Old thread (I presumed locked because of spam): http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273415

http://healthyliving.freedomblogging.com/2009/04/30/still-no-swine-flu-in-oc-health-officials-ready-with-antivirals/5557/
http://www.ocregister.com/articles/health-county-state-2384745-officials-swine

Seeing as a senior from University High, [i]the high school I attend[/i], has been affected, I have a personal interest in continuing intelligent discussion...

The article mentions that any spread of the disease would have presented itself by now, but my friend seems to have started to show flu-like symptoms today. Sure, it's an inaccurate diagnosis brought on by some sort of panic, but it's still not a good sign.

Maybe there's an overreaction to all of this, but to cite the example of the great flu pandemic in 1918, the disease started off mild and worsened, and even them, most cases were mild, despite the massive deaths across the globe.

Eh, I guess I have a greater reason to be worried now. Any feedback to this?

At this point, either the disease will stop on its own or nothing will stop it. Isolation of patients, school closings, and so on, cannot possibly stop the spread. They serve another purpose: hopefully spreading out the peak of the epidemic so as not to overwhelm medical and public health resources.

We were told that a dorm-living student at CSULB fell ill on Sunday, and as of Tuesday evening was listed as a "probable positive" for the swine flu. And that that student's case was fairly mild. I haven't heard one way or another about the further tests on that student. And I haven't heard anything else, and students aren't at this point rampantly disappearing from classes. I think we'll probably have a pretty good guess about the speed of further spread by next Monday.

It might stop on it's own because it's May and almost summer. That would still be something to potentially worry about, because the disease would still be there, widely disseminated, waiting for next winter to re-emerge.

There are two different aspects to an epidemic disease: infectivity and virulence. What always makes flu a concern is its very high infectivity - it's easy to catch and spread. And a completely novel strain - as this one appears to be - can have spectacularly high infectivity (as I recounted from my long-ago experience with the "Hong Kong flu" pandemic). As for its virulence - how bad a disease it causes - the Mexican data is spotty and contradictory, but in its spread outside of Mexico, its virulence doesn't seem to be outside the bounds of the endemic yearly flu. Of course, that's bad enough, estimated as killing more than 30,000 each year in the U.S. (mostly the very old, the very young, and the already-sick.) The H5N1 "bird flu" that's been around for a decade has nasty virulence - but seems to be lacking human-to-human infectivity.

The pork farmers made the WHO and CDC change the official name from "swine flu" to "H1N1 Type A Influenza Strain" because it was harming their sales. Hee hee... Does anyone know whether it does comes from from pigs or not? Like, if you didn't properly cook some pork and ate it, what would happen?

Not properly cooking pork is a bad idea because of many diseases. Flu isn't one of them; eating pork won't get you sick.

This strain does also infect pigs, and probably came from a factory farm. Lakes of feces and enormous numbers of pigs crowded together, all with swarms of flies, are a perfect environment for strains of the virus to mix and evolve. The flies spread it to nearby humans, and we get an outbreak.

[color=green]Moderator says: this post was made in response to a now-deleted post.[/color]

Let's put to rest that rumor:

[quote="mathcounts.org"]National Competition Update(4/30)

 In light of the news surrounding the Swine Flu in the United States and internationally, MATHCOUNTS would like to ensure all individuals planning to attend the 2009 Raytheon MATHCOUNTS National Competition that, at this time,  there are no plans to cancel or reschedule the event.  We will continue to monitor all regularly updated information and will follow the guidelines set forth by the Centers for Disease Control in order to provide as safe and healthy an environment as possible for all students, coaches and guests attending next week’s competition.

Should the status of the competition change, MATHCOUNTS will notify all attendees via e-mail and on the MATHCOUNTS Web site at http://www.mathcounts.org, as soon as possible.  In the meantime, individuals are encouraged to follow the CDC’s recommended actions to stay healthy. [/quote]

It's terrible that this disease has killed so many people so far, and it could become a real danger.  But let's put things in perspective.  So far, the flu is suspected to have killed 168 people worldwide, and only 16 of these deaths are confirmed, according to [url=http://en.wikipedia.org/wiki/2009_swine_flu_outbreak]Wikipedia[/url].  By contrast, more than 41000 people in the USA alone die in car crashes each year, according to the [url=http://www-nrd.nhtsa.dot.gov/Pubs/811017.PDF]National Highway Traffic Safety Administration[/url].  That's an average of more than 110 people every [i]day[/i].

Of course, it is quite possible that this illness will start claiming lives much more rapidly, but so far, it hasn't.  Remember how worried people were about bird flu a few years ago?  According to the [url=http://www.who.int/csr/disease/avian_influenza/country/cases_table_2009_04_23/en/index.html]WHO[/url], the epidemic has taken 257 lives worldwide to date.  That is a lot of people, but in all probability, more people died in traffic accidents in the past three days.  I hope I'm not treating human life too lightly, but I just think that right now, certain everyday things should scare us a lot more than swine flu.

The number of new cases has actually slowed down recently.

[quote="JBL"]Any future posts of the form, "I heard ..." without any supporting documentation get deleted.  Any posts of questions that could be answered by looking at the website of the [url=http://www.cdc.gov/h1n1flu/]CDC[/url] or [url=http://www.who.int/csr/disease/swineflu/en/index.html]WHO[/url] get deleted.  Reposts of posts violating the preceding sentences get deleted.  Complaints posted here about any of the preceding get deleted.

As Brut3Forc3 noted, there was a reason the previous thread got locked, and I don't plan to have this one taken over by middle schoolers explaining how much smarter they are than their little sisters, or whatever.  Informed, legible, on-topic posts are (as always) welcome.[/quote]

I'm sorry. I thought I didn't press submit. I truly didn't know. I typed it the first time at school and the school computers are sometimes glitchy (such as the times.). Sorry again!  :oops:  :blush:

Here's a really good overview of the "state of play" from the point of view of a public health professional:
http://scienceblogs.com/effectmeasure/2009/05/swine_flu_the_overreaction_ove.php

I just received an email from the president of my university in which we learn that the one suspected case of swine flu on my campus is now a confirmed case - and that all other tests so far performed of our students have been negative.

So this will show up in state and national statistics as a new confirmed case - but note that it is already old news. The student fell ill a week ago, and is reported to be "recovering."

There's no essential change from the message he sent Friday, which included, "Because no new cases have been identified, CSULB will operate on a business as usual basis, with classes continuing to be offered."

Read this article...
[url=http://www.cnn.com/2009/HEALTH/04/28/swine.flu.mexico/]Why only deaths in Mexico[/url]

Flus can't kill many people by themselves. What really kills people is the complication that follows the flu. By far no non-Mexican has died from influenza A (the single person who died is US was a Mexican immigrant). I am not an expert on this topic, but is the world overreacting do this stuff?

Despite common belief, this flu virus kills healthy, young people who has developed sufficient immune system. The main reason these people die from influenza A is that their inexperienced immune system overreacts and kills the normal body cells in an attempt to eradicate the virus. Young Mexicans have not been immunized sufficiently in comparison to Americans, resulting in a slaughter.

Also, please do not call this swine flu! A bunch of media people simply heard that the genome of the virus had a strand commonly found in pig viruses and labeled the virus "swine flu." Recent research has found no incidents of a pig becoming inflicted with the H1N1 Type A Virus. The flu also prefers human-to-human transmission-again no pigs involved. However, the media likes the erroneous name better because the official name feels "awkward." Silly, but that's how media earns money. All around the world, pig farmers are losing business because people think eating pork would make them sick. Let's at least try to help WHO and CDC in their efforts to clear misconception!

[quote="Aqua3993"]Recent research has found no incidents of a pig becoming inflicted with the H1N1 Type A Virus.[/quote][url=http://news.yahoo.com/s/ap/20090503/ap_on_he_me/med_swine_flu_people_to_pigs]False[/url]. Pigs can get it from pigs, humans can get it from pigs, humans can get it from humans, and pigs can get it from humans. At this point, it is both a swine flu and a human flu.

Still, the new evidence over the last few days indicates that we can relax: it appears to be harder to spread among humans than our usual seasonal strains, and merely as virulent. We're not likely to see large fractions of the population all get sick at the same time in the near future.

On that reaction, a few schools in Seattle closed last week in response to a few cases. They reopen tomorrow.

[quote="Aqua3993"]Still this flu should not be a reason not to eat pork or slaughter a bunch of pigs (happened in some middle eastern countries).[/quote]
[url=http://blogs.siliconindia.com/manojtiwari/Beware_of_Swine_flu_most_dangerous_flu_ever-bid-OVm0y1v334274924.html]Swine Flu Faqs[/url]
True; that, however, is a cultural case. Because of religous and cultural standards, most Middle Eastern countries already ban the consumption and sale of pork and pigs; to them, this 6th outbreak of swine flu is yet another reason to do so.
One should avoid eating pork regardless; pork is a very unhealthy and possibly dangerous food to begin with.
[url=http://en.wikipedia.org/wiki/Pork#Potential_health_risks]Pork-Potential Healthrisks[/url]

[quote="Eragon"]One should avoid eating pork regardless; pork is a very unhealthy and possibly dangerous food to begin with.
[url=http://en.wikipedia.org/wiki/Pork#Potential_health_risks]Pork-Potential Healthrisks[/url][/quote]
Sorry, but that's just wrong. Of the mentioned reasons, only one (!) isn't caused by uncooked meat...
And banning any kind of meal for religious reasons is stupid. But I accept a lot of other reasons like cruelty, unhealthyness, or economical/-logical reasons.

EDIT: I will no longer backseat mod. Sorry for being offensive if I seemed so to anyone :blush:

As well: First severe case of swine flu in Canada occurred to an Edmonton girl. She is recovering in hospital and breathing on her own.

[quote="Newanda"]Religious practices shouldn't be insulted.[/quote]
Are you joking? We have the right to criticize anything we want. Religion included.

Anyways: back on topic

Two students at a nearby college have it now. :( http://www.fairfield.edu/index_notice.html

[quote="Newanda"][quote]And banning any kind of meal for religious reasons is stupid.[/quote]

Please retract this statement. Religious practices shouldn't be insulted. I am sure many Muslims could explain to you why their cultural beliefs aren't "stupid."[/quote]

I'm waiting for their explaination, same for e.g. jewish people...
And I surely will not retract any statement for a reason like "religion must not be criticized".

[quote="Ihatepie"][quote="Newanda"]Religious practices shouldn't be insulted.[/quote]
Are you joking? We have the right to criticize anything we want. Religion included.[/quote]

We may have the right to criticize anything, but on a forum such as AoPS, mods have a right to, well, moderate. Don't criticize other religions. /backseat modding

In other news, [url=http://www.youtube.com/watch?v=W57pnVhEhjU]this[/url] and [url=http://www.youtube.com/watch?v=kW6FF63H4xc&feature=related]this[/url] give good reasons why swine flu control should happen now, not when it becomes a major pandemic.

[quote="crazypianist1116"]We may have the right to criticize anything, but on a forum such as AoPS, mods have a right to, well, moderate. Don't criticize other religions. /backseat modding[/quote]  Backseat moderating is fine if somewhat pointless when the moderator is in agreement with it.  When the moderator [i]doesn't[/i] agree with it, I'm not sure what to say about it, but it does come off as rather silly.  Religious beliefs aren't more special than any other kind of belief, and should be subject to exactly the same level of scrutiny -- in particular, there's no reason at all that they should be above criticism.  Absurd positions taken because of a professed religious belief aren't any better than absurd positions taken for any other reason, and objections to the name "swine flu" surely fall into the category of absurd positions.

No more discussion on this tangent in this thread, please.

I don't really hear news about the Swine Flu anymore.

I also have a friend that went to Mexico but didn't come back because of the border being closed..  :|

[quote="Sabien5"]I have the swine flu, it's really not that bad.[/quote]

 :o , what is the treatment ?, Im from Nebraska and a friend of my have H1N1..I'm really scare of that, because i was with him two days ago..

[quote="Sabien5"]he probably had a really high fever, like me-104F.  All i really did was sleep, eat and absorb vitamin c and drink warm water.WARM[/quote]
...and go on AoPS

I thought there was only, like 40,000 people infected in the whole world, and with something like 140 dead. Do you know where you got if from? I was traveling south and went through big, hub, crowded airports. I went outside all the time and I was in the downtown of a big city. I was careful not to touch things, like the railing on stairs, and washed my hands every flight/trip. It was scary going there.

I live in a city near the border of Mexico (it's not THAT close, but closer than Phoenix, Arizona)
Some of my teachers were talking about washing your hands after you use the bathrooms, etc. It doesn't seem like it's bad, but when your town is infected, it's a big deal. No one thought that it would hit my city/county, but it did.
It's really scary--I read it could come back even worse.

[quote="worthawholebean"]H1N1 is really no more dangerous to an individual than standard seasonal flu. Most of the hype is more mass hysteria than anything else.[/quote]
This does not seem to be true (if you just look at published mortality rates anyway). Nevertheless, worrying about this out of your own personal safety seems quite silly. If you are the parent of a very young child then taking precautionary measures might be in order (but so should you for ordinary colds).

@Kalle: What do you mean? If you can minimize contact with a very big crowd to prevent an illness that might cause death, YOU SHOULD! I come from a city of less than 70,000 people, and my state has only had a few cases.

What's so special about illnesses? You can die in a plethora of ways by leaving your home. Are you currently living your life in a way that optimizes life expectancy?

I think the cost of avoiding people is bigger than the (expected) cost of potentially dying of swine flu.

What I mean are the small things. Like, if you are climbing stairs, you don't have to drag your hand on the handrail. If you just shook hands with some people at a meeting and were about to eat, you could wash your hands first.

[quote="Kalle"]If you are the parent of a very young child then taking precautionary measures might be in order[/quote]  Well, the two things that make this flu interesting seem to be that (1) it is killing or hospitalizing people who are not just young children or the elderly, but rather teenagers and others who almost never die from seasonal flus, and (2) it is clearly not seasonal.  These are both far from normal behaviors, and my sense is that they have driven much of the concern from the epidemiological community.

Just because a news-media hysteria has died down does not mean serious concern was not warranted initially, nor even that serious concern is not warranted [i]now[/i].  The possibility of a flu that could kill hundreds of millions is not fantastical -- it was proven less than a century ago, in a world in which global spread was much more difficult than it is in our own.  That newspaper reports (and worse, anything that gets on cable TV "news") are invariably somewhat sensationalistic is not particularly surprising, but [url=http://scienceblogs.com/effectmeasure/]people who think about public health for their living[/url] are still concerned, and it has reached well over 70 countries and been declared a pandemic.

The cost of avoiding people depends an awful lot on your situation -- in particular, in a world in which a reasonably infectious and deadly version of the flu is floating about, it could quite possibly be more cost-effective for people who wake up with a headache and cough to stay home for a day rather than go into work or school (and in fact this may even be true in a world without pandemic flu).

I have virtually no incentive to post in this topic, since I am a home-schooled Upstate New Yorker.
I was wondering; theoretically, how fast can H1N1 mutate? If it can mutate into something deadly overnight, then I can understand WHO's concern.

Upstate New York is not exactly swine flu-free:
http://www.health.state.ny.us/diseases/communicable/influenza/h1n1/docs/state_h1n1_cases_by_county.pdf
(from the NYS DOH [url=http://www.health.state.ny.us/diseases/communicable/influenza/h1n1/index.htm]swine flu page[/url] -- for whatever reasons, Onondaga and Erie counties seem particularly hard-hit).  Also, it [url=http://www.cdc.gov/h1n1flu/update.htm]already is deadly[/url] in the usual meaning of the word: at least 45 people in the US have died.  (Admittedly, it seems that risk from being a [url=http://www.who.int/mediacentre/news/releases/2009/road_safety_report_20090615/en/index.html]bicyclist or pedestrian[/url] is quite a lot higher :).)  My sense (from reading, for example, in the link I left in my previous post) is that mutations for infectiousness or deadliness could happen at any time, but that this is an entirely random process of rather low probability (so it's not possible to predict when such a mutation might actually take place).  In fact, my girlfriend (an immunologist) suggests that it might be better to get the flu now, when it probably won't kill you, so your body can develop the antibodies to protect you if it does become more virulent --somewhat analogous to getting a flu shot.  (On the other hand, she also thinks I should avoid places with lots of infected people.)

A good FAQ page from the WHO is here: http://www.who.int/csr/disease/swineflu/frequently_asked_questions/levels_pandemic_alert/en/index.html

JBL, you may be right for all I know. It's just really hard to believe without numbers. So what would a statistical estimate of the fatality rate in healthy adults be? What is the risk that I will die, and how will this be influenced if I avoid railings? How does it stack up to the fact that I sometimes drive tired and that I am a smoker (every other factor of potential death becomes irrelevant as a smoker  :P)? To just say that a disease is potentially deadly (as all diseases are) just doesn't mean squat. It is my feeling that statistics are not presented in media because something like a confidence interval will be misinterpreted. Similarly, even though experts like your girlfriend might suggest washing your hands often and avoid shaking hands, I suspect this is just theoretical advice. The only research I can find into whether an increase in hand-washing helps to a significant degree are tested on groups of people not including me, like in very poor countries or in the health sector. My unscientific guess is that avoiding door knobs will not help you at all.

I am probably technically wrong, but I am worn out by the fact that in sweden one of the biggest source of news are single studies that show that you should not touch handles on carts in grocery shops, not eat this, not drink that, exercise in a particular way, blow your nose in the correct weather direction etc. Things have to be put in a much larger perspective than just a study that shows handles on carts have harmful microbes.

On that note, I was just sick for the first time in four years a few weeks ago, most likely with some type of influenza. I had 39+ Celcius body temperature for several days, and was to some extent sick for over a week. When I get the swine flu I will probably forget everything I just said and curse the gods.

@JBL: Yay my county only has one case!
@Kalle: Get better soon.

Another question: Would getting fit if I am not right now be a good idea?

Right now the only times that I may get H1N1 are the four tests I have to take today, tomorrow, and the next day, Welcome Weekend for Jack Kent Cooke Foundation, and AwesomeMath. The last two are approx. a month away.

[quote="Kalle"]It's just really hard to believe without numbers. So what would a statistical estimate of the fatality rate in healthy adults be? What is the risk that I will die, and how will this be influenced if I avoid railings? How does it stack up to the fact that I sometimes drive tired and that I am a smoker (every other factor of potential death becomes irrelevant as a smoker  :P)?[/quote]  Well, have you read any of the links I've posted?  In the US, there have been more than 17,000 confirmed cases and 44 deaths (they corrected the earlier 45 -- apparently no one is South Dakota has died from swine H1N1 after all), which means that the answers right now are "small," "small at present," and "small."  But these are all consistent with what I've said.  You are almost certainly shortening global life expectancy more by smoking (your own life) and driving (your own and those of all the pedestrians and bicyclists nearby) than you would by not washing your hands after you sneeze.

Something like 1-2 million people are killed yearly by drivers world-wide (from a WHO link in my previous post).  If the US death rate as recorded by the CDC numbers really is representative of the death rate world-wide, 1.5 million deaths is about what you would get if swine flu infected 1 in 10 people, which so far looks to me not incredibly likely.  But a slightly more contagious slightly more deadly version of the same thing could easily rack up tens or hundreds of millions of deaths.  I would much, much, much rather have a small amount of overreaction every time an infectious disease pops up that seems both contagious and deadly than I would have an underreaction when the next killer flu evolves.  I think that's the relevant perspective from a public health/potential pandemic view (which is quite different from the "personal health and wellness" view that you mention).

[quote="1=2"]Another question: Would getting fit if I am not right now be a good idea?[/quote]  Probably, but not for any swine flu related reason :)

[quote="JBL"][quote="1=2"]Another question: Would getting fit if I am not right now be a good idea?[/quote]  Probably, but not for any swine flu related reason :)[/quote]

Crap. How am I going to get dad to exercise?

I know that I've been asking more questions (2 or 4) than giving answers (0), but I just have a couple other miscellaneous questions:

Will any conspiracies be unearthed because of H1N1?
Who would stand to benefit from this pandemic if
(a) it dies out harmlessly?
(b) it kills many before it does out?
(c) it wipes out most of humanity?

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding your own.[/color]

The answer to c is of course the mass conspiracy of people who plot to kill off humanity so they can take over and rule it in their own way. That or the mole people.

does anyone know if there are cases in Manhattan? would it be safe to visit?

If every single person with swine flu piled into Manhattan it would be safe to visit. Don't worry about it.

Somewhere earlier in this thread I read not to say "swine flu" but say "H1N1" but I think that that that just makes people panic because it sounds scientfic and scary.  :mad:


greatest common divisor
function
algebra
floor function

1. I am thinking of a 4 digit number. My 4 dgit number is peculiar because when the sum of the sqaures of the dgits is subtracted from itself, the result is 2009. Find my 4-digit number.

2. The numerator of a fraction is 15 more than the 1/3 the denominator. If the GCF of the numerator and denominator is added to both of them, the resulting fraction is eqaul to 4/5. Find the orig fraction.

3. Find the value of GIF(1og 2009 base 2) +GIF(1og 2009 base 3)+ GIF(1og 2009 base 3)+.............+ GIF(1og 2009 base 2009). (greatest integer function)

[hide="1"]2054-i just tried all the numbers from 2010 up... (we know the answer couldn't be that large)[/hide]

Somewhat more systematically is to say $ 1000a \plus{} 100b \plus{} 10c \plus{} d \minus{} a^2 \minus{} b^2 \minus{} c^2 \minus{} d^2 \equal{} 2009$, where a,b,c,and d are the digits. It is easy to see that a must be 2, then b must be 0. From here it's brute force, we find c=5 d=4.

For 2, since the numerator and denominator are integers, the fraction is $ \frac {x \plus{} 15}{3x}$. We guess x=12 and find it works. To prove there are no other solutions, we know $ gcd(x \plus{} 15,3x) \equal{} k \le x \plus{} 15$, so $ \frac {4}{5} \equal{} \frac {x \plus{} 15 \plus{} k}{3x \plus{} k} \le \frac {2x \plus{} 30}{4x \plus{} 15} \implies x \le 15$, and none other up to 15 work.

2. [hide]
x/y

x-15=y/3
x/y=4/5, since the original fraction will be very close to 4/5

x= about 26

so, x must be 27 or 28...testing out 27 shows that it works

thus, the original fraction was 27/36[/hide]

3. [hide]2 gives you 11, 3 gives you 7, 4 gives you 6, 5-6 give you 5, 7-12 give you 4, 13-44 give you 3, 45-2008 give you 2, and 2009 gives you 1

so...

1*11
1*7
1*6
2*5
6*4
32*3
1964*2
1*1

add those up and you get 4083, our answer[/hide]

not with the greatest integer function...with the least integer function you get the answer you just said

also I believe your answer's off by 1 or 2...not too sure because I didn't actually do it out your way


Let G be a graph and let X and Y be subsets of V(G).
We say X and Y are linked in G if |X|=|Y| and there are |X| vertex disjoint (X,Y)-paths.
Let Z be a subset of V(G). Z is a (X,Y)- detaching set if G-Z has no (X,Y) paths.

[b]Lemma 1[/b]: Let d be the size of a minimum (X,Y)-detaching set in G. Then there exists a d-subset X' of X and a d-subset Y' of Y such that X' and Y' are linked.

Assuming that Lemma 1 is true, prove Menger's Theorem using Lemma 1 and prove Lemma 1 using Menger's Theorem.

Let x and y be vertices of G, then:
[b]Menger's Theorem[/b]: max{|P|: P is a set of inner disjoint xy-paths}= min{|S|: S is a vertex (x,y)-separator}


number theory
greatest common divisor
number theory solved

I  found this one on an incomplete paper which had only the year:1997.

the sequence: x_0=2,x_(n+1)=2.(x_n)^2 - 1 is considered.
prove that for all n in N we have:
                 gcd ( n , x_n )=1

It is very hard .
   We can if n=k is true then n=k+1 is true
  (n,x-k)=(n,[x-k] 2 )=(n,2[x-n] 2 )=1
   => (n,x-(k+1))=1  :D

[quote="Thanhliem"]It is very hard .
   We can if n=k is true then n=k+1 is true
  (k,x-k)=(n,[x-k] 2 )=(k,2[x-n] 2 )=1
   => (k+1,x-(k+1))=1  :D[/quote] .sorry :(

Well, this problem follows easily from the much harder problem in topic Sequence with strange property.

Yeah. Basicly, that problem states hat the primes which divide xn are greater than n.


function
real analysis
real analysis unsolved

[quote="puuhikki"]Does there exist a sequence of continuous functions $ (f_i)_{i\in \mathbb{N}}$, $ \mathbb{R}\to \mathbb{R}$ such that
1. $ \mathbb{R}\setminus\mathbb{Q} \equal{} \{x\in\mathbb{R}: \operatorname{sup}|f_i(x)| \equal{} \infty\}$?
2. $ \mathbb{Q} \equal{} \{x\in\mathbb{R}: \operatorname{sup}|f_i(x)| \equal{} \infty\}$?[/quote]

As stated this would imply $ \mathbb{R} \backslash \mathbb{Q} \equal{} \mathbb{Q}$.  I'm assuming there's some typographical error.

i think the sequence $ f_i$ can vary from one problem to the next :)

My mistake. :blush: There are two questions.

1. Does there exist a sequence of continuous functions $ (f_i)_{i\in \mathbb{N}}$, $ \mathbb{R}\to \mathbb{R}$ such that $ \mathbb{R}\setminus\mathbb{Q} \equal{} \{x\in\mathbb{R}: \operatorname{sup}|f_i(x)| \equal{} \infty\}$?
2. Does there exist a sequence of continuous functions $ (f_i)_{i\in \mathbb{N}}$, $ \mathbb{R}\to \mathbb{R}$ such that $ \mathbb{Q} \equal{} \{x\in\mathbb{R}: \operatorname{sup}|f_i(x)| \equal{} \infty\}$?

The questions are so similar that I thought it is useless to start two topics. So when solving 1., don't look at the question 2. and vice versa.


Please respect that the Round Table portion of this message board is meant for serious discussion.  The [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=139]Fun Factory[/url] board is the appropriate forum for jokes & basic socializing & lighter fare.

The moderators of this forum will move appropriate threads to the Fun Factory forum & will remove distracting posts from threads as necessary.

Knights of the Round Table, hearken to your moderators...

The Round Table is indeed a place for serious discussion. This does not mean that this board has to be deathly serious. Discussion can be easygoing - but no spamming or IM-ish conversation. Or ziggurats. Go to Fun and Games for that. Still, welcome everyone and post away. =)

New rule: Any whining will be ruthlessly edited or deleted at moderator discretion. You have been warned.

If I get four complaints about a particular post, set of posts, or thread, it will be deleted, editted, or locked, whichever is most appropriate and easiest.

Don't post several times in a row on any one particular thread, unless you are posting highly constructive things. If you post 3 or 4 times in a row when it could have easily been combined into one post, it will likely be deleted. Direct complaints about this policy to the edit button located near the bottom right hand corner of your posts.


A band has four trumpets, three trombones, three saxophones and two drums.  If each section has its own set of seats, in how many ways can the 12 band members be seated?

 \begin{tabular}{|c|c|c|} \hline
  trumpets & trombones & drums \\
  $ \Box$ $ \Box$ & $ \Box$ $ \Box$ $ \Box$ & $ \Box$ \\ \cline{2-2}
  $ \Box$ $ \Box$ & $ \Box$ $ \Box$ $ \Box$ & $ \Box$ \\
   & saxophones & \\\hline
 \end{tabular}

Um all we have to do is multiply how many rearrangements can be made or
4! * 3! * 3! *2 or 24*36*2 or $ \boxed{1728}$

[quote="isabella2296"]Just a note on factorials:

For an integer $ n$, $ n!$ is $ \boxed{n \cdot (n \minus{} 1) \cdot (n \minus{} 2) \cdot ... \cdot 2 \cdot 1}$.

So, for example, $ 4!$ would be $ 4 \cdot 3 \cdot 2 \cdot 1$, or $ 24$.

The $ !$ is called [b]factorial[/b]. $ 4!$ is pronounced $ 4 \text{ factorial}$.[/quote]

my suggestion for you is to

[b]not try to teach people about this stuff unless they ask for it[/b]

no offense, but this can get very annoying to some people, so keep it to yourself unless you are asked by someone to help them or somehtin like that  :wink:


what are the subjects above "passed" section in your report?

Those that you are currently most working on.

but for one of the subjects i only answered one and i got it right

one problem isn't really reliable to see if you passed or not...
when i first started out, a lot of subjects were really good, but were still in the not passed section
as you do more problems, alcumus will have a better idea of where your abilities are and then it will tell you if it's passed or not


and why do I have to expect something to happen?

Did you make it through? Hint: right click your way through to teleport the square.

what was that? that creepy face was creepy  :o  lol it was ez to get through though...


topology
function
analytic geometry
combinatorial geometry
Functional Analysis
real analysis
real analysis unsolved

$A$ is a compact set in a $\mathbb{R}$ normed vectorial space (infinite dimension).
Prove that the closure of $Conv(A)$ is compact.
(here $Conv(A)$ is the minimal convex set that contains $A$)
Is $Conv(A)$ always compact ?

I hope I'm not making any stupid mistakes, and the following provides a counterexample for the second part:

Take the space to be the space of real bounded sequences with the $\sup$ norm, for example, and $A$ to be the set of sequences having $\frac 1n$ on the first $n$ positions and $0$ everywhere else for $n\ge 1$, and the sequence consisting only of $0$. $A$ is a convergent sequence of elements in our space, together with its limit, and any set of this type in a metric space is compact.

The convex hull of $A$ consists only of sequences which are eventually $0$, but it's easy to see that the closure of the convex hull contains sequences which are not eventually $0$, so the convex hull of $A$ is not closed.

I am not good in topology, but I have one question: I managed to prove the result in Banach spaces, but I have already found it difficult. Is there any simple solution for this particular case?

:blush:  I forgot to mention that the result holds for Banach spaces. I don't know wether it is true or not in the general case.
Anyway what was your approach ?

I proved that the closure of the convex hull (which I will note K) is precompact first. Indeed, since $A$ is compact, it is precompact, thus given any epsilon there is a covering of A with a finite number of balls of radius epsilon. Let them be $B(x_1,\epsilon),..., B_k(x_k, \epsilon)$. Since the convex hull of $x_1,...,x_k$ is compact, it is precompact and thus we can cover it with some balls $ B(y_1,\epsilon),...,B(y_m, \epsilon)$. It is easy to show that $ B(y_1, 2\epsilon),..., B(y_m, 2\epsilon)$ is a convering of convA. Then the closed balls wil cover K and we have the claim.
   Take now a sequence $a_n$ of elements of K. Since K is closed and the space is complet, it suffices to show that we can extract a Cauchy sequence from $a_n$. Let $B_{n,1},...,B_{n,r_n}$ be a sequence of closed balls convering K and having radius $1/n$. It is easy to see that there exist strictly increasing functions $f_1,...,f_n,...$   such that there exists $b_n$ with $ a_{f_1(f_2(...(f_n(k)))...}$ is in $ B_{n,b_n}$ for all $k$/ It is clear that $a_{f_1(f_2(...f_k(k))))...}$ is a Cauchy sequence.

Does your friend know whether it remains true in any space? I could hardly believe that.

No, I don't think it's true in any space. Take the space of real sequences which are eventually $0$, and the same set $A$ as in my first post. It won't be true that any sequence of points in $\mbox{Conv}(A)$ has a subsequence converging to some point in the space (since we can take such a sequence converging to a point with infinitely many non-zero coordinates in the larger space of bounded real sequences).

It should be pointed out here that grobber and alekk are using different defintions of convex hull. arobber is taking it to mean the set of convex combinations of the given set, where a convex combination is a linear combination of some finite number of of the elements, with postiive coefficients that add up to 1. alekk defined a larger set: the smallest convex set containing the given set.

In grobber's original example, the closed convex hull - which is what alekk's question concerned - is compact, since it is closed and totally bounded.

But don't the two definitions of "convex hull" yield the same set? The set I defined is clearly contained in alekk's set, while the convex hull as I take it to be is clearly convex, so alekk's convex hull is contained in the one I used. Am I missing something here?

And in my first example, I agree that the closed convex hull is compact (by closed convex hull I understand the intersection of all [b]closed[/b] convex sets containing the set), but the aim was to prove that the convex hull is not compact, and I think it isn't. I was trying to answer alekk's second question, which is whether the convex hull (the word "closed" does not appear before "convex hull" here :)) is always compact.

I think our two definitions are equivalents ? isn't it?

Here's a different proof I wanted to post some time ago, but forgot :):

We want to prove that the closed convex hull $\overline{\mbox{co}(K)}$ of a compact subset $K$ of a Banach space $\mathcal B$ is compact.

Let $\mathcal B^{*}$ be the dual of $\mathcal B$. By the [url=http://planetmath.org/encyclopedia/AlaoglusTheorem.html]Banach-Alaoglu theorem[/url], the unit ball $\mathcal B^{*}_{1}$ of $\mathcal B^{*}$ is weak-$*$ compact. It's easy to see that the weak-$*$ topology on $\mathcal B^{*}$ is Hausdorff, so in this topology $\mathcal B^{*}_{1}$ is a compact Hausdorff space.

Every $x\in\mathcal B$ can be identified with a functional $\varphi_{x}$ on $\mathcal B^{*}$ defined by $\varphi_{x}(f)=f(x),\ \forall$. All functionals of the type $\varphi_{x}$ are weak-$*$ continuous, so their restrictions to $\mathcal B^{*}_{1}$ are continuous functions on a compact Hausdorff space. Since $x\mapsto\varphi_{x}|_{\mathcal B^{*}_{1}}$ preserves norms, $\mathcal B$ is identified isometrically with a (closed) subspace of the space $C(\mathcal B^{*}_{1})$ of continuous maps on $\mathcal B^{*}_{1}$.

The closed convex hull of $K$ in $C(\mathcal B^{*}_{1})$ coincides with $\overline{\mbox{co}(K)}$, so it suffices to prove the desired result for $\mathcal B=C(X)$, where $X$ is a compact Hausdorff space. This version, however, follows from Ascoli’s theorem, which in this particular situation says that a collection of continuous real valued maps on $X$ is compact in the uniform topology iff it is closed, bounded, and equicontinuous: $\overline{\mbox{co}(K)}$ is closed by definition, and its boundedness and equicontinuity are clearly inherited from $K$, so we’re done.


Hi! I'm BOGTRO. Some of you might have heard of me (through Fantasylover's big mouth?). Anyway, if [i]y[/i]ou applied, you probably want to know if you got accept[i]e[/i]d. If not, you might want to see who the player[i]s[/i] are.

So, [i]w[/i]ithout further ado, th[i]e[/i]

[url=http://www.youtube.com/watch?v=oHg5SJYRHA0]PLAYER LIST[/url]

:P

Behind you are 12 [i]c[/i]losed doors. 12 people have keys to these doors. They [i]a[/i]re waiting the sig[i]n[/i]al to come out. Behind:

[b][hide="DOOR 1"] DOJO!! [/hide][/b]

[hide="DOOR 2"] BUGI!! [/hide]

[hide="DOOR 3"] GAME.GUY!! [/hide]

[b][hide="DOOR 4"] HONGSQUARE!! [/hide][/b]

[b][hide="DOOR 5"] PYTHAG011!! [/hide][/b]

[b][hide="DOOR 6"] FANTASYLOVER!! [/hide][/b]

[b][hide="DOOR 7"] ZAPI2007!! [/hide][/b]

[hide="DOOR 8"] MZ94!! [/hide]

[hide="DOOR 9"] EPICFAILURE!! [/hide]

[b][hide="DOOR 10"] JOE10112!! [/hide][/b]

[hide="DOOR 11"] CIRCLESQUARED!! [/hide]

[b][hide="FINAL DOOR"] AIME15!!! [/hide][/b]

This group of 12[size=9]+1[/size], over the next few months, will be competing for a possibility to win a huge amount of money[size=9][hide="."]Some people might want to quote posts.[/hide][/size]

PLEASE CONFIRM. WE CANNOT START WITHOUT YOU CONFIRMING!

Its not too late to sign up! We have room for replacements, and MAYBE (hint hint) one more "player"!

Good luck to you all!!

{BOLD=CONFIRMED, not CONFORMED}

/conform

And don't click "Final Door"; it's a rickroll.

[b] RULES OF THE GAME [/b]

* I will require you to be VERY active. Prods will be sent after 48 hours of inactivity (no posts and no pms), and if the prod is not responded to by 48 more hours, you will be replaced.

* CONFIRMING DEADLINE IS FRIDAY AT 7:30 PM ET. IF YOU DO NOT CONFIRM BY THEN, I HAVE JUST AS GOOD REPLACEMENTS.

* Any discussion is allowed and encouraged, unless otherwise stated.

* Sharing quicktopics is strictly prohibited.

* If you are sent information intended for someone else... THAT'S MY PROBLEM, NOT YOURS. Go ahead and open it. EXCEPTION: IF YOU RECIEVE ANYTHING TITLED "Information for the mole only", delete it and tell me. If you open it and you are caught, you will be penalized by disqualification.

* Never decline help. It might come in odd ways, but never decline it.

* If you need to be replaced, say so in the thread.

* Final submissions for anything must be [b]BOLDED[/b]

* You should read the ENTIRE thread many times. [b]To make sure you read the rules, you must post "I lost" in your next post. SOME PEOPLE MISSED THIS?[/b]

[b] RULES OF THE MOLE [/b]

Congratulations! You have been accepted into the AoPS Mole! Let's make sure that you know what to do.

* First, you confirm. This means you post in thread.

* Tasks will be posted. Follow the directions in the thread.

* The [i]Mole[/i] attempts to sabotage the group, while keeping himself hidden. If you are the mole, confusion = Very good.

* Tasks are worth a certain amount of money, depending on how well you do.

* Execution quizzes occur every 3 tasks. The questions will be about who you think is the mole. You MAY NOT go back in the thread (can you go back in time?) during execution quizzes. I will permit going back in the thread during tasks, owing to the fact that this is a forum, and you won't nessacarily be here all the time.

* When 3 people are left, there will be a final quiz of 30 questions (normal is 10). If the pot is large enough, and you score high enough on the quiz, you win and take all the money. If the mole succeeds in stopping either the players from getting a certain pot, or preventing the players from getting a high enough score, the mole gets all the money.

* The mole does not have to answer the execution quizzes. Thus, he cannot be eliminated.

* Exemptions will be given throughout the game, usually in place of money. If you obtain an exemption, you do not have to do the next execution quiz.

* If you are "executed", you cannot post or talk to anyone regarding the mole.

If I forgot anything tell me.

/confirm

w00t door 10!

realize taht aime has 3 ! behind his name in final door.

also, pythag is in bold

maybe hints to find out the mole?

i am here

i like how door nine opens to epic failure  :lol:

Hint (which may appear throughout the game):

Scrutinize my first post very very carefully. You might find information you don't expect. Some things might... [i]change[/i]

/confirm... thanks bogtro! i'm excited :D

chance and change are italicized. Huh.

Also, can we look at the first few pages

1. After the quizzes
2. During the quizzes?

[quote=Rules]You MAY NOT go back in the thread [/quote]

Hmm...

You may not refer to stuff written before the last quiz.

And I think some people forgot to post that they lost.

[quote="BOGTRO"]So, without further ado, the

[color=red]PLAYER LIST [/color][/quote]

You suck... :P

Hopefully it wasn't a real rickroll...

[quote="Liar"](through Fantasylover's big mouth?)[/quote]

I never said anything about Mole 5... :wink:

Anyways, I confirm.

EDIT:

Ok ok I lost :wink:

Am I still a smart person if I just rickrolled myself?

Now it's underlined.

@BOGTRO: Depends on if you can solve USA TST 2005 #3, which is triival.

Apparently there remains interest in this game.

The task is starting over. NOTE THERE ARE RULE CHANGES:

GOOGLE IMAGE SEARCH:

The first person in line will guess an image. The second person in line must enter a search keyword of 3 words or less that has the picture on the first page of searching. (Note: the website is images.google.com).

The second person then posts an image, and the third person tries to guess, and so on. Each person has only 36 hours to complete his/her turn [Exception: You can request extra time or to be moved to the end of the line IF you have a good reason].

If your answer is wrong [Does not return the image in the first page], you are out of the round. If your deadline expires, you are out of the round.

In the first round, if more then 7 people survive, you win 10000. In the second round, if more then 2/3 of the people survive, you win another 10000.

The total possible is 20000.

Go.

[quote="zapi2007"]can you post a player list?[/quote]

Right, sorry.

Randomly:

1. AIME15 [Note: must guess last person's picture]
2. zapi2007
3. james4l
4. pythag011
5. algebra2000
6. Circlesquared
7. Joe10112
8. EpicFailure
9. hongsquare
10. Bugi
11. Dojo [Still posts picture, AIME15 must guess]

Go.

Get moving. This game will be declared player-abandoned very soon.

THIS GAME IS DECLARED PLAYER-ABANDONED. IF ALEKIM IS STILL INTERESTED, THE MOLE 6 SIGNUPS CAN BEGIN.

blah...probably because AIME15 just didn't post.

Let's try it right now!

[img]http://www.maximumpc.com/files/u46168/microsoft_0.jpg[/img]

no looking at the source code  :P

[quote="Bugi"]Who was the mole? BTW we were here... I would call this mod abandoned  :wink:[/quote]

Why? I've been trying to get this game going for over a month. Nobody is interested.

Out of interest, who did you guys think was the mole?

Come on, zapi wants to play, I want to play... Call this abandoned iff mole is inactive, ok?

how about new task.

Post a picture.

If you don't post a picture, you lose this game. (no not that game)

Then, one of the players remaining is randomly selected to be the mole.

We would need more than two players obviously, but...oh well.


i dont get these at all! i dont get adding them or anything. are there any good websites? i have tried over and over to figure this out but i cant!
thanks

Hmmm... any good websites?  I can't find any... oh wait, no wonder! I'm at one!

People here can probably explain it pretty well.  Do you understand the concept of bases other than base ten?

Hexidecimal is just base 16.  (I can explain more about different bases, or you can look at some old threads)

So, here is the basic idea:

We are working in hexidecimal, which is base 16.  In base 10, we have digits for 0, 1, 2, 3, ..., 9 -- everything up to 10.  In base 16, the same thing has to happen -- we need digits for 0, 1, 2, ..., 15 -- everything up to 16.  Because, in our normal system, we don't have single-digit symbols to represent the number 10, 11, 12, 13, 14 or 15, we need to create them.  Standard practice is to use letters -- A for 10, B for 11, ... and F for 15.

Now, for adding.  In our normal (decimal) system when adding numbers we carry [i]only when[/i] the sum of the numbers in a given place is 10 or greater.  Similarly, in hexidecimal, we will carry only when the sum of the numbers in a given place is 16 or greater.  Thus,
[tex]8 + 7 = F_{16}, 9 + 7 = 10_{16}[/tex].

Tell me if you need more.

Haha, that's really funny -- guess I'm the slow typist here.

Did you read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12354]this thread[/url], or are your problems because of it?

how do you know whether to write down 12 as c or to carry the 1 to the next slot?

That depends whether it is [tex]12_{10}[/tex] or [tex]12_{16}[/tex].  If you are adding and you convert to base 10, you then must convert back to base 16 before carrying.  Thus, if you are adding

[tex]B7_{16} + 15_{16}[/tex], you convert [tex]7_{16}[/tex] and [tex]5_{16}[/tex] to base 10, in which they become [tex]7_{10}[/tex] and [tex]5_{10}[/tex].  Then you add them to get [tex]12_{10}[/tex].  Then you convert back into hexidecimal to get [tex]C_{16}[/tex].

Note how I used the subscripts to keep track of what base I was in all the time.  This is very useful for making sure you know when to carry and so on.

Think of it as a counting system where you start with just making tally marks.  Eventually, you get tired of writing so many tally marks, so you decide to draw a square for every 16 tally marks (and erase those tally marks replaced).  Even so, you're ending up writing too many squares, so every 16 squares, you draw a triangle and so on.  Now let's say we have 5 triangles, 11 squares, and 15 tallies, and we're adding on 1 triangle, 4 squares, and 6 tallies.  First we lump them all together.  Then, seeing that we have a total of 6 triangles, 15 squares, and 21 tallies, we can just eliminate 16 of the tallies to get 1 square and 5 tallies left.  But wait, we have 16 squares now!  That means instead of writing them as squares, we could just write an additional triangle.  this gives us 7 triangles, 0 squares, and 5 tallies.

Now just replace all those shapes and such with digit places.

wow thats a really cool way of thinkin about it

Add the hexadecimal numbers A1E and 19D. how do i do that?

You show us how you think it should be done and we'll help you out if you get in trouble.

also, are there any good books on bases so i can further practice?

i AM in that class max lol and i dont get it....thats why i have all these questions. thanks! can you help on the addition?

[quote="eqn1190"]i dont know. i dont get adding letters. and then how do i write 142_10 as a hexadecimal?[/quote]

Hmmm... that's 142 tallies.  So that means 8 squares I guess.  8*16 = 128, so that means we have 14 tallies left.  We're done.  Now 14 = E, so we have 8 and then E = [tex]8E_{16}[/tex]


modular arithmetic
vector
linear algebra
matrix

Hello,

I'm wondering how I can efficiently solve the following problem:

Given $ n$ and $ p$ ($ n$ and $ p$ $ <$ $ 4000$) and $ p$ objects $ a_i$ ($ 1 < a_i < 500000$), I have to find the maximum sum $ S$ possible if I can take any subset of the $ p$ objects (each object can only be taken once). Also, the sum $ S$ must be divisible by $ n$.

I tried the classic dp, but I can't afford the time nor the memory... (around one second time).

I also tried taking the numbers modulo $ n$ when reading them then doing the dp on those numbers, and adding stuff in the original array at the end, but this doesn't always return the right answer, at least not the way im doing it, with d[i] = 1 if I can get to sum i, considering all numbers modulo $ n$, and last[i] = last number that's part of sum i.

Example:

n = 3, p = 5
1
2
3
3
4

The maximum sum is 2 + 3 + 3 + 4 = 12, which is maximum and also divisible by n = 3

Taking the numbers modulo n, we'd have:
1
2
0
0
1

The maximum sum we can make is 3 (1 + 2 or 2 + 1)
Adding the corresponding numbers in the original array, we'd get 1 + 2 or 2 + 4, both of which are correct, but neither of which are maximum.

I'd appreciate any ideas.

Assuming you can afford 16MB, you can try $ dp[i][j] \equal{} 1$ iff you can get sum $ i \pmod{n}$ using the first $ j$ numbers. Then $ dp[i][j] \equal{} dp[(i\minus{}a_j\plus{}n)\%n][j\minus{}1]$ since you can get $ i$ with the first $ j$ numbers iff you can get $ i \minus{} a_j \pmod{n}$ with the first $ j\minus{}1$. The dp array will have to be 4000 x 4000. But if you can't afford that memory, you can notice that in each step of the dp, e.g. increasing $ j$, you only need to know the previous row. So you can store it in a 4000 x 2 array instead, where one row is the previous row and the other is the current. It runs in $ O(np)$ time which should be very fast.

16 MB is fine.

This would still require I keep a vector with the initial numbers taken modulo $ n$, and your $ a_j$ is actually the $ j\minus{}th$ element taken modulo n, right?

Also, the answer is in a[0][p] right? How can I rebuild the solution? I think I can use a secondary matrix, but I can't really figure out what to store in it.

Yeah, sorry, I assumed you store each $ a_j \pmod{n}$. Also, I missed the part about the maximum sum, so instead of letting $ dp[i][j]$ be whether or not you can get that, you should instead have it be "maximum sum you can get which is $ i \pmod{n}$ using the first $ j$ numbers". Then the update step would be

$ dp[i][j] \equal{} \max\{dp[i][j\minus{}1], dp[(i\minus{}a_j \pmod{n}\plus{}n)\%n][j\minus{}1] \plus{} a_j\}$

if $ dp[(i\minus{}a_j \pmod{n}\plus{}n)\%n][j\minus{}1]$ was previously set (e.g. is non-negative assuming you initialize the entire table to -1). Note that I previously forgot to include the $ dp[i][j\minus{}1]$ term; if you can get the sum with the first $ j\minus{}1$ elements, you can definitely get it with the first $ j$. Then the answer will be $ dp[0][p]$.

If you want to rebuild the solution, just do the following. Let $ S$ be the maximum sum. If $ dp[0][p\minus{}1] \equal{} S$ then you know $ a_p$ is not used; otherwise $ a_p$ is used and the remaining sum is $ S \minus{} a_p$. Just iterate backwards over the table to get the rest of the numbers in the same way.

I did this, but this does not fit in 18 megs, which is the maximum the problem allows. To get around this limitation, I keep 2 vectors of size 4000, and work with those (basically the last two lines of the matrix). For rebuilding the solution, I keep a char helper[4000][4000] which fits in memory. However, this gets some TLEs. I precomputed modulos, but I think the problem lies in the fact that I have to copy the contents of each vector to the other at every step. This means 2*n*p operations, basically, and I'm not sure how I can get around doing this copying. Or maybe I could also speed it up by getting rid of the helper char matrix, but then I don't know how to rebuild the solution without a matrix...

Can we get rid of copying the two vectors at each step somehow?

Hmm, 2*n*p operations should run very fast, well under a second. But you shouldn't have to copy the contents of the vectors at all. Just swap which one is the current row and which is the previous row, e.g. have two variables cur and prev initialized to 0 and 1, respectively. Then you should always be updating dp[i][cur] = something that depends on dp[][prev]. With each iteration set cur = 1-cur and prev = 1-prev, so that you swap which one is the last row. That should eliminate copying.

Hello, sorry I took so long to reply, I wasn't home.

I did that and I'm getting a lot of wrong answers. Is there any reason why that might happen, or did I make a mistake somewhere?


Find the minimum possible least common multiple of twenty natural numbers whose sum is $801$.

I think it's $42 = 2*3*7$ with $3,42,...,42$

We must write $801$ as $\sum p_{ij}^a_{ij}i$  with $p_i$ prime and $max(p_{ij}^ {a_{ij}})$ minimum.

It's not possible to do so with only one prime factor to obtain less than $48$.

With a lcm like $p_1 * p_2$ is not possible either with $p_1 * p_2 < 48$ and only $20$ numbers
With a lcm like $p_1 * p_2 * p_3$ is possible with $p_1 * p_2 * p_3 < 42$
With a lcm like $p_1^{a_1} * p_2^{a_2}$ is not minimal (min value is $2^4*3$)
....

Since 801=20*40+1, by PHP there exist a number among the 20 which is >=41.
Thus lcm of the numbers>=41
For lcm to be 41, each of the numbers must be 1 or 41.
Let x of them be 1 and y of them be 41.
Then 41y +x=801 and x+y=20, whence 40y=781, which is not possible
Next we check 42
We note that 19*42+3=801, and thus 19 forty-twoes and one three makes l.c.m 42.
Thus the answer is 42
Bye

@sayantanchakraborty this is kind of hit and trial.. any better solution?

[u][b]Claim:[/b][/u] The answer is $42$.
Let the numbers be $x_1,x_2,\cdots x_{19}, x_{20}$. According to the problem, $\sum_{i=1}^{20} x_i=801$.
Let the $L.C.M$ of the numbers be $L$. Therefore, $\forall i \in \mathbb{N}, 1 \leq i \leq 20$, we get that $x_i \vert L \Rightarrow x_i \leq L$.
$\Rightarrow \sum_{i=1}^{20} x_i \leq 20L \Rightarrow 20L \geq 801 \Rightarrow L \geq  41$. Now we check cases (as we want the minimum value of the $L.C.M$, we start with the minimum cases, gradually increasing the $L.C.M$ and stopping whenever a case satisfies the condition).

Case $(i)$: If $L=41 \Rightarrow x_i= 1$ or $41$ $\forall$ $i$. Clearly, this case does not work out. That is because if all the $x_i's=41$, their sum is $820>801$, but if any one of them is equal to $1$, their sum becomes $780<801$.

Case $(ii)$: If $L=42 \Rightarrow$ the maximum possible value of $\sum_{i=1}^{20} x_i = 42*20= 840$, when $x_i=42$ $\forall$ $i$. Further, notice that if we change just one of the $x_i's$ to $3$, the sum decreases by $42-3=39$, and thus it becomes $840-39=801$. Therefore, $\boxed{L=42}$ works!

There is no point in checking any more cases, as this is the minimum value of the $L.C.M$ that can satisfy the given conditions. Thus, proved!


find (delta)y and dy if y=sin(x), and x changes from 0 to PI/6

$dy/dx = cos(x) \rightarrow dy = cos(x)dx \rightarrow dy = cos(pi/6)pi/6 \rightarrow \sqrt{3}pi/12$.
$\Delta y = \frac{sin(pi/6)-sin(0)}{pi/6 - 0} = 3/pi$.


What is the greatest number of pieces of pizza you can get if you cut a circular pizza with 5 straight cuts? What is the greatest number of pieces of pizza you can get if you cut the same pizza with 7 straight cuts? You must supply solutions to both questions! Yummy.

1 with 0 cuts
2 with 1 cut
4 with 2 cuts
7 with 3 cuts
11 with 4 cuts
16 with 5 cuts
22 with 6 cuts
29 with 7 cuts

so the first one passed through 0 points
the second one passed through 1 point
the third passed through 2 points
the fourth passes through 3 points


[quote="mr.danh"]Let a triangle ABC. The incenter touches sides $ BC,CA,AB$ at $ D,E,F$ respectively. $ M\in BE, N\in CF$ such that $ DM\parallel AC, DN\parallel AB$. Prove that $ MN\parallel EF$[/quote]
[hide="Solution"]
Notice that $ DM\parallel EC\implies \triangle BMD\sim \triangle BCE$, so $ \frac {DM}{BD} \equal{} \frac {EC}{BC}\implies MD \equal{} \frac {BD\cdot EC}{BC}$.  Similarly, we can see that $ DN \equal{} \frac {FB\cdot DC}{BC}$.  Yet, since $ FB \equal{} BD$ and $ CD \equal{} EC$, we have that $ DN \equal{} MD$.  Also $ AF \equal{} AE$.  Since $ MD\parallel AE$ and $ DN\parallel AF$, we have that $ \angle FAE \equal{} \angle MDN$, so $ \triangle MDN\sim \triangle EAF$, which implies that $ \angle AEF \equal{} \angle DMN$.  Furthermore, since $ MD\parallel AC$, we have that $ \angle DME \equal{} \angle MEA$, so \[ \angle FEM \equal{} \angle AEM \minus{} \angle AEF \equal{} \angle DME \minus{} \angle DMN \equal{} \angle EMN\] so $ MN\parallel EF$. [/hide]


How can I use $\delta, \epsilon$ to show that
$f(x,y) = \frac{xy}{x^{2}+y^{2}}$ if $(x,y) \ne (0,0)$ and $f(x,y) = 0$ if $(x,y) = (0,0)$ is discontinuous or continuous at $(0,0)$?

I tried the following but with no avail:

Fix $\epsilon > 0$. Find $\delta$ such that if $|\sqrt{x^{2}+y^{2}}| < \delta$ then this implies $|\frac{xy}{x^{2}+y^{2}}-0| < \epsilon$.

But we have $|x^{2}+y^{2}|<{\delta}^{2}$ so $|\frac{1}{x^{2}+y^{2}}| > \frac{1}{{\delta}^{2}}$.

I'm having trouble reducing this expression to something that is less than $\epsilon$. Do I need to substitute an $r = \sqrt{x^{2}+y^{2}}$ to make it easier?

Also, what about determining the continuity of
$f(x,y) = \frac{xy^{2}}{x^{2}+y^{2}}$ if $(x,y) \ne (0,0)$ and $f(x,y) = 0$ if $(x,y) = (0,0)$:  is it discontinuous or continuous at $(0,0)$?

Thanks.

For the original problem, the function is not continuous. Find points arbitrarily close to the origin with "large" values.

The substitution $r=\sqrt{x^{2}+y^{2}}$ does help (for both problems), especially if you go all the way into polar coordinates.

Can you explain your first advice a little more? How exactly do you put these "large values" in mathematical language?

For example, $f(t,t) = \frac{1}{t^{2}}\to \infty \ \textnormal{for}\ t \to 0$. Hence $f$ is not continuous at $(0,0)$

In this case, "large" merely meant "tending to something that isn't zero". The function is at least bounded.


a,b,c >=0
Find the minimun of

 (a+b)(b+c)(c+d)(d+a)/abcd- [(a+b+c)(b+c+d)(c+d+a)(d+a+b)/abcd]^¼[/list][/youtube][/b]

[quote="m.candales"]a,b,c >=0
Find the minimun of:
 $ P \equal{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)}{abcd} \minus{} [\frac {(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(c \plus{} d \plus{} a)(d \plus{} a \plus{} b)}{abcd}]^{\frac {1}{4}}$[/quote]
First, we will prove that:
$ \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)}{16abcd} \ge \sqrt [4]{\frac {(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(c \plus{} d \plus{} a)(d \plus{} a \plus{} b)}{abcd}}$
Or:
$ 3^4(a \plus{} b)^4(b \plus{} c)^4(c \plus{} d)^4(d \plus{} a)^4 \ge 16^4a^3b^3c^3d^3(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(c \plus{} d \plus{} a)(d \plus{} a \plus{} b)$
By AM-GM inequality, we have:
$ (a \plus{} b)(b \plus{} c) \equal{} ac \plus{} b(a \plus{} b \plus{} c) \ge 4\sqrt [4]{\frac {ab^3c(a \plus{} b \plus{} c)^3}{3^3}}$
Similarly, we have:
$ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} d)^2(d \plus{} a)^2 \ge 4^4\sqrt [4]{\frac {(abcd)^5(a \plus{} b \plus{} c)^3(b \plus{} c \plus{} d)^3(c \plus{} d \plus{} a)^3(d \plus{} a \plus{} b)^3}{3^{12}}}$
Or:
$ 3^4(a \plus{} b)^4(b \plus{} c)^4(c \plus{} d)^4(d \plus{} a)^4 \ge \frac {1}{3^2}\sqrt {(abcd)^5(a \plus{} b \plus{} c)^3(b \plus{} c \plus{} d)^3(c \plus{} d \plus{} a)^3(d \plus{} a \plus{} b)^3}$
So we need to prove;
$ \sqrt {(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(c \plus{} d \plus{} a)(d \plus{} a \plus{} b)} \ge 9\sqrt {abcd}$
Which is clearly true by AM-GM
Second, by AM-GM:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)}{abcd} \ge 16$
So we have $ P_{min} \equal{} 13$


inequalities
function
algebra
domain
vector
real analysis
real analysis unsolved

Does the following inequality
\[ \left\| f \right\|_{L^4 \left( \Omega  \right)}  \leqslant \left\| {\nabla f} \right\|_{L^2 \left( \Omega  \right)}
\]
hold where $ \Omega  \subset \mathbb{R}^N$ is open bounded set and $ N\equal{}1,2,3$.

[quote="alekk"]if not, take a constant ?!?[/quote]

A trivial counter-example  :) . If yes?...  :oops:

Consider a function $ f$, and let $ g(x)\equal{}f(x/c)$ for some $ c>0$. We get $ \nabla g(x)\equal{}\frac1{c^2}\nabla f(x/c)$, so $ \|g\|_4\equal{}c^{N/4}\|f\|_4$ and $ \|\nabla g\|_2\equal{}c^{N/2\minus{}2}\|\nabla f\|_2$.

For $ N<8$, choosing very large values of $ c$ makes $ \|g\|_4$ greater than $ \|\nabla g\|_2$; all we need is a large enough $ \Omega$. For $ N>8$, very small values of $ c$ make $ \|g\|_4$ greater than $ \|\nabla g\|_2$, and the inequality fails on any $ \Omega$.
This argument doesn't say anything for $ N\equal{}8$, but the inequality can't be true for any other $ N$.

beware that the domain is bounded. So even if $ f$ vanishes on the boundary, your function $ g$ will maybe not. (I think that the Sobolev embedding/inequality theorems could be useful)

I know that- which is why I said "all we need is a large enough $ \Omega$". We may have an inequality of the form $ \|f\|_4\le C(\Omega)\|\nabla f\|_2$, but $ C(\Omega)$ goes to $ \infty$ as the size of $ \Omega$ grows.

Quibble: $ \nabla g$ is the gradient, not the Laplacian, right? Then (this is from jmerry's first post), we'd have $ \nabla g(x)\equal{}\frac1c\nabla f(x/c)$ and $ \|\nabla g\|_2\equal{}c^{N/2\minus{}1}\|\nabla f\|_2.$

Then you would wind up replacing $ N<8,N>8$ in the argument that follows by $ N<4,N>4.$ The substance of the argument would remain, however.

OK. I don't deal with these symbols every day. The Laplacian is a little easier to deal with, since we don't have to worry about vector norms on the RHS.


Let $n$ be nonnegative integer, and let $p$ be a prime number that is congruent to $7$ modulo $8$.
Prove that $\sum_{k=1}^{p-1}\{ \frac{k^{2^{n}}}{p}-\frac{1}{2}\}= \frac{p-1}{2}$.


It is known that 3 sides of triangle are consecutive integers and the largest angle is twice the smallest angle. Find the perimeter of triangle.

[hide]
If A=2B, then $a^2=b^2+bc$.
Hope this helps.[/hide]

[hide]
so we have triangle ABC with
$\angle$ ABC = B   and    AB = x
$\angle$ BAC = 2A   and   CB = (x+2)
$\angle$ ACB = A  and   AC = (x+1)

Apply some 'cos rule' (Al-kashi theorem)

$\therefore \cos A = \frac{(x+1)^2+(x+2)^2-(x)^2}{2(x+1)(x+2)}$

$\therefore \cos A = \frac{(x+5)}{2(x+2)}$

and

$ \cos 2A = \frac{(x+1)^2+(x+2)^2-(x)^2}{2(x+1)(x+2)}$

$\therefore \cos 2A = \frac{x-3}{2x}$

now with some trigonometrical gymnastics,

$\cos 2A = \cos ^2A - \sin ^2A = \cos ^2A - (1- \cos ^2A) = 2 \cos ^2A-1$

$\therefore 2 \cos ^2A-1 = \frac{x-3}{2x}$

$\therefore  \cos ^2A = \frac {\frac {x-3}{2x}+1}{2}$

and we also have..

$ \cos A = \frac{(x+5)}{2(x+2)}$

$\therefore \cos^2 A = \frac{(x+5)^2}{(2(x+2))^2}$

$\therefore \frac{(x+5)^2}{(2(x+2))^2} = \frac {\frac {x-3}{2x}+1}{2}$

an algebraic slog reveals

$ x = -3 ,  - \frac{1}{2} , 4 $

We're looking for positive values here... so $ x = 4 $
$ \therefore $ the sides are, 4 , 5 and 6
$ \therefore $ the perimeter = 15 units
[/hide]


A permutation (a_1,a_2,...,a_k) of the integers (1,2,...,n) is said to be fantastic if a_1+a_2+...+a_k is divisible by k for each k from 1 to n.

Find all fantastic permutations.

do you mean combinations? :ewpu:

1 divides every number. so (1), (2), (3),.....,(n) are all fantastic permutations.

for divisiblity by 2, we should have two terms whose sum is even. so, we should select two odd terms or two even terms.

for divisibility by 3, we should have three numbers which sum to a multiple of 3.

for divisibility by n, n should be odd.


Let $P$ be a pedal-cevian point and let $Q$ be its isotomic conjugate. Is it true that also $Q$ is a pedal-cevian point?

Yes it is true.

We denote as D, E, F, the cevian traces (the traces of the lines AP, BP, CP) of a point P, on the sidelines BC, CA, AB respectively, of a given triangle ABC. If the lines through D, E, F and perpendicular to BC, AC, AB respectively, are concurrent at a point so be it P', then the point P, is called pedal-cevian point with respect to ABC.

If now, D', E', F', are the isotomic conjugate points of D, E, F respectively, with respect to BC, CA, AB respectively ( symmetrical with respect to their midpoints ), then we have that:

a) – The cevians AD', BE', CF', are concurrent at a point so be it Q, by Ceva’s theorem.

b) – The lines though D', E', F' and perpendicular to BC, CA, AB respectively, are concurrent at a point so be it Q',  by Carnot’s theorem.

Hence, the point Q, isotomic conjugate of P, with respect to ABC, is also a pedal-cevian point.

     Kostas Vittas.

Dear Mathlinkers,
an article (last of three) concerning the first synthetic proof of the PC-lines and the de Longchamps point has been put on my website
http://perso.orange.fr/jl.ayme vol. 6 Pedal-cevian lines go through the de Longchamps's point, p.43
Sincerely
Jean-Louis


taking it mod (n+2), we have 1^1995+2^1995+...+(-2)^1995. since 1995 is an odd power, everything except 1^1995=1 cancels out, meaning that whole thing is exactly one more than a multiple of (n+2)


algebra
polynomial
function
domain
combinatorics proposed
combinatorics

Prove that

\[ \sum_{k\equal{}1}^{n\minus{}1}(\minus{}1)^{k\minus{}1}\binom{n}{k}(n \minus{} k)^{n\minus{}2} \equal{} n^{n\minus{}2}\]

for integers $ n \geq 3$.

Tricky!  The identity fooled me into thinking about labeled trees.  I claim that for any polynomial $ P(k)$ of degree less than $ n$ it is true that

$ \sum_{k\equal{}0}^{n} (\minus{}1)^k {n \choose k} P(k) \equal{} 0$.

This is just a matter of evaluating finite differences and can be proven using the following two simple lemmas:

1)  The polynomial $ \Delta P \equal{} P(k\plus{}1) \minus{} P(k)$ has degree one less than the degree of $ P$.  In particular, if $ P$ has degree less than $ n$ then $ \Delta^n P$ is identically zero.

2)  $ \Delta^r P(0) \equal{} \sum_{k\equal{}0}^{r} (\minus{}1)^{r\minus{}k} {r \choose k} P(k)$.

The result then follows by setting $ P(k) \equal{} (n\minus{}k)^{n\minus{}2}$ and $ r \equal{} n$.

An alternative approach that makes at least some use of the special form of the given polynomial:  How many onto functions are there from $ \{1, 2, \dots, n\minus{}2\}$ to $ \{1, 2, \dots, n\}$?

Method 1: For any set $ S$ of size $ k$, there are $ (n\minus{}k)^{n\minus{}2}$ functions whose range fails to intersect $ S$.  By inclusion-exclusion applied to the $ n$ events "the number $ i$ is not in the range of our function", the number of onto functions is \[ \sum_{k\equal{}0}^n (\minus{}1)^k \binom{n}{k} (n\minus{}k)^{n\minus{}2}.\]

Method 2: The proposed range is larger than the domain, so there are no such functions.


calculus
integration
linear algebra
matrix
email
advanced fields

The international scientific-educational website EqWorld http://eqworld.ipmnet.ru presents extensive information on algebraic, ordinary differential, partial differential (mathematical physics), integral, functional, and other mathematical equations. Outlines exact solutions and some methods for solving equations, includes tables of solutions and interesting articles, gives links to math websites, lists useful handbooks, monographs, textbooks, journals, software, etc.

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Dear Friend:
                 Isaac Newton invented the science of 'Mathema-
tical Calculus' with its powerful 'equations.(Wilhelm Leibnitz did
invented 'independently,' his "integral calculus" late than New-
ton.) With his invention,the world of 'Gravitational Laws Of Phy-sics" with its mysteries was finally fathomed.Other 'application'
was later realized.

                Albert Einstein,did add 'mathematical spices' by
adopting the works of 'Lobachevsky,'and 'Riemann'which is the
'Non-Euclidean Geometry." The said German-Born American Citi-
zen "mathematical physicist"did also adopted both 'Tensor,'and
'Ricci Calculus' solving the problem of "Gravitational Field."(With
application on infinistissimal microphysics world.)

               Werner Heisenberg,another german "Theoretical Phy-
sicist,"applied other 'mathematical ideas' to his "Quantum World"
inquiry.The mathematics he used is:"MATRIX ALGEBRA"in its advance state.Same mathematics became the solid foundation
related to 'mathematical research application' during the late
1950's.

            Do you believed that with ever increasing sophistication
of 'Technological Invention'(Field Of Mathematical Physics.)in
today's science world,a must brand of "NEW MATHEMATICS"
must be invented to accommodate raising problem in "Quantum
Relativistic World?"(This same question was echoed by some
physicists.) "Equations,"mathematically speaking seems on the border line of "extinction" related to its usefulness in modern-day
mathematical physics world? Do you believed concerning my theoretical opinion that really mathematical equations are on the
"BRINK" of extinction?

                                          Respectfully,
                                            Mikhail

Dear Friends,
                  I know that some of you who made a "reply" to my
June 25th,2006 post message entitled:"A Must Mathematical
New Ideas" might wondered why i haven't reply to the message
posted.I had some problems with my "Password."Its only today that i was able to retrieved my lost past word.To be frank,my
password is usually "hacked"most of the time.Can you,or anyone
in this "art of problem solving"math forum member used my
alternate email address?(When it is convenient to the members.)
My alternate email:antonio65@vfemail.net.This more safer.I can
easily answer all possible reply for me.antonio65@vfemail.net is
very much "ENCRIPTED." So,its more safer.

                           Multitude Of thanks!

                                                    Respectfully,
                                                   Mikhail


Find the integral part of $ (\sqrt2\plus{}\sqrt3\minus{}\sqrt6)(\minus{}\sqrt2\plus{}\sqrt3\plus{}\sqrt6)(\sqrt2\minus{}\sqrt3\plus{}\sqrt6)$ without using calculator or knowing the approximations of the square roots.

[hide]6   because $ \sqrt2, \sqrt3, \sqrt6$ are all needed for the integral parts
and we have $ 6 \minus{} 6 \plus{} 6 \equal{} 6$[/hide]

Actually, that's incorrect. The answer is $ 4$, I believe.

Why $ 4$  :blush: ?

Oh! So the integral part doesn't mean the part without surds?

Well the approximate value of that thing is 4.11026, so I believe he meant the integer part of that, which is $ 4$ (unless I'm mistaken).

It is explicitly mentioned that the problem needs to be solved without using a calculator. I want to know how the problem is solved manually? Is is that -- We calculate the product of the 3 surds and pick out the integral part from that?

You could also try to find the highest possible integral lower bound to the resulting expression that you get after expanding.

Possibly adding or subtracting some very small quantity to cancel out the square roots?

Let $ a\equal{}\sqrt{2}$, and $ b\equal{}\sqrt{3}$. Then the product
\[ P\equal{}(a\plus{}b\minus{}a b)(a\minus{}b\plus{}a b)(\minus{}a\plus{}b\plus{}ab)\]

It is easy to compute that \[ (a\plus{}b\plus{}ab)P\equal{}23\]

So it suffices to show that $ \frac{23}{a\plus{}b\plus{}ab}\in[4,5)$ or
\[ \frac{23}{4}\ge a\plus{}b\plus{}ab>\frac{23}{5}\]

I dont' know if this helps but it may...

[quote="ifai"]Find the integral part of $ (\sqrt2 \plus{} \sqrt3 \minus{} \sqrt6)( \minus{} \sqrt2 \plus{} \sqrt3 \plus{} \sqrt6)(\sqrt2 \minus{} \sqrt3 \plus{} \sqrt6)$ without using calculator or knowing the approximations of the square roots.[/quote]

$ (\sqrt2\plus{}(\sqrt3\minus{}\sqrt6))(\sqrt2\minus{}(\sqrt3\minus{}\sqrt6)\equal{}2\minus{}(\sqrt3\minus{}\sqrt6)^2\equal{}2\minus{}(9\minus{}6\sqrt3)\equal{}6\sqrt3\minus{}7$

$ (6\sqrt3\minus{}7)(\minus{}\sqrt2\plus{}\sqrt3\plus{}\sqrt6)\equal{}
\minus{}6\sqrt6\plus{}18\plus{}18sqrt3\plus{}7\sqrt2\minus{}7\sqrt3\minus{}7\sqrt6\equal{}
18\plus{}11sqrt3\plus{}7sqrt2\minus{}13sqrt6$.

Thus the integral part is 18?

[quote="BOGTRO"]

$ (\sqrt2 \plus{} (\sqrt3 \minus{} \sqrt6))(\sqrt2 \minus{} (\sqrt3 \minus{} \sqrt6) \equal{} 2 \minus{} (\sqrt3 \minus{} \sqrt6)^2 \equal{} 2 \minus{} (9 \minus{} 6\sqrt3) \equal{} 6\sqrt3 \minus{} 7$

[/quote]

$ (\sqrt{3}\minus{}\sqrt{6})^2\equal{}3\plus{}6\minus{}2\sqrt{18}\equal{}9\minus{}6\sqrt{2}$ not $ \sqrt{3}$.

[quote="BOGTRO"]$ (\sqrt2 \plus{} (\sqrt3 \minus{} \sqrt6))(\sqrt2 \minus{} (\sqrt3 \minus{} \sqrt6) \equal{} 2 \minus{} (\sqrt3 \minus{} \sqrt6)^2 \equal{} 2 \minus{} (9 \minus{} 6\sqrt3) \equal{} 6\sqrt3 \minus{} 7$

$ (6\sqrt3 \minus{} 7)( \minus{} \sqrt2 \plus{} \sqrt3 \plus{} \sqrt6) \equal{} \minus{} 6\sqrt6 \plus{} 18 \plus{} 18sqrt3 \plus{} 7\sqrt2 \minus{} 7\sqrt3 \minus{} 7\sqrt6 \equal{} 18 \plus{} 11sqrt3 \plus{} 7sqrt2 \minus{} 13sqrt6$.

Thus the integral part is 18?[/quote]
Integral part of $ x$ is $ \lfloor x\rfloor$.


function
modular arithmetic
number theory
prime factorization
number theory proposed

Let $ \sigma(\cdot): \mathbb{N}_0 \to \mathbb{N}_0$ be the function from every positive integer $ n$ to the sum of divisors $ \sum_{d \mid n}{d}$ (i.e. $ \sigma(6) \equal{} 6 \plus{} 3 \plus{} 2 \plus{} 1$ and $ \sigma(8) \equal{} 8 \plus{} 4 \plus{} 2 \plus{} 1$).
Find all primes $ p$ such that $ p \mid \sigma(p \minus{} 1)$.
[i](Salvatore Tringali)[/i]

[hide="Solution"]

If $ n\equal{}\prod_{i\equal{}1}^k p_i^{x_i}$ is the prime factorization of $ n$, then
$ \sigma(n)\equal{}\prod_{i\equal{}1}^k \frac{p_i^{x_i\plus{}1}\minus{}1}{p_i\minus{}1}$

We know that $ \frac{t^{u\plus{}1}\minus{}1}{t\minus{}1}\equal{}t^u\plus{}t^{u\minus{}1}\plus{}\ldots\plus{}1$

So, in our problem, when we apply the above, we get that $ p$ must divide at least one of the factors. So there are $ a,b$ such that $ a^b\mid p\minus{}1,a\ge 2,b\ge 1,a$ is prime.

Also, $ A\equal{}a^b\plus{}a^{b\minus{}1}\plus{}\ldots\plus{}1\le a^b\plus{}\frac{a^b}{2}\plus{}\ldots\plus{}\frac{a^b}{2^b}\equal{}a^b\frac{2^{b\plus{}1}\minus{}1}{2\minus{}1}<2a^b$

Now, $ A\equal{}mp,m\in\mathbb{N}$
If $ m\ge 2$, then $ A\ge 2p>2p\minus{}2\ge 2a^b$, impossible, so $ A\equal{}p$

$ A\equiv 1\pmod a^b$, because $ a^b\mid p\minus{}1$

$ a(a^{b\minus{}2}\plus{}\ldots\plus{}1)\equiv 0\pmod a^b$, if $ b\ge 2$. Left side isn't divisible by $ a^2$ so $ b\equal{}1$

$ p\minus{}1\equal{}a$, $ p$ and $ a$ are primes, so $ p\equal{}3$, and $ a\equal{}2$

[/hide]

[quote="Bugi"]
Also, $ A \equal{} a^b \plus{} a^{b \minus{} 1} \plus{} \ldots \plus{} 1\le a^b \plus{} \frac {a^b}{2} \plus{} \ldots \plus{} \frac {a^b}{2^b} \equal{} a^b\frac {2^{b \plus{} 1} \minus{} 1}{2 \minus{} 1} < 2a^b$

[/quote]
This argument is not right, it should be
$ A\equal{}\frac{a}{a\minus{}1}a^b \minus{} 1/(1\minus{}a)\leq2a^{b}$


[quote="artofproblemsolving"]Prove or disprove that every positive rational number can be written in
the form
(a^2 + b^3)/(c^5 + d^7)
where a, b, c, d are positive integers.[/quote]
Yes .this problem is true . We will prove the more general problem . Let $ a,b,c,d$ be four positive numbers which satisfy  $ gcd(ab,cd) \equal{} 1$
then all positive rational number can represent in the form :
\[ \frac {x^a \plus{} y^b}{z^c \plus{} t^d}\]
where $ x,y,z,t$ are positive integers .
[b]Lemma 1[/b]  :All natural numbers can represent in the form $ \frac {x^a \plus{} y^b}{z^c \plus{} t^d}$ .
Proof : The first by the chinese remainder theorem ,exist an integer k such that :
\[ k\equiv 0 (\mod ab); k\equiv 1 (\mod c); k\equiv 1 (\mod d )\]
If we choose $ x \equal{} n^{\frac {k}{a}},y \equal{} n^{\frac {k}{b}},z \equal{} n^{\frac {k \minus{} 1}{c}},t \equal{} n^{\frac {k \minus{} 1}{d}}$ then we have
\[ \frac {x^a \plus{} y^b}{z^c \plus{} t^d} \equal{} \frac {n^k \plus{} n^k}{n^{k \minus{} 1} \plus{} n^{k \minus{} 1}} \equal{} n\]
Lemma 1 has been proven .
[b]Lemma 2 [/b]: Let q be an positive integer , then exist an integer $ m$ such that $ mq$ is c-power and d-power .We represent $ q$ in the form : $ \prod_{i \equal{} 1}^r p_i^{a_i}$. For each $ a_i$ exist an positive integer $ b_i$ such that
\[ b_i \plus{} a_i\equiv 0 (\mod cd)\]
Then we only need to choose $ m \equal{} \prod_{i \equal{} 1}^r p_i^{b_i}$ .
Ok ,now consider our problem . Consider a rational number $ \frac {p}{q}$ .By lemma 2 ,exist an positive integer $ m$ such that $ mq$ is a cd-power , so we can write it in the form $ u^{cd}$ . By lemma 1 ,exist x,y,z,t such that
\[ mp \equal{} \frac {x^a \plus{} y^b}{z^c \plus{} t^d}\]
Therefore ,we have :
\[ \frac {p}{q} \equal{} \frac {mp}{mq} \equal{} \frac {x^a \plus{} y^b}{(z.u^d))^c \plus{} (t.u^c)^d}\]
so if we chose $ z' \equal{} z.u^d,t' \equal{} t.u^c$ then $ \frac {p}{q}$ can represent in the form : $ \frac {x^a \plus{} y^b}{z'^c \plus{} t'^d}$ ,that ends the proof .

There exists an easier solution to the proposed problem;

[i]Proposition.[/i] Let $ a,b,c,d$ be four positive integers satisfying $ (ab,cd)\equal{}1$. Prove all rational numbers can be represented in the form
$ \frac{x^a \plus{} y^b}{z^c \plus{} w^d}$ where $ x,y,z,w$ are arbitrary integers.

Proof) We see by the conditions that $ (a,c) \equal{} (b,d) \equal{} 1$.
Let $ p_1 , p_2 , p_3 , p_4 , q_1 , q_2 , q_3 , q_4$ be positive integers satisfying
$ p_1 a \minus{} p_2 c \equal{} 1$
$ p_3 c \minus{} p_4 a \equal{} 1$
$ q_1 b \minus{} q_2 d \equal{} 1$
$ q_3 d \minus{} q_4 b \equal{} 1$
(It is trivial that such integers exist)
Let $ x \equal{} m^{p_1} n^{p_4}$, $ y \equal{} m^{p_2} n^{p_3}$, $ z\equal{}m^{q_1} n^{q_4}$, $ w\equal{}m^{q_2} n^{q_3}$.
Observe that by the conditions, $ \frac{x^a}{z^c} \equal{} \frac{y^b}{w^d} \equal{} \frac{m}{n}$
Therefore $ \frac{x^a \plus{} y^b}{z^c \plus{} w^d} \equal{} \frac{m}{n}$, as desired.


Guys, λίγη βοήθεια γιατί το μυαλό μου έχει γίνει χυλός και δε βλέπω μπροστά μου...

Ας θεωρήσουμε δύο σύνολα $ A, B$.

Aν υπάρχει συνάρτηση $ f: A\to B$ η οποία είναι 1-1, τότε είναι απλό να κατασκευάσεις μία συνάρτηση $ g: B\to A$ η οποία να είναι επί. Αν $ b\in B$ και $ b \equal{} f(a)$ για κάποιο $ a\in A$, απλώς θέτεις $ g(b) \equal{} a$; αντίθετα, αν $ b\notin f(A)$ διαλέγεις ένα $ a'\in A$ και θέτεις $ g(b) \equal{} a'\forall b\notin f(A)$. Κανένα πρόβλημα, η $ g$ είναι μία αριστερή αντίστροφη της $ f$: $ g\circ f \equal{} \textrm{id}_A$.

Τώρα, αν ξεκινήσουμε με μία συνάρτηση $ g: B\to A$ η οποία είναι επί, ένας τρόπος να κατασκευάσουμε μία δεξιά αντίστροφη της $ g$ είναι μέσω της απεικόνισης $ f: a\mapsto b \in g^{ \minus{} 1}(a)$, όπου $ b$ κάποιο στοιχείο του $ g^{ \minus{} 1}(a)$ που επιλέγεται αυθαίρετα για κάθε $ a\in A$. Έτσι, $ g\circ f \equal{} \textrm{id}_A$.

Εδώ είναι το ερώτημά μου: στο παραπάνω επιχείρημα, όταν λέω ότι "για κάθε $ a\in A$ επιλέγουμε $ b\in g^{ \minus{} 1}(a)$...", δεν επικαλούμαι σιωπηρά το αξίωμα της επιλογής? Υπό μία έννοια, καταλήγω να [i]κατασκευάζω[/i] ένα στοιχείο του $ \prod_{a\in A} g^{ \minus{} 1}(a) \equal{} \textrm{Maps}(A\to\cup_{a\in A} g^{ \minus{} 1}(a)) \equal{} \textrm{Maps}(A\to B)$, δεν είναι έτσι?

Ωστόσο, είναι δυνατό να χτυπάει το αξίωμα της επιλογής τόσο νωρίς... ή είμαι απλώς ηλίθιος?

Μάλλον το δεύτερο... αλλά δείξτε κατανόηση γιατί ο εγκέφαλος μου δε λειτουργεί ;) Ό,τι έχετε ευχαρίστηση... :)

Cheerio,

Durandal 1707

Μάλλον το πρώτο. Σε όσα βιβλία κοίταξα (όχι πολλά) το είχε ως συνέπεια του ΑΕ. Δες πχ. [url=http://www.math.ntua.gr/logic/set-theory/synola-2.pdf]εδώ[/url] σελίδα 26.

Kai omws tha peraseis opwsdipote meso tou aksiomatos tis epilogis  :wink:

Thanks guys, είχα κι εγώ καταλήξει στο προφανές με το που σταμάτησα να είμαι τυφλός και είδα γραμμένο το post μου... αλλά thanks και πάλι :)

Ωστόσο, κοιτάξτε ποια ήταν η πραγματική απορία μου (που εξαιτίας της αναρωτιόμουν για το παραπάνω)...

Για δύο σύνολα $ A,B$, μπορείς να ορίσεις ότι $ |A|\leq|B|$:
1. Αν υπάρχει 1-1 απεικόνιση $ A\to B$
2. Αν υπάρχει επί απεικόνιση $ B\to A$

Oι ορισμοί 1-2 είναι προφανώς ισοδύναμοι υπό AC (μάλιστα, αν δεν κάνω λάθος, ο 1 συνεπάγεται τον 2 υπό πολύ ασθενέστερα αξιώματα επιλογής (ADC, ACC, whatever...)).

Ωστόσο, αν δεν δεχτούμε το AC, η ισοδυναμία των 1-2 δεν είναι πια προφανής. Μπορεί να κατασκευαστεί ένα αντιπαράδειγμα όπου να μπορεί να αποδειχθεί ότι (χωρίς AC) οι δύο ορισμοί δεν οδηγούν απαραίτητα στο ίδιο συμπέρασμα?

Cheerio,

Durandal 1707

[quote="labradford"]Πρέπει να είναι ισοδύναμα με το AC. Δες στο "Axiom of choice" του Herrlich, σελίδα 13, άσκηση Ε4.
Η ισοδυναμία που ψάχνεις πρέπει να είναι η α <->β. Το βιβλίο μπορείς να το βρεις στο [url=http://www.eknigu.org/]eknigu[/url] για κατέβασμα![/quote]

Δεν μπόρεσα να κατεβάσω το βιβλίο, οπότε δεν είμαι και σίγουρος τι εννοείς :(

Nevermind πάντως γιατί, ούτως ή άλλως, την είχα λύσει την απορία μου :) Thanks though!

Cheerio,

Durandal 1707

Γιατί; Μια χαρά κατεβαίνει..! Είναι από τα λίγα σε pdf.
http://www.eknigu.org/lib/M_Mathematics/MA_Algebra/MAml_Mathematical%20logic/
έχει και μερικές ωραίες ασκησούλες, κοίταξέ το... :)

[quote="labradford"][url=http://rapidshare.com/files/145848246/Herrlich_H._Axiom_of_choice__LNM1876__Springer__2006__ISBN_3540309896__206s__MAml_.pdf]Να'το[/url]
Επανακυκλοφόρησε και πριν από ένα μήνα το γνωστό [url=http://www.amazon.com/Axiom-Choice-Thomas-J-Jech/dp/0486466248/ref=sr_1_1?ie=UTF8&s=books&qid=1221596056&sr=8-1]axiom of choice του jech[/url] στην τιμή των... 15 ευρώ... Πρέπει να είναι από τα πιο φτηνά βιβλία μαθηματικών εκεί έξω. Περιμένω τον ταχυδρόμο :)[/quote]

Thx for the heads up :)

Cheerio,

Durandal 1707


A metal sphere with diameter 6 cm is dropped into a cylinder of water so that the metal sphere is completely submerged inthe water. The radius of the cylinder is 5 cm, and the water level is 6 cm high before the metal sphere is placed into the container. How high is the water level of the cylinder after the metal sphere is placed into the container?

[hide]The volume of the water in the cylinder is $150\pi.$ The volume of the sphere is $36\pi.$ The combined volume is $186\pi.$

The base of the cylinder is $25\pi,$ so the water level is now at $\frac{186\pi}{25\pi},$ or $\boxed{7.44\text{ inches}}.$[/hide]

so you just basically find the total volume and put it in a fraction over the area of the base . gotcha  :P

what he was doing was using the formula base*height=volume, but instead of volume as the unknown, height was the unknown, and he made it height=volume/base. That's [u]why[/u]

[quote="Go Around the Tree"]you just basically find the total volume and put it in a fraction over the area of the base[/quote]


Welcome to Mafia Game 24.

On voting, up to two people can be on the block (accused and seconded) at once. On days where the vote is stalling, a deadline will be placed. Now the roles: (see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38004[/url]  for discriptions on [b]standard roles[/b])

It is a mafia vs. citizens vs. cultists game

[i]Mafia's side[/i]
The mafia wins if they are no longer outnumbered by the rest of the roles in the game.

3 Mafia-[b]Standard roles[/b] EXCEPT one of them is the mafia leader who must send the kill out at night. This mafia leader cannot be inducted into the cult (see below). If the mafia leader is killed, then a new mafia leader is chosen from the remaining loyal mafia.

1 Devil- [b]standard role[/b]

1 Thief-The thief can "rob" a person's action for a night. for instance, if the inspector inspects someone, and the thief "robs" him/her, the person will not be inspected. To take away a group's action, the thief must target the group's leader. Although the thief and mafia are on the same side, the mafia and the thief are unaware of each other's identities.

[i]Cultists side[/i]
The cult wins if their kind is the last remaining left in the game.
in the event of the mafia dying out and the # of citizens and cultists being the same, the cultists win.

Cult Leader-may induct one person into their cult every odd night. The person becomes part of the cult. The person still keeps his/her old night action and can pretend to still be that role. The Mafia leader and the mason leader (see below) cannot be inducted into the cult. Cult members only know their cult leader. No game-related communication is allowed between cult members and their leader at anytime.

2 cult members-members of the cult mentioned above.

[i]Civilian side[/i]
The Civilian side wins if their kind is the last remaining kind in the game.

5 citizens-[b]standard role[/b]. One of these is a miller, who shows up as mafia when inspected.

1 angel-[b]standard role[/b]

1 inspector-[b]standard role[/b]

1 priest-[b]standard role[/b]

1 magician-chooses two people in the night and switches them. Anything that would happen in the night to one person now happens to the other person.

1 agent-the agent inspects people in the night to see if they are cult members.

2 masons-one of these masons is the mason leader. This mason cannot be inducted into the cult and is the one who ultimately inducts people into the mason society. People who are inducted keep their old roles. For instance, if the mafia is inducted, the person is still loyal to the mafia.  Mason members know the identity of all the other masons and can talk to any other mason during night.
________

The following people will be playing:

Perfect628
Besttate
Fraxia
Slamchess
Weiqan
Pianoforte
Lena
Laughinghead505
dts
Fireemblem13
Peter VDD
Jashper
Jli
Jackdillon
Ars
Daine
Ignite168
Junggi
Mustafa
Koda

Day 1 has begun.

Just want to say, a 5 day limit is WAY WAY WAY too short. I seriously recommend making this at least 2 weeks, although a deadline should ideally only be placed when discussion is lagging, not all the time.

[quote="Koda"]21 people??
Another boring game is on sight.[/quote]

What's wrong with 21 people?

[quote="Lena"]when is it day in Game 24?

3 pm EST was 1 hour and 50 minutes ago[/quote]

It's already started I think.  :lol:

it is currently US Eastern Time 4:36 PM.

there should have been an announcement of that

This shouldn't start as i never recieved a role pm. when pming, is it case-sensitive? if so, it makes sense.

day has started. Perfect628, I accidently forgot to PM you because you were the last to sign up.

to clarify: masons know who else are masons; they do not know the other masons' other more important role.

Hey folks, glad to be here!  Can't wait to get going :)

Ars, did you even read the above post?

Try again, and pay closer attention this time... :wink:

I did your the one being ignorant

Apathy = no one cares which is not true in this game
I care but alone I can't do much

you made it into someone Mr. Thy

I switched myself and the person who was the largest source of apathy

try figuring out who

It was a joke post!

I was saying the game had been killed by apathy.

[quote="Peter"]I think many of us care, but since kheg is not doing his job, there's really not much we can do...[/quote]

I suspect "many" is an overestimation though...

[quote="Temperal"][quote="1=2"][quote="Peter"]Since kheg is not doing his job[/quote]

APATHY!!! Someone, excluding me, since I'm young and inexperienced, should replace him.[/quote]

Well, the good [b]do[/b] have to die young.

Who's next on the mod list?[/quote]
um how about we just screw this game?
it's unfair to give another person a messed up game and take away his/her right to start a new game

How about miyomiyo? He's on the end of the list, and he could just finish this game and mod 31 as well, in something like 10 years.

EDIT:  I can't take over; I'm going away to camp for a month on Saturday.  Good luck with the game, though!

Well, Lena the mafia is dead, so I don't see why not. I'll try to get the list of survivors.

Sorry for posting way too much in this game, I'm trying to get 26 up and running.

Er, I think I'm just going to end this game.

MOD-ABANDONED.


Circle $ A$ has radius $ 100$. Circle $ B$ has an integer radius $ r<100$ and remains internally tangent to circle $ A$ as it rolls once around the circumference of circle $ A$. The two circles have the same points of tangency at the beginning and end of circle $ B$'s trip. How many possible values can $ r$ have?

$ \textbf{(A)}\ 4 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ 50 \qquad
\textbf{(E)}\ 90$

[hide]since the radius of circle A is 100, its circumference is $ 200\pi$. the circumference of circle B must be an integer factor of this in order for the points of tangency to coincide after the rotation. however, this circumference must also be divisible by $ 2\pi$, so that the radius of circle B is an integer. this problem becomes a search for factors of 100. $ 100 \equal{} 2^2*5^2$. there are then 3 choices for the number of 2s (0, 1, or 2) and 3 choices for the number of 5s (0, 1, or 2). therefore there are $ 3*3 \equal{} 9$ factors of 100. however, one of them is 100 itself, and r<100. so there are $ 9 \minus{} 1 \equal{} 8$ possible values for r.

(B)[/hide]

[hide]We know the circumference of Circle $ A$ and $ B$ are $ r*2\pi$ and $ 100*2\pi \equal{} 200\pi$, respectively.
We want to find the values of $ r$ such that $ \frac {200\pi}{2r\pi} \equal{} \frac {100}{r}$ is an integer > 0.
The possible values of $ r$ which satisfy the condition are $ 1, 2, 4, 5, 10, 20, 25, 50, 100$. However, as specified in the problem, $ r < 100$. Hence, There are $ 10 \minus{} 1 \equal{} 9$ possibilities of r.[/hide]

Except, you counted wrong :)

It's $ 9\minus{}1\equal{}8$.

[hide]As everyone else said, you  have to find all the factors of $ 200 = 5^{2}*2^{3}$, giving $ 3*4 = 12$ possibilities.  However, $ r$ has to be an integer, and since we know that $ 2\pir$ has to divide $ 200\pi$, any odd factor of $ 200$ can't be counted.  This takes away $ 3$ possibilities, $ 5$, $ 25$, and $ 1$.  Also, we can't use $ 200$, since then $ r$ would equal $ 100$ (but $ r < 100$).  Therefore, there are $ 12 - 4 = 8$ possibilities.[/hide]

Sorry for the huge bump for a small question.

Can someone explain why we divide by $2\pi$?


Er, what if qualifying wasn't your goal?  I easily qualified, but I didn't make my goal.

The topic should be called "Did you qualify" lol

i wanted to make AIME through 10A, and i got 121.5. Although i wanted to get a 140+, i am still happy i made AIME.

Yeah I qualified, but got nowhere near my goal of 140+ on the 12.

I qualified, got a 133.5 on the 10A, but that was nowhere near my goal of 150. 1 stupid mistake, 1 misbubble and 1 feeling stupid, didn't want to solve.

[quote="archimedes1"]I qualify with a 132, I believe.  My goal after taking all the previous AMC 10s was to get a 144-150, but looking back I see that was a pretty stupid goal because things are always different in the test room for me, for some reason.[/quote]

thats exactly how i feel.

except i got 126 -_-

I'm going to use implied bragging like Hamster =)... I failed by about 12 points.  I almost got a 162 if I had seen those last two bubbles on the back ~_-.  :)

[quote="Hamster1800"]I qualified, but fell short of my goal by 6 points :( Yeah, my goal was 156. :P[/quote]

yeah well i got 127.5 above my goal which was -6.

I came, I saw, I fainted.

126 on the 12.

Ppl are talking about 150s and I didn't even qualify... :noo:  :stink:  :(

Oh, well, there's always a B. I better qualify on that one. *sniffle sniffle*

Well, my goal was to do better than last time, which i failed.  126 then 124.5 on the AMC 10A

My goal was around 130 on 12A, and I would have made it, but I made 3 stupid mistakes, which is about average for me. All the other freshmen at my school chickened out and took the 10A. Which they all failed miserably.

/me got 114 on 12A :(

My goal was a perfect. I failed to do that.

I don't see why people are complaining about missing their goals. My goal was a perfect, but I didn't even get to take it, so I got pretty much a 0. That is 150 below my goal. Somebody try to beat that without looking ridiculous.

I scored a 126 on the 10. Not too happy, but glad I still passed. And I hope to qualify for the USAMO, although quite unlikely. If I had failed, I wouldn't even make AIME this year because I can't take the B tests.


geometry
trigonometry
circumcircle
angle bisector
geometry proposed

In triangle $ABC$, $M$ is midpoint of $AC$, and $D$ is a point on $BC$ such that $DB=DM$. We know that $2BC^{2}-AC^{2}=AB.AC$. Prove that \[BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}\]

[color=darkred][b][u]Lemma.[/u] Let $\triangle ABC$ and the points $\{\begin{array}{c}M\in (AC)\ ,\ MA=MB\\\ D\in (BC)\ ,\ \widehat{DAB}\equiv\widehat{DAC}\end{array}$. Then[/b] $\boxed{\ DM=DB\Longleftrightarrow b=2c\ \ \vee\ \ 2a^{2}=b(b+c)\ }$.[hide="Proof of the lemma."] Apply the generalized Pythagoras' theorem to the common vertex $C$ in $\triangle ABC$ and $\triangle MDC$ : $\{\begin{array}{c}2ab\cos C=a^{2}+b^{2}-c^{2}\\\ DM^{2}=DC^{2}+MC^{2}-2\cdot DC\cdot MC\cdot\cos C\end{array}$.
Are well-known the relations $\frac{DB}{c}=\frac{DC}{b}=\frac{a}{b+c}$. Therefore, $DM=DB$ $\Longleftrightarrow$ $DB^{2}=DC^{2}+MC^{2}-2\cdot DC\cdot MC\cdot\cos C$ $\Longleftrightarrow$ $(DB+DC)(DB-DC)=MC^{2}-2\cdot DC\cdot MC\cdot\cos C$ $\Longleftrightarrow$ $a\cdot\frac{a(c-b)}{b+c}=\frac{b^{2}}{4}-b\cdot \frac{ab}{b+c}\cdot\frac{a^{2}+b^{2}-c^{2}}{2ab}$ $\Longleftrightarrow$ $4a^{2}c-4a^{2}b=b^{3}+b^{2}c-2a^{2}b-2b^{3}+2bc^{2}$ $\Longleftrightarrow$ $4a^{2}c-2a^{2}b=b^{2}c-b^{3}+2bc^{2}$ $\Longleftrightarrow$ $2a^{2}(2c-b)=b(2c^{2}+bc-b^{2})$ $\Longleftrightarrow$ $2a^{2}(2c-b)=b(b+c)(2c-b)$ $\Longleftrightarrow$ $b=2c$ or $2a^{2}=b(b+c)$.[/hide][/color]
[color=darkblue][b][u]Remark.[/u] Using the above simple lemma this problem becomes very easily and "painted nastily" ![/b][/color]

let $P$ be a point on line $AC$ and at side of segment $AC$ such that $AP=AB$

$2BC^{2}-AC^{2}=AB.AC \longleftrightarrow CM.CP=BC^{2}$ so circumcircle of $\triangle PBM$ is tangent to $BC$ at $B$

and cause $DM=BD$ , $DM$ is tangent to the circle ,too.

we have $\angle ABP=\angle MPB=\angle DBM=\angle DMB$ hence $\angle MDC=\angle BDM+\angle BMD=\angle A$ so

quadrilateral $MDBA$ is cyclic,hence $\angle DBM=\angle DMB=\angle CAD=\angle BAD$ so $AD$ is anglebisector of

$\angle A$ therfore $AD$ is perpendicular to $BM$ so $c=AM=\frac{b}{2}$
we have:

$CM.AC=CD.BC \rightarrow \frac{b^{2}}{2}=\frac{ab}{b+c}.a \Longrightarrow a^{2}=3c^{2}=\frac{3b^{2}}{4}$$\star$

so
$BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}$

My solution is a little bit similar as Amir.S 's:
Let $B'$ be the piont on the line $AC$ and $B'A=BA$ and $B',A,C$ lie in order.
Let $A'$ be the point on the line $BC$ with the properth $A'B=CB$ and $A'B,C$ lie in order.
then $2BC^{2}-AC^{2}=AB*AC-\to A,B,A',B$lie on a circle and $\triangle CAB \sim \triangle CA'B'$
and we easy to check that $\triangle CB'B\sim \triangle CBM,\triangle BMD \sim \triangle B'BA \triangle MDC\sim \triangle BAC$
in fact $A-\to A',B-\to B',M\to M'=B, D--\to D'=A$
then we have that $\frac{BD*DC}{B'A*AC}=(\frac{BC}{B'C})^{2}=\frac{AC^{2}AB}{2(AB+AC)}$

[quote="radio"][quote="Amir.S"]therfore $ AD$ is perpendicular to $ BM$ [/quote]

Why is this?[/quote]
yes,thanks. that was wrong ,but I didn't use it in rest of solution,Proving that $ AMDB$ is cyclic is enough to complete the solution.

How, Amir.S? So you won't get $ c\equal{}\frac{b}{2}$...

[quote="radio"]How, Amir.S? So you won't get $ c \equal{} \frac {b}{2}$...[/quote]
That's not nessecery also.
as i see know the problem is more easier than I thuaght,The relation $ CB^2 \equal{} CM.CP$ and the fact that D lies on angle bisector is enough to solve the problem.
this relation is equal to $ a^2 \equal{} \frac {b(b \plus{} c)}{2}\Longleftrightarrow \frac {a^2bc}{(b \plus{} c)^2} \equal{} \frac {b^2c}{2(b \plus{} c)}\Longleftrightarrow BD.DC \equal{} \frac {AC^2.AB}{2(AB \plus{} AC)}$cause:$ BD \equal{} \frac {ac}{b \plus{} c}\ ,DC \equal{} \frac {ab}{b \plus{} c}$
thanks again  :wink:

-Amir.S

A nice problem :wink:

Consider point $ K$ on ray $ CA$, in which $ AK\equal{}AB$,

Simply we can show that $ CM.CK\equal{}CB^2$

It means $ \triangle CBM$ and $ \triangle CKB$ are similar to each other,and:

$ \angle CBM\equal{}\angle AKB \Longleftrightarrow ABDM$is cyclic.

According to as inversion lemma we have:

$ A'B'\equal{}\frac{k.AB}{OA.OB}$

(which $ k$, as radius and $ O$,as the center of inversion,and$ A',B'$ the inverse points of $ A,B$respectively)

then: $ AB\equal{}\frac{\frac{AC^2}{2}.BD}{CD.\frac{AC}{2}}\equal{}\frac{AC.BD}{CD}\equal{}\frac{2CM.BD}{CD}$

and: $ AC\equal{}2CM$

according to these equalities,lets find out the value of $ CD.BD$:

$ CD.BD\Longleftrightarrow \frac{CD.CB.BD}{CB}\Longleftrightarrow \frac{CM.CA.BD}{CB}$

$ \Longleftrightarrow \frac{2CM^2.BD}{\frac{BC}{CD}.CD}$

$ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2.2CM(\frac{BC}{BD}).CD}$

$ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2.2CM(\frac{BD}{CD}\plus{}1).CD}$

$ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2(\frac{2CM.BD}{CD}\plus{}2CM).CD}$

$ \Longleftrightarrow \frac{(4CM^2)(2CM.BD)}{2(AB\plus{}AC).CD}$

$ \Longleftrightarrow \frac{AC^2.AB}{2(AB\plus{}AC)}$

and we proved that: $ CD.BD\equal{} \frac{AC^2.AB}{2(AB\plus{}AC)}$

hmmm,but I really don’t like it! :mad: