AMC
AIME
probability
email
AMC 12
geometry
algebra

Hello,

For those who have read/received my guides, what topics/strategies/areas would you want most help on? The only thing that I cannot do is USAMO... That will probably not be available until long time, but that's okay since there are excellent guides out there.

I'm open to suggestions and I'll consider them highly. Thanks.

An AMC level guide to Algebra, which I need, as well as a few of my friends.

It would be awesome to do AIME-level number theory. Also, I think it would help to put in a problem set of 15-20 problems relating to the subject and difficulty.

The Interm Geo guide is great btw!

hmmm i would appreciate a guide on number theory as well. combinatorics would be my second choice topicwise

Concerning there are 42 views and 3 replies, I'll go with the majority in the given. I don't think I will wait more... But any reply after this will be waitlisted.

Next guide will be on AIME-level number theory. To make AIME15's voice heard, I might insert algebra-NT problems as well.  :)

[quote="mathemonster"]hi silverfalcon,
i recieved the HD round 1 but not any of the guides even though i requested them. Could u send? AIME level probability would be nice.[/quote]

I'm not sure why you did not receive them. My guess is that either email didn't come through, or when they did, I simply replied that you go to my website and you didn't. All of them are available at:

http://www.hiddendiscoveries.com

[quote="AIME15"]I think there's no point for AIME level probability, because we have Intro to C&P and Intermediate C&P, which goes beyond AIME level...

And you copied my avatar :([/quote]

I don't have intermediate, because i don't think there are many on the aime. I have intro btw + V2. As for the avatar - sorry lol.

when will you be done with the next guide?

If you can include AMC12 level of Plane Geometry, I will greatly appreciate it.
Thanks.

Number theory or combinatorics would be the best. Personally I would prefer combinatorics but seeing how there's already an Intermediate C & P book you should do number theory instead. Medium hard AMC 12 - Medium AIME level if possible.

Next guide will probably take a long time. I'm not sure exactly how long.

AMC12 level of Plane Geometry can be done with Introductory to Geometry. Meanwhile, the first edition geometry techniques pretty much cover good deal of it provided that you do some problems and learn some theorems. It's funny that after making that guide, I look at AHSME geometry and they look incredibly easy (I used to do bad on just AMC geometry).

Combinatorics and probability/counting are two different disciplines that I want to write someday. Only problem is that they are pretty difficult to write since I'm horrible at it... Maybe someday.  :D

P.S. The problems for next year's HDMO are in process... I think it's pretty sinister  :roll:

Hi, I was just wondering if you stopped making the polynomial article because I thought there was a pole for your second article and polynomials got selected. (thanks for all your guides)

[quote="rOadToAIME"][quote="Silverfalcon"]

AMC12 level of Plane Geometry can be done with Introductory to Geometry. Meanwhile, the first edition geometry techniques pretty much cover good deal of it provided that you do some problems and learn some theorems. It's funny that after making that guide, I look at AHSME geometry and they look incredibly easy (I used to do bad on just AMC geometry).

[/quote]
Does that mean that you've already made introductory geometry guide?
If so, can I still get it from you?[/quote]

I made first edition for Intermediate Geometry Guide. It doesn't have any practice problem but I think with tips and guides from the guide (along with practices of course), you should be able to handle most of easy/medium AMC geometry. Later geometry is little bit too hard with just the guide but I think it's sufficient for the easy/medium ones. It's available at my website: http://www.hiddendiscoveries.com

Ah, with polynomial article.. Yes, I stopped and I don't think I will start that article anytime in the future. If I do, it'll be for 100% Olympiad level though.


probability
Pascal\u0027s Triangle
binomial coefficients

${4 \choose 2}= 6$, so we have $6 \cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{2}{3}\cdot \frac{2}{3}= \frac{8}{27}$.

Find the probability of getting a question wrong and a question right. We're getting two wrong and two right, so square each. Then take into account that we can miss any two questions and multiply by 6.

Well.. Basically, u can follow Bernoulli's Trial...
C(n,r) x p^r x q^(n-r) , where p = the probability of success, and q = the probability of failure...

From the question:
A multiple choice exam has three choices and four questions. What is the chance that a candidate who guesses all the answers randomly gets exactly two questions right?

It has 4 questions which they have 3 choices each. It has C(4,2) ways that the "Random guesser" guesses 2 questions right randomly.
Then, in this case, the probability of success = 1/3, failure = 2/3
According to Bernoulli's Trial, write the formula as i wrote....u will get the answer as MellowMelon :D

PS: Is that clear for u??? My English is not my native language...hoho..I try to make u understand... :rotfl:

[quote="bos1234"]How would you take into account of missing 2 questions?[/quote]
That's where multiplying by ${4 \choose 2}$ comes in. You can miss either 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, or 3 and 4, for a total of 6. But Pascal's triangle / binomial coefficients is much more efficient if, say, you were counting how many ways you could miss 5 problems out of 10.

In the interest of clarity:
The formula that FaBreGas-4 gave
$\binom{n}{r}p^{r}\cdot q^{n-r}$ represents a [url=http://en.wikipedia.org/wiki/Binomial_distribution]Binomial Distribution[/url]( $n =$ # of trials,$r=$ # of "successful" trials, $p=$ probability of success, $q=1-p$).  A binomial distribution consists of repeated[url=http://en.wikipedia.org/wiki/Bernoulli_trials]Bernoulli trials[/url].  There is also a Bernoulli distribution, which is a special case of the Binomial Distribution where $n=1$, i.e.
$p^{r}\cdot q^{1-r}$.

[quote="bos1234"]A multiple choice exam has three choices and four questions. What is the chance that a candidate who guesses all the answers randomly gets exactly two questions right?[/quote]
[hide="Tell me if I am wrong"]There are $\binom{4}{2}$ ways to choose 2 questions to get right. We must get two questions right and two wrong. The probability of us getting the chosen pair right is $\left(\frac{1}{3}\right)^{2}$. The probability of getting the other two wrong is $\left(\frac{2}{3}\right)^{2}$. So multiply these latexed numbers to get:
\[\binom{4}{2}\cdot \left(\frac{1}{3}\right)^{2}\cdot \left(\frac{2}{3}\right)^{2}\]

\[6\cdot \frac{1}{9}\cdot \frac{4}{9}\]

\[\frac{8}{27}\]
So the probability is $\frac{8}{27}$[/hide]

Edited.

[quote="vishalarul"][quote="bos1234"]A multiple choice exam has three choices and four questions. What is the chance that a candidate who guesses all the answers randomly gets exactly two questions right?[/quote]
[hide="Tell me if I am wrong"]There are $\binom{4}{2}$ ways to choose 2 questions to get right. We must get two questions right and two wrong. The probability of us getting the chosen pair right is $\left(\frac{1}{3}\right)^{2}$. The probability of getting the other two wrong is $\left(\frac{2}{3}\right)^{2}$. So multiply these latexed numbers to get:
\[\binom{4}{2}\cdot \left(\frac{1}{3}\right)^{2}\cdot \left(\frac{2}{3}\right)^{2}\]

\[6\cdot \frac{1}{9}\cdot \frac{4}{9}\]

\[\frac{12}{27}\]
So the probability is $\frac{12}{27}$[/hide][/quote]
Your wrong.  Chekc your math


logarithms
calculus
algebra solved
algebra

Find all a>0 s.t. there exists a positive x with the property                          log {base a} x=x.

Hey, it DOES look better that way! :D  :D  :D But the text which uses the logarithm includes the coindition of x being positive since a logarithm is defined on (0,infinity), that's why I posted it like that.

using calculus, we may find out 0<a<e^(1/e)..

Yeah, that's it. This was proposed for an older IMO, but I can't remember which one. Maybe it can be found in a shortlist or something.


2x^3+4x^2+6x+8=2468 and x is a positive real number, what is the value of x^3+9x^2+9x+7?

How would you do this problem?

[quote="8parks11"]$2x^{3}+4x^{2}+6x+8=2468$

divided the equation by 2 and u get
$x^{3}+2x^{2}+3x+4=1234$

you could find $x=10$ easily from here.
(although this might be difficult at first, 1234, and coefficients of 2,3 and a 4 should give you the hint)

now the second one would simply be
$1997$[/quote]
But there are multiple possible values of x.

10 is the only real value:

$x^{3}+2x^{2}+3x+4=1234$
$x^{3}+2x^{2}+3x-1230=0$
$(x-10)$ must be one of the factors, long or synthetic division to get
$(x-10)(x^{2}+12x+123)=0$

The discriminant, $B^{2}-4AC$ is $12^{2}-4(1)(123)$ which we can see is negative so the other two roots are not real.


The points Q(9,14) and R(a,b) are symmetric with respect to the point P(5,3). what are the coordinates of point R

well, q is just 4 over and 11 up from p, so r will be 4 back and 11 down from p, or[b] (1,-8).[/b]


In a magic square, the sum of the three entries in any row, column, or diagonal is the same value.  The figure shows four of the entries of a magic square.  Find $x.$

\[ \begin{tabular}{|c|c|c|}\hline x & 19 & 96\\ \hline 1 & & \\ \hline & & \\ \hline \end{tabular} \]

[hide="Hint"]First find the bottom left square (let it be $y$).[/hide]
[hide="Further hint"]$y+1=19+96$[/hide]
[hide="And if you are lazy, here is the answer"]$x=200$[/hide]

Sorry for reawakening this thread (for like the 10000th time) but can someone post a solution and explain, please?  :)

[hide]

Call the bottom left box $a$.  We see that $a+1=19+96\Rightarrow a=114$.  Since the common sum is $x+115$, using the diagonal from bottom left to upper right we see that the middle box is $x-95$.  From the middle column we see that if $b$ is the bottom entry in this column, $x-95+19+b=x+115\Rightarrow b=191$.  Using the bottom row and diagonal from upper left to bottom right, we see that $x+x-95=114+191\Rightarrow x=200$.

[/hide]

[hide="OK, here you go :) "]
We see that the sum of each row/column/diagonal is $x+115$. So the block below the $1$ is $114$. Then the middle block is $x-95$, and the block below it (next to the $114$) is $191$. Now we get that the bottom right corner is $x-190$.
Now simply solve for $x$:
$x+115=x+x-95+x-190$
Thus $x=200$[/hide]

Edit: Right... for some reason I did not see the solution above until after I posted mine... and I did not get another topic reply notification either. I guess his solution is a bit shorter than mine :)


linear algebra
matrix
geometry
3D geometry

Loosely inspired by the problem kunny posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294673]here.[/url]

Let $ A\equal{}\begin{bmatrix}1&1\\\minus{}1&\minus{}2\end{bmatrix}.$

Find an expression for $ A^n$ that involves familiar sequences and prove your answer is correct.

hello, i will start with
$ A^2\equal{}\begin{bmatrix}0&\minus{}1\\1&3\end{bmatrix},$

$ A^3\equal{}\begin{bmatrix}1&2\\\minus{}2&\minus{}5\end{bmatrix},$

$ A^4\equal{}\begin{bmatrix}\minus{}1&\minus{}3\\3&8\end{bmatrix}.$
Sonnhard.

Prediction:
For $ n\geq 2$

$ A^n\equal{}\begin{bmatrix}(\minus{}1)^{n\minus{}1}F_{n\minus{}2}&(\minus{}1)^{n\minus{}1}F_n\\(\minus{}1)^nF_n&(\minus{}1)^nF_{n\plus{}2}\end{bmatrix}$,where $ F_n$ is $ n$th Fibonacci number.

For $ n \equal{} 2$ it is true. Assume it is true for all $ k$ s.t. $ 2\leq k < n$. Now,

$ A^{n\plus{}1}\equal{}\begin{bmatrix}(\minus{}1)^{n\minus{}1}F_{n\minus{}2}&(\minus{}1)^{n\minus{}1}F_n\\(\minus{}1)^nF_n&(\minus{}1)^nF_{n\plus{}2}\end{bmatrix}\begin{bmatrix}1&1\\\minus{}1&\minus{}2\end{bmatrix}\equal{}$

$ \equal{}\begin{bmatrix}( \minus{} 1)^{n \minus{} 1}F_{n \minus{} 2} \minus{} ( \minus{} 1)^{n \minus{} 1}F_{n}&( \minus{} 1)^{n \minus{} 1}F_{n \minus{} 2} \minus{} 2( \minus{} 1)^{n \minus{} 1}F_{n}\\( \minus{} 1)^nF_{n} \minus{} ( \minus{} 1)^nF_{n \plus{} 2}&( \minus{} 1)^nF_{n} \minus{} 2( \minus{} 1)^nF_{n \plus{} 2}\end{bmatrix}\equal{}$

$ \equal{}\begin{bmatrix}( \minus{} 1)^nF_{n \minus{} 1}&( \minus{} 1)^nF_{n \plus{} 1}\\( \minus{} 1)^{n \plus{} 1}F_{n \plus{} 1}& ( \minus{} 1)^{n \plus{} 1}F_{n \plus{} 3}\end{bmatrix}$.

So it is true for every positive integer $ n$ s.t. $ n\geq 2$, by induction.

That's correct, Ognjen09.

If you reverse the recursion and accept that $ F_0 \equal{} 0, F_{ \minus{} 1} \equal{} 1,$ and $ F_{ \minus{} 2} \equal{} \minus{} 1,$ you can have these formulas be correct for $ n \equal{} 1$ and $ n \equal{} 0$ as well. In fact, you can push that further: $ A^{ \minus{} 1}$ is also an integer matrix, and Ognjen09's formulas should work for all $ n\in\mathbb{Z}.$

Diagonalization will help us calculate the power, and also would give a closed form formula for the Fibonacci in the process.

I created the problem by taking the usual generating matrix for the Fibonacci numbers, namely $ \begin{bmatrix}0&1\\1&1\end{bmatrix},$ and left-multiplying it by $ \begin{bmatrix}0&1\\\minus{}1&\minus{}1\end{bmatrix},$ which is a well-known matrix cube root of unity.


I'm moving out of the country to Japan in a few days, probably by the end of January 2008. I will not be attending a public school once I'm there, neither Japanese nor international. I've read the FAQ's on how to take the AMC's as a homeschooler, but the registration process was unclear to me. Since it's past December 15, I'm guessing the next deadline is Jan. 27.

Can I even take the AMC's in Japan anyway? The registration form on the AMC website seems to have North American countries/states listed as international, but no others. I also do not know yet what homeschooling organization I will enroll in, so am I allowed to turn in a registration form without a school name?

I'm aware of receiving pins/medals for acheiving the highest score in a school for two or more consecutive years. How does that apply to homeschoolers, if at all?

Thanks in advance.

how are u aware of winning those school winner pins and medals.

For this sort of question, you definitely want to talk to someone at the AMC.

As for the school winner question- homeschoolers are not included.

Your registration questions are too specfic to an individual and too detailed to answer on this forum.  Please contact the AMC office by any of the means listed on any of our printed material, or at our website.  However, as a general answer, yes it is possible to register as a homeschool participant outside the boundaries of the U.S., but in order to participate officially, you need to follow the rules including adminstration and proctoring of the contest as any other homeschool participant.

For the awarding of pins/medals, see the Frequently Asked Questions, especially General Question #7 at
http://www.unl.edu/amc/f-miscellaneous/faq.shtml
Our scoring and reporting programs only flag high-score pin/medal winners if the school has a team, that is three or more participants.  See the Teachers' Manual, page 5, Section IV and page 8, Section XI at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
That means singleton participants generally do not receive a pin/medal unless there are other circumstances.  Again, your case is too specialized to be addressed in this forum.

Steve Dunbar
MAA Director,  American Mathematics Competitions

all right, thank you both, jmerry and Mr. Dunbar, for your answers. I'll contact someone at the AMC.


In the beginning, the Administrator created FTW, and it was a roaring success right from the start. People flocked to the game, excited and anticipating great things, and they were not disappointed. But it was not long before things took a bad turn. A very bad, nasty turn. There will always be evil people in this world, and as such, there will always be evil people in FTW. And so they came, and they spammed and hacked and made it a very unenjoyable experience.

Soon the hundreds of FTWers dwindled down to the twenties, and then again to a mere sixteen. This brave band decided that if no one else would do the job, they would sort out which among them were the troublemakers, and ban them one by one. There will be mistakes - that is expected. But they desperately hoped they could remake FTW into the amazing game it once was, or else the Internet forever would be a barren wasteland.

Good luck.

The Rules(read them)
[list]
    [*] Votes must be in bold. If you do not bold your vote, it will not be counted.
    [*] Once a player has reached lynching state, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.
    [*] The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
    [*] Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post.
    [*] Don’t edit/delete previously submitted posts
    [*] Don’t quote any PMs from me. Neither true, nor false. Seriously.
    [*] If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough! No choice will be made.
    [*] Treat this game as a commitment. Be considerate – don’t leave us hanging.
    [*] Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/list]

The Living:

1. ernie
2. meewhee009
3. spaceguy524
4. PowerOfPi
5. zapi2007
6. pythag011
7. EggyLv.999
8. KING1
9. cf249
10. james4l
11. mathfreak123
12. westiepaw
13. Chompy
14. AIME15
15.SuperNerd123
16. AceOfDiamonds

The Dead:

None as of yet.

[quote="meewhee009"][b]Confirm[/b]
Haha... FTW... Nice Izzy.  :)[/quote]

You can't say you didn't see that coming. :P

Yeah...
I was expecting something 'Izzy' style.
This was on my list of possible themes.  :D

Yay self-referential mafia game.

See Mafia 39* for excellent example (not really) of a self-referential mafia game!
Oh... confirm!

*woohoo shameless self-advertising!

Awesome! I [b]confirm![/b] Cant wait! W00T! Yay!

@james4l
Yeah, I read your game...
Cult win.
Niice...
I was studying my Mafia, and like... I like reading threads :)

[quote="AIME15"][quote="ernie"]Town also made horrible lynches, a pattern which is now turning into an epidemic.[/quote]

Day 1, Spaceguy524 was, er, STUPID.  I maintain that the Day 1 lynch was mostly Spaceguy524's fault.[/quote]

Day 2 lynch was also mostly KING1's fault.

[quote="ernie"]@izzy: Can you give us the cop sanities?[/quote]

AIME15: naive
ernie: paranoid
pythag011: insane

[url=http://www.quicktopic.com/43/H/ZBHmQ2aN8wdz]Moderator QuickTopic[/url] (I would have never guessed that ernie and AIME15 were talking outside the thread)
[url=http://www.quicktopic.com/43/H/atRCahfRNeZ2X]Mafia QuickTopic[/url]

That was my first game as mafia, which was interesting. After the second night, I realized that my investigating abilities were useless, since everyone had a power role. For the mafia, my ability to see the mod QT wasn't very helpful except for my roleclaim and providing false facts. westiepaw's power was the critical advantage for the mafia in my opinion, and the siblings slightly balanced it.

For the town, there was a lot of power roles, most of which were very useful, but some of which weren't (like cf249's). I think that part of the advantage for the mafia was the fact that some key power roles where lynched early on, and the first non-pro-town was killed late in the game. I think meewhee's mistake lynched Chompy and ernie's mistake lynched meewhee (by mistake, I mean voting too early).

The overall game was a little bit towards the mafia.

well meewhee didn't post a good defense in my opinion.
she didn't realize that it was lylo(lynch or lose), and was going on about something like lynch me and lynch someone else if i'm town blah blah blah
this is very nooby play(making 'deals')

I think the setup was ok in terms of balance, but town played badly.

Quite a few of the early lynches were pretty bad...

spaceguy524, the Administrator, zapped by lightning, Day 1
james4l, the messenger, suffocated by a pillow, Night 1
pythag011, the moderator, defenestrated Night 1
KING1, the power user, thrown down a hatch Day 2
cf249, the outcast, drowned Night 2
SuperNerd123, the memmer, toppled off a mountain Day 3
AceOfDiamonds, the multi-account, poisoned Night 3
EggyLv999, the multi-account, scared to death Night 3
Mathfreak123, the serial killer, run over by an ambulance Night 3
Chompy, the doctor, killed by poison gas, Day 4
AIME15, the moderator, killed by a creepy kid, Night 4
Meewhee, the special user, committed suicide, Day 5

Spaceguy: Lynching him was really really stupid. At least give him time to roleclaim or ask for replacement.
KING1: Yeah, he was playing really really badly (refusing to restate your defense, HOW HARD IS IT TO MAKE 1 POST SAYING YOUR DEFENSE INSTEAD OF "GO LOOK FOR IT YOURSELF", EXTREMELY OVERREACTING TO EVERYTHING) but if you look at the interactions between him and other people as well as his meta he doesn't seem that lynch-worthing.

The town also decided day 3/4 that it had no evidence to work with even though it had A LOT OF EVIDENCE TO WORK WITH. There was lots of information that the town could have used, etc. Instead town decides to rush lynches based on not-that-much information.

Note: In the future, if you refuse to defend yourself from a case, I DONT CARE IF YOU ARE ALMOST CONFIRMED TOWNIE, POLICY LYNCH YOU ANYWAYS. (By almost confirmed, i mean QUITE POSSIBLY CONFIRMED but not necessarily. Godfathers, etc.)

At the end when we were in LYLO mode, and the thing that happened with the other lynched.... well I assumed the worse.
Anyways...
AHAHA!
WESTIEPAW WAS SCUM!
YOU DIDN'T LISTEN TO ME!
HAHAHA!
I'm sorry, I wanted to gloat :)

I agree that this game was a bit mafia-biased.  But I=mafia.  W00t!

And yes, the first several days were a DISASTER for town.  :P   This was my first mafia game, so it was interesting to play as a mafia.

Thanks for modding, izzy!  You had some of the most colorful flavortext I've ever seen.  :)

[quote="isabella2296"]xD Thanks all.

Sorry for the mafia-bias. :([/quote]

Its not a problem, it wasn't the mafia-bias that killed town, it was town's fault that they lost.

[quote="AceOfDiamonds"]as a townie, you shouldn't say 'if i turn up town, lynch blahblahblah.'
that is stupid and very scummy
how do you know they're scum for sure?
if you know for sure why not just state your reasons?
and anyway, you were already lynched at 2 votes so you can't do anything about it[/quote]
Yeah that's what I meant. I couldn't post a defense because I was already going to die. :( So that just came up.

@ernie: Yes, I know.  The townies made such wonderful lynches on the first several days!!!  :P

lastpost (this is used to preserve order of locked topic)


videos
geometry
geometric transformation

While chatting with colleagues from other countries, I've noticed that people in general have some weird prejudices about Bosnia... In addition to that, some of them, who think they know everything, got it all wrong...

What do you really know about Bosnia and Herzegovina?

[quote="bubka"]If that was directed at me, I have no weird prejudices about BH. I sympathize with it. It was one of the countires to bore the brunt of the cold war most badly.[/quote]

No, I wasn't thinking of you... In fact, some pepole from non-European countries on the IMO this year inspired me...

[quote="fredbel6"]What I know (correct or incorrect):

Capital is Sarajevo, its population consists of Bosniaks, ethnic Serbs and ethnic Croatians (who are respectively muslims, orthodox christians and catholics).
When Yugoslavia fell apart, these three groups clashed.  Initially Bosniaks and Croatians were somewhat allied against the ethnic Serbs but they also turned on each other (like in Mostar where the bridge was blown up).
Serbia (under Milosevic) sent troops on behalf of the ethnic Serbs in Bosnia.
In 1995 8000 men were killed by Mladic's troops in Srebrenica when the UN failed to properly protect their stronghold (and even assisted in the separation of the men from their families).
In the same year, a peace agreement was reached in Dayton, USA.
Serbian war criminals Mladic and Karadzic are still to be found.  The war has left Bosnia as one of the poorest countries in the former Yugoslavia.

So what is correct of this?[/quote]

You didn't skip your history classes...  :D Well done... Yes, everything is correct, but there is one thing to add. Not only that the troops from Milosevic's FR Yugoslavia participated, on the side of the Bosinan Serbs, but also the troops from Croatia participated (ofcourse, on the Bosnian Croats' side)...

i think it is unfair to prejudices a country or a nation, because  one or some idiots.

[quote="bubka"]Are you serb, croat, or bosniak?

By the way, I have been thinking about this since the world cup: Is Serbia nad Montenegro a country? And if so, where did it break away from?[/quote]

In 1992 Serbia and Montenegro became the Federal Republic of Yugoslavia, when Yugoslavia started to fell apart.
In 2003 they became Serbia/Montenegro (I guess it didn't make sense to keep referring to themselves as Yugoslavia)
Serbia/Montenegro was a country until a few months ago (Montenegro voted for a breakaway, one of the reasons is that on their own they have a better chance for EU membership than together with Serbia, which is accused of not handing over war criminals to the Tribunal in Den Haag).
Serbia is thus since a few months a (landlocked) country on its own.

[quote]You didn't skip your history classes  :D ...  Well done[/quote]
Actually, we've heard a lot of talk in high school and elementary school about the suffering, but we never studied Bosnia (or the Yugoslavia breakup) in school, so I never had a clue why they were fighting, or even what nations and groups were involved.  I just dediced to sort it all out on my own. :huh:

hsiljak, you might also be able to answer this question

I saw a video on television (yesterday!) which could prove war crimes, against Serbs.  They said it has caused unrest in Serbia, Croatia and Bosnia.
It was from 1995 (Operation Storm?) and it involved crimes against Serbs in Krajina.  They didn't show him, but they said their commander was from Saudi-Arabia.
So if Croats and Bosniaks turned against each other in 1992, what were Bosniaks (with an Arab leader) doing in Krajina helping Croats to kick out Serbs? :huh:

[quote="bubka"]Are you serb, croat, or bosniak?

By the way, I have been thinking about this since the world cup: Is Serbia nad Montenegro a country? And if so, where did it break away from?[/quote]

I'm a Bosniak.

And about the Serbia and Montenegro... The Socialist Federative Republic of Yugoslavia (disintegrated in 1992) was made up from 6 republics: Slovenia, Croatia, Bosina and Herzegovina, Serbia, Montenegro, Macedonia. In 1991 and 1992 Slovenia, Croatia, Bosnia and Herzegovina and FYR Macedonia claimed their independence. Serbia and Montenegro formed "the third Yugoslavia": Federal Republic of Yugoslavia. In the beginning of the XXI century, it changed its name to Serbia and Montenegro. After the Montenegrins claimed their independence in May/June this year, this Union dissapeared.

EDIT: Ooops, I see it was already answered, sorry...

[quote="fredbel6"][quote="bubka"]hsiljak, you might also be able to answer this question

I saw a video on television (yesterday!) which could prove war crimes, against Serbs.  They said it has caused unrest in Serbia, Croatia and Bosnia.
It was from 1995 (Operation Storm?) and it involved crimes against Serbs in Krajina.  They didn't show him, but they said their commander was from Saudi-Arabia.
So if Croats and Bosniaks turned against each other in 1992, what were Bosniaks (with an Arab leader) doing in Krajina helping Croats to kick out Serbs? :huh:[/quote][/quote]

Well, I'll try to explain:
The war between Croats and Bosniaks begun in 1993, and ended in 1994, when they formed some sort of pact, puting their armies under one Armed Forces Command. In the same time, the so-called Federation of Bosnia and Herzegovina was formed (a consequence of the "Washington agreement"). So, Bosniaks and Croats fought together until the end of the war (Dayton peace agreement).
In the 1995, the Serbs were holding their posts in Krajina, area both in Croatia and Bosnia. The UN protected enclave of Bihać (in bosnia) was surrounded. after the shameful Serbian conquest of the UN protected enclave of Srebrenica, everyone knew that Bihac must be liberated-fast! The forces of the 5th Corps (Bihac Corps) went in an offensive, and liberated a big portion of Bosnian part of Krajina. They met with the units from Croatia on the state border. Some of Bosnian troops fought day or two in Cratia, and vice versa.

Arabs in Bosnia are another important questions... They were volounteers, teaching Bosniak units about combat and so... It's often speculated about the numbers of Arabs in Bosnian Army. Some say that here were few thousands of Arabs, but that's not true. There couldn't be more then 300...

@x-and-y: I agree. But why do you say it in this context?

My answer to Bubka's question was only a "I think" answer, open for improvement.  You probably know your own country and the surrounding ones ten times better :roll:

[quote]n the 1995, the Serbs were holding their posts in Krajina, area both in Croatia and Bosnia. The UN protected enclave of Bihać (in bosnia) was surrounded. after the shameful Serbian conquest of the UN protected enclave of Srebrenica, everyone knew that Bihac must be liberated-fast![/quote]
I heard of the Krajina Serbian republic in Kroatia, but I didn't know it also had parts in Bosnia.
When you speak of the Bihac enclave in Bosnia, you mean it was a Bosniak enclave?

[quote]The forces of the 5th Corps (Bihac Corps) went in an offensive, and liberated a big portion of Bosnian part of Krajina. They met with the units from Croatia on the state border. Some of Bosnian troops fought day or two in Cratia, and vice versa. [/quote]

Bihac Corps = Bosniak (or Croatian?)

[quote]Arabs in Bosnia are another important questions... They were volounteers, teaching Bosniak units about combat and so... It's often speculated about the numbers of Arabs in Bosnian Army. Some say that here were few thousands of Arabs, but that's not true. There couldn't be more then 300... [/quote]
Is it true that the Arab militia were in the end dissatisfied with the rather secular lifestyle of the Bosniaks? (I heard Bosniaks don't wear veils, drink alcochol,etc..)  I heard that (apart from language maybe) in customs and appearance the people from different groups were practically the same.

So if understand correctly, the images I saw yesterday were Serbs falling victim to Bosniak troops (led by an Arab leader) in the Croatian part of Krajina?

I saw two movies about Bosnia recently :
Warriors (British) about 1992
[url]http://www.imdb.com/title/tt0119873/[/url]
De enclave (Dutch) about the events (and the consequences) in Srebrenica, and Dutchbat's shortcomings

[url]http://www.imdb.com/title/tt0309562/[/url]
Have you seen them?

[quote="bubka"]How did the Serbia and Montenegro play in the World Cup as Serbia and Montenegro?[/quote]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=101872[/url]
Bubka you may find this topic (that I started) relevant as well.

Bihac was a Bosniak enclave. 5th Corps was part of the Army of Bosnia and Herzegovina (Bosniak).
When I'm talking about Krajina, I'm talking just about the geographical term, not the so-called Republic of Serbian Krajina.
The republic was only in Croatia, but the Bosnian Krajina was also occupied by the Serbs.
And about the Arabs-yes, it is true that we're a bit more secular than them, so they did feel a bit strange :D
And the different groups here don't even differ in language. it's hard to say at glance who's who...

And the footage you've seen... I really can't say was it taped in Bosnia or Croatia-probably somewhere on the border.

The movies you saw: I've seen Warriors, and it's a great movie. But many people here (mostly the Croats) don't share this opinion. Because of this movie, the chief editor on the National TV had to resign (bacause he allowed it to be broadcasted...)-ofcourse, it has never been officially said that it was the reason, but...
I haven't seen the Duch one yet.

I would reccomend you to see the movie "No man's land", by Danis Tanović. It's an Oscar awarded film, a nice story about the war... it's full of metaphores, subtle, somethimes ironical picture of us, and the Europeans...

If I can be so insolent (and that is what I wonder about all ladies and gentlemen from the Balkan here) how old are you?  I mean, did you see much of the war?  :|

[quote="hsiljak"]
I haven't seen the Duch one yet.

I would reccomend you to see the movie "No man's land", by Danis Tanović. It's an Oscar awarded film, a nice story about the war... it's full of metaphores, subtle, somethimes ironical picture of us, and the Europeans...[/quote]

Well the Dutch one is very good.  It's about a translator who lost his family in Srebrenica.  He later marries one of the soldiers, while he works as translator in Den Haag. It's low budget, but as it was made from European perspective, it all seems very realistic (all the languages are correct, instead of everyone just speaking English). Many events like Karremans drinking with Mladic or drunk Dutch soldiers partying in Zagreb while Srebrenica is still happening, have been incorporated into the film.

No Man's land, I have seen some parts of that movie with interruptions (I remember the man on the mine), I probably should watch it again.

There is another dull film about Balkan conflict.  I don't know the name, but the plot sounded dumb : USAer loses family in Paris due to muslim fundamentalists' bombings, shoots random muslims in local mosque, then goes along with some Serbs to plunder? :huh:
And what did you think of "the Peacemaker" (angry bosnian runs around in Manhattan with NUCLEAR BOMB)

It's just that the first movies seem to give a somewhat realistic view, but I would like to know what the people themselves think.

Peacemaker, is, in my humble opinion, a pretty dumb movie... It has no connection to the reality whatsoever. It's the typical Americans-save-the-world-from-ugly-Russians-and-their-impersonators film. Behind Enemy lines (with Owen Wilson) is a bit better, but it should have been done more seriously. I'll watch the Dutch film, as soon as I find it... Thank you for recommending it. The film about the guy in Paris- I haven't seen it, but I remember something like that...

How old am I? 18. War begun when I was 3, ended when I was 7... So I remember some things...

Wow. Our ages are the same. I am 18 too. But no war in india started when i was 3 and no war ended when i was 7.

[quote]The film about the guy in Paris- I haven't seen it, but I remember something like that... [/quote]

I found it!  It is "the Savior".
[url]http://en.wikipedia.org/wiki/Savior_%28movie%29[/url]

It doesn't sound too good, is it good? :maybe:

[quote="bubka"]Where do you think I could download those movies?[/quote]

You naughty boy! :wink:   You must rent or buy them or watch them on television

Well as hsiljak already sai, "No Man's Land" is recommendable, and "Warriors" en "De enclave" are pretty good, but I hope you are becoming enough of a man to see the description of "the Peacemaker" and "the Savior" are pretty...dumb

Oh yes I have watched "No man's land". I was thinking of mentioning it. I got the theme and all (it was very moving) but the setting escaped me - except that it was serbia vs soem other army.

Doesn't B&H have a blue-and-gold diagonally seperated flag?

[quote="fredbel6"]Hey hsiljak (forgive my ignorance)
but allow me to ask this :

under Tito/communism (I know Tito didn't survive communism in Yugoslavia), in Bosnia:
- were people allowed to travel/live/work abroad, unlike in Warsaw Pact nations (I'm pretty sure they were allowed to travel.
- did the average family have a car? (did your parents have a car?)
- did the average family have a television?
- were films/music from the USA imported
- were there commercial newspapers/television stations, or was it state controlled?

Well, you don't have to answer all of these questions of course  :roll: , I'm a bit older(21) and quite frankly I too must admit communist Yugoslavia rather "escaped" me back then  :oops: .[/quote]

How do any of those matter really? None of them is a yardstick in measuring a communist government. I am not saying Yugoslavia under Tito was great, but if it wasn't, that's not for any of these reasons.

They matter to me  I respect your opinion, but that doesn't stop me from still awaiting hsiljak's answer.  You cannot answer instead of him or other Yugoslavs.
 :|

[quote="fredbel6"]Hey hsiljak (forgive my ignorance)
but allow me to ask this :

under Tito/communism (I know Tito didn't survive communism in Yugoslavia), in Bosnia:
- were people allowed to travel/live/work abroad, unlike in Warsaw Pact nations (I'm pretty sure they were allowed to travel.
- did the average family have a car? (did your parents have a car?)
- did the average family have a television?
- were films/music from the USA imported
- were there commercial newspapers/television stations, or was it state controlled?

Well, you don't have to answer all of these questions of course  :roll: , I'm a bit older(21) and quite frankly I too must admit communist Yugoslavia rather "escaped" me back then  :oops: .[/quote]

Hmmm...
Traveling, working, living abroad? Yes, ofcourse... A large portion of yugoslavs went to Western Europe in the 70s (esspecially West Germany), to work there (Gastarbeiters). When they would have made enough money for a cosy living in Yugoslavia, they would return here. Many people from here worked and lived in Lybia, Iraq... Government supported that... In that time they could travel more than we can now (Now, all countries descendants of Yugoslavia, except Slovenia have huge problems with visa requirements...). Some of my close relatives went in bussiness delegations to USA, for example...

My parents didn't have a car (they don't have it today...). But an average family could have afforded a car (Lada, Topolino, Polszky Fiat, Yugo, Zastava, PZ etc. were quite cheap cars...).

TV? Yes, an average family had a TV in the 70s.

Films from USA were imported. US music made also a huge impact on the popular Yugoslav music. Most of the rock bands here made their first songs listening to Rolling Stones, Pink Floyd, Led Zeppelin...

First commercial televisions appeared in the end of communist rule, but indipendent newspapers appeared much before.

Satisfied?

And the flag of B&H looks like this. First flag of indipendent B&H was quite different (dominantly blue coat of arms with golden fleur-de-lyses on white background).

Wow another similarity between me and hsiljak. We don't own a car as well. Cheap cars aren't available here. The cheapest cost the equivalent of 4000 dollars.

[quote="fredbel6"][quote="vedran6"]Sorry if I am late on this topic but I want to add one thing.In this recent horrible wae in B&H(1992-1995) the Army BiH was multi-ethnical because there were a lot of Bosnian Croats and Serbs in it which fought for independent and united Bosnia and Herzegovina.[/quote]

Against whom and at what time during the war?[/quote]

Against both the forces of HVO (Croatian Defence Council) and the Army of Republika Srpska. Army of bosnia and Herzegovina was multiethnic all the time during the war (there were more Crats and Serbs in the Army in the beginning though, before the Washington agreement).

[quote="fredbel6"]HVO , were they Croats from Croatia or from Bosnia (in general)?

So the Bosnian official army fought both Croats and Serbs?  But were there no Bosniaks in this army?  What about them?[/quote]

HVO are Croats from Bosnia in general. But some forces from HV (Croatian Army)-Armed Forces of Republic Croatia fought side by side with them.

Bosniaks were the majority in the official Bosnian army. But thre were also significant number of Serbs and Croats in ARBiH (Army of Republic Bosnia and Herzegovina).

[quote="futuremontrealian"]I have a question I bet you might be able to answer.  How do you pronounce Ljubljana (I have no idea if that's correct spelling)... as in capitol of slovenia?  Yes I realize it's not Bosnia but it's practically right next door, and I never really know if I'm saying (or spelling) the capitol right.[/quote]

I'll try to help you... It is in a way similar to the American pronunciation of the word Lloobllanah... Looks funny, right:) The ll (in original: lj) iss pronounced like ll in "llama". "Oo" is pronounced /u/, "anah" as /ana/. I hope that this helped you...

well, people from ljubljana pronounce it something like"Lublana".

this facts aren't well known in croatia, so i suppose they aren't known in the rest of the world:
in the beginning of the war croatian army (HV) was different:
there were muslims (from croatia) and some croatian serbs (not many) in it. people went to war 'cause it was their duty as civilians (like paying taxes). later, HV changed from state army (državna) into national army of croats (nacionalna) were was a great (great) mayority of croats (from croatia and from herzegovina as well). there were some changes in commanding structures (generals Tus and Domazet became less important, and generals like rojs and gotovina became important)

after operation oluja (storm) there HV went to liberate some positions in BiH which were under serbian army (in accordance with Army Bih- something signed in Split). so the Bihać and other were liberated. HV had some fights in BiH (HV even came 20 km from Banja Luka when UN stopped it from entering), and Army Bih had some fights in Croatia (e.g. Plitvice).
some people in croatia as well in bosnia think that BiH would be different if the UN hadn't stoped HV from entering Banja Luka (BiH wouldn't be separated into Federation of Bosnia and Herzegovina and Republic of Srpska)

all in all, the war is behind us.

btw. frederik I've read that there is some conflict (verbal one) in Belgium caused by Y. Leterme and something about learning languages (I remember u have said something before) so pm me.

I always thought it was like Liubliana (the j is replaced by an 'i' sound).

But isn't this grossly offtopic?

[quote="Andreas"][quote="fredbel6"]What I know (correct or incorrect):
In 1995 8000 men were killed by Mladic's troops in Srebrenica when the UN failed to properly protect their stronghold (and even assisted in the separation of the men from their families).
[/quote]

In this case I would say, it is incorrect or, at least, incomplete.
I suggest to read [url=http://www.eurasia-rivista.org/cogit_content/articoli/ArgomentoSrebrenica.shtml]this[/url] and [url=http://www.tibereide.it/articoli_dettaglio.asp?articolo_id=169&articolo_categoria=1]this[/url], in order to understand more about Srebrenica and in general about the defamatory campaign against the Serbs. (unfortunately I found it only in italian  :( )[/quote]

Incorrect? Incomplete? I would like to inform you that the International Crime Tribunal for former Yugoslavia in Den Haag has classified the "conquest" of Srebrenica as genocide, my dear friend. The Tribunal isn't pro Bosniak, nor pro Serbian, but it liberated the Bosniak army leader in Srebrenica Naser Oric, who was accused in one of the texts above for illing 3000 Serbs in Srebrenica...
Two texts which you mentioned are quite interesting, but also quite inconvenient and their auhors are a bit "aligned" towards the Serbs, don't you think? One of them says he had a conversation with Karadzic, but didn't find it important to talk to anyone from the other side...

You don't have an English, French, Dutch (or maybe German? :( ) translation of those texts?  Because I don't know any Italian...

[quote="hsiljak"]I would like to inform you that the International Crime Tribunal for former Yugoslavia in Den Haag has classified the "conquest" of Srebrenica as genocide, my dear friend. The Tribunal isn't pro Bosniak, nor pro Serbian, but it liberated the Bosniak army leader in Srebrenica Naser Oric, who was accused in one of the texts above for illing 3000 Serbs in Srebrenica... [/quote]
What do you mean?  Do you consider Naser Oric, personally, as a war criminal.  If so, then doesn't his liberation by the tribunal make the tribunal biased? :|

[b]hsiljak[/b], I wouldn't justify crimes committed by a part or those by the other one, I only would say that crimes against Serbs are, in my as in the opinion or a lot of non-aligned people, such as Yves Bataille, voluntarily omitted by the international comunity.

About genocide, there're a lot of qualified people such as Perez Pujol, director of the Anatomical Legal Institute of Cartagena, that put in discussion the effective number of murdered people in Srebrenica.

Also it is known that in Racak, for example, as in other places, in some presumed common graves were putted UCK rebels disguised as civils.

Finally about the seriousness of the International Crime Tribunal and in particular of Carla Del Ponte there would be a lot to discuss!

[quote="fredbel6"] If so, then doesn't his liberation by the tribunal make the tribunal biased? :|[/quote]

when you are found guilty of killing a single person (civil court) you get a bigger punishments than the ones found guilty in Den Haag.

a victim is always a victim. a crime is always a crime. (or it's the way it should be)

so the tribunal should give stricter punishments (remember just Biljana Plavshich. she is in a better place now (a nice swedish female jail) than she would be in her home on a freedom)

[quote]when you are found guilty of killing a single person (civil court) you get a bigger punishments than the ones found guilty in Den Haag. [/quote]
Or in other words : you don't feel the tribunal is doing a good job? :|

[quote="deimos"]well u could say so. they should be more efficient, faster and stricter (emphasis on faster)[/quote] Well it doesn't look like they will punish a lot of the big players.  Milosevic has died, and Karadzic and Mladic are not even arrested yet.  There are many conspiracists who believe Milosevic was either murdered or intentionally denied proper medical care in order to get him killed before he could expose certain things.  What do you think of this?

I have always found it weird that countries like the USA have taken such a strong pro-muslim and anti-Serb stance, in both the Bosnian war as the
Kosovo war?  Why? :|

Edit : people who wish to discuss the overthrow of Milosevic in 2000, and the role of foreign players like CIA in that, are definitely more than welcome in my [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105653]Flower Revolution topic[/url]
[hide]I admit it, I'm desparate to bring my topic back to life  :( [/hide]

[quote="bubka"]oh my god this tiopic is still going on?
but apparently hsiljak has given up.[/quote]
What do you mean?  There is nothing wrong with this topic.  What do you mean by given up?

[quote="bubka"]
but apparently hsiljak has given up.[/quote]

Nope, I'm still here... I just had some important things to do...

@fredbel6 (about Oric): I really don't know what to think. Maybe his forces did something that could be characterised as a war crime, but I doubt thos crimes were so massive as described in the official Hague accusation and in media.

(about American pro-Muslim stand): I don't think it was a strictly pro-Muslim stand. US goverment wanted to "balcanize" (divide) the area: smaller countries, smaller problems... Who knows when can Serbs (with Russians, ofcourse) make a problem for the NATO? If they live in a tiny country, the problem will be tiny too... It is just one of the reasons, ofcourse.


at 2:15 o clock, what will be the angle between the hour and minute hands of a clock?

[hide]22.5 degrees.

Well, you know that between 2 and 3 o'clock, there is 30 degrees. If the minute hand is on the 3, that means that the hour hand is 1/4 of the way between the 2 and the 3, since is is 1/4 of the way into the hour. Since there is still 3/4 of the distance between the 2 and the 3, you take 30 * 3/4 = 22.5 degrees. [/hide]

GOOD THINKING elementsofmathkid!!

[hide="hint"]the answer is not 30 degrees[/hide]

[hide]The minute hand is at 90 degrees and the hourhand is at 135/2 degrees

180/2 - 135/2 = 45/2 = 22.5 or 22 degrees 30 minutes.[/hide]

Many students fall into the trick of thinking that the angle is simply 30, but of course, the hour hand has the freedom to move too  :)


Romeo and Juliet are found dead on the floor in a bedroom. When they were discovered, there were pieces of glass and some water on the floor. The only furniture in the room is a shelf and a bed. The house in is a remote location, away from everything except for the nearby railway track. What caused the death of Romeo and Juliet?

I already told you once that these titles are not tolerable! ( http://www.mathlinks.ro/Forum/viewtopic.php?t=116637 )

PS: known ;)

Yeah, I have seen it before too.

[hide="answer"]They are fish. The fish bowl fell and broke because of the trains.[/hide]


function
complex numbers
complex analysis
complex analysis unsolved

find the number of zeros of the function $ f(z)\equal{}z^4\plus{}5z\plus{}1$ in the square $ 0<Re z<1, 0<Im z<1$ .

i tried to make a picture of what f does to the boundary of the square, so that i can compute the winding no around 0 but this is exactly where i have the problem. can someone give hints? thanks in advance.

[quote="kouboy"]
 can someone give hints? thanks in advance.[/quote]
It seems to me that you are using the [url=http://en.wikipedia.org/wiki/Argument_principle]argument principle[/url] instead use [url=http://en.wikipedia.org/wiki/Rouche%27s_theorem]Rouche's theorem[/url]. :maybe:

Personal question: Usually working on a disk is the most ideal set to work on. Suppose that we can map that square biholomorphically onto a unit circle. Can we now work on the unit circle instead. The problem is of course finding the proper mapping?

You'd better just parametrize the sides. Let's do the right vertical side for instance. It is $ z\equal{}1\plus{}it$, $ t\in[0,1]$. The function becomes $ (1\plus{}it)^4\plus{}5(1\plus{}it)\plus{}1$. Note that the real part of this expression is $ 1\plus{}t^4\minus{}6t^2\plus{}6> 0$ all the time, so there are no full turns around $ 0$ involved. Thus, all you need is to compute the values at the endpoints and subtract their arguments. The value at $ 1$ is $ 7$, the argument is $ 0$. The value at $ 1\plus{}i$ is $ 2\plus{}5i$ and the argument is $ \arctan 5/2$. So, the total increment of the argument over this side is $ \arctan 5/2$. You can do three other sides similarly :).

cant we use sylow theory's suggestion? i have a feeling that 5z+1 will dominate z^4 on the boundary. so there are are no zeros.

fedja's suggestion also seems to give the same answer since the value at i is also 2+5i...

[quote="kouboy"]i have a feeling that 5z+1 will dominate z^4 on the boundary. [/quote]
Yeap. Actually, not only on the boundary but inside as well. Indeed, $ |5z\plus{}1|> |5z|\ge 2^{3/2}|z|\ge |z|^4$ everywhere in the square, so the problem is much simpler than I thought :blush:.


The square
\[ \begin{tabular}{|c|c|c|}
\hline
50&\textit{b}&\textit{c}\\ \hline
\textit{d}&\textit{e}&\textit{f}\\ \hline
\textit{g}&\textit{h}&2\\ \hline
\end{tabular}
\]is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $ g$?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 25 \qquad
\textbf{(C)}\ 35 \qquad
\textbf{(D)}\ 62 \qquad
\textbf{(E)}\ 136$

[hide]
If the power of a prime $ p^n$ other than $ 2,5$ divides $ g$, then from $ 50 \cdot 2 e\equal{} 50dg$ it follows that $ p^n|e$, but then considering the product of the diagonals, $ p^{2n} |gec$ but $ p^{2n} \nmid 100e$, contradiction. So the only prime factors of $ g$ are $ 2$ and $ 5$.

It suffices now to consider the two magic squares comprised of the powers of $ 2$ and $ 5$ of the corresponding terms. These satisfy the normal requirement that the sums of rows, columns, and diagonals are the same, owing to our rules of exponents; additionally, all terms are non-negative.

The powers of $ 2$:
\[ \begin{tabular}{|c|c|c|}
\hline
1&\textit{b}&\textit{c}\\
\hline
\textit{d}&\textit{e}&\textit{f}\\
\hline
\textit{g}&\textit{h}&1\\
\hline
\end{tabular}\]So $ 1 \plus{} 1 \plus{} e \equal{} g \plus{} e \plus{} c \Longrightarrow g \equal{} 2 \minus{} c$, so $ g \equal{} 0,1,2$. Indeed, we have the magic squares
\[ \begin{tabular}{|c|c|c|}
\hline
1 & 0 & 2 \\
\hline
2 & 1 & 0 \\
\hline
0 & 2 & 1 \\
\hline
\end{tabular}, \quad \begin{tabular}{|c|c|c|}
\hline
1 & 1 & 1 \\
\hline
1 & 1 & 1 \\
\hline
1 & 1 & 1 \\
\hline
\end{tabular}, \quad
\begin{tabular}{|c|c|c|}
\hline
1 & 2 & 0 \\
\hline
0 & 1 & 2 \\
\hline
2 & 0 & 1 \\
\hline
\end{tabular},\] The powers of $ 5$:
\[ \begin{tabular}{|c|c|c|}
\hline
2&\textit{b}&\textit{c}\\
\hline
\textit{d}&\textit{e}&\textit{f}\\
\hline
\textit{g}&\textit{h}&0\\
\hline
\end{tabular}\]Again, we get $ 2 \plus{} e \equal{} g \plus{} e \plus{} c \Longrightarrow g \equal{} 0,1,2$. However, if we let $ g \equal{} 2, c \equal{} 0$, then $ e \equal{} d \plus{} e \plus{} f \Longrightarrow d \equal{} f \equal{} 0$, which obviously gives us a contradiction, and similarly for $ g \equal{} 0, c \equal{} 2$. For $ g\equal{}1$, we get \[ \begin{tabular}{|c|c|c|}
\hline
2 & 0 & 1 \\
\hline
0 & 1 & 2 \\
\hline
1 & 2 & 0 \\
\hline
\end{tabular}\]In conclusion, $ g$ can be $ 2^0 \cdot 5^1, 2^1 \cdot 5^1, 2^2 \cdot 5^1$, and their sum is $ 35\ \mathbf{(C)}$.
[/hide]

[hide="Alternative Solution"]From the table: $ \begin{tabular}{|c|c|c|}
\hline
50&\textit{b}&\textit{c}\\
\hline
\textit{d}&\textit{e}&\textit{f}\\
\hline
\textit{g}&\textit{h}&2\\
\hline
\end{tabular}$

We have $ 50dg\equal{}2gh \iff 25d\equal{}h$. Similarly, We have $ 50bc\equal{}2cf \iff 25b\equal{}f$. Then we have $ beh\equal{}100e \iff bh\equal{}100 \Rightarrow bd\equal{}4$. Now we can casework $ (b,d)\equal{}(4,1), (1,4), (2,2)$:
[b]
Case 1: $ b,d\equal{}(4,1)$.[/b]
Using $ 25d\equal{}h$ and $ 25b\equal{}f$, we have
$ \begin{tabular}{|c|c|c|}
\hline
50&4&\textit{c}\\
\hline
1&\textit{e}&100\\
\hline
\textit{g}&25&2\\
\hline
\end{tabular}$
Taking the columns, we have $ 50g\equal{}100e\equal{}200c \iff g\equal{}2e\equal{}4c$. Also, by the diagonals, $ 100e\equal{}ceg \iff 100\equal{}cg \iff 4g^2\equal{}100 \Rightarrow \boxed{g\equal{}5}$. (We also have $ e\equal{}10, c\equal{}20$.)

[b]Case 2: $ b,d\equal{}(1,4)$.[/b]
By symmetric reasoning, this will be the same as case 1 excepted reflected over a diagonal line, so we will have $ \boxed{g\equal{}20}$.

[b]Case 3: $ b\equal{}d\equal{}2$.[/b]
We have:
$ \begin{tabular}{|c|c|c|}
\hline
50&2&\textit{c}\\
\hline
2&\textit{e}&50\\
\hline
\textit{g}&50&2\\
\hline
\end{tabular}$
Similarly to above, we have $ 100c\equal{}100e\equal{}100g \iff c\equal{}e\equal{}g, ceg\equal{}100e \iff cg\equal{}100$. Clearly, $ \boxed{g\equal{}10}$.

Adding up all the cases, we have $ 5\plus{}10\plus{}20\equal{}\boxed{\textbf{(C)}\, 35}$.[/hide]

Sorry to revive, but there is a much simpler way (involving almost no number theory) than the solutions presented above. By equating the products of each row, column, and diagonal, we can systematically reduce the number of unknowns to one variable. For example, we can get everything in terms of $b$:

-----------------$50$ ---    $b$  ---     $\frac{20}{b}$

-----------------$\frac{4}{b}$ --- $10$ --- $25b$

-----------------$5b$ ---  $\frac{100}{b}$  ---  $2$

Since all the entries are integers, $b \in \{1,2,4\} \to g \in \{5,10,20\}$. So the answer is $5+10+20=35.$

[quote=viperstrike]Sorry to revive, but there is a much simpler way (involving almost no number theory) than the solutions presented above.[/quote]
\[ \begin{tabular}{|c|c|c|}
\hline
50&\textit{b}&\textit{c}\\ \hline
\textit{d}&\textit{e}&\textit{f}\\ \hline
\textit{g}&\textit{h}&2\\ \hline
\end{tabular}
\]
Let $p$ be the common product of each row, column, and diagonal. Note that the product of all the numbers in the grid is $p^3$.

Suppose we take the numbers on one diagonal, the numbers on the other diagonal, the numbers on the central row, and the numbers on the central column, and multiply all of them together (repeating the center square four times). We then get the value $p^4$. However, since this product counts each number once and the center square three extra times, we have $p^4 = p^3 \cdot e^3 \implies  p = e^3$.

But we know from the given numbers that $p = 50 \cdot e \cdot 2 = 100 e$, so $100e = e^3 \implies e = 10$. Thus the common product must be $50 \cdot 10 \cdot 2 = 1000$. Now we can easily express all the other entries in the grid in terms of $g$, as above.

In the first solution listed on the AOPS website, how do they get e=d+e+f?
Link: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_22


Prove that the set $ A\equal{}{ n,n\plus{}1,n\plus{}2,n\plus{}3,...,2n}$ can't be devides into two subsets which haven't same element and have same product of their elements.
Sorry for my pour English.

By Bertrand's Theorem, there exists a prime number $ p$ between $ n$ and $ 2n$, the result follows.


Hello,

I want have a proof of the generalization of inequality of Erdos Debrunner in space :

If we have a n-gone in space, and an other m-gone inscribed in this n-gone with 3=<m=<n, then

A(m-gone)>= at least the area of one of the other gone corner.

thanks[/hide]

Now, how do you define the area of a non-planar polygon? :? Also, what do you mean by "inscribed"?.


Geia. Psaxno na bro mia leptomeris apodeixi gia tin anisotita tis anadiataxis. Opois exei exei kati as to kanei ena link.  :)  :?:

ρε παιδια εχει καμια αποδηξη κανεις εκτος απο επαγωγη?

Δε σου αρέσει η επαγωγή δηλαδή; :lol: Έχω μια απόδειξη με μετασχηματισμό αλλά ουσιαστικά είναι σαν την επαγωγή, οπότε δεν ξέρω αν εννοείς αυτό.

Επειδή ζητάς [b]λεπτομερή απόδειξη [/b], δε σε παραπέμπω στο
   [color=red]<< Στεργίου - Σκομπρής , Ανισότητες >>[/color] , αλλά προτείνω να μου στείλεις με ένα PM τη διεύθυνσή σου , για να σου στείλω φωτοτυπία όποιες αποδείξεις βρω στην Αγγλική(έχω του Andy LIu και του Cθcera) ,  αλλά και άλλων.
  Δυστυχώς δεν έχω χρόνο να σου την πληκτρολογήσω στο LATEX.

[u]   Μπαμπης[/u]

καλα και η επαγωγη ειναι δεκτη :P .ειχα δει μια αποδηξη με την δημιουργια ενος αθροισματος. επειδη μου θυμισε και μια αλλη ασκηση δοκιμασα γραφηματα. μηπως μπορει κανεις να με βοηθησει? :oops:

Λοιπόν θα βάλλω μια απόδειξη αυτή που είναι σε εμένα τουλάχιστον η πιο γνωστή .
[color=darkblue]ΠΡΟΤΑΣΗ[/color]
To ;auroisma  $S=a_{1}b_{1}+...+a_{n}b_{n}$  eίναι μέγιαστο αν οι ακολουθίες $a_{1},...,a_{n}$ , $b_{1},.....,b_{n}$ είναι ομόνυμα διατεταγμένες .Το $S$ είναι ελάχιστο όταν οι δύο ακολουθίες είναι αντίθετα διατεταγμένες δηλαδή η μία κατά αύξουσα σειρά και η άλλη κατά φθίνουσα σειρά.

[color=red]ΑΠΟΔΕΙΞΗ[/color]
Έστω $a_{r}>a_{s}$ . Θεωρούμε τα αθροίσματα
$S=a_{1}c_{1}+....+a_{r}c_{r}+...a_{s}c_{s}+....+a_{n}c_{n}$
$S'=a_{1}c_{1}+....+a_{r}c_{s}+....+a_{s}c_{r}+.....+a_{n}c_{n}$
Το $S'$ προκύπτει από το $S$ αν αλλάξουμε την θέση των $c_{s}$ και $c_{r}$ .
Άρα $S'-S=(a_{r}-a_{s})(c_{s}-c_{r})<0$

[color=brown]Άσκηση[/color]
Αποδείξτε την ανισότητα ΑΜ-ΓΜ με αναδιάταξη

Ναι ουσιαστικά αυτό που κάνεις silouan είναι η επαγωγή κατά βάθος.

Όσο για την άσκηση:

Θεωρούμε κατ' αρχήν ότι όλα τα $a_{i}$ είναι θετικά, και κανένα μηδέν (αν είναι κάποιο μηδέν τότε είναι προφανής)

Έστω $G = \sqrt[n]{a_{1}a_{2}\cdots a_{n}}$. Θέλουμε να δείξουμε ότι $a_{1}+a_{2}+\ldots+a_{n}\geq n G$

Έστω $x_{i}= \frac{G^{i}}{\prod_{j\leq i}a_{j}}$ (παρ. ότι $x_{n}= 1$) και $y_{i}= \frac{\prod_{j\leq i}a_{j}}{G^{i-1}}$ και βλέπουμε ότι $y_{i}= \frac{G}{x_{i}}$ και άρα οι $(x), (y)$ είναι αντίθετα διατεταγμένες, άρα:

$x_{1}y_{1}+x_{2}y_{2}+\ldots+x_{n}y_{n}\leq x_{1}y_{2}+x_{2}y_{3}+\ldots+x_{n}y_{1}$
$G+G+\ldots+G \leq a_{2}+a_{3}+\ldots+a_{1}$
$nG \leq a_{1}+a_{2}+\ldots+a_{n}$

Ωραία cmad . :)  Ναι πράγματι είναι κάτι σαν επαγωγή απλά εβαλα τη λύησ για να υπάρχει.....


A hospital develops a test for a certain disease. If an individual has the disease, the test will come out positive $ 98\%$ of the time. If the individual does not have the disease, the test will only come out positive $ 10\%$ of the time. In the general population, $ .1\%$ of people have the disease. Joey is tested and the test comes out positive. What is the probability that Joey has the disease? Round your answer to the nearest tenth of a percent.

This is a great problem!  It's actually one that Howard Raiffa and I used to try to teach public policy students, including some cross-registered future doctors from the medical school some years back.

Here's the way I liked to explain the solution:  [hide]Consider a total general population of 1 million people who are all tested for the disease.

Among that 1 million, 1,000 people will actually have the disease and 999,000 will NOT have the disease.

Of the 1,000 who actually have the disease, 980 of them will test positive.

Of the 999,000 who do NOT have the disease, 99,900 will also test positive.

So the total number of people who will test positive is (980 + 99,900), of whom 980 actually have the disease.

So the odds of a true positive are only 980/99,900 or approximately 1.0%.[/hide]

@sophia
I like your solution better than the given one :D

copied from Solution page:
Problem:
A hospital develops a test for a certain disease. If an individual has the disease, the test will come out positive $98%
$ of the time. If the individual does not have the disease, the test will only come out positive $10%
$ of the time. In the general population, $.1%
$ of people have the disease. Joey is tested and the test comes out positive. What is the probability that Joey has the disease? Round your answer to the nearest tenth of a percent.

Solution:
Before the test, Joey had a probability of $.1%
$ of having the disease. Given this, the probability he has the disease and tests positive is $.1%\cdot98%=.00098
$. On the other hand, the probability that he does not have the disease yet tests positive is $99.9%\cdot10%=.0999
$. Since we know he tests positive, the probability that the first of these scenarios occurs is $\frac{.00098}{.00098+.0999}\approx.0097\approx\fbox{1.0\%}.
$.


is the normal of the plane 2x+4y-3z = 1 equal to 2i + 4j - 3k? can someone please explain the reason behind this?

Write that equation with a dot product. What does the difference between two points on the plane look like?

not sure I understand what you mean, can you please explain what you mean by write that equation with a dot product?

You can check it by picking two point lying in the plane, for example (x,y,z)=(0,0,-1/3)=P1 and  (x,y,z)=(1/2,0,0)=P2, and form the vector from P1 to P2: v=(1/2)i+(1/3)k
Now the dot product with the supposedly normal is 2(1/2)-3(1/3)=0. Indeed, the normal is by definition normal to any vector in the plane, which means the dot product must vanish.

Maybe you'll find [url=http://www.netcomuk.co.uk/~jenolive/vect13.html]this[/url] useful. It basically comes down to taking two non-parallel vectors in the given plane and form their cross product: the cross product is perpendicular to the two vectors, and hence is a normal.


Το $ABCD$ είναι ένα κυρτό τετράπλευρο στο οποίο το τρίγωνο $BCD$ είναι ισόπλευρο και $\angle DAB = 30^{\circ}$ .

Να δείξετε ότι : $(AC)^{2}= (AD)^{2}+(AB)^{2}$ .   :)

Έχει εύκολη λύση με διανύσματα:

[hide]
Έστω $O$ το συμμετρικό του $C$ ως προς τη $BD$, τότε το $A$ κινείται στον κύκλο $(O, R)$ με $R = BC = CD = DB$ (φυσικά όχι σε ολόκληρο τον κύκλο, αλλά στο μέρος που βρίσκεται στο εσωτερικό των προεκτάσεων $CB$ και $CD$).

$| \vec{AC}|^{2}= | \vec{OC}-\vec{OA}|^{2}= | \vec{OC}|^{2}+| \vec{OA}|^{2}-2 \vec{OA}\cdot \vec{OC}= 2R^{2}(2-\sqrt{3}\cos (\angle AOC))$

$| \vec{AD}|^{2}= | \vec{OD}-\vec{OA}|^{2}= | \vec{OD}|^{2}+| \vec{OA}|^{2}-2 \vec{OA}\cdot \vec{OD}= 2R^{2}(1-\cos (\angle AOC+\frac{\pi}{6}))$

$| \vec{AB}|^{2}= | \vec{OB}-\vec{OA}|^{2}= | \vec{OB}|^{2}+| \vec{OA}|^{2}-2 \vec{OA}\cdot \vec{OB}= 2R^{2}(1-\cos (\angle AOC-\frac{\pi}{6}))$

Και βλέπουμε ότι $| \vec{AD}|^{2}+| \vec{AB}|^{2}= | \vec{AC}|^{2}$

[/hide]

Με δεδομένο ότι το BCD είναι ισόπλευρο, αν θεωρήσουμε το συμμετρικό C', του C ως προς τη BD, τότε η κορυφή Α βρίσκεται σε κύκλο κέντρου C' και ακτίνας C'B = C'D ( για να είναι <DAB = 30º ).

     Από το τρίγωνο ABD => (AB)² + (AD)² = 2(AM)² + (BD)²/2, (1), όπου Μ, είναι το μέσον του ΒD.

     Από το τρίγωνο ACC'  =>  (AC)² + (AC')² = 2(AM)² + (CC')²/2,  (2).

     Άρα για να ισχύει  (AC)² = (AB)² + (AD)², (3),  από (1), (2), προκύπτει ότι αρκεί να

αποδειχθεί ότι ισχύει  (BD)²/2 = (CC')²/2 - (AC')², => (BD)² = (CC')² - 2(AC')², (4).

     Επειδή όμως BD = AC', η (4) γίνεται 3(BD)² = (CC')² = 4(CM)², η οποία αληθεύει.

     Άρα η (3) αληθεύει και η πρόταση έχει αποδειχθεί.

     Κώστας Βήττας

ΥΓ.   Κολλημένος σε ένα παιδαριώδες λάθος πράξεων, συμπέραινα ότι ήταν λάθος το πρόβλημα ( το οποίο αληθεύει και όταν το ABCD είναι μη κυρτό ). Ευχαριστώ θερμά τους Ramanujan και cmad, που μου άνοιξαν τα μάτια.

Στην περίπτωση του μή κυρτού τετραπλεύρου ABCD, το πρόβλημα αληθεύει μόνο όταν η κορυφή Α βρίσκεται πάνω στον κύκλο (c'), κέντρου C' και ακτίνας C'B = C'D.

     Αν θεωρήσουμε ένα σημείο A' στον κύκλο (c), κέντρου C και ακτίνας CB =CD, έχουμε πάλι <DA'B = 30º, αλλά η ισότητα
(A'C)² = (A'B)² + (A'D)², δεν ισχύει.

     Κώστας Βήττας.


[quote="Davron"]Let $a,b,c$ be postitive real numbers such that $a+b+c=1$. Show that inequality holds

${\frac{3}{2} \geq \sqrt{\frac{ab}{c+ab}}+ \sqrt{ \frac{bc}{a+bc}} + \sqrt \frac{ca}{b+ca}}$

Davron[/quote]
Yes, this is very easy. Jensen forever! (The Myth's trace.) :)

[quote="tchebytchev"] $\frac{1}{2}(\frac{a}{(b+c)} + \frac{b}{(a+c)}) \geq \sqrt{\frac{ab}{(b+c)(c+a)}}$[/quote]
Very nice!
I meant to: ${\sqrt{\frac{ab}{c+ab}}+ \sqrt{ \frac{bc}{a+bc}} + \sqrt \frac{ca}{b+ca}}=\sum_{cyc}\sqrt{\frac{ab}{(c+a)(c+b)}}=$
$=\sum_{cyc}\left(\frac{a+b}{2}\cdot\sqrt{\frac{4ab}{(a+b)^2(a+c)(b+c)}}\right)\leq\sqrt{\sum_{cyc}\left(\frac{a+b}{2}\cdot\frac{4ab}{(a+b)^2(a+c)(b+c)}\right)}=$
$=\sqrt{\frac{2(ab+ac+bc)}{(a+b)(a+c)(b+c)}}.$
But $\sqrt{\frac{2(ab+ac+bc)}{(a+b)(a+c)(b+c)}}\leq\frac{3}{2}\Leftrightarrow8(ab+ac+bc)(a+b+c)\leq9(a+b)(a+c)(b+c).$  :D


Can Somebody please explain mass points of triangle or direct me to a site that does?

Do you mean the centroid? If so, then: http://mathworld.wolfram.com/TriangleCentroid.html

Interesting, I've never heard that term before ;)

[quote="tetrahedr0n"]If you're wondering about the geometric problem-solving technique known as "mass point geometry," just search the forums, as there are tons of threads on it.[/quote]

Yes, in fact, [url=http://mathcircle.berkeley.edu/mpgeo/mpgeo.html]this[/url] article is very helpful.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39142]This[/url] is also excellent.  There are more threads concerning mass points on the forums...do a search to find them.

The main idea of mass point geometry is that you assign a certain weight to a vertex or a point according to the ratio of lengths.

For example, let's say $C$ is a point on the line segment $AB$ such that $AC: CB=2: 3$. Then we can assign a weight of $3$ to $A$ and a weight of $2$ to $B$. Also, for the weight of $C$, we simply add the weights of $A$ and $B$ to get $5$.

This idea is very useful for many cases that have to do with ratios. For instance, we can prove that the medians of a triangle divide each other in the ratio 2:1.
[hide="Proof"]
In $\triangle ABC$, let $D$ be the midpoint of $AB$, let $E$ be the midpoint of $BC$, and let $F$ be the midpoint of $AC$. Let $G$ be the centroid.

We assign weight $1$ to each of the vertices of the triangle.
(This is because $AD: DB=1: 1$, $BE: EC=1: 1$, and $CF: FA=1: 1$.)

Then, the midpoints $D$, $E$, and $F$ have weights $2$.

Now compare the mass of $C$ and $D$. $C$ has weight $1$ while $D$ has weight $2$. This implies that $CG: GD=2: 1$. In the same way, we can prove $BG: GF=2: 1$ and $AG: GE=2: 1$.
[/hide]

For more information, as tetrahedr0n said, search "mass points".
Also, see [url=http://www.math.uncc.edu/~hbreiter/problems/Orlando/MassPointsWeb.htm]this web site[/url] for some basic practice problems :)

djshowdown2 (I think that's his username, don't remember exactly :blush:) explain mass point geometry to me last year. You can PM him. :)

[quote="Altheman"]mass points are just a simplification of vectors[/quote]

or, equivalently, area ratios.


Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $ 4 : 3$. The horizontal length of a “$ 27$-inch” television screen is closest, in inches, to which of the following?
[asy]import math;
unitsize(7mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3));
fill((0,0)--(4,0)--(4,3)--cycle,mediumgray);
label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW);
label(rotate(90)*"Height",(4,1.5),E);
label("Length",(2,0),S);[/asy]$ \textbf{(A)}\ 20 \qquad
\textbf{(B)}\ 20.5 \qquad
\textbf{(C)}\ 21 \qquad
\textbf{(D)}\ 21.5 \qquad
\textbf{(E)}\ 22$

[hide="My Solution"]
The height, length, and diagonal are in the ratio $ 3: 4: 4$.  The length of the diagonal is $ 27$< so the horizontal length is $ \frac{4}{5}(27)\equal{}21.6$ inches, which corresponds to answer choice $ \textbf{D}$.
[/hide]

[quote="Corrina Chang"][hide="My Solution"]
The height, length, and diagonal are in the ratio $ 3: 4: 4$.  The length of the diagonal is $ 27$< so the horizontal length is $ \frac {4}{5}(27) \equal{} 21.6$ inches, which corresponds to answer choice $ \textbf{D}$.
[/hide][/quote]

Please delete message...I can't read.

[quote="crazypianist1116"][quote="worthawholebean"]The horizontal length of a “$ 27$-inch” television screen [b]is closest, in inches, to which of the following?[/b][/quote]
The message contains only quote and code tags. Please add some text of your own.[/quote]

Note to self:  Go back to kindergarten.

Don't worry, we've all made the same mistakes (grrrr 77 on psat math).

[hide="my solution"]

okay, so we know that the length = $ \frac43$ the height.

so, if we let the length=$ a$, the height = $ \frac34a$

$ a^2\plus{}(\frac34a)^2\equal{}27^2$

$ a^2\plus{}\frac{9}{16}a^2\equal{}27^2$

$ \frac{25}{16}a^2\equal{}729$

$ a^2\equal{}466.56$

$ a\equal{}21.6$

21.6 is closest to 21.5(yes 007math, [b]closest[/b].)
thus, the answer is Choice $ (\text D)$
[/hide]

corrina chang, your solution is good, but why did you say that
[quote]The height, length, and diagonal are in the ratio 3:4:4.[/quote]

[quote="nikeballa96"]
corrina chang, your solution is good, but why did you say that
[quote]The height, length, and diagonal are in the ratio 3:4:4.[/quote][/quote]

He definitely meant to say 3:4:5, which yields a much shorter solution (you know $ 27\equal{}5x$, so $ 4x\equal{}\dots$)


The radius of the first circle is $ 1$ inch, that of the second $ \frac{1}{2}$ inch, that of the third $ \frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:
$ \textbf{(A)}\ \frac{3\pi}{4} \qquad\textbf{(B)}\ 1.3\pi \qquad\textbf{(C)}\ 2\pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \text{none of these}$

[hide]
The areas are pi, pi/4, pi/16, pi/64, so on. Call the sum of the areas divided by pi x.
$ x \equal{} \displaystyle\sum^{\infty}_{a\equal{}0}{\frac{1}{4^{a}}} \equal{} 1 \plus{} \frac{1}{4} \plus{} \frac{1}{16} \plus{} ...$
Multiply by 4 to get
$ 4x \equal{} 4 \plus{} x$
Solve to get
$ x \equal{} \frac{4}{3}$

$ \boxed{\text{D}}$ [/hide]


by plugging $1$ you can get:

$f(1)+f(n)=f(1+2n)+f(1-2n)=f(1)+f(-n)$ hence $f(n)=f(-n)$ (it seems that you don't need $f(1)=f(-1)$.

also

$f(1)+f(n)=f(3n)+f(-n)=f(3n)+f(n)$ hence $f(1)=f(3n)$

now as if $n \not \equiv 0 \mod 3$ then one of $1+2n$, $2n-1$ $\equiv 0 \mod 3$ you can proceed easily to get $f(n)=f(1)$.

pluging m=0 directly yields
f(0)+f(n)=2f(0),
f(n)=f(0)=c is thus a solutin for any c from Z.

[quote="pooyan_seva"]find all functions $f: Z \longrightarrow Z$ such that $f(1)=f(-1)$ and for all $m,n\in Z$
               $f(m)+f(n)=f(m+2mn)+f(m-2mn)$[/quote]

dear pooyan-seva , you can find it in the mathemathic magazin No.20  ;)

thanks for your guide Ehsan. I saw the problem in the magazine , but it didn't have the solution!! :lol:

thank you for your nice solutions  ..
but the point is.... :blush:

the question I've posted wasn't correct....
i'm sorry about that .  the correct form is here:
find all functions$f: Z\longrightarrow Z$ such that :
$f(1)=f(-1)$ and for all$m,n\in Z, f(m)+f(n)=f(m+2mn)+f(n-2mn)$

sorry!! :lol:  ;)

as you see the $m$ is changed to $n$ .
by the way it's not that easy.

Well, we don't have a nice function... There are infinitely many ugly ones  :!:
(*)  $f(x)+f(y)=f(x+2xy)+f(y-2xy), \ \forall x,y\in \mathbb{Z}$
Denote $a=f(1)=f(-1)$.
For $y=1$ we get
(1)  $f(x)+a=f(3x)+f(1-2x)$
For $x=1,\ y=x$
(2)  $a+f(x)=f(1+2x)+f(-x)$
For $y=-1$
(3)  $f(x)+a=f(-x)+f(2x-1)$.
From (2) and (3) we get $f(2x+1)=f(2x-1),\ \forall x\in \mathbb {Z}$, hence
(4)  $f(t)=a$, \ for all odd $t\in\mathbb{Z}$.
It follows from (2) and (4) that
(6)  $f(x)=f(-x), \ \forall x\in {Z}$.

Now return to (*) and substitute $y=2k+1, \ k\in\mathbb{Z}$. By using (4), we obtain:
      $f(x)=f(x(4k+3)),\ \forall k\in \mathbb{Z}$.
For $x=2k+1$,  (*) and (6) yields to
      $f(y)=f(y(-4k-1))=f(y(4k+1))$
Since  every odd number $m$ is either $4k+3$, either $4k+1$, it results that
      $f(x)=f(xm)$, for all odd $m$ and $x$ in $\mathbb{Z}$.
So if $x=2^km$, where $k\in\mathbb{N}$ and $m$ is an odd integer, we'll have $f(x)=f(2^k)$.

Now, conversely, if we chose arbitrary integer values for $f(2^n), \ n\in \mathbb{N}$ and define $f(2^nm)=f(2^n)$ for every odd integer $m$ and $n$ in $\mathbb{N}$, it is easy to see that (*) is satisfied by all $x,y \in \mathbb{Z}$.


LaTeX
limit
analytic geometry
calculus
calculus computations

Define f (x, y) =(xy)/(x^2 +(y ln(x^2))^2) if x is not equal to 0, and f (0, y) = 0 if y is not equal to 0. Determine
whether lim(x,y)→(0,0) f (x, y) exists, and what its value is if the limit
does exist.

Everything is much more legible if you use [[LaTeX]] (see also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]).

$ f(x, y) \equal{} \frac {xy}{x^2 \plus{} (y \ln(x^2))^2}$ for $ x \neq 0$, $ f(0, y) \equal{} 0$ for $ y \neq 0$.  Determine whether $ \lim_{(x, y) \to (0, 0)} f(x, y)$ exists, and find its value if it exists.

(You can use the "quote" button to see the code for what I've written.)

It's clear that if the limit exists, it must be zero.  So, the work is really contained in the first part of the question.  One possible approach (not necessarily the best, but the one I thought of) is to switch to polar coordinates -- most of the $ r$s go away, and then you have to analyze the last term with an $ r$ in it.


I am not sure how successful it will be even if it is added because I see that the economics forum is not doing too well, but I was wondering if we can add a finance forum in the Science and Arts category and maybe a Business Administration Forum, or something of the sort. Not sure if it is proper to ask here, but whatever.

They would just clutter up the boards even more.  If no one uses the Economics forums, don't count on people using these.

I believe each of those topics can be discussed in the Economics forum.

But nonetheless, this is the wrong place. Moving topic...

Oops sorry. (tolken, when you read this...could you move it again?...my mouse slipped on the forum selection)

And now I'll move to the economics forum, which is where I would expect to find finance topics.


function USAMTS AMC AIME USA(J)MO USAMO geometry

I'm trying to write a program where you enter a number, y, and the program solves an equation for x in terms of y.  But if you cant, or it is too difficult to solve for x in terms of y, how can you go about this?  Is there a way to do this?  Cause on the ti-89 you can solve things like this, you can enter a y value and it'll tell you an x value.

Also, is there a program that will allow me to work with numbers larger than 2^32-1 that i can get for free?  Something like Maple?  Or how can I get beyond the 2^32-1 threshold?  Cause I wrote a program that works with large numbers, but it fails after it reaches that value, and I want to see if I can get it to run past that.

Thanks.

Well, the basic idea is that, since unsigned long can only be used up to a certain number, you have to represent large numbers as arrays. You would probably have to make a class and have member functions for things such as addition, subtraction, etc... Look around on the net for some sample classes if you need to.

Fierytycoon

I would advise not posting too much mathematical content as replies to this question since it is essentially the same as USAMTS #1, and it would be good if this site doesn't answer that question.

Oh, sorry, I didn't know that. Haven't checked out the problems yet.

Fierytycoon

I have to agree with ComplexZeta.  If you are writing this program to aid in solving USAMTS #1, that's fine, but keep it to yourself.  And if not...well, still keep it to yourself.  Sorry, but we don't solve problems for competitions that are currently going on.

yeah, me neither.  Sorry, I just looked at them, and it would answer the 1st question.  I'll edit the first message, but can someone answer this in general, how to get around solving for x if you really cant?  Or is there no way to do this.

And tycoon: what are classes?  I just started programming and i dont know too much.  Your explanation is probablly beyond what I am capable of doing right now.  But do you know any free programs similar to Maple that I can download?  Thanks.

[quote]what are classes? [/quote]

They are sorta like structs (not sure whether you've gotten to those yet), except with a few technical differences. They are pretty much used to represent objects. For example, you might have a class representing a math geek, and, within that class, have variables such as his/her AMC and AIME scores, and functions doing whatever. ok, that was a bad explanation, so just look on google and find some online tutorial that will explain it better.

[quote]how to get around solving for x[/quote]

you can, but seeing as how you said you are just beginning to learn programming, it would probably be too difficult a task for you at this point. I wouldn't attempt the programming task just yet.

Fierytycoon

[quote]actually i know what structs are. But i dont understand how that would help me with my problem. As far as I understand, there is no way using a C++ compiler to go beyond 2^32-1, right? So basically what im asking is, do you know any programs that will allow me to do this, while still using my code? The ti-89 can go much further than 2*32-1. But how?[/quote]

As I said in my first post, you have to code your own class representing large numbers. You would have an array of digits as a variable, and this array would represent the actual number. You would then have to overload the +, -, *, /, etc... operators for your class so that they will work with large numbers. So, for example, you would have to write a + operator that adds two arrays of digits.

Fierytycoon

oh ok, i think i understand now.   thanks.

The TI-89 solves equations by checking each number within a specific tolerance, I think. Not quite sure. Anybody know exactly how?

-Isaac

[quote]my math teachers said before that it keeps making guesses until it gets closer and closer to the answer, but im not sure. I think thats how i should proceed with my program if i cant solve for x. [/quote]

Hmm, coding this could be hard as well. First of all, how would you know how many roots there are in the equation? Also, how would you know which way to move to get "closer and closer to the answer"? With more complex functions with hills and loops and such, it would not always be closer to y=0.

Fierytycoon

yeah you're right.  Right now i have no clue how i would do it, but i think that will somehow be my approach.  First i'd have to keep it simple then maybe if i get that to work i'd try and make it more complex.  Right now I really dont have a specific problem i want to do, so i have some room to do whatever i want.

By the way, where did you learn to program?  Did you learn it in school?  I just took a course over the summer (it recently finished) and i want to make sure that i  dont forget the stuff that i learned.  Aside from making problems for myself to solve, do you know of any competitions or anything like that to help force myself to maintain what i have learned?
thanks.

If your function is continuous you could use the bisection method. For non-continuous functions its pretty impossible, as you could never know even if you were real real close to 0 at one point it may jump up to 1000 or something, you have no idea.

You probably know, but the bisection method is: (you have to rearrange so you get f(x)=0 first) choose any two points so one gives a positive value of function, one gives a negative (ie they are on opposite sides of the root) and check the average to see which side this is on, then repeat..

[quote]First i'd have to keep it simple then maybe if i get that to work i'd try and make it more complex. [/quote]

Definitely. Maybe just create a program that solves linear equations first. That wouldn't be so hard, but you'd be able to sort out the basic concept of having a variable and grouping expressions. For a start, perhaps have a binary tree that you can use to group each partitioned expression in order of importance. Hmm, that was a bit confusing too, but you should be able to find some stuff on google.

[quote]By the way, where did you learn to program? Did you learn it in school?[/quote]

Yeah, I did take the AP Computer Science course in school, but before that, I had already learned to program just because it was fun.

[quote]Aside from making problems for myself to solve, do you know of any competitions or anything like that to help force myself to maintain what i have learned?[/quote]

The main contest for high-schoolers would be USACO tests, which you need to take to qualify for the US team and participate in the IOI (international olympiad for informatics). Owever, I have never taken them since my school doesn't have a lot of kids interested in this sorta stuff :( There is a very nice practice site (made by the USAMO ppl) at http://ace.delos.com/usacogate . You can learn a lot there about algorithms.

Fierytycoon

You don't need people at your school taking the USACO to take it. I was the only one at my school to take it in 2002-2003, and you just take it on your own at your own home whenever you want within a window of 3 or 4 days. Some of the later rounds require you to have proctors. But remember that the tests are extremely challenging. I always did very poorly on the contests, but I never practiced. If you practice a lot you will probably do well, but it's like everything else. You won't be good if you don't work hard.

[quote]Some of the later rounds require you to have proctors. [/quote]

Yeah, the US Open does, and its the most important and required round. So did you just find a proctor at school?

Fierytycoon

No, I didn't do that round. I only did the first two mainly because it just happened that the rounds kept being the same weekends as math contests and when I got back from the math contests, I was too tired to work for a few hours on challenging programming problems.

Well, you have to do that round to qualify for the USACO summer camp (top 15).

Fierytycoon

Has anyone learnt any graphics for C/C++? I know that they're not included in the actual standard C thingies but you can get extra libraries and stuff for that. But it all seems way too confusing, it seems like nobody has had the idea to design a graphics library that lets you draw a rectangle in less than 100 lines or something ridiculous.

Well, there's a CMU graphic library from the Carnegie Mellon site, and of course the innate Win32 Graphics itself. And, for bigger projects, OpenGL and DirectX.

Fierytycoon

Well, go through the problems on the training site that I mentioned above, and perhaps get a nice algorithm book.

Fierytycoon

[quote="Idiot_without_a_village"]On the subject of USACO, what books did you guys use to prep for it??? I am trying to learn C write now but is that all I need for the USACO????[/quote]

Well, my programming ability is lousy at best, but I suppose I can answer some of this.  For the USACO, you only need to know one programming language (C, C++, or Pascal).  They might expand it to include another language or two sometime soon, but probably not Java.

So basically to prep, you'll want to learn the language well and then work on the training pages (http://ace.delos.com/usacogate).  I took a really cool data structures/algorithms class this past year, and we used a great book called Algorithms in C++.  I don't remember the author... it's probably easy to find on amazon.

This is also a good site to practice at:

http://acm.uva.es/problemset/

Fierytycoon

My preferred big integer libraries are GMP if you want speed, Miracl if you want class support, NTL if you want nice number theory built-ins, and FreeLIP if it's a quick and dirty job. Tom St. Denis (whose personality leaves much to be desired, alas) has an excellent portable bigint library that has a very similar calling interface to FreeLIP--I have forgetten its name.

I don't know if you were taught this in a calculus class, but this is the method used by TI graphing calculators to find roots and intersections.
X(n+1)=X(n)-Y(X(n))/Y'(X(n))
It has you make the first guess and set the range for the answer, but this is a feasible method for beginning/intermediate programmers trying to solve for values not easily obtained analytically.  Repeated many times, it can be extremely accurate.

That is Newton's method. It's also sometimes a good thing to know for competitions if you're running out of divine inspiration and think that some numerical insight will help you, since you can probably work out a few terms by hand to a significant degree of accuracy. Who knows? If you see 1.414, it's probably sqrt(2). You'd be amazed how many problems you can solve like that without knowing the ``correct'' way of solving it.


Be
$x^4-14x^2+24x-k$

Find K for P(x) has four differents roots

[quote="shyong"][quote="kunny"]The answer is $-16<k<11$[/quote]

Can you show how ?[/quote]

All right,shyong.

Let $f(x)=x^4-14x^2+24x$.
$x^4-14x^2+24x=k$ has four different roots $\Longleftrightarrow y=f(x)\ \text{has four interseption points}$ with $y=k$.
From $f'(x)=4x^3-28x^2+24=4(x-1)(x-2)(x+3)$, we can see that $f(x)$ has two minimal values $f(-3)=-21,f(2)=-16$ and
a maximal value $f(1)=11$. Thus by sketching thr graph of $y=f(x)$, the desired condition is $f(2)<k<f(1)$,yielding $-16<k<11$.

kunny

[quote="kunny"][quote="shyong"][quote="kunny"]The answer is $-16<k<11$[/quote]

Can you show how ?[/quote]

All right,shyong.

Let $f(x)=x^4-14x^2+24x$.
$x^4-14x^2+24x=k$ has four different roots $\Longleftrightarrow y=f(x)\ \text{has interseption points}$ with $y=k$.
From $f'(x)=4x^3-28x^2+24=4(x-1)(x-2)(x+3)$, we can see that $f(x)$ has two minimal values $f(-3)=-21,f(2)=-16$ and
a maximal value $f(1)=11$. Thus by sketching thr graph of $y=f(x)$, the desired condition is $f(2)<k<f(1)$,yielding $-16<k<11$.

kunny[/quote]

Dear kunny ,
By using your method , shouldnt $f(-3)=-117$ and $f(2)=8$ which gives us the interval for $k$ is $8<k<11$ ?  :blush:

Oops! I made a mistake. :blush: The right answer is $8<k<11$, shyong.

kunny


Take any prime p.  Prove that, when written in base p, there exists a Fibonacci number ending in 4 consecutive zeroes.

Set a finite upper bound (as a function of p) on how big n will have to be to be certain that such a number appears before F(n).

(The Fibonacci numbers are defined here by F(1) = F(2) = 1,
F(n + 2) = F(n) + F(n + 1) for all n :ge: 1.)

I thought you could use a periodicity argument to show that a Fibonacci number could be divisible by anything, let alone something of the form p^4.

Perhaps you can, perhaps not.  The only way to find out is to try to do it.

Yeah.. I think I posted something similar to this ages ago. Basically you just [hide]consider the Fibonacci sequence mod p^4 (or in fact anything you want). Then it must repeat somewhere (finite number of pairs, infinite length of sequence) and then you just trace the sequence back to the pair 1,1, where the previous number must be 0.[/hide]


limit
trigonometry
logarithms
algebra proposed
algebra

Find
$ \lim_{x \to 1^\minus{}} \arccos x \cdot \cos \left(\frac{1}{\ln x}\right)$

[quote="aktyw19"]Find
$ \lim_{x \to 1^ \minus{} } \arccos x \cdot \cos \left(\frac {1}{\ln x}\right)$[/quote]

$ \minus{}|\arccos x|\le\arccos x \cdot \cos \left(\frac {1}{\ln x}\right)\le |\arccos x|$

Hence $ \lim_{x \to 1^ \minus{} } \arccos x \cdot \cos \left(\frac {1}{\ln x}\right)\equal{}0$

hello, we have $ |\arccos(x)\cdot \cos\left(\frac{1}{\ln(x)}\right)|\le |\arccos(x)|$ and $ \lim_{x \to 1^{\minus{}}}|\arccos(x))|\equal{}0$ so the searched limit is zero.
Sonnhard.


number theory
greatest common divisor
number theory proposed

Let $ n$ and $ k$ be positive integers, $ n\geq k$. Prove that the greatest common divisor of the numbers $ \binom{n}{k},\binom{n\plus{}1}{k},\ldots,\binom{n\plus{}k}{k}$ is $ 1$.

Let prime p devide $ gcd$. Then $ p\le n$. If $ p\le k$, then exist $ 1\le i\le k$, suth that $ p|n\plus{}i$, therefore $ v_p(\binom{n\plus{}i\minus{}1}{k})\equal{}0$ - contradition. If $ k<p\le n$, then exist $ i\equal{}1,..k$, suth that $ p|n\plus{}1\minus{}i$, therefore $ p\not |\binom{n\plus{}k\plus{}1\minus{}i}{k}$ - contradition.

if $d$ is the greatest common divisor of these numbers, then $d$ must divide their consecutive differences, i.e $d|\dbinom{n}{k-1},...,\dbinom{n+k-1}{k-1}$, so it can be done by induction.


divide it into quadrants and just kinda try to count the ones on the edge?

While the quadrant strategy is what you should do, the key is noticing that the radius of the circle is 5 and the points where the circle passes through the points of intersection of the grid lines are where you would be making 3 - 4 - 5 right triangles.  There are two of those in each quadrant and it can show you exactly which squares fit wholly inside the circle.

Hope that makes sense,

frost