[url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Ito.html]Kiyoshi Itô[/url] passed away on 17th November 2008. "Kiyoshi Itô, a pioneer in probability theory, died on November 10 in Kyoto, Japan, at the age of 93. Itô's work has had a profound influence on the development of probability theory and has been applied across a wide range of areas. His work centers on the study of random processes revolving around Brownian motion, which can be seen in phenomena as varied as the diffusion of pollen in water and the fluctuation of stock prices. During the 1920s, the great mathematician Norbert Wiener showed Einstein's model of Brownian motion existed as a stochastic process, and some of its key features were established. But Wiener was unable to construct a real stochastic integral, which was an obstacle to progress because the classical tools of integration do not work for random processes like Brownian motion. Itô realized a clever opening in the problem and was able to develop a stochastic integral, which proved to be crucial to the study of the mathematics stemming from Brownian motion. His achievement is all the more impressive when one realizes he did this work in the isolation of Japan during World War II and was unaware of many spectacular developments taking place in probability theory, some of which could have facilitated his work." ([url=http://mahalanobis.twoday.net/stories/5336461/]Mahalanobis[/url])

Some obituaries: [url=http://www.ams.org/news/home-news.html#ito]AMS[/url], [url=http://agonist.org/stirling_newberry/20081124/kiyoshi_ito_creator_of_ito_calculus_dies]Agonist[/url] and [url=http://www.nytimes.com/2008/11/24/business/24ito.html?_r=1]New York Times.[/url]

1977:[url=http://en.wikipedia.org/wiki/Asahi_Prize]Asahi Prize[/url]
1987:[url=http://en.wikipedia.org/wiki/Wolf_Prize_in_Mathematics]Wolf Prize in Mathematics[/url], [url=http://en.wikipedia.org/wiki/Order_of_the_Sacred_Treasure]Order of the Sacred Treasure[/url]
1998:[url=http://en.wikipedia.org/wiki/Kyoto_Prize]Kyoto Prize[/url]in Basic Sciences
2006:[url=http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss_Prize]Carl Friedrich Gauss Prize[/url]

2003:[url=http://en.wikipedia.org/wiki/Person_of_Cultural_Merit]Person of Cultural Merit[/url]
2008:[url=http://en.wikipedia.org/wiki/Order_of_Culture]Order of Culture[/url]

[url=http://www.kurims.kyoto-u.ac.jp/~kenkyubu/past-director/ito/ito-kiyosi.html]Kiyosi Itô (RIMS)[/url]


Hello
[hide="Solution"]
2ooth positive odd number =1+(199)(2)=399
[/hide]
Thank u.

A simple way to answer this question would just to be multiply 200 by the smallest even positive integer, which is 2. This yields 400. Since integers alternate between odd, even, odd, even, etc., we can just subtract 1 from 400 to get the 200th odd integer, which is 399. 399 is the answer.


algebra
polynomial
arithmetic sequence
More Sequences

One member of an infinite arithmetic sequence in the set of natural numbers is a perfect square. Show that there are infinitely many members of this sequence having this property.

Let $a$ be the common difference and $an+b$ the sequence. If $ak+b=m^2$ for some $k,m$, then $\infty$ many squares are given by $(na+m)^2=n^2a^2+2nam+m^2=(k+an^2+2mn)\cdot a + b$.

This immeadiatly generalizes to any powers.

Here is the general of this problem
[b]Problem 1 [/b] If an arithmetic sequence contain infinite   p th power of an integer if and only  if contains infinite $ p^{th}$ power of an integer .
[u]proof[/u] (It is an fundamental of Polynomial congruent) because $ x_n$ is an arithmetic sequence then
$ x_{n \plus{} 1} \equal{} an \plus{} b$
Suppose exist an p th power of an integer in this sequence $ a_{n_0 \plus{} 1} \equal{} k^p$
Imply that $ k^p\equiv \minus{} b (\mod a)$
Consider the number $ (k \plus{} ta)^p$ $ t > 0$
Because $ (k \plus{} ta )^p\equiv k^p\equiv \minus{} b (\mod a)$ so $ (k \plus{} ta)^p$ is an term of the sequence .
So there are many p th power in this sequence.
[b]Problem 2[/b] If a arithmetic sequence contain an $ m,n$ power the it contains mn power . Prove by congruence . where $ \gcd(m,n)\equal{}1$


induction
calculus
calculus computations

\[ \left( 1\plus{}x^{2^0}\right)\left( 1\plus{}x^{2^1}\right)\left( 1\plus{}x^{2^2}\right)\cdots\left( 1\plus{}x^{2^n}\right) \equal{} 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\mbox{ for }n\equal{}0,1,2,\ldots\]
Let
 \[ P (n) : \left( 1\plus{}x^{2^0}\right)\left( 1\plus{}x^{2^1}\right)\left( 1\plus{}x^{2^2}\right)\cdots\left( 1\plus{}x^{2^n}\right) \equal{} 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\].
Then

[i]i[/i]. \[ P (0) : \left( 1\plus{}x^{2^0}\right) \equal{} 1\plus{}x^{2^{0\plus{}1}\minus{}1}\]
[i]ii[/i]. Assume that $ P(n)$ holds for some fixed $ n\geq 0$. Then,
\[ P (n) : \left( 1\plus{}x^{2^0}\right)\left( 1\plus{}x^{2^1}\right)\left( 1\plus{}x^{2^2}\right)\cdots\left( 1\plus{}x^{2^n}\right) \equal{} 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\]
\[ \Rightarrow \left( 1\plus{}x^{2^0}\right)\left( 1\plus{}x^{2^1}\right)\left( 1\plus{}x^{2^2}\right)\cdots\left( 1\plus{}x^{2^n}\right)\left( 1\plus{}x^{2^{n\plus{}1}}\right) \equal{} \left( 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\right)\left( 1\plus{}x^{2^{n\plus{}1}}\right)\]
\[ \equal{} \left( 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\right)\plus{}x^{2^{n\plus{}1}}\left( 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\right)\]
\[ \equal{} \left( 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\right)\plus{}\left( x^{2^{n\plus{}1}}\plus{}x^{2^{n\plus{}1}\plus{}1}\plus{}x^{2^{n\plus{}1}\plus{}2}\plus{}x^{2^{n\plus{}1}\plus{}3}\plus{}\cdots \plus{}x^{2^{n\plus{}1}\plus{}2^{n\plus{}1}\minus{}1}\right)\]
\[ \equal{} \left( 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\right)\plus{}\left( x^{2^{n\plus{}1}}\plus{}x^{2^{n\plus{}1}\plus{}1}\plus{}x^{2^{n\plus{}1}\plus{}2}\plus{}x^{2^{n\plus{}1}\plus{}3}\plus{}\cdots \plus{}x^{2\cdot 2^{n\plus{}1}\minus{}1}\right)\]
\[ \equal{} 1\plus{}x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^{2^{n\plus{}1}\minus{}1}\plus{}x^{2^{n\plus{}1}}\plus{}x^{2^{n\plus{}1}\plus{}1}\plus{}x^{2^{n\plus{}1}\plus{}2}\plus{}x^{2^{n\plus{}1}\plus{}3}\plus{}\cdots \plus{}x^{2^{n\plus{}2}\minus{}1}\]
\[ \Leftrightarrow P(n\plus{}1)\]

Therefore $ P(n)$ holds for $ n\equal{}0,1,2,\ldots$ by induction.

Also, the expoonents on the LHS can be multiplied out as in binary to create every number up to the sum of each. Hence the RHS is the expanded product, where
$ \sum_{i\equal{}0}^n 2^i \equal{} 2^{n\plus{}1}\minus{}1$.


For any natural number $ (a,b)$, $ a^2\plus{}ab\plus{}1$ is divisible by $ b^2\plus{}ab\plus{}1$. Prove that $ a\equal{}b$

We have the condition $ \Leftrightarrow a^2\plus{}ab\plus{}1|a^2\minus{}b^2\equal{}(a\plus{}b)(a\minus{}b)$
As $ gcd(a^2\plus{}ab\plus{}1, a\plus{}b)\equal{}1$, for $ a\minus{}b \neq 0$, we have $ a^2\plus{}ab\plus{}1| \ |a\minus{}b|<max\{a,b\}$, impossible, done.


topology
real analysis
real analysis unsolved

Let $Y$ be a closed subspace of a metric space $(X,d)$ and $g: Y\to\mathbb{R}$-a continious function. Prove that $g$ has a continious continuation $f: X \to \mathbb{R}$.


induction
rotation
function
analytic geometry
binomial coefficients

7.  Consider an infinite chessboard, with numbers in each square. Initially all of the numbers are 0, except for a single 1 on a black square. Go through a series of steps: at every step, replace each number with the sum of its four neighbors (all numbers get replaced simultaneously).

   1. After n steps, what is the sum of all the numbers on the board? What number is in the position of the original 1?
   2. Give as precise a description as you can of the arrangement of numbers on the board after n steps. (Even if you know the answer, you may find it hard to state clearly. Do your best: this is part of the challenge of the problem.)

The design of the chessboard became a (sort of) multiplication table for the nth row of Pascal's triangle. The sum of all the digits were $ 4^n$ and for the black square, I put that it as nC(n/2) if n/2 is an integer. Otherwise, it's zero.

What's this that I hear about a formula for the problem?

Err...the formula is that it's a multiplication table of two Pascal's triangles. More specifically, after [i]n[/i] steps, if a given square is [i]r[/i] diagonal moves upwards-right of the bottom square in the pattern and [i]s[/i] diagonal moves upwards-left, then, for integer [i]r[/i] and [i]s[/i], the value of that square is (nCr)(nCs).

Thus, the center square is actually (nC(n/2))^2.

I proved it using Induction + Pascal's identity:
[hide](b)  Note that at any step, either none of the white squares are filled (we will ignore writing in zeroes, and instead leave the square blank) or none of the black squares are filled, because white can not be reached in an even number of moves and black can not be reached in an odd number of moves, assuming (0,0) is black.  So to make things easier for us, we will work with a square instead of a diamond shape.  To do this we rotate the grid 45 degrees, so that the points with numbers on them are now parallel to the x-axis.  We shift this all so that the square that was originally at the bottom is now at (0,0).  To get an idea of what this actually means, here is a diagram for n = 3:
[code]
      1
    3   3                              1 3 3 1
  3   9   3                            3 9 9 3
1   9   9   1      becomes           3 9 9 3
  3   9   3                           1 3 3 1
    3   3
      1   [/code]
Each point $ (x,y)$ in the first representation becomes a point $ a,b$ in the second.  Specifically, $ (0, - n)$ in the first representation becomes $ (0,0)$ in the second, and $ (0,n)$ becomes $ (n,n)$.

Now, we make our conjecture:  For any point $ (a,b)$ in the second representation, the number on that point after $ n$ steps is $ \binom{n}{a} \binom{n}{b}$.  (We could put $ a$ and $ b$ in algebraic terms of the original $ x$ and $ y$ if we wanted to, but it's just an unnecessary pain.)

As the old saying goes, we proceed by induction on $ n$.  For $ n = 1$, the base case, clearly we have $ \binom{1}{0}\binom{1}{0} = 1$, $ \binom{1}{0}\binom{1}{1} = 1$, $ \binom{1}{1}\binom{1}{0} = 1$, and $ \binom{1}{1}\binom{1}{1} = 1$.  For all other points, the binomial coefficients will be zero, because $ a$ or $ b$ will be greater than $ n$.

Now we assume it is true for $ n = k$, and prove it true for $ n = k + 1$.  First of all, we pick some point $ (a,b)$ in the representation for step $ k$.  Consider the group of four points $ (a,b), (a,b + 1), (a + 1, b),$ and $ (a + 1, b + 1)$.  By the definition of a "`step"' in the problem statement, we know that the number on $ (a + 1,b + 1)$ at step $ k + 1$ will be the sum of the numbers on those four points at step $ k$.  Now we use our assumption that it works for $ n = k$:
\begin{eqnarray*} \binom{k}{a} \binom{k}{b} + \binom{k}{a} \binom{k}{b + 1} \\
+ \binom{k}{a + 1} \binom{k}{b} + \binom{k}{a + 1}\binom{k}{b + 1} & = & \left( \binom{k}{a} + \binom{k}{a + 1} \right) \left( \binom{k}{b} + \binom{k}{b + 1} \right) \\
& = & \binom{k + 1}{a + 1} \binom{k + 1}{b + 1}, \end{eqnarray*}
where the last step follows directly from Pascal's identity.  This is what we wished to prove, so the induction is complete!

[/hide]

Hahaha... On this problem I messed up the first step. I thought the sum was 5 rather then 4, and then the next was 20.... I found formula's for the messed up arrangements >_<

ARG. Maybe if I didn't mess this up I would have gotten in... dangz

Or you could have tried starting earlier. I wrote a 3-page solution to #1 that ended up being way off the mark (I used a repeating sequence which could be chopped off to form a ring, and erroneously concluded that any number of campers would do). A month later, I realized my mistake, and chopped that solution down to 1.25 pages (I tend to leave no room for thinking to the reader in my solutions...).

Yeah... Thats true... I guess I should start on the problems earlier next year... I was rushing to get them finished the night they were due. >_<

I think the generating function approach is rather elegant!

(a1) Shown above are the first two steps.  Each number spawns into four adjacent boxes, so the total quadruples with each step.  Since we start with a total of $ 1$, the total after $ n$ steps is $ 4^n$.

(a2) Consider the paths of length $ n$ that start from the "original $ 1$" and use only [i]right[/i], [i]left[/i], [i]up[/i], or [i]down[/i].  There are $ 4$ paths of length $ 1$ that each end on squares containing $ 1$'s after the first step.  On the second step, these paths may overlap (take $ LD$ and $ DL$ or $ UL$ and $ LU$) on a square containing a $ 2$.  Or, they could each bounce back to the middle square, which has now rought $ 4$.

As suggested, one can easily show inductively that on the $ n$th step, the number in a box coincides with the number of paths of length $ n$ from the middle square.  Therefore, the number in the middle square after $ n$ steps is the number of $ n$-words from the alphabet $ \{R,L,D,U\}$ with just as many $ R$'s as $ L$'s and $ D$'s as $ U$'s (the letters were chosen to indicate direction).  Now, consider the expansion of

\[ \left(R\plus{}\frac{1}{R}\plus{}U\plus{}\frac{1}{U}\right)^n\]

Let $ 1\le i\le n$.  Choosing $ R$ or $ U$ from the $ i$th factor is the same as moving right or up on the $ i$th step.  Similarly, choosing $ \frac{1}{R}$ or $ \frac{1}{U}$ on the $ i$th factor is the same as moving left or down on the $ i$th step.  Thus, to find the number of paths of length $ n$ that end on the middle square, we must find the constant term.  Notice

\[ \left(R\plus{}\frac{1}{R}\plus{}U\plus{}\frac{1}{U}\right)^n\equal{}\left(R\plus{}U\plus{}\frac{R\plus{}U}{RU}\right)^n\equal{}(R\plus{}U)^n\left(1\plus{}\frac{1}{RU}\right)^n\]

Each term in the expansion of $ \left(1\plus{}\frac{1}{RU}\right)^n$ will have the same power in $ R$ as in $ U$, but $ (R\plus{}U)^n$ will have no such terms unless $ n$ is even, in which case $ {n\choose n/2}R^nU^n$ is the only one.  In looking for the constant term when $ n$ is even, we multiply the central terms together,

\[ \left({n\choose n/2}R^nU^n\right)\left({n\choose n/2}\frac{1}{R^n}\frac{1}{U^n}\right)\equal{}\boxed{{n\choose n/2}^2}\]

If $ n$ is odd, the constant term is $ \boxed{0}$.

(b) Actually, the method used in (a2) extends naturally to (b).  Instead of finding the constant, find the coefficient of $ R^xU^y$, where $ (x,y)$ are the coordinates of the sought box.  Let $ a\equal{}\frac{n\minus{}y\plus{}x}{2}$ and $ b\equal{}\frac{n\minus{}x\minus{}y}{2}$.  If $ a$ and $ b$ aren't integers, the square $ (x,y)$ holds a $ \boxed{0}$.  If they are integers, the square holds

\[ \boxed{{n\choose a}{n\choose b}}\]

creating a multiplication table of the $ n$th row in Pascal's triangle.

Wow, that's very clever.  I think I gained a level in my algebra skill after reading that. :D


A right circular cone has height 20 and base radius of length 10. Find the radiuis of its circumscribed sphere

[hide="Solution"]The plane that cuts the cone in half vertically contains a great circle of the sphere, which is also the cricumcircle of the triangle formed at the intersection of the cone and the plane. The radius of this circle is $ 10\sqrt{5}$, which is also the radius of the sphere.[/hide]


$\cot 10 + \tan 5 =$

$\textbf{(A)}\ \csc 5 \qquad
\textbf{(B)}\ \csc 10 \qquad
\textbf{(C)}\ \sec 5 \qquad
\textbf{(D)}\ \sec 10 \qquad
\textbf{(E)}\ \sin 15$

As a sidenote, I remeber this being in AoPS Volume 2.

10000th User's solution is excellent. I have nothing to add.  :)

[hide]is it b? or c? im not sure[/hide]

[color=green][size=75]comment hidden by mod[/size][/color]

Work out everything you can. If you still can't find the answer, post what you've done (and please hide it so that anyone who hasn't done it yet and doesn't want hints can still try), and we can point you to what to do next.


FTW Harvard college AMC AIME Gauss blogs

This is the asme as the previous one that got locked. NOtice- THis is not Inappropriate things people say on FTW.

[11:52] mewto55555: now i can say i pwn
[11:52] FantasyLover: 1696
[11:52] FantasyLover: GOD
[11:52] mewto55555: no was
[11:52] superllamel: rm?
[11:52] mewto55555: i almost had coronary arrest
[11:52] mewto55555: no way
[11:52] superllamel: plz
[11:52] superllamel: in a bit then?
[11:52] THECHAMP: mewto cd?
[11:52] superllamel: if i get my rating higher?
[11:52] mewto55555: VENI VIDI PWNICI!
[11:52] mewto55555: I CAME I SAW I PWNED!

After my stunning defeat of superllamel...

[quote="1=2"][quote="levans"]For those of you who know how to behave, try to find ways to make the chat more useful.[/quote]

Huh[size=42]ii[/size][/quote]

define useful.......................................................

iin77 forbade me to make another one

so my uncle makes it?!?!?!?!?!?!?!?!?!??!

iin isnt even a mod/admin so what he says in that matter doesnt count

iin is a ftw mod =\

YEAH NOW THAT MAKES A LOTTA SENSE!

[quote="tinytim"][11:06] RacismFilter: Hey! No being racist! Next time you'll be kicked or banned![/quote]

Hah! Nice.

that was mine.

for usernames my length, type /me 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 and then a timestamp and whatever you want.

It will come out:

[color=blue]* splatyango 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[12:00] someone: blah whatever[/color]

lol i was telling iin about me gettin eaten

[19:50] [iin77]:
so uh
[19:50] [iin77]:
you should probably like... rub poison on yourself

bleh these quotes suck

except the racistfilter

that was pretty tasty

me celebrating...

[16:28] myyellowducky82: OMG
[16:28] myyellowducky82: I'M OVERLY HAPPY GUYS
[16:28] myyellowducky82: WAT IS WRONG W/ ME
[16:28] myyellowducky82: LOL
[16:28] myyellowducky82: IT TOOK ONE DAY
[16:29] myyellowducky82: WAT A MIRACLE
[16:29] myyellowducky82: LOL
[16:29] myyellowducky82: LOL
[16:29] myyellowducky82: IM SO FUNNY
[16:30] myyellowducky82: LOL
[16:30] myyellowducky82: IM STRANGE LIKE THAT
[16:30] myyellowducky82: ER
[16:30] myyellowducky82: OH WOW
[16:30] myyellowducky82: LOL
[16:30] myyellowducky82: im the only one talking
[16:31] myyellowducky82: lol
[16:31] myyellowducky82: i'll stop...

and

[16:29] myyellowducky82: i can talk!
[16:29] myyellowducky82: i can WALK!
[16:29] myyellowducky82: I CAN WALK MOMMY!

[quote="Squarerootofisoscelestria"][quote="mewto55555"][12:20] crazychinesemaniac: what does vulgar mean
[12:20] SoccerMan101: IDK
[12:21] Squarerootofisoscelestria: -.-
[12:21] crazychinesemaniac: sir the police are here to escort you
[12:21] Jongao: lol
[12:21] crazychinesemaniac: please go quietly
[12:21] Squarerootofisoscelestria: LMAO
[12:21] Squarerootofisoscelestria: NOOOOOOOOOOO
[12:21] crazychinesemaniac: or i will use the duct tape
[12:21] Squarerootofisoscelestria: I DISAPPEARED FOR NO APPARENT REASON!
[12:21] crazychinesemaniac: ur creating a scene as well
[12:21] Squarerootofisoscelestria: UM
[12:22] Squarerootofisoscelestria: I LIKE HOW THAT SOUNDS!!!!!!!!!!
[12:22] crazychinesemaniac: thank you officer...come again[/quote]

Lmao I guess I was getting pretty speechless at the end because I had no idea who he was talking to and why he was saying it.

PS.  How do these people not know what vulgar means?!   :rotfl:  :rotfl:  :rotfl:[/quote]

Acctually that wasnt directed at you  :D . I was talking to someone else, here was the  basic scenario. The person I was talking to (I forgot who) said they were hungry. Then I said that right next door was a Mc D's since I was at my moms work. And I pretended to give him an order. Then we pretended his credit card got rejected and he started making a scene. Finally, I called the cops and shot him if I remember correctly. The end.

[19:05] dannyhamtx: baseball failed
[19:05] (^_^): TU MADRE FAILED
[19:06] dannyhamtx: wat
[19:06] (^_^): TU MADRE IS WAT
[19:06] dannyhamtx: i thought we were twins

[quote="guineapiggies"][quote="mewto55555"][b][size=200]NOtice- THis is not Inappropriate things people say on FTW.[/size][/b][/quote]

umm yeah. i think this stuff is starting to KINDA starting to make this not what this topic is supposed 2 b[/quote]

i'm pretty sure that pg-13 is alright if it's, you know, actually funny, as opposed to "profanity! taboo material! counterculture beliefs! i'm witty!"

[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:10] Uprising42: leetl
[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:10] Uprising42: leet
[22:11] Uprising42: 1337
[22:11] pacman2812: Uprising
[22:11] Yoh2241: Lebg
[22:11] Uprising42: 1337
[22:11] Uprising42: 1337
[22:11] Uprising42: 1337
[22:11] pacman2812: Stop spamming.
[22:11] Uprising42: 1337
[22:11] Yoh2241: LEAA
[22:11] Uprising42: 1337
[22:11] Uprising42: 1337
[22:11] Uprising42: 1337
[22:11] pacman2812: I still have the mod complex
[22:11] Yoh2241: trapped game?
[22:11] Textangle: FloodingFilter: Uprising42 has been murdered for saying 1337!
[22:11] * pacman2812 have mod complex
[22:11] Textangle: yes trapped game
[22:11] Yoh2241: lol
[22:11] Textangle: whats mod complex?
[22:11] pacman2812: I think I'm a mod
[22:11] Yoh2241: what's w/ weird?
[22:12] Textangle: why are you not?

[19:07] mewto55555: the initails of my least favorite person are LE
[19:07] dannyhamtx: !!
[19:07] cherry_pi: hi levans
[19:07] Jongao: YO NO SPEAKO ITLAIONO
[19:07] cherry_pi: hi mewto
[19:07] guineapiggies: hi cherry!
[19:07] levans: mine ar MT
[19:07] cherry_pi: hi piggies
[19:07] mewto55555: levans?
[19:07] guineapiggies: what?
[19:07] dannyhamtx: mine would be DZ
[19:07] dannyhamtx: guess who that is
[19:07] guineapiggies: how does that work?
[19:07] mewto55555: arent your initals le?
[19:07] guineapiggies: yeah
[19:07] mewto55555: CHERRY!
[19:07] cherry_pi: MEWTO!
[19:07] guineapiggies: hmm
[19:07] dannyhamtx: ......
[19:07] cherry_pi: lol
[19:07] levans: mmmhmmm
[19:07] mewto55555: lol u sound like iin
[19:07] guineapiggies: i thought i cracked ur username thing
[19:07] guineapiggies: lol
[19:07] guineapiggies: who levans?
[19:07] * Please use control-enter for private messages
[19:07] mewto55555: do all mods/admins say mmhmm?
[19:07] dannyhamtx: i need to get hold of their conversation and write a mello drama
[19:07] cherry_pi: laaaaaaaa
[19:07] dannyhamtx: and earn billions
[19:07] guineapiggies: i say mhm...
[19:07] Jongao: stop sucking up to levans!
[19:08] guineapiggies: i'm not
[19:08] cherry_pi: laaaaaaaaaa
[19:08] levans: mewto wasn't
[19:08] guineapiggies: oh
[19:08] mewto55555: i wasnt?
[19:08] levans: i'm his least favorite person
[19:08] mewto55555: cool
[19:08] guineapiggies: nvm
[19:08] guineapiggies: lol
[19:08] Jongao: LOL
[19:08] mewto55555: can i be a mod now!
[19:08] guineapiggies: lol
[19:08] Jongao: NO
[19:08] mewto55555: =P jk
[19:08] guineapiggies: don't ask that!
[19:08] cherry_pi: lol
[19:08] guineapiggies: XD
[19:08] levans: Sure!
[19:08] Jongao: only smart people are mods!
[19:08] guineapiggies: whoa!
[19:08] mewto55555: i wont get out of bed for less than admin!
[19:08] mewto55555: whoa
[19:08] guineapiggies: lol
[19:08] Jongao: lol
[19:08] cherry_pi: lol
[19:08] levans: mod means mute, right?
[19:08] Jongao: lol
[19:08] guineapiggies: huh?
[19:08] cherry_pi: |()|
[19:08] guineapiggies: oh
[19:08] guineapiggies: lol
[19:08] dannyhamtx: I need to get hold of mewto and cherry's conversations and write a mello drama and earn billions
[19:09] guineapiggies: ok
[19:09] mewto55555: dude uncalled for
[19:09] dannyhamtx: ......
[19:09] * dannyhamtx died
[19:09] Jongao: HE WILL LOWER 359 PTS FROM YOUR RATING!
[19:09] guineapiggies: lol as long as danny doesn't come 2 my school
[19:09] * mewto55555 has been abused by levans [size=200]
[19:09] levans: That's Mr. dude to you[/size]
[19:09] * guineapiggies says "poor mewmew"
[19:09] guineapiggies: lol
[19:09] * Please use control-enter for private messages
[19:09] mewto55555: LOL
[19:09] mewto55555: copy-paste time
[19:09] guineapiggies: lol

[19:12] mewto55555: cherry?
[19:12] cherry_pi: ???????
[19:12] cherry_pi: sry
[19:12] guineapiggies: lol
[19:12] mewto55555: YOU HATE ME!!!!!
[19:12] guineapiggies: mewmew <3 cherry cherry<3 mewmew
[size=200][b][19:13] mewto55555: THATS MR. MEWMEW![/b][/size]

a bit later.....
[19:18] mewto55555: AND NEVER FORGET THE MISTER AGAIN!

[quote="mewto55555"][19:07] dannyhamtx: mine would be DZ
[/quote]

oh geez, i wonder who could that be... :lol:

[quote="dannyhamtx"][18:01] math154: kkk

Hmmmmmmmmmmmmmmmmmmmmm[/quote]

lmao  :rotfl:

[quote="27r"][11:50] jjx1: brb gotta watch curious george[/quote]

ROFLOL

[16:11] hurdler: i should go do somehting reproductive now

these weren't exact quotes, just paraphrased (and some may seems a little cooky or dreamy) :D

LHM: hi gauss
LHM: are you still muted
*guassy doesn't answer*
nick42: wait gauss is muted?
nick42: by whom?
LHM: i guess it's probably izzy or andrew
nick42: hey izzy!
nick42: hey andrew!
LHM: IZZYANDREW ANDRZY IZZDREW!!!!
*complete silence*

The next day...

gauss1181: hmmm
LHM whispers to me: yay you got unmuted!
gauss1181: ummm, thx?
*gauss1181 is feeling apathetic*
xoangieexo: GAUSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
xoangieexo: i think i found a picture of izzy
gauss1181: WHAT? TELL ME NOW!
*no answer*
gauss1181: TELL ME NOW YOU LITTLE FREAK
xoangieexo: ooo i typed isabella on google
xoangieexo: *link*
gauss1181: no way
xoangieexo: *link* this is guass
xoangieexo: shes hot
xoangieexo: i wish she was a model girl
gauss1181: IM NOT A MODEL GIRL!!!!!!!!!!!!!!!
gauss1181: n00bs
xoangieexo: *posts like a thousand more links*, *says a thousand more times: GAUSSY IS HOT
gauss1181: ewww!
xoangieexo: whats ur real name
LHM: *whispers* is your name _______?
gauss1181: how'd u know that?!?!?!?!
gauss1181: did stargirl tell u
LHM: yeah
LHM: i heard u were 6th place at *insert state here*
xoangieexo: fine then, this is gauss _-------->
nick42: shes too hot then
gauss1181: IM NOT HOT!!!
gauss1181: im not a man's chick

lololololll....

gauss1181: ..
isabella2296: GAUSSY!
gauss1181: um, hi?
*gauss1181 bites her lip*
gauss1181: did my blog hurt anyone?
isabella2296: just for fun
i stared at Gaussy's blog for 300 seconds
then i came to FTW
and my eyes were burning
and then i was like OOUUUUUCCCHHHH
and my mom was asking isabella what's wrong
gauss1181: did u tell your mother about ME?
isabella2296: i was like, "Gaussy's blog hurt my eyes

EDIT: ooops, sorry for reviving :blush:

[quote="gauss1181"][quote="xoangieexo"]FAIL
I never said you were hot
I said YOU SHOULD MODEL
I didn't say i found a pic of izzy.
I didn't say I WISH YOU WERE A MODEL GIRL
I SAID: MODEL POTENTIAL!
fail.[/quote]
that was just a ****ing paraphrase, i didn't remember exactly
didnt you even read the top of the post???? :mad:[/quote]

yes I did
I was showing how your memory FAILS.


quadratics algebra polynomial inequalities

Just read it from the last line to the 1st one.
It follows that the 1st inequality

$x^{6}-2(x^{5}-x^{3}+x)+1\ge 0$

is valid for all $x$.


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

Calculate
$ \int \frac{dx}{sin^2x\plus{}sinx\plus{}1}$

Let $ \sin x \plus{} \frac 12 \equal{} \frac {\sqrt {3}}{2}\tan t$.

[hide="Answer"]$ \frac {2}{\sqrt {3}}\ln \left|\tan ^ { \minus{} 1} \frac {2\sin x \plus{} 1}{\sqrt {3}}\right| \plus{} C.$[/hide]

Neglect the comment! :blush:

kunny

hello, setting $ z\equal{}\tan(x/2)$ i have got
$ \int\frac{2(1\plus{}z^2)}{1\plus{}2z\plus{}6z^2\plus{}2z^3\plus{}z^4}\,dz$
Sonnhard.

[quote="Dr Sonnhard Graubner"]
$ \int\frac {2(1 \plus{} z^2)}{1 \plus{} 2z \plus{} 6z^2 \plus{} 2z^3 \plus{} z^4}\,dz$
Sonnhard.[/quote]

What and then?

[b] kunny[b]
Can you show solution? [/b]

[quote="kunny"]Let $ \sin x \plus{} \frac 12 \equal{} \frac {\sqrt {3}}{2}\tan t$.

[hide="Answer"]$ \frac {2}{\sqrt {3}}\ln \left|\tan ^ { \minus{} 1} \frac {2\sin x \plus{} 1}{\sqrt {3}}\right| \plus{} C.$[/hide][/quote]

It is not so simple.... :ninja:

[quote="aktyw19"]Calculate
$ \int \frac {dx}{sin^2x \plus{} sinx \plus{} 1}$[/quote]

My solution:

Let $ \ I \equal{} \int \frac {1}{\sin ^2x \plus{} \sin x \plus{} 1}\ dx$

Let $ \ \sin x \equal{} \minus{}\frac {(2\minus{}\sqrt {3})t \plus{} (2 \plus{}\sqrt {3})}{t \plus{} 1}\implies \cos x\ dx \equal{} \frac {2\sqrt {3}}{(t \plus{} 1)^2}\ dt$

$ \implies I \equal{} 2\sqrt 3 \int \frac {t \plus{} 1}{\left((6 \minus{} 3\sqrt 3)t^2 \plus{} (6 \plus{} 3\sqrt 3)\right)\sqrt {(4\sqrt {3} \minus{} 6)t^2 \minus{} (4\sqrt {3} \plus{} 6)}}\ dx$

Now consider $ \ I_1 \equal{} \int \frac {t}{\left((6 \minus{} 3\sqrt 3)t^2 \plus{} (6 \plus{} 3\sqrt 3)\right)\sqrt {(4\sqrt {3} \minus{} 6)t^2 \minus{} (4\sqrt {3} \plus{} 6)}}\ dx$

and substitution $ \ (4\sqrt {3} \minus{} 6)t^2 \minus{} (4\sqrt {3} \plus{} 6) \equal{} z^2$ will reduced it to a well known integral.

And consider  $ \ I_2 \equal{} \int \frac {1}{\left((6 \minus{} 3\sqrt 3)t^2 \plus{} (6 \plus{} 3\sqrt 3)\right)\sqrt {(4\sqrt {3} \minus{} 6)t^2 \minus{} (4\sqrt {3} \plus{} 6)}}\ dx$

and first put $ \ t \equal{} \frac 1p$ after this substitution put $ \ (4\sqrt {3} \minus{} 6) \minus{} (4\sqrt {3} \plus{} 6)p^2 \equal{} u^2$ and we are done with the problem.
 :lol:

Sorry, Neglect my post above, that didn't make sense.

kunny

Just like what Sonnhard did, use $ z \equal{} \tan{\frac{x}{2}}.$ Using this substitution, we have

$ \sin{x} \equal{} \dfrac{2z}{1 \plus{} z^2}$ and $ dx \equal{} \dfrac{2dz}{1 \plus{} z^2}.$

Then we obtain

$ \int\frac{2(1\plus{}z^{2})}{1\plus{}2z\plus{}6z^{2}\plus{}2z^{3}\plus{}z^{4}}\,dz$,

but I guess this method is pretty much harder.


probability
algorithm
combinatorics unsolved
combinatorics

This isn't a test problem, it's real-world.  In days long gone by I could have easily solved it myself, but that was then, this is now, and my brain leaks.  The last statistics course I took was 20 years ago, and my textbook didn't survive my moves since.

Here's the issue:  A "modeling agency" that advertises heavily on the Internet is requiring everyone who applies for any modeling job through them to submit a fee for a "background check".  That is unique in my experience, and I'm convinced the "agency" is a scam, using the supposed background check only to generate up-front cash for themselves.  It has been discussed at some length on a modeling forum, and one of the victims of this group talked to the agency owner about it.  She reports to us, "[b]One thing that I can tell you is that they had a (fashion) show booked where 10 girls had been convicted of a felony and therefore couldn't work the show. "[/b]

This seems to me to be so exceedingly unlikely that it is impossible as a practical matter.  (First rule of fibbing:  try to say something that is at least feasible.)  It occurred to me to make a post on the modeling forum that gave the probability of a casting call with that result . . . but then I discovered that I'm old, forgetful, and don't know how to figure it out anymore.

Hence the question.  Since this is a real-world question the data is something of a variable, and although I could take the answer to a representative case or two, I'd prefer to have a formula to calculate the answer based on the input assumptions.

However, to set the stage:

1.  The average large fashion show has 15-25 models in it.  A representative number would be 20.

2.  We could consider that the population from which the fashion show models are chosen is the universe of all 5'10" skinny girls in the world (which is a very large number, effectively infinity for this purpose) or we could assume that the show is cast from a group of models who attended a casting call - typically 100 or so.

3.  We should assume that the distribution of felons among the applicants is random (that is, there is no external way to tell who is a felon when the choices are made).

4.  I have not been able to find a source of the percentage of convicted felons in (a) the general US population, or (b) among women age 16-25; perhaps someone else has a source.  However for purposes of this question, we should assume that the population is drawn from girls age 16-25, and assign (or find, if someone can) representative percentages of felons for that group.  I believe 2% would be a reasonable guess; 5% on the high side, 10% very high, and 20% unreasonably high, but for illustration purposes I'd like to assume a percentage somewhat higher than that in the general population of that type.

I believe that is all the information needed to solve the problem (or to construct an algorithm to solve the problem, and change the variables as need be).

Then the problem is: what is the probability that at least ten randomly selected models in a fashion show would be convicted felons?

If anyone can suggest the answer so that I can inform a group of aspiring models and, perhaps, save them from wasting some time and money, I would be most grateful.

Let me restate your question:  there is a fashion show with $n$ models.  Each model is a felon with probability $p$.  What is the probability that at least $m$ of those $n$ are felons?

The answer, which I'm not going to explain, is $\sum_{i = m}^n {n \choose i}p^i(1-p)^{n-i}$, which just sums the probabilities that $m$ of $n$ are felons, than $m + 1$ of $n$ are felons, and so on.  A computer can give you many values for this sum very quickly, but here is a table that summarizes them:

[code]            n = 15      n = 20      n = 25
p =  2%   2.8 E -14   1.6 E -12   2.5 E -11
p =  5%   2.3 E -10   1.1 E  -8   1.6 E  -7
p = 10%   1.9 E  -7   7.1 E  -6   7.1 E  -5
p = 20%   1.1 E  -4   2.6 E  -3   1.7 E  -2[/code]

So, if 20% of aspiring models are felons and there were 25 models in the show, there is a full 1.7% chance that 10 or more would be felons.  However, if only 2% of aspiring models are felons and there were only 15 in the show, the chance is only about 1 in 360,000,000,000,000.

Every state has a percentage of felons between 0.1% and 7.5%, the vast majority of those in the 4%-or-less neighborhood.  However, more than 90% of the American prison population is male, so we should expect no more than 10% * 7.5% = 0.75% of the American population to be female felons, which would give us about 1.5% of American women as felons.  So, I would say your 2% number is probably quite generous.  (We could also try to take into account various demographic factors.  I could only see them skewing this percentage lower, however.)

Thank you, kind sir.  I have passed it on.

Roger


Let $ A \in M_n(C),p \in N^{*}$ Such that $ A^{p\plus{}1}\equal{}A,Rank(A^{p})\equal{}1$ Prove that :
1) $ \det(I_n\minus{}A^{p})\equal{}0$
2) $ \det(I_n\minus{}A)\equal{}0$ if $ p$ is odd and $ A \in M_n{(R)}$
Here $ I_n$ is $ nxn$ identity matrix.

We solve 1),2) together.

From $ 0<Rank(A)\equal{}Rank(A^{p\plus{}1})\leq Rank(A^p)\equal{}1$ we conclude $ Rank(A)\equal{}1$. Let $ \bar{v}\not\equal{}\bar{0},\lambda\not\equal{}0$ with $ A\bar{v}\equal{}\lambda\bar{v}$. Then $ A^{p\plus{}1}\equal{}A$ implies $ \lambda^p\equal{}1\Longrightarrow A^p$ has eigenvalue $ 0\Longrightarrow\det(I\minus{}A^p)\equal{}0$. If $ A$ is real, we must have $ \lambda\in\mathbb{R}^*$ otherwise the complex conjugate $ \hat{\lambda}\not\equal{}\lambda$ would be another eigenvalue of $ A$ contradicting the rank. Since $ p$ is odd, we must have $ \lambda \equal{}1$, i.e. $ \det(I\minus{}A)\equal{}0$


analytic geometry
geometry
rectangle
conics
parabola
ratio
AMC

[quote="lightrhee"][quote="Sunny"]
EDIT:  BTW, Yiyi?  David?  How'd you find the AHSMC part II?  Kinda easy?  Maybe?  I had an hour left....I drew pictures.   :D[/quote]

I had an hour left too!  :D The first four problems were too easy and the last one wasn't hard either.
I was little bit nervous however, because easy problems meant that many people are going to get all(or most) of the problems. So a small mistake cost much more in this contest. :wallbash_red: I prefer harder problems.[/quote]

Yeah I know eh...its seems like the kinda contest everyone's going to do really well on.  Though I'm not really happy with my writeup for the last question...but oh well.  I drew pictures of dinosaurs. lol.

Well it was quite a bit easier than I expected, and i had an extra half hour by the end. I actually checked through the solution to avoid those 2*5=7 type mistakes, and only ended up correcting spelling errors and capital cases...lol

BTW, I do hope they won't mark on creativity. I did Q1 using coordinates.
Just because. :roll:  :blush:
(No offense to Richard...:rotfl: )

[quote="YY"]
BTW, I do hope they won't mark on creativity. I did Q1 using coordinates.
Just because. :roll:  :blush:
(No offense to Richard...:rotfl: )[/quote]

Bleh, coordinates suck.  lol.

I know, I am against bashing in general.
Just too lazy to think creatively...lol, didn't sleep well last night.

[quote="YY"]I did Q1 using coordinates.[/quote]
 :rotfl:

Sorry for this double post :spam:

I just found out that m+1 and n+1 could be substituted with m' and n' in #5. :wallbash_red:
That makes the expression so simple. Why didn't I see that during the competition? :(
______________________
EDITED: a type found :blush:

[quote="lightrhee"]Sorry for this double post :spam:

I just found out that m+1 and n+1 couble be substituted with m' and n' in #5. :wallbash_red:
That makes the expression so simple. Why didn't I see that during the competition? :([/quote]

Oh wow....lol no kidding.  Well...maybe if I actually thought about math for the last hour....then I would seen something.

I was talking to Dr. Sands, he said Dr. Liu wasn't around for the question making....maybe that's why it was too easy?

[quote="Sunny"][quote="jpark"]

lol, good luck with that.. ;)[/quote]

Just watch.   :P   I might need some help though, hehehehehe.

EDIT:  BTW, Yiyi?  David?  How'd you find the AHSMC part II?  Kinda easy?  Maybe?  I had an hour left....I drew pictures.   :D[/quote]

lol, speaking of which, Ian's proctoring me for APMO.. I could try to steal it..  :roll:

[quote="lightrhee"]Sorry for this double post :spam:

I just found out that m+1 and n+1 couble be substituted with m' and n' in #5. :wallbash_red:
That makes the expression so simple. Why didn't I see that during the competition? :([/quote]

UUUURGHHHHHHHHHHHHHH...I found out last night as well!
That's it! 2 days after contest is officially the best time for contest writing. :!:

And BTW, I do know a non-coordinate solution for Q1. :o

[quote="jpark"]
lol, speaking of which, Ian's proctoring me for APMO.. I could try to steal it..  :roll:[/quote]
Good luck with that. :D  I already got [url=http://www.artofproblemsolving.com/Forum/resources/files/china/65/China-National_Olympiad-2006-65-37.pdf]CMO 2006[/url].

[quote="Sunny"]David?  Yiyi?  Can one of you post it?

I wasn't allowed to keep my question sheet.   :(   And besides, there'd be a big giraffe on it anyways.  No idea why my principal wanted it.[/quote]

I don't have it with me either, though I can remember the questions...
Don't know why they didn't allow me to keep the sheet this year... :|
Mines doesn't contain any giraffes, but does have answers I got. I normally draw on scrap paper.

[quote="YY"]
Mines doesn't contain any giraffes, but does have answers I got. I normally draw on scrap paper.[/quote]

I usually draw on scrap paper too.  But I thought the giraffe woulda been more interesting chewing on the diagram for the first question.   :D

[quote="YawningLion"][quote="TheDreamer"]What is the AHSMC, and where did you guys write it?  Seems like our school doesn't have it.[/quote]

I think it's the Albertan high school contest, so if you want to write it you'd better move to Alberta.[/quote]

Hahaha, seems like Alberta is where all the action is.

But Jenn, where did you write it?

I was allowed to keep my paper.
1.
Suppose the square and the rectangle have the same area and the line containing the horizontal diagonal of the square passes through the centroid of the rectangle(see the picture below). Find [i]l[/i].

2.
A jacket with an original price of $\$100$ was put on sale and the price was reduced by an integer percentage. After a while, the price was reduced again by an integer percentage, this time to $\$79.17$. Find the percentages of the two price reductions.

3.
Let $S(n)$ denote the sum of the (base-ten) digits of a positive integer [i]n[/i]. Find all positive integers [i]x[/i] such that $x=13S(x)$.

4.
A rectangle ABCD has three of its corners lying on the parabola $y=x^2$. $A=(0,0)$, $B=(b,b^2)$ for some b>0, and $D=(d,d^2)$ for some d<0. The fourth corner C lies somewhere above the parabola.

(a) Show that C lies on the parabola $y=x^2+2$.
(b) Find a simple formula for the ratio AB/AD, in terms of b only.

5.
Let A be a set of positive integers such that for every $m, n \in A$, $|m-n| \geq \frac 1{20} (m+1)(n+1)$. What is the maximum number of elements that A can have?

As you can see, they were not hard.

[quote]1. Suppose the square and the rectangle have the same area and the line containing the horizontal diagonal of the square passes through the centroid of the rectangle(see the picture below). Find l.
[/quote]

I don't see the picture below..  :D

Thanks for practice.  The questions were AIME like.

[hide]

1.

Let the diagonal of the square be 2s.  We have $2l = 2s^2$.  If $A$ is the small angle made by the diagonal of the rectangle, then $\tan A = 2/l$, but it is also $s/(l/2 + s)$.  We now have 2 equations in two variables, whereby $l = 4$.

2.

Work in cents.  Say it was reduced by 100-x percent, and then 100-y percent.  The new price will be the old price multiplied by the factor $\frac{xy}{10000}$.  It implies $xy = 7917 = 3*7*13*29$ , which, combined with the constraints $x,y \in [0,100] \cap \mathbb{N}$, the percentages are 87 and 91.  So the price was reduced by 9% and then 13% (or the other way around obviously)

3.

For n, any number with k digits, we have $S(n) \le 9k$ and $n \ge 10^{k-1}$.  Then, if $10^{k-1} \le n = 13S(n) \le 117k$, it implies $10^{k-1} \le 117k$.  It is easy to show that for $k\ge 4$, it doesn't hold.  So for $13S(n) = n$, we must have $n = 100a + 10b + c$ for digits $a,b,c$ (which may equal zero.)  Then, we want $13(a+b+c) = 100a + 10b + c$, simplifying to 4c + b = 29a.  If $a\ge 2$, then the LS is at most 45, which doesn't work.  If $a = 0$ clearly $b=c=0$.  If $a = 1$, then we can have (5,9), (6, 5), (7,1) as solutions (c,b).  In total, the answers are 117, 156, 195.

4.

From vectors, $(b,b^2) \bullet (d,d^2) = 0$, implying $bd(bd+1) = 0 \Rightarrow bd = -1$.

a) Then, $C = (b+d, b^2 + d^2) = (b+d, (b+d)^2 + 2)$ as required.
b) $\frac{AB}{AD} = \sqrt{\frac{b^2 + b^4}{d^2 + d^4}} = \sqrt{\frac{b^2 + b^4}{\frac{1}{b^2} + \frac{1}{b^4}}} = b^3$.

5.

Let A be such a set.  Choose m to be largest.  Expanding, it comes to $m(19-n) \ge 21n + 1$.  If $n \ge 19$, the equation is unsatisfiable.  Thus, the second largest element is at most $18$.  Now choose m to be second largest; since that particular m is $\le 18$, we have $18 \ge \frac{21n+1}{19-n}$ whence $n \le 8$.  Continuing in the same fashion, 5, 3, 2, 1.  So the most # of elements we can have is 7.

[/hide]

[quote="billzhao"][quote="jpark"]I don't see the picture below..  :D[/quote]

The picture is near bottom of his post, as an attachment. Do you still not see it?[/quote]

oh, i see it now.. i'm beginning to think there's something wrong with my computer.. i could see more than 24 quotes too..  :?

[quote="lightrhee"]I just found out the result, but I don't think I can post it here until Monday. Very shocking :o  result
Anyway, I can tell you that I'll be at Calgary on April 1st for the Tournament of the Town and the award ceramony. :)[/quote]

Very stupid mistake.  I checked over my answers a few weeks ago. Wow i really need to work on this mental math thing. :wacko:


Perhaps there's already something like this, but I don't know.

For the math forums (especially High School forums), in addition to moderators, it would be nice if there was a group of "checkers," to look over the threads systematically to check validity of consistency of solutions, and to make sure all questions asked get answered.

I propose this because many users (including myself) intuitively link "0 replies" with "unanswered" and "(positive #) replies" with "answered" and miss that the "reply" was only a further asking for clarification or only answers to part of the question. In such a way, quite a few truly unanswered threads get missed. With "checkers," it would be insured that everyone is satisfied with response to their inquiry.

Logically, checkers would therefore be people who have a huge pool of mathematical knowledge and have a lot of time, so like, genii. It would of course be best if they could take complementary subjects (like, one checks algebra and one checks geometry and so on).

This is just a personal suggestion. Thank you for reading and considering.

--PhireKaLk6781, formerly r15s11z55y89w21

Being a "checker" in your sense does not require any special privileges. Anybody who is qualified can do it [b]if[/b] he has both time and desire (and some people both with and without moderator privileges are already doing things like that). So, I do not see why we should make them a special formal group.

On the other hand, some fora have a notion of "an accepted answer" and can list the questions to which there is no accepted answer. Something like that might be useful to try though it may take some thought to figure out in what form it might be really efficient here.


inequalities
function
calculus
derivative
integration
LaTeX
real analysis

The function $ f(x,y)$ has continuous partial derivaives and $ f(0,0)\equal{}0, |gradf|\leq1$ if  $ x^2\plus{}y^2\leq5$. Prove that $ f(1,2)\leq\sqrt{5}$.

Let $ p: [0,1] \to \mathbb{R}^2$ be defined by $ p(x) \equal{} (x,2x)$.

Then $ f \circ p$ is continuously differentiable and its derivative is $ (f \circ p)'(x) \equal{} \nabla f(p(x)) \cdot p'(x)$, which by the Cauchy-Schwarz inequality is $ \leq 1 \cdot \sqrt {5}$. Also,

$ f(1,2) \equal{} f(p(1)) \equal{} f(p(1)) \minus{} f(p(0)) \equal{} \int_0^1 f(p(x))dx$

so

$ |f(1,2)| \equal{} \left|\int_0^1 f(p(x))dx\right| \leq \int_0^1 \max_{x \in [0,1]}|f(p(x))|dx \leq (1 \minus{} 0) \sqrt {5}$.

The inequality cannot be improved on, as can be seen by letting $ f(x,y) = \frac {x + 2y}{\sqrt {5}}.$ For this function, $ |\nabla f| = 1$ everywhere, and $ f(1,2) = \sqrt {5}.$

This example is really just the flip side of FibonacciFan's proof. He used the Cauchy-Schwarz inequality, so we examine the equality case of that inequality.

$ \text{\LaTeX}$ tip for IOO: the appropriate symbol for the gradient is \nabla. And be sure to leave a space after it: \nabla f, not \nablaf. If you insist on using "grad", make it \text{grad}\,f. (\operatorname doesn't work on this site.)

Thanks,
I'm beginner here and my english, $ LATEX$ is poor. I'm trying to improve....


Let $ a,b,c,t$ be positive numbers such that $ abc\equal{}1$. Prove that

$ (a\plus{}\frac{1}{b}\minus{}t)(b\plus{}\frac{1}{c}\minus{}t)(c\plus{}\frac{1}{a}\minus{}t)\leq (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})(1\minus{}t)^2\plus{}4\minus{}3t$

Let $a$, $b$, $c$ be positive real numbers such that $abc=1$, then for any real number $t$, the inequality

$\left(b+\frac{1}{c}-t\right)\left(c+\frac{1}{a}-t\right)\left(a+\frac{1}{b}-t\right)$

$\leq (2-t)^3+s\left[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9\right]$

holds if and only if $s\geq\frac{2(t-1)^4}{4t^2-7t+4+\sqrt{4t^4-8t^3+9t^2-8t+4}}.$

$t=0$ see : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299899&start=1327

$t=3$ see : http://www.artofproblemsolving.com/Forum/viewtopic.php?t=406993

$t=\frac{1}{3}$ see : http://www.mathoe.com/dispbbs.asp?boardid=107&id=20783&star=3#84873


function
inequalities
integration
calculus
real analysis
real analysis unsolved

[quote="Cezar Lupu"]For any continuous convex at $[0, 1]$ function $f$ prove the inequality $\ds\frac25\int_{0}^{1}f(x)\, dx+\frac23\int_{0}^{3/5}f(x)\, dx\ge \int_{0}^{4/5}f(x)\, dx.$
(A. Prymak)[/quote]

Making in all the integrals the integration interval $[0,1]$ the given inequality can be restated in the following way :
$\int_{0}^{1}\frac25 f(t) dt+\int_{0}^{1}\frac25f(\frac35x) dt \ge \int_{0}^{1}\frac45f(\frac45t) dt.$, but due to convexity of $f$ we have that

$\frac25 f(t)+\frac25 f(\frac35t) \geq \frac45 f(\frac45t)$ which proves the given assertion


Show that every odd number can be expressed as four squares which two are same; in other words

    2k+1=x^2 + y^2 + 2(z^2) for every k (x,y,z in Z )

It is immediately followed by the fact that for all $ x\equal{}1(mod 4)$  $ x$ can be expressed as sum of two squares.

Write $ 4k\plus{}2$ as the sum of three squares, say $ 4k\plus{}2\equal{}a^2\plus{}b^2\plus{}c^2$
then we know that one is even and two are odd in a,b,c. Say a,b are odd and c is even
$ 2k\plus{}1\equal{}((a\plus{}b)/2) ^{2}\plus{}((a\minus{}b)/2)^{2}\plus{}2(c/2) ^{2}$

But can every 4k+2 can be expressed as a^2 +b^2 + c^2? I think I need some more clarification.....

Yeah. All positive integers not of the form $ 4^n(8k\plus{}7)$ are representable as the sum of three squares. I think the proof of this fact is not elementary. You can find a proof on this forum


count the number of pairs (a,b) in N, for which  1/a  +  1/b  =  1/2004

[hide]We can simplify  to.  This means that

I'm not exactly sure, but wouldn't that mean that ?[/hide]

Yes the classic problem resolves into just finidng the factors of (2004)^2.

Now an extension of the problem. (Hopefully it is not too difficult for the level here)

[b]given (1/a)+(1/b) = (1/c)

if a,b,c all are positive integers, and have NO common factors (for all the three numbers)  (In other words GCD of a,b.c is 1)

prove a+b is a square number.[/b]

1/a + 1/b = 1/c

ab = ac + bc

ab - c(a+b) = 0

(a-c)(b-c) = ab - c(a+b) + c 2 = c 2

(a-c)(b-c) = c 2

I think I'm stuck...

Attempt :

$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$
$\Rightarrow a + b = \frac{ab}{c}$

One can see that any prime divisor $p$ of $c$ divides either $a$ or $b$, but not both else $p$ is a common divisor of $a$, $b$, $c$.
           $p | c$ $\Rightarrow$ $p|a$ or $p|b$ but not both --------------- (1)

Now, assume that some prime divisor $p$ of $c$ also divides $ab/c$, then it also divides $a + b$.
           $p | (a + b)$                                     --------------- (2)

By (1) and (2), $p$ divides both $a$ and $b$, and hence $a$, $b$ and $c$ have a common divisor $p$.

Thus, no prime divisor of $c$ divides $ab/c$.

Now, consider any prime divisor $q$ of $ab/c$. It divides either $a$ or $b$. This prime $q$ cannot be a divisor of $c$ as argued before. But as $q$ divides
$ab/c = a + b$, so it divides both $a$ and $b$, and hence, $q^2 | (ab/c)$, as $q$ does not divide $c$.

Thus, if a prime divides $ab/c$, then its square also divides the latter.
As, $a + b = ab/c$, we conclude that $a + b$ is a square.

*******************************************************************************************************
Sambit


function
trigonometry
real analysis
real analysis unsolved

Let $ f: R\minus{}>[a,b]$ a continuous function.
Prove that for every bijective function $ g: R\minus{}>R$,the equation $ f(x)\equal{}g(x)$ has at least one real root.

[quote="xxxxtt"]Let $ f: R \minus{} > [a,b]$ a continuous function.
Prove that for every bijective function $ g: R \minus{} > R$,the equation $ f(x) \equal{} g(x)$ has at least one real root.[/quote]

Warning, the statement  can be  false without other conditions !   Take for exemple $ f(x) \equal{} arctan(x)$ and  $ \ g(x) \equal{} \minus{} x$ if $ x\notin \{0,1\}$ , $ g(0) \equal{} 1 , g(1) \equal{} 0$
 :cool:

Diogene , I believe that you are wrong .Since your function ,g is not injective ,hence not bijective .$ g( \minus{} 1) \equal{} g(0) \equal{} 1$.So you counter example is wrong. Can someone provide a solution to this ?

It isn't true, whether or not that counter-example is correct: let $ f(x) \equal{} \sin x$ and $ g(x) \equal{} \minus{}x$ except $ g(0) \equal{} 1$, $ g(\minus{}1) \equal{} 0$.


abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $ f(x) \equal{} x^n\minus{}1$ and let $ K$ be the splitting field of $ f(x)$ over $ \mathbb{Q}$. Prove that the Galois group $ Gal(K/\mathbb{Q})$ is an abelian group.

An element of the Galois group is uniquely determined by what it does to a primitive root.


[quote="sharkman"]Find the x and y intercepts for y = x^3 - 27.

MY WORK:

I set x^3 - 27 to = 0 and solved for x.

I found x = to be -3 and +3.

I then set x = 0 and solved for y.  I found y to be -27.

My intercepts are:

(-3, 0), (3, 0) and (0, -27)

The textbook tells me that (3, 0) is not an intercept in this case but does not give the reason for it.

Why is the point (3, 0) not an intercept in this case?[/quote]
How does x=-3? If it does, then you're saying that -27=27. Which on course is untrue. The x-intercept is 3 and the y-intercept is -27.

Well, according to the math textbook, the answers are:

Point (0, -27)

Point (3, 0)

and it also tells me that there is no symmetry.

Explain.

[hide]The y-intecept is obviously (0, -27) and the x intercept is (3, 0) only.  To find it, we make y=0.  So $0=x^{3}-27$.  So $x^{3}=27$, and $x=3$.  And to be even more sure, i just put it into my graphing calculator, and it does indeed only cross the x-axis one time.[/hide]


$ (\forall)m,n\in \mathbb{N}$ show that :

$ (\exists)x,y\in \mathbb{N}$ such that $ (m^4\minus{}m^2\plus{}1)(n^4\minus{}n^2\plus{}1)(m^4\plus{}3m^2\plus{}1)(n^4\plus{}3n^2\plus{}1)\equal{}x^2\plus{}y^2$

The terms on the LHS are of the form $ z^4\pm 2z^2\plus{}1\plus{}z^2\equal{}(z^2\pm 1)^2\plus{}z^2\equal{}(z^2\pm 1\plus{}iz)(z^2\pm 1\minus{}iz)$, i.e. if we implicitly define $ x\plus{}iy\equal{}(n^2\plus{}1\plus{}in)(n^2\minus{}1\plus{}in)(m^2\plus{}1\plus{}im)(m^2\minus{}1\plus{}im)\in\mathbb{Z}[i]$ we have $ x^2\plus{}y^2\equal{}(x\plus{}iy)(x\minus{}iy)\equal{}LHS$ and are done


Find the minimum value of x and maximum value of y so that the inequality is true:

$\sum a^{3}+x \sum a^{6} \geq yabc+6a^{2}b^{2}c^{2}$. (a,b,c are the sides of an acute triangle)

cheers! :D

Is this a joke? I mean, isn't it obvious that it is false?

maybe you forgot a 2 before the second sum? But in this case, the inequality becomes trivial.   :?

sorry, but it seems that I certainly forgot something, but I can't remember now what...  :blush:

The problem was like so:
Find the minimum value of x and maximum value of y so that the inequality is true:

$\sum a^{3}+x \sum a^{6} \geq yabc+6a^{2}b^{2}c^{2}$.(a,b,c are the sides of an acute triangle)

I am going to shoot myself!

This is what happens when you don't have Internet at home(for the moment). I just hurry to post.

sorry, again.

Still not far from trivial. Just take $ a=b=c$ and thus $ 3+3a^3x\geq 6a^3+y $ for all a. Thus, making a small and very large we find $ x\geq 2, y\leq 3$. It's clear that these values work.


inequalities
trigonometry
inequalities unsolved

[quote="Hogar"]Hokay, so I dunno whether someone else has posted this or not. I've searched half the forum and came up empty-handed so don't judge me too hard if it has already appeared.

The Problem: Basicly, you have a triangle ABC and a random point P inside it. Let X,Y,Z are the lengths of the bisectors of angles APB, APC and BPC.
(one end of the stretch with length X is P and the other lies on AB etc.). If AB=BC=AC=1 you need to prove that min{XY,YZ,ZX}=<1/12.

Good luck to everyone who undertakes this. A piece of advice: the author isn't a fan of calculations so there MUST be some kind of clever geometric solution, but I'm not that pretentious about it so ANYTHING would be helpful.[/quote]

Let : $ AP = p, BP = q,CP = r$ and $ \angle BPC = \psi_1,\angle CPA = \psi_2,\angle APB = \psi_3$ , $ (\psi_1 + \psi_2 + \psi_3 = 2\pi)$

$ l_a ,  l_b , l_c$ - lengths of the bisectors of $ \angle{BPC},\angle{CPA},\angle{APB}$

Then : $ l_a = \frac{2qr\cos(\psi_1/2)}{q+r}$ , $ l_b  = \frac{2rp\cos(\psi_2/2)}{r+p}$ , $ l_c = \frac{2pq\cos(\psi_3/2)}{p+q}$

$ l_a \le \frac{q+r}{2}\cos(\psi_1/2) \le \frac{1}{2}\cot(\psi_1/2)$  , ($ q+r \le \frac{1}{\sin(\psi_1/2)}$ )

So : $ l_a \le \frac{1}{2}{\cot(\psi_1/2)}$ , $ l_b \le \frac{1}{2}\cot(\psi_2/2)$ , $ l_c \le \frac{1}{2}\cot(\psi_3/2)$

$ \boxed{l_a*l_b + l_b*l_c+l_c*l_a \le \frac{1}{4}(\cot(\psi_1/2)\cot(\psi_2/2)+\cot(\psi_2/2)\cot(\psi_3/2)+\cot(\psi_3/2)\cot(\psi_1/2))= \frac{1}{4}}$

And Min($ l_a*l_b$  ,  $ l_b*l_c$  ,  $ l_c*l_a$ ) $ \le \frac{1}{12}$ :)


calculus
integration
inequalities
function
real analysis
real analysis solved

Let $f,g$ be $2$continuous functions defined over $R^+$,such that g is positive.If there exist $A>0$ such that$f(x)\le A+ \int ^x_0 {f(t)g(t)}dt$ then $f(x)\le A e^{ \int ^x_0 {g(t)}dt}$

The name is Gronwall lemma, there is some application to show that solution of
some linear differential equation is bounded.

Looks like "Gronuol's" inequality (sorry, I couldn't find the right english writing)

If we write our inequality in form $\frac{f(x)g(x)}{A+\int_{0}^{x}f(t)g(t)dt}\leq g(x)$ and integrate it we obtain
$ln(A+\int_{0}^{x}f(t)g(t)dt)\leq lnA+\int_{0}^{x}g(t)dt$. Hence $f(x)\leq A+\int_{0}^{x}f(t)g(t)dt \leq Ae^{\int_{0}^{x}g(t)dt}$


may I ask how you came across this in the first place?

Just surfin on you tube.  Then I wanted to see Fergie's music video, then this thing came up on the right side of the screen and it looked funny.

[quote="math92"]Then I wanted to see Fergie's music video,[/quote]
why, oh god, why?

[quote="junggi"][quote="math92"][quote="spix"]The key words here are "see" and "Fergie", I  don't think it has anything to do with the music part.[/quote]

Hehe.     :D[/quote]
ehh don't you have a girlfriend(lizzy or w/e)?[/quote]
We're still friends, but not in that way anymore.

Also, they're at it again:
http://youtube.com/watch?v=n7Kh859EarE


analytic geometry
trigonometry
algebra proposed
algebra

Consider four points $P_1,\ P_2,\ P_3$ and $P_4$ on the coordinate plane with origin $O$ which satisfy the condition \[ \overrightarrow {{OP}}_{n-1}+\overrightarrow {{OP}}_{n+1}=\frac{3}{2} \overrightarrow {{OP}}_{n},\ (n=2,3). \]

(1) If $P_1,\ P_2$ lie on the curve $xy=1,$ then prove that $P_3$ doesn't lie on this curve.

(2) If $P_1,\ P_2$ and $P_3$ lie on the circle $x^2+y^2=1,$ then prove that $P_4$ lie on this circle as well.

Hello, how much time is permitted to do these entrance exams ? Is it the same time for all universities ?  The exam of tokyo university seem to be longer than the T.I.tech exam, and what about the exam of kyoto university (I see that you posted 11 problems in 2 exams with the same date ??) is that what we call "math course 1 and course 2"   :?:

$\overrightarrow{OP_{1}}+\overrightarrow{OP_{3}}={3\over 2}\overrightarrow{OP_{2}}\implies \overrightarrow{OP_{3}}=-\overrightarrow{OP_{1}}+{3\over 2}\overrightarrow{OP_{2}}$

$\overrightarrow{OP_{2}}+\overrightarrow{OP_{4}}={3\over 2}\overrightarrow{OP_{3}}\implies \overrightarrow{OP_{4}}=-\overrightarrow{OP_{2}}+{3\over 2}\overrightarrow{OP_{3}}=-{3\over 2}\overrightarrow{OP_{1}}+{5\over 4}\overrightarrow{OP_{2}}$

(1) Put $\overrightarrow{OP_{1}}=\langle a,{1\over a}\rangle, \overrightarrow{OP_{2}}=\langle b,{1\over b}\rangle, ab\neq 0$. Then

$\overrightarrow{OP_{3}}=\langle-a+{3\over 2}b,-{1\over a}+{3\over 2b}\rangle$.

In order for $P_{3}$ to lie on $xy=1$, we must have $\left({3\over 2b}-{1\over a}\right)\left({3b\over 2}-a\right)=1$, but that simplifies into $2b^{2}-3ab+2a^{2}=0$, which has no solutions other than $(0,0)$ as its discriminant is negative.

(2) Put $\overrightarrow{OP_{1}}=\langle\cos\alpha,\sin\alpha\rangle, \overrightarrow{OP_{2}}=\langle\cos\beta,\sin\beta\rangle$. Then

$\overrightarrow{OP_{3}}=\langle-\cos\alpha+{3\over 2}\cos\beta,-\sin\alpha+{3\over 2}\sin\beta\rangle$ and we also have

$(-\cos\alpha+{3\over 2}\cos\beta)^{2}+(-\sin\alpha+{3\over 2}\sin\beta)^{2}=1$, which simplifies into $\cos\alpha\cos\beta+\sin\alpha\sin\beta={3\over 4}$

Now $\overrightarrow{OP_{4}}=\langle-{3\over 2}\cos\alpha+{5\over 4}\cos\beta,-{3\over 2}\sin\alpha+{5\over 4}\sin\beta\rangle$, giving

$|\overrightarrow{OP_{4}}|^{2}={9\over 4}-{15\over 4}(\cos\alpha\cos\beta+\sin\alpha\sin\beta)+{25\over 16}={9\over 4}-{15\over 4}\cdot{3\over 4}+{25\over 16}=1$


trigonometry
algebra
polynomial
function
algebra unsolved

Let a function :
 $ f(x)= a_1cos(x+\alpha_1)+a_2cos(x+\alpha_2)+ ... + a_ncos(x+\alpha_n) $
 ( $ \alpha_i , a_i $ are real constants )
 Prove that : If there exists $ x_1, x_2 $ satisfying : $ x_1-x_2 $ is different from $ m\pi $ for all $ m\in Z $  and $ f(x_1)=f(x_2)=0 $ , then $ f(x) = 0 $ for all $ x $

$f(x)$ can be expand to form:
$f(x)=a\sin{x}+b\cos{x}=\sqrt{a^2+b^2}sin(x+t)$
Then If exist $x_1,x_2 ,\sqrt{a^2+b^2}=0->a=b=0$.
And we are done.


If you are interested in coming to HMMT  PM E^(pi*i)=-1!  The deadline is FEBUARY 1ST!

If you decide to join, I need your name, nickname, and the individual tests you want take.

Also, BOGTRO, I prefer not to have my (father's) e-mail address on a public forum, so please edit it out.  PM me if you want to join.


I stumbled across this one today in my never-ending journey to the end of the internet.  Positively BRILLIANT!

[url]http://www.flickr.com/photos/marnanel/211728740/[/url]

LOL.
at least it was a Politically Correct #5 and arm band.   :lol:


Let $f$ be the function defined by $f(x)=ax^2-\sqrt2$ for some positive $a$. If $f(f(\sqrt2 ))=-\sqrt 2$, then $a=$

$\text{(A)} \ \frac{2-\sqrt2}{2} \qquad \text{(B)} \ \frac12 \qquad \text{(C)} \ 2-\sqrt2 \qquad \text{(D)} \ \frac{\sqrt{2}}{2} \qquad \text{(E)} \ \frac{2+\sqrt2}{2}$

[hide]Nothing special here. You can even work backwards to find that the answer's d.  :P  :rotfl: [/hide]

I got $\boxed{D}$ and [hide="heres how:"]

$f(\sqrt{2})=2a-\sqrt2$
$f(2a-\sqrt2)=2a^2-a\sqrt{2}-\sqrt{2}$
We know that is equal to $-\sqrt{2}$.
$2a^2-a\sqrt{2}-\sqrt{2}=-\sqrt{2}$
$a(2a-\sqrt2)=0$
Hence $a=\frac{\sqrt{2}}{2}$.[/hide]


calculus
limit
logarithms
calculus computations

Calculate:

$ \mathop {\lim }\limits_{x \to  \plus{} \infty } {\left( {\log x} \right)^{\frac{{ \minus{} 2}}{x}}}$

Thanks, because I d'ont find solution for this.

Antonio

$ \lim_{x\to \infty} (\log x)^{\frac{\minus{}2}{x}}$
$ \equal{}\lim_{x\to \infty} e^{\frac{\minus{}2}{x}\ln (\log x)}$
Because both $ \ln$ and $ \log$ are both strictly increasing, as well as both being less than x,
$ \log x <x\longrightarrow \ln (\log x) < x$
Thus:
$ \lim_{x\to \infty} \frac{\minus{}2\ln(\log x)}{x}\equal{}0$
And finally:
$ \lim_{x\to \infty} e^{\frac{\minus{}2}{x}\ln (\log x)}\equal{}e^0\equal{}1$

[quote="EastyMoryan"]$ \log x < x\longrightarrow \ln (\log x) < x$
Thus:
$ \lim_{x\to \infty} \frac { \minus{} 2\ln(\log x)}{x} \equal{} 0$[/quote]  This implication does not hold (although the consequence is true).  But the fact that $ \lim_{x \to \infty} \frac{\log x}{x} \equal{} 0$ is well-known, and so $ \lim_{x \to \infty} \frac{\log\log x}{x} \equal{} \lim_{x \to \infty} \frac{\log\log x}{\log x} \cdot  \frac{\log x}{x} \equal{} 0$, as well.


evaluate: $x+2x^2+3x^3+4x^4+\dots+nx^n$. in terms of $x$

i know how to do this if it is an infinite sum and $|x|<1$.
but how would you go about solving if it is a finite sum and $x$ can have any real value. :?

i posted this a while back

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50853[/url]

Note that $S(1 - x)$ is a regular geometric series.

To elaborate more on what Kalle is saying:

$S = x + 2x^2 + 3x^3 + \cdots$
$Sx = x^2 + 2x^3 + 3x^4 + \cdots$
$S(1-x) = x + x^2 + x^3 + \cdots$


Welcome to this forum, Chemrock.

The reaction of carbonyl coumpunds (ketones) with peracids is know as the Baeyer-Villiger reaction (or rearrangement) and the product is an esther. In the mechanism of this reaction there is a migration of one of the groups attached initially to the carbonyl carbon to an electron deficient oxygen atom, and the group migrates with its electrons, i e., as an anion. The global result of the Baeyer-Villiger reaction is the insertion of an oxygen atom between the group with greatest migratory aptitude and the carbonyl carbon. Experimental studies on the migratory aptitudes of several groups on this (and related reactions) have showed that we have the following order

$ H > C_6H_5\minus{} > R_3C\minus{} > R_2CH\minus{} > RCH_2\minus{} > CH_3\minus{}$

(note that hydrogen, or more correctly the hydride ion, has the greatest migratory aptitude, and this is why the reaction of peracids with aldehydes is not usefull). On the other, as it should perhaps be expected, electron righ groups migrate more readily compared with electron deficient groups: that is why benzene rings with electron donating substituents (RO-, R2N-, alkyl-) migrate more easily than benzene rings with electron withdrawing substituents (NO2-, RCO-, NC-).
So, for you reaction 1. the product will be MeO-C6H4-O-CO-C6H5. We also know that the nitro group is a more deactivating compared with the cyano group, so in the second reaction we would get NO2-C6H4-CO-O-C6H4-CN.


calculus
integration
number theory
Putnam
college contests

Could someone read over my solution? I think I did it wrong.

[hide="Problem"]Prove that the expression
$\frac{\gcd{(m,n)}}{n}\binom{n}{m}$
is an integer for all pairs of integers $n \ge m\ge 1$.[/hide]
[hide="Solution"][u]Case 1[/u]: $n=m$

Our expression equals $\frac{\gcd{(n,n)}}{n}\binom{n}{n}=1$, which is an integer.

[u]Case 2[/u]: $n>m\ge1$

Note that $\frac{\binom{n}{m}}{n}$ is an integer for integers $n,m$ with $n>m\ge1$. So $\frac{\gcd{(m,n)}}{n}\binom{n}{m}$ is also integral, as $\gcd{(m,n)}$ is an integer.

-We have covered all the cases, so we are done. QED.[/hide]

[quote="1234567890"][u]Case 2[/u]: $n>m\ge1$

Note that $\frac{\binom{n}{m}}{n}$ is an integer for integers $n,m$ with $n>m\ge1$.[/quote]

$n = 4, m = 2$.

There are really two ways.  One requires no thinking, but a lot of plodding - given a prime $p_i$ in the prime decomposition of $m$ or $n$, prove that $p_i$ has a positive exponent in that quantity.

The other way requires a theorem from number theory; given integers $m,n$, there exist integers $a$ and $b$ such that $am+bn=\gcd(m,n)$.  What happens if you make this substitution?

OK, thanks. I get how to solve it now. That was a stupid assumption. We have (6,3), (14,7)..... Yeah, I was wrong.  :D


Consider random walk on $ \mathbb{Z}^2$ started at the origin.what is the probability ro hit $ (a,b)$ for $ a,b\in \mathbb{Z}$ given.

you may know that a symmetric random walk in $ \mathbb{Z}^2$ comes back infinitely many times at $ 0$, say. Moreover, if $ n \equal{} |a| \plus{} |b|$, it is clear that there is a non null probability that, starting from $ 0$ at time $ t \equal{} 0$, the random walk will hit $ (a,b)$ before time $ t \equal{} n$. Because the random walk comes back infinitely many times at $ 0$, it implies that the probability of hitting $ (a,b)$ is $ 1$.

Now what about in $ \mathbb{Z}^3$ where the situation is quite different ..

[quote="alekk"]... Moreover, if $ n \equal{} |a| \plus{} |b|$, it is clear that there is a non null probability that, starting from $ 0$ at time $ t \equal{} 0$, the random walk will hit $ (a,b)$ before time $ t \equal{} n$ ...[/quote]
can you please explain this more?
Thanks.

for example the path that goes $ |a|$ times in the direction of $ (a,b)$ along the $ x$-axis, and then reaches $ (a,b)$ along the $ y$-axis after the $ |b|$ remaining steps. There is a non-null probability that the random walk will take this path (exactly $ \frac{1}{4^n}$ more precisely).

I wonder if this kind of problem could be dealt with the technic of embedding  of the random walk into a Two (or three as Alekk suggests) dimensional Brownian Motion.
I think that this kind of embedding is treated in in Volume 1 of the Book of Rogers and Williams.

ok, I accept that we visit origin infinite times and there is a positive possibility to visit $ (a,b)$.
but I didin't get why its possibility is 1 ?
why after visiting $ (a,b)$ we hit origin by possibility 1?
 :maybe:

suppose that you throw a dice infinitely many times: what is the probability of getting a $ 6$ ? it is $ 1$ ! this is the same reasoning.

But why are we sure after visiting $ (a,b)$  we'll visit origin again?
sorry if it's a stupid question.

is there any calculation way??

well if you agree that we hit the origin infinitely many times, and if you think we hit (a,b) even once, then we must hit IT infinitely many times as well.

let T be the stopping time for the first time we hit (a,b) then consider the random walk $ S'_n\equal{}\sum_{T\plus{}1}^{T\plus{}n} X_j$ starting at this point (which hits the origin infinitely many times)

another way to say this is that the set of accessible points is equivalent to the set of recurrent points for random walks on Z and $ Z^2$

hope this helps


calculus
derivative
geometry
rectangle
function
real analysis
real analysis unsolved

Suppose the partial derivatives of a mapping $ f : U \to \mathbb{R}^n$ defined in an open cell $ U \subset \mathbb{R}^m$ are bounded. Then prove that the mapping is uniformly continuous in $ U$. (It does not have to be differentiable.)

I'll do this for $ n\equal{}2.$ The notation is awkward, but it should be clear how to extend this proof to $ n$ dimensions.

Assume that $ \left|\frac{\partial f}{\partial x_1}\right|\le M$ and $ \left|\frac{\partial f}{\partial x_2}\right|\le M$ on $ U,$ which is an open rectangle (we'll need that shape).

Consider $ f(y_1,y_2)\minus{}f(x_1,x_2)\equal{}f(y_1,y_2)\minus{}f(x_1,y_2)\plus{}f(x_1,y_2)\minus{}f(x_1,x_2)$

$ |f(y_1,y_2)\minus{}f(x_1,x_2)|\le|f(y_1,y_2)\minus{}f(x_1,y_2)|\plus{}|f(x_1,y_2)\minus{}f(x_1,x_2)|$

By the mean value theorem, $ f(y_1,y_2)\minus{}f(x_1,y_2)\equal{}(y_1\minus{}x_1)\frac{\partial f}{\partial x_1}(\xi,y_2)$ so we can say that

$ |f(y_1,y_2)\minus{}f(x_1,y_2)|\le M|y_1\minus{}x_1|$

By a similar MVT argument, $ |f(x_1,y_2)\minus{}f(x_1,x_2)|\le M|y_2\minus{}x_2|$

Putting those together,

$ |f(y_1,y_2)\minus{}f(x_1,x_2)|\le M(|y_1\minus{}x_1|\plus{}|y_2\minus{}x_2|)\equal{}M\|y\minus{}x\|_1\le\sqrt{2} M\|y\minus{}x\|_2$

That's a Lipschitz condition, which implies (but is stronger than) uniform continuity.

Sorry, how can we extend this to $ n$ dimensions? Both the argument and the value of the function are in several dimensions, and I don't see how to apply the chain rule here.

Oh, I didn't notice that the codomain was in several dimensions. In that case, make this exact argument for each component of $ f$ and add the results together.

As for the argument being in more than two dimensions, do this:

$ f(y)\minus{}f(x)\equal{}f(y_1,y_2,\dots,y_n)\minus{}f(x_1,y_2,\dots,y_n)\plus{}f(x_1,y_2,\dots,y_n)$
$ \minus{}f(x_1,x_2,y_3\dots, y_n)\plus{}\cdots\plus{}f(x_1,x_2,\dots ,x_{n\minus{}1},y_n)\minus{}f(x_1,x_n,\dots,x_n)$

Then break that up into a sum of $ n$ differences and estimate each one by the MVT.

The conclusion should still be that $ f$ satisfies a Lipschitz condition on $ U.$


Use the change of base formula so you get
[hide]$\frac{\log3}{\log2}\cdot\frac{\log4}{\log3}\cdot\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}\implies\frac{\log8}{\log2}\implies\boxed{3}$[/hide]

[quote="ckck"]Evaluate the product $(log_{2}3)(log_{3}4)(log_{4}5)(log_{5}6)(log_{6}7)(log_{7}8)$.

How would you do this problem? Logrithm confuses me, as I have no idea what to do.[/quote]

[hide]

Use the change of base property several times. The first one is shown below.

$\log_{2}{3}= \frac{\log{3}}{\log{2}}$

Many logs will cancel out, leaving $\log_{2}{8}$.

We can rewrite this as $\log_{2}{2^{3}}= 3$.

[/hide]

[hide]Chain rule of logs, $(\log_{a}b)(\log_{b}c)=\log_{a}c$, so the expression becomes $\log_{2}8=3$.[/hide]

[quote="pianoforte"]Who is Lazy Jim? I don't get it.[/quote]
That's the name of the identity, it's not very well known.
The key for this and many other log problems is using the change of base property, particularly when writing logs in terms of other logs.

[quote="bpms"][quote="rnwang2"][hide]
Since $(\log_{a}b)(\log_{b}c)=\log_{a}c$, the product becomes $\log_{2}8=\boxed{3}$.

[b]Note[/b]: In general, $(\log_{a}b)(\log_{c}d)=(\log_{a}d)(\log_{c}b)$.
[/hide][/quote]
This general property is called the Lazy Jim log property by the way. It is one of the more obscure, yet powerful log identities.[/quote]
Really? Neither mathworld nor wikipedia has that as an article or whatever. I couldn't find something mathrelated when I googled that. So I don't think that Lazy Jim is it's name(I never even heard of this till now)
Also, I don't think that this property is all that obscure......unless by obscure you mean "many school textbooks don't have it."

[quote="Kamil Witek"]oh this question is from aops v.2 chapter 1 right?[/quote]

Yes this was from the first chapter.

In AOPS vol. 2 ch. 1 they say it may sometimes be called the "chain rule for logarithms". Personally, I would find this confusing, I mean, people could misinterpret it as using the conventional chain rule on a logarithmic function?

[quote="mathgeniuse^ln(x)"][quote="pianoforte"]bpms, where did you hear it called "Lazy Jim log property". I could not find that term either. I do remember the property from aops vol 2.[/quote]

I bet anything he made it up.[/quote]
Well you lost anything because [i]I[/i] didn't make it up.

It sounds like your teacher or something taught it to you under that name.  It's not a common name for it, and I don't like calling it the chain rule either, since it's such a direct consequence of the base-change formula.

Just thought I might add, it becomes really obvious if you set $d = b^{f}$ for some $f$


topology
geometry
real analysis
real analysis unsolved

assume that $F$ is borel set in $\mathbb{R}^{n}$ such that $H^{s}(f)=\infty$ then prove that there exists compact subset $E\subset F$ such that $0<H^{s}(E)<\infty$

You can find a complete solution in
P. Mattila, Geometry of Sets and Measures in Euclidean Spaces, Cambridge University Press, 1999.

Sam-n, do you have a short proof for this nice result ? The one in Mattila is not that short ..


[quote="levi"]Let ABC a nonisosceles triangle. For any $X\in {\{O,L,\Gamma\}}$
we will note $X'\in GX \cap BC,\ r(X)=\frac{X'B}{X'C}$. Prove that

$r(O)=\frac{c^2-a^2}{a^2-b^2}\cdot \frac{a^2+c^2-b^2}{a^2+b^2-c^2}$ $;\ \ r(\Gamma)=\frac{c-a}{a-b}\cdot \frac{s-b}{s-c}$ $;\ \ r(L)=\frac{c^2-a^2}{a^2-b^2}.$[/quote] ([b]L[/b] - [u]Lemoine[/u]'s point; $\Gamma$ - [u]Gergogne[/u]'s point; [b]O[/b] - center of circumcircle).


probability
combinatorics unsolved
combinatorics

I am having a hard time solving this problem. May be one of you guys can solve it.

Each of the 2n members of a committe flips a fair coin in deciding whether or not to attend a meeting of the committe; a committe member attends the meeting if an H appears. What's the probability that a majority will show up for the meeting (i.e. at least n+1 committe members present)?

-Swapnil

The Coin is flipped $2n$ times. The Probability for being in or not is each $\frac{1}{2}$ so with Bernoulli's Formula I get:

$P(X \geq n+1) = 1 - P(X \leq n) = 1 - \sum_{k=0}^n \binom{2n}{k} \frac{1}{2^k} \frac{1}{2^{2n-k}}$
$= 1 - \frac{1}{2^{2n}}\sum_{k=0}^n \binom{2n}{k} = 1 - \frac{1}{2^{2n+1}}\left(2^{2n} + \binom{2n}{n}\right)$
$= \frac{1}{2} - \frac{1}{2^{2n+1}}\binom{2n}{n}$


Solve in non- negative intergers the equation $ \displaystyle\frac{a!\plus{}b!}{n!}\equal{}3^n$.

a=1,b=2,n=1 is a solution (and the only one i suppose :maybe: )

Clearly, we have $ n>0$ and let $ M\equal{}\max\{a,b\}$ and $ m\equal{}\min\{a,b\}$. We must have $ n\leq m$, since for $ n>M$ we get the contradiciton $ 3\leq 3^n\leq \frac{2}{M\plus{}1}\leq 2$ and for $ m < n \leq M$ the ratio $ \frac{a!\plus{}b!}{n!}\equal{}\frac{m!\plus{}M!}{n!}$ is not an integer. We rewrite the original equation to get $ 3^n\equal{}\frac{m!}{n!}\left(1\plus{}\frac{M!}{m!}\right)$ which yields $ m\in\{n,n\plus{}1\}$ and $ M\in\{m\plus{}1,m\plus{}2\}$

Case 1: $ m\equal{}n,M\equal{}n\plus{}1$ yields $ 3^n\equal{}n\plus{}2$, i.e. $ n\equal{}1$
Case 2: $ m\equal{}n,M\equal{}n\plus{}2$ yields $ 3^n\equal{}n^2\plus{}3n\plus{}3$, which requires $ 3|n$, but then the RHS is not divisble by 9
Case 3: $ m\equal{}n\plus{}1,M\equal{}n\plus{}2$ yields $ 3^n\equal{}(n\plus{}1)(n\plus{}3)$ which is impossible
Case 4: $ m\equal{}n\plus{}1,M\equal{}n\plus{}3$ yields $ 3^n\equal{}(n\plus{}1)(n^2\plus{}5n\plus{}7)$ requiring $ n\equal{}3k\minus{}1$ with $ k>0$ and substitution gives $ 3^{3(k\minus{}1)}\equal{}k(3k^2\plus{}3k\plus{}1)$ which has no solution.

Putting all together the only solution is $ n\equal{}1$ and $ (a,b)\in\{(1,2),(2,1)\}$

Note because the original question says $ \text{non\minus{}negative}$ so when $ a\equal{}1$ or $ b\equal{}1$ then $ a\equal{}0$ and $ b\equal{}0$ are also solutions. As $ 1!\equal{}0!$.

But a and b cannot be simult. 1
See the eqn properly
edit : IGNORE THIS POST

[quote="madness"]But a and b cannot be simult. 1
See the eqn properly[/quote]

What I mean is that:

$ (a,b,n) \equal{} (0,2,1), (2,0,1)$ are also solutions.


geometry
3D geometry
tetrahedron
pyramid
octahedron

What does a regular tetrahedron look like?

It is just a pyramid with an equilateral triangle for a base, and three equilateral triangles for faces.  So basically 4 congruent equilateral triangles!

can't describe it mucj=h better than that.

I was just wondering. Is there a formula for the volume of a regular totetrahedron based on side length?

I think that it would make sense. Just plug some parts into 1/3bh

I got $\frac {x^3 \sqrt 2}{12}$

Where $x$ is the edge length. I probably messed up =o

that's what I got, therefore a square pyramid is that times 2, and an octahedron is a tetrahedron by 4


algebra
polynomial
complex numbers
topology
real analysis
real analysis unsolved

Suppose that $P$ is a complex polynomial of degree at least 2 and look at the set of complex numbers $x$ such that if we start with $x$ and iterate $P$ we obtain a bounded sequence. Is this set compact?

Yes, of course. Choose $R$ so that $|P(x)|>|x|$ when $|x|>R$; this is always possible. Then the set of $x$ for which $|P^k(x)|$ is unbounded is $\bigcup_k P^{-k}(U)$, where $U=\{z: |z|>R\}$ and $P^{-k}$ is the inverse of $P$ iterated $k$ times. Each $P^{-k}$ is open by continuity of $P^k$, and therefore the union is open. The complement is therefore closed; since it is contained in the disk of radius $R$, it is compact.

Sorry, I don't understand your solution: what I meant was that the sequence $x_{n+1}=P(x_n)$ and $x_1=z$ is bounded. Now, what does this have to do with the fact that $P^k(x)$ is unbounded? I really don't understand;

He's just going about the problem by discussing the complement of your set.

Since $P$ is polynomial, eventually it gets monotone (outside a (possibly rather large) ball), so the set you care about must be bounded.

Then, if the sequence diverges for some $z$, we know eventually $x_n$ is in the region where $P$ is monotone. So jmerry just looks at the inverse image of the complement of that ball around the origin under $k$ iterations of $P$. If some number is not in the union for any $k$, then obviously it is a $z$ for which the sequence is bounded. If it is in one of those inverse images, then obviously it is one for which the sequence is not bounded, because once a term of the sequence is outside that ball, it can't get back in -- it's now monotone and diverging like $x^j$ for some $j \ge 2$.

Finally, all those inverse images are open, so their union is open. That's the set of points for which teh sequence diverges. Complement of open set is closed! :D

Now that I write this, I am not sure how much better an explanation it is than what jmerry said anyway. Hopefully it is a little more clear.

Oh, sorry, I understood it. Thanks a lot.  :blush:  :oops:


Prove that twice the length of an inner diagonal of a rectangular box is always greater than the sum of its length, width, and height.

[hide]
Note that the length of the inner diagonal of the box, $d$, is $\sqrt{l^{2}+w^{2}+h^{2}}$ where $l,w,h$ are the length, width and height of the box respectively (this follows from 3-dimensional Pythagoras).

By Cauchy-Schwartz:
$(1^{2}+1^{2}+1^{2})(l^{2}+w^{2}+h^{2}) \geq (1l+1w+1h)^{2}$.
$4(l^{2}+w^{2}+h^{2}) > 3(l^{2}+w^{2}+h^{2}) \geq (1l+1w+1h)^{2}$.
$2\sqrt{l^{2}+w^{2}+h^{2}}> (l+w+h)$.
$2d > l+w+h$.
[/hide]


would the answer be $ 7$?

Imnot sure I just amde geometric points out of them and had the shortest distance.

I seriously doubt thats right but Ithink you might sub one variable into the other or plot it out geometrically (I think thats what you do)

[hide="Short"] By Minkowski's inequality,

$ \sqrt{x^{2}+y^{2}}+\sqrt{(1-x)^{2}+(-y)^{2}}+\sqrt{x^{2}+(y-1)^{2}}+\sqrt{(3-x)^{2}+(4-y)^{2}}\ge \sqrt{ (x+(1-x)+x+(3-x))^{2}+(y+(-y)+(y-1)+(4-y))^{2}}= \boxed{5}$

This is equivalent to the geometric interpretation of minimizing a sum of distances.[/hide]

WOw..

It was as simple as recognising that $ (1-x)^{2}=(x-1)^{2}$.

Then, just plot the geometric points out of it......

Thanks!!

Would the point in consideration be the intersection of the diagonals of the quadrilateral whose vertices are (0,0), (1,0), (0,1), and (3,4)?

Yes.  Speaking of which, my manipulation was invalid, so let me correct myself.

[hide="Correction"] Minkowski's as I have used it only holds for the sum of two distances.  That's because for more than two distances, the equality case may not be attainable.  Thus the correct usage would've been

$ \sqrt{(x-1)^{2}+y^{2}}+\sqrt{(-x)^{2}+(1-y)^{2}}\ge \sqrt{1^{2}+1^{2}}= \sqrt{2}$
$ \sqrt{x^{2}+y^{2}}+\sqrt{(3-x)^{2}+(4-y)^{2}}\ge \sqrt{3^{2}+4^{2}}= 5$

So the correct answer is $ \boxed{5+\sqrt{2}}$, which is attainable by pda's argument. [/hide]


Is there a website that lists all the nations that won the IMO beginning from 1959.

//http://olympiads.win.tue.nl/imo/index.html
P/S: I must move this topic.  :wink:

http://imo-official.org/ lists a lot more (even the contestants, and often their scores as well).

[quote="sylow_theory"]Interesting. So USSR was the best overall naition in IMO  :maybe:[/quote]
After China. But this year USA might overtake it. (And let's hope Hungary as well :D )

I'm doing research on the international maths olympiads and need a lot of data... the more the better of course : )

Now for the IMO, the mentioned website is very informative yet I would like to have the data in a different format because downloading all the XML Spreadsheets is really tyring. I've sent an email to the webmaster of imo-official already but so far haven't received any reply.

Does anyone have an idea where to get the whole bunch of imo-official data as a SQL dump or a *.csv?

HELP PLEASE

Cheers
l8ty


Find all functions$ f: Q^\plus{}\to Q^\plus{}$ satisfy:
$ f(x)\plus{}f(\frac 1{x})\equal{}1$
$ f(2x)\equal{}2f(f(x))$

Still no proof but it's clearly

f(x)=x/(x+1)

Okay, so I have a proof but it uses more advanced math than would be expected on one of the Ireland olympiads... can you site the source a little more clearly? Like, which contest precisely and which problem number? I have reason to believe that this is not the source you claim.


calculus
derivative
function
trigonometry
calculus computations

Find all functions $ f(x)$ such that $ \frac{d}{dx}(\arctan f(x)))\equal{}\frac{d}{dx}(\arctan x)$.

$ \begin{array}{l}
 \arctan f(x) \equal{} \arctan x \plus{} c \Leftrightarrow  \\
 \arctan f(x) \minus{} \arctan x \equal{} c \Leftrightarrow  \\
 \arctan \left( {\frac{{f\left( x \right) \minus{} x}}{{1 \plus{} xf\left( x \right)}}} \right) \equal{} c \Leftrightarrow \frac{{f\left( x \right) \minus{} x}}{{1 \plus{} xf\left( x \right)}} \equal{} \tan c \Leftrightarrow  \\
 f\left( x \right) \equal{} \frac{{\tan c \plus{} x}}{{1 \minus{} x\tan c}} \\
 \end{array}$
for appropriate x

[quote="mathxl"]$ \begin{array}{l} \arctan f(x) \equal{} \arctan x \plus{} c \Leftrightarrow \\
\arctan f(x) \minus{} \arctan x \equal{} c \Leftrightarrow \\
\arctan \left( {\frac {{f\left( x \right) \minus{} x}}{{1 \plus{} xf\left( x \right)}}} \right) \equal{} c \Leftrightarrow \frac {{f\left( x \right) \minus{} x}}{{1 \plus{} xf\left( x \right)}} \equal{} \tan c \Leftrightarrow \\
f\left( x \right) \equal{} \frac {{\tan c \plus{} x}}{{1 \minus{} x\tan c}} \\
\end{array}$
for appropriate x[/quote]

Perfect mathxl  :)  :)  :)


inequalities
geometry
trigonometry
3D geometry
tetrahedron
triangle inequality
geometry unsolved

In a triangle $ABC$, let $A_{1}$, $B_{1}$, $C_{1}$ be the points where the excircles touch the sides $BC$, $CA$ and $AB$ respectively. Prove that $A A_{1}$, $B B_{1}$ and $C C_{1}$ are the sidelenghts of a triangle.

[i]Proposed by L. Emelyanov[/i]

Namdung mentioned a "very simple geometric proof" in http://www.mathlinks.ro/Forum/viewtopic.php?t=46088 . However, apparently nobody else has seen this proof until now, so let me post my proof, neither simple nor geometric, but actually showing some stronger inequalities. Note that it is rather long since I tried to avoid massive computations, but it is rather a typical inequality proof than a geometry proof, as you will surely see.

First I rename some points:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and let its A-excircle, its B-excircle and its C-excircle touch its sides BC, CA and AB at the points X, Y and Z, respectively. Prove that the segments AX, BY and CZ are the sidelengths of a triangle.[/color]

[i]Solution.[/i] We will show that $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$. This will yield $\left(BY+CZ\right)^2=BY^2+CZ^2+2\cdot BY\cdot CZ>BY^2+CZ^2>AX^2$, so that BY + CZ > AX, and similarly CZ + AX > BY and AX + BY > CZ. Thus, the segments AX, BY and CZ are the sidelengths of a triangle, and moreover, since $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$, this triangle must be acute-angled, so we thus show even more than the problem asks us for!

So, it remains to prove the inequalities $BY^2+CZ^2>AX^2$, $CZ^2+AX^2>BY^2$ and $AX^2+BY^2>CZ^2$. We will only prove $BY^2+CZ^2>AX^2$, since the other two inequalities will then follow by cyclic translation.

Let $s =\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Remember the half-angle formula $\cos^2\frac{B}{2}=\frac{s\left(s-b\right)}{ca}$. Since $s =\frac{a+b+c}{2}$, we have $s-b=\frac{a+b+c}{2}-b=\frac{c+a-b}{2}$, and thus, this formula becomes $\cos^2\frac{B}{2}=\frac{\frac{a+b+c}{2}\cdot\frac{c+a-b}{2}}{ca}=\frac{\left(c+a+b\right)\left(c+a-b\right)}{4ca}=\frac{\left(c+a\right)^2-b^2}{4ca}$, so that $\left(c+a\right)^2-b^2=4ca\cos^2\frac{B}{2}$. Now, applying this formula to the triangle ABX, we get $\left(AB+BX\right)^2-AX^2=4\cdot AB\cdot BX\cdot\cos^2\frac{\measuredangle ABX}{2}$. Now, we have AB = c, and since X is the point where the A-excircle of triangle ABC touches its side BC, we have BX = s - c. Thus, AB + BX = c + (s - c) = s. Finally, < ABX = B. Hence, this equation becomes $s^2-AX^2=4\cdot c\cdot\left(s-c\right)\cdot\cos^2\frac{B}{2}$. But since $\cos^2\frac{B}{2}=\frac{s\left(s-b\right)}{ca}$, this transforms into

$s^2-AX^2=4\cdot c\cdot\left(s-c\right)\cdot\frac{s\left(s-b\right)}{ca}=\frac{4s\left(s-b\right)\left(s-c\right)}{a}$,

and thus

$AX^2=s^2-\frac{4s\left(s-b\right)\left(s-c\right)}{a}=s\left(s-\frac{4\left(s-b\right)\left(s-c\right)}{a}\right)$.

Now, denote s - a = x, s - b = y and s - c = z (note that these numbers x, y, z are then positive). Then, y + z = (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, and x + y + z = x + (y + z) = (s - a) + a = s. Thus, this becomes

$AX^2=\left(x+y+z\right)\left(x+y+z-\frac{4yz}{y+z}\right)$.

Similarly,

$BY^2=\left(x+y+z\right)\left(x+y+z-\frac{4zx}{z+x}\right)$;
$CZ^2=\left(x+y+z\right)\left(x+y+z-\frac{4xy}{x+y}\right)$.

Hence, the inequality we want to prove, $BY^2+CZ^2>AX^2$, is equivalent to

$\left(x+y+z\right)\left(x+y+z-\frac{4zx}{z+x}\right)+\left(x+y+z\right)\left(x+y+z-\frac{4xy}{x+y}\right)>\left(x+y+z\right)\left(x+y+z-\frac{4yz}{y+z}\right)$.

Upon division by x + y + z, this becomes

$\left(x+y+z-\frac{4zx}{z+x}\right)+\left(x+y+z-\frac{4xy}{x+y}\right)>x+y+z-\frac{4yz}{y+z}$.

Equivalently,

$x+y+z+\frac{4yz}{y+z}>\frac{4zx}{z+x}+\frac{4xy}{x+y}$.

Now, we can directly prove this inequality by expanding, but this is a bit uncreative and requires computations. Let's prove this inequality in a slightly different way, by showing a stronger [i]symmetric[/i] inequality. In fact, first we rewrite this inequality by adding $\frac{4yz}{y+z}$ to both sides:

[color=green][b](1)[/b][/color] $x+y+z+\frac{8yz}{y+z}>4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}\right)$.

Now, since (z + x) (x + y) = x (x + y + z) + yz > x (x + y + z), we have $\frac{8yz\left(z+x\right)\left(x+y\right)}{y+z}>\frac{8yzx\left(x+y+z\right)}{y+z}=8\frac{xyz\left(x+y+z\right)}{y+z}$, so that $\frac{8yz}{y+z}>8\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$. Thus, the inequality [color=green][b](1)[/b][/color] would immediately follow from the following symmetric inequality:

[color=green][b](2)[/b][/color] $x+y+z+8\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\geq 4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}\right)$.

Now, though this inequality be destroyed by expanding, it's quite sharp. Here is a halfways "synthetic" proof of this inequality:

By subtracting $4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$ from both sides of the inequality [color=green][b](2)[/b][/color], we bring it into the form

$x+y+z+4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\geq 4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$.

But this can be proven as follows:

$x+y+z+4\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}=\left(x+y+z\right)\cdot\left(1+\frac{4xyz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$
$=\left(x+y+z\right)\cdot\frac{4xyz+\left(y+z\right)\left(z+x\right)\left(x+y\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$
$=\left(x+y+z\right)\cdot\frac{x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$
$=\frac{\left(x+y+z\right)\cdot\left(x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$
$\geq\frac{\left(\sqrt{x\cdot x\left(y+z\right)^2}+\sqrt{y\cdot y\left(z+x\right)^2}+\sqrt{z\cdot z\left(x+y\right)^2}\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$           (by Cauchy-Schwarz)
$=\frac{\left(x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}=\frac{4\left(yz+zx+xy\right)^2}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$
$=4\left(yz\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}+zx\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}+xy\frac{yz+zx+xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$
$=4\left(\left(\frac{yz}{y+z}-\frac{x^2yz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)+\left(\frac{zx}{z+x}-\frac{y^2zx}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)+\left(\frac{xy}{x+y}-\frac{z^2xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)\right)$
$=4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{x^2yz+y^2zx+z^2xy}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$
$=4\left(\frac{yz}{y+z}+\frac{zx}{z+x}+\frac{xy}{x+y}-\frac{xyz\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\right)$.

And [color=green][b](2)[/b][/color] is proven, so the problem is solved.

  Darij

In http://www.mathlinks.ro/Forum/topic-59415.html

   Darij elegantly convinced us that Medians triangle is always possible.

   Nagelians are isotomics of Gergonnenians, so they should  always be on the 'right'

   side of Medians and thus they should not struggle for triangle  existence.

   Thank you.

   M.T.

I don't understand what you are saying   :(

Medians of a triangle can form a triangle of their own, they have enough length in

   them.  If this is true (Darij  gave a convincing argument, so it is not iffy as it was before

   he delivered the proof),  then cevians of Nagel point

   are even 'more triangulable' then medians, if you consider position of Nagel point

   w.r.t. Centroid and Gergonne point.

   Salute and thank you.

   M.T.

[quote="Arne"]I don't call that a proof, armpist...[/quote]

 It is perfectly all right, it is not your fault, it happens with my proofs from time to time,

 they have that deceiving first impression.  Yet, I suspect

 that you did not spend enough time trying to fathom the

 underlying original and fruitful  idea. You can and probably should for your own good

 get back to it after the Holidays,  when you will be able to concentrate exclusively on it.

  Happy New Year to All !

  Thank you.

   M.T.

Dear Perfect_radio
 where is "Tuymaada"?

take a look at [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=98]this[/url].

Consider the triangle $A'B'C'$ whose medial triangle is $ABC$ and let $D$, $E$, $F$ be the tangency points of the incircle of $ABC$ with $BC$, $CA$, $AB$ respectively. Then $A'D=AA_1$, $B'E=BB_1$ and $C'F=CC_1$.

Assuming WLOG that $\angle A\le90^\circ$ fold the triangles $ABC'$ and $ACB'$ along the sides $AB$ and $AC$ until $B'$ and $C'$ coincide in a point $P$. If $\angle A<90^\circ$ there are two possibilities for $P$: one above the plane of $ABC$ and one below, mirror images, so it does not matter which one we choose. In any case the triangle $PEF$ is formed and $PE=BB_1$, $PF=CC_1$. If $\angle A=90^\circ$ then $P$ is unique and it is in the plane of $ABC$.

By triangle inequality $PE+EF>PF$ and $PF+EF>PE$, but $EF$ is a chord of the incircle so $EF$ is smaller than an altitude of $ABC$, and so smaller than $AA_1$, hence $PE+AA_1>PF$ and $PF+AA_1>PE$, i.e., $BB_1+AA_1>CC_1$ and $CC_1+AA_1>BB_1$. Folding along the other acute angle we obtain the remaining inequality.

Observe that if the triangle is acute then we can fold the three triangles and obtain a nice tetrahedron. If the triangle is rectangle we obtain a degenerate tetrahedron when folding along the right angle. When the triangle has an obtuse angle then it is impossible to form a tetrahedron by folding the three triangles around $ABC$.


Expressed in simplest form, $(3x^{3})(2y)^{2}(4x^{4})$ equivalent to what?

[hide]Just multiply it out and get:
$48x^{7}y^{2}$[/hide]

Note that whenever we have two things multiplied together inside parenthesis, and then raised to a power, like this:

$(a \cdot b)^{2}$

It can be rewritten as $a^{2}\cdot b^{2}$.  It's a simple rule of exponents you should always know.

True. Using that, you got that $(3x^{3})*(2y)^{2}*(4x^{4})=3x^{3}*4y^{2}*4x^{4}=48x^{7}y^{2}$.

Another important thing to know is that $x^{m}*x^{n}=x^{m+n}$, and that $\frac{x^{m}}{n^{n}}=x^{m-n}$.

Please, always remember to post not only your answer, but your solution too (to everyone).

combine the coefficents and combine the variables to get:

48 (x^7) (y^2) :D