ok im really sorry i dont know ANY french at all  :(  i took spanish at school...
 but i wanted to know if any of you guys are actually french, and if you are, do you know who nikkfurie (or nikk furie, or nick furie....!) is? i have had a REALLY hard time trying to find his music here in the states (i cant even order anything off amazon). any help you could give me about where i could find a cd or something would be GREAT!

merci
 :lol:

Hello,

I'm actually french but I don't know who you're talking about. However I can ask friends who play music.

Is Nick Furry a guitarist? a singer?

its uh... letsee, i think he's a DJ (in the most technical sense of the term- he mixes music). It's very Urban/r&b... kind of electronica. the song he mixed for ocean's 12 (te a la menthe) was originally done by "la caution".... any info on that group would be awesome as well. danke.... i mean, gracias... er.......... merci!

I'm sorry :( ,

I can't help you: I don' know DJs very much...

Maybe somebody else...

Je crois que "Thé à la menthe" ne sera pas sur l'album de La Caution mais sur l'album instrumental de Nickffurie, qui est prêt depuis bien longtemps (ms aucune date, il devait être distribué mondialement si je me souviens bien).
"Thé à la menthe est également utilisé par Kourtrajmé dans un de leur court métrage et présent sur le cd de la bande son du dvd livré avec la réédition.
http://www.infratunes.com/forum/message.php?sid=3932

http://www.kourtrajme.com/accueil.htm

:)  merci!
Je voudrais trouver un copie de l'album instrumental de Nickffurie.  Ou est ca?  Est-ce qu'il y a un place qui n'est pas en l'internet?

(I hope this is right.)

Je ne sais pas du tout...
tout ce que je sais sur Nickfurrie Google me l'a appris...

Donc cherche sur Google !

J'essaye de comprendre ce que vous dites sur l'album instrumentale.  Qu'est que c'est "ms" ("ms aucune date")?  Je pense que vous disiez que cette album de Nickfurrie a libere, n'est pas?  C'est tres difficile a trouver l'information sur Google, parce que tous les pages sont en Francais, que je ne comprend pas. (Mon amie a ecrit pour moi.  Ell n'ecrit pas bien en francais et elle est desolee pour le grammaire et vocabulaire qui n'est pas correct.) :blush:

again, merci...   :)

ms est mais, il a enlevé les consones. C'est le style SMS, on dit salut: slt

je suis désolé...

je ne comprends pas.

ah.... this is as far as my freelance-French goes...
Can you please try to put that into English? I am very sorry that I do not understand what you are saying. :blush:

Thank you!!!!
merci!

Sometimes we write "ms" instead of "mais" because it's shorter, "ms" isn't a french word. "ms aucune date" = "mais aucune date" = "but no dates"

[size=200]OOOOOOOOH![/size]

 :blush:  :)
THANK YOU!!!!!
*g*


Let $a,b,c$ be positive real numbers and $a+b+c=1$.prove that this inequality holds:
$ \sqrt{\frac{ab}{ab+c}}+\sqrt{\frac{ac}{ac+b}}+\sqrt{\frac{bc}{bc+a}} \geq \frac{3}{2}$

obviously of the reversed sign. The inequality can be proved by amgm on $\frac{a}{c+a}$ and $\frac{b}{c+b}$ and so on and sum them up.

Let me post a full solution :  :)
 $ \sqrt{\frac{bc}{bc+a}}= \sqrt{\frac{bc}{bc+1-b-c}}= \sqrt{\frac{bc}{(a+b)(a+c)}} \leq \frac{1}{2}. (\frac{b}{a+b}+\frac{c}{a+c}) $
  Write 2 similar ineqs and add them all --->Q.E.D

[quote="blackcode"]May be you have something wrong[/quote]
I don't think so. :?:

This can be deduced from the following 'well known' ineq:
$\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}} \le \dfrac{3}{2}$
where $x,y,z$ are positive real numbers with $x+y+z=xyz$.

very nice solutions thanks.you are cool. :D

This was already posted in http://www.mathlinks.ro/Forum/viewtopic.php?p=135067

thank you but i think there is no matter if we think about it again,and we won't lose anything. :D

i think the inequalities is wrong.
it should be (ab/(ab+c))^(1/2)+(bc/(bc+a))^(1/2)+(ca/(ca+b))^(1/2)<=3/2.[/list][/quote][/code]


limit
trigonometry
integration
real analysis
real analysis unsolved

Proove the following limits

$ \lim_{n\to \infty }\, \left(\sum_{b=1}^{n}\sum_{a=1}^{b n}\frac{1}{a^{2}+n^{2}}\right)=\frac{\pi }{2}$

$ \lim_{n\to \infty }\, \left(\sum_{b=1}^{n}\sum_{a=n}^{b n}\frac{1}{a^{2}+n^{2}}\right)=\frac{\pi }{4}$

This is a little bit hard. it took me some minutes to solve and I am sure of the solution. the double series increass very and very slowly and the limit of the first sum is zero. so we cannot change the order.


Let $a,b,c$ be non-negative numbers such that $a^{2}+b^{2}+c^{2}=3$. Prove that
$1+4abc \geq 5\min\{a,b,c\}$.

:D [quote="toanhocmuonmau"]If $a,b,c \ge 0,a=\min(a,b,c)$, then
\[bc \ge a\sqrt{b^{2}+c^{2}-a^{2}}=a\sqrt{3-2a^{2}}\]
:)
[/quote]
The main difficulty of the problem is to find a nice proof to
$1+4a^{2}\sqrt{3-2a^{2}}\geq 5a$. :D

[quote="silouan"][quote="Vasc"] The main difficulty of the problem is to find a nice proof to
$1+4a^{2}\sqrt{3-2a^{2}}\geq 5a$. :D[/quote]

This is equivalent to show that
$32a^{5}+32a^{4}-16a^{3}-16a^{2}+9a-1\geq 0$ for a$\in$[0,1][/quote]

I think you're wrong, silouan.

[quote="silouan"]This is equivalent to show that
$32a^{5}+32a^{4}-16a^{3}-16a^{2}+9a-1\geq 0$ for $a\in[0,1]$
[/quote]

Ok, you meant to say $a\in[\tfrac{1}{5}, 1]$ [b]not[/b] $a\in[0,1]$.

I think your solution is nice only for $a \geq \frac 1{2}$.
But also for $a\leq \frac 1{2}$ there is a nice solution. :P

[quote="toanhocmuonmau"]ok, could you please show me your solution?[/quote]
For $a \leq \frac 1{2}$, we have $\sqrt{3-2a^{2}}\geq \sqrt{\frac 5{2}}$, and hence $1+4a^{2}\sqrt{3-2a^{2}}\geq 1+4a^{2}\sqrt{\frac 5{2}}\geq 5a$.


function
quadratics
algebra
floor function

I made the PDF document for this round for the solutions as well so enjoy looking at them.

If you have any questions or comments (I'll appreciate word "thank you!" too! :D ), please post here.

Thanks

Silverfalcon

[quote="eryaman"]Is there any way to do problem number 3 without just adding them all up?[/quote]

notice the pattern in the differences between each term (after the greatest integer function is performed)

0,0,1,2,3,5,7,9,12,15....

difference b/w each term: 0,1,1,1,2,2,2,3,3....

although in the end, you have to add the terms up, if you aren't allowed to use a calculator, this will save time in finding each term

I don't think there are any easy way.

I made that problem so I can see who is making careless mistakes.

But luckily, most people got right on that problem. :)

I thought I qualified ... why am I not on the qualifiers list?

[quote="Silverfalcon"]Nice and neat.

Qualified. :)[/quote]

Unless that was meant for someone else.

Fixed.

Sorry about it. This COP has been very active one :D and I'm starting to prepare for final as well.  Also, I went to tennis match yesterday so I was tired..

Thanks for pointing out. :)

Bleh, I can't believe I made such a careless mistake on #5, $\frac{81\pi}{6} = \frac{27\pi}{3}$  I've done hundreds of those type of problem argh.

[quote="white_horse_king88"]Darn - ...

I knew I should have sent in the answer yesterday night instead of today morning.

I couldn't get access to a computer in the morning - complicated story.

Sorry Silverfalcon.[/quote]
Dont worry, I believe there is a round 1 B.  :lol:

[quote="chess64"][quote="eryaman"]Is there any way to do problem number 3 without just adding them all up?[/quote]

I made a calculator program to do it in like 15 seconds.  :D[/quote]I did that too...but wasn't sure if that was against the rules, so I did it by hand also.

I don't know how to put that in calculator.

I used calculator when I was adding after taking greatest integer function though.

But from now on, I'll add that the putting the equation into the calculator and calculating the value is not allowed.

[quote="Silverfalcon"]

But from now on, I'll add that the putting the equation into the calculator and calculating the value is not allowed.[/quote]

That's not what I did. I noticed that the pattern was quadratic equation, so I wrote it as one. I made a program that took the greatest integer function of each $y$, based on the $x$'s, as the $x$ increased each time. At the end, it automatically summed them for me.  :?

[quote="spoudyal"]When will I be able to participate if I did not this time.[/quote]

Read about COP B


linear algebra
matrix
rotation
geometry open
geometry

[quote="flip2004"]Let $ (x_k) , (y_k) $ defined by  $ { x_{k+1}\choose y_{k+1}} ={a\; \; \;  b \choose c\; \; \; d}{x_k \choose y_k}\; \; ,\; \; k=0,1,2,..., $ where  $ x_0,y_0,a,b,c,d $ are real numbers. In plane let the points $ M_k =(x_k,y_k) , k=0,1,... . $
  [i]Question:[/i] [b] Find the ,,start-point(s)" $M_0$ and  all matrices  $ {a\; \; \; b\choose c\; \; \; d} $  for which :
 [i] i)[/i]  $\; \; \;  x_0<x_1<x_2<... $ , and
  [i] ii)[/i]  the polygonal line  $  M_0M_1M_2...M_n... $  is convex .[/b]

[b]Note:[/b] ,,semi"-open problem. I hav'nt a complete solution.[/quote]

I doubt this is open.  The dynamics of a 2x2 matrix are well known and make this problem obvious, unless I completely misunderstand the statement.  If there are
complex eigenvalues the orbits will lie on some type of spiral and eventually rotate more than 360 degrees, so nothing convex.  If the eigenvalues are real then after some affine transformation the orbits are on curves $|x^a| = |y^b|$ for real $a,b$.  For convexity you need the orbit to lie on just one branch of the curve, which (I think) means the eigenvalues are positive, or maybe it's enough if they have the same sign.
I think the curve is automatically convex, it looks like $y = x^u$.

The other condition, that x1,...  be increasing is just some (possibly ugly) condition on the orientation of the orbit-curve relative to the x and y axes.  It may be enough to just check the location of the eigenlines, or that the starting position is in the correct quadrant between the eigenlines.

Anyway, nothing complicated is needed to solve this.


A gardener plants three maple trees, four oak trees, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. What is the probability that no two birch trees are next to one another?

[hide]Consider the sequence MMMOOOO.  If we find the number of ways this can be arranged, then multiply this by $\binom{8}{5}$, since we are choosing 5 of the 8 blanks in _M_M_M_O_O_O_O_ for the location of the birch trees, we can find the total number of ways without two birch trees next to each other.  So MMMOOOO can be arranged in $\binom{7}{3}$ ways, and the total number of ways the trees can be ordered is $\frac{12!}{3!4!5!}$.  So our answer is $\frac{\displaystyle\binom{8}{5}\binom{7}{3}}{\frac{12!}{3!4!5!}}=\frac{\displaystyle\binom{8}{5}\binom{7}{3}3!4!5!}{12!}=\frac{7}{99}$.[/hide]

No. of ways of planting the trees is 12!/(3!4!5!)=27720
XMXMXMXOXOXOXOX,  X is the possible positions for the birch trees when they are not together.
No. of ways of planting trees so that no 2 birch trees are together is [7!/(3!4!)]*[8C5]=35*56=1960

Hence probability=1960/27720

I think thats the answer.[/code]

varun if you had bothered to reduce you'd find your answer identical to the other one.

here's another way: ignore the oak and maple, and just take the ratio of the number of ways to plant birch w/o adjacent ones to the total number of ways to plant them. (because the ways to plant oak and maple are symmetric in all cases and cancel out in ratios anyway)


geometry
trapezoid
geometric transformation
geometry proposed

$ S$ is the area of a trapezoid which its height $ \equal{}4$,2 diagonal lines are perpendicular.
Find $ 15S$ when length of one line is $ 5$.

This problem was one of practices suggested by my math teacher, Mr.Cho, for KMO preparation. So, pretty crude translation(Kor->Eng) -_-;

Assume $ AD \parallel BC$, $ AC\equal{}5$. By translation $ T(\overrightarrow{AD})$, we have $ C \to C'$. And $ \triangle{ADB}\equal{}\triangle{CC'D}$. So $ \triangle{BDC'}\equal{}S_{ABCD}$. Hence we have:

$ 5 \cdot BD \equal{} 2S \equal{} 4BC' \equal{} 4 \sqrt{5^2\plus{}BD^2}$

which yields $ BD\equal{}\frac{20}{3}$. Hence $ 15S\equal{}15 \cdot \frac12 \cdot 5 \cdot \frac{20}{3}\equal{}250$

Before describing my one, i appreciate your reply, shobber.
definitions: $ y\equal{}\overline{AB},$ $ x\equal{}\overline{BC},$ $ z\equal{}\overline{BF}$ (Refer to the image)
sol.
Assume the line $ \overline{EC} \equal{} 5$ that is the "one line" and $ \overline{AC}\parallel\overline{EF}$
then $ \overline{BE} \equal{} 5\minus{}x$
(1)$ y: z\equal{}x: (5\minus{}x)$(because $ ABC$ is similar to $ FBE$)
Hence $ xz \equal{} y(5\minus{}x)
S_{AEFC} \equal{} S_{AEF}\plus{}S{AFC}
                  \equal{} \frac{1}{2}(y\plus{}z)(5\minus{}x) \plus{} \frac{1}{2}(y\plus{}z)x
                  \equal{} \frac{1}{2}(y\plus{}z) * 5$
(2)$ x: y\equal{}3: 4$(because $ ABC$ is similar to $ EDC$)
Hence $ y\equal{}\frac{4}{3}x$, and $ z\equal{}\frac{20}{3} \minus{} \frac{4}{3}x
y\plus{}z \equal{} \frac{4}{3}x \minus{} \frac{4}{3}x \plus{} \frac{20}{3}$
In conclusion, $ 15S_{AEFC} \equal{} \frac{1}{2}(y\plus{}z) * 5 * 15
                                          \equal{} \frac{1}{2} * \frac{20}{3} * 5 * 15
                                          \equal{} 250$


$f(n)$ is defined for all positive integer $n$ and take on non-negative integer values such that $f(2)=0,f(3)>0,f(9999)=3333$.
Also, for all $m,n$,it is given that $f(m+n)-f(m)-f(n)=0$ or $1$
Determine $f(2005)$

[quote="shobber"]if $f(m+n)=f(m)+f(n)$ then $f(x)=kx$. but since $f(2)=0$, so $f(x) \neq kx$, hence we must have $f(m+n)=f(m)+f(n)+1$.[/quote]

I interpreted the given statement as "All $m$ and $n$ satisfy one of $f(m+n)-f(m)-f(n)=0$ or $f(m+n)-f(m)-f(n)=1$". If $f(m+n)=f(m)+f(n)+1$ for all $m$ and $n$, then $f(1+1)=f(1)+f(1)+1$ implies $f(1)=-\frac 12$, clearly violating the definition of $f$.

$f(n) = \left\lfloor\frac n3\right\rfloor$?
If so, then $f(2005) = 668$.

Well we have

$f(3) - f(1) = 0$ or $1$
$f(4) - f(3) - f(1) = 0$ or $1$
$f(5) - f(4) - f(1) = 0$ or $1$
...
$f(9999) - f(9998) - f(1) = 0$ or $1$.

If we add all those together we have $f(9999)-9997f(1) = 3333-9997f(1) \in \{0,1,...,9997\}$, but since $f(1)$ is an integer, it must be $0$.

So $f(1) = 0 \Rightarrow f(3) = 1$. But then $f(4) = 1$ or $2$ and you can keep building $f(k)$ up to $2005$ but from what I see there can be lots of different values...

The floor function given by octagram seems to work, but I'm still not sure that $668$ is the only value possible.

Actually, once you have $f(3)=1$, you have $f(3n)\ge n$. Then $3333=f(9999)\ge f(2004)+f(7995)\ge f(2004)+2665$, so $668\ge f(2004)\ge 668$. Now, since $f$ is nondecreasing, $f(2005)\ge 668$. A similar argument at 2007 shows that $f(2007)=669$ and $f(2005)\le 669$. On the other hand, $f(6015)=2005$, so $f(2005)\le\frac{2005}3$ and $f(2005)=668.$

The statements about the value of $f$ for small $n$ are redundant; we could prove them knowing only the value at 9999.


hi im new to this forum but my friends told me about this site and how its great for helpin ppl in math..but i need to be helped to help tutoring this kid from church..she just asked me these 5 questions out of no where and its above my level. but shes nice and has respected me greatly so im wantin to help her out. thx.

1)
If adding consecutive integers starting with 1 on a nonscientific calculator with an eight-digit display, how many integers would have to be added to exceed the capacity of the calculator?

2)
A semicircle is inscribed in an equilateral triangle as shown. Express the radius in temrs of the side S.
it shows the circle's diameter on the base of the triangle and the circle's edge seems to be 1/2 way up the tip with the other bottom half of the circle not shown. (sry no pic)

3)
Arrange the digits 0 through 9 so that the first digit is divisible by 1, the first two digits are divisible by 2, and the first three digits are divisible by 3. Continue until the first nine digits are divisible by 9 and the ten-digit number is divisible by 10.

4)
What is the maximum number of acute angles that any convex polygon can have?

5)
Find a possible next three terms in the following pattern:
1,1,1,3,5,9,17,31,__,__,__,......

id really kno how to do it so i can explain...sry not that good with these stuff im only in geometry and this girl from algebra is asking...its optional to help. just wanna help ppl out. thx

[hide="hint for number 5"]Think about the Fibonacci sequence[/hide]

[hide]
1. Formula for addition of n consecutive integers is $\frac{(n)(n+1)}{2}$
So we're looking at $\frac{(n)(n+1)}{2}>100000000$
Which solves to be $14142$.

2. We see that the area of the triangle is $\frac{s^{2}\sqrt{3}}{4}$, but also $sr$, so $r=\frac{s\sqrt{3}}{4}$

5. Each term past the 3rd is the sum of the 3 previous terms. [/hide]

[hide]The first thing i would do in number 4 is set an upper bound; that is, n where an n-gon CAN'T have acute angles, and work down.[/hide]

[quote="Ubemaya"][quote="Hoborobo"][hide]
2. We see that the area of the triangle is $\frac{s^{2}\sqrt{3}}{4}$, but also $sr$, so $r=\frac{s\sqrt{3}}{4}$
[/hide][/quote]

There's a semicircle inscribed, not a circle...[/quote]

So? Draw it out. You'll see what I'm talking about.

I'm pretty sure the semicircle would be able to have a larger radius.

An acute angle is an angle such that it's measure is $\frac{\pi}{2}-\epsilon$, where $\epsilon$ is positive. Just use that for the acute angle problem.

[hide="Question1"]Question one is asking for what number does the sum of all integers less than it exceed 99,999,999
The sum of consecutive integers can be given by $\frac{n(n+1)}{2}$
We want this to be$\ge 99,999,999$.
$\frac{n(n+1)}{2}\ge 99,999,999$
$n(n+1) \ge 199999998$
$n^{2}+n-199999998 \ge 0$
By the quadratic equation, $n \ge \frac{(-1 \pm \sqrt{1+799999992})}{2}$

Work out the root of 799999992 on a calculator, and find that it's roughly $28284.27$
We're only interested in positive solutions, obviously, so take $-1+28284.2713$ and divide by two.
You should get that $n \ge 14142.6357$, but looking at that, you see you need to have the one before that. So you'd need to add $\boxed{14142}$ consecutive integers to overload the calculator.
[/hide]
[hide="Question2"]
If the equilateral triangle has side length S, and one whole diameter of the circle is touching S, then you can see that the diameter = S. A radius is half of a diameter, so the radius is equal to $\boxed{\frac{S}{2}}$
[/hide]
[hide="Question3"]
The ten digit number MUST have the zero at the end to be divisble by ten.
Also, the first digit may be anything, because anything is divisible by one.
For something to be divisible by three, it's digits must add to a multiple of three. See if you can work it out with these hints.

Also, we've already used the 0, and for a number to be divisible by five, it must end in a zero or a 5. Thus, the fifth number is 5, the tenth is zero.
[/hide]
[hide="Question4"]
Think of a regular, convex polygon with 89 degree angles. Now think of one with 89.5 degree angles. As the angle measure gets closer to 90, the sides get closer to a square. Thus, the most acute angles possible are from a triangle (three angles).
[/hide]
[hide="Question5"]
This is like a fibonacci series, but adding the past three terms instead of one. There's a complicated means of doing something called determining the characteristic polynomial, but for now, suffice it to say this:
The fourth term is the sum of the 1st, 2nd, and 3rd terms.
The fifth term is the sum of the 2nd, 3rd, and 4th terms.
The sixth term is the sum of the 3rd, 4th, and 5th terms.
The next three terms are: $\boxed{57, 105, 193}$
[/hide]

He never said that the diameter WAS the base, but that it was ON the base. He said that the semicircle was inscribed, so Hoborobo is right.

True. I made a reading error. I think hoborobo is correct as well.


algebra
polynomial
Vieta
function
algebra proposed

Using the cubic formula, this problem isn't very hard. So, do you want us to factor it? Thanks

One of metods is find factor $(y^{3}+ay+1)|(y^{15}+15y^{10}-198y^{5}+1)$. Then $y_{1}+y_{2}+y_{3}=0$, were $y_{i}=\sqrt[5]{x_{i}},x_{i}$ are roots $x^{3}+15x^{2}-198x+1=0$. But my calculations don't give factor. May be mistake.

We can prove that if the roots of the:
$x^{3}-3x+1=0$
equation are $x_{1}$, $x_{2}$, $x_{3}$ then the roots of the
$x^{3}+15x^{2}-198x+1=0$ equation are $\sqrt[5]{x_{1}}$, $\sqrt[5]{x_{2}}$ and $\sqrt[5]{x_{3}}$
From the first equation using Vietes formulas:
$x_{1}+x_{2}+x_{3}=0$
So:
$\sqrt[5]{x_{1}}+\sqrt[5]{x_{2}}+\sqrt[5]{x_{3}}=0$

My congratulations, Thaakisfox!
Try the following problem:
$x_{1},$ $x_{2}$ and $x_{3}$ are roots of equation $x^{3}-16x^{2}-57x+1=0.$
Prove that $\sqrt[5]{x_{1}}+\sqrt[5]{x_{2}}+\sqrt[5]{x_{3}}=1.$  :wink:
Thank you for the interest.

See also this one  :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?t=69828

[quote="arqady"][quote="me@home"]How did you come up with this problem?[/quote]
As always: from the end. :wink:  :)[/quote]

Yeah, I think this sort of problems are much easier to construct than to solve.
For example start with a polynomial $ x^{3} \minus{} px \plus{} q \equal{} 0$ such that there are three real roots $ x,y,z$. By Vieta we have formulas for $ \sum x,\sum xy$ and $ xyz$ in terms of $ p,q$. From there we get e.g. $ \sum x^{2} \equal{} 2p,\sum_{sym}x^{2}y \equal{} 3q,\sum x^{3} \equal{} \minus{} 3q$ and so on for all symmetrical functions of $ x,y,z$ of higher degrees.

By the way, I wonder if for any "starting polynomial" $ \sum_{i \equal{} 0}^{n}a_{i}x^{i}$ with $ a_{n} \equal{} 1$, all the symmetrical functions of its roots, being polynomials of the $ a_{i}$, have [i]integer[/i] coefficients. I remember a very nice book by I.G.Macdonald  "Symmetric Functions and Hall Polynomials" which probably has an easy proof of this using ring theory  :roll:

arqady, Please show your solution if you can !


limit
continued fraction
number theory proposed
number theory

Let $x \in \mathbb{R}$. Prove that $x$ is irrational if and only if there exist sequences $\{p_n\}$ and $\{q_n\}$ of positive integers such that \[ \lim_{n \rightarrow + \infty} \left(q_nx - p_n\right) = 0  \] and such that \[ x \neq \frac{p_n}{q_n},\ \forall n \in \mathbb{N}.  \]

If $x$ is irrational, then the sequence of fractions given by the continued fraction of $x$ will do.

Otherwise, assume that for $x=\frac ab$ such a sequence is given.
Then also $\lim_{n \to \infty} aq_n - b p_n = 0$, but $aq_n - b p_n$ just gets integer values different from $0$, so contradiction.


helo
solve in IR :
$ x^2\minus{}\sqrt{x}\minus{}14\equal{}0$

[quote="sami.d"]helo
solve in IR :
$ x^2 \minus{} \sqrt {x} \minus{} 14 \equal{} 0$[/quote]

[hide="Hint"]Set $ z\equal{}\sqrt x$ and solve the quadratic equation.[/hide]

[hide]
Substitue $ \sqrt{x}$ with $ y$,where $ y \geq 0$, then the equation becomes
$ y^4\minus{}y\minus{}14\equal{}0$
Observe $ y\equal{}2$ is a root of this equation, then by synthetic division (or long division), it's factored as
$ (y\minus{}2)(y^3\plus{}2y^2\plus{}4y\plus{}7)\equal{}0$
Because $ y \geq 0$, the cubic is always greater than or equal to $ 7$, therefore has no zero.
Thus, the only root is $ 2$.[/hide]


!این جا برای مساله های هندسه حالا دیگه لطف کنید سوالهاتون رو با همون سیستم بفرستید دیگه

سلام  خب بذارید از همون اول خفن شروع کنیم.

Dar mosalase $ABC$ dayeri $(\omega)$ delkhah az ro'oose $B,\ C$ migozarad .dayeriye $C$ bar $AB\ \text{va}\ \omega$dar$P,Q$momas ast.agar $I$ markaze dayere mohati mosalase $ABC$ bashad.sabet konid cheharzelyie $PQIB$ mohatiye.

lotfan kalame tanha nanevisid agar hal kardid rahe kamel benevisid.(ghabl tavajoh Ashegh joon)

daste shomaa dard nakone che soale ghashangi bood ;)
khob; baayad saabet konim $\angle PQI=180-\angle PBI$ yaa in ke $\angle PQB+\angle BQI=180-\frac{\angle B}{2}$
amma baa tavajoh be yek masaleye besiaar maroof: $2\angle BQI=\angle BQC$
pas $\angle BQI=\frac{\widetilde{BC}}{4}$ va $\angle B=\frac{\widetilde{XYC}}{2}$
hala baayad saabet konim $\angle PQB=180-\frac{\widetilde{BC}+\widetilde{XYC}}{4}$
yaa in ke $\angle PQB=90+\frac{\widetilde{XB}}{4}$
va dar nahaayat baayad saabet konim $PQ$ nimsaaze khaarejiye zaaviyeye $\angle XQB$ ast; ke in ham baa yek zaaviye baaziye kochak dar mosalase $\triangle XQB$ saabet mishavad... :|

rahe hale ghashangi bood ;)   vali agar ooon mas'aleye maroofo(man ta hala nadideboodamesh ;) ) nemidoonestid chejoori esbat mikardid.man khodam ba Enekas bad az 2 saat toonestam halesh konam.
vali hala oon mas'aleye maroof ro chejoori esbat mikoni. :?:

[quote="amir2"]rahe hale ghashangi bood. ;) [/quote]

akhe Ramtin jan, emsal medale tala gereftan

Bale rahehale ghashangiye vali fekrkonam ghalate.

[quote="ramtin madani"]va dar nahaayat baayad saabet konim $PQ$ nimsaaze khaarejiye zaaviyeye.... $XQB$[/quote]

baghyasho nagereftam mishe kamel tozih bedi

khob age $QS$ momaas moshtarake do daayere baashad; dar aan soorat $\angle QSB=\angle QXB-\angle QBX$ va chon midaanim $QS=SP$ pas $\angle PQS=\frac{\angle QSB}{2}=\frac{\angle QXB-\angle QBX}{2}$
va dar nahaayat $\angle PQX=\angle PQS+\angle SQX=\frac{\angle QXB+\angle QBX}{2}$
pas $QB$ nimsaaze khaarejiye $\angle XQB$ ast.
oon masale kheyli maroofe; mige ke saabet konid nimsaaze $\angle BQC$ az $I_{\triangle ABC}$ migozarad; esbaatesham baa enekaase :P (man balad nistam).

akhe amir joon to chera inghadr toopi? :D

in masale ba $2$ ta enekas e besiar ziba ,kheily khoshghel hal mishe.

ageh vaght shod rahe hallo post mikonam... ;)

Vali man ba ye enekas halesh kardam :D  :D

khob yeki raahe hale enekaasi ro bezaare dige :|

in ham rahe hal taghdim be agha Ramtin :

[b][color=blue]Lemma[/color][/b].dar mosalase $ABC$ agar $I$ markaze dayere mohati dakheli va $I_a$ markaze dayere mohati kharegi nazire rase $A$bashad darim
$AB.AC=AI.AI_a$
ba tashaboh rahat mitoonid sabet konid.

khob hala bar migardim sare soal be markaze $B$ va be shoa'e$AB.AC$ enekasi mizanim ba tavajoh be lemm:

[color=darkblue]khate $AC$ ----> dayereye mohiti mosalas jadid
dayereye delkhah ke az ro'oose $B,C$ migozarand---->khati delkhah ke az $C$ migozarad
$I$--->$I_b$(ba tavajoh be lemm)[/color]

baraye inke sabet konim chehar noghteye $B,I,P,Q$ roye yek dayereand pas az enekas bayad sabet konim $P,Q,I_b$ roye yek khat hastand

farz mikonim mahale barkkhord $PQ$ ba nimsaz khareji zaviye $C$ ,$X$ bashad,Sabt mikonim $X$ haman $I_b$  ast.(bedoone joziyat)
ba tavajoh be shekl $P'$ vasate kamane $ACB$ ast.choon $T$ markaz tajanoose do dayere ast pas $PQ\parallel P'Q'$dar natije $\angle PQT=\angle TQ'P'=\angle TCX$ dar natije chehar zelyei $TCXQ$ mohatiye.ma midoonim ke agar $I_b\equiv X$angah $P'A=P'X$ va hamchenin darim $AP'^2=P'T.P'P$ pas bayad sabet konim
$P'X^2=P'T.P'P$ ke oonham choon $\angle TXP'=\angle TQP'=\angle QPT$ esbat mishavad.
Dar natije $X\equiv I_b$ :D  :D

[quote="pooyan_seva"]aghayoone asatid bebakhsid ... man ye soale fanni dashtam:
che joori shekl mikeshin?
yani koja rasmeshoon mikonin??
  :D   :roll:  mersi.[/quote]
man(fekr konam hame!!!) too paint windows mikesham. bad attach mikonam.

ma mamoolan sheklhaye India3 (hendese) ro ba dast mikeshim agar bekhahim ghashang beshe dige az pargaro khatkesh ham estefade mikonim.

ye soale asoone dige:
chegoone mitavan mosalas ra ba dashtane toole se miane an rasm kard??(lotfan rahe halletoono bezarid)

ye soale asoone maroofe dige:
az noghte R khati rasm konid ke khat mafroozi ra dar D va yek dayere mafrooz ra dar E va F ghat konad be tori ke RD=EF
khaheshan rahetoono bezarido nagid koja boode ya inke mikhore az chi hal beshe
merC

in sola kheili khafane(soal Courte) mesle inke ba khatkesh va pergar ham nemishe rasm kard.

:o  :o m mage to halesh kardi ? ya inke az koja midooni soale asoonie?

kheili doooost daram rahe haleto bebinam.

yek masaleye dige
neshan dahid ke khotoote oulere se mosalasi ke as yek mosalase mafroiz tavasote azlaye mosalase padake an
joda mishavand yak noghteye moshtarak darand vain noghte roye dayereye no noghteye mosalase mafruz gharar darad.
in yeki ro ba bordar hal kardan kare hasrate fil ham nist!
man khodam ino ba menelaoos hal kardam.

 lemma
agar az se rase mosalasi se khate movasi rasm konim va in se ta ra nesbat be nimsaze motanazereshan
gharine konim se khate jadid ham res hastand.
 dar zemn midoonim khotoote ouler se mosalas ra agar nesbat be nimsaze motanazereshan
az zaviyeye mosalase asli gharine konim se khat movazi khate oelere mosalase asli mishavand.

ina rahnamayiye khoobi bood.  :idea:

amir jan baraye etelaet bayad begam ke mesle inke in soalo faghat vakili too iran hal karde
vali fekr konam ba enekas hal beshe vali chon vaght nadaram esbatesho be ohde shoma mizaram
merC

babak jan vase etelaet arz mikonam ke mesle inke insoalro Ostad Vakili ham natooneste hal kone.
va sabet shode ke ba khat kesh va pargar rasm nemishe kard.

BaBaK Jan agar mishe vaght dashte bash va rahe hal ba enekas ro befrest va in masooliate sangin ro az dooshe ma bardar :D  :D

valla amirm jan ta oonjai ke man midoonam vakili tabestoon goft ke faghat oon hallesh karde
badesh man ye soali dashtam mikhastam bedoonam ketabe khoob baraye tabdilato abzare khoob too hendese (adad mokhtalet,hendese bordari,...) chi hast??
merC :)

nemidoonam chera hichki javabe soale avale mano nadade shayad kheili asoone vali dar har soorat man halesho minivisam:
farz konid dar mosalase $ABC$ $AM_1,BM_2,CM_3$ se miane mosalas bashand
hal az noghte $M_1$ be vasate pare khate $BG$ ke $G$ markaze seghle mosalas mibashad,vasl konid va vasate $BG$ ra $K$ benamid
hal midanim ke $KG=1/3BM_2,GM_1=1/3AM_1,M_1K=1/3CM_3$
pas mosalase $KGM_1$ ghabele rasm ast
hal biaid pare khate $GK$ ra dar nazar begirid va in pare khat ra az dotaraf be andaze khodash emtedad dahid ta noghate $B,M_2$ (ke $B$ dar samte $K$ va $M_2$ dar samte $G$ vaghe bashad)
hamchenin pare khate $M_1G$ra az samte $G$ be andaze $2/3$ $M_1G$emtedad dahid ta noghte $A$ be vojood ayad
sepas noghte $M_2$ ra be $A$ vasl konido be andaze khodash az samte $M_2$ emtedad dahid ta noghte $C$ be vojood ayad
hal noghate $A,B,C$ra motavalian be ham vasl konid
ta mosalase $ABC$ be vojood ayad
hala aya mitavan mosalas ra ba dashtane $3$ nimsaze dakhelie an keshid??

[quote="mohamadhosein"][quote="mohamadhosein"]yek masaleye dige
neshan dahid ke khotoote oulere se mosalasi ke as yek mosalase mafroiz tavasote azlaye mosalase padake an
joda mishavand yak noghteye moshtarak darand vain noghte roye dayereye no noghteye mosalase mafruz gharar darad. :rotfl:[/quote][/quote]

in masale ziaadam sakht nist. kheyli maroofe :roll:
va agar aan noghteye hamresi P baashad  va ras haaye mosalase paadak X va Y va Z baashand; jame toolhaaye dotaa az se paare khat $PX,PY,PZ$ baa sevomi baraabare.

[quote="amir2"]ye soal shabihe hamin ham hast ke mige:
agar $F$ noghteye Fermaye mosalase $ABC$ bashad .khotoote Euler 3 mosalase $ABF\ ,ACF\ ,BCF$ hamresand.

([color=blue]Noghteye ferma[/color]: noghteyi ke 3 ta zele mosalasro ba zaviye $120^{\circ}$ mibinand)[/quote]

Fekr konam too markaze seghl $ABC$ hamresand :roll:

kasi mitoone ye seri az vigegihaye noghteye ferma ra post kone?


calculus
integration
trigonometry
function
derivative
algebra
linear equation

Can someone do a full solution for this integral. I used integration by parts but it just doesn't seem to work. $ \int e^{2x}\sin x\;dx$

More generally

Take $ \int e^{ax}\sin(bx)\,dx$ and apply integration by parts.

Yes but when I do Integration by parts I keep getting an integral with the natural exponent and a sin or cos. Should I just apply an integration formula or is there another way

Well, since I was just teaching this very thing yesterday, I'll have mercy on you and show you the trick. This is standard, and is explicitly shown in most calculus books that I know about.

In an integration by parts, we integrate one factor and differentiate the other. We'd like a plan under which the resulting integral is simpler than what we started with. But, as you've pointed out, exponentials, sines and cosines, are "indestructible": integrate them or differentiate them and they stay basically the same - the integral gets neither simpler nor more complicated. In the end, we take advantage this very unchanging nature.

Let $ u\equal{}\sin x,dv\equal{}e^{2x}\,dx.$ Then $ du \equal{}\cos x\,dx$ and $ v\equal{}\frac{1}2e^{2x}.$

$ \int e^{2x}\sin x\,dx\equal{}\frac{1}2e^{2x}\sin x\minus{}\frac{1}2\int e^{2x}\cos x\,dx$

Again: $ u\equal{}\cos x,dv\equal{}e^{2x}\,dx,du\equal{}\minus{}\sin x,v\equal{}\frac{1}2e^{2x}.$

$ \equal{}\frac{1}2e^{2x}\sin x\minus{}\frac{1}2\left[\frac{1}2e^{2x}\cos x\plus{}\frac{1}2\int e^{2x}\sin x\,dx\right]$

Distributing, we get

$ \int e^{2x}\sin x\,dx\equal{}\frac{1}2e^{2x}\sin x\minus{}\frac{1}4e^{2x}\cos x\minus{}\frac{1}4\int e^{2x}\sin x\,dx$

The unknown integral appears on both sides. Treat this as an equation - a linear equation - for that unknown.

$ \left(1\plus{}\frac{1}4\right)\int e^{2x}\sin x\,dx\equal{}\frac{1}2e^{2x}\sin x\minus{}\frac{1}4e^{2x}\cos x\plus{}C$

$ \int e^{2x}\sin x\,dx\equal{}\frac{2}5e^{2x}\sin x\minus{}\frac{1}5e^{2x}\cos x\plus{}C$

---

In my own class, I will return to this problem later on this semester after I've shown them the complex exponential. Oh, and by the way, if you want to know where $ 5$ comes from, $ 5\equal{}2^{2}\plus{}1^{2}.$

---

Yet another method that can be used in the method of undetermined coefficients as taught in a first differential equations course.

ahhh thank you. I got that same integral and realized that it couldn't be integrated completely. I didn't even think of transferring the integral to the other side. Thanks again

Note that the choice of which function to make $ u$ and which to make $ dv$ was quite arbitrary - it would have worked either way. But once you make that choice, you have to make essentially the same choice the second time. It's essentially the same amount of work each way. If there's a choice to make, it would be for human error reducing reasons - like letting $ u$ be the trig function because you tend to make sign errors when you integrate trig functions.

Answer

$ \int e^{2x}\sin{x}\, dx$

$ \equal{} Im\left[\int e^{2x}e^{ix}\, dx\right]$

$ \equal{} Im\left[\int e^{(2\plus{}i)x}\, dx\right]$

$ \equal{} Im\left[\frac{1}{2\plus{}i}e^{(2\plus{}i)x}\plus{}c\right]$

$ \equal{} Im\left[\frac{1}{5}e^{2x}(2\minus{}i)(\cos{x}\plus{}i\sin{x})\plus{}c\right]$

$ \equal{}\frac{1}{5}e^{2x}(2\sin{x}\minus{}\cos{x})\plus{}c$

Euler's relationship is beautiful, I still can't seem to get my head around it fully. Not only is it a prime example of mathematical elegance, it is also of great practical importance. For instance, in alternating current circuits with resistors, inductors and capacitors, the relationships which hold for complex voltages and currents etc. are analogous to those for direct current circuits. By converting to phasors and calculating complex impedances, equations can be obtained which give the current in any branch at any time and allow voltage gains to be calculated. I pity those people who never will know or understand the implications of this beautiful relationship.

sludgethrower, that [i]is[/i] in my lecture plans for this semester, about 6 weeks from now. I'm just not ready for it quite yet. And when I get there, I'll throw in this problem just for good measure:

$ \int xe^{2x}\sin x\,dx$

$ \int e^{ax}\sin(bx)\,dx$

Let's set $ u = e^{ax}\implies du = ae^{ax}\,dx$ & $ dv =\sin(bx)\,dx\implies v =-\frac{\cos(bx)}{b}$, so the integral becomes to

$ \int e^{ax}\sin(bx)\,dx =-\frac{e^{ax}\cos(bx)}{b}+\frac{a}{b}\int e^{ax}\cos(bx)\,dx$

Now we apply integration by parts to the remain integral. We take $ u = e^{ax}\implies du = ae^{ax}\,dx$ & $ dv =\cos(bx)\,dx\implies v =\frac{\sin(bx)}{b}$, therefore

$ \int e^{ax}\cos(bx)\,dx =\frac{e^{ax}\sin(bx)}{b}-\frac{a}{b}\int e^{ax}\sin(bx)\,dx$

Back substitute

\begin{eqnarray*}\int{e^{ax}\sin (bx)\,dx}& = &-\frac{{e^{x}\cos (bx)}}{b}+\frac{a}{b}\left({\frac{{e^{ax}\sin (bx)}}{b}-\frac{a}{b}\int{e^{ax}\sin (bx)\,dx}}\right)\\ & = &-\frac{{e^{x}\cos (bx)}}{b}+\frac{{ae^{ax}\sin (bx)}}{{b^{2}}}-\frac{{a^{2}}}{{b^{2}}}\int{e^{ax}\sin (bx)\,dx}\\ b^{2}\int{e^{ax}\sin (bx)\,dx}& = &-be^{ax}\cos (bx)+ae^{ax}\sin (bx)-a^{2}\int{e^{ax}\sin (bx)\,dx}\\ \left({a^{2}+b^{2}}\right)\int{e^{ax}\sin (bx)\,dx}& = & e^{ax}\left({a\sin (bx)-b\cos (bx)}\right)\\ \int{e^{ax}\sin (bx)\,dx}& = &\frac{{e^{ax}\left({a\sin (bx)-b\cos (bx)}\right)}}{{a^{2}+b^{2}}}+k,\,\forall k\in\mathbb R\end{eqnarray*}

Well, now you find the constraints for $ a,b$

About Kent's integral, looks like it is

[hide="¿?"]
$ \int{xe^{2x}\sin x\,dx}=-\frac{1}{{25}}e^{2x}\left({\left({5x-4}\right)\cos x+\left({3-10x}\right)\sin x}\right)+k$[/hide]

The method of undetermined coefficients, as it applies here:

We predict that the answer will be of the form $ ae^{2x}\sin x\plus{}be^{2x}\cos x\plus{}C$; this comes from the "indestructible" nature of trig and exponential functions. Knowing what to predict is also the trickiest part of the method.

Now, we differentiate our proposed answer; its derivative is $ 2ae^{2x}\sin x\plus{}ae^{2x}\cos x\plus{}2be^{2x}\cos x\minus{}be^{2x}\sin x$
$ \equal{} (2a\minus{}b)e^{2x}\sin x\plus{}(a\plus{}2b)e^{2x}\cos x$.
For this to work, we need $ 2a\minus{}b \equal{} 1$ and $ a\plus{}2b \equal{} 0$; this leads to $ a \equal{}\frac{2}{5},b \equal{}\minus{}\frac{1}{5}$ as in the other solutions.

From my working above we can see that:

$ \int e^{2x}\sin{x}\, dx \equal{}\frac{1}{5}e^{2x}(2\sin{x}\minus{}\cos{x})\plus{}c$

By simply taking the real part of the complex integral I used above we get:

$ \int e^{2x}\cos{x}\, dx \equal{}\frac{1}{5}e^{2x}(\sin{x}\plus{}2\cos{x})\plus{}c$

Now we wish to find the following integral:

$ I \equal{}\int x e^{2x}\sin{x}\, dx$

Let $ u \equal{} x\Rightarrow u' \equal{} 1$

Let $ v' \equal{} e^{2x}\sin{x}\Rightarrow v \equal{}\frac{1}{5}e^{2x}(2\sin{x}\minus{}\cos{x})$

$ \Rightarrow I \equal{}\frac{1}{5}x e^{2x}(2\sin{x}\minus{}\cos{x})\minus{}\frac{1}{5}\int e^{2x}(2\sin{x}\minus{}\cos{x})\, dx$

$ \equal{}\frac{1}{5}x e^{2x}(2\sin{x}\minus{}\cos{x})\minus{}\frac{2}{25}e^{2x}(2\sin{x}\minus{}\cos{x})\plus{}\frac{1}{25}e^{2x}(\sin{x}\plus{}2\cos{x})\plus{}c$

$ \equal{}\frac{1}{25}e^{2x}((10x\minus{}3)\sin{x}\minus{}(5x\minus{}4)\cos{x})\plus{}c$

Thanks for the [i]really[/i] detailed explanations everyone.  :lol:

$ \{\begin{array}{c}S=\int e^{ax}\sin bx\ dx\\\\ C=\int e^{ax}\cos bx\ dx\end{array}\|$ $ \implies$ $ \{\begin{array}{ccccc}bS=\int e^{ax}(-\cos bx)'\ dx & = &-e^{ax}\cos bx+a\int e^{ax}\cos bx\ dx &\implies &\boxed{\ bS-aC=-e^{ax}\cos bx\ }\\\\ bB=\int e^{ax}(\sin bx)'\ dx & = & e^{ax}\sin bx-a\int e^{ax}\sin bx\ dx &\implies &\boxed{\ aS+bC=e^{ax}\sin bx\ }\end{array}\|$ $ \implies$

$ \{\begin{array}{c}S=\frac{1}{a^{2}+b^{2}}\cdot e^{ax}\cdot|\begin{array}{cc}-\cos bx &-a\\\ \sin bx & b\end{array}|\implies\boxed{\ S=\frac{1}{a^{2}+b^{2}}\cdot e^{ax}\cdot (a\sin bx-b\cos bx)\ }\ \ \ \blacksquare\\\\ C=\frac{1}{a^{2}+b^{2}}\cdot e^{ax}\cdot|\begin{array}{cc}b &-\cos bx\\\ a &\sin bx\end{array}|\implies\boxed{\ C=\frac{1}{a^{2}+b^{2}}\cdot e^{ax}\cdot (b\sin bx+a\cos bx)\ }\ \ \ \blacksquare\end{array}\ \ \bigodot$


If $a_{n+1}=a^2_n-na_n+1$ and $n\in N^+$, prove that if $a_n \geq 3$, then:
\[a_n \geq n+2\]

By induction.  It is true for n = 1.  Suppose it is true for n = k.  (that $a_k - k \ge 2$)

Then $a_{k+1} = a_k^2 - ka_k + 1 = a_k(a_k - k) + 1 \ge 2a_k + 1 \ge 2k + 5 > k+3$ as required.


[quote="miyomiyo"]There is a hole in my ceiling of radius 4 and I have a bowl of radius 2.  Every 5 seconds, a drip of water falls through the hole in the ceiling.  After this happens, I move the bowl so that it is centered at the most recent drip of water.  Overall, what percentage of water drops will I catch?[/quote]
well
i would guess
[hide]1/4[/hide]
but with the moving the bowl thing i'm not sure if that affects the answer..if it does i have no clue.

Is this a self-written problem?  Because it looks to me very unlikely to have a non-calculus solution -- the area of intersection of the two circles will usually be unpleasant, and then we have to take an integral to average the resulting probability.  (Also, of course, we need a distribution for the locations of the water drops; if they always fall in the same place, the question is much easier then if the distribution is uniform.)

A simpler question to ask would be for a square hole and bucket, provided the sides of the bucket are always parallel to the sides of the hole.


Today was my math exam on school. This ubercool problem was asked to us, I think it is quite beautiful, though it's not hard.

Given a triangle $ABC$ and a line that intersects the sides of the triangle in $D,E$ and $F$.
Construct $G$, the harmonic conjugate of $D$ with respect to $A$ and $B$. Similarly, construct points $H$ and $I$, who are harmonic conjugates of $E$ and $F$, with respect to $A,C$ and $B,C$ respectively.

Prove that $CG, BH$ and $AI$ are concurrent.

Wow, that's strange  :o  How come he asked that on an exam? I can't really relate it to a subject we saw this year. Did you make some exercises about harmonic conjugates in class? Because, to be honest, I don't even know what that is, a harmonic conjugate  :P

[hide]Definition of harmonic conjugate for the points $D, E, F$ and then Menelaos' on the transversal $DEF$.[/hide]

Kurt, I guess we just have a superteacher :D We studied projective, affine and Euclidean geometry, so that explains this question.

For harmonic conjugates: Here, we can just say that two points, say $C$ and $D$ are harmonic conjugates with respect to $A$ and $B$ if they satisfy: $\left(ABC\right) =-\left(ABD\right)$.
It has a strong connection with polars of points with respect to two intersecting lines...

Andreas, that's the most straightforward way, and this solution makes it to an easy straightforward problem. But I found a much cooler solution this morning, and I used no calculation,[hide="I just used..."]polars and Desargues[/hide]

I think the solution me and probably some others gave is much more beautiful.

Nobody interested in the problem? Andreas posted a solution, but there's another very cool solution (At least I think it is very cool  :) )

I'll post it soon if nobody finds it. It involves polars.

It's pretty obvious what you're hinting at. We first prove that, for example, $GH \cap BC = \left\{ F \right\}$ (one way or the other). Then, by Desargues Theorem applied to the triangles $ABC$ and $GHI$, we are done.

Of course you are right, that's indeed what I meant.  :)

To show that $F$ is the intersection point you were talking about, you can use polars.

Maybe it's because it's the first problem I've ever seen to be solved with Desargues, or maybe it's the polars, but I thought it was a nice method.
I think I'm just still very impressed by the power and elegance of projective geometry.  :lol:

In fact the problem is trivial with menalaus and ceva theorems.

Out of personal curiosity, is this course you speak of an university, high school, or Olympiad course?  It must be an interesting course.

It is a High School course, this came right out of the first chapter on projective geometry.
Projective geometry is definitely one of the most beautiful things I ever learned about, and it's applications in e.g Olympiad geometry are endless.

Wow, a high school course!!!! That's amazing, I wish I had those oppurtunities...

Well, you can find plenty of courses on Projective geometry on the net, so it's really not that exclusive...
The course also has a very analytical approach, but sometimes, the synthetic methods are also discussed and they are very beautiful and powerful.

If you are really kean on learning about projective geometry, H.S.M Coxeter wrote some nice books on this subject, and you can find plenty of free courses also. (Cfr. Kiran Kedlaya's Geometry Unbound)


i have an exam in 3 hours
i would want to ask because i see in my textbook
1)the sequence of terms of a convergent series converges to 0
2)if proposition 1) is not true than the series is divergent

it really contradicts what i've been reading for the last 6-7 hours or
so..im really disoriented
please help me

This is a one-sided implication. Are you trying to read a nonexistent converse into it?

All convergent series have terms which tend to zero.
Some series with terms that tend to zero do not converge.


floor function
induction
inequalities
algorithm
algebra proposed
algebra

Suppose the sequence of nonnegative integers $a_1, a_2, \ldots, a_{1997}$ satisfies
\[ a_i + a_j \leq a_{i+j} \leq a_i + a_j + 1  \]
for all $i,j \geq 1$ with $i + j \leq 1997$.  Show that there exists a real number $x$ such that $a_n = \lfloor nx \rfloor$ (the greatest integer $\leq nx$) for all $1 \leq n \leq 1997$.

copied and pasted...

Note that in order for $a_n = [nx]$, we must have $x \in [\frac{a_n}{n}, \frac{a_n+1}{n}]$. Thus, there will exist such an $x$ for all $1997$ of the $a_n$ iff none of these intervals are disjoint, i.e. $\forall m, n, \frac{a_n+1}{n} \ge \frac{a_m}{m}$. We will prove this by induction on $m+n$. The base case will be $m+n = 2$, i.e. $m=n=1$, for which the inequality is obvious. Now, we do the inductive step. First, if $m=n$, the inequality is obvious, so forget that case. Now, if $n>m$, use the division algorithm to find $n = mq+r$. Note that $r$ is STRICTLY less than $n$, so $m+r < n+m$ and we can apply the inductive hypothesis: $\frac{a_r+1}{r} \ge \frac{a_m}{m}$ $\Rightarrow a_{mq+r} + 1 \ge qa_m + a_r + 1$ $= qa_m + r \cdot \frac{a_r+1}{r} \ge$ $qa_m + r \frac{a_m}{m} = n \frac{a_m}{m} \Rightarrow \frac{a_n+1}{n}$ $\ge \frac{a_m}{m}$, as desired. If $n<m$, use the division algorithm to find $m = nq+r$, and as $r$ is strictly less than $m$, we have $r + n < m + n$, and we can again use the inductive hypothesis: $\frac{a_r}{r} \le \frac{a_n+1}{n} \Rightarrow na_m = na_{qn+r} \le n(qa_n + a_r + q)$ $\le nqa_n + nq + nr \frac{a_r}{r} \le$ $nqa_n + nq + nr \frac{a_n+1}{n}$ $= nqa_n + nq + ra_n + r = m(a_n+1)$ $\Rightarrow \frac{a_m}{m} \le \frac{a_n+1}{n}$, as desired.

really not sure this is right... is it?

[hide="Solution"]If $a_n=\lfloor nx \rfloor$, $a_n \leq nx < a_n+1 \rightarrow \frac{a_n}{n} \leq x < \frac{a_n+1}{n}$.
If there is truly some $x$ that satisfies $a_n=\lfloor nx \rfloor$ for all $n$, then $[\frac{a_1}{1}, \frac{a_1+1}{1}) \cap [\frac{a_2}{2}, \frac{a_2+1}{2}) \cap \cdots \cap [\frac{a_1997}{1997}, \frac{a_1997+1}{1997}) \neq \{\}$.

Thus, for all $m$ and $n$, $\frac{a_m}{m} \leq \frac{a_n+1}{n}$.

We use induction to prove that for all $n$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq n$.

First the base case, $n=1$. It is obvious that $\frac{a_1}{1} \leq \frac{a_1+1}{1}$, so $\frac{a_1}{1} \leq \frac {a_j+1}{j}$ for all $j \leq 1$

Assume that for all $n \leq k$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq k$.

We must show that for all $n \leq k+1$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq k+1$.

$\frac{a_{k+1}}{k+1} \leq \frac{a_b+a_{k+1-b}+1}{k+1}$ by the given condition.
$\frac{a_b+a_{k+1-b}+1}{k+1} \leq \frac{\frac{a_j+1}{j}+\frac{a_j+1}{j}+1}{k+1}=\frac{2a_j+2+j}{j(k+1)}$ for all $j \leq k$ by the induction hypothesis.

$\frac{2a_j+2+j}{j(k+1)} \leq \frac{a_j+1}{j}$ iff $2a_j+2+j < (k+1)(a_j+1) \rightarrow (k-1)a_j > j-k+1$.
This is true if $j > k-1$. If $j=k$, then $(k-1)a_k > 1$, which is obvious.

Also, $\frac{a_{k+1}}{k+1} \leq \frac{a_{k+1}+1}{k+1}$, so the induction hypothesis implies that for all $n \leq k+1$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq k+1$.

Lastly, to complete the induction step we need to show that for all $n \leq k$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq k+1$.
By the induction step, for all $n \leq k$, $\frac{a_n}{n} \leq \frac {a_j+1}{j}$ for all $j \leq k$. Thus, $
Thus for $1 \leq n \leq 1997$, there exists a real number $x$ such that $a_n = \lfloor nx \rfloor$.
$Q.E.D.$[/hide]

[quote=calc rulz][hide="A different approach"]

As above, we wish to show $ \frac{a_{m}}{m}<\frac{a_{n}\plus{}1}{n}\forall m,n\in\{1,2,...,1997\}$

We first prove that we can add as many terms as we like to the sequence so that it still satisfies the above property. We show that given a sequence $ a_{1},a_{2},...,a_{k}$ such that $ a_{i}\plus{}a_{j}\le a_{i\plus{}j}\le a_{i}\plus{}a_{j}\plus{}1$ for all $ i,j\ge 1$ with $ i\plus{}j\le k$, we can add a term $ a_{k\plus{}1}$ at the end so that it still satisfies the same property.

Any such $ a_{k\plus{}1}$ must satisfy $ a_{i}\plus{}a_{k\plus{}1\minus{}i}\le a_{k\plus{}1}\le a_{i}\plus{}a_{k\plus{}1\minus{}i}\plus{}1$ for all $ i$ with $ \frac{k\plus{}1}{2}\le i\le k$, so the intervals $ [a_{i}\plus{}a_{k\plus{}1\minus{}i},a_{i}\plus{}a_{k\plus{}1\minus{}i}\plus{}1]$ cannot be disjoint, which is equivalent to showing that $ a_{i}\plus{}a_{k\plus{}1\minus{}i}\le a_{j}\plus{}a_{k\plus{}1\minus{}j}\plus{}1$ and $ a_{j}\plus{}a_{k\plus{}1\minus{}j}\le a_{i}\plus{}a_{k\plus{}1\minus{}i}\plus{}1$ where $ 1\le k\plus{}1\minus{}i < k\plus{}1\minus{}j\le j < i\le k$.

For the first inequality we note that $ a_{k\plus{}1\minus{}i}\plus{}a_{i\minus{}j}\le a_{k\plus{}1\minus{}j}$ and $ a_{i}\le a_{j}\plus{}a_{i\minus{}j}\plus{}1$ and add these. For the second we note that $ a_{j}\plus{}a_{i\minus{}j}\le a_{i}$ and $ a_{k\plus{}1\minus{}j}\le a_{k\plus{}1\minus{}i}\plus{}a_{i\minus{}j}\plus{}1$ and add these.

It thus follows that we can pick $ a_{k\plus{}1}$ satisfying all the necessary inequalities. We can now add terms past the $ 1997th$ term and still have a sequence that has the desired conditions, meaning that $ a_{i}\plus{}a_{j}\le a_{i\plus{}j}\le a_{i}\plus{}a_{j}\plus{}1$ is true even when $ i\plus{}j > 1997$.

Now let us consider $ a_{m}$ and $ a_{n}$. We wish to show that $ \frac{a_{m}}{m}<\frac{a_{n}\plus{}1}{n}$ or equivalently $ n a_{m}< m a_{n}\plus{}m$.

Next we prove that for a positive integer $ r$, $ r a_{m}\le a_{rm}\le r a_{m}\plus{}r\minus{}1$ using induction. The base case is $ r \equal{} 1$ which is clearly true. Assume it's true for $ r$. Then we have that $ (r\plus{}1) a_{m}\le a_{m}\plus{}r a_{m}\le a_{m}\plus{}a_{rm}\le a_{rm\plus{}m}\le a_{m}\plus{}a_{rm}\plus{}1\le a_{m}\plus{}r a_{m}\plus{}r\minus{}1\plus{}1\le (r\plus{}1) a_{m}\plus{}r$ so it holds true in general.

Now we have that $ n a_{m}\le a_{mn}\le m a_{n}\plus{}m\minus{}1 < m a_{n}\plus{}m$ and we are done.[/hide][/quote]

The problem I see in this solution is that you proved these intervals are pairwise non disjoint, but you haven't proved that all of these intervals have a common element. Though Helly's theorem makes this insignificant.

Actually we don't need $a_i\ge 0$. Assumption they are integers is sufficient.

Solution from [i]Twitch Solves ISL[/i]:

We are trying to show there exists an $x \in {\mathbb R}$ such that \[ \frac{a_n}{n} \le x < \frac{a_n+1}{n} \qquad \forall n. \] This means we need to show \[ \max_i \frac{a_i}{i} < \min_j \frac{a_j+1}{j}. \] Replace 1997 by $N$. We will prove this by induction, but we will need some extra hypotheses on the indices $i,j$ which are used above.

[b][color=red]Claim:[/color][/b]  	Suppose that 	[list] 		[*]Integers $a_1$, $a_2$, \dots, $a_N$ satisfy the given conditions. 		[*]Let $i = \operatorname{argmax}_n \frac{a_n}{n}$; 		if there are ties, pick the smallest $i$. 		[*]Let $j = \operatorname{argmin}_n \frac{a_n+1}{n}$; 		if there are ties, pick the smallest $j$. 	[/list] 	Then \[ \frac{a_i}{i} < \frac{a_j+1}{j}. \] 	Moreover, these two fractions are in lowest terms, 	and are adjacent in the Farey sequence of order $\max(i,j)$.
[i]Proof.[/i]  	By induction on $N \ge 1$ with the base case clear. 	So suppose we have the induction hypothesis 	with numbers $a_1$, \dots, $a_{N-1}$, 	with $i$ and $j$ as promised.
Now, consider the new number $a_N$. 	We have two cases: 	[list] 		[*]Suppose $i+j > N$. 		Then, no fraction with denominator $N$ 		can lie strictly inside the interval; 		so we may write for some integer $b$ 		\[ \frac bN \le \frac{a_i}{i} 			< \frac{a_j+1}{j} \le \frac{b+1}{N}. \] 		For purely algebraic reasons we have 		\[ \frac{b-a_i}{N-i} \le \frac bN \le \frac{a_i}{i} 			< \frac{a_j+1}{j} \le \frac{b+1}{N} 			\le \frac{b-a_j}{N-j}. \] 		Now, 		\begin{align*} 			a_N &\ge a_i + a_{N-i} 				\ge a_i + (N-i) \cdot \frac{a_i}{i} \\ 			&\ge a_i + (b-a_i) = b \\ 			a_N &\le a_j + a_{N-j} + 1  				\le (a_j+1) + (N-j) \cdot \frac{a_j+1}{j} \\ 			&= (a_j+1) + (b-a_j) = b+1. 		\end{align*} 		Thus $a_N \in \{b,b+1\}$. 		This proves that $\frac{a_N}{N} \le \frac{a_i}{i}$ 		while $\frac{a_N+1}{N} \ge \frac{a_j+1}{j}$. 		Moreover, the pair $(i,j)$ does not change, 		so all inductive hypotheses carry over.
[*]On the other hand, suppose $i+j = N$. 		Then we have 		\[ \frac{a_i}{i} < \frac{a_i + a_j + 1}{N} < \frac{a_j+1}{j}.  \] 		Now, we know $a_N$ could be either $a_i + a_j$ or $a_i + a_j + 1$. 		If it's the former, then $(i,j)$ becomes $(i,N)$. 		If it's the latter, then $(i,j)$ becomes $(N,j)$. 		The properties of Farey sequences ensure that 		the $\frac{a_i + a_j + 1}{N}$ is reduced, either way. 	[/list] $\blacksquare$

[i]A different Inductive solution: [/i]

We will solve the more generally when $1997$ is reaplaced by any natural number $\alpha$, by using strong induction on $\alpha$. Base case $\alpha = 1$ is trivially true. Assume the assertion holds for all $\alpha < \beta$ (for some $\beta \ge 2$) and we consider the case when $\alpha = \beta$. So we have numbers $a_1,a_2,\ldots,a_{\beta}$.

[b][color=#f00]Claim: [/color][/b] We may WLOG assume $a_1 = 0$.

[i]proof:[/i] Suppose $a_1 \ge 1$. Then by induction $a_i \ge i$ for all $i$. Then we can change each $a_i$ to $a_i - i$ and $x$ to $x-1$ and note that the condition originally were satisfied if and only if they are now satisfied. So whenever $a_1 \ge 1$, we do this until we get $a_1 = 0$. This proves our claim. $\square$

So now we have $a_1 = 0$. Then for any $i$,
$a_i \le a_{i+1} \le a_i + 1$. So the sequence $a_1,a_2,\ldots,a_{\beta}$ will look something like
$$0,0,\ldots,0 ; 1,1,\ldots,1 ; 2,2,\ldots,2 ; \ldots$$
Assume the number $i$ appears $b_i$ times in $a_1,a_2,\ldots,a_{\beta}$ and let $k$ be the largest such that $b_k$ in non-zero. Note that all of $b_0,b_1,\ldots,b_k$ must be non-zero and $b_0 + b_1 + \cdots b_k = \beta$. Define
$$S_m = \sum_{i=0}^{m-1} b_i  ~ \forall ~ 1 \le m \le k+1 $$

[b][color=#f00]Claim: [/color][/b] for all $i,j \ge 1$ with $i+j \le k+1$ we have
$$S_i + S_j \le S_{i+j} \le S_i + S_j + 1$$

[i]proof: [/i] We have
\begin{align*}
& a_{S_{i+j} + 1} = i +j = a_{S_{i} + 1} + a_{S_{j} + 1} \le a_{S_i + S_j + 2} ~ \implies S_{i +j} + 1 \le S_i + S_j + 2 \\
& a_{S_{i+j}} =i+j-1 = (i-1) + (j-1) + 1 = a_{S_i} + a_{S_j} + 1 \ge a_{S_i + S_j} ~ \implies S_{i+j} \ge S_i + S_j
\end{align*}
This completes the proof of our claim.  $\square$

Now we want to prove that the following system of inequalities has a solution:
$$x \cdot S_i < i \le x \cdot(S_i+1) ~ \forall ~ i = 1,2,\ldots,k+1 \qquad (1)$$
If $k+1 = \beta$, then we must have $b_0 = b_1 = \cdots = b_k = 1$ which would mean $a_i = i-1 ~ \forall ~ i$, and that case is easy. So now assume $k+1 < \beta$. Consider the numbers
$$S_1,S_2,\ldots,S_{k+1}$$
which satisfy
$$S_i + S_j \le S_{i+j} \le S_i + S_j + 1 ~ \forall ~ i,j \ge 1 \text{ with } i+j \le k+1$$
By our induction hypothesis, there exist a real number $y$ satisfying
\begin{align*}
S_i \le y \cdot i < S_i + 1 ~ \forall ~ i =1,2,\ldots,k+1 \\
\implies \frac{1}{y} \cdot S_i \le i < \frac{1}{y} \cdot (S_i + 1) ~ \forall ~ i=1,2,\ldots,k+1
\end{align*}
So if we take $x = \frac{1}{y} + \epsilon$ for sufficiently small $\epsilon > 0$, then $(1)$ will be satisfied. \\
[b]This completes the proof of the problem. [/b]

Replace $1997$ with $N$ and induct on $N$. At $N=1$, the result is trivial.

Now, remark that if $a_i+a_j$ with $i+j=N$ is always one of $\lfloor Nx\rfloor$ and $\lfloor Nx\rfloor - 1$, meaning we are done unless $a_i+a_j$ is fixed because then $a_{i+j}$ would have to be $\lfloor Nx\rfloor$.

Now first suppose that $a_i+a_j = \lfloor ix\rfloor + \lfloor jx\rfloor = (i+j)x - [ix] - [jx]$ is always equal to $\lfloor Nx\rfloor = Nx - [Nx]$ where $[\bullet]$ is the part of $\bullet$ after the decimal point for nonnegative $\bullet$. Hence $[Nx] = [ix]+[jx]$. In particular, this means that $[ix]\le [Nx]$ for all $i$. We would like to show that $x+\frac{1-[ix]}{i}=\frac{a_i+1}{i} > x$ is larger than $\frac{\lfloor Nx\rfloor + 1}{N} = x + \frac{1-[Nx]}{N}$. Indeed, this is clear because $[ix]\le [Nx]$ and $1>[Nx]$.

Now suppose that $a_i+a_j=(i+j)x-[ix]-[jx]$ were always equal to $\lfloor Nx\rfloor - 1 = Nx-[Nx]-1$ so $[ix]+[jx] = [Nx]+1$. It is clear that all $[ix]$ are greater than $[Nx]$. We analogously have $\frac{\lfloor Nx\rfloor}{x} = x - \frac{[Nx]}{N} > x-\frac{[ix]}{i} = \frac{a_i}{i}$, so it is possible to shift the choice of $x$ downward to accommodate the conditions.

It's not hard to prove inductively that \[ia_1\le a_i\le ia_1+i-1\]
So, if we denote $b_i=a_i-ia_1$, we have $0\le b_1\le i-1$ and
\[b_i+b_j\le b_{i+j}\le b_i+b_j+1, \forall i+j\le 1997\]
However, \[a_n=\lfloor nx\rfloor\Leftrightarrow a_n\le nx< a_n+1\Leftrightarrow nx-1< a_n\le nx\Leftrightarrow x-a_1-\frac{1}{n}< \frac{b_n}{n}\le x-a_1\]
So it is enough to prove the existence of some $y$ satisfying \[\frac{b_n}{n}< y\le \frac{b_n+1}{n}\]
If we manage to prove that $\frac{b_m}{m}\le \frac{b_n+1}{n}, \forall m,n$, then the intervals $\left(\frac{b_i}{i}, \frac{b_i+1}{i}\right]$ are not disjoint and we can take $y$ from their intersection.
\\ Proceed by induction on $m+n$. The base case is trivial, since $b_1=0$. Suppose the statement true for any $(i,j)$ with $i+j<m+n$.
\\ If $m<n$, then write $n$ as $m\cdot c+r$, with $r<m$. We know, by the induction hypothesis that \[\frac{b_m}{m}\le \frac{b_r+1}{r}\]
On the other hand, we also know that $b_r+cb_m\le b_n$. So we have \[b_n+1\ge b_r+cb_m+1=cb_m+r\cdot \frac{b_r+1}{r}\ge cb_m+r\cdot \frac{b_m}{m}=n\cdot \frac{b_m}{m}\Rightarrow \frac{b_n+1}{n}\ge \frac{b_m}{m}\]
If $m>n$, then write $m$ as $n\cdot c+r$, with $r<n$. We know that \[\frac{b_r}{r}\le \frac{b_n+1}{n}\]
and also that $b_m\le b_r+cb_m+c$. So we have \[b_m\ge r\cdot \frac{b_n+1}{n}+c(b_n+1)=\frac{b_n+1}{n}\cdot m\Rightarrow \frac{b_n+1}{n}\ge \frac{b_m}{m}\]
If $m=n$, the inequality is obvious. So the induction is done and the problem is now finished.


Solve:

$ \frac{3}{\sqrt{2}}\sec \theta\minus{}\sqrt{2}\csc \theta\equal{}1$

[hide]
I know that it can be easily seen that $ \theta\equal{}45^{\circ}$ is a solution but how to find by solving the equation
[/hide]

hello, why did you posted the same problem twice?
Sonnhard.

Very sorry there was some server problem. :omighty:


ratio
absolute value
real analysis
real analysis unsolved

When does the series $\sum \frac{1}{1+z^{n}}$ (for $z \in \mathbb{C}$) converge and when does it diverge? How do I prove it?

Two useful tests:
1) If the series converges, its general term tends to $0$.
2) If the ratio $\frac{a_{k+1}}{a_{k}}$ tends to some limit that is less than $1$ in absolute value, then $\sum_{k\ge 1}a_{k}$ converges absolutely.
These two are enough to answer your question.


[hide="hint"] If Andrew doesn't know what Tom has, there must be two (positive) numbers that are consecutive to Andrew's.[/hide]

Hint:

[hide]For Andrew to say ' I don't know....' for the first time => his number is not 1
For Tom to say 'I don't knw ...' for the first time ==>  his number is not 1  or 2

etc...[/hide]

I don't know if this necessarily is a mathcounts problem (it's more like a trick or logic question) but here's my answer.
[hide="Solution"]Assuming that I didn't switch anything around, the first statement means that Andrew doesn't have a 1, Tom doesn't have a 2, and so forth. On his eleventh statement, Andrew proves that he has a 21, and that shows that Tom has a 22 (as he showed he didn't have a 20 in [i]his[/i] 10th statement). Therefore the answer is 22.[/hide]
Don't look at this until you have tried the problem... and don't look at the giveaway hints either... just THINK..

[quote="236factorial"]I seriously hate this problem, but I seriously need help on it too.
The problem goes:

Pete chose two consecutive positive integers, and wrote them down on two separate sheets of paper. He gave one of the numbers to Andrew and gave the other number to Tom, who both knew that the numbers they were given are consecutive positive integers. Then the following highly intellectual conversation took place between Andrew and Tom:

Andrew: " I do not know what number you got."

Tom: "I don't know what number you got either."

Andrew: " I do not know what number you got."

Tom: "I don't know what number you got either." ....

Andrew and Tom both said the same sentence 10 times. Then Andrew's 11th sentence sounded like:
"Now I know what number you got."

What number did Tom get?[/quote]

[hide]Andrew cant have a one so Tom cant have a two
Andrew cant have a 3 so Tom cant have a 4 and so on for 10 times

Andrew can't have a 21 so Tom can't have 22 so the answer is 22?[/hide]

that's one real tough problem, let me think about it...

Ah, I give up on that problem. It's really confusing.

I don't really know what the answer really is, because the answer key was not avaliable. This problem was intended for 5th and 6th graders.

Maybe 22 is correct...


$ a,b,c$are positive numbers,prove that
\[ (\frac{a^2\minus{}ab\plus{}b^2}{a\plus{}b})^7\plus{}(\frac{b^2\minus{}bc\plus{}c^2}{b\plus{}c})^7\plus{}(\frac{c^2\minus{}ca\plus{}a^2}{c\plus{}a})^7\geq\frac{a^7\plus{}b^7\plus{}c^7}{2^7}\]

And can we generlize like this:for $ k\geq1$
\[ (\frac{a^2\minus{}ab\plus{}b^2}{a\plus{}b})^k\plus{}(\frac{b^2\minus{}bc\plus{}c^2}{b\plus{}c})^k\plus{}(\frac{c^2\minus{}ca\plus{}a^2}{c\plus{}a})^k\geq\frac{a^k\plus{}b^k\plus{}c^k}{2^k}\]

[quote="manjil"][quote="Jianxing113725"]$ a,b,c$are positive numbers,prove that
\[ (\frac {a^2 \minus{} ab \plus{} b^2}{a \plus{} b})^7 \plus{} (\frac {b^2 \minus{} bc \plus{} c^2}{b \plus{} c})^7 \plus{} (\frac {c^2 \minus{} ca \plus{} a^2}{c \plus{} a})^7\geq\frac {a^7 \plus{} b^7 \plus{} c^7}{2^7}
\]
And can we generlize like this:for $ k\geq1$
\[ (\frac {a^2 \minus{} ab \plus{} b^2}{a \plus{} b})^k \plus{} (\frac {b^2 \minus{} bc \plus{} c^2}{b \plus{} c})^k \plus{} (\frac {c^2 \minus{} ca \plus{} a^2}{c \plus{} a})^k\geq\frac {a^k \plus{} b^k \plus{} c^k}{2^k}
\]
[/quote]

I think the generalization is not possible![/quote]

Can you give an example?And how about the first inequlity...that'a true


For $ a,b,c > 0$ such that $ abc \equal{} 1$

$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq (a \plus{} b \plus{} c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$

$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq(1/a \plus{} 1/b \plus{} 1/c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$

P.S. I cannot prove them. :blush:

[quote="manlio"]For $ a,b,c > 0$ such that $ abc \equal{} 1$

$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq (a \plus{} b \plus{} c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$    (1)

$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq(1/a \plus{} 1/b \plus{} 1/c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$  (2)

P.S. I cannot prove them. :blush:[/quote]

We have $ (1) \Leftrightarrow 3(a^2 \minus{} a \plus{} 1)(b^2 \minus{} b \plus{} 1)(c^2 \minus{} c \plus{} 1) \ge a \plus{} b \plus{} c$

              $ (2)\Leftrightarrow 3(a^2 \minus{} a \plus{} 1)(b^2 \minus{} b \plus{} 1)(c^2 \minus{} c \plus{} 1) \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$

and $ (1)\Leftrightarrow (2)$ because if put $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}$  $ (xyz \equal{} 1)$
then  $ (2)\Leftrightarrow 3(x^2 \minus{} x \plus{} 1)(y^2 \minus{} y \plus{} 1)(z^2 \minus{} z \plus{} 1) \ge x \plus{} y \plus{} z$

Now.we prove (1)
              $ 3(a^2 \minus{} a \plus{} 1)(b^2 \minus{} b \plus{} 1)(c^2 \minus{} c \plus{} 1) \ge a \plus{} b \plus{} c$
$ \Leftrightarrow 3 \plus{} 3\sum a^2b^2 \plus{} 3\sum a^2 \ge 3\sum ab(a \plus{} b) \plus{} \sum a$  (Note $ abc \equal{} 1$)

Put  $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca$

We need to prove
$ 12 \plus{} 3(q^2 \minus{} 2p) \plus{} 3(p^2 \minus{} 2q) \ge 3pq \plus{} p \Leftrightarrow 3p^2 \plus{} 3q^2 \plus{} 12 \ge 7p \plus{} 6q \plus{} 3pq$
:)

Applying [b]AM-GM Inequality[/b] (note that $ p^2 \ge 9 ,q^2 \ge 9$),we obtain

$ 3p^2 \plus{} 3q^2 \plus{} 12 \equal{} (\frac {1}{3}p^2 \plus{} \frac {8}{3}p^2) \plus{} (\frac {1}{2}q^2 \plus{} \frac {5}{2}q^2 )\plus{} 12 \ge \frac {8}{3}p^2 \plus{} \frac{5}{2}q^2 \plus{} \frac {39}{2} \equal{} 7(\frac {p^2 \plus{} 9}{6}) \plus{} (q^2 \plus{} 9) \plus{} \frac {3}{2}(p^2 \plus{} q^2) \ge 7p \plus{} 6q \plus{} 3pq$ [b](Q.E.D)[/b]

Equality holds if only if $ a \equal{} b \equal{} c \equal{} 1$  OK  :wink:


Linguistic background aside, which language do you think is inherently the most difficult? i.e. duh for English speakers Indo-European languages are easier than Sino-Tibetan languages but for say, an infant with no experience whatsoever...

And how does English rank in difficulty among other languages?

So, you're wondering which language is the most difficult for infants to acquire?

There are so many different aspects of language acquisition that ranking languages by difficulty of acquisition would be highly subjective (and probably meaningless).  But, which aspects of language infants learn earlier or later are universal across languages.  For example, infants learn to produce certain sounds before others. Vowels are easy, and so are so-called "stop consonants" like p, b, d, t, g, k.  Fricatives like s, f, and th are harder, and so are approximants like y and l.  Some of the hardest consonants to produce are tongue clicks (like are found in some African languages) and the English "r" sound, and children often do not master these until they are grade-school age.  Surprisingly, tones are extremely easy for infants to pick up, and infants can learn to mimic tones accurately in the first stages of language acquisition.

If you're interested in this subject, you might want to read a book on child language acquisition, like Charles Yang's book [i]The Infinite Gift.[/i]

Perhaps you can't "sound" out the word, but you can often work out what it means, since the characters have radicals which relate to the meaning.

But, AznTiger and vallon22, you're confusing [b]literacy[/b] with [b]fluency[/b].  You can speak Chinese as a native speaker without being able to read and write it, and fluency as a native speaker is a better measure of knowledge of a language (for linguistic purposes) than literacy.

That said, it is true that it takes several years longer for children to become fluent in reading Chinese or Japanese than children learning alphabetic writing systems.

[quote="Healker"]Definitely, mandarin chinese, it's very hard!!!!

That's why I want to learn (someday) japanese, it's much easier than mandarin.  :lol:

(Besides, the symbols are nicer! No ofense, of course.)[/quote]

Well, some Japanese writing is easier, but quite often [i]kanji[/i] (Chinese characters) are scattered throughout sentences, so you need to know how to read those as well.

[quote="Healker"]Definitely, mandarin chinese, it's very hard!!!!

That's why I want to learn (someday) japanese, it's much easier than mandarin.  :lol:

(Besides, the symbols are nicer! No ofense, of course.)[/quote]

I'd have to disagree with that.

In Chinese, to compensate for tonal difficulties and a difficult writing system, the grammar is almost nothing compared to other languages. For example, there is nothing of following in Chinese:
-verb conjugation (the difference between be, am, is, are)
-noun declension (gender, number (rules about plurality...), etc.)
-irregular verbs (there not being any conjugation, how are verbs to be irregular?)
-other inflection

...and why Mandarin? I feel that people say "Mandarin" because that might be the most technical sounding way they know to say "Chinese"... Mandarin is just the most common dialect of Chinese... it's like saying, "Hmm... the most widely used language is British English." (And Mandarin is arguably not the hardest dialect of Chinese - Cantonese has 2 to 4 more tones than Mandarin (depends on whose counting).)

And by the way, Japanese does actually have some sort of grammar. To give you an idea, here are the Wikipedias for Japanese grammar [url]http://en.wikipedia.org/wiki/Japanese_grammar[/url] and Chinese grammar [url]http://en.wikipedia.org/wiki/Chinese_grammar[/url]. Now just compare the size of those scroll bars!

(And in defense of Japanese, it is way easier to pronounce than Chinese - in fact, it's even easier than Spanish.)

And as for the writing system - it's fine if you think it's nicer, but being able to recognize Chinese characters, I'd say that at least - at [i]least[/i] - a third of the symbols in Japanese writing are kanji, that is, borrowed Chinese characters. In fact, I can make out the general topic of each section just by reading the kanji. For a sample of Japanese writing, see the Japanese Wikipedia about Japanese language at [url]http://ja.wikipedia.org/wiki/%E6%97%A5%E6%9C%AC%E8%AA%9E#.E6.96.87.E6.B3.95[/url].

[quote="vallon"]
I would have to say chinese
there are thousands of characters that have to be memorized
there is no way to "sound" out the word, like in english or some other languages
you have to know every single character in order to know what a sentence means (ok, maybe if you didn't know one, you could still figure it out, but...)
also, you can't sound a word out when you "spell" it[/quote]

In Chinese, true, you might come up against a word you didn't know, and not be able to pronounce it, but maybe be able to deduce the meaning of the word (and actually quite often the pronunciation as well - 90-95% of Chinese characters are 1/2 meaning and 1/2 pronunciation clues in them). But think - what would happen if you came across an English word that you didn't know? You'd be able to pronounce it, or maybe there were 2-3 possible ways that came in to your head on how to pronounce it, and you maybe could deduce the meaning from the context and also from the word itself through affixes and recognized roots.

Was there that big a difference?

Many will compare the thousands of characters to the 26 letters of the English alphabet. But is this what should be compared? In terms of functionality, a character says way more than a letter. Instead, a closer counterpart for the character would be, say, a small word or affix. (Longer words in English might correspond to multiple characters in Chinese, for example.) And the number of characters that are needed to be memorized (which the word "thousands" is a bit misleading - it gives the impression that one would need to memorize around 10,000 characters), which is at most 3500 I'd say (most places will say that to read a newspaper you need to memorize 2500 or so), is comparable to the number of short words and affixes.

---------

Now I'm not saying that Chinese is a super-easy language to learn, I'm just pointing out some things to which I believe the majority are uninformed.

The above point about grammar is definetely true.  I learned Chinese as a first language, but living in America, it has quickly deterioated.  I learned English very young, so the grammar mostly comes intuitively, but the more I learn, the more frustrating I get.  As a foreign language, taking French is so annoying, verbs and adjectives have like a million conjugations, and everything varies on masculine feminine.  Gramatically, Chinese is much much easier.

Chinese, definitely.

I'm taking spanish at school, with no prior knowledge of spanish except for numbers from 1-10, amigo, hola, and adios, and it's a breeze.

I'm taking chinese classes, with some prior knowledge of speaking.  It's totally hard.

Chinese is hard...

I would say that the hardest language for a child to learn fluency-wise (not literacy-wise) would be a language with a lot of inflection.  In English, grammar-wise as long as you know subject-verb-object, you can make yourself understood.  In a language like Latin, an immense amount of knowledge is required to form even the simplest sentences.

The Economist has an article on this subject, and Chinese is actually relatively easy compared to some languages.

For sound complexity, one language stands out. !Xóõ, spoken by just a few thousand, mostly in Botswana, has a blistering array of unusual sounds. Its vowels include plain, pharyngealised, strident and breathy, and they carry four tones. It has five basic clicks and 17 accompanying ones. The leading expert on the !Xóõ, Tony Traill, developed a lump on his larynx from learning to make their sounds. Further research showed that adult !Xóõ-speakers had the same lump (children had not developed it yet).

...

With all that in mind, which is the hardest language? On balance The Economist would go for Tuyuca, of the eastern Amazon. It has a sound system with simple consonants and a few nasal vowels, so is not as hard to speak as Ubykh or !Xóõ. Like Turkish, it is heavily agglutinating, so that one word, hóabãsiriga means “I do not know how to write.” Like Kwaio, it has two words for “we”, inclusive and exclusive. The noun classes (genders) in Tuyuca’s language family (including close relatives) have been estimated at between 50 and 140. Some are rare, such as “bark that does not cling closely to a tree”, which can be extended to things such as baggy trousers, or wet plywood that has begun to peel apart.

Most fascinating is a feature that would make any journalist tremble. Tuyuca requires verb-endings on statements to show how the speaker knows something. Diga ape-wi means that “the boy played soccer (I know because I saw him)”, while diga ape-hiyi means “the boy played soccer (I assume)”. English can provide such information, but for Tuyuca that is an obligatory ending on the verb. Evidential languages force speakers to think hard about how they learned what they say they know.

The most difficult languages to learn could possibly be [url=http://www.naclo.cs.cmu.edu/]here.[/url] But this is coming from someone with no background of linguistics except for one competition, so be warned.

Are we assuming this is for english speakers?

As some people that have defended chinese as "not too hard" claim, they are native speakers. So, it makes sense that it wouldn't be too hard to them. But to others, whose native language is something like English, French, or Klingon (i wonder if that's actually true...), it proves to be difficult.

For native speakers of english, I would Japanese is one of the hardest to learn. While Chinese may have more characters (roughly 4000 to be considered "fluent", twice that of the 2000 joyo kanji), Japanese has actual grammar.

Also, the japanese (like the koreans) decided to use chinese characters, even though chinese is fundamentally different from their language. As a result, the kanji are even more difficult to master as the chinese characters, because chinese characters have 2 parts:
1) Meaning
2) Reading or Pronunciation.

But Japanese Kanji have 2+ reading, the common ones having 10.

Also, Japanese has an honorifics system where the entire order and conjugation of a sentence, as well as expressions you are allowed to use, changes depending on the person you are talking to.

yeah, i think east asian languages are some of the hardest. but so is navajo and icelandic, especially navajo because the sounds are very hard for english speakers

Simplest language in the world: 한글! Guess the language without searching it up

[quote=del-math]I've noticed that a lot of Indian languages, like Hindi and Tamil and so on, use thick sounds that English speakers cannot reiterate.[/quote]

True, now i may be biased, but i feel that for an infant, Indian languages are easy to learn. The alphabet itself is very organised in Indian languages as compared to any western language. Some of them do have loads of grammar but almost none of them have any form of 'silent letters' which you may find in western languages. That is, when you read a word, usually there is one and only one pronunciation for it. This is particularly nice.

Many of you guys are missing a point. [b]Pronunciation doesn't matter, except in certain circumstances, like !Xoo.[/b] After a little wikipedia surfing, watching videos by the youtube channel Al Muqaddimah, and a day of practice, you can pronounce the pharyngeal sounds with relative ease. So, its better to choose a language by its grammar.

[quote=sachima]I agree with Mycroft.

Ignoring the writing aspect, Chinese is not really as difficult as most people say. Of course, it is still hard because of its tones, but I don't think it would top the list of most difficult languages. Grammatically, it's quite simple, with no verb conjugations, cases, or really any inflection. But being Chinese, I might be a little biased...

Also ignoring the writing aspect, Japanese is harder than Chinese because there are actually verb tenses. In addition to that, formality is a very important part of their speech, which affects verb conjugation and pronouns and just makes things more complicated. But if you throw the writing part into play, Japanese is easier to read, but they do use some kanji, which borrowed from Chinese. Japanese is also relatively easy to pronounce.

Some may even say English is very hard to learn because of its irregularity and gigantic vocabulary. But then again, grammar in English is more flexible than other languages (German, for instance), and the languages has also lost most of its cases.

To me, it seems like Russian, Icelandic, and Finnish are very hard to learn because their grammar is complicated. But that's just from my limited knowledge of them d:[/quote]

I disagree. Japanese is not harder than chinese (i am learning japanese) - though they are both grammatically easy. You do need kanji for japanese, and kanji is chinese, so being chinese helps learning japanese. (I'm not so its harder for me). I think the hardest language is a language which has a writing system and alphabet not similiar to any other and has wild grammar.

If you address the original post, in which it says, "Which is the hardest language for a baby to learn," then all languages would be equally easy.

[quote=AznTiger]this may seem biased but i think chinese. There are over 5000 characters which must be painstakingly memorized. And even the smallest mistake can change the meaning. i.e. ni can mean dirt or you[/quote]

I mean, sometimes that's the same with Spanish but I don't disagree.

Some languages in Papua New Guinea have over 700 verb forms!

Wow, really?

I would argue that Chinese (all dialects) is the hardest, because it is both a tonal and character based language, and English has neither of those.

Probably something in the Uralic family. Those can get insanely complex. Like Nenets, where you give every word an ending based on its "importance" to the sentence, on a scale of 1 to 16.

as a native chinese + english speaker, so i'd say chinese definitely isn't the hardest, which proves that difficulty is subjective. lol


Find all functions $ f: \mathbb R^{+}\rightarrow \mathbb R$ such that
\[f(x)f(y)=f(xy)+2005\left(\frac{1}{x}+\frac{1}{y}+2004\right),\]
for all $ x,y>0$.


advanced fields
advanced fields unsolved

I haven't done relations for a while, so when I went back to my textbook to 're-learn' what they were I got slightly confused.

It says a relation on A is defined of subsets of AxA.

I then go to the definition of reflexive, which says R is reflexive on A iff for all X that is an element of A, xRx.

Ok consider A = {x,y}

Then are the following different relations reflexive? I don't see why not

R1=(x,x)
R2=(y,y)
R3=(y,y) (x,x)
R4=(y,y) (x,x) (x,y)
R5=(y,y) (x,x) (y,x)
R6=(y,y) (x,x) (y,x) (y,x)

So when it says all a such that (a,a) is in R to be reflexive, does it mean all A in the relation, or all A in the initial set? Thanks

[quote="Jrb599"]...
for all x [b]that is an element of A[/b], xRx.
...
does it mean all A in the relation, or all A in the initial set?[/quote]


geometry
incenter
HMMT
circumcircle
Euler
geometric transformation
homothety

Let $G$, $H$, and $I$ denote the centroid, orthocenter, and incenter of triangle $ABC$ respectively. If $HI = 2$, $IG = 3$, and $GH = 4$, then compute the value of $\cos(A) + \cos(B) + \cos(C)$.

Still nobody answers it! I guess Geometry will be a fun subject at HMMT next year. ;)

I'll guess [hide]$\frac{5}{4}$[/hide] for kicks.

ah, now I have a method. and it isn't too ugly. eh, I still don't really want to work through it though

ok, I bravely slogged through it, and got

[hide][hide]Use standard notation (O is the circumcenter, N the nine-point center, R the circumradius, r the inradius). We know 2 OI^2 = 2R^2 - 4Rr (*) and that cos A + cos B + cos C = 1 + r/R (^). We also know that the incircle and the nine-point circle are tangent, so IN = R/2 - r, or 4*IN^2 = R^2 - 4Rr + 4r^2 (**). Now, we know NOHG (pronounced NOHG) are collinear on the Euler line, and that the homothecy about G with factor -1/2 takes H -> O -> N, so GO = 1/2 * GH = 2, GN = 1/2 * GO = 1. Now, use Stewart's theorem to find IN, IO. We get 2^2 * 1 + 3^2 * 3 = 3 * 1 * 4 +  4 IN^2 => 4 IN^2 = 19 (!), and 3^2 * 6 + 2^2 * -2 = 4 * 6 * -2 + 4 IO^2 => 47 = 2 IO^2 (!!). Now, consider 47(!) - 19(!!). We get 47 (4 IN^2) - 19 (2 IO^2) = 0. We plug in (*) and (**) to find 47 (R^2 - 4Rr + 4r^2) - 19 (2R^2 - 4Rr) = 0 => 9R^2 - 112Rr + 188r^2 = 0 => 188(r/R)^2 - 112(r/R) + 9 = 0 => (2 * (r/R) - 1) * (94 * (r/R) - 9) = 0. The triangle's not equilateral (otherwise all these fun distances would be 0), so we ignore the first factor => r/R = 9/94, so from (^), we get an answer of [/hide]103 / 94.[/hide]

Excellent work. That is correct. I had a bit less calculation, using the fact that $-\frac{2r}{3}\left(R - 2r\right) = \vec{IG} \cdot \vec{IH}$, which conveniently appears in the Law of Cosines on $\angle{HIG}$, dividing this expression by $R(R-2r)=OI^2$, etc.

Can you post a proof for this

\[2 OI^2 = 2R^2 - 4Rr \]

It's called Euler's theorem.

See more [url=http://mathworld.wolfram.com/EulerTriangleFormula.html]here.[/url]

Here is a quick proof:

Let $O$ and $I$ be the circumcenter and incenter, respectively, of triangle $ABC$, and extend $AI$ past $I$ to $A'$ on the circumcircle of $\triangle{ABC}$. Let $P$ denote the projection of $I$ onto $\overline{AC}$. Let $2\alpha, 2\beta$, and $2\gamma$ denote the measures of $\angle{A}, \angle{B}$, and $\angle{C}$ respectively. We have $m\angle{AA'C} = 2\beta$ and $m\angle{A'CI} = \alpha + \gamma$ so that $m\angle{CIA'} = \alpha+\gamma$; hence, $A'I = A'C$. The power of $I$ with respect to the circumcircle of $ABC$ is $R^2 - OI^2 = AI \cdot IA' = AI \cdot A'C$. But as $IP = r$, simple trigonometry gives $AI = \frac{r}{\sin(\alpha)}$ and $A'C = 2R\sin(\alpha)$ so that $R^2 - OI^2 = 2Rr$. The result now follows.


How many binary connectives are there in 3-valued logic? How is this determined?

Let's count:
Binary connectives which acts on two statements $A$ and $B$
There are three values for each $A$ and $B$, so there are $3^3=27$ possible outcomes of truth values
For each of the $27$ logical outcomes, a binary connective would assign one of the $3$ truth values, so there are $3^{3^3}=27^3=19683$ different binary connectives we have to consider.

I really doubt the usefulness of multi-value logic, for what could be proven in multi-valued logic could also be proved by 2-valued logic :D

Thank you for your answer.

One other question, how many binary and unary connectives are there in k-valued logic? How is this determined?


Try using the Chinese Remainder theorem.

It's enough to calculate $2005^{2005}\mod 7$ and $2005^{2005}\mod 5$.  The second one is obviously zero, and the first one is $3$ via $FLT$.

So we want a number that's a multiple of $5$ and $3\bmod 7$.  By the Fundamental Theorem of Guessing Small Numbers That Might Work, we get that the answer is just $10$.

[quote="devious_"]blahblahblah, how did you get that the first one is 3? Is it because $2005 \equiv 3 \mod 7$?[/quote]

$2005\equiv 3\pmod 7$

$2005^{2005}\equiv 3^{2005}\equiv 3^{4\cdot 501+1}\equiv 3^1\pmod 7$

[quote="3X.lich"]
$2005^{2005}\equiv 3^{2005}\equiv 3^{4\cdot 501+1}\equiv 3^1\pmod 7$[/quote]
I know that $2005^{2005}\equiv 3^{2005} \mod 7$, but I don't really understand the significance of writing $3^{2005}$ as $3^{4\cdot 501+1}$ and how congruence to $3 \mod 7$ follows from that...

By $FLT$, $3^6=1\bmod 7$.  Now, just use the laws of exponents.


The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.

We know that there are 14 two person games, therefore there are a total of 28 players (some may be repeated). There are 20 members of the club, so 8 members are repeated. 12 members, therefore, are not repeated, so there is a set of six games with 12 distinct players.

[quote="Arvind_sn"]12 members, therefore, are not repeated, so there is a set of six games with 12 distinct players.[/quote]

The 12 non-repeated members do not (and most likely will not) have to play against each other.  :wink:

Does the problem specify that they have to be playing against each other though? >.<

Yes, we can't jump from "there are 12 non-repeating members" to "there are a set of 6 games with 12 non-repeating members".

Though, I did the same thing when I first tried this problem.    :)

Ohhhh,

that makes more sense then. Because this was suspiciously easy lol

Thanks! :D

Actually, if you take away a game where there is at least one player who plays more than once, then repeat for a total of at most 8 times, you'll be left with 6 games with 12 players who each play once. I think...

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356634#356634

This is linked to from the contests section. Next time, just revive the topic with your question. (same goes for your other one)


geometry
3D geometry
pyramid
LaTeX
function
calculus
integration

I dont remember exact wording but this will be enough to be working on.

1. What is the smallest positive factor of 2005 which has at least two digits?

2. A company is selling same jelly in two kinds of jar.
     Jar J has j ounces jelly and contains 25% more than jar E.
     Jar E has e ounces of jelly and contains 20% less than jar J.
     If j and e are relatively prime, what is ordered pair (j,e)

3.  Find smallest n which is multiple of 9, containing 0,2,5 at least once as its digit,
     and no other digit.

4.  for positive a, y=x^3 + x^2 + ax meets x-axis at exactly two points. Find a.

5. four digit number POOQ is pooq when P,Q are distinct prime numbers and O = 0,
    and  exactly one of P and Q divides POOQ.
    Find the number of pooq.

6. Find the volume of regular pyramid where base is hexagon of area 12 and each
    isosceles triangle is also area of 12.

can somebody else post 7 and 8?
well, only four people got 8, but it was very very sevenable.

based on my (poor) memory and my scrap paper...
7. $a_0 = 4, a_1=9, a_2=200, a_3=5, a_{n}=a_{n-1}-a_{n-2}+a_{n-3}-a_{n-4}$ for all positive integer $n\ge 4$.  find $a_{2005}$

i don't really know how to use Latex,although i know i should....
anyhow,might not be exact words,but understandable,at least for me.....

8.x,y,z are all real numbers,0<(should be less than and equal to,same thing for the next several)x^2+y^2<4,and 0<z<[x^2+y^2+1](all less/greater than and equal to,and this bracket should be great integer function),let S denote the volume of solid formed by x,y,z,find S

it's strange there's only 4 people get 8,because personally thinking,none of them is really hard,especially 8,and it's very unlikely to happen that most people made careless mistakes...........

well, for one thing i blanked out as soon as i saw a cylinder minus the stepped pyramid that i was supposed to find the area of for number 8. i think it's just the fact that not many of us are familiar with 3-d geometry...

Tai Ho of the Y team did a double integral for 8 because he overlooked the floor function. So sometimes it's the other way around. :) The answer he got was actually pretty close. 7 is still an amazing score.


The following sequence of reactions was used to prepare compound Q:

$ 2,3,4-tribromocyclopentanone+(C_{2}H_{5})_{3}N\, (2 eqs.) \longrightarrow A$

$ A+CH_{3}OH/H^{+}\longrightarrow B$

$ B+h \nu \longrightarrow C$

$ C+H_{3}O^{+}\longrightarrow D$

$ D+NaHO(aq)/heat \longrightarrow E$

$ E+H_{3}O^{+}\longrightarrow Q$.

Compound Q has molecular formula $ C_{10}H_{8}O_{4}$, and the main feature of its infra-red spectrum is a strong, broad band centered at about 3000 cm-1. Ascertain the structure of each intermediate compound A-E and of compound Q.

sorry for reviving an old topic but im pretty jobless rite now and was looking for some tough org questions. :roll:
in the first step you get 2 bromo cyclopentadiene
which is unstable(antiaromatic) and dimerizes(III step) to give endo product
second step protects the keto group
and fourth step I think selectively deprotects  one of the carbonyl groups.
Now fifth is obviously a favorskii rearrangement and u get a monocarboxlic acid with an acetal, and the last step I assume is a deprotection of the keto group and hydrolysis of br.
Is dat rite??
Im not gr8 at spectroscopy, especially IR so the IR data doesn’t mean much to me.

[quote="VARUNARASU"]in the first step you get 2 bromo cyclopentadiene[/quote]

Wrong.

[quote="VARUNARASU"]which is unstable(antiaromatic) and dimerizes(III step) to give endo product [/quote]

Wrong.

[quote="VARUNARASU"]second step protects the keto group[/quote]

Right.

[quote="VARUNARASU"]and fourth step I think selectively deprotects one of the carbonyl groups.[/quote]

Right.

[quote="VARUNARASU"]Now fifth is obviously a favorskii rearrangement [/quote]

Right.

[quote="VARUNARASU"]and u get a monocarboxlic acid with an acetal, and the last step I assume is a deprotection of the keto group and hydrolysis of br.[/quote]

Wrong.

[quote="VARUNARASU"]Im not gr8 at spectroscopy, especially IR so the IR data doesn’t mean much to me.
[/quote]

Then study it.

sorry first one was a typo - bromo cyclopentadienone which dimerizes giving endo product
i've bin thinking abt 3rd step. maybe its a cycloaddition[2+2] reaction as dimerisation of cyclopentadienone is spontaneous

i just went thru IR. and i think the last step gives a dicarboxlic acid, coz if there was hydroxyl group too, you wud probably have a sharp,(3590-3650) or broad(3200-3550) depending on the conc.


function
trigonometry
induction
algebra unsolved
algebra

1. Find all function $f: \mathbb R \to \mathbb R$, satisfying $f(x+f(y))=f(x)+ \sin y$ for all real numbers $x$ and $y$.

2. Find all function $f: \mathbb N \to \mathbb N$ such that $f(f(f(n))) +f(f(n)) +f(n) = 3n$ for all $n \in \mathbb N$.

[quote="MathVietNam"]1.Find all function f: R into R, satisfying: f(x+f(y))=f(x)+ Siny, for all real numberx,y[/quote]

No such function exists :

Let $ P(x,y)$ the assertion $ f(x\plus{}f(y))\equal{}f(x)\plus{} \sin y$

1) $ f(\mathbb R)\equal{}\mathbb R$
If $ a\in f(\mathbb R)$, $ \exists b$ such that $ a\equal{}f(b)$. Then  $ P(b,y)$ $ \implies$ $ a\plus{}\sin y\in f(\mathbb R)$, and so $ [a\minus{}1,a\plus{}1]\subset f(\mathbb R)$
An immediate induction shows that $ f(\mathbb R)\equal{}\mathbb R$ and $ f(x)$ is a surjection.

2) $ f(\mathbb R)\neq\mathbb R$
Let $ a\in\mathbb R$. Since $ f(x)$ is a surjection, $ \exists b$ such that $ f(b)\equal{}a$
Then $ P(0,b)$ $ \implies$ $ f(a)\equal{}f(0)\plus{} \sin b$ $ \in[f(0)\minus{}1,f(0)\plus{}1]$
So $ f(\mathbb R)\subseteq[f(0)\minus{}1,f(0)\plus{}1]$

Since 1. and 2. are in contradiction, no such function exists.

[quote="MathVietNam"]2.Find all function f :N into N ,Such that: f(f(f(n))) +f(f(n)) +f(n) = 3n[/quote]

$ f(n)$ is injective, so if $ n_0$ is the least non fixed point we would have lhs>rhs  :wink:

ps. beautiful part 1 on the image of f! :D

For n = 1, we easily get f(1) = 1. Suppose that, for n < k, we have f(n) =n. We prove that f(k) = k. if p = f(k) < k then by the induction
hypothesis f(p) = p = f(k), and this contradicts the injectivity of f. If f(k) > k, then f[f(k)] >= k.
If we had f(f(k)) < k, then, as before, we would get the contradiction
Similarly, we have f{f[f(k)]} > k. Hence, f{f[f(k)]} + f[f(k)] + f(k) > 3k, which contradicts the original condition. Thus f(k) = k.


[quote="GameBot"]There are 4 separate large boxes, and inside each large box there are 3
separate small boxes, and inside each of these small boxes there are 2
separate smaller boxes. How many boxes, counting all sizes, are there
altogether?[/quote]

We have 4 large boxes, and $ 4 \times 3 \equal{} 12$ small boxes, and $ 12 \times 2 \equal{} 24$ smaller boxes, for a total of $ 4 \plus{} 12 \plus{} 24 \equal{} 40$.


Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3$
Prove that:
\[ \frac{1}{{{a^2}}} \plus{} \frac{1}{{{b^2}}} \plus{} \frac{1}{{{c^2}}} \ge {a^2} \plus{} {b^2} \plus{} {c^2}\]

[quote="leviethai"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that:
\[ \frac {1}{{{a^2}}} \plus{} \frac {1}{{{b^2}}} \plus{} \frac {1}{{{c^2}}} \ge {a^2} \plus{} {b^2} \plus{} {c^2}\]
[/quote]

WLOG, assume $ a \ge b \ge c$.

if $ a \le 1\plus{}\sqrt{2}$, then we have

$ \frac{1}{a^2}\ge (a\minus{}2)^2 \equal{} a^2\minus{}4a\plus{}4,$

so we get

$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2} \ge a^2\plus{}b^2\plus{}c^2\minus{}4(a\plus{}b\plus{}c)\plus{}12 \equal{} a^2\plus{}b^2\plus{}c^2$;

if $ a > 1\plus{}\sqrt{2}$, then we have $ b\plus{}c<2\minus{}\sqrt{2}$, and we get

$ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}>\frac{1}{b^2}\plus{}\frac{1}{c^2} \ge \frac{8}{(b\plus{}c)^2} > \frac{8}{(2\minus{}\sqrt{2})^2}>9\equal{}(a\plus{}b\plus{}c)^2>a^2\plus{}b^2\plus{}c^2$.

done.

[quote="leviethai"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that:
\[ \frac {1}{{{a^2}}} \plus{} \frac {1}{{{b^2}}} \plus{} \frac {1}{{{c^2}}} \ge {a^2} \plus{} {b^2} \plus{} {c^2}\]
[/quote] we have $ \sum \frac{1}{a^2} \ge \sum \frac{1}{bc} \equal{}\frac{a\plus{}b\plus{}c}{abc} \ge \frac{27}{(ab\plus{}bc\plus{}ca)^2}$  we prove that $ \frac{27}{(ab\plus{}bc\plus{}ca)^2} \ge a^2\plus{}b^2\plus{}c^2$
by AM-GM $ (\sum a^2)(\sum bc)^2 \le (\frac{\sum a^2\plus{} 2\sum bc}{3})^3\equal{}27$
we done

[quote="nguyenvanhuyen"][quote="leviethai"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that:
\[ \frac {1}{{{a^2}}} \plus{} \frac {1}{{{b^2}}} \plus{} \frac {1}{{{c^2}}} \ge {a^2} \plus{} {b^2} \plus{} {c^2}\]
[/quote] we have $ \sum \frac {1}{a^2} \ge \sum \frac {1}{bc} \equal{} \frac {a \plus{} b \plus{} c}{abc} \ge \frac {27}{(ab \plus{} bc \plus{} ca)^2}$  we prove that $ \frac {27}{(ab \plus{} bc \plus{} ca)^2} \ge a^2 \plus{} b^2 \plus{} c^2$
by AM-GM $ (\sum a^2)(\sum bc)^2 \le (\frac {\sum a^2 \plus{} 2\sum bc}{3})^3 \equal{} 27$
we done[/quote]

well done...
 :)

[quote="leviethai"]Let $ a,b,c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$
Prove that:
\[ \frac {1}{{{a^2}}} \plus{} \frac {1}{{{b^2}}} \plus{} \frac {1}{{{c^2}}} \ge {a^2} \plus{} {b^2} \plus{} {c^2}\]
[/quote]
$ \Leftrightarrow \sum \frac{1}{a^2}\plus{}2\sum bc \ge \left ( \sum a \right )^2\equal{}9$
By $ AM\minus{}GM$
$ \sum \frac{1}{a^2}\plus{}2\sum bc \ge \frac{3}{abc}\plus{}2\sqrt{3abc(a\plus{}b\plus{}c)}\equal{}3\left ( \frac{1}{abc}\plus{}\sqrt{abc}\plus{}\sqrt{abc} \right )  \ge 9$

[b]But[/b]
$ \frac{a\plus{}b\plus{}c}{abc}\ge\frac{27}{(ab\plus{}bc\plus{}ca)^{2}}$
Can you give more details on your prove?Thank you!

[quote="wxd686"][b]But[/b]
$ \frac {a \plus{} b \plus{} c}{abc}\ge\frac {27}{(ab \plus{} bc \plus{} ca)^{2}}$
Can you give more details on your prove?Thank you![/quote]

very easy
$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)^2\ge \ 3\sqrt[3]{abc}.9\sqrt[3]{a^2b^2c^2}\equal{}27abc$

[quote="wxd686"][b]But[/b]
$ \frac {a \plus{} b \plus{} c}{abc}\ge\frac {27}{(ab \plus{} bc \plus{} ca)^{2}}$
Can you give more details on your prove?Thank you![/quote]
$ \frac {a \plus{} b \plus{} c}{abc} \equal{}\frac {9}{abc(a\plus{}b\plus{}c)} \ge \frac {27}{(ab \plus{} bc \plus{} ca)^{2}}$
Because $ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)$


An exhibit hall in a convention center measures 20 by 30 yards.
A square area of 9 square yards must be reserved in one corner
for registration. How many square yards will remain for exhibit
space?


function
real analysis
real analysis unsolved

Please give me the link if the following was  posted.
Let $f$ be a function with the property that every point of discontinuity is a removable discontinuity and f may be discontinius at some (even inifinety) numbers $x$
We define $g(x)=lim_{y\rightarrow x}f(y)$

 Prove that $g$ is continious

It seems obvious but could anyone post a clear and rigorous mathematical proof ?Thank you

|g(x) - f(y)| < epsilon, |y - x| < delta
Let y->t ( for any t in [x-delta, x + delta])
then
|g(x) - g(t)|<= epsilon, |t - x|<=delta


function
integration
real analysis
real analysis unsolved

Let $f : [0,1] \to \mathbb R$ be a continuous function such that $\int_{0}^{1}f(x) \, dx = 0$.
Prove that if $n \in \mathbb N$, $n$ odd, then there is $c \in [0,1]$ such that $\int_{0}^{c}f(x) \, dx = \left( f(c) \right)^{n}$.

Presumably, we can find a class of functions $\mathcal G$ such that the conclusion says that there is $c \in [0,1]$ such that \[\int_{0}^{c}f(x) \, dx = g(f(c)) .\]

[quote="perfect_radio"]Let $f : [0,1] \to \mathbb R$ be a continuous function such that $\int_{0}^{1}f(x) \, dx = 0$.
Prove that if $n \in \mathbb N$, $n$ odd, then there is $c \in [0,1]$ such that $\int_{0}^{c}f(x) \, dx = \left( f(c) \right)^{n}$.

Presumably, we can find a class of functions $\mathcal G$ such that the conclusion says that there is $c \in [0,1]$ such that
\[\int_{0}^{c}f(x) \, dx = g(f(c)) . \]
[/quote]Let $g: \mathbb{R}\to \mathbb{R}$ be continuous such that $\xi g(\xi) > 0$ whenever $\xi \ne 0$ (and thus $g(0) =0$). Then for any continuous $f$ with $\int_{0}^{1}f(x)dx = 0$ there is $c \in [0,1]$ such that
\[\int_{0}^{c}f(x) \, dx = g(f(c)) . \]
Here's why: If $f(0) = 0$, set $c = 0$. Otherwise we may assume $f(0)>0$, and $f$ must change its sign on $[0,1]$. Consider $h(\xi) = \int_{0}^\xi f(x) dx-g(f(\xi))$, and let $x_{0}= \inf \, \{x | f(x) = 0\}$.  Thus $f > 0$ on $[0,x_{0})$.

Then $h(0) =-g(f(0)) <0$ and $h(x_{0}) = \int_{0}^{x_{0}}f(x)dx-0 > 0$. Hence $h$ must vanish at some $c \in (0,x_{0})$. :)