Princeton college probability topology search function Harvard

I have the opportunity to do the Budapest Semesters in Mathematics program this spring. I know a few people who have done it, and they all seemed to have had a great time. However, I'm still not sure and I'd like some advice, especially from anyone who has done it or knows anything about the program/university education in Hungary.

Pros:
- Everyone says it's really fun.
- Studying abroad is a good experience.
- I'm a British citizen as well as one of the USA, so I can pretend to be British so people will hate me less. Also EU membership probably helps a lot.

Cons:
- Although people tell me that the math department there is really good, I'm at Princeton, which is arguably the best in the world (I am interested in analysis; you might disagree if you're in a different field). Why shouldn't I savor all 8 chances to study with people like Daubechies and Fefferman and Stein?
- By the end of the upcoming fall semester, I'll have taken courses up to the intro graduate level in real and complex analysis and probability theory. I don't know know anything about the graduate program there and whether appropriate classes and/or research opportunities would be available in analysis. I could just go and take courses in algebra and topology and stuff like that, but I feel I would be unhappy doing no analysis for an entire semester.
- Although tuition, room, and board are all much cheaper for this program than for Princeton, I'm not exactly sure how financial aid works; it's possible I'll end up paying the same amount or more than if I stayed at Princeton.
- I'm not too interested in learning Hungarian. If I learn another language, I'd much prefer it to be German or Chinese. I'd hate to half-way learn Hungarian by staying there for a semester, and then just one year later have to start learning German for grad school and keep getting confused between my memories of Hungarian and the German I'm learning. I'm not that great with languages...

Anyway, any advice would be most welcome. Feel free to try to convince me in either direction or to just clarify my views on any of the above points.

The short version: yes.  ;)

When I was there, there were good functional analysis and analytic number theory classes among the harder group of courses.  (Some courses are easier/for less experienced people; you won't take any of those if you go.)  I personally strongly recommend the graph theory and combinatorics classes, and the problem-solving-y class (I forget its name) is reputed to be most excellent.

I have to run right now, but you might search through past posts to see what I and others have said about BSM before -- I'll try to address your comments more specifically, probably sometime tomorrow if I remember.

Oh yeah here's the link in case anyone else is interested. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68721
There's lots of useful information there. Unfortunately it doesn't really address my particular concerns... Mainly I am concerned that I have already done too much math and the classes won't be hard enough (except the ones in fields that I'm not terribly interested in, but maybe it's good for me to explore other stuff a bit), and that the other students won't be at the same level as I am. I even have that problem at Princeton -- there are certainly plenty other people as good at math or better than I am in some general sense, but I really have no one to talk to about the analysis topics I'm interested in, which is sometimes rather depressing.

I looked at the list of profs on their website and it seemed there were a few who might be interested in topics that I'd want to work on. (Note: I have to do at least one piece of junior independent work that semester for Princeton.) It seems like most of the profs are really into combinatorics and graph theory. But some of them work on the sort of number theory problems I was interested in lately, like the arithmetic progressions in the primes problem. And there was one guy who seemed like he might do some geometric analysis...

I also looked at the course offerings for Spring 07. There are definitely classes that cover topics that I don't know, but the most interesting ones by title (Topics in Analysis, Mathematics of Fractals, Probability Theory, Functional Analysis, Real Functions and Measures) all either intersect nontrivially with or are contained in what I've already done.

edit: the analytic number theory class sounds pretty exciting.

[quote="tokenadult"]How much time have you spent in other countries? If not much, you should go for that reason alone.[/quote]
The first year and a half of my life, and none since then. But there's a good chance I'd study in Europe after I graduate (either for grad school or with one of those English things that gives me 1 or 2 years to study there between undergrad and grad school), and I'll be going to Bulgaria this summer for a week and all over Europe for several weeks next summer on vacation. So it's not like there won't be other opportunities...

Going from the USA to Europe for graduate studies in mathematics? I see no reason why... :maybe: The "English thing" looks like a better idea (although I don't really know what it is), because starting a grad school a year later will probably work to your advantage.

Short thoughts on your specific concerns:

[quote]- Although people tell me that the math department there is really good, I'm at Princeton, which is arguably the best in the world (I am interested in analysis; you might disagree if you're in a different field). Why shouldn't I savor all 8 chances to study with people like Daubechies and Fefferman and Stein?[/quote]
Because the teaching will be better.  (Actually, I have no idea who any of those people are, so this is a guess.  Certainly, the teaching is better in Budapest than at Harvard.)  Also, because Budapest is much more interesting than Princeton, NJ :-p

[quote]- By the end of the upcoming fall semester, I'll have taken courses up to the intro graduate level in real and complex analysis and probability theory. I don't know know anything about the graduate program there and whether appropriate classes and/or research opportunities would be available in analysis. I could just go and take courses in algebra and topology and stuff like that, but I feel I would be unhappy doing no analysis for an entire semester.[/quote]
There is no "graduate program" -- BSM isn't a university, and its professors are not permanently affiliated with it.  There are courses at a variety of levels.  Better-prepared students (which usually includes those from the "big name" schools, i.e. Harvard, MIT, Stanford, etc.) take the harder classes.  I don't think you'll have to worry too much about finding "hard enough" things to do or other smart students (everyone there has, after all, enrolled in a study abroad program in order to study mathematics) -- you can always just take many classes, and I think there's always at least a handful of really talented people -- but finding hard enough things in your particular area of interest may be tricky, especially since you've taken several courses in analysis.  It might be worthwhile to talk to someone at the program about this.

[quote]- Although tuition, room, and board are all much cheaper for this program than for Princeton, I'm not exactly sure how financial aid works; it's possible I'll end up paying the same amount or more than if I stayed at Princeton.[/quote] You'll have to talk to them and talk to Princeton about this.  If you're paying less than 10,000 USD to Princeton total, my guess is that you'll more-or-less break even.  If you're paying more like 0 to 5,000 USD to Princeton, you might end up paying more to go there, and I don't think there's much way around this.

[quote]- I'm not too interested in learning Hungarian. If I learn another language, I'd much prefer it to be German or Chinese. I'd hate to half-way learn Hungarian by staying there for a semester, and then just one year later have to start learning German for grad school and keep getting confused between my memories of Hungarian and the German I'm learning. I'm not that great with languages... [/quote]  No helpful advice here, except that Austria and Germany are both relatively close by, so you could always head there to practice your German. :)  Hungarian is hard and most of us just learn a little bit and then forget it all.  It is also possible that Princeton will force you to take Hungarian.

As a second-language speaker of Chinese (who has also studied German from time to time, and can read German in my subjects of study), I would say that you SHOULD be able to pick up enough radio broadcasts in German while living in Hungary to help your German along a good bit. Listening to the radio (NOT watching TV) can be quite helpful for developing reading fluency, as it gets your brain used to following along with the language at native speaker speed of speech.

Of course, it also ought to be possible to listen to Internet radio in almost any language from your computer in Princeton, NJ, so this is a fairly minor reason to travel to central Europe.

For more on language learning in general, see

http://learninfreedom.org/languagebooks.html

Can you tell I'm a big fan of study abroad experiences?

[quote="statistical"]Knowledge of German would allow one to operate relatively easily in Hungary.  As far as I know it is the most popular foreign language for Hungarians to learn, although I have heard that there is some increase in the popularity of English.[/quote]

If you have some questions regarding learning German you may consult me.  :D

[quote="JBL"]Because the teaching will be better.[/quote]
I wouldn't be surprised if the teaching is better on average. But the teaching at Princeton doesn't suck. It's not like I'm dying to get away from it.

[quote="JBL"]Also, because Budapest is much more interesting than Princeton, NJ :-p[/quote]
This is a good reason. ;-)

[quote="JBL"] If you're paying less than 10,000 USD to Princeton total, my guess is that you'll more-or-less break even.  If you're paying more like 0 to 5,000 USD to Princeton, you might end up paying more to go there, and I don't think there's much way around this.[/quote]
Unfortunately, next year I'm expecting to pay somewhere in the range of -2500 to 2500 USD. The cost is actually a pretty serious concern for me, as you should be able to infer from that aid estimate.

Thanks again for all of your advice, everyone. I think I'm going to let my decision rest on the financial aspect. I'll talk to Princeton and see what kind of aid I'll get. I can also wait to find out my 07-08 aid award; the spring application deadline isn't until November and I should know my aid within a month or so.

I would have expected Princeton to routinely provide aid for study abroad, and at any rate this would be the kind of thing that it might be possible to apply for a special grant for. Bon voyage.

I don't have any idea how widely German is spoken in Hungary, but it is not very difficult to navigate Budapest* with English and extremely limited Hungarian -- indeed, this is what most people who attend BSM do.  It is my experience in other countries (e.g. Slovenia) that German is the foreign language of the older portion of the population and that the younger set (certainly including college-age students) prefer English by a substantial margin.

Also, good luck with funding!

* It is worth keeping in mind that this comment is relative to me -- for someone with little experience in big cities, for example, the combination of a large city and not speaking the language might be quite daunting.

My travels in Hungary were not as recent as yours, but a little German went much farther than a lot of English at that time.  I also have heard from Hungarians that the younger generation (at least among the educated classes who might want to go abroad) prefer to learn English. Because so much of the tourism and money, and I guess also television and radio, comes in from German-speaking countries, the merchants, cab drivers, etc have very little choice but to know some German.   There is some residual familiarity with Russian from the old days but I don't know if as many people speak that as French.

As a native Hungarian, I'd have to say that knowing German is better right now than knowing English, although it's starting to catch up. The Hungarian language itself...good luck learning it without soliciting a few laughs :D I was lucky enough to have it as my first language.


advanced fields
advanced fields unsolved

Could you give an example of two non-isomorphic Hilbert spaces.

$ \ell^2(\mathbb Z)$ and $ \ell^2(\mathbb R)$, where $ \mathbb R$ is used as an index set. The second space is not separable: its canonical basis is uncountable.


rules
-a person posts a 5-10 letter word and everyone else has to find anagrams of it.
-when posting an anagram, it must come with the word used in a sentence
-you get points for the number of letters in the anagram and a bonus 3 points if you use all the letters in the original word
-if you cannot think of another anagram to post, post FORFIET. the last person to post an anagram gets the total number of anagrams listed for that word
-the angram must be a rearrangement of the letters(no roots). for example, from computer, you may not use compute.
-post all previously used anagrams
-names are not allowed
-abbreviations are not allowed

EX:
WORD: compact
a
am
map
cat
act
camp
tap
FORFEIT
whoever posted a gets 1 point
whoever posted am gets 2 points
whoever posted map, cat, act, or tap gets 3 points
whoever posted camp gets 4 points
whoever posted tap gets 7 points since there are 7 words listed.

yes i know these are not all the anagrams of compact
i left out the sentence part

post in the form
WORD:*word*
*1st word*
*2nd word*
..
..

lets begin

WORD: circular

thanks abcdeafg

Aren't you supposed to let people join?

Circular: Curl

@Saszs: You're supposed to answer his first.

Palientologist: Toenail

Trigonometry

No, you guys. You stick with 1 word until someone forfeits, then you add the next word. Also, you need a sentence.

Word: CIRCULAR
Rail

I slid on the [b] rail.[/b]

Word: CIRCULAR
rail
rural

[b]Rural[/b] regions tend not to have cities.

Hey! I put: Curl the first time I posted in this thread!

I curl my hair often.

So I post: Crail

I carry my crail with me to picnics.

ha ha

arc

the arc of a semicircle is 180 degrees.

car

I will not be eligible to drive a [b]car[/b] anytime soon.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car

[b]Liar[/b]

A [b]liar[/b] does not always tell the truth.

This should go in the Fun Factory, not the Games section. Mods, please move this.

This, I suppose, is relatively organized discussion, so is possibly non-frivolous?

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial

It is crucial that you keep playing this game.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural

The crural bone is your shinbone.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil

Uracil is a base in RNA that pairs with adenine.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A :P

I bought [b]a[/b] pair of shoes.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A
Circa

Circa 2009 this game was created.

Isn't the point of the game to get as many points as possible? (so I have 18 pts as of now, which I think is higher than anyone else)

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A
Circa
Air

I breathe [b]air[/b].

I have 12 points.

Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A
Circa
Air
Lira

Italy used to use the Lira.

I have 22 points.

Isn't that a proper noun?

Anyway:
Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A
Circa
Air
Lira
Curari

Curari is dangerous in the circulatory system.

18 pts.

Some capitalize it, but others don't. Think: do we consider the dollar a proper noun?

Anyway:
Word: CIRCULAR
Rail
Rural
Curl
Crail
Arc
Lair
Car
Liar
Crucial
Crural
Uracil
A
Circa
Air
Lira
Curari
Curial

I would not live in a curial government system.

28 pts

http://www.mathlinks.ro/viewtopic.php?search_id=1681996282&t=203861

For some reason we start at circular both times

EDIT: 2500th Post, newest Birch-Swinnerton Dyer member!


Determine the number of ways to arrange the letters of the word ALABAMA

Yes; here's how to solve it.

We have $ 7$ letters, and $ 7!$ is 5040. However, we have $ 4$ A's, so we divide by $ 4!$ or $ 24$.

$ \frac{5040}{24} \equal{} \boxed{210}$.

Can't you do $ \frac{7!}{4!}$ and simplify to get $ 3!$ and get $ 6$?

no because 7!=7*6*5*4*3*2*1 and 4!=4*3*2*1

if you divide, it equals 7*6*5, which does NOT equal 3!


induction
search
combinatorics unsolved
combinatorics

Assume that we have "m" coins(m > 2), they're similiar except one of them which has different weight(lighter or heavier)from others. Furthermore we have weighting machine(with two pans).
now fix a natural number like "k", find the largest number of "m" that we can find that coin with different weight with "k" times using the machine!
please solve, and reply it to me
Good luck

It is not so easy, but a famous problem.
Each weighing has 3 possible outcomes (Left < Right, Left > Right, Left = Right).
There are $2m$ possible answers to the problem (which is the bad coin (m) and whether it is light or heavy (2)).
Therefore, $3^k \geq 2m$.
However, not all possible sequences of weighings allow determination of the coin.  If  Left = Right for each weighing, we might identify the false coin but cannot determine whether it is light or heavy.
Also, if Left < Right each time or Left > Right each time (where we have some definite convention for choosing which of two given sets of $i$ coins to place on the Left pan), then we can't tell whether the coin is light and in one group of coins, or heavy and in the other.  So the information-theoretic lower bound is
$3^k - 3 \geq 2m$.   You can reach the bound by at each stage dividing the coins into 3 equal or nearly-equal size subsets, weighing two against each other, and thus reducing from 6 initial possibilities (light/heavy coin in subset 1/2/3) to 2, and repeating (at each stage reducing the number of possibilities by 3).

But here is a point : when two parts are equal you gain some good coin
I finde a better bound : 4*(3^(k-2)) + 1    (this is lower bound)
At first show that you can find it in 12 coin, then show you can finde it in 13 with 3 times.
then by inductoin show this
Goodbye

There is a strong and a weak version of the search problem.
The weak version is to only find the coin.
The strong version is to find the coin and determine whether it is lighter or heavier.
The information-theoretic bound that I described applies to the strong problem.
For 3 weighings (the famous "12 coin problem") the strong version has a bound of 12 and the weak version has a bound of 13, as you mention.

Yes! of course and the problem that Aligator said is the weaker one


geometry
geometric transformation
reflection
similar triangles
geometry proposed

[hide="Solution"]
Let $ X$ and $ Y$ be the feet of the perpendiculars from $ S$ and $ T$ to $ BC$.  Furthermore, allow the reflection of $ D$ across $ M$ be $ E$.  Notice that $ BU \equal{} 2BD$ and $ BC \equal{} 2BM$.  It follows that\[ DM \equal{} BM \minus{} BD \equal{} \frac {BC}{2} \minus{} \frac {BU}{2} \equal{} \frac {UC}{2}\]so $ UC \equal{} 2DM$, $ \frac {UC}{DM} \equal{} 2 \equal{} \frac {BC}{BM}$.  Thus, $ \frac {UC}{BC} \equal{} \frac {BM}{DM}$.  Now, it is well-known that $ AE\parallel MS$ (in fact, $ MI$ is the midline of $ \triangle ADE$, look at Lemma 1 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218243), so $ \frac {AS}{SB} \equal{} \frac {EM}{BM} \equal{} \frac {DM}{MB} \equal{} \frac {BC}{UC}$. Notice that $ SX\perp BC$, $ AL\perp BC$, and $ TY\perp BC$.  It follows that $ SX\parallel AL\parallel TY$.  This gives us that $ \frac {LY}{LX} \equal{} \frac {AT}{AS}$ and we obtain similar triangles $ \triangle SBX\sim \triangle TBY$.  Thus, $ \frac {TY}{SX} \equal{} \frac {BT}{SB}$.  From $ AU\parallel TC$, we see that $ \frac {AT}{BT} \equal{} \frac {UC}{BC} \equal{} \frac {AS}{SB}$.  Thus, $ \frac {LY}{LX} \equal{} \frac {AT}{AS} \equal{} \frac {BT}{SB} \equal{} \frac {TY}{XS}$.  This, combined with the fact that $ \angle SXL \equal{} 90 \equal{} \angle TYL$ gives that $ \triangle TLY\sim \triangle SLX$, so $ \angle LSX \equal{} \angle LTY$.  $ AL\parallel SX\parallel TY$, so $ \angle ALS \equal{} \angle LSX \equal{} \angle LTY \equal{} \angle ALT$, as desired.    [/hide]

[quote="Virgil Nicula"][color=darkred]Let $ \triangle ABC$ with the incircle $ w \equal{} C(I,r)$ . Denote : the midpoint $ M$ of $ [BC]$ ; $ S\in MI\cap AB$ ; $ L\in BC$ so that

$ AL\perp BC$ ; $ D\in BC\cap w$ ; the reflection $ U$ of $ B$ w.r.t. $ D$ ; the point $ T\in AB$ for which $ CT\parallel AU$ . Prove that $ \widehat {ALS}\equiv\widehat {ALT}$ .[/color][/quote]

[color=darkblue][b][u]Remark.[/u][/b] Prove easily that the points $ S$ , $ T$ are harmonical conjugate w.r.t. the

points $ A$ , $ B$ ,  i.e. $ \frac {SA}{SB} \equal{} \frac {TA}{TB} \equal{} \frac {|b \minus{} c|}{a}$ . Since $ LA\perp LB$ results $ \widehat {ALS}\equiv\widehat {ALT}$ .[/color]

To prove the lemma in the QuattoMaster's link, I think there is a faster way, namely using another
 lemma ( 2 ):
“ Two cevians BP and CQ concur on AD iff ÐADQ = ÐADP”, which can be proved by extending BP and CQ until they cut the parallel through A to BC in P’ and Q’ respectively and finding AP’ = AQ’

Now, for the problem: extend AD to D’, on (ABC), see that DM||D’C, hence
ÐADM = ÐAD’C = ÐABC.
According to the lemma (2), ÐADU = ÐABC, therefore DU and AB are perpendicular; similarly for DV, and BCVU is cyclic, done.

Method 2
Let AD cut (ABC) 2nd time at D’. By construction, since HD=DD’, there exists  AM/MC=AD/DD'. Apply Ceva for AD, BM and CU concurrent at E and get  AU/BU=AM.CD/MC.BD and, by the previous equalities we get AU/BU=(AD/BD)^2 , which shows DU and AB perpendicular, similarly for DV and AC, so BCVU is cyclic, done.

Remark: See that, this way, ÐUDN = ÐVDC = |ÐB - ÐC|.

(Here the sign Ð stand for 'angle sign').

Best regards,
sunken rock


I have done a few of the old AIME's from the 80's. I want to start doing more AIME practice.

My question: should I work thourgh old AIME's and THEN do the Mock AIME's, or should I do all the Mock AIME's, and save the regular ones for when I've practiced more?

I don't see what's the difference as long as you get both things done. :maybe:

[quote="t0rajir0u"]Go for the difficult mocks.  Then when you do the real thing you'll think, "Wow, these are so much easier than Mildorf's problems!"   :wink:[/quote]

Awesome.  :)

Who's Mildorf?

He was a IMO participant who made 5 mock AIMEs and like a bazillion mock USAMOs.

[quote="rzsolt"][quote="Iversonfan2007"]yea if you average 9ish on real aimes, you'll average like 5ish on mildorf aimes :wink:[/quote]

lol, mildrof is gonna kill me. I'm kind of excited.  :D[/quote]

yes i get excited when i'm about to get killed too


For all nonnegative reals $ x,y,z,$ we have
\[ x^2(x^2 \minus{} y^2)(x^2 \minus{} z^2) \plus{} y^2(y^2 \minus{} z^2)(y^2 \minus{} x^2) \plus{} z^2(z^2 \minus{} x^2)(z^2 \minus{} y^2)\geq 8(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2.\]

[quote="Ji Chen"]For all nonnegative reals $ x,y,z,$ we have
\[ x^2(x^2 \minus{} y^2)(x^2 \minus{} z^2) \plus{} y^2(y^2 \minus{} z^2)(y^2 \minus{} x^2) \plus{} z^2(z^2 \minus{} x^2)(z^2 \minus{} y^2)\geq 8(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2.
\]
[/quote]
After mixing vars (x-z,y-z,0) LHS decrease RHS const, so we must prove it only for z=0
so
$ x^4(x^2 \minus{} y^2)\plus{} y^4(y^2 \minus{} x^2)\geq 8y^2x^2(x \minus{} y)^2$
$ (x^2\plus{}y^2)(x\plus{}y)^2(x\minus{}y)^2\ge8x^2y^2(x\minus{}y)^2$ but $ (x^2\plus{}y^2)(x\plus{}y)^2\ge 8x^2y^2$ by AM-GM.
Equality when x=y=z or x=y z=0

[quote="dima ukraine"][quote="arqady"][quote="dima ukraine"]
After mixing vars (x-z,y-z,0) LHS decrease [/quote]
Why?[/quote]
z=min(x,y,z)[/quote]
I don't see it. :(  Explain please, why $ LHS$ decreases? Thank you!

[quote="arqady"][/quote][quote="dima ukraine"][quote="arqady"][quote="dima ukraine"]
After mixing vars (x-z,y-z,0) LHS decrease [/quote]
Why?[/quote]
z=min(x,y,z)[/quote][quote="arqady"]
I don't see it. :(  Explain please, why $ LHS$ decreases? Thank you![/quote] $ x\ge y\ge z$
$ z^2(z^2 - x^2)(z^2 - y^2)\ge0$
$ x^2(x^2 - y^2)(x^2 - z^2) +y^2(y^2 - z^2)(y^2 - x^2)=(x^2-y^2)^2(x^2+y^2-z^2)=(x-y)^2(x+y)^2(x^2+y^2-z^2)=(x-y)^2(x+y)^2(x^2+(y-z)(y+z))$

Yes! Now I see it!
Very nice proof, dima ukraine!

[quote="Ji Chen"]For all nonnegative reals $ x,y,z,$ we have

$ x^2\left(x^2 \minus{} y^2\right)\left(x^2 \minus{} z^2\right) \plus{} y^2\left(y^2 \minus{} z^2\right)\left(y^2 \minus{} x^2\right) \plus{} z^2\left(z^2 \minus{} x^2\right)\left(z^2 \minus{} y^2\right)$

$ \geq 8(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2.$[/quote]After dima ukraine's nice proof, we stick down an effective method:

$ \sum{x^2\left(x^2 \minus{} y^2\right)\left(x^2 \minus{} z^2\right)} \minus{}8\prod{(y\minus{}z)^2}$

$ \equiv F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} s \plus{} t)$

$ \equal{} 4\left(s^2\plus{}st\plus{}t^2\right)x^4\plus{}2(2s\plus{}t)\left(s^2\plus{}st\plus{}6t^2\right)x^3$

$ \plus{}t^2\left(s^4\plus{}2s^3t\plus{}45s^2t^2\plus{}44st^3\plus{}13t^4\right)x^2\plus{}2(2s\plus{}t)\left(8s^2\plus{}8st\plus{}3t^2\right)x$

$ \plus{}t^4\left(6s^2\plus{}6st\plus{}t^2\right)\geq 0$,

which is clearly true for $ x\leq y\leq z.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=6558


function
integration
LaTeX
calculus
geometry
real analysis
real analysis unsolved

let $f : [-2,2]\to R$,  $f$ is $C^3[-2,2]$ function such that:

                 $\int_0^2 f(x)dx -\int_{-2}^0 f(x)dx= f(2)-f(-2)$.

    Prove that there is $c$ in $(-2,2)$ such that $f'''(c)=0$.

[b] Moderator edit: Edited for $\LaTeX$ [/b]

just use $f(x)=f(0)+xf'(0)+\frac{x^2}{2}f^2(0)+\frac{x^3}{6}f\"'(a)$
and partly integer,right?
nice problem ;)

can u eleborate? I think there's something wrong here.

It's wrong.Where did you use the relation given for this problem?

Let's proceed by absurd:
$\not \exists c \in (-2,2)$ such that $f'''(c) = 0$
Thus $f'$ is either convex or concave. Let's assume that is convex. For concave case the

proof is the same.

Let $g: [-2,2] \rightarrow \mathbb{R}$, $g = f(x) - f(-x)$
$g$ is odd, so $g'$ is even. And $g \in C^3[-2,2]$. Since $g$ i(s odd, $g(0) = 0$.

We can rewrite the integral:
\[f(2) - f(-2) = g(2) = \int_0^2 f(x)dx - \int_{-2}^0 f(x) dx = \int_0^2 g(x) dx\]
This represent the area of the triangle of verteces $(0,0) ; (0,2) ; (2, g(2))$, so the area

delimited by the function $h(x) = {g(2) \over 2} x$.

Now suppose $g(x) \neq h(x)$ in $(0,2)$.
Thus $\exists x_1,x_2 \in (-2,2)$ such that $g(x_1) > h(x_1)$ and $g(x_2) < h(x_2)$. Suppose

$x_1 < x_2$. It's nearly the same for $x_2 < x_1$.

In this case we have:
$\exists a,b,c \in (0,2)$ with $0 < a < b < c$ such that:
\[f'(a) = {g(x_1) \over x_1}\]
\[f'(b) = {g(x_2) - g(x_1) \over x_2 - x_1}\]
\[f'(a) = {g(2) - g(x_2) \over 2 - x_2}\]
It's easy to see that $g'(a) > g'(b) < g'(c)$. Since $g'$ is even, we also have $g'(-b) < g'(a)$. But this is impossible for $g'$ convexity.

Thus $g(x) = h(x)$. In this case $g'''(x) = 0 \;\;\forall x \in (-2,2)$
It follows that $f'''(x) + f'''(-x) = 0 \;\Rightarrow f'''(x) = -f'''(-x) \;\; \Rightarrow f'$ is odd. And if it is odd, $f'''(0) = 0$. But this is absurd since we have supposed $f''' \neq 0 \forall x\in (-2,2)$.

Thus we have that $\exists c \in (-2,2)$ such that $f'''(c)=0$.


Distinct points $ A$, $ B$, $ C$, and $ D$ lie on a line, with $ AB\equal{}BC\equal{}CD\equal{}1$.  Points $ E$ and $ F$ lie on a second line, parallel to the first, with $ EF\equal{}1$.  A triangle with positive area has three of the six points as its vertices.  How many possible values are there for the area of the triangle?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 6 \qquad
\textbf{(E)}\ 7$

[hide]
The two parallel lines are a fixed distance away from each other, and that's the altitude, so we only care about the base. There are only three possible lengths for the base, so our answer is $ \textbf{(A)}$.
[/hide]

I personally counted 6 distinct triangles, let's assume line segment EF is far to the left, so we can form $\triangle  AEF,\triangle EBF,\triangle EDF,\triangle ACF\triangle ADF\triangle ACE \triangle ADE$ I am going to post an image soon


[hide]Since $-t^2 + 60t + 700$ is a parabola with $a < 0$ the point with maximal altitude is the vertex $V$.
[/hide]

[hide]Rewrite the equation as $t^2-60t-700=t^2-60t+900-1600=(t-30)^2-1600$. Since we multiplied by -1, The highest value is 1600, when t=30. [/hide]

[quote="thebrokenlight"]I assume, just as with the AMC, these problems shouldn't be solved using Calculus?[/quote]

This definitely does NOT need calculus :rotfl: read the last few posts ;)

[quote="4everwise"]A ball is thrown upward from the top of a tower. If its height is described by $-t^2+60t+700$, find the greatest height the ball attains.[/quote]

[hide]we need to maximize the function $f(t)=-t^2+60t+700$.
$\displaystyle\frac{b}{2a}=\frac{60}{2}=30$
We know that the maximum occurs at 30.
f(30)=1600
The maximum is 1600.
[/hide]


Find [size=150]one[/size] pair of functions \[ f(x): (0,1)\to [0,1]  \] and \[ g(x): (0,1)\to [0,1]  \] that satisfy  for each a belongs to \[ [0,1]  \]There is only one x and infinitely many y that satisfy f(x)=g(y)=a
I'm poor in English.Maybe I've typed something wrong.

[quote="wuhao_jshj"]
I'm poor in English.Maybe I've typed something wrong.[/quote]
這裡可以打中文的.

试定义两个函数f(x),g(x),使其均以（０，１）为定义域，以［０，１］为值域，且对任意的a属于［０，１］，f(x)=a有唯一解,g(x)=a有无数解．

我想是存在的，
比如 先把（0，1）的任一个数写成无穷位数的小数。
（比如0.1=0.1000000000……，）
然后取出这些数位的第1，第3，第5，第奇数为（如0.1024000000……，取出后为0.120000000）
可以知道所有由这类新取出的数对应了所有（0，1）的数，对每个（0，1）的数都有无数个x和他们对应。
对于0，1两点，因为x属于（0，1）已经有无穷多个x和他们对应了，因此可以把任一个x中的无穷多个分一点无穷多给0和1

据说是复旦大学保送生暨自主招生考试题
我已在下面网页有一种构造，但大家齐心协力，看看有没有更多构造
http://www.aoshoo.com/bbs1/dispbbs.asp?boardID=14&ID=2156&page=1 :D   :P
建议大家加入此论坛，确实非常不错！
http://www.aoshoo.com/bbs1/index.asp


modular arithmetic
Pascal\u0027s Triangle
number theory theorems
number theory

It's known that if we paint even numbers of pascal triangle with one color and odd ones with other color, we'll see a nice fractal
http://en.wikipedia.org/wiki/Sierpinski_triangle
one of it's properties is that $ 2|\binom{n}{k}$ for all $ k \equal{} 1,2,\dots,n \minus{} 1$ iff $ n$ is a power of 2.

will it be true with 2 replaced by 3 ?
or maybe the general result:
$ p|\binom{n}{k}$ for all $ k \equal{} 1,2,\dots,n \minus{} 1$ iff $ n$ is a power of $ p$ ?

another observation: in every row there are more numbers which $ \equiv 1\pmod{3}$ then numbers which $ \equiv 2\pmod{3}$
how to prove it?

The theorem for such kind of problem is Lucas' theorem ( http://en.wikipedia.org/wiki/Lucas%27_theorem ).

Thanks, interesting theorem, it solves the first problem (even the general form)
But what about the second one?

It solves the second one as well.  The possible coefficients that appear in Lucas' theorem $ \bmod 3$ are

$ {2 \choose 2} \equiv 1 \bmod 3$
$ {2 \choose 1} \equiv 2 \bmod 3$
$ {2 \choose 0} \equiv 1 \bmod 3$
$ {1 \choose 1} \equiv 1 \bmod 3$
$ {1 \choose 0} \equiv 1 \bmod 3$
$ {0 \choose 0} \equiv 1 \bmod 3$

and the remaining terms are congruent to $ 0 \bmod 3$.  In other words it is very unlikely for factors congruent to $ 2 \bmod 3$ to appear, and even then there must be an odd number of them in order for the product to be congruent to $ 2 \bmod 3$.


find all prime numbers(p) that m and n are integer and p=m^2+n^2 ,p|m^3+N^3 - 4.

This problem was given on the fifth mock test in Project Z.  My solution from that test, verbatim:

It is well known that there exist integers $ m, n$ such that $ p \equal{} m^2 \plus{} n^2$ iff $ p \equal{} 2$ or $ p \equiv 1 \bmod 4$.  Since $ p \equal{} 2$ does not satisfy the problem conditions, $ p > 2$ and therefore $ p$ is odd.  Let $ u \equal{} mn, v \equal{} m \plus{} n$.  We want to find $ p$ such that $ m^3 \plus{} n^3 \equiv 4 \bmod p$.  Note that $ p \equal{} v^2 \minus{} 2u$; therefore,

$ u \equiv \frac{v^2}{2} \bmod p \Leftrightarrow$
$ m^3 \plus{} n^3 \equal{} v(m^2 \minus{} u \plus{} n^2) \equiv \minus{}uv \equiv \minus{} \frac{v^3}{2} \Leftrightarrow$
$ (v \plus{} 2)(v^2 \minus{} 2v \plus{} 4) \equiv 0 \bmod p$.

[b]Case:[/b]  $ p | v \plus{} 2$.  AM-QM gives

$ \sqrt{ \frac{m^2 \plus{} n^2}{2} } \ge \frac{v}{2} \implies \sqrt{2p} \ge v$

so that we can only have $ v \equal{} p\minus{}2$, and in fact this can only occur when $ \boxed{ p \equal{} 5 }$.  When $ p > 5$, we have $ p \minus{} 2 > \sqrt{2p}$.

[b]Case:[/b]  $ p | v^2 \minus{} 2v \plus{} 4$.  But $ v^2 \minus{} 2v \plus{} 4 \equiv 2(u \minus{} v \plus{} 2)$, hence $ p | u \minus{} v \plus{} 2 \equal{} (m\minus{}1)(n\minus{}1) \plus{} 1$.  Because $ m, n \ge 1$ we know that $ (m\minus{}1)(n\minus{}1) \plus{} 1 \ge 1$, but because $ v \ge 2$, we have

$ u \minus{} v \plus{} 2 \le u \le \frac{p}{2}$

by AM-GM, hence we never have $ (m\minus{}1)(n\minus{}1) \plus{} 1 \equiv 0 \bmod p$.

Zuming Feng's year-round Olympiad training program.  Or something along those lines; I was never really clear on the details.


Let $ ABC$ a triangle, where $ \overline{AB}\equal{}6$, $ \overline{AC}\equal{}9$ and $ \overline{BC}\equal{}11$. If $ JL//BC$, find $ \overline{AL}$.

[hide]AL = 6

It's 1:30 a.m. so I won't try to put my solution into words, but in the end you just use $ |AI| \equal{} \frac{a\plus{}c\minus{}b}2.$[/hide]

[hide]
$ AJ \equal{} s \minus{} b \equal{} \frac {6 \plus{} 9 \plus{} 11}{2} \minus{} 9 \equal{} 4$
$ AL \equal{} \frac {AJ}{AB}\cdot AC \equal{} \frac {4}{6}\cdot 9 \equal{} 6$
* $ s$ is the half perimeter of the triangle.[/hide]


I really just heard this term somewhere and I don't know what it means. :(
Could someone explain what "Roots of unity filter" means :)

is it concerned with computers or maths coz it has meanings in both of them!

The roots of unity filter is a sum that extracts the sum of every $ n^{th}$ coefficient in a polynomial $ P(x)$.  For example, $ P(1)$ is the sum of every coefficient and

$ \frac{P(1) \plus{} P(\minus{}1)}{2}$

is the sum of every even coefficient (why?).  The roots of unity filter is the sum

$ \frac{P(1) \plus{} P(\zeta) \plus{} ... \plus{} P(\zeta^{n\minus{}1})}{n}$

where $ \zeta$ is an $ n^{th}$ root of unity.  (Similar sums exist to extract the coefficients of index congruent to $ a \bmod n$ for each residue $ a$.)

Now how exactly is this concept useful in solving combinatorics problems . Can it be used to find the coefficient of a particular power of a variable in a product like in the AMC 12 A 2008 problem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=178172413&t=189177]AMC 12 2008A #19[/url] can indeed be solved by a roots of unity filter; its role is to isolate coefficients.  The roots of unity filter is also particularly well-suited to finding sums like $ {2007 \choose 0} \plus{} {2007 \choose 3} \plus{} {2007 \choose 6} \plus{} ... \plus{} {2007 \choose 2007}$, which is just

$ \frac{2^{2007} \plus{} (1 \plus{} \omega)^{2007} \plus{} (1 \plus{} \omega^2)^{2007}}{3}$.

This is the question of how many subsets of a set with $ 2007$ elements have cardinality divisible by $ 3$.  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215429]IMO 2008 #5[/url] can be attacked using the 2nd roots of unity filter.


For the 1966 AHSME, does anyone have a list of only answers. Thanks.

If you search 1966 AHSME, you will get a bunch of topics with question numbers, and solutions. Some don't have question numbers though, not all of the questions are there, and it would be kind of inconvenient for you.

I'm editing the 1966 AHSME in the AoPSWiki for problems 1 and 3-10 now. Any solutions would be appreciated.

P.S. I'm reviving because there was no answer to this question, and some people might be looking for the 1966 AHSME.


function calculus derivative geometry trigonometry

Find the range of the function: $y=\sqrt{x^2-2x+5}-\sqrt{x^2+6x+10}$.

It is easy to find the upper bound of the range, but what is the lower bound?

WARNING: TEDIOUS METHOD AHEAD.

differentiating give:

$\frac{x-1}{\sqrt{x^2-2x+5}} - \frac{x+3}{\sqrt{x^2+6x+10}}=0$

simplifying gives:

$\frac{(x-1)\sqrt{x^2+6x+10} - (x+3) \sqrt{x^3-2x+5}}{whatever}=0$

thus:

$(x-1)\sqrt{x^2+6x+10} = (x+3) \sqrt{x^3-2x+5}$

squaring both sides:

$x^4+4x^3-x^2-14x +10 = x^4+4x^3 +2x^2 +12x+45$
=$3x^2+26x +35=0$
(x+7)(3x+5)=0

sub in x=7 and x=-5/3 to find max and min

Thank you. But maybe differentiation is a little difficult for a senior 2 student. Can you find a more elementary solution?

My way to find the upper bound is to change $y$ into the difference between the distance of two segments. But I can't find the lower bound of the function by this method.

i am actually very weak in other ways of finding maxima and minima, thats what i'm working towards to learning now. Sorry, cant help you here.

Anywayz, do any schools teach other areas of mathematics other from ie basic trigonometry calculus etc?

[quote="Extra"]Anywayz, do any schools teach other areas of mathematics other from ie basic trigonometry calculus etc?[/quote]
Huh? What do you mean by that?

There are several problems with what you've done:

First, -5/3 is a false root: the derivative of the function at -5/3 is -8/5, not 0.  So, at best, you have one of the extrema.

Second, you've assumed that the points of zero derivative will necessarily be extrema.

Third, you assumed the function will [i]have[/i] extrema.  Try graphing the function for $x > 0$ -- it will be clear fairly quickly why you have been unable to find the minimum.


inequalities
triangle inequality
inequalities unsolved

(a+a+b+c)(a+b+b+c)(a+b+c+c) \geq (a+a+a+c)(b+b+b+a)(c+c+c+b)

Work out, simplify, Muirhead... done :P

Does anyone have a nicer sol? (less straightforward)

How often will I have to say this: Muirhead does [b]not[/b] work for cyclic (for Peter: and non-symmetric) inequalities.

Let's prove the inequality: We have to show that

$ \left( 1\plus{}a\right)\left( 1\plus{}b\right)\left( 1\plus{}c\right)\geq\left( 3a\plus{}c\right)\left( 3b\plus{}a\right)\left( 3c\plus{}b\right)$.

In fact,

$ \left( 1\plus{}a\right)\left( 1\plus{}b\right)\left( 1\plus{}c\right)\minus{}\left( 3a\plus{}c\right)\left( 3b\plus{}a\right)\left( 3c\plus{}b\right)$
$ \equal{} 2\cdot\left(\left( b\plus{}c\right)\left( a\minus{}b\right)^{2}\plus{}\left( c\plus{}a\right)\left( b\minus{}c\right)^{2}\plus{}\left( a\plus{}b\right)\left( c\minus{}a\right)^{2}\right)\geq 0$.

Proof of identity left to the reader.  :D

  Darij

[quote="manlio"]For $a,b,c$ positive reals such that $a+b+c=1$ prove the ineq

$(1+a)(1+b)(1+c) \geq (3a+c)(3b+a)(3c+b)$[/quote]

Substituting x=1+a, y=1+b, z=1+c we get that the inequality is equivalent to:

xyz >= (2x-z)(2y-x)(2z-y)

Does this help at all?  Because all I'm able to do from here is the identical substitution to that of darij's.

[quote="manlio"]Nice substitution

[quote="blahblahblah"]

Substituting x=1+a, y=1+b, z=1+c we get that the inequality is equivalent to:

xyz >= (2x-z)(2y-x)(2z-y)

Does this help at all?  Because all I'm able to do from here is the identical substitution to that of darij's.[/quote]

If you put $x=p+q,y=q+r,z=r+p$ you can solve $xyz >= (2x-y)(2y-z)(2z-x)$ very easily by AM-GM.

Try it!!![/quote]

But... can't you only do that substitution (with positive p,q,r) if a,b,c satisfy the triangle inequality?


Given a rectangular piece of paper and a pencil and [i]nothing else[/i], construct a 60 degree angle.
Hint: With these conditions it is also possible to construct a 30, 15, even 1 degree angle.

What I'd do is draw fold teh paper in a manner where I can draw a 1 line width line perpendicular to its base. Then I'd fold teh paper again in a manner that I can draw a line with 2 lines' length touching the base of the pernedicular line. It satisfies a 30-60-90 triangle and so it is 60 degrees.  This is for a lined piece of paper. :P

Masoud Zargar

I've seen paperfolding books that show how to fold equilateral triangle and even regular pentagon. Too bad I don't own one. :(

can you rip the paper? I measured out distances and I ripped it to put the paper in the right position...

I used a ripped-off corner of the paper as a straightedge. I used the right angle on the paper's corner to make a perpendicular. Then I picked an arbitrary length for one of the legs and put the edge of the ripped off part up to it ad marked off the distance, and then marked it off again, so I had twice the distance. Then I just used that peice of paper and turned it around so that the ends of the distance I had markedmet up with both legs, and drew taht, and had a 60 degree angle. I don't see how to construct 15 or 1 degree angles though...

I suppose that works. :)

Here was my method:

First, align the paper like a landscape. Imagine its points are labeled ABCD.
Then imagine that the top edge has a midpoint P. There are also two other points on the bottom edge Q and R such that CR = QD. Angle QPR should be close to 60 degrees, just by eyeballing. Now fold the paper along PQ such that AP lines up with PR, and at the same time fold the paper along PR [i]backwards[/i] such that PB lines up with PQ. Crease and unfold, and you have a 60 degree angle.

The 30 can be made by taking the 60 and folding it in half, and the 15 by folding the 30 in half.

And come to think of it, I don't think the 1 is actually possible... mainly due to the paper thickness :blush:

1. I would first fold in half length wise on the blue line.

2. Then I would fold the short side on the red line, such that one corner of the side stays in place, while the other corner ends up on the blue line. Then I would fold the green line.

The green line is the same length as the short side. If we repeated step 2 going the other way, we would then have an equilateral triangle. So the short side and the green line make a 60 degree angle.

You can then construct a 30 degree and a 15 degree by halving the 60 degree angle.

[quote="lingomaniac88"]How would you construct a 1-degree angle?[/quote]
I don't think thats possible.  If it is, then it would be really hard.

This problem (and others like it) were the basis of the 2006 ARML Power question...

I hated that power round, it was difficult to tell what was rigorous and what was not. I thought it vague.

Allow me to paraphrase B5 and B6:
You can make the fold, unless there is no fold that satisfies those conditions.
Well, DUH!

Found a instruction on how to fold a regular pentagon.
[url]http://www.zingman.com/origami/oriPics/roses/fivefold_rose_diagrams.swf[/url]


Hey,

I'm having trouble understanding double sums. Can someone provide examples or problems?

You mean like $ \sum_{n\equal{}1}^{\infty}\sum_{i\equal{}1}^{\infty}a_{n,i}$?  This just means that you sum over the first index, and then the second one.  It really is nothing more than a double sum:
$ \sum_{n\equal{}1}^{\infty}\sum_{i\equal{}1}^{\infty}a_{n,i}\equal{}a_{1,1}\plus{}a_{1,2}\plus{}a_{1,3}\plus{}\cdots\plus{}a_{2,1}\plus{}a_{2,2}\plus{}\cdots$.

[url=http://mathworld.wolfram.com/DoubleSeries.html]This[/url] page has some examples, as does [url=http://people.revoledu.com/kardi/tutorial/BasicMath/Sum/Double%20summation.htm]this one[/url], which shows it from a matrix perspective.


When all possible 4-digit numbers containing each of the digits 1, 2,
3, and 4, are listed in ascending order, which number is in the twelfth position?

We can simply list them out in order:

$ 1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431$

So $ \boxed{2431}$.

An easier way would be to take the thousands digit.  Numbers with thousands digit $ 1$ would be numbers 1-6 among 24 in order, and numbers with thousands digit $ 2$ would be numbers 2-12 among 24.  Thus, the greatest possible number with only 1, 2, 3, and 4 and thousands digit $ 2$ is $ \boxed{2431}$.


function
logarithms
inequalities
calculus
inequalities proposed

Show that $\sum_{k=1}^n \left (k \cdot lnk \right) \geq \frac{n(n+1)}{2}ln \left (\frac{n+1}{2} \right )$.

jianuovidiu, I think I have found something! :)

Consider the function $F(x)=xlnx$ and observe that $f''(x)=\frac{1}{x}>0$
so f is convex!
Using Jensen's ineq we get
$\displaystyle \sum_{k=1}^n \left (k \cdot lnk \right) \geq nf(\frac{1+2+3+...+n}{n})=nf(\frac{n+1}{2})=\frac{n(n+1)}{2}\ln\frac{n+1}{2}$
and we are done! :D  :lol:  :)

Jeane,I don't know if your solution is the same as the one given,but I think Chebyshev's inequality is  working here.
Did you prove it with Chebyshev? ;)

ok. What is your solution? did you use calculus or elementary math?

my first attempt was using Cebasev's ineq but I did not get to the
wanted result...... :(

[quote="cezar lupu"]Sorry,I realised Chebyshev is too weak in this case. :oops:[/quote]

i used a lemma. i'll post later, i aspect more sollution here :P

Well this is my solution:

LEMA: Let $a_k > 0 ,k=1,2,...,n$. Then we have: $\prod {a_k}^{a_k} \geq \left ({\frac {a_1+a_2+...+a_n}{n} \right)}^{a_1+...+a_n}$
The proof o this lema is: We take the logarithm and the ineq. is equivalet with $\frac {\sum {a_klna_k}}{n}$ $\geq {\frac {\sum a_k}{n}}ln\left ( \frac {\sum a_k}{n} \right )$ which can be prooved by using Jensen tu the function $f(x)=xlnx$.

Take now $a_k=k$ and use LEMA and we get the desired result.

q.e.d.

Well, i see that is almost the same sollution gave by socrate.....and if i look carefuly is the same  :D  :P  :roll:


superior algebra
superior algebra unsolved

Let $ K$ be a field of characteristic $ p$. Let $ \beta$ be a root of $ f(x)\equal{}x^p\minus{}x\minus{}a\in K[x]$. Show that $ K(\beta)/K$ is a normal extension.

In fact, the roots of $ f$ are exactly $ \beta, \beta\plus{}1,\ldots \beta\plus{}p\minus{}1$.

Oh wow. That was dumb. Since if $ n\in\mathbb{Z}$, then $ (\beta \plus{}n)^p\minus{}(\beta\plus{}n)\minus{}a\equal{}\beta^p\plus{}n^p\minus{}\beta\minus{}n\minus{}a\equal{}\beta^p\minus{}\beta\minus{}a\equal{}0$.


Vietnamese Mathlinkers : nice to see you.

HI dude ,

thanks for your hearty message


Euler
function
number theory unsolved
number theory

Let $\varphi(n)$ is the Euler function and we define $\delta(n)$ is the sum of the divisors of $n\in N$. Prove that if $\varphi(n)\mid n-1$ and $n+1\mid \delta(n)$ so that $n$ is a prime.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=lehmer&t=14533

Pierre.


Find the equations of all lines tangent to the graph of [img]http://www.sosmath.com/CBB/latexrender/pictures/0cce27b69701fe52c780bf3bdddf7bf5.gif[/img] that also contain the point [img]http://www.sosmath.com/CBB/latexrender/pictures/d334b8221235e6bd3283aea4dc703c49.gif[/img].

The procedure:
Find the tangent line $ y\equal{}ax\plus{}b$ at the point $ (t,f(t))$ ($ f$ is our polynomial).
Write down the equation $ 5a\plus{}b\equal{}9$; the tangent line goes through the point (not on the curve).
Solve this cubic equation.

The last step is the killer- unless the problem was rigged very carefully, the answer or answers will be extremely ugly.


search
function
topology
set theory
group theory
advanced fields
advanced fields unsolved

Show that every nonempty set admits a binary operation that makes it into a group.

My first impression is that some advanced set theory is required. That's why I posted it here :)
(I don't know if it has been discussed before; the search function doesn't give me any results)

uhm.. i don't think anything advanced is needed: if this set is finite, you can put it in bijection with some cyclic finite group; otherwise, a direct sum of $\kappa$ copies of $\mathbb{Z}/2\mathbb{Z}$ has cardinality $\kappa$.

the most advanced thing used here is that finite subsets of an infinite cardinal are equipotent to the cardinal itself, which follows from axiom of choice, in the form $|X\times X| = |X|$ for $X$ infinite.

[quote="ma_go"]a direct sum of $\kappa$ copies of $\mathbb{Z}/2\mathbb{Z}$ has cardinality $\kappa$.[/quote]

I don't think so: $\kappa$ copies of ${\bf Z}/2\bf Z$ gives you $2^\kappa$ elements, no?

That depends on how you understand the words "direct sum". You may choose to understand it either as the group of all functions from a set $U$ of cardinality $\kappa$ to $\mathbb Z/2\mathbb Z$ (and then you are right) or as the group of functions whose values are $0$ at all but finitely many elements of $U$. Ma_go obviously had the latter interpretation in mind. It is like the story with the product topology when the number of spaces is infinite.

The term "direct sum" always means the latter. owk

Yes, this is the standard convention, but how would you call the first object? :?

I remember a book on group theory in which the larger object was called "Cartesian product" [which it is], and the smaller one was "direct product". :) Which kind of makes sense when multiplicative notation is used.


function calculus derivative algebra domain limit logarithms

This is the problem:
Find and describe all local extrema of f(x)=x^(5/3)-5x^(2/3). Indicate on which regions the function is increasing and decreasing.

So far, this is what I have:
[hide]
I have gotten the minimum. Basically, I set up the derivative of f(x) to 0 and solved. I did the second derivative test on it and it turned out to be a minimum. BTW, the min is at 2.
[/hide]

Now, I just need to show where the function is increasing and decreasing. I kinda cheated by using a graphing calc. I wasn't supposed to.   :oops:   Now, I need to show it without using my calculator.

Any help would be appreciated.

$ f(x)\equal{}x^{\frac{5}{3}}\minus{}5x^{\frac{2}{3}}$

$ f'(x)\equal{}\frac{5(x\minus{}2)}{3x^{\frac{1}{3}}}$

$ f''(x)\equal{}\frac{10(x\minus{}1)}{9x^{\frac{4}{3}}}$ isnt reall that important

critical points are at 0 and 2,

f'(-1)>0, f'(1)<0, f'(3)>0

so increasing from $ (\minus{}\infty, 0) \cup (2, \infty )$
decreasing (0,2)

[quote="kenn4000"]critical points are at 0 and 2[/quote]
To be pedantic about it, there isn't a critical point at $ 0.$ Rather, that's a point that is not in the domain of the function.

The point of making that distinction it that $ f$ cannot have a maximum or minimum at a point that is not in the domain. On the other hand, you must recognize $ 0$ as a point to be dealt with and explained; else you can't really make sense of what the graph should look like. In particular, you should understand $ \lim_{x\to 0^\plus{}}f(x)$ and $ \lim_{x\to 0^\minus{}}f(x)$ before you sketch the graph.

This is a prime example of something I usually get stuck on.
Graphing the function yields all $ x<0$ are inadmissible. But why is this?

Actually, $ 0$ is in the domain of $ f$- those are positive powers. Since they're thirds, it's even in the interior of the domain and is a true critical point. It's a vertical cusp and a local maximum.

Euclid's Turtle: your calculator doesn't know how to take non-integer powers of negative numbers.

[quote="Kent Merryfield"][quote="kenn4000"]critical points are at 0 and 2[/quote]
To be pedantic about it, there isn't a critical point at $ 0.$ Rather, that's a point that is not in the domain of the function.[/quote]  Not in the domain of the derivative, but the original function is defined over all of $ \bf R$.

Sorry about that, guys - I was reading the wrong line of kenn4000's post when I wrote that. I was looking at one of the derivatives rather than the original function. And yes, as you can clearly read from what kenn4000 said, there is a local maximum at $ 0.$

hello, the function $ f(x)\equal{}x^{\frac{5}{3}}\minus{}5x^{\frac{2}{3}}$ is defined for all
real $ x$ with $ x \geq0$. The point $ x\equal{}0$ isn't in the domain of the derivative of $ f(x)$, but $ x\equal{}0$ gives a local maximum of $ f(x)$.
Sonnhard.

to $ \text{jmerry}$

I don't think it is my calculator... graphing it with several devices yields the doman to be only positive reals.
Also i tried  $ (\minus{}1)^{\frac{5}{3}}$ and got $ \minus{}(\minus{}1)^\frac{2}{3} \approx .5000000001\minus{}.8660254037i$  :(

That [b]is[/b] a calculator and/or software issue, and it comes from treating a fraction like $ \frac23$ as a real number (or approximate real number) rather than as an exact fraction with an odd denominator.

The underlying calculator/software issue appears to be that of defining $ a^b$ as $ e^{b\ln a}.$ If $ a>0,$ this leads to unambiguous results, but for $ a<0,$ we must deal with the meaning of the logarithm of a negative number.

The simple response to that would be to simply return an error message; if you stick to the real numbers, a negative number does not have a logarithm. The more complicated response would be to pick an arbitrary branch of the complex logarithm and use that. The problem with that is that the answer depends on the branch of the logarithm.

For instance, there are three possible values for $ (\minus{}1)^{\frac23}: 1$ and $ \minus{}\frac12\pm\frac{i\sqrt{3}}2.$ (Note the roundoff error in the post by Euclid's Turtle.)

There is an available unambiguous real-number-only definition for $ a^{\frac pq}$ for $ a<0$ and rational number $ \frac pq$ as long as $ \frac pq$ is in lowest terms and $ q$ is odd. To get that definition out of a piece of software, you must persuade the software to treat $ \frac pq$ as an exact rational number, and that at minimum requires a symbolic rather than numerical processor.

Alternatively, you can wake up and realize that you're smarter than your calculator. In my teaching, I always prefer this.


Mike starts at one corner of a tetrahedron.  In a move, he walks along any edge to a different corner.  In how many ways can he end up where he started after 6 moves?

[quote="math154"]Let the number of ways to return to the point after $ n$ moves given that Mike is on the point be $ a_n$, and the number of ways to return to the point after $ n$ moves given that Mike is not on the point be $ b_n$. Then we want to find $ a_6$. We also have the recursions $ a_n \equal{} 3b_{n \minus{} 1}$ and $ b_{n \plus{} 1} \equal{} a_n \plus{} 2b_n\implies a_n \equal{} b_{n \plus{} 1} \minus{} 2b_n$, so $ 3b_{n \minus{} 1} \equal{} b_{n \plus{} 1} \minus{} 2b_n\implies b_{n \plus{} 1} \equal{} 2b_n \plus{} 3b_{n \minus{} 1}$. Clearly, $ b_0 \equal{} 0$ and $ b_1 \equal{} 1$, so $ b_2 \equal{} 2\implies b_3 \equal{} 7\implies b_4 \equal{} 20\implies b_5 \equal{} 61\implies a_6 \equal{} \boxed{183}$.

Alternatively, there are $ 3^6 \equal{} 729$ possible paths in total, and $ 4$ total points that he can end up in, so there are about $ \frac {729}4 \equal{} 182.25$ ways per point. But the starting point has to be "special," and we see that there are $ \boxed{183},182,182,182$ ways for each point. :P[/quote]

thanks for the solution, but in solution 1, i do not understand the two recursions you given.. sorry..

$ a_n\equal{}3b_{n\minus{}1}$ because if we are at the original point, then the next move has to be to one of the $ 3$ other points. Then it is like we have $ n\minus{}1$ more moves and are on another point, so there are $ b_{n\minus{}1}$ ways to proceed, for a total of $ 3b_{n\minus{}1}$, and $ b_{n\plus{}1}\equal{}a_n\plus{}2b_n$ because if we are on another point and have $ n\plus{}1$ moves left, then we can go to the original point in $ 1$ way, and then it is like having $ n$ more moves on the original point, which has $ a_n$ ways to go on, or we can go to one of the $ 2$ other non-original points, after which we have $ b_n$ ways to proceed, for a total of $ a_n\plus{}2b_n$.


inequalities
geometry
inequalities proposed

Yes, easy. I rewrite your inequality denoting the variables by x, y, z instead of a, b, c, since I will later use a, b, c for the substitution variables.

So we have to prove:

[i]For any three positive reals x, y, z, we have

$\frac{x^2+yz}{\left(y+z\right)^2}+\frac{y^2+zx}{\left(z+x\right)^2}+\frac{z^2+xy}{\left(x+y\right)^2} \geq \frac32$.[/i]

[b]Proof.[/b] Upon multiplication by 2, this inequality becomes

$\frac{2\left(x^2+yz\right)}{\left(y+z\right)^2}+\frac{2\left(y^2+zx\right)}{\left(z+x\right)^2}+\frac{2\left(z^2+xy\right)}{\left(x+y\right)^2} \geq 3$.

Now, to prove this inequality, set a = y + z, b = z + x, c = x + y. Then, some calculation yields that $2bc-ca-ab+a^2=2\left(x^2+yz\right)$, what, together with a = y + z, results in

$\frac{2\left(x^2+yz\right)}{\left(y+z\right)^2} = \frac{2bc-ca-ab+a^2}{a^2} = \frac{2bc-ca-ab}{a^2}+1$
$= \frac{b\left(c-a\right)-c\left(a-b\right)}{a^2}+1 = \frac{b\left(c-a\right)}{a^2} - \frac{c\left(a-b\right)}{a^2} + 1$.

Similarly,

$\frac{2\left(y^2+zx\right)}{\left(z+x\right)^2} = \frac{c\left(a-b\right)}{b^2} - \frac{a\left(b-c\right)}{b^2} + 1$;
$\frac{2\left(z^2+xy\right)}{\left(x+y\right)^2} = \frac{a\left(b-c\right)}{c^2} - \frac{b\left(c-a\right)}{c^2} + 1$.

Thus, the inequality we have to prove,

$\frac{2\left(x^2+yz\right)}{\left(y+z\right)^2}+\frac{2\left(y^2+zx\right)}{\left(z+x\right)^2}+\frac{2\left(z^2+xy\right)}{\left(x+y\right)^2} \geq 3$,

simplifies to

$\left(\frac{b\left(c-a\right)}{a^2} - \frac{c\left(a-b\right)}{a^2} + 1\right) + \left(\frac{c\left(a-b\right)}{b^2} - \frac{a\left(b-c\right)}{b^2} + 1\right)$
$+ \left(\frac{a\left(b-c\right)}{c^2} - \frac{b\left(c-a\right)}{c^2} + 1\right) \geq 3$.

This is clearly equivalent to

$\left(\frac{b\left(c-a\right)}{a^2} - \frac{c\left(a-b\right)}{a^2}\right) + \left(\frac{c\left(a-b\right)}{b^2} - \frac{a\left(b-c\right)}{b^2}\right) + \left(\frac{a\left(b-c\right)}{c^2} - \frac{b\left(c-a\right)}{c^2}\right) \geq 0$.

After a little rearrangement of terms, this becomes:

$\left(\frac{a\left(b-c\right)}{c^2}-\frac{a\left(b-c\right)}{b^2}\right)+\left(\frac{b\left(c-a\right)}{a^2}-\frac{b\left(c-a\right)}{c^2}\right)+\left(\frac{c\left(a-b\right)}{b^2}-\frac{c\left(a-b\right)}{a^2}\right) \geq 0$.

But this inequality is clearly true, since after some algebraic manipulations inside the brackets it becomes

$a\left(b+c\right)\left(\frac{1}{b}-\frac{1}{c}\right)^2+b\left(c+a\right)\left(\frac{1}{c}-\frac{1}{a}\right)^2+c\left(a+b\right)\left(\frac{1}{a}-\frac{1}{b}\right)^2 \geq 0$.

$\blacksquare$

  Darij

Ok,but this ineq is still true:

$\frac{4a^2+5bc}{(b+c)^2}+\frac{4b^2+5ac}{(a+c)^2}+\frac{4c^2+5ac}{(a+b)^2}\ge \frac{27}{4}$

Is your way useful for it,Darij?

In my opinion, the following stronger inequalities are valid:

 (A) $\frac{a^2+2bc}{(b+c)^2}+\frac{b^2+2ac}{(c+a)^2}+\frac{c^2+2ab}{(a+b)^2} \ge \frac{9}{4}$;

(B) $\frac{a^2+bc}{b^2+bc+c^2}+\frac{b^2+ac}{c^2+ca+a^2}+\frac{c^2+ab}{a^2+ab+b^2} \ge 2$.

You forgot factor $\frac{1}{3}$ ;)
Anyway, your solution is very nice and short. Especially, if we compare it with Darij's one...  :) Без в??ких обид :)

Yeah, you are right,[b] Myth[/b], I forgot it , but I don't want to edit it (no difference).

PS. Actually, [b]Myth[/b], [b]darij[/b]'s solutions(especially in geometry) are not always so long, but I don't know why,they become very long in his write up.    :D

[quote="Vasc"]
 (A) $\frac{a^2+2bc}{(b+c)^2}+\frac{b^2+2ac}{(c+a)^2}+\frac{c^2+2ab}{(a+b)^2} \ge \frac{9}{4}$;[/quote]

  Do you think $2$ is the best constant.I'm not sure about it.

[b]To Vasc:[/b]

[quote="Vasc"]In my opinion, the following stronger inequalities are valid:

 (A) $ \frac{a^{2}+2bc}{(b+c)^{2}}+\frac{b^{2}+2ac}{(c+a)^{2}}+\frac{c^{2}+2ab}{(a+b)^{2}}\ge \frac{9}{4}$;

(B) $ \frac{a^{2}+bc}{b^{2}+bc+c^{2}}+\frac{b^{2}+ac}{c^{2}+ca+a^{2}}+\frac{c^{2}+ab}{a^{2}+ab+b^{2}}\ge 2$.[/quote]

Thanks, Vasc, for these two nice and - I think, exceptionally (for you) - easy inequalities! I have proven both of them using one and the same idea, and this idea can also be used to find alternative proofs of the original inequality of this thread.

I begin with a cool theorem of Valentin:

[b]Lemma 1 (generalized Schur inequality / Vornicu-Schur inequality).[/b] Let a, b, c, x, y, z be six nonnegative reals such that $ a\geq b\geq c$, and such that either $ x\geq y\geq z$ or $ x\leq y\leq z$. Then,

$ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.

[hide="Proof of Lemma 1"]Actually, this is just a rephrase of one of Valentin's extensions of the Schur inequality posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=1166 . For the sake of completeness, here is a [i]proof of Lemma 1:[/i]

First consider the case $ x\geq y\geq z$. In this case, we have the inequalities $ x\geq y$ and $ a-c\geq b-c$ (that's because $ a\geq b\geq c$); note that all four sides of these two inequalities are nonnegative (in fact, both a - c and b - c are nonnegative since $ a\geq b\geq c$), so that we can multiply these inequalities and get

$ x\left(a-c\right)\geq y\left(b-c\right)$.

Since $ a-b\geq 0$ (again from $ a\geq b\geq c$), we can multiply this with a - b and get

$ x\left(a-b\right)\left(a-c\right)\geq y\left(b-c\right)\left(a-b\right)$,

so that

$ x\left(a-b\right)\left(a-c\right)-y\left(b-c\right)\left(a-b\right)\geq 0$;

in other words,

$ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)\geq 0$.

Also, $ a\geq b\geq c$ yields $ c-a\leq 0$ and $ c-b\leq 0$, so that $ \left(c-a\right)\left(c-b\right)\geq 0$, and since $ z\geq 0$, we get

$ z\left(c-a\right)\left(c-b\right)\geq 0$.

Adding this to the inequality $ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)\geq 0$, we get $ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$, as intended to prove.

Now remains the case $ x\leq y\leq z$. This time, have the inequalities $ y\leq z$ and $ a-b\leq a-c$ (that's because $ a\geq b\geq c$); note that all four sides of these two inequalities are nonnegative (in fact, both a - b and a - c are nonnegative since $ a\geq b\geq c$), so that we can multiply these inequalities and get

$ y\left(a-b\right)\leq z\left(a-c\right)$.

Since $ b-c\geq 0$ (again from $ a\geq b\geq c$), we can multiply this with b - c and get

$ y\left(a-b\right)\left(b-c\right)\leq z\left(a-c\right)\left(b-c\right)$,

what rewrites as

$ -y\left(b-c\right)\left(b-a\right)\leq z\left(c-a\right)\left(c-b\right)$.

Hence,

$ y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.

Also, $ a\geq b\geq c$ yields $ a-b\geq 0$ and $ a-c\geq 0$, so that $ \left(a-b\right)\left(a-c\right)\geq 0$, and since $ x\geq 0$, we get

$ x\left(a-b\right)\left(a-c\right)\geq 0$.

Adding this to the inequality $ y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$, we get

$ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.

And thus Lemma 1 is proven.[/hide]

A corollary of Lemma 1 is:

[b]Lemma 2.[/b] For any three positive reals a, b, c, we have

$ \frac{\left( a-b\right) \left( a-c\right) }{\left( b+c\right)^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{\left( c+a\right)^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{\left( a+b\right)^{2}}\geq 0$.

[hide="Proof of Lemma 2"]
[i]Proof of Lemma 2.[/i] Since the inequality in question is symmetric, we can WLOG assume that $ a\geq b\geq c$. Then, $ a+b\geq c+a\geq b+c$; hence, $ \left( a+b\right)^{2}\geq \left( c+a\right)^{2}\geq \left( b+c\right)^{2}$, or, equivalently, $ \frac{1}{\left( a+b\right)^{2}}\leq \frac{1}{\left( c+a\right)^{2}}\leq \frac{1}{\left( b+c\right)^{2}}$, what rewrites as $ \frac{1}{\left( b+c\right)^{2}}\geq \frac{1}{\left( c+a\right)^{2}}\geq \frac{1}{\left( a+b\right)^{2}}$. Thus, if we denote $ x=\frac{1}{\left( b+c\right)^{2}}$, $ y=\frac{1}{\left( c+a\right)^{2}}$ and $ z=\frac{1}{\left( a+b\right)^{2}}$, then we have $ x\geq y\geq z$. Now, since the numbers a, b, c, x, y, z are all nonnegative and satisfy $ a\geq b\geq c$ and $ x\geq y\geq z$, we can apply Lemma 1 and get

$ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.

Remembering that $ x=\frac{1}{\left( b+c\right)^{2}}$, $ y=\frac{1}{\left( c+a\right)^{2}}$ and $ z=\frac{1}{\left( a+b\right)^{2}}$, we immediately see that this is equivalent to

$ \frac{\left( a-b\right) \left( a-c\right) }{\left( b+c\right)^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{\left( c+a\right)^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{\left( a+b\right)^{2}}\geq 0$,

and Lemma 2 is proven.[/hide]

Now, we have $ a^{2}+2bc=\left(a-b\right)\left(a-c\right)+\left(bc+ca+ab\right)$, $ b^{2}+2ca=\left(b-c\right)\left(b-a\right)+\left(bc+ca+ab\right)$ and $ c^{2}+2ab=\left(c-a\right)\left(c-b\right)+\left(bc+ca+ab\right)$; hence, Vasc, your first inequality

$ \frac{a^{2}+2bc}{\left( b+c\right)^{2}}+\frac{b^{2}+2ca}{\left( c+a\right)^{2}}+\frac{c^{2}+2ab}{\left( a+b\right)^{2}}\geq \frac{9}{4}$

is the sum of the inequality

$ \frac{\left( a-b\right) \left( a-c\right) }{\left( b+c\right)^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{\left( c+a\right)^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{\left( a+b\right)^{2}}\geq 0$,

which is our Lemma 2, with the inequality

$ \frac{bc+ca+ab}{\left( b+c\right)^{2}}+\frac{bc+ca+ab}{\left( c+a\right)^{2}}+\frac{bc+ca+ab}{\left( a+b\right)^{2}}\geq \frac{9}{4}$,

which is a rephrase of the Iran 1996 inequality

$ \frac{1}{\left( b+c\right)^{2}}+\frac{1}{\left( c+a\right)^{2}}+\frac{1}{\left( a+b\right)^{2}}\geq \frac{9}{4\left( bc+ca+ab\right) }$.

Proving your second inequality will require another lemma:

[b]Lemma 3.[/b] For any three positive reals a, b, c, we have

$ \frac{\left( a-b\right) \left( a-c\right) }{b^{2}+bc+c^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{c^{2}+ca+a^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{a^{2}+ab+b^{2}}\geq 0$.

[hide="Proof of Lemma 3"]
[i]Proof of Lemma 3.[/i] Since the inequality in question is symmetric, we can WLOG assume that $ a\geq b\geq c$. Since

$ \left( c^{2}+ca+a^{2}\right)-\left( a^{2}+ab+b^{2}\right) =\left( a+b+c\right) \left( c-b\right)$,

from $ c\leq b$ we conclude that $ c^{2}+ca+a^{2}\leq a^{2}+ab+b^{2}$, so that $ \frac{1}{c^{2}+ca+a^{2}}\geq\frac{1}{a^{2}+ab+b^{2}}$. Similarly, $ \frac{1}{b^{2}+bc+c^{2}}\geq\frac{1}{c^{2}+ca+a^{2}}$, so that $ \frac{1}{b^{2}+bc+c^{2}}\geq\frac{1}{c^{2}+ca+a^{2}}\geq\frac{1}{a^{2}+ab+b^{2}}$. Thus, denoting $ x=\frac{1}{b^{2}+bc+c^{2}}$, $ y=\frac{1}{c^{2}+ca+a^{2}}$ and $ z=\frac{1}{a^{2}+ab+b^{2}}$, we have $ x\geq y\geq z$. Because of this, and also because $ a\geq b\geq c$ and since a, b, c, x, y, z are nonnegative reals, we can apply Lemma 1, and get

$ x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.

Remembering that $ x=\frac{1}{b^{2}+bc+c^{2}}$, $ y=\frac{1}{c^{2}+ca+a^{2}}$ and $ z=\frac{1}{a^{2}+ab+b^{2}}$, we immediately see that this inequality rewrites as

$ \frac{\left( a-b\right) \left( a-c\right) }{b^{2}+bc+c^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{c^{2}+ca+a^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{a^{2}+ab+b^{2}}\geq 0$,

what proves Lemma 3.[/hide]

Since $ a^{2}+bc=\left(a-b\right)\left(a-c\right)+a\left(b+c\right)$, $ b^{2}+ca=\left(b-c\right)\left(b-a\right)+b\left(c+a\right)$ and $ c^{2}+ab=\left(c-a\right)\left(c-b\right)+c\left(a+b\right)$, your second inequality

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

turns out to be the sum of the inequality

$ \frac{\left( a-b\right) \left( a-c\right) }{b^{2}+bc+c^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{c^{2}+ca+a^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{a^{2}+ab+b^{2}}\geq 0$,

which is our Lemma 3, with the inequality

$ \frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}+\frac{b\left( c+a\right) }{c^{2}+ca+a^{2}}+\frac{c\left( a+b\right) }{a^{2}+ab+b^{2}}\geq 2$,

which looks to me like a rather interesting and not too simple inequality (and my proof of this inequality is not really elegant:

$ \frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}+\frac{b\left( c+a\right) }{c^{2}+ca+a^{2}}+\frac{c\left( a+b\right) }{a^{2}+ab+b^{2}}-2$
$ =\frac{bc\left( b-c\right)^{2}\left( b^{2}+bc+c^{2}\right)+ca\left( c-a\right)^{2}\left( c^{2}+ca+a^{2}\right)+ab\left( a-b\right)^{2}\left( a^{2}+ab+b^{2}\right) }{\left( b^{2}+bc+c^{2}\right) \left( c^{2}+ca+a^{2}\right) \left( a^{2}+ab+b^{2}\right) }$

is clearly $ \geq 0$, so the inequality must hold).

Now, I have two new proofs of the initial inequality

$ \frac{a^{2}+bc}{\left( b+c\right)^{2}}+\frac{b^{2}+ca}{\left( c+a\right)^{2}}+\frac{c^{2}+ab}{\left( a+b\right)^{2}}\geq \frac{3}{2}$.

[b]First proof:[/b] Adding your first inequality

$ \frac{a^{2}+2bc}{\left( b+c\right)^{2}}+\frac{b^{2}+2ca}{\left( c+a\right)^{2}}+\frac{c^{2}+2ab}{\left( a+b\right)^{2}}\geq \frac{9}{4}$

with the inequality

$ \frac{a^{2}}{\left( b+c\right)^{2}}+\frac{b^{2}}{\left( c+a\right)^{2}}+\frac{c^{2}}{\left( a+b\right)^{2}}\geq \frac{3}{4}$

(which follows from Nesbitt and AM-QM), we get

$ \frac{2a^{2}+2bc}{\left( b+c\right)^{2}}+\frac{2b^{2}+2ca}{\left( c+a\right)^{2}}+\frac{2c^{2}+2ab}{\left( a+b\right)^{2}}\geq 3$;

dividing by 2 yields

$ \frac{a^{2}+bc}{\left( b+c\right)^{2}}+\frac{b^{2}+ca}{\left( c+a\right)^{2}}+\frac{c^{2}+ab}{\left( a+b\right)^{2}}\geq \frac{3}{2}$,

qed..

Similarly, one can deduce the initial inequality from your second inequality.

[b]Second proof:[/b] Since $ a^{2}+bc=\left(a-b\right)\left(a-c\right)+a\left(b+c\right)$, $ b^{2}+ca=\left(b-c\right)\left(b-a\right)+b\left(c+a\right)$ and $ c^{2}+ab=\left(c-a\right)\left(c-b\right)+c\left(a+b\right)$, the inequality

$ \frac{a^{2}+bc}{\left( b+c\right)^{2}}+\frac{b^{2}+ca}{\left( c+a\right)^{2}}+\frac{c^{2}+ab}{\left( a+b\right)^{2}}\geq \frac{3}{2}$

turns out to be the sum of the inequality

$ \frac{\left( a-b\right) \left( a-c\right) }{\left( b+c\right)^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{\left( c+a\right)^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{\left( a+b\right)^{2}}\geq 0$,

which is simply our Lemma 2, with the inequality

$ \frac{a\left( b+c\right) }{\left( b+c\right)^{2}}+\frac{b\left( c+a\right) }{\left( c+a\right)^{2}}+\frac{c\left( a+b\right) }{\left( a+b\right)^{2}}\geq \frac{3}{2}$,

which is clearly equivalent to the Nesbitt inequality

$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$.

Hence, we have not only proved the initial inequality, but also shown that it is weaker than Nesbitt.

[b]To Hungkhtn:[/b]

[quote="hungkhtn"]Ok,but this ineq is still true:

$ \frac{4a^{2}+5bc}{(b+c)^{2}}+\frac{4b^{2}+5ac}{(a+c)^{2}}+\frac{4c^{2}+5ac}{(a+b)^{2}}\ge \frac{27}{4}$

Is your way useful for it,Darij?[/quote]

Well, I don't think so (not checked anyway), but your inequality is just Vasc's first inequality $ \frac{a^{2}+2bc}{\left( b+c\right)^{2}}+\frac{b^{2}+2ca}{\left( c+a\right)^{2}}+\frac{c^{2}+2ab}{\left( a+b\right)^{2}}\geq \frac{9}{4}$ multiplied by $ \frac52$ plus the inequality $ \frac{a^{2}}{\left( b+c\right)^{2}}+\frac{b^{2}}{\left( c+a\right)^{2}}+\frac{c^{2}}{\left( a+b\right)^{2}}\geq \frac{3}{4}$ multplied by $ \frac32$.

[b]To Armo:[/b]

[quote="Armo"]PS. Actually, [b]Myth[/b], [b]darij[/b]'s solutions(especially in geometry) are not always so long, but I don't know why,they become very long in his write up.    :D[/quote]

I know that my solutions are long, actually I try to be as detailed as possible and this makes my posts so long and boring. However, when I try writing in a less detailed way, my solutions quickly become unclear, so I prefer writing more than less. Sorry for my style anyway. BTW, your proof of the initial inequality was very nice!

[b]To Hungkhtn:[/b]

[quote="hungkhtn"][quote="Vasc"]
 (A) $ \frac{a^{2}+2bc}{(b+c)^{2}}+\frac{b^{2}+2ac}{(c+a)^{2}}+\frac{c^{2}+2ab}{(a+b)^{2}}\ge \frac{9}{4}$;[/quote]

  Do you think $ 2$ is the best constant.[/quote]

Both of Vasc's inequalities are tight in the sense that equality holds for a = b = c.

  Darij

[quote="darij"]Sorry for my style anyway[/quote]
Solution is solution, long or short. :) Writing your solution in more details you are making our work(I mean to understand your solution) much easier, so, thanks.  :)

Ok,My solution quite same with Daiji.

We have
$\sum\frac{a}{b+c}+\sum\frac{ab}{a+b)^2} \ge \frac{9}{4}$

And to proof $\sum\frac{a^2+2bc}{(b+c)^2} \ge \frac{9}{4}$

We only solve:

$\sum\frac{a^2}{(b+c)^2}+2(\frac{9}{4}-\sum\frac{a}{b+c}) \ge \frac{9}{4}$

$(\sum\frac{a}{b+c}-\frac{3}{2})^2-2\sum\frac{ab}{(a+c)(b+c)}+\sum\frac{a}{b+c} \ge 0$

But
$\sum\frac{a}{b+c}\ge 2\sum{ab}{(a+c)(b+c)}$

$\leftrightarrow \sum{a^3}+(\sum{a})(\sum{ab}) \ge 2\sum{ab(a+b)}$

It true by Schur!

Very special your solution Darij to the inequalities

(A) $\frac{a^2+2bc}{(b+c)^2}+\frac{b^2+2ac}{(c+a)^2}+\frac{c^2+2ab}{(a+b)^2} \ge \frac{9}{4}$;

(B) $\frac{a^2+bc}{b^2+bc+c^2}+\frac{b^2+ac}{c^2+ca+a^2}+\frac{c^2+ab}{a^2+ab+b^2} \ge 2$.

In fact, you have proven (A) by summing the inequalities

$\displaystyle \frac{\left( a-b\right) \left( a-c\right) }{\left( b+c\right) ^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{\left( c+a\right) ^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{\left( a+b\right) ^{2}}\geq 0 $
and
$\displaystyle \frac{bc+ca+ab}{\left( b+c\right) ^{2}}+\frac{bc+ca+ab}{\left( c+a\right) ^{2}}+\frac{bc+ca+ab}{\left( a+b\right) ^{2}}\geq \frac{9}{4}$,

and (B) by summing the inequalities

$\displaystyle \frac{\left( a-b\right) \left( a-c\right) }{b^{2}+bc+c^{2}}+\frac{\left( b-c\right) \left( b-a\right) }{c^{2}+ca+a^{2}}+\frac{\left( c-a\right) \left( c-b\right) }{a^{2}+ab+b^{2}}\geq 0$
and
$\displaystyle \frac{a\left( b+c\right) }{b^{2}+bc+c^{2}}+\frac{b\left( c+a\right) }{c^{2}+ca+a^{2}}+\frac{c\left( a+b\right) }{a^{2}+ab+b^{2}}\geq 2$.

I proven (A) using the following nice inequality

$\frac{a^2+2bc}{b+c}+\frac{b^2+2ac}{c+a}+\frac{c^2+2ab}{a+b} \ge \frac{3(a+b+c)}{2}$.

Can someone to find a nice proof for it?

To prove (B), I used Cauchy-Schwarz in a special variant, then Schur.

Hungkhtn, I think that the best inequalities of this type are the following

(C) $\frac{2a^2+5bc}{(b+c)^2}+\frac{2b^2+5ac}{(c+a)^2}+\frac{2c^2+5ab}{(a+b)^2} \ge \frac{21}{4}$;

(D) $\frac{2a^2+3bc}{b^2+bc+c^2}+\frac{2b^2+3ac}{c^2+ca+a^2}+\frac{2c^2+3ab}{a^2+ab+b^2} \ge 5$.

But, I haven't solution to these inequalities, yet.

What do you mean by

[quote]To prove (B), I used Cauchy-Schwarz in a special variant [/quote]

Which is the special variant?

What do you mean by

[quote]To prove (A) I used
$\displaystyle \frac{a^2+2bc}{b+c}+\frac{b^2+2ac}{c+a}+\frac{c^2+2ab}{a+b} \ge \frac{3(a+b+c)}{2}$ [/quote]

And then how do you use it to complete (A), you cannot use Chebychev?

Can you please detail better your two proofs?

I don't understand. :(

I wait (a week :P ) somebody find all these answers.

[quote]Hungkhtn, I think that the best inequalities of this type are the following

 $\frac{2a^2+5bc}{(b+c)^2}+\frac{2b^2+5ac}{(c+a)^2}+\frac{2c^2+5ab}{(a+b)^2} \ge \frac{21}{4}$;

But, I haven't solution to these inequalities, yet.[/quote]

Oh,I'm very interested in this inequality.I also think it's true and if it's true,$\frac{5}{2}$ is the best constant.I will check it,but It's really very difficult.

[quote="manlio"]Dear Vasc, when will you post your solution for

$\displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p} $

I hope less than a month  ;)[/quote]

did anyone solve it?

or at least one of the four nice ineqs? they are really tough  :blush:

yes four last ineq is solved already :P

look for vasc_nice ineq 1,2,4,5,6 this was posted by manlio
some of the general one was solved by siuhochung last 3 month :D

[quote="perfect_radio"][quote="siuhochung"]for this general ineq, I think I solved it for $p \ge 0$, but I don' t know what to do with $p \le 0$[/quote]

did you post it on the forum? only thing i could find were the many particular cases of it[/quote]
I haven't, because I haven't found the solution for $p \le 0$. For $p \ge 0$, it's the old trick about vornicu-schur and millions of expansion.

Here's my solution. (I think this is the shortest one)

$\Rightarrow 3\sum_{cyclic} \frac{a^2+bc}{(b+c)^2} \geq (\sum_{cyclic}a^2 + bc)(\sum_{cyclic} \frac{1}{(b+c)^2})$ (by $cheby-shev$)

$= \frac{1}{2} (\sum_{cyclic}(a+b)^2 )(\sum_{cyclic}\frac{1}{(a+b)^2 }) \geq \frac{9}{2}$ (by $Cauchy$)

$Q.E.D.$

Sorry for not detailed explain.

WLOG $a\geq b \geq c$

hmm First, $\frac{1}{(b+c)^2} \geq \frac{1}{(c+a)^2 } \geq \frac{1}{(a+b)^2}$

and $a^2 + bc\geq b^2 + ca \Leftrightarrow a+b \geq c$ is OK!

and $ b^2 + ca \geq c^2 + ab \Leftrightarrow b+c \geq a $

If $a>b+c$ , $a<\sqrt{\frac{3}{2}}(b+c)$.  (cuz from $\frac{a^2}{(b+c)^2})$

Let $a=k(b+c)$ ($1<k<\sqrt{\frac{3}{2}}$)

$\frac{a^2+bc}{(b+c)^2 } + \frac{b^2+ca}{(c+a)^2} + \frac{c^2+ab}{(a+b)^2}  \geq k^2 + \frac{2bc + b^2 + c^2 + k(b+c)^2 }{((k+1)(b+c))^2} = k^2 + \frac{1}{k+1} \geq \frac{3}{2}  $

Then is it ok?

hmm; then this is not the shortest one.  :D

[quote="Chang Woo-JIn"]Here's my solution. (I think this is the shortest one)

$\Rightarrow 3\sum_{cyclic} \frac{a^2+bc}{(b+c)^2} \geq (\sum_{cyclic}a^2 + bc)(\sum_{cyclic} \frac{1}{(b+c)^2})$ (by $cheby-shev$)
[/quote]
why?
i think they may not be similarly sorted.

The following statement are valid:
If $a,b,c$ are non-negative real numbers and $p>-2$, then
$\displaystyle \frac{2a^2+(1+2p)bc}{b^2+pbc+c^2}+\frac{2b^2+(1+2p)ca}{c^2+pca+a^2}+\frac{2c^2+(1+2p)ab}{a^2+pab+b^2} \ge \frac{3(3+2p)}{2+p}$

I proven this inequality in two cases:
1) Case $p\ge -1$. The inequality is equivalent to
$A(b-c)^2+B(a-c)^2+C(a-b)^2 \ge 0$,
where
$A=(2+p)(b+c)(pa+b+c)(b^2+pbc+c^2)-(1+p)(c^2+pca+a^2)(a^2+pab+b^2)$,...
This is true, since for $a \ge b \ge c$ we have $B \ge 0$, $C\ge 0$ and $A+B \ge 0$.
2) Case $-2<p\le -1/2$. The inequality can be proven using Cauchy-Schwarz Inequality in a special form.
First, we write the inequality in the form
$\sum \frac{f(a,b,c)}{b^2+pbc+c^2} \ge 6$,
where $f(a,b,c)=(4+2p)a^2-(1+2p)(b-c)^2$.
Then, apply C-S Inequality.
Finally, we arrive at
$2(2+p)\sum a^2(a-b)(a-c)+3\sum bc(b-c)^2 \ge 0$,
which is clearly true. :)

Did anyone look over Vasc's soln?

I did, but I got stuck at $B,C \geq 0$ (I can't prove it when $p<0$). Can anyone help me?

[Edit] Don't mind this post. I finally proved those results after thinking again, but I'm still missing the other crucial ones... I'll probably never find out how to prove them  :D

[quote="Vasc"]...
For $p=-1/4$, we get the nice inequality

$\sum \frac{1}{4b^2-bc+4c^2} \ge \frac{9}{7(a^2+b^2+c^2)}$.

And this last inequality implies the very nice inequality

$\sum \frac{1}{5b^2-bc+5c^2} \ge \frac{1}{a^2+b^2+c^2}$.

I wait nice or ugly solutions to these inequalities. ;)[/quote]

How about extremely ugly? Expansion reduces the very nice inequality to

\[\sum_{sym} \frac{25}{2}a^6-5a^5b-25a^4b^2+\frac{15}{2}a^3b^3+8a^4bc+6a^3b^2c-4a^2b^2c^2 \ge 0 \]

Which is evident by summing (with appropriate weights) the following

\[
\sum_{sym} a^6 - 2a^5b + a^4bc \ge 0 \\
\sum_{sym} a^6 - 9a^4b^2+8a^3b^3 \ge 0 \\
\sum_{sym} a^6-2a^4b^2+a^2b^2c^2 \ge 0
\]

and then applying Muirhead.

The inequality that resulted first is equivalent to
\[\sum_{sym} 56a^6-28a^5b-128a^4b^2  +44a^3b^3+\frac{95}{2}a^4bc + 31a^3b^2c-\frac{45}{2}a^2b^2c^2 \ge 0 \]

Which is considerably more difficult to "dumbass" but can be realized by adding (with suitable weights) the following

\[
\sum_{sym} a^6 - 2a^5b + a^4bc \ge 0 \\
\sum_{sym} a^5b - 4a^4b^2+3a^3b^3 \ge 0 \\
\sum_{sym} a^4b^2-a^4bc-a^3b^3+2a^3b^2c-a^2b^2c^2 \ge 0 \\
\sum_{sym} a^4bc - 2a^3b^2c+a^2b^2c^2 \ge 0
\]

and then applying Muirhead.

We used the inequality Iran96:
$ \sum\frac{a}{b+c}+\sum\frac{bc}{(b+c)^2}\ge\frac{9}{4} $
We have:
$ LHS=\sum\frac{a^2}{(b+c)^2}+\sum\frac{bc}{(b+c)^2}\ge\frac{1}{3}(\sum\frac{a}{b+c})^2+\frac{9}{4}-\sum\frac{a}{b+c}\ge\frac{3}{2}\Leftrightarrow $
$ \Leftrightarrow (2\sum\frac{a}{b+c}-3)^2\ge 0 $
that is true!
________
Marin Sandu


Find all positive integers such that the sume of the cubes of the digits of $N$ equals $N$.

[hide="Hint"] Find an upper bound. [/hide]

[hide]$N$ can have at most $4$ digits.
This is a correction to my earlier post when I incorrectly stated that $N$ can have at most $2$ digits.
[/hide]

[quote="M4RI0"]$s(n) \leq n$, where $n$ is the sum of digits

Does that help :?:[/quote]

I'm not sure what you mean  :huh:

[hide="The upper bound to which I was referring"] Let $n$ be a $d$-digit number.  The sum of the cubes of the digits of $n$ has the property

$S_{3}(n) \le 9^{3}\cdot d$ (can you see why?) [/hide]

Oopps. What I meant was $n \leq a(n)$, where $a(n)$ is the product of digits :blush:


A = {1, 2, 3, 4, 5, 6} is the set of first six positive integer...

[b]Find the all permutations of A such that the difference of each two adjoined elements are at least 2...[/b]

What happens if A = { 1, 2, 3, ...., n }

[b]Is there any relation of the solution with fibonacci numbers ?[/b]

[quote="maxal"]See http://www.research.att.com/~njas/sequences/A002464[/quote]

this link does not contain the solution and also does not contain the exact answers to my questions... Is there anyone else who has comprehensive solution ?

Read carefully - it's exactly the answer to your question:
%C A002464 Permutations of 12...n such that none of the following occur: 12, 23, ..., (n-1)n, 21, 32, ..., n(n-1).

[quote="AutumnRose"]The following article describe a more general problem with a lot of interesting result:[/quote]

thank you and also thanks to maxal, but my problem still continues... :(

I understood the solutions that you sent me...but in my set A = { 1,2,3,4,5,6}
question is not only the permutations with 6 elements, but also the permutations of A with 5 elements or 3 elements.  and in this condition situation changes for example a 4-element permutation of A is { 1,2,3,5} or { 2,3,5,6} so if it stays just {1,2,3,4} there will be no problem only "n" changes and similar procedure can run, but all elements of permutation are not consequtive...

So what can be done now?

please detailed solution...

[quote="maxal"]The [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]inclusion-exclusion principle[/url] will help to solve the problem for arbitrary size permutations.
The solution should not be much different from the case of full-size permutations.[/quote]

CAN YOU WRITE HERE THE DETAILED SOLUTION I COULDN'T SOLVE IT...

Let's count number of permutations of $ \{1,2,\dots,n\}$ of size $ k$ that do not satisfy any of the properties $ P_s \equal{}$ "$ s$ and $ s \plus{} 1$ are adjacent in a permutation".
The inclusion-exclusion principle implies the expression:
\[ \sum_t ( \minus{} 1)^t \sum_j \sum_{x_1 \plus{} \dots \plus{} x_j \equal{} t} \sum_{y_0 \plus{} \dots \plus{} y_{j \plus{} 1} \equal{} n \minus{} (t \plus{} j)} 2^j \binom{n \minus{} (t \plus{} j)}{n \minus{} k} (k \minus{} t)!\]
where
$ t$ stands for the number of properties that permutations satisfy;
$ j$ is the number of chains they form (i.e., if $ s$ and $ s \plus{} 1$ are adjacent and $ s \plus{} 1$ and $ s \plus{} 2$ are adjacent, then $ s, s \plus{} 1, s \plus{} 2$ form a chain);
$ x_i\geq 1$ is the number of properties in the $ i$-th chain, the length of this chain is obviously $ x_i \plus{} 1$, so that the total length of all chains is $ t \plus{} j$;
$ y_i\geq 0$ is the number of single elements between $ i$-th and $ i \plus{} 1$-st chains;
$ 2^j$ is the number of orientations of all chains (factor of 2 per each chain);
$ \binom{n \minus{} (t \plus{} j)}{n \minus{} k}$ is the number of ways to remove $ n \minus{} k$ elements from $ n \minus{} (t \plus{} j)$ single elements (so that $ (k \minus{} t \minus{} j)$ single elements remain);
$ (k \minus{} t)!$ is the number of ways to permute $ j$ chains and $ (k \minus{} t \minus{} j)$ remaining single elements (note that $ j \plus{} (k \minus{} t \minus{} j) \equal{} k \minus{} t$).

The above formula simplifies to
\[ \sum_{t \equal{} 0}^k ( \minus{} 1)^t (k \minus{} t)! \sum_{j \equal{} 0}^{\min\{t,n \minus{} t\}} \binom{t \minus{} 1}{t \minus{} j} \binom{n \minus{} t}{j} \binom{n \minus{} t \minus{} j}{n \minus{} k} 2^j\]
In particular, for $ n \equal{} 6$ and $ k \equal{} 0,1,\dots,6$, we have 1, 6, 20, 48, 90, 124, and 90 permutations respectively.

P.S. As one may notice, this solution is just a slight extension to the case $ k\equal{}n$ which I gave at http://www.mathlinks.ro/viewtopic.php?t=65890 awhile ago.


geometry
linear algebra
matrix
integration
logarithms
inequalities
function

Lets start the pointless math thing

Given $\triangle ABC$ has an area of $2e$ and $\triangle ABC$ is an equilateral triangle, what is its height?

yo I'm not sure but I'll guess sqrt(2esqrt(3))

actually the answer should be pointless unless the answer is quite obvious.
Also you should post another problem after you solve it.

$\left[\begin{array}{c}6\16\16\16\end{array}\right]\times[2]=$

Please post another problem after you solve it.

$\begin{bmatrix}1&2&3&4\\5&6&7&8\end{bmatrix}\cdot\begin{pmatrix}1&2\\3&4\\5&6\\7&8\end{pmatrix}=$

[quote="sapphyre571"]$\begin{bmatrix}I&could\\care&less\end{bmatrix}$

What is the product of the reciprocal and additive inverse of $M$ if $M$ is the sum of the phone numbers of all your relatives divided by the product of their ages?[/quote]

$\pi?$ aka (-1)

Bob is a bus driver and is $x$ years old, where $x$ is an integer. One day, when he drove the bus he decided to count the number of people who got on and got off of the bus. At his first stop, 5 men and 6 women got onto the bus (assume the bus was empty before this stop). Then at the second stop half of the women left the bus, while 1 women got back on the bus. Also 1 man got off the bus. Three hours later, the bus driver arrived at his third stop where all the men got off the bus, and 10 more women got on the bus. Then, at the fourth stop, 5 men got on the bus. He then counted the number of men and women on the bus. What is the name of the bus driver?

Man thats tough...I'm guessing eggs!

If $P^{3}=N_{2}$ where does $\Theta$ go?

Since no one posted an answer yet I shall post one.

This can be rewritten as $P\cdot P\cdot P=N_{2}$. Since $O=P-1$ it equals
$O\cdot P+1\cdot P-1=N_{2}$

$\int_{\infty-\Phi^{2}}^{\alpha}\mu =$

No one is posting here.

$\int_{\infty-\phi^{2}}^\alpha\mu\equiv\mathbb{ALPHA\%Error. "CENts" is a bad command.
}$

[quote="hunter34"]actually the answer should be pointless unless the answer is quite obvious.
Also you should post another problem after you solve it.

$ \left[\begin{array}{c}6 \
16 \
16 \
16\end{array}\right]\times[2] =$[/quote]

haha i just learned this [yes, i'm dumb, but i already know that, thank you very much.]
[32
32
32]
yay. xD

 :whistling:
great, i have to post another problem...

Okay, here goes:
What is the sum of:
1+2+3+...+24

88 because i say so
who is the president of yugoslavia

There isn't one...and I mean one
...hmmm...
what are all the digits of [i]e[/i]

hey guys:

If ra5249 eats 5 tacos evryday, then Where Does He Buy them???

If any1 (but ra5249) can solve, they get a prize!

Police

[hide]jkjk guys :rotfl: [/hide]

Every degenerate triangle is a line. Every line is of length 42, since 42 is the answer to everything. Thus, the two segments that make up this triangle sum to a number less than or equal to 42. However, since 42 is the answer to everything, they sum to exactly 42. Thus, we have two numbers $a$ and $b$, WLOG $a\leq b$, such that $0 < a, b < 42$. However, if $a$ and $b$ are not both 21, they can be extended and/or shrunk so that they have length 21. Thus, $a = b = 21$. Since the side lengths of the triangle are 21, 21, and 42, this triangle is isosceles, so we are done.

Show that the AMC Forum will always be locked.

Well, facts are that 42 is the answer to everything, the AMC forum is currently locked, and the usual stuff (1=2, 9001 is > 9000, etc.)

Thanks.
Being locked forever would mean that the lock time would increase linearly. more specifically, each hour we increase in time, we add an hour to the amount of time spent locked. hence, the lock time follows the function $f(x)=x$ which is it's own inverse. The inverse is $f^{-1}(x)=42$, and the range is amount of time spent locked. this line will forever be going up from y=0, and up in this situation means more time being locked. Hence, the AMC forum will be locked forever. (sorry for not making much sense, but basically since a function is congruent to the inverse, and the inverse of the function of the time spent locked is infinite, the lock time will be equal to the inverse which is infinite)

*sorry, Lord.of.AMC, but you're never going to be able to rule you're forum from now on.

Prove that there exists exactly 42 irrational positive integers with $\text{Im}(-\pi)$ in the interval between 1337 and $9000(2-0.\overline{9})$

@cire_il I know right.

First of all, note that $9000(2-0.\overline9) = 9000(2 - 1) = 9000$. Having established this, we can continue with the proof.

By the Pigeonhole Principle, we know that there must be at least 183 numbers within 1337 and 9000 with same imaginary residue mod 42, since $(9000-1337)/42 = 7663/42 = 182.452... \approx 183$. Thus, since irrational numbers are the norm of real numbers (from the real numbers, it just makes sense that we pick an irrational number with probability 100% and a rational number with 0%), and that since 42 is the answer to everything, there must be at least 42 irrational numbers in the subset of 183 real numbers, so we are done.

[quote="cire_il"]Darn, it's kind of sad how this is the most active thread in the Texas forum[/quote]
It's a tad depressing.

*sigh...

Find x if y=mx+b, b=mx+y and x=3.
Now that you found x tell me if
you are driving a car that is traveling along a cartesian plane at 70.e units per second and you throw a triangle out the window and a policeman notices, how many purple sheep can Bob buy if Sally Sold Seashells by the Salty Sea Shore for 5 human souls :oops:

Huh?
I HEARD THE WORD "REVIVING" AND THE WORD "POINTLESS"! I'M HERE!

Could you guys take this to another thread at least? I'd rather not have this showing up in my past posts feed anymore. :)


What's the largest number of elements that a set of positive integers between $1$ and $100$ inclusive can have if it has the property that none of them is divisible by another?

We can write each positive integer $ n$ in a unique way as $ n \equal{} 2^k\cdot q$ with $ k,q\in\mathbb{N}$ and $ q$ odd.
If two numbers have the same $ q$, one is divisibly by another. Since there are only $ 50$ odd numbers in $ [1,100]$, the requested number is at most $ 50$.
But the set $ \{51,52,53,...,100\}$ is an example of such a set with $ 50$ elements, so the maximum number of elements is $ 50$.

I think, by the way, that the source of this problem is Erdös.


trigonometry
calculus
derivative
calculus computations

Show that:

$ sin^{ \minus{} 1}\frac {x}{a} \equal{} \minus{} cos^{ \minus{} 1}\frac {x}{a} \equal{} \frac {1}{\sqrt {a^2 \minus{} x^2}}$

[hide="What I did"]y=$ sin^{ \minus{} 1}\frac {x}{a}$

$ siny \equal{} \frac {x}{a}$

$ x \equal{} asiny$

$ \frac {dx}{dy} \equal{} acosy$

\\\ly for $ y \equal{} \minus{} cos^{ \minus{} 1}\frac {x}{a}$

$ \frac {dx}{dy} \equal{} asiny$

But how can I show that both these expressions are $ \frac {1}{\sqrt {a^2 \minus{} x^2}}$[/hide]

For a good reason too. They aren't equal to their expression, but their derivatives are =]


Prove that $ BQ$ and $ AR$ lines intersect on the altitude corresponding to vertex $ C$.
($ ACPQ$ and $ CBRS$ are external squares on the sides of the triangle $ ABC$ - See figure.)

See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=39101

And notice that the solutions of this problem can help us to prove a very nice and hard problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=79051

It means that [b]Huyen Vu[/b]'s solution is still work with our problem here, and we can say that our property is still true when two similar rectangles are constructed inwardly the given triangle.


The angle-bisectors $ AA_1,BB_1,CC_1$ of a triangle $ ABC$ with permeter $ p$ meet $ B_1C_1,C_1A_1,A_1B_1$ respectively at $ A_2,B_2,C_2$. The line through $ A_2$ and parallel to $ BC$ meets $ AB,AC$ at $ A_3,A_4$. Construct the points $ B_3,B_4$ and $ C_3,C_4$ in a similar way. Prove that
$ AB_4\plus{}BC_4\plus{}CA_4\plus{}BA_3\plus{}CB_3\plus{}AC_3\leq p$


Rid-a-Spot comes in a 1-quart container. There are 4 cups in  1 quart. To wash windows, 2 cups of Rid-a-Spot are to be mixed  with 5 cups of water. How many quarts of this mixture can be made  from the 1-quart container of Rid-a-Spot? Express your answer as a decimal to the nearest tenth.

1 quart of RAS is 4 cups.

That allows 2 mixes of 7 cups total liquid each.

That gives 14 cups of total solution, or $ \boxed{3.5}$ quarts.


Let $ a,b ,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} k$. Find the minimum value of the following expression:
$ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}$

[quote="daigiaga1994"]Let $ a,b ,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} k$. Find the maximum value of the following expression:
$ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}$[/quote]

Let $ a\to0,b\to0,c\to{k},$ then

$ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}\to {\plus{}\infty}.$

But we have

$ [\frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}]_{min}\equal{}\frac{3\sqrt{3}}{2\sqrt{k}}.$

[quote="fjwxcsl"][quote="daigiaga1994"]Let $ a,b ,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} k$. Find the maximum value of the following expression:
$ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}$[/quote]

Let $ a\to0,b\to0,c\to{k},$ then

$ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}\to { \plus{} \infty}.$

But we have

$ [\frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }}]_{min} \equal{} \frac {3\sqrt {3}}{2\sqrt {k}}.$[/quote]
I'm so sorry.  I've edited my post


Hello

could someonepost the problems of schweitzer 2007 ?

or put a file in PDF on the forum ?

thanks

Isn't problem 4 just the restatement of Cauchy-Davenport??!


I'm going over systems of equations in my old textbook I used this past school year, and as I was solving some problems, I noticed this one.

$ 5.$
$ y\equal{}x\plus{}1$
$ y\equal{}2x\minus{}1$

I don't know how you can solve these with elimination or substitution! Please help!!!

Multiply one of the equations by -1 then add the equations together. Solve for x. Plug x value back in to solve for y.

Since the problems have y= in them, I would say it's easier to substitute. So, you let x+1 and 2x-1 equal each other. Then, solve 2x-1=x+1 to get x=2. Plug x back in to solve for y, which is 3.

Latexified

Since they both equal y, they'd be equal to each other, so
$ x\plus{}1\equal{}2x\minus{}1$

$ x\equal{}2$

Plug in

$ y\equal{}2\plus{}1\equal{}3$

$ y\equal{}4\minus{}1\equal{}3$

And if you have a calculator, they are already in slope-intercept form, so you can graph them and find the intercept.

(Note: I am not saying that this is the best method; in fact, I would recommend one of the algebraic solutions. It is also not the fastest. It's just a useful tool to know for later, because more complicated systems can usually be solved most quickly by graphing.)

If you come across a question like this, and you're not sure how to solve it, just rearrange the equations and solve from there. It's easier to do the method people above mentioned, but if you have a test or something and you forget, just rearrange them.  :wink:

[quote="myyellowducky82"]I'm going over systems of equations in my old textbook I used this past school year, and as I was solving some problems, I noticed this one.

$ 5.$
$ y \equal{} x \plus{} 1$
$ y \equal{} 2x \minus{} 1$

I don't know how you can solve these with elimination or substitution! Please help!!![/quote]
Substitute x+1=y into the second equation
x+1=2x-1
x=2
substitute into first equation
y=3
(2,3)


$ x$, $ y$ and $ z$ are positive odd integers. What is the remainder when $ x^2\plus{}y^2\plus{}z^2$ is divided by 4?

The square of x, y, and z are all odd. So if you add them all together, and divide by 4, you have a remainder of 3.

Let $ n$ be an odd integer, then

$ n \equiv 1 \mod 4$

OR

$ n \equiv 3 \mod 4$

We find that

$ n^2 \equiv 1 \mod 4$

for both cases. So

$ x^2 \plus{} y^2 \plus{} z^2 \equiv 1\plus{}1\plus{}1 \equiv \boxed{3} \mod4$

[hide="Alternate Solution"]The way that modularmarc did it is what I think is the best way, because you prove it and you can fully understand it.  However, you may be able to do it slightly faster this way, but it's cheap.  You can imply from the way that the question is asked that there is only $ 1$ answer, and that [b]any[/b] odd value(s) for $ x$, $ y$, and $ z$ will get the same answer.  So, if we simply assume $ x \equal{} y \equal{} z \equal{} 1$, we get $ 1^2 \plus{} 1^2 \plus{} 1^2\equiv1 \plus{} 1 \plus{} 1\equiv\boxed{3}\text{ (mod } 4)$[/hide]

[hide="Or, fastest way for countdown"]
Plug in $1,3,5$. \[9+25+1=35\]. Remainder when divided by $4$ is $\boxed{3}$ [/hide]