke.  I did do the whole princeton review book for it though, but there were still some things on there that tripped me up.

i took it freshman year after honors bio, got 770 without feeling very well prepared for it. i know someone though who took it after AP bio and didn't do too well. :( it depends on your preparation for the test inside AND outside of the class

As for me, I took it 4 years ago, cold turkey.  I did absolutely no studying for it, took Bio-M, and emerged with a 790.

The only reason I didn't take E, which is supposedly easier, is that I hated ecology with a vengeance and like playing around with molecules.

For advice, I'd say AP Biology is unnecessary.  Regents Biology, coupled with much outside reading, should do the trick.

Heh, yea I ended up getting 760, better than I expected.

I'm most likely going to take the October 8 Bio SAT II. What do you think is the best way to study for it? I have been using Kaplan's but I've found that the explanations are a bit sketchy sometimes.

I'm not the sort of person that can just hardcore memorize everything. For me, there needs to be a reason for something to occur. For that reason, I'm not sure if I should use Campbell's for SAT IIs or if it's too much overkill.

I just got the book today and it's around 1200 pages so I can make it if I read a little over 20 pages a day.

What do you guys suggest?

20 pages a day in campbells is good. overkill? yeah, but you will know more about the wonderful science of biology for it.

Sounds good :) 20 pages a day it is (and I actually read more than that today lol)

Good! And it's not like Campbell's is boring to read either; I've always found it to be very addictive. Then again, I am a bioaddict. lol

yeah, the thing is, bio's definately got a harder curve than physics- though, there's not much to it, just make sure you're not making the stupid mistakes they want you to

i just took it june, and it wasnt very bad at all.

some questions are a little "wierd" though, until you give them some thought.

but, preparation-wise?  I think any first-year biology text should do... you needn't go into AP


Find the number of ordered triples of sets $ (A,B,C)$ such that
$ A \cap B \cap C \equal{} \phi , A \cap B \not\equal{} \phi , A \cap C \not\equal{} \phi$
and $ A,B,C \subset  \left\{\,\right. {1,2,...,n}\left.\,\right\}$

The first equation is not needed.

This is my reasoning:

If $ \#A \equal{} k$, than there are $ \binom{n}{k}$ possibilities for $ A$. Because $ A \cap B \equal{} \emptyset$ and $ A \cap C \equal{} \emptyset$ we have
$ \#B \equal{} \binom{n \minus{} k}{m}, \quad \#C \equal{} \binom{n \minus{} k}{p}$ with$ 0 \leq m,p \leq n \minus{} k$.

Is this already good ?

edit: I misread the problem  :blush:

make a table of $ n$ rows, one for each of the elements $ 1, 2, ..., n$, from the top down, and 3 columns, one for each of the sets $ A, B, C$, from left to right. put a 1 in a cell of the table if the element represented by that cell's row is in the set represented by that cell's column, a 0 otherwise. note that each table corresponds to a choice of $ (A, B, C)$, and vice versa. then the given conditions translate to: 1) no row has all 1's, 2) at least one row is $ 110$ 3) at least one row is $ 101$.

the number of tables satisfying condition (1) is $ 7^n$. of these tables, the ones NOT satisfying (2) OR (3) are the ones where each row is only one of the following: $ 000$, $ 100$, $ 010$, $ 001$, $ 011$. there are $ 5^n$ such tables. the tables satisfying (1) and (2) but not (3) are those where each row is one of $ 000$, $ 100$, $ 010$, $ 001$, $ 011$, $ 110$, and at least one is $ 110$. the number of such tables is $ 6^n \minus{} 5^n$. similarly, the number of tables satisfying (1) and (3) but not (2) is also $ 6^n \minus{} 5^n$.

so the answer is $ 7^n \minus{} 5^n \minus{} 2(6^n \minus{} 5^n) \equal{} 7^n \minus{} 2\cdot 6^n \plus{} 5^n$.


integration
real analysis
real analysis unsolved

yes, it's true and i have a long solution with [b]squeezing lemma[/b] for it  ;)

[quote="P.H"]Is it true ?
the limit of the 1^k + 2^k +....+n^k when n gets bigger and bigger is n^k+1 /k+1.
thank you.[/quote]

$1^k+...+n^k = n^k\left((\frac{1}{n})^k+...+1\right) = n^{k+1}\frac{1}{n}\left((\frac{1}{n})^k+...+1\right)$

We see that $\frac{1}{n}\left((\frac{1}{n})^k+...+1\right) \sim \int_0^1 x^k dx = \frac{1}{k+1}$ (for $k \neq -1$) because it's riemann sum.

QED


[quote="Tiago Guerreiro"]hi :)

A king decided to give to a prisoner the chance to get free. He took him to a room, with two exit doors called $A$ and $B$, each one with a guard. He said to the prisoner: " One of the two doors leads to freedom, while the other door leads to death; besides, one of the guards allways speaks truth, while the other guard allways lies. You can ask just one thing to one of the guards and pick a door to get out." The prisoner thought a little; then, he asked to one of the guards: " If I ask to your partner which door leads to freedom, what would he answer me? "; "$A$" the guard said.
"Thanks", said the prisoner, and went out through $B$

What appened to the prisoner ? Explain why.

I'm sorry for my bad english.  :blush:[/quote]The prisoner was alive and free. Assuming the chosen guard was the liar then he would lie about which door his partner would pick. Thus the truth-teller would pick $B$. Otherwise, if he had chosen the truth-telling guard then he would answer truthfully that his partner, the liar, would say $A$. Thus the correct choice is still $B$.

[quote="Tiago Guerreiro"]hi :)

A king decided to give to a prisoner the chance to get free. He took him to a room, with two exit doors called $A$ and $B$, each one with a guard. He said to the prisoner: " One of the two doors leads to freedom, while the other door leads to death; besides, one of the guards allways speaks truth, while the other guard allways lies. You can ask just one thing to one of the guards and pick a door to get out." The prisoner thought a little; then, he asked to one of the guards: " If I ask to your partner which door leads to freedom, what would he answer me? "; "$A$" the guard said.
"Thanks", said the prisoner, and went out through $B$

What appened to the prisoner ? Explain why.

I'm sorry for my bad english.  :blush:[/quote]
Bad English? It's good. :D

B. because
a) If he was lying then B would be correct.
b)If he was telling the truth then his partner would lie (So when his partner says A, it is a lie) then also B is true

I love these type. Here's another one:

At the dinner table, Jeff was next to Anne. But Jeff was not next to Jack. If Jack was not next to Sandy and there were only 4 chairs at the table, who is Sandy next to?

[hide="My Solution"]

He got freed.

Solution:

If the guard was lying, then the other guard would have said B, and the other is telling the truth, so B.

If the guard was telling the truth, then the other guard would have said A, but would have also been lying, so B.

B either way.

The prisoner was smart.[/hide]

This should be moved to puzzles and brainteasers (I think).

This is like the one where the guy was caught by native americans and they said if you say a true statement we will feed you to lions and if you say a false statement we will throw you off of a cliff. What did the man say? Also like knights and knaves...

[quote="mustafa"]Answer to the second problem

[hide]Jeff[/hide]

I agree that this should be moved to puzzles and brainteasers[/quote]

Yep, ur right. I am sorry I don't know this site well enough, i didn't even know that the puzzles and brainteasers section EXISTS!! lol  :blush:  :blush:
Will go there.... :)

Hey there's another answer to this 'one door leading to freedom and other door leading to death' riddle. The answer is go to any guard and ask him: "If my friend came and ask you which door leads to freedom, what would you tell him?" Then take the door that the guard tells you the take.

Think about why this question works.

This definitely belongs in Puzzles and BrainTeasers.

Moved.

[quote="bpms"][quote="i_like_pie"][quote="mathnerd314"]Extension:

Same situation, but there are three doors. One of them is guarded by a person who gives completely random answers.[/quote]
If you're only allowed to ask one question, then no matter what you ask, if you get the random guy, you're in trouble.[/quote]
He didn't say you are only allowed to ask 1 question...[/quote]

He said "same situation"

[quote="mustafa"]He said "same situation"[/quote] Well, since 1 question will not solve the problem, we should consider 2 or more questions.  :lol:

You could ask one guy the same question over and over again...

[quote="mathnerd314"][quote="mustafa"]Wouldn't that be impossible? If you are unlucky and talk to the guy with the random answers, then you would be dead. Are you sure that you can only ask one person one question?[/quote]

:oops_sign:

Challenge: Find the minimum number of questions needed to guarantee that you can correctly deduce the right door.

[hide]Reason: I don't remember the original problem :roll:[/hide][/quote]

since this is random, you can apply the law of random average thingie (uhh maybe not a law but...)...you know, that in random events, the sums of the random consequences of events each approach close to 1/2

(i.e. flip a coin 1000 times. your results will be close to 500 heads and 500 tails. not exactly, but fairly close.)

ask any one guy 4 questions (same questions) to figure out the identity of the guy you're interrogating (if u hit the random guy, then the chances he'll blab either yes or no will be 1/2--therefore, you would know to skip the guy and find your way out). if the guy answers the same answer, then just ask the question about "if i ask your partner about which door is the way out..."

[quote="perfectnumber628"]but if he says things that are consistently either true or false, you won't know for sure if he's the random guy or not.

maybe you ask them each one question that you know the answer to.  if 2 give false answers and the other one gives a true answer, then you know he's the true one so ask him.  if 2 give true answers, then you know for sure the other guy is false, so just ask him which door, and then go in the other one.  oh, but since there's 3 doors, you might have to ask him more than once.[/quote]

Actually, I don't think you would have to ask him twice; you know it's not his door, so you wouldn't go through it. You would ask him which of the two other doors leads to freedom and go through the other one. Also, if two are false and one is true, you can just go through his door. So I guess the answer is 4 at most?

The liar might have the freedom door.....

I agree with perfectnumber's solution, it is much better than anything I could come up with. To give a definitive number,

[hide]You would need to ask each guard one question, (What is $2+2$?). If two people say 5, and one says 4, then ask the one that said 4 which door to go through, taking 4 questions. If two people say 4, and one says 5, go the the one that says 5, and ask him which door will not lead to freedom, and go through the door that he indicates, also giving an answer of 4! Therefore, the answer is 4![/hide]


un problema sencillo  :blush:  :

sea $G$ el baricentro del triangulo $ABC$ . demostrar que si $AB+GC = AC+GB$ entonces ABC es isosceles

Ola!!
Mi primer intento fue poner a GC y GB en terminos de los lados, pero me keda algo feo... Me imagino k al efectuar todas las cuentas llegas a la solucion, pero sin embrago, buskare una solucion bonita...
(Buen problema) :lol:


algebra
polynomial
number theory
prime numbers
number theory unsolved

Problem:  Prove that : forall positive integer $a$, there exist $p>q$ are primes such that. $q|ap-1$

Let's take two numbers $a$ and $a-1$ these numbers are coprime, which means that fron Dirichlet's theorem, in sequence $k(a-1)+a$ there exists infinitely many prime numbers...in particular there is a prime number $p>a$ in this sequence. We have then, $p=k(a-1)+a$ for some $k$. We obtain from this \[ (a-1)|p-a  \] ,which means that \[ (a-1)|(pa-a^2)  \] but of course we also have \[ (a-1)|(a^2-1)  \] so by adding we have: \[ (a-1)|pa-1  \] and as $p>a$ also $p>(a-1)$ Let $q$ be any prime which divides $a-1$ Then \[ q|pa-1  \] and $q\leq (a-1)<p$ so we found our two numbers $p,q$. This proof won't work for $a=1$ but then our statement becomes rather obvious :) this won't also work for $a=2$ cos' then $a-1=1$ but then we need to find an example... something like $q=3$ and $p=5$ should be "good"

Thank you, jastrzab!.

Here is my solution:
Case 1: $a\ge 3$
Because in the sequence $k(a-1)+1$ there exists infinitely many prime numbers. Thus, there is a prime number $p>a$ in this sequence. We have $a-1| p-1$ and $a-1|p(a-1)$ .By adding we have $a-1|p(a-1)+p-1=pa-1$.

@: Can we prove it without using [b]Dirichlet's theorem[/b] ?.

And is the following problem is true ?

Let $P(x)$ (not constant) be a polynomial whose polynomial are integers such that $P(0)=-1$. Show that we can find $p>q$ are prime numbers such that $q|P(p)$.


Just continuing on from the :rt2:^(:rt2:^... post..
When its the square root of 2, that converges. When its a little bit higher, it doesn't. Whats the biggest x so that x^(x^(x^... converges?

Everyone from SF Bay Area who went to ARML 2003 had better know this now. This number (I don't want to give it away) was the theme of our ARML T-shirt in 2003.

let me guess, whats a number between 1 and 2 and famous enough to get on an ARML t-shirt?  [hide]golden ratio[/hide]?

Nope, not the golden ratio. We had that as our theme in 2001 though.

Wait, are you serious -- it's a number between 1 and 2 and not :phi:?  I'm not sure I believe you . . .

[hide]e^(1/e)[/hide]?  You had a shirt devoted to [hide]e^(1/e)[/hide]?

I'll do this.

Suppose xxx... = a is a real number.

Then xa = a, or x = a1/a.

So we are looking for the maximum of f(t) = t1/t.

f'(t) = (1 - ln t)/t2 e(ln t)/t

f'(t) = 0 <=> 1 - ln t = 0 <=> t = e.

So we have an extremum for t = e.

Hence result : x = e1/e.

Interesting that it's so close to :rt2:.  Go figure.

e^(1/e) is right. What about the lower bound of convergence? Does anyone want to tackle it? If you have seen it before, don't post please.

Well, definitely every positive number works.  I would argue that the function definitely doesn't converge at 0, since we get 0^x = x, which means either x > 0, in which case we get x = 0, or x < 0, in which case we get x is undefined, or we get x = 0, in which case x = 0^0 which is usually accepted to be 1.  So, that doesn't work.  Negative numbers I'd have to look at, though.

There are positive numbers <1 for which the tower doesn't converge.

I think this prb is somehow related to the following (proposed for an IMO, a long time ago, can't remember the year):

Find all positive a for which the equation log {a} x=x has a solution in positive numbers. The solution involves e^(1/e) (a can have any value >0 and <=e^(1/e)).

By the way, the lower bound for convergence is e^-e.


Let $x,y,z$ be positive real numbers such that $x+y+z=3$.Prove that:
$\frac{\sqrt{x+y^{2}z^{2}}}{y+z}+\frac{\sqrt{y+z^{2}x^{2}}}{z+x}+\frac{\sqrt{z+x^{2}y^{2}}}{x+y}\geq \frac{3\sqrt{2}}{2}$

My inequality from : http://www.mathlinks.ro/Forum/viewtopic.php?t=148011  :D
Do you like it? :P


Levi had 10,000 points to exchange for prizes. Each prize
he selected required half the number of points he had at the
time he selected it. How many points did he have left after
choosing four prizes?

Correct me if I'm wrong:

After $ 4$ exchanges, his money will be $ 2^4$ times less than it was at the start.  Thus he has $ \frac{10000}{16}\equal{}625$.

How do you put a box around 625?

your answer is correct

in order to put a box around something, use "\boxed{x}" where x is what you want to put inside the box

also, you have to take away the quotes and put dollar signs, but you know that because you can write in Latex

so...to put 625 in a box, you write \boxed{625} (with dollar signs around it) to get $ \boxed{625}$


what is the formula for the area of a sphere

$ \frac{4\pi*r^3}{3}$ where $ r$ is the radius.


quadratics
algebra
polynomial
quadratic formula

I need a few tips on this. I just completed the intro series. Now, i am doing more advanced books but when i try any of the problems i just freeze and don't know how to start but when i look at some of the solution then i can proceed very quickly with the solution. How do i get better at starting the problem?
-------
thanks

Step 1: Write down the question.
Step 2: Plug in some numbers.  Do a much easier case first.  I always like to say it like this:

If you can't immediately prove it in general, then at least try to prove by example!! ;].

Usually, once you plug in some things, you see the pattern or you see what the problem is trying to get at.

I had a problem once that I had to find a recursive formula for something regarding permutations, and I had to write out all permutations of 1,2,3,4,5 before I saw how the recursion worked.  Took a while, but I was the only one who came up with the solution.

This should really be moved to Other Problem Solving Topics.

Generally, most problems have an underlying mathematical idea that is learned in school. The trick is to identify what this concept is (which could be something as simple as the quadratic formula, which I will show below) and how you can manipulate it to the problem.

Take the following problem:

[hide="Example"]
[quote="Problem"]Given that $ 1 \pm \sqrt {3}$, $ 0$, and $ 3$ are solutions to $ f(x)$, find $ f(x)$.[/quote]

One can easily solve the problem the long method, by multiplying $ [(x \minus{} 1 \minus{} \sqrt {3})(x \minus{} 1 \plus{} \sqrt {3})(x)(x \minus{} 3)]$. However, if you note the root $ 1 \pm \sqrt {3}$, you'll realize that it's partially in the form of the quardratic formula solution. Therefore, we realize that $ a \equal{} 1$, $ b \equal{} \minus{} 1$, and we can simply solve for $ c$ from $ b^2 \minus{} 4ac \equal{} 3$. And from our values of a,b, and c, we can determine a second degree polynomial. From there, we see that $ 0$ and $ 3$ are roots, so we multiply the second degree polynomial by $ x$ and $ x \minus{} 3$, respectively.
[/hide]

Simple things like that will save you a lot of time, and in the long run, you'll view problems differently.


Point M lies in the interior of triangle ABC such that $ \hat{APB}\equal{}\hat{CPB}\equal{}\hat{APC}$;The two lines PC and AB intersect at E and  the two lines PB and AC intersect at F Prove that 4EF$ \leq AB\plus{}AC$

[quote="tranquoc"]Point M lies in the interior of triangle ABC such that $ \hat{APB} \equal{} \hat{CPB} \equal{} \hat{APC}$;The two lines PC and AB intersect at E and  the two lines PB and AC intersect at F Prove that $ 4EF\leq AB \plus{} AC$[/quote]
Let the ray BF meet the circumcircle of AFC again at $ D'$.
Since $ \angle AFC \equal{} 120^o, \angle AD'C \equal{} 60^o. \angle AFD \equal{} 180^o \minus{} \angle AFB \equal{} 60^o$.
So $ \angle ACD' \equal{} \angle AFD' \equal{} 60^o$. Thus $ AD'C$ is equilateral. Let M be the midpoint of AC and extend $ D'M$ to meet the circle again at N. Since $ AD'C is equilateral$ we have $ \frac {DM}{MN} \equal{} 3$. Let $ M'$ be the foot of the perpendicular from F to AC.
Since M is the midpoint of AC, we have MN ≥ M'F. So $ \frac {D'D}{DF} \equal{} \frac {D'M}{M'F} \geq 3$. Hence $ \frac {D'F}{DF} \geq 4$. Define $ E'$ similarly. A similar argument shows that $ \frac {E'F}{EF} \geq 4$.
Now take $ D"$ on the ray FD such that $ \frac {D"F}{DF} \equal{} 4$ and similarly take $ E"$ on the ray FE such that $ \frac {E"F}{EF} \equal{} 4$. Then $ D"E" \equal{} 4DE$. The triangle E'D"D' has $ \angle E'D"D' \geq\angle EFD > 90^o$, so $ D'E' > D"E'$. Similarly $ \angle EE"D" \geq \angle EFD$, so $ D"E' \geq D"E"$ and hence $ D'E' \geq 4DE$
$ \Rightarrow AB \plus{} AC \equal{} AE' \plus{} AD' \geq D'E'\geq 4DE$


there is a 12 letter word made up of four $ X$s , four $ Y$s and four $ Z$s

how many different 12 letter words can be made so that there are no $ X$s in the first 4 letters, no $ Y$s in the second 4 letters and no $ Z$s in the last four letters?

Solution

[hide]If we look at the word $ XXXXYYYYZZZZ$ (absolutely cannot happen), we see that the $ X$'s, as they move, will displace other letters and basically define a word.  Consider if all $ X$'s displace all $ Y$'s.  This only can occur in the word $ ZZZZXXXXYYYY$ by the third condition, giving us one word.  Noticing the symmetry when all $ X$'s displace all $ Z$'s, we will multiply by two later.  If one $ X$ displaces one $ Y$, then three $ X$'s must displace three $ Z$'s.  Our word in general form becomes $ YYYZXZZZXXXY$.  There are four ways to arrange each group of four letters, giving us $ 64$ new words.  We currently have $ 2 \cdot 65\equal{} 130$ words.  If there are two $ X$ displacing two of each letter, it is symmetric and we cannot multiply by two.  The word has a form of $ YYZZXXZZXXYY$.  There are 6 ways to arrange each group of four letters, which gives us $ 216$ new words.  Thus, there are $ 346$ words.[/hide]

Good.  What if we replaced the number "4" in the problem with "$ n$"?

For those who just want to know the answer, it is sequence [url=http://www.research.att.com/~njas/sequences/A000172]A00172[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

My reasoning -:
$ ( \dashrightarrow \mbox{indicates words obtained with the given 4 letter\minus{}groups rearrangement} )$

Starting with
$ ZZZZ \ XXXX \ YYYY \dashrightarrow \ 1$
$ YZZZ \ ZXXX \ XYYY \dashrightarrow \ 64$
$ YYZZ \ ZZXX \ XXYY \dashrightarrow \ 216$
$ YYYZ \ ZZZX \ XXXY \dashrightarrow \ 64$
$ YYYY \ ZZZZ \ XXXX \dashrightarrow \ 1$

Hence a total of $ 346$ words

[quote="JBL"]What if we replaced the number "4" in the problem with "$ n$"?[/quote]

Using the same methodology as above to solve the most general case ( $ m$ different alphabets $ n$ times in the given word ) , i think the number of possible words should be   $ \  \sum_{i \equal{} 0}^{n} \binom{n}{i}^m$

I like your generalization from 4 to $ n$ -- I'm not so sure I like your generalization from 3 to $ m$.  Or, at least, you're going to have to state more clearly what you think this generalization is.  (Thing to consider: try to apply your claim for $ n \equal{} 1$ and $ m \geq 4$.)  There is a question that has as its answer the expression you have, but I don't think it's the question you're thinking of.


calculus
integration
algebra
partial fractions
calculus computations

substitute x-1/x=u and the integral becomes
$ \begin{array}{l}
 \int {\frac{{du}}{{{u^2} \plus{} 3}}} \mathop  \equal{} \limits_{du \equal{} \sqrt 3 t}^{u \equal{} \sqrt 3 t} \int {\frac{{\sqrt 3 }}{3}}  \cdot \frac{1}{{{t^2} \plus{} 1}}dt \equal{} \frac{{\sqrt 3 }}{3}\arctan t \plus{} c \equal{} \frac{{\sqrt 3 }}{3}\arctan \left( {\frac{u}{{\sqrt 3 }}} \right) \plus{} c \equal{}  \\
  \equal{} \frac{{\sqrt 3 }}{3}\arctan \left( {\frac{{x \minus{} \frac{1}{x}}}{{\sqrt 3 }}} \right) \plus{} c \\
 \end{array}$

Thanks, mathxl.

Alternatively, we could factorise $ x^{4} \plus{} x^{2} \plus{}1$ as $ (x^{2}\plus{}x\plus{}1)(x^{2}\minus{}x\plus{}1)$

and then use partial fractions to get

$ I\equal{} \frac{1}{2}\int \frac{1}{x^{2} \plus{} x\plus{}1} \plus{} \frac{1}{x^{2}\minus{}x\plus{}1}\ dx$

and then complete the squares in the denominators to get

$ \frac{1}{2} \int \frac{1}{\frac{3}{4} \plus{} (x\plus{}\frac{1}{2})^{2}} \plus{} \frac{1}{\frac{3}{4} \plus{} (x\minus{}\frac{1}{2})^{2}}\ dx$.


If $ a, b, c, d$ real numbers, then
$ \prod(2a^4\minus{}2a\plus{}1)\geq0$

Why did you take 4 numbers?^^

Let $ f(x)\equal{}2x^4\minus{}2x\plus{}1$. We have $ f'(x)\equal{}8x^3\minus{}2$ and $ f''(x)\equal{}24x^2 \ge 0$. Hence $ f$ gets minimal if $ f'(x)\equal{}0$. That is, if $ x\equal{}\frac{1}{\sqrt[3]{4}}\equal{}x_0$. Hence $ f(x_0)\equal{}\minus{}\frac{3}{2\sqrt[3]{4}}\plus{}1\equal{}\frac{2\sqrt[3]{4}\minus{}3}{2\sqrt[3]{4}}>0$.


Hello people! Let me know how my new sig is, for this si the first tiem I have changes it. Also feel free to suggest suggestions.

Cheers,
bubka

I don't understand...you could put an explanation not that you need to...because i'm a bit strange

Just in case I meant the sig because I'm strangely used to people typing like that when they eat acid pops...

Ok someone likes Harry Potter too much.

bubka: is it you?

Yes, the human in the picture is with almost all certainty Bubka.

Isn't there any place where Bubka can host those pictures without that annoying thing at the bottom giving the filesize? :o

I am one of the two in the picture  :P
Guess which one?  :rotfl:

Huh? Which do you think is the foreground then?

[quote="Treething"]poor doggy and poor people to the right and roughly in the background of the picture[/quote]
 :|   Treething, what are you saying?  Who is poor??

Hmmm ... are we talking fiscal standards here?

EDIT: ok financial whatever

Looks cool :)

I think Treething means the people that we cannot see in the photo, but, if they were standing there, would be kicked by bubka.

haha... never mind. i don't mean poor in the money way though

what's the name of your dog? (that's the dog isn't it?)

I can only say that my sister is called [url=http://extreme.atheist.googlepages.com/life,...]Pushpita[/url].

I realised that I like your sig it shows us your presence rather more quickly than usual...

lol i have no idea what they're talking about

*sigh*
There was an old feature on the forums that got removed where you could have a signature
a few users still sort of copypaste a sig in but idk

how come they took out signatures?

they sound cool


Given two piles of stones. An automatic machine choose the pile, which contain an even number of stones (if both piles contains even numbers of stones then machine selects an arbitrary pile), and put half of these stones to another pile. If both piles contains an odd number of stones then machine stops. Find all pairs of initial numbers of stones in the piles s.t. the machine will necessarily stop.

Let 2^a be the largest power of 2 that divides the number of stones in the first pile.
Let 2^b be the largest power of 2 that divides the number of stones in the second pile.

The robot terminates for sure, if and only if a=b holds.

It is correct!  :)
Are you waiting someone will write solution? ;)

If those two exponents are equal, at each step, they remain equal and decrease, so eventually the machine wil reach a position in which it will have two odd piles. On the other hand, if they are not equal, at each step there are two possibilities: if $a=b+1$, then the machine can break up the column with $b+1$, which will now have exponent $b$, and the column with initial exponent $b$ will have exponent $b+1$, so we simply switch $b,b+1$ between the two columns (which means we never stop). On the other hand, if $a>b+1$, then it also breaks up the one with exponent $a$, which will now go to $a-1$, while the other column will keep having exponent $b$. The machine can continue this way until it gives the pile with initial exponent $a$ an exponent $b+1$, after which it continues as above.

Too complicated for such problem...
Just note that if $m+n$ is odd then machine never stops ($m$ and $n$ are number of stones). For common case unite $2^a$ stones into one big STONE (I assume $a<b$).


algebra
polynomial
function
algorithm
geometry
3D geometry
number theory

Does anyone know any website, know the proof or even better, have their own proof that

a^3 + b^3 = c^3 has no integer solutions for a b c

If you do, please tell me the proof or the website.
thanks

how much number theory do you know, I think this can be shown using properties of Z[ (-1+ :sqrt: -3)/2]

Dinky question:

What do people mean when they refer to the field Z^2 or Z^3 or, in this case, Z[Sqrt(-3)]?

first, Z^2 is just ZxZ, which is pairs of integers of the form (a,b) for a,b in Z.  So you can think of the lattice points in the plane as Z^2, similarly Z^3 = ZxZxZ = (a,b,c) a,b,c in Z.

but Z[a] is something different, an element of Z[a] would be b0*a0 + b1*a1 + b2*a2...
where bi is in Z.

an example is Z[x] which is just polynomials with integer coefficients, or Z[i] is guassian integers = a+bi with a,b in Z and i = :sqrt: -1.  This branch of number theory is called Algebraic Number Theory and has many applications.  For example, the proof that Grobber gave for an n-gon in the lattice relies on properties of Z[i] and its unique factorisation.

in a ring like Z[i] you have the norm function, so that N(a+bi) = (a+bi)*(a-bi) = a^2+b^2, so it is a way of mapping gaussian integers to non-negative integers.  You can prove some nice properties of the norm function, like N(a*b)=N(a)*N(b), and with the use of the norm function, you can prove a division algorithm in Z[i]. and build up and prove unique factorization into guassian irreducibles.  This post is getting kinda long, but yea, algebraic number theory is pretty fascinating stuff that has many applications.

If you would like to learn more about algebraic number theory, apply here: http://www.math.ohio-state.edu/ross/ or feel free to post any follow up questions.

Well, I don't know much about algebraic number theory, but I think I get the idea now. Thanks for the response.

if you go to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10102
somebody asked about x^3+y^3=2z^3, the proof for x^3+y^3=z^3 is very similar, just with more cases and longer.

zscool, was that a typo or did you answer a different question?  He wanted to know about the Fermat expression x :^3: + y :^3: = z :^3:.

(I don't know if "Fermat expression" really means anything, but the point is that it's a case of Fermat's Last Theorem.)

Hm...is it just me or...I've seen some mathematicians getting triples for x^3+y^3=z^3.

-interesting_move

If you mean positive integer triples, it's safe to say that you're wrong.  Since no such triples exist, after all.  (Fermat's Last Theorem, n = 3.)

Hmm.. well either they were finding non-integer ones, or they made a mistake somewhere :D
Though maybe you meant a^3 + b^3 + c^3 = d^3? That has solutions..

That (what MMM said re: 3 cubes to a cube) is a possibility.

If they don't have to be positive, a fun one is 13+123-93=103.

http://homepages.cwi.nl/~dik/english/mathematics/flt3.html gives FLT for n = 3 without algebraic NT.


Hi i am new in the forum and i was hoping you could help me with this problem:

Prove that the determinant of the matrix A:
2a+p  2b+q  2c+r
2p+x  2q+y  2r+z
2x+a  2y+b  2z+c

is 9 times the determinant of the matrix B:
a  b  c
p  q  r
x  y  z

Well i am really confused with this problem, i dont know where is better to start so any help is welcome.
Thank you.

Note that $ A\equal{}\left(\begin{array}{ccc}2&1&0\\0&2&1\\1&0&2\end{array}\right)\left(\begin{array}{ccc}a&b&c\\p&q&r\\x&y&z\end{array}\right)$, so by the product rule for determinants you just need to show that the first matrix in the above product has determinant 9.

Ok thank you very much,  your answer is great and helped me a lot, i just never imagined that i could express de matrix A as the product of matrix B and other matrix.


A point is rotated x degrees counter-clockwise from the point (1,0) to the point (-.6,.8) on the unit circle. What is the value of x?

They say that the value of the y coordinate is equal to sin(number of degrees rotated). When I do $\sin^{-1} .8$, I get something like 53.13 degrees. Do I then use the supplements theorem because the cosine value is negative? Does x equal 127 instead of 53?

I'm quite new to trigonometry, so be nice about this  :)

I'd guess it would be 127 not 53, just looking at the two points on a graph.


A park is in the shape of a regular hexagon 2 km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of 5 km. How many kilometers is she from her starting point?

$\text{(A)} \ \sqrt{13} \qquad \text{(B)} \ \sqrt{14} \qquad \text{(C)} \ \sqrt{15} \qquad \text{(D)} \ \sqrt{16} \qquad \text{(E)} \ \sqrt{17}$

[hide="Answer"]Simple geometry. Draw the hexagon, draw the right triangle. It has short leg $1$, obviously, and it has longer leg $2\sqrt{3}$, so the hypotenuse is $\sqrt{1+(2\sqrt{3})^2}=\sqrt{13}\Rightarrow \boxed{A}$.[/hide]


x1 , x2 , ... , xn  are  reals  in  the  interval  [a,b] . Let  M=(x1+x2+...+xn)/n , V=(x1^2+x2^2+....+xn^2)/n . Show  that
M^2>=4Vab/(a+b)^2
Vietnam  1981
Thanks!

Classical result:

(xi-a)(xi-b)<=0 => V - (a+b)M + ab <=0
=> V/M^2 <= -ab/M^2 + (a+b)/A = -ab(1/A-(a+b)/2ab)^2 + (a+b)^2/4ab <= (a+b)^2/4ab and we have done.

Namdung


If $ x,y,z\in \mathbb{R}$ and $ x^2\plus{}y^2\plus{}z^2\equal{}14$ , find the maximum value of :

$ P\equal{}2|xy\plus{}yz\plus{}zx|\plus{}M\minus{}3m \ ,\ (M\equal{}\max\{x^2,y^2,z^2\}\ ,\ m\equal{}\min\{x^2,y^2,z^2\})$

Clearly the inequality achieves its maximum iff $ x$, $ y$, $ z$ are all positive or negative. So let $ x \ge y \ge z \ge 0$. Hence the inequality is equialent to $ 28\minus{}(x\minus{}y\minus{}z)^2\minus{}(y\minus{}2z)^2$, which is $ 28$ iff $ x\equal{}3$, $ y\equal{}2$, $ z\equal{}1$.


Sa se calculeze:

$\int\frac{x*arctan{\sqrt{x^{4}+2x^{2}}}}{\sqrt{x^{2}+2}}dx\;,\;\;x\in(0,\infty)$

[quote="Paul V."]Sa se calculeze:

$\int\frac{x\cdot\arctan{\sqrt{x^{4}+2x^{2}}}}{\sqrt{x^{2}+2}}dx\;,\;\;x\in(0,\infty)$[/quote]
mai frumos asa... :lol:


The numbers $1$ throgh $1993$ in some order are written in a row. the following operation is performed: if the first number in the row is $K$, then the first $k$ numbers are reversed. Show that after several such operation number $1$ will occur on the first place.

Consider the largest positive integer $n_0$ which ever reaches the first position. It's clear then that after performing the move which takes $n_0$ to the $n_0$'th position, it will never reach the first position again. Then consider the largest positive integer $n_1<n_0$ which ever reaches the first position, and so on. This process has to end, and it can only end when $1$ reaches the first position.

Another of those useless remarks: The problem was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=62215 , with a solution being posted by... guess whom.
In fact it would have been the easiest problem of the BWM 2002 second round, if not about every solution shorter than a page would have been graded as incomplete or not detailed enough.

  Darij


linear algebra
matrix
linear algebra unsolved

let A be any matrix with complex entries.show I+AA* is nonsingular(A*=(A bar)transpose)) :rotfl:

Hint: $\langle AA^{*}x,x\rangle=\langle A^{*}x,A^{*}x\rangle\ge 0$.


[quote="ductrung"][quote="pvthuan"]I have not seen this one anywhere. That's good inequality. Can you give any hint?[/quote]
Hello Thuan,
$\sum_{cycl}\sqrt{\frac{b+c}{a}}\ge\sqrt{6}\frac{a+b+c}{\sqrt{ab+bc+ca}}$
Enjoy.[/quote]

This one is still that hard. Can you be more instructive?

[quote="phamduyhiep"][quote="ductrung"]For positive reals $a,\ b$ and $c$, prove that
\[ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}} \ge 2\ \sqrt{3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}  \][/quote]
It's easy but very nice!
$2\sum \sqrt{ \frac{b+c}{a}. \frac{c+a}{b} } \geq 2\sum \sqrt{ \frac{4c}{ \sqrt{ab}}} \geq 12$
and $\sum \frac{b+c}{a} =\sum a(\frac{1}{b}+\frac{1}{c}) \geq \sum \frac{4a}{b+c}$.[/quote]
Your solution is nice!

[quote="phamduyhiep"]It's easy but very nice!
$2\sum \sqrt{ \frac{b+c}{a}. \frac{c+a}{b} } \geq 2\sum \sqrt{ \frac{4c}{ \sqrt{ab}}} \geq 12$
and $\sum \frac{b+c}{a} =\sum a(\frac{1}{b}+\frac{1}{c}) \geq \sum \frac{4a}{b+c}$.[/quote]
Nice, Hiep! I found it this way
\[ \sum_{cycl}\sqrt{\frac{b+c}{a}}\ge \sqrt{6}\frac{a+b+c}{\sqrt{ab+bc+ca}}\ge 2\cdot\sqrt{3+\sum_{cycl}\frac{a}{b+c}} \]


vector
linear algebra
matrix
graph theory
combinatorics unsolved
combinatorics

Let $ G$  be the complete graph with $ 2n$ vertices, and suppose that $ S_1,S_2,...,S_k$ are subgraphs of $ G$ with the following properties:

(a) Every $ S_i$ is a complete bipartite graph;

(b) Every edge of $ G$ is contained by an odd number of subgraphs $ S_i$.

Show that $ k\geq n$.

Nice one! Sketch of a solution:

Consider the $ \mathbb{F}_2$-vector space $ \mathbb{F}_2^{2n}$. Let $ e_j$ be the $ j$-th basis vector for every $ j\in\left\{1,2,...,2n\right\}$. Enumerate the $ 2n$ vertices of $ G$ by $ 1$, $ 2$, ..., $ 2n$.
For every $ i\in\left\{1,2,...,k\right\}$, the graph $ S_i$ is a complete bipartite graph, so there are two disjoint subsets $ L_i$ and $ R_i$ of $ \left\{1,2,...,2n\right\}$ such that $ V\left(S_i\right) \equal{} L_i\cup R_i$ and $ E\left(S_i\right) \equal{} \left\{\left\{\lambda,\rho\right\}\mid \lambda\in L_i\text{ and }\rho\in R_i\right\}$.
For every $ i\in\left\{1,2,...,k\right\}$, define two vectors $ \ell_i\in\mathbb{F}_2^{2n}$ and $ r_i\in\mathbb{F}_2^{2n}$ by $ \ell_i \equal{} \sum_{\lambda\in L_i}e_{\lambda}$ and $ r_i \equal{} \sum_{\rho\in R_i}e_{\rho}$.
Then, condition (b) says that the $ 2n\times 2n$ matrix $ \sum_{i \equal{} 1}^kL_iR_i^T \plus{} \sum_{i \equal{} 1}^kR_iL_i^T$ equals to the matrix $ E \minus{} I_{2n}$, where $ E$ is the $ 2n\times 2n$ matrix with every entry equal to $ 1$, and $ I_{2n}$ is the $ 2n\times 2n$ identity matrix. But $ \det\left(E \minus{} I_{2n}\right) \equal{} 1$ (remember, we are working over $ \mathbb{F}_2$), so that $ \mathrm{Rank}\left(E \minus{} I_{2n}\right) \equal{} 2n$. But on the other hand, $ E \minus{} I_{2n}$ is the sum of $ 2k$ matrices of rank $ \leq 1$ each (since $ E \minus{} I_{2n} \equal{} \sum_{i \equal{} 1}^kL_iR_i^T \plus{} \sum_{i \equal{} 1}^kR_iL_i^T$, and every matrix of the form $ UV^T$ for two vectors $ U$ and $ V$ has rank $ \leq 1$), and thus has rank $ \leq 2k$ (since $ \mathrm{Rank}\left(U_1 \plus{} U_2 \plus{} ... \plus{} U_v\right)\leq\mathrm{Rank}U_1 \plus{} \mathrm{Rank}U_2 \plus{} ... \plus{} \mathrm{Rank}U_v$ for any matrices $ U_1$, $ U_2$, ..., $ U_v$). Thus, $ 2n\leq 2k$ and therefore $ k\geq n$.

  darij


linear algebra
matrix
Princeton
college
articles
Support
geometry

Hello, all!
How was your performance at the JEE 2009?

1st paper was pretty easy
2nd maths was a bit tough  :mad:
but 2nd paper section 4 was real goldmine esp for phy and chem

[hide]
i may get 280 to 320

i did not say anything  :P
[/hide]

[quote]
but 2nd paper section 4 was real goldmine esp for phy and chem
 [/quote]
I  agree !!
btw,when are the solutions of different coaching institutes coming out?
i am waiting to see how much i'll get !
And i guess cut off's will be really high this time ! :P

some nondescript prerna class has already put up solns

i already can spot a few mistakes in them  :rotfl:

we shud wait for fitjee ones

i also found the paper easy :P   :rotfl:  (is that possible)
aakash insti will release sol by 11pm (got an sms)

[quote]
some nondescript prerna class has already put up solns
[/quote]
can you put up the link?

i think its not a new thing

but do u even kno where is this prerna classes??

they have relesed dolutions of paper1 only

and must be wrong coz i m getting 159 :rofl:

an awesome 74/80 in physics (and that also only last ques wrong)

what abt u???

not bad at all

but certainly some answers must be wrong (i wud be happy if they r not) :blush:

how can i get so much...me...not even an engineering aspirant :o

talking abt  last ques in physics...

isnt $ B \equal{} 0$ in "s"
and $ E not equal to 0$ in "r"???

i wud b jumping if i get even half of it in Paper2

meanwhile...last >287 implies <500 rank
wat do u guys think abt this yr??

One of my original questions, the brown colouration one made it to the paper.howbout that  :D

me getting 154 :(
tht matrix match thing?the first one is faulty?rite?how is the gravitational potential energy continously increasing in s and t?
O_O
and mechanical energy doesnt do decrease in s?
v=(2g'h)^0.5
g'<g
so mgh>0.5mv^2
:(
i screwed up chem 37 in chem :( 52 and 65 in physics and math
:D
i hardly get 50% in my coaching classes in math :D math was so dead easy
its chem tht i goofed up :(

hello every1.

[b]@ rituraj[/b]

how was ur PMT i suppose u r going 2 cum first... :D

[quote="sumanth0811"]
@nuclear
pron ones!/quote]
I loooove seafood!  :rotfl:
Do I get a free demonstration of the 'cooking' procedure for prawns?

[quote="nuclear_alchemist"][quote="sumanth0811"]
@nuclear
pron ones!/quote]
I loooove seafood!  :rotfl:
[/quote][/quote]

i luv live [hide="fish"]mermaids  :P [/hide]

no i wud say prons r better :rotfl:

@tuhin..well well well..i'll get first rank but from bottom

chem ended up all chances i had to get into top 50 :(

I blew my chance of qualifying for olympiad luck has crushed me :mad:

Hey everyone has forgotten abt this thread...

About 24 hrs for the IIT results to be out :D
Could anyone post the link where the official results will be available..?

@ Conjurer
Most probably 08:00 AM on 25th May.. But actually it may come a bit earlier too.. Last year the results came 20 min. before.. So, we need to stick to comp much early..   :roll:

@ _MM_
Results will be out at every IIT site.. Though the timings for each site to display the results won't be same..  :!:

Results will be available at 8 AM sharp on 25th at following official JEE 2009 sites of 7 IITs:

http://www.jee.iitb.ac.in/main.php
http://www.jee.iitd.ac.in/
http://www.iitg.ac.in/jee/
http://www.jee.iitk.ac.in/
http://www.iitkgp.ernet.in/jee/
http://jee.iitm.ac.in/
http://jee.iitr.ernet.in/

Though last year, few people told me that at one of the IIT site, the result was available at 12 mid-night the day earlier. I will be trying tonight.

The current thread had a life post JEE and till results are out. From tomorrow, let us use new thread created 'Post IIT JEE 2009 Result'.

results out on iitkgp.ernet.in/jee/resultstatus.php but there is some error....
wtf!!!

Nitin Jain - AIR 1 (IIT Delhi Zone)
Shubham Tulsiani - AIR 2 (IIT Mumbai Zone)

No surprises! :) Bansal will never go out of business.

its awesome that topper is not  frm bansal

Delhi rocks!!! :D

last year our mod was AIR-1[hide]in seeing the result...he checked out his result at 11.45 or so ..one day earlier :rotfl: [/hide]


[quote="kysuke"][quote="27r"][quote="kysuke"]cool invitational we had today huh?[/quote]

Ehh.. crazy team round problems. Not to mention the awards ceremony.

Otherwise it was okay.[/quote]

If you're talking about precalulus, I sure loved the arithmetic.[/quote]

I'm in Alg 2, so I don't know about precalc, but our team round was crazy. The top team round score was 52.

[quote="Poincare"]I MEAN COME ON
GIVE ME SOME REASON TO LIVE IN FLORIDA[/quote]
What in the world? I feel deja vu:

http://www.playworkz.com/swingset/maoforums/index.php?showtopic=1153777708

I have come! To save this forum and get unmuted from FTW!


What percent of $ 5x$ is equal to $ x/5$?


I am considering homeschooling my son next year, and he is about ready to learn calculus and physics.  I'm looking for a good book that teaches them simultaneously.  I can find a lot of physics books that assume you have had calculus already, and calculus books that assume that math is a stand-alone subject, but I've not yet found a book that introduces the concepts simultaneously.

It seems to me that co-teaching physics and calculus is such an obvious idea (calculus was initially developed in order to do physics, after all), that I'm surprised I'm having such trouble find texts for co-teaching them.

Does anyone have any suggestions?  If not a single book, then a physics book and a calculus book that can be used together?

Two titles to consider are:

1. The Mechanical Universe

2. Integrated Calculus and Physics, Volumes I and II

Thanks.

By "The Mechanical Universe" I assume you mean the telecourse filmed at Caltech in 1985.  That looks like a useful pointer for physics.

By "Integrated Calculus and Physics" I assume you mean the books by Andrew Rex and  Martin Jackson.  They look like the sort of book I had in mind.  Are they any good?   How do they compare with other approaches to teaching calculus or physics?

I read Vol. I of [u]Integrated Calculus and Physics[/u] after I finished Frank Loxley Griffin's [u]Introduction to Mathematical Analysis[/u] and I still found it demanding.

I'm only familiar with the [u]Advanced Edition of The Mechanical Universe[/u]. It would need to be supplemented with a calculus text while the ICP is a combination calculus and physics text.

I never did find an integrated calculus and physics book.  Instead we ended up using the AoPS online calculus course (which was excellent) and the [i]Matter and Interactions[/i] textbook for physics [url]http://matterandinteractions.org/[/url], which was a good fit, given its use of computational models.


3.  If α is a real root of $x^5 - x^3 + x - 2 = 0$, show that $[a^6] = 3$.

$P(x) = x^5 - x^3 + x - 2$

$P'(x) > 0$ so $P(x)$ is strictly increasing and $\alpha$ exists and is unique.

$P(0) = -2$ so $0 < {\alpha}$ and equation can be rewritten $\alpha^6 = 2(1+ \alpha^2)/ \alpha) - 1 > 3$
Finally $P(4^{1/6}) > 0$ so ${\alpha}^6 < 4$

[quote="Diogene"]Already on this site : http://www.mathlinks.ro/Forum/topic-43943.html
:cool:[/quote]

$P(1.201) < 0 \Longrightarrow a > 1.201 \Longrightarrow a^6 > 3$

Sure but how do you guess and prove that without using a computer (= in competition conditions) ??


logarithms
floor function
inequalities
induction
probability
expected value
combinatorics unsolved

for n \geq k \geq 3 assume X={1,2,...,n} and Fk is a family of subsets with k members such that any two subsets in Fk has at most k-2 members in common. show the existance of Mk (subset of X) with at least  [log(n) (in base 2)]+1 members such that has no common members with Fk.([x] means integer part of x)

I have made readable form, but I have some doubts about statement. What is actually $M_k$?

For $n \geq k \geq 3$ assume $X=\{1,2,...,n\}$ and $F_k$ is a family of subsets with $k$ members such that any two subsets in $F_k$ has at most $k-2$ members in common. Show the existance of $M_k\subset X$ with at least  $[\log_2 n]+1$ members such that has no common members with $F_k$. ($[x]$ means integer part of $x$).

I do not understand your question. $M_k$ is a subset of $X$ which has to be constructed (that's the question  :P ).
And, as usual for sam-n, it looks like a probabilistic method one.

Pierre.

sniff... I don't want to talk with you, but problem in current state is meaningless, sinse $F_k$ can cover entire $X$.

[quote="pbornsztein"][quote="Myth"]sniff... I don't want to talk with you,[/quote]

 :P

[quote]
 but problem in current state is meaningless, sinse $F_k$ can cover entire $X$.[/quote][/quote]

Ok, here is the statement that I have :
For $n \geq k \geq 3$, let $X = \{1,2, \cdots , n \}$ and let $F_k$ be a family of $k$-elements subsets from $X$ such that any two subsets in $F_k$ have at most $k-2$ common elements. Show that there exists a subset $M_k$ of $X$ with at least $[ \log_2 (n) ] + 1 $ elements containing no subset in $F_k$.

That does not mean that $M_k$ has no common element with the subsets in $F_k$....

Pierre.

As Myth remarked, the problem, as stated, is absurd. I happen to know, however, the correct formulation:

[b]Ammended Problem.[/b] Let $F_k$ be a family of $k$-element subsets of $[n]$ such that every two of them have at most $k-2$ elements in common. Show that there exists a subset $M_k \subset [n]$ with at least $\lfloor \log_2 n \rfloor + 1$ elements such that no element of $F_k$ is a subset of $M_k$.

And yes, Pierre, the original proof is probabilistic.  :lol:

--Vesselin

But your turn to write complete solution Vess  :D

Pierre.

:D  :D  :D

It's an easy probabilistic argument. Let $m = \lfloor \log_2 n \rfloor  + 1$. It is sufficient to show that the expected number of elements of $F_k$ that are contained in a random $m$-element set $X \subset [n]$ is less than $1$. But this expected number equals $|F_k|\binom{m}{k}\Big/\binom{n}{k}$, so we have to show that
\[
|F_k| < \binom{n}{k}\Big/\binom{m}{k}.
\]
This is where we'll use that every two sets in $F_k$ have at most $k-2$ elements in common. This means that each $(k-1)$-element subset lies in at most one member of $F_k$; since, in addition, each member of $F_k$ contains $k$ elements (and therefore has $k$ $(k-1)$-element subsets), it follows that
\[
|F_k| \leq \frac{1}{k}\binom{n}{k-1} = \frac{1}{n-k+1}\binom{n}{k},
\]
so we are left to prove that $\binom{m}{k} > n-k+1$ for  $m=\lfloor \log_2{n} \rfloor + 1$, which is a triviality.

--Vesselin

[quote="pbornsztein"][quote="vess"]
so we are left to prove that $\binom{m}{k} > n-k+1$ for  $m=\lfloor \log_2{n} \rfloor + 1$, which is a triviality.

[/quote]

Hmmm... I wouldn't say that...

Pierre.[/quote]

Sorry, we actually need the opposite inequality,
\[
\binom{m}{k} < n-k+1,
\]
which of course isn't true in general. However, we may assume that $k \leq m$ (otherwise the problem is trivial), in which case it suffices to show that
\[ (*)
\binom{m}{k} < n-m+1.
\]
An easy induction shows that $\binom{m}{k} \leq 3 \cdot 2^{m-3}$ for $m \geq 3$, and now (*) becomes quite obvious.  :)


Does anyone have a solution guide? I don't even think there is one...at least not that I know of :? Does anyone on this forum even need one?

A search of the forum shows the answer to your question [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=83348&highlight=zeitz+solutions#p83348]here[/url] as well as a couple of other places -- there is a solution manual, but it is not available to the public, and at least one member of the forum is in possession of one.

[quote="Mario_KaRt"]Does anyone on this forum even need one?[/quote]Yup :) There are over 5800 members on this site, but as you may guess not everyone is a genius. We are all here to learn from each other and to increase our problem solving abilities, mathematical abilities, and some of us just like to *hang out* with their friends (yeah I know kinna *gecky* :P).

This issue has been previously discussed in this post:
http://www.mathlinks.ro/Forum/viewtopic.php?t=19879
so maybe you'll have a look there too :P

[quote="Mario_KaRt"]Does anyone on this forum even need one?[/quote]

I do  - there are some pretty tough problems there.  :)

Edit: Thanks to those of you who have sent me a copy - I meant the above note as a 'those problems are so hard I can't solve them all in a reasonable amount of time'.  I know Paul & he sent me a copy a while back, though I do appreciate the efforts of those who sent me copies!

Yea I wouldn't mind purchasing one either.


integration
calculus
calculus computations

Compute $ \int \frac{\sqrt[4]{x^3}}{x^7 \plus{} 1}\,dx$.


abstract algebra
superior algebra
superior algebra unsolved

1. $ G_1=Aut(G),\cdots......G_n=Aut(G_{n-1})$   what kind of G will yield $ G_n=G_{n+1}=\cdots$

2. similarly, $ V$ is a $ R-module$  $ E_1=End(_{R}V)$  $ E_2=End(V_{E_1})$  $ E_3=End(_{E_2}V) \cdots$ what kind of V and R will yield $ E_n=E_{n+2}$ or simply $ E_n=E_{n+1}$.

for example when $ G=S_n$ all $ Aut(G_n)=S_n$

The automorphism-tower problem seems hard. It is proven, that for every finite, centerless group (i.e. $ Z(G) \equal{} 1$) the automorphism tower becomes stationary (Wielandt) in a finite number of steps. The same holds for centerless Chernikov-groups and some other types of groups.
For arbitrary groups the automorphism tower becomes stationary too but this could take an infinite number of steps. (At least this paper claims that: http://arxiv.org/abs/math/9808014)

And by the way: $ Aut(S_6)\neq S_6$ (index of $ S_6$ in $ Aut(S_6)$ is 2) and $ 1 \equal{} Aut(S_2) \neq S_2$. The equality holds exactly for $ n\neq 2 \wedge n\neq 6$.

The endomorphism-ring-problem is easy. For each and every module we have $ E_3 \equal{} E_1$ and $ E_4 \equal{} E_2$ so the sequence is always periodic. For commutative $ R$ one has in fact $ E_2 \equal{} E_1$.


Let $ f: [a,b]\rightarrow R$ be a continuous function. If there exists $ c\in [a,b]$ so that $ f(c)>1$, prove that there exists $ m,n\in [a,b]$ so that $ f(x)>1\forall x\in [m,n]$.

I think it's a direct conclusion from the Darboux property...  :wink:

It's much simpler than that; we can work straight from the definition of continuity at $ c$.

Isn't this really trivial? Continuity at c implies there exists a neighborhood around c such that f(x) is "sufficiently close" to c -- define sufficiently close to be the different between f(c) and 1.

Uhm...well actually it was just a test to see how many people know the definition of continuity. :lol:  :blush:  Or maybe not but when in 12th grade one forgets a lot about the concepts learned a year ago.


number theory
prime numbers
prime factorization

Is it possible to decide if $2*3*5*7*11*13*17*19*23*29+1$ is a prime number without testing? And how can you decide that?

Well notice these are all the prime numbers, up to 29.

Thus, it is not divisible by any prime number, up to 29.

So we can see that this number is prime!

In other words...

Let $p_{i}$ be the $i^{th}$ prime.

$p_{1}p_{2}p_{3}\cdots p_{n}+1=p_{x}$

where $x>n$

edit: I just corrected it...

[quote="1=2"]In other words...

Let $p_{i}$ be the $i^{th}$ prime.

$p_{1}p_{2}p_{3}\cdots p_{n}=p_{x}$

where $x>n$[/quote]

to elaborate...

$p_{x}+1$, therefore, is NOT divisible by $p_{1}, p_{2}, p_{3}...$ etc.
so its prime. which is kind of what i said earlier, but...

Close, but not quite.... it is either prime OR has a prime factorization that uses primes that you didn't use in the factorization, it's no always prime. For example:

2*3*5*7*11*13 + 1  = 30031
30031 = 59*509

hm... good point.  :blush:
i confused this with the other theorem... the one that says $n$ is prime if it is not divisible by any prime $<\frac{n}{2}$.


Let $ a,b,c$ be three side lengths of a triangle. Find the greatest constant $ k$ such that the following inequality holds:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\ge k\left(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\minus{}\frac{3}{2}\right)$

I hope ı found correct :maybe:
http://www.imagehack.eu/uploads/3baa1917b3.gif

I hope ı found correct   :maybe:
[url=http://imageshack.us][img]http://img167.imageshack.us/img167/7435/114mathlinksroig2.gif[/img][/url]


geometry
geometric transformation
rotation
vector
real analysis
real analysis unsolved

Find all continuous curves $M(t)$ on the real plane such that  the length of the segment whose endpoints are $M(t_1), M(t_2)$ depends only on $|t_1-t_2|$.

[b]Answer:[/b] a circle, a line, or a point (or their part).

Here is a sketch of my proof:

Let us take two distinct points $A = M(t_A)$ and $B= M(t_B)$ (say $t_A< t_B$). Pick $c > 0$ so that $M$ is defined on the neighborhoods of $t_A$ and $t_B$.
Define $A_t = M(t_A + t)$, and $B_t = M(t_A + t)$. Let $R_t$ be the isometry that maps $A$ to $A_t$, $B$ to $B_t$, and preserves the orientation of the plane.

It is easy to see that the map $t \mapsto R_t$ is continuous. It is "almost" a homomorphism from the additive group $({\mathbb R}, +)$ to the group of plane isometries:
\[ R_a \circ R_b = R_{a+b}  \]
if $a,b, a+b \in (-c, c)$.

The isometry $R_0$ is the identity transformation. We may assume that $c$ is small enough s.t. $R_t$ is close enough to the identity transformation for $t\in (-c, c)$.
Now every plane isometry (that preserves the orientation) is either a rotation or a translation. Consider the sequence
\[ R_{c/2}, R_{c/4}, R_{c/8}, \dots  \]
Each is a square root of the previous. Moreover since all of them are close enough to the identity, the square root is defined uniquely. We get that for $t=\frac{c}{2^k}$ either
i) $R_{t}$ is the rotation around a fixed center $O$ by the angle $t\phi_0$, or
ii) $R_{t}$ is the translation by the vector $t v_0$.

Since the map $t \mapsto R_t$ is "almost" a homomorphism, this also holds when $t$ is a binary rational. By continuiuty this holds for all reals.

Now the curve is an orbit of $R_t$. So it is either a circle or a line.


Using the x- and y-coordinates from the unit circle and the equation: cosθ=(sq.root of 2)/(2) , the angle θ is

PLS HELP

I do not know what the rest of the hypothesis such as (using x and y coordinates from the unit circle) is useful for.

Finding the initial answer which is $\theta = \frac{\pi}{4}$ presents no difficulty, but since $\theta$ is multivalued, we get the same answer by adding $2n\pi$ where $n = 0,1,2,3,..$
So $\theta = \frac{(1+8n)\pi}{4}$, where $n =0,1,2,3,4,5,...$

That's only half of the solutions- there are two solutions in each period, not 1.

[quote="akech"]Thanks jmerry;-
Okay, the complete solution should be
$\theta = \frac{(1+8n)\pi}{4}$, where $n = 0,\pm1,\pm2,\pm3,\pm4,\pm5,...$[/quote]

You just added negative integers to n for your  answer

but...
Cosine is also $\frac{\sqrt2}{2}$ at  $\theta = \frac{-\pi}{4}$

Thus...
$\theta = \frac{(-1+8n)\pi}{4}$, where $n = 0,\pm1,\pm2,\pm3,\pm4,\pm5,...$

If your teacher isn't picky (or isn't very well educated)
You can just put
[hide]
(sqrt2/2,sqrt2/2)
for xy coordinates

[/hide]

but if your teacher is well educated
then just listen to what others posted

You would just add 2pi for the next coordinate i think

spriggin, are you related to inteval as well?

 :|

$\cos{\beta}= \frac{\sqrt{2}}{2}\leftrightarrow \theta = \pm \frac{\pi}{4}\pm 2k\pi \leftrightarrow (x,y) = \left(\frac{\sqrt{2}}{2},\pm\frac{\sqrt{2}}{2}\right)$