Support algebra function domain \/closed

Is it just me, or is this site becoming progressively slower by the day since the merge with mathlinks? It used to be rather fast (pages loaded within a second or two), but now it's taking in excess of 1 min 10 sec (I just stopwatched it) for pages to fully load, which is rather painful. I have a 2 megabit connection and other pages have been loading at their normal speed, so I have a feeling that the problem is not on my end.

Weird - I'm not seeing this at all, but a few students in class had this problem today.  How widespread is this?

I noticed it also but I thought it was because I had spyware or something.

Is this just today, or over the last week, or what?

It's been happening to me too...for about a week or so.

By the way, shouldn't this thread be in Support rather than in Questions and Suggestions? I just tried deleting my browser (Mozilla) cache, closing my browser, and starting over. The whole site, including images, loaded in ten seconds with a cable modem connection. I logged in instantly, and then I got here right away. No noticeable slow-down here. I get cable modem pool latency problems once in a while, when a lot of other subscribers are online at the same time as I, but don't notice any particular slowness to this AoPS site on a consistent basis.

Now it takes my computer 4-5 seconds instead of 2-3.  On my computer there isn't much of a difference.

Is it the whole site that is a problem, or just the message board?

Are there others besides me and tokenadult who are *not* seeing any difference?

Also, for those of you experiencing slowness - what browser/OS are you using?

I am using IE 6 and I have Windows XP.  Most of the time the site is fine but every once in a while it will take a large amount of time to load something.

I've been having the same problems, along many pages not loading and tex not working sporadically.

[quote="rrusczyk"]Please note how long you've been having the problems, as well as your internet connection type and browser...[/quote]

I'm not complaining.  But since you are asking...  The site has been noticeably slower for me, since the merge.  I can't honestly quantify the amount.  It just feels slower.  [Maybe half as fast, on average, if I had to make a guess.]

Sometimes -- perhaps two or three times a week -- pages will consistently take a minute or more to load.  I think I've always seen this behavior in the wee hours of the morning (USA).  The site is essentially unusable when this occurs.  Most times it seems to clear up fairly quickly (under ten minutes?)  A few times I've just given up for the night.  This never occured prior to the merge.  I think I was at least as active on this web site prior to the merge, so I think this is a new behavior.

I see both of these behaviors with IE5, IE6, and NS4.7x.

[quote="rrusczyk"]Please note how long you've been having the problems, as well as your internet connection type and browser you are using - this will help us diagnose the problem, as we've made several server changes over the last couple weeks.  Knowing exactly when the problems happen will help us zero on which change caused them.[/quote]
The speed change coincided precisely with the merge. Some days it has been a bit better, and some a bit worse, but it seems to be going slowly downhill on average since then.

My computer, etc:
Win XP Home
Mozilla 1.6
2 megabit down / 128 kilobit up cable connection through Charter Communications in Traverse City, MI
More than capable hardware to run the computer acceptably

Windows XP Pentium 4, AOL or Microsoft Internet Explorer.

Tex images aren't loading and whenever I click on basically anything, it says "Bad Request."  I can get to the forums in two ways: either going to the homepage and clicking the forum button, or hitting the browser's back button.  I couldn't click the button now.

[quote="confuted"]The speed change coincided precisely with the merge. Some days it has been a bit better, and some a bit worse, but it seems to be going slowly downhill on average since then.[/quote] Please report any abonormality in the functionality of the site as soon as it appears. The merge took place a couple of weeks ago. By reporting quickly we can operate faster in order to solve the problem.

On the other hand, it is strange, as only some people are affected by this behaviour, which seems to indicate that it is not a server problem. However you all have this since the merge and this is clue pointing that it is related closely to the server ...  :?

[quote="Valentin Vornicu"]Please report any abonormality in the functionality of the site as soon as it appears. The merge took place a couple of weeks ago. By reporting quickly we can operate faster in order to solve the problem.[/quote]
I actually complained about it the day of the merge to Mr. Crawford, who reassured me:

[quote="Mathew Crawford"]The site will get faster. We're adding RAM on the server end and much of the slowness is due to all the little quirks of combining so many things at once. Ultimately the site will probably get faster soon over the next couple of years because the influx of members means a greater influx of capital with which to make improvements.

I hope you enjoy the new international base of students.[/quote]

So I held off on further complaints for a while.

[quote="rrusczyk"]Is it the whole site that is a problem, or just the message board?[/quote]
Excuse the double post, but it's the entire site, not just the message board.

[quote="rrusczyk"]Has it always been the entire site?  To the degree I see a slow-down, I only see it on the forum, and the merge didn't touch anything else on the site (though we have made server changes since doing the merge that might affect the whole site).[/quote]
I don't know; I don't often visit the rest of the site. I come directly to the message board when I visit, without going through the "main page."

RC-7th: How do you connect to the internet?

Through a network, either home or work. I use cable or DSL

Are all pages always slow?  Sometimes slow?  Have you tried deleting your cookies?

(As for the east coast - our server is on the east coast, so I don't think geography is the problem)

Also, what are you calling slow?  5 seconds?  10 seconds?  They're loading in around 1 second on cable for me, 5 seconds on dialup.

[quote="RC-7th"]To me the whole site is slow, whatever browser or computer I use, but I live on the east coast, maybe that has something to do with it. Interestingly, to me the mathlinks.ro website is much faster.[/quote] the only real difference between the two domains is the DNS addressing. Other than that, which should not affect this, both sites should have the same speeds.

Did you try changing templates? Do you have the problem on the SubSilver template?


Let $ G$ be a simple graph with vertex set $ V\equal{}\{0,1,2,3,\cdots ,n\plus{}1\}$ .$ j$and$ j\plus{}1$ are connected by an edge for $ 0\le j\le n$. Let $ A$ be a subset of $ V$ and $ G(A)$ be the induced subgraph associated with $ A$. Let $ O(G(A))$ be number of components of $ G(A)$ having an odd number of vertices.
Let
$ T(p,r)\equal{}\{A\subset V \mid 0.n\plus{}1 \notin A,|A|\equal{}p,O(G(A))\equal{}2r\}$ for $ r\le p \le 2r$.
Prove That $ |T(p,r)|\equal{}{n\minus{}r \choose{p\minus{}r}}{n\minus{}p\plus{}1 \choose{2r\minus{}p}}$.

[hide="hint"]
use lattice path. start from 0,0. at any step if next no. is taken go right, else go up.
next use two specific operations on the path:
1.delete 1 right step from every odd lengthed horizontal strip to make it even lengthed.
2.all consequtive right steps are halved,and other half is discarded.
 now find the inverse process of above operations.
[/hide]

Sorry ith_power, could you please elaborate your hint ?

[hide]the first term is no of paths possible when both operations are done. i.e. without any restriction other that $ 0,n\plus{}1 \notin A$.
the second term is no of paths $ P$ from which we can arive at same path after first operation.[/hide]


induction
combinatorics unsolved
combinatorics

There are $ n\geq 5$ people in a party. Assume that among any three of them some two know each other. Show that one can select at least $ \frac{n}{2}$ people and arrange them at a round table so that each person sits between two of his/her acquaintances.

Use induction on $ n$ :

Basis cases are straightforward.

Assume the result holds for $ n$ and let's consider a collection of $ n\plus{}2$ people.
If any two persons are acquainted the result holds trivially.
Assume that people $ A$ and $ B$ are not acquainted, and using the induction hypothesis to the group of the $ n$ other people, we may choose at least $ \frac n 2$ of them to be disposed as desired. Let $ M$ be any of these people sitting around the table, then $ M$ is acquainted to $ A$ or $ B$, so that $ A$ or $ B$, say $ A$, is acquainted to at least half of the people around the table. If to of the people around the table, sau $ C$ and $ D$ are acquainted with $ A$, then just insert $ A$ between $ C$ and $ D$.
If $ A$ is not acquainted to two neighbours around the table, then $ \frac n 2$ is even and the people around the table are $ M_1P_1M_2P_2 \cdots M_kP_k$ such that $ A$ is acquainted to each of the $ M_i$'s, and not to any of the $ P_i$'s thus $ B$ is acquainted to each of the $ P_i$'s. Note that $ n \geq 5$ implies that $ k \geq 2$.
If $ k\equal{}2$ then we may use the arrangement $ M_1AM_2P_1BP_2M_1$.
If $ k \geq 3$ then, some two among $ P_1P_2P_3$ are acquainted, say $ P_1$ and $ P_2$. Then, instead of $ M_1P_1M_2P_2$ then use $ M_1AM_2P_1P_2$ which gives an arrangement of $ 2k\plus{}1$ persons, as desired.

Pierre.

if we mention the friends of A( one person in n people) group B and the others group C then because one of group C is not friends with A, each two people in C are friends and if the number of people in C is more than n/2 it's solved.
so we mention that each person has more than n/2 friends so if we start from A and choose one of its friends and so on then after on less n/2 times (if get cycle we have to come more than n/2 times) we get to a person if it was a friend of A than it's solved if not it's from the group C and because it has more than n/2 friends it has on less one friend which is also friend with A.


algorithm
number theory
Euclidean algorithm

hey ppl ...help me with dis
this is the given problem

Given two vessels, one of which can accommodate a litres of water and the other - b litres of water, determine the minimum number of steps required to obtain exactly c litres of water in one of the vessels.

At the beginning both vessels are empty. The following operations are counted as 'steps':

    * emptying a vessel,
    * filling a vessel,
    * pouring water from one vessel to the other, without spilling, until one of the vessels is either full or empty.

Input

An integer t, 1<=t<=100, denoting the number of testcases, followed by t sets of input data, each consisting of three positive integers a, b, c, not larger than 40000, given in separate lines.
Output

For each set of input data, output the minimum number of steps required to obtain c litres, or -1 if this is impossible.

in c++
i have done  a bfs on the possible states ...and i am storing the distance in a map corresponding to every state.. checking if the state has occured previously in the map ..if not then updating distance

now this approach times out ..time limit 1 sec..
what optimisation can i make

best regards
vasu

[quote="TripleM"]As far as I can see, theres only two possibilities (this is a well-known problem, after all).
Either:
fill A, pour A->B, empty B, pour A->B. Repeat.
Or the same, but starting with B.

Just run them both, and see which one is shorter.[/quote]
That's an NP algorithm. There is a better way. I can transfer water from one container to another. I can also compute $x-(Y-y)$ where $Y$ is the capacity of a container, and $y$ the amount it holds. These are the operations the Extended Euclidean Algorithm uses. So now we have to translate back to pours. This is pretty easy.

Yeah, that would be the faster (algorithmically) way to do it, but with such small bounds on the sizes of the two vessels, either way works fine. (my way got AC in 0.33 seconds, and thats in java!)

i just got a TLE trying to run the same algorithm as stephen's....probably i am doing the some stupid mistake
just tel me if i have got it right(ur algo that is)
we run thru the set of states starting with(a,0) and pouring from 1st to 2nd always..and emptying 2nd if it is full..filling 1st if its empty

and then run thru set of states starting with(0,b) and pouring from 2nd to 1st always..emptying 1st when its full...and filling 2nd when its empty

am i right??
and one more thing ..just chekd spoj status..does the test data really require the huge amt of memory ur program uses..or is it because of java...(this might sound like a dumb question:) ...anyways)

thx

Yup, thats all I did.
And yeah, java uses a huge amount of memory for all sorts of weird stuff, its not really the program.

Can you elaborate more on that answer?

I deeply apologize for my incompetence, but I don´t understand what you meant, starting by the meaning of the x.


Prije par dana sam obrisao sa našeg podforuma ovakvu temu

http://www.mathlinks.ro/viewtopic.php?t=248100

u dogovoru sa ostalim moderatorima (direktiva s vrha je da se ovakvi crankovi brišu). Ko želi da pogleda rad, na gore navedenom threadu se nalazi link. Zašto ovo pišem:

Nisam znao kakvoj naučnoj veličini se zamjerim! http://uncyclopedia.wikia.com/wiki/Tsolkas  :rotfl:

Zanimljivo...
Ali, 2. link svako moze da edituje i da stavi svoj tekst, pa ne dokazuje tacnost, a 1. link treba detaljno procitati i onda se moze zakljuciti da li ima greske, ili je dokaz tacan.

2. link je poprilično duhovit (mada Uncyclopedia ima mnogo smješnijih članaka, dakako), dok je prvi prevelik čitalački zalogaj za mene. Naime, pročitao sam rad, ali nisam našao previše smisla (priznajem, nisam se ni trudio baš previše da ga nađem). Na svakom naučnom forumu postoje ljudi koji su sigurni da su dokazali da je Einstein pogriješio, da imaju dokaz FLT na pola strane, dokaz Riemannove hipoteze na dvije strane i sl (društvo sa elite securityja ili math.e foruma će se sjetiti Spreče). A ovaj Grk je dodatno nanervirao moderatore postajući svoj link po cijelom forumu, te ulazeći u svađu s pojedinim moderatorima... Ovaj thread sam postavio tek toliko da se ne osjetite zakinutim u odnosu na ostale nacionalne podforume na kojim thread (još uvijek) nije obrisan.

Vi vjerojatno niste upoznati s Gezom... autorom, pogledao sam, 3417 postova na forum.hr. E, uglavnom, taj je (nažalost, naišao sam na njegove postove tek nakon što je prestao objavljivati) imao sve moguće teorije... to trebate pročitati. No, posebnu je zainteresiranost pokazivao za broj 4 za koji je zaključio da upravlja prirodom, što li.

Naravno, volio je tetraedre (napisao je i knjigu o njihovim svojstvima) i, po njemu, postoje četiri spola čovjeka.

Od svoje daljnje djelatnosti, tvrdio je da je 1:2 oznaka za dvije polovine i da je to strogo veće od 1. Mislim da je imao i ideju da matrice zapravo u sebi nužno imaju neko dublje značenje (i barem četiri dimenzije). Volio je velike riječi i pisao ovakve izjave: "[i]... DOVOLJAN JE MOJ PRIMJER. ŠTO GOD SAM PAMETNO NAPISAO KAO PAMET SVJETSKE ZNANOSTI POPLJUVANO JE OD REDUKCIONISTA I NEZNALICA ŠTO JE ZNANOST.
TEK KADA ZNANOST POSTANE SLOBODNA UMNOST TEK TADA JE MOGUĆ POVRATAKA ZNANOSTI. SVAKO HERMETIZIRANJE, SHEMATIZIRANJE I TVRĐENJE DA POSTOJI ZNANSTVENA ISTINA DO KOJE SE DOLAZI SAMO EKSPERIMENTOM JE TUŽALJKA ZNANOSTI.[/i]"

Još ću vas ostaviti s ovom njegovom misli: (na pitanje kakvi su ljudi kao on) "[i]Visoki, elegantni, kao muškarci lijepi, naočiti, obrazovani, samoobrazovani, druželjubivi, pokretljivi, dobrotvori, plemeniti, po prirodi istraživačii i kreatori, u zvanju i karijeri uspješni, imaju veliki radni i kretaivni kapacitet, dobri su ronioci, ribari, skiperi, plesaći, ljubavnici, pijanci, vlo dobro kuhaju, razumiju se praktički i sve vještine ...., jedina im je velika mana što nemaju sluha.[/i]"

Uostalom, upišite u Google geza site:forum.hr pa se smijte. Ja se smijem već dvije godine.

Sve gornje izjave mogu se pronaći na gorespomenutom sajtu.


calculus
integration
real analysis
real analysis unsolved

[quote="flip2004"]  $ J_n(f)  = \int\limits_{0}^1 |f(x)|^n dx$
$  J_0(f):=\frac{1}{2}$ [/quote]

With the definition of $J_n(f)$ we have $J_0(f)=\int\limits_{0}^1 1dx=1$

$J_n$ has the same property of Wallis integral
$J_{n+1}\leq J_n$,  $J_{2n}=\frac{(2n)!}{(2^n.n!)^2}J_0$, $J_{2n+1}=\frac{(2^n.n!)^2}{(2n+1)!}J_1$, $(n+1)/(n+2)=\frac{J_{n+2}}{J_n}\leq \frac{J_{n+1}}{J_n}\leq1$ we get $J_{2n+1}/J_{2n}$ tends to $1$ when n tends
$+\infty$ with Stirling formulae we get $J_1$

The definiton of $ J_n(f) $ as an integral is valid  for $ n\ge 1 $ . However, thanks for remark.Alex


Can any of you find f(x,y,z) that the following ineq is right where x,y,z≥0:
$ (x^2\plus{}xy\plus{}y^2)(y^2\plus{}yz\plus{}z^2)(z^2\plus{}zx\plus{}x^2)$≥f(x,y,z)
I found that f(x,y,z)=$ \frac{(x\plus{}y\plus{}z)^2(xy\plus{}yz\plus{}zx)^2}{3}$
I am needing another f(x,y,z) that is stronger than this one.Let me know your idea

It is rather hard to understand what exactly you are looking for. Why, for example, not to take $ f(x,y,z)$ to be just the left hand side? :?


Think of one word for each of the following properties.

1) This 8-letter word has five vowels in a row.
2) This 9-letter word has exactly one vowel (this includes y)
3) This 10-letter word can be typed on the top row of a standard QWERTY typewriter.
4) This word has all of the vowels in order (a,e,i,o,u).
5) This word has all of the vowels out of order (a,e,i,o,u), but is only 7 letters long.
6) This ten letter word has 1 of one letter, 2 of another letter, 3 of another letter, and 4 of another letter.

*** Spoilers Ahead ***
.......partial list of answers
>
>
>
>
.
.
.
.
.>
>
>
>
>
1) queueing

2) strengths

3) typewriter

4) facetious

5)

6)
>

On Today, at 10:31 am ; junggi wrote:
[quote]I'm sure queueing isn't a word[/quote]

True. According to dictionary.com queueing is not a proper word.

My attempt at solving the problem is given below.

[hide]PARTIAL ANSWERS:

1) Not found a solution. Solution with 7 letters: queuing.

2) strengths

3) typewriter, proprietor, repertoire

4) abstemious,  abstemiously,  abstentious,  acheilous,  acheirous,  acleistous,  affectious,  aleikoum,  anemious,  annelidous,  arsenious,  arteriosus,  arterious,  bacterious,   facetious,  facetiously,  fracedinous,  majestious,  parecious,  pareciously,  tragedious

5) Sequoia, Oiseaux

Note:  Till date, I have not been able to solve (6)[/hide]

[quote="junggi"]I'm sure queueing isn't a word[/quote]
However, [url=http://www.google.com/search?hl=en&q=queueing&btnG=Google+Search]Google[/url] and [url=http://en.wikipedia.org/wiki/Queuing_theory]wikipedia[/url] don't seem to have much of a problem with it. In fact, wikipedia even redirects "queuing theory" to "queueing theory".

(6) is fun.

randomdragoon wrote:

[quote](6) is fun[/quote]

Solution to (6):

[hide]SLEEVELESS -----> This word contains precisely one 'V', two 'L's, three 'S's and four 'E's.[/hide]

The instructions at the top of the first post include:

"Think of one word..."

For any entry of 1) - 6) you should have THAT disqualified from counting for NOT following directions.  I don't care if more than one word fits the other criteria; it's a puzzle or game, and there are rules to be upheld.

K. Sengupta, for each numbered entry where you entered more than one word, that numbered entry should not count.

Following directions is part of puzzle solving.

box of rocks, are you [b]joking[/b]?

Another fun but slightly easier one:
Find a word (any length) with 6 consecutive consonants. (where y counts as a vowel).

On #3, "typewriter" is not a coincidence- it's marketing. The QWERTY keyboard was originally designed to slow down typists so that keys wouldn't get stuck. It's impractical with modern equipment, but persists because everyone's familiar with it.

To "box of rocks": Don't be so obnoxious. In this context, there's nothing wrong with finding multiple answers.

Not a real answer to #6: "these teeth"

tim1234133,

no, I'm not joking.  The user who calls my correct answer wrong will learn about following directions.
------------------------------------------------------------

jmerry,

you have an insecurity about following instructions too?

There is absolutely nothing obnoxious about what I've done.  It's called reading the directions and adhering to them.  You're
the person having trouble with that.  There is no "in this context."  You give one word or you don't.

When I make up a puzzle, I don't want multiple answers given with "one answer."  That's plain.  If I want multiple answers, THEN I'll type that.

Don't affix the name "obnoxious" to a user; it shows you are being a name-caller yourself, which is rude.

[quote="mathnerd314"]Please stop accusing and actually get back to the problem:

[quote="tim1234133"]Find a word (any length) with 6 consecutive consonants. (where y counts as a vowel).[/quote]

[hide]Rhythms, dysrhythmic, dysthymia...[/hide][/quote]y counts as a vowel?

box of rocks wrote:
[quote]K. Sengupta, for each numbered entry where you entered more than one word, that numbered entry should not count. [/quote]

My opinion is at direct variance with the foregoing observation.
It would be observed that  if one is asked to write just one number satisfying x^2-5x+6 = 0; then x=2,3 is as valid as a response as x=2 or x=3. Accordingly,for any given query possessing multiple answers, it is NOT improper to write more than one answer.

Consequiently, I do not agree at all with the foregoing observation since my response is fully in conformity with the provisions governing this sub-forum.

Also, remember that if you write three words, you have also written one word. I did not use the word "exactly".

[quote="randomdragoon"]I can't get the six consecutive consonants... but here is another one

Find a 7-letter word that has exactly one vowel (where y counts a vowel), but also has no "s".[/quote]

scratch


inequalities
geometry
circumcircle
inequalities proposed

Prove that in any triangle $ ABC$ exists the inequality

$ \boxed {\sqrt {\frac {R \plus{} 2r}{R}}\le\frac {OH}{OI}\le\sqrt {\frac {3(3R \minus{} 2r)}{R}}}\ .$

[quote="Virgil Nicula"]Prove that in any triangle $ ABC$ exists the inequality $ \boxed {\frac {OH}{OI}\ge \sqrt {\frac {R + 2r}{R}}}\ .$[hide="Hint."]
${ \frac {OH}{OI}\ge \sqrt {\frac {R + 2r}{R}}}\Longleftrightarrow R^2\le OH^2 + 4r^2\Longleftrightarrow p^2\le 4R^2 + 4Rr + 3r^2\ .$ This means that

the power of the orthocenter $ H$ w.r.t. the circumcircle of $ \triangle ABC$ is at least equal to $ - 4r^2\ .$[/hide][/quote]
I think it's Gerretsen inequality, we can prove it if uses fomula: $ IH^2=(4R^2+3r^2+4Rr)-p^2 \ge 0$

[quote="Virgil Nicula"] [color=darkred]Prove that in any triangle $ ABC$ exists the inequality $ \boxed {\sqrt {\frac {R + 2r}{R}}\le\frac {OH}{OI}\le\sqrt {\frac {3(3R - 2r)}{R}}}\ .$[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] The relations $ \|\begin{array}{c} a^2 + b^2 + c^2 = p^2 + r^2 + 4Rr \\
 \\
OI^2 = R^2 - 2Rr \\
 \\
OH^2 = 9R^2 - (a^2 + b^2 + c^2) \\
 \\
IH^2 = 4R^2 + 4Rr + 3r^2 - p^2 \\
 \\
9\cdot IG^2 = p^2 + 5r^2 - 16Rr\end{array}$ are well-known. Thus,

${ \frac {OH}{OI}\ge \sqrt {\frac {R + 2r}{R}}}\Longleftrightarrow R^2\le OH^2 + 4r^2\Longleftrightarrow p^2\le 4R^2 + 4Rr + 3r^2\ .$

The power of $ H$ w.r.t. the circumcircle of $ \triangle ABC$ is at least equal to $ - 4r^2\ .$

[b][u]Remark.[/u][/b] Prove easily that $ \boxed {R\cdot OH^2 = 2R\cdot IH^2 + (R + 2r)\cdot OI^2}\ .$

Now the proposed left inequality is evidently !

Apply the [b]Stewart's theorem[/b] to the cevian $ [IG]$ in $ \triangle HIO$ :

$ 3\cdot IH^2 + 6\cdot OI^2 = 9\cdot IG^2 + 2\cdot OH^2\ .$  Using and the upper identity eliminate :

$ \|\begin{array}{ccc} OH\ \nearrow & \implies & 4\cdot (R - r)\cdot OI^2 = 9R\cdot IG^2 + R\cdot IH^2 \\
 \\
IH\ \nearrow & \implies & 3(3R - 2r)\cdot OI^2 = 18R\cdot IG^2 + R\cdot OH^2 \\
 \\
OI\ \nearrow & \implies & 4(R - r)\cdot OH^2 = 3(3R - 2r)\cdot IH^2 + 9(R + 2r)\cdot IG^2\end{array}$

Thus, $ \|\begin{array}{c} \frac {IH}{IO}\le2\cdot\sqrt {1 - \frac rR} \\
 \\
\frac {IG}{IO}\le\frac 23\cdot\sqrt {1 - \frac rR} \\
 \\
\frac {HI}{HO}\le \sqrt \frac {4(R - r)}{3(3R - 2r)}\end{array}$ and $ \sqrt {\frac {R + 2r}{R}}\le\frac {OH}{OI}\le\sqrt {\frac {3(3R - 2r)}{R}}\ .$

[b]Remark.[/b] Prove easily that $ 9\cdot(IG^2+IH^2-GH^2)=-12r(R-2r)\le 0$,

i.e. $ \boxed {m(\widehat {GIH})\ge 90^{\circ}}$ and $ \frac {2\cdot IO^2+IH^2}{2\cdot GO^2+GI^2}=3\ .$[/color]


If $x_1 ,x_2, \ldots ,x_n$  are the roots of the equation \[x^n\plus{}p_1 x^{n\minus{}1}\plus{}p_2 x^{n\minus{}2}\plus{} \cdots \plus{}p_{n\minus{}1}x\plus{}p_n\equal{}0,\] where $p_1,p_2, \ldots, p_n \in \mathbb R$. Prove that \[ (1\plus{}x_1 ^2)(1\plus{}x_2 ^2)\cdots (1\plus{}x_n ^2)\equal{}(1\minus{}p_2\plus{}p_4\minus{}p_6\plus{}\cdots)^2\plus{}(p_1\minus{}p_3\plus{}p_5\minus{}p_7\plus{}\cdots)^2.\]

[quote="jackie-chan01"]$ If x_1 ,x_2,...,x_n \ are \ nth \ roots \ of \ equation \ x^n \plus{} p_1 x^{n \minus{} 1} \plus{} p_2 x^{n \minus{} 2} \plus{} ... \plus{} p_{n \minus{} 1}x \plus{} p_n \equal{} 0 \ where \ p_1,p_2...p_n \in R$ .
Prove that $ (1 \plus{} x_1 ^2)(1 \plus{} x_2 ^2)..(1 \plus{} x_n ^2) \equal{} (1 \minus{} p_2 \plus{} p_4 \minus{} p_6 \plus{} ...)^2 \plus{} (p_1 \minus{} p_3 \plus{} p_5 \minus{} p_7 \plus{} ..)^2$[/quote]

Let $ P(x)\equal{}\prod_{k\equal{}1}^n(x\minus{}x_k)\equal{}x^n\plus{}p_1x^{n\minus{}1}\plus{}p_2x^{n\minus{}2}\plus{}...\plus{}p_n$

$ P(i)\equal{}i^n(1\minus{}ip_1\minus{}p_2\plus{}ip_3\plus{}p_4\minus{}ip_5 ...)$
$ P(\minus{}i)\equal{}(\minus{}1)^ni^n(1\plus{}ip_1\minus{}p_2\minus{}ip_3\plus{}p_4\plus{}ip_5...)$

So $ 1\minus{}p_2\plus{}p_4\minus{}p_6...\equal{}\frac{P(i)\plus{}(\minus{}1)^nP(\minus{}i)}{2i^n}$ and $ p_1\minus{}p_3\plus{}p_5\minus{}p_7...\equal{}\frac{P(i)\minus{}(\minus{}1)^nP(\minus{}i)}{(\minus{}2i)i^n}$

So $ (1\minus{}p_2\plus{}p_4\minus{}p_6..)^2\plus{}(p_1\minus{}p_3\plus{}p_5\minus{}p_7...)^2$ $ \equal{}\frac{(P(i)\plus{}(\minus{}1)^nP(\minus{}i))^2}{4(\minus{}1)^n}$ - $ \frac{(P(i)\minus{}(\minus{}1)^nP(\minus{}i))^2}{4(\minus{}1)^n}$

So $ (1\minus{}p_2\plus{}p_4\minus{}p_6..)^2\plus{}(p_1\minus{}p_3\plus{}p_5\minus{}p_7...)^2$ $ \equal{}P(i)P(\minus{}i)$ $ \equal{}\prod_{k\equal{}1}^n(i\minus{}x_k)\prod_{k\equal{}1}^n(\minus{}i\minus{}x_k)$ $ \equal{}\prod_{k\equal{}1}^n(1\plus{}x_k^2)$

Q.E.D.


geometry
circumcircle
perpendicular bisector
geometry proposed

According to the estimated number of participants who gave a correct solution, this was the hardest (!) problem from today's paper. So here is this great German killer - be warned!

Given a circle k and three pairwisely distinct points A, B, C on this circle. Let h and g be the perpendiculars to the line BC at the points B and C. The perpendicular bisector of the segment AB meets the line h at a point F; the perpendicular bisector of the segment AC meets the line g at a point G.

Prove that the product $BF\cdot CG$ is independent from the position of the point A, as long as the points B and C stay fixed.

The actual problem behind the problem: Why on hell should the points B and C stay fixed?

  Darij

Wow, the German olympiads are SO tough these days :D
[That's called sarcasm, for those who didn't realize that ;)]

It is really obvious that BF = cR/b with R the radius of the circumcircle, and CG = bR/c, so that the product equals R2  (for all possible positions of A, B, C).


function
topology
real analysis
real analysis unsolved

for every closed set $ E$  on real line, is there a differentiable function $ f$ which satisfies $ f^ {\minus{} 1}\{0\} \equal{} E$?

I think SouthStar interpreted the question to be "Is there a differentiable function $ f: E\to \mathbb{R}$, ...". Then $ f\equiv 0$ works. I'm assuming the question is actually $ f: \mathbb{R}\to\mathbb{R}$.

Ooooh. Couldn't have been that easy. It was 2 am. :)

I have seen some results for infinitely differentiable functions but not for this general case.

Define $ \varphi(x)\equal{}\begin{cases}x^2(1\minus{}x)^2& x\in(0,1)\\0&\text{otherwise}\end{cases}.$

Note that $ \varphi$ is differentiable and that $ \varphi(x)\ne0$ precisely on $ (0,1).$

Any open set in $ \mathbb{R}$ can be written as a countable disjoint union of open intervals, so let $ \mathbb{R}\setminus E\equal{}\bigcup_{k\equal{}1}^{\infty}I_k.$

If there are any infinite intervals there, pull them aside to be treated separately. Write $ I_k\equal{}(a_k,b_k).$

Define $ f(x)\equal{}\sum_{k\equal{}1}^{\infty}2^{\minus{}k}(b_k\minus{}a_k) \varphi\left(a_k\plus{}\frac{x\minus{}a_k}{b_k\minus{}a_k}\right)$

If we have an interval of the form $ (a,\infty),$ then add $ (x\minus{}a)^2$ to $ f(x)$ for $ x>a.$

If we have an interval of the form $ (\minus{}\infty,b),$ then add $ (b\minus{}x)^2$ to $ f(x)$ for $ x<b.$

The coefficients $ 2^{\minus{}k}(b_k\minus{}a_k)$ were added to assure that the series for $ f'(x)$ would converge uniformly so that $ f$ would be differentiable. We have a differentiable function $ f$ which is zero on $ E$ and strictly positive on the complement of $ E.$

We can see how to adjust this proof to get a $ C^{k}$ function - all that requires is a modification of the definition of $ \varphi.$ With just a few more adjustments, both to $ \varphi$ and to the coefficients of the sum, we can make a $ C^{\infty}$ function.

But this proof doesn't generalize so easily to $ \mathbb{R}^{n}$ for $ n>1,$ since I used a specifically $ \mathbb{R}^1$ characterization of open sets.

[quote="Kent Merryfield"]
But this proof doesn't generalize so easily to $ \mathbb{R}^{n}$ for $ n > 1,$ since I used a specifically $ \mathbb{R}^1$ characterization of open sets.[/quote]
Actually, it does: of course, it has been used that every open set is a [i]countable union[/i] of open intervals, but the fact that those intervals are [i]disjoint[/i] hasn't been really used anywhere ;).


Let $S=\{1,2,\cdots,40\}$. Find the minimum value of $n$ such that in any $n$ element subset $P$ of $S$ there exists distinct $a,b,c,d$ in $P$ with $a+b=c+d$.

i discovered a general formula but i'm not sure if its correct. because if we take 20 instead of 40 my formula gives n=9, but i think that it should be n=8. if you want i can post my formula.

:blush: sorry my formula had some mistakes. now it gives n=8 in case of 20. why hasn't anyone posted a solution yet? is it too easy or too hard? will i post my solution? say something...


calculus
integration
logarithms
real analysis
real analysis unsolved

Calculate the following integrals: (in the first dx is omited)

what about the second? I'm not sure if in the first integral it is not: $ ln(e^{x}\plus{}1)$ instead of $ ln(e^{x\plus{}1})$. I found these integrals in Internet and I show them here, because I think they are interesting.

[hide="Number 2"]Integrating by parts,

$ \int \frac{\ln(1 \plus{} \sqrt{x})}{(1\plus{}x)^2}\,dx \equal{} \underbrace{\minus{} \frac{\ln(1 \plus{} \sqrt{x})}{1\plus{}x}}_{\equal{} A} \plus{} \underbrace{\frac{1}{2} \int\frac{dx}{\sqrt{x}(1 \plus{} \sqrt{x})(1\plus{}x)}}_{make\,\,x \equal{} t^2} \equal{}$

$ \equal{} A \plus{} \int\frac{dt}{(1\plus{}t)(1\plus{}t^2)} \equal{} A \plus{} \frac{1}{2} \int \left(\frac{1}{1\plus{}t} \minus{} \frac{t}{1\plus{}t^2} \plus{} \frac{1}{1\plus{}t^2}\right)\,dt \equal{}$

$ \equal{} A \plus{} \frac{1}{2} \ln(1\plus{}t) \minus{} \frac{1}{4} \ln(1\plus{}t^2) \plus{} \frac{1}{2} \arctan t \plus{} C$.

You can finish from here.[/hide]


For each positive integer number n let A be set of all positive integers x such that gcd(x,n)>1
n is called good if for any x,y in A then x+y also in A. Find all good number.
       ( Sorry,my English is bad  :| )

[quote="Nbach"]For each positive integer number $n$, let $A_{n}$ be the set of all positive integers $x$ such that $\gcd(x,n)>1$: $n$ is called good if, for any $x,y \in A_{n}$, then also $x+y \in A_{n}$. Find all good number.[/quote]
Obviously $n = 1$ isn't good. So suppose $n \ge 2$. If $n = p^{k}$, where $p \in \mathbb{N}$ is prime and $k$ is any positive integer, then $\gcd(x,n) > 1$ means that $p \mid x$, whenever $x \in \mathbb{N}$. Hence $p \mid (x+y)$, if $x, y \in A_{n}$, proving that any prime power is good. Now suppose that $n$ has at least two distinct prime factors $p, q \in \mathbb{N}$. Then there exists a minimal $m \in \mathbb{N}^{+}$ such that $\gcd(mq+p,n) = 1$, which eventually implies that $n$ isn't good.


$ a_1,a_2,...,a_n$ is a sequence of real numbers. $ a_{i \plus{} j} \le a_i \plus{} a_j$ for every natural $ i,j; i \plus{} j \le n$  .prove that  $ a_1 \plus{} \frac {a_2}{2} \plus{} \frac {a_3}{3} \plus{} ... \plus{} \frac {a_k}{k} \ge a_k$ for every natural $ k \le n$


How do you think Armaggedon will be?
I'm looking for different scenarios so don't be afraid to post your ideas.

It will be terrifying with a nice scenario, loud music and many people crying  :(
Hiiiiiiiii, Haaaaaaa.

BUT !! Don't worry man, USA president, Schwarzie, will save all of us with his big muscles  :D

I understand you foretell a sudden, but expected, end for mankind. Something like an asteroid or comet.

hey the movie is quite good (armaggedon :D).

The movie was good, the end was good, but Bruce Willis(with his big muscles, and he is not even president  :D ) could not have missed the chance to save the world.

Of course not, but I don't uderstand why couldn't Ben Affleck have saved the day.

Three world superpowers will rise (Asia(Asian Empire), Russia+Germany(Russian Empire), and Europe+America(Allied Forces)). They will fight a huge nuclear war(World War 3) which will destroy near all the planet. The Russians have the best tacticians in the world(chess players like Karpov and Kasparov) and the best machines from German car companies. The Russians will win. Iceland will be searched, when Bobby Fischer is found he will be executed and the Russians will be the Best at Chess Forever. The Russians will create a huge superpower spanning over the entire world, everyone will be schooled, live, and die Russian. The Russians will have a Communist Government. Now that the Caribbean, Taiwan, Japan, and all the Islands north of Australia sank, South America is no more, most of China is a feast for scavengers, and whats left of America and Europe are the two biggest trophies, Russia will have full dominance. This will leave three occupied continents: Russia(the habitable part of Asia), Europe, and North America. Then they will split and have another war(World War 4) which will destroy all but Russia. Then the sun will explode and the world will be lost forever....

Here is a good (and hilarious) scenario of how the world will end:
http://www.albinoblacksheep.com/flash/end.php

i haven't seen any chuck norris bs yet

WHAT IS WRONG WITH YOU PEOPLE

[hide="my prediction"]A chinese Hitler will gain control of China, start shipping out poisonous stuff, like toothpaste, or Water guns, and Chaos will ensue. Then China will just go out there, and capture all of the world, and then Communism will ensue, And communism=bad. Then some random person will realize "OMG!!!!!!!!!" Then the 7th year of the Tribulation will ensue.[/hide]

Has anyone else noticed that this thread is about 4 years old?

lol I was about to wonder how guests can post  :D

lol apprently its not really called "armageddon" in scientific terms.  its actually global killer (or something like that).  i took a science pre test for some disasters unit and i put armageddon as my answer and it was wrong...


Let $n \geq 1$ be a natural number. And let $a_n$ be the number of different possibilities to fill a  $2 \times 2 \times n$ cuboid by means of $1 \times 1 \times 2$ cuboids which are positioned in a axis-parallel way. Determine $a_n$.

[quote="grobber"]Isn't the recursion straightforward?

$a_{n+2}=2a_{n+1}+5a_n$, I think (there might be something I'm missing, of course :)).[/quote]

I think you got wrong. Please rethink.  :)  And be so kind and find an explicit formula of $a_n$.

Yup, I realized what I did wrong. Sorry about that, I'll rethink :).

How's this: $a_{n+2}=2a_{n+1}+a_n+4\sum_{i=0}^n a_i$, where we define $a_0=1$. Still wrong?  :blush:

[Edit: actually, from that recurrence formula, we get the nicer one $a_{n+3}=3a_{n+2}+3a_{n+1}-a_n,\ \forall n\ge 0$ by subtracting $a_{n+2}$ from $a_{n+3}$]

I give you another try.  ;)  A hint the recursive formula is something like:

\[a_n = u \cdot a_{n-1} + v \cdot a_{n-2} + w \cdot a_{n-3}\] with $u,v,w \in \mathbb{Z}$. This formula is valid for most cases. Two special case considerations are necessary.  :)

I edited my post before seeing the hint :D. Besides, orl, the fact that a recurrence formula is different from the one you had in mind when you initiated the topic, doesn't necessarily make that particular recurrence formula incorrect.  ;)

You are right. I must say: I am impressed. 4000 postings and most of them really on mathematics. That is really an :alien: performance. !!! Keep up the good work.  :)

Thank you, that's very kind of you :).

If I am not mistaken (again :D), the closed form should be something like $a_n=\frac{2(-1)^n+(2-\sqrt 3)^{n+1}+(2+\sqrt 3)^{n+1}}6$. Even if this is not exactly accurate due to computational mistakes, it can be obtained in the usual manner: $a_n=\sum_{i=1}^3 \alpha_i\cdot r_i^n$, where $r_1,r_2,r_3$ are the three roots of $x^3-3x^2-3x+1$, the characteristic equation of the recurrence. From the first three terms, we compute $\alpha_i$, and I found them to be $\frac 13,\frac{2-\sqrt 3}6,\frac{2+\sqrt 3}6$ for $r_1=-1,r_2=2-\sqrt 3,r_3=2+\sqrt 3$.

Your expicit formula is perfectly right. Then we just would like to know how you obtained the recurrence formula.  :lol:

How any recurrence formula is obtained :). Just consider how the first few pieces "at the bottom" of the box can be arranged. You'll obtained the first one, that ugly one with $\sum_{i=0}^n$. Then subtract two consecutive such expressions to get the short and nice one :).


hi,
i cant seem to get the answer when i try to find a discount from an original price tag or find the original price when discount has already been giving.

eg:

what does  ring originally sell for if it has been given 20% discount and is on sale for 149.99

dont you find the discount price of 149.99 using 20% then add it to 149.99?

No, that would be finding 20% of the sale price, and then adding it, which doesn't give you anything.  20% off is the same as 80% of the original.  So
$ .8x\equal{}149.99\Rightarrow x\equal{}187.49$

The thing to be careful of here is of what you are taking the percentage. You can think from the shopkeeper's perspective - how would she have changed the price?

Or instead of thinking in terms or the discount, think in terms of the sale price.

So if there's a 20% discount, think of it as 80% of the normal price.

If there's a 6% tax, then think of it as 106% (1.06) of the original price.


AMC
AIME
AMC 10
USA(J)MO
USAMO
percent
AMC 12

Does anybody think the floor score could possibly be a $ 7$? I'm a sophomore, and I think I got a 7 maximum (unless I guessed right...). I thought this time's AIME was a bit easier.

Maybe. I'm a sophomore too, and I got an 8 or 9, depending on 12. It might be a 7. Or an 8, if you're really unlucky.

Last year it was a 6. Maybe something like that'll happen again.

yea im a sophomore and i got a 8

but anyways,

i would compare this test as the same difficulty as the 2006 AIME

if you look at the first 10, they are about the same difficulty of  the first 10 that year.

that year's cutoff was 8 and that was limited expansion.

im betting this year, with a full expansion, is going to be 7 (someones gonna get like 144/7), and they will probably make the cutoff like with a 193 index...

WHOO! Following the weird pattern here, I'm a sophomore and got an 8, a huge improvement over a 4 from last year.  This is crazy, because i barely made the AIME, getting a 117 on the AMC 10A

I feel pathetic in terms of improvement...I've had a 7-7-9 the last three years (including this year).

If the floor is an 8 this year, I'll be so mad, because I made stupid mistakes.

[quote="E^(pi*i)=-1"]I think it is certainly possible that the floor will be $ 7$, but I think the high AMC/low AIME profile won't be as prevalent this year as in the past.  I would guess $ 8$ until I see a counterexample.  Sorry about that (I got $ 7$ on the 2006 AIME, which had an $ 8$ floor, so I feel your pain).[/quote]

i know someone that got a 144/7 (well he thinks he got) if he didnt guess one right

Was that $ 144$ on the AMC 12 or AMC 10?  If it was on the 10, it doesn't affect floor score placement, as far as I know.

compare this contest to the difficulty of the 2006 AIME where the floor was 8

[quote="the future"]i know someone that got a 144/7 (well he thinks he got) if he didnt guess one right[/quote]
Did he take the AMC 10 or the 12?

i think 12...
but i dont know if 1 person can decide it..

if he does then it will be problaby with a really high index

so

either 8 floor/no index required

or 7 floor/200 index

which means some people with 117/8 or 118.5/8 will be excluded

[quote="miller4math"]im not sure about anything else, but my brother did get a 145.5/ 8 on the amc 12..
which im pretty sure passes him.

I wrote down my amc 10 score, will they count that for my index?[/quote]

that wasnt suppose to be for anything but identification...

so it doenst really matter if you wrote down 12 or 10, they should already have it in the database, the AMC score you wrote down for the AIME should only be for identifying you...i think

The score you write down has no bearing on what they use for qualification. It's just an identity check.

I would guess 8, personally, as this AIME seemed easier than the past ones. Of course, having gotten a 9, I'm alright with that, but it would suck to be someone who got a 7.

And in terms of improvement: 1-5-9.  :D

[quote="unimpossible"]james4l did you take AIME I or II?[/quote]
I took AIME II. I also took AMC10. My score isn't too good, so I am on the edge of being bumped off  :(

[quote="the future"]Uhh unimpossible, he took the AMC 10/AIME (hes only in 6th grade) so the 144/7 is irrelevant (but i do know someone without an AoPS account that got a 144/7 on the 12/AIME).

But seriosuly, you guys should stop worrying.

Theres no way the cutoff will be greater than 7.  And theres no way the index will be greater  than 196.[/quote]

I hope you're right  :)

On a different note, when do they announce whos made USAMO.

Ive read somewhere that the AMC Director said that the AMC would be sending out on April 11, 2008 the scores for the AIME, the solutions for the AIME, and a list of all the USAMO qualifiers to each school.

However, I have spring break all next week.  So do they also email every person on whos made USAMO (i know they did last year), and would that also be on April 11, or would that be on an earlier date?

Thanks.

I pretty sure that it says on the teachers manual that invitations for USAMO would be sent out on April 6-11. I think the solutions and results for the AIME I have been sent out already, but I'm not sure about the AIME II...

[quote="AMCDirector"]AIME results for AIME I and AIME II will be mailed out starting approximately April 11.  This is because we include a list of USAMO qualifiers along with the results, the answer pamphlet and the certificates of participation.  See the AIME Teachers' Manual, Section V, page 5.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

This is what the AMC Director wrote but thats when they are mailed out.  So I was just wondering when do we acutally find out by email (if we do).

It seems from what he wrote, the AIME I and AIME II results are mailed out at the same time.  So I dont think anyones gotten their I results back yet...

Well, the AIME I results were supposed to come out about 10 days after the test, and there's already been a thread asking about what people's official scores were. I took the AIME I, but I don't think my school's received the scores yet...

Yeah I created that thread when I thought that the results were already being sent out.  The AMC director corrected me, saying that the results won't be mailed out until the 11th.  We can just revive the thread around the 11th.

[quote="physisnu"]Hmm... How about this?

I am an 8th grader...

[b]AMC 10: 150.0/150.0

AMC 12: Didn't take.

AIME II : 8/10[/b]

USAMO: ?/9+

OMG I wrote 502 for #8....

OMG I wrote 120 for #10....

OMG I could've guessed that 8 moves would make some sort of a pattern for #9....

But I dind't.... :(

Silly, silly mistakes.....I shouldn't make any in next year's AIME II!!!
[b]
Anyway, my index is 230. A good possibility of taking the USAMO?[/b][/quote]

No question, you will make it--don't stress it. Btw, AIME is out of 15.

I missed #1 (along with two other mistakes) on the AIME last year, which got me a 5, missing the floor by 1 (which was low enough; I shouldn't complain).
I missed #1 again this year, which got me a 6. If it misses a floor of 7, I'm seriously going to grind my head into the wall  :wallbash_red:

[quote="worthawholebean"][quote="mathcrazed"]Just kidding =P...  Honestly, the first 7 or 8 questions of the AIME II were pretty simple.  #8 was slightly harder, but other than that... Probably just me though.[/quote]2 was very hard if you're talking percentage of people who answered correctly.[/quote]

Is it just me, or do the AMC people just like messing with people on #2s?

I think people just don't pay enough attention to #2's. They're like, okay, I better not miss #1 because that would look stupid, but then focus their attention on the later problems.

Like me.  I got an 11 on AIME last year as a sophomore, and this year I missed #1 and #2 but still scraped out an 8.  With an almost decent AMC12B score I'm right on the bubble.  It's sad because, excluding 8th grade, my USAMO index has steadily dropped since freshman year.b

Actually, I would say that #2 on last year's AIME I and this year's AIME II were just downright tricky. Especially last year.


calculus
integration
logarithms
Putnam
algebra
partial fractions
calculus computations

Evaluate

$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{m^{2}n+mn^{2}+2mn}$

[hide]
so far the only ideas i have are, partial fractions and finding some clever way to turn it into a double integral. neither of which have gotten me very far.
[/hide]

I got, if my calculations are correct, that the value of your sum is $\frac{7}{4}$.

could you please provide a proof for your answer.

I would try the integral representation
\[\frac{1}{m+n+2}= \int_{0}^{1}x^{m+n+1}\, dx \, . \]

[quote="didilica"]The sum equals
$\int_{0}^{1}x(\ln(1-x))^{2}dx=\frac{7}{4}$ by Ravi`s idea, however I have another method for calculating this sum based on a, let`s say , long computational method.
Didi[/quote]

how did you get that integral. i follow what Ravi did but i dont see how they are related.

Write the sum as $\sum_{n=1}^\infty \sum_{m=1}^\infty \int_{0}^{1}\frac{x^{m}}{m}\cdot \frac{x^{n}}{n}\cdot x \, dx$.
Now interchange the sums with the integral and use the Taylor expansion of $\log$ around $1$.

PS: I'm not sure what steps to do in order to justify that interchange... But this is the idea, I think.

to interchange i think it just has to be a sequence of only positive terms. but i could be wrong.

Here is a nice generalization:

Let $k\geq 2$ be a natural number.

Calculate,

$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{nm^{2}+n^{2}m+kmn}$.

This problem has been discussed in detail in the Putnam section, if I am not wrong.

i am pretty sure that it was. however i was unable to find it.

solution involving partial fractions is quite ugly:

$S=\sum\frac{1}{n}\sum\frac{1}{m(m+n+2)}$
$=\sum\frac{1}{n(n+2)}[1+\frac{1}{2}+...+\frac{1}{n+2}]$

do another partial fractions, then combine terms.

answer is indeed 7/4.


I think it's very well-known , but i just can't find it out .

$a , b , c > 0$ with $abc = 1$ , prove that

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

actually this problem if i am not mistaken was discovered by darij and it was without the condition of $abc=1$:

$a^{2}+b^{2}+c^{2}+2abc+1 \geq  2 (ab+bc+ca)$ , for all positive reals.

there are a lot of ways to solve it, one of them is with schur, $2abc+1 \geq  \frac{9abc}{a+b+c}\geq  2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})$.

actually the "stronger" one holds, and it is equivalent with the one with the condition $abc=1$ : $a^{2}+b^{2}+c^{2}+3 \sqrt[3]{a^{2}b^{2}c^{2}}\geq  2 (ab+bc+ca)$, and it can be solved in the same way, or directly by popoviciu for $f(x)=e^{2x}$

As for the initial problem i think it could be solved by mixing variables.


29 seems to be a pretty straightforward. Let x be the weight of the tank when empty and y be the weight of the fuel needed to completely fill up the tank. Then you can express both statements in the problem in x and y, so you just have to solve the simultaneous equations.

Fierytycoon

The numbers given in problem 29 make it especially nice. This will happen often in math contests. Note that the difference between 25% and 75% is 50%, so half a tank of fuel (without the container) is 1200 tons. Thus 25% is 600. Thus the weight of the empty tanker is 2500-600=1900 tons.

I don't understand question 30. Is the original rectangle ABCD?

Well I have run out of paper within reaching distance of this seat so I'll just start this and get someone else to finish  :D
First of all you know that AB = BC because thats how we've folded it. So AB=5 and B to the other top corner is 3. Now that right angled triangle is a 3,4,5 triangle so C is 4 from the top. Now we got a heap of right angled triangles.. the line connecting A to C must have length :sqrt:80, dividing that in half gives :sqrt:20 somewhere, and you can finish it off from there, I think it comes out to 5:sqrt:5 or something?


Are you sure it is sum, and not product?

Because $ \prod_{k\equal{}2}^n \frac {k^3\minus{}1} {k^3\plus{}1} \equal{} \left ( \prod_{k\equal{}2}^n \frac {k\minus{}1} {k\plus{}1} \right ) \left ( \prod_{k\equal{}2}^n \frac {k^2 \plus{} k \plus{} 1} {k^2 \minus{}k \plus{} 1} \right ) \equal{}$ $ \frac {2} {n(n\plus{}1)}\cdot \frac {n^2\plus{}n\plus{}1} {3}$, while the sum does not seem to telescope that well.


I would like to participate in USAMTS this year. I am a resident of the United States but I am not a citizen of the United States. Would I be able to participate in USAMTS?

I think so, because I'm only a resident too.

Yes, as long as you are a resident (in other words, have a US mailing address), you may participate.


[quote="pvthuan"]Năm nay Kỳ thi SMO do Hội toán học Singapore tổ chức, dành cho học sinh các nước trong khu vực sẽ được tổ chức vào ngày 29 tháng 5 năm 2007. Dưới sự điều phối của hội toán học Hà Nội, tính đến nay kỳ thi này đã được tổ chức hai lần tại Việt Nam.

Năm nay có sự tham gia rộng rãi của nhiều tỉnh hơn những năm trước.

Có em nào đã được lọt vào vòng loại thì lên tiếng nhé.[/quote]

Có em :)
Bây giờ em mới có đề các năm 2004, 2005 và 2006 nếu ai có đề năm nào khác thì cho em với :D

[quote="pvthuan"]Năm nay khối cấp 2 thì Hanoi Ams được 5 em, Nguyễn Trường Tộ được 1 em, Hải Dương có 4 em. Chúc mừng nhé. Vậy còn 10 vị trí nữa sẽ thuộc về các tỉnh Vĩnh Phúc, Phú Thọ, ....[/quote]
Thầy có danh sách không ạ, share cho em với :)

Sao ở miền trung thiệt thòi thế nhỉ? Chẳng bao giờ được tham gia mấy cái như thế này!

Chắc người ta nghĩ các bạn không giỏi English đấy thôi :D

[quote="N.T.TUAN"]Chú tìm thêm [hide="ở đây"] http://sms.math.nus.edu.sg/Competitions/CompetitionHomePage.aspx[/hide] xem sao? Cứ search trên Google là ra cả đống ấy mà :D[/quote]

Làm gì có đề ở đó anh ơi! Chỉ có 2 cái đề Senior in trên THTT 2 năm vừa rồi thui  :lol:

[quote="tunganh"][quote="hieuchuoi@"]Khối Senior Hải Dương được 4 bạn  :D Còn Junior cũng được 4 hả thầy Thuận? Ơ sao đều thế nhỉ  :D

Chán quá cái đề vòng loại của Senior bài BDT còn bị sai đề  :( làm em mất bao thời gian, tưởng tính sai (ngố  :blush: ) Mà sao thi loại vẫn được dùng máy tính ạ? Thi thật có được không ạ  :D (thấy trong đề ghi là not allowed) Mà em thấy cần phải cho giấy nháp thoải mái một chút  :D hôm thi thầy Thủy bảo là thi thật nháp luôn cả vào bài làm cho người ta chấm  :D[/quote]
Đi thi không được dùng máy tính đâu anh, còn đề thi thật đến 2,3 trang (tức 4-6 mặt), nháp vô tư mà anh :D[/quote]
Em thấy xin thêm nháp vẫn được mà, chẳng qua người ta muốn đánh giá kỹ năng của mình qua cách nháp đó. Vậy ai có đề Junior các cuộc thi khác cũng được share cho em với

Theo yêu cầu của một số bạn không có điều kiện về Hà Nội, tôi trích dưới đây một số bài toán tập huấn đã được sử dụng.

Ui thay` ơi ,cho em hỏi thế đá có danh sách ruì ạ ,thầy có thể post danh sách lên đc ko ạ ?đến giờ mà chưa đc thông báo nghĩa là... :(

Khối Senior ở Hà Nội ko đc ai hả thầy?Nếu có danh sách thầy post qua cho bọn em coi  thầy nhá

Theo tôi biết thì ở Hanoi có 4 em sau đây

1) Pham Thuy Duong
2) Dang Dung Ha
3) Truong Ba Thang
4) Ha Khuong Duy

Bốn em này đều là học sinh của Khối chuyên KHTH Hanoi. Số còn lại thì tôi không biết.

[quote="quangpbc"]Không biết qua các năm thì năm nay thành tích được đánh giá là tăng hay giảm anh Tuân nhỉ  :([/quote]
Giảm so với các năm trước. Nói chung mấy cái trò này anh chả quan tâm.  :lol:

[quote="Trịnh Lưu Ly"]Hê!Vô lý.Làm gì có chuyện không cho dùng máy tính,nếu gặp số to thì làm thế nào?! :wink:  :(  :huh:[/quote]
Không được dùng máy tính mà bạn! Thằng bạn mình đi thi về bảo vậy, đây là thi kiến thức, chứ có phải thi máy tính đâu! Mà thấy nó bảo, lúc thi 4 đứa TH được ngồi cạnh nhau, ko biết từ TA nào có thể hỏi nhau được, thế mới hay!  :wink:

[quote="NguyenDungTN"] Mà thấy nó bảo, lúc thi 4 đứa TH được ngồi cạnh nhau, ko biết từ TA nào có thể hỏi nhau được, thế mới hay!  :wink:[/quote]
Cái này đừng nói trên mạng nhe em . :P

[quote="pvthuan"]Vang[b] rội[/b] .[/quote]

 :D Sai roi` tha^y` MrThuan oi .

Ma`sao ko cho cac tinh? khac' thi anh TUAN nhi?. Ba^t' co^ng qua'  :(

Theo như một thày nói thì do là các tỉnh khác không chịu cấp kinh phí , thế thôi. Chứ thi cái này ai chẳng thi được. Mà blog của anh ngon rồi đấy, vào comment mau. :P

Comment rồi ông anh ạ. Mà đi nói chuyện này ở đây mới khổ chứ :wacko:  :wallbash:  :whistling:

[quote="pvthuan"]Vang rội nhất vẫn là năm 2005. Tôi nhớ năm đó là độ Vĩnh phúc có 20 em dự thi thì có 18 HCV, 2 HCB. Năm đó Vĩnh phúc tập trung đội tuyển khá sớm (trước 1 tháng) và tập huấn khá nhiều (riêng Mr Thuận đã khoảng 10 buổi). Năm nay có Hanoi Ams cũng rất mạnh, ra quân 5 thì đều được 5 HCV.[/quote]
Nói chung hai năm nay Hanoi ams năm nào cũng hầu như toàn vàng (đấy là THCS, còn THPT năm ngoái thi thì chỉ có bạc thôi)

cho em hoi the bao gio`,giay chung nhan gui ve` truong` a,em muon biet vi` can co' 1 chut viec :)

Thường thì Hội toán học Hà Nội sẽ tổng kết và trao chứng nhận cho các em tham gia vào dịp đầu năm học mới hoặc tháng 11. Em nào tham gia thì trực tiếp hỏi hiệu trưởng của trường đó về việc này.

Ngày 28, 29 tháng 11, năm 2007, nhân Hội thảo Toán Giải tích và Ứng dụng, Hội Toán Học Hà Nội sẽ trao giải của cuộc thi SMO 2007. Các em tham gia, đoạt giải, đã biết thông tin này chưa?

(sẽ có ít nhất là 3 em không chứng kiến giây phút này vì đang ở Singapore rùi)  :)

T__T sao bon e ko dc bao' a?
Muon nhan giai thuong thi` lay o dau a? hicc

Chào các anh chị em. Hì, h mới biết có chỗ để mình viết Tiếng Việt trên Mathlinks. Mong mọi người giúp đỡ.


The $64$ squares of an $8 \times 8$ chessboard are filled with positive integers in such a way that each integer is the average of the integers on the neighbouring squares. Show that in fact all the $64$ entries are equal.

where do you get these problems?(you know,you send a lot of problem) :D

Take the smallest integer among the ones written on the chessboard. Then all neighbouring integers must be equal to this one. Continue repeating this process and you easily get that they are all equal. Too easy, and too well-known.

Proof: by contradiction. Suppose that every square has a different integer value. Then, without loss of generality, we will pick an arbituary square and say this has the lowest value. This square, though, is equal to the arithmetic mean of its neighbours. This implies that one of its neighbours has a lower value than it. But we specified the square to have  the lowest valu. Contradiction! Thus, all squares have equal values.

Q.E.D

we can also introduce a maximal element..
(the proof is still the same as above)

Choose the square (any one) which is filled up with the smallest of all the positive integers filled in the 64 squares. It any of the neighbors is filled with positive integer greater then the smallest positive integer, then the average of all the integers in the squares occupying the neighbor’s positions will be greater than the said smallest integer. This is contradiction !Hence all the neighboring squares are filled with same smallest integer continuing is this way till all the squares are filled in finite number of steps

If you know $Geogebra$ and $\LaTeX$$,$ please use them neelanjan. Your solution will look good. :)
BTW, your solution is completely wrong - from the beginning to the end.

[quote=eshan]If you know $Asymptote$ and $\LaTeX$$,$ please use them neelanjan. Your solution will look good. :)[/quote]

[quote=neelanjan]atfirst , i do not know asymptote . you will see only my solutions ! , see it and tell me where i am wrong . :mad:[/quote]

i think neelanjan is right, it is the depth of the solution you need to consider, not the way it is presented. Everybody is not supposed to know everything!


algebra
polynomial
Rational Root Theorem

Please, help me to solve this problem:
"Define the number of positive roots of the equation:
12x^5 + 6x^4 – 4x^3 – x – 34 = 0"

The solutions are going to be of the form $ \frac{a}{b}$, where $ a|12$ and $ b|34$.
What are the divisors of 12 and 34?
$ a = {\pm1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12}$
$ b = { \pm 1, \pm 2, \pm 17, \pm 34}$

Since we are looking for positive roots, $ a$ and $ b$ must have the same sign.
Thus our answers fall in this list:
$ \frac{a}{b} = {1, \frac{1}{2}, \frac{1}{17}, \frac{1}{34}, 2, \frac{2}{17}, 3, \frac{3}{2}, \frac{3}{17}, \frac{3}{34}, 4, \frac{4}{17}, 6, \frac{6}{17}, 12, \frac{12}{17}}$

Note that it is not all of numbers that are going to satisfy your equation. But you know where to look now for your answers and that was from a theorem known as Rational Root Theorem which you should study.

[quote="akech"]The solutions are going to be of the form $ \frac {a}{b}$, where $ a|12$ and $ b|34$.
What are the divisors of 12 and 34?
$ a = {\pm1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12}$
$ b = {\pm 1, \pm 2, \pm 17, \pm 34}$

Since we are looking for positive roots, $ a$ and $ b$ must have the same sign.
Thus our answers fall in this list:
$ \frac {a}{b} = {1, \frac {1}{2}, \frac {1}{17}, \frac {1}{34}, 2, \frac {2}{17}, 3, \frac {3}{2}, \frac {3}{17}, \frac {3}{34}, 4, \frac {4}{17}, 6, \frac {6}{17}, 12, \frac {12}{17}}$

Note that it is not all of numbers that are going to satisfy your equation. But you know where to look now for your answers and that was from a theorem known as Rational Root Theorem which you should study.[/quote]
You'd still need to look for irrational positive roots, also.

You could use [url=http://www.artofproblemsolving.com/Wiki/index.php/Polynomial#Descartes.27_Law_of_Signs]Descartes' Rule of Signs[/url] to find the possible numbers of positive roots.


How many ways can you arrange 7 green bottles and 8 blue bottles in a row, if exactly $1$ pair of green bottles is side by side?

[hide]We have six blocks of green bottles, one containing two bottles and five containing one bottle each.  There are 6 ways to order these blocks.  Then let $x_{n}$ be the number of blue bottles between the $(n-1)$th and $n$th blocks of green bottles.  We have $x_{1}+x_{2}+\cdots+x_{7}=8$.  Here $x_{1}$ and $x_{7}$ are nonnegative integers, and the rest are positive.  Let $x_{1}=y_{1}-1$ and $x_{7}=y_{7}-1$.  Substituting, we obtain $y_{1}+x_{2}+x_{3}+\cdots+x_{6}+y_{7}=10$, where all variables are positive integers.  There are $\binom{9}{6}$ solutions to this equation, so the answer is $6\binom{9}{6}$.[/hide]

[quote="calculuslover800"]Are the bottles distinct?[/quote]
No, they generally aren't for these types of problems.

Although, you could solve it so that they are distinct.  :D


function
search
algebra
functional equation
algebra proposed

Find all functions $f: \mathbb N\cup\{0\}\rightarrow \mathbb N\cup\{0\}$ that \[f(f(n))+f(n)=2n+3k\] which $k\in\mathbb Z$

Omid,

edit out the word "easy."
I told you I would come back to check up on this.
It's not mathematical AT ALL.

I apologize for spamming.
[i]To A cute angle :[/i]
If you search the topics that has the word "easy" you will see a great list.

Please finish it and don't tell me what to do :mad:

Stating something is easy (or hard) is wrong.  I expect it out of elementary school children, but not serious people who call themselves mathematicians.

1) You still decided not to make a change in the word.

2) An apology for spamming from you is EMPTY; it is actions insofar as NOT typing out said words in the future that matter.

3) Argumentation/Logic -- Don't change the subject on me.  We are talking about you and your words.  I am focusing my attention on one person (user) at a time.  Do not bring in other users for an attempted comparison.

The word "easy" in titles is indeed not the best, especially when not anything else is given.
Reason: it's impossible to know what the thread is about before opening -> searching is hard for it, and finding even more.
This changes if there is more information given, e.g. that it is a functional equation; but it would be even better to give "parts" of the problem in the title, like the equation itself.

But more on what Pleaseread(/A cute angle/sohcohtoa/all the other names he uses) refers to:
I agree that "easy" is subjective; but there are easier and harder problems (here "easy" is defined on how hard the average mathematicians solving it considers it), and when the author considered this one to be easy, then it is at least not one of the hardest problems he knows (otherwise he wouldn't say so). These kind of statements are for example usefull if you are looking for not-to-hard problems; you just have to remember who stated this (e.g. an "easy" from vess is normally not what most people call easy ;) ), but then there is no problem in using this terms.

And it is much more childish to use a couple of accounts instead of one; but I think you will never stop this...

Conclusion:
@all: try not to use "easy" in titles, especially without more specific terms on the problems.
But telling that one thinks it is easy is indeed ok.
@Pleaseread: stop doing this finally!
@moderators: delete this in some days please.

hallo,
i think the problem is nice.
$f(n)=n+k$ is this right ?

Yes, that's the only solution. But post your solution.


geometry
3D geometry
sphere
modular arithmetic
pigeonhole principle

(1) Given $ n$ integers. Prove that there are some integers which sum is divisible by $ n$.
(2) Put 10 points in a circle with radius of 5. Prove that there are 2 points which distance is less than 2.

Not sure if I have misunderstood this, but for "given n integers" we will always have n 1s or n 2s or n 3s or more generally n ms. Summing them you will have nm, which is divisible by n.

For the second one, you could evenly spread out all the points on the sphere, even when you do that, the points will always be just over 3 apart.

One valid set of $ n$ integers is $ 1, 2, 3, \ldots, n\minus{}1, n\plus{}1$, in which case it is not as trivial as you have made it out to be. Obviously in this case $ 1 \plus{} (n\minus{}1)$ is one sum that works, but proving there is always such a sum is slightly trickier.

[hide="1"]Let the numbers be $ k_1, k_2, \ldots k_n$. Consider each of the following sums modulo $ n$: $ k_1$, $ k_1 \plus{} k_2$, $ \ldots$, $ k_1 \plus{} k_2 \plus{} \ldots \plus{} k_n$. If any of them are $ 0 \pmod{n}$, then that is the desired sum. If none of them are $ 0 \pmod{n}$, then they are one of the following $ \mod n$: $ 1, 2, 3, \ldots n\minus{}1$. By the pigeonhole principle, at least two of the sums have the same remainder ($ n$ sums and $ n\minus{}1$ possible remainders). Then their positive difference is $ 0 \pmod{n}$, giving the desired sum.[/hide]


Find the dimension of the subspace of those linear operators $\varphi$ on $M_{n}(\mathbb{R})$ for which the identity $\varphi(A^{T})=(\varphi(A))^{T}$ holds for every matrix $A\in M_{n}(\mathbb{R}).$
(A. Oliynyk)

sorry, I think it should be $\frac{n^{2}(n^{2}+1)}{2}$. May I know if i'm right?

Define $A_{ij}$ to be matrix that is 0 except the ij-th position, where it is 1.

These $n^{2}$ matrices constitutes a basis for $Mn(R)$.

Thus we only need to consider $\phi$ on these matrices.

For $A_{ii}$, $\phi(A)$ must be symmetric.
For $A_{ij}$, $i < j$, $\phi(A)$ is arbitrary.

Thus the answer is $n \frac{n(n+1)}{2}+\frac{n(n-1)}{2}n^{2}= \frac{n^{4}+n^{2}}{2}$

seems more acceptable
Would you post a complete proof ?
thank's

[quote="Soarer"]Thus the answer is $n \frac{n(n+1)}{2}+\frac{n(n-1)}{2}n^{2}= \frac{n^{4}+n^{2}}{2}$[/quote]

check this ?


Given a natural number $n,$ find all polynomials $P(x)$ of degree less than $n$ satisfying the following condition \[ \sum^n_{i=0} P(i) \cdot (-1)^i \cdot \binom{n}{i} = 0. \]

Just use Lagrange Interpolatin formula for P(x).


x,y are rational, solve this equation:
x^n+y^n=x^m+y^m

**m,n are natural

?

you sure you didnt make any mistake in the qn?


Let $ a_1, a_2, \dots , a_{99}$ be positive integers all less than 100 (but not necessarily distinct). If the sum of any two or more of them is never a multiple of 100, show that they all must be equal.

Yeah, I fail at pigeonhole.

[hide="Part of a solution"]
So define $ s_n$ to be the cumulative sums of $ a_n$, i.e., $ s_n \equal{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n$.
Define $ s_0 \equal{} 0$.
Then if $ s_i \equiv s_j \bmod{100}$ and $ i \neq j$, we have $ s_{i\plus{}1} \plus{} s_{i\plus{}2} \plus{} \ldots \plus{} s_{j} \equiv 0 \bmod{100}$.
(Since no $ a_n$ can be 0 or 100, there is no way the above sum could consist of just one number.)
So, assume that $ s_0, s_1, s_2, \ldots, s_{99}$ are all distinct.
The above holds if you permute the elements of $ a_n$ in any way.
[/hide]

I keep feeling that the last step I'm missing is either obvious or trivial, and possibly both.


Does there exist a continuous function which is not monotonic on any interval?

$ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {\cos(3^{n}x)}{2^{n}}$ has the property stated.

Edited: $ x$ is installed back where it was missing upon a request by an acute reader

Another approach: for any $ 0\le a<b\le 1$ the set of all functions $ f\in C[0,1]$ that are monotone on $ [a,b]$ is closed and has no interior points. Taking the union over intervals with rational endpoints, we conclude that the set of functions that are monotone on some interval is of first category in $ C[0,1]$.

The classical Weierstrass function $ f(x)\equal{}\sum_k 10^{\minus{}k}\cos(100^k x)$ is the most famous example. To see that it works, just consider $ f(\pi(n\plus{}1)/100^k)\minus{}f(\pi n/100^k)$ with odd and even $ n$. What is interesting is that you can even find a differentiable (though, of course, not continuously differentiable) function with this property ;)

Edit: mlok and akech were faster :)

[quote="bookworm_vn"][quote="akech"]$ f(x) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {\cos(3^{n})}{2^{n}}$ has the property stated.[/quote]

there is no $ x$ in the expression...  :oops:[/quote]

Thanks, it was too obvious to include;-


[quote="Arvind_sn"]I need help on this area problem, please. You have a regular hexagon with area 36. Find the side length.[/quote]
[hide]The area of a hexagon is $\frac{3x\sqrt{3}}{2}$. Solving the equation $\frac{3x\sqrt{3}}{2}=36$, $x\sqrt{3}=24$ and $\boxed{x=8\sqrt{3}}$. [/hide]

[quote="merajatartofprobsolving"]I think area of regular hexagon is ((3)pow(1/2)*3)/2*side*side[/quote]

It is, so chinesechess, you should solve for x in this:$\frac{3x^2\sqrt3}{2}=36$.

Oops. Yeah, you're right...that was weird. [hide]$x=2\sqrt[4]{12}$ i think. [/hide]

Or you could divide the hexagon into 6 congruent equilateral triangles from the centre... If the area is 36, the area of a triangle is 6, and therefore each side is 2....

the answer should be 2 the power of 3 over 4

$[ABCDEF]=6\frac{a^2\sqrt{3}}{4}$

we have: $[ABCDEF]=36$, that's why $a^2=8\sqrt{3}\iff a=\sqrt{8}\sqrt[4]{3}=2\sqrt{2}\sqrt[4]{3}$  :)

EDIT: We can generalize it to regular $n$ polygon and compute with letters, not with $36$.. :P

[quote] I think you divide that 36 by 6 but dont take my word 4 it k? get it checked afterwards k? [/quote]

I don't think it works that way...in doing that you split the hexagon into 6 pieces, each with area 6, but that doesn't get you much closer to finding the side length

Here is my approach:
[hide]Since a hexagon consists of 6 equilateral triangles, each triangle has an area of 6. Area of an equilateral triangle is $\frac{(s^2)\sqrt{3}}{4}$. So 24=$s^2\sqrt{3}$, so $s^2=8\sqrt{3}$. Therefore $s= 2\sqrt[4]{12}$.[/hide]

Can a mod please insert the symbols since it is kind of messy like this and I don't know how to insert them? Thank you.
[size=75][color=green]
Mod note: Latexed as much as I could [/color][/size] :)

[quote="life=tennis+math"]Here is my approach:
[hide]Since a hexagon consists of 6 equilateral triangles, each triangle has an area of 6. Area of an equilateral triangle is $\frac{((s^2)\sqrt{3})}{4}$. So $24=(s^2)\sqrt{3}$, so $s^2=8\sqrt{3}$. Therefore $s= 2\sqrt[4]{12}$.[/hide]

Can a mod please insert the symbols since it is kind of messy like this and I don't know how to insert them? Thank you.[/quote]
Fixed :)

the main approach is that when the opp. vertices of a hexagon are joined the diagonals are concurrent. thus 6 equilateral triangles are formed with each side being each side of the hexagon.

Dude... this thread is 6-7 years old.... don't revive it....


geometry
LaTeX
AMC
USA(J)MO
USAMO
number theory
prime numbers

The problems I'll be posting in this topic are just some problems from previous Canadian contests.  Anyone is welcome to try them.

@mods : If these questions do not suit the level of Getting Started, please move it to the forum of your desire.

1. A square whose area is [tex]100cm^2[/tex] is inscribed in a semicircular region.  The radius of the semicircle, in cm, is...

a)[tex]  \sqrt{50}[/tex]  b) [tex]\sqrt{75}[/tex] c)[tex]  \sqrt{125}[/tex] d) [tex] \sqrt{200}[/tex]  e)[tex]  \sqrt{500}[/tex]

2.  If [tex]\sqrt{x} + \sqrt{y}=\sqrt{135}[/tex], where x and y are positive integers, then the value of x+y is...

a) 135 b) 75 c) 45 d) 30 e) 15

3.  [tex]\frac{2^{1990}-2^{1989}}{2^{1990}+{2^{1989}}}[/tex], equals

a) 0 b) [tex]2^{-3978}[/tex] c) [tex]\frac{1}{3}[/tex] d) [tex]\frac{1}{2}[/tex] e) [tex]2^{3978}[/tex]

4  If [tex]a,b,$ and c[/tex] are prime numbers with [tex]a+b=c[/tex] and [tex]1<a<b<c<30[/tex], then the number of possible sets [tex]{a,b,c}[/tex] is...

a) 0 b) 2 c) 3 d) 4 e) 5

5.  If 630 is expressed as the product of three positive integers with a sum of 26, then the smallest of these integers is...

a) 2 b) 3 c) 5 d) 6 e)7

I will post more once these are completed.

1.[hide]The diagnol of the square is the radius of the smeicirle.

The side is 10, so the diagnol is  :sqrt: 10 * :sqrt:2 = :sqrt: 200 or d[/hide]

5.[hide]The prime factorization of 630 is 2*32*5*7.

Just by looking at the possible way there can be 3 numbers, we have 10, 9, 7.  The smallest number in this set is 7, or 7 [/hide]

And btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?

Oh no.  These are some problems from the Grade 10 Math Contest in Canada.  It's called Cayley and I'll be taking this this year =D.  So I'm preparing for it as well.

The CMO is different.  There are around 25 problems, similar to the AMC.  I don't have much information about the CMO as I have not yet taken it yet but I will be taking it this year.

(4)
Since c>b>a, and all must be prime, the least c is (2+3) or 5. Hence, every value of c is odd since the only even prime is 2. An odd can only be written as a sum of an odd and an even, so we know a must always be 2. So, the number of ways such that a + b = c where all are primes between 1 and 30 is simplythe number of twin primes between 1 and 30. So, there are four twins, (3,5), (5,7), (11,13), and (17,19). ;)

Good job everyone.  I don't have the answers but they all look correct to me since they do correspond to one of the multiple choice.

Next set of questions.

6.  The distance from the point (-1,-2) to the midpoint of the line segment joining points (2,3) and (-4,3) is...

a) [tex]\sqrt{20}[/tex] b) [tex]\sqrt{24}[/tex] c) 1 d) 4 e) 5

7. If [tex]x^2 + y^2 = -2xy[/tex], then [tex]x^3 + 2y^3[/tex] equals...

a) [tex]y^3[/tex] b) [tex]y^6[/tex] c) [tex]3y^3[/tex]  d) [tex]7y^3[/tex]  e) [tex]9y^3[/tex]

8. The base of a triangle is increased by 10%.  For the area to remain unchanged, the altitude must be decreased by...

a) 9% b) 9/1/11% c) 10% d) 11%
e) 11/1/9%

Note : Does anyone know how to work percentage in Latex

9. The sum of four numbers is 64.  If 3 is added to the first number, 3 is subtracted from the second, the third is multiplied by 3, and the fourth is divided by 3, then all four results will be equal.  The difference between the largest and the smallest of the original numbers is...

a) 27 b) 32 c) [tex]\frac{128}{3}[/tex] d) [tex]\frac{130}{3}[/tex] e) 52

10.  If [tex](x-a)(x-10) + 3 = (x+b)(x+c)[/tex] is true for all [tex]x[/tex] and if [tex]a,b,$ and c[/tex] are integers, then the number of possible values of [tex]a+b[/tex] is

a) 0 b) 2 c) 3 d) 4 e) 6

(6) is very very straightful...
the midpoint of (-4,3) and (2,3) is (-1, 3).
the distance from (-1,2) to (-1,3) is 5.

(7) $x^2+y^2=-2xy \Longrightarrow x^2+y^2+2xy=0 \Longrightarrow (x+y)^2=0$ therefore $x=-y$, hence $x^3+2y^3 = (-y)^3+2y^3 = y^3$

(8) suppose it decreases by $x$. and the original base and altitude are $a$ and $h$.
$\frac{1}{2}ah = \frac{1}{2}\cdot 1.1a \cdot h(1-x)$
we get $x=1-1/1.1 = 1/11$.

use \% to implement $\%$.

9.[hide]

{w+x+y+z}={64}

 [/hide]

I know I went wrong somwhere, since [hide]256[/hide] isn't one of the answer choices..

Oops I think I messed up

9.

[hide] We have a+3=b-3=3c=d/3

a+b+c+d=64, so dividing by 4 we have each equation must equal 16.  Solving gives a=19, b=13, c=48, d=16/3.  c is the largest and d is the smallest, so 144/3-16/3=128/3 C[/hide]

that's the same thing I did at first too

[quote="h_s_potter2002"]
And btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?[/quote]

Erm.... correct me if I'm not mistaken, but the only CMO I know of is the Canadian Mathematical Olympiad... which is the Canadian equivalent of the USAMO and is FAR harder than these questions. It is also 5 questions long, however, and people qualify through the COMC.

Unless there is a different CMO I don't know of....

Could someone explain number 3? I can't seem to solve it.

On #1 isnt the radius the diagonal of half the square?

I believe so

In which case, the answer would be root 125

[quote="Lucky707"][quote="h_s_potter2002"]
And btw, is this representative of what would be on the first exam for the CMO?  Like if I got 4/5 on this one, then I would get 4 problems out of every 5 problems on the exam the CMO has?[/quote]

Erm.... correct me if I'm not mistaken, but the only CMO I know of is the Canadian Mathematical Olympiad... which is the Canadian equivalent of the USAMO and is FAR harder than these questions. It is also 5 questions long, however, and people qualify through the COMC.[/quote]

I agree; the COMC is, to my knowledge, the 25-question multiple-choice test that kool_dudy was referring to.  The CMO is definitely olympiad-style; I believe it's 5 questions long.

However, these questions aren't from either test - they're from one of {Fermat, Cayley, Hypatia, etc.}.  (I can't keep track of all of them...)  And there are many other CMOs - China, for one.  But the Chinese olympiad problems are nowhere near as hard as these.   :D  (I jest.)


vector
linear algebra
matrix
linear algebra unsolved

So far, we are in the second week of class and last class we talked about linear independence and linear dependence..

without further ado, here is a problem assigned to us by the prof.:

Let V be a vector space over the reals
Let u and v be distinct vectors in V.
Prove that {u,v} is linearly independent if and only if {u+v,u-v} is linearly independent.

So... basically, how do you approach this problem? All I have written down is:
Assume {u,v} is linearly independent.
I'm trying to prove the if first, and then the only if second... and yet i'm still completely stuck.

Hints would be greatly appreciated.. thanks.

Generalization: Suppose $A$ is an $n\times m$ matrix whose rows are independent vectors in $F^{m}$ and suppose $\{u_{1}\,u_{2},\dots u_{m}\}$ is an independent set in $V,$ which is a vector space over $F.$ For $i=1,2,\dots,n$ let $v_{i}=\sum_{j=1}^{m}a_{ij}u_{j}.$ Then $\{v_{1},v_{2},\dots,v_{n}\}$ is independent.

As a corollary, if $A$ is an $n\times n$ matrix with independent columns (that is, an invertible matrix), then $\{u_{1}\,u_{2},\dots u_{n}\}$ is independent if and only if $\{v_{1},v_{2},\dots,v_{n}\}$ is independent.

10000th user's insistence that the field not have characteristic $2$ is tantamount to insisting that $\begin{bmatrix}1&1\\ 1&-1\end{bmatrix}$ be invertible. In his post, he has in effect used $A^{-1}$ and the $2$ that appears in the denominator of his fractions is simply $-\det(A).$

Proof of the claim: Suppose $\sum_{i=1}^{n}\alpha_{i}v_{i}=0.$

Then $\sum_{i=1}^{n}\alpha_{i}v_{i}=\sum_{i=1}^{n}\left(\sum_{j=1}^{m}a_{ij}u_{j}\right)= \sum_{j=1}^{m}\left(\sum_{i=1}^{n}\alpha_{i}a_{ij}\right)u_{j}=0.$

Since the $u_{j}$ are independent, this means that for each $j,$ $\sum_{i=1}^{n}\alpha_{i}a_{ij}=0.$

But that then means that a linear combination of the rows of $A$ is the zero vector in $F^{m}.$ Since the rows of $A$ are independent, that means that $\alpha_{i}=0$ for each $i.$

Hence we have proved that the $v_{i}$ are independent.

[quote="10000th User"]$a_{1}\mathbf{u}+a_{2}\mathbf{v}=\left(\frac{a_{1}+a_{2}}2\right)(\mathbf{u}+\mathbf{v})+\left(\frac{a_{1}-a_{2}}2\right)(\mathbf{u}-\mathbf{v})$

Note: $\mathsf{V}$ is over a field with characteristic other than 2.[/quote]

I was wondering, how did you come up with that solution? I know it works out  algebraically, but I'm still trying to arrive at that solution

[quote="Kent Merryfield"]Gilbert0000, what is the inverse of $\begin{bmatrix}1&1\\1&-1\end{bmatrix}\ ?$[/quote]

.5 .5
.5 -.5

still.. im not exactly following how i'm supposed to use this in my proof..

 :(

[hide="Spoiler inside"]
Write $a_{1}\mathbf{u}+a_{2}\mathbf{v}=b_{1}(\mathbf{u}+\mathbf{v})+b_{2}(\mathbf{u}-\mathbf{v})$

$a_{1}\mathbf{u}+a_{2}\mathbf{v}=(b_{1}+b_{2})\mathbf{u}+(b_{1}-b_{2})\mathbf{v}$

$\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}$

Solve for $b_{1},b_{2}$
[/hide]
Nice generalisation Prof. Merryfield :) Is this true?--> $A$ is invertible iff the vectors are linearly independent

if you mean ALL the row vectors ( or equivalently column vectors), yes. otherwise no


inequalities
Euler
algebra
polynomial
number theory
prime numbers
absolute value

I saw the following problem in Komal and I don't have any idea of a solution. Can someone help me, please?
 Does there exist an integer n such that every prime factor of (2^n)-1 is at most (2^(n/1993))-1?

And what is AMM?

Let n1,...,ns are pairwise coprime numbers. It is well known (2ni-1,2nj-1)=1.

Consider n=n1...ns. We have 2ni-1 | 2n-1 ==> (2n-1) /  (2n1-1)...(2ns-1) = m is an integer number and m < 2s.

So our task to find s and ni (i=1..s) such that ni < n1...ns/1993 and s < n1...ns/1993.
For example, s=1993, ni=p1...p1993+pi where pi are successive prime numbers 2,3,5,7 and so on.

Kmal informatation can be found at:

http://komal.elte.hu/info/bemutatkozas.e.shtml

AMM (American Mathematical Monthly):

http://gort.ucsd.edu/newjour/a/msg02973.html

It were grateful if you could (please) be more explicitely in your solution.

My solution is rigorous. Do you not understand some details?

To be sincerely, I don't understand how you proved that every prime factor of 2^n-1 is smaller than 2^(n/1993)-1.

We obtained 2n-1 = m(2n1-1)...(2ns-1) and each factor is less than 2n/1993-1 ==> each prime divisor is less than 2n/1993-1.

Please, write private message next time.

I thought I saw an error. Your last statement is false. Indeed, I will prove that the factor m is not smaller than 2^(n/19993)-1. It is obvious that prod (2^n_i-1) is strictly smaller than prod 2^n_i, which is 2^(n_1+...+n_s). Let us denote for simplicity 2^(n/1993)=a. Then it is clear that 2^n-1 is strictly greater than a^1992*(a-1). So proving that m>a-1 reduces to proving that a^1992>2^(n1+...+n_k). This is equivalent to proving that 1992 prod n_i>1993 sum n_i. Since n_i are different, it is clear that sum n_i/ prod n_i is smaller than (1+2+...+s)/s!, which is obviously smaller than 1992/1993. So, I believe that the whole argument of the solution is incorrect.

You are wrong.

[quote]So proving that m>a-1 reduces to proving that a^1992>2^(n1+...+n_k). This is equivalent to proving that 1992 prod n_i>1993 sum n_i.[/quote]
It is a gross error: m>a-1 reduces to prooving 1992 \sum ni > 1993 \sum ni !!!

I use obvious inequality 2ni-1 \geq 2ni-1 ==> m=(2n1+...+ns-1) / (2n1-1)...(2ns-1) \leq 2n1+...+ns / 2n1-1...2ns-1 = 2s.
All numbers ni are greater than 1993 ==> n/1993 = (n1+...+ns)/1993 > 1993 = s ==> m \leq 2s-1 < 2n/1993-1.

Well, I'm a bit confused by your argument. You have said first that you take n=n_1*n_2*...n_s and noe you tell me that n=n_1+n_2+...+n_s. What should I believe? If you take n=n_1*n_2*...*n_k, then my argument ( which is good-try to look very well) shows that m>2^(n/1993)-1, while if you take n=n_1+...+n_k, why is it neccesarily that 2^n_i-1|2^n-1. And I wonder why doesn't any other meber interfere and say anything?

Sorry for my poor english, but I think that it's time someone different from myth and harazi read the proofs given by myth and harzi and completely close this useless dialog.

So I should understand that everyone believes Myth's solution is corect? Ok then, I won't insist anymore. But I must say that I'm not satisfied with that solution and that, in my opinion, the problem is not solved yet.

Do you read private messages, harazi?

I wrote to you that I really was wrong and my solution is incorrect. Execuse me for my self-confidence.

Then I must apologise. Please, excuse me.

Let Q_k(x) denote the k th cyclotomic polynomial. We have deg(Q_k(x))= phi(k), where phi is the Euler totient function.
We have (2^n)-1= Product Q_k(2) where k takes divisors of n. Since Q_k(2) is an integer for all k, we have for all prime divisor of (2^n)-1 is a divisor of Q_k(2) for some divisor k of n. Thus we have p<(n^2)*2^phi(n)
(because all the coefficent of P_k(x) is smaller than n). Now we only have to choose n such that (n^2)*(2^phi(n)) < 2^(n/1993). This can easily done for n^2< a^n for any a>1 and n sufficently large and phi(n) <n/c for any given c>0 if we choose n=p(1)*p(2)*...*p(l) for sufficent large ( here p(i) denotes the i th prime number.).
In fact by choosing n=p(1)*p(2)*...*p(m) for sufficent large m, and using more sophisticated method to estimate phi(n), we can choose n such that for all prime divisors p of (2^n)-1 we have p<n^d for some constant d.

I don't know cyclotomic polinomial's theory. Can anyone verify this slution?

I think the solution Ihvietbao posted is corect. There is something I don't understand though:
[quote](because all the coefficent of P_k(x) is smaller than n)[/quote]

I think that a proper form of this should be: because all the coefficient of Q_k(x) are smaller that n (or k).
Could you, Ihvietbao, please explain it to me why this is true?

I think that a proper form of this should be: because all the coefficient of Q_k(x) are smaller that n (or k).
Could you, Ihvietbao, please explain it to me why this is true?

Well, I can't prove that statement, so I have a correction:
We don't need to estimate the coefficent of P_k(x) that "sharp".
We only have to use a weaker estimation, which is almost obvious:
the absolute value of the coefficent of x^j in P_k(x) doesn't exceed jCphi(k) (0<=j<=k) (uCv is the binomial coefficent), because all the roots of P_k(x) has modul 1.
Thus we have abs(P_k(2))<= 2^k*(\sum jCphi(k)) = 4^phi(k) <= 4^phi(n) (because k is a divisor of n). (in the above sum j runs from 0 to phi(k))

So for each prime divisor of (2^n)-1 we have p<=4^phi(n), and now the proof continues as posted and we are done.
(At first I thought that all the coefficents of cyclotomic polynomials are either -1,1 or 0, but in fact, P_105(x) has a coefficent -2!)


Prove this inequality for [tex] a,b,c >0 [/tex]
[tex] \sqrt{\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}}+\sqrt{\frac{2(ab+bc+ca)}{a^2+b^2+c^2}} \geq 1+\sqrt{2} [/tex]  :coool:

I think I have a very funny solution.

If $\sqrt{a},\sqrt{b},\sqrt{c}$ are sidelengths of a triangle, then the inequality is trivial, because $2(ab+bc+ca) \ge a^2+b^2+c^2$

If not, let me use a lemma,
Assume x,y,p,q are positive, then if $x \ge p \ge y$, $q \ge y$, $xy \ge pq$, then $x+y \ge p+q$
Proof:
Assume $xy=kpq$ where k is a constant > 1. Then, $x=\frac{kpq}{y}$.
$x+y \ge p+q$
iff $kpq+y^2 \ge y(p+q)$
iff $(k-1)pq + (p-y)(q-y) \ge 0$
Since $k-1 >0$ and $q-y \ge 0$, $p-y \ge 0$, the inequality is true.

Now, let $x = \sqrt{\frac{\sum a^4}{\sum a^2b^2}}$, $y = \sqrt{\frac{2(\sum ab)}{\sum a^2}}$, $p=1$, $q=\sqrt{2}$
It's easy to see that $x \ge p$, $q \ge y$.
$p \ge y$ because $\sqrt{a},\sqrt{b},\sqrt{c}$ does not form a triangle.
To prove $xy \ge pq$, we need to prove
$(\sum a^4)(\sum ab) \ge (\sum a^2b^2)(\sum a^2)$
This resolves to $(5,1,0) \succ (4,2,0)$ and $(4,1,1) \succ (2,2,2)$ and muirhead.
So we can apply our lemma and the question is done.

[quote="Anh Cuong"]Very nice approach  :) , but I still don't understand why it is trivial when [tex] \sqrt{a},\sqrt{b},\sqrt{c} [/tex] are sidelenghs of a triangle  :( .[/quote]

oops, I think I must have something wrong in my mind... let me think again..  :blush:

I tried this inequality but haven't finished yet
My approach is to prove two inequalities:
$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}\ge\frac{a^2+b^2+c^2}{ab+bc+ca}$
[tex] \frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}} \geq 3 [/tex]

My solution was not completed, can anyone give me a hand?

[quote="silouan"][quote="keira_khtn"]Yes,nice![/quote]
Do you have a aolution for the ideas of Minhkoa?[/quote]
Yes,
1/my proof for first lemma:
After expandind we get the equivalent ineq:
$(a^4bc+b^4ca+c^4ab-3a^2b^2c^2)+\sum(a^5b+b^5a-a^2b^4-a^4b^2)\geq 0$
which is obviously true by Cauchy and Muirhead
2/Solution of second lemma:Using Cauchy ineq for 3 expression:
$LHS\geq 3 \sqrt[3]{\frac{(a^4+b^4+c^4)(ab+bc+ca)^2}{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)^2}}$
It remains to prove:
$(a^4+b^4+c^4)(ab+bc+ca)^2 \geq (a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)^2$
Which is equivalent to:
$abc^2(a-b)(a^3-b^3)+ab^2c(c-a)(c^3-a^3)+a^2bc(b-c)(b^3-c^3)+abc[a^5+b^5+c^5-abc(a^2+b^2+c^2)] \geq 0$
********
Return the given ineq,Multiplying both two hands we get:
$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}+2\sqrt{\frac{2(a^4+b^4+c^4)(ab+bc+ca)}{(a^2b^2+b^2c^2+c^2a^2)(a^2+b^2+c^2)}}\geq 3+2\sqrt2$
Since Minhkhoa's two lemma and it follows


function
inequalities
real analysis
real analysis unsolved

Determine, with proof, all functions $f: %Error. "matbbl" is a bad command.
{R}\longrightarrow %Error. "matbbl" is a bad command.
{R}$ that map every closed interval onto a closed interval of the same length.

[hide]
i dont have a proof but i think that the answer is $f=x+b$ for some constant $b$.
[/hide]

Since $[x,y]$ is sent to an interval of length $x-y$, $|f(x)-f(y)|\le |x-y|$ for all $x,y$. This gives continuity. Moreover, $|f(a)-f(b)|=|x-y|\ge |a-b|$ for some pair $a,b\in [x,y]$. We must have $\{a,b\}=\{x,y\}$ for this pair of inequalities to hold, and $|f(x)-f(y)|=|x-y|$ for all $x,y$. For each $x$, either $f(x)=f(0)+x$ or $f(x)=f(0)-x$. By continuity, the same choice must be made for all positive $x$ and the same choice must be made for all negative $x$. A simple check ($f(x)\neq f(-x)$) shows that either $f(x)=f(0)+x$ for all $x$ and $f(x)=f(0)-x$. Making the choice $f(0)=c$, both choices work.


AwesomeMath
summer program
HCSSiM
Mathcamp
Ross Mathematics Program

Sorry if this has been asked before - I did not spend an hour searching on the forums for this question.

Can you really apply to more than one math summer camp? (i.e. I am refering to ones like Awesomemath, HCSSiM, mathcamp, ross program, etc) All these camps don't mind if you are accepted and then choose to not go?

Thanks.

Of course.  You can apply to more than one college, can't you?

are the apps to any of these programs open yet?

I am trying to decide whether I should plan to set aside time to do some of the awesome math early registration problems, but they won't be up for another 3 weeks.

another question: if I am accepted to more than one camp, about when would I need to decide which one I want to go to? I realize this would probably vary from camp to camp. Also, if I apply early to AwesomeMath, do I have to go?

Thanks.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate these:
$ \int\frac{x^4\plus{}1}{x^6\plus{}1}dx$
$ \int ln(lnx)dx$

[quote="dulldrake"]
$ \int\frac {x^4 \plus{} 1}{x^6 \plus{} 1}dx$

[/quote]

Try:

$ (\cos x)^{\frac{1}{4}} \equal{} t$

.

hello, to your first integral, note that
$ \frac{x^4\plus{}1}{x^6\plus{}1}\equal{}\frac{1}{3}\frac{x^2\plus{}1}{x^4\minus{}x^2\plus{}1}\plus{}\frac{2}{3(x^2\plus{}1)}$.
Sonnhard.

hello, to your second integral, set $ \ln(x)\equal{}t$, then we have to solve $ \int e^t \cdot \ln(t)\,dt$.
Sonnhard.

hello, this is ok, Maple 12 gives us the following answer
$ \int\ln(\ln(x))\,dx\equal{}\ln  \left( \ln  \left( x \right)  \right) x\plus{}{\it Ei} \left( 1,\minus{}\ln
 \left( x \right)  \right)$
Sonnhard.


Pascal\u0027s Triangle
Pythagorean Theorem
geometry

im rather confused on how there are so many results attributed to china, arabia, and india,
(such as, they attribute pascal's triangle, the bernolli numbers, the pythagorean theorem, the base 10 system, etc) but the people who invented/discovered these are never named in the books i read asides from ramanujan, but perhaps that is because he went under the wing of a western european mathematician.Al kwarizi is only known (did i spell that rihgt?) because we coined algebra after his name. perhaps it is because i live in america, so the resources around me are completely biased for west european mathematicians

does anyone have any names?

The reason (in my personal opinion) is that we are from the west, and for centuries the west misjudge as poor and non-important the science in the east. It is just rescently (not more than two centuries) that the west realize that there are as many geniuses as in the West. How could a genius not to be European, or descendent of europeans in the case of America. Just a personal opinion. Al Kwarizmi is just an exception to my argument, but, well, all non-mathematical rule has its exceptions.

So, just for the record, Al Kwarizmi is the name from which we derive the word "algorithm."  The word algebra comes from the Arabic [i]al-jabr[/i], used in the title of one of his works.  As for the actual content of your question, we also don't know many particular ancient Babylonian, Egyptian, or Mayan mathematicians, but they certainly must have existed.  Very early mathematics was largely not about actually studying mathematics but instead about such practical tasks as calculating boundaries between lands (surveying) and astronomical calculations.  As such, many of the people doing mathematics weren't doing it in the way that gets your name attached to a theorem.


$\color{blue} \boxed{\textbf{SOLUTION}}$

By $\textbf{Vieta's Formula,}$
$$x_1+x_2=a, x_1x_2=a $$

So, ${x_1}^2+{x_2}^2=a^2 -2a$
And, $2(x_1+x_2)=2a$

We need to show,
${x_1}^2+{x_2}^2 \ge 2(x_1+x_2) \implies a^2-2a \ge 2a \implies a \ge 4$

Which is true because $\triangle \ge 0 \implies a^2-4a \ge 0 \blacksquare$

[quote=qwedsazxc][quote=lian_the_noob12]Here is a mistake in the Question, It should be $a \ge 4$

$\color{blue} \boxed{\textbf{SOLUTION}}$

By $\textbf{Vieta's Formula,}$
$$x_1+x_2=a, x_1x_2=a $$

So, ${x_1}^2+{x_2}^2=a^2 -2a$
And, $2(x_1+x_2)=2a$

We need to show,
${x_1}^2+{x_2}^2 \ge 2(x_1+x_2) \implies a^2-2a \ge 2a \implies a \ge 4$ as desired $\blacksquare$[/quote]

No problem w/ the question. As you know the problem is to prove that $a^2-4a\geq0$, which is true since $x_1,x_2$ being real leads to $a\leq0$ or $a\geq4$.[/quote]

Oh Yes, Sorry I didn't notice this, So for real roots
$$\triangle \ge 0 \implies a^2-4a \ge 0$$

[quote=moldovan]Let $ a$ be a real number. Prove that if the equation $ x^2\minus{}ax\plus{}a\equal{}0$ has two real roots $ x_1$ and $ x_2$, then: $ x_1^2\plus{}x_2^2 \ge 2(x_1\plus{}x_2).$[/quote]

Note that $x_1+x_2=x_1x_2=a$, so the inequality is equivalent to $a^2-4a\geq 0$ which is true because this is the discriminant.


geometry
circumcircle
ratio
inequalities
geometry proposed

Let $ ABC$ be an acute triangle. Points $ D$, $ E$, and $ F$ lie on segments $ BC$, $ CA$, and $ AB$, respectively, and each of the three segments $ AD$, $ BE$, and $ CF$ contains the circumcenter of $ ABC$. Prove that if any two of the ratios $ \frac{BD}{DC}$, $ \frac{CE}{EA}$, $ \frac{AF}{FB}$, $ \frac{BF}{FA}$, $ \frac{AE}{EC}$, $ \frac{CD}{DB}$ are integers, then triangle $ ABC$ is isosceles.

If i have understood the problem, $AD,BE,CF$ are concurrent and the point of concurrency is the circumcenter of
$\triangle ABC.$
If we consider four following cases then we are done.(Other cases are symmetric)
$\boxed {case (i):-}$
$\frac {BD}{CD}$ and $\frac {CD}{BD}$ are both integers then $BD=CD \implies AB=AC$
$\boxed {case (ii):-}$
$\frac {AF}{BF}$ and $\frac {AE}{CE}$ are both integers.
$\implies \frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}$ and $\frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}$ are integers.
$\implies \frac {b^2(c^2+a^2-b^2)}{a^2(c^2+b^2-a^2)}- \frac {c^2(b^2+a^2-c^2)}{a^2(c^2+b^2-a^2)}=\frac {b^2-c^2}{a^2}$ is an integer.
Since $\triangle ABC$ is acute, So $-1 < \frac {b^2-c^2}{a^2} < 1$
hence $AB=AC$
$\boxed {case (iii):-}$
$\frac {AF}{BF}$ and $\frac {CE}{AE}$ are integers.
$\implies \frac {CD}{BD}$ is an integer.
Now similarly as previous we get $CA=CB$
$\boxed {case (iv):-}$
$\frac {BF}{AF}$ and $\frac{CE}{AE}$ are integers.
[hide]Someone for this case please ??? [/hide]

Yeah, I can help with the last case ($\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)}$ and $\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)}$ are integers). Denote $a^2=x,b^2=y, c^2=z$ and $x,y,z$ are the sidelenghts of a triangle, so we can write $x=v+w, y=u+w, z=u+v$. Now, we know that $\frac {a^2(c^2+b^2-a^2)}{b^2(c^2+a^2-b^2)} \ge 2$ and $\frac {a^2(c^2+b^2-a^2)}{c^2(b^2+a^2-c^2)} \ge 2$ (in fact, if we have both of them $2$, we are done since $D$ is the midpoint of $BC$, case already treated). If you are patient enough to multiply the paranthesis, you will get $xz+2y^2 \ge x^2+xy+2yz, xy+2z^2 \ge x^2+2yz+xz$. We have 3 cases:
1. $x \ge y \ge z$ , when $xy+2z^2 <x^2+2yz+xz$, impossible (the equations are symmetric in $y$ and $z$).
2. $y \ge z \ge x$ which implies $ u \ge w \ge v$ and the two inequlities become $uw \ge uv+2vw ; uv \ge uw+2vw$, obviously impossible.
3. $y \ge x \ge z$ which implies $ w \ge u \ge v$ and the two inequlities become $uw \ge uv+2vw ; uv \ge uw+2vw$, obviously impossible.

So, at least one of the ratios is $1$ and we already solved this case.


From what I have learned of complex numbers, it seems to me that any number's nth root yields $n$ complex roots.

So for example, $8$ must have $3$ 3rd roots.

Which should be:

$2 cis 0$
$2 cis 120$
$2 cis 240$

So say your solving the simple equation: $x^3 = 8$

Should one include all 3 roots in their answer or just the real roots? I was taught that if  the equation is $x^2 = 4$, then the answer must be $\pm2$

ask your teacher what he/she wants, on contests give all solutions unless it says real roots, etc

And sometimes (in lower classes, Algebra 1 and such) they only ask for the single real positive root. If you're learning De Moivre's theorem, then the assignment would probably to provide all roots.

[quote="breez"]it seems to me that any number's nth root yields $n$ complex roots.[/quote]

This depends strongly on exactly what you mean.  The equation $x^n = a$ has $n$ complex roots - however, strictly speaking, $x = \sqrt[n]{a}, a \in \mathbb{R}$ is defined to produce only the principal value - the single positive real root.

So if you wanted to find the solutions to $x^3 = 8$, you would include all the roots.  If you wanted to evaluate $\sqrt[3]{8}$, you would include only $2$.  Strictly speaking.


The sum of three different positive unit fractions is $ \frac78$. What is the least number that can be the sum of the denominators of these fractions?

Rather obviously, it's $ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}$, where the sum of the denominators is $ 2\plus{}4\plus{}8\equal{}\boxed{14}$.

It's relatively easy to see that $ frac12\plus{}\frac14\plus{}\frac18\equal{}\frac78$, and since no smaller values work, $ 2\plus{}4\plus{}8\equal{}\boxed{14}$.

What if there was an extension to this problem, like if 3 unit fractions add to 37/81?

Is there a way to solve this systematically?


Bag A contains four yellow marbles, two green marbles and three blue marbles.  	Bag B contains four blue marbles and four green marbles.  If you choose one marble from each bag, what is the probability that you end up with one blue marble and one yellow marble?  Express your answer as a common fraction.

There are no yellow marbles in Bag B, so they have to come from Bag A. Thus our answer is $ \frac{4}{9}\cdot\frac{4}{8}\equal{}\boxed{\frac{2}{9}}$.


What is the significance of the Art of Problem Solving logo?

It's an optical illusion. There are different amounts of little cubes depending on how you look at it.

That's not the only significance.  Imagine that the colors were not highlighted and the tiling was arbitrary within the same hexagon.  Can you think of a math problem that might be asked about the figure?

I don't know if I see it correctly, but it looks like it could be an area/volume question. Clarification?

Indeed, that problem is the inspiration for the logo.