Four positive integers a, b, c, and d have a product of 8! and satisfy

ab+a+b = 524
bc+b+c = 146
cd+c+d = 104.

What is a-d?

Cats -- it's worth pointing out that those aren't necessarily the only solutions to the equations, just the only set that satisfies abcd = 8!.  There are a few other cases you have to eliminate.


Να υπολογιστεί το όριο:

${ \lim_{n\to\infty}n^{2}\left(\int_0^1\sqrt[n]{1+x^{n}}dx - 1\right)}$

$ \alpha=\frac{1}{n}$ και $ y=x^{\frac{1}{\alpha}}$ για ευκολία, αναπτύσσεις σε σειρά και, αν δεν κάνω κάπου λάθος:
${ \lim_{n\to\infty}n^{2}\left(\int_0^1\sqrt[n]{1+x^{n}}dx - 1\right)} = \sum_{k=1}^\infty \frac{(-)^{k-1}}{k^2} = \frac{\pi^2}{12}$.

Δεν το περίμενα το αποτέλεσμα που βγήκε, nice :)

[hide="Disclaimer"]
Χρειάζεται βέβαια να κάνεις λίγο τετριμμένο book-keeping για να ανταλλάξεις όρια, σειρές και ολοκληρώματα, αλλά το DCT και το συμπαγές $ [0,1]$ απλοποιούν πολύ τα πράγματα. Ωστόσο, αν από υπερβολική αισιοδοξία έχασα κάτι στη χαμαλοδουλειά και είναι εκεί το ζουμί της άσκησης, πες το μου για να την κοιτάξω με μεγαλύτερη λεπτομέρεια.
[/hide]

Cheerio,

Durandal 1707


probability
percent
blogs
expected value
probability and stats

I don't know what this falls under, so I posted here.

The odds of winning the Florida State Jackpot are almost 1 in a 23,000,000 (23 million) chance of winning.

If I play 5000 different numbers, what are my overall odds?

Each ticket has a 1 in 23 million chance of winning.

I have 5000 tickets, 5000 different numbers.

What are my total odds?

I don't know how to calculate it. More than 2 answers please, incase someone miscalculates.

I'd say it's $ \frac {5000}{23000000}$. This belongs to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=498]Probability and Statistics[/url] forum.

If my odds are 1 in 23 million with 1 ticket, how can my odds be 1 in 2.17 with 5000 tickets?

1 in 2.2 Million?

Would that be correct?

so 10,000 tickets = 1 in 1.2 Million

50,000 numbers = 1 in 217,000 chance of winning 20 to 300 Million

200,000 numbers = about 1 in a 86,000 chance of winning 20 to 360 Million.

If the odds are correct, and the lottery is up to 300 million, and some millionair spends 35Million on 35 Million different numbers, they would be guaranteed to win 300 million dollars and only have spent 35Million....

Playing exactly 23 million numbers still kind of leaves the possibility you won't win, 35 million numbers to 45 million numbers is a guarantee.

Someone explain why no millionairs with 500 and 600 and 700 million dollars don't play 50 million numbers when the lottery gets up to 250-350Million dollars.

It is a guaranteed win, and guaranteed profit in the MILLIONS over night.

Even if 2 other people win, which is not all that uncommon, they still make millions more than they spent.

oh, I see.

They would have to buy about 1.4Million tickets every hour for 24 hours straight in order to play 35 million different numbers.

Impossible unless they have over 100 people buying tickets at all hours for 24 hours.

Actually, it is less than 24 hours time that they have to buy 35 million tickets.

It can still be done, and it would be world history.

I'm going to contact a millionair and place an offer, for free, without asking for anything in return.

If they make a habit of it, it would change lottery laws and become illegal - even though it isn't cheating.

Interesting.

There was a funny story in the news not so long ago: a large number of people all got five out of six numbers (or some such) and won the first prize (not grand prize) in a lottery near me.  Initially, there was worry that it had to do with some sort of insider scam, but then they discovered that all of those people just played whatever lucky numbers their fortune cookie gave them.  Also, the lottery did much worse than it would have if all those players had all six numbers: the first prize is a fixed dollar amount (several hundred thousand, perhaps?) for each winner while the jackpot gets split between multiple winners.  (In fact, because this is the structure of most lotteries, buying a ticket when no one has won in several weeks and the jackpot has grown unusually large [i]can[/i] have positive expected value.  It just doesn't happen very often.)

[quote="blahblahblah"]this is an incredibly silly post.  you are not going to find a lottery which has only 30 million combinations of tickets which pays out 300 million dollars but somehow has a ticket price of less than 10 dollars.

lottery corporations employ people who are capable of doing these very basic calculations so that they don't lose money.

at a bare minimum, if there are x combinations of tickets in a lottery and the ticket price is y, then the total payout is going to be less than xy.  This means that each individual ticket has a negative expected value.  But the expected value of the combination of tickets is the sum of expected values of each of the individual tickets.  A sum of negative numbers is negative, and so you lose money no matter how many tickets you buy.

If you want to try a similar experiment to see why this is a bad strategy but lose less money, go to a casino, play roulette, and put a bet on every single number.[/quote]

The odds are posted on the national and state lottery pages loser.

http://www.flalottery.com/inet/lottoMain.do

"1:22,957,480" <--the actual odds in Florida.

Scroll to the bottom.

Just like a scratch off ticket says the total odds are 1:3.5 (some got different odds)

so 1 in every 3.5 tickets are a winner on average/total.

That is why if you go buy 10 scratch off tickets you will have at least 1 winner, sometimes you'll have 3-5 winners from 10 tickets at 1 each, 2 each, 5 each, 10 each, and 20-25 each.

--^ I tried it over 3 dozen times with over 400 dollars.

So if the lotto is up to 300 million, the odds of 1:22,957,480 don't change.

Lotta don't get that high very often though, I guess.

Yes, the next estimated jackpot is 3,000,000 rather than 300,000,000. When the jackpot grows into hundreds of millions, it receives more media coverage, and the number of people buying lottery tickets increases as well. This negatively affects your expected payout because the jackpot may be divided between multiple winners.

Don't forget to read the fine print:
[quote]The advertised jackpot is an estimate of the amount the Lottery will pay a single jackpot ticket in 30 annual payments... You can choose to receive your portion of the FLORIDA LOTTO jackpot in a one-time Cash Option payment... (approximately half of the advertised jackpot amount.) Appropriate federal income taxes will be withheld from both Cash Option and Annual Payments at the time payments are made.
[/quote]

I don't know about other states, but the proposition that established the California lottery specified that $ 50\%$ of the gross revenue should go to prizes. You can work backwards from that number to get an expected value: assuming the game designers are doing their jobs (which they are), the expected value of a one dollar ticket is about 50 cents.

Rolling jackpots do cause that to fluctuate somewhat, but the effect that mlok describes flattens out that fluctuation somewhat. The long-term average return remains the same.

I don't understund why people play such games, the probability of wining is very low,

In Islam, such games are forbidden  :P

Many Christians also object to gambling, which leads to the weird patchwork of laws here in the USA.

Most states ban gambling (Nevada is the big exception) except for state-run lotteries and a few things like horse racing. On the other hand, there's a loophole which allows for full-scale casinos anyway- many Native American tribes have their own land, on which they're not subject to state laws. In recent years, this autonomy has been exploited, and we have tribe-run casinos scattered everywhere.

Incidentally, casino games tend to return much more than lotteries- most of the standard casino games return about $ 90\%$ to the players.

[quote="jmerry"]
Incidentally, casino games tend to return much more than lotteries- most of the standard casino games return about $ 90\%$ to the players.[/quote]

most of them return more, actually, but it doesn't really make a difference what the rate of return is if you play repeatedly.

[quote="blahblahblah"]this is an incredibly silly post.  you are not going to find a lottery which has only 30 million combinations of tickets which pays out 300 million dollars but somehow has a ticket price of less than 10 dollars.[/quote]The Florida lottery jackpot has been a multiple of $ \$$23M a number of times, with a ticket price equal to $ \$$1. The history is [url=http://www.flalottery.com/inet/downloads/Top30Jackpots.pdf]here[/url]. The number of winners depends on the number of tickets sold.

It appears from the data that if the jackpot is several times the number of combinations in dollars, the expected value of a 1$ \$$ ticket is larger than 1$ \$$. This may be different for record jackpots ($ \$$80M and above).

Having read the official rules of the Florida lottery, I found no restriction on the number of tickets that can be bought by one person. Apparently, one can buy $ \binom{53}{6}\approx 23\cdot 10^6$ tickets when the jackpot is large enough for this to be profitable. However, the investment is not risk-free because of the possibility of multiple winners. And there may be logistic obstacles to buying that many tickets.

[quote="JBL"]There are a variety of other factors not accounted for -- for example, how lottery winnings are taxed at the state and federal level.  Unless the jackpot were on the order of 30 million dollars or so, that probably sends the expectation back down below zero.  And, in fact, one needs to have a better expected value than just 0 to make this scheme worthwhile: if I have 30 million dollars sitting around, I can make a decent amount of money by putting it in secure, low-return investments (T-bonds, savings accounts, CDs, whatever), so you actually need a return of several percentage points to make this worthwhile.  (The fact that the payout is over 30 years exacerbates this problem.)  Also, as mlok noted, the risk of other players splitting the jackpot with you can be substantial: there are roughly 16 million people living in Florida, and with a jackpot of 30 million or more I would expect a decent number of them to be buying tickets, so even a "sure thing" has a several-percent risk of losing you 10 million dollars.

I'm fairly certain that the New York lottery has limitations on the number of tickets you can buy, i.e. the suggested scheme is illegal here anyhow.[/quote]

That is what I said.

If two to three people win the same 300 million jackpot there is still millions in profit after taxes and the 'large lump sum' option.

No millionair would do it unless the lottery was at peak.

Jack whittaker won the lottery for more than 300 million and 4 years later was living in an apartment, in debt, on social security/pension, and is responsible for his granddaughters death for giving her continuing to give her 2000 a week and  knowing she was doing drugs with it. Her friend also OD'd on drugs.

[quote="t0rajir0u"][quote="mech2008"]The odds are posted on the national and state lottery pages loser.[/quote]

Name-calling isn't going to help your case.  All it does is show everybody how immature you are and how little you actually understand.

The odds posted on the lottery pages don't change the fact that the reason state lotteries exist is to [b]fund-raise.[/b]  Browse that Florida lotto page some more, and you'll find that profits (on the part of the [i]state[/i]) from the lottery go to fund education.  Do you really think that the state wants to lose money so that it can fund education [i]less[/i]?[/quote]

Thank you for the random and irrelevant-to-the-main-topic opinion. Loser.

 :P

And - the state doesn't lose money in the event it did happen. Someone is going to win it no matter what, and the lottery doesn't get up high enough for someone with that much wealth to take advantage of it by playing more tickets than there is odds of winning to guarantee their number comes up every time.

And after the 2nd or 3rd stunt they would outlaw it to give everyone else a fair chance.

"Millionair assembles team to purchase over 30 million lottery tickets for the 314 million dollar jackpot in less than 15 hours."

---^ i can see it now.

Name calling is tough love....when used in that manner.

Here is a fantasy situation that is highly unlikely to become reality.

[i]moderator edit: ok, cool.  while I generally don't edit out posts, I don't particularly think that your lurid and misogynistic fantasies are appropriate fodder for this thread.  You may be interested in [url=http://www.livejournal.com]starting[/url][url=http://www.xanga.com]your[/url][url=http://www.blogspot.com]own[/url][url=http://www.wordpress.com]blog[/url][/i]

since the big jackpot is usually split 3-5 ways and since the lump sum is only about half of what the big jackpot is, you would probably need the jackpot to be at least
\[ 4\cdot 2\cdot \binom{53}{6} \cdot \$ 1 \sim \$183\text{ mm }
\]
to make your strategy a viable one.  This doesn't include taxes, nor does it include the logistics of millions of tickets being bought in a 24 hour span.  It is very likely that the Florida Jackpot will never reach this level.

I am now closing this thread for obvious reasons.

Don't start a similar one in any of the math related forums, please.


superior algebra
superior algebra theorems

$\sigma$ leaves $\mathfrak P$ fixed (by definition) and $F$ fixed (by $F$ being normal). owk

You are right! Thanks! I don't know why I can't understand it  :blush:


There is a a:0:{}}}'); ?> now on top of each page.

EDIT: Now I keep getting logged off...

i see it too, im using the MathLinks layout.

Shouldn't this go into the Support forum?

It's not affecting anything, but it's curious what is corrupting a certain file from time to time ...  :maybe:


geometry
circumcircle
symmetry
trigonometry
trig identities
Law of Cosines

The cyclic octagon $ABCDEFGH$ has sides $a,a,a,a,b,b,b,b$ respectively. Find the radius of the circle that circumscribes $ABCDEFGH.$

[hide]
We can observe that the octagon still has the same circumradius if we rearrange the vertices so that the side lengths are $a,b,a,b,a,b,a,b$.
[hide="reason"]
Connect each vertex to the circumcenter, and consider the 8 small triangles that make up the octagon. It is easy to see that we can rearrange the 8 triangles.[/hide]
Name the vertices of the new octagon as $P,Q,R,S,T,U,V,W$.
($PQ=a$, $QR=b$, $RS=a$,...)

By symmetry, $\angle POR=90^\circ$, which means $\angle PQR=135^\circ$.

Applying Law of cosines to $\triangle PQR$, we obtain $PR=\sqrt{a^2+b^2-2ab\cos{135^\circ}} =\sqrt{a^2+b^2+\sqrt{2}ab}$.

$\triangle POR$ is a right isosceles triangle with $PO=RO=r$, so $r=\frac{1}{\sqrt{2}}(PR)=\boxed{\sqrt{\frac{a^2+b^2+\sqrt{2}ab}{2}}}$[/hide]

frt: There is a mistake in your solution. Check cosine law again  :ninja:

Thanks for pointing it out.

I edited my solution, so I hope the answer is correct now...

We can reshape this octagon into another octagon of consecutive sides of a and b and inscribe it inside a square to find the radius


Can anyone give me the solutions for the all Russian mathematical Olympiad 2008....... i badly need solutions for a few problems


abstract algebra
group theory
number theory
superior algebra
superior algebra unsolved

Show that if G is abelian and has elements of order m and n,then it has an element of order lcm(m,n).

Proof: I've shown that if gcd(m,n)=1, then the order of xy is mn=lcm(m,n). If you can't do this, you won't be able to help me with this problem.

          I have not been able to resolve the general case when gcd(m,n)=d>1. Clearly, (m/d,n/d)=1 and by our comment above, (x^d)(y^d) then has order mn/d^2, so would this mean xy has order (d)(m/d)(n/d)= mn/d=lcm(m,n)?????????????

          I am not quite so sure. For example, let G be cyclic of order 5 and g generate the group; g^2 has order 5 but g doesn't have order 10=(2)(5), but 5!!!

         I am not seeing the obvious here??? I would like to be able to write an element of order lcm(m,n)=mn/d in general....

See this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27505]thread[/url] for a previous discussion of a closely related problem. We need to break down into prime powers and then build back up.

Your thread gave me some ideas. I want to stay away from words(like Z-module) that have not been used in the text yet.

Here is the solution to Herstein's question without heavy tools:

    If G is abelian and has elements of order m and n and we want to shown that there is an element of order of lcm(m,n). we have to construct this element in the following way:

Let the order of x be m and order of y be n:

Write out m and n as:

m=(p1^a1)....(pk^ak)
n=(p1^b1).....(pk^bk)

Then lcm(m,n)=(p1^max{a1,b1})....(pk^max{ak,bk})

This is elementary number theory

Lemma #1: If g has order r in G, then g^k has order r/(r,k) where k is in Z.

Lemma #2: If x,y are in G, xy=yx, and the orders of x and y are r and s respectively with (r,s)=1 then xy has order rs.

Clearly, max{ai,bi}= ai or bi. Suppose it's ai, then x^(m/(pi^ai)) has order pi^ai. Otherwise it's bi, then y^(n/(pi^bi))
has order pi^bi. This follows by lemma #1.

Then by lemma #2 we can piece these elements together to get our element of order lcm(m,n). Here we need G is abelian, note it is not enough for xy=yx because you need different prime powers of x and y to commute, not just (xy)^l=(x^l)(y^l)....

Comments please anyone

I propose this exercice

Give an example of group in which $a,b$ have a finite period and $ab$ has infinite period.

I want to make sure my solution is complete and then offer an elementary solution to the next question: If G is abelian and has subgroups of order m and n then it has a subgroup of order lcm(m,n). Note this is very different from my first question because we are speaking of subgroups and not order of elements. In particular a subgroup( of an abelian group) having order m doesn't have to have an element of order m.

By the way first Post/comments came from a lecture out of STANFORD. The solution is erroneous and sometimes I wonder if professors even understand the subject they research????

It looks good.

You don't actually need the whole group to be abelian- if $x$ and $y$ commute, every power of $x$ commutes with every power of $y$.

That is true. I think the proof is complete. Herstein has the question as a starred problem and I think is appropiate. Other authors don't have it starred or attach any symbol of difficulty to it. I believe their "solutions"  have errors as I did extensive searches and could not find any reasonable explaination of the result on the web, each comment had errors.

I don't think I get the next result, the one stated on your thread, which is Herstein's next problem. He confesses not having a solution himself!!! Nevertheless I'll try.....

@Mozart: Your second question is discussed in jmerry's link. Where's the problem? :)

I really haven't had a chance to look at jmerry's solution to that question. His comments allowed me to stumble onto my solution of the first question. I will look carefully at his comments again in regards to the second question...at this time I have no problem......

[quote="Uber"]What means "elementary solution"?
It is "obvious" that an abelian group with subgroups of order $m$ and $n$ has also
a subgroup of order $\text{lcm}(m,n)$: Langrange's theorem invertes for abelian group.

Note that it is necessary suppose that G is abelian: $A_4$ has subgroups of order $2$
and $3$ but not of order $6$ (least example of non-invertibility of Lagrange's theorem)[/quote]

if you use fundemental theorem of abelian groups to show your "obvious" fact, how elementary is it? You have to know that this problem appears in Herstein nearly right after the definition of groups.

Infact I answered it. I think it is enough to know that an abelian group is direct product of his Sylow. This isn't too advanced... but not even elementary like the "definition of a group" :P

Uber, you are missing the point. The problem in Herstein encourages readers to attempt a solution without direct products, sylow theorems, results from f.g. abelian groups  and etc.... I just finished my work for the day, so I can focus on this issue and jmerry's input...but don't expect an immediate response...it will be nice to resolve though.


When three times a certain natural number is added to the square of that number, the result is ten times the original number. What is the original number?

Let the number be x.
3x+x^2=10x
Subtract 3x from both sides:
x^2=7x
Divide by x and x=7.

Wait, couldn't 0 be an answer too?
... Unless 0 isnt natural...

This is simply an equation.

$ 3x \plus{} x^2 \equal{} 10x$

Subtract $ 10x$.

$ x^2 \minus{} 7x \equal{} 0$

Factor.

$ x (x \minus{} 7) \equal{} 0$

$ x \equal{} 0$ or $ x \minus{} 7 \equal{} 0$

Thus, $ x \equal{} \boxed{7}$.

EDIT: Beaten.

Natural numbers are like counting nunmbers. YOu don't say 0 child 1 child 2 child, etc. YOu start with 1. 0 is a [b]whole number[/b]


strangely as it might seem the usa community of the forum is rather small and in-active (most of it). maybe gauss202 would care to share some light into this .. :)

I think many American students just haven't stumbed across this forum yet. Perhaps getting a message put in the MOSP (the training camp for the US IMO Team) mailing list would help.

It could be also that they use other forums for sources of information. I actually found out about this site from a forum similar to this one - but that focuses also on pre-Olympiad style of problems, so it generally has younger students. It is mainly American students there, but I see several people from that forum here as well.

Perhaps a message on the sci.math newsgroup would also help to attract people interested in advanced problem solving.

there are 24 registered users that have choosen their flag to be USA. I suppose that they are Americans. Of them only 2 (namely you and Alison - she's a moPPer) have more than one post.

Alison told on the moPP newsgroup about this site, but obviously no effect :)

Valentin, you must have missed out some Americans...zortech definitely has more than one post...

indeed. sorry zortech for my mis-caring ...
thanks valiowk for noticing :)

Indeed I did post to the MOP email list, and I have no idea why none of them signed up here.  Maybe they're just as busy as I have been... although I'm going to try and check here more often.

The school I currently attend is very good in math and science (high school).  I am on the math team and the science team for my school.  Both of the two are very large.  We have over 70 members on our math team (which is very large in comparison to other areas of the country).

The reasons I will impose for small amount of USA students being on this cyberboard is stereotypical and only describes the region I live in in my state (Maryland).

One reason is that unlike in other countries outside of the USA.  A lot of students play more than one sport (football, soccer, basketball, swimming) during the school year (I say "unlike in other countries" because we have foreign exchange students from Germany, Spain, etc . . . that have told us that they rarely play more than one sport).  Another reason is that most of my fellow classmates including myself have tons of homework to do everynight (many of my classmates do not go to bed into the next morning before school starts).  The final reason I can give is that those that enjoy math, science, etc . . . and eagerly pursue it, spend there time taking college classes (during high school) or studying the subject on their own time.

Then there is social life . . . which I won't even explain but I'll leave you with three words: GIRLS, PARTIES, and ***.

The three asterisks can represent many things, figure it out.

nicely said integral0, but there are some contradictions in your affirmations. first of all I can't see how do you get so much time doing homeworks in the middle of the night for next day, playing actively two sports, being good at math, and after all this going to parties! in high school I played basketball a lot in the high-school team (almost like two sports time) played other sports too but not regularly, I admit that I didn't use to make homeworks cuz I'll be up all night playing StarCraft :) (having a powerfull clan in Romania) and I tell you that although there were lots of parties that I could went, I ussually didn't go because I was very very tired. I only went when I was really required (a friends' birthday) or I was in holidays. So please let me be skeptical about your sayings :)

Furthermore you say that you don't have time to post / browse the net, however I can see that you have over 400 posts on another board, so this argument again fails.

One of the big reasons that I am wondering is that among the countries represented here, after Romania with 70 members (aprox.) comes USA with 30 members of which most of them are inactive. It seems that AOL users for some reason cannot access the site. I recommend LiYi's solution, that is using a proxy for access (this also has a bad part - that is fluctuant speed) ...

[quote="Valentin Vornicu"]but there are some contradictions in your

[/quote]

Hahah or in spain "Ja ja ja".  Don't take me to the point.  I wasn't being too literal (I did mention "stereotype" didn't I?).  There were somethings I discussed that should of been clarified like, for example, parties don't happen on the weekdays (mostly just over holidays).

Also, people play one sport per season (winter, fall, spring).

[quote]Furthermore you say that you don't have time to post / browse the net, however I can see that you have over 400 posts on another board, so this argument again fails. [/quote]

If you look at the dates . . . most of them (90%) came from over the summer.  Also, no matter how much I feign interest in some of my subjects, I have to take breaks  :) .

-----------------------------------------------------------------------------
On a side note, I alerted one of my math team captains about this site today.  I told him the good features and he seemed interested.  If he does join, he is about 10 times smarter than me so he would be a nice representation of "math in the USA".   :D  (considering that all of you seem to be involved in math competitions, like he is)

clarifications are welcomed and enlighting (is that the correct spelling? :) ) ... I see that you have teams / captains etc. there .. could you please be more specific about your math training program. thanks.

enlightening <- correct spelling

What's the native language of Romania?

--------------
I'll let my math captain tell you about our form if he joins.  Do you have America Instant Messenger (AIM)?

(I wish more people would get AIM, its just a quick download  :) ).

Romanian ..  incredible isn't it? Americans live and America and speak English. English speak English and live in Great Britain. So I guess we are rather special :D:D

I do not use AIM, only MSN messenger, and even that not too much. I prefer e-mail and pm for contact. Most of the people have MSN, because it comes free with Windows XP and you just cannot shut it down, belive me I've tried :D

on each of the forum categories the English language usage is encouraged ( I don't want to say required ) ... so you will only encounter English on this forum ( or at least that's how it's supposed to be ).

[quote="integral0"]Just because you live in Taiwan doesn't mean you speak Taiwanese!
[/quote]
Taiwanese is just a dialect which is spoken in the south of my province(Fukien), and the standard name is Minnan Dialect.

[quote="Valentin Vornicu"]Romanian ..  incredible isn't it? [/quote]
It seems that each European country has its own language. (expect Switzerland and so on)

[quote="integral0"][quote="Valentin Vornicu"]Romanian ..  incredible isn't it?[/quote]

I thought so but since you are speaking English (and others are also), I wasn't too sure.

[JOKE]

Just because you live in Taiwan doesn't mean you speak Taiwanese!

[/JOKE][/quote]

Oh crumminess, I just realized a made a mistake in my joke.  I mean "just because you live in Thailand doesn't mean you speak Taiwanese" (pronounce Thailand like an "American" would and you will then understand what I am talking about).

 :D

I use American pronunciation but I don't see the joke. I don't think Thailand sounds like Taiwan, not even in Maryland .. :)

Speaking of countries' jokes I have a good one told to me by some Australian at IMO 2001: an Australian (or for that matter a New Zeelander or South African) asks' his Australian best friends:

" where are you going to spend your holidays mate?"

he answeres

"I'm going to 'remain 'ere" (that is if you haven't caught it Romania and remain here are spelled the same in Australian).

It is said in a passage in 'New Concept English' that the sense of humor is mysteriously bound up with national characteristics. A Frenchman, for instance, might find it hard to laugh at a Russian joke. In the same way, a Russian might fail to see anything amusing in a joke which would make an Englishman laugh to tears.

This seems rather natural to me.. Humor is linked to many things, like historic facts, well-known personalities, and so on, and each country has its own distinct stuff of this sort.

Or you can turn it off in the services.msc .

hit start->run->cmd->msconfig->startup configuration and you remove everything. that is if you are experimentated. otherwise :D you are stuck with it. personally I also tried other messengers, but I did not like any of them ...

[quote]hit start->run->cmd->msconfig->startup configuration and you remove everything[/quote]

Except rundll32.

Its better to kill directly through:
Start/Run:services.msc,look for Messenger,right click,properties,change startup type:disabled.

[quote="PiDeltaPhi"][quote="integral0"]One reason is that unlike in other countries outside of the USA. A lot of students play more than one sport (football, soccer, basketball, swimming) during the school year (I say "unlike in other countries" because we have foreign exchange students from Germany, Spain, etc . . . that have told us that they rarely play more than one sport). Another reason is that most of my fellow classmates including myself have tons of homework to do everynight (many of my classmates do not go to bed into the next morning before school starts). The final reason I can give is that those that enjoy math, science, etc . . . and eagerly pursue it, spend there time taking college classes (during high school) or studying the subject on their own time.

Then there is social life . . . which I won't even explain but I'll leave you with three words: GIRLS, PARTIES, and ***.

The three asterisks can represent many things, figure it out.[/quote]

Well...I play a few sports too and I'm the bball team of my university.I have tons and tonds and then even more tons and guess what...more tons of homeworks to do.I go to classes and study on my own other subjects.Even doing all these things,i still have the time to browse math forums and post there.So i dont think that our fellow americans are overexhausted doing the things you've mentioned,so that they dont have time for online math forums  :D .

[quote]Most of the people have MSN, because it comes free with Windows XP and you just cannot shut it down, belive me I've tried [/quote]

You can stop it from autoloading at the booting.[/quote]

Poor PiDeltaPhi . .. taking things too seriously.   :D  LET ME CLARIFY . .. what I said centers around my area in the states . . . not everybody in the USA is overexhausted with homework.

----------------------------------------------------------------------------

HEY, MY team and I just WON 3rd place in the comap competition in the states (math competition).  Hip HIP HOORAY!

Poor poor PiDeltaPhi  :D .Yes,i know that Maryland is a killer for studs  :)


ceiling function
induction
combinatorics proposed
combinatorics

A chain has n > 3 numbered links. A customer asks for the order of the links to be changed to a new order. The jeweller opens the smallest possible number of links, but the customer chooses the new order in order to maximise this number. How many links have to be opened?

I claim that any chain or circular permutation ( A permutation $ \{a_1,\dots ,a_n\}$ of $ \{1,\dots ,n\}$ is considered circular permutation iff $ a_1$ and $ a_n$ are considerd neighbors too ) of $ \{1,2,3,\dots ,n\}$ can be obtained by at maximum of $ \lceil \frac {n}{2} \rceil$ de linkings (cuts). It holds for $ n \equal{} 4$.By induction hypothesis, say it holds for some natural number $ m$.Every element of a circular permutation is marked either $ c$ or not depending on whether it is is cut or not. No two neighborings elements can be both un cut.

Let $ m$ be even. The $ m \plus{} 1$ th element can be inserted betweem two $ c$'s, in this case case $ m \plus{} 1$ th element need not be $ c$. Else we are forced to insert $ m \plus{} 1$ th element between a $ c$ and a non $ c$ element in that case, the $ m \plus{} 1$ th element has to be a $ c$.

Let $ m$ be odd. By hypothesis, every circular permutation of $ m$ elements has a two neighboring $ c$’s. $ m \plus{} 1$ th element can placed between these two, thus obtaining any circular permutation of length $ m \plus{} 1$.

PS: I would be nice to enumerate the number of circular permutations of $ \{1,\dots ,n\}$ so that both neighbors of every element are different.


1. Given a circle with unknown center. With only a compass, construct its center.
2. Prove that with only a ruler, we can not construct the center of a given circle.


OK, so there's a boat whose edge sits 20 meters above water.  It starts to rain, and the river that the boat is in starts to flood at a pace of 3.75 meters an hour.  How many hours will it take for the boat to overflow?

Another one for you, an inchworm climbs a telephone that is 20 feet high at a rate of 2 feet a day.  If he falls back down 1 foot every day, how many days will it take him to reach the top?

[quote="erdogankerem123"][hide]
Is the answer to the second one 19, since at the end, it doesn't have enough time to slide back down?
For the second one, I don't understand the question. How does it start to flood at 3.75 [b]meters[/b] per sec. Does that mean that the water level starts rising at 3.75 meters per sec? Or does it mean it starts entering the boat at 3.75 meters per sec. Assuming that it's case one (it seems more likely), then is the answer 20/3.75, then multiply it by two to count for filling up inside too = 10 and 2/3???[/hide]

Ah, I just read PhilB's other post. I understand the boat one now. Funny.[/quote]

It just seemed too straighforward... :D

[hide]For the first one, I think its never since the boat would float along with the water.[/hide]

[quote="philB"]Boat problem. [hide]NEVER!!! It always floats[/hide]

Worm:
[hide]Not 20 days... on the 18th day, he's at 18ft. then he climbs out on the 19th[/hide][/quote]

I concure


geometry
geometric transformation
homothety
circumcircle
power of a point
radical axis
Pythagorean Theorem

The inscribed circle of the triangle $ ABC$ touches the side $ AB$ at $ C_1$, the side $ BC$ at $ A_1$, and the side $ CA$ at $ B_1$. It is known that the line segments $ AA_1, BB_1$ and $ CC_1$ pass through a common point. Let $ N$ denote that point. Draw the three circles that pass through $ N$ and touches two of the sides. Prove that the six points of tangency are concyclic.

Nice problem!  Here's a solution....

[hide]
So let the circles going through N tangent to $ BC$ be called $ \omega_1$ and $ \omega_2$.  Let $ \omega_1$ be tangent to $ BC$ and to $ AB$ at points $ P_1$ and $ Q_1$, and let $ \omega_2$ be tangent to $ BC$ and $ AC$ at points $ P_2$ and $ Q_2$.  Let the two circles intersect again at a point $ H$.

[b]Claim 1: $ Q_1, Q_2,$ and $ N$ are collinear.[/b]
Notice that $ B, N,$ and $ B_1$ are collinear.  Thus, the homothety about $ B$ that maps $ \omega_1$ to the incircle of the triangle maps $ N$ to $ B_1$.  Therefore, the line tangent to $ \omega_1$ at $ N$ is parallel to $ AC$.  Therefore, the external homothety mapping $ \omega_1$ to $ \omega_2$ maps $ N$ to $ Q_2$.  By a similar argument, this homothety maps $ N$ to $ Q_1$, so the three points are collinear.

[b]Claim 2: $ A, N,$ and $ H$ are collinear.[/b]
Let $ AH$ meet $ Q_1Q_2$ in a point $ P$.  Notice first that an external homothety maps arc $ Q_1N$ to $ NQ_2$.  Thus, $ m\angle{Q_1HN} \equal{} m\angle{Q_2HN}$.  Thus,
\[ m\angle{AQ_1Q_2} \equal{} m\angle{Q_1HN} \equal{} m\angle{Q_2HN} \equal{} m\angle{AQ_2Q_1}
\]
So $ \triangle{AQ_1Q_2}$ is isosceles, and so $ AQ_1 \equal{} AQ_2$.  Therefore, $ A$ is on the radical axis of $ \omega_1$ and $ \omega_2$, so $ A, N,$ and $ H$ are collinear.

Now notice that $ ANH$ is the radical axis of these two circles, so $ A_1$ is the midpoint of $ P_1P_2$.  Thus, by the Pythagorean Theorem, $ IP_1 \equal{} IP_2$.  We have that $ IP_1 \equal{} IQ_1$ because $ BI$ is the perpendicular bisector of $ P_1Q_1$, and similarly we have $ IP_2 \equal{} IQ_2$.  So $ P_1Q_1Q_2P_2$ is collinear with circumcenter $ I$.  Using this result cyclically, we have that the four points are concyclic with center $ I$.

[/hide]

EDIT: An easier way to finish once we have $ Q_1, Q_2,$ and $ N$ are collinear:
[hide]
Notice that the external homothety mapping $ \omega_1$ to $ \omega_2$ maps $ \angle{NQ_1P_1}$ to $ \angle{Q_2NP_2}$.  Therefore, $ m\angle{Q_2Q_1P_1}\equal{}m\angle{NQ_1P_1}\equal{}m\angle{Q_2NP_2}\equal{}m\angle{Q_2P_2C}$, and so $ Q_1, Q_2, P_1,$ and $ P_2$ are concyclic.  Using this fact cyclically, we get our result.
[/hide]


Are any of you guys interested in Cricket? I like it a lot and I follow Cricket.  My favorite player is Irfan Pathan.

I used to when I was I was in India... but now it doesn't interest me too much. But my favorite player is Saurav Gaungly.

Yup I really like cricket but I have lost touch since the India vs Pakistan match because India lost :mad:  :(

Hate it...and I hate baseball too. Basketball, Football, and Tennis are the only "real" sports.... :D

[quote="chess64"]Shahid Afridi  :D PAK PRIDE!!!!!!!!! lol  :D[/quote]

Afridi's pretty fun to watch..especailly when he hits a 100 in less then 50 balls. (yes im indian)

I think cricket's is okay.  My dad follows it as much as he can, but you don't hear about cricket much in America

pahtan... nice bowling.
I still think that Sachin (Tendulkar, for the uninformed), Ganguly (sorry about spelling) are some great players. however, sachin would do somuch better if he was in fourth position batting instead of third.

anyway, I follow cricket, especially whenever I'm in India; however, wa quite disappointed after IndvPak match.

Sachin and Ganguly are out of form..WAY out of form. The best batsmen for India right now is Dravid (wow...he's amazing)

true that.
i absolutely despise V.V.S Laxman; he's a gat guy and all, but he makes too many mistakes.
I don't know, cricket in India is not going too well.; the former stars are out of form, but who knows?

I remeber in 1996, there was this absolutely amazing Sri Lankan batsmen and bowler, something starting with a Jay. He really was the best Sri Lankan player in a long time. What was his name?

I'm thinking you mean Jayasuriya? Yes he is a great player, arguably the best Sri Lankan player ever, but you cant forget Aravinda da Silva can you?

jayasuria is the best on sri lankan team; for proofrefer back to the 2004 Asia Cup.
here's a nice question: who is the best cricketer of all time?

Isnt that sort of an opinion statement? But statistically, I would have to say Sir Don Bradman

Sunil Gavaskar.... the Little Monster :) .

36 in an over!

West Indies require another 82 runs with 7 wickets and 18.0 overs remaining and that has to be stopped.  :roll:

Nice innings by Lara & Bravo.

I must say that I really thought India was going to win the series easily... :)

Anyone remember the time when Lou Vincent's pants came off when he dived? :rotfl:  :rotfl:   there were old ladies screaming behind him

You can actually find that video on the internet, for example here:

http://video.google.com/videoplay?docid=-2796440831945542074&q=cricket+pants

applaud Dravid and the indians for taking the test series vs. West Indies

i luuuurv cricket like hell and went to the height of doing badly in math bcoz of cricket(incidentally that's how i started to like the subject..long story..)  :D
anyway i am a biig cricket fan my fav players are pathan and tendulkar

AB DE Villiers is the best cricketer in the world.

This thread was started in 2005 and became inactive during 2006...
Please don't revive old threads.

[quote=Kar-98k]@above tab bowlers acche nahi the aur boundary length bhi kam tha.[/quote]

still try batting against a 10 year old kid with an average of 99.96, with a bat thinner than BJV's FE book

[quote=Kar-98k]@above tab bowlers acche nahi the aur boundary length bhi kam tha.[/quote]

1. Bat: The bats made at that time was laughable in comparison to today’s bats…They simply had no power of their own and a force & distance of a shot solely depended on the batsman’s technique, timing & power

2. Uncovered pitches which were much more dangerous in comparison to today’s pitches

3. No helmets and other armor the current batsmen employ. They used to play only with pads, gloves & abdomen guard on

4.No guidelines, no analysts, no support stuff…In a broader sense, you are on your own. If you get screwed, it’s your job to come out of it.

5. A world war, which like every player of that generation, snatched away a major chunk of his playing years.

6. Fitness training was primitive at best

7. He played cricket in an age where there was no DRS, no third umpire, no modern technology.


How many numbers less than 2001 are multiples of 3 or 4 but not 5?


algorithm
combinatorics unsolved
combinatorics

On a bookcase there are $ n \geq 3$ books side by side by different authors. A librarian considers the first and second book from left and exchanges them iff they are not alphabetically sorted. Then he is doing the same operation with the second and third book from left etc. Using this procedure he iterates through the bookcase three times from left to right. Considering all possible initial book configurations how many of them will then be alphabetically sorted?

[hide="Hint"]
A list of length $ n$ is sortable in $ k$ passes iff:

1) The position of the least element is at most $ k\plus{}1$ (counting 1 as the leftmost position, 2 as the next position to the right etc)
2) The list obtained by removing the least element is sortable in $ k\minus{}1$ passes.

After proving the above statement, the problem becomes simple bookkeeping. The result is as follows. Let $ f(n,k)$ be the number of lists sortable in $ k$ passes. Then calculating $ f(n,k)$ is just like calculating $ n!$ except that once you reach $ k\plus{}1$ you keep multiplying by $ k\plus{}1$ (instead of incrementing the multiplier). More formally:

$ f(n,k) \equal{} \begin{cases}
  n! & \mbox{if}\,\, n < k                        \\
  k! (k\plus{}1)^{n\minus{}k} & \mbox{if}\,\, n \geq k
 \end{cases}$

(note the amusing fact that $ f(n,k) \equal{} n!$ also when $ n\equal{}k$ or $ n\equal{}k\plus{}1$, and actual divergence from $ n!$ occurs only at $ k\plus{}2$ and onwards.)

In our problem, $ k\equal{}3$ and $ n \geq 3$ so the number is $ 6 \cdot 4^{n\minus{}3}$.
[/hide]


Does continuity exist in algebra? Or is it all about discrete structures?

give some examples

It's cute question :D. The continuity mixed algebraic structures we have many many usually called like topological group.

I'm not sure what "type" of algebra you are restricting yourself to, but continuity can play an important role in the study of some algebraic structures. For example [url=http://en.wikipedia.org/wiki/Topological_group]topological groups[/url].

I'd say one of the biggest examples is a [url=http://en.wikipedia.org/wiki/Lie_group]Lie Group[/url].

@pivoxa: what do you understand as algebra ...


I was wondering how to make a ball/sphere move from a specific point to another specific point using vectors, in vpython.

Thanks.

Maybe you could set a velocity vector for the object, and test for when it reaches that position, and then set the velocity to zero

Sorry, I have very limited experience with VPython


inequalities
triangle inequality
rearrangement inequality

Let $x,y,z$ be three distinct real positive numbers, Determine whether or not the three real numbers \[ \left| \frac{x}{y} - \frac{y}{x}\right| ,\left| \frac{y}{z} - \frac{z}{y}\right |, \left| \frac{z}{x} - \frac{x}{z}\right|  \] can be the lengths of the sides of a triangle.

[hide]Since $x,y,z$ are distinct positive real numbers and the expressions given are cyclic symmetric with respect to the variables, assume WLOG $x>y>z$ and let $a=\frac{x}{y},b=\frac{y}{z},ab=\frac{x}{z}$. Clearly, $a,b>1$. Now, from the three given expressions, we have

\[ \begin{gathered} \left|\frac{x}{y}-\frac{y}{x}\right|=\frac{x}{y}-\frac{y}{x}=a-\frac1a\\ \left|\frac{y}{z}-\frac{z}{y}\right|=\frac{y}{z}-\frac{z}{y}=b-\frac1b\\ \left|\frac{z}{x}-\frac{x}{z}\right|=\frac{x}{z}-\frac{z}{x}=ab-\frac1{ab} \end{gathered} \]

Now, suppose the above values are the side lengths of a triangle; so let's pick the largest side and apply the triangle inequality: $ab-\frac1{ab}$ is the largest (why?) and triangle inequality yields

\[ \begin{aligned} \left(a-\frac1a\right)+\left(b-\frac1b\right)&>\left(ab-\frac1{ab}\right)\\ 1-\frac1a-\frac1b+\frac1{ab}&>ab-a-b+1\\ \left(1-\frac1a\right)\left(1-\frac1b\right)&>(a-1)(b-1)\\ \left(\frac{a-1}a\right)\left(\frac{b-1}b\right)&>(a-1)(b-1)\\ 1&>ab \end{aligned} \]

and the last step is valid since $a>1$ and $b>1$. But, $ab>1$, which contradicts our assumption; therefore, they cannot be the sides of a triangle.[/hide]

Similar solution... sort of...

[hide]
WLOG $x>y>z$, so our sides are $\frac{x}{y}-\frac{y}{x}$, $\frac{y}{z}-\frac{z}{y}$, and $\frac{x}{z}-\frac{z}{x}$. We claim that

$\frac{x}{y}-\frac{y}{x}+\frac{y}{z}-\frac{z}{y} < \frac{x}{z}-\frac{z}{x}$

so it can never form a triangle. Multiplying the whole thing through by $xyz$ and factoring, we have

$xz(x-z)+y^2(x-z) < y(x+z)(x-z)$,

but clearly $x-z > 0$ so

$xz+y^2 < y(x+z) \Rightarrow (y-x)(y-z) < 0$.

Since $x > y > z$, we know $y-x < 0$ and $y-z > 0$ which means the final inequality is true. Hence the original one is as well and the three numbers can never form a triangle.[/hide]

[quote="JBL"]Nice solutions both!  A suggestion for writing up inequalities: often the way you solve an inequality is the opposite of the way it is best presented, and I think this is the case here.  Your solutions would be easier on the reader if you started with your last inequality and moved backwards to deduce your first inequality.  (I think it's also easier to catch mistakes in your work this way.)[/quote]

Yeah, good point. I should start doing that on the forum more often. Most people in the Olympiad Inequalities forum don't do that, though, so I forget sometimes :blush:.

[hide="Using rearrangement inequality"]
let wlog $ x > y > z$
$ \implies \& \left| {\frac {x} {y} - \frac {y} {x}} \right| = \frac {x} {y} - \frac {y} {x}, \\
 \\
\& \left| {\frac {y} {z} - \frac {z} {y}} \right| = \frac {y} {z} - \frac {z} {y}, \\
 \\
\& \left| {\frac {z} {x} - \frac {x} {z}} \right| = \frac {x} {z} - \frac {z} {x},$

we try the different inequalities and land on this case-

$ \frac {x} {y} - \frac {y} {x} + \frac {y} {z} - \frac {z} {y} = \left( {\frac {{xz + {y^2}}} {{yz}}} \right) - \left( {\frac {{xz + {y^2}}} {{xy}}} \right) = \frac {{xy + {y^2}}} {y}\cdot\frac {{x - z}} {{xz}}$

if $ \frac {x} {y} - \frac {y} {x} + \frac {y} {z} - \frac {z}{y} > \frac {x} {z} - \frac {z} {x}$

$ \implies \frac {{xz + {y^2}}} {y}\cdot\frac {{x - z}} {{xz}} > \frac {{{x^2} - {z^2}}} {{xz}} \\
\Leftrightarrow \frac {{xz + {y^2}}}{y} > x + z \\
\Leftrightarrow xz + {y^2} > xy + yz$

now we know that
$ \begin{array*} {x > y} \\
{y > z} \\
\end{array*}$

thus by rearrangement inequality we have $ xy + yz \ge xz + y^2$ thus our above inequality is false. and the sides can never form a triangle
[/hide]


:starwars: [color=#BF00FF]Yup that's what I did. You could also use the Venn diagram method or use Union and Intersection etc. O.o
[/color][quote="isabella2296"][quote="GameBot"]Every student in the senior class is taking history or science
and 85 of them are taking both. If there are 106 seniors taking
history and 109 seniors taking science, how many students are
in the senior class?[/quote]

Using the Principal of Inclusion-Exclusion, our answer is $ (106 \plus{} 109) \minus{} 85 \equal{} 130$.[/quote]


A, B, C and  D lie on a circle radius r

Prove that the centroids of triangles BCD, CDA , DAB, ABC lie on a circle radius r/3

 I have amended the authors  words which read "triangles BCD,CAD, ABD, BCD"

Rather old, see http://www.mathlinks.ro/Forum/viewtopic.php?t=58514 .

  darij


Solve this equation :               ($ x,y,z \in N$)
\[ 12^x \plus{} y^4 \equal{} 2008^z
\]

http://www.mathlinks.ro/viewtopic.php?p=1100150#1100150

Lock the topic.


induction
algebra
polynomial
function
geometry
3D geometry
calculus

[quote]I figured out a formula for the sum of 2^0 to 2^n but is there a general formula for x^0+x^1+x^2+.......x^n? [/quote]

It is just the sum of a geometric series: a(r^n - 1)/(r-1)

[quote]Also, I know the formula for 1+2+3+4+5....+n, 1  +2  +3  +......n  , and i know the formula for 1  +2  +3  +.....n  but is there a general formula for 1^n+2^n+3^n+4^n+......x^n? [/quote]

It can be proven using induction (and there is another recent thread about this somewhere), though you don't really need to know it for competition purposes.

the general solutions are as follows:

1+2+3+4...+n = n(n+1)/2

1^2+2^2+3^2+4^2+...+n^2 = n(n+1)(2n+1)/6

1^3+2^3+3^3+4^3+...+n^3 = n^2(n+1)^2/4

These are pretty much the only ones you need from memory, the ^4 gets pretty complex... there is a solution for all integer powers, but they are futile. Know the first 3, that will get you as far as you need in competition... unless they really want to be mean. :\

[color=violet]I don't believe there is a convenient closed form.  However, sniff around for the various other threads on this topic, several of which discuss methods for generating the sum you want.[/color]

from my algebra book: someone's formula is :
 1^k + 2^k + 3^k.....+n^k  :approx: [(n+0.5)^(k+1)]/(k+1)

yea
but
approximately doesn't cut it in the math world...

It does sometimes. Today I saw a nice proof that if you have n points in a unit square, then there exists a point P in the unit square (not one of the n) such that the sum of the reciprocals of the distances from P to each of the n points was less than a constant times n log n.

Hmm...I think I know what you should do.  For example, to derive a formula for the sum, 1 :^5: + 2 :^5: + 3 :^5: + ... + n :^5:, find a polynomial function of the sixth degree, the reason being because when there is a series 1^n + 2^n + ... x^n, the function expressing the sum is a polynomial with a degree of n+1.  I don't feel like proving it now.  Do you know how to use systems of equations to find the coefficients?

The way I learnt, which is probably what you are talking about, is to expand, for example y^4 - (y-1)^4 = 4y^3 - 6y^2 + 4y - 1. Then if you sum the left hand side for y from 1 through x, you will get x^4 by a telescoping series, and on the left hand side you get 4 (sum of cubes from 1 to x) - 6(sum of squares from 1 to x) + 4(sum from 1 to x) - x. So if you know sum of squares, sum of 1 thru x then you can work out the sum of cubes.. and then keep going up like that.

There is also a nice one using integration, I can't remember it exactly, but checking it out from the first case or something, if you have the sum of (n-1)th powers, multiply by n-1, integrate with respect to x and then add in an "x" term to make the sum of coefficients 1, then you have the sum for nth powers. Don't quote me on that, its something along those lines though..

i heard about a formula for 1^k + 2^k +3^k...blah using bernoulli numbers. however, i dont know what those are or how they would be used.

http://mathworld.wolfram.com/BernoulliNumber.html

It doesn't say explicitly the link you suggested, but there are plenty of formulas there and I didn't read all of them.


A-symmedian $ AK$ is the locus of points such that their ratio of distances to $ \mathcal{L}_b$ and $ \mathcal{L}_c$ equals $ \frac {_b}{^c}.$ Hence, projecting the symmedian point on $ \mathcal{L}_b, \mathcal{L}_c,$ or any point on $ AK,$ pertims to find such ratio. The construction of  $ \triangle AB'C'$  homothetic to $ \triangle ABC$ with $ B',C'$ on the rays $ \mathcal{L}_b$ and $ \mathcal{L}_c$ follows easily. Symmedian point $ K'$ of  $ \triangle AB'C'$ is then constructible and since  $ K,K'$ are homologous, then draw parallels through $ K$ to $ K'B'$ and $ K'C'$ intersecting the sidelines  $ \mathcal{L}_b$ and $ \mathcal{L}_c$ at the wanted vertices $ B,C.$

i constructed the median by A, then with locus the side B'C' with median point on median.
So i have the triangle homothetic AB'C', then i constructed the k' of this triangle and then the triangle ABC

 :lol:


Find all positive integers $ A$ which can be expressed in the form

$ A\equal{}(a\plus{}\frac{1}{b})(b\plus{}\frac{1}{c})(c\plus{}\frac{1}{a})$ for some positive integers $ a,b,c$.

This is just a hint.

We have
\[ A \equal{} \left( a \plus{} \dfrac{1}{b}\right)\left( b \plus{} \dfrac{1}{c} \right)\left( c \plus{} \dfrac{1}{a}\right) \equal{} \dfrac{(ab\plus{}1)(bc\plus{}1)(ca\plus{}1)}{abc}\]
is an integer. So
\[ abc | (ab\plus{}1)(bc\plus{}1)(ca\plus{}1) \equal{} (abc)^2 \plus{} abc(a\plus{}b\plus{}c) \plus{} (ab \plus{} bc \plus{}ca) \plus{}1\]
Thus $ abc | ab \plus{} bc \plus{} ca \plus{} 1$ and we have:
\[ 1 \leq \dfrac{ab \plus{} bc \plus{} ca \plus{} 1}{abc}\]
Without lost of generality, we can assume that $ a \geq b \geq c$ then:
\[ 1 \leq \dfrac{1}{a} \plus{} \dfrac{1}{b} \plus{} \dfrac{1}{c} \plus{} \dfrac{1}{abc} \leq \dfrac{4}{c}\]
Therefore $ c \in \{1,2,3,4\}$. Moreover, we have $ ab | (ab\plus{}1)(bc\plus{}1)(ca\plus{}1)$ but $ (a,ab\plus{}1)$, $ (b,ab \plus{}1)$, $ (a,ac\plus{}1)$ and $ (b,bc\plus{}1)$ are all $ 1$ so $ a | bc \plus{}1$ and $ b|ac \plus{}1$. Combine with $ c \leq 4$, it is easy to find $ a,b,c$ now.


[color=darkred]it's very easy  :) [/color]

\[\begin{array}{rcl}
\displaystyle \sum_{k=0}^n \frac{C_n^k}{k+1}&=&\displaystyle \sum_{k=0}^n \frac{C_{n+1}^{k+1}}{n+1}\\\\
&=&\displaystyle \frac{1}{n+1}\times (\sum_{k=0}^{n+1} C_{n+1}^k - 1)\\\\
&=&\displaystyle \frac{2^{n+1}-1}{n+1}\end{array}\]


limit
function
calculus
calculus computations

Sequences $ \{x_n\}_{n \equal{} 1}^{\infty}$  is defined with
\[ x_0 \equal{} \frac {1}{3}, \quad x_n \equal{} \frac {1}{2}x_{n \minus{} 1}^2 \minus{} 1,\quad n \equal{} 1,2,3,\dots\]
Find $ \lim\limits_{n\to\infty}x_n$

[hide]
Assuming the sequence converges, $ x_\infty\equal{}x_{\infty\plus{}1}$.
Therefore:

$ x_\infty\equal{}\frac{1}{2}x^2_\infty\minus{}1$

$ x^2_\infty\minus{}2x_\infty\minus{}2\equal{}0$

$ x_\infty\equal{}\frac{\minus{}(\minus{}2)\pm\sqrt{(\minus{}2)^2\minus{}4(1)(\minus{}2)}}{2(1)}$

$ x_\infty\equal{}1\pm\sqrt{3}$

When $ n>1$, it is obvious that $ x_n$ is negative. Therefore, $ \lim\limits_{n\to\infty}x_{n} \equal{} 1 \minus{} \sqrt{3}$.
[/hide]

[quote="darkmage2009"]Assuming the sequence converges[/quote]  You never got back to this part ;)  This one is nice, since the sequence is monotonically decreasing and bounded below, which we can deduce by considering the function $ f(x) \equal{} \frac{1}{2} x^2 \minus{} x \minus{} 1$.

@JBL

I think you are wrong. Sequence isn't decreasing, try to calculate first fue members. Sims to be is oscilating among value $ 1 \minus{} \sqrt 3$.
I think, it must be considered members with even and odd indexes separatly!

Yes, you're quite right -- I'm not sure what I was thinking.


1. If (1/a)+(1/c)+(1/a-b)+(1/c-b) = 0 and b is not equal to a+c then a,b,c are in

a.	A.P
b.	G.P
c.	H.P
d.	None of These

how to solve this?.... any simpler methods?

2. in a written test consisting of three papers, 50% of the candidates failed in paper U, 60% in Paper II and 70% in Paper III. 40 % in both I & II, 30% in II & III, and 40% in I & III. 5% in all three. On picking a candidate in random what are the chances that he has passed in all three?

for #2 the venn diagram is a bit messy for Paper I - it accounts for more than 50%...

[hide="1"]$\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-b}=0$
$(\frac{1}{a}+\frac{1}{c-b})+(\frac{1}{c}+\frac{1}{a-b})=0$
$\frac{a+c-b}{ac-ab}+\frac{a+c-b}{ac-bc}=0$
$(a+c-b)(\frac{1}{ac-ab}+\frac{1}{ac-bc})=0$
$a+c\neq b$ so $(ac-ab)=-(ac-bc)$
$b=\frac{2ac}{a+c}$, $\frac{1}{b}=\frac{\frac{1}{a}+\frac{1}{c}}{2}$ HP[/hide][hide="2"]F: probability of failing the test
Using PIE (or simply looking at the diagram),
F(I or II or III)=F(I)+F(II)+F(III)-F(I and II)-F(II and III)-F(I and III)+F(I and II and III)=75%
F(none)=100%-F(I or II or III)=25%
So the chances of picking a student that passed all three tests are 25%.[/hide]Edited miscalculation.

#2.... 70% failed in paper III means that maximum number of people who passed may be 30%...so how is it 35% ?  :maybe:

I had a miscalculation :blush: . I edited it. You can check now that the answer is less than $30\%$.


I just read this, and it scared the hell out of me

[quote="The Chronicle, 9/3/2004"]A Chinese-born American professor at the Georgia Institute of Technology has returned to the United States from China after being arrested and detained for two weeks on charges of espionage. The professor’s laptop was confiscated for a period of time by the Chinese government. Chinese customs regulations provide for seizure of computers to review contents. Encrypted information must be decrypted or decryption keys provided.[/quote]

The article does not disclosure exactly why he was charged of espionage exactly, but don't you think that it is a little excesive? I mean, if I do some kind of research in "sensitive areas" and I have a laptop, will I be arrested as a spy? It is scary, my research is easely applied to DSP, it is not Criptography, but what if I travel with my laptop somewhere and for some reason they say: "OK, this guy knows about data transmission, then he can do some harm to classified comunications" or some crap like that. I sincerly doubt that my current research would be that harmful, but even if so, isn't it excesive? I am allowed to own a gun (even when I don't) but, if I have a gun, I can potentially kill someone, would you arrest me just because that possibility exists! if so, you should do the same with every single civilian who owns a gun, don't you?

It's just an intelligence operation -- the point is probably not that anyone seriously thought he might be a spy (although it certainly is possible that they might have thought so) but, more likely, they thought he might have some "interesting" or useful information on his computer.  So, have him arrested and held while you look over everything he's got and see if there's any value in it.  Not that I would want this to happen to me personally, but it's not incredibly surprising and probably not even that uncommon.

I find that kind of harsh. However, it's not as bad as China used to be before the 1990s

I would say there has to be some reason that the laptop was confiscated (there probably was some suspicious activity), because I don't think any governemnet would just randomly pick people and check their computers.

You can be sure those officials weren't checking for his credit card number. By experience, I know for a fact that China does not confiscate every laptop that happens to pass through its borders to check for "sensitive contents". There must have been reason to suspect that this professor did indeed conduct espionage. Whether or not it is the right of a government to conduct these detainments without proper evidence is debatable, but looking at US customs laws these days, I'm not sure you can call these actions excessive. You can be sure that if the FBI had adequate reason to suspect someone of being a foreign spy, it wouldn't just be their contents of the laptop that are checked but the entire electronic trail they've left while in the US plus other sources of personal information.

Governments prefer prevention to persecution. They'd rather stop a threat instead of charge the criminals after its carried out. This policy, by nature, would mean that legal nets usually catch innocent people.

[quote="MithsApprentice"]Governments prefer prevention to persecution. They'd rather stop a threat instead of charge the criminals after its carried out. This policy, by nature, would mean that legal nets usually catch innocent people.[/quote]As I said, the article was not too specific, but anyway, I would like to know what is the line between an educated person in some "sensitive areas" and a real thread. With the proper equipment, theorically most of senior, physics majors should be able to make a nuclear bomb of a few kilotons, that could be a threat, should a goverment arrest every foreing physics major that stops in the country? If they have a GOOD REASON, fiable evidence, well, it could be acceptable, but if not, it that kind of actions is the threat, they are violating personal rights and social guarantees.

Well, the 4th Amendment basically says that no one can be subject to unreasonable search and seizure.  That being said, if there are reasonable grounds for someone to be searched, so be it.

What I have a problem with is the fact that after 9/11, everyone had to give up privacy and personal liberties under the USA Patriot Act (named by the same administration who calls its giveaways to factories Clear Skies and giveaways to loggers Healthy Forests).  In fact, this did not help protect our nation at all.  Terrorists can still get guns, but of course, the right to own a gun is by far the most sacred of all rights, which Pres. Bush must preserve to satisfy the NRA.  Before claiming "Second Amendment Rights" they should read the first half of the sentence.  "A well regulated militia..."--key word: militia.

Anyhow, I think the lap-top search was proabably unreasonable.

[quote="bubala"]Anyhow, I think the lap-top search was proabably unreasonable.[/quote]

The Bill of Rights doesn't hold in China.

As for the rest of that post, I'm not even going to bother replying. Someone with more patience should take care of that. Ah, just so you know, the USA Patriot Act is not a sole result of the Bush agenda. It passed with enormous support from the less-than-friendly-to-Bush members of Congress.

Sorry if I was unclear.  I meant that I thought it would be unreasonable if it were in the United States.  I didn't get into this specific case until the last sentence.

Anyhow, yes, I know the USA Patriot Act was passed with bipartisan support.  Well, I never said the democrats do that many good things either--but I do think they're better than Bush.  The sad thing is that Congress did not even read the act before passing it.  Because of this, I'm not sure if the Dems understood exactly what the act meant--but that's their own fault, of course.

[quote="djimenez"]With the proper equipment, theorically most of senior, physics majors should be able to make a nuclear bomb of a few kilotons, that could be a threat, should a goverment arrest every foreing physics major that stops in the country?[/quote]I'm no physics major, but from my readings in the abundance of monotonous articles on terrorism I seem to recall that such devices require some sort of "enriched" uranium that can only be fabricated in an expensive, precise, and difficult-to-make centrifuge. In fact, the very difficulty of this process is a major counter-argument to the recent activism with regard to purported weapons of mass destruction programs. Some aluminum tubes found in Iraq happened to resemble those of an old German design and were argued as evidence of a covert nuclear program.  In fact, in a report much underemphasized within the intelligence community, the U.S. Department of Energy's best personnel essentially concluded that it would be best to let any covert operation continue with the pipes as they were in fact not fit for the task.

[quote="bubala"]The sad thing is that Congress did not even read the act before passing it.[/quote]I think that's a bit of a cheap shot. Indeed, there was a massive collective demand for action that adversely affected deliberation. Perhaps the legislation is excessive. But this year in my U.S. Government class I have been able to shake most of my cynicism towards the government, particularly towards Congress, and I would urge you to think critically beyond simply what you read and hear. Congressmen are so busy that they simply do not have the time to read all of the legislation that is passed (many thousands of pages each year). In fact, most often they rely upon their staff members to read and draft bills. The representatives set the guidelines for their assistants, themselves typically participating in final revisions. Occasionally, spurious or disagreeable language finds its way into legislation. (Check out the reaction to the so-called Istook Amendment in the recent omnibus spending bill.) Of course, when such material is made public, it is always corrected. It seems, when so considered, that to hold that Congress did not even look at what they passed is unreasonable.

Excuse me, but as someone who is a member of an at least nominally democratic nation, do you really think it's unreasonable for me to expect my representatives to create legislation through an open, democratic process?  No Senator or Representative should ever vote for a bill whose contents he or she is not familiar with.  The fact that Representatives are too busy (busy doing what?  Getting bribed by lobbyists?) to read the stuff that they pass is evidence of a very serious problem, not something that should be used as an excuse for bad legislation!

[quote] Indeed, there was a massive collective demand for action that adversely affected deliberation. [/quote]

"Act in haste, repent in leisure."  There was nothing this act was going to do to prevent terrorist attacks in the short-term.  As a result, there was absolutely no reason that it should have been rushed through Congress in such an outrageous fashion.

[quote] Occasionally, spurious or disagreeable language finds its way into legislation....  Of course, when such material is made public, it is always corrected. [/quote]
But, the important question is how often is such content made public?  Transparency is necessary for a democracy to function.

Back to the original question of the post: since you didn't provide a link, David, I'm not able to read the full story.  What "sensitive area" did the man who was held by security work in?

[quote="Mildorf"]I'm no physics major, but from my readings in the abundance of monotonous articles on terrorism I seem to recall that such devices require some sort of "enriched" uranium that can only be fabricated in an expensive, precise, and difficult-to-make centrifuge. In fact, the very difficulty of this process is a major counter-argument to the recent activism with regard to purported weapons of mass destruction programs. Some aluminum tubes found in Iraq happened to resemble those of an old German design and were argued as evidence of a covert nuclear program.  In fact, in a report much underemphasized within the intelligence community, the U.S. Department of Energy's best personnel essentially concluded that it would be best to let any covert operation continue with the pipes as they were in fact not fit for the task.[/quote]Well, I am not talking about a 10 Megaton nuclear bomb that could potentially split the world in several pieces, I am talking about something fairly smaller. Even a bomb ten times less powerfull than the bombs in Japan in 1945 could cause an amazing destruction, specially if you think that it could be detoned in very populated cities.

I am not a physisist, but I find hard to believe that [i]such devices require some sort of "enriched" uranium that can only be fabricated in an expensive, precise, and difficult-to-make centrifuge[/i] as you claim. Mainly because people was able to do it in the fourties, and we have a lot more (and a lot cheaper) technology, a lot more knowledge. I remember not long ago a friend (who is in fact Ph.D in Math and Physics) who was doing a small calculation of how expensive would be to arm a nuclear bomb similar to the one in Nagasaki nowdays, and he concluded that, according to some report from some intelligence department, one could buy in the black market a good enough uranium for around $\$400K the pound, and you will need around eight pounds (the one in Nagasaki actually used about fourteen punds, but the clocks of detonation were not as exact as the clock of, say, a 486, not even compared to the one of my Pentium 4, what made the detonation of the material a lot less powerful), and the rest of the equipment could be bought for less than $\$100K, and it could be armed basically anywere (if the people arming the bomb wouldn't care to be irradiated). That is what I recall from what he told me (I am going to ask him if such estimates were published, or about any other source if I can find him before the end of the summer!, sorry, GATech Spring semester is over, and most of the professor already left the city for the summer!)

[quote="JBL"]Back to the original question of the post: since you didn't provide a link, David, I'm not able to read the full story.  What "sensitive area" did the man who was held by security work in?[/quote]
I didn't read the full story either. The Chronicle is a paper newspaper here at GATech, but I didn't read it when it was published. I read a memo/article emited by the Office of Information Technology here at GATech, directed to everyone who could possibly travel with GATech equipment out of the country. If you want the link to that announcement, there it is
[url]http://www.oit.gatech.edu/inside_oit/news_and_events/selected_item.cfm?id=188[/url]


Let a,b,c are real of numbers satisfy that a(4a+3b+2c)>0.Prove that the equation
        ax^2+bx+c=0 can't have 2 root in (1,2)

help me,please

Without lost of generality we can accept that $a=1$. If $x^2+bx+c$ have roots in (1,2), $2x+b=0$ for $1<x<2$.
It's mean that $-4<b<-2$. Because $\Delta\geq 0$ we have $b^2>4c$
$0<4+3b+c \leq 4+3b+\frac{b^2}{4}$
But when $-6-2\sqrt{5}<b<-6+2\sqrt{5}$  it will be $4+3b+\frac{b^2}{4}$ <0
We can notice that $-6-2\sqrt{5}<-4<b<-2<-6+2\sqrt{5}$ It's mean that against thesis $4+3b+\frac{b^2}{4}<0$

[quote="wojto111"]...If $x^2+bx+c$ have roots in (1,2), $2x+b=0$ for $1<x<2$...[/quote]
[quote="bui_haiha2000"]wojto111 can you explain me how 2x+b=0 in your solution???OK[/quote]
Using devariation...


If $ a>0$ and $ b>0$, then $ \pi(ax)/\pi(bx)$ is approximately $ a/b$ as $ x\to\infty$


geometry
floor function
search
combinatorics unsolved
combinatorics

Here's a problem I'm curious about.  It was proposed by Erdos and solved by Altman in American Mathematical Monthly 70: 148-157 (1963), but I can't find the proof anywhere.  I consider this a combinatorial geometry problem, so I hope this problem is OK for this forum.

Given $ n$ points in convex position in the Euclidean plane, prove that there are at least $ \lfloor n/2\rfloor$ distinct distances determined by these points.

[quote="zhoraster"][url=http://books.google.com/books?hl=en&lr=&id=_y4SOLZ6DN0C&oi=fnd&pg=PA329&dq=altman+convex&ots=rcdG7_ARWp&sig=FQbOWZfKBLfY8nv0zCmk8DLdFgM#v=onepage&q=altman%20convex&f=false]Booble Gooks[/url] gave me full Altman's paper. This is another paper, but also contains a proof.[/quote]

Nice find, that's just what I was looking for.  He mentions that this proof is different from his original proof of this statement.  I wonder if the original proof is much different.

Albanian Eagle: I think that's also what I was looking for, but it won't let me look at it unless I pay.  :(

Thanks guys.

You're at a university (or will be shortly), right?  Your school library almost certainly has an online subscription -- try poking around the library website, or go talk to a librarian at whichever library has the math books.  They'll certainly be able to get it for you.

This problem was proposed at the 2007 Brazilian National Olympiad. I didn't know it was Erdos's. So it is not a surprise that I couldn't solve it. :lol:

Anyway, the problem is discussed here: http://www.mathlinks.ro/viewtopic.php?p=957993#957993


function
number theory unsolved
number theory

Let $ (a,b,c)\in\mathbb{Z}^{3}$, define $ f(a,b,c)\equal{}(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)$.
Find all integer triples $ (a,b,c)$ such that $ f(f(a,b,c))\equal{}(a,b,c)$


topology
real analysis
real analysis unsolved

Let $K$ be a compact topological space, $X$ a metric space, and $f \in C(K \times X)$. Define
	\[g(x) = \max_{s \in K}f(s,x).\]
	Must $g$ be continuous on $X$?

The answer is yes, for any topological space $X$. Fix $x\in X$ and $\epsilon>0$. For any $s\in K$ we have $f(s,x)\le g(x)$, hence there is a product neighborhood $U_{s}\times V_{s}$ of $(s,x)$ in which $f<g(x)+\epsilon$. The set $K$ is covered by finitely many of $U_{s}$. Let $V$ be the intersection of the corresponding $V_{s}$. It is clear that $g\le g(x)+\epsilon$ in $V$. This proves that $g$ is lower semi-continuous. Upper semi-continuity is easy.


function
calculus
integration
algebra
domain
polynomial
complex analysis

Prove or disprove: if $ f$ and $ g$ are analytic on $ \Omega$ and $ f(z) g(z) \equal{} 0$ on $ \Omega$, then either $ f(z) \equal{} 0$ on $ \Omega$ or $ g(z) \equal{} 0$ on $ \Omega$.

If $ f$ is not 0 at one point, it is not 0 in a neighborhood. On that neighborhood g is all 0, which implies g is 0 on the whole region.

it came to my attention that the ring of analytic functions is an integral domain.

Is the field of fraction the ring of all meromorphic functions? Or what is the field of fraction? Is it strictly larger? What can be said about such things?

[quote="Jacobson"]
Is the field of fraction the ring of all meromorphic functions? Or what is the field of fraction? Is it strictly larger? What can be said about such things?[/quote]
It is true for entire functions. That is, the field of quotients of entire functions is the meromorphic functions on $ \mathbb{C}$. This is seen from the Weierstrass product factorization.

For other domains, I don't know, but the only possibility is that the meromorphic functions are a larger ring. Any element in the field of quotients is a meromorphic function obviously. For example, for a disc, for any meromorphic function we would like to construct an analytic function that has zeros where the meromorphic function has its poles $ (a_k)$. One approach would be to take Blaschke product, but for it to have the right properties we must require that $ \sum (1 \minus{} |a_k|) < \infty$, and I don't think this is true in general.

On the torus $ \mathbb{C}/(\mathbb{Z}\plus{}i\mathbb{Z})$, all analytic functions are constant, and the field of meromorphic functions is larger. I'm not sure about subsets of $ \mathbb{C}$.
Meromorphic functions on a connected surface are always a field.

There are several ways to construct a function with prescribed zeroes. If you don't mind having extra zeroes, the $ \overline\partial$-technique is the easiest approach. Otherwise you have to construct an infinite product which amounts to finding analytic in the domain functions $ f_k$ each of which has only one zero at the $ k$-th pole and such that $ f_k\to 1$ uniformly inside the domain. For the disk, it is a simple exercise in polynomial approximation: just approximate $ \minus{}\log(z\minus{}z_k)$ in $ |z|<|z_k|\minus{}\delta_k$ by a polynomial $ P_k$ and put $ f_k(z)\equal{}(z\minus{}z_k)e^{P_k(z)}$. For multiply-connected domains, the life is just a tiny bit more complicated. Try to figure out what happens, say, for the annulus $ 1<|z|<2$ yourself. Once you have done it, the general case shouldn't give you much trouble. :)


Does any one here listen to and/or like Asian music such as:

JPop (Japanese Pop) or JRock,
CPop, Kpop or anything else?

Please feel free to share

Yes, I'm listenning to Yuming's songs now. :lol:

We're forced to listen to some (not modern music) Japanese music in English.

Just some folk music with a "shamisen" (says Wiki).

[quote="1=2"]eh... I like traditional Asian.

You know, something mid-1070.[/quote]

So do I.

I also like Jay Chou.

hmmm

for chinese, i like jay chou!
yeah!

i listen to some japanese music...
aqua timez.

most of their songs are the openings to my anime, so im forced to listen to them every single time.
it grows on you.

Mainstream Asian music is more horrendous than ordinary white power mainstream music in my eyes. It's like an unintelligible cross between the worst elements of pop and rock.

K-pop is the worst.


The equation $ x^2 \plus{} ax \plus{} b \equal{} 0$ has two distinct real roots. Prove that the equation $ x^4 \plus{} ax^3 \plus{} (b \minus{} 2)x^2 \minus{} ax \plus{} 1 \equal{} 0$ has four distinct real roots.

Dividing by $ x^2$ ,we get

$ x^2\plus{}\frac{1}{x^2}\plus{}a(x\minus{}\frac{1}{x})\plus{}b\minus{}2\equal{}0$

$ (x\minus{}\frac{1}{x})^2\plus{}a(x\minus{}\frac{1}{x})\plus{}b\equal{}0$

Put $ t\equal{}x\minus{}\frac{1}{x}$ $ \Rightarrow$  For every real t there will exist two real x,since Discriminant of $ x^2\minus{}tx\minus{}1\equal{}0$ is

positive we get

$ t^2\minus{}at\plus{}b\equal{}0$

Since $ a^2>4b$,$ t$ will be real.Also,$ t$ can assume all real values except when x=0 and by putting x=0 in the equation

$ x^4 \plus{} ax^3 \plus{} (b \minus{} 2)x^2 \minus{} ax \plus{} 1 \equal{} 0$

we can see that $ 0$ is not the root .

Therefore,for every real $ t$ we get two real $ x$.Since we will get 2 real $ t$ therefore we will get 4 different real $ x$.Hence

the quartic has 4 distinct real roots


The three sides of a right triangle have integer lengths. The hypotenuse is 1 foot longer than the longer leg, and the shorter leg is 9 feet. What is the number of feet in the perimeter of the triangle?

There is a general form for any odd integer $ k$ as a primitive pythagorean triple : $ k$, $ \frac{k^2\minus{}1}{2}$, $ \frac{k^2\plus{}1}{2}$. For $ k\equal{}9$, we have the triple $ (9,40,41)$. The perimeter is thus $ 9\plus{}40\plus{}41\equal{}\boxed{90}$.

i think ur thinking of 5,12,13     9, 12,13 doesn't satisfy pythag


The value of $ \frac{1}{16}a^0\plus{}\left (\frac{1}{16a} \right )^0\minus{} \left (64^{\minus{}\frac{1}{2}} \right )\minus{} (\minus{}32)^{\minus{}\frac{4}{5}}$ is:

$ \textbf{(A)}\ 1 \frac{13}{16} \qquad
\textbf{(B)}\ 1 \frac{3}{16} \qquad
\textbf{(C)}\ 1 \qquad
\textbf{(D)}\ \frac{7}{8} \qquad
\textbf{(E)}\  \frac{1}{16}$

[hide="Click for solution"]
$ x^0\equal{}1$ for any number $ x$ such that $ x \neq 0$ and $ \minus{}32\equal{}\minus{}2^5$.   Thus, our expression equals $ \frac{1}{16}\plus{}1\minus{}\frac{1}{\sqrt{64}}\minus{}\frac{1}{(\minus{}2^5)^{4/5}}\equal{}\frac{1}{16}\plus{}1\minus{}\frac{1}{8}\minus{}\frac{1}{16}\equal{}\frac{7}{8}$, or $ \boxed{\textbf{(D)}}$.
[/hide]


The point $ P$ is a fixed point on a circle and $ Q$ is a fixed point on a line. $ R$ is a variable point on the circle such that $ P,Q,R$ are not collinear. The circle through $ P,Q,R$ meet the line again at $ V$. Show that the line $ VR$ passes through a fixed point.

I think this problem has been posted before, but I am not quite sure.  Either way, here is the solution (I think it works...):
[hide="Solution?"]
Take two points on the fixed circle, call them $ R$ and $ R'$.  Say that the circumcircle of $ \triangle PQR$ and the circumcircle of $ \triangle PQR'$ meet the fixed line at $ V$ and $ V'$ respectively.  Now, allow $ PQ$ to meet the fixed circle at $ M$, $ RV$ to meet it at $ N$, and $ R'V'$ to meet it at $ N'$.  We notice that $ PRVQ$, $ PR'V'Q$, $ MNRP$, $ NN'R'R$, and $ MN'R'P$ are cyclic.  Therefore, $ \angle NMP \equal{} 180 \minus{} \angle PRN \equal{} 180 \minus{} \angle PQV\implies MN\parallel QV$.  Similarly, we can see that $ MN'$ parallel to $ QV'$.  Thus, we have that $ MNN'$ is parallel to $ QVV'$, so $ M$, $ N$, and $ N'$ are collinear.  However, they all lie on a circle.  Hence, they must all be the same point.  Therefore, $ PQ$, $ RV$, and $ R'V'$ meet on the fixed circle, so their intersection is fixed.  This gives us that $ RV$ must pass through a fixed point. [/hide]


linear algebra
matrix
algebra
polynomial
integration
linear algebra unsolved

Let $A\in M_{3}(\mathbb{C})$. If $|det(A+zI_{3})|\leq 1$, for any $z\in\mathbb{C}$ with $|z|=1$, then

                                                   \[ det(A+zI_{3})=z^{3}, \]
 for any $z\in\mathbb{C}$ with $|z|=1.$ ;)

Now, it's just an ultra-classical problem on polynomials painted with a little bit of matrix stuff. Just the kind of "linear algebra" problem to be avoided.

Harazi, you always have something concerning all problems i post. I really enjoyed solving this problem.
To tell you the truth this problem is due to Dan Stefan Marinescu. :)  ;)

I have to dissapoint you: it's not Dan Marinescu who found it, it's just the first author that you probably saw in a magazine. And I'm not saying things only about your problems, but about problems with no value. I apologize for this kind of attitude, but I'm a little bit sick to see 5 times the same idea painted on some different walls and then exposed as a super-brilliant one. I like you enjoyed solving the problem, but that won't convince me too much. And please, when I say it's a super classical idea, I usually know what I'm saying and therefore I do not need biographic and auto-biographic informations about that problem or idea.

$det(A+zI_3)=det(A)+az+bz^2+z^3$

k any non-zero integer

$|2k\pi det(A)|=|\int_{0}^{2k\pi}(det(A)+ae^{ix}+be^{2ix}+e^{3ix})dx|\leq \int_{0}^{2k\pi}(|det(A+e^{ix}I_3)|) dx\leq 2\pi$

for any k non zero integer $|det(A)|\leq \frac{1}{|k|}$ so det(A)=0

for z such that |z|=1
$|det(A+zI_3)|=|z||(a+bz+z^2)|=|a+bz+z^2|\leq 1$
integrate again like above you will have a=0

$|det(A+zI_3)|=|z||(b+z)|=|b+z|\leq 1$ integrate again like above b= 0

Hi! I think that these exercices are classicals if you have already seems them. It is not the case of everybody and personnaly, I happy when I discover the Cezar's questions...  :roll:  :P

(Excuse me for my pitiful English).


[hide]Small has area of 64pi, so large has area of 128pi, which gives an radius(not area) of 8 :sqrt: 2. I'm sure you could do it better, but... Oh yes, I need to answer the question!:rotfl:  :blush:  :oops: The answer  is 10. :)  [/hide]

Billy

You did not answer my question, however.

what do you mean by  "where c is not divided by squares of prime." and isnt his answer correct?

Where c is not divisible by the square of any prime

I think Billy's answer is right.

His answer is right now but before it, he didn't find a+c.

By phrase "c is not divided by squares of prime," it means that c can't be simplified using square root.

See, dividing by at least one square of prime means that it can be simplified. :)

Hope that was clear enough.

He means that c is in simplest radical form: ie. it is not divisible by any number of the form n2.

[hide]Concentric circles where the area between big circle=little circle, little circle radius=8 so area is 64(pi) and the area between is 64(pi).
Therefore, the big circle is 128(pi).  Because the area is twice as big, the radius must be sqrt(2) times as big.  So the radius is 8sqrt(2).  THerefore, a+c=10.[/hide]

[quote="Silverfalcon"]Wording has been changed.

Consider two concentric circles. The larger circle has radius R and the smaller has radius of 8. The area between those two circles is equal to the area of the smaller circle. Radius R is equal to:

$a \sqrt c$ where c is not divided by squares of prime. Find a+c.[/quote]

[hide]the area of the smaller circle is 64 pi

the other circle is 128 pi

sqrt 128=8 sqrt 2

2+8=10[/hide]

[hide]The area of one is 64pi. They're concentric. meaning they share a common senter.
SO the other must be 128pi in area so its 8 :sqrt:2.
THey want a+b,
8+2=10
>>10<<[/hide]

[hide] 10. sqrt of 128= 8* sqrt 2.[/hide]


[b][u]The author of this posting is: romano.[/b][/u]
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Let a triangle ABC . Erect isoceles triangles : $BCA_1, CAB_1, ABC_1$ ( at $A_1, B_1, C_1$ ) . M , N , P are the midpoints of BC , CA, AB . d1 is the line through M , perpendicular to $B_1C_1$ . d2 is the line through N , perpendicular to $A_1C_1$ . d3 is the line through P , perpendicular to $B_1A_1$ . Prove that : d1 , d2 , d3 are concurrent

[b][u]The author of this posting is: grobber.[/b][/u]
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Actually, it's kind of easy if you've heard of orthologic triangles before:

This relation is true iff the perpendiculars through $A_1, B_1, C_1$ to $NP, PM, MN$ respectively concur, and this is obvious, since these perpendiculars are precisely the perpendicular bisectors of $BC, CA, AB$ respectively, which concur at the circumcenter of $ABC$.

[b][u]The author of this posting is: romano.[/b][/u]
_____________________________________________________________

Can you tell me about orthologic triangles ? Now , I can't understand why "This relation is true iff the perpendiculars through A1 , B1 , C1 to NP , MP , NM respectively concur"

[b][u]The author of this posting is: grobber.[/b][/u]
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This is a theorem (it may be called [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160533]the Orthologic Triangles Theorem[/url]):

Let $ ABC, MNP$ be two triangles. Then the perpendiculars through $ M, N, P$ to $ BC, CA, AB$ are concurrent iff the perpendiculars through $ A, B, C$ to $ NP, PM, MN$ are concurrent, in which case the triangles $ ABC, MNP$ are called orthologic triangles.

You could look for more info on the Internet.


ARML
linear algebra
matrix
probability
expected value

Suppose I have a bag with 12 slips of paper in it.  Some of the slips have a 2 on them, and the rest have a 7 on them.  If the expected value of the number shown on a slip randomly drawn from the bag is $ 3.25$, then how many slips have a 2?

Well... there's an easier way to do this.

First we note that this is essentially an algebra problem in disguise. Upstate 2009 ARML team, think back to Coach Reuter's hilarious lampooning of word problems. In particular, this is a "I have k dimes and p quarters" problem hidden very sneakily, but in an obvious way.

We can first note that if the expected value is 3.25, by the definition of expected value the average of the numbers of the slips must also be 3.25.

So we set up a system of two equations, where k is the number of slips of paper with a 2 on it and p is the number of slips of paper with a 7 on it-

k + p = 12
2k + 7p = 39 (Since we can multiply both sides of the average by 12.)

After this, have at it- you can go for matrix algebra if you like (GAH) or you can just rearrange (1) to be p = 12 - k, then plug that into (2).

Well, we can solve it easier i think(though its easy enough as it is).
Take 2 off any of the numbers and divide by 5, making them 0 & 1 instead of 2 & 7. apply the same on 3.25, and we get 0.25 = 1/4. so, 1/4 of the 12 had 7 on them, which means 3/4*12=9 had 2 on them.

You could also just say that you have twelve slips and 12*3.25 is 39. Then just guess and check your way thought it. It won't work on larger, more complicated probs, but for this it works just fine.

[quote=franzliszt]The easiest way to do this is read Intro to C+P Chapter 11.3. (Sorry for the bump, but why does Alcumus use problems from the text???)[/quote]

True. This is also in Intermediate C&P Chapter 1, Review of Counting and Probability Basics.

[hide=my sol]Let $x$ denote the number of 2-slips. Thus, there are $12-x$ 7-slips. Then, the expected value is $\frac{2x+7(12-x)}{12}.$ We know this is equal to $3.25,$ so now we have the equation $$\frac{2x+7(12-x)}{12}=3.25.$$ Solving gives $x=\boxed9.$

[hide=Waiting]
3.25 - 2 = 1.25
3.25 - 7 = 3.75
Thus there are 3 times as many 2s.
This gives 9.
[/hide]


algebra
polynomial
function
algebra unsolved

Find all polinomial function $ f(x) \in \mathbb{R}[X]$ s.t. $ \forall (a,b,c) \in \mathbb{R}^3$ with $ ab\plus{}bc\plus{}ca\equal{}1$, the following equation holds:
$ f(a\minus{}b)\plus{}f(b\minus{}c)\plus{}f(c\minus{}a)\equal{}2f(a\plus{}b\plus{}c)$

good work  :wink:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=99448#99448]IMO 2004 problem 2[/url]

EDIT: Sorry, didn't notice the $ 1$ and $ 0$ switching.

MellowMelon, i already know IMO2004, in fact this problem seems similar..  :wink:
[edit: don't worry..]


abstract algebra
superior algebra
superior algebra unsolved

Suppose [img]http://lh4.ggpht.com/_miqRA1rlzTU/ST3ktnjsdiI/AAAAAAAAACQ/1FPLxuOS9Yw/s144/cd.PNG[/img] is a commutative diagram of $ R$-modules and $ R$-module homomorphisms.
(a) Suppose that $ \phi$ is injective. Prove that the map $ \phi_0 : \text{Ker } f \rightarrow \text{Ker } g$ is injective.

(b) Suppose that $ \psi$ is surjective. Prove that the map $ \bar{\psi} : \text{Coker } f \rightarrow \text{Coker } g$ is surjective.


We have just learnt some trig identities in my math class and i want to know what they are and what they are used for.

$\displaystyle\sin{x}=\frac{2\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}$

And there are two more identities similar to that one, they express $\cos{x}$ and $tg{x}$ in terms of $tg\frac x2$

Yes. We can express $\sin x$, $\cos x$ and $\tan x$ in terms of $\tan\frac{x}{2}$.

The formulas are:

[hide]
$\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$

$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$

$\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}$

[/hide]

I use this formulas in order to obtain algebraicaly identities by substituting $\tan\frac{x}{2}$ with a constant ($t$)  ;)

ok thanks Slizzel.
And, do the three formulas have any other names??

[quote="Psycho"]normally, they're referred to as the "t-formulas"[/quote]

what does the "t" stand for??

I would guess the t represents tangent. Probably t = tan(x/2) in these cases. I have seen this before, with the tan(x/2)'s replaced by t's, so the formulas are just simple functions of t.


Λίγη βοήθεια στην παρακάτω άσκηση: Να αποδείξετε ότι στον αλγόριθμο του ευκλείδη τα υπόλοιπα τουλάχιστον υποδιπλασιάζονται μετά από δύο βήματα, δηλαδή ισχύει:
$ r_{i\plus{}2}<\frac{1}{2}r_{i}$

Ίσως να είναι εύκολη αλλά δεν μπορώ να σκεφτώ κάτι... :blush:

Έλεγξε τι γίνεται αν $ r_2 > r_1/2$.

Τότε υποχρεωτικά  θα είναι $ r_{3}<r_{1}/2$
Ευχαριστώ :lol:


I just forgot how to prove this. I am unsure if there is a better way:

Prove that $ \lim_{x\rightarrow\infty}\frac{1}{x} \equal{}0$.

(Using the $ \delta$-$ N$ definition of such a limit).

Now I'm not sure, but say we choose $ N\equal{}\max\left( 0,\frac{1}{\epsilon}\right)$.

Thus $ x>\frac{1}{\epsilon}$ and $ x>0\implies 0<\frac{1}{x}<\epsilon$. So $ \left| \frac{1}{x}\minus{}0\right|<\epsilon$, so we're done.

The thing that I am unsure about is if you are allowed to take such an $ N$.

Is this proof correct? It is also a but unclear to me what functions are allowed for $ N$ and what functions are not allowed.

You can take such a function, but since $ \epsilon>0$ by standard assumption, $ N\equal{}\max\left(0,\frac1{\epsilon}\right)$ is silly. Just say $ N\equal{}\frac1{\epsilon}.$  Except that that probably isn't an integer. So we actually say this: choose $ N$ to be an integer such that $ N>\frac1{\epsilon}.$ We know we can do that because of the Archimedean property of the real numbers.


logarithms
probability
geometry
geometric transformation
reflection
probability and stats

Let $ \xi_1,\xi_2,...$ be independent, identically distributed random variables with distribution \[ P(\xi_1=-1)=P(\xi_1=1)=\frac
12 .\] Write $ S_n=\xi_1+\xi_2+...+\xi_n \;(n=1,2,...),\ \;S_0=0\ ,$ and \[ T_n= \frac{1}{\sqrt{n}} \max _{ 0 \leq k \leq n}S_k .\] Prove that $  \liminf_{n \rightarrow \infty} (\log n)T_n=0$ with probability one.

[i]P. Revesz[/i]

reflection principle says that $ P[max_{k \leq n} S_k \geq a] \equal{} 2 P[S_n \geq a]$, hence Borel-Cantelli lemma gives the conclusion in no time.


superior algebra
superior algebra unsolved

is there a monic multilinear map $f : \mathbb{R}^\mathbb{N}\to \mathbb{R}$, i.e. for all $\lambda , a_{i}, a'_{i}\in \mathbb{R}$ we have

$f(1,1,...) = 1$ (or weaker: $f \neq 0$)
$f(a_{1}, ... , a_{n-1}, a_{n}+a'_{n}, a_{n+1}, ...) = f(a_{1}, ... , a_{n-1}, a_{n}, a_{n+1}, ...)+f(a_{1}, ... , a_{n-1}, a'_{n}, a_{n+1}, ...)$
$f(a_{1}, ... , a_{n-1}, \lambda a_{n}, a_{n+1}, ...) = \lambda f(a_{1}, ... , a_{n-1}, a_{n}, a_{n+1}, ...)$ ?

in particular, we would have $f(2,2,...,2,1,1,...)=2^{n}$, so that $f$ can't be continuous. i guess the existence of such a $f$ would employ zorn's lemma, but i have no idea :-)

Can't we just set $f(a_{1}, a_{2}, ...) = a_{1}\cdot a_{2}\cdot ...$ iff only finetely many $a_{i}$ are $\neq 1$ and $f(a_{1}, a_{2}, ...)=0$ otherwise¿

yes! unbelievable that's is so simple :-D