Vista just got released, so are you gonna buy it? Poll above ^^

I am not.

How about...I am not going to buy it because I do not need a new operating system at the moment

I know a guy who has it (he was a beta tester I think). It looks pretty cool and it has a lot of neat features. But I am fine with Windows XP and he said i shouldn't rush out to buy Vista, but if i am able to get a good deal on it then it would be a good idea to get it

I know what catches people - the graphics.

If you seriously want the graphics, go to Vista Transformation Pack and get a vista look completely. Choose system file override etc and it'll be as good as vista itself.

Vista is just a bunch of graphics.

I have the latest public beta and then deinstalled it back to XP because I like XP better but I am now on Suse 10.2. GET LINUX!

I have beta version and I don't like it. It seems to me that XP is more stable.

[quote="jackdillon"]I am pretty much fine with xp, but I would consider getting it if it continues getting good reviews.[/quote]

Mind linking to the site that you find mostly good reviews of Vista? As far as I know most sites don't give Vista good reviews...

I have heard lots of bad things about Vista, but yesterday I went into Best Buy and tested it for myself until the salespeople started giving me dirty looks.  I liked it a lot, and it has some pretty cool upgrades for people who want a few more advanced tools.

However, I talked to people who said they had upgrading issues.  The general consensus seemed to be don't put it on an old computer with lots of installed programs, but go ahead and put it on a recently-purchased computer, or better yet, buy a new computer with Vista.

Go to CompUSA. They won't give you bad looks because teh more you are there, they think you're interested and they work on commission so they think you'll buy it.

As I said, why not linux? Whats up with cool graphics? There's [url=http://www.softpedia.com/get/System/OS-Enhancements/Vista-Transformation-Pack.shtml]Vista Transformation Pack[/url] which makes your computer look completely like vista (Choose sytem files and it looks like vista)

And you can have even better candy with linux :P.

Maybe, but not now. I will consider it after (at least) half a year so that I can know more about it.

I most probably won't buy it unless I get it really really really cheap...
Personally, I feel that since graphics software like AERO,Flip 3D and the new desktop bar(DOCK),as well as IE7 are available on the net for free and since the security has not been as good as promised(I read somewhere that Vista challenged all hackers worldwide last year to crack it's advanced security, and they managed to do so, calling it child's play!!) it's kinda like a waste of money. I can definitely wait.. :D

remember the days that XP kept getting bad reviews?

I saw one that went like "Look at XP sideways doesn't that look funny!"

And now it's the golden standard.  Soon Vista may be.

I don't know though.  It needs improvements.

And also, what worries a lot of people is the compatibility.  Even though it has a compatibility option called "Windows XP SP2"  which doesn't really work well...

I have mixed feelings.

Halo 2 online PC with Vista!!!!

but other than that there is no reason to buy it im probably gonna wait until prices start dropping...

If you want Vista, wait until they come up w/ a "Service Pack 2" so that it's more "secure".

However, Linux is better!

ok well i HAVE a vista. it's okay. it has nice graphics but it's really annoying because that it always tells you it has important updates it needs to install and the notice keeps popping up every 10 minutes. otherwise it's fine

I've been using Vista for 2 months on my personal computer and 1 month at the AoPS office. I can tell you that at this moment I do not recommend the upgrade. Way too many programs don't work, or behave erratically. It's so much of a hassle, plus that it needs enourmous hardware resources, not to mention the costs of upgrading (because you're not only upgrading the OS, you also have to upgrade some of the programs you use for their new versions, compatible with Vista).

I am a mac freak, and while I like cool looking graphics (total pwnage) I am most certainly not going to get vista (it hurts my pride just have xp) :lol:

Well, I wouldn't replace my trusty xp with vista but if I got a new computer I would get vista (but do you really have a choice? unless you get a mac...macs are icky...but windows is icky too and so is linux...lol)

I have a thing for getting a free Vista upgrade, but I didn't take because Vista needs to be more stable. Other than that it should be fine.
Don't uninstall Vista Transformation pack. It will mess up your computer.

In my opinion - Vista should never be installed on an old computer as an upgrade. Its too new and you'd waste its features by buying it like that. I really like my XP/Kubuntu computer (Its totally skinned out and junk xD) at the moment. I would never buy a new computer with XP though.

why at all vista?
ubuntu+Beryl(now Compiz Fusion)+ubuntu community+Repositories+Open Source>>>>>>>>>>Vista :D

Vista sounds good, and I tested it and it was okay. However, I'm not getting it any time soon.

[quote="Anaxerzia"]Kubuntu>Ubuntu[/quote]

yes may be kubuntu looks beautiful but it really isn't that stable as a GNOME environment it really crashes if u don't handle properly ...really difficult to use...doesn't have desktop effects of it's own(though KDE 4 may have one)....community support is high for ubuntu..is well supported...has desktop effects...beryl...very stable..

well anyways u can always install either environments in the respective linuxes :wink:

[url=http://ubuntuforums.org/attachment.php?attachmentid=36282&d=1182682967]here's[/url] what beryl+ubuntu can do to ur desktop all that for free

Gentoo > *ubuntu xD

Naw, ubuntu's good (and for the record, xubuntu, kubuntu, ubuntu, and any other *ubuntu are the same linux - just with different window managers x->xfce, k->kde, normal->gnome), but when I switched to gentoo, I could tell that it was much faster, and I prefer it. Personally, I use fluxbox. I haven't used beryl for quite a while. I prefer the simple black screen.

Vista looks nice. The graphics are good. The downsides are 1) It takes about 2GB of ram to actually get all the eye-candy, 2) nothing has been ported yet, 3) it's very expensive.

Personally, if I got a new computer, I'd probably get rid of the windows partition altogether. I haven't done it with the computers I have because I started with windows, but there's really not all that much use for it. (I mainly use it for starcraft)

[quote="#H34N1"]I know what catches people - the graphics.

If you seriously want the graphics, go to Vista Transformation Pack and get a vista look completely. Choose system file override etc and it'll be as good as vista itself.

Vista is just a bunch of graphics.
[/quote]

I set my XP computer to Windows Classic ('98) style, and I'd do the same on Vista.

[quote="mgao"]remember the days that XP kept getting bad reviews?
I saw one that went like "Look at XP sideways doesn't that look funny!"
And now it's the golden standard.  Soon Vista may be.[/quote]

I still think that xp is a load of rubbish.. (But new computers were sold with it so it soon became widespread.) That is why i use 2003, which i think to be the best windows today. Perfect successor of 98. It's much faster and the design is better. Yes, win98 design is better because it's simple and faster (and isn't ugly at all). And i haven't had compatibility problems except that there is no windows media player for this. (mind that i don't really play games so i don't really have experiences about win2003's gaming capabilities)

MAC OSX 10.5=MAC OSX 10.4 =10*WindowsXP=$ \infty$ * WindowsVista.

Linux fits in there somewhere... :huh:

My guess is MAC OSX 10.4 =2*Linux in general

I only had vista for 5 days....XP is more stable...and if you know working on LINUX, don't hesitate....LINUX is the best...

I know it's been a while, but in response to 1=2's favor of mac over linux, I'd like to mention that mac is actually a flavor of unix. Personally, I feel that the lack of openness on apple's part is a problem (and for the record, it's a bigger problem with microsoft). The nearly complete openness of linux (the closed components are pretty much just hardware drivers for graphics cards and such) allows for a much larger selection of packages.

Also, with regard to the linux distribution argument (which I contributed to a while back), many linux kernels are very similar, all being based on the same thing. What you are really choosing is 1) the package manager, 2) the community, and 3) the documentation.

Like it except for the fact it is annoying about all the security measures. I have it already.

Since this post is already revived, I'm going to ask a new question:
Do you think Windows 7 will be better?

since this post has been revived, i will respond that i hope windows 7 has the speed and convenience of xp and the graphics of vista

@revival:

The only reason I keep Windows on my computer is because I need it to run certain applications, and because certain applications run faster in Windows. Otherwise, I don't have much use for it.

i might consider it but probably not.

my dad is like all over computers so he would probably buy it for his computer in his office.

Vista has a nice layout and graphics (better than XP), but the system can be confusing.
They alert you of installations and updates, so it can be annoying...

[quote="Elixir10"]I have Vista on my laptop. It isn't half as bad as everyone makes it out to be, if your computer is powerful enough.

User Account Control annoys me, though.[/quote]

don't revive


Helo
Proof this:
$ 1\times2\times3\times...\times n\ge2^{n\minus{}1}$

[quote="sami.d"]Helo
Proof this:
$ 1\times2\times3\times...\times n\ge2^{n \minus{} 1}$[/quote]
[hide="Proof"]
This is for $ n$ in the set of counting numbers.
I'll use induction.
$ 1 \times2\times3\times...\times n \ge2^{n \minus{} 1}$ (how do u do the question mark above $ \ge$ ?)
Works for $ n \equal{} 1$: $ 1\ge2^0$
Let $ k \equal{} n$
$ 1 \times2\times3\times...\times k \ge2^{k \minus{} 1}$
Prove For $ k \plus{} 1$
$ 1 \times2 \times3 \times... \times k \times (k \plus{} 1) \ge 2^{k}$
$ 2^{k \minus{} 1} \times (k \plus{} 1) \ge 2^{k}$
$ \frac {2^{k \minus{} 1} \times (k \plus{} 1)}{2^{k \minus{} 1}} \ge \frac {2^{k}}{2^{k \minus{} 1}}$
$ (k \plus{} 1) \ge 2$
$ k \ge 1$
this is true for all counting numbers
[/hide]

We want to show that $ n! \geq 2^{n\minus{}1}$

If $ n\equal{}0$, we have $ 1 \geq \frac{1}{2}$, and if $ n\equal{}1$, then we have $ 1 \geq 1$

If $ n \geq 2$, the statement is equivalent to $ \prod_{j\equal{}2}^n j \geq \prod_{j\equal{}2}^n 2 \iff \prod_{j\equal{}2}^n \frac{j}{2} \geq 1$, which is true since each factor on the left is at least 1.

For $ n\geq 4, \ n! \equal{} 1\cdot 2\cdot 3\cdot 4\cdot \cdots n.$
$ \ \ \ \ \ \  2^{n} \equal{} 2\cdot 2\cdot 2\cdot 2\cdot \cdots 2.$
$ \therefore n!\geq 2^n.$ Done!


Let $ a,b,c > 0$ satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \sum_{cyc} \left[\frac {a}{bc}(a^2 \minus{} b^2)\right] \plus{} 4 \geq 2\sum_{cyc}\left(\frac {b^2 \plus{} c}{c}\right)$
Hope you will like this; dear friends. :)
[edited]; {thanks to nguoivn.}

Dear Polta, you should check again your problem. It's equivalent to:
$ \sum\ a^4\plus{}2abc(a\plus{}b\plus{}c) \ge \sum\ a^2b^2\plus{}2\sum\ a^3b$
But this ineq is not trues, try $ a\equal{}b\equal{}1; c\equal{}0$  :wink:

[quote="nguoivn"]Dear Polta, you should check again your problem. It's equivalent to:
$ \sum\ a^4 \plus{} 2abc(a \plus{} b \plus{} c) \ge \sum\ a^2b^2 \plus{} 2\sum\ a^3b$
But this ineq is not trues, try $ a \equal{} b \equal{} 1; c \equal{} 0$  :wink:[/quote]

note that $ a \plus{} b \plus{} c \equal{} 1$, although $ a$, $ b \rightarrow \frac {1}{2}$, $ c\rightarrow 0$ is a counterexample.  :lol:

@earldbest: nguoivn's inequality is homogeneous, so $ a\equal{}b\equal{}1$, $ c\equal{}0$ is valid (except the fact, that $ c>0$, but that doesn't matter).

Sorry for my mistake in calculation. the inequality should be:
[quote="Potla"]Let $ a,b,c > 0$ satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \sum_{cyc} \left[\frac {a}{bc}(a^2 \minus{} b^2)\right] \plus{} 4 \geq 2\sum_{cyc}\left(\frac {b^2 \plus{} c}{c}\right)$
[/quote]
 :(

This is new one is wrong with $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$:

$ \sum_{cyc}\left[\frac{a}{bc}(a^{2}\minus{}b^{2})\right]\plus{}4\geq 2\sum_{cyc}\left(\frac{b^{2}\plus{}c}{c}\right)$

Seems odd that there is a $ \frac{b^2\plus{}c}{c}$, as $ \frac{c}{c}\equal{}1$. Perhaps something else was meant?

Yes, I understand... I did not start from a correct point , so that the whole inequality is wrong. Sorry about that.

OK, applying the correct hint I have arrived at this new (and correct, I think....) inequality:
For positive reals $ a,b,c$ satisfying $ a\plus{}b\plus{}c\equal{}1$ prove that :
\[ \boxed{12\minus{}\sum_{cyc}\left[\frac{a}{bc}(13b^2\minus{}3a^2)\right]\geq 6\sum_{cyc}\left[\frac{b^2\plus{}c}{c}\right]}\]
Hope this one is all right.....
Sorry to you all for posting a wrong problem before... :(

[quote="Potla"]OK, applying the correct hint I have arrived at this new (and correct, I think....) inequality:
For positive reals $ a,b,c$ satisfying $ a \plus{} b \plus{} c \equal{} 1$ prove that :
\[ \boxed{12 \minus{} \sum_{cyc}\left[\frac {a}{bc}(13b^2 \minus{} 3a^2)\right]\geq 6\sum_{cyc}\left[\frac {b^2 \plus{} c}{c}\right]}\]
Hope this one is all right.....
Sorry to you all for posting a wrong problem before... :([/quote]
Since this post is remaining unanswered for such a long time, I will post  a hint.
[hide="Hint"]Rewrite the inequality in a simpler form, and use Vasc's inequality:
$ a,b,c,p,q,r\in\mathbb R$; $ 3(1\plus{}r)\geq p^2\minus{}pq\plus{}q^2$; then we have
\[ \sum a^4 \plus{} r\sum a^2b^2\plus{} (p\plus{}q\minus{}r\minus{}1)abc\sum a\geq p\sum a^3b\plus{}q\sum ab^3\]
That really kills the problem..... :( [/hide]


Let x,y,z be positive reals. Prove that \[ \frac {x}{x \plus{} 2} \plus{} \frac {y}{y \plus{} 2} \plus{} \frac {z}{z \plus{} 2} \ge \frac {(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2 \plus{} 2x \plus{} 2y \plus{} 2z}\]

[quote="pythag011"]Let x,y,z be positive reals. Prove that
\[ \frac {x}{x \plus{} 2} \plus{} \frac {y}{y \plus{} 2} \plus{} \frac {z}{z \plus{} 2} \ge \frac {(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2 \plus{} 2x \plus{} 2y \plus{} 2z}
\]
[/quote]

This is a very easy inequality.

This inequality is the same as

$ \frac{x^2}{x^2 \plus{} 2x} \plus{}\frac {y^2}{y^2 \plus{} 2y} \plus{} \frac{z^2}{z^2 \plus{} 2z} \ge \frac {(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2 \plus{} 2x \plus{} 2y \plus{} 2z}$

But this is a direct consequence of Cauchy's lemma

Cauchy's lemma states:

$ x^2/a \plus{} y^2/b \plus{} z^2/c \ge (x\plus{}y\plus{}z)^2/a\plus{}b\plus{}c$

$ Q.E.D$

[quote]Can I have a link to Cauchy's lemma?[/quote]

[color=red]Cauchy's lemma [/color]or [color=red]Andreescu inequallity[/color]

if $ x_1,x_2,...,x_n \in \Re$ and $ a_1,a_2,...,a_n$ are positive reals then:

 $ \frac {x_1^2}{a_1} \plus{} \frac {x_2^2}{a_2} \plus{} ... \plus{} \frac {x_n^2}{a_2} \geq \frac {(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}$

[hide="proof"]aply B-C-S for the n-tuples $ (\frac{x_1}{\sqrt {a_1}},\frac{x_2}{\sqrt {a_2}},...,\frac{x_n}{\sqrt {a_n}})$ and $ (\sqrt {a_1},\sqrt {a_2},...,\sqrt {a_n})$ [/hide]

PS:this inequallity is very useful to proove other inequallities.In the solution above (that befuddlers done) there is a clasic mothod for using cauchy's lemma
another apllycation of cauchy's lemma is nesbitt's inequality:[url]http://www.artofproblemsolving.com/Wiki/index.php/Nesbitt%27s_Inequality[/url]

if you have any questions we are here  :)


Find all real $x,y,z$ that \[\left\{\begin{array}{c}x+y+zx=\frac12\\ \\ y+z+xy=\frac12\\ \\ z+x+yz=\frac12\end{array}\right.\]

[hide="Probably the ugliest (but brave...) solution.... :P"]

find $x$ in the first expression and then substitute it in second and third expressions. Then find $y$ in the third and substitute in the second. Now we have a polynomial in $z$ that is:

$P(z)=4z^{5}+2z^{4}-20z^{3}-5z^{2}+11z-2$

Now we have that if $x=y=z= \pm \sqrt{ \frac{3}{2}}+1$ we have solution. thus $2z^{2}+4z-1$ must factor $P(z)$. In fact:

$P(z)=(2z^{2}+4z-1)(2z^{3}-3z^{2}-3z+2)=(2z^{2}+4z-1)(2z-1)(z-2)(z+1)$

And so we're able to find all the solution:

$(x,y,z)=(\pm \sqrt{ \frac{3}{2}}+1,\pm \sqrt{ \frac{3}{2}}+1,\pm \sqrt{ \frac{3}{2}}+1)$

$(x,y,z)= (\frac{1}{2},-1 , 2)$ and cyclics.

[However when we find $x$ in the first expression we have to divide per $z+1$ so we have to check this, but in the final polynomial is obvious that we consider it.]

[/hide]

Just a quick question. I did the question as well but I started with the $ \text{Rational Roots Theorem}$  and got $ x\equal{}\minus{}1$ as a root.
Etc...
I am just curious how you got the $ x\equal{}y\equal{}z\equal{} \pm \sqrt{ \frac{3}{2}}\plus{}1$ from the polynomial $ P(z)\equal{}4z^{5}\plus{}2z^{4}\minus{}20z^{3}\minus{}5z^{2}\plus{}11z\minus{}2$.

Thanks ^_^

Also I keep seeing it around how do you make the words blue and one needs to click on the words to make stuff appear? The link thing.

[hide]He saw that by setting $ x \equal{} y \equal{} z$, then $ z \equal{} 1 \pm \sqrt {\frac {3}{2}}$ is a solution in the original system. Thus $ z \equal{} 1 \pm \sqrt {\frac {3}{2}}$ must be a solution in $ P(z)$ as well.
Since $ 1 \pm \sqrt {\frac {3}{2}}$ are the roots of $ 2z^2 \plus{} 4z \minus{} 1 \equal{} 0$, $ 2z^2 \plus{} 4z \minus{} 1$ must be a factor of $ P(z)$.

To make some words blue such that one needs to click on the words to make stuff appear, type:
[code][hide=words that you want to be blue]
stuff that you want to appear when one clicks the blue words
[/hide][/code][/hide]

Note that
(1) $ (x \minus{} y)z \equal{} \minus{} (y \minus{} z)$,
(2) $ (y \minus{} z)x \equal{} \minus{} (z \minus{} x)$, and
(3) $ (z \minus{} x)y \equal{} \minus{} (x \minus{} y)$.

[b]Case 1[/b]: $ x$, $ y$, and $ z$ are distinct.

   Hence, $ xyz \equal{} \minus{} 1$.  Therefore,
\[ xy \plus{} y^2 \equal{} \frac {y}{2} \plus{} 1\,,
\]

\[ yz \plus{} z^2 \equal{} \frac {z}{2} \plus{} 1\,,
\]
and
\[ zx \plus{} x^2 \equal{} \frac {x}{2} \plus{} 1\,.
\]
   Call the three equations above (*).  Thus, if $ s \equal{} x \plus{} y \plus{} z$, then
\[ s^2 \minus{} xy \minus{} yz \minus{} zx \equal{} \frac {s}{2} \plus{} 3\,.
\]
   However,
\[ 2s \plus{} (xy \plus{} yz \plus{} zx) \equal{} (x \plus{} y \plus{} zx) \plus{} (y \plus{} z \plus{} xy) \plus{} (z \plus{} x \plus{} yz) \equal{} \frac {3}{2}\,.
\]
   Therefore,
\[ s^2 \plus{} 2s \minus{} \frac {3}{2} \equal{} \frac {s}{2} \plus{} 3\,.
\]
   Consequently, $ s \equal{} \minus{} 3$ or $ s \equal{} \frac {3}{2}$.  If $ x \plus{} y \plus{} z \equal{} s \equal{} \minus{} 3$, then $ xy \plus{} yz \plus{} zx \equal{} \frac {15}{2}\,.$   Note that
\[ x^2 \plus{} y^2 \plus{} z^2 \equal{} s^2 \minus{} 2(xy \plus{} yz \plus{} zx) \equal{} \minus{} 6\,.
\]
   Thus, $ x$, $ y$, or $ z$ is not real.  Therefore, $ x \plus{} y \plus{} z$ must be $ \frac {3}{2}$.  In this case, $ xy \plus{} yz \plus{} zx \equal{} \minus{} \frac {3}{2}$.  Therefore, $ x,y,z$ are the roots of
\[ t^3 \minus{} \frac {3}{2}t^2 \minus{} \frac {3}{2}t \plus{} 1 \equal{} 0\,.
\]
   Hence, $ (x,y,z)$ is a permutation of $ \left(2, \minus{} 1,\frac {1}{2}\right)$.

 *PS: Alternatively, if $ s \equal{} \minus{} 3$, from (*),
\[ (z \minus{} x)y \equal{} \left(y \plus{} z \minus{} \frac {1}{2}\right)(y \minus{} z) \equal{} \minus{} \left(x \plus{} \frac {7}{2}\right)(y \minus{} z)\,,
\]

\[ (x \minus{} y)z \equal{} \left(z \plus{} x \minus{} \frac {1}{2}\right)(z \minus{} x) \equal{} \minus{} \left(y \plus{} \frac {7}{2}\right)(z \minus{} x)\,,
\]
and
\[ (y \minus{} z)x \equal{} \left(x \plus{} y \minus{} \frac {1}{2}\right)(x \minus{} y) \equal{} \minus{} \left(z \plus{} \frac {7}{2}\right)(x \minus{} y)\,.
\]
Hence,
\[ \minus{} 1 \equal{} xyz \equal{} \minus{} \left(x \plus{} \frac {7}{2}\right)\left(y \plus{} \frac {7}{2}\right)\left(z \plus{} \frac {7}{2}\right)\,.
\]
Thus,
\[ 1 \equal{} \frac {343}{8} \plus{} \frac {49}{4}(x \plus{} y \plus{} z) \plus{} \frac {7}{2}(xy \plus{} yz \plus{} zx) \plus{} xyz\,,
\]
or $ 1 \equal{} \frac {343}{8} \plus{} \frac {49}{4}( \minus{} 3) \plus{} \frac {7}{2}\left(\frac {15}{2}\right) \minus{} 1 \equal{} \frac {251}{8}$, which is a contradiction.  Thus, $ s \equal{} \frac {3}{2}$.

***********

[b]Case 2[/b]: Two of $ x$, $ y$, and $ z$ are equal.

Then, (1), (2), and (3) ensure that $ x \equal{} y \equal{} z$.  In this case, they are the roots of
\[ t^2 \plus{} 2t \minus{} \frac {1}{2} \equal{} 0\,.
\]
That is, $ x \equal{} y \equal{} z \equal{} \minus{} 1\pm\sqrt {\frac {3}{2}}\,.$


Hello. Can anybody send me/post the packing list? I just realized that I should do that and I'm not sure what to bring.

Here's the "official" packing list, for students of course (on page 2).


For any $n,T \geq 2, n, T \in \mathbb{N}$, find all $a \in \mathbb{N}$ such that $\forall a_i > 0, i = 1, 2, \ldots, n$, we have

\[\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k},\] where $S_k = \sum^k_{i=1} a_i.$

My Solution, please mention if im wrong.

First we have, $T^2\ge 4$ then, $$\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k}=  \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k} < 4\sum_{k=1}^{n}\frac{1}{a_k}\Rightarrow \frac{1}{2}\left( an(n+1)\sum_{k=1}^{n}\frac{1}{S_k}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}\right)< 4\sum_{k=1}^{n}\frac{1}{a_k} $$

\[ \Rightarrow \left( an(n+1)\sum_{k=1}^{n}\frac{1}{S_k}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}\right)< 8\sum_{k=1}^{n}\frac{1}{a_k} \Rightarrow \left( an(n+1)+\frac{a^2}{2}\right)\left( \sum_{k=1}^{n}\frac{1}{S_k}\right)< 8\sum_{k=1}^{n}\frac{1}{a_k}  \]

Now we calculate this finite sums. Let's try, \[ \frac{\frac{1}{S_1}+\frac{1}{S_2}+\dots+\frac{1}{S_n}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}=\frac{\frac{1}{S_1}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}+\dots+\frac{\frac{1}{S_n}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}} \]

Then,
\[ \Rightarrow \frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}=  \frac{\sum_{sym}\left(\prod_{i=1}^{n-1}a_i\right)}{\prod_{i=1}^{n}a_i}\]

\[ \Rightarrow \frac{\prod_{i=1}^{n} a_i}{\left( \sum_{j=0}^{n}(n-j)a_j\right) \left( \sum_{sym} \left(\prod_{i=1}^{n-1}a_i\right)\right)}\left( \frac{(a^2+2an(n+1)}{2}\right)< 16 \]

If we choose $n\rightarrow \infty$ then $\frac{\prod_{i=1}^{n} a_i}{\left( \sum_{j=0}^{n}(n-j)a_j\right) \left( \sum_{sym} \left(\prod_{i=1}^{n-1}a_i\right)\right)} \rightarrow 1$ quickly.

We assume it has a value equal $1$ also its easy to see $2\le n\le 3$. After trying, solution for $a$ is $1$ and $2$ which is obvious. So we are done
! :oops:

[quote=mathmaths]I can't see in the second row:$$ \Longleftrightarrow \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}<3\sum_{k=1}^{n}\frac{1}{a_k}$$.
    For $T^2\ge 4$,why this line is true?[/quote]

Cause

$$ \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}= \sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} <3 \sum_{k=1}^{n}\frac{1}{a_k}<T^2 \sum_{k=1}^{n}\frac{1}{a_k} \Longleftrightarrow T^2\ge 4>3$$   :maybe: i think


integration
calculus
function
inequalities
real analysis
real analysis solved

I?m going to post some more of Cristinel Mortici?s problems (they are taken from a book of his: ?600 problems?) and I?m going to name them P1, P2 etc. in separate posts.

P1

Given a>0 and f:[0,1]->R an integrable function s.t. :
a)	0<=f(x)<=a^2/3
b)	integral from 0 to 1 of f(x)dx=a
Show that integral from 0 to 1 of (f(x))^1/2 dx >= a^2/3

We use Cauchy-Schwarz inequality
(\int h.g) 2  \leq \inth 2 .\intg 2

(*)   (\int f 1/4f 3/4) 2 \leq \int f 1/2 \int f 3/2

The LHS is a 2

\int f 3/2 = \intf. \sqrt f \leq a 1/3\intf=a 1/3 a

(*)  =>

a 2  \leq a 1/3 a\int \sqrt f

2-1/3 -1 = 1- 1/3 = 2/3

a 2/3  \leq  \int \sqrt f

Nice inequality

The solution that is in the book goes like so:

 \int  sqrt(f(x)) dx=  \int f(x)/sqrt(f(x)) dx >= \int f(x)/a^(1/3) dx =
= 1/a^(1/3) \int f(x) dx=a^(2/3).

this is the "official" solution .. don't ask me why it is like that .. I don't know integrals .. not yet.. next year! :D :D :D

cheers!

The solution from the book, is nice, very short.

But we write f/ \sqrt f,  \sqrt f must be different than 0.
I don't see why f(x) cannot be 0 in some point of [0;1]

Cheers

It took me some time to remember my soln:

 \int {0 to 1} (f(x))^1/2*((f(x))^1/2-a^1/3) dx <= 0 since the function we integrated is <= 0 from the first conditiond. Now we multiply and we get  a= \int {0 to 1} f(x) dx <= a^1/3* \int {0 to 1} (f(x))^1/2 dx. We divide it by a^1/3 and we get exactly what we want. It's also short and it avoids division by (f(x))^1/2.

Bye! :D


slove for x
x=x+1
where x is a complex number
is there any?[/i]

Square both sides
$ x^2\equal{}x^2\plus{}2x\plus{}1$
$ x\equal{}\minus{}\frac{1}{2}$
Just kidding  :D
Of course there are no complex solutions, because a complex number is the sum of its real and imaginary part, and that just adds 1 to the real part while making the imaginary part stay the same.

hello, the equation $ x\equal{}x\plus{}1$ has no solutions in complex numbers.
Sonnhard.

there is no complex solution! and x<x+1 always

@anuonhu: for complex nos. there is no such thing as x<x+1 (as far as I know)
maybe you meant magnitude (modulus) of x < magnitude of (x+1) ?
but even that is not necessary
E.g.  $ x\equal{} \minus{}3\plus{}2i  (\equal{}> x\plus{}1\equal{} \minus{}2 \plus{} 2i)$

anyway,
@ericlou: there is no complex no. x such that $ x\equal{}x\plus{}1$

$ x \equal{} x \plus{} 1$ if and only if $ 1 \equal{} 0$, which isn't true.  This isn't unique to the complex numbers; it's true (well, by definition) in every ring.


[quote="fourierseries"]{$a_{n}$} is a sequence.

If $a_{n}< \sum^{2n+2006}_{k=n}\frac{a_{k}}{k+1}+\frac{1}{2n+2007},$

Prove $a_{n}<\frac{1}{n}$ for n=1,2,3,...[/quote]

Something is not correct, because the inequality can be satisfied with unbounded $a_{n}$.  For example, $a_{k}= k+1$  for all $k$, will work.


email
induction
combinatorics unsolved
combinatorics

Given an integer $n$  greater than 2,show that there exists an integer $H(n)$ such that if there are $H(n)$ points on the plane,no three of which are collinear,then one can always choose $n$ points among them to form a convex n-gon.
BTW:my english is too low,should I say [color=red]on the plane [/color]or [color=red]in the plane[/color]?

if you tell me your email i can send you a whole article on this problem. it is known as the erdos-sekerez happy ending problem i think...

I'm aware of the fact that it's been posted before ans solved (by Pierre, I think), but I've never tried to solve it so I'd like to give it a shot, since I find it very enjoyable :).

We can prove it by induction on $n$. Suppose the assertion is true for all naturals less than $n$, and the aim is to prove it for $n$.

We construct a hypergraph having the points as vertices, and as "sides" the $n-1$-size sets which form convex $n-1$-gons. If we prove that we can find a complete sub-hypergraph on $n$ points, we're done: it's easy to see that if every $n-1$ among $n$ points form a convex $n-1$-gon, then all the points must form a convex $n$-gon.

So the problem is now this: given $H(n-1)$, we can find some $H(n)$ s.t. for every $H(n)$-element set with the property that every $H(n-1)$-subset contains a colored $n-1$-subset, we can find an $n$-element subset with all its $n-1$-subsets being colored.

The above is an easy consequence of Ramsey's Theorem for hypergraphs.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=26701 and http://www.mathlinks.ro/Forum/viewtopic.php?t=21981 .

  darij


function
modular arithmetic
number theory
totient function
relatively prime

How, using modular arithmetic, would you determine the last two digits (tens and units) of $ 2^{100}$ ?

Maybe...
[hide]From Euler's Totient function (the word of the week!), $ 2^{40} \equal{} 2^{100\varphi} \equiv 1 \pmod {100}$.  This implies that $ 2^{80} \equiv 1 \pmod {100}$, which goes away because of multiplication by $ 2^{20}$.  We either brute force $ 2^{20}$ or break it up.  $ 2^{20}\equal{}(2^{10})^2 \rightarrow 2^{10} \equiv 24 \pmod {100} \rightarrow 2^{20} \equiv 76 \pmod {100}$.  I hope the answer is $ 76$.[/hide]

You can look in the forum here to find extra help.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216082[/url]

But by the Euler's function : $ a^{\varphi x}\equiv 1 \pmod {x}$ if $ gcd(a,x)\equal{}1$
am I wrong?

Dang...

$ 2^{10} \equiv 24 \pmod {100}$
$ 2^{20} \equiv 76 \pmod {100}$
Fortunately, $ 76^5 \equiv 76 \pmod {100}$.

So I used the wrong solution to get the right answer.

You can "split" the mod 100 into its prime components mod 25 and mod 4.

For this problem:
Clearly, $ 2^{100}\equiv0 \pmod 4$. By Euler's Totient function, $ 2^{20}\equiv1 \pmod {25}$, so $ 2^{100}\equiv 1 \pmod {25}$. Therefore, $ 2^{100}\equiv 76 \pmod {100}$. (Test out 1, 26, 51, and 76 - the existence and uniqueness of a corresponding solution is guaranteed by the Chinese Remainder Theorem.)

ok I found a solution to $ 2^{999}$  :lol: :
by the Euler's: $ 2^{1000}\equiv 1 \pmod {25}$
$ 2*2^{999}\equiv 1 \pmod {25}$
also note that $ 13*2\equiv 1 \pmod {25}$ and $ 2^{999}\equiv 0 \pmod {4}$
$ 2*13*2^{999}\equiv 13 \pmod {25}$
 $ 13 \pmod {25}$ are $ 13,38,63,88$.
Only 88 is divisible by 4 so the answer is $ 88$.

Applying the same methods, its last [i]four[/i] digits are $ 4688$.

When the exponent is relatively prime to $ 10$, it can be a lot easier to use the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url].  In particular, $ \lambda(100) \equal{} 20, \lambda(1000) \equal{} 100, \lambda(10000) \equal{} 500$, and generally $ \lambda(10^k) \equal{} 5 \cdot 10^{k\minus{}2}, k > 4$.  By comparison, the value of $ \phi$ is $ 8$ times larger.


function
real analysis
real analysis unsolved

Let $f: [0,+\infty] \to [0,+\infty]$-continiously differentiable function such that $lim_{x\to+\infty}\frac{f'(x)}{f(x)}=-\infty$.
a).Prove that for all $A \in \mathbb{R}$, there exists $N \in \mathbb{R}$ such that $\forall n\geq N$, $\forall p\in \mathbb{N}$, $f(n+p) \leq f(n)e^{pA}$.
b).Prove thet $\sum_{n\geq1}f(n)$ converges.
c).Give an equivalen of $R_n=\sum_{n\geq k}f(n)$

a) first of all note that $f$ decreases on a neibourghood of infinity.
if $A>0$ then $f(n+p)\le f(n)\le f(n)e^{Ap}$
if $A<0$ it is sufficient to prove that $e^{-Ax}f(x)$ decreases on a neibourgeooh of infinity. Differentiating and using the fact that $\frac{f'(x)}{f(x)}\to -\infty$ we get that the derivate of  $e^{-Ax}f(x)$ is ngative on a neibourghood of infinity.

b)Fix an $A<0$. Then we have that $\sum_{n\ge 1}f(n)=S(N)+\sum_{n\ge N}f(n)<S(N)+f(N) \sum_{n\ge N} e^{A(n-N)}<S(N)+f(N)\frac{1}{1-e^{A}}$. So, the sum converges by comparision.


Here is a challenge. Probably really easy for most people but I found this interesting.  :D

Mr. Keterson found an interesting problem for his class. It said:

If \[\displaystyle (3x-1)^{7} = a_{7}x^{7} + a_{6}x^{6} + ... + a_{1}x + a_{0}\] for all \[x\], then \[a_{0}+a_{1}+....+a_{6}+a_{7}\] equals what?

What was the answer to Mr. Keterson's this nice problem?  :lol:

Here is my solution. I hope it's right.

[hide]
\[(3x-1)^{7} = \sum_{i=0}^{n} \binom {n}{i}x^{n-i}y^{i}\]
To find our need, we need to work on the part of the binomial, so:
\[= \sum_{i=0}^{7} \Rightarrow \binom {7}{0} \binom {7}{1} \binom {7}{2} \binom {7}{3} \binom {7}{4} \binom {7}{5} \binom {7}{6} \binom {7}{7} = 128\]   [/hide]

You forgot to put '+' signs between all of those terms.  Also, there is a much more elgant approach to this problem.  See if you can find it.


geometry
rectangle
linear algebra
matrix
AMC
USA(J)MO
USAMO

On a chess table $n\times m$ we call a [i]move [/i] the following succesion of operations

 (i) choosing some unmarked squares, any two not lying on the same row or column;
 (ii) marking them with 1;
(iii) marking with 0 all the unmarked squares which lie on the same line and column with a square marked with the number 1 (even if the square has been marked with 1 on another move).

We call a [i]game [/i]a succession of moves that end in the moment that we cannot make any more moves.

What is the maximum possible sum of the numbers on the table at the end of a game?

You say:
(iii) marking with 0 all the UNMARKED squares which lie on the same line and column with a square marked with the number 1 (even if the square has been MARKED with 1 on another move).

I see a contradiction between the two adjectives
Can you explain it to me one more time?

I see no contradiction. You did not read carefully the problem's text (as some of the contestants did!). I shall give you an example:

if at your first move you choose two different 1s, say the first two on the main diagonal (up-left corner, + one below right and down), then you mark with 0 the other 2 squares with which these 1s form a rectangle.

then in your second move, if you choose only one unmarked square, say the one in the right-down corner, you have to mark with 0, the other two corners of the rectangle, because those squares lie on the intersection of the colum on which the last 1 chosen lies, and one of the first two 1s lies. So you must color in your second move exactly 4 squares (in this beginning of a game that I have chosen).

[quote="iura"]You mea what is the greter number of ones one can put in a m*n matrix s.t. for any 2 ones, there is no other 1 lying in a row with the first and in a column with another?
It looks like some problem from USAMO[/quote] yes I belive you can re-phrase the problem in this manner.

If the problem is so, I know the solution.
Consider the sides of rectangle and consider the number of unit squares on border the 1's project to.
If we can't take two squares lying in distinct columns and rows then all 1's lie in a row or column so we get at most max{m,n} 1's.
If we can, this squares project into 4 unit squares.
The next squares add at least 2 new squares on the border ( otherwise this 1 is in a row and column with anether 1's contradiction).
As there are 2(m+n) squares in the border, we get at most m+n-2 squares.
This bound is easily achieved so desired number is m+n-2

oh my god! this problem has been so confusing!!!

let me explain what i've understood of it:

at first the board has all the fields unmarked.

step 1 choose some UNMARKED fields, any two not on the same line/column;

step 2 consider ALL fields containing "1" and for any UNMARKED field that lies at intersection of a line that contains a "1" AND a column that contains a "1" mark that field with "0"...  so the remark

[quote]even if the square has been marked with 1 on another move [/quote]

simply specifies that we consider ALL the "1"'s... recent and old

Iura, if you understand what i say you see that now you CAN have 3 fields that are verices of a right triangle... here are a series of moves for 2x2

put "1" in (1,1) ... no "0"'s
put "1" in (2,1) ... no "0"'s (is it obvious?)
put "1" in (1,2) ... field (2,2) is on the same line with (2,1) and on the same column with (1,2) so mark it with 0...

that's the end.

Then I still don't understand the problem!
What squares are MARKED?

Now I think MARKED squares are those in which there is written a 0 or 1.
Then my answer is m+n-1.
The proof is similar to my previous:
Take projections on sides.
First square projects on 4.
Next squares add at least 2.
So we get most m+n-1 which is achievable (Kappa's example for 2*2)

Yup... that's right!  i forgot to mention that m+n-1 can be reached, so i got 6/7 :blush:


Please help me with this one:
   The rate at which an infectious disease spread throught the population directly proportional to the fraction of the population, not yet expose to the disease
   Let y is a fraction of the population affected to the disease, t is the time in month. Find the differential equation.

I still don't understand what my teacher asks for. In my thought, from "The rate at which an infectious disease spread throught the population directly proportional to the fraction of the population" :
 I write $\frac{d_d}{d_t}$ as the rate the infectious dease spread so
$\frac{d_d}{d_t}=ky$ But I don't understand what my teacher means by "not yet expose to the disease".
  Please help me!

[quote="Gibbenergy"] But I don't understand what my teacher means by "not yet expose to the disease".[/quote]

Say that you have a population $P$, of which $I(t)$ members of the population have been exposed.  Then the "number of people not yet exposed to the disease" is $P - I(t)$, and the differential equation is

$\frac{dI}{dt} = k(P - I)$

For a constant $P$.  Hope this helps :)


analytic geometry
Support
inequalities
triangle inequality

THe points A(0,1) B(1,1) C(1,0) and D(0,0) are in a plane.  P is also a point.  What is the mnimum value of PA+PB+PC+PD

[hide="???"]
Wouldn't just be $2+\sqrt{2}$?
[/hide]

[hide]
It's a square, so the smallest value is at the point where the diagonals intersect, so the answer would be the sum of the diagonals, $2\sqrt{2}$[/hide]

[quote="alkjash"]It's a square, so the smallest value is at the point where the diagonals intersect[/quote]

Can you support this claim?

[hide="With, possibly..."] Triangle inequality? [/hide]

alright let me do a bit of a different question that will have to make you actually use the triangle inequality instead of intuition.

A=(2,0) B=(2,4) C=(0,2) D=(0,1) what is the shortest distance PA+PB+PC+PD if P is some point.


limit
inequalities
real analysis
real analysis unsolved

Hi everyone,

I'm having quite a hard time with the following problem:

Let $ 0 < p < q\leq\infty$, where $ 0 < p < 1$. Find a sequence of continuous linear mappings $ \{\Lambda_{n}\}$ into $ \mathbb C$ such that

$ \lim_{n}\Lambda_{n}(f) \equal{} 0$ for every $ f\in l^{p}$
and there exists a $ g\in l^{q}$ such that $ \{\Lambda_{n}(g)\}$ does not converge (to a finite number).

I have been fighting this one for quite a while now so all help is most appreciated  :oops: .

-Laura

PS. The case where both $ p$ and $ q$ are $ < 1$ is the most important one for me
and here $ l^{p}$ means the sequence space (as usual).

For the unimportant case $ q > 1$: let $ \Lambda_{n}(f) \equal{}\sum_{i \equal{} n}^{2n}f_{i}$. For any $ f\in\ell^{p}$ we have
$ |\Lambda_{n}(f)|\le\sup_{n\le i\le 2n}|f_{i}|^{1\minus{}p}\sum_{i \equal{} n}^{2n}|f_{i}|^{p}\le\sup_{i\ge n}|f_{i}|^{1\minus{}p}\sum_{i \equal{} 1}^{\infty}|f_{i}|^{p}\to 0$ as $ n\to\infty$. On the other hand, the sequence $ g_{i}\equal{} 1/i$ is in $ \ell^{q}$ for all $ q > 1$ and $ \Lambda_{n}(g)\not\to 0$.

Have you tried $ \Lambda_{n}(f) \equal{} n^{\alpha}\sum_{i \equal{} n}^{2n}f_{i}$ with a suitably chosen $ \alpha$? Holder's inequality should tell you which $ \alpha$ to use.

The thing is that as far as I know Hölder's inequality requires that $ p>1$ (maybe there is an extension to this?). I can't come up with a way to estimate the functional for the $ l^{p}$- functions. But  thanks anyway mlok!

The best I have managed to come up with so far is that if $ f\in l^{p}$ , then we must have
$ \liminf_{j}\frac{|f(j)|}{j^{\minus{}1/p}}\equal{} 0$,
because otherwise there would exist a $ K\in\mathbb N$ and $ c > 0$ such that
$ \sum_{k > K}|f(p)|^{p}\geq c\sum_{k > K}j^{\minus{}1}\equal{}\infty$.

On the other hand setting $ g(j): \equal{} j^{\minus{}\frac{1}{2}(\frac{1}{p}\plus{}\frac{1}{q}) }$
we have $ g\in l^{q}\setminus l^{p}$ and $ \liminf_{j}\frac{|g(j)|}{j^{\minus{}1/p}}\equal{}\infty$.

So using that could work, except of course that $ \liminf$  is not linear in addition. And I haven't found any way to overcome this. So maybe it is not the right approach.

Noteworthy to anyone trying to solve this is the known fact that for every $ 0 < p < 1$
$ (l^{p})^{*}\equal{} l^{\infty}$,
and so every functional on $ l^{p}$ is of the form
$ \Lambda_{n}(f): \equal{}\sum_{j}g_{n}(j) f(j)$, where $ g_{n}\in l^{\infty}$.

-Laura

What I had in mind was $ \Lambda_{n}(f) \equal{} n^{\alpha}\sum_{i \equal{} n}^{2n}f_{i}$ with $ q^{\minus{}1}\minus{}1 <\alpha < p^{\minus{}1}\minus{}1$. Then the $ \ell^{q}$ sequence $ g_{i}\equal{} 1/i^{\alpha\plus{}1}$ satisfies $ \Lambda_{n}(g)\not\to 0$. Also, $ \Lambda_{n}(f)\to 0$ if $ f\in\ell^{p}$ is a [b]nonincreasing[/b] sequence (in absolute value).

I'm afraid that the above is the best one can do, and here is why.

[b]Lemma 1.[/b] Let $ 0 < p <\infty$. Let $ \Lambda_{n}(f): \equal{}\sum_{j}g_{n}(j) f(j)$, where $ g_{n}\in\ell^{\infty}$ for each $ n$. If $ \Lambda_{n}(f)\to 0$ for any $ f\in\ell^{p}$, then $ \sup_{n}\|g_{n}\|_{\infty}<\infty$.

Proof. Suppose not. Then there is a sequence $ (n_{k})$ such that $ \|g_{n_{k}}\|_{\infty}\ge 2\|g_{n_{k\minus{}1}}\|_{\infty}$. Choose $ i_{k}$ so that $ |g_{n_{k}}(i_{k})|\ge 0.9\|g_{n_{k}}\|_{\infty}$. Define $ f$ by $ f(i_{k}) \equal{}\mathrm{sign}(g_{n_{k}}(i_{k}))/k^{2/p}$ and $ f(j) \equal{} 0$ for other $ j$. Clearly $ f\in\ell^{p}$. However, $ \Lambda_{n_{k}}(f)\ge 0.9\cdot 2^{k\minus{}1}/k^{2/p}\to\infty$.

[b]Lemma 2.[/b] Under the assumptions of Lemma 1 we have $ \max_{i\le m}|g_{n}(i)|\to 0$ as $ n\to\infty$, for any fixed $ m$.

Proof: consider finite sequences.

Combining Lemma 1 and Lemma 2, you should be able to conclude that $ \Lambda_{n}(g)\to 0$ for any $ g\in\ell^{1}$.

Thanks a lot mlok!!!

I feel stupid that I didn't have the courage to question the existence of such a $ \{\Lambda_{n}\}$   :oops: .

-Laura


Let $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$ . Show that :

$ \frac{x\plus{}y\plus{}1}{x^4\plus{}y^4\plus{}1}\plus{}\frac{y\plus{}z\plus{}1}{y^4\plus{}z^4\plus{}1}\plus{}\frac{z\plus{}x\plus{}1}{z^4\plus{}x^4\plus{}1} \leq \frac{1}{xyz}$


probability
geometry
3D geometry
sphere
puzzles

A meatball has four fly droppings on it. These droppings, are in random places and you must calculate the probability that you are able to cut a semi-sphere with all four droppings on one side.

[hide]you can verify your solution here http://www.projecteureka.org/problem/question/397 [/hide]

You are visiting your girlfriend and she orders pizza. Your evil girlfriend has perfect eyesight and notices fly droppings in three places on the pizza. She is seeking revenge on you for refusing to babysit her poodle and proceeds to cut the pizza. Assuming that the droppings occurred independently and in random places, what is the probability that she will be able to cut a half-circle pizza slice with all three droppings in one half?

Can you solve the simpler variant ?

[hide]http://www.projecteureka.org/problem/question/382[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Hi everybody,

I am having some difficulties with the following problem:

Let p>2 be prime and $ \chi$ a primitive dirichlet character modulo $ p^n$. Show that there is a $ k \in \mathbb{N}$, such that $ \chi(k)$ is a primitive (n-1)th root of unity in $ \mathbb{C}^{*}$.

Thanks in advance :-) Flo

That cannot be correct, for example take $ 5^4$: all it's characters have orders dividing $ 500$, being coprime to $ 3$.

Hm, thanks for the hint. I did type it in correctly.. Hmm. What could be ment? (p-1)th root instead of (n-1)th?

Yes, that would be true. Or more generally a $ \varphi(p^n)$-th primitive root of unity.

Okay, and how would I prove this? :-)

Its clear that every $ \chi(k)$ is a $ \varphi(p^n)$-th root of unity, but why is there also a [b]primitive[/b] $ (p\minus{}1)$-th root?

Every finite abelian group is (non-canonically) isomorphic to it's group of Dirichlet characters. Especially, there is an element of order $ p\minus{}1$ in the characters, so at least one of it's values must have multiplicative order $ p\minus{}1$.

I am not sure if I get that.

You are saying that the group of dirichlet characters mod $ p^n$ is isomorphic to $ \mathbb{Z}_{p^n}$. Okay. You also say that one of the characters has order p-1 in that group, okay, I agree. But why does this imply the existence of a k for EVERY character, such that $ \chi(k)$ is a (p-1)-th root of unity (and why should it be primitive)? I am also wondering where you used that $ \chi$ is primitive?

As stated, $ \chi$ is fixed, so we can choose $ k$ depending on $ \chi$. Well, for that other version, take a $ k$ such that $ k$ is a primitive $ (p\minus{}1)$-th root of unity $ \mod p^n$ (for simplicity, just take $ k\equal{}a^{p1n}$). Now note that characters are determined by their values at a primitive root $ a \mod p^n$, and a primitive character needs to have a primitive $ \varphi(p^n)$-th root as value at $ a$. Now just note that $ a$ is a $ sp^{n\minus{}1}$-th power of $ a$, $ \gcd(s,p\minus{}1)\equal{}1$ (or just $ s\equal{}1$ if choosen $ k$ carefully) and conclude.


Show that there exist infinitely many pairs $(m,n)$ of integers such that $4mn-m-n+1$ is a perfect square.


calculus
integration
calculus computations

As mathematical consultant to the Happy Pumpkin Candy Factory, you must design the production schedule for their wildly popular chocolate covered-aspargus Halloween candies. It is now 200 days before Halloween, and the factory must have 2 million candies produced by Halloween. Past research shows that to produce candies at a rate of R candies per day costs the factory$ 50 \plus{} R/1000$cents per candy. To store $ A$ candies costs $ A/2$
cents per day (to rent storage space, power the refrigerators, hire security guards, etc.)

Question:
Let $ t$ be the time (in days) from today, and let $ C(t)$ be the number of candies produced by
time $ t$ (thus, the total output from today to time $ t$). Find a definite integral over $ t$ for the total
production cost from now until Halloween in terms of $ C(t)$. Slice the time interval $ [0, 200]$into
pieces; figure out the cost to the factory from time $ t$ to time $ t \plus{} dt$.

the cost $ z(n)\equal{}(50\plus{}\frac{R(n)}{1000})\plus{}\frac{C(n)}{2}$ on day t
so the total cost is $ \sum_{k\equal{}1}^n z(k)\equal{}50n\plus{}\frac{\sum_{k\equal{}1}^n R(k)}{2}\plus{}\frac{1}{2}\sum_{k\equal{}1}^n C(k)\equal{}50n\plus{}\frac{C(n)}{2}\plus{}\frac{1}{2}\sum_{k\equal{}1}^n C(k)$

if we assume further that $ R(t)\equal{}R$ is constant.. then

$ C(n)\equal{}nR, z(n)\equal{}50\plus{}\frac{R(1\plus{}500n)}{1000}, \sum_{k\equal{}1}^n z(k)\equal{}\frac{200\plus{}R(n\plus{}3)}{4}n$

the way you describe it, the only integration i can see would be with the counting measure.  unless of course you charge instantaneously based on the amount of the day that something is in the storage... which sounds impossible?


Let $ S$ be a set of $ 16$ distinct positive integers, all less than $ 60$. Show that there exist
four pairwise distinct elements $ a, b, c, d \in S$ such that $ a \plus{} b \equal{} c \plus{} d$.

[hide]There are $ \binom{16}2 \equal{} 120$ different ways to choose two distinct elements from $ S$, but there are less than $ 120$ possible sums (in fact, only 115). Hence, by Pigeonhole it follows that there must be two pairs $ (a,b)$ and $ (c,d)$ such that $ a \plus{} b \equal{} c \plus{} d$. Furthermore, $ a \equal{} c$ implies $ b \equal{} d$, in which case $ (a,b)$ is identical to $ (c,d)$, contradiction, and vice versa, so it follows that $ a,b,c,d$ are pairwise distinct.[/hide]

But do we need $ 16$? Can we get the same result for a smaller number! Maybe ... think, just $ 14$ would be enough ... (with a different, more arduous proof).


number theory
greatest common divisor
number theory unsolved

Let $ S$ be a set of all 6-tupils $ (a,b,c,d,e,f)$ of positive intgers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\equal{}f^2$. Consider the set $ T\equal{}<abcdef: (a,b,c,d,e,f)\in S>$. Find the gcd of all numbers in T.

The answer is 24. The 6 tupil (1,1,1,2,3,4) belongs to S since $ 1\plus{}1\plus{}1\plus{}4\plus{}9\equal{}16$ and the product of its elements is 24.
Suppose 3 does not divide any of the numbers  a-f. Looking modulo 3 we  get then $ 1\plus{}1\plus{}1\plus{}1\plus{}1\equal{}1$. Contradiction. So at least 1 number is divisible by 3.
Suppose all numbers are odd. Then modulo 8 we get $ 1\plus{}1\plus{}1\plus{}1\plus{}1\equal{}1$ which is contradiction. So there is at least one even number. Since there are 6 numbers, there are at least 2 even numbers. If the two even numbers are not divisible by 4 we get (again modulo 8)
$ 1\plus{}1\plus{}1\plus{}1\plus{}4\equal{}4$ or $ 1\plus{}1\plus{}1\plus{}4\plus{}4\equal{}1$ either case is a contradiction. So the product is divisible by 8.


algebra
polynomial
function
logarithms
inequalities
number theory
Diophantine equation

Find all the solutions of the following equation in the set of positive integers:

$x^{y}=y^{x}-(x+y)$

Related to some similar problems, I have a conjecture (probably stupid question, but who knows): for any polynomial in two variables $P(x,y)$ such that $P(x,x)$ is not zero, the equation $x^{y}=y^{x}+P(x,y)$ has only finitely solutions in natural numbers. Can anyone prove it?

the polynomial will have finite degree and $x^{y}-y^{x}$ grows more rapidly than any polynomial.
with these idea we can formulate a rigorous proof I think :maybe:

the problem would be devising a method for finding those solutions (in some special cases at least) :)

I think (2,5) is the only solution because if you consider the function
f(x,y)=x^y-y^x+x+y the integer values are going to be very large or small beyond a certain value of x and y. After this value the least positive integer value will occurr when x=y giving f(x,y)=2x. This still is not rigorous however.

Albanian eagle, I think it's not so easy as you think.  :wink:

Harazi I never thought it was easy! :o  (it was jus a vague idea)
but I think I managed to complete a proof which is a bit long and so I'm going to post it as I turn back from IMO. :)

Thanks for your comments guys, but you only talked about the solvability of general equations not the solutions of this specific equation that was in the first post...

[quote="harazi"]I talked about the general one since the one you posted is very simple.  :wink:[/quote]
:huh: So can you solve it?

I have corrected the flaw in the proof I posted two days ago. The following proof should be flawless:

[b]The Diophantine equation[/b]

$(1) \;\; x^{y}-y^{x}+x+y = 0$

[b]where $x$ and $y$ are natural numbers, has only the solution $(x,y)=(5,2)$.[/b]

[u][b]Proof:[/b][/u] First we consider the cases with $\min(x,y) \leq 2:$

* $x = 1 \;\; \Rightarrow \;\; 0y+2 = 0 \;\; \Rightarrow \;\; \mbox{no solution.}$

* $x = 2 \;\; \Rightarrow \;\; 2^{y}= y^{2}-y-2.$

By inspection we find that there is no integer solution for $y \leq 3$. Futhermore, since $2^{y}\geq y^{2}$ when $y \geq 4$, the number of solutions is zero in this case.

* $y = 1 \;\; \Rightarrow \;\; x = 0 \;\; \Rightarrow \;\; \mbox{no solution because}\;\; x>0.$

* $y = 2 \;\; \Rightarrow \;\; (2) \;\; 2^{x}= x^{2}+x+2.$

Trying with $x=3, 4 , 5$ in (2) and using the fact that $2^{x}> x^{2}+x+2$ for $x \geq 6$, gives us only one solution of (2), namely $x = 5.$

Next assume $\min(x,y) \geq 3$. Then  by (1)

$x^{y}\;<\; y^{x}\;\; \Rightarrow \;\; \ln(x^{y}) \;<\; \ln(y^{x}) \;\; \Rightarrow \;\;y \cdot \ln x \;<\; x \cdot \ln y \;\; \Rightarrow \;\;{\textstyle \frac{\ln x}{x}}\;<\;{\textstyle \frac{\ln y}{y}}.$

The function ${\textstyle \frac{\ln t}{t}}$ is strictly decreasing when $t > e.$ Seeing that $\min(x,y) \geq 3 > e,$ the last inequality gives $x > y.$ Thus $x = y^{t}$ for a real number $t>1$. Then by (1)

$y^{ty}-y^{y^{t}}+y^{t}+y \;=\; 0$

$y^{y^{t}}\;=\; y^{ty}+y^{t}+y$

$y^{y^{t}}\;<\; y^{ty}+2y^{t}$ (divides the inequality by $y^{ty}$)

$(3) \;\; y^{y^{t}-ty}\;<\; 1+2y^{t(1-y)}$

Here the exponent $t(1-y) <-2$ since $t>1$ and $y \geq 3$. Therefore $y^{(1-t)y}\: <\: y^{-2}\: \leq \: 3^{-2}\: =\:{\textstyle \frac{1}{9}}$. Consequently

$(4) \;\; 1+2y^{t(1-y)}\;<\; 1+{\textstyle \frac{2}{9}}\;=\;{\textstyle \frac{11}{9}}\;<\; y.$

Combining the inequalities (3) and (4) we obtain $y^{y^{t}-ty}< y,$ which implies $y^{t}-ty < 1$. Hence $t < 2,$ giving $x = y^{t}< y^{2}$.

Now $x>y$, so $x = y+n$ for a natural number $n$. According to (1)

$y^{x}\; = \; x^{y}+x+y \;<\; x^{y}+2x \;<\; x^{y}+2y^{2}$

$y^{x-y}\; = \; ({\textstyle \frac{x}{y}})^{y}\:+\:{\textstyle \frac{2y^{2}}{y^{y}}}$

$y^{n}\; = \; ({\textstyle \frac{y+n}{y}})^{y}\:+\: 2y^{2-y}\; \leq \; (1+{\textstyle \frac{n}{y}})^{y}\:+\: 2 \cdot 3^{2-3}\; < \; e^{n}\:+\:{\textstyle \frac{2}{3}}$

$y \;<\; \sqrt[n]{e^{n}+{\textstyle \frac{2}{3}}}\; < \; 4.$

Thus the only possibility is $y=3$. In this case we obtain the equation

$(5) \;\; 3^{x}= x^{3}+x+3.$

Since $3^{3}-3^{3}-3-3 =-6 \neq 0$ and $3^{x}> x^{3}+x+3$ when $x \geq 4$, equation (5) has no solution. Hence the only solution of equation (1) is in fact $(x,y) = (5,2)$. [b]q.e.d.[/b]

[quote="harazi"]I have a conjecture (probably stupid question, but who knows): for any polynomial in two variables $P(x,y)$ such that $P(x,x)$ is not zero, the equation $x^{y}=y^{x}+P(x,y)$ has only finitely solutions in natural numbers.[/quote]As stated, the conjecture is false, because when $x = 1$ or $y = 1$, there could be infinitely many solutions.  But if you ignore those cases, then I think Albanian Eagle's approach does work.  Unfortunately, I am feeling too lazy to post a solution.

For $x=1$, the equation is $1=y+P(1,y)$, which will be a polynomial in $y$, and hence have only finitely many solutions (positive integers or otherwise). The same can be said when $y=1$.

Unless I am misreading it, these cases are not exceptions to the conjecture.

What if $P(x, y) = 1-y$?  Won't that lead to an infinite number of solutions when $x = 1$?

Its very easy.
Wlog  x>y, x=y+t where t is natural number.

y^(y+t)=〖(y+t)〗^y-(2y+t)
y^t=〖(1+〗_y^t)y – 2/y^(y-1)   - t/(y^y)<e^t<3^t
Thus y<3. Since y integer, y=1 or 2.
When y=2 then x=5.


In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest possible number of men who are married, have their own telephone, own their own car, and own their own home?

I don't get either of your solutions.

Can you explain yours, jhredsox?

[hide]I think what Jhredsox did was he started with the 70 people with telephones. Then you can choose 100-85 = 15 of those people to not be married, and choose 100-75 = 25 of those people to not have a car, and then same for the homes. Then you would only be left with 10 people who have all four, which you can't decrease any more[/hide]

[quote="alex1337"][hide="solution"]Add up the number of people that have each trait and you get 310.  Then subtract all 100 of them having 3 of the 4 traits you get 10.  Those ten traits have to be filled with 10 people having all 4 traits.  [b]10[/b][/hide][/quote]

I don't understand.  :maybe:

Why do you have to subtract 100 three times?


Given$ a,b,c$ be none-gative numbers, no two of which are zero.Prove that:
$ 12(\frac{a^2}{(b\plus{}c)^2}\plus{}\frac{b^2}{(c\plus{}a)^2}\plus{}\frac{c^2}{(a\plus{}b)^2})\plus{}\frac{40abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 8(\frac{a^3}{(a\plus{}b)^3}\plus{}\frac{b^3}{(b\plus{}c)^3}\plus{}\frac{c^3}{(c\plus{}a)^3}) \plus{}11$
I hope you will like it,my friends  :)

I can't fix my big problem.any body pleasez help me? :!:

[quote="Allnames"][quote="tuandokim"]let :$ \frac {a}{b \plus{} c} \equal{} x$
so $ \sum x^2 \plus{} 2xyz \equal{} 1$
so we have to prove that:
$ 1 \plus{} 16xyz \geq 8\sum x^3$
(can be killed easily by Lagrange Multipliers)[/quote]
It should be $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$,tuandokim.[/quote]
oh ,i see
thanks
 :blush:

[quote="tuandokim"]let :$ \frac {a}{b \plus{} c} \equal{} x$
so $ \sum x^2 \plus{} 2xyz \equal{} 1$
so we have to prove that:
$ 1 \plus{} 16xyz \geq 8\sum x^3$
(can be killed easily by Lagrange Multipliers)[/quote]
Are you sure? $ RHS\equal{}8 \sum \ \frac{a^3}{(a\plus{}b)^3}\plus{}11$
Regards $ \frac{a}{a\plus{}b}$ not$ \frac{c}{a\plus{}b}$,...
You wrote $ 1\plus{}16xyz \ge 8\sum x^3$
If so $ x^3\plus{}y^3\plus{}z^3 \ge \frac{3}{8}$ but$ xyz \le \frac{1}{8}$ :oops:
Now, try it :wink:

Yes,This ineq is hard and strong.Given $ a,b,c>0$.prove:
$ 12( \sum\frac{a}{b\plus{}c} . \frac{b}{a\plus{}c} ) \plus{} \frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \le 8( \sum\frac{a^3}{(a\plus{}b)^3} ) \plus{}11$
This ineq is easier than ineq of quykhtn.

[quote="nguoivn"]It's a weak and not nice ineq with me  :maybe: . I'm sure that the author's solution can't solve the stronger:
$ \sum\ \frac {a^2}{(b \plus{} c)^2} \plus{} \frac {25abc}{6(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {5}{6}. \sum\ \frac {a^3}{(a \plus{} b)^3} \plus{} \frac {23}{24}$  :)
An old and similar question: (#5)
http://www.mathlinks.ro/Forum/viewtopic.php?t=267158[/quote]
I don't know how to solve this kind of inequality.can nguoivn or quykhtn-ga1 tell me the way to solve this ineq? :?:

[quote="nguoivn"]It's a weak and not nice ineq with me  :maybe: . I'm sure that the author's solution can't solve the stronger:
$ \sum\ \frac {a^2}{(b \plus{} c)^2} \plus{} \frac {25abc}{6(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {5}{6}. \sum\ \frac {a^3}{(a \plus{} b)^3} \plus{} \frac {23}{24}$  :)
An old and similar question: (#5)
http://www.mathlinks.ro/Forum/viewtopic.php?t=267158[/quote]
Thanks my brother  :lol: .Yes,this inequality very strong and now, I cansolve it :wink:
With my inequality,I have a nice solution.

[quote="secrets"]I dont' have experience for this problem,help me![/quote]
This inequality not hard.I believe you can solve it ,my friend :)

[quote="nguoivn"]It's a weak and not nice ineq with me  :maybe: . I'm sure that the author's solution can't solve the stronger:
$ \sum\ \frac {a^2}{(b \plus{} c)^2} \plus{} \frac {25abc}{6(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ \frac {5}{6}. \sum\ \frac {a^3}{(a \plus{} b)^3} \plus{} \frac {23}{24}$  :)
An old and similar question: (#5)
http://www.mathlinks.ro/Forum/viewtopic.php?t=267158[/quote]
I can solve it,my brothers :lol: .this inequality nice and hard
@nguoivn:[hide]anh doan dc y tuong cua em roi,phai o ?em se post loi giai sau(use Pham kim hung) :) [/hide]

Yes, using the well-known:
$ \sum\ \frac{a^3}{(a\plus{}b)^3} \plus{} \frac{5abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 1$
We only need to prove:
$ \sum\ \frac{a^2}{(b\plus{}c)^2} \geq \frac{1}{8} \plus{} \frac{5}{6}. \sum\ \frac{a^3\plus{}b^3}{(a\plus{}b)^3}$
How'll you solve this problem, quykhtn-qa1 ?
I solved it by my old result  :)

[quote="nguoivn"]Yes, using the well-known:
$ \sum\ \frac {a^3}{(a \plus{} b)^3} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 1$
We only need to prove:
$ \sum\ \frac {a^2}{(b \plus{} c)^2} \geq \frac {1}{8} \plus{} \frac {5}{6}. \sum\ \frac {a^3 \plus{} b^3}{(a \plus{} b)^3}$
How'll you solve this problem, quykhtn-qa1 ?
I solved it by my old result  :)[/quote]
Yes,It is equivalent to:
$ \frac{2a^2\plus{}5bc}{(b\plus{}c)^2}\plus{}\frac{2b^2\plus{}5ca}{(c\plus{}a)^2}\plus{}\frac{2c^2\plus{}5ab}{(a\plus{}b)^2} \ge \frac{21}{4}$ :)
It can solve by SOS :)
you wrote:''I solved it by my old result'': I want know it ,my brother :)

I used my old result: $ \sum\ \frac{a^2}{(b\plus{}c)^2} \plus{} \frac{1}{2} \geq\ \frac{5}{4}. \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}$
And now, SOS completed the proof  :)
Besides:
[quote="quykhtn-qa1"]Yes,It is equivalent to:
$ \frac {2a^2 \plus{} 5bc}{(b \plus{} c)^2} \plus{} \frac {2b^2 \plus{} 5ca}{(c \plus{} a)^2} \plus{} \frac {2c^2 \plus{} 5ab}{(a \plus{} b)^2} \ge \frac {21}{4}$ :)
It can solve by SOS :)
[/quote]
Using the old result above, we obtain an well-known (stronger than Iran96)
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{} 2\sum\ \frac{ab}{(a\plus{}b)^2} \geq\ \frac{5}{2}$
I had had 2 proofs for it  :)

[quote="nguoivn"]I used my old result: $ \sum\ \frac {a^2}{(b \plus{} c)^2} \plus{} \frac {1}{2} \geq\ \frac {5}{4}. \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$
And now, SOS completed the proof  :)
Besides:
[quote="quykhtn-qa1"]Yes,It is equivalent to:
$ \frac {2a^2 \plus{} 5bc}{(b \plus{} c)^2} \plus{} \frac {2b^2 \plus{} 5ca}{(c \plus{} a)^2} \plus{} \frac {2c^2 \plus{} 5ab}{(a \plus{} b)^2} \ge \frac {21}{4}$ :)
It can solve by SOS :)
[/quote]
Using the old result above, we obtain an well-known (stronger than Iran96)
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 2\sum\ \frac {ab}{(a \plus{} b)^2} \geq\ \frac {5}{2}$
I had had 2 proofs for it  :)[/quote]
Thanks, my brother :lol: .Please post 2 proofs you spoke :)


$ AD \cap BC, AE \cap BF$ (one formula, comma in the middle) shows as AD^BC, AE^BF. Should be AD n BC, AE n BF.
$ AD \cap BC$, $ AE \cap BF$ (two formulas, comma between) shows as AD n BC, AE^BF. Should be AD n BC, AE n BF.

On the other hand,
$ AD \cup BC, AE \cup BF$  (one formula, comma in the middle) shows correctly
$ AD \cup BC$, $ AE \cup BF$  (two formulas, comma between) shows correctly

I think it started yesterday (May 18 2009)
Using Win2000, WinXP, IE 6.0, FireFox 3.0, various PCs.

Fixed.  Let me know if there are other commands that are broken, or if there are other cached images with \cap that I need to delete from the server (as I did the two you noted above).

Thank you for a quick reaction. The only formula still giving trouble (cached on the server) is $ P \equiv AD \cap BC,$ (comma included at the end, just this particular combination of letters), which still shows as P = AD^BC. I replaced it with  $ P \equiv AD \cap BC$, (comma outside of dollar signs), which shows correctly. I did not see any other problem.

Got that one, too.  Thanks for pointing it out.

This is a different problem, but I'll post in in this thread on the off chance that they're related.

I ran into it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277804]here.[/url]

The norm symbol \| isn't working properly - it's showing up a single bar. That is

$ \|x\|$ looks the same as $ |x|$ which it shouldn't; it should be $ \parallel{}x\parallel{}$ but with better spacing.

And it is showing up correctly on this post but not on the one I link. Huh?

Edit: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277944]here[/url] for further elaboration. It makes a difference what you put after it?

As a workaround you can use the synonym \Vert for\|
$ \Vert x\Vert_{2}^{2}\quad\left\Vert\frac{x}{\|x\|}\right\Vert \equal{}1\quad\left\Vert\frac{x}{\|x\|_{\infty}}\right\Vert_{\infty}\equal{}1$


trigonometry
limit
calculus
function
derivative
calculus computations

$\displaystyle\lim_{x\to 0}\frac{\sin{3x}}{5x} = \frac{3}{5}$
[color=red][i]
mod edit: moved from Intermediate[/i][/color]

what is l'hopitals?

EDIT: ive read it in MathWorld, but how does one prove that?

L'Hop(i with a weird sign on top)tal is basically when the limit of f(x)/g(x) is in the form 0/0 or $\frac {\infty} {\infty}$, the actual limit is equal to f'(x)/g'(x), as x approach the same number.

So lim x-> 0 sin3x / 5x = lim x->0 3 cos 3x / 5 = 3/5

or....

lim x->0 sin3x / 5x = lim x->0 sin3x / 3x (3x/5x) = 1*3/5 = 3/5

[quote="xxreddevilzxx"][quote="AntonioMainenti"][quote="xxreddevilzxx"]$\displaystyle\lim_{x\to 0}\frac{\sin{3x}}{5x} = \frac{3}{5}$[/quote]This follows immediately from the very well-known $\lim_{x\to0}\frac{\sin{x}}{x}$. Do you want us to prove that?[/quote]

Yea... is that well known related to l'hopital?[/quote]Er.. I can't remember it now. It's a geometrical demonstration that also uses a fundamental theorem of limits.

It's not related to L'Hospital's rule, by the way.

You can use the squeeze theorem.  I would google it because usually $\lim_{x\to 0}\frac{\sin x}x$ is a very, very common example and there should be at least a few proofs of it online.

Btw, limits are usually considered calculus.  Please post limit problems in the calculus forums from now on.

Thanks

L'Hopital's rule is overkill. $\lim_{x\to 0} \frac{\sin(x)}{x}$ can be proved by the Sandwich Theorem and we can get limits of this form simplified. I will generalize this for your case:

$\lim_{x\to 0}\frac{\sin {nx}}{mx}$, $m,n\in \mathbb{R}$

$\frac{1}{m}\lim_{x\to 0}\frac{\sin {nx}}{x}$

Multiplying the limit by $n$ we get:

$\frac{1}{m}\lim_{x\to 0}\frac{n\sin {nx}}{nx}$

$\frac{n}{m}\lim_{x\to 0}\frac{\sin {nx}}{nx}$

$u=nx$, $\lim_{u\to 0}\frac{\sin {u}}{u}$ by Sandwich Theorem

Therefore $\lim_{x\to 0}\frac{\sin {nx}}{mx}=\frac{n}{m}$ Q.E.D.

Using L'hopital's rule here leads to difficulties, because it is circular reasoning if you subscribe to the school of thought that the trigonometric functions are defined in terms of their relation to the unit circle.  If, conversely, you believe that $\sin$ is defined in terms of power series, then the conclusion can be immediately read off of the series.

for evaluating $\lim_{x\rightarrow 0} \frac{\sin{x}}{x}$, use the fact:
$\sin{x}=\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}$

the result follows ...

[quote="blahblahblah"]Using L'hopital's rule here leads to difficulties, because it is circular reasoning if you subscribe to the school of thought that the trigonometric functions are defined in terms of their relation to the unit circle.[/quote]Could you explain this? I'm assuming it wasn't just a pun on "circular" :D

The limit $\frac {\sin x}{x}$ is needed to prove that the derivative of $\sin x$ is $\cos x$.


Suppose we have many ideal resistors of resistance exactly 1 ohm each. Construct a scheme connecting only those resistors (with ideal, zero-resistance wires, of course;-) that has resistance of PI ohms between two of its points, with accuracy of 3,141592 plus/minus 0,000001 (one millionth) Ohm. (That means, the resistance must be in interval 3,141591 to 3,141593.) We cannot divide a resistor or do similar dirty tricks.

What is the minimal number of resistors we must use?

(26 is quite good, I have found a better solution, but I am still not sure it is the best.)

$\pi$ can be approximated to the desired accuracy using a 26-resistor network that is purely series-parallel. The only special feature of $\pi$ I use is that it can be approximated very accurately using a rather simple fraction.

Attached is a a schematic diagram.

The basic idea I used to obtain this circuit is as follows:
Find the continued fraction representation of the resistance. Here, we can approximate $\pi$ using the fraction $355/113$. (this is the fourth convergent in the continued fraction representation of $\pi$).

Let's say the continued fraction representation is
$a_1 + 1/(a_2+1/(a_3+1/(a_4+...$

Let the convergents be $c_1, c_2, c_3, c_4, ...$
Then to go from $c_n$ to $c_{n+1}$
If $n$ is odd, add a series chain $n$ resistors, in parallel to the parallel chain added in the previous step.
If $n$ is even, add of a parallel chain of $n$ resistors, in series to the series chain added in the previous step.

Looks good...

(I've found a solution using less than 26 resistors, but i'm not sure it is the best possible solution.)

[b]kubus:[/b] Could you please post  your solution?

I don't know where did you find this one  :lol: ,I'll post better solutioon with only 10 resistors.See the figure,7 parallel rezistors and 3 series.If we add them,

$\frac{1}{7}+3=\frac{22}{7}$

$\frac{22}{7}$ is the the original value of $\pi$.

Read the original post. $\frac{22}{7}\approx3.142857\ldots$ and its precision is too low!
What do you mean by original value of $\pi$?


logarithms
trigonometry
calculus
calculus computations

yes, OF COURSE, you can simplify #1, $\frac{1}{\sqrt{x^2+a}}$
2
forgot the negative sign
$\frac{-\sin{(2x)}}{\sin^4{x}}$

6, typo
$\frac{-\cos{x}-\sin{x}-1}{(1+\sin{x})^2}$

[quote="amirhtlusa"]yes, OF COURSE, you can simplify #1, $\frac{1}{\sqrt{x^2+a}}$ :)
2
forgot the negative sign
$\frac{-\sin{(2x)}}{\sin^4{x}}$ :)

6, typo
$\frac{-\cos{x}-\sin{x}-1}{(1+\sin{x})^2}$ :( [/quote]

[quote="dheardie"][quote="kunny"][quote="ambierona"]6, i think
$\displaystyle \frac{-\sin{x}-1}{(1+\sin{x})^2}$[/quote]

Thank you for your reply, ambierona.

You can simplify. :)

kunny[/quote]

$-\frac{1}{1+\sin(x)}$ :) [/quote]

hmmm
i think my answer for 4, is correct, but it might simplify or something...

Haha I suck at simplifying.. Wow...  :blush:


Let $ \Gamma$ be a circumcircle. A circle with center $ O$ touches to line segment $ BC$ at $ P$ and touches the arc $ BC$ of $ \Gamma$ which doesn't have $ A$ at $ Q$. If $ \angle {BAO} = \angle {CAO}$, then prove that $ \angle {PAO} = \angle {QAO}$.

Two hints:

[hide="It is a known fact that..."]$ PQ$ meets the circumcircle again at the midpoint of arc $ BC$ that contains $ A$[/hide]

[hide="Then prove that..."]$ APOQ$ is cyclic[/hide]

let $ M$ be the midpoint of arc $ BC$ not containing $ A$, so $ A,O,M$ are collinear...

let $ O'$ be the center of $ \Gamma$, so $ O', O, Q$ are collinear, and $ OP\parallel{} O'M$...

then, $ \angle QAO\equal{}\angle QAM\equal{}\frac{1}{2}\angle QO'M\equal{}\frac{1}{2}\angle POO'\equal{} \angle QPO$...

this implies that $ APOQ$ is cyclic, and since $ PO\equal{}OQ$ we conclude that $ \angle PAO\equal{}\angle QAO$... :D

Change the supposition we have problem:
(O) is a circumcircle of $ \Delta ABC$ and $ AD,BM,CN$ are 3 bisectors. $ (I),(J),(K)$ touch to line segment $ BC,CA,AB$ at $ P,E,F$ and touch the arc BC which doesn't have A, the arc CA which doesn't have B,the arc AB which doesn't have C at $ Q,E,F$. Prove that  $ AQ,BE,CF$ are concurrent  :D

Hey who can solve my problem?  :D  :wink:

[quote="dgreenb801"][quote="cyshine"]
[hide="It is a known fact that..."]$ PQ$ meets the circumcircle again at the midpoint of arc $ BC$ that contains $ A$[/hide]
[/quote]
How do you prove this?[/quote]
Let K be the intersection of $ \Gamma$ and PQ,I is the circumcenter then $ \frac {OP}{IK} \equal{} \frac {OQ}{IQ}$
So $ OP//KI$ or $ IK\perp BC$
Therefore K is the midpoint of arc BC  :wink:

[quote="livetolove212"]Hey who can solve my problem?  :D  :wink:[/quote]
It very easy :)

Hi dgreenb801,

Sorry I didn't answer your question before. I know livetolove212 have already proved it, but I have a different proof.

Consider the homothety with center $ Q$ that takes the circle with center $ O$ to the circumcircle. It also takes $ BC$ to a line tangent to the circumcircle parallel to $ BC$, which touches it in the midpoint of the arc $ BC$ (just do some angle chasing). This tangency point is the image of $ P$ under such homothety, so it belongs to line $ PQ$.

$\textbf{Solution}$

Let $R$ denote the midpoint of arc $BC$ not containing $A$ and $T$ denote the midpoint of arc $BC$ containing $A$. Let $S$ denote the center of circle $\Gamma$ and $\Omega$ denote the circle with center $O$. From the condition that $\angle{BAO}=\angle{CAO}$, we have that $R$, $O$ and $A$ are collinear. Further it is important to note that $T$, $S$ and $R$ are collinear.

Extend $PQ$ to intersect the circle at $T'$. By the homothety mapping $\Omega$ to $\Gamma$ with center $Q$, the tangent to circle $\Gamma$ at $T'$ is parallel to $BC$. Therefore $T$ is the midpoint of major arc $BC$ and $T'=T$.

Since homothety maps $P$ to $T$ and $O$ to $S$, it follows that $ST \| PO$. Hence,

\[\angle{OPQ}=\angle{STQ}=\angle{RTQ}=\angle{QAR}=\angle{OAQ}\]
Therefore $POQA$ is cyclic. Denote its circumcircle as $\omega$. Since $OP=OQ$, $O$ is the midpoint of minor arc $PQ$ of circle $\omega$ and therefore $\angle{PAO}=\angle{QAO}$.

Here's a nice one...

Let $D,E$ be the midpoints of the arc $BC$ not containing $A$ and arc $BC$ containing $A$ respectively, and let $QF$ be a diameter of $\gamma$. It is clear that $Q,P,E$ are collinear. Suppose to the contrary that it is not true. Then if $G$ is the intersection of $AP$ with $\gamma$, then it is clear that $FG$ is not parallel to $DE$, so they intersect at a point $H$. By Pascal's theorem, $P,O,H$ should be collinear. But clearly $PO//DE$, so they can't be collinear, since $F\in DE$, absurd. Q.E.D.

This could also be solved by using harmonic division.
Let M, N be the midpoint of arc ABC, BQC.X be the circumcenter. Then O(P,M,X,N) is harmonic.

Extend $AO$ and $QP$ to meet $\Gamma$ at $T$ and $M$ respectively. Denote $O_\Gamma$ as the center of $\Gamma$.

Since $Q$ is the point of tangency of the smaller circle, which we'll call $w$, to $\Gamma$, we have that $Q$, $O$ and $O_\Gamma$ are collinear. Now consider the homothety about $M'$, the midpoint of arc $BC$, mapping $O_\Gamma$ to $O$. Since $\triangle{M{'O_\Gamma}P} \sim \triangle{M'OQ}$, we have that $P$ is mapped to $Q$, hence $Q$, $P$ and $M'$ are collinear, implying that $M'=M$.

Since $T$ and $M$ are both midpoints of their respective sides of arc $BC$, we have that $TM \perp BC$, implying that $TM \parallel OP$. It follows that $\triangle{SOP} \sim \triangle{STM}$, implying that quadrilateral $APOQ$ is cyclic. Note that $\triangle{QOP}$ is isosceles. We now have that $\angle{PAO}=\angle{PQO}=\angle{QPO}=\angle{QAO}$. $\square$

[hide=Solution]
Let $\omega$ be the circle centered at $O$ passing through $P$ and $Q$. Note that $\sqrt{bc}$ inversion fixes $\omega$, as arc $BQC$ gets mapped to segment $BC$ and vice-versa, and line $AO$ is fixed, so since there is only one circle centered on $AO$ which is tangent to $BC$ and internally tangent to $\Gamma$, we have that $\omega$ is fixed. Thus, $P$ and $Q$ map to each other, so $AP$ and $AQ$ are isogonal in $\angle{BAC}$. This implies that $\angle{PAO}=\angle{QAO}$.
[/hide]

[quote=franzliszt]Let the circumcircle of $\triangle ABC$ be $\Omega$ and let $AO\cap \Omega=L$ and $QP\cap \Omega=M$. By Fact 5, $L$ is the midpoint of $\widehat{BC}$ not containing $A$ since $O$ is on the $A$-angle bisector, and by the Shooting Lemma (I think that's what it's called? Lemma 4.33 in EGMO.), $M$ is the midpoint of $\widehat{BC}$ containing $A$. Thus, $L$ and $M$ both lie on the perpendicular bisector of $BC$ and $LM\perp BC$.

Let the circle centered at $O$ be $\omega$. Since $BC$ is tangent to $\omega$ at $P$, se have $OP\perp BC$. Combining this with $LM\perp BC$, we find that $OP\parallel LM$.

Now we prove the following

$\textbf{Claim:}$ Quadrilateral $APOQ$ is cyclic.

$\textit{Proof.}$ This is just angle chasing. Write $$\measuredangle OPQ=\measuredangle LMQ=\measuredangle LAQ=\measuredangle OAQ$$ and the claim is proven. $\square$

To finish, note that $\triangle OPQ$ is isosocles so $$\measuredangle PAO=\measuredangle PQO=\measuredangle OPQ=\measuredangle OAQ$$ and we find that $P$ and $Q$ are isogonal as required. Done. $\blacksquare$[/quote]

Yeah, I believe the Shooting Lemma is cited for the angle bisector condition and length condition (see USAMO $2015$/$2$).

[hide=Solution]
[asy]
unitsize(1cm);
pair A, B, C, O, P, Q, M, N;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
O=(3.749016,-1.113632);
P=(3.749016,0);
Q=(3.670988,-2.224527);
M=(4,-sqrt(5));
N=(4,16sqrt(5)/5);
draw(A--B--C--A--M--O--P--O--Q--N);
draw(circumcircle(A, B, C));
draw(circle(O, distance(O, P)));
label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$O$", O, E);
label("$P$", P, NE);
dot("$Q$", Q, SW);
dot("$M$", M, SE);
label("$N$", N, dir(90));
[/asy]

Consider the homothety centered at $Q$ mapping the small circle to $\Gamma$. Then, $P$ maps to $N$, the midpoint of arc $BC$ containing $A$, so $Q$, $P$, and $N$ are collinear. Let $AO$ intersect $\Gamma$ again at $M$. Since $\angle BAM=\angle MAC$, this means that $M$ is the midpoint of arc $BC$ not containing $A$.

Now, we claim that $APOQ$ is cyclic. We have
\begin{align*}
\angle AQP&=\angle AQN\\
&=\angle ABN\\
&=\angle ABC-\angle NBC\\
&=\angle ABC-\angle NMC\\
&=\angle ABC-90^{\circ}+\frac{\angle BAC}2\\
&=\angle ABC+\angle BAM-90^{\circ}\\
&=\angle AOP.\end{align*}

Therefore, $APOQ$ is cyclic. Since $\angle PQO=\angle OPQ$, this means that $\angle PAO=\angle QAO$.

[hide=Solution]We perform a $\sqrt{bc}$ inversion centred at $A$ followed by a reflection across the angle bisector
We see that $$BC \iff \odot(ABC)$$ And the circle with centre $O$ will map to another circle that is tangent to the inverses of $BC$ and $\odot(ABC)$
But these two swap which means we still get a circle tangent to $BC$ and $\odot(ABC)$

Now we also know that under this inversion (followed by reflection) $A$,$O$ and the centre of the new circle are collinear (Since $O$ lies on the angle bisector of $\triangle ABC$, the reflection does not change anything after the inversion)

These two facts combined together give us that the new circle is the same as the old circle which means $P$ and $Q$ which are the tangency points must swap
But now it is obvious that a line passing through the centre of Inversion will map to its isogonal conjugate which means that $\angle PAO = \angle QAO$ since line $AO$ is nothing but the angle bisector in $\triangle ABC$ which finishes the proof  :D $\blacksquare$[/hide]
Yes, a proof mostly with words  :rotfl:
Nvm there are 2 inversion solutions in the thread (Which are the same as mine  :laugh:  Saw this after writing the post  :rotfl: )

Let $QP \cap \odot{ABC}=M$. It is well known that $MB=MC$. Let $\angle{OAB}=x, \angle{OPQ}=y$.
Note that $\angle{BCM}=90-x$, $\angle{BQM}=90-x$, so $\angle{QBM}=x-y$. Now $\angle{QAC}=x-y$, so $\angle{QAO}=y=\angle{QPO}$. Consequently, $QAOP$ Is cyclic, and we are done.

Note that $O$ lies on the angle bisector of $\angle A$

We perform the following transformation: Invert at centre $A$ with radius of inversion $\sqrt{bc}$ and follow it by reflection across the angle bisector
Firstly note that $B \leftrightarrow C$
Also, it is easy to see that $BC \leftrightarrow$ arc $BC$

We claim that this transformation takes $P$ to $Q$
Since inversion preserves intersections, the image of the circle with Centre $O$ must still be tangent to the images of line $BC$ and arc $BC$

But since these 2 transform into each other, the new circle must also be tangent to both line $BC$ and arc $BC$
This along with the fact that $O'$ lies on $AO$ and therefore the centre of the new circle lies on $AO$ gives us that the circle transforms into itself and thus $P'=Q$ since there can only be one circle that is tangent to both $BC$ and arc $BC$ and centred on $AO$

Hence we get $\angle PAO = \angle QAO$

The main claim is that

[b][color=#f00]Claim.[/b][/color] $OPAQ$ is cyclic.

It is not hard to see how this claim implies the result. In particular, we will then have $\measuredangle QAO = \measuredangle QPO = \measuredangle OQP = \measuredangle OAP$ by inscribed angles since $OP=OQ$.

[i]Proof.[/i] Extend $AO$ to meet $(ABC)$ at $M_1$. By Lemma 1.18, we know that $M_1$ is the arc midpoint of $\widehat{BC}$ not containing $A$. Similarly, extend $PQ$ to meet $(ABC)$ at $M_2$. By Lemma 4.33, we also know $M_2$ is the arc midpoint of $\widehat{BC}$ containing $A$. It follows that $M_1M_2\perp BC$, so $OP \parallel M_1M_2$ by tangent properties.

To finish, angle chase suffices; $$\measuredangle AOP = \measuredangle AM_1M_2 = \measuredangle AQM_2 = \measuredangle AQP,$$ since $AM_2M_1Q$ is given to be cyclic. Thus, the claim is proven. $\blacksquare$

By reasons described above, this implies $\angle PAO = \angle QAO$, as desired.

Let $QP$ meet $ABC$ at $S$. It's well known that $S$ is midpoint of arc $BAC$. Let $G$ be circumcenter of $ABC$ and $SK$ be diameter of $ABC$.
$\angle OAQ = \angle KAQ = \angle KSQ = \angle SQG = \angle PQO$ so we need to prove $APOQ$ is cyclic. Note that $OP || GK$ cause both are perpendicular to $BC$.
$\angle AQP = \angle AQS = \angle AKS = \angle AOP$ so $APOQ$ is cyclic and $\angle PAO = \angle PQO = \angle OAQ$.
we're Done.

[b][color=#00f]Solution one[/color][/b]
[hide=one liner(and a half, may be :P )]The image of the reflecion of the inversion $\Psi(A,\sqrt{bc})$ wrt. the bisector of $\angle A$ swaps $B$ with $C$ and $AP$ with $AQ$. Hence $AO$ is the bisector of $\angle PAQ$ as desired.[/hide]

[b][color=#00f]Solution two[/color][/b]
[hide=without inverting]By shooting lemma(lemma 4.33 in EGMO), we have that $AO$ and $QO$ pass through different midpoints $M,N$ of the arc $BC$. Consider \[\angle QAO=\angle QNM=\angle QPO\] gives $A,P,Q,O-$concyclic. Since $OP=OQ$, we have that $\angle QAO=\angle PAO$ as desired.[/hide]

Suppose the circle with center $O$ is $\omega$. Let $PQ\cap \Gamma = M$ and $AO\cap \Gamma=N$.

[color=#f00][b]Claim - [/b][/color] $M$ is the midpoint of $\overarc{BC}$ that contains $A$.
[b]Proof - [/b] The homothety centered at $Q$ mapping $\omega$ to $\Gamma$ also maps $BC$ to $B'C'$ such that $B'C'$ is tangent to $\Gamma$ & $B'C'||BC$ and $P$ to $M$. Using all these information, we can conclude that $MB=MC$ thus minor arcs $MB$ and $MC$ are equal.

[color=#f00][b]Claim - [/b][/color] $N$ is the midpoint of $\overarc{BC}$ that does not contain $A$.
[b]Proof - [/b] As $\angle{BAN}=\angle{BAO}=\angle{CAO}=\angle{CAN}$, $BN=CN$ thus minor arcs $BN$ and $CN$ are equal.

Therefore, $MN$ is the diameter of $\Gamma$. Also, $OP||MN$.

[color=#f00][b]Claim - [/b][/color] $APOQ$ is cyclic
[b]Proof [/b] $\angle{AQP}=\angle{AQM}=\angle{ANM}=\angle{AOP}$.

As $OP=OQ$, we have $\angle{QAO}=\angle{OAP}$. $\blacksquare$

Let $N$ be the midpoint of arc $BC$ containing $A$ and let $M$ be the midpoint of arc $BC$ not containing $A.$ Let $O_1$ be the center of $\Gamma.$ By the Circles Inscribed in Segments Lemma, $P,Q,$ and $N$ are collinear. Hence, $$\angle QAO=\angle QNM=\angle O_1QP=\angle QPO$$ so $APOQ$ is cyclic. Since $OP=OQ,$ $$\angle PAO=\angle OQP=\angle OPQ=\angle QAO.$$ $\square$

We define a few points first. Let $X$ be the midpoint of the arc $BC$ that doesn't contain $A$. Let $X_1$ be the midpoint of the remaining arc $BC$. Define $X_3$ to be the center of $\Gamma$.
[i]Claim:[/i] The quadrilateral $OPAQ$ is a cyclic quadrilateral.
[i]Proof:[/i] Firstly, by the circle inscribed in segments lemma we find that the points $A,O,M$ are collinear. Now we angle chase to get the following \[\angle OPQ = \angle DX_1Q = \angle XX_1Q = \angle XAQ = \angle OAQ \] This proves that the quadrilateral $OPAQ$ is a cyclic quadrilateral. $\square$
And since we have $OP = OQ$ we get $\angle PAO = \angle QAO$

Let $M_1$ be the midpoint of arc $BAC$,  and let $M_2$ be the midpoint of $BC$. Clearly we see that $BC \perp M_1M_2$, since $M_1M_2$ is the diameter of $\Gamma$.  Since $OP \perp BC$ and $M_1M_2 \perp BC$, we know that $OP || M_1M_2$.

Note that by Lemma 4.33(From EGMO) we know that $Q, P, M_1$ are collinear, and trivially by incenter properties $A, O, M_2$ are collinear.

[b]Claim: [/b] We claim that $APOQ$ is cyclic.
[b]Proof: [/b] Note that $$\measuredangle QAO = \measuredangle QAM_2 = \measuredangle QM_1M_2 = \measuredangle QPO.$$ Thus we have proved that $APOQ$ is cyclic.

Since $APOQ$ is cyclic, we know $\measuredangle PAO = \measuredangle PQO$. But since triangle $\triangle{OPQ}$ is isosceles due to $OP = OQ$, we know $$\measuredangle PAO = \measuredangle PQO = \measuredangle OPQ = \measuredangle QAO$$ as desired.

[hide = Solution] Denote by $M_A$ as the midpoint of arc $\widehat{BC}$ not containing $A$, and $M_{BC}$ the midpoint of arc $\widehat{BC}$ containing $A$. Let $O_{\Gamma}$ be the center of $\Gamma$.

Consider the homothety at $Q$ sending the circle with center $O$ to the circle with center $O_{\Gamma}$. Then we have that $Q, P, M_{BC}$ and $Q, O, O_{\Gamma}$ are collinear. Hence $\overline{OP} \parallel \overline{O_{\Gamma}M_{BC}}$. We now make the following claim:

[b][u]Claim[/u]: [/b] $A, P, O, Q$ are concyclic points.

\begin{align*}
\angle QPO &= \angle QM_{BC}M_A \\
&= \angle QAM_A \\
&= \angle QAO.
\end{align*}

Thus we have $A, P, O, Q$ are concyclic and as $\angle QPO = \angle OQP$, we have that indeed $\angle QAO = \angle PAO$, as desired. $\blacksquare$ [/hide]

Let $D$ be the midpoint of minor arc $BC$ and let $D'$ be its antipode.
Note that since $OP = OQ$ it follows that $\measuredangle OPQ = \measuredangle PQO$, and thus if $APOQ$ is cyclic then the result follows.
Note that $Q, P, D'$ are collinear.
Thus, we then have that \begin{align*} \measuredangle AOP &= \measuredangle DOP = \measuredangle CPO + \measuredangle DCP + \measuredangle ODC = 90^\circ + \measuredangle DAB + \measuredangle ABC \\ &= 90^\circ + \measuredangle DAB + \measuredangle ACB + \measuredangle BAC = 90^\circ + \measuredangle BAD + \measuredangle ACB \end{align*} and that \[ \measuredangle AQP = \measuredangle AQD' \]
By considering arcs, the result follows.


geometry
parallelogram
trigonometry
MATHCOUNTS
trig identities
Law of Cosines

In the parallelogram shown, what is the number of inches in the sum of the squares of the lengths of the diagonals?
[img]http://i202.photobucket.com/albums/aa68/MBarracade/Scraps/trapezoid.jpg[/img]

and

Two different circles and a triangle lie in the same plane. What is the greatest number of points that can belong to at least two of the three figures?

[b]Question 1[/b]
First draw the diagonals.
[asy]pair A = (0, 0);
pair B = (8, 0);
pair C = (11, 6);
pair D = (3, 6);
draw(A--B--C--D--cycle);
label("$x$", A--D, W);
label("$y$", D--C, N);
label("A", A, SW);
label("B", B, SE);
label("C", C, NE);
label("D", D, NW);
draw(A--C);
draw(B--D);
label("$d_1$", (3, 1));
label("$d_2$", (5, 5));[/asy]
Now, let $ x$ and $ y$ be the side lengths and let $ d_1$ and $ d_2$ be the diagonals. Then it is a known fact that
\[ d_1^2 + d_2^2 = 2x^2 + 2y^2
\]
Or in other words, the sum of the squares of the diagonals is equal to the sum of the squares of the four sides.

[hide="Proof - Hidden because it uses trig and is not MATHCOUNTS material"]
Let $ m\angle ABC = m\angle CDA = \theta$. Then it follows that $ m\angle BCD = m\angle DAB = \pi - \theta$ (or $ 180 - \theta$ if you prefer degrees), because those two angles are supplementary. Now, using the law of cosines,
\[ d_1^2 = x^2 + y^2 - 2xy\cos \theta
\]
and
\begin{align*}d_2^2 & = x^2 + y^2 - 2xy\cos (\pi - \theta) \\
& = x^2 + y^2 + 2xy\cos \theta\end{align*}
(because $ \cos (\pi - \theta) = - \cos \theta$). Then it follows that
\begin{align*}d_1^2 + d_2^2 & = (x^2 + y^2 - 2xy\cos \theta) + (x^2 + y^2 + 2xy\cos \theta) \\
& = 2x^2 + 2y^2,\end{align*}
as desired.
[/hide]

So the answer, plugging in, is $ 2(6)^2 + 2(8)^2 = 2 \cdot 36 + 2 \cdot 64 = \boxed {200}$.

[b]Question 2[/b]
What do you mean by [i]belong[/i]? The way I see it, the answer is infinity because there are an infinite number of points that lie in a region, no matter how small.

I would assume that he means intersection points.

 [geogebra]4446f248fa1a156315edd391a656824bf4f15c47[/geogebra]

A circle and a line can intersect in at most 2 points. Since a triangle is essentially 3 line segments, at most 6 points.

There are 2 circles, so 12 points.

Now the circles can intersect with each other in 2 points, so a total...

of[b] 14 [/b]points.

Soo....

If a circle, square, and triangle are all drawn on a plane, and none of the square's sides are collinear with the sides of the triangle, then the largest number of points that can belong to at least two of the three figures is 8+6=14?

I think...

circle to square = 8
circle to triangle = 6
square to triangle = 6
Total: 20 [if I am not mistaken]

I think the problem is that you haven't considered the square to triangle case.

[geogebra]e058f3bed240a150ed050e9ad7898f608f16f54b[/geogebra]

That is correct, james4l.

@M_M: for these problems, remember to consider ALL pairings.


I doubt that this is the right place to post this but,

What is the largest number that can be obtained using 4 2s?

Be warned, it is harder than it looks. :!:

I remember seeing this on an NJML contest.
[hide]It's $ 2^{2^{22}}$.[/hide]

yeah that is the one. It requires quite a bit of thought to actually work it out though if you don't know the answer doesn't it?

What about tetration? (placing 2's to the upper left)

but in the last 2 numbers, $ 2^2$ evaluates to only 4 while 22 is alot bigger than 4
i assume by tetration you mean $ 2^{2^{2^{2}}}$

[quote="stevenmeow"]
i assume by tetration you mean $ 2^{2^{2^{2}}}$[/quote]

Yes, but what you wrote is equal to $ ^{4}2 \equal{} ^{^{2}2}2$. Suppose we let this be $ x$. Then the nested tetration with four $ 2$'s is $ ^{^{^{2}2}2}2 \equal{} ^{x}2$, which is going to be a lot greater than $ 2^{2^{2^{2}}}$.

This [i]was[/i] on a Math League contest a few months ago.
[hide="The answer is"]
$ 2^{2^{22}}$.
[/hide]

Could someone explain tetration, or post a link? (I just need a simple explanation.)

Also, what's wrong with $ 2{}^2{}^2{}^2$? Because 2*22=88, which is less than 222...

The inequality holds if they are powers of $ 2$.
$ 2^{22} \ln 2 > 222 \ln 2 < 2^{2^{2}} \ln 2$.

EDIT Lazarus posted first....

Fishythefish, I'll make it clear and simple: $ 2^{22}$ is not equal to $ 2\cdot22$, it's equal to

$ 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$

Also, a tetration is an iterated exponent: for example $ ^{4}2 \equal{} 2^{2^{2^2}} \equal{} 2^{\left(2^4\right)} \equal{} 2^{16} \equal{} 65,\!536$

If tetration is a allowed operator(I'm hearing about it for the first time), I'm $ 99.9\%$ sure this is gonna be the biggest...insane. :P
[hide]
$ ^{^{22}2}2$

To understand properly:
$ ^{22}2 \equal{} 2^{2^{2^{2^{2^{.^{.^{.^.}}}}}}}$($ 2$ appears $ 22$ times)

and $ ^{^{22}2}2 \equal{} 2^{2^{2^{2^{2^{.^{.^{.^.}}}}}}}$($ 2$ appears $ ^{22}2$ times)

I'm announcing $ \$1000000000000000000000000000000000000000000000000000$ :P  for the person who writes the number in latex correctly and posts it here... In full decimal expansion. :P

Offer valid till end of this april. :P[/hide]

[quote="andersonw"]$ 2^{22}$ is not equal to $ 2\cdot22$, it's equal to
$ 2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$
[/quote]

Ive not read the entire thread but that isnt $ 2^{22}$ or is it? :D :P :maybe:

Oh...my bad. I thought they meant $ (2{}^2){}^2{}^2$
It makes sense now...sorry guys. :blush:
How do you raise powers to powers, and why can't I use more than one digit for each caret?

Oh, and I'm not new to powers or anything, I know what 2^22 is.I just kind of messed up.  :roll:

[quote="jvenezuela716"][quote="ProtestanT"]What about tetration?[/quote]

An even better way:

$ ^{2^{22}}2$[/quote]

No. Mine was better.

I'm pretty sure this problem is done being solved and discussed.  Locked.

Also, for future reference, post brainteasers and such in [url=http://www.artofproblemsolving.com/Forum/index.php?f=337]the puzzles & brainteasers forum.[/url]


This has been puzzling me for a while:

Assume Jim takes a spaceship and travels to a star that is 9.5 LY away and back at .95c.

The time that has passed on the earth during this trip is approximately 20 years. Applying relativity, we note that Jim should have aged only about 6 years.

a) Am I applying relativity correctly (I am pretty sure I am, but thought I should ask before the next question)

b) In what sense has Jim not gone faster than the speed of light? He has covered 19 LY in 6 years, and by extension if he was going .999c he could have covered much more.

If I understand your question correctly, you are referring to the paradox that Jim has exceeded light speed travel. This is not the case however. To Jim he has not traveled 19 cy. Due to length contraction his observed round trip distance was approximately 5.93 cy.

Thanks. So I can see in what sense he hasn't gone faster than the speed of light.

I guess I just find it quite surprising/interesting that one could travel very far (i.e. nearest non-solar system planet) within seconds given the correct equipment.