integration
logarithms
geometry
trapezoid
LaTeX
floor function
ceiling function

Dertermine whether the series converges conditionally,converge absolutely or diverge.

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{\sqrt{n}}}{n}$

First of all it would be nice to figure out what you mean by $ (\minus{}1)^{\sqrt n}$. Is it $ e^{\pi i\sqrt n}$ or something else? :?

[quote="azatkgz"]Dertermine whether the series converges conditionally,converge absolutely or diverge.

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{\sqrt {n}}}{n}$[/quote]
I'm sure that you mean :
$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{[\sqrt {n}]}}{n}$
this is not so simpl problem and is interesting  :)
also here is generalization:
$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{[\sqrt {n}]}}{n^{p}}$
it was already posted at this forum . I only have posted hint, but not full solution.

This is $ \sum_{k=1}^{\infty}(-1)^n\sum_{n=k^2}^{(k+1)^2-1}\frac1n=
\sum_{k=1}^{\infty}(-1)^kb_k$ where $ b_k=\sum_{n=k^2}^{(k+1)^2-1}\frac1n.$

$ \int_{k^2}^{(k+1)^2}\frac1x\,dx<b_k<\int_{k^2-1}^{(k+1)^2-1}\frac1x\,dx.$

${ \ln(\frac{(k+1)^2}{k^2}})<b_k
<\ln(\frac{(k+1)^2-1}{k^2-1}).$

Since these logarithms tend to zero, that shows that $ b_k\to0.$ We also need to know that $ b_k$ is decreasing. It is, but the estimate above isn't strong enough to show that.

Once we establish that $ b_k$ is decreasing, we will have that the series converges by the alternating series test.

A combination of the trapezoid rule and midpoint Riemann sums, using the fact that $ \frac1x$ is concave upwards, gives us

$ \frac12\left(\frac1{k^2}\minus{}\frac1{(k\plus{}1)^2}\right)\plus{}\ln\left(\frac{(k\plus{}1)^2}{k^2}\right)
<b_k<\ln\left(\frac{(k\plus{}1)^2\minus{}\frac12}{k^2\minus{}\frac12}\right).$

That might be enough to prove that $ b_k$ is decreasing; I haven't checked to be sure.

The relevent commands in LaTeX are \lfloor and \rfloor: \lfloor \sqrt{n}\rfloor gives $ \lfloor \sqrt {n}\rfloor$.  As with many other such symbols you can make them fit around larger things by using \left and \right, i.e. $ \left\lfloor \sum_{i \equal{} 0}^\infty f(i) \right\rfloor$, which looks nicer than $ \lfloor \sum_{i \equal{} 0}^\infty f(i)\rfloor$.  \lceil and \rceil are the commands for the ceiling function.

[quote="Kent Merryfield"]${ \ln(\frac {(k + 1)^2}{k^2}}) < b_k < \ln(\frac {(k + 1)^2 - 1}{k^2 - 1}).$[/quote]Even if we have trouble proving that $ b_k$ is decreasing, the left-hand bound is definitely decreasing.  Furthermore, the difference between the left and right sides is $ O(\frac {1}{k^2})$.  So we can replace $ b_k$ by the left-hand side; the error terms form an absolutely-convergent series.

I agree with
Kent Merryfield. you need that $ b_{k}$ is decreasing.
more simple way:
$ \sum ( \minus{} 1)^{n}a_{n}$ convergent conditionally if $ a_{n}$ is monotonic and tends to $ 0$ (Leibnit's Theorem).
in this case you must prove that $ b_{k}$ is decreasing. I note that this is main moment at this problem
  :)

Ravi B: point taken. Yes, that's the right way to look at it.

Here is my solution.  We want to show that the series is monotonically decreasing and that its limit is zero.  I will
use the form stated eariler, then instead of using integrals to find bounds, use a discrete method with the Harmonic numbers

$ H_n \equal{} \sum_{k \equal{} 1}^{n}\frac1k$

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{\sqrt {n}}}{n} \equal{} \sum_{k \equal{} 1}^{\infty}( \minus{} 1)^n\sum_{n \equal{} k^2}^{(k \plus{} 1)^2 \minus{} 1}\frac1n$

$ \sum_{k \equal{} 1}^{\infty}( \minus{} 1)^n\sum_{n \equal{} k^2}^{(k \plus{} 1)^2 \minus{} 1}\frac1n \equal{} \sum_{k \equal{} 1}^{\infty}( \minus{} 1)^kb_k$ where $ b_k \equal{} \sum_{n \equal{} k^2}^{(k \plus{} 1)^2 \minus{} 1}\frac1n \equal{} H_{(k \plus{} 1)^2 \minus{} 1} \minus{} H_{k^2} \equal{} H_{k^2 \plus{} 2k} \minus{} H_{k^2 \minus{} 1}$

To prove its limit is zero, use the Euler-Maclaurin integration Formula identity for the Harmonic series (this I will not show, but use $ f_k \equal{} 1/k$ with the forumla on the link [url]http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html[url]

$ H_n \equal{} Ln(n) \plus{} \gamma \plus{} \frac{1}{2n}  \minus{} \frac{1}{12n^2} \plus{} \cdot\cdot\cdot$ where the rest of the terms have negative powers of n.

$ \lim_{k \minus{}> \infty} b_k \equal{} \lim_{k \minus{}> \infty} H_{k^2 \plus{} 2k} \minus{} H_{k^2} \equal{} \lim_{k \minus{}> \infty} Ln(k^2 \plus{} 2k) \minus{} Ln(k^2 \minus{} 1)$
$ \lim_{k \minus{}> \infty} Ln(\frac{k^2 \plus{} 2k}{k^2 \minus{} 1}) \equal{} 0$ by using L'Hospital's Rule

Now to show it is monotonically decreasing, we will examine the difference $ b_k \minus{} b_{k \plus{} 1}$ which, if positive, will imply that $ b_k$ is decreasing, and we would be done.

$ b_k \minus{} b_{k \plus{} 1} \equal{} \left(H_{k^2 \plus{} 2k} \minus{} H_{k^2 \minus{} 1}\right) \minus{} \left(H_{(k \plus{} 1)^2 \plus{} 2(k \plus{} 1)} \minus{} H_{(k \plus{} 1)^2 \minus{} 1}\right)$
$ b_k \minus{} b_{k \plus{} 1} \equal{} H_{k^2 \plus{} 2k} \minus{} H_{k^2 \minus{} 1} \minus{} H_{k^2 \plus{} 4k \plus{} 2} \plus{} H_{k^2 \plus{} 2k}$
$ k^2 \plus{} 2k \minus{} (k^2 \minus{} 1) \equal{} 2k \plus{} 1$  implies that subtracting the first difference makes
$ H_{k^2 \plus{} 2k} \minus{} H_{k^2 \minus{} 1} \equal{} \sum_{n \equal{} k^2}^{k^2 \plus{} 2k}\frac1n$, and similarly for the second,
$ k^2 \plus{} 4k \plus{} 2 \minus{} (k^2 \plus{} 2k) \equal{} 2k \plus{} 2$
$ H_{k^2 \plus{} 2k} \minus{} H_{k^2 \plus{} 4k \plus{} 2} \equal{} \minus{}\sum_{n \equal{} k^2 \plus{} 2k \plus{} 1}^{k^2 \plus{} 4k \plus{} 2}\frac1n$
Thus
$ b_k \minus{} b_{k \plus{} 1} \equal{} \sum_{n \equal{} k^2}^{k^2 \plus{} 2k}\frac1n \minus{}\sum_{n \equal{} k^2 \plus{} 2k \plus{} 1}^{k^2 \plus{} 4k \plus{} 2}\frac1n$
where the positive series has $ 2k \plus{} 1$ terms and the negative has $ 2k \plus{} 2$ terms.
At this point, we will break up the two series and intermingle them term by term, leaving the last one out on its own.

$ \sum_{n \equal{} k^2}^{k^2 \plus{} 2k}\frac1n \minus{}\sum_{n \equal{} k^2 \plus{} 2k \plus{} 1}^{k^2 \plus{} 4k \plus{} 2}\frac1n
\equal{} \left(\frac{1}{k^2} \minus{} \frac{1}{k^2 \plus{} 2k \plus{} 1}\right) \plus{} \left(\frac{1}{k^2 \plus{} 1} \minus{} \frac{1}{k^2 \plus{} 2k \plus{} 2}\right) \plus{} \cdot\cdot\cdot \plus{} \left(\frac{1}{k^2 \plus{} 2k} \minus{} \frac{1}{k^2 \plus{} 4k \plus{} 1}\right) \minus{} \frac{1}{k^2 \plus{} 4k \plus{} 2}$
$ \equal{} \sum_{n \equal{} 0}^{2k}\left(\frac{1}{k^2 \plus{} n} \minus{} \frac{1}{k^2 \plus{} 2k \plus{} 1 \plus{} n}\right) \minus{} \frac{1}{k^2 \plus{} 4k \plus{} 2}$

$ \equal{} \sum_{n \equal{} 0}^{2k}\frac{2k \plus{} 1}{(k^2 \plus{} n)(k^2 \plus{} 2k \plus{} 1 \plus{} n)} \minus{} \frac{1}{k^2 \plus{} 4k \plus{} 2}$
$ > \frac{(2k \plus{} 1)(2k)}{(k^2 \plus{} 2k)(k^2 \plus{} 2k \plus{} 1 \plus{} 2k)} \minus{} \frac{1}{k^2 \plus{} 4k \plus{} 2} \equal{} \frac{4k^2 \plus{} 2k}{(k^2 \plus{}2k)(k^2  \plus{} 4k \plus{} 1)} \minus{}\frac{1}{k^2 \plus{} 4k \plus{} 2} \equal{} \frac{3k^4 \plus{} 12k^3 \plus{} 7k^2 \plus{} 2k}{(k^2 \plus{} 2k)(k^2 \plus{} 4k \plus{} 1)(k^2 \plus{} 4k \plus{} 2)} > 0$

$ b_k \minus{} b_{k\plus{}1} > 0$

Thus $ b_k$ is decreasing, hence it converges conditionally.

yes you are right.  :lol:.
I solved by approxymation which also works. solution like your is also in some books.


What's the probability of rolling a multiple of 3 with a fair pair of dice? Express your answer as a common fraction.

You must roll either atleast one 3 or 6, out of the 36 possiblities.

There is a 2/6=1/3 chance of rolling a 3 or 6 on the first roll. Out of the remaining 2/3 chance, on the second roll, there is a 2/6=1/3 chance of getting a 3 or 6.
2/3*1/3=2/9

$ 2/9\plus{}1/3\equal{} 2/9\plus{}3/9\equal{} 5/9$

There are $ 6\cdot6$ total outcomes. We want $ 3, 6, 9, 12$. To get $ 3$, there are $ 2$ ways, $ (1,2), (2,1)$. To get $ 6$ there are $ 5$ ways, $ (1,5), (5,1), (2,4), (4,2), (3,3)$. To get $ 9$ there are $ 4$ ways, $ (6,3), (3,6), (4,5), (5,4)$. Finally to get $ 12$, there is only $ 1$ way, $ (6,6)$. We add them up \[ 2 \plus{} 5 \plus{} 4 \plus{} 1 \equal{} 12\] As there are a total of $ 36$ outcomes, our answer is $ \boxed{\frac{1}{3}}$.

That's what I did at first also. Read the problem closely- They are not asking for the sum of the 2 outcomes, rather just the probability there is atleast 1 multiple of 3.

but, the answer is 1/3. i believe akhil0422 is right, though the question is oddly worded. could someone explain to me how to do it right (unless akhil0422 is right...)?


If you are given $x<y$ and $x+y<z$ where x, y and z are real numbers, which of the following can be proved to be true?
a. $x^{2}<y^{2}$
b. $z>2x$
c. $3x+y<2z$
d. $(x-y)(x+y-z)>0$
Don't only tell the answer, prove it.

[hide="a"]see althemans[/hide]

[hide="b"]$x<y$
$2x<y+x<z$
$2x<z\implies z>2x$[/hide]

EDIT:Ya i didnt see that one i think i sorta forgot about 0..

[hide="c"]
$x<y$
$3x<y+2x$
$3x+y<2y+2x<2z$
$3x+y<2z$[/hide]
[hide="d"]$x<y$
$\rightarrow x-y<0$

$x+y<z$
$\rightarrow x+y-z<0$

Therefore

$(x-y)(x+y-z)>0$[/hide]

[hide]d)

If $x<y$ then $x-y<0$
If $x+y<z$ then $x+y-z<0$

So multiplying the two inequalities on the right side, we get that

$(x-y)(x+y-z)>0$, so there.[/hide]


number theory
greatest common divisor
relatively prime
number theory proposed

We are given 4 integers with the property that the square of the sum of two is divisible by the product of the other two. prove that there are 3 of the integers that are equal.  :D

so you can divide out by the gcd and the property still holds so that WLOG the gcd is 1, let the four numbers be a, b, c, d. Suppose a, b had some common prime factor p. Then from ad|(b+c)^2, we know p|b+c, so p|c, and similarly p|d, which can't be true, so the numbers are in fact pairwise coprime. Now, for some prime p dividing a, from ab|(c+d)^2, we have p|c+d, so c + d = 0 mod p. Similarly b + d = b + c = 0 mod p, so we can find 2b = 2c = 2d = 0 mod p. So if p is odd, we get p|b, c, d, which isn't true. Thus p is even, so all numbers must be in fact powers of 2. Since we already deduced they are pairwise relatively prime, if there is more than one nontrivial power of two, these two would have a common factor (2), which isn't true, so at least three of them are 1. (the fourth is 1, 2, or 4). So, upon multiplying by the gcd (which we divided out by at the very beginning), the numbers are x, x, x, and (x, 2x, or 4x).


[quote="Vasc"]If $ a,b,c,d\in [\sqrt\frac 3{5},\sqrt\frac 5{3}]$, then

$ 2(\frac a{b} \plus{} \frac b{c} \plus{} \frac c{d} \plus{} \frac d{a})\ge \frac b{a} \plus{} \frac c{b} \plus{} \frac d{c} \plus{} \frac a{d} \plus{} 4$.[/quote]

Isn't this inequality true for all $ a,b,c,d\in [\sqrt\frac 1{2},\sqrt 2]$ ?  :huh:
If we write the inequality in the form:

$ \dfrac{2a \minus{} c}{b} \plus{} \dfrac{2b \minus{} d}{c} \plus{} \dfrac{2c \minus{} a}{d} \plus{} \dfrac{2d \minus{} b}{a}\ge 4$

After we apply the Cauchy-Schwartz inequality, we get that:

$ LHS \equal{} \dfrac{(2a \minus{} c)^2}{b(2a \minus{} c)} \plus{} \dfrac{(2b \minus{} d)^2}{c(2b \minus{} d)} \plus{} \dfrac{(2c \minus{} a)^2}{d(2c \minus{} a)} \plus{} \dfrac{(2d \minus{} b)^2}{a(2d \minus{} b)}\ge \dfrac{(a \plus{} b \plus{} c \plus{} d)^2}{ab \plus{} bc \plus{} cd \plus{} da}$

We're left to prove that $ (a \plus{} b \plus{} c \plus{} d)^2\ge 4(ab \plus{} bc \plus{} cd \plus{} da)\iff (a \minus{} b \plus{} c \minus{} d)^2\ge 0$

[quote="Inequalities Master"] [quote="Vasc"]If $ a,b,c,d\in [\sqrt\frac 3{5},\sqrt\frac 5{3}]$, then

$ 2(\frac a{b} \plus{} \frac b{c} \plus{} \frac c{d} \plus{} \frac d{a})\ge \frac b{a} \plus{} \frac c{b} \plus{} \frac d{c} \plus{} \frac a{d} \plus{} 4$.[/quote]

Isn't this inequality true for all $ a,b,c,d\in [\sqrt\frac 1{2},\sqrt 2]$ ?  :huh:[/quote]
Yes, it is true,  :P   and your proof is very nice. :lol:

Actually, I wanted to post the following inequality.

If $ a,b,c,d\le \frac 5{3}$ are positive real numbers satisfying $ abcd\equal{}1$, then

$ 2(a\plus{}b\plus{}c\plus{}d)\ge \frac 1{a}\plus{}\frac 1{b}\plus{}\frac 1{c}\plus{}\frac 1{d}\plus{}4$.


Okay, so I was playing around with different things and I noticed that both $\verb|\parskip 1em|$ and $\verb|\setlength{\parskip}{1em}|$ do the same thing.
Which is the better "programming" practice? I am thinking if (La)TeX's equivalent of W3C put together a standard which one would fall under "strict" and which under "transitional." And why would that be so?

Ben

\\Edit - I assume that if there is a more correct way known that that will be thrown out, but I thought I might tack it on explicitly. So if there is a more correct way than either of those please share.

Firstly, as far as \parskip is concerned, using \setlength is the usual method as it can cope with more complicated commands like \setlength{\parskip}{1ex plus0.5ex minus0.2ex} which means that the normal exta line spacing is 1ex but can be between 0.8 and 1.5 acording to circumstances.

There are lots of similar examples which work in simple cases but you have to be a lot more careful in more complicated examples such as x^2 which should be x^{2} - the latter is vital for x^{2y} for obvious reasons.

LaTeX (or rather TeX) is now 30 years old so it is bound to develop alternative ways of doing things. There was a major change when LaTeX2.09 became LaTeX 2e so although, for example, [i]\it text[/i] for italics was retained for backwards compatibility, the LaTeX2e method is to use [i]\textit{text}[/i].

Packages are written by lots of individuals who may use their own styles and can clash with one another. On the other hand, the amsmath packages were a huge advance introducing new commands and replacing deficient commands like \eqnarray with a better version \align (see [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url]).

If you want to find good coding practice that works then the nearest you'll get to a standard reference is [url=http://www.amazon.com/Companion-Addison-Wesley-Tools-Techniques-Computer/dp/0201362996/sr=1-1/qid=1161947729/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]The LaTeX Companion (2nd ed)[/url].

Yeah I found it pretty awesome how [url=http://www.tlhiv.org/vitae/vitae.ltx]Troy Henderson[/url] goes like:

[code]
\documentclass{article}
%%%%%%%%%%%
% My name %
%%%%%%%%%%%
\def\name{Troy Lee Henderson, IV}
\def\cv{Curriculum Vitae}
\title{\cv\ of \name}

%%%%%%%%%%%%%%%%%%%%%%
% My contact address %
%%%%%%%%%%%%%%%%%%%%%%
\def\address{
	Department of Mathematical Sciences\\
	252 Thayer Hall\\
	United States Military Academy\\
	West Point, NY  10996}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% My contact phone numbers %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\def\phone{
	Phone: (845) 938-5649\\
	Fax: (845) 938-2409}
...
[/code]

Is this is a standard thing to do, or is it just him?

[quote="chess64"]Is this is a standard thing to do, or is it just him?[/quote]I think this is the work of someone who is comfortable with LaTeX and has a good understanding of it. Pointers are the use of \def rather than \newcommand and the use of .ltx rather than .tex file extension. Many users not as familiar with LaTeX would also use a pre-defined class for their letters and CVs.

[quote="stevem"]. . . commands like \setlength{\parskip}{1ex plus0.5ex minus0.2ex} which means that the normal exta line spacing is 1ex but can be between 0.8 and 1.5 acording to circumstances.[/quote]
How do plus and minus affect the spacing according to circumstances? I haven't seen those commands before.
Thanks for the explanation though. Unfortunately, as a broke college student, I won't be able to purchase a book to help me along any time soon.

You may need a little bit of leeway to prevent having a single line of a paragraph at the bottom of a page or at the top of a new one. Maybe the text has to be fitted into a box or two columns balanced so the bottom of each column are on the same line.

See 6.3.2 of [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url] which is probably the standard online LateX introduction and is regularly updated. Incidentally you'll find a lot more online resources in the Documentation and Help section of the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html]TeX FAQ[/url] page

Sweet, I can't wait to play with that. Maybe I will have to take a look at that Not so short intro after all. The two have already read didn't mention any such thing. Thanks, Stevem!

Ben


It seems that PROMSY is not as popular as the other two. I wonder why. Which one will better prepare me for AIME?

[quote="beta"]I want to go to PROMYS but I don't think they are going to accept me.[/quote]

I think they are.  You have lots of time for the questions, and looking at what you've posted, you should have no problem with the problems.

[quote="Nukular"]
But two weeks before June 1st, as I found out last year, means you better have all 8 problems answered.[/quote]

I don't have complete solutions to all 8 of them.


algebra
polynomial
Putnam
algebra unsolved

let P and Q 2 polynomials wich have the same roots, and lets suppose that P - 1 and Q - 1 have the same roots
prove that P = Q

[hide="solution"]
If they have the same roots, then one is a constant multiple of the other, or Q(x)=kP(x).
By the next condition, P(x)-1 and kP(x)-1 have the same roots. So kP(x)-1 is a multiple of P(x)-1, hence k must be 1.[/hide]

sorry, but they have the same roots, not the same multiplicities of roots !

It followed from abc theorems for polinoms.

This problem comes from Putnam 1956 (B#7).

Pierre.


Suppose that $a,b,c>0$. Prove that \[\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \frac{3}{1+abc}\] Can anyone use the SOS method to solve it ?


I'm pretty sure a rectangle can be cut into only acute triangles, but I can't remember how.

[quote="junggi"][quote="i_like_pie"][quote="wigianto"]Can a rectangle be cut into acute triangles?[/quote]
Plain and simple: no.[/quote]
why can't it be if a square can be?[/quote]
For some reason, I thought the problem said "2 acute triangles."

Yes, it can be. :blush:


Recently I have found a section called "Olympiad resources", but I couldn't figure out how to get there from the Index page... It's not under the Olympiad section... Is there a direct link from the Index page to the Olympiad resources page? :?

Sorry if it's a stupid question... Thanks in advance!

It's quite a strange question indeed, but anyway, here it is:

[url]http://www.mathlinks.ro/Forum/resources.php[/url] :D

[quote="frt"]Recently I have found a section called "Olympiad resources", but I couldn't figure out how to get there from the Index page... It's not under the Olympiad section... Is there a direct link from the Index page to the Olympiad resources page? :?[/quote]

Look at the upper right corner of any ML page. There is a button called "resources". Admittedly, few people noted this, as the button formerly just linked to the Download Section.

  darij

PS. My provider seems to be strangely dysfunctioning today, so I had to trick it in order to access ML. Let's see whether this post appears...

[quote="frt"]I see... thanks. :oops:

Is there a way to get there from[url=http://www.artofproblemsolving.com/Forum/index.php]AoPS Index[/url]? I usually use AoPS, not Mathlinks...  :(

Sorry, but thanks a lot! :)[/quote]

There will be very soon.  I plan to make a button for it Friday.

There should be some sort of announcement with a clarification on this page later. (Valentin will have the timeframe on that.)  :)

There is now a 'Contests' button atop the AoPS view of forum (next to Blogs).  This will take you to the contest collection page.


Could someone please explain the sp3d1 and sp3d2 hybridizations and their effects in bonding? Thanks!

Well, as I'm sure you know, we can think about $sp^{3}$ orbitals, for example, as mixture of and $s$ and 3 $p$ orbitals.  This gives us a total of 4 equivalent hybrid orbitals, which allow us to make 4 bonds (including lone pairs).  The same exact thing happens when we start mixing in $d$'s.  Five $sp^{3}d$ orbitals will be present in a molecule with five bonds (inc. lone pairs) to the central atom (e.g. $PCl_{5}$), making a trigonal bipyramidal shape, and six $sp^{3}d^{2}$ orbitals will be present in a molecule with six bonds to the central atom (e.g. $XeF_{6}$), making an octahedral shape.

Incidentally, all of the above is a lie.  If you look back into the history of chemistry, I believe it was Linus Pauling who developed the idea of these "hybrid orbitals," but these are merely an approximation of reality.  I believe that the hybridization model usually works fairly well up to $sp^{3}$, but once we start mixing in $d$ orbitals to account for expanded octets, calculations using the valence bond model start to break down.

Off the top of my head, one qualitative difference between the hybridization picture and the molecular orbital (a much more general and applicable model) of these molecules is that hybridization implies that row 2 elements cannot bond to beyond 4 atoms because they do not have $2d$ orbitals, while in reality, the problem is that these elements are too small to accomodate more than four atoms.


inequalities
quadratics
search
analytic geometry
graphing lines
slope
symmetry

[hide="Approach 1"] By Jensen's $(x-a)^{2}+(b-x)^{2}\ge 2 \left( \frac{b-a}{2}\right)^{2}$.  The minimum occurs when $x-a = b-x \implies x = \frac{a+b}{2}$.  [/hide]
[hide="Approach 2"] By QM-AM we also happen to have $(x-a)^{2}+(b-x)^{2}\ge 2 \left( \frac{b-a}{2}\right)^{2}$.  Then proceed as above. [/hide]
[hide="Approach 3"] $y = 2x^{2}-2(a+b)x+a^{2}+b^{2}= 2 \left( x-\frac{a+b}{2}\right)^{2}+a^{2}+b^{2}-\frac{(a+b)^{2}}{2}$ which clearly has its minimum when $x = \frac{a+b}{2}$. [/hide]
[hide="Approach 4"] Let $f(x) = (x-a)^{2}+(x-b)^{2}$.  Now, $2 \sqrt{f(x)}$ is the sum of a distances of a point $(x, x)$ from $(a, b)$ and from $(b, a)$.  By triangle inequality the minimum clearly occurs when $(x, x)$ is between the two points, that is, at $\left( \frac{a+b}{2}, \frac{a+b}{2}\right)$. [/hide]

[quote="da ban man"]Let y=(x-a)^2+(x-b)^2 with a and b constats.  For what value of x is y a minimum in terms of a and b?[/quote]

[hide="Completing the square"]$y = (x-a)^{2}+(x-b)^{2}$

$y = (x^{2}-2ax+a^{2})+(x^{2}-2bx+b^{2}) = 2x^{2}-2(a+b)x+(a^{2}+b^{2})$

$y = 2\left( x-\frac{a+b}{2}\right)^{2}+\textrm{unimportant term}$

Now, it is clear that the minimum will occur when $x = \frac{a+b}{2}$[/hide]

[quote="p4fn2w"][quote="t0rajir0u"]
[hide="Approach 2"] By QM-AM we also happen to have $(x-a)^{2}+(b-x)^{2}\ge 2 \left( \frac{b-a}{2}\right)^{2}$.[/hide]
[/quote]

Question please - can you explain more about the quadratic mean, QM? After doing a quick search, I find that QM = $\sqrt{\frac{\sum a_{i}^{2}}{n}}$, so shouldnt QM-AM show that (Im assuming that AM$\ge$QM):
$\frac{(x-a)^{2}+(b-x)^{2}}{2}\ge \sqrt{\frac{(x-a)^{4}+(x-b)^{4}}{2}}$?
How did you get from this to the elegant equation that you have? Is this a case of clever variable manipulation (or am I completely wrong about QM)?[/quote]I think you got the other way around.... should be $\text{QM}\ge\text{AM}$....

Well, you don't necessarily need to throw in what you have into the AM side.

He did $(x-b)^{2}=(b-x)^{2}$, and then said, for quantities $x-a$ and $b-x$, it must be true that:

$\frac{(x-a)+(b-x)}{2}\leq \sqrt{\frac{(x-a)^{2}+(b-x)^{2}}{2}}=\sqrt{f(x)/2}$

The x's cancel out on the left, and you get:

$\frac{b-a}{2}\leq \sqrt{f(x)/2}$

$\left( \frac{b-a}{2}\right)^{2}\leq f(x)/2$

$2\left( \frac{b-a}{2}\right)^{2}\leq f(x)$

Which is what he has

I'm not going to present this in a fully rigorous way, but its good for fast thinking if it's a step towards solving a larger problem under time pressure, so its worth mentioning. It looks long, but it actually goes through one's head very, very quickly.

[hide="Approach 6"]
Plot (not necessarily draw out, but consider the graphs of) $A=(x-a)^{2}$ and $B=(x-b)^{2}$ (suppose for now that b>a), which intersect at $x=\frac{a+b}{2}$. Note that the system is totally symmetric about this axis. Therefore, at this point, the rate at which A is increasing is equal to the rate at which B is decreasing. Say the slope of A is $k$ and the slope of B is $-k$.

If you move to the right, A gets steeper, and its rate of increase is higher: call this $k+m$ for some positive $m$. Also, the rate of decrease of B decreases, since B gets shallower. Let's call the rate of increase of B $-k+n$ for some positive $n$.

Now, the net rate of increase of the sum has risen from $k+(-k)=0$ to $(k+m)+(-k+n)=m+n$, and thus the sum is increasing to the right of the symmetry point.

If you do a similar case for the left of the symmetry point, you find that the sum is decreasing to the left.

Overall, the sum decreases till it gets to the symmetry point, and the increases again. Clearly, the minimum value is at the symmetry point.

[hide="More"]You can more accurately quantify the gradient of increase of the functions with calculus, but at a glance, it's not necessary in symmetric situations. The same argument can be used to find the minimum of, say, $\log_{2}(4^{x}+4^{-x})$

Since $\log$ is strictly increasing, we have to minimize the inside, and by similar argument, the minimum happens at the symmetry point, $x=0$. The minimum is $\log_{2}(4^{0}+4^{-0})=\log_{2}(1+1)=1$

Be careful to make sure that the system is in fact symmetric and that the concavities match before using this.[/hide]
[/hide]


At the end of $ 1994$, Walter was half as old as his grandmother. The sum of the years in which they were born was $ 3838$. How old will Walter be at the end of $ 1999$?

$ \textbf{(A)}\ 48 \qquad \textbf{(B)}\  49\qquad \textbf{(C)}\  53\qquad \textbf{(D)}\  55\qquad \textbf{(E)}\ 101$

[hide="hint"]Try writing two equations with the years in which the two people were born as the variables.[/hide]

[hide]We know that the sum of 1994 subtracted by walter age and 1994 subtracted by grandma's age should be 3838;

3838 = (1994-w) + (1994-g)
3838 = 3988 -w-g
150 = w+g
We also know 2walters = 1grandma:
2w = g
150 = w + 2w
50 = w
Now we add 5 so the answer is D[/hide]

[hide="Another way"]Let W be the year in which Walter was born and G be the year in which his grandmother was born.

The 2 equations are $1994-W=\frac12(1994-G)$ and $W+G=3838$. Thus we have $G=3838-W$, which gives $994-W=997-1919+\frac12W$, which results in $W=1944$. At the end of 1999, Walter will be $1999-1944=55\implies\boxed D$[/hide]


[quote="Armo"]In a triangle $b+c=2R\sqrt{3}$.Prove that $\frac{\sin{2A}}{\sin{3A}}>\frac{2}{3}$.
Hereby b, c are the sides of the triangle, R is circumradius.[/quote]

From b+c=2R :sqrt: 3 it derives sinB + sinC=  :sqrt: 3.Therefore 2cos(A/2)cos[(B-C)/2]=  :sqrt: 3<= 2cos(A/2)
so cos(A/2)>=  :sqrt: 3/2
cosA= 2cos 2 (A/2) - 1>= 2*3/4 - 1= 1/2 so 1/2<=cosA<1
We get Sin2A/sin3A = 2cosA/(4cos 2 A - 1) ... :)
...try to finish it yourself :P
...there's not so much to do
Artiona

$\frac{\sin{2A}}{\sin{3A}}=\frac{2\cos{A}}{4\cos^{2}A-1}$.From $b+c=2R\sqrt{3}$ we get $A\in(0;\frac{\pi}{3})$. Consider the function $f(x)=\frac{x}{x^{2}-1}$ which is decreasing ... :lol:


Consider the sequence of numbers $a_{1}, a_{2},...$ defined such that $a_{1}=a_{2}=1$, $a_{3}=251$ and for all $n \ge 3$

$a_{n+1}={{2007+a_{n}a_{n-1}}\over{a_{n-2}}}$.

Prove that all of the numbers in this sequence are integers.

What is this? I don't know.
[quote="rjt"]

$a_{n+1}={2007+a_{n}a_{n-1}}\over{a_{n-2}}$.

[/quote]

Pretty sure that's what he meant.

[hide="guess"] The sequence appears to follow the recursion

$a_{2n}= 9 \cdot a_{2n-1}-a_{2n-2}$

$a_{2n+1}= 252 \cdot a_{2n}-a_{2n-1}$. [/hide]


For any real numbers x, y and z prove inequality:
$ (x^2 \plus{} y^2 \minus{} z^2)(x^2 \plus{}
\plus{} z^2 \minus{} y^2)(y^2 \plus{} z^2 \minus{} x^2) \le (x \plus{}y\minus{}z)^2(x\plus{}z\minus{}y)^2(y\plus{}z\minus{}x)^2 .$

[quote="Erken"][quote="NguyenDungTN"]For any real numbers x, y and z prove inequality:
$ (x^2 \plus{} y^2 \minus{} z^2)(x^2 \plus{} \plus{} z^2 \minus{} y^2)(y^2 \plus{} z^2 \minus{} x^2) \le (x \plus{} y \minus{} z)^2(x \plus{} z \minus{} y)^2(y \plus{} z \minus{} x)^2 .$[/quote]
This nice inequality comes from Poland NO of some year and it was discussed on ML for sure...
I'll try to find a link.
Edited:
Here is the link:
http://www.mathlinks.ro/viewtopic.php?search_id=68269355&t=18908[/quote]

Thank you, it was appeared firstly in Kvant 1977! ! :) 15 years later, it were Poland MO, too!


Fie a>0 si $f:[ \frac{1}{a},a] \to [ \frac{1}{a},a]$ bijectiva.Daca $f^{-1}(x)= \frac{1}{f(x)}$,pentru orice $x \in [ \frac{1}{a},a]$,aratati ca:
a)f are cel putin un punct de discontinuitate;
b)daca f este continua in 1,atunci f are o infinitate de puncte de discontinitate;
c)exista o functie f care verifica conditile din enunt si are un numar finit de puncte de discontinuitate.

a)Daca presupunem ca f este continua,avem ca f monotona.
Dar din enunt avem faptul ca $f$ si $f^{-1}$ au monotoni diferite.Contadictie
Astept  ajutor pt. b).c).

silviu,pentru b) se aplica punctul a). ;)


conics
ellipse
geometry
3D geometry
sphere
calculus
calculus computations

What are the semimajor and semiminor axes of the ellipse that contains two unit circles in its interior and has:
1) Minimum area?
2) Minimum arclength?
What are the semi-axes of the ellipsoid that contains two unit spheres in its interior and has:
3) Minimum volume?
4) Minimum surface area?


Two of the four interior angles of a particular parallelogram are 130 degrees each. What is the number of degrees in each of the other two interior angles?

Interior angles adjacent to obtuse angles in a parallelogram are always supplementary with them.

So each of the other two interior angles measures $ 180\minus{}130\equal{}\boxed{50}$ degrees.

parellelogram is 360 degrees 2 angles are 130 two together =260  subtract 360-260=100/2=50

[hide="the parellelogram is here or is it ?"]50 degrees it is here !! : )[/hide]

[size=50][hide="can any of yall read this " "][color=#FFFFFF][size=200]YOU CANT SEE ME OR CAN YOU ????? HA HA HA : )[/size][/color][/hide][/size]

[hide="rating"][size=200][color=#408000][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH[/u][/i][/b][/color][/size][/hide]

:weightlift: --> :bruce:-->  :starwars:-->  :stretcher:

you excercise -->you get strong --> got fight--> and come back dead : (


1.Determine the value of the expression $ \sqrt {5 \plus{} \sqrt {5^2 \plus{} \sqrt {5^4 \plus{} \sqrt {5^8 \plus{} ...}}}}$.
2. We are given a regular tetrahedron ABCD all of sides who have length four. A plane passes through this tetrahedron, intersecting edges AC, AD, BC, and BD at points W, X, Y, and Z. If AW=2, AX=3, and BZ=1, then determine BY.

[hide="hint for 1"]Let that expression equal $ x$.  Then $ x \equal{} \sqrt{5 \plus{} x\sqrt{5}}$.[/hide]

Argh. How did I not notice that?

Ok, so I solved #1, just need #2 now.

Uh...bump? Since this hasn't gotten a reply (for #2), should I repost in Pre-Olympiad?


The problem is that you have to find how many solutions and what they are to the following problem.

abc
+def
____
ghi

Where (a,b,c,d,e,f,g,h,i) represent one of the integers from 1 to 9 distinctly.

Ex. 735
+246
_____
981

Each integer 1 to 9 is used once. Is there a way to solve this problem without brute forcing it? I'm pretty stuck... Thanks for the help.

@ [b]barce11[/b]: Please refrain from posting the same question in multiple forums. One question is enough. If you are not getting the answer out of this forum, try somewhere else. We are here to help and help is a choice, not an obligation.

@ everyone else: Continue discussion in the other topic in HSB [url]http://www.mathlinks.ro/viewtopic.php?t=177525[/url]

Topic should be [u]Locked[/u]


Sorry about my question.
I would like to know what is the Gergonne point.

Hmm, as I remember(maybe I am wrong) if we have a triangle ABC and it's incircle touches BC,CA,AB at D,E,F then Gergonne's point is the intersection point of AD,BE and CF.

Yeah, that's right.  (A quick proof of the concurrence follows directly from Ceva's & equal tangents.)

[quote="Armo"]Strange question. :?[/quote]
i have seen few problems related to gergonne point...

But also there is "[b][color=darkblue]Gergonne's relation[/color][/b]"
[color=darkblue][b]In a triangle $ ABC$ consider 3 ceviens, $ AA',BB',CC'$ that are joining in the same point $ P$,$ A' \in BC$,$ B' \in AC$,$ C' \in AB$.Prove that: \[ \frac{A'P}{AA'}\plus{}\frac{B'P}{BB'}\plus{}\frac{C'P}{C'C}\equal{}1\][/b][/color]

[color=red][b]Lemma(well known)[/b][/color] [b][color=darkblue]Let $ ABC$ be a triangle and $ M$ a point in the interior .Let $ d_1,d_2,d_3$ be the projections of $ M$ on $ BC,AC,AB$ and $ h_1,h_2,h_3$ the projections of points $ A,B,C$ on the sides $ BC,AC,AB$.Prove that \[ \frac{d_1}{h_1}\plus{}\frac{d_2}{h_2}\plus{}\frac{d_3}{h_3} \equal{}1\][/color][/b]
 \[ \frac{[AMC]}{[ABC]}\equal{}\frac{d_2 \cdot AC}{2} \cdot \frac{2}{h_2 \cdot AC}\equal{}\frac{d_2}{h_2}\]
\[ \frac{[BMC]}{[ABC]}\equal{}\frac{d_1 \cdot BC}{2} \cdot \frac{2}{h_1 \cdot BC}\equal{}\frac{d_1}{h_1}\]
\[ \frac{[AMB]}{[ABC]}\equal{}\frac{d_3 \cdot AB}{2} \cdot  \frac{2}{h_3 \cdot AB}\equal{}\frac{d_3}{h_3}\]

_________________________________________
$ 1\equal{}\frac{d_1}{h_1}\plus{}\frac{d_2}{h_2}\plus{}\frac{d_3}{h_3}$

[color=darkblue][b]Proof of Gergonne's relation[/b][/color]
Let $ d_1,d_2,d_3$ and $ h_1,h_2,h_3$ be the segments with the same property as in [color=red][b]Lemma[/b][/color]
In the triangle $ A'AB$,$ d_1 \parallel{} AB \Rightarrow \frac{d_1}{h_1}\equal{}\frac{A'P}{A'A}$
Similar, $ \frac{d_2}{h_2}\equal{}\frac{CP}{C'P}$ and $ \frac{d_3}{h_3}\equal{}\frac{B'P}{B'B}$


Hi everyone,

Could someone please help me on this?

The problem says "Find a statement form, S, equivalente to -(P V Q) and show that it is logically equivalent by constructing the truth table for "S if and only if -(P V Q)" and showing that this statement form is tautology."

My attempt:

P      Q         -P V-Q

T      T           T
T      F           F
F      F            F
F      T           T

The statement is "I'll take you to the movies if and only if you do not get bad grades or do not do your homework"

Would this work?

Thank you in advance

There are some ambiguities here. The statement $ S$ does not have $ \neg (P\vee Q)$ in it. It has $ \neg P\vee \neg Q$ if $ P$ is "You get bad grades" and $ Q$ is "You do your homework."

A tautology for $ \neg (P \vee Q)$ is $ \neg P \wedge \neg Q$ and a tautology for $ \neg P \vee \neg Q$ is $ \neg (P \wedge Q)$.

To show tautologies with a truth table, you need the two in question in the table, and there truth values should match for every combination of $ P$ and $ Q$ 's values (which you set the first to columns up fine).

[quote]A tautology for  is  and a tautology for  is .

To show tautologies with a truth table, you need the two in question in the table, and there truth values should match for every combination of  and  's values (which you set the first to columns up fine).[/quote]

Isn't a tautology the final column in the truth table that consists of all T's? Then are those two just called equivalents? Would I then do the following:

P               Q                 -(P V Q)       -P V -Q

T               T                  T                   T
F               F                   T                   T
T               F                   T                   T
F               T                   T                    T

Is this correct? Thank you very much

OK, so you could do it either way. If you match truth values then you have shown equivalence and not tautology, this is true. But you can construct the tautology in the final column by making it "column 3 if and only if column 4," then you will have T's all down that last column.

Note: the table you gave should have for column 3: FTFF, and for column 4: FTTT.


geometry
trigonometry
arithmetic sequence
trig identities
Law of Cosines
Law of Sines

Let $abc$ be a triange.
suppose $a^2,b^2,c^2$ are in arithmetic progression.
Show that $\text{cotan A, cotan B, cotan C}$ are also in arithmetic progression

[hide]
using law of cosines and law of sines,
\begin{eqnarray*} \cot A &=& \frac{\cos A}{\sin A}\\ &=& \displaystyle \frac{\displaystyle \frac{b^2+c^2-a^2}{2bc}}{\displaystyle \frac{a}{2R}}\\ &=& \displaystyle \frac{-a^2+b^2+c^2}{\displaystyle \frac{abc}{R}} \end{eqnarray*}
Similarly,
$\cot B = \displaystyle \frac{a^2-b^2+c^2}{\displaystyle \frac{abc}{R}}$
$\cot C = \displaystyle \frac{a^2+b^2-c^2}{\displaystyle \frac{abc}{R}}$

Since $\frac{abc}{R}$ is a fixed value, it is sufficient to prove that the numerators are in arithmetic progression.

Let $a^2=x-k$, $b^2=x$, $c^2=x+k$
Then
$-a^2+b^2+c^2=x+2k$
$a^2-b^2+c^2=x$
$a^2+b^2-c^2=x-2k$

QED

[/hide]


A number $ x$ is 3 larger than its reciprocal.  What is the value of $ (x\minus{}\frac{1}{x})^4$?

Since the problem says so, $ x\minus{}\frac{1}{x}\equal{}3$.  Thus, we have $ 3^4\equal{}\boxed{81}$

$ x-\frac{1/x}$ is 3 (stated from the first sentence), so $ (x-\frac{1}{x})^4=3^4=81$.


Hi, I need a help with some conditional probability...

Let P(X|Y) denote a probability of the phenomenon X supposing that the phenomenon Y has already occured. Prove that there aren't any phenomenons A, B and descompositions of the set of all basic phenomenons C_1, ..., C_m and D_1, ..., D_m  such that for each i = 1, ..., m  we have P(A|C_i) < P(B|D_i) and P(A) > P(B).

Thanks.

Did I miss something or is your statement untrue?

Consider the following experience : You throw a fair cubic dice. If you get 1,2,3,4 or 5 you choose a ball into an urn U_1 which contains 1 black ball and 2 white balls (with uniform probability). If you get 6 then, you choose a ball into an urn U_2 which contains only black balls.

Now consider the event A 'the chosen ball is white' and B 'the chosen ball is black'.
Let's partition the universe into C_1 = D_2 = 'the ball comes from U_1' and C_2 = D_1 = 'the ball comes from U_2'.

We then have : p(C_1) = p(D_2) = 5/6 and p(C_2) = p(D_1) = 1/6.
p(A/C_1) = 2/3 and p(B/D_1) = 1.
p(A/C_2) = 0 and p(B/D_2) = 1/3.
Thus p(A/C_1) < p(B/D_1) and p(A/C_2) < p(B/D_2).

In another hand :
p(A) = p(C_1)p(A/C_1) + p(C_2)p(A/C_2) = 5/9
and
p(B) = p(D_1)p(B/D_1) + p(D_2)p(B/D_2) = 1/6 + 5/18 = 4/9
so that p(A) > p(B).

Pierre.

Thanks, you're right and the question is wrong. But if someone doesn't know that the question is wrong, it isn't so easy to find a contraexample as you - it isn't so obvious..., so I've been trying to prove it and mostly it's very difficult to prove that some untrue theorem is valid... :)

Thanks.

Well, I didn't know too before trying to see how it could be true for m = 2...

Pierre.


[hide] Firstly, multiply everything by 5.  We have that $20logm+5logn=logw-log32.$  Since $20logm=log(m^{20}), 5logn=log(n^{5})$, and $logw-log32=log(\frac{w}{32})$, we have that $log(n^{5})+log(m^{20})=log(\frac{w}{32}).$  Since logm+logn=logmn, we have that $log(n^{5})+log(m^{2}0)=log(m^{20}*n^{5})=log(\frac{w}{32})$.  This means that $m^{20}*n^{5}=w/32\Rightarrow w=32*m^{20}*n^{5}$.  [/hide]

[quote="Ignite168"]If $4log(m)+log(n)=\frac{log(w)-log(32)}{5}$ express w in terms of m and n.[/quote]

[hide]$5 log m^{4}n=log \frac{w}{32}$

$log m^{20}n^{5}=log \frac{w}{32}$

$w=32m^{20}n^{5}$[/hide]

[hide]
I haven't opened other hide tags yet so it might repeat :roll:
20log m +5 log n =log 32 = log w, w=32 n^52^20[/hide]

edit: yup exactly the same


[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative real numbers.Prove that:
$ 2(a^2 \plus{} b^2 \plus{} c^2) \ge a\sqrt [3]{7b^3 \plus{} c^3} \plus{} b\sqrt [3]{7c^3 \plus{} a^3} \plus{} c\sqrt [3]{7a^3 \plus{} b^3}$[/quote]
Actually, the more general inequality holds:

If $ a,$ $ b,$ $ c,$ $ k$ are nonnegative real numbers, then
$ a^2 \plus{} b^2 \plus{} c^2 \ge a\sqrt[3] {\frac {kb^3 \plus{} c^3}{k \plus{} 1}} \plus{} b\sqrt[3] {\frac {kc^3 \plus{} a^3}{k \plus{} 1}} \plus{} c\sqrt[3] {\frac {ka^3 \plus{} b^3}{k \plus{} 1}}.$

[quote="can_hang2007"][quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative real numbers.Prove that:
$ 2(a^2 \plus{} b^2 \plus{} c^2) \ge a\sqrt [3]{7b^3 \plus{} c^3} \plus{} b\sqrt [3]{7c^3 \plus{} a^3} \plus{} c\sqrt [3]{7a^3 \plus{} b^3}$[/quote]
Actually, the more general inequality holds:

If $ a,$ $ b,$ $ c,$ $ k$ are nonnegative real numbers, then
$ a^2 \plus{} b^2 \plus{} c^2 \ge a\sqrt {\frac {kb^3 \plus{} c^3}{k \plus{} 1}} \plus{} b\sqrt {\frac {kc^3 \plus{} a^3}{k \plus{} 1}} \plus{} c\sqrt {\frac {ka^3 \plus{} b^3}{k \plus{} 1}}.$[/quote]

Actually, the more general inequality holds:

If $ a,$ $ b,$ $ c,$ $ k$ are nonnegative real numbers, then
$ a^2 \plus{} b^2 \plus{} c^2 \ge a\sqrt[3] {\frac {kb^3 \plus{} c^3}{k \plus{} 1}} \plus{} b\sqrt[3] {\frac {kc^3 \plus{} a^3}{k \plus{} 1}} \plus{} c\sqrt[3] {\frac {ka^3 \plus{} b^3}{k \plus{} 1}}.$

?

[quote="can_hang2007"]Sorry hedeng. I got mistake when typing the general result. I have edited my post.[/quote]

Can you upload what you do for solution?

thank you very much.

xzlbq played a Chinese name for you,灿红.

 :D

[quote="can_hang2007"]If $ a,$ $ b,$ $ c,$ $ k$ are nonnegative real numbers, then
$ a^2 \plus{} b^2 \plus{} c^2 \ge a\sqrt [3] {\frac {kb^3 \plus{} c^3}{k \plus{} 1}} \plus{} b\sqrt [3] {\frac {kc^3 \plus{} a^3}{k \plus{} 1}} \plus{} c\sqrt [3] {\frac {ka^3 \plus{} b^3}{k \plus{} 1}}.$[/quote]
Using Holder's Inequality, we have
$ \left( \sum a\sqrt [3]{kb^3 \plus{} c^3}\right)^3 \le \left[ \sum a(kb^3 \plus{} c^3)\right]\left( \sum a \right)^2 .$
It suffices to prove that
$ \left(k\sum ab^3 \plus{} \sum a^3b \right) \left( \sum a\right)^2 \le (k \plus{} 1)\left( \sum a^2\right)^3.$
Using Vasile's Inequality, we have
$ \sum ab^3 \le \frac {1}{3}\left( \sum a^2\right)^2, \quad \sum a^3b \le \frac {1}{3}\left( \sum a^2\right)^2.$
Also, by the Cauchy-Schwarz Inequality,
$ \left( \sum a\right)^2 \le 3\sum a^2.$
Combining these inequalities, the conclusion follows. Equality holds if and only if $ a \equal{} b \equal{} c.$
[quote="hedeng123"]
xzlbq played a Chinese name for you,灿红.
 :D[/quote]
I used a translator to translate this and it gives 灿红 = Can Red? Actually, the name "Can" of mine has its meaning in English is "meticulosity". ;)


I was watching

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#layout

and I was wondering, what's the standard layout?

Wat do you mean by standard? Different classes like article, report, book clearly use different default layouts and of course different countries use different paper sizes and layouts. Have a look at [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=geometry]the geometry package manual[/url], particularly 6 Default Settings. The geometry package is the best way of setting page layout.


Hello,

Find all positive integers $ x,y,z$ satisfying simultaneously two conditions.

i) $ \frac{x\minus{}y\sqrt{2009}}{y\minus{}z\sqrt{2009}}$ is a rational number .

ii) $ x^2 \plus{} y^2\plus{}z^2$ is a prime .

Let $ n$ be the integer for which $ \frac{x\minus{}y\sqrt{2009}}{y\minus{}z\sqrt{2009}} \equal{}n$. So $ x\minus{}y\sqrt{2009}\equal{}ny \minus{} nz\sqrt{2009}$, $ x\minus{}ny \equal{} (y\minus{}nz)\sqrt{2009}$. LHS is rational, so $ y\minus{}nz\equal{}0$, $ x\equal{}ny$. So we have $ y\equal{}nz,x\equal{}n^2z$.

$ x^2\plus{}y^2\plus{}z^2\equal{}n^4z^2\plus{}n^2z^2\plus{}z^2\equal{}(n^2\minus{}n\plus{}1)(n^2\plus{}n\plus{}1)z^2$. Thus $ z\equal{}1,n^2\minus{}n\plus{}1\equal{}1$, i.e., $ n\equal{}1$.

$ \boxed{x\equal{}y\equal{}z\equal{}1}$


analytic geometry
combinatorics proposed
combinatorics

Given 37 integer points in $\mathbb{R}^3$, no three of them are collinear. Prove that there are three points, those centroid is an integer point.

[i](2 points)[/i]

There are $5$ points with equal first and secound coordinates modulo $3$ respectively, and among these $5$, there are $3$ whose last coordinate sum a multiple of $3$, they work!, because among five integers, we have $3$ different residues modulo $3$, or $3$ equal residues modulo $3$.


abstract algebra
superior algebra
superior algebra unsolved

give example $ R$-module $ M$ where $ M$ don't contain maximal submodules

Your answer can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275696


Find the Wronskian $ W(t)$ of two solutions of the following differential equation without solving the equation
$ x^{2}y''\plus{}xy'\plus{}(x^{2}\minus{}v^{2})y\equal{}0$.

[quote="projektnemesis"]Find the Wronskian $ W(t)$ of two solutions of the following differential equation without solving the equation
$ x^{2}y'' \plus{} xy' \plus{} (x^{2} \minus{} v^{2})y \equal{} 0$.[/quote]

Use Liouville-Ostrogradsky's formula.

Which is to say, as I mentioned in another thread, that $ W$ satisfies the differential equation $ W'\plus{}\frac1x W\equal{}0.$


number theory
relatively prime
AIME
AIME 2000

A sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50}=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

[hide]$x_{k}= \sum x_{i}-x_{k}-k \Rightarrow 2x_{k}= \sum x_{i}-k$. Summing the realtions over all $k$ between $1$ and $100$ we get $2\sum x_{i}= 100\sum x_{i}-\frac{101 \cdot 100}{2}\Rightarrow \sum x_{i}= \frac{2525}{49}$. Hence $x_{50}= \frac{\frac{2525}{49}-50}{2}= \frac{75}{98}$ and so $m+n = 173$.[/hide]

my solution is the same as Andreas!! Nice!

Let $s$ be the sum. Clearly, $x_k=(s-x_k)-k$ so $$x_k=\frac{s-k}{2}.$$ Summing this over all $k$ gives $$s=50s-2525\rightarrow s=2525/49.$$ Finally we have $$x_{50}=\frac{s-50}{2}=\frac{75}{98}\rightarrow 173.$$

Let $x_1+x_2+\cdots+x_{100}=s$.
Then, we can write $x_k=(s-x_k)-k\Rightarrow x_k=\frac{s-k}{2}$, for $1\le k\le100$.
We add up all the hundred equations obtained. On left hand side we get $s$, and on right hand side we get $50s-\frac{1+2+\cdots+100}{2}=50s-2525$.
So, $s=50s-2525\Rightarrow s=\frac{2525}{49}$.

This gives $x_{50}=\frac{s-50}{2}=\frac{\frac{2525}{49}-50}{2}=\frac{2525-2450}{98}=\frac{75}{98}=\frac{m}{n}$.

This gives $m=75,n=98\Rightarrow m+n=75+98=\boxed{173}$.

[hide=AIME PYQ]This is also AIME I 2000 P10.[/hide]


28. Each point on the four sides of a 1m  1m square is coloured one of  colours so that no two points that are exactly 1m apart are coloured the same. What is the smallest  for which such a colouring can be made?

30. A positive integer is equal to the sum of the squares of its four smallest positive divisors. What is the largest prime that divides this positive integer?

You already posted number 30 in getting started. Why are you posting it again?


function
calculus
derivative
analytic geometry
algebra
domain
integration

Hi all, hehe embarassing that my first post needs to be a cry for help but, I do need help, appreciate it if someone could take a look:

[img]http://math.hartford.edu:8200/webwork2_course_files/M144Austin/tmp/gif/nkern-4807-setchap_4sec3prob1image1.png[/img]

The graph above is the derivative f(prime) of a function f .
Give the x-coordinate of the point where f has a maximum:
Give the x-coordinate of the point where f has a minimum:
Give the x-coordinate of the second inflection point of f

Follow the graph of $f'$ from left to right:
If $f'(x)$ turns from positive to negative when $x$ passes $a$, then $f$ has a maximum at $a$.
If $f'(x)$ turns from negative to positive when $x$ passes $a$, then $f$ has a minimum at $a$.
If $f'(x)$ just touches $0$ without going to the other side, that's an inflection point for $f$.

I don't know about [b]second[/b] inflection point, though.  :huh:

The graph of $f$ should have inflection points at $x=2,4,$ and $6:$ the max and min values of $f'.$ The one at $x=4$ is the coincidence of an inflection point coinciding with a critical point; the other two are more typical "on the slant" inflection points.

I thought that the inflection point is a point when a curve changes its direction of concavity, so wouldn't (20,2) be about the first inflection point? and (20,6)

and um...what would i define as [img]http://www.mathlinks.ro/Forum/latexrender/pictures/0cc175b9c0f1b6a831c399e269772661.gif[/img] ?

i thought that the answer would be:
20
0
20
but this seems to be incorrect

thanks btw for the help :)

Well, I got 2 out of 3 right. Should be enough for D+ in Calc 1.

Note that $f(2)\ne 20.$ Well, to be exact, we don't know what $f(2)$ is, but whatever it is, it's certainly not the same as $f(6).$

We know that $f'(2)=f'(6)=20.$

We do have inflection points at $2,4,6.$ About all we can say for sure about the values of $f$ as those points is that $f(2)<f(4)<f(6).$ $f$ has a minimums at $x=1$ and a maximum at $x=7.$ We can add that $f(1)<f(2)<f(4)<f(6)<f(7).$ In fact we can add that $f$ is increasing on the interval $[1,7].$ But we still don't know any values of $f.$

Note: when mlok said "$a$", he wanted you to consider each of $2,4,$ and $6$ in turn as possible values of $a.$

Oooh Oooh!!! I understand now, thanks alot :)

hehe and one more quick question while on the topic...

does:

[img]http://math.hartford.edu:8200/webwork2_course_files/M144Austin/tmp/gif/nkern-3082-setchap_4sec3prob3image1.png[/img]

even have an inflection point?

Edit: shown is the graph of [img]http://www.mathlinks.ro/Forum/latexrender/pictures/7c32e30c61841f413089ffe3ab242947.gif[/img]

No, at least not within the domain shown. Apparently concave up throughout.

At least that's the answer if you're showing me the graph of $f.$ If you're showing me something else, then the answer would change.

[quote="Kelth"]Oooh Oooh!!! I understand now, thanks alot :)

hehe and one more quick question while on the topic...

does:

[img]http://math.hartford.edu:8200/webwork2_course_files/M144Austin/tmp/gif/nkern-3082-setchap_4sec3prob3image1.png[/img]

even have an inflection point?

[color=red]Edit: shown is the graph of [/color][img]http://www.mathlinks.ro/Forum/latexrender/pictures/7c32e30c61841f413089ffe3ab242947.gif[/img][/quote]

Yes, the graph has a local minimum at $x\approx 0.3$, which means that $f^\prime$ has a horizontal tangent at $x\approx 0.3$, i.e. we know that the derivative of $f^\prime$ is zero there, that is we know that $f^{\prime\prime}(0.3) =0$ which is an inflection point (IP) of $f$. And since $f^\prime$ goes from decreasing to increasing there, we know that the IP the $f$ goes from concave down (CD) to concave up (CU) there. Here is a plot I made at [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=graphs&s2=equations&s3=basic]www.quickmath.com[/url] thought was pretty close the the one you have of $f^\prime$ ([color=red]red[/color]) and which also includes $f^{\prime\prime}$ ([color=blue]blue[/color]) and a possible $f$ ([color=green]green[/color]), namely, the one for which the constant of integration is 7:

[url=http://ezimagecenter.com/out.php/i188_MaxMinofafunctionfromthederivativeofafunction.JPG][img]http://ezimagecenter.com/out.php/i188_MaxMinofafunctionfromthederivativeofafunction.JPG[/img][/url]

You'll get it  :showoff:


The focus of a parabola is also a focus of the ellipse defined by $4x^{2}+9y^{2}=36$. One end of the latus rectum of the parabola coincides with the vertex of the ellipse that is nearest the common focus. What is the distance between the focus of the parabola and the vertex of the parabola?

[hide]

looking at the ellipse
$a=3$ and $b=2$

$c^{2}= a^{2}-b^{2}= 5$

we will focus on the positive side (quadrant 1 of the ellipse)
$c = \sqrt{5}$

yielding the focus at point $(\sqrt{5},0)$
now, $(3,0)$ is the closest vertex of the ellipse

we now have a parabola whose focus is at $(\sqrt{5},0)$ and whose latus rectum starts at $(3,0)$
the length of the latus rectum is $2(3-\sqrt{5})$

the focal length is quarter of that, or $\frac{3-\sqrt{5}}{2}$

I hope I got that terminology right
[/hide]


are we supposed to have received mop hw yet?  :|

I just got an email reminding me to bring my homework to camp but I never got the homework :(?

[quote] MOSP Student,

Please remember to bring you MOSP Homework solutions with you to camp.

Happy problem solving.

Rachelle [/quote]

uhhh...

2008 MOSP Homework set  (pdf format, cover page, 1 instruction page , and 4 pages of problems) is being sent from the AMC office to each MOSP participant today, (Tues, June 3, 2008).  Watch for it at the email address you gave to the AMC for MOSP.

Steve Dunbar
MAA Director, American Mathematics Competitions

this might sound like a really stupid question but i just want to make sure

so if im in red group, i only do the red group problems right?  why is there no black group set?

god wow
with the number of projects i have due this week
im gonna have to cut school to finish this set

oh man that would be horrible now wouldn't it...

I'm probably going to sound pretty clueless, but what does the notation on the first red problem mean, where it's f: Q -> Q?

It means f is a function from the rationals to the rationals.

[quote="jb05"]Oh man 6 day notice for 15 problems. Exciting stuff.[/quote]

slightly more exciting because I am not even in the country right now and don't have a pdf reader. yeah.

[quote="SamE"]There's probably not a black group hw because we already got like 150 problems from Zuming over the last couple weeks. You should feel lucky just to have 15 problems. In 2005, they gave red and blue 28 problems each. I don't know what they did the last two years, though.[/quote]

But they gave you more than a week. I know in 06 we got about that, but we had a full month, so it was quite comfortable.

And @not_trig: I'm sure everybody knows you're away by now.

[quote="Sean Gu"]So far I've only gotten 3 problems,,,does it matter how much u did by 6/10? :D[/quote]

Don't panic: MOP Homework is not supposed to be like homework for a school class.  We don't expect you to be able to solve all, or even most, of the problems: the purpose of the homework is to give everyone a challenge before MOP.  My first year at MOP, I could only solve a couple of the homework problems.  Ultimately, it really doesn't matter how many you solve: what matters is how much you learn in the process of trying to solve them and then discussing them at the beginning of MOP.  It sounds like you've been putting a good amount of effort into working the problems, which is great!

Some past ones, like  2005 and 2006 MOP Homework, have been posted on AoPS... Search the Downloads forum.

[quote="pythag011"]Can we get MOP homework even if we are not in MOP?[/quote]
Continuing this, would it be okay for someone to post a link of the MOP homework or something?
Or you could email it to me... there should be a button right below my post... :D

EDIT: Never mind, thanks <person whose name I am not sure if I should mention>! :D

If you are tricky enough, you can probably find all the problems on AoPS.

this is non-collaborative right? or no...

i think it's non-collaborative in the sense that you should work on each problem for a few hours on your own, then if not, ask other people -cough- altheman -cough- for hints.  :wink:


Prove that there exists c belonging to (a,b) such that f(c)f'(c)=c, given that f is differentiable on (a,b) and satisfies f^2(a)-f^(b)=a^2-b^2.

If we also know that f is continuous on [a,b] we can apply rolle to g(x)=[f(x)]^2-x^2 on [a,b]


trigonometry
geometry
circumcircle
geometry proposed

Let $ABC$ be the isosceles triangle ,$BC=a , AB=AC=b$ and $M \in AC , N \in AB$ that changes such that $a^2\times AM \times AN=b^2\times BN \times CM$ if $P=(BM\cap CN)$. Find the locus of $P$.

[quote="ehsan2004"]... $a^2\times AM \times AN=b^2\times BN \times CN$ ...[/quote]
I think there is a typo and it should be $a^2\times AM \times AN=b^2\times BN \times CM$. Assume this is the case.

Using the sine theorem for the triangles $\triangle ABM, \triangle ACN, \triangle BMC, \triangle CNB$,

$\frac{AN}{AC} = \frac{\sin \widehat{ACN}}{\sin \widehat{ANC}},\ \ \frac{AM}{AB} = \frac{\sin \widehat{ABM}}{\sin \widehat{AMB}},\ \ \frac{BN}{BC} = \frac{\sin \widehat{BCN}}{\sin \widehat{BNC}},\ \ \frac{CM}{CB} = \frac{\sin \widehat{CBM}}{\sin \widehat{CMB}}$

By the problem condition,

$\frac{AN \cdot AM}{AC \cdot AB} = \frac{AN \cdot AM}{b^2} = \frac{BN \cdot CM}{a^2} = \frac{BN \cdot CM}{BC \cdot CB}$

$\frac{\sin \widehat{ACN} \cdot \sin \widehat{ABM}}{\sin \widehat{ANC} \cdot \sin \widehat{AMB}} = \frac{\sin \widehat{BCN} \cdot \sin \widehat{CBM}}{\sin \widehat{BNC} \cdot \sin \widehat{CMB}}$

Since $\angle ANC = 180^\circ - \angle BNC,\ \angle AMB = 180^\circ - \angle CMB$, we have $\sin \widehat{ANC} = \sin \widehat{BNC},\ \sin \widehat{AMB} = \sin \widehat{CMB}$ and

$\sin \widehat{ACN} \cdot \sin \widehat{ABM} = \sin \widehat{BCN} \cdot \sin \widehat{CBM}$

$\frac{\sin \widehat{ACN}}{\sin \widehat{BCN}} = \frac{\sin \widehat{CBM}}{\sin \widehat{ABM}}$

Since in addition, $\angle ACN + \angle BCN = \angle CBM + \angle ABM = \angle B = \angle C$, it follows that

$\angle ACN = \angle CBM,\ \ \angle BCN = \angle ABM$

But then the triangles $\triangle MBC \sim \triangle MCP$ are similar, having the angle at the vertex M common and the angles $\angle CBM = \angle ACN \equiv \angle MCP$ equal. Likewise, the triangles $\triangle NCB \sim \triangle NBP$ are similar, having the angle at the vertex N common and the angles $\angle BCN = \angle ABM \equiv \angle NBP$ equal. It follows that their remaining angles are also equal, i.e., $\angle CPM = \angle BCM \equiv \angle C$ and $\angle BPN = \angle CBN \equiv \angle B$. As a result, $\angle BPC = 180^\circ - \angle B = 180^\circ - \angle C$. Thus the locus of the points P is a circle tangent to the lines AB, AC at the points B, C.

I note $m(\widehat {ABM})=x,\ m(\widehat {BCN})=y$

$[m(\widehat {ACN})=C-y,\ m(\widehat {CBM})=B-x]$.

$\frac{MA}{MC}=\frac ba\cdot \frac{\sin x}{\sin (B-x)},\ \frac{NA}{NB}=\frac ba\cdot \frac{\sin (C-y)}{\sin y}\Longrightarrow$

$\frac{b^2}{a^2}=\frac{MA}{MC}\cdot \frac{NA}{NB}=\frac{b^2}{a^2}\cdot \frac{\sin x \sin (C-y)}{\sin (B-x)\sin y}\Longrightarrow$

$\sin x\sin (C-y)=\sin (B-x)\sin y\ (C=B)\Longrightarrow$

$\cos (B-x-y)-\cos (B+x-y)=$

$=\cos (B-x-y)-\cos (B-x+y)\Longrightarrow x=y.$

Therefore, the geometrical locus of the point $P$ is the circumcircle

which is tangent in the points $B,\ C$ to the lines $AB,\ AC$ respectively.


Hi I'm currently in high school taking 4x. We're at the unit on logorithms and I'm heaving trouble with expressing logorithms.

For example: Log(15)=x Log(50)(base 20)=y
Express Log(9)(base 40) in terms of x and y

does anyone know where I can find more practice problems? Maybe a particular book or something? Thanks.

Hi! Logg :)

Take a look!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=logarithm&t=82841[/url]

thanks for those, I tried em and got same answers as others. But I'm wondering where you got these, did you make them up? Because I was wondering if there was any books on this kind of problems, thanks.


calculus
derivative
calculus computations

How do you solve this problem?

In the figure below, PQ represents a 40 foot ladder with end P against a vertical wall and end Q on level ground.  If the ladder is slipping down the wall, what is the distance RQ at the instant when Q is moving along the ground $\frac{3}{4}$ as fast as P is moving down the wall?

Let $RQ=x, PR=y$ where $x,y\in (0,\infty)$. Then we have the relation\[ x^2+y^2=1600  \]by Pythagoras' Theorem. We take implicit differentiation w.r.t $t$ (time).\[ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0  \]\[ \Rightarrow x\frac{dx}{dt}=-y\frac{dy}{dt}  \]Substitute $\frac{dx}{dt}=\frac{3}{4}\frac{dy}{dt}$,\[ y=-\frac{3}{4}x  \](We have a negative sign here because $P$ is moving towards $R$ while $Q$ is moving away from $R$.)
Substitute $y=-\frac{3}{4}x$ back into the original equation,
$x^2+\frac{9}{16}x^2=1600$
$\Rightarrow x=32$
 :D

Thanks for your help -- I see what I did wrong.


induction
advanced fields
advanced fields unsolved

Show that an infinite well-orderable set has nonisomorphic well-orderings.

well ... just swap two elements in the order. generalization: http://www.mathlinks.ro/Forum/viewtopic.php?t=190094

Swapping two elements doesn't suffice, the orderings $ 1<2<3<...$ and $ 2<1<3<...$ are isomorphic.

ah, thank you.

the claim is equivalent to: for every infinite ordinal number a, there exists an ordinal number b, which is equipotent to a but not equal to a. this is clear, put b=a+1.

to be explicit: take an infinite set $ X$ with a well-ordering $ <$. adjoin an element $ \infty$ and define a well-ordering on $ X' \equal{} X \cup \{\infty\}$ so that $ x < \infty$ for all $ x \in X$. now, $ X$ and $ X'$ are not isomorphic: if $ f : X \to X'$ is an isomorphism, we have a largest element $ s \in X$, which is mapped to $ \infty$. assume $ x < s$ and that we know $ f(y) \equal{} y$ for all $ y < x$. then $ f(x) \equal{} \inf \{X \backslash \{f(y) : y < x\}\} \equal{} x$. thus by transfinite induction, we have $ f(x) \equal{} x$ for all $ x < s$. but then $ s \in X$ has no preimage under $ f$, a contradiction. since $ X$ and $ X'$ are equipotent, $ X$ has a well-ordering which is isomorphic to the one on $ X'$, thus not isomorphic to the given one on $ X$.


algebra
polynomial
Irreducible
irreducible polynomial

$ n,m$ are positive integers and $ n \ge m$ $ \epsilon _i \in \{\minus{}1,1 \}$ $ i\equal{}1,2$
Now for what $ (n,m)$   $ f(x)\equal{}x^n\plus{}\epsilon_1 x^m\plus{} \epsilon_2$ is irreducible and for what $ (n,m)$   $ f(x)$ is reducible.
I think this is a result of Ljunggren's.Can somebody give the result and the proof?
Thank you :)

Actually Ljunggren in his "On the irreducibility of certain trinomials and quadrinomials", Math Scand. 8, (1960), 65-70 proved the following theorem:

Let $ n \ge 2m$, $ d \equal{} (m, n), n1 \equal{} n/d$ and $ m1 \equal{} m/d$. Then the polynomial
        $ g(x) \equal{} x^n \plus{} ax^m \plus{} b'$ where a, b = 1 or -1
is irreducible. The exceptional cases are the cases where n1+m1  divedes by 3 and
       1) n1 and m1 are odd and a = 1
       2) n1 is even and b = 1
       3) m1 is even and b = a

Thank you for the result :) Can you post [b]On the irreducibility of certain trinomials and quadrinomials[/b] here,Namdung?I want to see the proof of it.


irrational number
inequalities unsolved
inequalities

(without a calculator)
Determine which of $ \left(\frac{1}{2}e\right)^{\sqrt{3}}$ and $ \left(\sqrt{2}\right)^{\frac{1}{2}\pi}$ is greater.

I did some REALLY rough approximations and determined that the one on the right is indeed slightly larger.  If you approximate each irrational number as a rational one (for example, take $ e\approx 2.7$), then you can see that the one on the right is a little larger.

Both numbers are transcendental.  Yay.


hello again ! :D

can you sum this series:

$ \sum_{m\equal{}0}^\infty\;\left\{\dfrac{(\minus{}1)^m\;(2m)!}{2^{4m\plus{}3}\;(m!)^2}\cdot\dfrac{(2m\plus{}1)}{(m\plus{}2)(m\plus{}1)}\right\}$

[hide]
$ answer: \qquad \boxed{13\minus{}4\left(1\plus{}\sqrt 5\right)}$
[/hide]

Do you want to actually post how to solve any of these?

This is just another manipulation of $ \frac{1 \minus{} \sqrt{1 \minus{} 4x}}{2x} \equal{} \sum_{m \geq 0} \frac{(2m)!x^m}{(m \plus{} 1)!m!}$.

Is there a way to actually figure that out?

some comment :
suppose you need to find sum $ \sum R((2m)!, (m!)^{2},m!...)a_{n}$
at first you must to think about something like this $ f((1 \plus{} \alpha x^{p} )^{\beta})$ and after simple algebraic manipulation you can get sum.

JRav: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192445 contains part of the analysis.  This falls into the category "facts about the Catalan numbers that one picks up reading about combinatorics, generating functions, etc."


logarithms
integration
inequalities
function
calculus
floor function
Euler

You just won a math contest and you get to choose between 2 prizes:

1) For every day for the rest of your life, you will get $\frac{1}{10}$th of the money you recieve the previous day starting with $\$900,000$

2) For every day for the rest of your life, you will get $\$ \frac{10}{n}$ every day, where $n$ is the $n$th day after you won the prize

Which do you choose? Why?

Well assuming that there is no infaltion and I am imorrtal I'd immediatly pick the second one.

In reality, I would probably pick the first one. I do not think that dollars will be still in use in $e^{90000}$ years

second one i think... first one converges and second one diverges... geometric series with $|r|<1$ and harmonic series respectively

I think I would choose the first. I would rather have a million dollers in capital to role into the right venture when it comes along, than chance it and die from a heart attack at 60 or something, and never got a chance to use the money.

True, the second series diverges, however it diverges wayyyyyyyyy too slowly.
After only $5$ days with the first choice, the winner would already have $900,000+90,000+9,000+900+90+9=\$99,999$.  After about two months (and [u][b]grossly[/b][/u] overestimating, the second choice would award $\$10(60*\frac{1}{2})=300$.  Realistically, you wouldn't want (nor be able) to wait around long enough to reap the benefits of the second option.  Plus, what would you be saving up for anyway?  Option 1 awards enough money on the first day to take and invest with, enabling you to earn money at an exponentially larger rate than the second way.  Choice 1 makes more sense.[/u]

True, the second series diverges, however it diverges wayyyyyyyyy too slowly.
After only $5$ days with the first choice, the winner would already have $900,000+90,000+9,000+900+90+9=\$99,999$.  After about two months (and [u][b]grossly[/b][/u] overestimating, the second choice would award $\$10(60*\frac{1}{2})=\$300$.  Realistically, you wouldn't want (nor be able) to wait around long enough to reap the benefits of the second option.  Plus, what would you be saving up for anyway?  Option 1 awards enough money on the first day to take and invest with, enabling you to earn money at an exponentially larger rate than the second way.  Choice 1 makes more sense.[/u]

QUESTION:
So, how long would it take for the second prize to give you $\$1000000$? :D

A long, long time.  It increases slower and slower the longer it goes too...

$\sum_{k=1}^{x}k^{-1}$ compared to $\sum_{k=1}^{x}900000*10^{-k}$

The first is $\frac{1-10^{-x}}{.9}$
The second is approximatly equal to $10(ln(x)+\int_0^1ln(ln(x^{-1}))dx)$

It would take around e^99999.4 days to get 1million dollars with the second option.

[quote="Chocoboom"]uhhh im i missing something? i seems to me that the 2nd option starts small and gets smaller. first day, you would get 10/1 so 10 dollars. On the hundredth day, wouldn't you get 10/100 so 10 cents?? im confused. :?[/quote]

you may not know this, but

$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...$

is called the harmonic series and the sum diverges (i.e. the sum is infinity), so if you had enough time, you would get infinite money, but you would die by then...

Is this a maths question or a philosophical question?  :huh:

Well obviously the 1st option is far more enticing because you would reach $1000000$ in a week and a day, you would reach $1000000$

$900000+90000+9000+900+90+9+0.9+0.09 = 999999.99$

Im not sure about America, but Australia round up/down every 5 cents so that would infact be 1000000  :D even though it never techinically reaches 1000000.

and for the $\frac{10}{n}$ thats sorta stupid

when $n\geq400 \implies \frac{10}{n} \leq 0.025$

and anything less than 0.025 dollars will be rounded to 0  :lol: so in fact you dont get anything.

Question: how much money will you have made when you die at age 80? [assuming you chose option 2 and you decide to live off the earnings from that only and you do not have a job and you dont find any money on the ground or steal or get it from family or freinds, etc. as well as assuming money is not rounded up or down]

[quote="Chocoboom"]uhhh im i missing something? i seems to me that the 2nd option starts small and gets smaller. first day, you would get 10/1 so 10 dollars. On the hundredth day, wouldn't you get 10/100 so 10 cents?? im confused. :?[/quote]Not all decreasing series converge.

$\frac11+\frac12+(\frac13+\frac14)+(\frac15+\frac16+\frac17+\frac18)+\cdots >1+\frac12+\frac12+\frac12+ \cdots$

[quote="Chocoboom"]no wait i still dont get it, after a thousand days, then wouldn't you barely get a penny a day? and as you keep going, after 10,000 days, you would get 0.1 cents which is nothing... how can you get an infanite amount?? :huh:  how can it diverge?[/quote]

notice that your argument is just intuitive...it has no mathematical basis, look at other posts for actual math showing that the harmonic series diverges

[quote="Chocoboom"]no wait i still dont get it, after a thousand days, then wouldn't you barely get a penny a day? and as you keep going, after 10,000 days, you would get 0.1 cents which is nothing... how can you get an infanite amount?? :huh:  how can it diverge?[/quote]

Diverging means that the sequence does not approach a fixed amount. You may be receiving .1 cents a day, but cumulatively, this will add up to an amount that will keep getting larger without bound.

As in 4everwise's example, look at
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots$

If you group them as 4everwise did, this expression is greater than
$1 + \frac{1}{2} +\frac{1}{2}+\frac{1}{2}+\cdots=1 + \frac{n}{2}$

However, we cannot evaluate $n$ since there will be an infinite amount of one-halfs, and thus $n$ will simply approach infinity. Therefore, [i]cumulatively[/i], you will receive an infinite amount. However, you may be receiving close to nothing on any particular day.

Yeah, it keeps increasing, just by less and less each time...

Well, the second one will ultimately reach infinity, while the first one will never go any higher than $\$1,000,000$.

But after 100 years (and I probably won't even live that long), I would only make about $\$110.82$ with number 2.  I'd take option number 1.

EDIT: I looked at a later post, and I guess Excel might have some rounding errors.  I don't know, I didn't test it out to be sure for myself (I would have used a TI-89 if I had one).  But still, I won't live long enough to see the second prize grow to be more than the first.

[quote="Chocoboom"]no wait i still dont get it, after a thousand days, then wouldn't you barely get a penny a day? and as you keep going, after 10,000 days, you would get 0.1 cents which is nothing... how can you get an infanite amount?? :huh:  how can it diverge?[/quote]

It's very hard to see intuitively.  But the idea is that $\frac{1}{n}$ decreases, but [b]too slowly[/b] for it to converge.

If you want to get a sort of computational feel for it, verify for some large values of $n$ the following inequality:

$1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \ge \ln (n + 1)$

In fact, simply consider the sequence $a_n = \ln (n+1) - \ln n = \ln \frac{n+1}{n}$.  The sequence gets smaller and smaller, but the [b]total sum[/b] of all the values in the sequence is obviously $\ln (n+1)$, so it goes to infinity!

Don't worry if you don't get it, you'll study this sort of thing later in high school (possibly college) :)

As regards the original problem,

$\text{Which reward to choose} \neq \text{Which reward ultimately generates more money}$

:)

Economically speaking, the first is far superior.

Yes. You have to consider things like discount rates and expected utility and all that.

huh... i see

yea just take the first one, put it in the bank once you get 99999 and the interest you get will rise more then choice 2

[quote="jelly119"][quote="Altheman"]yes...the first one you get about $\$ 10^6$ within a few days, the second one is the harmonic series (multiplied by 10), and you approximately get $10*\log_2 n$ total after the $n$th day, and even if you live 100 years, $10*\log_n (100*365 )\approx 151.56$, which is clearly less[/quote]

With Excel I got $\sum_{n=1}^{36500}\frac{10}{n}=110.822......$ Far away from 151.56. Something wrong??[/quote]
Well, using the better aproximation $\sum_{n \leq x } \frac{1}{x} \approx \log_e(x)+\gamma$ ($\gamma = 0.57...$), you get: $\$ 110,82$ ;)

PS: But why do you use Excel for that¿?¿
It's a shame on math when this is used for doing math!

It doesn't seem like a good idea to use excel in this case since the rounding errors add up and are probably significant.

Also does $\gamma=\int^1_0ln(-ln(x))dx$? If this is true how do you prove it?

One sometimes defines it by the formula I used (one has to show then that it is indeed a constant). But the proof I know uses $\gamma : = 1-\int_1^\infty \frac{\{t\}}{t^2} dt$ and partial summation. For a proof one will need calculus....
But see http://mathworld.wolfram.com/Euler-MascheroniConstant.html for more identities.

No. I have seen quite a few though.

My favorite is
$4tan^{-1}(\frac{1}{5}) - tan^{-1}(\frac{1}{239})$

To prove it with $\gamma : = 1-\int_1^\infty \frac{\{t\}}{t^2} dt$:

Usefull Lemma:
Let $f,g$ be functions with $g$ having a continous derivation on $[a,b]$, $a,b \geq 0$. Define $F(x) : = \sum_{n \leq x } f(n)g(n)$. Then:
\[ \sum_{a < n \leq b} f(n)g(n) = F(b)g(b)-F(a)g(a)-\int_{a}^b F(t) g'(t) dt \]

Proof: some simple algebraic manipulation and using $g(x+1)-g(x) = \int_x^{x+1} g'(t) dt$ (that is all calculus needed!), thus $F(n)(g(n+1)-g(n)) = \int_n^{n+1} F(t)g'(t) dt$ for integers $n$ since $F$ os constant in between integers.

The previous one used with $a=1 , b=x >2$ gives after another algebra:
\[ \sum_{n \leq x} f(n)g(n) = F(x)g(x)-\int_{1}^x F(t) g'(t) dt \]

Now set $f(x)=1 , g(x) = \frac{1}{x}$, giving with the last one above:
$F(x)=\sum_{n \leq x} 1 = \left[ x \right]$ with $\left[x\right]$ being the greatest integer $\leq x$ and $\{x\} = x - \left[x\right]$ in the upcomming stuff.
$\sum_{n \leq x} \frac{1}{n} = \sum_{n \leq x} f(n)g(n)= \frac{\left[x\right]}{x} + \int_{1}^x \frac{\left[t\right]}{t^2} dt=$
$= \frac{x-\{x\}}{x} + \int_{1}^x \frac{t}{t^2} dt - \int_{1}^x \frac{\{t\}}{t^2} dt= \log(x)+\gamma+h(x)$
with $h(x) = \int_x^\infty \frac{\{t\}}{t^2} dt - \frac{\{x\}}{x}$.

Now when we look at $h$:
$\int_x^\infty \frac{\{t\}}{t^2} dt < \int_x^\infty \frac{1}{t^2} dt = \frac{1}{t}$
which gives with the trivial part $\frac{\{x\}}{x} < \frac{1}{x}$ that $|h(x)| < \frac{1}{x}$.
But this is what we need.

In any event, you would get pwned by inflation in the second one :(

sure the second option diverges...but only if we go for $\infty$ days...so lets be realistic, thats never going to happen.
all so does $\frac{10}{9745651316541}$ maintain its actual value or will it be rounded down (which is most likely what the people awarding the prize will do) to $0$

How about this choice?

$\$$44,000 immediatly
or
$\$$2,000 a year forever

lets assume that you will live for more than 22 years...so option 2 is the obvious choice...unless i am missing something.

Yeah, that one is definitely choice 2, no question...

Actually, I'd probably still go with choice 1 for this one.  There's a reason your non-mathematical instinct tells you to go for the immediate reward.  It's because immediate rewards are definite rewards.  You can enjoy the utility of the entire $\$ 44,000$ immediately, whereas in the second choice you would have to wait a long time for the overall utility to be larger.  And, of course, you could always invest the $\$ 44,000$.

[quote="t0rajir0u"][quote="0714446459923"]Well I tohught it was defined as the limit of the difference of the harmonic series and the logarithm. I was just curious how it was possible to turn it into an expression that doesn't use floor or rounding functions or anything similar to that.[/quote]

Have you seen all the different expressions for $\pi$?  :)

Altheman, why $log_2$?   :huh:[/quote]

i suppose that i do not have a rigirous proof, and it seems that the natural log is better, but i was thinking about the proof when you show that the harmonic series is divergent by doing the comparison test with $\frac{1}{2}+\frac{1}{2}$

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{2}$
$\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}>\frac{1}{2}$

then you have that from $2^n+1$ to $2^{n+1}$, you have another $\frac{1}{2}$...but this is not that good an approximation

[quote="Altheman"]
i suppose that i do not have a rigirous proof, and it seems that the natural log is better, but i was thinking about the proof when you show that the harmonic series is divergent by doing the comparison test with $\frac{1}{2}+\frac{1}{2}$

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{2}$
$\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}>\frac{1}{2}$

then you have that from $2^n+1$ to $2^{n+1}$, you have another $\frac{1}{2}$...but this is not that good an approximation[/quote]

This one is a little better:

$\int_{1}^{n} \left( \frac{1}{ \lfloor t \rfloor } - \frac{1}{t} \right) \, dt = 1 + \frac{1}{2} + ... + \frac{1}{n} - \ln (n+1) \ge 0$

(In fact, the limit of the integral is finite, and is called $\gamma$, or the Euler-Mascheroni constant.)


Keto esters are bifunctional reactive molecules and are important synthons for the
synthesis of aliphatic and heterocyclic compounds.

a. Two isomeric keto esters X and Y have the same molecular formula C5H8O3.
Deduce their possible structures
Each ester is first reacted with benzyl bromide in the presence of CH3ONa, and the
resulting products are treated with 1 or 2 equivalent of a strong base (such as lithium
diisopropyl amide, LDA) followed by 1 equivalent of CH3I.

The products at the end of the second step are then hydrolysed by aq.HCl.

b. Write down the reaction sequences involved.

c. At the end of the reaction, the final product of keto ester X is a neutral
compound (molecular formula C11H14O) whereas keto ester Y, gives a keto
acid (molecular formula C12H14O3). Explain.
d. Keto ester X gives different products depending upon the amount of LDA
used. Explain what happens when
i. 1 equivalent of LDA is used.
ii. 2 equivalents of LDA are used.

[hide="Answer"]There are three possible compounds: A, OHC-CH2CH2-CO2CH3; B, CH3COCH2CO2CH3; and  C, CH3CH2COCO2CH3. It's clear from the description given in part c that the product from keto esther X undergoes decarboxylation, and this can only happen for a beta ceto acid. Then, keto esther X is compound B, and the product is forms is CH3COCH(CH3)CH2Ph. Keto esther Y is very likely compound C, since A could give several side products under the basic conditions used.

[quote="VARUNARASU"]Keto ester X gives different products depending upon the amount of LDA
used.[/quote]

Of course: when 1 eq. is used, only carbon 2 is deprotonated, while with 2 eqs. both carbon 2 and carbon 4 are deprotonated, and with reaction with an electrophile it is carbon 4 that reacts (why?).[/hide]

Correct :10:  :10: ( ‘cept the first part A would be Ch3COCOOCH2CH3 coz OHC-CH2CH2-CO2CH3 is an aldehyde)
[quote="carcul"]
Of course: when 1 eq. is used, only carbon 2 is deprotonated, while with 2 eqs. both carbon 2 and carbon 4 are deprotonated, and with reaction with an electrophile it is carbon 4 that reacts (why?).
[/quote]
i think thats bcoz carbon 4 is a better nucleophile as there is lesser delocalization

[quote="VARUNARASU"]‘cept the first part A would be Ch3COCOOCH2CH3[/quote]

No, that cannot be the case because the base used was sodium methoxide. In reactions of keto esthers whose esther part is -CO2R the base used must be RONa, because if it were R'ONa then part of the base would be consumed in a transestherification to yield molecules with -CO2R'. Therefore, structure assignement of the product would be ambiguous. So in your problem we can only have methyl esthers.

[quote="VARUNARASU"]i think thats bcoz carbon 4 is a better nucleophile as there is lesser delocalization[/quote]

Yes: if a proton is less acidic, then the corresponding conjugate base is a stronger nucleophile.

But transesterification being a reversible reaction would take place only when the base is in high concentration. But here the concentration of base(not mentioned),  I suppose would be less as the enolates formed being more stable(less basic) than the base forward reaction is favored. Am I rite in saying this??

P.S this problem was one of the preparatory problems of the 2001 chem Olympiad and im giving the answers from the official solutions.

The concentration of base cannot be low: it must be at least the same as the keto esther. And transestherification would certainly occur, I am sure of that.


algebra
functional equation
algebra unsolved

My teacher recently mentioned Abel's functional equation which has form
$ f(\alpha(x))\equal{}f(x)\plus{}h(x)$ for the given h

I'd like to know what are the standard approaches for such equations.

General solution to this equation:

$ f(x) \equal{} \sum^{\infty}_{n \equal{} 0} {h[\alpha^n(x_0)] \minus{} h[\alpha^n(x)]}$

if the infinite sum converges.

Bluntly I dont know how do you obtain this, i've found it in "Functional equations and how to solve them" if u want sth more on the subject u may want to read it.


geometry
analytic geometry
rotation
geometry unsolved

[quote="udyam.bnghs"]hi friends, i'm a 17-year old indian guy...  i've a problem on geometry......

     two non-overlapping  squares of sides a and b placed inside a squares of side 1. prove that maximum of (a+b) is 1.[/quote]
The maximum can clearly be reached, when you just place the squares next to each other. Assume $ a\plus{}b > 1$, then $ b > 1\minus{}a$. Assume that the square's sides are parrallel to the the $ x$ and $ y$ axis, such that the left most and lowest point are $ (0,0)$. Let $ x_a,y_a$ be the coordinates of the left most and lowest point of the square, with side $ a$. And likewise with $ x_b,y_b$. Wlog assume $ x_a < x_b$. Since $ a\plus{}b > 1$, we see that the height of the sqaures put on top of each other are too big. So $ x_b \ge x_a \plus{} a$. But the right most point of the square $ b$ must be $ x_b \plus{} b \ge x_a\plus{}a\plus{}b \ge a\plus{}b > 1$, which clearly doesn't belong to the square. Contradiction. Hence our assumption was wrong and $ a\plus{}b>1$ can never occour.

[quote="udyam.bnghs"]but the right most point of the square b may be less than x(b)+b. (if the sides of this square are not parallel to the axes!).  then?[/quote]
Oh yeah. I didn't realise that one could rotate the squares. :oops:

i have just fought with the problem. and ultimately get failure!!!


my school just got the result today

i got 73, not too bad though

it's a very good mark, are you the school champion?

of course cynthia did better than me
i am the 2nd

no.....
actually she hasn't known her score yet
i gonna tell her tmr
now i will not disclose her score..

i guess her 3c solution wasn't good enough, although her solution was like 3-pages long

I see. I don't know my score yet, I will bug my teacher to tell it to me tomorrow.
Nice job on COMC and good luck on CMO!

Does this mean that all schools have gotten their results back already?

if you look at http://www.cemc.uwaterloo.ca

You see that it says School Results are posted, so I believe yes..

In fact, I just found out that I got 79/80.

80/80  :)  A bit surprising since I did omit some trivial cases on 3c.

[quote="vincent_five"]u guys have ur school # and password?
anyways congratz[/quote]

no teacher emailed me

77.  Which, considering the two mindless mistakes I made, was the best I could have hoped for.

Congratulations to everyone else!

[quote="british8985"]sounds like you all have school id and password... :([/quote]

our teachers do, yeah

I got 80 and was pleasantly surprised that no marks were taken off for 3c.

congratz to everyone!

btw when will the official result be posted?

[quote="mathgeniuse^ln(x)"][quote="ddziabenko"]I got 8 and was pleasantly surprised that no marks were taken off for 3c.[/quote]

I believe you mean 80, right.[/quote]

 Right :oops_sign:

I got a  81 in this COMC..

I found a fatal error on the 3c, and then they had to give me a bonus mark for that

because i gave them the alternative solution.

The known solution only works only up to the prime number smaller than 2^2000. but after that, you need a different

strategy to attack it.

The markers admitted their mistakes, and then adjusted my mark so that my great work would not be depicted

as same as other 'ordinary works'

Hey!
I found that too haha. Just thought I'd make an account just to tell you that! I blew it thought by making a dumb mistake on 8b where I put the answer for ALL the paths rather than just the path of 14. They're probably gonna edit the solutions on the site at the same time they release the results and standings. I had to get a fricken email from them telling me they were "further investigating" my solution. Arghhhh. They never told me if I was right though... and definitely not bonus points.... wow..... you're pretty lucky. Definately a historical groundmark in COMC history.

actually, i saw the same problem with the 3c

I thought there was an arternative solution, but i just decided to stick with what I had as a solution, cuz it started getting too complicated

so I got 80 (79 + 1) cuz i got 9 on part II #4 (i still don't know why)

so my score is 40 + 10 + 10 + 11 + 9 = 80

i hope waterloo doesn't make any mistakes from now on

:rotfl:  :spam: At least this wasn't the "help with this problem on my math test" spam.

the official result has been posted

the cutoff for cmo is 72

there are a number of american kid got 72 or above

so the cutoff may be 71 :)

I just got an email for the "repechage" problem set, which basically means I get another chance to qualify for CMO!!! Yay!!!

Also, does anyone know the difficulty level of the problems? CMO? APMO? USAMO? COMC? AIME? :D  :D

does anyone know the cut off for the recharge problem set?

i got 73, but i haven't recieved any invited letter, how are about you?

no, they don't send you if you make it do they


inequalities
calculus
derivative
integration
inequalities proposed

Let $ a,b,c>0$ prove that

\[ a^2\ln(b\plus{}c)\plus{}b^2\ln(c\plus{}a)\plus{}c^2\ln(a\plus{}b)\ge \frac{1}{2}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\]

Um... $ a \equal{} b \equal{} c \equal{} \frac {1}{4}$??? LHS is negative, RHS positive...

The inequality is reversed, isn't it...

Ok, I can prove the opposite inequality:

First of all, $ x/2 > ln(x)$ for all $ x > 0$. (true because $ ln(2) < 2/2$ and the derivative of $ ln(x)$ is more than $ 1/2$ when $ x<2$ and the derivative of $ ln(x)$ is less than $ 1/2$ when $ x > 2$.)

Then the LHS $ < 1/2(a^2(b\plus{}c) \plus{} b^2(a\plus{}c) \plus{} c^2(a\plus{}b)) <$ RHS.

Thank actually I make it by integral and Nesbit  :P

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge\frac{3}{2}$

$ \int (\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b})da\equal{}\frac{1}{2}\frac{a^2}{b\plus{}c}\plus{}b\ln(c\plus{}a)\plus{}c\ln(a\plus{}b)$

Two more times with $ db,dc$ we get  $ \sum a^2\ln(b\plus{}c)\minus{}\frac{1}{2}\sum ab(a\plus{}b)$ from RHS we get $ \frac{3}{2}abc$ so

$ \sum a^2\ln(b\plus{}c)\minus{}\frac{1}{2}\sum ab(a\plus{}b)\ge \frac{3}{2}abc$

Actually I known this working is not right  :P  and maybe there are coefficient for the equality... :)


geometry
3D geometry
prism
tetrahedron
octahedron
pyramid
dodecahedron

[hide]The formula for diagonal is $s\sqrt{3}$, and the formula for surface area is $6s^{2}$, so the answer to your question is $6s^{2}= 6(7^{2}) = 294$.[/hide]

its not so much "knowing the formula", its knowing how to derive it... the way you derive something like this is somewhat easy geometry (but can get messy). Anyway the reason why its $s\sqrt{3}$ is that by pythagoras the length from one corner to another on the same face is $\sqrt{2}$. Now we use pythagoras again, this time with one leg of the triangle being the edge pointing in the direction we haven't used yet, and the other leg being the diagonal of the perpendicular face. In this way, we have a distance formula in n-dimensions: keep using the 2-d pythagoras over and over, in particular the space diagonal of an n-cube is $\sqrt{n}$. Sorry to have posted so much..

[quote="fireforce"]Anyone konw the formula for other space shapes other than a cube or rectangular prism?
(ex. icosahedroid)[/quote]

I know the formula for a tetrahedron with side s: the diagonal (which either doesn't exist or coincides with the side) is also of length s. :roll:

Sorry, I'm really sleep-deprived...

With an octahedron, you would treat it as two separate pyramids and then find the height (again, using Pythagorean). Then multiply by two.

And with dodecahedrons and icosahedrons, it would get really messy.

[quote="mr. math"]how would u do this problem:

A cube has a space diagnol of 7(square root of 3). what is the surface area of the cube?[/quote]

Space diagonal=d.The formula is 2 d^2.

[hide]I don't know off the top of my head formulae for certain polyhedra, but it is possible to derive it.

Anyway for the problem,

$s\sqrt{3}=7\sqrt{3}$.

$s=7$, so surface area is $6*7^{2}=294$

$294$[/hide]

For the original problem:

[hide]The space diagonal is $7\sqrt{3}$ so the sidelenth is 7. The area of one face is $7^{2}=49$, so the total surface area is $49*6=294$.[/hide]

maybe maximize the surface area? or since you have a^2+b^2+c^2 (If I were writing the problem) I would do fun stuff with Cauchy...

[quote="Allan Z"][quote="AdorableAddict"]New Problem:
The space diagonal of a rectangular parallelepiped is $7\sqrt{3}$. Find its surface area.[/quote]

I dont think this is sufficient to solve the problem...
lets assume the dimension of this retangular parallelepiped is [b]$a*b*c$[/b], then $a^{2}+b^{2}+c^{2}=(7\sqrt{3})^{2}=147$, this is not enough to solve $2ab+2bc+2ac$  :maybe:[/quote]

[hide="Maximization"] $a^{2}+b^{2}+c^{2}\ge ab+bc+ac$ by Cauchy or Rearrangement. $\boxed{294}\ge2(ab+bc+ac)$ [/hide]

[quote="lotrgreengrapes7926"][quote="mathnerd314"]With an octahedron, you would treat it as two separate pyramids and then find the height (again, using Pythagorean). Then multiply by two.[/quote]
The answer is $\frac{s\sqrt2}{2}$.[/quote]

Umm, I keep getting the formula for the space diagonal of an octahedron to be $\frac{2s\sqrt{6}}{3}$

The centroid of an equilateral triangle is $\frac{s\sqrt{3}}{3}$ units away from each of the vertices, so using a right triangle to find the height you get : \[s^{2}=\left(\frac{s\sqrt{3}}{3}\right)^{2}+h^{2}\]  \[s^{2}=\frac{3s^{2}}{9}+h^{2}\]  \[h^{2}=\frac{6s^{2}}{9}\implies h=\frac{s\sqrt{6}}{3}\] Multiply this by 2 and you get $\boxed{\frac{2s\sqrt{6}}{3}}$

hahaha [size=200][color=green]"Moo with him!"[/color][/size]
jk  :o  :maybe:  :rotfl:

ya.. you can get the max suface area by cauchy  :P :. I didnt think of finding the max area since AdorableAddict didnt specify it   :wink:

[quote="lotrgreengrapes7926"][quote="DanK"]The centroid of an equilateral triangle is $\frac{s\sqrt{3}}{3}$ units away from each of the vertices, so using a right triangle to find the height you get :[/quote]
There are no equilateral triangles in an octahedron. It's 2 regular square pyramids put together at their bases.[/quote]

 :blush: The first signs that school is taking too much of a toll on you, you try to use tetrahedrons rather than square pyramids for an octahedron... :oops:

I am pretty sure that $\frac{s\sqrt{2}}{2}$ is only the height of 1 pyramid, though, which would make the space diagonal just $s\sqrt{2}$...

Btw, what is Cauchy?

Cauchy, sometimes called Cauchy-Swarz Inequality is a special case of Holder's Inequality. Find info about it [url=http://www.artofproblemsolving.com/Wiki/index.php/Cauchy-Schwarz_Inequality]here[/url].

[quote="lotrgreengrapes7926"]Oh, yeah. :blush:

Actually, there are 2 types of space diagonals on an octahedron. Let's label the 2 vertices $A$ and $A'$. And the other 4 points $B,C,D,E$. There is the diagonal $AA'$ or $BD$ (also $CE$). Surprisingly, I think they have the same length. $AA'=s\sqrt2$ and $BD=CE=s\sqrt2$.[/quote]

Not as surprising, since the octahedron is symmetrical--all the sides have the same length, so it would look the same regardless of which points you choose to be the vertices. ;)

Yeah, as long as it's a regular octahedron that's what you would expect.  Now, if you get into irregular ones, none of these formulas work anyway, so...


Hello,

I have a few legal questions to ask, maybe (and surely) there are a few people here who will be able and willing to help.

Imagine I have discovered a very important security breach on a company's or organization's website or product, and imagine this breach would matter very much to this company.

1) Is it lawful to try to sell the breach to this company (by disclosing its effect but revealing the way to patch the problem, if any, only after getting paid) ?

2) Is it lawful to pubicly disclose a security breach if it can imply huge losses for the company, but still can be of educational interest to other people.

3) If you propose a company to buy your solution to a security breach the company wouldn't be able to resolve, can it be interpreted as blackmailing ?

Most important question: What country are you in?

We can't know anything about legality if we don't know your country. Morality on another hand, is a different matter.

Ignita said that someone in the position you propose needs to engage competent counsel that knows which country you are in, where the company you are approaching is in, which intermediate countries might have jursidiction, how you came to have the knowledge you propose to sell.

However, in general, if you obtained the information without violating any rules, agreements or laws, you generally have a right to distribute information that might cause harm to another.  In another context, slander is damaging someone reputation by saying something that is not true.  If it is true, there is no liability.

People often sell the legal right to do something legal to someone who doesn't want the first person to exercize their rights.  Selling the right to sue someone, and agreeing to a settlement, falls into this catagory.

I find it hard to believe that you have, without doing anything improper, come into some knowledge that is so dangerous to a large organization that would pay to get the information.  Why hasn't someone else already found and resolved what you are talking about?  It could happen, but you could also win the lottery.

One good way to approach this might be to propose that you start working for this company, perhaps with a contingency that you save them money to keep your job or your company's position as a consultant to them.  If you found one without inside info or access how much better will you and your friends be with inside info and access??   Rather than getting a one time payment that might get you blackmail charge, why not get a career, if you really are so good?

I know, this is all conjecture and none of it really happened, but this might be the best way to leverage the situtation.  Get a good attorney to negotiate a contract where they pay for results, then produce the reults and get the money for providing a service.  It's far better than skating on the edge of illegality.

Hope this works out for you and the one with the web site issue.

If you have a fault with a company's product, you most certainly have the right to reveal the information. I don't know of a country without product liability laws. In fact, if you are in some way damaged because of this fault (which, if you've discovered it, you probably are damaged) you can sue.

If the information concerns breaching the security of a company's website, there is a good chance you can be sued. Cyber law is something that is still developing, especially in common law. Something US courts (and many international courts) consider very heavily is intent. In this case, corporate attorneys may very well be able to assert that your actions are "unconscionable", in which case, they will find many torts under which to sue you.

In the case of something like a bank, where the information behind encrypted online servers could be very valuable, you're committing both criminal and civil offenses by illegally stealing the information or distributing methods by which such information can be obtained. You probably would, however, be able to offer your services in fixing the problem.

A lot of what you've asked (and in fact, a lot of law in general) depends not only on what is done, but how it's done, the intent of the parties, the context, the magnitude of foreseeable consequences, etc.

Remember that you're on very thin ice. It's like watching someone open their safe and remembering the combination. The observation in itself is not illegal (in most situations). But taking any actions with that information could get you in trouble, especially since you are knowingly using harmful information.