If you could have a playlist consisting of any songs you choose, with the limit only to have 10 songs on it, each song having any length,

[b]what would you put on that list?[/b]

[hide="My List"]
[i]Drunken_Math's Personal Playlist
[/i]
Carmen Suite No. 1 - Bizet
Carmen Suite No. 2 - Bizet
Sugar, We're Going Down - Fall Out Boy
In The End - Linkin Park
Collide - Howie Day
Pirates of the Carribean (entire soundtrack medley) - Bedalt
When I Come Around - Green Day
Wake Me Up When September Ends - Green Day
Boulevard of Broken Dreams - Green Day
Letterbomb - Green Day
[/hide]

[quote="math92"][hide="some good songs i have"]
Usher ft..Lil Jon & Ludacris - Yeah
Black Eyed Peas - Don't Phunk With My Heart
Black Eyed Peas - Lets Get it started
Cold Play - Speed Of Sound
Game - Raps MVP
George Clinton - Atomic Dog
Green Day - American Idiot
Green Day- Boulevard Of Broken Dreams
Greenday - Holiday
Ludacris- Get Back
Ludacris - Number One Spot
The Longest Yard Soundtrack - 12 - D12-Eminem
U2 - Vertigo	5,924 KB
Weezer - Beverly Hills	[/hide]
 :)[/quote]

Dude, he said [b]10[/b] songs.

Hmm..I like your list Drunken_Math...all decent songs..

[hide="My random songs"]
1)LoTR RotK Soundtrack
2)Desert Rose-Sting
3)Jupiter-from the Planets
4)In the End-Linkin Park
5)Pirates of the carrabean
6) Dragostea din tea-romanian pop song- really funny
7)amish paradise
8)Its my life...(forgot name)
9)Walking in your footsteps-The police
10)A thousand miles-Sting
11) (strangly enuff) Phantom of the opera
CD's
1)A beetoven one
2) the other one as a part of that pack
3) a set of relaxing classics
4)LoTR RotK Sound track
5)The first linkin park CD
6)the planets
7)a weird al cd
8)Brand new day
9) the best of the police
10)not really a cd, but I like the music in StarCraft and the secret tune: Radio Free Zerg.
[/hide]

I like.. a lot of songs.. btw "Its My Life" is by Bon Jovi. <from new jersey, go jersey!>

[hide]I can't really decide, but here are some good songs:

1) 3 doors down- When I'm Gone
2) Creed - One
3) Green Day - Time of Your Life (i'm a sucker for classics)
4) Linkin Park - In the End
5) Led Zeppelin - Stairway to Heaven (along with its freaky backwards satanic messages..)
6) Ludacris - Stand Up
7) Metallica - Enter Sandman
8) Nickelback - Someday
9) Sublime - What I Got
10) Disturbed - Prayer (or Shout 2000.. but only sometimes)

Those arent all my favorite songs, but some really good ones.
As an 11, though many will say i'm an idiot for including this.. but maybe Baby Got Back?  :lol:[/hide]

[quote="math92"]U2 - Vertigo	5,924 KB[/quote]
That's relevant.

Green day sucks.

Deep Purple>Green Day.
Pink Floyd>Green Day.
Ramones>Green Day.
White Stripes>Green Day.
Queen>Green Day.
Van Halen>Green Day.
Rolling stones>Green Day.
Pearl Jam>Green Day.
The Doors>Green Day.

[quote="nr1337"]Oh, such a hard limit...

[hide]
Pink Floyd - Shine On You Crazy Diamond
Pink Floyd - Welcome To The Machine
Pink Floyd - Have A Cigar
Pink Floyd - Wish You Were Here
Pink Floyd - Echoes
Pink Floyd - Fearless
Pink Floyd - See Emily Play
Pink Floyd - Sheep
Pink Floyd - Comfortably Numb
Pink Floyd - Brain Damage/Eclipse
[/hide]

There are so many more that I want to put in there....[/quote]
We have a winner!!!

[hide]
Coldplay - Speed of Sound
Creed - My Sacrifice
Creed - One Last Breath
Goo Goo Dolls - Here Is Gone
Mathchbox Twenty - Unwell
Snow Patrol - Run
U2 - One
U2 - Walk On
U2 - Yahweh
Yellowcard - Only One
[/hide]

[quote="nr1337"]Oh, such a hard limit...

[hide]
Pink Floyd - Shine On You Crazy Diamond
Pink Floyd - Welcome To The Machine
Pink Floyd - Have A Cigar
Pink Floyd - Wish You Were Here
Pink Floyd - Echoes
Pink Floyd - Fearless
Pink Floyd - See Emily Play
Pink Floyd - Sheep
Pink Floyd - Comfortably Numb
Pink Floyd - Brain Damage/Eclipse
[/hide]

There are so many more that I want to put in there....[/quote]

I wonder who your favorite band is...

@PenguinIntegral, First of all, thats some interesting music for a penquin. Second, dont be so harsh in disaproving in someones interest. BTW, I think green day is okay, but not that great...

[quote="PenguinIntegral"]Deep Purple>Green Day.
Pink Floyd>Green Day.
Ramones>Green Day.
White Stripes>Green Day.
Queen>Green Day.
Van Halen>Green Day.
Rolling stones>Green Day.
Pearl Jam>Green Day.
The Doors>Green Day.[/quote]

There must be something wrong with my browser, those "less than" signs are coming up as "greater than" signs. :P

i bet you more people will disagree with penguin

[quote="jmadsen"]There must be something wrong with my browser, those "less than" signs are coming up as "greater than" signs. :P[/quote]

Nothing wrong with your browser, since mine shows the same thing. But perhaps his keyboard is broken. I'd offer to fix the post...but eh... ;)

but that would be against the code of modness you should just delete his post for being false :D

Erm, are you guys implying that Green Day is better than the Doors? Because that's just plain BS. Sure, the Doors were pretty screwed up, but still, they were one of the top 15 rock bands ever. Green Day can't even compare to them at all, since they're music is typical of the so-called 'rock' of today which has dropped off so much from groups in the 60s-80s.

This is irrelevant but my hyper 60 year old English teacher listens to Bon Jovi and Simple Plan ( she does step aerobics too.) She plays "departing music" in class.

I like some rock but i listen to mostly pop(unlike it seems every one here)

i hate green day and so does everyone else that lives in the Binghamton area. ever since they came here to do a show all you hear on the radio is green day I HATE GREEN DAY!! :mad:

what state is binghamton in...and i dont think you can generalize everyone in the area thats almost like saying everyone in texas is conservative which is almost true except for like austin,texas and some other random people like me

It's not so much that they are bad, but that they are bad and people think they are gods.

Hold on penguin just because ou guys have difference in opinion doesnt mean either of you are right so just do whatever and not debate this

Well you can't really say that Green Day sucks because most of you can't sing either.

[quote="PenguinIntegral"][quote="math92"][quote="PenguinIntegral"]Green day sucks.
[/quote]
I think they have some good music.[/quote]
What?[/quote]
i like most of their songs like:
Holiday
American Idiot
Boulavard of Broken Dreams


Quadrilateral $ ABCD$ such that $ BD \perp AC$. Midline of $ AB$ intersect midline of $ CD$ at $ O$( such that $ O$ is inside $ ABCD$).Prove that $ ABCD$ is cyclic $ \Leftrightarrow S_{ABO}=S_{CDO}$ ($ S$ denotes area).

We denote as $ M,$ $ N,$ the midpoints of the segments $ AB,$ $ CD$  respectively and let $ P$ be, the intersection point of $ AC,$ $ BD.$

$ \bullet$ When $ ABCD$ is cyclic, it is easy to show that $ \angle MBO = 90^{o}-\angle MOB = 90^{o}-\angle ACB = \angle CBP = \angle NOC$ $ ,(1)$

From $ (1)$ we conclude that $ (MBO) = (NOC)$ $ \Longrightarrow$ $ ABO) = (CDO)$ $ ,(2)$

$ \bullet$ When $ (2)$ is true, we draw the circumcircle of $ \bigtriangleup ABC$ which intersects the segment line $ BD,$ at a point so be it $ D'.$

By the same way as before we can prove that $ ABO) = (CD'O)$ $ ,(3)$

From $ (2),$ $ (3)$ $ \Longrightarrow$ $ CDO) = (CD'O)$ $ ,(4)$

From $ (4)$ $ \Longrightarrow$ $ DD'\parallel CO$ $ \Longrightarrow$ $ AC\perp OC$ $ ,(5)$

Because $ (5)$ is a false result, we conclude that $ D'\equiv D.$

That is the quadrilateral $ ABCD$ is cyclic and the proof is completed.

Kostas Vittas.


calculus
derivative
trigonometry
calculus computations

Hi guys,

    Can you help solve the deriv. of the following:

$\sin(x)\cos(x)-\cos(x)$

Im guessing I can use the product rule? and let

$u = \cos(x), v = \sin(x)$

= $\frac d{dx} \cos(x)\sin(x) - \frac d{dx} \cos(x)$,

= $\frac d{dx} \cos(x)\sin(x) + \sin(x)$

=$\cos(x) \frac d{dx} \sin(x) + \sin(x) \frac d{dx} \cos(x) + \sin(x)$,

= $\cos(x)^2 + \sin(x) \frac d{dx} \cos(x) + \sin(x)$,

= $\cos(x)^2 - \sin(x)^2 + \sin(x)$

My calculator gives me the following answer:

$2\cos^2(x)+\sin(x)-1$

Can my answer be simplified to give me that?  If it can be, can you please explain it?  Thanks in advance guys...
Chris

Yes, you're right, use the identity $\sin^2 x = 1-\cos^2 x$, then you have

$\cos^2 x - (1-\cos^2 x)+\sin x = 2\cos^2 x + \sin x -1$


prove that for all [img]http://i44.servimg.com/u/f44/11/94/14/08/412.gif[/img][img]http://i44.servimg.com/u/f44/11/94/14/08/4510.gif[/img]

The following inequality is true too:
$ \sum_{cyc}(a^3\plus{}a^2d\plus{}3abc\minus{}a^2b\minus{}a^2c)\geq0$ is true too. :wink:

[quote="arqady"]The following inequality is true too:
$ \sum_{cyc}(a^3 \plus{} a^2d \plus{} 3abc \minus{} a^2b \minus{} a^2c)\geq0$ is true too. :wink:[/quote]

Don't you mean $ \sum_{cyc}(a^3 \plus{} a^2d \plus{} 6abc \minus{} a^2b \minus{} a^2c)\geq0$ ?

[quote="y-a-s-s-i-n-e"]thank's arqady
but how we can prove your inequality[/quote]
I think, the way $ b\equal{}a\plus{}x,$ $ c\equal{}a\plus{}y$ and $ d\equal{}a\plus{}z,$ where $ a\equal{}\min\{a,b,c,d\}$ helps here.

thank's again arqady.

but is there another solution  :ninja: ?


invariant
combinatorics
Chessboard
algorithm
IMO Shortlist

A natural number is written in each square of an $ m \times n$ chess board. The allowed move is to add an integer $ k$ to each of two adjacent numbers in such a way that non-negative numbers are obtained. (Two squares are adjacent if they have a common side.) Find a necessary and sufficient condition for it to be possible for all the numbers to be zero after finitely many operations.

It is possible iff the sum of the number in white squares is the same as the sum in white square.

[i]It is possible -> Sums are equal: [/i]
It is obvious that each move change of the same quantity the sum in black and in white. Then, cause at the  end the two sum are both equal, they had to be equal also initially.

[i]Sums are equal -> it is possibile:[/i]
[b]Lemma[/b] Given three adjacent squares with $a,b,c$ it is possible to trasform them in $0,x,y$  doing only allowed moves.
[hide="Proof"]If $a<b$ I simply add $-a$ to the first two squares, obtaining $0, b-a, c$. Otherwise I add $a-b$ to the last two squares, and then add $-a$ to the first two squares obtaining$ 0,0,a+c-b$.[/hide]
Applying the lemma on each row (left to right) I succed in having all squares with zeroes apart from the last two columns.
I repeat the same operation on both the last two columns (up to down) and now I have all zeroes except in the 2x2 square at bottom right corner.

Now I only have to prove it for 1x1, 2x1,1x2,2x2 (the assumption that the sums are equal remains because doing the allowed operation doesn't change the difference between the sums).
Since they are all easy cases the problem is finished.

p.s. Sorry for my terrible english

I'm going to rephrase dario2994's solution, as well as cleaning it up some.

Color the board with alternating black and white colors like a chessboard. The claim is that it's possible to turn the whole board into zero if and only if the sum of numbers on black cells and the sum of numbers on white cells are equal. This is true even if there is no restriction that numbers must remain non-negative throughout.

First, the proof that this is necessary: each time we do such move, we change both black and white sums by an equal amount, so their difference is an invariant. At the end they are equal (to zero), so at the beginning they must also be equal.

Now, the proof that this is sufficient:

Given any three squares A, B, C with B being adjacent to both A and C (note that A and C don't have to be on the same row/column), we can make A to be 0. Let $a, b, c$ be the contents on A, B, C respectively. Then add $a$ to both B and C, and subtract $a$ from both A and B, to obtain $0, b, a+c$ respectively. Note that the addition clearly doesn't break the rule that there is no negative cell, and the subtraction will also be fine because we have added $a$ to B.

Now, there exists a Hamiltonian path across the grid (e.g. just take a "snake" pattern that covers all cells). Using the above, we can repeatedly apply it from an end to make all but two squares zero. (If the squares are $A_1, A_2, A_3, A_4, \ldots$, first make $A_1$ zero by applying the above to $A_1, A_2, A_3$; then make $A_2$ zero by applying $A_2, A_3, A_4$, and so on.) This leaves at most two nonzero squares at the other end of the path. Let the last two squares of the path be X and Y, with numbers $x,y$ on them. Since they are adjacent squares on the path, they are of different colors. By our assumption, black sum is equal to white sum; since all other numbers are zero, we must have $x = y$ in order for the sums to be equal. This means we can just subtract $x$ from both to have all zero.

Apply the checkerboard colouring on the m by n chessboard.  Let $ S_b$  and  $S_w$ denote sum on white and black respectively. We claim that the only restriction needed is $S_b=S_w$ Suppose otherwise, notice that $S_b-S_w$ is invariant since adjacent squares are 1 black and 1 white. So if we were to reach all 0's we would need $S_b- S_w=0 $ contradiction.
Now we shall provide a construction that works.
Consider a random row. Take the topmost box in that row, suppose it contains the number $a$ and a number $b$ directly below it. Operate such $(a,b)->(0,b-a)$ continue repeating this process down the same column, until all the numbers in the column except for the bottommost box is 0. Do the same for the remaining columns. Now we should have that all rows are only 0's except for the last row. Now repeat what you did for the columns with the last row. This will leave us with all boxes 0 except for the bottom rightmost box. However, note that $S_b- S_w=0$ condition already gives us that the final box must be 0, so we are done.

Cool problem. Let's tag the squares as $a_1, a_2, \ldots, _n$ The necessary conditions are:

                1. $\sum a_i = 2k$, in other words the summation of the numbers must be even.
		2. Let's color the board white and black. Let the summation of numbers in black and white cells be $S_b$ and $S_w$ respectively. Then,
		\[S_b = S_w\]

	First of all, let the total sum be $X$. After each move $X$ becomes, $x - 2k$. So, if $X$ is odd and $X$ can't be $0$, as desired

	Now, $S_b = S_w$. Suppose, $a, b, c$ be three adjacent squares and $c$ is closest to the left corner. If $b \geq c$, then we will remove $c$ from each.
	\[\left(a, b, c\right) = \left(a, b - c, 0 \right)\]
	If $c \geq b$,
	\[\left(a, b, c \right) \to \underbrace{\left(a + c - b, c, c \right)}_{\text{Adding $c - b$ to $a$ and $b$}} \to \left(a + c - b, 0, 0 \right)\]
	So, we will get $0$ in every squares, expect the most right two columns. We will do the same in those columns and will be left with $2$ adjacent corners, $x$ and $y$. And, they will be of different color, obviously. As, $S_b = S_w$, it is simple that $x = y$ and we are done.

Nice !
Clearly $S_b=S_w$ since we are adding an integer to neighbouring cells. We claim this is sufficient as well.
We proceed via induction on $m+n$. The base case can be easily handled. Take any board with perimeter $m+n+1$, and $S_b=S_w$. We want to prove that we can make all it’s entries zero at some point. Look at its first column. If $S_b=S_w$ in that column, we are done by our induction hypothesis. If not, WLOG $S_b > S_w$. Simply add $S_b-S_w$ to any white cell and the cell adjacent to it in the same row. Now we have $S_b=S_w$ in the first column, and hence the remaining board too. Applying the induction hypothesis separately, we are done :D

First ISL Combo! (Although it's pretty old)
-----
Call $S_{b}$ and $S_{w}$ the sum of the black and white squares, respectively. Clearly, $S_{b}-S_{w}$ is an invariant. We have $S_{b}-S_{w}=0$ at the end, so we also need $S_{b}=S_{w}$ at the beginning.

[b][color=#f00]Claim:[/color][/b] The above condition is sufficient.
[i]Proof:[/i] Our algorithm is as follows: Let $a_{1}, a_{2}, a_{3}$ be numbers in consecutive cells, with $a_{1}$ adjacent to $a_{2}$ and $a_{2}$ adjacent to $a_{3}$. We take cases based on which of $a_{1}$ and $a_{2}$ is greater.
[u]Case 1[/u]: $a_{1} \geq a_{2}$.
In this case, we can add $a_{1}-a_{2}$ to $a_{2}$ and $a_{3}$.
[u]Case 2[/u]: $a_{1} \leq a_{2}$.
In this case, we can add $-a_{1}$ to $a_{1}$ and $a_{2}$.

Applying this to each row, we have everything canceling except the last two entries. However, these last two numbers are clearly equal, since $S_{b}=S_{w}$. $\square$

Since we have shown that this is sufficient, we are done. $\blacksquare$


At the time I am posting this, problems 24 and 25 on the AMC 2005 A are identical.

In fact, my sources tell me that #25 should look like this:

25. Let $S$ be the set of all points with coordinates $(x,y,z)$ where $x$, $y$, and $z$ are each chosen from the set $\{0,1,2\}$. How many equilateral triangles have all their vertices in $S$?
(A) $72$ (B) $76$ (C) $80$ (D) $84$ (E) $88$

And the answer is for you to figure out...  :harhar:

I guess Silverfalcon accidentally posted #24 for #25 as well...I posted #25 in the intermediate forum over the weekend I think.

the answer is:
[hide]80, C[/hide]
it has already been discussed:
take a look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=368861#p368861]here![/url]

Fixed.

It's really hard to do this when you have to edit problems sent by some other people. And chess64, the main reason that I confused 24 and 25 were because you posted that problem twice.  :? Next time, search the forum.. and try to remember which problems you posted.  :D

[quote="chess64"]....

:rotfl: I did? Which one? 24? I searched for that one but couldn't find it.[/quote]

Okay, well..

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367585#p367585
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357250#p357250

Oh, oops :blush: and it was only about two weeks ago, too :oops:


1. In $ \triangle ABC$, $ \angle ACB\equal{}90^0$. The circumcenter is $ O$. $ CD$ is the altitude to $ AB$, then we make a circle with $ D$ as the center, and $ DK$ is the radius, where $ K$ is the midpoint of arc $ AB$ and $ DK$ intersects $ AB$ at $ M$. If $ AB\equal{}c$, define $ CM$ in terms of $ C$

2. In $ \triangle ABC$, $ AB\equal{}AC$. $ M$ is the midpoint of $ BC$, $ O$ is on the extension of $ AM$ such that $ OB\perp AB$. $ Q$ is on $ BC$ ($ BQ>QC$), $ E$ is on $ AB$ and $ F$ is on the extension of $ AC$ such that $ EQF$ are collinear. Prove that $ OQ\perp EF$ if and only if $ QE\equal{}QF$

3. In $ \triangle ABC$, $ AC\equal{}BC\equal{}13$ and $ AB\equal{}10$. $ P$ is on $ BC$ such that $ BP: PC\equal{}1: 3$; $ Q$ is on $ AC$ such that $ AQ: QC\equal{}5: 1$. $ AP$ and $ BQ$ intersect at $ M$, find the value of $ cos\angle AMQ$

Sorry for my bad english,  :)


geometry
perpendicular bisector
combinatorics unsolved
combinatorics

$ 2000$ points on the plane such that no $ 8$ of them are collinear
prove that there exists $ 250$ segments with different length

I know this is posted before but I can't find a solution :?:

[url]http://www.mathlinks.ro/viewtopic.php?p=457182#457182[/url]
I think it was given at a Romanian olympiad but I'm not sure,anyway:

assume that the segments have $ k$ different length $ x_1,x_2,\ldots ,x_k$ we want to show that $ k\geq 250$.
color each segment of length $ x_i$ with color $ i$ ($ 1\leq i\leq k$).
now let $ S$ be the number of monotonic paths with length $ 2$,i.e. the number of segments with the same color that share a common endpoint.
we know that for every two points $ X,Y$ there are at most $ 7$ points on their perpendicular bisector,let $ V$ be a point on the perpendicular bisector of $ XY$ now note that $ |VX|\equal{}|VY|$ so the color of $ VX$ and $ VY$ must be the same,so for every point on the bisector of $ XY$ we have a monotonic path of length $ 2$,so the number of monotonic paths of length $ 2$ is at most:

$ S\leq 7\binom{2000}{2}$ (*)

now let $ n_{ij}$ be the number of segments of color $ j$ that are connected to point $ i$ ($ 1\leq i\leq 2000$),so we have:

$ S\equal{}\sum\binom{n_{ij}}{2}$

now note that:

$ S\equal{}\sum\binom{n_{ij}}{2}\geq 2000k\binom{\frac{1999}{k}}{2}$

thus according to (*) we have:

$ 7\binom{2000}{2}\geq 2000k\binom{\frac{1999}{k}}{2}$

$ \Rightarrow k\geq 250$

QED


[quote="qntty"]How do you pronounce it?[/quote]

It's pronounced "ARM-ULL". Like for example... uh... rhymes with "carmel".

While I agree with you on how to pronounce "ARML", I don't agree that it rhymes with either Carmel (the place) or caramel (the candy). Accent matters.

[quote="Ihatepie"]I just say Ay-ar-em-ell just like I say Ay-eye-am-ee and you-ess-ay-em-o.[/quote]

I sometimes do the same, but was informed that the last two are typically pronounced "Amy" and "Yoo-Sam-Oh", however I'm not sure if IMO does the same thing (Eye-Moh doesn't sound all that different from Eye-Em-Oh)

What I was saying is I don't really care how you are supposed to say it, I just say it letter by letter.

I say all acronyms that don't sound like jibberish when you try to say them (such as AMC; amik 12?). Therefore I pronounce: AIME, USAMO, MOsP(mop), PUMaC, NYSML, and ARML. I don't pronounce: HMMT, AMC, IMO.

[quote="JBL"][quote="BOGTRO"]rhymes with "carmel".[/quote]  I would just like to note that this is among the worst possible ways to explain how to pronounce something, since the pronunciation of the word "caramel" is varies (I don't know if it's regional variation or just individual).  For example, my girlfriend pronounces the second "a" in "caramel," and I do not.[/quote]

Ok... I didn't know how else to discribe it. Its pronoucned "ARE-MULL" I guess.

... which is another way of writing the same thing that has already been said.

In IPA: [ɑɹmɫ̩].  Or maybe [aɹmɫ̩].  I'm never quite sure of the difference.

Just say it as aaaaaaarrrrrrrrrrrrrrrrrrrmmmmmmmmmmmmmmmallllllllllllllllllllllllllllllllllllllllllllllllll

You have to be screaming when you say it.

Hmm... so who had the "THIS IS ARML!!!" shirts last year at Iowa?


I have a very nice solution for the second part . Induction !! :wink:
As April proved it is true for n=1 and n=2 . Assume that it is true that
$\sum\frac{a^{n-1}}{b^{n}}\geq \sum\frac{a^{n-2}}{b^{n-1}}$
and we will prove the desired .

By Cauchy  and the induction hypothesis we have
$\sum\frac{a^{n}}{b^{n+1}}\sum\frac{a^{n-2}}{b^{n-1}}\geq (\sum\frac{a^{n-1}}{b^{n}})^{2}\geq \sum\frac{a^{n-1}}{b^{n}}\sum\frac{a^{n-2}}{b^{n-1}}$

so $\sum\frac{a^{n}}{b^{n+1}}\geq \sum\frac{a^{n-1}}{b^{n}}$

QED

Silouan, I think you missed the fact that $n \in \mathbb R_{\geq 1}$ :ninja:

Anyway, the second inequality is equivalent to $\sum_\text{cyc}a^{n+1}b^{2n+1}\geq \sum_\text{cyc}a^{2n}b c^{n+1}$. We prove this by $AM-GM$, using the equivalent condition found by [url=http://mathlinks.ro/Forum/viewtopic.php?p=585871&search_id=1417354003#585871]Sung-yoon Kim[/url] (the part in bold at the end). Not very nice, but efficient :D


This is definitely not classroom math level.  If you want to try the forum and $ \text{\LaTeX}$, go to the test forum (near the bottom of the forum list.)

***Mods: Please move to Intermediate/Pre-Olympiad.***

[quote="Duy_VN"]Prove that : $ \frac {1}{(1 \plus{} x)^3} \plus{} \frac {1}{(1 \plus{} y)^3} \plus{} \frac {x^3y^3}{(1 \plus{} xy)^3} \ge \frac {3}{8} (x,y > 0)$[/quote]

[hide="first step"]
dunno, this is a pretty complicated inequal
i think the first step is to change the 3rd term to $ \frac{1}{(1\plus{}\frac{1}{xy})^3}$

[/hide]

[quote="The QuattoMaster 6000"][quote="Duy_VN"]Prove that : $ \frac {1}{(1 \plus{} x)^3} \plus{} \frac {1}{(1 \plus{} y)^3} \plus{} \frac {x^3y^3}{(1 \plus{} xy)^3} \ge \frac {3}{8} (x,y > 0)$[/quote]
[hide]
By GM-HM and AM-GM, we have that
2 + (x + \frac {1}{x}) + (y + \frac {1}{y}) + (xy + \frac {1}{xy})\le 2 + 2 + 2 + 2 = 8

[/hide][/quote]

This inequality sign should be switched; I'm not sure if that affects the solution, but it looks like it does.

[hide="nevermind"]Divide both sides by 3. From AM-GM, we get

$ (LHS/3)\ge \frac {xy}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} xy)}$ and we need to show that the RHS is $ \ge 1/8$... nevermind, it isn't strong enough.[/hide]

EDIT: Still 100% on posting incorrect solutions to inequalities. :D

[hide="Wrong"]Determine that $ f(x) \equal{} \frac {1}{(1 \plus{} x)^3}$ is convex. Write the third term of the inequality as $ \frac {1}{(1 \plus{} \frac {1}{xy})^3}$ by dividing the numerator and denominator by $ x^3y^3$. Then the inequality is

$ f(x) \plus{} f(y) \plus{} f(\frac {1}{xy}) \geq \frac {3}{8}$

$ \frac {f(x) \plus{} f(y) \plus{} f(\frac {1}{xy})}{3} \geq \frac {1}{8}$

By Jensen's,
$ \frac {f(x) \plus{} f(y) \plus{} f(\frac {1}{xy})}{3} \geq f(\frac {x \plus{} y \plus{} \frac {1}{xy}}{3}).$

By AM-GM on the terms inside the $ f$,
$ f(\frac {x \plus{} y \plus{} \frac {1}{xy}}{3}) \geq f(x \cdot y \cdot \frac {1}{xy}) \geq f(1) \equal{} \frac {1}{8}$

as desired.[/hide]

[quote="ghjk"]The final step is wrong, it should be<= not >=
Hint: The key idea is proving:1/(1+x)^3+1/(1+y)^3>=1/(1+xy)^3(just by expansion because it maybe wrong! :wink: )[/quote]

mathlink's students are so stupic  :rotfl:

Because of self-esteem, this will be the last post I have posted in your topic, Duy-VN! :mad:
Here's my solution:
We have the following inequality: a^3/(a+b)^3+b^3/(b+c)^3+c^3/(c+a)^3>=3/8(a,b,c>0)(This is the Vietnamese-inequality in national competition)
Now just let b/a=x,c/b=y we will get the LHS in your inequality(DONE!)
PS: Be respectful to others, Duy_VN! If you think you are good, compete in Olympiad in the section-inequality!

***Mods: Please delete the last post by Duy_VN in this topic. It has no beneficial purpose, and only insulted fellow AoPSers.***

[quote="Duy_VN"][quote="ghjk"]The final step is wrong, it should be<= not >=
Hint: The key idea is proving:1/(1+x)^3+1/(1+y)^3>=1/(1+xy)^3(just by expansion because it maybe wrong! :wink: )[/quote]

mathlink's students are so [b]stupic[/b]  :rotfl:[/quote]

Well, you're not one to lead by example.

a) We usually refer to ourselves as AoPSers
b) Since it WAS in the Vietnamese National Olympiad, this is better suited for the Pre-Olympiad forum, where the problem would be easily beasted.
c) Since when was beasted a verb?


$ \frac {AC}{AB} \equal{} \frac {AB}{AD}$
$ AD \times AC \equal{} AB^2$
$ 143 \equal{} AB^2$
$ AB \equal{} \sqrt {143}$
$ DB^2 \equal{} AB^2 \minus{} AD^2$
$ DB^2 \equal{} 143 \plus{} 121 \equal{} 264$
$ DB \equal{} \sqrt {264} \equal{} 2\sqrt {66}$

actually, no.

$ AC$ $ \equal{}$ $ 13$ and $ DC$ $ \equal{}$ $ 2$, so $ AD$ $ \equal{}$ $ 11$.

Then we can use the altitude to hypotenuse geometric mean.

So $ \frac {AD}{DB}$ $ \equal{}$ $ \frac {DB}{DC}$
$ \frac {11}{DB}$ $ \equal{}$ $ \frac {DB}{2}$
$ DB^2$ $ \equal{}$ $ 22$
$ DB$ $ \equal{}$ $ \boxed{\sqrt {22}}$


Tom bought a $ 30 jacket at a 20 percent discount and a$20 shirt at a 30 percent discount. What is the single percentage discount he received on his total purchase?

[hide="Solution"]

$ 100(1\minus{}\frac{4/5\cdot20\plus{}7/10\cdot20}{30\plus{}20})\equal{}40$.

[/hide]

Please preview before submitting.


Given a,b,c are 3 possitive real numbers prove that
$\frac{a^3}{a^2+b^2+a^2b}+\frac{b^3}{b^2+c^2+abc}+\frac{c^3}{c^2+a^2+a^2c}\ge \frac{a+b+c}{a+2}$

[quote="Megus"]It doesn't look correct: what about $a \longrightarrow 0$ and $b=c=\frac{1}{2}$ ? Then left side is something about $\frac{3}{4}$ while right side is about $1$.[/quote]
are you sure?  the right side is about $1/2$

sorry, my mistake - eh... i can't count  :blush:  :blush:  :blush:

this ineq is nice after all - i can't complete my proof but here's what might be useful:

we can rewrite it as:

$\frac{a^2}{b (\frac{a}{b} + \frac{b}{a}+a)}+\frac{b^2}{c(\frac{b}{c}+\frac{c}{b}+a)}+\frac{c^2}{a(\frac{c}{a}+\frac{a}{c}+a)} \geq \frac{a+b+c}{a+2}$

and it is obvious that $\frac{c}{a}+\frac{a}{c}+a \geq a+2$ though I dont know how to use it in this case.

\[
\begin{array}{l}
 a^2 b \le \frac{1}{2}a\left( {a^2  + b^2 } \right) \\
 abc \le \frac{1}{2}a\left( {b^2  + c^2 } \right) \\
 a^2 c \le \frac{1}{2}a\left( {a^2  + c^2 } \right) \\
 \end{array}
\]

use this inequalities to get the result.

are you sure?I think it will be:
$\frac{{a^3 }}{{a^2 + b^2 + a^2 b}} + \frac{{b^3 }}{{b^2 + c^2 + abc}} + \frac{{c^3 }}{{c^2 + a^2 + a^2 c}} \ge \frac{2}{{a + 2}}\left( {\frac{{a^3 }}{{a^2 + b^2 }} + \frac{{b^3 }}{{b^2 + c^2 }} + \frac{{c^3 }}{{c^2 + a^2 }}} \right) $
sorry for my not have the ability to complete your solution?
I think it is nice.but still have a problem.

Sorry for making a mistake there.
This proof should be ok now:
\[
\frac{{a^3 }}{{a^2  + b^2 }} \ge \frac{{2a - b}}{2}
\]

[quote="siuhochung"]
$\frac{{a^3 }}{{a^2  + b^2 }} \ge \frac{{2a - b}}{2}
\$][/quote]
yes.It is ok now.I think it for half an hour :)


lol i dont think so, its just two lines, and it has no asymptotes or whatever they are called? im not sure ;)

[quote="ZetaX"]No it's not a true hyperbola (there are no foci), but a degenerated conic intersection.[/quote]

ok, shweet :D thanks


Does anyone know where there previous AMTNJ competitions?


parameterization
calculus
calculus computations

Please someone help me to verify my answer. I think my answer is right but most of my classmates and my teacher purposed different answer. I've think of it, but i still find my answer is right. Really annoying they didn't accept my argument  :( . This is about Gamma distribution.

[b]Problem[/b]
Show that in life testing, the waiting time until the death happen has Gamma distribution.

We are all agree that it has Gamma distribution, but I have different parameter.
Here my answer (please verify this???):
Let the  random variable $W$ be the time until the death happen. Then the distribution of $W$ is \[ G(w) = \text{Pr}(W \leq w) =1- \text{Pr}(W > w)  \]
Look at   $\text{Pr}(W > w)$, if the event $W > w$ happens, it means that the time until the death happen is greater than $w$. This is equivalent with the event, that before $w$  the death doesn't happen or there is no death before $w$. This is in relation with Poisson distribution, that is there is no death in time interval of length $w$. If  random variable $X$ is the number of changes in an interval of length $w$, that has Poisson distribution then \[ Pr(W>w) = Pr(X=0) = e^{- \lambda w}  \]
So $G(w) = 1 - e^{- \lambda w}$, and the p.d.f of $W$ is \[ g(w) = G'(w) = \lambda e^{- \lambda w}  \]
Since the p.d.f of Gamma type with parameter $\alpha$ and $\beta$ is \[ f(x) = \frac{1}{\Gamma (\alpha) {\beta}^{\alpha}} x^{\alpha -1} e^{- \frac{x}{\beta}}  \] So $g(w) = \lambda e^{- \lambda w} = \frac{1}{\Gamma (1) \left(\frac{1}{\lambda} \right)^{1}} w^{1 -1} e^{- \frac{w}{\frac{1}{\lambda}}}$
So it has Gamma distribution with parameter $\alpha = 1$ and $\beta = \frac{1}{\lambda}$.

ALL of my classmates who also solve this  said that i was wrong and also my teacher  :(  :mad: . they come with parameter $\alpha = k$ and $\beta = \frac{1}{\lambda}$.  BUT I think they answer can be right IF the the problem is modified to " $k$ death " not "the death happen" .
So Please.......... It is REALLY annoying that no one in my side. If i were wrong just tell me.
I must say many thanks to smeone who give me a reply. :)

[quote="Beginner"]
ALL of my classmates who also solve this  said that i was wrong and also my teacher  :(  :mad: . they come with parameter $\alpha = k$ and $\beta = \frac{1}{\lambda}$.  BUT I think they answer can be right IF the the problem is modified to " $k$ death " not "the death happen" .
[/quote]
Hmm... What is $k$? There seems to be no $k$ in the problem as you stated it. Most likely, you, indeed, just solved different problems.  :?

I'm also confuse with that one. Why there is $k$?
the problem just say "until the death happen". NOT "Until $k$ death happen". I've read the problem a few time.

Any other comment/reply,  please?
am I right ?

I found this in  [i]introduction to mathematical statistics, 5th edition[/i]  [i]Hoog & Craig[/i].
[quote="introduction to mathematical statistics, 5th edition . Hoog & Craig,"] Let the random variable $W$ be the time that is needed to obtain excactly $k$ changes (possibly death), where $k$ is a fixed positive integer. Then the distribution function of $W$ is \[ G(w)= \text{Pr}(W \leq w) = 1 - Pr(W >w).  \] However, the event $W>w$, for $w>0$, is equivalent to the event in which there are less than $k$ changes in a time interval of length $w$. That is, if the random variable $X$ is the number of changes in a time interval of length $w$, then \[ \text{Pr}(W>w) = \sum_{x=0}^{k-1} \text{Pr}(X=x) = \sum_{x=0}^{k-1} \frac{(\lambda w)^x e^{- \lambda w}}{x!}.  \] It is left as an exercise to verify that \[ \int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{(k-1)!} \, dz = \sum_{x=0}^{k-1} \frac{(\lambda w)^x e^{- \lambda w}}{x!}.  \] If , momentarily, we accept this result, we have, for $w>0$, \[ G(w) = 1 - \int_{\lambda w}^{\infty} \frac{z^{k-1} e^{-z}}{\Gamma(k)} \, dz = \int_{0}^{\lambda w} \frac{z^{k-1} e^{-z}}{\Gamma(k)} \, dz,  \] and for $w \leq 0$, $G(w)=0$. If we change the variable of integration n the integral that defines $G(w)$ by writing $z= \lambda y$, then \[ G(w) = \int_{0}^{w} \frac{\lambda^k y^{k-1} e^{-\lambda y}}{\Gamma (k)} \, dy , \qquad w>0  \] and $G(w)= 0$, $w \leq 0$. Accordingly, the  p.d.f of $W$ is \[ \begin{ eqnarray*} g(w) = G'(w) = \frac{\lambda^k y^{k-1} e^{-\lambda y}}{\Gamma (k)} \, dy , \qquad w>0  \] \[ =0 \qquad \text{elsewhere} \end{eqnarray*}  \] That is $W$ has gamma disribution with $\alpha =k$ and $\beta= \frac{1}{\lambda}$  [/quote]

I think what my friends do is the problem like in this book. But the actuall problem is different, here we don't  deal about "$k$ death occur" but only  the first death occur.
Reply please.....

[quote="Beginner"]Any other comment/reply,  please?
am I right ?
I think what my friends do is the problem like in this book. But the actuall problem is different, here we don't  deal about "$k$ death occur" but only  the first death occur.
Reply please.....[/quote]
I do not see anything wrong with your argument. What I meant is why don't you ask your classmates or your teacher what $k$ is in their answer? They definitely should have an answer to that and I believe that this answer will clarify everything right away. As I said, the most probable thing is that you and your classmates solved two different problems. Most likely, you are also right that the problem they solved was about $k$ deaths instead of one. But the only way to find that out for sure is to ask them a question.

I had a situation like this in my class once when one student came up with a different answer from everybody else. That time the cause turned out to be that he had a different edition of the textbook and the problem in his (older) edition was slightly different from the same number problem in the newer edition everybody else was using (so, he was right after all, but it took some time to figure that out). I suspect that something like this happened to you but I can only guess here.

Thanks for the reply, but now we are on holiday.
Its okay when I'm right.
Again Thanks for the reply.


number theory
Additive combinatorics
Additive Number Theory
counting
IMO Shortlist

The integer $ 9$ can be written as a sum of two consecutive integers: $ 9 \equal{} 4\plus{}5.$ Moreover, it can be written as a sum of (more than one) consecutive positive integers in exactly two ways: $ 9 \equal{} 4\plus{}5 \equal{} 2\plus{}3\plus{}4.$ Is there an integer that can be written as a sum of $ 1990$ consecutive integers and that can be written as a sum of (more than one) consecutive positive integers in exactly $ 1990$ ways?

Yes.
From $n=(x+1)+\dots+(x+1990)=1990x+1989\cdot 995$ we get $n$ - odd and $n$ is a multiple of $995$.
Suppose $n=(x_{i}+1)+\dots+(x_{i}+k_{i})$, for $i=\overline{1,1990}$ and one of the $k_{i}$'s is 1990. Also $k_{i}>1$.
Then $2n=k_{i}(2x_{i}+k_{i}+1)$. For any $k_{i}|n$, $k_{i}<\sqrt{n}$ there is an integer $x_{i}=\frac{1}{2}\left(\frac{n}{d}-k_{i}-1\right)$.
So $2n$ has exactly $1990$ divisors which are $>1$ and $<\sqrt{n}$, which means that $2n$ has $2\cdot1991=2\cdot11\cdot181$ divisors.
Since $2n$ has already $3$ prime divisors: $2,5,199$ we conclude that these are the only prime divisors of $2n$. Now since $n$ is odd, it is clear that $n$ must be $5^{10}\cdot199^{180}$ or $5^{180}\cdot199^{10}$. So, such $n$ exists, but these are the only values $n$ can take.

yes $ 5^{10}\cdot199^{180}$ has exactly 1990 divisors, but no all are neccessery $ <\sqrt{n}%
$

I didn't exactly understand your question, but I did make some typos in the solution:

[quote="freemind"] Then $ 2n \equal{} k_{i}(2x_{i} \plus{} k_{i} \plus{} 1)$. For any $ k_{i}|n$, $ k_{i} < \sqrt {n}$ there is an integer $ x_{i} \equal{} \frac {1}{2}\left(\frac {n}{d} \minus{} k_{i} \minus{} 1\right)$.
So $ 2n$ has exactly $ 1990$ divisors which are $ > 1$ and $ < \sqrt {n}$, which means that $ 2n$ has $ 2\cdot1991 \equal{} 2\cdot11\cdot181$ divisors.
[/quote]

should be:

___
Then $ 2n \equal{} k_{i}(2x_{i} \plus{} k_{i} \plus{} 1)$. For any $ k_{i}|n$, $ k_{i} < \sqrt {2n}$ there is a positive integer $ x_{i} \equal{} \frac {1}{2}\left(\frac {2n}{k_i} \minus{} k_{i} \minus{} 1\right)$.
So $ 2n$ has exactly $ 1990$ divisors which are $ > 1$ and $ < \sqrt {2n}$, which means that $ 2n$ has $ 2\cdot1991 \equal{} 2\cdot11\cdot181$ divisors.
___

Hope this helps.

Let $n$ be a number with the desired properties.

By the first condition we get $n=x+(x+1)+\cdots+(x+1989)=995(2x+1989) \Rightarrow 5\mid n,199 \mid n$ and $n$ is odd.

By the second condition there exists $1990$ pairs of $(j,k)$ such that $n=j+(j+1)+\cdots+(j+k)$,or $2n=(k+1)(k+2j)$.It is also easy to see that $k+1<k+2j$ so $\frac{d(2n)}{2}-1=1990 \Rightarrow d(2n)=3982$.Now it is natural to set $2n=2\times 199^{y_1} \times 5^{y_2} \times {p_1}^{y_3} \times \cdots {p_m}^{y_m}$.Then $3982=2\times 11\times 181=d(2n)=2(y_1+1)(y_2+1)\cdots(y_m+1)$.We also know that $y_1,y_2 \ge 1$.This forces us to conclude that either $y_1+1=11,y_2+1=181$ or $y_1+1=181,y_2+1=11$ and the rest $y_i=0$Thus $(y_1,y_2)=(10,180),(180,10)$ and the numbers $5^{180}\times199^{10}$ and $5^{10}\times199^{180}$ satisfy our conditions.So the answer to the problem is:Oh,Yes!

Proof (sketch):

The first condition implies $n \equiv 995$ (mod 1990). So 5 and 199 appear in $n$'s prime factorization. Also note that n has to be odd.

If n can be decomposed into an even number 2k, then $n \equiv k$ (mod 2k). If n can be decomposed into an odd number, d, of consecutive integers, then d $\mid$ n. We can combine these two statements to get an iff for the number of numbers in which odd n can be decomposed: All divisors of 2n work.

Also note that $\sqrt{2n}$ provides the upper bound of divisors. This is convenient, since it just divides the divisors of 2n by 2.

So we just need an n that prime factors 5 and 199 with 1991 divisors. This gives us the two solutions $5^{10} * 199^{180}$ and $5^{180} * 199^{10}$.

I constructed a number $x=LCM[3,5,7,....,2653]$. It probably satisfies both the conditions.
Because it can be written as the sum of $1990$ consecutive integers. All can be written as the sum of $n$ consecutive integers where $n$ is any number which is not divisible by $4$ and $2\le n\le 2654$, that is exactly $1990$. Please  PM me or post here if there is any mistake.


trigonometry
calculus
calculus computations

[quote="vazk"]
\[ \begin{gathered} f\left( x \right) \equal{} \sin ^3 \left( {\sin ^2 \left( {\sin x} \right)} \right) \hfill \\
whats{\text{ }}f'\left( x \right)? \hfill \\
\end{gathered}\]
I got some problems with it[/quote]

We have :

$ f^{'}(x)\equal{}6\, \left( \sin \left(  \left( \sin \left( \sin \left( x \right)
 \right)  \right) ^{2} \right)  \right) ^{2}\cos \left(  \left( \sin
 \left( \sin \left( x \right)  \right)  \right) ^{2} \right) \sin
 \left( \sin \left( x \right)  \right) \cos \left( \sin \left( x
 \right)  \right) \cos \left( x \right)$


calculus
integration
logarithms
calculus computations

[quote="EquinoX"]how do I solve the differential equation of dS/dt = 600-60S(2*10^6) ??

here's what I've did so far:

integral of dS/(10-S/(2*10^6)) = integral of 60 dt
2*10^6 integral of dS/(2*10^7-S)) = integral of 60 dt
2*10^6 ln(2*10^7-S) = 60t + c
ln(2*10^7-S) = 3*10^5t +c
2*10^7-S = e^(3*10^5t +c)
S = 2*10^7 - e^(3*10-5)t + c

is this right?? my feelings say that there is something wrong with this[/hide][/quote]

Is the problem supposed to be:
1. $\frac{dS}{dt}= 600-60S(2*10^{6})$
or
2. $\frac{dS}{dt}= 600-\frac{60S}{(2*10^{6})}$
In either case, you can solve it using either separation of variables or integrating factor method. You have used the first method and if the problem is 1, then your first line is not correct.

it's suppose to be the second one. and I use separation of variables to solve it..but I don't know where I made a mistake

These problems are tough, I've always done them in the case where an initial condition is given to find $C$.  Your error was your brought $C$ to the outside when it was exponentialized when really you should of just tacked it in from of your exponent.  See here (I think this is right, it gets spotty with the $\pm$ due to having no initial condition):

[hide]$\frac{dS}{dt}= 600-\frac{60S}{(2*10^{6})}$
$\frac{dS}{dt}= 600-\frac{3S}{(10^{5})}$
$\frac{dS}{dt}= \frac{6 \cdot 10^{7}-3S}{(10^{5})}$
$10^{5}\cdot \frac{dS}{6 \cdot 10^{7}-3S}= dt$
$10^{5}\int \frac{dS}{3S-6 \cdot 10^{7}}=-\int dt$
$10^{5}\cdot \frac{1}{3}\cdot \ln |3S-6 \cdot 10^{7}|=-t+C$
$\ln |3S-6 \cdot 10^{7}|=-\frac{3}{10^{5}}\cdot t+C$
$|3S-6 \cdot 10^{7}|= e^{(-\frac{3}{10^{5}}\cdot t+C)}$
$|3S-6 \cdot 10^{7}|= e^{(-\frac{3}{10^{5}}\cdot t)}\cdot e^{C}$; Because $e^{C}$ is just another constant, we can write this again as $C$.
$|3S-6 \cdot 10^{7}|= C \cdot e^{(-\frac{3}{10^{5}}\cdot t)}$; This is where you would test your initial condition to find $C$.
$3S-6 \cdot 10^{7}= \pm C \cdot e^{(-\frac{3}{10^{5}}\cdot t)}$
$S=2 \cdot 10^{7}\pm \frac{1}{3}\cdot C \cdot e^{(-\frac{3}{10^{5}}\cdot t)}$
$S=2 \cdot 10^{7}+C \cdot e^{(-\frac{3}{10^{5}}\cdot t)}$; The $\frac{1}{3}$ can just be taken in with the $C$ as can the $\pm$.

$\boxed{S=2 \cdot 10^{7}+C \cdot e^{(-\frac{3}{10^{5}}\cdot t)}}$[/hide]


geometry
3D geometry
incenter
circumcircle
combinatorial geometry
algebra
factorization

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=140150
Find the least positive integers $m,k$ such that
a) There exist $2m+1$ consecutive natural numbers whose sum of cubes is also a cube.
b) There exist $2k+1$ consecutive natural numbers whose sum of squares is also a square.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=140151
If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then
\[AI+BI+CI\leq 3R \]
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=140152
Let $M, N$ be points inside the angle $\angle BAC$ usch that $\angle MAB\equiv \angle NAC$. If $M_{1}, M_{2}$ and $N_{1}, N_{2}$ are the projections of $M$ and $N$ on $AB, AC$ respectively then prove that $M, N$ and $P$ the intersection of $M_{1}N_{2}$ with $N_{1}M_{2}$ are collinear.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=140153
Consider five points in the plane, no three collinear. The convex hull of this points has area $S$. Prove that there exist three points of them that form a triangle with area at most $\frac{5-\sqrt 5}{10}S.$

Thanks to iura  :wink:


Solve in real numbers:

$ x^5\equal{}5y^3\minus{}4z, y^5\equal{}5z^3\minus{}4x, z^5\equal{}5x^3\minus{}4y$

http://www.om.edu.pl/zadania/om/om59_1r.pdf
Try that link again...
Just normal casework :wink:


projective geometry
ratio
vector
geometry solved
geometry

Could anybody show me some propeties about tripolar centroid??? what's it about?

thanks a lot

José Carlos

[i]The definition:[/i] Let ABC be a triangle and P a point in its plane. Let the lines AP, BP, CP meet the lines BC, CA, AB at the points A', B', C', respectively. Let the lines B'C', C'A', A'B' meet the lines BC, CA, AB at the points A", B", C", respectively. Then, the centroid of triangle A"B"C" is called the [i]tripolar centroid[/i] of the point P with respect to triangle ABC.

[i]Remark.[/i] It is well-known (and follows from Desargues, applied to the triangles ABC and A'B'C') that the points A", B", C" lie on one line (the so-called [i]tripolar[/i] of the point P with respect to triangle ABC), so that the triangle A"B"C" is degenerate. But still, we can speak of a centroid of this triangle. In fact, if X, Y, Z are any three points, then we can always define the centroid of triangle XYZ as the point which divides the segment between the point X and the midpoint of the segment YZ in the ratio 2 : 1, or, equivalently, as the point P satisfying the vector equation $\overrightarrow{PX}+\overrightarrow{PY}+\overrightarrow{PZ}=\overrightarrow{0}$. Such a point always exists and is unique, even if triangle XYZ is degenerate. Thus, the tripolar centroid is well-defined.

Initially the notion of tripolar centroids was introduced by myself to produce new triangle centers (which would, however, mostly be very far from interesting). Later, Wilson Stothers, an English geometer, came up with some properties of tripolar centroids. See [url=http://groups.yahoo.com/group/Hyacinthos/messages/8809?viscount=34]Hyacinthos messages #8809, #8811, #8813, #8814, #8844[/url].

  Darij


algebra
polynomial
inequalities
absolute value
algebra proposed

Let $a,n$ be integer numbers, $p$ a prime number such that $p>|a|+1$. Prove that the polynomial $f(x)=x^n+ax+p$ cannot be represented as a product of two integer polynomials.

[i]Laurentiu Panaitopol[/i]

Since it's been unanswered for so long, here's a quick proof, albeit non-elementary:

Let $f(z)=z^n+az+p,\ g(z)=z^n+p$. We have $|f-g|=|az|$. For $z$ on the unit circle, this is $=|a|<-1+p=-|z^n|+p\le|z^n+p|=|g|$. By Rouche's Theorem, $f$ and $g$ must have the same number of roots inside the unit disk, and since $g$ has none, all the roots of $f$ are outside the closed unit disk. This proves the irreducibility of $f$, since if it were reducible, one of its factors would have to have free term equal to $\pm 1$, so it would have to have a root with absolute value $\le 1$.

We can also use triangular inequality to prove the fact that all roots are outside the unit disk. This makes it more elementary.

[quote=nmd27082001]Can we use Rouche theorem in olympiad?[/quote]

Should be proved I assume


geometry
rectangle
ratio
geometric transformation
rotation
trigonometry
analytic geometry

[quote="bodan"]We are given a rectangular grid such that the sides of the rectangles are in ratio $ 1:\sqrt{3}.$ Prove that for any lattice points $ X$ and $ Y$ there exist lattice points $ A,B$ and $ C$ such that triangles $ XAB$ and $ YAC$ are equilateral.[/quote]I'm afraid that it's not true for any lattice points $ X,Y...$

In the complex plane let $ O = 0$ and $ P = a\sqrt 3+b i$, where $ a,b\in\mathbb{Z}$

After rotation by 60 degrees left, $ P$ goes to $ Q = x\sqrt 3+y i$

$ \left[\begin{array}{c}x\sqrt 3\\ y\end{array}\right] =\left[\begin{array}{cc}\cos 60^{\circ}&-\sin 60^{\circ}\\ \sin 60^{\circ}&\cos 60^{\circ}\end{array}\right]\left[\begin{array}{c}a\sqrt 3\\ b\end{array}\right]\Rightarrow$

$ \left[\begin{array}{c}x\sqrt 3\\ y\end{array}\right] =\frac{1}{2}\left[\begin{array}{cc}1 &-\sqrt 3\\ \sqrt 3 & 1\end{array}\right]\left[\begin{array}{c}a\sqrt 3\\ b\end{array}\right]$

So we get $ x\sqrt 3 =\frac{1}{2}(a-b)\sqrt 3\Rightarrow\boxed{x =\frac{1}{2}(a-b)}$ and $ \boxed{y =\frac{1}{2}(3a+b)}$

Similarly, if we rotate by $ -60^{\circ}$, we get $ \boxed{x =\frac{1}{2}(a+b)}$ and $ \boxed{y =\frac{1}{2}(-3a+b)}$

At any case, we can see that, $ Q$ is lattice point $ \iff a+b\equiv 0\mod 2$
We'll say that the coordinates of $ Q$ [color=brown]are eliminated under addition in [/color]$ \mod 2$
$ \hline$

Now, let $ X(0,0)$ and $ Y(m\sqrt 3, n)$

For any (lattice) point $ A(a\sqrt 3, b)$, if we want to form equilateral triangles $ XAB,YAC$, it is necessary that the coordinates of the vectors $ \overline{XA},\overline{YA}$ [color=brown]are eliminated under addition in [/color]$ \mod 2$

$ a+b\equiv 0\mod 2\ \ (1)$
and
$ a-m+b-n\equiv 0\mod 2\iff a+b\equiv m+n\mod 2\iff$
$ m+n\equiv 0\mod 2\ \ (2)$

If $ (2)$ holds, then we can find infinite points $ A$, it's enough that $ a,b$ satisfy $ (1)$

But if $ (2)$ doesn't hold [u][b]there is no solution[/b][/u]

Your arguments are true and can be restated as [i]the rectangular grid should be replaced by a regular trianglular one.[/i] (the grid consists of equilateral triangles). In this case $ m$ and $ n$ are always of the same parity.

This is because I derived the problem from the following algebraic one:

[color=darkred][b]Algebraic Problem.[/b] Prove that for any integers $ x,y\in\mathbb{Z}[\omega]$ the system of equations \[ x\equal{}\omega a\plus{}\omega^{2}b\quad\quad\quad\quad y\equal{}\omega a\plus{}\omega^{2}c\] has a solution $ a,b,c\in\mathbb{Z}[\omega]$ (Here $ \omega\equal{}\frac{\minus{}1\plus{}i\sqrt{3}}{2}.$)[/color]

The set $ \mathbb{Z}[\omega]$ is exactly this triangular grid, but when I first expressed it geometrically, I somehow got the rectangular grid.

Try to prove the following more general problem, which is (probably) not as trivial as the previous one:

[color=darkred][b]Problem.[/b] We are given a regular triangular grid and fixed lattice points $ X_{1},\ldots,X_{n\minus{}1}$ and a fixed line $ l$ of the grid. Prove that there exist lattice points $ A_{1},\ldots,A_{n}$ such that triangles $ \triangle X_{1}A_{2}A_{3},$ $ \triangle X_{2}A_{3}A_{4},$ $ \ldots$ $ \triangle X_{n}A_{1}A_{2}$ are all equilateral for some point $ X_{n}\in l.$[/color]

In other words, prove that we can pick a point $ A_{1}$ so that the image we obtain after rotating it $ n\minus{}1$ times along fixed points belongs to some very small set depending on $ A_{1}$ and a fixed line.


modular arithmetic
number theory proposed
number theory

Let $p$ be an odd prime. Prove that:
\[\displaystyle\sum_{k\equal{}1}^{p\minus{}1}k^{2p\minus{}1} \equiv \frac{p(p\plus{}1)}{2} \pmod{p^2}\]

https://artofproblemsolving.com/community/c6h265462

[color=#FF0000]mod edit: see here also:
https://artofproblemsolving.com/community/c6h5001
https://artofproblemsolving.com/community/c6h231922
https://artofproblemsolving.com/community/c6h156047
[/color]

  darij

Sorry to revive but for the sake of posterity here's a solution.

This looks scary, because of the $p^2$, but is it really that bad? We rearrange the hypothesis to \[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} + (p-k)^{2p-1} \equiv p \pmod{p^2}\].  Now observe that by truncation, we can obtain that \[\displaystyle k^{2p-1} + (p-k)^{2p-1} \equiv p(2p-1)k^{2p-2} \pmod{p^2}\] (Truncation is fancy for the binomial theorem) Hence, we need to show that \[\displaystyle\sum_{k=1}^{p-1} (2p-1)k^{2p-2} \equiv 1 \pmod{p}\]. By [b]Fermat's little theorem[/b], \[\displaystyle k^{2p-2} \equiv (k^{p-1})^2 \equiv 1 \pmod{p} \] Now \[\displaystyle\sum_{k=1}^{p-1} (2p-1)k^{2p-2} \equiv  \sum_{k=1}^{p-1} -1 \equiv 1 \pmod{p}\], so we're done.

This is a fantastic problem! I'll post in spite of the above solution essentially being the same as mine  :P

[hide=solution]
[b]Lemma:[/b] $k^{2p-1}+(p-k)^{2p-1} \equiv -pk^{2p-2} \pmod{p^2}$.
Proof: By the Binomial Theorem,
\[ (p-k)^{2p-1} \equiv (-k)^{2p-1}+\dbinom{2p-1}{1}(-k)^{2p-2}p \equiv -pk^{2p-2}-k^{2p-1} \pmod{p^2} \]
and adding $k^{2p-1}$ yields the desired result.

Now by FLT, we have $\sum_{k=1}^{p-1} k^{2p-2} \equiv p-1 \equiv -1 \pmod{p}$.
Therefore,
\[ \sum_{k=1}^{p-1} -k^{2p-2} \equiv 1 \pmod{p} \implies \sum_{k=1}^{p-1} -k^{2p-2} = jp+(p+1) \]
for integer $j$. Multiplying both sides of the equation by $p$ and taking$\pmod{p^2}$ yields that
\[ \sum_{k=1}^{p-1} -pk^{2p-2} \equiv p(p+1) \pmod{p^2} \]

However, by our Lemma, this means that
\[ \sum_{k=1}^{p-1} k^{2p-1}+(p-k)^{2p-1} = 2\sum_{k=1}^{p-1} k^{2p-1} \equiv p(p+1) \pmod{p^2} \]
Since $p$ is an odd prime, $2^{-1}$ exists$\pmod{p^2}$, so we may conclude.
[/hide]


$ {PQR}$ is an equilateral triangle. The point $ {U}$ is the midpoint of $ {PR}$. Points $ {T}$ and $ {S}$ divide $ {QP}$ and $ {QR}$ in the ratio $ {1: 2}$. The point of intersection of $ {PS}$, $ {RT}$ and $ {QU}$ is $ {X}$. If the area of $ \triangle{QSX}$ is $ {1}$ square unit, what is the area, in square units, of $ \triangle{PQR}$

From menelaus' theorem

$ \dfrac{QR}{RS} \times \dfrac{SX}{XP} \times \dfrac{PT}{QT}\equal{}1$

then $ \dfrac{SX}{XP}\equal{}\dfrac{1}{3} \Leftrightarrow SX: XP\equal{}1: 3$

And then,the area of $ \bigtriangleup QSX$ is $ 1$, so $ \bigtriangleup$ the area of $ \bigtriangleup XSR$ is $ 2$.($ QS: SR\equal{}1: 2$)
And the area of $ \bigtriangleup XPU$ is $ 6$ and the area of $ \bigtriangleup XTP$ is $ 3$.($ SX: XP\equal{}1: 3$)

Therefore,$ \bigtriangleup ABC\equal{}\bigtriangleup QSX\plus{}\bigtriangleup XSR \plus{} \bigtriangleup XTP \plus{}\bigtriangleup XPU$=$ 12$

It is easy to see that

$ \bigtriangleup PXU \equiv \bigtriangleup UXR$
$ \bigtriangleup PTX \equiv \bigtriangleup XSR$
$ \bigtriangleup QXT \equiv \bigtriangleup QSX$

So they also have equal area.

 Since $ \bigtriangleup XSR$ and $ \bigtriangleup QSX$ share the same height on sides $ QS$ and $ SR$ so the area of $ \bigtriangleup XSR$ is $ 2$. ( because $ QS: SR\equal{}1: 2$ )

Since the area of $ \bigtriangleup PQS$ equals the area of
$ \bigtriangleup PXT \plus{} \bigtriangleup XTQ \plus{} \bigtriangleup XQS \equal{} 2 \plus{} 1 \plus{} 1 \equal{} 4$  ( we use the equalites above)

and since $ \bigtriangleup PQS$ is exactly one third of $ \bigtriangleup PQR$ (because $ QS: SR\equal{}1: 2$ ) the area of $ \bigtriangleup PQR$  equals $ 12$.

Diagram for above solution:

[geogebra]ad206d211ba0454f2819d67e4f061f47068a5568[/geogebra]

It is well known that $ \frac{XS}{SP}\plus{}\frac{XT}{RT}\plus{}\frac{XU}{QU}\equal{}1$. This is true in any triangle.
Now use Menelaus to find $ \frac{PX}{XS}\equal{}3$, then $ \frac{XS}{SP}\equal{}1/4\equal{}\frac{XT}{TR}$, so using the above formula, $ \frac{XU}{QU}\equal{}1/2$. So the answer is $ 1 \cdot 2 \cdot 3 \cdot 2\equal{}12$.


function
induction
LaTeX
strong induction

I came across a problem that asked to prove the following result using induction:
$ f_r(x)\equal{}(1\plus{}x\plus{}x^{\minus{}1})(1\plus{}x^3\plus{}x^{\minus{}3})...(1\plus{}x^{3^r}\plus{}x^{\minus{}3^r})$,
then
$ f_r(x)\equal{}x^a\plus{}x^{a\minus{}1}\plus{}....\plus{}x^2\plus{}x\plus{}1\plus{}x^{\minus{}1}\plus{}...\plus{}x^{\minus{}a}$ where $ a\equal{}\frac{3^{r\plus{}1}\minus{}1}{2}$.

My only problem in using induction is picking my base case. Should I show the base case for $ r\equal{}0$ and use that to show assuming our result holds for $ r\equal{}n$ implies it holds for $ r\equal{}n\plus{}1$, or should I show it holds for $ r\equal{}1$ as the base case and proceed as in the last sentence? Or is it necessary to prove the base case on both $ r\equal{}0,1$ and use a kind of "weak", strong induction? I doubt the former is necessary.
Thanks for any help!

As for the base case, you probably should choose $ r\equal{}0$.  In general, we want to choose the least $ r$ that still works.  If we were to choose $ r\equal{}1$, then technically we'd still have to prove the case of $ r\equal{}0$ separately (which is sort of a waste of your time, though it's somewhat trivial).

It doesn't really matter whether you want to use the strong or weak form of induction.  Since
$ f_{n\plus{}1}(x) \equal{} f_n(x) (1 \plus{} x^{3^{n\plus{}1}} \plus{} x^{\minus{}3^{n\plus{}1}})$
I would probably just do weak induction.

Start with $ r = 0$. Weak induction is fine since you don't need $ r = 0$ through $ r = n$ to prove $ r = n + 1$. You just need $ r = n$.

This proves that every integer between $ - a$ and $ a$ can be distinctly represented in balanced ternary using $ r + 1$ digits.

Edit: $ \text{\LaTeX}$

Yeah, I was just worried that if I proved the $ r\equal{}0$ base case, using my inductive assumption, I wouldn't have a valid base case to complete the induction. In other words, metaphorically speaking, just because i've proved all the dominoes will fall once I knock one over, I haven't shown they'll all fall 1 by 1. Sorry if that's a little bit confusing.

Think about the relationship between this problem and the ability to uniquely express integers in [url=http://en.wikipedia.org/wiki/Balanced_ternary]balanced ternary[/url]


Να λύσετε στους θετικούς ακεραίους την εξίσωση:
$ (x\plus{}2)(y\plus{}2)(z\plus{}2)\equal{}(x\plus{}y\plus{}z\plus{}2)^2$.

Meta apo kapoies prakseis kateliksa sto na vroume tis thetikes akeraies lyseis tis eksisosis:

                                               $ x^2\plus{}y^2\plus{}z^2\minus{}xyz\equal{}4$ alla kollisa edw.

Any ideas pws mporoume na synexisoume apo edw gia na vroume tis lyseis?
thanks.

~Kostas

Φίλε Κώστα πού ακριβώς βρίσκεσαι στο Cambridge γιατί κι εγώ εκεί είμαι... Αν επιτρέπετε τι ηλικία έχεις. Σπουδάζεις?

Eimai junior sto Harvard kai spoudazw economics.Afto to eksamino den eimai sto Cambridge though :P
But anyways,exeis kamia idea gia afto to provlimataki gt exw kollisei. hehe  :help:
Thanks

~Kostas

[quote="new_member"]Meta apo kapoies prakseis kateliksa sto na vroume tis thetikes akeraies lyseis tis eksisosis:

                                               $ x^2 \plus{} y^2 \plus{} z^2 \minus{} xyz \equal{} 4$ alla kollisa edw.

Any ideas pws mporoume na synexisoume apo edw gia na vroume tis lyseis?
thanks.

~Kostas[/quote]

Ας σκεφτούμε το z σαν σταθερό. Για z=1 βρίσκουμε τις λύσεις (2,1,1) και (1,2,1) και για $ z\equal{}2$ τις λύσεις $ (x,x,2)$. Για $ z > 2$ η εξίσωση γράφεται

$ x^2 \minus{} (z^2 \minus{} 4) \left(\frac{2y\minus{}xz}{4\minus{}z^2} \right) \equal{} 4$

Για κάθε $ z>2$, αυτή η εξίσωση έχει άπειρες λύσεις.  Π.χ. κάθε λύση της $ a^2 \minus{} (z^2\minus{}4)b^2 \equal{} 1$, δίνει την λύση $ (2a,b(4\minus{}z^2)\plus{}az,z)$. Η εξίσωση $ a^2 \minus{} (z^2\minus{}4)b^2 \equal{} 1$ είναι εξίσωση Pell και έχει άπειρες λύσεις.

Δεν ξέρω αν μπορούμε να πούμε κάτι περισσότερο.

[quote="new_member"]Dimitri,exeis lysi gia afti?
Dystyxws exei ksexastei :(

~Kostas[/quote]

Όχι δυστυχώς δεν έχω κάποια λύση αν και την είχα προσπαθήσει αρκετά.....


I don't get how to do these problems...

EDWARD
x        R
--------
MURROW

?

Where did u get this nonsense?!
it doesn't work out (unless repeats are allowed)
R obviously can't be 1 or 0
R also can't be 9 beccause if it is E has to be 1 and D has to be 1 or 0 but it can't be one since E already is and if can't be 0 since that means W has to be 0 as well
If R is 8, E is 1, D is 2, M is 9 and you work it out and it turns out it doesn't work
R can't be 5 because it would mean tens digit of r*r, which is 2 + something *r equals to 5, which doen't work
R can't be 3  because r*r + tensdigit of d*r must be greater than 10 and when it is, a is 4, but that means w also has to be 4.
7 doesn't work because E has to be 1, so d is greator than 1, so 7*7 +tens digit of d*7 has a tens digit of 5, which is sent over to the hundreds digit of the multiplication, which makes A=6, but that means somenumber*7+4=7, which doesn't work!
!!!!!!!! (im getting pissed off :blush:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad:  :mad: )

2 more to go

4 cannot be the value of r because the carry over from the tens digit of multiplication is 1 and there is no number x for 7x+1=4 (mod 4)

YAY one more
im gonna be so mad if this is the right answer
lets see...
.......... it doesn't work and i leave it to you to find out WHY
just to clear some parts in the explanation, by tens digit and all that sh*t im refering to the carry over in the multiplication (is that even the right term)

my explanation for this is either u copieid it wrong and wasted my time or its in a different base (base 230948092 maybe we'll never know)


algebra
polynomial
geometry
perimeter
analytic geometry
function
parameterization

=A subset $ S$ of $ \mathbb R^2$ is called an algebraic set if and only if there is a polynomial $ p(x,y)\in\mathbb R[x,y]$ such that
\[ S \equal{} \{(x,y)\in\mathbb R^2|p(x,y) \equal{} 0\}
\]
Are the following subsets of plane an algebraic sets?
1. A square
[img]http://i36.tinypic.com/28uiaep.png[/img]
2. A closed half-circle
[img]http://i37.tinypic.com/155m155.png[/img]

[quote="Third Edition"]Neither nor.

1. A polynomial vanishing on a set $ A\times B$ with $ A,B$ infinite subsets of $ \mathbb R$ vanishes everywhere.

2. Assuming that the half circle line belongs to the unit disk (wlog), $ S$ is contained in the zero-set of $ f(X,Y)\equal{}X^2\plus{}Y^2$. This implies that of power of $ f$ divides $ p$. Since $ f$ is irreducible, $ f$ divides $ p$. Thus $ p$ must vanish on the complete circle line which is assumed not to be.[/quote]

Well, it's no clear from context if with "an square" you mean the solid figure or its perimeter, and the same with "circle" or "circumference".
More important, the problem does not say anything about the coordinates X, Y being aligned with any of the figures, they may be slanted.

Anyway, if the author of the problem refers to "solid figures", the poin (1) above solves both cases.
If the problem is about the perimeters, then both a straight lines -the ones containing the sides of the square- and circumferences have rational-function parameterizations:

$(at+b, ct+d)$

and

$(x_0+r\frac{2t}{1+t^2},y_0+r\frac{1-t^2}{1+t^2})$

being $t$ the parameter and $a$, $b$, $c$, $d$, $x_0$, $y_0$, $r$ real number constants.

So we have polynomial equations in $t$ with an infinite number of solutions, so should be identically null. Id est:
- a polynomial subset of the plane containing a segment of line should contain the whole line
- a polynomial subset of the plane containing an arc of circumference should contain the whole circumference.

Q.E.D.


$ ABC$ are vertices of the equilateral triangle with sideleght $ 8$ , we draw a semicircle with diameter $ BC$ outside the triangle . A line $ l$ is tangent to the semicircle such that the distance from $ A$ to $ l$ is $ 9$ . Compute the distances from the remaining vertices to $ l$

Let $ A(0,4\sqrt{3}), B(\minus{}4,0),C(4,0)$.

For a line $ p: y\equal{}kx\plus{}n$ to be a tangent of the circle $ x^2\plus{}y^2\equal{}16$, we must have

$ n^2\equal{}16(1\plus{}k^2)\qquad(1)$

The distance of $ A$ from $ p$ is

$ d_A\equal{}\frac{|4\sqrt{3}\minus{}(k\cdot 0\plus{}n)|}{\sqrt{1\plus{}k^2}}$

which yields

$ (4\sqrt{3}\minus{}n)^2\equal{}81(1\plus{}k^2)\qquad(2)$

Dividing $ (2)$ into $ (1)$ we get

$ {4\sqrt{3}\over n}\minus{}1\equal{}\pm{9\over 4}\iff n\in\left\{{16\sqrt{3}\over 13},\minus{}{16\sqrt{3}\over 5}\right\}$

Plugging those values in $ (1)$, we find that the first one doesn't yield a real value of $ k$. Hence $ n\equal{}\minus{}{16\sqrt{3}\over 5}\implies k\equal{}\pm{\sqrt{23}\over 5}$

Since the problem is intrisically symmetric, it doesn't matter which value for $ k$ we choose.

Therefore $ p: y\equal{}{\sqrt{23}\over 5}x\minus{}{16\sqrt{3}\over 5}$

Now $ d_B\equal{}\frac{\left|0\minus{}\left(\minus{}4\cdot{\sqrt{23}\over 5}\minus{}{16\sqrt{3}\over 5}\right)\right|}{\sqrt{1\plus{}{23\over 25}}}\equal{}4\plus{}\sqrt{{23\over 3}}$

Similarly, $ d_C\equal{}4\minus{}\sqrt{{23\over 3}}$

Let's call $D, E, F$ the midpoint of $BC$, the contact point of $l$ with the semicircle (i.e. projection of $D$ onto $l$) and the projection of $A$ onto $l$. Let $MN\parallel EF, M\in DE, N\in AF$. Take also $m(\angle BDM)=\alpha\implies m(\angle BAN)=60^\circ-\alpha$. We get $DM=4cos\  \alpha, AN=8cos\ (60^\circ-\alpha)\ (\ 1\ )$. Because $ME=NF, AF-DE=9-4=5$, we finally get $8cos\ (60^\circ-\alpha)-4cos\ \alpha=5$, or $sin\ \alpha=\frac{5}{4\sqrt{3}}$, a.s.o.

Best regards,
sunken rock

[color=darkred][b][u]PP[/u].[/b] Let $\triangle ABC$ be an equilateral with sidelength $l$ . We draw a semicircle with the diameter $BC$ outside of triangle. A line $d$ is tangent

to the semicircle such that the distance from $A$ to $d$ is $a$ . Compute the distancies $b\ ,\ c$ from the remaining vertices $B\ ,\ C$ to the line $d$ .[/color]

[b][u]Proof[/u].[/b] Denote : the midpoint $D$ of $[BC]$ ; the point $X\in BC\cap d$ and [u]suppose[/u] w.l.o.g. that $C\in (BX)$ ; $T\in d$ so that $DT\perp d$ ;

$P\in d$ so that $AP\perp d$ ; $\phi =m\left(\widehat{DAP}\right)$ . Observe that $AD=\frac {l\sqrt 3}{2}$ and $a=\frac l2+\frac {l\sqrt 3}{2}\cdot\cos\phi$ , i.e. $\cos\phi =\frac {2a-l}{l\sqrt 3}\implies$

$\sin\phi =\sqrt{\frac {2\left(l^2+2al-2a^2\right)}{3l^2}}$ . Thus, $\{b,c\}=\left\{\frac l2\cdot\left(1\pm\sin\phi\right)\right\}\iff$ $ \left\{b,c\right\}=\frac l2\pm\sqrt {\frac {l^2+2al-2a^2}{6}}$ .

[b][u]Compatibility[/u] test[/b] : $l\ \le\ 2a\ \le\ (1+\sqrt 3) l$ .


LaTeX
integration
induction
calculus
calculus computations

Find the sum of the convergent series, $ \sum_{n \equal{} 0}^{\infty}\left[ \left(\frac {2}{3}\right)^n \minus{} \frac {1}{(n \plus{} 1)(n \plus{} 2)}\right]$

Sorry for the bad LaTeX!  :blush:

I can't figure out how to make the brakets/parenthesis longer.
[color=green][As moderator, I just gave you a little editing help. - K.M.][/color]

$ \sum_{n \equal{} 0}^\infty \left(\frac {2}{3}\right)^n \equal{} \frac {1}{1 \minus{} \frac {2}{3}} \equal{} 3$
$ \sum_{n \equal{} 0}^\infty \frac {x^{n\plus{}2}}{(n \plus{} 1)(n \plus{} 2)} \equal{} \sum_{n \equal{} 0}^\infty \int_0^x \int_0^y z^n dzdy$
$ \equal{} \int_0^x \int_0^y\sum_{n \equal{} 0}^\infty z^n dz dy \equal{} \int_0^x \int_0^y \frac {1}{1 \minus{} z}dz$
$ \int_0^x  \ln(1 \minus{} y) dy\equal{}  1 \plus{} (1 \minus{} x)(\ln(1 \minus{} x) \minus{} 1)$

taking limits as $ x\to 1^ \minus{}$ we get $ 1$

so $ 3 \minus{} 1 \equal{} 2$

That's a ridiculously complicated way to sum the telescoping series $ \sum_{n \geq 0} \frac{1}{(n \plus{} 1)(n \plus{} 2)} \equal{} \sum_{n \geq 0} \frac{1}{n \plus{} 1} \minus{} \frac{1}{n \plus{} 2}$.

math92: there are lots of commands like \Big, \Huge, \LARGE, etc., but there's also \left and \right as in the first equation in kenn's post.

[quote="Tyl"]We have $ \sum\limits_{n \equal{} 0}^\infty r^n \equal{} \frac1{1 \minus{} r}$ for $ |r| < 1$, and divergence of the series otherwise.

This can be proven by noting through induction (or telescoping) that $ \sum\limits_{n \equal{} 0}^k r^n \equal{} \frac {1 \minus{} r^{k \plus{} 1}}{1 \minus{} r}$ and then taking the limit $ k \to \infty$. (For $ r \equal{} 1$, note that the terms of the series are constantly $ 1$, easily showing divergence to $ \infty$.)[quote="kunny"][quote="kenn4000"]$ \sum_{n \equal{} 0}^\infty \left(\frac {2}{3}\right)^n \equal{} \frac {1}{1 \minus{} \frac {2}{3}} \equal{} 3$[/quote]

How about $ \sum_{n \equal{} 0}^\infty r^n\ (r\in\mathbb{R})$? :wink:[/quote][/quote]

Yes, that's right.


modular arithmetic
topology
combinatorial geometry
real analysis
real analysis unsolved

Given $A$ a set of points in the plane.
 Assume that $A$ is not convex.
 Let $B$ be the union of the segments whose endpoints belong to $A$, namely, if $M,N$ are in $A$
then all the points of the segment $MN$ are in $B$. Prove or disprove $B$ is convex provided that
  a) $A$ is path-connected
  b) $A$ is connected
 I would be very appreciate if anyone can devise any weaker condition  :lol:

I think the answer is positive in the first case :?

Take any point $P$ belonging to the convex hull of $A$. This point must then be the convex linear combination of a few points in $A$, meaning that it lies inside a convex polygon $A_1A_2\ldots A_n$, with vertices among the points of $A$. For each $i\in\overline{1,n}$, let $\mathcal L_i$ be a continuous path connecting $A_i,A_{i+1}$ (the indices are $\pmod n$).

Now consider the case when there is an $i$ s.t. there is a homotopy $H_i(x,t)$ taking $\mathcal L_i$ to $A_iA_{i+1}$ with the property that $H_i(x,t)=P$, for some $x,t$, and the case when there are no such homotopies for any $i$.

The results seem very intuitive, but I think the proofs should not be that easy, since we deal with continuous curves, and we know what a pain these can be (think about the Jordan Curve Theorem).

[quote="grobber"]To Myth: I understood it to be the usual definition of connectedness: there are no two sets $A_1,A_2$ s.t. $A_1\cup A_2=A$ and $\overline{A_1}\cap A_2=\overline{A_2}\cap A_1=\emptyset$. Or, equivalently, there's no proper subset of $A$ which is both open and closed in the topology induced by the usual topology on $\mathbb R^2$.

Maybe dickchimeny means something else?[/quote]

 Yes, it is the connectedness!!

I didn’t manage to make it very concise, but I’ll try.

We have to show that each point in the convex hull belongs to the set $B$. Take such a point $P$. Since $P$ is in the convex hull, we can interpret this by saying that no matter which line through $P$ we choose, there are points of the set $A$ in both half-planes determined by the line (disregard the otherwise easy to treat limit cases when $P$ lies on the boundary of the convex hull). On the other hand, since we assume that $P$ does not lie on a segment with endpoints in $A$, given a line through $P$, there is a point from $A$ on precisely one of the rays belonging to the line and starting from $P$.

For simplicity, identify each ray starting from $P$ with $x\in[0,1)$ s.t. the ray has the direction of $e^{2\pi ix}$.

If there are no points from $A$ on the rays $x$ and $y>x$, then there must be no points from $A$ in one of the angles determined by the two rays, namely the smaller one (otherwise we could split the set $A$ in two sets, one inside each angle determined by the two rays, and we would contradict the connectedness). This will show fairly easily that there are, infact, no points from $A$ on the rays corresponding to $x\in$an entire interval of length $\frac 12$, closed at one end and open at the other one, and this will contradict the fact that $P$ is in the convex hull.

It’s a bit unclear, and I don’t think I have all the details worked out, but even if it’s flawed, it should give a starting point for further investigations :).

Thanks for the proof!! But can you make it clearer??
 For example, at each step you indicate what you're going to prove?
 By this problem, I'm asking , not quizing! So feel free to explain it clearly for me :D

I'll help, but which part should I explain better?

[quote="grobber"]I didn’t manage to make it very concise, but I’ll try.

If there are no points from $A$ on the rays $x$ and $y>x$, then there must be no points from $A$ in one of the angles determined by the two rays, namely the smaller one (otherwise we could split the set $A$ in two sets, one inside each angle determined by the two rays, and we would contradict the connectedness). This will show fairly easily that there are, infact, no points from $A$ on the rays corresponding to $x\in$an entire interval of length $\frac 12$, closed at one end and open at the other one, and this will contradict the fact that $P$ is in the convex hull.

.[/quote]

- This one. What you want to prove in this part?
- You proved b), right?

Yes :).

I have to apologize about that. Before writing it, I had another message saying "I think [b]b)[/b] works as well (blahblahblah)". After that, I erased my initial message, and posted this one, but forgot to mention what I was proving :). Sorry again. Indeed, that message is concerned with [b]b)[/b].

So correct!! :D  :D
 Thanks, infact b) is the part I need!!  :)

b) is, of course, stronger than a), so we have the problem secured pretty good now :). What weaker conditions did you have in mind?

I have no idea
 But, as you said, the problem is pretty well-secured now. ;)


Let m,n be positive integers with 1 \leq n \leq m-1. A box is locked with several padlocks, all of which be opened to open the box, and all of which have different keys. Each m people have keys to some of the locks. No n people can open the box but any n+1 people can open the box. Find the smallest number l of locks and in that case find the number of keys that each person has !

Let N be the total number of keys.

- If a key C has been distributed to more than m-n people then, by the pigeon-hole principle, each group of n people may open the padlock associated to C. It follows that this padlock is useless...
Thus, if a minimal configuration does exist, each key has not been distribured to more than m-n people.
It leads to N  \leq L(m-n)   (1), where L is the number of padlocks.

- Let P be one of the people.
There are t = C(m-1,n) groups of n people not containing P. Let G_1,...,G_t be these groups. Each of these groups miss at least one key to open the box. Since, for each i, the group G_i \cup {P} is able to open it, it follows that P possess these missing keys.
Let 1 \leq i<j \leq t. Let C(i) (resp. G_i) be a missing key for G_i (resp. G_j)? Since G_i =/= G_j, there exists at least a person Q which belongs to G_i but not to G_j. Since the group G_j \cup {Q} can open the box, it follows that Q has the key C(j). Thus C(j) is not a missing key for G_i. Therefore P has at least one different key for each of the groups G_i's.
Since P is an arbitrary person, we deduce that the total number of distributed keys is N  \geq m*C(m-1,n).  (2)

From (1) and (2), we deduce that L  \geq C(m,n).

To see that this is the minimum we are looking for, it suffices to construct such a situation.
Let k = C(m,n). According to above, it is easy to procced as follows :
Let G_1,...,G_k be all the groups with n people. For each i, we install a padlock S_i, and give the associated key C_i to every person not belonging to G_i. It is easy to verify that in that case the statement of the problem is satisfied.

Thus, the minimal number of padlocks is L = C(m,n). And, in that case, each person possess exactly C(m-1,n) keys since (1) and (2) have to be equalities).

Pierre.


[quote="mdk"]To take off from the ground, an airplane must reach a sufficiently high speed. This is known as the takeoff velocity.  A plane accelerates from rest at a constant rate of 5.00m/s^2 along a runway that is 1800 m
long. Assume that the plane reaches the required takeoff velocity at the end of the runway.

What is the distance traveled by the plane in the last second before taking off?

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway?[/quote]

[hide="Solution"]
To find the answers you need to use the kinematic equations:
$ {v(x)}^2 = v_{0}^2 + 2ax$, and

$ x(t) = x_0 + v_0 t +\frac{1}{2} a t^2$

You know several things, $ v_0 = 0,\, a = 5.00,\, x_{\text{max}} = 1800$

From which you can determine,

$ v_{\text{takeoff}} = \sqrt{2ax_{\text{max}}} =\sqrt{2\cdot 5.00\cdot 1800}$

$ v_{\text{mid-point}} =\sqrt{2a(\frac{1}{2}x_{\text{max}})} =\sqrt{ax_{\text{max}}}$

$ \frac{v_{\text{mid-point}}}{v_{\text{takeoff}}} =\sqrt{\frac{ax_{\text{max}}}{2ax_{\text{max}}}}=\sqrt{\frac{1}{2}}$

$ \Delta x = x\text(T) - x(T-1) = \frac{1}{2}a(T^2 -(T-1)^2) = \frac{1}{2}a(2T - 1)$

Determine $ T$ from $ x_{\text{max}} =\frac{1}{2}aT^2$,

$ T = \sqrt{\frac{2x_{\text{max}}}{a}}$

$ \therefore \Delta x =\frac{1}{2}a(2\sqrt{\frac{2x_{\text{max}}}{a}}-1)$

[/hide]


calculus
integration
calculus computations


$ a_n\equal{}\sum_{i\equal{}1}^n\frac{1}{i^s}$
$ y_n\equal{}\frac{a_n^v}{n}$
Find all $ v$ such that
$ \{y_n\}$ is converge at a point different from 0.

v exist for $ s<1$ and $ v\equal{}\frac{1}{1\minus{}s}$.

[quote="Rust"]v exist for $ s < 1$ and $ v \equal{} \frac {1}{1 \minus{} s}$.[/quote]
You are wrong.For example  try $ s\equal{}2$
It true for all $ s\in R$ except $ s\equal{}1$

If $ s > 1$ we get $ a_n\to const>1$, therefore $ a_n^v\to C>0, y_n\to 0$.

[quote="Rust"]If $ s > 1$ we get $ a_n\to const > 1$, therefore $ a_n^v\to C > 0, y_n\to 0$.[/quote]
I am sorry.
But can you post your solution? You always post without full solution..

When s<1 $ a_n\equal{}\frac{1}{1\minus{}s}n^{1\minus{}s}(1\plus{}o(1))$. It give solution $ v\equal{}\frac{1}{1\minus{}s}.$


[quote="insung0705"]if x> y then [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/be8828a15a5f32eecfc9ee10bfc9bee1.gif[/img]
if x=y then we can see that
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2530fc6258d876683a0054b54139a9f8.gif[/img]
and so 2n is a square[/quote]

you missed x<y

[quote="Parkdoosung"]Assume that positive integer $x,y,n$ satisfy $(n+1)x^{2}-(n-1)y^{2}=n$.
Prove that $2n$ is a sqaure.[/quote]

2n= [(y-x)^2+(y+x)^2] / [(y-x)(y+x)+1]
let 2n=k. y-x=a,x+y=b. a>=0 because if a<0, then n<0 contradiction
if a=0: 2n=4x^2 let, a be a positive integer
k=a^2+b^2/ab+1. if k is integar, k is square. this is imo 1988, problem 6


uh i think theres someone named Iris Aliaj with 16777194 posts??

if you click his name he has 5 stars that says moderator just like admin??

is this a hack??


Actually, if $\{a,b,c,d\}\subset(-1,1)$ then $ab+ac+ad+bc+bd+cd+4>0.$ ;)

[quote="arqady"][quote="riemann abdellah"]but,do u know the solution?or,was it posted before that?[/quote]
O.K.
[quote="riemann abdellah"]let a,b&c integers such that:
  -1<a<1   -1<b<1   -1<c<1
    prove that:
    ab+ac+bc+1>0[/quote]
Since $\{a,b,c\}\subset\mathbb Z\cap(-1,1)$ then $a=b=c=0.$
Hence, $ab+ac+bc+1=1>0.$ :([/quote]

lol , I guess maybe a,b,c should be real  :P

[quote="riemann abdellah"]i didn't mean inetgers,but reels,so,can u solve it?[/quote]
Let $a>0,b>0,c<0.$
Hehce, $ab+ac+bc+1>ab-a-b+1=(1-a)(1-b)>0.$  :)


search
function
complex numbers
number theory unsolved
number theory

Exuse if it has posted before .Prove that if $ab+1|a^2+b^2$ then $\frac{a^2+b^2}{ab+1}$ is a perfect square.

Yes, this has been posted before several times, and yes, it's easy to find this question using the search function (or google for it, or go to the Kalva website).

how can u google or search for this kind of problem. i mean one can hardly just enter "ab+1|a^2+b^2" or similar. so what are the search arguments for searching such a problem?

IMO 1988 Problem 6, perfect square, Engel, ...

these words can be used to search for it :)

That's true, Arne, but if you don't know it comes from IMO, you cannot find it easily...

i had a nice but not findable solution with complex numbers,
I will try to remember it (it was too odd)

To vakiliyan=A little persain please:
avalan ,harfe ketabe bineshe dabirestano copy nakon :D (Aya anan ke midanand ba ana  ke nemidanan barabarand)
2voman_In ayeye ghorane ,bayad benevisi ke kojaye ghoran omade :)

[quote="lomos_lupin"]avalan ,harfe ketabe bineshe dabirestano copy nakon :D   (Aya anan ke midanand ba ana ke nemidanan barabarand)
2voman_In ayeye ghorane ,bayad benevisi ke kojaye ghoran omade  :) [/quote]

I agree with Mr Tavakolli

lomos would U say your solution with Complex number.

I have tired to remember it.
BUt i cant even remember how it begin. :blush:
I am sorry.


jWe3 are given distinct positiveintegers a(1).........,a(20) .The set of pairwise sum {a_i+a_j/ 1 <= i <= j <= 20} contains 201 elements .What is the smallest possibled number of elements in the set {/a_i-a_j/ with 1 <= i < j <20} ,the set of positive differencee between the integers

same problem as: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=201+elements&t=3947 thus topic is locked. please continue discussion there.


trigonometry AMC AIME USA(J)MO USAMO geometry USAMTS \/closed

So, I just looked at the WOOT program that is offered here. I'm considering  taking it next year, but I have a few questions:

Background - How much background knowledge is necessary in order to understand the classes? I think I'd be set for algebra and trigonometry, but aside from that i only know material in the "Intro" series. Would a knowledge of the "Intermediate" books / Aops V2 be sufficient to understand the matieral?

Workload - how much homework is assigned for WOOT? Is it comparable to something like an AP course, or a regular school course, etc..? Will I be able to take another Aops class while doing WOOT (e.g. the number theory seminar)?

My dilemna is that I will be a senior this year, so itll be my only chance to take an interesting class like WOOT(I think its highschool and under, right?) However, I'm probably going to have to work for a good part of the summer in order to pay for it, and I don't want to spend all this time working for a class I'd be totally lost in.

Overall, my goal is really just to learn elementary math and problem solving/proof writing more indepth before I go to college. I've found that the classes offered here are a great way to do that(I learned more in one hour of Aops than in 1 month of school...), although I don't really care about math competition scores. So, I'm not sure if it would be better for me to take "subject" classes here, or to go with WOOT.

So, if anyone could answer some of these questions, or give some recommendations, that would be great. Thanks in advance.

My answers probably aren't too reliable but...

Background knowledge-wise, if you can score moderately well on AIME (maybe 6+) and if you can do most stuff from intro books, it should be okay, I think...

Workload: Well, for each class, you get a handout before it, and you get some problems afterward to work on and discuss. It's not a lot of work. You also once in a while get contests and problem sets to work on.

I think if you're really interested in math, and you like doing challenging problems, you should take this class. I took it this year, and found it quite hard, but I still enjoyed it.

Ye, it was a tough class but very interesting.
Actually, I joined the class this year with a 5 on the 2007 AIME. But I had done AoPS 2 between AIME and September last year, so I suppose Carbon57 has got it right, more or less.
Workload is definitely variable. If you don't have much time, don't do so many problems on the handout. If you do, by all means try and do all of them. You also get 4 Practice Olympiads and 3 Problem Sets (I think) throughout the course, these you should do. (Even though I didn't submit answers to Problem Set 3; extremely busy time for me.)

Finding time to do the problems is almost undisputably the most annoying part of WOOT. I guess it depends on how much other stuff you have to do, but as you'll be a senior next year, there shouldn't be much to worry about in that regard after you've finished apps  :P . But always keep in mind that WOOT is a lot more interesting, more important, and worth your time than English class.

I agree with the previous posters in that WOOT requires lots of time outside of class.  An important difference from school/AP courses is that, for me at least, the time you spend outside need to be in chunks of at least an hour or two if you're going to get any worthwhile problems done.  I was in your position in terms of mostly Intro-level knowledge before the class, and the biggest thing I noticed was that the problems in WOOT take a [i]lot[/i] of time compared to many intro-level problems.  So you might plan on most likely not being able to fully digest every single problem presented during the program, including problem sets, post-class problems, and handout problems.

Background you can pick up; a good indicator is the ability to score reasonably well on the AIME.

Workload is probably going to be the more important consideration for you as a senior.  What they tell you about WOOT is that what you get out of it is what you put into it; the more of the work you do, the more you benefit from the class.  If you plan on doing all the work, you'll have to periodically set aside hours (as archimedes1 says) for the problem sets and the practice olympiads, and as a senior, this just might not be feasible for you.  If you don't plan on doing a lot of the work, it's not really worth the price tag.

[quote="ejr322"]My dilemna is that I will be a senior this year, so itll be my only chance to take an interesting class like WOOT(I think its highschool and under, right?) [/quote]
Although the majority of people who take it are high schoolers (since it's olympiad prep), I don't think there are any rules banning older students from taking it.

tim: you will benefit from it right now.

[quote="timwu"]How high is "reasonably good" for AIME? In terms of USAMO, how high a score is good enough, or, how many problems do you need to be able to solve (not taking account the actual writing)? I scored around 9 in my practice AIMEs (about 3-4) but only got a 6 on this year's. I did one year's USAMO (2005) for preparation and solved 3 not under time limit and my actual score was far worse, 1 on this year's (with one solved problem and some partial). How much do I need to improve in order to benefit from this class?[/quote]

The only thing about the AIME that keeps it from being a better proxy for the fact that you don't need to be able to bubble your answers correctly to do well on USAMO.
I think that you're ready for WOOT. When I took it as a junior, I found it to be extremely hard, and I definitely didn't understand all of the material, but I benefited from it. The great thing about the class is essentially they provide you with enough material to keep you busy no matter how much time you have on your hands. Take my advice with a grain of salt, since I was busy with a lot of things that weren't math the year I took WOOT, but I felt like I was doing maybe a quarter of the work, but I was still spending a lot of time on it and getting a lot out of it. It was pretty much how I prepared for USAMO that year, which was really nice since it meant I didn't need to waste time looking for resources or trying to find good mock tests or anything like that- I just had it all there. It is pricy, but it seems like most things are these days.

Just read my signature to see how I did this year. (0 on USAMO  :help: )
Am I ready? (and my high school, I think, is a high-workload school.  :( )
I've done, like, half of AoPS V2.  :D
And I'm going to AwesomeMath.

blackbelt, you're right in the target audience for WOOT.

Looking at these posts, I'm so happy I did WOOT in 8th grade. Something tells me I'm going to hate all the high school workload along with USAMO practice this year...

on the AIME I, during the test, i correctly solved 7 questions, but my official score was slightly lower and due to my not so good 126 on AMC 10, i didn't make USAMO. before AIME i was at algebra 1 level, and now i have finished geometry and algebra 2/pre-calc and i have worked through Aops volume 2. and recently on the AIME II (taken unofficially) i scored an 8. and on recent AIME's i can score anywhere from 7-9 (except 2007 AIME I where i scored a 5), and on the AIMEs in the 90's (significantly more difficult ones) i score within 4-7, usually around 5 or 6 though, and i get around the double digits on the early ones. i am going into 9th grade, so i am qualified to take WOOT?

and the WOOT brochure was sent to me, does that mean anything, or is it sent to everyone?

cognos - yes, you're coming to WOOT at the right time.  It should be a fit for you.

[quote="diophantient"][quote="cognos599"]and the WOOT brochure was sent to me, does that mean anything, or is it sent to everyone?[/quote]
I think it's sent to everyone who is on the mailing list for things like USAMTS and other things like that, and maybe like most of the people who bubbled "yes" on the AIME for the mailing list thing.

@the AoPS administrators: I heard that WOOT has gotten slightly less difficult (in terms of advancedness of the material) over the years, is this true?[/quote]

It sure would seem that way to a student taking the class multiple years in a row -- the student gets better!

would it be too hard for me if i only know things from the "intro" series?

Probably, yes.  If you don't know anything that is in, say the Intermediate Algebra book, then most of WOOT would be a blur.

ok. i'll try to get some intermideate books when i'm ready for them

[quote="cognos599"]so its okay if i have little skill in USAMO-like tests (if today i were to take USAMO, i might score 0-2), but enough skill to most likely qualify for USAMO.[/quote]

Absolutely.  You'll get plenty of practice with USAMO like problems in WOOT.

Also, try the USAMTS (http://www.usamts.org) this year to get some practice writing solutions.

i actually did round 1 of USAMTS.... i really liked the problems and i plan to do it again this year (except i will do all 4 rounds). i have 1 last question, that is kind of unrelated to this topic but if i register say today, and say in a month (before the class starts) i decide that i am unable to do it, would i be able to cancel the order. the probability of that happening is very close to the probability of winning the lottery and then on the same day get struck by lightning 21 times, but i am just checking.

You can drop by the end of September and receive a full refund.

(Not sure if this is the right place for this, but I didn't want to make another WOOT thread.)
Typically, how many students do you have in WOOT each year?

Sorry about reviving this topic, but is there a "Are you ready" thing for WOOT? Like on a practice USAMO I can solve problem 1 almost all of the times and problem 4 partially. On the AIME II this year I solved 9 of the problems but made a lot of stupid mistakes.

Sounds like you're ready.  The AIME and USAMO are essentially "Are You Ready" tests.  If you feel you can typically handle 9 problems on the AIME, then you're definitely ready.


calculus derivative trigonometry function integration percent limit

okay, until the AoPS Calculus classes start, these are likely going to be trivial. Objective: keep calculus fresh in my mind, especially the slightly more advanced BC stuff the AoPS class won't cover.

I feel like volumes today. Easy:

Find the volume of the solid of revolution produced by rotating the curve $ y \equal{} x^4$ for $ 0\leq{x}\leq3$ around the $ y$-axis.

*cough* contribute to calculus tidbits thread because I'm caught up with two coursework's worth of study.. :D

[quote="cognos599"]blah, i don't know that much calculus.... In calculus class, we are on chapter 2 derivatives.

I, however, know derivatives (power, trig functions, sum/diff/prod/quotient, and maybe others that i cannot think of), I also know basic integrals, ..... but not volumes. ARGH, you make me feel so stupid.... :)[/quote]

what book are yall using

we're still doing delta-epsilon prooofs :dry:

the average score for chapter 2 test (lim is a 55 :|

our book combines calculus AB and BC so our course is called Calculus AB-BC (its an AP course). our book is called CALCULUS: seventh edition. Its by Larson, Hostetler, and edwards.

ps. i HATE the epsilon delta definition of the limit..... ALOT! we did not spend alot of time on it however... (thank goodness).

wait 55 percent???? ouch, what did you get? and what grade is the course supposed to be for (if you didn't skip any courses, when would you take it)...

I don't remember the delta-epsilon business.. do I need it?

I don't use a book XD it's an online applet with blindingly small font that I spend hours decoding. I'm glad it's over.

Okay we use Calculus by Stewart.

I used Calculus: eighth edition for AB Calc. Larson tangentially mentioned the $ \delta\minus{}\epsilon$ definition of a limit, while Stewart, on the other hand, went quite deep. We have "Problems-Plus" at the end of our section, consisting of around 10 problems which are extremely difficult. We have like 5 days to complete 6 of those problems.

We learned how to do really weird $ \delta\minus{}\epsilon$ proofs like proving that $ \lim_{x\to 0} x\left(\dfrac{\lfloor x \rfloor}{x}\right)\equal{}1$ etc.

Um we were supposed to prove that for HW and I haven't proved it yet. :dry:

[quote="undefined117"]Find the volume of the solid of revolution produced by rotating the curve $ y \equal{} x^4$ for $ 0\leq{x}\leq3$ around the $ y$-axis.[/quote]
I'll use cylindrical shells.

$ \int_0^32\pi{x}\cdot{}x^4\;dx$

What we're doing is getting the circumference of the shell, $ 2\pi{x}$, and multiplying it by the height $ y\equal{}x^4$ and integrating over $ x\in[0,3]$. We get $ \boxed{243\pi}$ as our answer.

Next question:

Find: $ \int{e^{\minus{}x^3}\;dx}$.

[hide="Hint"]Use Taylor series[/hide]

$ \displaystyle\int^3_0 e^{\minus{}x^3}\,dx$ and we know that $ \displaystyle\int e^{x} \,dx \equal{} e^x$

so $ \displaystyle\int^3_0 e^{\minus{}x^3}\,dx \equal{} e^{\minus{}3^3} \minus{} e^{0} \approx \minus{}1$

what is wrong with that??? i'm almost sure it is wrong but what is wrong with it?

$ \displaystyle\int^3_0 (1 \plus{} \frac{\minus{}x^3}{1!} \plus{} \frac{(\minus{}x^3)^2}{2!} \plus{} \ldots)\,dx$

now don't you just integrate each of those terms separately.... i guess... i don't know integral calculus yet, i'm just going with what i know.... if im totally off, tell me...

You need to use the chain rule, but it won't work here.

Using Taylor series, we know that $ e^x = \sum_{n = 0}^\infty\frac {x^n}{n!}$. So

\begin{align*}
&\int_0^3e^{-x^3}\;dx\\
&=\int_0^3\sum_{n=0}^\infty(-1)^n\frac{x^{3n}}{n!}\;dx\\
&=\left[\sum_{n=0}^\infty(-1)^n\frac{x^{3n+1}}{(3n+1)n!}\right]_0^3\\
&=\boxed{\sum_{n=0}^\infty(-1)^n\frac{3^{3n+1}}{(3n+1)n!}}
\end{align*}

okay sorry this was a bad problem. We could evaluate that if we wanted, but eh, with Maxima:
[code](%i1) float(sum((-1)^n*3^(3*n+1)/((3*n+1)*n!),n,0,1000));
(%o1) 0.89297951156918[/code]
Which is close (first 11 digits) to the actual value of the integral (calculated with a graphing calculator).

is the chain rule the same for differentiation and integration... ??

No. It's not
Take $ \sin(2x^2)$ for example. You can apply "chain rule" to $ \sin2x$ however, since you're only dividing by a constant which is allowed in integration. You can't divide by variables or functions. Same goes for product rule in differentiation. Usually these problems are solved with integration by parts or by substitution.

An attempt to keep the Indiana forum/this thread from dying:

A cone with base radius 6 and height 10 is stood point-down and filled halfway (in terms of height) with water. By how many degrees must it be tilted before water starts to spill out? (Use a calculator if you want.)

i don't think indiana responds to your level of enthusiasm for calculus  :wink:

UH what is the chain rule for antiderivatives

Example?
Some cannot be solved with easy methods.

[quote="undefined117"]*cry* but calc can be so much fun!

@ryanryan91: solve this. :)[/quote]

ryanryan91 = EPIC BEAST @ Calculus and math at general...... RYANRYAN91, post something.....

Uhh I guess its basically just itnegration by substitution you are talking about?

[quote="#H34N1"]
We learned how to do really weird $ \delta \minus{} \epsilon$ proofs like proving that $ \lim_{x\to 0} x\left(\dfrac{\lfloor x \rfloor}{x}\right) \equal{} 1$ etc.[/quote]

Excuse my intrusion into the Indiana Forum, but I just wanted to comment that the above limit is wrong. The limit does not exist because $ \lim_{x\to 0^\minus{}} x\left(\dfrac{\lfloor x \rfloor}{x}\right) \equal{} \minus{}1$ and $ \lim_{x\to 0^\plus{}} x\left(\dfrac{\lfloor x \rfloor}{x}\right) \equal{} 0$.

To see this, imagine putting in positive x values very close to zero. $ \lfloor x \rfloor$ will be 0, making the value of the entire expression zero. Then imagine putting in negative x values very close to zero. $ \lfloor x \rfloor$ will be -1, making the value of the entire expression equal to -1. Note that the expression itself is undefined at x=0, but this fact does not effect the value of the limit.

[quote="moogra"]what are you talking about?
try integrating sin(x^2) with elementary methods[/quote]

It would be pretty hard (more precise, impossible), as it doesn't have elementary antiderivative (it's a variation on a Fresnel integral).

Yes, that was the point. There is no "chain rule" with integration.

i only know chain rule for differentiation.... guess what, we are starting integration soon in calculus....... i'm so excited.... should i not be??? blah


limit
real analysis
real analysis unsolved

The sequence $\{a_n\}$ satisfies the following recursive formula.

$a_1=1,\ a_{n+1}=\frac{a_1+2a_2+\cdots +na_n}{n+1}\ (n\geqq 1)$

Find $\lim_{n\to\infty} \sqrt[n] {a_{n+1}}$.

$\displaystyle(n+1)a_{n+1} = \sum_{k=1}^{n} ka_k = \sum_{k=1}^{n-1} ka_k + na_n = 2na_n$, so we have $a_n = \frac{2^{n-2}}{n} \; \; \; \forall n>1$.
obviously, its limit is $2$.


arithmetic sequence
combinatorics proposed
combinatorics

Let $ B_i\equal{}\{A_{i1},A_{i2},...\}$ be the set that partitions the set of positive integers where
(i)every element $ A_{ij}$ is a infinite arithmetic progression (with non zero common difference).
(ii) $ B_i$ is in one to one correspondence with $ \mathbb{N}$.

$ B_i$ and $ B_j$ are said to be distinct iff $ \exists A_{ir}\in B_i,A_{js} \in B_j$ so that $ A_{ir} \neq A_{js}$.

Prove that $ C\equal{}\{B_1,B_2,\dots|B_i$ and $ B_j$ are distinct $ \quad  \forall i,j\}$ is in one to one correspondence with $ \mathbb{R}$

Of course we can encode any such arithmetic progression $A$ by two integers meaning that we can encode any $B$ by a sequence of integers so that $C$ has [b]at most[/b] the cardinality of the continuum. So the main issue is to prove that $C$ is uncountable.
It clearly suffices then to construct an [b]injective [/b]mapping from, say, $\{2,3\}^n$ to $C$.
So we start with a sequence $(a_1,a_2,\dotsc)$ with $a_i \in \{2,3\}$ and construct $B=\{A_1,A_2,\dotsc\}$ as follows.
Start with $A_1=a_1\mathbb{N}+1, A_2=a_1\mathbb{N}+2,\dotsc,A_{a_1-1}=a_1\mathbb{N}+(a_1-1)$. The part that remains is $a_1\mathbb{N}$.
We then continue by partitioning this into $A_{a_1}=a_1a_2\mathbb{N}+a_1, A_{a_1+1}=a_1a_2\mathbb{N}+2a_1,\dotsc,A_{a_1+a_2-1}=a_1a_2\mathbb{N}+(a_2-1)a_1$. The part that remain is $a_1a_2\mathbb{N}$ and it should be pretty obvious how to iterate this step.
It is also clear that the sequence $B$ constructed this was is a countable partition of $\mathbb{N}$ as desired and that this mapping is injective, so that we are done. (Of course we could have replaced $\{2,3\}$ by any set of at least two positive integers, excluded the number $1$ which would destroy injectivity).