analytic geometry
calculus
calculus computations

Show that any collection of pairwise disjoint discs in the plane is countable. What happens if we replace discs by circles?

I think any disc in $ \mathbb{R}^2$ contains a rational pair of coordinates - and by injecting these into the set $ \mathbb{Q} \times \mathbb{Q}$ we have an injection into a countable set. Is this right?

Also, with circles we can just consider the circle with radius 1 centered at the origin. All possible circles smaller than this one also with center at the origin have radii in the range $ (0, 1)$ and can be mapped by $ r \mapsto \frac{1}{r} \minus{} 1$ to $ (1, \infty)$. I think this is a bijection to an uncountable set - is this right?

is it possible that we are solving your homework?  :P

i wonder if they're in the same course; they posted the same problem within 2 days of each other. if they're cheating on their homework, they should feel very very guilty and go to a confessional *right* now :wink:

This is indeed my problem sheet for the Christmas holidays - Pythagorean12 is probably taking the same course as me. At least I try out the problems myself before asking for hints or help :P Thanks!

Here's a follow-up question for you: Let $ I$ be a collection of pairwise disjoint circles in the plane with distinct centers. Must $ I$ be countable?


Well if you took out john it would be alternating three letter names and names with double letters (ie ll in bill).

No; the order of the words does not matter.

[quote="moogra"][quote="E^(pi*i)=-1"][hide]
Dan does not belong.  All of the others are common words, not just names.
[/hide][/quote]

Nice, but what does john, and joe mean?[/hide][/quote]
[hide]john can be used to mean toilet, joe can be used to mean sweetheart or coffee.[/hide]

[quote="randomdragoon"][quote="sapphyre571"]harry spelled "harry" doesn't mean anything else
hairy does

natty doesnt belong because it's not a common name[/quote]
harry means to harrass or to annoy.
http://dictionary.reference.com/search?q=harry&x=0&y=0

Natty is a common name; it is short for Nathaniel. However, this puzzle doesn't have anything to do with names because none of the words were capitalized.[/quote]
wait is E^(pi*i)=-1 right
b/c
dan
–noun Martial Arts.
a degree of expertise in karate, judo, tae kwon do, etc., usually signified by the wearing of a cloth belt of a particular color; level: a sixth-degree dan.
[url]http://dictionary.reference.com/browse/dan[/url]

I think joe does not belong because that word has 2 vowels while all the others only have one (not including "y"s). :P

is it "john" because it has a silent letter: "h"

[quote="randomdragoon"]which of the following does not belong?

bob bill dan harry john joe natty pat[/quote]

[hide="guess"]dan can be anagramed into D.N.A. but the rest can't be anagramed into an abbreviation.[/hide]

[quote="junggi"][quote="K81o7"][quote="junggi"]wait is E^(pi*i)=-1 right
b/c
dan
–noun Martial Arts.
a degree of expertise in karate, judo, tae kwon do, etc., usually signified by the wearing of a cloth belt of a particular color; level: a sixth-degree dan.
[url]http://dictionary.reference.com/browse/dan[/url][/quote]

Wow...
Dictionary.com doesn't actually know what dan means.  If they don't know Japanese, they shouldn't have Japanese words...[/quote]
uh it's not really "Japanese"
I'm sure it's actually from Chinese, but used in other countries like Korea or Japan.
And that IS the correct meaning(except, of course, it's not exclusively for karate, blabla, but for other stuff as well. And for other stuff they don't wear belts)[/quote]

First of all, dan IS a Japanese word.  Perhaps it was taken from Chinese long ago, but it is an actual Japanese word.

And second, dan has nothing to do with belt color.  Belt color is more of an American idea...
Dan has a much more general meaning than that...they could at least make the distinction between dan and kyu, because that's where much of the meaning of dan is contained: a dan is one who has mastered the basics and can pass them on to a kyu, or a beginner.

[quote="K81o7"][/quote][quote="junggi"][quote="K81o7"][quote="junggi"]wait is E^(pi*i)=-1 right
b/c
dan
–noun Martial Arts.
a degree of expertise in karate, judo, tae kwon do, etc., usually signified by the wearing of a cloth belt of a particular color; level: a sixth-degree dan.
[url]http://dictionary.reference.com/browse/dan[/url][/quote]

Wow...
Dictionary.com doesn't actually know what dan means.  If they don't know Japanese, they shouldn't have Japanese words...[/quote]
uh it's not really "Japanese"
I'm sure it's actually from Chinese, but used in other countries like Korea or Japan.
And that IS the correct meaning(except, of course, it's not exclusively for karate, blabla, but for other stuff as well. And for other stuff they don't wear belts)[/quote][quote="K81o7"]

First of all, dan IS a Japanese word.  Perhaps it was taken from Chinese long ago, but it is an actual Japanese word.

And second, dan has nothing to do with belt color.  Belt color is more of an American idea...
Dan has a much more general meaning than that...they could at least make the distinction between dan and kyu, because that's where much of the meaning of dan is contained: a dan is one who has mastered the basics and can pass them on to a kyu, or a beginner.[/quote]
Well so in Japanese, dan and kyu are not a "a degree of expertise"?
Like in Korea, we say how much someone is better at something than someone else, based on their dan. (like in go, 9dan is much stronger than 7dan.) This definition you give is probably one of the definitions of the words dan in Japanese, but in Korea or U.S, it's much more commonly used as "a degree of expertise," so I would say dictionary.com KINDA knew what they were talking about.(the "[only ]in karate, judo, tae kwon do, etc., usually signified by the wearing of a cloth belt of a particular color; level: a sixth-degree dan." is wrong, i know).
Also, I'm sure Japanese use dan as a rating of some player's strength in go?
And I brought up "uh it's not really "Japanese"" because it sounded as if you were saying "it originated from Japan, it's only a Japanese word,"etc, which it clearly not true.I'm fairly sure it originated from Chinese(in fact most of Japanese words originated from China, didn't they?).
And now that I think about it, the word dan as the definition presented in dictionary.com is used in U.S. very commonly, and the other definitions we know of is almost never used. So the word blended into the U.S. language, and the dictionary gives us the definition used in U.S. And it's an American dictionary, isn't it?

natty , because it's not a good name  :lol:

actually not

[b]K81o7:[/b]  Without giving any corresponding written form, it is hard to say that there is only one word that is pronounced "dan" in Japanese or in Chinese.  The definition given by dictionary.com certainly accords with one usage of "dan" in rankings in go, and I would not be surprised if the martial arts usage were also common.

Even if we are discussing the same word, English is not the only language in which the same word has multiple meanings :!:

[b]junggi:[/b]  The Japanese spoken language evolved independently of Chinese.  It is Kanji, an aspect of their written language, that was taken from the Chinese system of writing.  Some words, like "ai" (love), do have both the same pronounciation and character in Japanese and Chinese, but I would rather you backed claims like that with evidence rather than speculation.

Certainly your point that an American dictionary focuses on the American usages of words is valid.

Edit:  Hey, evidence is good.  [i]Right below the definition from dictionary.com is this gem:[/i]

[quote="dictionary.com"][Origin: 1940–45; < Japn < MChin, equiv. to Chin duàn step, grade] [/quote]

So that settles it.  It is a Japanese word with a Chinese equivalent from which it probably derived.

I'm not sure where the opinion came from that words borrowed from other languages don't really count as words of a given language, since any non-Germanic English word was borrowed from French, and many American English words were borrowed from Native American languages.

eh ok

randomdragoon, could you post the answer to the problem?

Simple answer: dan is the only word that is not in the Scrabble dictionary.

Cheers  :lol:

[quote="junggi"]Well so in Japanese, dan and kyu are not a "a degree of expertise"?
Like in Korea, we say how much someone is better at something than someone else, based on their dan. (like in go, 9dan is much stronger than 7dan.) This definition you give is probably one of the definitions of the words dan in Japanese, but in Korea or U.S, it's much more commonly used as "a degree of expertise," so I would say dictionary.com KINDA knew what they were talking about.(the "[only ]in karate, judo, tae kwon do, etc., usually signified by the wearing of a cloth belt of a particular color; level: a sixth-degree dan." is wrong, i know).
Also, I'm sure Japanese use dan as a rating of some player's strength in go?
And I brought up "uh it's not really "Japanese"" because it sounded as if you were saying "it originated from Japan, it's only a Japanese word,"etc, which it clearly not true.I'm fairly sure it originated from Chinese(in fact most of Japanese words originated from China, didn't they?).
And now that I think about it, the word dan as the definition presented in dictionary.com is used in U.S. very commonly, and the other definitions we know of is almost never used. So the word blended into the U.S. language, and the dictionary gives us the definition used in U.S. And it's an American dictionary, isn't it?[/quote]

True...yet I just wanted to stress that they have an extremely narrow definition there.  It's not only a degree of expertise: for example in go, a "dan" player (or just a dan) signifies on who has a dan rating, but also one who is experienced and is one to take example from.  The cloth belt is completely unrelated...the definition they give is thus incomplete.  Even when it's used in the US, the word "dan" seems to have that meaning.  (I know this only because I play go, have been learning karate for 6 years, and have been learning Japanese for 3-4 years...in those settings, the word "dan" is analogous to sensei.  To kyus, it doesn't matter what dan the sensei is.  Of course, I guess that might be different among dans.)

And yeah, I wasn't saying that the word originated from Japan...it just sounded like you were saying it was a word in only the Chinese language.

To t0rajir0u: That's true...I suppose we just were referring to the most common (in my experience) use of the word.  By the same reasoning, dictionary.com's definition of "kami" is incorrect. :D


Let $ (x_n)$ be a convergent sequence of real numbers with limit $ L$. Prove that $ \frac{x_1\plus{}x_2\plus{}...\plus{}x_n}{n}\to L$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167852]here[/url] or your previous thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=138902]here[/url]. This has been asked many times before, and the result is frequently used as a lemma.


I'm running TeXnicCenter with MiKTeX on my computer.

I get the error code "TeX capacity exceeded, sorry [0.2=200000]"  .

I understand I should try to increase the TeX capacity, but how do I do that in MiKTeX?

Thanks!

That's a really misleading message, left over from the old days when memory was in short supply. Nowadays that is extremely unlikely to be the cause of the problem. Almost certainly it is due to an error in the document (which could cause recursion for example). It could be as simple as a missing }

So see if you can find the source of the error. Try cutting down the document until you find where the error is happening and if necessary come back here with a minimal document that's causing the error.

I understand the problem now.

Apparently I was trying to input a huge image file (according to latex standards) and the program was simply unable to perform such a command -- or something like that, I'm nost sure (the image itself was 500 px by 500 px).  After resizing the image to 200px by 200 px, the error goes away.

Ahh -- wasted about 1.5 hours trying to search for ways to increase tex capacity off the web.  Thanks though for your insight, it was very much appreciated.


According to observers in the frame F, the following events occurred in the xy plane.  A singly charged positive ion which had been moving with constant velocity v = 0.6c in the +y direction passed through the origin at t = 0.  At the same instant a similar ion which had been moving with the same speed, but in the -y direction, passed the point (2, 0, 0) on the x axis.  The distances are in cm.

(a) What is the strength and direction of the electric field, at t = 0, at the point (3, 0, 0)?

(b) What is the strength and direction of the magnetic field at the same place and time?

[hide](a) $\frac{25}{18}e$ in the $+x$ direction[/hide]


A quadrilateral has sides length 24,25,26,and 27, and area 180x, where x is a real number. Find $ x^2$

The side lengths of a quadrilateral do not uniquely determine its (shape, nor its) area.  Is the quadrilateral also cyclic?

[u][b]Answer: 663[/b][/u]
[b][u]Formula: Read the post[/u][/b]
I'm going to have to disagree with you t0rajir0u. Look into Bretschneider's formula which works for ANY quadrilateral and into Brahmagupta's formula which generally deals with cyclic quadrilaterals.

Well onto the problem:
1) I will assume the quadrilateral is in fact cyclic, just for the sake of easy math (sorry!).
2) [img]http://upload.wikimedia.org/math/c/5/e/c5eb0f06ae3ffb12a5338b9ffc93e76d.png[/img] Where T is the semi perimeter, (p + q + r + s)/2, and p, q, r, s are the side lengths.
3) Plugging in your values gives you 180x = sqrt(51*27*26*25*24).
4) 180*180*x^2=51*27*26*25*24
5) x^2=51*13
6) x^2=663

Bretschneider's formula doesn't tell you anything.  The diagonals aren't uniquely determined by the sides.  (Consider the case when $ a \equal{} b \equal{} c \equal{} d$; the quadrilateral (which is a rhombus) can have any area from $ a^2$ to $ 0$.)  You are correct in the case that the quadrilateral is given to be cyclic; that's why I asked.

How can a rhombus have an area of zero? Reread my question and just wanted to point out that it is a serious question. Wouldn't a rhombus not be a polygon by definition if it didn't enclose an area?

Strictly speaking, no.  We call such shapes [i]degenerate.[/i]  My point still stands.

Gealgebra, t0rajir0u is trying to say that you cannot [i]assume[/i] that the given quadrilateral is cyclic unless you can show it is cyclic.

[quote="Gealgebra"][u][b]Answer: 663[/b][/u]
[b][u]Formula: Read the post[/u][/b]
I'm going to have to disagree with you t0rajir0u. Look into Bretschneider's formula which works for ANY quadrilateral and into Brahmagupta's formula which generally deals with cyclic quadrilaterals.
[/quote]

If you looked up the first formula that you cited, you would notice that the formula is also dependent on the sum of the opposite angles of the quadrilateral...That is you have to plug in 5 values to that equation, not four.

Isn't it six? I recall the last term is $ \cos\left(\frac {A \plus{} C}{2}\right)$

$ \cos \left( \frac {A \plus{} C}{2} \right) \equal{} \cos \left( \frac {B \plus{} D}{2} \right)$, so this is really only one new independent parameter.  In some sense, it measures the extent to which a quadrilateral fails to be cyclic.


N is the product of five distinct odd primes, of the form abcab, where a,b,c are digits, and 4<a<8. Find one value of N.

hmm...calculators allowed? :P only way I can think of...

Assuming calculators are allowed, I do a little trial and error [hide]That repeated pattern of abcab makes me think of divisibility by 1001. 1001=7*11*13. I multiply by 3 and 5 (the smallest two primes). That gives 15015. Hmmm too small. I replace 5 with 17 (the next smallest prime after 13) and I get 51051. YAAAAAY. a=5, b=1, and c=0. [/hide]

[color=cyan]Calculators are not allowed on these contests, sorry.

As a starting point, try to set caps on the size of the prime numbers involved.[/color]

yea, mystic, no calcs allowed, but ur answer is one of hte correct answers.

It seems that everyone in my class did this question using guess and check (i didn't do this problem, cuz i hate guess and checking), which makes me think this problem is stoopid.

[color=cyan]What was Linda's solution?[/color]

she just nodded to the guess and check one

You only have to find one value. Just assume c=0, because there are like 48312742734984672384923674 values of N with c=0 that satisfy the property, and they're all equal to 7*11*13*p*q. So you just find two primes that multiply to a two digit number between 40 and 89.


What is the value of $ 3^5 \minus{} 3^4$?

Factor out the $ 3^{4}$ to get $ 3^{4}(3\minus{}1)\equal{}3^{4}\cdot 2\equal{}81\cdot 2\equal{}162$.


let a,b,c are real number
      find k is real number smallest such as with all a,b,c we have:
       (a+b+c)/3<=k.max(/a/,/b/,/c/)+(abc)^(-1/3)

hello, do you meant $ \frac{a\plus{}b\plus{}c}{3}\le k\cdot \max(|a|,|b|,|c|)\plus{}(abc)^{\minus{}\frac{1}{3}}$?
Sonnhard.


vector
linear algebra
matrix
modular arithmetic
abstract algebra
linear algebra unsolved

Panurge has 2009 sheep. If I remove a sheep (it does not matter which ; the property is valid for each choice of sheep) then there exists a partition of the 2008 sheep remaining in two parts such as the sum of the weights of the sheep of a part is equal to the sum of the weights of the sheep of the other part.
   What can you say about the weights of the sheep ?

Hmm... our white wolf chose to post this in the linear algebra section - in what way is this a linear algebra problem?

Let $ x$ be the $ 2009\times 1$ column vector consisting of the weights of the sheep. The one obvious restriction is that we should have $ x_j>0$ for all $ j.$

The condition is that there exists a $ 2009\times 2009$ matrix $ A$ with diagonal entries of $ A$ zero and all off-diagonal entries equal to $ \pm1$ such that $ Ax\equal{}0.$

If this had been $ 2008$ sheep instead of $ 2009,$ we would have ruled this out as impossible. If we compute the determinant of $ A$ mod $ 2$ we find that it equals $ m\minus{}1\pmod{2},$ where $ m$ is the size of the determinant. So if $ m$ is even, $ A$ would have to be invertible and we couldn't have a notrivial $ x$ with $ Ax\equal{}0.$ But $ m$ was odd here, so we can't rule out the possibility that cheaply.

So far I can prove that in the case of $ m\equal{}3$ (three sheep), the sheep must be exactly the same weight. (That is, $ x$ must be a scalar multiple of the vector of all ones.) I haven't worked up to the case of arbitrary odd $ m$ yet.

[hide="oops"]It took me some googling to realize what our white wolf was alluding to in the title...

[quote="Kent Merryfield"]If this had been $ 2008$ sheep instead of $ 2009,$ we would have ruled this out as impossible. If we compute the determinant of $ A$ mod $ 2$ we find that it equals $ m \minus{} 1\pmod{2},$ where $ m$ is the size of the determinant. So if $ m$ is even, $ A$ would have to be invertible and we couldn't have a notrivial $ x$ with $ Ax \equal{} 0.$ But $ m$ was odd here, so we can't rule out the possibility that cheaply.[/quote]

Well, it would be somewhat surprising if you could show that all sheep have zero weight...
Basically, you have solved the problem. Just remember that you don't want to show that the kernel of $ A$ over $ \mathbb{F}_2$ is zero, but you want to show that it has dimension $ 1$. And an $ m\times m$ matrix has rank $ \geq k$ if and only if it has a $ k\times k$ minor with nonzero determinant.

This is very a famous problem. For instance:
http://www.mathlinks.ro/viewtopic.php?t=27219
http://www.mathlinks.ro/viewtopic.php?t=18134
and some more.

  darij[/hide]

Hi Darij,
here it is not the same problem. My subsets have not necessarily $ 1004$ elements each.
Remark: the parity of n=2009 has nothing to do here. It is sufficient to assume $ n\geq{3}$.

But the parity does matter - there is no nontrivial solution for an even number of sheep. See my previous post, about the invertible matrix mod 2.

Update: a colleague of mine whom I told the problem tells me that the positivity matters: if you don't have $ x_j>0$ for all $ j,$ you get other answers.

Ouch, not again.  :oops:

This will most likely be another humiliation, but let's try: Isn't positivity completely irrelevant? We can replace every sheep of weight $ x$ by a sheep of weight of $ \left|x\right|$, and the condition of the problem doesn't change, because if a list $ \left(x_1,x_2,...,x_k\right)$ of integers can be partitioned into two sub-lists with equal sum, then so can the list $ \left(\left|x_1\right|,\left|x_2\right|,...,\left|x_k\right|\right)$.
So we seem to have arrived at http://www.mathlinks.ro/viewtopic.php?t=224623 . It remains to find out whether TTsphn's solution is correct (which I cannot tell without trying to understand his English, what needs some better concentration that I am able to at the moment).

  darij

A nontrivial example with $ n \equal{} 5:$ the weights of the sheep could be $ 3, 3, 1, 1, 1.$

Darij is right: positivity is basically irrelevant. My colleague gave me a matrix whose kernel is spanned by $ \minus{} 1, 3, 1, 1, 3$ but his example was easily altered.

Here's the matrix $ A$ for this example, with some blocks that I noticed. Note that the lower right $ 3\times 3$ block is essentially the only solution for the $ n\equal{}3$ case.

$ \left(\begin{array}{cc|ccc}0&1&\minus{}1&\minus{}1&\minus{}1\\1&0&\minus{}1&\minus{}1&\minus{}1\\\hline 1&\minus{}1&0&1&\minus{}1\\1&\minus{}1& 1&0&\minus{}1\\1&\minus{}1&1&\minus{}1&0\end{array}\right).$

Hi Kent,
I am stupid. $ n$ is necessarily odd.
Darij has simplified the problem:
Now we may assume the weights to be only non zero during the calculations.
If $ n\equal{}5$ I think that there exists only 2 non trivial solutions:
$ \{1,1,1,3,3\},\{1,1,3,3,5\}$.
If the TTsphn's solution is correct then, for all odd number $ \geq{7}$, there exist solutions with pairwise distinct (positive ?)weights.
Example for $ n\equal{}7$:
$ \{3,31,49,19,21,17,23\}$.


Please tell me i'm not the only person on here who didn't make nats in their 8th grade year

Every person who's in 8th grade who didn't make nats didn't make nats. And alot of them are on here, such as... mathcountsfreak, 27r, AIMEman (sorry for using you guys, but you're probably the only people I actually know), cognos599, blah. There, feel better? xD

mathcrazed, you forgot to list your best friend...the "funny kid"
but OMG I didnt make nats in my 8th grade year either...dang bummer...so sad...crying...
Infact, I didnt even CD at our state.

But then again, I was sitting 2-3 spots away from 2 6th graders who also didnt make CD that year and both have a legit shot at winning Nationals this year.

oh yea and mathcrazed, your question that you posted about Bill,emily... is kinda weird,did you forget something?  because it seems that they can go at any speeds.

mwpl11(, undefined117, MathAndKnowledge, MustItAlwaysBeMe, the future, and not to mention COGNOS599. all of whom are really good at math (maybe with the exception of mustit.... and cognos599) but didn't make nats.

i didn't make nats. and i'm in 8th grade.

HIGH FIVE!!! w00t!!

You can put me on the list I guess...I placed 8th (ugh, lack of sleep). At least I made USAMO in my first year of doing AMC's.

who DIDNT make it?

person from our school who got 1st school, 1st chapter, and got..... 16th state.

Also, the 1st - 4th places from every other chapter in the state, (except our 4th place state, 1st chapter guy.)

A brief summary of those who are going to nats:

1st place 7th grader who got 5th chapter
2nd place 8th grader who got 6th or 7th chapter
3rd place (ME!) 7th grader who got 5th chapter
4th place 8th grader who got 1st chapter...

pretty weird.

I didn't make it, and I had a score of 35! I wish I was in an easy state like oklahoma...

me too, partially because I was a alternate...

Nooooo! Two points away from Nats! Noooo!  :rotfl:

I didn't make nationals! :D
I was emo for quite some time. I screwed myself on target...

...Meh. I can't expect to compete with this kid who got first in School, Chapter, State...but I suspect he might not do so well at Nationals. :maybe:

I thought this years target was a little easy but i got owned on it anyway.....I sucked at sprint (i dunno if we can say scores yet though) i can tell you right now that i barely made double digits :blush:

I didn't make nats. I'm in seventh grade, got 6th in state. I hope i dont fail again in 8th grade. :wink:

Same here. Except I got fifth (after tiebreakers  :(... nope, I'm still not over that!). But yeah, Chapter this year wasn't a great indicator of how you were gonna do at State, since it was so easy. I completely phailed Sprint at State... *sigh* at least I'm only a lowly seventh grader!

I failed like the first set on the target, and somehow got the last one, but missed the third set - Yea - I died. Actually, this year's target was a joke, maybe even a good diss.

[quote="mathcrazed"]You're just lucky this year's AIME was easy distracted =P. But... since it's your first year, OMG GENIUS RUN.[/quote]
You made USAMO, smart one. lol, i was bored, so i looked at the list. i didnt see bobby though...
I didn't make nats in my 7th grade year. WHAT NOW

I didn't even get close to nats... 31st. Why does TX have to be so hard????? I'm only in 6th, though. I have plenty of time.

Thats what I said 2 years ago... You better start studying....


It's not realistic, but here's a strip detailing exactly how Windows Vista screws over even the best of computers:

Why does everyone say vista is so bad?

i mean what if a son/daughter of a dad who works for windows saw this?

that would be insulting.


Do you have a role model? Post here what this person is.

My two role models are Michael Jackson and Eliot Spitzer

Ermm do you know who those people are? :huh:

i think he does, but that's sorta odd

just in case, they're both (alleged) molesters...  :lol:

my role model is...

er, not sure, but probably my parents

michael jackson was a damn good artist and a kickass children
he just liked to touch little boys on the side but other than that he was a pretty cool dude

my role model is Asakura Hao! he's the main antagonist/villian in Shaman King, and he wants to kill all humanity, but it's not his fault, since people had killed him in the past unjustly, so he wants revenge. (he reincarnated himself. ) He's really srong, never gives up, and repects no one but himself  :lol:

Mr. Crawford? james4l? Mr. R (hehe, not even going to bother to try to spell it)? Mr. Vornicu? Mr. Patrick? Eh, no idea.

Hmmm...
Yi Sun, graduated from Harker in 2004 I think, Ipho and IMO gold medalist, Intel STS second prize, played piano too :D

1. My biology teacher. He shows that you can be a nerd, a husband and father, a football player in HS, as well as a gamer all at once.  :o

2. t0rajir0u of course, who makes AoPS seem like 2nd grade math.

my sis: she just makes me happy and she is always someone i can talk to. I can talk to her and confess and do anything to her and she will take it lightly

my parents: well.. mostly my mom b/c she told me to never give up.

my friends: i could talk to them about stuff that i don't want my family to know.

ANDREW ARDITO

and the AoPS staff.

And all IMO qualifiers who are famous on AoPS, and Sherry Gong.

And Darryl Wu.

ASAKURA HAO!!! (once again) HAO'S THE BEST!!!!!! WHOOOHOOO!!!!

Pete Townshend
Elizabeth Meckstroth
Michael Piechowski
hmmm...
Oh, My brother! =P

Me
Matt Damon
Chewbacca
Richard Ruscyk
David Patrick
Steve Sigur
Me
Myself
You (just kidding!!!)
ME

I'd have to say:

Andrew Wiles, who sort of inspired me to think that nothing

is unprovable.

Richard Rusczyk, who has been a great teacher, a great writer,

and who has taught me SO much. Thanks!

David Patrick, an amazing instructor, who really helped me in

Counting & Probability.

All the other employees of AoPS, for too many reasons to list.

Many of the AoPS members.

Previous IMO participants, including the 2 girls, who have inspired

me to make it. Soon.

And others, too numerous to mention all of them.

Micheal Jackson
My Bro
The great mathematicians.
My cat

other than Hao:
me
myself
I
"this one"
dreambold22

and.........

my best buddy Fuyuki-chan.

My parents
My calculator
Steve Hawkings
Me (jk)
J.R.R. Tolkien
t0rajir0u
Arnav Tripathy
Miles Dillon Edwards
Every yellow person on the face of this Earth ftw :lol: (a bit racist, I know)


algebra
function
domain
floor function
calculus
calculus computations

Find the domain and set of points of discontinuity of the function
$ f(x)\equal{}[x]sin(\frac{\pi}{[x\plus{}1]})$
where $ [x]$ denotes the greatest integer function.

hello Pankaj
[hide="Domain"]
$ R-[-1,0)$ as $ \left floor\lfloor{x+1}\right floor\rfloor$ should not equal to zero.
[/hide]

[hide="Continuity"]
The function is discontinuous at x belongs to $ N$ where $ N$ is an integer in the provided domain. [/hide]
thank u


How many cubic feet are in the volume of a round swimming pool
which is 16 feet in diameter and 4 feet deep throughout? Express
your answer in terms of $ \pi$.

the volume of this pool (assuming that it is cylindrical) is equal to the base times the height. Because the diameter of the base is 16, the radius is 8, so the base is $ 64\pi$ square feet. Multiplying this by the depth, 4 feet, gives the volume, $ \boxed{256\pi}$ cubic feet.

[quote="gaussintraining"]Since this is a cylinder, we have $ \pi (8^2)(4) \equal{} \boxed{256 \pi}$.

Edit: Beaten[/quote]

haha by one minute :D


geometry
number theory
least common multiple

[color=cyan]Here are 4 problems of mine.  Have fun.

1. 99 can be expressed as the sum of 99 consecutive integers.  Compute their product.

2. What is the smallest positive integer that, when divided by 2, 3, 4, 5, 6, and 7, leaves remainders of 1, 2, 3, 4, 5, and 6, respectively?

3. Compute the sum: 1/1 + 1/(1 + 2) + 1/(1 + 2 + 3) + . . . + 1/(1 + 2 + 3 + . . . + 100)  in simplest form.

4.  In triangle ABC, BC = 39, the length of the altitude to AB is 15, and the length of the median to AB is 25.  Compute both possible values for the area of the triangle.[/color]

Quote:1. 99 can be expressed as the sum of 99 consecutive integers. Compute their product.

[hide]99 = -48+ -47 + ... + 47 + 48 + 49 + 50

Product = (48!)^2 * 49 * 50

Is that what you wanted?[/hide]

No, it's not what he wanted. Be very careful. The problem reminds me of another one:

What is the expanded version of the product

(x-a)(x-b)(x-c)...(x-z)?

Read the question a bit more carefully. Is 1+2+3+...+k equal to k(k+1)?

Quote:4. In triangle ABC, BC = 39, the length of the altitude to AB is 15, and the length of the median to AB is 25. Compute both possible values for the area of the triangle.

[hide]The altitude is 15. Apply Pythagorean a few times, and, depending on the location of the median with respect to the altitude, AB = 112 or 32. Thus the area is either 840 or 240.[/hide]

[hide]

2.

It's 1 less than a multiple of 2,3,4,5,6, and 7. The LCM of all of them is

2:^2:*3*5*7 = 420

420 -1 = 419

[/hide]

TheSouthPaw:

I don't think question #2 says that it is 1 less than a multiple of 2, 3, 4, ..., but 1 less than amultiple of 2, 2 less than a multiple of 3, etc...

yup...same way we discussed, with slightly different notation.
make sure you take the LCM of all of those, not the product, btw.

My bad, interpreted your "less" as "more"

yeah, forgot about the lcm part, and i delete it. lol. sorry

[quote]fiery, u made a mistake in #1.
There is a zero in that sum, so the product is just 0.[/quote]

You're right, my bad, the answer is 0.


calculus integration trigonometry logarithms function limit inequalities

Prove

$ \sum_{n\equal{}1,3,5,...} \text{Ci}(n\pi)\equal{}\frac12 \ln 2\minus{}\frac14$

[quote="Dr. Doe"]Prove

$ \sum_{n \equal{} 1,3,5,...} \text{Ci}(n\pi) \equal{} \frac12 \ln 2 \minus{} \frac14$[/quote]
Whta is def of $ \text{Ci}$?
you mean $ \text{Ci}(t)\equal{}\int_{0}^{t} \frac {1}{\cos (x)}dx$?

Cosine Integral is defined as $ \text{Ci}(x)\equal{}\minus{}\int_x^\infty \frac{\cos t}{t} \, dt$  .

Similarly $ \int_0^\infty e^{-n\pi x}\, \frac{x}{1+x^2}\, dx=(-1)^{n-1}\, \text{Ci}(n\pi) $ .

So $ \sum_{n=1}^\infty \text{Ci}(n\pi)=\int_0^\infty \sum_{n=1}^\infty (-1)^{n-1}\, e^{-n\pi x}\, \frac{x}{1+x^2}\, dx=\int_0^\infty \frac{1}{e^{\pi x}+1}\, \frac{x}{1+x^2}\, dx $ .

Can you show that this integral it equal to $ \frac12\, \left(\log 2-\gamma\right) $ ?

[quote="Doe John"]We need the formula

$ \int_0^\infty e^{-sx}\,\frac{x}{a^2+x^2}\, dx=\sin(as)\,\left(\frac{\pi}{2}-\text{Si}(as)\right)-\cos(as)\,\text{Ci}(as) $ .

$ \Rightarrow \, \int_0^\infty e^{-(2n+1)\pi x}\, \frac{x}{1+x^2}\, dx=\text{Ci}((2n+1)\pi) $

Thus $ \sum_{n=0}^\infty \text{Ci}((2n+1)\pi)=\int_0^\infty \sum_{n=0}^\infty e^{-(2n+1)\pi x}\, \frac{x}{1+x^2}\, dx  $

$ =\int_0^\infty \frac{e^{-\pi x}}{1-e^{-2\pi x}}\, \frac{x}{1+x^2}\, dx=\frac12 \int_0^\infty \frac{1}{\sinh \pi x}\, \frac{x}{x^2+1}\, dx $

$ =\frac12 \, \left(\log 2-\frac12 \right)=\frac12 \log 2-\frac14 $ . ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=164412&hilit=ln+sinh]Abel's integral[/url])[/quote]

In the proof you linked to, the contour goes through the pole at the origin.  Don't you want to avoid that?

[quote="Doe John"]The function $ f(z)=\frac{z}{(1+z^{2})\sinh\pi z} $ doesn't have a pole at $ z=0 $, since $ \lim_{z\to 0} f(z)=\frac{1}{\pi} $ .[/quote] Thanks. I should have noticed that it was a removable singularity.

And how do we know that the other three sides of the contour evaluate to zero as $N \to \infty$ without explicitly using the ML inequality?

sos440

How do you derive that series for csch(x)?

Just consider the logarithmic differentiation of
\[ \frac{x}{2} \frac{\cosh (\frac{x}{2})}{\sinh (\frac{x}{2})} = \prod_{n=1}^{\infty} \left( 1+ \frac{x^2}{\pi^2 n^2} \right)^{(-1)^{n-1}},\]
which is easily verified from
\[ \sinh x = x \prod_{n=1}^{\infty} \left( 1 + \frac{x^2}{\pi^2 n^2} \right), \text{ and } \cosh x = \prod_{n=1}^{\infty} \left( 1 + \frac{x^2}{\pi^2 ( n - \frac{1}{2} )^2} \right).\]

Can you prove the following formulas?

$ 2\sum_{n=1}^\infty \text{Ci}(n\pi x)=\psi\left(1+\left\lfloor\frac{x}{2}\right\rfloor\right)-\log\frac{x}{2} $

$ 2\sum_{n=1}^\infty (-1)^n\, \text{Ci}(n\pi x)=\psi\left(\frac12+\left\lfloor\frac{x+1}{2}\right\rfloor\right)-\log\frac{x}{2} $

Maybe the hint $ \int_0^\infty e^{-sx}\, \frac{a\,\sin as-x\cos as}{1+x^2}\, dx=\text{Ci}(as) $ is helpful.

$ \int_0^\infty e^{-n\pi\alpha x}\, \frac{a\,\sin(n\pi\alpha)-x\cos(n\pi\alpha)}{1+x^2}\, dx=\text{Ci}(n\pi\alpha) $

$ \int_0^\infty \frac{\sin\alpha\pi-x\left(\cos\alpha\pi-e^{\alpha\pi x}\right)}{\cosh \alpha\pi x-\cos\alpha \pi}\, \frac{dx}{1+x^2}=2\sum_{n=1}^\infty \text{Ci}(n\pi\alpha) $

We can deduce the second formula from the first one.

$ 2\sum_{n=1}^\infty (-1)^n\, \text{Ci}(n\pi x)=2\cdot 2\sum_{n=1}^\infty \text{Ci}(2n\pi x)-2\sum_{n=1}^\infty \text{Ci}(n\pi x) $

Another interesting formula could be

$ \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \, \text{Ci}\left((2k+1)\pi x\right)=(-1)^{\lfloor x+\frac12 \rfloor}\cdot \frac{\pi}{4}\log \pi x  $

$+\frac{\pi}{4}\log\frac{\pi}{4}-\pi\log\left(\frac{\pi}{\Gamma\left(\frac14\right)}\right)+\frac{\pi}{2}\sum_{k=0}^{\lfloor x-\frac12 \rfloor} (-1)^k\, \log\left((2k+1)\frac{\pi}{2}\right) $ .

$ \frac{4}{\pi} \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \cos\Big((2k+1)\pi t\Big)=(-1)^{\lfloor t+\frac12 \rfloor} \quad \text{for all real} \quad t\notin\Bbb{Z}+\frac12 $

$ R:=-\frac{4}{\pi}\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\, \text{Ci}\Big((2k+1)\pi x\Big)=\int_x^\infty (-1)^{\lfloor t+\frac12 \rfloor}\, \frac{dt}{t} $

$ =\int_x^{\lfloor x\rfloor +\frac12} (-1)^{\lfloor t+\frac12 \rfloor}\, \frac{dt}{t}+\sum_{n=\lfloor x\rfloor+1}^\infty \int_{n-\frac12}^{n+\frac12} (-1)^{\lfloor t+\frac12 \rfloor}\, \frac{dt}{t}$ $=(-1)^{\lfloor x+\frac12 \rfloor} \, \log\frac{\lfloor x \rfloor +\frac12}{x}+\underbrace{\sum_{n=\lfloor x\rfloor+1}^\infty (-1)^n \, \log\frac{n+\frac12}{n-\frac12}}_{S} $
where the last series $ S=2\log\left(2\cdot \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)-\sum_{n=1}^{\lfloor x \rfloor} (-1)^n \, \log\frac{n+\frac12}{n-\frac12} $

in turn $ -\sum_{n=1}^{\lfloor x \rfloor} (-1)^n \, \log\frac{n+\frac12}{n-\frac12}=\sum_{n=1}^{\lfloor x\rfloor} (-1)^n \, \log\left(n-\frac12\right)-\sum_{n=1}^{\lfloor x \rfloor} (-1)^n \, \log\left(n+\frac12\right) $

$ =-\sum_{n=0}^{\lfloor x \rfloor-1} (-1)^n \, \log\left(n+\frac12\right)+\log\left(\frac12\right)-$ $\sum_{n=0}^{\lfloor x \rfloor-1} (-1)^n \, \log\left(n+\frac12\right)-(-1)^{\lfloor x \rfloor}\log\left(\lfloor x \rfloor+\frac12\right) $.

$ S=2\log\left(\sqrt{2}\,\, \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)-2 \sum_{n=0}^{\lfloor x\rfloor-1} (-1)^n \log\left(n+\frac12\right)-(-1)^{\lfloor x \rfloor}\log\left(\lfloor x \rfloor+\frac12\right) $

$ R=-(-1)^{\lfloor x+\frac12 \rfloor}\log x+2\log\left(\sqrt{2}\,\, \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)-$ $2\sum_{n=0}^{\lfloor x\rfloor-1} (-1)^n \log\left(n+\frac12\right)+\underbrace{\left((-1)^{\lfloor x+\frac12\rfloor}-(-1)^{\lfloor x \rfloor}\right) \log\left(\lfloor x\rfloor+\frac12\right)}_{T} $

where $ T=\left((-1)^{\lfloor x+\frac12\rfloor-\lfloor x \rfloor}-1\right) (-1)^{\lfloor x \rfloor} \log\left(\lfloor x\rfloor+\frac12\right) $ gives an additonal summand iff $ \left\lfloor x-\frac12\right\rfloor > \lfloor x \rfloor-1 $.

Thus $ R=-(-1)^{\lfloor x+\frac12 \rfloor}\log x+2\log\left(\sqrt{2}\,\, \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)-$ $2\sum_{n=0}^{\lfloor x-\frac12 \rfloor} (-1)^n \log\left(n+\frac12\right) $.

$ \frac{4}{\pi}\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\, \text{Ci}\Big((2k+1)\pi x\Big)=(-1)^{\lfloor x+\frac12 \rfloor}\log x-2\log\left(\sqrt{2}\,\, \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)+$ $2\sum_{n=0}^{\lfloor x-\frac12 \rfloor} (-1)^n \log\left(n+\frac12\right) $

$ =(-1)^{\lfloor x+\frac12 \rfloor}\log(\pi x)-2\log\left(\sqrt{2\pi}\,\, \frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right)+$ $2\sum_{n=0}^{\lfloor x-\frac12 \rfloor} (-1)^n \log\left((2n+1)\frac{\pi}{2}\right) $


algebra
polynomial
quadratics
AMC
USA(J)MO
USAMO
quadratic formula

[quote="darkquantum"]Solve for all x:

x^4+2ix^3-4x^2-4ix+4=0.

where i^2=-1.[/quote]

(x:^2:-2):^2:+2ix(x:^2-2)=(x:^2:-2)(x:^2:+2ix-2)  The first factor gives x=:pm::rt2:. Using the quadratic formula on the second factor, we find x=-i:pm:(:sqrt:(-4+8)/2=-i:pm:1.

Yea, that would work, but how did u see the factoring?

Just try to deal with imaginary and real parts separately, and then you can get that factorization.

Ahhh,, ok.

Anyone do it the dividing way?

Separately dealing with real and imaginary parts is fine as part of your whole "brainstorming" process, I guess. But that doesn't actually result in a proof, because x is a complex variable.

yeah...I realized that...still, it gave me a factorization.

Yes, but if you are asked to come up with factors on the USAMO (not sure why that would be), you can just pull them out of a hat and say ``Hey look! They work.," and they have to give you full credit.


Let $x,y,z>0$. Prove that:
$\frac{(x+y)(y+z)(z+x)}{xyz}+\frac{xyz}{(x+y)(y+z)(z+x)}\geq\frac{x+z}{y+z}+\frac{y+z}{x+y}+\frac{1}{2}$

[quote="Hawk Tiger"][quote="Yu-Dong Wu"]In fact, we can prove the inequality as follows.

$\frac{(x+y)(y+z)(z+x)}{xyz}+\frac{xyz}{(x+y)(y+z)(z+x)}\geq\frac{x+z}{y+z}+\frac{y+z}{x+y}+\frac{49}{8}$

Proof. For $\frac{(x+y)(y+z)(z+x)}{xyz}+\frac{xyz}{(x+y)(y+z)(z+x)}-\frac{x+z}{y+z}-\frac{y+z}{x+y}-\frac{49}{8}$

$=(8y^4zx+16xyz^4-17xy^2z^3-9x^2y^3z-17xy^3z^2+8x^4yz-17x^3yz^2-17x^2yz^3-9x^3y^2z-42x^2y^2z^2+16x^3z^3+8x^4y^2+8x^2z^4+8y^2z^4+16y^3z^3+8y^4x^2+8y^4z^2+16x^3y^3+8x^4z^2)/[8xyz(x+y)(y+z)(z+x)]$,

Now we only need to prove the following inequality.

$8y^4zx+16xyz^4-17xy^2z^3-9x^2y^3z-17xy^3z^2+8x^4yz-17x^3yz^2-17x^2yz^3-9x^3y^2z-42x^2y^2z^2+16x^3z^3+8x^4y^2+8x^2z^4+8y^2z^4+16y^3z^3+8y^4x^2+8y^4z^2+16x^3y^3+8x^4z^2\geq 0$

It's obvious that the above inequality is symmetric. Hence, there is no harm in supposing $x \geq y \geq z$. Setting $x=z+u+v$, $y=z+u$($u\geq 0$,$v\geq 0$), then the above inequality is equivalent to

$378z^3u^3+410z^2u^4+190zu^5+126z^4u^2+126z^4uv+110v^2z^4+102z^3v^3+24v^4z^2+428zu^3v^2+615v^2z^2u^2+393v^2z^3u+475zu^4v+820z^2u^3v+567z^3u^2v+167zu^2v^3+205uv^3z^2+24zuv^4+32u^6+96u^5v+8u^2v^4+48u^3v^3+104u^4v^2\geq 0$.

The last inequality holds obviously.
Thus, the proof is completed.[/quote]How many does this solution take you?
how patient you are![/quote]

I do the calculation with the famous mathematics software Maple 9.0.


Express $ 3.\bar{7}$ as a common fraction.

Let's take care of the $ \bar{7}$ part first, shall we? Well, we know that such a case of $ \bar{x}$ represents $ \frac{x}{9}$, so the fractional part is $ \frac{7}{9}$. Now that we know that our denominator would be $ 9$, we find what $ 3$ $ 9$ s would be, for the whole part. $ 3\times9\equal{}27$, so we have $ 27$ $ 9$ s in the whole part. Adding the two numerators together, $ 7\plus{}27\equal{}34$, making our final answer $ \frac{34}{9}$.

NOTE: If someone would clarify on why $ \bar{x}\equal{}\frac{x}{9}$, that would be very helpful of you. I forgot why.  :blush:

Let x be .aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa where a is a digit.

$ x\equal{} \bar{a}$

Multiply by 10:

$ 10x\equal{} a. \bar{a}$

This works because there are an inifnite number of a's.
Subtracting the first equation from the second:

$ 9x\equal{}a$

$ x\equal{}\frac{a}{9}$

Alternatively, recall that in general, $ \frac {\overline{a_1a_2\ldots a_mb_1b_2\ldots b_n} \minus{} \overline{a_1a_2\ldots a_m}}{\underbrace{99\ldots99}_n\underbrace{00\ldots00}_m} \equal{} 0.a_1a_2\ldots a_m\overline{b_1b_2\ldots b_n}$ (It can be proven many ways, including one with the technique nt811581147 mentioned above).

Thus, $ 3.\overline7 \equal{} 10\cdot0.3\overline7 \equal{} 10\cdot\frac{37 \minus{} 3}{90} \equal{} \boxed{\frac {34}9}$.


Three distinct lines, all contained within a plane, separate that plane into distinct regions. What are all of the possible numbers of distinct regions of the plane that could be separated by any such three lines?

For $ n$ distinct lines, what is the largest number of distinct regions of the plane separated by those lines?

you can induct and see that the largest is $ \frac{n(n \plus{}1}{2} \plus{} 1$. The smallest occurs only if all the lines are parallel, making $ n \plus{} 1$ regions

This was from Engel.
[hide="Key lemma"]
Each intersection point is the highest point of exactly 1 region (if not so, you can rotate the whole figure so it is so). There is 1 more step to the problem.[/hide]
[hide="Rest of solution"] Draw a horizontal line above the intersections of all points, from here we find there's n+1 regions without a highest point, so the answer is $ {n \choose 2} \plus{} n\plus{}1 \equal{}$[/hide]

I much prefer the solution in Engel, for
[hide="two reasons..."] It encodes in its form the information that the first few terms are powers of $ 2$ (for obvious reasons), and it generalizes in the most obvious possible way to $ d$ dimensions (whereas the inductive approach is no help at all).  That is, if you get the answer in the nice form, which is
[hide="this"] $ {n \choose 2} \plus{} {n \choose 1} \plus{} {n \choose 0}$ [/hide] [/hide]

[quote="dgreenb801"]This was from Engel.
[hide="Key lemma"]
Each intersection point is the highest point of exactly 1 region (if not so, you can rotate the whole figure so it is so). There is 1 more step to the problem.[/hide]
[hide="Rest of solution"] Draw a horizontal line above the intersections of all points, from here we find there's n+1 regions without a highest point, so the answer is $ {n \choose 2} \plus{} n \plus{} 1 \equal{}$[/hide][/quote]
What is "highest point", exactly?

In an appropriate coordinate system, it means "point with the highest $ y$-coordinate."  The only subtlety that can arise is when a line is parallel to the $ x$-axis, but you can change coordinates so that this is not the case for any of a finite set of lines.

I see (how this is true from my drawing of case n=4)... but how do we prove the lemma and the fact that there are n+1 regions if we draw a line above all the intersections?
Also, what's the generalization to the d-dimension (is it $ \sum_{k\equal{}0}^d \binom{n}{k}$? if so, proof?)? Thanks.

Yes, it is.

Outline of an inductive argument: In the $ d^{th}$ dimension, we start with $ 1$ region. Then, after drawing $ k \minus{} 1$ objects, when we draw the $ k^{th}$ object, it will divide each previously drawn object into $ \sum_{i \equal{} 0}^{d \minus{} 1}\binom{k \minus{} 1}i$ $ (d \minus{} 1)^{th}$ dimension objects, each giving one more region, so the total number of regions is $ 1 \plus{} \sum_{k \equal{} 1}^n\sum_{i \equal{} 0}^{d \minus{} 1}\binom{k \minus{} 1}i \equal{} 1 \plus{} \sum_{i \equal{} 0}^{d \minus{} 1}\sum_{k \equal{} 1}^n\binom{k \minus{} 1}i \equal{} 1 \plus{} \sum_{i \equal{} 0}^{d \minus{} 1}\binom n{i \plus{} 1}\equal{}\sum_{i \equal{} 0}^d\binom ni$ (where we have used the inductive hypothesis and the Hockeystick Identity).

[quote="dgreenb801"]This was from Engel.

Each intersection point is the highest point of exactly 1 region (if not so, you can rotate the whole figure so it is so). There is 1 more step to the problem.
 Draw a horizontal line above the intersections of all points, from here we find there's n+1 regions without a highest point, so the answer is $ {n \choose 2} \plus{} n \plus{} 1 \equal{}$[/quote]

Same question: how do you prove there's n + 1 regions without a highest point? What is a horizontal line above the intersections of all points? how do points intersect?

for 3D, we are dividing by planes, and for d dimensions, we are dividing by d - 1 dimensional objects right? (and I think thats what math154 meant by objects?).

Again, the trick is to use a coordinate system such that none of the lines in question are parallel to a coordinate axis.  That means that all $ n$ lines extend upward, so a horizontal line above every intersection point intersects each of the $ n$ lines exactly once.  The $ n \plus{} 1$ regions between these lines open upward infinitely, hence have no highest points.  It is much easier to convince yourself of this via a diagram.

Note that [i]we have now reduced this to the $ 1$-dimensional problem.[/i]  Of these $ n \plus{} 1$ regions, all but one has a "rightmost side."  This suggests how to approach the $ d$-dimensional problem by inducting on $ d$ instead of on $ n$.

How about the maximum number of intersections?

[quote="hurdler"]each intersection is uniquely determined by a pair of lines. There is a one to one correspondence between the number of intersections and the number of pairs of lines, so we have $ \dbinom n2$ intersections at most.[/quote]
Oh, I meant in dimension $ d$... the formula wouldn't be always the same, would it?

It's precisely $ {n \choose d}$.  In dimension $ d$, it is $ d$ generic $ d\minus{}1$-hyperplanes that determine a point (for example, $ 3$ planes in dimension $ 3$).


ceiling function
inequalities
combinatorics unsolved
combinatorics

There are $ 1993$ towns in a country, and at least $ 93$ roads going out of each town. It's known that every town can be reached from any other town. Prove that this can always be done with no more than $ 62$ transfers.

We will prove more general result. Let $ G$ be a connected graph with the minimum degree of vertex at least $ d$ and the longest distance between two vertices at least $ D$ (the diameter of $ G$). Then
$ |G| \geq \lceil \frac{D\plus{}1}{3} \rceil (d\minus{}2) \plus{} D \plus{} 1$.
Indeed, consider the two vertices with distance $ D$ and the path connecting them. It's clear that if we pick two vertices of this path which are at least $ 3$ edges apart then they don't have a common neighbour because otherwise we could've shorten the path. For the same reason no two non consecutive vertices of this path are connected. So if we name the vertices in path by $ 1, 2, ... D\plus{}1$ then every vertex $ 1, 4, 7, 10, ...$ gives us at least $ d\minus{}2$ new vertices which don't belong to the path. After adding the vertices in the path our inequality follows.

Suppose that there are two towns which require at least $ 63$ transfers, i.e. $ D \geq 63$. Since $ d \geq 93$ we obtain the bound:
$ 1993 \geq \lceil \frac{63\plus{}1}{3} \rceil (93\minus{}2) \plus{} 63 \plus{} 1 \equal{} 22 \cdot 91 \plus{} 64 \equal{} 2066$
which clearly can't hold.

It seems to me that we can lower $ d$ to $ 90$ instead of $ 93$...

[quote="TomciO"] For the same reason no two non consecutive vertices of this path are connected. [/quote]

I assume you don't have to use this, right?

I have to use this because otherwise some of that $ d\minus{}2$ vertices which are produced by $ 1, 4, 7, ...$ could appear somewhere in the path.


[hide="Outline"] Using the remainder theorem, plug in a $ 2$ for all $ x$'s. If you get 0 for your answer, then $ x\minus{}2$ is indeed a factor since it doesnt have a remainder when it is divided by the polynomial. If you get any other number but 0, then x-2 is not a factor[/hide]

[quote="salahcool"]Use the factor theorem to decide if (x - 2) is a factor of

f(x) = x^5 - 2x^4 + 3x^3 - 6x^2 - 4x + 8.

salah cool

Regards[/quote]

Sinc your factor is $ (x\minus{}2)$, take $ 2$ out of it and plug it into $ f(x) \equal{} x^5 \minus{} 2x^4 \plus{} 3x^3 \minus{} 6x^2 \minus{} 4x \plus{} 8$ to get $ f(x) \equal{} (2^5 )\minus{} (2*2^4) \plus{} (3*2^3) \minus{} (6*2^2) \minus{} (4*2 \plus{} 8)$ so... $ f(x) \equal{} (32)\minus{} (32) \plus{} (24) \minus{} (24) \minus{} 4*2 \plus{} 8$ =$ 0$ so $ (x\minus{}2)$ is a factor.

Alterantivly you could use synthetic division

On a more general note,

For any non-constant function $ f(x)$, it can be written as

$ f(x) \equal{} (x\minus{}a)q(x) \plus{} r(x)$

where $ deg \ q > deg \ r$.

Clearly,

$ f(a) \equal{} (a\minus{}a)q(a) \plus{} r(a) \equal{} r(a)$

So if $ f(a) \equal{} 0$, then $ x\minus{}a$ is a factor of $ f(x)$

That might be the remainder theorem: the remainder when $ p(x)$ is divided by $ x\minus{}a$ is $ p(a)$

[quote="salahcool"]Use the factor theorem to decide if (x - 2) is a factor of

f(x) = x^5 - 2x^4 + 3x^3 - 6x^2 - 4x + 8.

salah cool

Regards[/quote]

if $ x\minus{}2$ divides $ f(x)$, 2 will be a root of $ f$. Testing it you´ll see that it indeed is a root.Hence it is a factor.  :wink:

I agree with AwesomeToad, I don't know what the factor theorem is.

Factor theorem and Remainder theorem are very closely related. The factor theorem just states, that if $ f(a) \equal{} 0$, then $ x\minus{}a$ is a factor of $ f(x)$.

[quote="modularmarc101"]$ deg \ q > deg \ r$.[/quote]Not quite correct. What if $ \deg(q)\equal{}n\minus{}1$? Then $ \deg(r)$ can be equal to $ n\minus{}1$?
Here is a link to one of my favorite proofs that I've written when I was young: [url=http://www.mathlinks.ro/viewtopic.php?p=344713#344713]Link[/url]


[quote="tiddlewink0"]How long in hours, after 12 noon, will it take for the hour hand of a clock to lie directly over its minute hand for the first time?[/quote]

This is a classic problem.
I bet you could use the search function and find like 3 entries.

[hide="Anyway, click here for the solution"]It's obvious that the desired time is between 1 and 2 o'clock.

So let the time be $ 1: x$, for some $ x\in [0,60]$. For every minute the minute hand moves,

the hour hand moves $ 0.5^{\circ}$
the minute hand moves $ 6^{\circ}$

Note that the hour hand has a $ 30^{\circ}$ "head start" since it begins at 1 and the minute hand begins at 12.

Now, we may set up our equation:

$ 30+0.5x = 6x$

$ 30 = 5.5x$

$ 60 = 11x$

$ x = \frac{60}{11}= 5\frac{5}{11}$

Hence, our desired time is $ 1: 5\frac{5}{11}$, so you need to wait $ \boxed{1\textbf{ hour and }5\frac{5}{11}\textbf{ minutes}}$[/hide]

EDIT: Didn't read the question.  :oops:

For some reason I got just 1 hour and 5 minutes.

That would be correct if the hour hand just jumped to the next hour after the minute hand completed a revolution, but by the time the minute hand has reached the 1, the hour hand has moved 2 and a half degrees. Vishalarul's answer iwould be right, if it was in hours.

grrrrrrrrrr

use the search button

the answer is 12/11 hours


algebra
polynomial
number theory proposed
number theory

Find all natural numbers $ (x,n)$ satisfying

$ x^{4}+21x+10=2^{n}$


invariant
geometry
geometric transformation
reflection
symmetry
projective geometry
geometry solved

first we define $(ABCD)=-1$ it means that $A,B,C,D$ Are harmonic points.in other word it means $\frac{AC}{AD}=\frac{CB}{BD}$.

let $\ell$ be a given line with $A,B,C$ three fixed points on it.
let $A^`,B^`,C^`$ be points such that $(AA^`BC)=-1, (BB^`AC)=-1,(CC^`AB)=-1$.prove:
$(AA^`B^`C^`)=-1, (BB^`A^`C^`)=-1,(CC^`A^`B^`)=-1$

ugly problem and ugly sulotion ,too.

let $AB=a,BC=b$.

$(CC'AB)=-1$ , $\frac{AC'}{C'B}=\frac{AC}{CB}=\frac{a+b}{b}$ therefore:$\frac{AC'}{a}=\frac{a+b}{a+2b},\frac{a}{C'B}=\frac{a+2b}{b}\rightarrow AC'=\frac{a(a+b)}{a+2b},C'B=\frac{ab}{a+2b}$

$(AA'BC)=-1$ , $\frac{AB}{BA'}=\frac{AC}{CA'},\frac{AB}{AC}=\frac{BA'}{CA'}=\frac{a}{a+b}$ therefore :$\frac{b}{CA'}=\frac{2a+b}{a+b},\frac{BA'}{b}=\frac{a}{2a+b} \rightarrow BA'=\frac{ab}{2a+b},A'C =\frac{b(a+b)}{2a+b}$

$(BB'AC)=-1$ , $\frac{AB}{BC}=\frac{AB'}{B'C}=\frac{a}{b}$ therefore $\frac{a-b}{b}=\frac{a+b}{B'C},\frac{a}{a-b}=\frac{AB'}{a+b}\rightarrow B'C=\frac{b(a+b)}{a-b}$

          $\frac{AC'}{C'A'}=\frac{\frac{a(a+b)}{a+2b}}{\frac{3a^2b+3ab^2}{(a+2b)(2a+b)}}$

          $\frac{AB'}{B'A'}=\frac{\frac{a(a+b)}{a-b}}{\frac{3ab(a+b)}{(a-b)(2a+b)}} $

therefore:$\frac{AC'}{C'A'}=\frac{AB'}{B'A'}=\frac{2a+b}{3b}$  IT MEANS THAT $ (AA'B'C')=-1$

the rest is like this. :D

The problem is very nice and I have three nice solutions. Here is the simplest one.

[color=blue][b]Problem.[/b] Let A, B, C be three points on a line l. Let A', B', C' be the harmonic conjugates of the points A, B, C with respect to the segments BC, CA, AB. Prove that the points A, B, C are the harmonic conjugates of the points A', B', C' with respect to the segments B'C', C'A', A'B'.[/color]

[i]Solution.[/i] Since harmonic conjugation is a projectively invariant notion, we can undertake a projective transformation which will map the point C to infinity. So we can simply assume that the point C is an infinite point. Then, the point C', being the harmonic conjugate of this point C with respect to the segment AB, will be simply the midpoint of this segment AB. On the other hand, since the point A' is the harmonic conjugate of the point A with respect to the segment BC, the point B must be the harmonic conjugate of the point C with respect to the segment AA'; since the point C is an infinite point, its harmonic conjugate B must therefore be the midpoint of the segment AA'. Hence, the point A' is the reflection of the point A in the point B. Similarly, the point B' is the reflection of the point B in the point A.

Now, the symmetry with respect to the midpoint C' of the segment AB maps the point A to the point B and the point B to the point A; thus, it also maps the reflection A' of the point A in the point B to the reflection B' of the point B in the point A. Thus, the point C' is the midpoint of the segment A'B'. On the other hand, the point C is the infinite point of the line l. Hence, the points C' and C divide the segment A'B' harmonically. In other words, the point C is the harmonic conjugate of the point C' with respect to the segment A'B'.

Similarly, by projecting the points A and B to infinity, we prove that the point A is the harmonic conjugate of the point A' with respect to the segment B'C', and that the point B is the harmonic conjugate of the point B' with respect to the segment C'A'.

Problem solved.

  Darij


This is kind of hard to explain so feel free to ask questions...

Whenever I get board, I look at the clock and try to find patterns.

For example, at 1:00 1*0=0

You can also use the fact that there are 3 numbers in 1:00, 1:01 and so on... if it is used in your equation, please bold it.

You can use addition, subtraction, multiplication, division, exponents and so on.

In each post, you should have at least 5 times and please don't double post.

i'll start it off... (going from 1:00 to 24:59)

1:00...1*0=0
1:01...1-1=0
1:02...[b]3[/b]-2-0=1
1:03...3^0=1
1:04...[b]3[/b]+1+0=4
1:05...5^0=1

..... I don't get it either... :( for a moment i thought it was using numbers in the digits to make an equation but that doesn't fit for all your cases. I have no idea then... and why did you bold the 3's?

[quote="Barnacle"]3:45 is my favorite time... can you figure out why?[/quote]

pythagorean triples

i kinda get it. so

1:16 = 6/[b]3[/b]=1+1 ????

or.. of course 1:16 could be 1=1^6 but thats just boring..

i kinda get it...

[quote="ToBe"]

1:16 = 6/[b]3[/b]=1+1 ????
[/quote]
I think that you have to use the numbers in the order that they are given.  You can't switch 6 and 1 around.  And i also think that you have to use the numbers given, you can't add an extra 3.  Am i right jli?

[quote="math92"]i get it now!!!

[quote="ToBe"]

1:16 = 6/[b]3[/b]=1+1 ????
[/quote]
I think that you have to use the numbers in the order that they are given.  You can't switch 6 and 1 around.  And i also think that you have to use the numbers given, you can't add an extra 3.  Am i right jli?[/quote]

he can use the 3 because there are 3 numbers

[quote="math92"]Oh, that makes sense...this is still confusing[/quote]

sorry about that


number theory
greatest common divisor
algebra solved
algebra

Find all infinite bounded sequences a_1,a_2,... of positive integers such that for all n > 2, a_n = (a_(n-1)+a_(n-2))/gcd(a_(n-1),a_(n-2)).

the only such sequence is (2,2,2,2........) let
g_n = gcd [a_n,a_(n+1)].then g_(n+1) divides both a_(n+1) and a_(n+2),
so it divides g_n*a_(n+2) - a_(n+1) = a_n as well.thus g_(n+1) divides both a_n and a_(n+1) so g_(n+1) divides g_n as well.
thus g_i form a non increasing sequence of positive integers and eventually a constant = gat this point a_i satisfy the recursion
ga_n = a_(n-1) + a_(n-2)
if g=1 then a_n=a_(n-1) + a_(n-2) > a_(n-1), so the sequence is increasing and unbounded .
if g>3 a_n =[a_(n-1) + a_(n-2)]/g <1/2{ a_(n-1) + a_(n-2)} < max[a_(n-1),a_(n-2)]
similarly  a_(n+1) < max[a_(n-1),a_n]  \leq  max[a_(n-2),a_(n-1)]
so that max[a_n,a_(n+1)] < max[a_(n-2),a_(n-1)]
therefore the maximum values if successive pairs of terms form an infinite decreasing sequence of positive integers .contradiction
thus g=2 and eventualy we have 2a_n = a_(n-1) + a_(n-2) or
a_n - a_(n-1) = -1/2[a_(n-1) - a_(n-2)] implies the a_i-a_(i-1) converges to zero and hence the a_i are eventually constant and this constant must be 2 as per the recurrence
now if a_n=a_(n+1)=2 for n>1 then gcd{a_(n-1),a_n} =gcd[a_(n-1),2] either equals 1 or 2.now
2= a_(n+1) = {a_n + a_(n-1)}/(gcd[a_(n-1),2)
implying that either a_(n-1) = 0 [impossible] or a_(n-1)=2
thus all a_i are equal to 2 as claimed
hope the solution is correct


probability
integration
function
probability and stats

The life times of a large batch of electric light bulbs are independently and identically distributed. The probability that the life time, $ T$ hours, of a given light bulb is greater than $ t$ hours is given by:

$ P(T>t) \equal{} \frac{1}{(1\plus{}kt)^{\alpha}}$,

where $ \alpha$ and $ k$ are constants, and $ \alpha >1$. Find the median $ M$ and the mean $ m$ of $ T$ in terms of $ \alpha$ and $ k$.

Nine randomly selected bulbs are switched on simultaneously and are left until all have failed. The fith failure occures at 1000 hours and the mean life time of all bulbs is found to be 2400 hours. Show that $ \alpha \approx 2$ and find the approximate value of $ k$. Hence estimate the probability that, if a randomly selected bulb is found to last $ M$ hours, it will last further $ m\minus{}M$ hours.

what have you done ?

the median $ M$ is satisfies $ P[t>M]\equal{}\frac{1}{2}$ and the mean $ m$ is given by $ \int_{t\equal{}0}^{\infty} P(T>t) \, dt$. Hence you can express $ m$ and $ M$ as functions of $ k$ and $ \alpha$. Then you can take the approximation $ M\equal{}1000$ and $ m\equal{}2400$, which gives you $ 2$ equations and $ 2$ unknowns: you can thus work out the corresponding values of $ \alpha$ and $ k$ - done -

More interesting should be to ask yourself how precise is your estimate of $ \alpha$ and $ k$ only based on $ 9$ observations ...


trigonometry
ARML
geometry
3D geometry
ratio
number theory
relatively prime

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=301536]Refer here for the full set of problems[/url]. Note, there are two forms (some problems appear in only one).

- Let $ \displaystyle f\left({\sqrt [3]{54 \plus{} 2\sqrt {675}} \minus{} \sqrt [3]{10 \plus{} \sqrt {108}}}\right) \equal{} a \plus{} \sqrt {b}$. Calculate $ a \plus{} b$ if $ f(x) \equal{} 2x \plus{} \frac {1}{x \plus{} f(x)}$.

- A spaceship is at $ (0,0,0)$. Every time it teleports, it moves to a lattice point one unit away. Let $ m$ be the number of paths the spaceship can take so that it takes 12 teleportations to reach $ (4,4,4)$, while avoiding the black holes at $ (1,1,1),(2,2,2),(3,3,3)$. Compute the sum of the digits of $ m$.

- Let $ \displaystyle z_k \equal{} \cos\left(\frac {2k\pi}{9}\right) \plus{} i\sin\left(\frac {2k\pi}{9}\right)$ for integer $ k$. What is $ (3 \minus{} z_2)(3 \minus{} z_4)\cdots(3 \minus{} z_{16})$?

- Consider a sequence defined as follows:
$ F_1 \equal{} F_2 \equal{} 1$
$ F_{n \minus{} 2} \plus{} 2F_{n \minus{} 1} \equal{} F_n \text{ for } n > 2$
Let $ \displaystyle{\sum_{n \equal{} 1}^{\infty}\frac {F_n}{10^n}} \equal{} R/G$ for relatively prime integers $ R$ and $ G$.  Compute $ R \plus{} G$.

- $ ABCDEF$ is a regular hexagon with side length 1, with point $ G$ randomly located on the hexagon's boundary or in its interior. What is the maximum value of $ AG^2 \plus{} BG^2 \plus{} CG^2 \plus{} DG^2 \plus{} EG^2 \plus{} FG^2$?

- Examine the figure in the pdf. $ AD$ is the angle bisector to side $ BC$ of triangle $ ABC$ and $ E$ is the midpoint of $ AD$. If $ AC \equal{} 10$, $ AB \equal{} 6.5$, and $ BC \equal{} 10.5$, compute the length of $ CE^2$.

- Consider the cubic $ x^3 \minus{} x^2 \plus{} 3x \plus{} 4 \equal{} 0$  with roots $ a, b, c$.  Let $ w \equal{} \tan^{ \minus{} 1}{a} \plus{} \tan^{ \minus{} 1}{b} \plus{} \tan^{ \minus{} 1}{c}$. Suppose the positive value of $ \sec(w) \equal{} \frac {\sqrt {m}}n$ where $ m$ is not divisible by the square or any prime. Calculate $ m \plus{} n$.

- If $ 6\tan^{ \minus{} 1}x \plus{} 4\tan^{ \minus{} 1}(3x) \equal{} \pi$, compute $ x^2$. (Yes, this is from ARML)

- Let $ a,b,c$ be positive integers with $ a$ odd. Determine $ a \plus{} b \plus{} c$ if
$ 2b \minus{} 6a \plus{} 8c \plus{} a \plus{} 2b \minus{} 5ab \equal{} \minus{} 396$
$ 8bc \plus{} 2b^2 \minus{} 3a^2 \plus{} 4ac \equal{} 887$

[quote="Caelestor"]- Consider a sequence defined as follows:
$ F_1 \equal{} F_2 \equal{} 1$
$ F_{n \minus{} 2} \plus{} 2F_{n \minus{} 1} \equal{} F_n \text{ for } n > 2$
Let $ \displaystyle{\sum_{n \equal{} 1}^{\infty}\frac {F_n}{10^n}} \equal{} R/G$ for relatively prime integers $ R$ and $ G$.  Compute $ R \plus{} G$.[/quote]
[hide="Solution"]
\[ S \equal{} \frac {F_1}{10^1} \plus{} \frac {F_2}{10^2} \plus{} \frac {F_3}{10^3} \plus{} \cdots\implies 20S \equal{} 2F_1 \plus{} \frac {2F_2}{10^1} \plus{} \frac {2F_3}{10^2} \plus{} \frac {2F_4}{10^3} \plus{} \cdots\]\[ \implies 21S \equal{} 2F_1 \plus{} \frac {F_3}{10^1} \plus{} \frac {F_4}{10^2} \plus{} \cdots\]
Also, $ 100S \equal{} 10F_1 \plus{} F_2 \plus{} \frac {F_3}{10^1} \plus{} \frac {F_4}{10^2} \plus{} \cdots$.

Hence, $ 100S \minus{} 21S \equal{} 79S \equal{} 9 \implies S \equal{} \frac {9}{79}$.

Answer: 88[/hide]

[quote="Caelestor"]- Consider the cubic $ x^3 \minus{} x^2 \plus{} 3x \plus{} 4 \equal{} 0$  with roots $ a, b, c$.  Let $ w \equal{} \tan^{ \minus{} 1}{a} \plus{} \tan^{ \minus{} 1}{b} \plus{} \tan^{ \minus{} 1}{c}$. Suppose the positive value of $ \sec(w) \equal{} \frac {\sqrt {m}}n$ where $ m$ is not divisible by the square or any prime. Calculate $ m \plus{} n$.[/quote]
[hide="Solution"]Letting $ \arctan a \equal{} x, \arctan b \equal{} y, \arctan c \equal{} z$, we have $ \tan x \equal{} a, \tan y \equal{} b, \tan z \equal{} c$.

We have $ \tan w \equal{} \tan(x \plus{} y \plus{} z)$, and applying tangent sum formula twice, $ \tan(x \plus{} y \plus{} z) \equal{} \frac {a \plus{} b \plus{} c \minus{} abc}{1 \minus{} (ab \plus{} bc \plus{} ca)} \equal{} \minus{} \frac {5}{2}$.

Noting that $ \sec^2 w \equal{} 1 \plus{} \tan^2w$, $ \sec w \equal{} \sqrt {1 \plus{} \frac {25}{4}} \equal{} \frac {\sqrt {29}}{2}$.

Answer: 31[/hide]

Hints:

- Let $ \displaystyle f\left({\sqrt [3]{54 \plus{} 2\sqrt {675}} \minus{} \sqrt [3]{10 \plus{} \sqrt {108}}}\right) \equal{} a \plus{} \sqrt {b}$. Calculate $ a \plus{} b$ if $ f(x) \equal{} 2x \plus{} \frac {1}{x \plus{} f(x)}$. [b]Can you simplify the cubic roots?[/b]

- Examine the figure in the pdf. $ AD$ is the angle bisector to side $ BC$ of triangle $ ABC$ and $ E$ is the midpoint of $ AD$. If $ AC \equal{} 10$, $ AB \equal{} 6.5$, and $ BC \equal{} 10.5$, compute the length of $ CE^2$.  [b]You might want to draw an extra line in the figure.[/b]

- Let $ a,b,c$ be positive integers with $ a$ odd. Determine $ a \plus{} b \plus{} c$ if
$ 2b \minus{} 6a \plus{} 8c \plus{} a \plus{} 2b \minus{} 5ab \equal{} \minus{} 396$
$ 8bc \plus{} 2b^2 \minus{} 3a^2 \plus{} 4ac \equal{} 887$
[b]Substitutions could make this problem considerably easier.[/b]

[quote="Ihatepie"]I'm almost sure I got that first one right, but it apparently wasnt. What is the real answer?[/quote]

[hide="Answer"]
Agree that the expression inside the parentheses is just 2 (simplify the cube roots to see this). From there, it's easy to solve the infinite fraction, and we have $ 1\plus{}\sqrt{10} \implies \boxed{11}$.
[/hide]

[quote="Caelestor"]Examine the figure in the pdf. $ AD$ is the angle bisector to side $ BC$ of triangle $ ABC$ and $ E$ is the midpoint of $ AD$. If $ AC \equal{} 10$, $ AB \equal{} 6.5$, and $ BC \equal{} 10.5$, compute the length of $ CE^2$.  [b]You might want to draw an extra line in the figure.[/b][/quote]
Hehe, I figured there was something like that, but I went ahead and spent 15 minutes Stewart bashing anyway.

Stewart's on an AMC 12? Doubt it. We're not as brilliant as you  :oops:  :P

[quote="Caelestor"][quote="Ihatepie"]I'm almost sure I got that first one right, but it apparently wasnt. What is the real answer?[/quote]

[hide="Answer"]
Agree that the expression inside the parentheses is just 2 (simplify the cube roots to see this). From there, it's easy to solve the infinite fraction, and we have $ 1 \plus{} \sqrt {10} \implies \boxed{11}$.
[/hide][/quote] That's what I sent in, and you marked it right. For some reason I thought you marked it wrong  :D

Let $ \displaystyle z_k \equal{} \cos\left(\frac {2k\pi}{9}\right) \plus{} i\sin\left(\frac {2k\pi}{9}\right)$ for integer $ k$. What is $ (3 \minus{} z_2)(3 \minus{} z_4)\cdots(3 \minus{} z_{16})$?

[hide="Solution"]

Consider $ f(x) \equal{} (x\minus{}z_2)(x\minus{}z_4)...(x\minus{}z_{16})(x\minus{}z_{18})$

The specifications that are placed on $ z_i$ make them the 9th roots of unity. Thus, $ f(x) \equal{} x^9 \minus{} 1$

Plugging in 3 for x yields

$ f(3) \equal{} (3\minus{}z_2)(3\minus{}z_4)...(3\minus{}z{16})(3\minus{}z_{18}) \equal{} 3^9 \minus{} 1 \equal{} 19682$

But f(x) is $ (3\minus{}z_{18}) \equal{} 2$ times the answer we want.

$ \frac{19682}{2} \equal{} \boxed{9841}$

Darn, I would not have gotten this problem had I not seen another just like it.
[/hide]

[hide]Generally, when you are going to square stuff when the same sum, you want to maximize one of the values. ($ 2^2$ is more than $ 1^2 \plus{} 1^2$, for example.

In this case, that means having G be on one of the points, lets say A, for example.

So, $ AG \equal{} 0$, $ BG$ and $ FG \equal{} 1$ because they are both along an edge. $ DG \equal{} 2$ when you realize that the hexagon is made up of equalateral triangles. You can use the pythagorean theorem and equilateral triangle to find that $ CG^2$ and $ EG^2 \equal{} 3$

So, $ 0 \plus{} 2(1) \plus{} 4 \plus{} 2(3) \equal{} 12$

I believe that this is the maximum.[/hide]

For the geometry one with the angle bisector and the bisector of the angle bisector, the way I did it was draw a parallel line through the midpoint of AD, creating a similar triangle with ratio 1/2 to ABC.  I then created a right triangle by drawing a line perpendicular to BC, and letting it intersect our parallel line at a right angle.  The rest is just Heron's and then the Pythagorean Theorem.  The computations involved take about 30 seconds to a minute.

Its funny to see everyone talk about these questions...

For number 24 on V.1 (its 25 on V.2), I added both equations together and actually managed to factor the entire left side of the equation with all the variables, getting $ (3a\minus{}b\plus{}4c\minus{}1)(a\plus{}2b\plus{}2) \equal{} (163)(3)$. Since they are all positive integers, I figured $ a\plus{}2b\equal{}161$ and $ 3a\minus{}b\plus{}4c\equal{}4$. So I thought maybe I can add $ x(a\plus{}2b) \plus{} y(3a\minus{}b\plus{}4c) \equal{} k(a\plus{}b\plus{}c)$ for some integer $ x, y,$ and $ k$, but there doesn't exist any integers $ x$ or $ y$ that does this.

Was I on the right track at all?

[quote="Elefont"]For number 24 on V.1 (its 25 on V.2), I added both equations together and actually managed to factor the entire left side of the equation with all the variables, getting $ (3a \minus{} b \plus{} 4c \minus{} 1)(a \plus{} 2b \plus{} 2) \equal{} (163)(3)$. Since they are all positive integers, I figured $ a \plus{} 2b \equal{} 161$ and $ 3a \minus{} b \plus{} 4c \equal{} 4$. So I thought maybe I can add $ x(a \plus{} 2b) \plus{} y(3a \minus{} b \plus{} 4c) \equal{} k(a \plus{} b \plus{} c)$ for some integer $ x, y,$ and $ k$, but there doesn't exist any integers $ x$ or $ y$ that does this.

Was I on the right track at all?[/quote]

Close. You're right that adding them up can create an easy factorization; unfortunately, yours is a tad off.

Ok so I think it actually factors into $ (\minus{}3a\plus{}b\plus{}4c\plus{}1)(a\plus{}2b\plus{}2)\equal{}(29)(17)$ but my original method still doesn't work and I don't know how to determine which factor equals $ 17$ and which equals $ 29$.

Can I have a hint?  :blush:

[quote="Elefont"]Ok so I think it actually factors into $ ( \minus{} 3a \plus{} b \plus{} 4c \plus{} 1)(a \plus{} 2b \plus{} 2) \equal{} (29)(17)$ but my original method still doesn't work and I don't know how to determine which factor equals $ 17$ and which equals $ 29$.

Can I have a hint?  :blush:[/quote]

How about $ (b\minus{}3a\plus{}4c)$ instead? :D


There is an interesting thread "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11090]0 a perfect square?[/url]" that topper posted on the Mathcounts forum. The issue he ran into was answering a Mathcounts problem, in which the definition of "perfect square" seemed to be at issue.

In my HarperCollins Dictionary of Mathematics, "perfect square" is defined as "an integer that is the square of another integer, such as 1, 4, 9, 16, . . . " The definition also has a cross-reference to the term "figurate number." I wonder if one basis for considering zero NOT to be a perfect square, etymologically, is that it does form a figure. (Hmmm . . . . )

The textbook Elementary Number Theory by Gareth A. Jones and J. Mary Jones defines a perfect square as "an integer of the form m=n^2, where n is an integer." No examples are given. Some of the discussion of factorization following that definition would seem to imply that zero is NOT a perfect square, but that is not stated explicitly one way or the other.

What do the sources you have at hand say?

according to http://en.wikipedia.org/wiki/Perfect_square

they say that a perfect square is a positive integer, which agrees with mathworld.

Hmmm.....interesting observation....My textbooks say the exact thing that you've mentioned in your post..A perfect square is a number in the form a such that b * b = a , where b is an integer.......

If I remember correctly it is actually a topic of dispute among mathematicians.  I believe, however, that the general concensus is that it is not a perfect square.

0 is not technically an integer; it's in its own category, even though it's included under integers. So, when you square 0, you aren't squaring an integer, and the result is also not an integer, which is why it isn't included under perfect squares.

This topic is like a year old. I don't know why you revived them but don't do it! ;)

It's like spamming.

[quote="JesusFreak197"]0 is not technically an integer[/quote]Why do you say that?

Sorry, a correction and rephrasing:

0 is classified as an integer, but it's different than any other integers, since it is both non-negative and non-positive. Since I now know that it's an integer, I'm a little less certain on whether or not it's considered a perfect square, but I still don't think it is...

Maybe what defines a perfect square is that it is the square of both a positive and negative integer? That would eliminate zero. What actually seems more likely is simply that a perfect square must be positive, and zero is not positive. So, even though $0^2$ is defined, since it is not positive, it is not considered a perfect square.


logarithms
search
number theory
greatest common divisor
least common multiple
number theory theorems

Can anyone provide an outline or some ideas of the proof (or a reference available on the net in which I can find such details) for the following relation, established by Matiyasevitch in 1985:

$ \pi \equal{} lim_{n\to\infty}\sqrt {\frac {6logF_1F_2...F_n}{log[F_1,F_2,...,F_n]}}$

Where the log is the logarithm to the base 10 and [x,y] is the Least Common Multiple of x and y, the $ F_n$ obviously are the fibonacci numbers.
Thanks for any information.

Yes, thanks.
Actually, I knew about this one, but it's not available online (or at least I can't find it), I was hoping somebody could save me a trip to the library.
It looks like it can be written as:

$ \frac{\pi^2}{6}\equal{}1\plus{}\frac{log(F_1,F_2,...,F_\infty)}{log[F_1,F_2,...,F_\infty]}$

with (x,y)=gcd of x and y, which I find even cooler than the previous version.

[quote="nsato"]Cool formula!  A search on Google turned up the following reference:

Matiyasevich, Yuri V. and Guy, Richard . "A new formula for $ \pi$" The American Mathematical Monthly 93.8 (1986): 631-635[/quote]

 is this article avaliable online or someone has it and could share ?


convert glutaric acid to 5-hydroxypentaniocacid.

is my method right? :?:

on heating the calcium salt of glutaric acid, do you get cyclobutanone or cyclopentanone?
[hide]or or you dont get either of them?  :maybe:  :o [/hide]

oops . I'm sorry , .
It's my mistake.
I'll correct it.
one mo.
Can we change , pentandioic acid to 5-hydroxy pentandioic acid.?

Your mechanism is incorrect. The first step should be lactone formation, not a cycloalkanone.

But isn't Ca in the medium ? our sir told . Ca C bond is significantly covalent and hence favours homolytic cleavage . so ,it results in the fromation of , CACO3 and the cyclic ketone . Lactone is formed if we use , Na salt of acid . Is this right? :?:

In that case, why don't you use the sodium salt right from the start? You would save one step.

I'm sorry , I'm thinking rubbish , Na salts don't give lactones.

 :blush:  .

when heated with CaO , and NaOH , alkane is produced , na? I was thinking some dum thing.

laugh out loud .

I'm thinking that I should get valero lactone and add water to get teh  result .
but getting it is difficult. I've tried different ways . I'll post them tomorrow. everywhere , so ,step becomes a problem.


algebra
system of equations
geometry unsolved
geometry

Each side of a polygon is extended to twice its length at one of its ends going around the polygon in a counterclockwise direction.If the newly obtained ends of segments form a regular polygon, prove that the original polygon is also regular.

Nice problem...I can't really produce synthetic solution :) Let $F_1$ be our initial n-gon, and $F_2$ the regular n-gon we've obtained. Consider this configuration on a complex plane. Let's put the center $O$ in the center of $F_2$ and axis such that vertices of $F_2$ have form $\epsilon, \epsilon^2,...,\epsilon^n$ where $\epsilon$ is the n-th root of unity. Now let $F_1$ have vertices in $z_1,z_2,...,z_n$ (we label them counterclockwise) Now assume we have  a side $z_1z_2$, the vertice created by dobling this side will have then the form $z_2+(z_2-z_1)=2z_2-z_1$ consecutively we may write system of equations: \[ 2z_2-z_1=\epsilon \] \[ 2z_3-z_2=\epsilon^2 \] \[ ....... \] \[ 2z_1-z_n=\epsilon^n \] this system has a unique solution in $C$. This means that for our $F_2$ there exists only one $F_1$ which "doubled" gives $F_2$. O.K.... but if $F_2$ is regular we are able to find such regular $F_1$ which "doubled" gives $F_2$ (easy to prove) and as for given $F_2$ , $F_1$ is unique then $F_1$ has to be regular :)


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$ Prove that:
\[ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} ac} \plus{} \frac {1}{1 \minus{} bc}\geq\frac {9}{3 \minus{} ab \minus{} ac \minus{} bc}
\]

One step Cauchy?? ab,bc, and ac are all less than 1.

[quote="arqady"]$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$
[/quote]
Is this condition necessary? This inequality is obvious using Andreescu.

Yes, but you have to have ab, bc and ac less than 1 to use Andreescu (or Cauchy, whichever you call it). The denominators can't be negative. If $ a^2\plus{}b^2\plus{}c^2\equal{}1$, then a,b, and c are all positive and less than 1.

Yes, it's obvious. Here is much better inequality from A. Khrabrov.
Let $ \{a,b,c,d\}\subset[1,2].$ Prove that:
\[ 16(a^2\plus{}b^2)(c^2\plus{}d^2)\leq25(ac\plus{}bd)^2\]

[quote="Bovy11"]Excuse me but what is Andreescu??Titu Andreescu?
I used Google but I'm not sure.[/quote]
$ \frac{a_1^2}{x_1} \plus{} \frac{a_2^2}{x_2} \plus{} ... \plus{} \frac{a_n^2}{x_n} \geq \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)^2}{x_1\plus{}x_2\plus{}...\plus{}x_n}$

for $ x_1,x_2,...,x_n$ reals and $ a_1,a_2,...,a_n$ positive reals

[quote="basted"]
$ \frac {a_1^2}{x_1} \plus{} \frac {a_2^2}{x_2} \plus{} ... \plus{} \frac {a_n^2}{x_n} \geq \frac {(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2}{x_1 \plus{} x_2 \plus{} ... \plus{} x_n}$

for $ x_1,x_2,...,x_n$ reals and $ a_1,a_2,...,a_n$ positive reals[/quote]
Are you sure? :wink:

I thought $ x_1,x_2,x_3...x_n$ had to be positive because once allnames gave a proof and can hang said it was wrong because they weren't positive.
After looking it up, it seems I'm right, the denominators have to be positive but the numerators can be anything.

I used Cauchy

$ (a \plus{} b \plus{}c)^2 \equal{} \left (\sum_{cyc} a\sqrt{1\minus{}ab} \cdot \frac{1}{\sqrt{1\minus{}ab}} \right )^2 \le \left ( \sum_{cyc} a^2(1\minus{}ab) \right ) \left ( \sum_{cyc} \frac{1}{1\minus{}ab} \right )$
$ \sum_{cyc} \frac{1}{1\minus{}ab} \ge \frac{(a\plus{}b\plus{}c)^2}{\sum a^2 \minus{} \sum a^3b}$
$ \frac{(a\plus{}b\plus{}c)^2}{1 \minus{} (a^3b\plus{}b^3c\plus{}c^3a)} \ge \frac{9}{3 \minus{} (ab\plus{}bc\plus{}ca)}$
$ \frac{1\minus{}2(ab\plus{}bc\plus{}ca)}{1 \minus{} (a^3b\plus{}b^3c\plus{}c^3a)} \ge \frac{9}{3 \minus{} (ab\plus{}bc\plus{}ca)}$

But now i don't know how to go on...any idea?

Better idea (maybe)

$ (a \plus{} b \plus{} c)^2 \equal{} \left (\sum_{cyc} a\sqrt {1 \minus{} bc} \cdot \frac {1}{\sqrt {1 \minus{} bc}} \right )^2 \le \left ( \sum_{cyc} a^2(1 \minus{} bc) \right ) \left ( \sum_{cyc} \frac {1}{1 \minus{} ab} \right )$
$ \sum_{cyc} \frac {1}{1 \minus{} ab} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum a^2 \minus{} \sum a^2bc}$
$ \frac {(a \plus{} b \plus{} c)^2}{1 \minus{} abc(a \plus{} b \plus{} c)} \ge \frac {9}{3 \minus{} (ab \plus{} bc \plus{} ca)}$
$ \frac {1 \minus{} 2(ab \plus{} bc \plus{} ca)}{1 \minus{} abc(a \plus{} b \plus{} c)} \ge \frac {9}{3 \minus{} (ab \plus{} bc \plus{} ca)}$

Maybe using pqr

$ (1 \minus{} 2q)(3 \minus{} q) \ge 9 (1 \minus{} pr)$

and $ p^2 \minus{} 2q \equal{} 1$ and $ q \le 1$...


the numbers are defined by recurrence:$a_1=1,a_{n+1}=(n+1)a_n+\sum_{k=1}^{n-1}a_k.a_{n-k}$
a)show that $b_{n+1}\leq b_n + \frac{4b_n^2}{n(n+1)}   if    b_n=\frac{a_n}{n!}$
b)prove that there exists a constant $\alpha$ such that :$Lim_{n\rightarrow \infty}\frac{a_n}{n!}=\alpha$

[quote="sam-n"]the numbers are defined by recurrence:$a_1=1,a_{n+1}=(n+1)a_n+\sum_{k=1}^{n-1}a_k.a_{n-k}$
[/quote]

Hi sam-n,

With $f(x)=\sum_{n\geq1}a_nx^n$ and
the recurrence relation it seems we can an equation with $f(x)$


I just figured out this problem, but I thought that it would be fun to post here.

Triangle $ABC$, inscribed in a circle, has $AB=15$, and $BC=25$. A tangent to the circle is drawn through $B$, and a line through $A$ and parallel to this tangent intersects $BC$ at $D$. Find $DC$.

[hide="Answer"]We can assume that it is a right triangle with $AC=20$, since that is a possibility and no specifications were given. Hence, we have $\triangle DAC\sim \triangle ABC$ because the tangent is perpendicular to $BC$, and $AD$ is parallel to the tangent, so $AD\perp BC$. Therefore, $\frac{DC}{AC}=\frac{AC}{BC}\Rightarrow 25\cdot DC=(20)^2\Rightarrow \boxed{DC=16}$.[/hide]


The page is: [url]http://www.artofproblemsolving.com/Forum/faq.php?mode=attach&f=139[/url]

You can find it while posting and clicking on "Add an Attachment" then go to "Allowed Extensions and Sizes"

I have only tried it on Opera 9.64.

The page doesn't seem to load correctly, I tried refreshing and it still didn't work.

I took a glance at the source and found

[code]
</body>
</html>

<!-- left side bar-->
<!-- must appear AFTER body-->

[/code]

I think that the "</html>" could be causing the problem but I'm not sure.

Edit: Looking at the source again, I found that "<html>" was only put once but the html close command was put twice, which is probably causing the problem.

I use the MathLinks board style and it doesn't have the side links. It comes normal for me.

EDIT: I tried it with the AoPS board style and it is like the picture.

2nd EDIT: Oh, yeah, and I use IE.

Confirmed in Firefox. That page has broken spacing.

This has been mentioned before (I forgot where though); I guess we can still live with it until it is fixed.

We don't really care that much about that page, so we're just going to leave it be until we move on to the AoPS 2.0 version :)

AoPS 2.0 Version???

When is it estimated to be done?

Hmm.. I searched around a bit and couldn't find any other place where it has been announced so it's has to be soon or a very well kept secret?

We said we're working on a couple new features for the site, and an improved look a couple of times now.
We have no idea when we're going to be done or what features will make it or not. It's not a secret it's just a thing that we're not sure about the timetable ourselves.


limit
real analysis
real analysis unsolved

Prove that $ \lim_{n\rightarrow\infty}\sqrt[n]{n}  \equal{}1$.

use natural log and it should unwrap from there

\[ (1 \plus{} x)^n \equal{} 1 \plus{} nx \plus{} \binom{n}{2} x^2 \plus{} \cdots > 1 \plus{} nx \plus{} \frac {n(n \minus{} 1)}{2}x^2 \ \ (x>0)\]
So, let $ x \equal{} \sqrt {\frac {2}{n}}$,
\[ \left(1 \plus{} \sqrt {\frac {2}{n}}\right)^n > 1 \plus{} n\sqrt {\frac {2}{n}} \plus{} (n \minus{} 1) \equal{} n \plus{} \sqrt {2n} > n\]
Therefore,
\[ \left(\left(1 \plus{} \sqrt {\frac {2}{n}}\right)^n\right)^{\frac {1}{n}} > n^{\frac {1}{n}} ( > 1)\]
Thus,
\[ \lim_{n \to \infty}\sqrt [n]{n} \equal{} 1\]

\[ (1 \plus{} x)^n \equal{} 1 \plus{} nx \plus{} \cdots > 1 \plus{} nx \ \ (x>0)\]
So, let $ x \equal{} \sqrt {\frac {1}{n}}$,
\[ \left(1 \plus{} \sqrt {\frac {1}{n}}\right)^n > 1 \plus{} \sqrt{n} > \sqrt {n} \equal{} n^{\frac{1}{2}}\]
Therefore,
\[ \left(\left(1 \plus{} \sqrt {\frac {1}{n}}\right)^n\right)^{\frac{2}{n}} > \left(n^{\frac{1}{2}}\right)^{\frac{2}{n}} (>1) \iff  \left(1 \plus{} \sqrt {\frac {1}{n}}\right)^2 > n^{\frac{1}{n}} (>1)\]
Thus,
\[ \lim_{n \to \infty}\sqrt [n]{n} \equal{} 1\]

Let $ \sqrt[n]{n} = 1 + \alpha_n \ \ (\alpha_n > 0)$.

So,
\begin{align*}
n
&= (1 + \alpha_n)^n \\
&= 1 + n \alpha_n + \binom{n}{2} \alpha_n^2 + \cdots \\
&> \binom{n}{2} \alpha_n^2 \\
&= \frac{n(n-1)}{2}\alpha_n^2
\end{align*}
Therefore,
\[ 0 < \alpha_n < \sqrt{\frac{2}{n-1}} \ \Rightarrow \ \lim_{n \to \infty} \alpha_n = 0\]
Thus,
\[ \lim_{n \to \infty}\sqrt [n]{n} = \lim_{n \to \infty} (1 + \alpha_n) = 1\]


If the diameter of the circle is decreased by 50%, the area is decreased by:

a. 25% b. 50% c. 75% d. 100% e. None of these

the diameter is $ \ x$ lets say. so now it's $ \ 50%x
$
we take out pi from both sides, then multiply by 4.
so now it's $ \ x^{2}$ and $ \ (.5x)^{2}=.25x$
So i believe it's c.?
Correct me please if i'm wrong.

Call the diameter $ d$.

$ \frac{(\frac 12 d)^2 \pi}{d^2 \pi} \equal{} \frac{1}{4}$

Answer is $ \frac 34$, or C


Prove that for positive integers $x_{1},...,x_{n}$, we have

$\prod_{1\leq i<j\leq n}(j-i)|\prod_{1\leq i<j\leq n}(x_{j}-x_{i})$

[hide="solution"]It is not Pre-Olympiad

http://www.mathlinks.ro/Forum/viewtopic.php?t=4858[/hide]


floor function
number theory
prime numbers
number theory unsolved

Find all odd prime numbers p such that

          1^(p-1)+2^(p-1)+...+2004^(p-1)    is divisble by p.

Bomb

They are all prime numbers for which $2004-\lfloor\frac{2004}{p}\rfloor$ is divisible by $p$ but I haven't found quick way of finding them all

Enumerate !  :D
(only 304 case to check)

[quote="cyberjoe"]Enumerate !  :D
(only 304 case to check)[/quote]
 :D  :D

Computer checking shows that [tex]p = 17[/tex] and [tex]p = 2003[/tex] are the only result. That seems a bit odd to me.

Dudes, Ondrob is obviously right... all of the terms are 1 mod p except those divisible by p.

Not that anyone doubted, just pointing it out just in case.

Well, it's clear enough that ondrob is right !
Homever his result doesn't lead directly to a solution.
BTW : My C program can't handle numbers like 2004^2003  :)

use the fermat's theoreme , and it's not very difficult !

:mad:
How did you found primes being the solution ? (without enumerating)
(and *gosh* it's obvious that one should use fermat's theorem !)


Two equal particles A and B, joined by a taut inelastic string ,lie at rest on an a rough horizontal table.

A force is applied to B in the plane of the table making an angle $\Theta$ with AB where $\Theta$ < $\pi/4$. The equilibrium is limiting so that B is on the point of  moving in a direction making an angle $\Phi$ with AB.

1.Show that $\Phi$=2$\Theta$.

2.Also show that  if  $\pi/4$ <=$\Theta$ < $\pi/2$ then $\Phi$ =$\pi/2$

Let f be the limiting friction force, $\,\mu\,.N\,$ (N being the normal contact force), F the applied force (both acting in B) and T the string’s tension, all in magnitude (or could I say, modulus ?).

Due B’s equilibrium, subjected to the tension T in BA direction, F making an angle $\;\Theta\;$ with AB (actually with it’s extension) and  friction force, f, making an angle $\;\Phi\;$ with BA (cause opposite to the direction of B’s imminent movement), must have their vector sum null.

We’ll ask for nulls projection in AB’s and in his orthogonal directions equal to zero.

$\qquad \large F\,cos\,\Theta\,=\,f\,+\,T\,cos\,\Phi\,$

$\qquad \large F\,sin\,\Theta\,=\,f\,sin\,\Phi\,$

By division and simplifying we arrive to $\;T\,=\,f\,.\,{\,sin\,(\Phi\,-\Theta) \over{ sin\,\Theta\,}}$;

In 1. for $\,\Theta\,<\,{\pi \over{ 4}}$, particle A will also be about moving and its friction force f, so T = f, for A’s equilibrium. Substitution in the expression above leads to $\,sin\,(\Phi\,-\Theta)\,=\,sin\,\Theta\,\qquad \Rightarrow \qquad \Phi\,=\,2\,\Theta.\,$

2. For $\;\;{\pi \over{ 4}}\,<\,\Theta\,<\,{\pi \over{ 2}}$, in the above expression, $\;\;\Phi\;\;$ would be greater than $\;\;{\pi \over{ 2}}\;\;$ and it wouldn’t be tension at all.

 So we’ll have $\;\;\Phi\,=\,{\pi \over{2}}\;\;$ and $\;\;T\,=\,f.\,cotg\,\Theta\;\;\;$ since A is no more about to move.

Friction force acting on the particles is
$F=\mu N= \mu m g$
1.Force diagrams for the ball B
$F \cos\Theta=\mu N \cos \Phi+T$
$F \sin\Theta=\mu N \sin\Phi$
$T$ is the force on the string.
For the ball A
$T= \mu N$
When we combine thes equations we get
$F \cos\Theta=\mu N(1+\cos\Phi)$
$F \sin\Theta=\mu N\sin\Phi$
Dividing these equations side by side
$\tan\Theta=\frac{\sin\Phi}{1+\cos\Phi)}=\tan(\frac{\Phi}{2})$
And $\Phi= 2\Theta$
2.For $\frac{\pi}{4} \leq \Theta \leq \frac{\pi}{2}$, in the above expression, $\;\;\Phi\;\;$ would be greater than $\frac{\pi}{2}\;\;$ and there wouldn’t be tension any in the string and ball A cannot move. So the limiting value is $\Phi= \frac{\pi}{2}$


I would like to see the code for the enumerate environment, mainly so that I can modify it to add a fifth level of nesting.  Where is the LaTeX source code located in a) Miktex, and b) tetex?

I think you totally misunderstand TeX and LaTeX.
1. MiKTeX and teTeX are just ways of installing LaTeX so they are no different.
2. You never change TeX itself unless you wish to re-compile it. But even then you would have to have read and understood [url=http://www.amazon.com/TeXbook-Computers-Typesetting-Donald-Knuth/dp/0201134489]The TeXBook[/url] before you could even start. That would be a major challenge. [url=http://www.eijkhout.net/tbt/]TeX by Topic[/url], which is downloadable, will give you some insight into the technical details. It's not a trivial read.
3. You normally modify the way TeX works by using packages or macros in your document.

So the first thing is to look on [url=http://tug.ctan.org/search.html]CTAN[/url] for a package that will do what you want. Unfortunately in this case it looks as if you won't be successful. The [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=toodeep]TeX FAQ[/url] explains that your only choice is to rewrite your document. [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/fd45f0b66e4dad06/ff04cd375fa67129]5th item level[/url] tells you a little more.

It is very rare to say something isn't possible in LaTeX; this is one of those occasions I'm afraid.

Perhaps I wasn't clear enough.  I'm not quite so foolhardy as to attempt any serious direct modification of the LaTeX source code.  I wanted to know where the source code was so that I could use it as a reference to attempt to create a package that allows deeper nesting.  I had assumed that it might be easier to know where to find the source code than to continually ask what the code is for individual macros.

It seems that I would need to redefine the enumerate environment (in a package), which currently uses the following syntax:

[code]
\def\enumerate{%
  \ifnum \@enumdepth >\thr@@\@toodeep\else
    \advance\@enumdepth\@ne
    \edef\@enumctr{enum\romannumeral\the\@enumdepth}
[/code]

The main problem that I'm running into is what the variants of numbers using @'s are.  I know that one is \@ne and three is \thr@@, but I need to know the @ variant of four so that I can change the \thr@@ in the above code to it.  Is there anything else that I would need to do to make this work (other than define enumv)?

[url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=ltxabbrv]LaTeX internal “abbreviations”, etc.[/url], page 345 of The TeXBook and in yourtexpath/tex/plain/base/plain.tex.

A trawl of the comp.text.tex produced [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/52985fafa99587a1/7f13fb1d091eb04e]this thread[/url] which is interesting for two reasons:
1. it contains suggested code;
2, it says that version 2 of the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url] allows you to set the depth. The manual has a new command
[code]\newlist{<name>}{<type>}{<max-depth>}[/code](It's simple to update enumitem by replacing enumitem.sty with the latest version).

Of course you could still use these two sources to help write your own package.

Thanks for the tip on enumitem.  It works perfectly!


inequalities
geometry
analytic geometry
triangle inequality
Cauchy Inequality

Any 2 sides of a triangle added together is greater than the third side

Is there a proof for this?

[quote="i_like_pie"]It can be proven using the Cauchy-Schwarz Inequality.[/quote]

The Cauchy-Schwarz Inequality is beyond the level of high-school basics, but if you are interested:

[url=http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality]Cauchy-Schwarz Inequality on the wiki[/url]

[quote="Anaxerzia"][quote="i_like_pie"]It can be proven using the Cauchy-Schwarz Inequality.[/quote]

What?? Why would you need that? Just think about what a triangle would look like if it had two sides shorter the nor equal to than the third, If equal its basically a line - and its impossible to have shorter...[/quote]
Intuition is most definitely not a proof. The Cauchy-Schwarz is, in fact, not needed, but some calculation is required..

Thats pretty basic stuff...its just what the TRIANGLE INEQUALITY SAYS!!

The sum of any two sides of a triangle is greater than the third side.

You probably (normally) learn this in like 7th/8th grade...

but in other words, its pretty elementary for high school

a very simple proof simply by definition of a line segment let $ A$ $ B$ $ C$ be  the vertices of the triangle then clearly the shortes path between $ A$ and $ B$ is the line segment $ AB$ hence $ AB < AC\plus{}CB$  :wink:

This is the simplest case of what is known as [b]Minkowski's inequality[/b].

Consider the triangle $ A(0, 0), B(a, b), C(a\plus{}c, b\plus{}d)$.  Minkowski's inequality states that

$ \sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{c^{2}\plus{}d^{2}}\ge\sqrt{(a\plus{}c)^{2}\plus{}(b\plus{}d)^{2}}$

which is simply the statement $ AB\plus{}BC\ge AC$.  Minkowski's inequality is, in fact, related to the Cauchy-Schwarz inequality.

[quote="gauss1181"]Thats pretty basic stuff...its just what the TRIANGLE INEQUALITY SAYS!!
...
its pretty elementary for high school[/quote]
Naming does not constitute a proof.  Your comment is both presumptuous and useless.

[quote="pardesi"]a very simple proof simply by definition of a line segment let $ A$ $ B$ $ C$ be  the vertices of the triangle then clearly the shortes path between $ A$ and $ B$ is the line segment $ AB$[/quote]
I'm not familiar with how Euclidean geometry is built up axiomatically, but is there a standard way to define distance without respect to a Cartesian coordinate system?

i don't think you should use a higher level theorem to prove a lower one.

I'm not sure how to do this but I would try drawing a triangle ABC with AB the longest side. construct a point P on BC such that AP=AB+AC.  Then, you assume that point P is between points B and C. and then somehow(i don't know) by contradiction, P cannot lie within points B and C. Thus, AB+AC>BC.

i'm new to proofs and i can't find an exact solution, but i think the general direction is correct.

[quote="unimpossible"]i don't think you should use a higher level theorem to prove a lower one.[/quote]
This is usually a valid concern, but to my knowledge the modern axiomatic construction of Euclidean geometry starts with defining distance in terms of an [url=http://mathworld.wolfram.com/InnerProduct.html]inner product[/url], the dot product, and the triangle inequality then follows by Cauchy, so really, an algebraic inequality is at the heart of this method of defining plane geometry as well.

Again, I'm not familiar with how Euclidean geometry is built up axiomatically without reference to a Cartesian coordinate system, and with reference to a Cartesian coordinate system the fastest way to rigorously establish the validity of the triangle inequality is algebraically.

[quote="t0rajir0u"]
I'm not familiar with how Euclidean geometry is built up axiomatically, but is there a standard way to define distance without respect to a Cartesian coordinate system?
[/quote]
well what i have done is simply used the definition of a line segment.

[quote="Temperal"]Oh, really? Just because that the shortest path between A and B is a line does not imply that AC>AB+BC[/quote]
Well, hold on there.  For one thing, the statement you're not trying to imply is untrue, and for another what pardesi's trying to say is that the path that goes from $ A$ to $ C$ to $ B$ is not the shortest path (because the shortest path is the straight line from $ A$ to $ B$, by definition), and therefore that it must be longer (or the same length); ergo $ AC\plus{}CB\ge AB$.

The issue I have with this proof is that it assumes a definition for a straight line that I don't find rigorous, or at least not sufficiently well-explained in the context of what I know about Euclidean and non-Euclidean geometry.

[quote="Anaxerzia"][quote="i_like_pie"]It can be proven using the Cauchy-Schwarz Inequality.[/quote]

What?? Why would you need that? Just think about what a triangle would look like if it had two sides shorter the nor equal to than the third, If equal its basically a line - and its impossible to have shorter...[/quote]
[quote="t0rajir0u"]...the triangle inequality then follows by Cauchy...[/quote]

That cauchy inequality is a bit advanced. I done some browsing and found this good site

http://books.google.com/books?id=KAJ8g5zuFdoC&pg=PA20&lpg=PA20&dq=any+2+sides+of+a+triangle+added+together+is+greater+than+the+third+side&source=web&ots=N3JznlM3hF&sig=AClcpdHsathSBpL_JNPFs5w3nOQ

see the [url=http://en.wikipedia.org/wiki/Line_(mathematics)]definition[/url] of straight line here.
[b]@Whoever rated my post spam[/b]  :mad:   :mad:  Remeber subtility is in the heart of mathematics definitions and subtle matters may be missed if u don't go through ur fundamentals well enough.
My proof was no joke
Try to realize the importance of a definition and the wonderful formulation.


Let be $ x\equal{}m^3\plus{}1$ and $ y\equal{}m^3\minus{}1\ ,\ m\in \mathbb{Z}$ . Show that if $ 7$ doesn't divides $ m$ then $ 7$ divides $ xy$ .

$ 7\not|m\Longrightarrow m^3\equiv\pm 1\bmod{7}\Longrightarrow 7|(m^3\plus{}1)(m^3\minus{}1)$