geometry
analytic geometry
vector
trigonometry
trapezoid
calculus
integration

Triangles $ ABC$ and $ ADE$ have areas $ 2007$ and $ 7002,$ respectively, with $ B \equal{} (0,0),$ $ C \equal{} (223,0),$ $ D \equal{} (680,380),$ and $ E \equal{} (689,389).$ What is the sum of all possible x-coordinates of $ A?$

$ \textbf{(A)}\ 282 \qquad \textbf{(B)}\ 300 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 900 \qquad \textbf{(E)}\ 1200$

There's a nicer solution with vector cross products being 1/2 the area of the trapezoid.

I'll leave someone to do the actual solution posting, but I solved this one very quickly.

Also, the correct solution isn't C, but [hide]E[/hide]

[quote="DiscreetFourierTransform"]There's a nicer solution with vector cross products being 1/2 the area of the trapezoid.
[/quote]

Hm, I think shoelace is just vectors in disguise.

[quote="DiscreetFourierTransform"]Yeah, it is, but the raw solution is elegant, powerful, and flexible.

It really makes "the ART of problem solving" take on a new meaning

(sorry for any perceived arrogance, by posting on the internet my sincerity is lost with the bytes.)[/quote]

Hm I would like to see that...I know basic vectors (dot, cross products) but I would need to review fancy stuff.  How complicated is the solution?  I would like to see it if someone could post it.

I'll post mine up tomorrow if nobody else has one.
I need to focus on actual HW, but I'll work on getting one up for you.

[quote="Altheman"][quote="mdk"]what did you get on the test altheman?[/quote]

118.5   :|

yea...i sucked it up...i did not do so well with the time pressure...i kind of spent too long on some problems, and i did not choose to do elegant solutions...i also had 3 careless errors that killed my score[/quote]

That interesting, you always seem to find the most elegant solutions on this forum. Maybe you're saving those for the aime.

A way of doing this problem consists of determining the heights of the triangles given their areas and base lengths and setting equations for the lines in which the bases of the triangles are contained. For triangle ADE, the base (DE) has a length of 9sqrt(2) and is contained in the line y = x-300. Dividing the area of the triangle (7002) by the 1/2*the length of the base, we get its height as 1556sqrt(2)/2. Similarly, for ABC, the base (BC) has a length of 223 and is contained in the line y = 0, and the height can be calculated to be 18. Now, for triangle ABC, no manipulation needs to be done with the height since it is already parallel to the y-axis. However, for ADE, the height is at a 45-degree angle from the vertical. Therefore, we must set up a right triangle with a right angle at the intersection of the line y = x-300 and a perpendicular line, and we get the y-component of the distance between point A and the line of DE to equal 1556. Now, the different possible values for y include 1556-18, 1556+18, 18-1556, -18-1556. These numbers all add up to 0. However, solving for x, we get x = y+300. Since there are four values of y we are adding up, our solution is 4*300 = [b]1200[/b], or [b]E[/b].

In my personal case, I had almost solved this problem when time was running out, and I made a quite egregious silly mistake by multiplying 300 by 2 instead of by 4, giving me an answer of C.  :wallbash_red:

yes yes yes. This is what realized when looking at it hours later.

Personally, although I got this one right, I used the shoestring method.  It was horrible brute force and got me a bit nervous because it was only #19.  (I spent 5-10 minutes on it  :blush: )

The computations on this problem didn't take me that long.  Once I figured out that $y = \pm 18$ it wasn't hard to calculate that the four $x$-values were given by $x = 300 \pm 1576 \pm 18$ (or something of that form), the sum of which is clear.

sorry for reviving old forum but it seems the vector solution was never actually posted...

mysmartmouth:
for three-space vectors $ a$ and $ b$, we know,
$ \parallel{}a \times b\parallel{} \equal{} \parallel{}a\parallel{}\cdot \parallel{}b\parallel{}\cdot \sin{\theta}$
where $ \theta$ is the angle between the two vectors.

and this is known to give area of trapezoid defined by the vectors (informally you can see this based upon the formula for area of triangle $ \frac{ab\sin{C}}{2}.$)

So in this case, you just had to view all of the sides as three-space vectors with k component of 0.
Discreet: is this ^ sort of what you did?

As for shoelace, i had to prove it in a vector-cal class using line integrals and such. it might be related to the cross-product characterization.

[quote="shake9991"]

As for shoelace, i had to prove it in a vector-cal class using line integrals and such. it might be related to the cross-product characterization.[/quote]

This proof is very nice!

It is based on a tricky use of green's theorem. In particular $ \int\int_R 1\equal{}\int_{\partial R} x\, dy$...

[quote="Altheman"][quote="shake9991"]

As for shoelace, i had to prove it in a vector-cal class using line integrals and such. it might be related to the cross-product characterization.[/quote]

This proof is very nice!

It is based on a tricky use of green's theorem. In particular $ \int\int_R 1 \equal{} \int_{\partial R} x\, dy$...[/quote]
Yes, and if you guys want, please start a topic in calculus forum about this discussion of line integrals and continue there, it is interesting and at the same time follow the guideline of this forum (no calc in Intermediate) ;)

[hide="Solution"]The y-coordinate of $A$ is $\pm 18$. Let $A=(a, b)$ where $b=\pm 18$.

Then the area of $\triangle{ADE}$ is the length of the altitude from $A$ to $\overline{DE}$ times the length of $\overline{DE}$ divided by 2.

The equation of line $DE$ is $x-y-300=0$ so the length of the altitude is the distance from $A$ to $\overline{DE}$ which is given by

$\frac{|a-b-300|}{\sqrt{1^2+1^2}}=\frac{|a-b-300|}{\sqrt2}$

so the area of $\triangle{ADE}$ is $\frac{|a-b-300|}{\sqrt2}\cdot DE=\frac92{|a-b-300|}=7002$ which gives $|a-b-300|=1556$.

If $b=18$ then $a=318\pm1556$ and if $b=-18$ then $a=282\pm1556$. Therefore the sum of all possible $a$ is $1200$ and $E$.[/hide]

@ahaan: thanks

[quote="AwesomeToad"]
$\frac{|a-b-300|}{\sqrt{1^2+1^2}=\frac{|a-b-300|}{\sqrt2}$
[/quote]You were missing a '}'. Here: $\frac{|a-b-300|}{\sqrt{1^2+1^2}}=\frac{|a-b-300|}{\sqrt2}$

[hide="Question"]
How did you get [quote="AwesomeToad"]The y-coordinate of $A$ is $\pm 18$. [/quote] ^^^ that? I'm sorry if I'm missing something obvious.
[/hide]

[hide="@your question"]The base ($\overline{BC}$) is given (since $B$ is at the origin and $C$ is $(223, 0)$)

Then the altidude $h$ is such that $h\cdot223/2=2007$ and thus $h=18$, which means $A$ lies 18 units from the x-axis.[/hide]

[quote=muti66][quote=Altheman]WOA MAN...IF WE CONSIDER THE LOCUS OF A, it is the interection of two pairs of parallel lines, 4 points, whose centroid is the intersection of BC and DE, =(300,0), so multiply by 4 to get 1200

SNAP[/quote]

Wait how does this work?[/quote]

The condition $[ABC] = 2007$ forces $A$ to lie on one of two lines, [b]which are both parallel to and equally spaced from $BC$[/b]. The condition $[ADE] = 7002$ forces $A$ to lie on one of two lines, [b]which are both parallel to and equally spaced from $DE$[/b]. Thus, taking both restrictions into account, we see that there are $4$ possible locations for $A$, which form the vertices of a parallelogram; to find the sum of their $x$-coordinates, we can find $4$ times the $x$-coordinate of its center. But this center is simply the intersection of $BC$ and $DE$ because of the bolded text, which is obviously $(300, 0)$, and thus the answer is $4 \cdot 300 = 1200$.

Pretty nice short-answer problem :P

Can someone please explain to me in Cantonmathguy's solution how you can just conclude it must be 4x the x-coordinate of the center? What is the logic behind this?
Thanks!

Sorry but /bump

Any input would be appreciated!

I think the center is the average of all the coordinates.

Can somebody explain how exactly we find the center's coordinates? How do we "obviously" get (300,0)?


Since $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}xz))$ we have $ 1\minus{}xyz\equal{}9\minus{}3(xy\plus{}yz\plus{}zx)$, which is equivalent to $ (x\minus{}3)(y\minus{}3)(z\minus{}3)\equal{}\minus{}8$, implying that $ x,y,z\in\{\minus{}5,\minus{}1,1,2,4,5,7,11\}$ for sure.

If we have $ \minus{}5$, then the others are both $ 2$ or both $ 4$, this gives the solution $ (x,y,z)\equal{}(\minus{}5,4,4)$ and its permutations.
If we don't have $ \minus{}5$: we can't have $ \minus{}1$, so $ x,y,z\ge 0$ and from the equality case of the power-mean inequality ,we get that $ x\equal{}y\equal{}z\equal{}1$.

W.L.O.G. $ x\geq y\geq z$. If $ x<0$, then $ x^3\plus{}y^3\plus{}z^3\equal{}x\plus{}y\plus{}z<0$ which contradicts. Since $ x^3\geq 3x\minus{}2\Longleftrightarrow (x\minus{}1)^2(x\plus{}2)\geq 0$, yieilding $ x^3\plus{}y^3\plus{}z^3\geq 3(x\plus{}y\plus{}z)\minus{}6,$ the eqaulity holds only for $ x\equal{}y\equal{}z\equal{}1.$

I however do think that $ 4^3\plus{}4^3\plus{}(\minus{}5)^3 \equal{} 4\plus{}4\plus{}(\minus{}5)\equal{}3$...

You prove your ineq for $ x$, but you use that $ x\ge\minus{}2$. That is true for $ x$ but not for $ y$ and $ z$, so you can't sum them up afterwards. ;)

Just a sidenote: finding all $ x,y,z \in \mathbb Z$ such that $ x^3 \plus{} y^3 \plus{} z^3 \equal{} 3$ is an open problem.

Another Method : from identity $ x^3\plus{}y^3\plus{}z^3 \equal{} (x\plus{}y\plus{}z)^3\minus{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)$
$ 3 \equal{} 27\minus{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)$ so $ (x\plus{}y)(y\plus{}z)(z\plus{}x) \equal{} 8$
thus $ (3\minus{}x)(3\minus{}y)(3\minus{}z) \equal{} 8$ which is easy to get $ (x,y,z) \equal{} (1,1,1)$

[quote="MWIT-19"]$ (3\minus{}x)(3\minus{}y)(3\minus{}z) \equal{} 8$ which is easy to get $ (x,y,z) \equal{} (1,1,1)$[/quote]
It's not right because as Peter said, $(x,y,z)=(4,4,-5)$ also works...

[quote="Peter"]Find all $(x,y,z) \in {\mathbb{Z}}^3$ such that $x^{3}+y^{3}+z^{3}=x+y+z=3$.[/quote]
We observe that $x^3+y^3+z^3=(x+y+z)(x^2+y^2+x^2-xy-yz-zx)+3xyz=x+y+z$
Writing $3$ as $x+y+z$  and factorizing them up we get after a few steps
$(x+y+z)^{2}-3(xy+yz+zx) +xyz=1$
Again writing $x+y+z=3$ we get,
$3^2 -3(xy +yz+zx)+xyz=1$
=>$8=3(xy+yz+xz)-xyz$ Now factorizing this up we get
=>$8=(3-z)xy +3z(x+y)$ now we know $3-z=x+y$
Therefore $8 = (x+y)( xy+3z)$ ,rewriting $(3-z)$ instead of $x+y$ we get
=>$8=(3-z)(xy+3z)$  .Now factorize $8$ and find the values of $x$,$y$ and $z$ which we get as
$(x,y,z) =(4,4,-5)${and all permutations of it} ;$(1,1,1)$


algebra
polynomial
superior algebra
superior algebra unsolved

Find all polynomials $f$ for which there exists a polynomial $h$ such that

$f(2t)=h(f(t))$.

This is probably easy but I can't seem to do it.  There is an $h$ for any monomial $f$ but I don't know if there exists $h$ for polynomial $f$. Thanks.

With consideration of the degree of the equation, $degh(x)\leq 1$, and I think the rest of the problem is just to write the equation explicitly and solve it using violence?

i agree with victor. I think its you can have binomial as long as one of the terms is constant.


Marsha borrowed one mystery, one comedy and one adventure video from a friend. If the friend had five mysteries, seven comedies, and six adventures from which to choose, how many different sets of tapes could Marsha have chosen?

This is simply $ 5 \cdot 7 \cdot  6 \equal{} 210$.


Recently started Alcumus.
So far, it seems as if the topics discussed are mostly constrained to counting and probability.
Are there other subjects which will become available when I obtain more points?

See the FAQ:

Will there be other subjects included in Alcumus?
Yes. We are developing material for Alcumus that supplements the rest of our texts, beginning with Introduction to Algebra. There is currently no available timeframe estimate as to when the Algebra problems will be ready.


Given $n$ is 3600, $k$ is the sum of the factors of $n$, and $l$ is the sum of the factors of $7n$, what is $\frac{l}{k}$?

[hide="Hint"]
What can we say about $l$'s relation to $k$?
[/hide]

[quote="prototypex24"][hide]$l=n+7n=8n$
$\frac{l}{n}=8$
[/hide][/quote]
Please read the problem carefully. He is looking for l over k, not l over n.


Caution:

In practice, working on this will be faster compared to the reading/digesting through these many following details IMO:

You live in a town which is on the grid of the xy-plane.  It is all of the real points (x,y) which simultaneously satisfy the following:  They lie to the right of x = 1, to the left of
x = 7, above y = 1, and below y = 7.

There are unit lattice points inside this square-shaped town, where x and y are positive integers.  There are unit long streets
that join certain pairs of lattice points.  For instance, (2,2) to (2,3) is one or (5,3) to (6,3) is another.  A unit long street also exists if it extends to the border around the town [The border being: (1,1) to (1,7) to (7,7) to (7,1), back to (1,1)].  For example,
(1,5) to (2,5) is a unit long street extending to the border.

You are in charge of putting in a rail line INSIDE of the town limits.

Rules:

1) It must be a quadrilateral.

2) The rail line cannot touch or cross the border of the town.
    (Certain unit long streets do.)

3) No part of the quadrilateral need coincide with any of the
    unit lattice points.  That is optional for you.

4) It must contain exactly 4.5 (that is, four and a half) square
    units of area.

5) When choosing placement for your (rail line) quadrilateral
    inside of the town, in reference to 3), you may allow x and y
    to take on any positive real numbers, so long as you keep
    them in their earlier mentioned limits.

6) A. Make your quadrilateral (rail line) cross as many streets
    as possible.  For example, if one of the sides of the  quadrilateral (rail line) were from (2.5,2.5) to (4.5,3.5), then
the line segments (streets) crossed would be:  (3,2) to (3,3),
(3,3) to (4,3), and (4,3) to (4,4).

 B. However, a quadrilateral side that coincides with either a vertical or a horizontal street is defined as NOT crossing that street.  For instance, the side of the quadrilateral (rail line) from (2,4) to (2,7) does not cross any streets.  (It lies right over unit long streets, and it seems to go through "intersections."  But all unit long streets end, and new unit long streets begin in the same orientation.)

*** What is the largest number of crossed streets by the rail line (quadrilateral) of 4.5 square units possible inside this town?

I would think the quadrilateral would have to be a

[hide]
square in the middle?[/hide]

Usually answers are perfect...

mgao,

no, your "hidden content" is not true, BUT I DID want
users to get steered toward what you offered, in order to throw them off from their expectations.  Also you did not state the number of
unit long streets specifically crossed with your shape so that I would know that you know you're crossing correctly.  That, really, is what I am getting after.  The shape backs it up is true.

mgao, good effort so far...

If I were allowed to, anyone who had given me your answer,
would have gotten partial credit.
>
>

*** EVERYONE, please keep working on this problem.


I am interested in chemical engineering right now, but I am having a hard time deciding to which school I should eventually apply. Could someone tell me the differences between MIT and Caltech? I know that Caltech is very small and has a rigorous curriculum, but what about MIT's curriculum? Also Caltech allows students access to world class laboratories, but again, what about MIT?

Please provide me with lots of information so I can have a better idea of which shool I should apply.

I'm not sure, and I will also be applying to both schools, but not to major in chemical engineering.

On the other hand, someone on my ARML team (GA) is going to caltech and he intends to study chemical engineering. He's only a 9th grader, but he won the AP State Scholar Award (most number of 4's and 5's on AP tests in the entire state). I think he took an average of about 6 a year.

Personally, I would chose MIT simply because Boston is a much nicer place than LA.

[quote=drunner2007]
On the other hand, someone on my ARML team (GA) is going to caltech and he intends to study chemical engineering. He's only a 9th grader, but he won the AP State Scholar Award (most number of 4's and 5's on AP tests in the entire state). I think he took an average of about 6 a year.[/quote]
He plans to go to caltech you mean?  Did he take ap exams in middle school?

On topic, just apply to both.  Then revisit this issue if you get accepted to both.  MIT is one of the top ranked math (top 5) and science (not sure of a number) schools up there with caltech.  Going to MIT has the added benefit of being able to cross-register at Harvard.

I agree with the advice to deal with this issue initially by applying to both. You can learn to adapt to either school if you are admitted--that's what the admission officers think if they admit you.

Several AoPSers were in the happy position this year of being admitted to both Caltech and MIT. I don't think they all chose the same school to attend. Both schools arrange opportunities to visit campus for admitted students, and attending a visit like that in each place may help you decide.

Good luck in your applications.

[quote="quantum leap"]
...MIT is one of the top ranked math (top 5) and science (not sure of a number) ...[/quote]
Top 5 in compsci as well.


calculus
derivative
algebra
polynomial
calculus computations

I was told that if $N(x)=\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)$ are polynomials, then there's a formula for $N'(x)$ in terms of $f$, $g$, $f'$ and $g'$. Can someone tell me what that formula is? Thanks.

Take a look.

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


[quote="mr. math"]i need help with this problem:

when drawing two marbles from a bag containing 4 green, 7 orange, and 10 black marbles, what is the probability of getting a green and an orange.

and this one:

when two cards are drawn from a standard 52-card deck, what is the probabiality that they are from diffeernt suits?[/quote]

[hide="1"]$\frac{4*7}{\binom{21}{2}}=\boxed{\frac{2}{15}}$[/hide]

[hide="2"]$1-4(\frac{\binom{13}{2}}{\binom{52}{2}})=\boxed{\frac{13}{17}}$[/hide]

[quote="mr. math"]i need help with this problem:

when drawing two marbles from a bag containing 4 green, 7 orange, and 10 black marbles, what is the probability of getting a green and an orange.

and this one:

when two cards are drawn from a standard 52-card deck, what is the probabiality that they are from diffeernt suits?[/quote]

Do you replace the marbles and cards or not?

[quote="Totally Zealous"]Do you replace the marbles and cards or not?[/quote]

from the wording of the problem, I don't think you do

I think it could go either way, it does not say anything in the problem about that.

[hide="Without Replacement"] $2\cdot\frac{4\cdot7}{21^{2}}=\boxed{\frac{8}{63}}$

$\boxed{\frac{3}{4}}$. :D [/hide]

How did you go from $2\cdot\frac{4\cdot7}{21^{2}}=\frac8{63}$ to $\frac34$??

They're answers to the two different problems.

Should have known that  :oops:

We still need to know whether replacement is allowed or not.

[quote="mr. math"]splashD was right, but can anyone tell me why he did what he did?[/quote]
[hide="1"] I don't know how to explain what SplashD did, but I'll do it a different way. First we consider choosing green and then orange. The second case is choosing orange then green. Each of these cases are equivalent to $\frac{4}{21}\frac{7}{20}=\frac{1}{15}$. Since there are 2 cases, we can multiply by 2. $2\cdot\frac{1}{15}=\frac{2}{15}$. [/hide]
[hide="2"] SplashD used complimentary counting: $1-\text{the probability that you choose 2 of the same suit}$.

Another way is to say you can choose any first card, for a probability of 1. For the second card, there are $13\cdot3$ possible cards out of $51$, for a probability of $\frac{13\cdot3}{51}=\frac{13}{17}$. [/hide]

[quote="Totally Zealous"]In SplashD's solution, what did the big parentesis with 2 numbers in them mean?

The 21 on the top and 2 on the bottom in his 1st problem solution.[/quote]
That's a binomial. $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is the formula. That's how many ways you can choose $k$ objects from a group of $n$ objects.


probability
Alcumus
logarithms
AMC
AMC 8
MATHCOUNTS
AMC 10

I have intro to probability and counting but I can't afford any other books. So is working through past amc exams  enough (and other problems)? I'm trying to do the problems but I often encounter things that I have never heard of (such as logs or modulus) and I frequently do not understand the solution. My goal is to get 100 on amc 10 (heh, a pathetic goal but I don't think I'll ever make aime. I'm just doing the math for fun. I only got through 7 problems last year  :( ). Any free resources/tips? Thanks.

The Art of Problem Solving website is a good resource. You can try [url=http://www.artofproblemsolving.com/Alcumus/Landing.php]Alcumus[/url] for extra counting and probability practice. You can post questions on the forum if you don't understand something.

I think you should work through problems, then do research (online) on anything you didn't understand.

When you come across those concepts you don't understand, books aren't completely necessary. Resources are available all over the web - just Google whatever you don't know. For example, "logarithm" gives:

http://www.sosmath.com/algebra/logs/log1/log1.html

http://www.math.utah.edu/~pa/math/log.html

http://www.themathpage.com/aPreCalc/logarithms.htm

oakroadsystems.com/math/loglaws.htm

http://www.purplemath.com/modules/logs.htm

And that's all I saw on just a first glance.

http://www.agmath.com has a bunch of free AMC lessons.  I recently found them, and I'm working through them.

[hide="The biggest secret to learning anything"]
LEARN HOW TO USE GOOGLE
Seriously, you can get any information you want. [size=59]Get a few free books in the process![/size]
Check this out: http://www.googleguide.com/
Go through it, understand it. Also, get a few books from a library. Notice I say few, not one. Most of the textbooks you will find at the library are 90% junk and 10% gold. Go through concepts you don't understand in each one, or one you think explains it the best.
[/hide]

adithyaganesh how good is that site (the amc lessons on number theory, counting, and probability) for preparing for amc10/12, or is it geared mostly towards amc8 and mathcounts

It is more geared towards high middle school level, but it can be useful on the AMC 10.

Most of the AMC 10 is basically extending what you learned in middle school.  (There are a few exceptions.)

so which lesson would be most useful to start out with then? (aops vol 1 is arriving in a week or two)

Merry xmas (if thats your thing)

Not sure, I started out with Number Theory, because it was top on the list.

Merry Christmas to you too.

My old high school also has a website with tons of material, which you might want to look at: http://activities.tjhsst.edu/vmt/. "Lectures" and "Wiki" are the relevant sections.

I'm sure there's plenty of books in .djvu form on Rapidshare or something like that. Not legal - I don't think - but it's always an option. Alternatively, try buying used off Amazon.


I thought  that a data base of phishing mails would be useful and will protect us from get phished. Here is my contribution :)
-------------------------------------------------------
Dear one
 I am writing this letter with due respect and
heartful of tears since we have not known or met
ourselves previously.

I am asking for your assistance and I will be so glad if you
can allow and lead me to the right channel towards your
assistance to my situation now.

I will make my proposal well known if I am given the
opportunity. I would like to use this opportunity to introduce
myself to you. well, i am Michael Dikko 20years old young man
and I know that this proposal might be a surprise to you
but do consider it as an emergency.

In nutshell, My (late) father Mr Morris Dikko
was a very wealthy gold and cocoa merchant who based
in Accra and Abidjan respectively. But he
was killed along side with my mother during last year's
Rebel attack and all his properties was totally
destroyed.

However, after their death I managed to
escape with a very important document (DEPOSIT
CERTIFICATE  Two million Five hundred
thousand U.S Dollars deposited by my late father in the bank
which i am the next of kin.

Meanwhile,i am saddled with the problem of
securing a trust worthy foreign personality to help me
transfer the money over to his country and into his
possession pending my arrival to meet with him.

Furthermore,you can contact the bank for
confirmation and i will issue a letter of authorization
on your name,that will enable the bank to
deal with you on my behalf.

I am giving you this offers
as mentioned with every confidence on your acceptance
to assist me or take me as your child and manage the
money.  I am inclined to offer you 15% of the total sum
as a mode of compensation for your effort after the
 successful transferring of these fund to your
nominated account overseas

Conclusively,i wish you send me a reply
immediately as soon as you receive this
proposal,please write me back with this email
address () Until then,i remain with the best regards.

Michael.
---------------------------------------------------------

There are so many of those. It's funny to think about how many clueless people will just throw out their bank information to this guy.

(Is that considered phishing? Normally phishing is just someone posing as ebay or something.)

I got one once from the widow of an Iraqi general who had misappropriated a fortune. It seems that she's now dying of prostate cancer and is looking for help in making things right ...

[quote="nutz_for2.718281828"]What is phishing? Is is fishing for phi?[/quote]
I hope you weren't trying to be funny, seriously.
Read, and the answer will come to you. *Ohm*
[quote="xtrememath8"]There are so many of those. It's funny to think about how many clueless people will just throw out their bank information to this guy.

[b](Is that considered phishing? Normally phishing is just someone posing as ebay or something.)[/b][/quote]

it's not really phishing, it's more of a Nigerian e-mail scam.

anyway it could become phising (I didn't try to know for sure :) )


Okay, there is this place called Tooly Land, and there is only one rule in Tooly Land, you guys have to figure out that rule based on the clues I give. I will keep adding more and more clues until somebody figures it out.

Clue #1;

Okay, because of the rule, there can be robbers in Tooly Land, but there aren't any murderers.

Note: This MAY be the answer, so dont look if you want to think about it more.

[hide]Double letters?[/hide]

Okay, that was it. If you still want to play, then don't look at Jongao's post.

Here are the clues;

1. There are robbers, but not murderers.
2. There are schools, but there are isn't any math.

FYI this is much more commonly referred to as things that can go through the Green Glass Door.

It's more fun playing with the red plastic window. :P

I think the point is that Green Glass Door can fit with this puzzle very well...

He means to say that this game is commonly called [b]"The Land of the Gr[u]ee[/u]n Gla[u]ss[/u] D[u]oo[/u]rs"[/b], which is a more telling (and honestly, less annoying-sounding) name.

A very obvious hint that a friend came up with: [i]You can take a poop but you can't take a s***.[/i]

Alright, this post is basically dead. It's kind of stupid anyway.  :|


integration
logarithms
limit
calculus
calculus computations

$\int_{0}^{+\infty }\frac{xlnx}{(1+x^{2})^{2}}$
 this intergration is convege or disconve ?

The only troublesome points are $0$ and $\infty$:

$\lim_{x \to 0^{+}}\frac{x \ln x}{\left( 1+x^{2}\right)^{2}}= \lim_{x \to 0^{+}}x \ln x =-\lim_{x \to 0^{+}}\frac{\ln \left( \frac1{x}\right)}{\frac1{x}}= 0$;

$\lim_{x \to \infty}\frac{\frac{x \ln x}{\left( 1+x^{2}\right)^{2}}}{\frac1{x^{2}}}= 0$.

Hence, $ \int_{0}^\infty \frac{x \ln x}{\left( 1+x^{2}\right)^{2}}\, dx$ converges.

Substitute $ x = \frac1{y}, \, dx =-\frac1{y^{2}}\, dy$:
\begin{eqnarray*}\int_{0}^\infty \frac{x \ln x}{\left( 1+x^{2}\right)^{2}}\, dx &=& \int_\infty^{0}\frac{\frac1{y}\ln \left( \frac1{y}\right)}{\frac{\left( y^{2}+1 \right)^{2}}{y^{4}}}\cdot \frac{-1}{y^{2}}\, dy \\ \ &=& \int_\infty^{0}\frac{y \ln y}{\left( y^{2}+1 \right)^{2}}\, dy .\end{eqnarray*}
Since we have proved earlier that $ \int_{0}^\infty \frac{x \ln x}{\left( 1+x^{2}\right)^{2}}\, dx$ converges, it is safe to conclude that the integral is actually $ 0$.


floor function
number theory
relatively prime

Let $ a$ and $ b$ be integers and $ n$ a positive integer.  Prove that
\[ \frac{b^{n-1}a(a+b)(a+2b) \cdots (a+bn-b)}{n!}\]
is an integer.

not too sure about my solution, any comments?
[hide]Let $ p$ be a prime dividing $ n$ and let $ k$ be the number of factors of $ p$ in $ n!.$
If we show that the numerator is divisible by $ p^{k}$ for all $ p,$ we are done.
Note that
$ k=\lfloor\frac{n}{p}\rfloor+\lfloor\frac{n}{p^{2}}\rfloor+...+\lfloor\frac{n}{p^{j}}\rfloor<\frac{n}{2}+\frac{n}{2^{2}}+...$
$ =n(\frac{n}{2}+\frac{n}{2^{2}}+...)=n$ where $ j$ is the maximum integer such that $ p^{j}|n.$
So $ k<n\implies k\le n-1.$

Case 1: $ p$ divides $ b.$ Then we are done, because $ p^{k}|b^{k}\implies p^{k}|b^{n-1}$ since $ k\le n-1.$

Case 2: $ p$ does not divide $ b.$
Let $ m_{i}$ be the maximum number such that $ m_{i}p^{i}<n.$ Then $ m_{i}=\lfloor\frac{n}{p^{i}}\rfloor$
Since $ p$ is relatively prime to $ b,$ there exist an integer $ x_{i}$ such that $ a+x_{i}b=0\mod p^{i}$ and $ 0<x_{i}<p^{i}.$
It follows that $ a+(x_{i}+cp^{i})b=0\mod p^{i},$ for $ c$ an integer.
Since $ x_{i}+(m_{i}-1)p^{i}<p^{i}+(m_{i}-1)p^{i}=m_{i}p^{i}<n,$ all $ a+(x_{i}+cp^{i})b$ where $ 0\le c\le m_{i}-1$ are factors of the numerator, and thus there are $ m_{i}$ factors of $ p^{i}$ in the numerator.
Summing the factors of $ p^{i}$ from $ 1$ to $ j,$ we get that there are $ m_{1}+m_{2}+...+m_{j}=\lfloor\frac{n}{p}\rfloor+\lfloor\frac{n}{p^{2}}\rfloor+...+\lfloor\frac{n}{p^{j}}=k$ factors of $ p$ in the numerator, so we are done.

[/hide]

sum(i=1,infiny)E(n/p^i)=vp(n!)
Are you sure ???  vp(12!)=9 end sum...=10[/code]


I am needing some help with some college algebra if anyone can assist me please let me know!!! :lol:

Post your problems in correct subforum and you will have a large chance to obtain solutions.


geometry
rectangle
perimeter
number theory
relatively prime

the rectangle shown is split into 9 squares. If the side lengths of the rectangle are relatively prime, what is the perimeter of the rectangle?

Do you just express everything in terms of the 2 smallest side lengths?

yeah. just assume that the two smallest square lengths have sides a and b, and then you can find the dimensions of the rectangle.

wow that solution probably took a lot of time.

thanks!!!  :lol:


Find all polynomials $ p(x)$ satisfying the equation: $ (x\minus{}16)p(2x)\equal{}16(x\minus{}1)p(x)$ for all $ x$.

[hide]$ (x \minus{} 16)P(2x) \equal{} 16(x \minus{} 1)P(x)$

$ x \equal{} 1$ yields:
$ ( \minus{} 15)P(2) \equal{} 16(0)P(1)$
$ P(2) \equal{} 0$

$ x \equal{} 2$ yields:
$ ( \minus{} 14)P(4) \equal{} 16(1)P(2)$
$ P(4) \equal{} 0$

$ x \equal{} 4$ yields:
$ ( \minus{} 12)P(8) \equal{} 16(3)P(4)$
$ P(8) \equal{} 0$

$ x \equal{} 8$ yields:
$ ( \minus{} 8)P(16) \equal{} 16(15)P(8)$
$ P(16) \equal{} 0$

Since $ x \equal{} 2,4,8,16$ are roots of $ P(x)$, we have $ P(x) \equal{} (x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16)Q(x)$ for some polynomial $ Q(x)$.

Plugging in this into the equation yields:
$ (x \minus{} 16)(2x \minus{} 2)(2x \minus{} 4)(2x \minus{} 8)(2x \minus{} 16) Q(2x) \equal{}$ $ 16(x \minus{} 1)(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16) Q(x)$

$ 16(x \minus{} 1)(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16) Q(2x) \equal{}$ $ 16(x \minus{} 1)(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16) Q(x)$

$ Q(2x) \equal{} Q(x)$

If $ r$ is a root of $ Q(x)$, then $ 2r, 4r, \ldots , 2^nr , \ldots$ are also roots of $ Q(x)$.
If $ r \neq 0$, then $ Q(x)$ has an infinite number of distinct roots, a contradiction.

Thus, $ r \equal{} 0$ is the only possible root of $ Q(x)$. Thus, $ Q(x) \equal{} Cx^n$ where $ n \in \mathbb{N}_0$.

Therefore:
$ Q(2x) \equal{} Q(x)$
$ C 2^n x^n \equal{} C x^n$
$ (2^n \minus{} 1)(Cx^n) \equal{} 0$
$ n \equal{} 0$ OR $ C \equal{} 0$

Thus, $ Q(x) \equal{} C$, and $ P(x) \equal{} C(x \minus{} 2)(x \minus{} 4)(x \minus{} 8)(x \minus{} 16)$ for some constant $ C \in \mathbb{R}$. [/hide]

My solution from WOOT:
Let the leading term of $p$ be $ax^n$ with $a\ne0$ (otherwise $p\equiv0$ which is included in the solution set). We obtain $xa(2x)^n=16ax^{n+1}$, so $n=4$.
$x=16\Rightarrow p(16)=0$
$x=1\Rightarrow p(2)=0$
$x=8\Rightarrow p(8)=0$
$x=2\Rightarrow p(4)=0$
By the FToA, we have $\boxed{p(x)=a(x-2)(x-4)(x-8)(x-16)},a\in\mathbb R$, which works.


AMC
AIME
ARML
USA(J)MO
USAMO
HCSSiM
summer program

Hi,

since most people here have had the experience of going all across the country doing math and going to camps, I was just wondering which camps are very useful.

I mean as in, when you come back from the camp, you really feel like you have learned a lot and know what to work on. This is all in terms of things like AMCs, and stuff like ARML.

if it helps
last year i got a 140 on amc 10, 2 on aime
(really bad amc score, barely qualified) and 7 on the AIME w/ 3 stupid mistakes so i expect about a 7-10 on AIME.
I got a 3 at the individual at ARML like 2 days ago.

I wanna shoot for 11-13 next year on the AIME and do decently at ARML.

Also, I was wondering if WOOT would be any good for me now?

I aim to do well on USAMO next year. (hoping i qualify  :oops: )

Are there any camps or things that you would recommend to me?

I live in the midwest, Louisville, KY, to be exact. I am not really concerned w/ the travel, just the math i suppose..The duration doesn't really matter either. It's better if it's later in July and early August, but I am open to suggestions.

Thanks for all replies and suggestions, they are all greatly appreciated.

havent been but I heard awesome math was good. You definitely missed all the deadlines by now though  :|

I went to [url=http://www.hcssim.org]HCSSiM[/url] last summer; I highly recommend it.  It is about exploring interesting math for its own sake, rather than contest preparation, though.  There is no exact deadline (we were passing out flyers at ARML Friday night), but if you want to go you should apply as soon as possible.

Here are some other nationally known math summer programs:
[list]
[*] [url=http://www.mathcamp.org]Mathcamp[/url]
[*] [url=http://www.math.ohio-state.edu/ross/]the Ross program[/url]
[*] [url=http://www.promys.org]PROMYS[/url]
[*] [url=http://www.awesomemath.org]Awesomemath[/url][/list]
All their deadlines have passed, though.  The deadlines for Ross and PROMYS were recent though; I don't know how lenient they are with late applications.

WOOT would be a fit for you in 2008-09.

You should also be sure to spend plenty of time on Olympiad problems during the year, rather than solely focusing on the AIME.


trigonometry
parameterization
parametric equation

I'm not sure if this is in the right place..
A bendable closet door is shaped so that it pivots at one end and the other end is attached to a straight line. Following is a rough diagram:[asy]size(300);for(int i=0; i < 10; ++i){draw((0,0)--(5*cos(pi/20*i),5*sin(pi/20*i))--(10*cos(pi/20*i),0));}for(int i=0; i < 50; ++i){draw((15,0)--(15+5*cos(pi/100*i),5*sin(pi/100*i))--(15+10*cos(pi/100*i),0));}[/asy]How do you calculate the equation of the outside boundary of the graph? (shaped like a cos graph) Thanks  :)

[quote="minsoens"]I'm not sure if this is in the right place..
A bendable closet door is shaped so that it pivots at one end and the other end is attached to a straight line. Following is a rough diagram:
How do you calculate the equation of the outside boundary of the graph? (shaped like a cos graph) Thanks  :)[/quote]

$ 1).$ Mathematical point of view
On a mathematical point of view, you are looking for the envelope of the right part of the bendable closet, whose equation is $ y \equal{}\minus{}x\tan(\theta)\plus{}2L\sin(\theta))$ where $ L$ is the length of one door and $ \theta$ the angle of the first door with the straight line continning the other end.

A simple way to compute the equation of the envelope of a group of curves is to compute the intersection of two of them and check what is the limit of this intersection when the two curves are nearly the same.

Intersection between positions for $ \theta_{1}$ and $ \theta_{2}$ is immediate :
$ x\tan(\theta_{1})\minus{}2L\sin(\theta_{1}))$ $ x\tan(\theta_{2})\minus{}2L\sin(\theta_{2}))$ and so $ x \equal{} 2L\frac{\sin(\theta_{1})\minus{}\sin(\theta_{2})}{\tan(\theta_{1})\minus{}\tan(\theta_{2})}$ and $ y \equal{}\minus{}x\tan(\theta_{1})\plus{}2L\sin(\theta_{1}))$.

When $ \theta_{2}\rightarrow\theta_{1}$, we have :
$ x \equal{} 2L\cos^{3}(\theta_{1})$
$ y \equal{} 2L\sin^{3}(\theta_{1}))$
And this is the parametric equation of the curve you are looking for.

You can also write $ y \equal{} ((2L)^{\frac{2}{3}}\minus{}x^{\frac{2}{3}})^{\frac{3}{2}}$

or also $ (\frac{x}{2L})^{\frac{2}{3}}\plus{}(\frac{y}{2L})^{\frac{2}{3}}\equal{} 1$

As you prefer $ \ldots$

$ 2)$ Correction for physical point of view.
If you draw the previous curve, you will not find your graph for the following reason : you drawn the envelope of segments and not of lines.
So as soon as the envelope point is left to the top of the closet, the drawing changes.
When does this occur ? : when $ 2L\cos^{3}(\theta_{1}) < Lcos(\theta_{1})$, so when $ \cos^{2}(\theta_{1}) <\frac{1}{2}$, so $ \theta_{1}>\frac{\pi}{4}$

For $ \theta_{1}>\frac{\pi}{4}$, the curve you are looking for is no longer the envelope, but the curve described by the top of the closet, that's to say a circle.

So, on a physical point of view, the equation is :

$ 2.1).$ Parametric equation :
For $ \theta\in[0,\frac{\pi}{4}]$ :
$ x \equal{} 2L\cos^{3}(\theta)$
$ y \equal{} 2L\sin^{3}(\theta))$

For $ \theta\in[\frac{\pi}{4},\frac{\pi}{2}]$ :
$ x \equal{} L\cos(\theta)$
$ y \equal{} L\sin(\theta))$

$ 2.2).$ Classical equation :
For $ x\in[0,\frac{L}{\sqrt{2}}]$ : $ y \equal{}\sqrt{L^{2}\minus{}x^{2}}$
For $ x\in[\frac{L}{\sqrt{2}},2L]$ : $ y \equal{} ((2L)^{\frac{2}{3}}\minus{}x^{\frac{2}{3}})^{\frac{3}{2}}$

[quote="pco"]A simple way to compute the equation of the envelope of a group of curves is to compute the intersection of two of them and check what is the limit of this intersection when the two curves are nearly the same.[/quote]I see. Thank you for your explanation  :lol:


vector
geometry
geometric transformation
rotation
trigonometry

[quote="Shyamkumar"]the torque of any of these forces about any axis perpendicular to the plane of the forces (ie parallel to the z-axis) is equal[/quote]

The torque of a force about an axis depends on the distance between the force line of action and the axis. So if we consider different axes with different distances to that line of action the torque for the same force will differ. Perhaps you are trying to say something else?

perhaps you're talking about a a pair of forces...

hey..
i am sorry about the typo....i meant 'net torque' instead of ' torque of each of the forces'....now,how's it done??  :maybe:
Shyam

$ \Sigma \ \ (P_i -O^') \times \vec F_i = \Sigma \ \ (P_i -O) \times \vec F_i+ \Sigma \ \ (O -O^') \times \vec F_i = \\ =\Sigma \ \ (P_i -O) \times \vec F_i + (O-O^') \times \Sigma\vec F_i \\$
 But $ \ \ \Sigma \vec F_i =0 \ \ \textrm{so} \ \  \Sigma \ \ (P_i -O^') \times \vec F_i =  \Sigma \ \ (P_i -O) \times \vec F_i$

[quote="shyamkumar"]Suppose that the net force on a body is 0.Also suppose that all forces (vectors) lie in the same plane (say the x-y plane).Then can u prove that the torque of any of these forces about any axis perpendicular to the plane of the forces (ie parallel to the z-axis) is equal (ie independent of the axis chosen)?[/quote]

Zero net force on a body implies translational equilibrium. However, translational equilibrium does not imply rotational equilbrium.

Consider a broomstick lying on a frictionless table top. (There is zero net vertical force on the broomstick, because the normal force of the table top balances the weight of the broomstick.)  Now imagine you attach two frictionless pullies, one to each side of the table top. Next run inextensible cords over each pully, and attach the ends of the cords which are lying on the table top to the broomstick at arbitrary points on the broomstick. Finally, hang equal weights from the free ends of the cords.  In such a case you would expect no translation, i.e. the center of mass of the broomstick will not move from its location; however, it should rotate, rather abruptly, until the ends of the broomstick come into alignment with the pullies.

[quote="genericme"]That's not the question, though...[/quote]

The answer to the question, is "no, you cannot." The experiment I just described is a counter-example.

[quote="shreyas_patankar"]Dr. No, about which two axes in your experiment is the net torque unequal?
I feel that the explanation given by Immanuel Bonfilis is satisfactory, if $ O$ and $ O'$ are points where two axes intersect the plane and forces $ F_i$ are applied at points $ P_i$[/quote]

The net torque would be non-zero for any arbitrary axes parallel to z-axis provided that the moment arms for each force are unequal. I would expect the broomstick to rotate only until $ r_1\sin \theta_1 \equal{} r_2\sin \theta_2$ at which point the rotation would stop because the broomstick would come into rotational equilibrium.

The problem is two show that net torque (zero or not) is equal about any axis that satisfies the given conditions.

[quote="Dr. No"]In such a case you would expect no translation, i.e. the center of mass of the broomstick will not move from its location[/quote]

For this to happen, you explicitly require that $ \sum \vec{F_i}\equal{}\vec{0}$, but the net torque may be non-zero (they can add up).

yes i could just roattae the broom about it's Centre of mass

[quote="shreyas_patankar"]The problem is two show that net torque (zero or not) is equal about any axis that satisfies the given conditions.[/quote]

No, the problem was can one prove that there is zero net torque about any axis given that there is zero net force on the would be rotating body? The answer is that zero net force does not in general imply zero net torque -- as my example shows -- and therefore the answer to the original question is no.

[quote="Dr. No"]In such a case you would expect no translation, i.e. the center of mass of the broomstick will not move from its location[/quote]

[quote="shreyas_patankar"]For this to happen, you explicitly require that $ \sum \vec{F_i} \equal{} \vec{0}$, but the net torque may be non-zero (they can add up).[/quote]
This is exactly what I have been saying.

[quote="shreyas_patankar"][quote="shyamkumar"]Suppose that the net force on a body is 0.Also suppose that all forces (vectors) lie in the same plane (say the x-y plane).Then can u prove that the torque of any of these forces about any axis perpendicular to the plane of the forces (ie parallel to the z-axis) [b]is equal [/b](ie independent of the axis chosen)?[/quote]
No mention of Zero.[/quote]

That is even easier to disprove than the way I was approaching the problem in the first place.

To begin with we have from the problem statement: $ \mathbf{F}_1 \equal{} \minus{}\mathbf{F}_2 \equal{}\mathbf{F}$.

Now suppose there exists an axis such that the torques produced by $ \mathbf{F}_1$ and $ \mathbf{F}_2$ about it are equal.

It follows that with respect to said axis $ \mathbf{r}_1\times\mathbf{F}_1 \equal{} \mathbf{r}_2\times\mathbf{F}_2$.

If we then substitute $ \mathbf{F}$ into this equation, we obtain

$ \mathbf{r}_1\times\mathbf{F}\equal{}\minus{}\mathbf{r}_2\times\mathbf{F}$

$ \Rightarrow \left(\mathbf{r}_1 \plus{} \mathbf{r}_2\right)\times\mathbf{F} \equal{} 0$

This can only be the case if the direction of $ \mathbf{F}$ is parallel to that of the vector sum $ \mathbf{r}_1 \plus{} \mathbf{r}_2$, or if $ \mathbf{r}_1 \equal{} \minus{}\mathbf{r}_2$.  Neither of these conditions must be the case in general.


Let
\[ S_n \equal{} \sum_{k \equal{} 1}^n \binom{n}{k}k^n.\]
1. Prove $ n^3|S_n$ if $ n$ is odd.
2. Prove $ \mathrm{ord}_2\left(S_n\right) \equal{} \mathrm{ord}_2\left(n!\right)$ for any $ n$, where $ \mathrm{ord}_2\left(n\right)$ is defined as the integer $ k$ such that $ 2^k|n$ and $ 2^{k \plus{} 1}\not |n$.


limit
LaTeX
real analysis
real analysis unsolved

show that if $ sum(u_n)$ converge then $ sum((u_n)^(n/n\plus{}1))$ converge

We can use Cauchy's criterion.

Let $ S_m = u_1 + ... + u_m$, then consider $ \frac {u_n}{n + 1} + ... + \frac {u_m}{m + 1}$ because
$ \frac {nu_n}{n + 1} + ... + \frac {(m + 1)u_m}{m + 1} = S_m - S_n - \frac {u_n}{n + 1} - ... - \frac {u_m}{m + 1}$

$ |\frac {u_n}{n + 1} + ... + \frac {u_m}{m + 1}|$
$ = |(S_n - S_{n - 1})\frac {1}{n + 1} + ... + (S_m - S_{m - 1})\frac {1}{m + 1}|$
$ \leq \frac {|S_{n - 1}|}{n + 1} + \frac {|S_{m}|}{m + 1} + |S_n| \frac {1}{(n + 1)(n + 2)} + ... + |S_{m - 1}|\frac {1}{m(m + 1)}$
which goes to 0 if n,m are large enough because $ S_n$ are bounded and $ \sum \frac {1}{n(n + 1)}$ is a convergent series.

So $ |\frac {nu_n}{n + 1} + ... + \frac {(m + 1)u_m}{m + 1}| \leq |S_m - S_n| + |\frac {u_n}{n + 1} + ... + \frac {u_m}{m + 1}|$ goes to 0 if n,m are large enough because $ S_n - S_m$ goes to 0 by Cauchy criterion of convergent series $ \sum u_n$

$ \lim_{n \to \infty} \frac {\frac {n u_n}{n \plus{} 1}}{u_n} \equal{} \lim_{n \to \infty} \frac {n}{n \plus{} 1} \equal{} 1$

and so both series have the same nature. Also by Abel's test, as sum u_n converges and n/(n+1) is a monotonic and limited sequence, then sum n u_n/(n+1) also converges.

I don't think either of these answers the original question; I think that was meant to be $ \sum u_n^{n/(n + 1)}$, with the assumption that $ u_n$ is positive.

(To bellahsayn: Use {} rather than () to enclose things like exponents in $ \text{\LaTeX}$.

[quote="jmerry"]I don't think either of these answers the original question; I think that was meant to be $ \sum u_n^{n/(n \plus{} 1)}$, with the assumption that $ u_n$ is positive.
[/quote]
I think so too. It's not a bad sum really! Let $ v_n\equal{}u_n^{n/(n\plus{}1)}$ and $ A\equal{}\{n \in \mathbb{N}: \ v_n \leq 2u_n\}.$
Then over $ \mathbb{N}\minus{}A$ we have $ v_n < \frac{1}{2^n}.$ Thus $ \sum v_n \leq 2\sum u_n \plus{} \sum \frac{1}{2^n} < \infty.$ Q.E.D.

[quote="Carcul"][quote="Jmerry"]I don't think either of these answers the original question;[/quote]

That should be the original poster to tell us.[/quote]

But this is what he was trying to tell us!  If you quote his text this is what you will see.

[code]
[quote="bellahsayn"]show that if $sum(u_n)$ converge then $sum((u_n)^(n/n + 1))$ converge[/quote]
[/code]

I agree that it was misleading.

Exactly. You wouldn't have had any problem with it if the dollar signs had been left out- the only problem was a $ \text{\LaTeX}$ mistake.

Missing the assumption that $ u_n$ is positive, on the other hand, is a mathematical mistake.


For a set of positive $\gamma_k=0 , 1, 2...,$ we have

${ \gamma_k}^2-\gamma_{k-1}\gamma_{k+1} \geq 0$ for $k=1, 2...,$

Is this not equivalent to $n^2 > (n+1)(n-1)$?

I think I'm missing the point... :(


A square of area 6 square units is split into rectangles of areas 3, 2 and 1 square units, as shown. What is the ratio of the length of the longer side of the gray rectangle to the length of the shorter side of the gray rectangle? Express your answer as a common fraction.
[asy]fill((2/3,1)--(2/3,.5)--(1,.5)--(1,1)--cycle,gray(.7));
draw(unitsquare);
draw((0,.5)--(1,.5));
draw((2/3,.5)--(2/3,1));
label("1",(5/6,3/4));
label("3",(.5,.25));
label("2",(1/3,.75));[/asy]

Since the square has a total area of $ 6$ square units, a $ 3$ unit portion, such as the bottom rectangle in the diagram, would resemble half the square. Also, since a square's area is $ s^2$, a side of this square is $ \sqrt {6}$ units long. Half of this is $ \frac {\sqrt {6}}{2}$. Thus, the longer side of the gray rectangle is $ \frac {\sqrt {6}}{2}$ units long. Also, since the two upper rectangles have the same length, the ratio of their widths most equal the ratio of their areas. Thus, the width of the $ 2$-square unit rectangle would be $ \frac {2}{3}$ the length of the square's side, and the width of the $ 1$-square unit rectangle would be the leftover $ \frac {1}{3}$ of the square's side. Thus, the shorter side of they gray triangle is $ \frac {\sqrt {6}}{3}$ units long. The ratio of $ \frac {\sqrt {6}}{2}$ to $ \frac {\sqrt {6}}{3}$ is $ \fbox{3/2}$.


Probability - Think About It Part 3

We simplify a probability problem involving walking along edges of a cube by thinking about the problem in a different way.

Can you someone please post the solution to the problem extension?
Thanks,
AlphaBetaTheta

I didn't do much thinking and I got 5^3/13^3

if we extend the diagram, we find that there are 13 possible paths all together. only 5 of them end in a 0, i.e. are successful, so the probability should be 5/13, right? what do you think guys?

I will work hard!
Hello!
everyone I had just arrived

[quote="math-rules"][hide="My Solution"]
If we extend the tree diagram that was an the video, we find out that after 6 minutes, we are either at 0 ar 2. The total number of ways is 13, and 5 of them end at 0, so the probabiliity is $\boxed{\dfrac{5}{13}}$.
[/hide][/quote]

The issue with this solution is that not all of the paths have the same probability of happening. For example, when you are on a vertex one step away from the beginning, the probability that you arrive at a vertex two steps away is $\frac{2}{3}$ while the probability you arrive back at the beginning is only $\frac{1}{3}$.

I just did the problem out again using paths and probabilities and got $\frac{61}{243}$. I'll try to make my work neater and post a picture later.

Edit: Here's my work. I tried to make it neat, but it was a little difficult, not to mention I had to retrace everything with pen because pencil didn't show up well enough.
The full photo is [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0298.JPG]here[/url] and zoomed in pictures can be found [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0299.JPG]here[/url], [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0300.JPG]here[/url], [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0301.JPG]here[/url], [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0302.JPG]here[/url], [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0303.JPG]here[/url], and [url=http://dl.dropbox.com/u/24673237/Probability%20-%20Think%20About%20It%20Part%203%20Solution/IMG_0304.JPG]here[/url].


calculus
integration
trigonometry
search
real analysis
real analysis unsolved

Calculate:
$ \int_{0}^{2\pi}\sin^nxdx\equal{}?$

$ \int_{0}^{2\pi}\cos^nxdx\equal{}?$

if n is odd then 0.
if n is even then $ 2 B(\frac{n\plus{}1}{2},\frac12).$

A link to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=89630878&t=246576]reduction formulas[/url]. Searching for that will get you some other topics as well.


function
inequalities
algebra
functional equation

Fie $f:[0,\infty ) \to [0,\infty )$ o functie continua cu proprietatea $f(x + f(x)) = x,\forall x \ge 0.$
Sa se arate ca f este monotona.

Let $a=f(0)$. Then $f(a)=f(0+f(0))=0$. Then $0=f(a)=f(a+f(a))=a$. Thus $f(0)=0$.
Consider the function $g(x)=x+f(x)$. We know that $g(x)$ is continuous, $g(0)=0$ and $g(x)\geq x$. Moreover, it is clear that $g$ is injective! It implies that $g$ is monotonuous and increases from 0 to $+\infty$. It means $g$ has inverse continuous increasisng function $g^{-1}$ s.t. $g^{-1}(0)=0$. Since $x=f(g(x))$ we conclude $g^{-1}(x)=f(g(g^{-1}(x)))=f(x)$. So $f$ is an increasing function.

[quote="Myth"][quote="fleeting_guest"]Is there an explicit example of such $f$?[/quote]
Is it a joke? $f(x)=kx$, where $k+k^2=1$.[/quote]

Anything nonlinear or nontrivial?

[quote="Myth"]Do you have some ideas?[/quote]

I hope you are not being sarcastic.

I have a boring standard proof that continuous $f(x)$ is linear, and a more interesting but false argument for constructing nonlinear solutions.

The proof starts from $g(x)$ and that $f(x)$ is a two-sided inverse of $g$.
From this,  $f(f(x)) = x - f(x)$.   (Note that $g(g(x)) = x + g(x)$ does not require continuity or any inverses, it is just equivalent to the functional equation of $f$.)
Iteration produces $f^n(x) = a_n f(x) + a_{n-1}x$
where $a_n$ is the negative continuation of the Fibonacci sequence $... 5,-3,2,-1,0,1,1,2,3,...$, and $a_0 = 0, a_{-1} = 1$.  The inequalities $f^n(x) \geq 0$ produce upper and lower bounds on $f$, converging to $f(x) = kx$  with $k^2 + k = 1$.

The reason for my questions earlier in the thread, however, was the interesting false idea, to let $f$ be a map that sends $F_s$ to $F_{s-1}$ where $F_s$ is a continuous solution of the Fibonacci recurrence, $F(s+2) = F(s+1) + F(s)$.   The iterates $g^n(x)$ satisfy this recurrence for any $g$, so it is natural to think of $g$ as $F_s \to F_{s+1}$.

This problem is perhaps indicating that there is no analogue for the real numbers, of the partition of integers into Fibonacci sequences.

[quote="fleeting_guest"][quote="Myth"]Do you have some ideas?[/quote]I hope you are not being sarcastic.
[/quote] Of course, not. But I really thought you have such an example.

[quote] Of course, not. But I really thought you have such an example.[/quote]

I am still puzzled by the continuous Fibonacci idea.
e.g., for which $a$ there exists a nonlinear solution of $g(g(x)) = g(x) + x$  on $[a,\infty)$ or $(a,\infty)$.   Continuous or discontinuous version of the problem are both interesting.


[quote="dondigo"]$x,y,z,a$ are non-zero real numbers.
Prove that if:
$x+y+z=a$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}$ then
one of numbers $x,y,z$ is equal to $a$.
Here is a part of my solution:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$
$\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=\frac{1}{x+y+z}$
$(x+y+z)(yz+xz+xy)=xyz$
$xyz+x^2z+x^2y+y^2z+xyz+xy^2+yz^2+xz^2+xyz=xyz$
$xyz+x^2z+x^2y+y^2z+xyz+xy^2+yz^2+xz^2=0$
$x^2(y+z)+x(y^2+2yz+z^2)+yz(y+z)=0$
$x^2(y+z)+x(y+z)^2+yz(y+z)=0$
$(y+z)(x^2+xy+xz+yz)=0$
So $y+z=0$, then $x=a$ or $(x^2+xy+xz+yz)=0$
I can't prove, that the second case is not possible[/quote]
$(y+z)(x^2+xy+xz+yz)=0$
$\Rightarrow (y+z)(z+x)(x+y)=0$  :D

[quote="dondigo"]So $y+z=0$, then $x=a$ or $(x^2+xy+xz+yz)=0$
I can't prove, that the second case is not possible[/quote]

[hide]$(x^2+xy+xz+yz)=0 \Leftrightarrow$

$x(x+z)+y(x+z)=0\Leftrightarrow$

$(x+z)(x+y)=0\Leftrightarrow$

$(a-y)(x+y)=0 \rightarrow y=a$

or

$(x+z)(a-z)=0 \rightarrow z=a$[/hide]


pm if you think its right and copy paste the qestion.... if i did 25 posts then i will rate you

6.
Find the smallest four-digit number that is divisible by each of the four smallest prime numbers.

remember i edit this posts

note to self:mz94 got 5 and 6 right

almighty bob got it right
pi guy got it wrong the answer was 80 not 50

*why* aren't you posting all the problems? It's much better if people can see your previous problems too [and see your answers].

he did, and then he took them away as people answered them

i quit.... but ill make another game later


What score do you need at the MC chapter competition to advance to state?  My first practice test I got a 41, missed 3 on the Sprint and 1 on the Target.  Its pretty bad, but will it be enough?  My goal this year is to atleast make state and perform competitively.

It depends on how strong your chapter is. For some chapters a 45 will not make it, and in other chapters, a 20 can make it to states.

[quote="the cliu"]It depends on the chapter you are in. Some chapters are a lot worse than others. One chapter in particular that I heard about needed a 17 to make states. My chapter needed like a 40 last year to qualify for states. Some people in chapters in beast states like Texas and California don't even make states with a perfect score.[/quote]

Well, my chapter had a bunch of epic schools, and its in Texas.  I hope I make it!  And how many individuals from each chapter move on?

Generally, 2 or 3 teams of 4 move on from chapter, and 4 individuals, for a total of 12-16. Of course, this depends on the size of your chapter.

The scores needed to advance to state vary from year to year and from chapter to chapter. For example, in Bergen County, New Jersey last year, the scores were 46, 45, 44, 43 (All from the same "school", which I was with, although you won't find this in any results because of a registration error), while in other chapters that I've heard of, a 23 was enough to move to state, and a 26 enough for national.

However, there are also situations ont he other extreme, where a 45 might not make it, or even in very rare cases a 46 might not make it because of countdown.

In Texas, where you say you are situated, I would guess that a 42/43 would be enough to move to state, but it depends on how good you are at countdown, and if countdown is official in Texas.

[quote="the cliu"]It depends on the chapter you are in. Some chapters are a lot worse than others. One chapter in particular that I heard about needed a 17 to make states. My chapter needed like a 40 last year to qualify for states. Some people in chapters in beast states like Texas and California don't even make states with a perfect score.[/quote]
:cough cough:  Yeah, that's my chapter.......still don't get why lake county's so pathetic.  Like really......what makes our chapter so much worse than the beastly Indiana school systems :cough :cough:

But for 90% of chapters 41/46 is good enough.  Even in states like Indiana and Washington it's probably good.......but if you live in California or Texas you need 46.......

[quote="Maybach"]but if you live in California or Texas you need 46.......[/quote]

The score required in the most competitive CA chapter was 45 :wink:

Heh, yeah it all depends on where you live.

Our chapter sent 40-50 to states. (But we should be able to send more, since our state is really bad [19 made cd in 2009] and 12-14 of the 16 countdowners @ state are from our chapter)

[quote="the cliu"]I think about 15 or so from each chapter move on. Texas is a beast state.[/quote]

that changes too, our chapter sends 10 (more if tie)

EDIT: more if they're on a team


algebra
function
domain
algebra unsolved

Denote $D=\{(x,y,z)\in\mathbb{R}^{3}|x>0,y>0,z>0;xyz=1\}$. Let function $f: D\to\mathbb{R}$ defined by $f(x,y,z)=\frac{xy+yz+zx}{x+y+z}$. Find set $\{f(x,y,z)|(x,y,z)\in D\}$.

$\{f(x,y,z)|(x,y,z)\in D\}\subseteq \mathbb{R}_{+}$ is clear
and $f(x,x,{1\over x^{2}})={x^{2}+{2\over x}\over 2x+{1\over x^{2}}}={x^{4}+2x\over 2x^{3}+1}$$\mathop{\longrightarrow}_{x\to 0}0$ and $\mathop{\longrightarrow}_{x\to \infty}\infty$.
So $\{f(x,y,z)|(x,y,z)\in D\}= \mathbb{R}_{+}$


how can i find "mathematical olympiads treasure" with pdf

1) It should be moved
2) I don't know
3) Look at http://www.mathlinks.ro/index.php?f=319 and http://www.mathlinks.ro/index.php?f=121


limit
trigonometry
calculus
calculus computations

Evaluate $\lim_{x\searrow 0}{\frac{xe^\frac{1}{x}}{\tan^{2}(x)}}$What's wrong with this limit? I don't find at this hour any way to get it :|

The $\tan^2 x$ is comparable to $x^2$. The $e^{\frac1x}$ is very small- it goes to zero faster than any power of $x$. The overall limit is zero.

sorry the arrow was $\searrow$ :) But my book says that it's 1. I reduced that thing too but $\lim_{x\searrow 0}{\frac{{e}^{\frac{1}{x}}}{x}}=?$

[quote="jmerry"]The $e^{\frac1x}$ is very small- it goes to zero faster than any power of $x$.[/quote]

 :?

Ah- you just reversed the meaning. Now that $e^{\frac1x}$ goes to $\infty$ faster than any power of $x$, and the limit is $\infty$.


1) Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
\[\frac a{\sqrt{a+bc}}+\frac b{\sqrt{b+ca}}+\frac c{\sqrt{c+ab}}\leq\frac32.\]
2) Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
\[\frac 1{\sqrt{a+bc}}+\frac 1{\sqrt{b+ca}}+\frac 1{\sqrt{c+ab}}\geq\frac92.\]
3) Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Find the minimum of
\[f=\frac 1{\sqrt{a+bc}}+\frac 1{\sqrt{b+ca}}+\frac 1{\sqrt{c+ab}}.\]

Problem 1) has two nice solutions while problems 2) and 3) are unsolved.

For the first one .
$2\frac{a}{\sqrt{a+ab}}=2\frac{a}{\sqrt{(a+c)(a+b)}}=2\sqrt{\frac{a}{a+c}\cdot\frac{a}{a+b}}\leq \frac{a}{a+c}+\frac{a}{a+b}$

Adiing the similar ones we get the desired inequality .

[quote="pvthuan"]
2) Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
\[\frac 1{\sqrt{a+bc}}+\frac 1{\sqrt{b+ca}}+\frac 1{\sqrt{c+ab}}\geq\frac92. \]

\]
Problem 1) has two nice solutions while problems 2) and 3) are unsolved.[/quote]

2.using cauchy then its sufficient to prove $2 \geq \sum \sqrt{a+bc}$ then $\iff 3\sqrt{\frac{\sum a+\sum ab}{3}}, \sum ab \leq \frac{1}{3}$ done

Have I made a major mistake or something it does not seem so hard

Sorry, I mean the second as follows:

Prove or disprove that
\[\frac1{\sqrt{a+b^{2}}}+\frac1{\sqrt{b+c^{2}}}+\frac1{\sqrt{c+a^{2}}}\geq\frac92.\]

[quote="toanhocmuonmau"][quote="pvthuan"]1) Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
\[\frac a{\sqrt{a+bc}}+\frac b{\sqrt{b+ca}}+\frac c{\sqrt{c+ab}}\leq\frac32. \]
2) Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
\[\frac 1{\sqrt{a+bc}}+\frac 1{\sqrt{b+ca}}+\frac 1{\sqrt{c+ab}}\geq\frac92. \]
3) Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Find the minimum of
\[f=\frac 1{\sqrt{a+bc}}+\frac 1{\sqrt{b+ca}}+\frac 1{\sqrt{c+ab}}. \]
Problem 1) has two nice solutions while problems 2) and 3) are unsolved.[/quote]
3) is equivalent to my inequality
\[\sum\sqrt{\frac{a}{a^{3}+bc}}\ge 2\sqrt{2}\]
if $a,b,c \ge 0,a^{2}+b^{2}+c^{2}=1.$ :P[/quote]

Problem (3) is of mine of course. In the book "Sang Tao Bat Dang thuc, 2006", it is placed in Open Section (but now I proved it). In Mathlink, we have many discussion about it (at least two discussion I think). As I know, Ji Chen has a solution to this one, but my own solution is not easier any more.

[quote="pvthuan"]2) Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove or disprove that
\[ \frac1{\sqrt {a \plus{} b^{2}}} \plus{} \frac1{\sqrt {b \plus{} c^{2}}} \plus{} \frac1{\sqrt {c \plus{} a^{2}}}\geq\frac92.\][/quote]
$ a \equal{} 0,b \equal{} \frac {2}{3},c \equal{} \frac {1}{3}$:

$ \sum_{cyc}\frac1{\sqrt {a \plus{} b^{2}}} \equal{} \frac {3}{2} \plus{} \frac {3}{\sqrt {7}} \plus{} \sqrt {3} \equal{} 4.365\cdots < \frac {9}{2}$.

By the way, the following inequalities are true :

If $ x,y,z$ be positive reals such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \frac {1}{(x \plus{} y^{2})^2} \plus{} \frac {1}{(y \plus{} z^{2})^2} \plus{} \frac {1}{(z \plus{} x^{2})^2}\geq\frac {243}{16}$,

$ \frac {1}{\sqrt {x \plus{} y^{2}}} \plus{} \frac {1}{\sqrt {y \plus{} z^{2}}} \plus{} \frac {1}{\sqrt {z \plus{} x^{2}}} > \sqrt {19}$,

$ \frac {1}{\sqrt {x \plus{} 8y^{2}}} \plus{} \frac {1}{\sqrt {y \plus{} 8z^{2}}} \plus{} \frac {1}{\sqrt {z \plus{} 8x^{2}}} \geq \frac{9}{\sqrt {11}}$,

$ \frac {1}{\sqrt {3x \plus{} 2y^{2}}} \plus{} \frac {1}{\sqrt {3y \plus{} 2z^{2}}} \plus{} \frac {1}{\sqrt {3z \plus{} 2x^{2}}} \geq \frac{9}{\sqrt {11}}$.

[quote="hungkhtn"][quote="toanhocmuonmau"][quote="pvthuan"]1) Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac a{\sqrt {a \plus{} bc}} \plus{} \frac b{\sqrt {b \plus{} ca}} \plus{} \frac c{\sqrt {c \plus{} ab}}\leq\frac32.
\]
2) Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac 1{\sqrt {a \plus{} bc}} \plus{} \frac 1{\sqrt {b \plus{} ca}} \plus{} \frac 1{\sqrt {c \plus{} ab}}\geq\frac92.
\]
3) Let $ a,b,c$ be positive real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Find the minimum of
\[ f \equal{} \frac 1{\sqrt {a \plus{} bc}} \plus{} \frac 1{\sqrt {b \plus{} ca}} \plus{} \frac 1{\sqrt {c \plus{} ab}}.
\]
Problem 1) has two nice solutions while problems 2) and 3) are unsolved.[/quote]
3) is equivalent to my inequality
\[ \sum\sqrt {\frac {a}{a^{3} \plus{} bc}}\ge 2\sqrt {2}
\]
if $ a,b,c \ge 0,a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1.$ :P[/quote]

Problem (3) is of mine of course. In the book "Sang Tao Bat Dang thuc, 2006", it is placed in Open Section (but now I proved it). In Mathlink, we have many discussion about it (at least two discussion I think). As I know, Ji Chen has a solution to this one, but my own solution is not easier any more.[/quote]
The inequality of mine is not yours, Hung. Your Inequality is
\[ \sum \frac{1}{\sqrt{a^2\plus{}bc}} \ge \frac{2\sqrt{2}}{\sqrt{ab\plus{}bc\plus{}ca}}\]
it can be proved by the AM-GM. :)

[quote="manlio"][quote="can_hang2007"]
\[ \sum \frac {1}{\sqrt {a^2 \plus{} bc}} \ge \frac {2\sqrt {2}}{\sqrt {ab \plus{} bc \plus{} ca}}
\]
it can be proved by the AM-GM. :)[/quote]

Dear can_hang2007, I know this inequality, but I have never seen a solution. Can you please post a proof if possible ?

I would be very happy to see it.  :)

Thank you very much.[/quote]
Never mind, Manlio.
Here is my solution with AM-GM :)
Problem. Let $ x,y,z$ be nonnegative real numbers. Prove that
\[ \frac{1}{\sqrt{x^2\plus{}yz}} \ge \frac{2\sqrt{2}}{\sqrt{xy\plus{}yz\plus{}zx}}\]
Solution.
Setting $ a\equal{}1/x,b\equal{}1/y,c\equal{}1/z$, the inequality becomes
\[ \sum \sqrt{\frac{a(a\plus{}b\plus{}c)}{a^2\plus{}bc}} \ge 2\sqrt{2}\]
By the Am-GM, we have
\[ \sum \sqrt{\frac{a(a\plus{}b\plus{}c)}{a^2\plus{}bc}} \ge  2\sqrt{2}\sum \frac{a(a\plus{}b\plus{}c)}{2(a^2\plus{}bc)\plus{}a(a\plus{}b\plus{}c)}\]
It suffices to prove
\[ \sum \frac{a(a\plus{}b\plus{}c)}{2(a^2\plus{}bc)\plus{}a(a\plus{}b\plus{}c)} \ge 1\]
By expandind, it is equivalent to
\[ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}6abc(a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}a^2b^2c^2 \ge 0\]
which is trivial. :)

Thank you very much. :)

Very nice and elegant solution for a hard inequality.
Congratulations.


Given that x, y and z be positive real numbers such that x^2 + y^2 + z^2 = 1

Prove that x^2yz +xy^2z and xyz^2=<1/3

$ 1 \equal{} (x^2\plus{}y^2\plus{}z^2)^2 \equal{} x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2$.

By AM-GM, $ x^4 \plus{} y^2z^2 \geq 2x^2yz$ and $ \frac{x^2y^2 \plus{} y^2z^2}{2} \geq xy^2z$. In this way we get $ 6$ inequalities, if we add them we get
$ 1 \geq 3x^2yz \plus{} 3xy^2z \plus{} 3xyz^2$, which is wat we wanted to prove.


I use the fact that $ k^2 \minus{} kl \plus{} l^2\geq 0$.
$ x^3 \minus{} 3x^2 \plus{} 5x \equal{} 1, x^3 \minus{} 3x^2 \plus{} 3x \minus{} 1 \equal{} \minus{} 2x, (x \minus{} 1)^3 \equal{} \minus{} 2x$. Similarly $ (y \minus{} 1)^3 \equal{} 4 \minus{} 2y$. Set $ x \minus{} 1 \equal{} k$,$ y \minus{} 1 \equal{} l$,$ x \plus{} y \equal{} a$. Adding the two previous equations you get $ (a \minus{} 2)(k^2 \minus{} kl \plus{} l^2) \equal{} 4 \minus{} 2a$. First solution: $ a \equal{} 2$. If $ a\neq 2$, then $ k^2 \minus{} kl \plus{} l^2 \equal{} \minus{} 2$, which is impossible. So, $ x \plus{} y \equal{} 2$.

[quote="equilibrium"]x^3 - 3x^2 + 5x = 1

y^3 - 3y^2 + 5y = 5

Find   x + y = ?[/quote]
[size=150][color=red][b]solution with cardan method[/b][/color][/size]
[img]http://www.mathmontada.net/vb/uploaded/17_1257632921.gif[/img]
with my best wishes
mrmohamed


dose anyone know any online resource on basic quantum mechanics . prefebrebly PDF.

I have the link(ftp) for feynman's lecture book. will post it soon.

also if anyone got the url of feynman's tips on physics I will be very glad.

[color=blue]Have a look at these books

[url=http://djvu.504.com1.ru:8019/WWW/2357215bd7a3283c99273a42a697fee3.pdf]1. Quantum mechanics - Concepts and applications 1999[/url]

[url=http://djvu.504.com1.ru:8019/WWW/524cdc1a61b3b826261737157d23bcf0.pdf]2. Quantum mechanics 2000[/url]

[url=http://djvu.504.com1.ru:8019/WWW/0db5e7c743fdadf9905259aaebe5df71.pdf]3.Introduction to quantum mechanics 2003[/url]

I hope that it can help you[/color]

sorry but the link is unavilable any other resource ???


How exactly do I integrate  :sqrt: (r :^2: - x :^2:) ?

[hide]trig substitution, a description can be found in any calc book ever written... if you don't understand it after looking it up, come back and we'll explain.[/hide]

Thanks... I found it online (I already handed in my calc book for the year).  I hadn't been able to find it before, but looking for the trig substitutions made it easier.


Find all polynomials $p(x,y)\in\mathbb R[x,y]$ such that for each $x,y\in\mathbb R$ we have
\[ p(x+y,x-y)=2p(x,y). \]

$P(x,y)=2P(\frac{x+y}2,\frac{x-y}2)=4P(\frac x2,\frac y2)$
But since a polynomial identity is also an identity in its coefficents we have that all terms of $P$ are of total degree $2$.
So $P(x,y)=ax^2+bxy+cy^2$.
So $P(x+y,x-y)=(a+b+c)x^2+2(a-c)xy+(a+c-b)y^2$.
So we must have $a+b+c=2a$ and $2(a-c)=2b$ and $a+c-b=2c$ but all of these equations are equal. So the answers are
\[ P(x,y)=(a+b)x^2+axy+by^2 \]

i've solved this one with another way(it is not a better way.).
see that easily$p(2^kx,2^ky)=4^kp(x,y)$
so for  as infinite as many $n$ we have:$q(n)=p(nx,ny)=n^2p(x,y)$(fix $x,y$ )
so always:$p(tx,ty)=t^2p(x,y)****$.
so:
$p(0,0)=0, p(x,0)=x^2p(1,0) ,p(0,x)=x^2p(0,1)$
so:from the above$p(x.y)=ax^2+by^2+cxy(h(x,y))$
if you use$****$ we have:$h(tx,ty)=h(x,y)$(for all $t$)
so $h$ is constant.
but as a bad chance i.ve mistakefully had the answer wrong on my sheet.(hop god help me :) )

i've solved it in asimilar way to mr ahmadi pour.
but i think mr dashmiz proof was deeper.


Q.

The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the calss. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.

Chandra and Bob, who each have a copy of the book, decide that they can save time by 'team reading' the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through they will tell each other about the part they read. What is the last page that handra should read so that she and Bob spend the same amount of time reading the novel?

A. 456

I have no idea how to solve this...

Help me!!!

[hide]We can ignore Alice, since she is irrelevant. Bob reads a page in 45 seconds, and Chandra 30 seconds. We can see that Chandra reads 1.5 times as fast as Bob (consider the interval of 90 seconds). There are 760 pages total, so, letting $ c$ be Chandra's # of pages (which is also her last page, since she starts at page 1):

\begin{align*}
\dfrac{1.5}{1.5+1}&=\dfrac{c}{760}\\
c=\fbox{456}
\end{align*}
[/hide]


If 74 hens lay 74 dozen eggs in 74 days and if 37 hens eat 37 kilograms of wheat in 37 days, how many kilograms of wheat are needed to produce 1 dozen eggs?

i don't get it

incase it's correct though:...[hide]1[/hide]

jorian

[hide]2?[/hide]

haha darn i see a huge discrepancy of answers

Yes I need a SOLUTION, however.

[hide]I had 1 originally, which was wrong. I now think it's 2... but don't really know why[/hide]

Thanks to everyone that's posted so far...

I also get $\boxed{2}$.

Try calculating amount of chicken-days per egg/wheat.

[quote="frostmourne"][hide] i got 2

74 hens lay 74 dozen eggs in 74 days, so one hen lays 1 dozen eggs in 74 days

37 hens eat 37 kilos of wheat in 37 days, so one hen eats one kilo of wheat in 37 days, or 2 kilos of wheat in 74 days

if one hen lays one dozen eggs and eats 2 kilos of wheat in 74 days, then 2 kilos of wheat is needed for 1 dozen eggs [/hide]

im not really sure this is right though :huh:[/quote]

Thats what I think too...  :wink:

[hide]i got the same answer using direct and inverse variation

for example: take the # of eggs to be the dependent variable.

# hens is directly proportional to the # of eggs (more hens, more eggs)

#days is also directly proportional to the # of eggs (more days, more eggs)

so you end up with $\frac{\#eggs}{\#hens\times\#days}= k$

so then you can plug in the information to find $k$. when you have $k$, you can plug in information to find how many days it takes for one hen to lay one dozen eggs...

then u do the same for the amount of wheat...sure it's longer, but it's way more fun  :D  and you'll [i]know[/i] it's right...[/hide]

Haha the wording owned me. I divided through by 74 to get 1 chicken lays 1 dozen eggs in 1 day -_- originally

Thanks frost


probability
function
integration
probability and stats

define Random():
a =Normal(u1, o1^2)
b=Normal(a, a*(1-a))
return b

What is the pdf of the output of Random()?

take a test function $ f$ and compute $ E[f(b)]$:
$ E[f(b)] \equal{} \int_{a \in \mathbb{R}}  \frac{\exp( \minus{} \frac{(a\minus{}u_1)^2}{2o_1^2})}{\sqrt{2 \pi u_1^2}} \int_{y \in \mathbb{R}} f(y) \frac{\exp( \minus{} \frac{(y\minus{}a)^2}{2 |a(1\minus{}a)|})}{\sqrt{2 \pi |a(1\minus{}a)|}} dy da \equal{} \int_y f(y) D(y) dy$
where you can easily find a closed form of $ D(y)$. Then $ D(y) \equal{}  \int_{a \in \mathbb{R}}  \frac{\exp( \minus{} \frac{(a\minus{}u_1)^2}{2u_1^2})}{\sqrt{2 \pi o_1^2}}  \frac{\exp( \minus{} \frac{(y\minus{}a)^2}{2 |a(1\minus{}a)|})}{\sqrt{2 \pi |a(1\minus{}a)|}} da$ is the distribution of $ b$. (I assumed that the variance is $ |a(1\minus{}a)|$. Is that what you meant ?)

[quote="alekk"]take a test function $ f$ and compute $ E[f(b)]$:
$ E[f(b)] \equal{} \int_{a \in \mathbb{R}} \frac {\exp( \minus{} \frac {(a \minus{} u_1)^2}{2o_1^2})}{\sqrt {2 \pi u_1^2}} \int_{y \in \mathbb{R}} f(y) \frac {\exp( \minus{} \frac {(y \minus{} a)^2}{2 |a(1 \minus{} a)|})}{\sqrt {2 \pi |a(1 \minus{} a)|}} dy da \equal{} \int_y f(y) D(y) dy$
where you can easily find a closed form of $ D(y)$. Then $ D(y) \equal{} \int_{a \in \mathbb{R}} \frac {\exp( \minus{} \frac {(a \minus{} u_1)^2}{2u_1^2})}{\sqrt {2 \pi o_1^2}} \frac {\exp( \minus{} \frac {(y \minus{} a)^2}{2 |a(1 \minus{} a)|})}{\sqrt {2 \pi |a(1 \minus{} a)|}} da$ is the distribution of $ b$. (I assumed that the variance is $ |a(1 \minus{} a)|$. Is that what you meant ?)[/quote]

Yes it is, thank you!


function
integration
calculus
logarithms
limit
linear algebra
matrix

The 11th International (15th National) Scientific Olympiad in Mathematics (for 3rd and 4th year university students) will be in Iran on 13-16 July 2010.

Schedule:
Tuesday 13 July 2010 - Mathematical Analysis, Numerical Analysis
Wednesday 14 July 2010 - Algebra/Operations Research, Linear Algebra

http://olympiad.sanjesh.org/en/calendar.html

please past to last olympiad problems and solutions or only problems

Program: http://olympiad.sanjesh.org/en/Program%20of%20Activities.html

In English:
ISOM 2010 event in Facebook: http://www.facebook.com/event.php?eid=137300312955574

In Russian:
ISOM 2010 event in vKontakte: http://vkontakte.ru/event18750000

Team of Ukraine

Students:
Ivan IURCHENKO, Kyiv Taras Shevchenko National University (Mathematics)
Ievgen MAKEDONSKYI, Kyiv Taras Shevchenko National University (Mathematics)
Maksym TANTSIURA, Kyiv Taras Shevchenko National University (Mathematics)
Oleg KOVALENKO, Dniepropetrovsk Oles Gonchar National University (Mathematics)
Andrii ARMAN, Kyiv Taras Shevchenko National University (Mathematics)
Bogdan MAREKHA, Kharkov V.N.Karazin National University (Chemistry)
Mykhailo VYBORNYI, Kyiv Taras Shevchenko National University (Chemistry)

Leaders: Dmytro MITIN, Kyiv Taras Shevchenko National University (Mathematics); Oleksandr POZHARYTSKYI, Odessa State Agricultural University (Chemistry); Kateryna MYKHAILYK, Igor NEKRASSA, Ministry of Education and Science of Ukraine

Team of Turkmenistan

students:
Gurbanov Pirmyrat ;Magtymguly Turkmen State University (Mathematics)
Tohtayev Rufat ; International Turkmen-Turkish University (Mathematics)
Shukurov Jemsit ;International Turkmen-Turkish University (Mathematics)
Muradov Halmurat ; International Turkmen-Turkish University (Mathematics)
Hojamuhammedov Sohbet;Magtymguly Turkmen State University (Chemistry)

Leaders:
Hajimamed Soltnov Magtymguly Turkmen State University (Mathematics),
Geldiniyazov Mametmyrat Magtymguly Turkmen State University (Chemistry)

Best regards

Problems of the 1st Day (Mathematical Analysis and Numerical Analysis) : http://rapidshare.com/files/406826274/day1.rar
(10 downloads only, megaupload seems to be banned here in Iran).

please anybody past to solutions of math analys and numeral analys

Problems of the 2nd Day (Algebra/Operation Research and Linear Algebra) : http://rapidshare.com/files/407100256/day2.rar
Each of the problems in Algebra and Linear Algebra is worth 25 points.

Mirror:
1st Day: http://www.mediafire.com/?iwqm2k3yenznxd2
2nd Day: http://www.mediafire.com/?iz1zoojdumnfvum

Results: http://mfi.re/?6th3kgv29zzb5ds

Ivan IURCHENKO - gold medal (Mathematics)
Ievgen MAKEDONSKYI -silver medal (Mathematics)
Maksym TANTSIURA - bronze medal (Mathematics)
Oleg KOVALENKO - bronze medal (Mathematics)
Andrii ARMAN - bronze medal (Mathematics)
Bogdan MAREKHA - bronze medal (Chemistry)
Mykhailo VYBORNYI - gold medal (Chemistry)

[quote="dmitin"]Results: http://mfi.re/?6th3kgv29zzb5ds
[/quote]

Link doesn't work - at least for me.

[quote="hsiljak"]Link doesn't work - at least for me.[/quote]

What about http://www.mediafire.com/?6th3kgv29zzb5ds
?

Scanned version of problems is attached to this posting.

Results are at the olympiad web-site: math http://olympiad.sanjesh.org/en/Result/11th.html
chemistry http://olympiad.sanjesh.org/en/Result/CH3rd.html

[b]The $15^{\text{th}}$ International Scientific Olympiad for University Students
13-16 July 2010, Shahid Beheshti University, Tehran, Iran
Mathematics[/b]

[b]First Day, 13 July 2010[/b]

[b]1. Mathematical Analysis (Pure and Applied Mathematics)[/b]
120 minutes, 100 points

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357907]1.[/url] (15 points) Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $(f\circ f\circ f)(x)=x$ for all $x\in\mathbb{R}$. Show that $f$ is the identity function.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357908]2.[/url] (15 points) Let $C(\Delta)$ be the set of all continuous complex valued functions on the closed unit disc $\Delta=\{z\in\mathbb{C}:\lvert z\rvert\le1\}$.
Let $d$ be the uniform metric on $C(\Delta)$ defined by
\[d(f,g)=\lVert f-g\rVert_{\infty}=\sup_{z\in\Delta}\lvert f(z)-g(z)\rvert \quad (f,g\in C(\Delta)).\]
If
\[A=\{f\in C(\Delta):\ f(0)=0\text{ and }\lvert f(z)-f(w)\rvert\le\lvert z-w\rvert^{1/2},\text{ for }z,w\in\Delta\},\]
show that $A$ is compact in $C(\Delta)$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357910]3.[/url] (30 points) Let $I=[0,1]$ and $X=C^1(I)$ be the space of all continuously differentiable real valued functions on $I$. Let $\lVert f\rVert_{\infty}=\sup_{x\in I}\lvert f(x)\rvert$ for $f\in X$ and define three metrics on $X$ as follows:
\[d_0(f,g)=\lVert f-g\rVert_{\infty},\ d_1(f,g)=\lvert f(0)-g(0)\rvert+\lVert f'-g'\rVert_{\infty}\text{ and }d_2(f,g)=\lVert f-g\rVert_{\infty}+\lVert f'-g'\rVert_{\infty},\]
for $f,g\in X$.
(i) (9 points) Is $(X,d_0)$ a complete metric space? Why?
(ii) (8 points) Is $(X,d_1)$ a complete metric space? Why?
(iii) (5 points) Show that $(X,d_2)$ is a complete metric space.
(iv) (8 points) Are the metrics $d_1$ and $d_2$ equivalent on $X$? Why?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357911]4.[/url] (15 points) Let $f$ be a continuous real valued function on $[0,\infty)$ such that
\[\sup_{p\ge0}\biggl|\int_0^pf(x)\,\mathrm{d}x\biggr|<\infty.\]
Show that for each $a,b>0$ the following integral exists and is equal to $f(0)\ln\frac{b}{a}$
\[\int_0^{\infty}\frac{f(ax)-f(bx)}{x}\,\mathrm{d}x.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357912]5.[/url] (25 points) For a continuous function $f$ on $[0,1]$ and $j\in\mathbb{N}$, define
\[\omega_j(f)=\sup\Bigl\{\lvert f(x)-f(y)\rvert:\ x,y\in[0,1],\ \lvert x-y\rvert<\frac{1}{j}\Bigr\}.\]
Let the functions $f$ and $f_n$, $n=1,2,\ldots$ be continuous on $[0,1]$ and let $\mathbb{Q}$ be set of rational numbers such that
(i) $\lim_{n\to\infty}f_n(r)=f(r)$ for each $r\in\mathbb{Q}\cap[0,1]$,
(ii) $\lim_{n\to\infty}\omega_j(f_n)=\omega_j(f)$ for each $j\in\mathbb{N}$.
Show that $\lim_{n\to\infty}f_n=f$ uniformly on $[0,1]$.

[b]2. Numerical Analysis (Pure and Applied Mathematics)[/b]
120 minutes, 100 points

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357913]1.[/url] (25 points) (i) (15 points) The iteration defined by $x_{k+1}=\frac{1}{2}(x_k^2+c)$, where $0<c<1$, has two fixed points $\xi_1,\xi_2$, where $0<\xi_1<1<\xi_2$. Show that
\[x_{k+1}-\xi_1=\frac{1}{2}(x_k+\xi_1)(x_k-\xi_1),\quad k=0,1,2,\ldots,\]
and deduce that $\lim_{k\to\infty} x_k=\xi_1$ if $0\le x_0<\xi_2$.
(ii) (5 points) How does the iteration behave for other values of $x_0$?
(iii) (5 points) Assume that $f(x)=8x^3+ax^2+bx-1$. Find $a$ and $b$ such that $f(\frac{1}{2})=0$ and the order of convergence of Newton's method, for finding root $x=\frac{1}{2}$ with suitable $x_0$, be exactly one.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=357914]2.[/url] (20 points) Let $A$ be an $m\times n$ matrix, $\mathbf{b}$ an $m$-vector, and $\alpha>0$.
Using the Euclidean norm, define
\[F(\mathbf{x})=\lVert A\mathbf{x}-\mathbf{b}\rVert_2^2+\alpha\lVert\mathbf{x}\rVert_2^2.\]
(a) (15 points) Prove that $F(\mathbf{x})$ is a global minimum when $\mathbf{x}$ is a solution of the equation
\[(A^{T}A+\alpha I)\mathbf{x}=A^{T}\mathbf{b}.\]
(b) (5 points) Prove that for $\mathbf{x}$, a global minimizer of $F$, we have
\[F(\mathbf{x}+\mathbf{h})=F(\mathbf{x})+\lVert A\mathbf{h}\rVert_2^2+\alpha\lVert\mathbf{h}\rVert_2^2.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=357917]3.[/url] (30 points) (i) (15 points) Let $f\in C^4[a,b]$, $F(x)=\int_0^xf(t)\,\mathrm{d}t$, for $x\in[a,b]$, $h=\frac{b-a}{n}$, $x_i=a+ih$, $i=0,1,\ldots,n$, $n\ge2$.
Explain how one can estimate values of $F(x_i)$, $i=2,3,\ldots,n$, with errors of order $h^4$, $O(h^4)$, using [u]only[/u] values of $f(x_i)$, $i=0,1,2,\ldots,n$, and suitable quadrature rules.
(ii) Let $w_i$ and $x_i$ be such that the following formula is exact for all polynomials of degree less than or equal to $2n-1$,
\[\int_{-1}^1f(x)\,\mathrm{d}x\approx\sum_{i=1}^nw_if(x_i).\]
Show that:
1) (5 points) $\displaystyle\sum_{i=1}^nw_i=2$,
2) (10 points) $\displaystyle\int_{-1}^1L_i(x)\,\mathrm{d}x=\int_{-1}^1(L_i(x))^2\,\mathrm{d}x$, where $\displaystyle L_i(x)=\prod_{\substack{j=1\\ j\ne i}}^n\frac{(x-x_j)}{(x_i-x_j)}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=357918]4.[/url] (25 points) The Bernoulli polynomials are defined by
\[\sum_{k=0}^n\binom{n+1}{k}B_k(t)=(n+1)t^n,\text{ with }\binom{n+1}{k}=\frac{(n+1)!}{k!(n+1-k)!},\]
where for any given $n$, the functions $B_0,\ldots,B_n$ are the Bernoulli polynomials.
Prove the following properties:
(i) (5 points) $B_n'=nB_{n-1}$, $n\ge1$.
(ii) (5 points) $B_n(t+1)-B_n(t)=nt^{n-1}$, $n\ge2$.
(iii) (5 points) $\displaystyle B_n(t)=\sum_{k=0}^n\binom{n}{k}B_k(0)t^{n-k}$.
(iv) (10 points) $B_n(1-t)=(-1)^nB_n(t)$.

[b]Second Day, 14 July 2010[/b]

[b]3. Algebra (Pure Mathematics)[/b]
120 minutes, 100 points

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=357919]1.[/url] (25 points) Let $G$ be a finite group and $H$ be a subgroup of $G$ with odd order and $[G:H]=2^n$, where $n\in\mathbb{N}$. Prove that all elements of $G$ with odd order are in $H$ if and only if $H\trianglelefteq G$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=357920]2.[/url] (25 points) Let $G$ be a finite group and $N<G$ be a maximal subgroup of $G$. Assume that $N$ is abelian, and $[G:N]=p^n$ for some prime $p$, $n\in\mathbb{N}$, and $N$ contains no non-identity normal subgroup of $G$. Show that $p$ does not divide $\lvert N\rvert$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=357921]3.[/url] (25 points) Prove that every ring of order $p^2$ with identity is commutative, where $p$ is a prime number.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=357924]4.[/url] (25 points) Let $R$ be a commutative ring with identity and $I$, $J$ be the ideals of $R$.
(a) If $R$ is a PID and $R=I+J$, prove $IJ=I\cap J$.
(b) Show that none of the two conditions in (a) can be dropped in order to obtain the same conclusion.

[b]3. Operations Research (Applied Mathematics)[/b]
120 minutes, 100 points

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&t=357927]1.[/url] (25 points) Let $S=\{\mathbf{y}\in\mathbb{R}^m:\ A^T\mathbf{y}\le\mathbf{c}\}$, where $A\in\mathbb{R}^{m\times n}$ with rank $m<n$, $\mathbf{c}\in\mathbb{R}^n$ and $K=\{\mathbf{x}\in\mathbb{R}^n:\ A\mathbf{x}=\mathbf{0},\ \mathbf{x}\ge\mathbf{0}\}$.
Consider the LP problem, $\min \mathbf{c}^T\mathbf{x}$ subject to $\mathbf{x}\in K$.
(i) (15 points) Show that $S$ is nonempty if and only if the optimum value of LP is zero.
(ii) (5 points) If there exists $\mathbf{y}\in S$ with $A^T\mathbf{y}<\mathbf{c}$, show that the unique minimal solution of LP is $\mathbf{x}=\mathbf{0}$.
(iii) (5 points) Can the LP problem be unbounded? Give reasons for your answer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&t=357928]2.[/url] (30 points) Consider the linear program (P) of the form,
\[\min \mathbf{q}^T\mathbf{z}\]
\[\text{s.t. }M\mathbf{z}\ge\mathbf{q}\quad (\text{P})\]
\[\mathbf{z}\ge\mathbf{0},\]
where $M$, $\mathbf{q}$ and $\mathbf{z}$ are properly partitioned to be:
\[M=\begin{bmatrix}O&-A^T\\A&O\end{bmatrix},\quad\mathbf{q}=\begin{bmatrix}\mathbf{c}\\-\mathbf{b}\end{bmatrix},
\quad\mathbf{z}=\begin{bmatrix}\mathbf{x}\\ \mathbf{y}\end{bmatrix}.\]
(i) (20 points) Show that (P) has an optimal solution if and only if (P) is feasible.
(ii) (5 points) Can (P) be infeasible? Give reasons for your answer.
(iii) (5 points) Can (P) be unbounded? Give reasons for your answer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&t=357929]3.[/url] (30 points) Consider the following linear program,
\[\min \mathbf{b}^T\boldsymbol{\lambda}\]
\[\text{s.t.}\ A^T\boldsymbol{\lambda}\le\mathbf{c},\quad(\text{P})\]
where $A$, $m\times n$, has full row rank.
Assume that $\overline{\mathbf{x}}$ is feasible to the dual of (P). Show that the program (P) is equivalent to the following program:
\[\max \overline{\mathbf{x}}^T\mathbf{z}\]
\[\text{s.t.}\ P_A\mathbf{z}=P_A\mathbf{c}\]
\[ \mathbf{z}\ge\mathbf{0},\]
where, $P_A=I-A^T(AA^T)^{-1}A$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&t=357930]4.[/url] (15 points) Show that in a simplex tableau when a basic variable is determined to be removed from the basis, it can not be immediately chosen to be a candidate for entering the basis in the next simplex tableau.

[b]4. Linear Algebra (Pure and Applied Mathematics)[/b]
120 minutes, 100 points

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=357932]1.[/url] (25 points) Let $A$ be a real $n\times n$ matrix such that its characteristic polynomial has no repeated roots. Let $B$ and $C$ be real $n\times n$ matrices which commute with $A$, prove $BC=CB$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=357933]2.[/url] (25 points) Let $A=(a_{ij})$ be a real $n\times n$ matrix such that $\sum_{j=1}^n\lvert a_{ij}\rvert<1$, for all $1\le i\le n$. Prove that the matrix $I-A$ is invertible.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=357934]3.[/url] (25 points) Let $V$ be a finite dimensional vector space over the field of rational numbers. Suppose that $T$ is a non-singular linear transformation on $V$ such that $T^{-1}=T+T^2$. Prove that 3 divides the dimension of $V$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=357935]4.[/url] (25 points) $M_r(\mathbb{C})$ denotes the vector space of $r\times r$ matrices over the field of complex numbers $\mathbb{C}$. Let $S$ and $T$ be linear transformations of $M_r(\mathbb{C})$ into $M_r(\mathbb{C})$ such that $S^r=T$ and $T^r=S$. If all the eigenvalues of $S$ are different, prove that $S$ and $T$ are similar matrices.

[quote="amparvardi"]Thank you!
Could you please post the problems in resources ?
It'll be very good.[/quote]

Do you mean [url=http://www.artofproblemsolving.com/Forum/resources.php]here[/url]?
How to do this?


[quote="ElChapin"]Let $ x$, $ y$ and $ z$ be postive real numbers such that $ x_1x_2x_3\ge 6$ and $ x_1\plus{}x_2\plus{}x_3\le 6$.

Determine the maximum value of $ x^7\plus{}y^7\plus{}z^7$.

:)[/quote]
For $a=b=1+2\sin10^{\circ}$ and $c=4-4\sin10^{\circ}$ we obtain:
$a^7+b^7+c^7=4327.085...$


It is possible to factor $2x^{2}-x-3$ as follows:
$2x^{2}-x-3$
In this step, we take the two and move it to the three, multiplying them.
$x^{2}-x-6$
$(x-3)(x+2)$
Now we "put back" the two.
$\frac{(2x-3)(2x+2)}{2}$
$(2x-3)(x+1)$
Prove that this is completely legal, and can be done for any factorable trinomal $ax^{2}+bx+c$ with $a,b,c\in\mathbb{Z}$ and $a\ge2$.

[hide="Proof"]Let the factorization of $x^{2}+bx+ac$ be $(x-r_{1})(x-r_{2})$, giving $r_{1}+r_{2}=-b, r_{1}r_{2}=ac$. Then by the described rule, we transform this into $\frac{(ax-r_{1})(ax-r_{2})}{a}=a(x-{r_{1}\over a})(x-{r_{2}\over a})$. By Vieta's formulas, ${r_{1}\over a}+{r_{2}\over a}=\frac{r_{1}+r_{2}}{a}=\frac{-b}{a}$ and ${r_{1}\over a}\cdot{r_{2}\over a}=\frac{r_{1}r_{2}}{a^{2}}=\frac{ac}{a^{2}}={c\over a}$. Therefore, the latter factorization indeed corresponds to $ax^{2}+bx+c$.[/hide]


Something is wrong with alcumus.

For one thing, my rating was originally about 84, and then dipped to 82 fo no apparent reason.
Another thing is that when I visit the Get Problem section, no problem shows up.
If you can offer an explanation, I would be glad to hear it.

Please check the other announcements. If you continue to do the same question given already, even though it may look like you score will go higher, in the end it drops back down. Also, if there is no problem showing, that means that there is no problem available at this time. If you are, however, doing the 'Mastery of Counting Fundamentals Level 5', I am pretty sure it is there to stop furthermore subjects, as the developers input in more subjects for Introduction to Algebra, to be released sometime in the near future.
Alcumus has been twiddled a little. Please check the other post on this forum regarding that to answer your questions, so that the forum won't get that messy (same topics coming up all the time).

[quote="AdithyaGanesh"]I know rating dips because other people get higher.  The rating is actually a number based on relative performance of everyone.[/quote]

Yes. Alcumus is almost 100% ranking based. Your score is determined by how others do.

The Alcumus score is based on your ranking relative to others.

Are you still having the problem that Alcumus is not giving you problems?

Yes, I am still having the problem about not getting the problems.

The glitch has been fixed now. Other than the subject 'Principle of Inclusion-Exclusion', you will find you can get new problems. I think today, Alcumus was given many more problems for you to do.


algebra
polynomial
calculus
integration
modular arithmetic

Let $ p$ be prime number and $ a_0,\ a_1,\ \cdots,\ a_{n\minus{}1}\ (n\geq 2)$ be integers.
Consider a polynomial with degree $ n$:$ f(x)\equal{}x^n\plus{}pa_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}pa_i x^{i}\plus{}\cdots \plus{}pa_0.$

(1) If the equation $ f(x)\equal{}0$ has integral solution $ \alpha$, then prove that $ \alpha$ is divisible by $ p$.

(2) If $ a_0$ isn't divisible by $ p$, then prove that the equation $ f(x)\equal{}0$ doesn't have integral solution.

(1) $ f(\alpha) \equiv \alpha^n \pmod{p}$, so $ p \mid \alpha$.

(2) $ f(\alpha) \equiv pa_0 \pmod{p^2}$, so $ p \mid a_0$.

Can we have a solution without using modular arithmetic?

[quote="kunny"]Let $ p$ be prime number and $ a_0,\ a_1,\ \cdots,\ a_{n \minus{} 1}\ (n\geq 2)$ be integers.
Consider a polynomial with degree $ n$:$ f(x) \equal{} x^n \plus{} pa_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} pa_i x^{i} \plus{} \cdots \plus{} pa_0.$

(1) If the equation $ f(x) \equal{} 0$ has integral solution $ \alpha$, then prove that $ \alpha$ is divisible by $ p$.

(2) If $ a_0$ isn't divisible by $ p$, then prove that the equation $ f(x) \equal{} 0$ doesn't have integral solution.[/quote]

[hide="1"]

If the equation $ f(x) \equal{} 0$ has integral solution $ \alpha$, then $ f(\alpha) \equal{} 0$
i.e.
$ \alpha^n \plus{} pa_{n \minus{} 1}\alpha^{n \minus{} 1} \plus{} \cdots \plus{} pa_i \alpha^{i}\plus{} \cdots \plus{} pa_0\equal{}0$

$ \implies  a_{n \minus{} 1}\alpha^{n \minus{} 1} \plus{} \cdots \plus{} a_i \alpha^{i}\plus{} \cdots \plus{} a_0\equal{}\frac{\minus{}\alpha^n}{p}$

Since $ a_{n \minus{} 1}\alpha^{n \minus{} 1} \plus{} \cdots \plus{} a_i \alpha^{i}\plus{} \cdots \plus{} a_0$  represents some integer , then $ p\mid (\minus{}\alpha^n)$
$ \therefore p\mid \alpha$ [/hide]

[hide="2"]
Suppose $ p\nmid a_0$  and the equation $ f(x) \equal{} 0$  has  integral solution $ \alpha$,

From (1)
$ \alpha^n \plus{} pa_{n \minus{} 1}\alpha^{n \minus{} 1} \plus{} \cdots \plus{} pa_i \alpha^{i}\plus{} \cdots \plus{} pa_1\alpha\plus{}pa_0\equal{}0$
Let $ \alpha\equal{}pd$, $ d$ is some integer.

$ (pd)^n \plus{} pa_{n \minus{} 1}(pd)^{n \minus{} 1} \plus{} \cdots \plus{} pa_i (pd)^{i}\plus{} \cdots \plus{} p^2da_1\plus{}pa_0\equal{}0$

$ \implies d^np^{n\minus{}2} \plus{} a_{n \minus{} 1}d^{n \minus{} 1}p^{n\minus{}2} \plus{} \cdots \plus{} a_i d^{i}p^{i\minus{}1}\plus{} \cdots \plus{} da_1\equal{}\minus{}\frac{a_0}{p}$

L.S represents for some integer , but $ p\nmid \alpha$.

Hence, it is contradictive.
[/hide]

Thank you!

Yes, that's standard solution that Japanese high school students can understand well.

Needless to say, mavropnevma's solution is correct as well.


dear orl, the rest of question is disappeared!!!

[quote="orl"]Given $n$ positive numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ such that $a_{1}\cdot a_{2}\cdot ...\cdot a_{n}=1.$ Prove $
\left( 1+a_{1}\right) \left( 1+a_{2}\right) ...\left(1+a_{n}\right) \geq 2^{n}.
 $[/quote]

Here it is. If there are more questions like this, let me know via PM, we'll see what can be done.

I'm sorry for not using Latex. By   Holder corollary (1+n-th root a1)(1+n-th root a2)...(1+n-th root an)>=(1+n-th root(a1*a2*...*an))^n, we have LHS>=(1+n-th root(a1*a2*..*an)^n=(1+1)^n=2^n.

$(1+b)(1+c) \geq (1+\sqrt{bc})^{2}$, which is equivalent to $b+c \geq 2\sqrt{bc}$. By Sturm's method minimum of $\left( 1+a_{1}\right)\left( 1+a_{2}\right) ...\left(1+a_{n}\right)$ is obtained if $a_{1}=...=a_{n}=1$. So $\left( 1+a_{1}\right)\left( 1+a_{2}\right) ...\left(1+a_{n}\right) \geq 2^{n}$.

[hide=Probably Wrong Solution]Note that
\begin{align*}
(1+a_1)(1+a_2)\cdot\cdot\cdot(1+a_n)&=1+\sum_{i=1}^{n-1}\left(\sum_{1\leq b_1,b_2\ldots,b_i\leq n,|\{b_1,b_2\ldots,b_i\}|=i}\left(\prod_{k=1}^{i}a_{b_{i}}\right)\right)+\prod_{i=1}^{n}a_i\\
&=2+\sum_{i=1}^{n-1}\left(\sum_{1\leq b_1,b_2\ldots,b_i\leq n,|\{b_1,b_2\ldots,b_i\}|=i}\left(\prod_{k=1}^{i}a_{b_{i}}\right)\right)\\
&\geq2+\sum_{i=1}^{n-1}\left(\binom{n}{i}\left(\prod_{k=1}^{n}a_{k}\right)^{\frac{n}{i\binom{n}{i}}}\right)\\
&=2+\sum_{i=1}^{n-1}\left(\binom{n}{i}\right)\\
&=2+2^n-\binom{n}{0}-\binom{n}{n}\\
&=2^n,
\end{align*}
as desired.[/hide]

From AM-GM $\begin{aligned} &\left(\frac{1+a_1}{2}\right)\left(\frac{1+a_2}{2}\right) \ldots\left(\frac{1+a_n}{2}\right) \geqslant \sqrt{a_1 a_2 \ldots a_n} \\ \Rightarrow & {\left[\left(1+a_1\right)\left(1+a_2\right) \ldots\left(1+a_n\right)\right] \frac{1}{2^n} \geqslant \sqrt{1} } \\ \Rightarrow &\left(1+a_1\right)\left(1+a_2\right) \ldots\left(1+a_n\right) \geqslant \sqrt{2^n} \end{aligned}$

@above wow

i overcomplicated this problem :noo:

$f(x)=\ln(1+e^x)$ is convex.

Let $b_i=\ln a_i$ so that $b_1\cdots+b_n=0$. Then by Jensen, $$\ln(1+e^{b_1})+\cdots \ln(1+e^{b_n})\geq n\ln(1+e^0)=n\ln 2.$$ Exponentiating gives the desired result.

[quote=orl]Given $n$ positive numbers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ such that $a_{1}\cdot a_{2}\cdot ...\cdot a_{n}=1.$ Prove \[ \left( 1+a_{1}\right) \left( 1+a_{2}\right) ...\left(1+a_{n}\right) \geq 2^{n}.\][/quote]

Use Carlson.
[quote=Algie]By Sturm's method minimum of $\left( 1+a_{1}\right)\left( 1+a_{2}\right) ...\left(1+a_{n}\right)$ is obtained if $a_{1}=...=a_{n}=1$.[/quote]

What is this method?

Sturm just destroys this problem, but simple AM-GM for reach bracket also easily finishes the problem

Sturm method is a method where you basically fix a sum/product/sum of squares, etc for a list of numbers, and then try to keep this sum/product the same while changing the numbers and looking on the other hand side. For example here we have the product fixed, so I can change a1,a2 -> a1/x, a2x and look how the hands of the inequality change.

[quote=Peter]Rewrite it as $\left(\frac{1+a_{1}}{2}\right)\cdot\left(\frac{1+a_{2}}{2}\right)\cdots\left(\frac{1+a_{n}}{2}\right) \geq 1$ and this follows immediately from AM-GM[/quote]

This is the best method!


Hey, I'm currently making a document, and I want one header for the first page, and another header for all the other pages. How can I do this? I'm using \lhead, \chead, and \rhead for the headers.

Just change \lhead, \chead and \rhead on the second page.

Where would I put that? I do not know where the second page starts, because I might add or remove things from the first page.

Just move them if necessary in the final document. You'll also find advice in [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr's documentation[/url] in section 18.