geometry
perpendicular bisector
national olympiad

Today 5th Bangladesh math olympiad has started.I am sending a problem for under 10th grade students.

Two parallel chords of a circle have length 10 and 14.The distance between them is 6 unit.The chord parallel to these chords and half way between them has length /sqrt(a).Then find a

try to solve
>moon

[hide="My attempt"]
Let $x$ be the distance from the center of the circle to the perpendicular bisector of the two chords.

Therefore $r^{2}=7^{2}+x^{2}$ and $r^{2}=5^{2}+(6+x)^{2}$  Setting these two equations equal yields $x=1$.

Then the length of the chord is $a$ where $r^{2}=(3+x)^{2}+(\frac{a}{2})^{2}$

This leads to $a=\sqrt{132}$
[/hide]

Good move!
But i think that u have done a stupid mistake.So try again.
>moon

[img]7344[/img]

(Click on it twice and it won't be distorted...)
So I get 46


geometry
rectangle
real analysis
topology
real analysis unsolved

In $\mathbb{R}^{2},$ let the phrase "dyadic rectangle" mean any set of the form $\left[\frac{j}{2^{m}},\frac{j+1}{2^{m}}\right] \times \left[\frac{k}{2^{n}},\frac{k+1}{2^{n}}\right]$ for $j,k,m,n\in\mathbb{Z}.$ We will use $|E|$ to mean the two-dimensional Lebesgue measure of the set $E.$

Let $\Omega$ be an open set in $\mathbb{R}^{2}$ and let $\mathcal{R}$ be the set of all dyadic rectangles $R\subset\Omega.$ If $|\Omega|<\infty$ and $\beta > 1,$ then there exists a constant $C_{\beta}$ which does not depend on $\Omega$ such that
\[\sum_{R\in\mathcal{R}}|R|^{\beta}\le C_{\beta}|\Omega|^{\beta}\]

[Note: this is quite false for $\beta=1.$]

I would hate to spoil the fun, so I'll not post a complete solution right away. But, since the problem is not obvious, here is a series of hints:
[hide="Hint 1"]Look at the rectangles of one fixed area first[/hide]
[hide="Hint 2"]If the area of each rectangle is, say, $c|\Omega|$ then there are not more than $c^{-1}$ rectangles of the same shape.[/hide]
[hide="Hint 3"]Two rectangles of the same area and very different shapes are "morally" disjoint[/hide]

So, where is it from? ;) :lol:


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Show that the following ring is absolutely flat.
$ A \equal{} K[X_1,X_2,.......]/(X_1^2 \minus{} X_1,X_2^2 \minus{} X_2,........)$

I am trying to do it using the equivalent criterion that any principal ideal is idempotent, but not much luck.

since flatness is a local property, absolut flatness is also a local property. furthermore, fields are local absolut flat rings (the converse also holds, but we don't need that). thus we have to check that $ A_m$ is a field for every maximal ideal $ m$ of $ A$, which means that $ \forall a \in m \exists b \in A \backslash m : ab \equal{} 0$.

now $ A$ is the colimit of the rings $ k[X_1,...,X_n] / (X_1^2 \minus{} X_1 , ... , X_n^2 \minus{} X_n)$, so we may assume that $ A$ is such a ring. a maximal ideal of $ k[X_1,...,X_n]$ is the kernel of some evaluation $ X_i \mapsto a_i$ where $ a_i \in \overline{k}$. thus $ A$ has exactly $ 2^n$ maximal ideals, namely $ (\overline{X_1} \minus{} e_1, ...., \overline{X_n} \minus{} e_n)$ where $ e_i \in \{0,1\}$, which are annihilated by $ (\overline{X_1} \minus{} (1 \minus{} e_1)) \cdot ... \cdot (\overline{X_n} \minus{} (1 \minus{} e_n))$.

Would you please explain your solution.I did not quite understand it.


[quote="abacadaea"]i dont know if im right but i can get this started:

factor out $ \frac{1}{2^{2}}$ which gets us $ \frac{1}{4}(1/1^{2}+1/2^{2}+1/3^{3}.......)$

thats as far as i can get[/quote]

That's a good start, but don't feel bad that you can't get any farther.  As far as I can tell, there is no way to get farther without the use of advanced mathematics (calculus.)  You might want to look at this page from Harvey Mudd's Fun Facts at http://www.math.hmc.edu/funfacts/ffiles/10001.3-5.shtml, which states (without proof) the cool and useful fact that:

$ \(1/1^{2}+1/2^{2}+1/3^{2}.......=\pi^{2}/6$

The proof of this assertion depends on calculus (Euler is credited with the original proof) and does not belong in this forum, but if we take that fact as a given, then the answer to the original problem is $ \pi^{2}/24$.

Edited to correct typos pointed out by t0rajir0u and Gyan--thanks!

That should be $ \frac{\pi^{2}}{24}$.  And no, it's not an elementary fact at all.

For those who want to learn more about the proof that $ \sum_{n=1}^\infty \frac{1}{n^{2}}= \frac{\pi^{2}}{6}$, there's a Wikipedia article: [url=http://en.wikipedia.org/wiki/Basel_problem]The Basel Problem[/url]

(edit: Oops, haha, thanks Gyan.)

You may like to correct a  typo in the above post,  The value  is $ \pi^{2}/6$ and not $ \pi^{2}/4$


hi,
just had my yr 11 maths exam today (non calculator, calculator one next week) and was wondering whether u peeps could help me with a question we had.

the question was height (y, given value h) over time (x, given value t), and u had to figure out the time when the height was 86, and then when it was 83 ( 2 separate questions, ie. evaluate when h=86, evaluate when h=83)

the equation was a 'sin' equation, so the answers to be given were the first value in the cycle.

cus i cant remember the equation frm the beggining, ill explain what i did. i replaced Y (as in Y=...) w/ 86 further on w/ 83 (so they were 86=...and 83=...) to work out the x valuefor each.

i swapped the numbers over the = sign (minusing when the number was being added, etc.) and eventually got to

sin('pi't)=1 (i dunno how to do pi on a keyboard) for the h=86 question

and

sin('pi't)=1/2 for the h=83 question

i didnt know what to do w/ the pi in the brackets (ie. move over the equals sign) so that was my last step and i left my answer like that

remember i had to do this WITHOUT a calculator, and it would have to be given in whole numbers (so i would think making pi=3.14.... would be too hard and wouoldnt suit)

so plz plz plz help me out. im stressing over this question and a little curious

thanks

$\sin(\pi t)=1$
$\pi t=\sin^{-1}(1)$
$t=1/2$

$\sin(\pi t)=1/2$
$\pi t=\sin^{-1}(1/2)$
$t=1/6$

$\sin^{-1}(x)$ is the inverse function of $\sin(x)$
therefore, $\sin^{-1}(\sin(x))=x$
example: $\sin^{-1}(0)=$ the values of $x$ where $\sin(x)=0$

Btw: $\sin^{-1}$ can be denoted as $\arcsin(x)$


Prove the inequality \[\frac{ac+c+1}{ac+3c+1}-\frac{ab+b+1}{ab+3b+1}\leq\frac{bc+c+1}{bc+3c+1}-\frac{b^{2}+b+1}{b^{2}+3b+1}\] for all positive real numbers $a,b,c$ such that $a\geq b\geq c.$

[quote="RDeepMath91"]$LHS-RHS \geq 0$
$\Leftrightarrow \sum_{cyc}(\frac{a}{a^{2}+3a+1}-\frac{a}{ca+3a+1}) \leq 0$
$\Leftrightarrow \sum_{cyc}\frac{a^{2}}{(a^{2}+3a+1)(ca+3a+1)}(c-a) \leq 0$

Anyone can continue ?? Maybe with rearrangement or something ...[/quote]
Please clearly!!!!! :wink:

[quote="Fitim"]Dont have solutions this problems??? :D[/quote]

stop spamming, please! save the time, was it?? save YOUR time, and you'll save the others.

[quote="bodan"]Prove the inequality \[\frac{a^{2}+a+1}{a^{2}+3a+1}+\frac{b^{2}+b+1}{b^{2}+3b+1}+\frac{c^{2}+c+1}{c^{2}+3c+1}\geq \frac{ab+b+1}{ab+3b+1}+\frac{bc+c+1}{bc+3c+1}+\frac{ca+a+1}{ca+3a+1}\] for all positive real numbers $a,b,c.$[/quote]

I'm afraid it's not true.
Take for example $a=1$, and $b=c=10^{-3}$


A wire of length x is bent into the shape of a square.

(a) Express the perimeter of the square as a function of x.

(b) Express the area of the square as a function of x.

The perimeter of the square is x.

The area of the square is $(\frac{x}{4})^{2}$.

Sorry, but can you post problems that require a little more thought?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1108892508&t=130891[/url]

yeah, ur not interval.

Please dont post the same problem twice, interval/sharkman.

Interval must be a student in the same course.

what is this course? and what is this test ur studying for? it would seriously help all the people ur having explain everything to u if they knew that.

If you are taking Algebra II and want to go by formulas and have a TI-83+ series calculator, then [url=http://www.ticalc.org/archives/files/fileinfo/255/25555.html]this[/url] might be of help.  I would not recommend it to anyone who wants to really learn how to solve problems, though.

ok...

1) i already told you that you cant post super easy problems here anymore

2) why dont you just admit that ur Interval.  what are the chances that you post the same exact problem??? i would seriously bet my life, cross my heart hope to die stick a finger in my eye, watever

anyways, just look at the other topic you posted.

locked.


ratio
trigonometry
calculus
calculus computations

Please, help with finding the radius of convergence of the series with term:

a_k= ((k^3).(sink).(x^k))/(6^k)

without using the Cauchy-Adamar fourmula ( 1/R = lim-sup (k-th root of |a/k|).

I am looking for an absolute convergence, of course.

Thank you for your time and help.

[quote="gauss202"]Try to apply an Absolute Comparison Test to $ \sum_{k \equal{} 1}^\infty k^3 x^k 6^k$, followed by Ratio Test. Then show for any x outside of that range, the $ \sin k$ can't "save it".[/quote]

Thank you - you pushed me further. In fact I have almost did it for the outside showing that there is a sequence of integers going to infinity for which sin is 1/2 so showing that in that case the term does not go to zero.


certo che ci sono italiani!!!!!
Io ad esempio sono italiano........

oh ciao....anche se con ritardo  :oops: .....BENVENUTO!


search
combinatorics proposed
combinatorics

Let n >= 2 be an integer and Dn the set of points (x; y) in the plane
such that x; y are integers with |x|; |y| <= n.
(a) Prove that if each of the points in Dn is colored in one of three
colors, there exist two points of Dn in the same color such that
the line through them passes through no other point of Dn.
(b) Show that the points of Dn can be colored in four colors so that
if a line contains exactly two points of Dn, those two points have
different colors.

[quote="carlosbr"]Latex added:

Let $n \geq 2$ be an integer and $D_n$ the set of points $(x,y)$ in the plane
such that $x,y \in \mathbb{Z}$ and $|x| \leq n, |y| \leq n$.

$(a)$ Prove that if each of the points in $D_n$ is colored in one of $3$
colors, there exist $2$ points of $D_n$ in the same color such that
the line through them passes through no other point of $D_n$.

$(b)$ Show that the points of $D_n$ can be colored in $4$ colors so that
if a line contains exactly $2$ points of $D_n$, those points have
different colors.[/quote]

Part $(b)$ is much too easy:
color $A$ when $x$ odd, $y$ odd
color $B$ when $x$ even, $y$ odd
color $C$ when $x$ odd, $y$ even
color $D$ when $x$ even, $y$ even
Contradiction follows by itself.

Part $(a)$ is a bit cooler luckily, i'll leave open for you to search on.  ;)

I wanted to solve this problem but its late and I am sleepy. So I am going to bed, but not until I will put an additional question:
Is it true that the only valid coloring for point b) is Peter's?

Ok. I just woke up. I solved a) while I was falling asleep yesterday. This is it:
Suppose there is a 3-coloring. Take the points $A(n-1,n), B(n-2,-n)$ and one by one the points $(n,n-2k+1), k\in\{1,n\}$.  $A$ and $B$ are different, and different from those points, so they have all the colors. Let's say A is red, B is blue, and the others are yellow. The points $(n,n-2k), k\in\{1,n-1\}$ can't be red(take them with A). So they are:
1) blue. The point $C(n-1,-n+1)$ is red (on line  with $(n,n-1)$ and $n,n-2$)). Now just take the point $D(n-3,n)$. It is on the same line with $D,B$ and $(n,n-1)$, and thus it can't be colored.
2)yellow. The point $E(n-1,n-1)$ cannot be blue or yellow so it's red. Now take $(n-3,n)$ with $E,B$ and $(n,n-1)$. Contradiction.

In fact, now I see that cases 1) and 2) are almost the same, and I did the same thing in 1) asI did in 2). :D


We just wanted to remind interested students that the regular enrollment deadline is tomorrow, August 31.

There are already over 100 students from numerous countries enrolled in the inaugural [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT program[/url].

[quote="ifai"]what is WOOT?[/quote]

Worldwide Online Olympiad Training.

Whoa the deadline is today.

That specific page is [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]here[/url].

[quote="Arne"]An olympiad training programma.

Please use the search function if you want to get relevant information, or look at the specific WOOT page, etc.[/quote]
thank you for your suggestion :blush:

The class lectures will vary in content.  The schedule is being updated and we will soon announce many of them to WOOTers.

[quote="manuel"]I already enrolled in WOOT but i am a bit lost.. when does it start? which are the classes?[/quote]

You should now see the WOOT message board when you are logged in.  You'll find that the students have already started discussing some challenging problems there.


If I have:

$\frac{\frac{p(x)}{g(x)}}{\frac{r(x)}{q(x)}}$

Then none of $g(x), r(x), q(x)$ are allowed to equal 0, right?

Whereas in the similar expression $\frac{p(x)}{g(x)}* \frac{q(x)}{r(x)}$
(Which is essentially equivalent to the first expression), neither $r(x)$ nor $g(x)$ can equal 0  but here, $q(x)$ can?

Thanks.

I'll assume you meant $\frac{p(x)}{q(x)}* \frac{g(x)}{r(x)}$ .  Yes, those are actually two different expressions, and the reason you can transform it to the all-fractional form is because you are assuming that $p(x)$ and $g(x)$ are not equal to zero.

Edited to what I meant to say, I made a few mistakes.
So in the second expression, q(x) CAN equal 0, right? But it can't in the first?

in
$\frac{\frac{p(x)}{g(x)}}{\frac{r(x)}{q(x)}}$
which is equivalent to
$\frac{p(x)}{g(x)}\div \frac{r(x)}{q(x)}$ or $\frac{p(x)}{g(x)}\cdot \frac{q(x)}{r(x)}$

this means that $g(x), r(x), q(x)$ cannot be equal to zero

its just like saying

$f(x)=\frac{x}{x}=1$

they are different


[quote="7h3.D3m0n.117"]does the following summation

$\sum_{i=0}^{\infty}\frac{1}{n!}$ $=e$?????

i worked it out and im pretty sure but my math teacher wasn't as sure as i was[/quote]

this is just the taylor expansion of e^x at x=1

In fact, that's the definition of $e$ that Euler originally used when defining it.

Yeah, more of a Calculusish problem, but I'll let it slide.  :)

[quote="boostage"]Euler originally used when defining it.[/quote]
I think it was Napier. Euler lived after Napier. Euler numbers can be defined as follows: Let $|x|<\pi/2$. Then $\text{sec }x$ has the series expansion
\[\text{sec }x-1=\sum_{i=1}^\infty \frac{E_{i}x^{2i}}{(2i)!}.\]
$E_{i}$ is the $i$th Euler number. $\text{Euler numbers}\ne \text{Napier's constant}.$

Of course, your summation symbol should be $\sum_{n=0}^{\infty}$ instead.  A few equivalences:

$e^{x}= \lim_{n \to \infty}\left( 1+\frac{x}{n}\right)^{n}= \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$

[quote="t0rajir0u"]your summation symbol should be $\sum_{n=0}^{\infty}$ instead.[/quote]
Are you sure? In [url=http://mathworld.wolfram.com/EulerNumber.html]Mathworld[/url] the definition is of the same form I gave.

You give $i$ in the summation constant but use $n$ in your expression.  Technically speaking,

$\sum_{i=0}^{\infty}\frac{1}{n!}= \frac{1}{n!}+\frac{1}{n!}+... = \infty$

Edit:  Oh, sorry, I was referring to [b]7h3.D3m0n.117[/b]'s original post, not yours.

[quote="7h3.D3m0n.117"]oops  :blush:  i'll fix it...i'm supposed to wait like a couple hours or something

and i thought summation and the Euler number was part of Algebra II???????[/quote]
Summation is Algebra II. Euler number is introduced in Algebra II. The two taken together is calculus...


ABC  triangle  x.y.z>0 ===> (x+y+z) 2 R 2  \geq xyc 2 +yza 2 +zxb 2

We easily prove
x 2 +y 2 +z 2  \geq -2xycosC-2yzcosB-2xzcosA by vector(1)
<=>(x+y+z) 2  \geq 2 \sum xysin 2 C
<=>(x+y+z) 2 R 2  xyc 2 +yza 2 +zxb 2
You try proving (1)
Note  \sum a(vectorIA)=vector0


Star polygons are created as follows: n points are equally distributed around the circumference of a unit circle. An initial point P(subscript)o is chosen and each of the remaining points are then consecutively labled clockwise P(ss)1,P(ss)2,....,P(ss)n-1. Segment P(ss)oP(ss)m is constructed where 0<m<n; then a chord of the same length is constructed beginning as P(ss)m and ending at the mth point counting clockwise from P(ss)m. This procedure is repeated until arriving at P(ss)o. The figure is a star polygon if and only if all n points were used to construct the segments. How many distinct star polygons are there for n=36?

[hide="solution"]
First off, note that the only way a star polygon will be formed is when $ m$ is relatively prime to $ 36$.  Thus, we compute $ \phi (36)\equal{}12$.  However, $ m\equal{}1$ and $ m\equal{}35$ will generate the same star polygon (verify this yourself), so we divide by 2, since there's always a pair that generates the same polygon, so our answer is $ \boxed{6}$.
[/hide]

Hint: Subscripts in Latex are P_o or P_{morethanonecharacter}. $ P_oP_m$. $ P_{more than one character}$


I tried to install virtualpdfprinter but kept getting an error msg:" this program requires windows version 5.0 or later." I have windows ME. Is there any ways to get around it? please help. Thanks.

Windows 5.0 means Windows 2000. From their website it seems that they support W95/98/ME only through networking, meaning you have to have the virtual printer installed on a computer with W2K or above; then you have to share the printer and print from W95/98/ME to the shared printer.


trigonometry
number theory
least common multiple
greatest common divisor
modular arithmetic
complex numbers
relatively prime

The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity.  The set $C = \{zw : z \in A \ \text{and} \ w \in B\}$ is also a set of complex roots of unity.  How many distinct elements are in $C$?

[quote="arkmastermind"]this problem seems to have come up several times in the forums, but oh well  :)[/quote]It's ok. He's an administrator.  :P

yeah its all good, people will still benefit from working the problem again or seeing the solution.

Let me attempt this problem...

[hide]
$z^{18}=1$ can be written in the form: $(r)(cis\theta)$.
The first root is $r=1;\theta=0$.
From there on, we see that we have $(1)(\cos(\frac{0+360n}{18})+i\sin(\frac{0+360n}{18}))$ for $n=0,1,2,3...17$ by De Moivre's?
So, we see we have the $\theta=0,20,40,60...320$

$w^{48}=1$ can be written in the form: $(r)(\cos(\theta)+i\sin(\theta))$
The first root is $r=1;\theta=0$.
From there on, we see that we have $(1)(\cos(\frac{0+360n}{48})=i\sin(\frac{0+360n}{48}))$ for $n=0,1,2,3...47$.
So, we have $\theta=0,\frac{15}{2},15,\frac{45}{2},30....\frac{705}{2}$

[/hide]

And then...I'm stuck. This roots of unity stuff is new to me (just learned it this year)...so yeah :/. Although I didn't get the answer, is this the correct approach? (or...did I even find the right $\theta$ values?)

I'm bad with roots of unity. Bah. :stink:

[hide="my guess"]I tried interpreting this geometrically. The roots of unity all form regular polygons. There are always $n$ solutions to $z^n=1$.

Therefore, there are 18 things in set A and 48 in set B. Howevern when you graph them, some overlap, so we want to LCM of $18$ and $48$, which is $\boxed{144}$.

Kind of random, but I think it works.[/hide]

[quote="chess64"]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=complex&t=73489[/quote]Cool! I guessed it correctly. :P

[quote="Zellex"]Why would you do the lcm to find the # of distinct products?[/quote]I have no idea.  :blush:

The previous guy is on the right track.

[hide]
We have one set of angles $0, 20, 40, \dots 340$, and another set $0, 7.5, 15, \dots 352.5$.  When we multiply complex numbers we add the angles and multiply the magnitude, which in both cases are 1.  so the question becomes how many angles are there that results from adding all the angles in set $A$ with all the angles in set $B$.  set the angles to radians we have set $A$ equals $\frac{n\pi}{9}$ and set $B$ equals $\frac{k\pi}{24}$.  adding gives $\frac{(8n + 3k)\pi}{72}$.  Since $3$ and $8$ are relatively prime, the maximum unreachable number is $24 - 8- 3 = 13$.  Experimenting we have $1, 2, 4, 5, 7, 10, 12, 13$ 8 unreachable numbers.  However, we can wrap around to $2*72 + 1$ and so on, which is achievable.  (Since we consider angles mod $2\pi$).  so all numbers are achievable, and 72 implies there are $2*72 = 144$ such valid angles
[/hide]

Don't ask me why the LCM worked, I have no idea...

Pkerichang, that's exactly the same as my solution in the other thread :lol:

Uh....I still don't get why it's LCM....it'd be helpful if someone could tell me...

Maybe it's just a coincidence. You could try it with different numbers to check.

[quote="scorpius119"]No, the lcm always works. [hide]If $w^a=1$, $z^b=1$, and $m$ is the lcm of $a$ and $b$, then $w^m=1$ and $z^m=1$ so $(wz)^m=1$. This limits the set to $m$ values.

All $m$ of these values would work if the number $\exp \frac{2\pi i}{m}$ were in the set. The set $A$ consists of numbers of the form $\exp \frac{2x\pi i}{a}$ and $B$ consists of numbers of the form $\exp \frac{2y\pi i}{b}$, where $x$ and $y$ are integers.

If $d$ is the gcd of $a$ and $b$, $a=d\alpha$, and $b=d\beta$, then since $ab=md$, $\exp \frac{2\pi i}{a}=\exp \frac{2\beta \pi i}{m}$. Similarly, $\exp \frac{2\pi i}{b}=\exp \frac{2\alpha \pi i}{m}$.

It follows that if $\beta x+\alpha y\equiv 1\pmod{m}$ for some integers $x$ and $y$, then $\exp \frac{2x\pi i}{a}\cdot \exp \frac{2y\pi i}{b}=\exp \frac{2\pi i}{m}$. This is true because $\alpha$ and $\beta$ are relatively prime, so $\beta x+\alpha y=1$ has a solution. Therefore there are always $m$ values in $C$.[/hide][/quote]Cool.  :)

Sorry for reviving, but I found a cool way to solve this, that I don't think anyone else has posted yet, so this is not spam.

[hide]$z^{18} = 1 \Rightarrow z = 1^{\dfrac {1}{18}}$
$w = 1^{\dfrac {1}{48}}$.
$wz = 1^{\dfrac {1}{18} + \dfrac {1}{48}} = 1^{\dfrac {11}{144}}$.
Since the denominator of the exponent is 144, the answer is 144.[/hide]

I guess I should post mine, it's quite instructive.

[hide="Solution"]Clearly $z=e^{\frac{2ai\pi}{18}}$ and $w=e^{\frac{2bi\pi}{48}}$, with $a\in \{1,2,...,18\}$ and $b\in \{1,2,...,48\}$. So $zw=e^{\frac{2\pi i(8a+3b)}{144}}$ (just use exponent laws and simplify), meaning each $zw$ is a $144^{th}$ root of unity. Furthermore, by Chicken McNugget, every integer greater than $8\cdot 3 - 8 - 3=13$ can be achieved by $8a+3b$. Now since adding a multiple of $2\pi$ to the numerator doesn't change the value of $zw$ (i.e. roots of unity are periodic with period $2\pi$), and since $8a+3b$ can achieve $\equiv \{1,2,...144\}\pmod {144}$ (because $8a+3b$ can be any value $>13$ from above), it follows that $zw$ can be any distinct $144^{th}$ root of unity. So, the answer is just $\boxed{144}$.[/hide]

Just to make sure, $8a+3b = 1$ is not possible, but $8a+3b=145$ is indeed possible, and $e^{\frac{2\pi i \cdot 145}{144}}$ is also a root of unity (just not reduced, it's the same thing as $e^{\frac{2\pi i \cdot 1}{144}}$), right?


What is the remainder obtained through dividing [b] 472005 + 292005[/b] by  [b]19[/b] ?

[quote="a_vakilian"]you can solve it easier:
$47^{2005}+29^{2005}=(47+29)(47^{2004}-47^{2003}.29+...-47.29^{2003}+29^{2004})$and we knwo that $19|(47+29)=76$. ;)[/quote]

Using the following general result which u actually showed in ur solution, u get the problem pretty fast.

 an  +  bn  is divisible by a+b if n is an odd integer.

 Puttin a = 47,b=29, we get a+b = 76 = M(19) .Also 2005 is odd . So an  +  bn is div. by a+b i.e. 76 and hence by 19


[quote="The Sigma"]I took the AMC 12 and I need help with this problem.

Here it is:

Joe and JoAnn each bought 12 oz of coffee in a 16 oz cup. Joe drank 2 oz of his coffee and then added 2 oz of cream. JoAnn added 2 oz of cream, stirred the coffee well, and then drank 2 oz. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?[/quote]
[hide]
After Joe drinks, he has 10 oz of coffee, and adds 2 oz of cream, making the amount of cream in his cup 2 oz. JoAnn adds 2 oz of cream (ratio 2 to 14) and since the ratio is $2: 14$ and 2 oz is drank, the amount of cream drank is $\frac{2\times2}{14}=\frac{2}{7}$ and the total cream minus cream drank is $2-\frac{2}{7}=\frac{12}{7}$ and the ratio between Joe's cup with cream and JoAnn's cup with cream is $\frac{2}{\frac{12}{7}}=\frac{14}{12}=\boxed{\frac{7}{6}}$...you might want to recheck that though.[/hide]

use the search function?....I think this was posted several times


geometry
rectangle
geometric transformation
rotation
reflection
combinatorics open
combinatorics

A rectangle m*n is divided into figures:
***
***   1)
***
***

o***
o*** 2)
***o
***o
(with their rotations and reflections).
Prove (or disprove) that the total number of figures 2) (with theiк rotations and reflections) in this division is 0.

There're lots of counter-examples. This is the one that I found:

a true statement would be: show that the
***o
***o
o***
o***
figurine does not generate any options the rectangle would not have had

and basically it's just the same as imo problem 3 :)


What is the largest integer $ k$ such that $ k^6 < 2^{20}$?

PoP, I think you made a little mistake. $ 3^3 \equal{} 27$, not $ 81$.  So $ 9$ works.  This is my solution:
We have that $ k^6 < 2^{20}$.  If we take the square root of both sides, we are left with $ k^3 < 2^{10} \equal{} 1024$.  We know that $ 1000 \equal{} 10^3 < 1024 < 11^3 \equal{} 1331$, so $ k \equal{} \boxed{10}$

I disagree.

This is what I did:

Take the sixth root of both equations.  We should get $ k < 2^{\frac {20}{6}}$.  That simplifies to $ k < 2^{\frac {10}{3}}$.  We know from heart that $ 2^10 \equal{} 1024$; therefore, the greatest number less than the cube root of $ 1024$ is $ \boxed{10}$.

Check it with a calculator.  It works.

Grr.  jxl28 beat me to it again.


The right side was already posted and for the left side you can use Vasc´s inequality twice and reduce it to
$a^{4}+b^{4}+c^{4}\geq a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$ :)

[quote="Yosh..."]What is Vasc's inequality?
Thx..[/quote]

I think he's talking about
\[(x^{2}+y^{2}+z^{2})^{2}\geq 3(x^{3}y+y^{3}z+z^{3}x) \]
$\forall x,y,z \in \mathbb R$

Any idea ?  :maybe:
My solution didn't use Vasc's inequality :D


geometry
AMC
AIME
conics
ellipse
USAMTS
3D geometry

I think it's harder than any AIME geometry problem I've ever seen. But keep in mind that you have like a month to do it.

On the other hand, I made the stupidest mistake ever.

I used the formula c^2 = a^2 - b^2 where c is the focal length, a is the semimajor radius, b is the semiminor radius.

I had all the correct data up to this point. So I put 4^2 = 5^2 - b^2 -> b=3 and thus area = pi*3*4 = 12pi.

The part where I screwed up was that c is the distance from the center to the focus, not the distance from the long edge to the focus. If I would plug in c=1 instead of c=4 in that case, I would have gotten the correct answer.

How many points do you think they will take off for that?

1 point I think; it's a minor stupid mistake. And you had everything up to there right, so it's not really a big mistake. At least that's how they'll grade #4.... (though I made non-trivial mistakes...)

Ah well, I botched that problem.  Hopefully I'll get 2 or 3 points for figuring out the major axis...

No, $12\pi$ is not the correct answer. According to the Math Jam, it's $10\sqrt 6 \pi$

[quote="zanttrang"][quote="pkerichang"][quote="zanttrang"]Argh... I understand the solution now.  If only I hadn't made such a stupid assumption when solving the problem.  Do you think I could get 3 points for figuring out the major axis of the ellipse and then botching the minor axis?[/quote]

Strangely I solved this problem the first day I got it....correctly too (maybe I tend to be good at geometry).

Btw, zanttrang, I bombed the third one...How much points to they give to completely written WRONG solution??? :huh:[/quote]
It depends on what sort of thought and effort you put into it - post your solutions and I'll take a look at it (and I'll post mine so you can make fun of my solution to  #5 :P).  Looks like our USAMTS scores might not be matching after all...[/quote]

I did in the other thread.  But If you get 3 pts for getting the major axis, and They somehow give my three points for trying, then we will still have the same score :lol:

Zanttrag, how did you completely botch it?

I did the same thing, got the semimajor axis and bombed the semiminor, and I got $\frac{35\pi \sqrt{2}}{2}$ as an answer.

What I did with the semiminor axis was I said that the line segment parallel to the base of the cone, with endpoints on the cone, and that goes through the midpoint of the semimajor axis would be equal to the semiminor axis. Apparently I was wrong...

[quote="djshowdown2"][quote="ben12345_5"]I was wondering...

How does problem 5 (round 4) compare to geometry problems that typically show up on problems 11-15 in the AIME?

 :?:[/quote]

I actually thought it was comparable to an AIME 3D geometry problem.  The hard part for me was formally writing it up, but I didn't think the answer was difficult to come by.  AIME (and most 3D geo) likes cross sections, and this could be solved taking two simple cross sections.  I suppose there was a good amount going on in this problem that it might have been a little long for an AIME question, but not much.[/quote]

I thought much the same thing.  As far as geometry goes, 3D geo is probably my weakest point (maybe collinearity/concurrency is worse...), yet I had little trouble with actually solving the problem.  The hard part was the write up for three reasons.

1.  Describing 3-D manipulations is always nasty, even with diagrams to help you.

2.  I had trouble proving that the intersection of the plane with the sphere was actually on the cross section used in the math jam

3.  I had trouble proving that the major axis was actually 10, (the line in the cross section) and not some other freakish line through the cone.

Maybe I was just being obsessive in my proof, since both #2 and #3 are pretty obvious intuitively, but it still took time to prove them.

Also, condolences to Philip on the simple mistake.  In any case, you should get a 4 on that one if everything else was correct, leaving you with a 97.  Still gold ;).


Bin $ A$ has one white ball and four black balls.  Bin $ B$ has three balls labeled \$1 and one ball labeled \$7.  Bin $ W$ has five balls labeled \$8 and one ball labeled \$500.  A game is played as follows: a ball is randomly selected from bin $ A$.  If it is black, then a ball is randomly selected from bin $ B$; otherwise, if the original ball is white, then a ball is randomly selected from bin $ W$.  You win the amount printed on the second ball selected.  What is your expected win?

$ \frac{1}{5} \cdot \frac56 \cdot 8 \plus{} \frac{1}{5} \cdot \frac16 \cdot 500 \plus{} \frac45 \cdot \frac34 \cdot 1 \plus{} \frac45 \cdot \frac 14 \cdot 7\equal{}\boxed{20}$

[quote=GameBot]Bin $ A$ has one white ball and four black balls.  Bin $ B$ has three balls labeled $1$ and one ball labeled $7$.  Bin $ W$ has five balls labeled $8$ and one ball labeled $500$.  A game is played as follows: a ball is randomly selected from bin $ A$.  If it is black, then a ball is randomly selected from bin $ B$; otherwise, if the original ball is white, then a ball is randomly selected from bin $ W$.  You win the amount printed on the second ball selected.  What is your expected win?[/quote]

Fixed it for Gamebot :-D

Wow! The founder of AoPS edited Gamebot's post!

Anyway, here's my solution. The chance of getting a white ball from bin A is $\frac{1}{5}$, and a black one $\frac{4}{5}$. From bin B only, the chance of getting $\$1$ is $\frac{3}{4}$, and getting $\$7$ is $\frac{1}{4}$. We multiply this by the black ball percentage to find the chance of getting these values of money. For the chance of getting $\$1$, it is $\frac{3}{5}$, while for $\$7$, it is $\frac{1}{5}$. We multiply the chances by the values, to see the expected amount of money made from just those two, and add. This is a lame $\$2$. Then we do the same thing for bin W. $\$8$ is $\frac{5}{6}$, and $\$500$ is $\frac{1}{6}$. We multiply this by the chance of getting a white ball ($\frac{1}{5}$). The chance for getting the $\$8$ from all the events is  $\frac{1}{6}$, and the $\$500$ is $\frac{1}{30}$. Again, we multiply the chance by the amount of money potentially made, and sum them. This is $\$18$. We add this to the original $\$2$ to get $\boxed{\$20}$


How poeple are selected to participate in math competitions in Iran?
Do you have special schools for talented youth in math.
And how are you prepared for competitions?
Thanks to respond.
senouy.

Iranian team is selected with some competitions. There are not olympiad schools but people who are best in the will be teached for he Team for IMO.  Our country is not rich. With our little but good Team prepration courses we obtain good results. We want to advance and our country needs to know the well-talented students.


Let m is integer . Prove that the polynomial :
   P(x) = x^{4} - 1994x^{3} + (1993+m)x^{2} -11x +m
has integer root

It is not true. There are finite set of integer values of m, when this polinom had integer root.

Sorry the equation has max one integer root

If $P(a)=0, \ a\in Z$ then \[m=\frac{a^{4}-1994a^{3}+1993a^{2}-11a}{a^{2}+1}=-a^{2}+1994a-1992-\frac{1983a-1992}{a^{2}+1}.\] It give finite values of a and finite values of m, when this polinom had integer root.


My AMC coordinator told me that the AMC 10A result never been sent to her so we don't get our scores.  Since AIME is approaching but i can't prove that i qualified it cause i don't have score report back yet, i was wodering should i just give up AIME A and wait patiently for the score and hopefully take the AIME B test?

I think you should ask your math teacher about the result.

You'll know your score before the AIME I test date.

[quote]the AMC 10A result never been sent to her so we don't get our scores.[/quote]

This forum is the incorrect location to find your school's scores.  Your teacher should contact the AMC office.  We can check to make sure the answer forms were received.  We can check to see if the report has been generated.  We can check to see if the report has been sent, and determine the email address to which it has been sent.  We can resend the report.
Nobody here can do any of those checks.

By the way, have your teacher check the spam filter on his/her email account.  Many schools have highly aggressive spam filters that trap our reports.

Steve Dunbar
MAA Director, American Mathematics Competitions


Let $ ABCD$ be a convex quadrilateral inscirbed in circle which has center at $ O$.Let $ P\equal{}AB \cap CD$.Line $ PO$ intersects segment $ AC$ and $ BD$ at $ E$ and $ F$ respectively.Prove that $ [ACQ]\equal{}[BDQ]$
([...] denotes the area)

Really sorry for my mistake.I forgot to type the last sentence.
EDIT:
Let $ ABCD$ be a convex quadrilateral inscirbed in circle which has center at $ O$.Let $ P \equal{} AB \cap CD$.Line $ PO$ intersects segment $ AC$ and $ BD$ at $ E$ and $ F$ respectively.Let $ Q$ be a point such that $ QE\perp AC$ and $ QF\perp BD$.Prove that $ [ACQ] \equal{} [BDQ]$
([...] denotes the area)

$ S_{AQC}\equal{}S_{DQB}\Leftrightarrow \frac{AC}{BD}\equal{}\frac{QF}{QE}\equal{}\frac{sin \angle QEF}{sin \angle QFE} (1)$
Let $ T, U$ be the midpoints of $ AC,BD$ then $ OT\perp AC, OU\perp BD.$
We have $ \frac{sin \angle QEF}{sin \angle QFE}\equal{}\frac{sin \angle TOE}{sin \angle UOE}$
By the law of sine:
$ \frac{sin \angle TOP}{TP}\equal{}\frac{sin \angle OTP}{OP} , \frac{sin \angle UOP}{UP}\equal{}\frac{sin \angle OUP}{OP}$
$ \Rightarrow \frac{sin \angle TOE}{sin \angle UOE}\equal{}\frac{TP}{UP}.\frac{sin \angle OTP}{sin \angle OUP} (2)$
On the other side, $ \Delta ACP\sim \Delta DBP \Rightarrow \Delta ATP\sim \Delta DUP$
$ \Rightarrow \frac{TP}{UP}\equal{}\frac{AT}{DU}\equal{}\frac{AC}{BD} (3)$
And $ \angle BUP\equal{}\angle CEP$
Hence $ \angle OUP\equal{}\angle OTP (4)$
From $ (1)(2)(3)(4) \rightarrow$ QED


advanced fields
advanced fields unsolved

If $X$ is normal and $y$ is a point of $\beta(X)-X$ ($\beta$ = Stone-Cech-Compactification), then $y$ is not the limit of a sequence of points of $X$.

:maybe: Consider a sequence $\{x_{n}\}$ with no accumulation points in $X$. It must have a subsequence that converges in $\beta X$, and the limit of the subsequence cannot be an element of $X$.


What editor do you thinks is the best one and why:
mcedit, joe, vim, kate, kwrite, kdevelop, emacs, rhide?

good ol emacs mann...dont ask me why... :D

you people use linux?

i use windows, and in windows i think visual C is good one!

mahbub: you know there will be no linux in IOI this year, right?

I used to use Windows too, but now I'm starting to use Linux. That's why I am asking which editor is the best: I want to know which one I should learn how to use.

[quote]
mahbub: you know there will be no linux in IOI this year, right? [/quote]
I think you meant windows.

-------------------------------------

I think VIM is good editor, especially for programming. After setting up your preferences and extra commands, your work becomes extremely fast. Also, you don't have to take your hands off the keyboard at all! The only problem is that it takes some time to get used to vim.

But that's just my 5 cents....


If anyone lost the sheet or was not at the SC Team Practice that week, it is found at [url]http://www.geometer.org/mathcircles/indprobs.pdf[/url]

[hide="Problem 1 Answer"]
Show that $ 3^n \ge 2^n$

Note: I am assuming that, as all proofs should be inductive, a proof for all non-negative integers is expected.

First, the inequality is true for $ n = 0$, as $ 1 \ge 1$.

Second, assuming $ 3^n \ge 2^n$, the inequality $ 3^{n + 1} \ge 2^{n + 1}$ needs to be shown as true. This can be done by rewriting this step as $ (3)3^{n} \ge (2)2^{n}$.  As both $ 3 \ge 2$ and $ 3^n \ge 2^n$ are true (the second because it is an assumed hypothesis), the inequality relating the product of the two is also true.  Thus, for n is equal to all non-negative integers:

$ 3^n \ge 2^n$[/hide]
[hide="Problem 3 Answer"]
Prove that for any $ n > 0$:
\[ 1^2 + 4^2 + 7^2 + 10^2 + \ldots + (3n - 2)^2 = n(6n^2 - 3n - 1)/2
\]
First, show that the equation is true for $ n = 1$:
\[ 1^2 = 1(6 - 3 - 1)/2 = 1
\]
Next, using the equation to be solved, show that:\begin{eqnarray*} 1^2 + 4^2 + 7^2 + 10^2 + \ldots + (3n - 2)^2 + (3(n + 1) - 2)^2 & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
n(6n^2 - 3n - 1)/2 + (3n + 1)^2 & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
\frac {6n^3 - 3n^2 - n + 2(3n + 1)^2}{2} & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
\frac {6n^3 + 15n^2 + 11n + 2}{2} & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
\frac {(n + 1)(6n^2 + 9n + 2)}{2} & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
\frac {(n + 1)((6n^2 + 12n + 6) - 3n - 3 - 1)}{2} & = & (n + 1)(6(n + 1)^2 - 3(n + 1) - 1)/2 \\
\end{eqnarray*}
And we are done.
[/hide]
I know my answers are not the most formal, but please let me know if one's absolutely wrong...

They look good to me. After you do a couple of the basic type of induction problems you should pretty much get it. So once you get the basic idea behind induction you can skip the "easier" ones if you like and just try the ones that aren't really obvious how you'd do them.

In linear algebra, there are some important theorem such as the replacement theorem that require the use of induction to solve.  Also, the Vandermonde determinant requires a sort of induction.

For #2, you can set $ \frac {1}{n^2} < \frac {1}{n(n \minus{} 1)}$ and the series can easily telescope from there. However, I have yet to find an inductive proof :(.

You could make what you've done into an inductive proof by showing that $ \sum_{k \equal{} 2}^n\frac {1}{k(k \minus{} 1)} < 1$ is true by induction.

I'm not a big fan of proving things by induction when it could easily be proven some other way. A problem like this is just for practice with induction for people who are new to it.

Hmmm.. I'm not sure how thats inductive... you just telescope into $ \frac{1}{k\minus{}1}\minus{}\frac{1}{k}$ and everythinhg cancels in the series except $ 1\minus{}\frac{1}{n}$, which is obviously less than 1. You don't need the domino effect of induction...

EDIT: Looking at the solution in that packet, I see that they do the same thinhg, though instead of telescoping they use induction :)

[hide="Problem 4 Answer"]
Show that if $ n$ lines are drawn on the plane so that none of them are parallel, and that no three lines intersect at a point, then the plane is divided by those lines into $ (n^2 + n + 2)/2$ regions.

First, 1 line splits a plane into 2 regions, and sure enough, $ (1+1+2)/2=2$.

Next, given $ n$ lines, the $ (n+1)$th line will intersect every other line for a total of $ n$ new intersections, guarenteed by the fact that no line is parallel and no three lines intersect at one point.  There is one region the line runs through 'before' each point it intersects and 1 'after' all of the new intersections.  The line will split each into two regions, adding $ n+1$ regions to the previous case.  Taking the hypothesis for $ n$ lines and adding the $ n+1$ regions, we have:

\begin{eqnarray*}
\frac{n^2 + n + 2}{2} + n + 1&&\\
\frac{n^2 + 3n + 4}{2}&&\\
\frac{(n^2 + 2n + 1)+n+3}{2} &=& \frac{(n+1)^2 + (n+1) + 2}{2}
\end{eqnarray*}
[/hide]
Problem 5 I've seen in [i]The Art and Craft of Problem Solving[/i].
[hide="Problem 5 Answer"]
Given the requirements in Problem 4, show that it is possible to color the resulting regions with two colors such that no two adjacent regions share the same color.

First, with 1 line the plane is split into only two regions, so color them differently.

Next, assume a coloring is possible for $ n$ lines.  Draw the $ (n+1)$th line and switch the coloring of all regions on one side of your line.  Doing so satisfies that all regions adjacent to another across the $ (n+1)$th line are colored differently, and all other edges must separate differently colored regions for our the coloring to be possible with $ n$ lines.  Thus, such a coloring is possible with $ n+1$ lines.
[/hide]


a.b,c >0 satisfy a+b<c+1 , a+c<b+1 , b+c<a+1
prove that  a^2 + b^2 + c^2 < 2abc + 1

Assume that $ a\leq b\leq c$ we have $ b\plus{}c\minus{}1<a\leq b\leq c<1$
Let $ f(a)\equal{}a^2\minus{}2bc.a\plus{}b^2\plus{}c^2\minus{}1$ then $ f(a)\leq max \{f(b\plus{}c\minus{}1);f(b)\}$
But
$ f(b)\equal{}b^2\minus{}2b^2c\plus{}b^2\plus{}c^2\minus{}1\equal{}(c\minus{}1)(c\plus{}1\minus{}2b^2)<0$ because $ c\plus{}1\minus{}2b^2>c\plus{}1\minus{}2b>0$
$ f(b\plus{}c\minus{}1)\equal{}(b\plus{}c\minus{}1)^2\minus{}2bc(b\plus{}c\minus{}1)\plus{}b^2\plus{}c^2\minus{}1\equal{}\minus{}2(b\minus{}1)(c\minus{}1)(b\plus{}c)<0$
Then OK!

[quote="tdl"]Assume that $ a\leq b\leq c$ we have $ b \plus{} c \minus{} 1 < a\leq b\leq c < 1$
Let $ f(a) \equal{} a^2 \minus{} 2bc.a \plus{} b^2 \plus{} c^2 \minus{} 1$ then $ f(a)\leq max \{f(b \plus{} c \minus{} 1);f(b)\}$
But
$ f(b) \equal{} b^2 \minus{} 2b^2c \plus{} b^2 \plus{} c^2 \minus{} 1 \equal{} (c \minus{} 1)(c \plus{} 1 \minus{} 2b^2) < 0$ because $ c \plus{} 1 \minus{} 2b^2 > c \plus{} 1 \minus{} 2b > 0$
$ f(b \plus{} c \minus{} 1) \equal{} (b \plus{} c \minus{} 1)^2 \minus{} 2bc(b \plus{} c \minus{} 1) \plus{} b^2 \plus{} c^2 \minus{} 1 \equal{} \minus{} 2(b \minus{} 1)(c \minus{} 1)(b \plus{} c) < 0$
Then OK![/quote]
then $ f(a)\leq max \{f(b \plus{} c \minus{} 1);f(b)\}$ why so :?:  :?:

we know $ \prod_{cyc}(a \plus{} b \minus{} c) \leq abc$
so the given inequality is $ \equiv$ to proving
$ \sum_{cyc}a^{2}\leq 2\prod_{cyc}(a \plus{} b \minus{} c) \plus{} 1$ ------------*

set,
$ a \plus{} b \minus{} c \equal{} u$
$ b \plus{} c \minus{} a \equal{} v$
$ c \plus{} a \minus{} b \equal{} w$
so,
the inequality * $ \equiv 8uvw \plus{} 4 \geq \sum_{cyc}(u \plus{} v)^{2}$
where $ 0 \leq u,v,w \leq 1$
set ,$ f(x) \equal{} (x \plus{} u)^{2} \plus{} (x \plus{} v)^{2} \plus{} (u \plus{} v)^{2} \minus{} 8uvx \minus{} 4$
$ f^{''}(x) \equal{} 4 > 0$
so f attains maximum in it's boundary .so we have to check just 4 values now
namely (d,e,f) where d,e,f are either 0 or 1.
so we have by checking max$ [f(u,v,w)] \equal{} f(1,1,1) \equal{} 0$
and we are done.

[quote="gauravpatil"]
$ f(a)\leq max \{f(b \plus{} c \minus{} 1);f(b)\}$ why so :?:  :?:[/quote]
Because graph of function $ f(a)$ is a parabol have shape $ \cup$ (efficient of $ a^2$ is $ 1 > 0$)
Hope you can understand my idea!

[quote="tdl"][quote="gauravpatil"]
$ f(a)\leq max \{f(b \plus{} c \minus{} 1);f(b)\}$ why so [/quote]
Because graph of function $ f(a)$ is a parabol have shape $ \cup$ (efficient of $ a^2$ is $ 1 > 0$)
Hope you can understand my idea![/quote]
i still cant
b and c are not constants and hence i still dont get the argument :?: :huh:  :wacko:
but i think i got the soln from your idea showing that the discriminant for the function is negative for numbers less than 1thanks :D

I am giving a solution which is quite similar to yours...(logically equivalent)
But this is for madness who wanted the proof of Corner Point Method....

Consider the scalar function
f(a,b,c) = a^2 + b^2 + c^2 - 2abc - 1

The constraint functions are:
a>=0,b>=0,c >=0,a+b-c-1 =< 0 , a+c-b-1=<0 , b+c-a-1=<0
Needless to say that we have three planes (actually six)
which form a closed region...

If a point of maxima is attained within the region
gradient f = 0
=> (a-bc)(i) + (b-ac)(j) + (c-ab)(k) = 0
which yields (a,b,c) =(0,0,0) or (1,-1,-1) or (1,1,1) and their permutations...

Its obvious that none of these points lie within the region(they lie on the boundary)...
A contradiction...

So maxima is not attained within the region but on the boundary....
(We are assured of the presence of a maxima by the continuity of 'f'...)
On the boundary,
if the maxima occurs on one of the planes then gradient is perpendicular to the plane...

(Rest part I will post in next post as this has become quite long)

[quote="gauravpatil"]
i still cant
[/quote]
Another way: With $ f(a)$ have $ 2$ roots $ a_1<a_2$ then $ f(a)<0\Leftrightarrow a_1<a<a_2$
In this $ f(b\plus{}c\minus{}1)<0;f(b)<0$ then $ b\plus{}c\minus{}1;b\in(a_1;a_2)$ then $ a\in(a_1;a_2)$, it mean $ f(a)<0$

[quote="tdl"]
Another way: With $ f(a)$ have $ 2$ roots $ a_1 < a_2$ then $ f(a) < 0\Leftrightarrow a_1 < a < a_2$
In this $ f(b \plus{} c \minus{} 1) < 0;f(b) < 0$ then $ b \plus{} c \minus{} 1;b\in(a_1;a_2)$ then $ a\in(a_1;a_2)$, it mean $ f(a) < 0$[/quote]
thanks a lot  :D  :lol: now i understand

We know that the normal to the planes are...
i + j - k , i - j + k , - i + j + k

So we have a set of equations like
a-bc = b-ca = ab-c and the permutations...(according to the normal)
This yields a=b, c=1 and its likes...

Also the corner points(intersection of two planes give similar points(a,a,1)
f(a,a,1)=0... So
maxima of the function is 0...

Actually we have are yet to check whether what we have got is maxima or minima...
This can be trivially done by choosing a point (a,b,c) and evaluating 'f' which is sure to give less than 0....
(check by yourself with a particular example....)

Hence we conclude that the maximum of f(a,b,c) is zero and it is attained iff
a=b,c=1 and its permutations

which gives our inequality.
a^2 + b^2 + c^2 <= 2abc - 1
(equality holds iff a=b,c=1 and its permutations)

Pardesi's boundary maximum approach is actually quite similar to what Bhargav did in the actual test. Another slick solution given by Shalex in the inequalities forum goes like this:
[quote]Assume that $ a \leqslant b \leqslant c$, with
$ b \plus{} c \leqslant 1 \plus{} a \leqslant 1 \plus{} b$
we have $ c \leqslant 1.$

1) If $ a \leqslant bc$, we get
$ (a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} (2abc \plus{} 1) \\
\equal{} a^{2} \plus{} (b \plus{} c)^{2} \minus{} 2(1 \plus{} a)bc \minus{} 1 \\
\leqslant a^{2} \plus{} (1 \plus{} a)^{2} \minus{} 2(1 \plus{} a)bc \minus{} 1 \\
\leqslant 2(1 \plus{} a)(a \minus{} bc) \leqslant 0$

2) If $ a > bc$ , we get
$ (a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} (2abc \plus{} 1) \\
\equal{} (a \minus{} bc)^{2} \minus{} (1 \minus{} b^{2})(1 \minus{} c^{2}) \\
\leqslant (b \minus{} bc)^{2} \minus{} (1 \minus{} b^{2})(1 \minus{} c^{2}) \\
\equal{} (1 \minus{} c)[b^{2}(1 \minus{} c) \minus{} (1 \minus{} b^{2})(1 \plus{} c)] \\
\equal{} (1 \minus{} c)(2b^{2} \minus{} c \minus{} 1) \\
\leqslant(1 \minus{} c)(2b \minus{} b \minus{} b) \equal{} 0$ [/quote]
 If I remember correctly, Arijit also gave a solution like this. Can anyone find a solution using trigonometric substitution or some geometric method?


IMO 2004 will take part in Greece in few days. Tonight, Greece Team, leading by King Otto and Captain Zagorakis will fight for title. I wish them success. And even if the lose the match, they are the heros and I hope greece mathematical team will show the best perfomance this year.

Namdung

Congrats to the Greeks here on the Euro victory. :)

My favorite is Czech but they lost.  :(

NO congrats to Greece  :P

Greece, the victory of humbleness and seriousness against vanity and presumption

I refer to Italy, a team of rich and presumptuous players.....a bad example for sport :(

Congratulation , Greece !!!!!!!!!!!!!!!!!!!   :)

I posted previous msg but I didn't realize to be log out.

Sorry

Yeah congrats Greece. I was in Athens a couple of days ago when Greece qualified in the Finals (meeting with Nickolas) and it was great ... the parade and all the people were gathering in the center of the city, it was really fantastic. I can only imagine what fiesta happened yesterday.

They will probably be still celebrating when we come up to Greece. :):)
Congrats!

We calculated our team leader must arrive to airport exactly when the greek footballists do.
It will be funny for him.

Unfortunatelly Greece is no longer land of maths but land of athlets

[quote="thatin"][quote="fredbel6"]I think you just broke the record of longest gap between two posts in one topic. :D[/quote]

are you refering to me? what do you mean?[/quote]
Well, yeah since you are the first person to post in this topic in TWO YEARS.
He means that yhou revived a dead topic and the gap between the last post and your post was, like, impossibly long.

[quote="thatin"]oh :blush:  is there any problem then?[/quote]
No probably not, since this is games and fun factory. But make sure you realize what you are posting in- it seems kind of odd how you are talking about how well a team will do from something that happened to years ago.
I would advise you not to revive topics in other forums, though.

i said nothing about the team! they were brilliant! i was just talking about the benefits athletes receive (which, in my opinion, they should have received!)


AMC AIME USA(J)MO USAMO inequalities USAMTS geometry

How do people go from doing AIME to USAMO or IMO? The USAMO and IMO are so much harder. Is it even possible to study for these?

You just do lots of old problems.  It also helps if you have someone to teach you basic techniques and theorems that are often used, such as basic inequalities for example.

[quote="topper"]Maybe there's a program somewhere you can take, plus there might be past problems.... Ask Mr. Rusczyk[/quote]

Although you don't [i]have[/i] to take a special class, to be sure.  Some people study on their own.

The single best strategy (and I can't say that it works, because I haven't done that well on USAMO in my one year of it, but still...) is to just do a lot of problems.  And read solutions after you're finished solving/trying to solve the problems, so that you can compare your work to the "official" work.  Of course, if your answers are slightly different, it doesn't mean they're wrong...it's just that you get a general idea of how one goes about answering olympiad-style problems by reading what other people have done to solve them.

funny, i find USAMO easier than AIME, probably because in USAMO i have 1.5 hours for each question and in AIME i have less than 15 minutes for each question!!

But bugzpodder, you're not from the US, right? Does your country have pre-olympiad short-answer contests as well?

COMC (Canadian Open Mathematics Competition) is used as the primary factor in determining the prospective students to participate in IMO.

Does CMO matter? I honestly don't know...

Tom, I dont believe that's the case.  To my knowledge, COMC (full short answer) and OLYMON (like USAMTS) is used to determine invitees to wintercamp (similar to MOSP in USA)  and high-scorers in COMC gets to write APMO and CMO, with ppl from wintercamp gets to write USAMO (well, unofficially because USA would not recognize the scores of anyone that is not an us citizen or immigrant)  the scores from these three olympiads determine the IMO team

We also have the Euclid contest, which is again, short answer (ie partial marks allotted) which is used for scholarship selection in Canadian universites i believe

we do have multiple choice tests in grade 9,10,11 but not grade 12, and they are not used to select IMO team.

Ah, yes yes.

Oh well, I guess I'm out of luck then. =|

This is a big step - I remember I was devastated the first time I looked at olympiad problems as a student.  Thought I'd never understand them, much less solve them.  It takes practice and patience.  We are developing courses to help students bridge this gap - next spring we hope to offer classes in geometry and inequalities, and our intermediate counting class also covers some of this area.

There are several decent books - the Zetiz book is very good, and the Engel book has a mountain of problems (can read more about them in our resources pages).  Also, anything by Andreescu is pretty good.  Drawback of Zeitz - very few solutions (but you have this board for that).  Drawback of Engel - pretty intimidating for the beginner.  Zeitz starts a good deal more slowly, I think.  Most of Andreescu's work that I've seen is more problem-oriented than text-oriented, and might best serve as a supplement to the books above.

Mostly, it's all about problems, and asking yourself when you read solutions 'How would I have thought of that in the first place?'  If you can't answer that question, then you haven't really learned the solution (and you should post here and one of us will help guide you to it).

It's also about asking questions - you have on this site a large stack of people who can answer your specific & general questions.  Use this resource.

[quote]Drawback of Engel - pretty intimidating for the beginner. [/quote]

That is true, at least from my experience. However, after taking in some of the sections slowly, you'll find that most of it isn't that bad at all, even for someone only done with AoPS 2 (actually, not even, still have a couple chapters).

I find Andreescu (Mathematical Olympiad Challenges) more helpful than Engel.. firstly because almost all the problems in the Olympiad book are accurate representations of what an Olympiad question is like as opposed to Engel, where the problems fall within a wide range of difficulty so that improvement is hard to measure.

Secondly, the problems in the olympiad book focus on very small sections of a topic that are repeatedly drilled until the reader becomes very familiar with the theme, while Engel is divided into huge chapters so it's easy to get lost and much harder to retain all of the information.

Both are good, though... Engel is somewhat easier, and focuses more on the strategies behind problem solving (invariance, extremal principle) as opposed to Andreescu, where entire sections are devoted to specific identities.

The problem with Andreescu/Gelca is that it deals with a lot of extremely specialized problems that almost never show up on olympiads. It's a great book if you want to work on tough problems on a very specialized topic, but I wouldn't recommend it as a full study course.

Zeitz is for beginners, serves as an intro of Olympiads, but the fact that it has no solutions is really disappointing
Engel is my fav Olympiad book (in general)
and Andreescu's book just plain sux -- waste of $$$

did someone say they had a solution manual?  Someone from mathlinks forum?

how did you go about getting it, and how can i go about getting it?

I thought someone said before that someone had it as a pdf from the mathlinks forum.

I got it from Paul Zeitz. I just asked him for it, and he gave it to me. I don't know if that is repeatable though.

Do you know Professor Zeitz personally?  I once asked him for the solution manual and he never responded....  :cry:

I do know him personally, but when I got it from him, I don't know if he knew who I was, although I had met him before.

is it an electronic copy?  if so can you upload it?

i could have sworn that someone said that someome from the mathlinks forum had an electronic copy...


conics ellipse geometry calculus integration derivative trigonometry

Is there a formula to find the "circumference" of an ellipse?  I know the formula for area, but don't know if there's one.

could you just write the equation of the curve and then use calc to get the length?  something like the integral of 1- the  derivative under a square root... don't remember that one offhand, though.

yeah, it's very easy to find the arc length of the graph
requires a little bid Calc...
i can post the proof though, if anyone interested.. :)

ok, here is one way:
assume you have an ellipse, with major axis of $2b$, and minor axis of $2a$, where $b>a$.
then the Parmetric equation for such ellipse is:
$x=a\cos{\theta}$
$y=b\sin{\theta}$
now the arclength ( the parameter) is find by:
$p=\int_{0}^{2\pi} \sqrt{a^2\sin^2{\theta}+b^2\cos^2{\theta}} d\theta$

maybe people posting here should know this  :roll:
[quote="DPatrick"]Please do not post calculus problems in this forum.

We have an entirely separate forum set up just for calculus:
[url]http://www.artofproblemsolving.com/Forum/index.php?f=66[/url]

Calculus does not appear on any of the high-school math contests.  It certainly is not an "Intermediate" high-school subject by any reasonable definition of the word "Intermediate".

Any posts involving calculus will be summarily moved to the Calculus Forum.[/quote]
just reminder thou, no offense, ok?  :gleam:

i posted just because one asked....
 :huuh:

[quote="amirhtlusa"]now the arclength ( the parimeter) is find by:
$p=\int_{0}^{2\pi} \sqrt{a^2\sin^2{\theta}+b^2\cos^2{\theta}}\, d\theta$[/quote]
That is correct, and that integral could be manipulated into any of several equivalent forms. Numerical methods work unusually well on this integral. However, unless $a=b$ (circle) or $ab=0$ (degenerate case), this integral cannot be expressed in closed form using the usual elementary functions.


Let $n\in N^*$ and the real numbers $x_k,\ k\in \overline {1,n}$ for which $\sum_{k=1}^n x_k =na$ and $\sum_{k=1}^n x^2_k =nb.$

Prove that for any $k\in \overline {1,n}$ there is the relation $ |x_k-a|\le \sqrt {(n-1)(b-a^2)}.$

i think the bound Arne gives is stronger.
In fact by cauchy schwarz inequality
$\sum_{j <> i} (x_j-a)^2 \ge \frac{1}{n-1}(na-x_i-(n-1)a)^2 = \frac{1}{n-1}(x_i-a)^2$
which gives the desired result.


AMC AIME MATHCOUNTS trigonometry calculus integration

This tournament will consist of 4 rounds. People will be in teams of 4. The first two rounds will be about 8-10 questions each, and after these two rounds are done, roughly 1/2 of the teams will advance to the next round. The 3rd and 4th rounds will be relay, and I will post more details about them when the time comes.
Rounds 1 and 2: Teams will be given a set of problems, and each member of the team will submit solutions to the problems. [b]A problem will be counted as wrong automatically if no solution is given, and no partial credit will be given, except on the last problem in each round. Absolutely no communication between the members of a team is allowed.[/b] The last problem in each round will require a proof, and will be worth 7 points, while the others are worth one point.

Deadline to sign up is [b]Thursday, March 19, at 5 PM PST (8 PM EST)[/b], but I might extend it.

Teams Joining:

Currently No Team:
funtwo

Team A:
AIME15
BOGTRO
pythag011?
ghjk?

Round 1 and 2 will be about AMC 10-12 level, maybe a little harder. The relays will be something like AIME problems 1-8.

Hai.

MATHCOUNTS concepts right?  Or is there trig/logs/that stuff?

Anyway, join.

I'd like to be with pythag011 and Bugi if possible :D

There might be a few problems involving trig/logs, but most of them don't involve them. (Only 1 problem per round at maximum will involve trig or logs.)

EDIT:

Change "I'd like to be with pythag011 and Bugi if possible "

to

"I'd like to be with pythag011 and Bugi and ghjk if possible"

:D

ghjk team?

wth i dont wanna be w/ bogtro

jk

that's ok

SO

[b]team a:[/b]

aime15
bogtro
(hopefully) pythag
(hopefully) ghjk

ill play.
does that mean i have to start another team?

I'm in.

Request: funtwo, Bugi, nikeballa96/FantasyLover

Teams Joining:

Team A:
AIME15
BOGTRO
pythag011?
ghjk?

Team B:
isabella2296
nikeballa96
funtwo
dragon96

Team C:
Mathmaticious

Actually, no pythag? (@ bogtro)

mieh he doesn't like to do tourney probelms

ill be on team B.
BTW, i will be gone march 19-23, os that ok?

i guess ill be on Team C to complete the 4th if thats ok with everyone else on that team?

Teams Joining:

Please Select a Team Name:
AIME15
BOGTRO
ghjk
pythag011

This. Is. How. We. Pwn.:
isabella2296
nikeballa96
funtwo
dragon96

Team C:
Mathmaticious
007math
yaofan
ReseesPieces

Deadline to sign up is over. Round 1 will be posted before Saturday.

I like our team name.  Let's keep it at:

Please Select a Team Name

or i suppose the

We Make Trivial Errors On AIME

Here is round 1.

Round 1:
This round consists of 7 problems, and each problem will be worth 1, point, except the last, which is worth 7 points. Please include a solution for problems 1-6, and proof for problem 7. Calculators are allowed, but not necessary.

1. Let ABC and ABD be two triangles such that angle ACB = angle ADB. Suppose E is the intersection of AD and BC, and AE=10, ED=2,and CE=4. Find BE.

2. Suppose p, q are primes. How many integral values of x and y are there such that $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{pq}$?

3. Find the last three digits of $ 57^{799}$.

4. What is the smallest positive real number N, such that $ \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} ... \plus{} \frac {a_999}{a_1000} \plus{} \frac {a_1000}{a_1}\le N$, for any real $ a_1,...,a_1000$?

5. In the figure below, one circle has radius 6 and the other has radius 8, and the distance between the centers is 12. Also, Q, P, and R are collinear, and QP=PR. Find PR.  [geogebra]fbdc55ea9ec89728b21e3c444a4809f4c2f43117[/geogebra]

6. A 5-digit number abcde is called good if a=1(mod 3), b=2(mod 3), and c=d+e(mod 3). Find the number of 5-digit good numbers.

7. Let $ a_1, a_2,...$ be a strictly increasing infinite positive integer sequence. Prove that there are infinitely many $ a_s$ such that $ a_r \equal{} p\cdot a_s \plus{} q\cdot a_t$,where p and q are positive integers.

Deadline is Sunday, March 29. PM me solutions. Sources will be posted once the round is over.

I don't think #4 has a maximum value. Just let $ x_1$ approaches infinity, $ x_2\equal{}1$, then your problem is unreasonable. Should you switch the sign and add more restrictions between $ x_1$,$ x_2$,...,$ x_n$?

oops, sorry, here is the correct problem 4:

What is the largest positive real number $ N$ such that $ \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} ... \plus{} \frac {a_{999}}{a_{1000}} \plus{} \frac {a_{1000}}{a_1}\ge N?$, if $ a_1,...a_{1000}$ are all positive real numbers.

Deadline extended one week, since only one person submitted so far. New deadline is next Sunday. If no one else submits, this tournament has to be canceled.

This tournament is canceled due to lack of submissions.

hee hee haa haa lack of submissions i'll submit if you explain the problem more clearly, so please do in any future tournaments you run

You know you just revived a really old post and it is canceled anyway.

wow... 22 day old topic. you didn't read powerofpi's post...  don't revive old topics

[quote="Farmstate"]Yaofan your name is stupid.[/quote]
[i][u][b]PLEASE STOP SPAMMING[/b][/u][/i] (BTW, what if that is his/her real name?)


inequalities
calculus
derivative
integration
real analysis
real analysis unsolved

let $ f: [0,1]\rightarrow \mathbb{R}$ be continiously derivative and $ f(0)\equal{}0$ prove:
$ \parallel{}f\parallel{}^2\le \int_0^1(f'(x))^2dx$

(which $ \parallel{}f\parallel{}\equal{}\sup{|f(t)|: 0\le t \le 1}$)

An other one : $ \sup_{0 < t \leq a }  |\frac {f(t)\minus{}f(0)}{\sqrt t}|^2 \leq \int_0^a (f'(x))^2dx$
 :cool:


See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=239265]this thread.[/url] Please do not post unnecessary topics in the forum.  :D


Let R is a principle ideal domain. Is projective R-module which has finite number of generators, a free module? Give an example...

It is [b]false[/b].  I'll give you the  example and you should check all the properties  :wink:

- Define $ R\equal{} \mathbb{Z} \oplus \mathbb{Z}$; then, $ R$ is a PID since $ \mathbb{Z}$ is.

- Define multiplication operation by $ (m,n) \cdot k \equal{} mk$.  Then, $ \mathbb{Z}$ is an $ R$ module.

- $ \mathbb{Z}$ is finitely generated and projective (as an $ R$ module) but it is not free.

Your example isn't a PID: it isn't a domain.
In fact, the statement is true.


something is prime if it has only 2 factors, one and itself

with that known, the numbers we are looking for can't be the product of two numbers if one isn't 1
ex: the number we are looking for can't be 5 times 7 because neither is 1

anyway
let's write out 30!
30*29*28*27*26*25*24*23*22*21*20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

let's think about it
since we won't get a prime if we multiply two numbers, then our answer is within these numbers

let's count the number of primes in this list

29, 23, 19, 17, 13, 11, 7, 5, 3, 2

there are 10 primes, so our answer is 10

Actually all you have to do is count the number of primes from 1 to 30... Because, in 30!, every number less than or equal to thirty is a divisor...

So the answer is 10...

Note: i am only 99.99 % sure... there is still a possibility that i'm wrong... And the reason would be.. "I don't understand the problem"... Because, if the way I understand it is correct, my answer is surely correct... :|

Yes, you are both right.

In general, the number of different prime factors of $ n!$ is the number of primes less than or equal to $ n$.

You didn't waste my time, have too much time on my hands, but i don't  :lol:
Anyways, if you prime factored it, the remaining numbers are 2, 3, and 5. 2 times 3 times 5 is 30, so that means if you pair them, you can get numbers that multiply up to 30. But there are also decimal ones, so then you just look for all the prime numbers up to 30. So yeah...

Yeah...in some problems you just have to recognize their transparency; you would not expect them to expect you to mulitply $ 30!$ out.  So, you look for the factors by writing $ 30!$ out which, when you do, makes you see immediately the answer.  Common sense is part of the game sometimes.


Find all solutions to: $\sqrt{x+5} - \sqrt{x-2} +1=0$

Indeed, now replace the "+1" with a "-1", and answer the same question. ;)

Anyway, this should go to the Intermediate section I think.

[quote="pieterminate"]Yeah, this isn't Olympiad, hardly unsolved.

Now, for the 2nd one, I found $x=11$ after squaring a lot.  And looking at the squares for differences of 5-(-2)=7 gave 16-9=7 as the only possibility.[/quote]
You're right. (On both statements  :P )

[quote="pieterminate"]None since $\sqrt{x+5}-\sqrt{x-2}>0$ :)  :roll:[/quote]

Ohh thanks....

I think: Find all solutions to: $\sqrt{x+5} - \sqrt{x-2} =1$.
It gve $x\ge 2 \ x+1=\sqrt{(x+5)(x-2)}\Longrightarrow x=11.$

[hide]Since $(x+5)-(x-2)$ is always 7 and $\sqrt{x+5}-\sqrt{x-2}=1$ is given, divide the two equations to get $\sqrt{x+5}+\sqrt{x-2}=7$.

Adding this last equation to the original one and dividing by 2, $\sqrt{x+5}=4\Rightarrow x=\boxed{11}$.[/hide]


calculus
integration
function
derivative
algebra
polynomial
real analysis

Hi guys, I haf some maths qns here. Hope all the maths expert can help me
may i know

1) What is the purpose of differential eqns??
2) How many types of differential eqns out there?? I heard of ODE, PDE, LP etc
2) Given a 2nd Order Differential Eqn like this

d^2y / dx^2 - 4y = 3e^-(2x)

how come I derive the Particular integral ??

Thank u

The first question is too phylosophical to give a unique and complete answer too. The standard answer is that they naturally arise when one tries to describe almost any process in nature from celestial mechanics to heat conduction to the dynamics of rabbit and fox population in a forest to...

The second question is also hard to answer because there are many classifications based on various principles (how many independent variables, how many unknown functions, how many derivatives, how close to linear, are there any preserved quantities, and so on, and so forth).

The third question is much easier  :). Since the characteristic polynomial is $r^2-4$ and $-2$ is a simple root of the characteristic polynomial, one should look for a particular solution of the form $A x e^{-2x}$ where $A$ is some real number (to find $A$, just plug this expression into the equation).


function
calculus
integration
quadratics
complex numbers
algebra
algebra unsolved

Prove that there is no $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = x^2-2$.

sorry, i made a mistake... i've modified the problem.

More generally, you may prove that if $a,b,c$ are real numbers such that $(b-1)^2 > 4(ac+1)$ then there is no function $f $ from $R$ onto $R$ such that $f(f(x)) = ax^2 +  bx +c$.

If necessary, I will post a proof later when I will have some time.

Note that the result above is not best possible, but the proof of the following is much harder :
Let $g(x)=ax^2+bx+c$ where $a,b,c$ are real numbers and $a \neq 0$. Then :
- If $(b-1)^2 - 4ac \leq 1$ then for each integer $r \geq 2$ there exists a function $f$ such that $f^r = g$.
- If $(b-1)^2 - 4ac > 1$ then for each integer $r \geq 2$ there is no function $f$ such that $f^r = g$.

Moreover, if $a,b,c$ are complex numbers and $a \neq 0$, then for each integer $r \geq 2$ there is no function $f$ from $C$ onto $C$ such that $f^r (z) = az^2 + bz +c$.

Reference :
R.E. Rice, B. Schweizer, A. Sklar, 'When is $f(f(z)) = az^2 + bz + c?$', A.M.M., 1980, p.252-263.

Pierre.

(1) :( Bouh Pierre have a french flag but not me ! how have a flag ?

(2) :D  This exercice is generalisable

Rufus isaacs and Meger give $g(x)=\frac b{1+\sqrt{a}} +x\sqrt{a}$
with $g(g(x))=ax+b$

if f is a function C1 for x>=0
f(0)=0 f(x)>0 for x>0 and $|f'(x)-k| \leq \mu x^\lambda$
with $0<k<1$ and $\lambda >0$ $x_n=f_n(x)$ with $f_{n+1}=f\circ
f_n$

then a,b>0 exist b<1 with $0\leq x<a$ and $x_n\leq b^n x$

$g_n(x)=k^{-n} f_n(x)$ CVU and $g'_n$ CVU   $g_n $ have $g$ for limit

$g\circ f=kg$

for $t>0$ put $g(z)=tg(y)$ give a function $z=\phi_t(y)$

Then (calculs) exist h $h\circ h=f$

References
(1) Probleme agregation 1949 Problems d'analyse maitrises de math Par Lelong viallard  p 72-74
(2) Berge  (For Menger) Graphes and Hypergraphes p 37

(3) for agregation 1949 Calcul differentiel et integral par Delachet PUF p 53-61

hello 20 7 04 07h49  :)

I join a fichier in tex (but in french) generalize the question  :D

Hmmm... Scientific Workplace generates a bad tex and it is sometimes unreadable for those who do not have SW.
Moreover Vidiani, the day and the hour of the moment you sent your post do appear in the head of the post, so it is useless to mention them  :D

Come on, still growing up little beettle :D

Pierre.

in plus you can use tex on the board too :) with the classical dollar signs :)

Let's give a proof of the not best possible result but still more general than the one proposed (see my first post on this topic) :

Let $f,g$ be two real functions such that $f^2 = g$ (where $f^0 = id$, and $f^{n+1} = fof^n$.
Clearly, $f$ and $g$ commute.
Let $(S)$ be the system with unkwon $x$) $g^2(x) = x$ and $g(x) \neq x$.

It is not difficult to prove the following lemmas in turn:

Lemma 1 :
If $a$ is a real number which is a solution of $(S)$, then :
i) the numbers $a,f(a),f^2(a),f^3(a)$ are four pairwise distinct solutions of $(S)$.
ii) For each nonnegative integer $i$ and $r \in \{0,1,2,3 \}$, we have $f^{4i+r}(a) = f^r(a)$.

Lemma 2 :
If $a,b$ are two soloutions of $(S)$, let $A = \{a,f(a),f^2(a),f^3(a) \}$ and $B = \{b,f(b),f^2(b),f^3(b) \}$.
Then $A \cap B = \emptyset$ or $A = B$.

Lemma 3 :
Let $S$ be the set of the solutions of $(S)$. If $S$ is not infinite, then $|S| = 0$ mod[4].

Now, let's assume that there exists a function $f$ such that $f(f(x)) = ax^2 + bx + c = g(x)$ for all real numbers $x$, where $a \nq 0$ and $(b-1)^2 > 4(ac+1)$.

Then for all $x$, we have :
$g^2(x) - x = (g(x)-x)(a(g(x) - x)+ 2ax + b + 1)$.
Thus the system $g^2(x) = x$ and $g(x) \neq x$ is equivalent to $a(g(x)-x) + 2ax+b+1=0$ (*) and $g(x) - x \neq 0$.
In turn, (3) is equivalent to  $a^2x^2 + ab+1)x + ac+b+1 = 0$, which is a quadratic equation in $x$, with discriminant $ \Delta = a^2((b-1)^2 - 4(ac+1)) > 0$, thus it has two distinct real roots, say $\alpha , \beta$.
Moreover, if $x$ satisfies (*) and $g(x)=x$, then $x = - \frac {b+1} {2a}$, thus there is at most one such real number. It follows that in our case, $(S)$ has one or two solutions, which is a finite number but not divisible by 4, which contradicts the lemma 3. Hence result.

Pierre.

[quote="VIDIANI"]:(  :blush:  :maybe:  :(
I do not understand your lamentation (pIerre and Valentin)
My tex is good plain Tex ![/quote] what I was trying to say is that you need not attach a tex file, but simple post in tex directly, as you see people doing it above :)

:D But my tex is vere very long (6 pages) with specific macros ! :lol:

[quote="harazi"]Sorry guys, can anyone have mercy and make this readable? I mean, I cannot read it not because of the french but because of the error that appears when I'm trying to open the file. Thanks a lot.[/quote]Actually, it's Tex, not Latex. I attached a pdf.


vector
linear algebra
linear algebra unsolved

Let $ X$ be a finite set of vectors of a vectorspace $ V$, and $ \Omega$ the linearly independent subsets of $ X$.

Show that $ (X,\Omega)$ is a matroid.

finite dimensional vector space? Which definition of matroid are you using?

If you are using the first definition given in http://en.wikipedia.org/wiki/Matroid then you'd only have to check for the augmentation property. This follows from the fact that every basis of a vector space has the same cardinality.

[hide="Matroid"]If $ X$ is a finite set, and $ \Omega$ is a set of some subsets of $ X$. Then $ (X,\Omega)$ is called matroid if
1. If $ S\in\Omega$, then every subset of $ S$ is also in $ \Omega$.
2. If $ U,V\in\Omega$ and $ |U| < |V|$, then there exists $ v\in U\V$ such that $ U\cup\{v\}\in\Omega$
 [/hide]

Ok. So 1) is trivial. For the second, note that every set of linearly independent vectors can be extended to a base. If 2) wasn't satisfied then $ U$ is a basis. But $ |V|>|U|$ guarantees a basis larger than $ U$ which gives a contradiction.


trigonometry
induction
inequalities proposed
inequalities

Consider a,b,x,y positive real numbers.If a>30 and b>30 and -1<=x,y<=1 and p(n) is the number of primes from 0 to n proove that:
     \[ p(a+b+x+y)<\sqrt{\frac{a^2+b^2}{2}}+\frac{x+y}{2}-5x\sqrt{1-y^2}-5y\sqrt{1-x^2}  \]
                                                                                Luca V.Iliesiu

$\sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}2 \ge 30$
$\frac{x+y}2 \geq -1$
Now for the other we can assume $x=cos(s)$ and $y=cos(t)$ therefore the other part of the right side becomes $5x\sqrt{1-y^2}+5y\sqrt{1-x^2}=5[sin(t)cos(s)+sin(s)cos(t)]=5sin(s+t) \leq 5$ therefore RHS $\geq 34$ and if we prove that the left side is $\leq 34$ then the problem is solved..I think..I may have done some mistakes..if so,then excuse me,but I am in a hurry,because I have homework to do :blush: If some mistakes are made,I will corect them as soon as I can.

Of course the left side is arbitraly big.

Yes, Luca, please correct the problem.It isn't correct in this form.

You may want to use $\pi(n)<\frac{1.25506n}{\ln(n)}$.

The problem is ok...Try to think about it...Hint:use mathematical induction

I see no body gave a solution...I'll give another hint(the most important one)...
If n a positive integer and n>55 then:
\[ \pi(n)<\frac{n}{2}-11  \]
P.S:This can be demonstrated by MI


How many perfect squares less than 100 have a tens digit of 6?

List perfect squares:
1,4,9,16,25,36,49,64,81.
The only perfect square with tens digit of 6 is 64.
The answer is 1.

Quicker way: Obviously 8^2=64.  Because 7^2 and 9^2 are both out of 6x, there is only one solution.

THe only perfect square the fits the example has a tens digit of 6 and only 64 is a perfect square in that category.


Suppose you have a triangle ABC in three dimensional coordinates.  Define the vectors $ \vec{a}$, $ \vec{b}$, $ \vec{c}$ (eh...I don't know how to get arrows on top), as Origin to A, Origin to B, Origin to C.  Which points can be represented as $ d_1\vec{a} \plus{} d_2\vec{b} \plus{} d_3\vec{c}$ for some combination of $ d_1$, $ d_2$, $ d_3$, which are all positive reals (a,b,c are the vectors, not scalar quantities)?  Can every point inside the triangle be represented as such?  Is this representation unique?

Edited in vectors :).

about the arrows, there is such a page that help a lot in LaTex: http://www.mathlinks.ro/Wiki/index.php/LaTeX:Symbols

About the problem i don't understood how we can get aa point with sum of the scalar vectors... :roll:

Well, not scalars, sorry, I meant that the $ a$, $ b$, $ c$ are vectors (pretend they have the arrow above them).

edit: arrows there now.

Three vectors are linearly dependent in $ \mathbb{R}^2$, so what do you think?

The condition that makes this idea nice is to specify $ d_1 \plus{} d_2 \plus{} d_3 \equal{} 1$; the keywords you want are [url=http://en.wikipedia.org/wiki/Convex_combination]convex combination[/url] and [url=http://mathworld.wolfram.com/BarycentricCoordinates.html]barycentric coordinates[/url] (otherwise known as - wait for it! - mass points).


Though in imaginary math world, Russia somehow sent [hide]4[/hide].

[hide]Israel sent 6-x delegates

Russia sent x delegates

U.S. sent 7-x delegates

Palestine sent 15-[(6-x)+x+(7-x)]=15-(13-x)=2+x delegates.

If Israel sent 4, x would equal 2 and Palestine would also send four, which can't be.

If the U.S. sent 4, x would equal 3 and both Israel and Russia would send 3, which can't be.

If Palestine sent 4, x would equal 2 and we've been over that already.

If Russia sent 4, nothing crashes or burns. Russia sent 4 delegates.[/hide]

MCrawford wrote:A total of fifteen delegates from Israel, Palestine, Russia and the United States meet at a conference.  Each of these countries has sent a different number of delegates (with each country sending at least one delegate).  Israel and Russia have sent a combined total of six delegates.  Russie and the United States have sent a combined total of seven.  One country has sent four delegates.  Which country was it?

[hide]i = number of delegates from Israel

p = number of delegates from Paelstine

r = number of delegates from Russia

u = number of delegates from the U.S.

i+p+r+u = 15

i+r = 6

r = 6-i

r+u = 7

6-i+u = 7

u = 7-6+i

u = 1+i

i+i+1 = 7

2i+1 = 7

2i = 6

i = 3

r = 6-i

r = 3

u = i+1

u = 4

i+r+u+p = 15

3+3+4+p = 15

p = 5

So the answer is the U.S.

[/hide]

h_s_potter2002 wrote:i+i+1 = 7
2i+1 = 7
2i = 6
i = 3

Spoiler:

[hide]But i+r=6 so then r=3.  But we can't have two numbers the same so r can't equal 3 also.[/hide]

MCrawford wrote:A total of fifteen delegates from Israel, Palestine, Russia and the United States meet at a conference.  Each of these countries has sent a different number of delegates (with each country sending at least one delegate).  Israel and Russia have sent a combined total of six delegates.  Russie and the United States have sent a combined total of seven.  One country has sent four delegates.  Which country was it?

[hide]I + P + R + U = 15

I + R = 6

R + U = 7

Therefore U = I + 1

I + P = 15 - (R + U) = 15 - 7 = 8

So P = R + 2

If R = 4, then P = 6, I = 2, and U = 3, which satisfies the given equality.

If U = 4, then R = 3, P = 5, and I = 3, but then I = R, which is impossible.

If P = 4, then R = 2, U = 5, and I = 4, but then I = P, which is impossible.

If I = 4, then U = 5, R = 2, and P = 4, but then I = P, which is impossible.

Hence, Russia sent four.

[/hide]

[After comparing answers]: looks like everyone favors the U.S. sending 4 for some reason.

[quote="LynnelleYe"]Currently it's two for the U.S. and four for Russia. The problem is I can't find the flaw in h_s_potter2002's reasoning, even though it produces an incorrect result. Weird...[/quote]

I don't see how he got to i+i+1=7...  this is probably where the flaw occured.

Yeah, that's what's it.  I tried to find a way where you solve one of the variables for another variable, but everything kept cancelling out and stuff, and I ended up getting stuff like u=u.


I noticed that there isn't a button to press to stay logged into the usamts site. Is this on purpose? Not that it bothers me, i'm just curious.

I've been wondering that too...

I don't think there is one, but then again, there's not that much point.


A 3-dimensional cross is made up of 7 cubes, one central cube and 6 cubes that share a face with it.
The cross is inscribed in a circle with radius 1. What's its volume?

If we denote the side by $s$, then we get the relation

$(\frac{s}{2})^2+(\frac{s}{2})^2+(\frac{3s}{2})^2=1$.  Solve for s.

Sorry, I think I hit submit too quickly.

Therefore $s=\frac{2}{\sqrt(11)}$,

and $V=7s^3=\frac{56}{11\sqrt{11}}$


geometry
3D geometry
MATHCOUNTS
probability
Pythagorean Theorem
algebra
difference of squares

[quote="Zenith Xan"]NOTE: A calculator is not needed for any of these problems, but probably recommended for at least one of them.

1. Two people are assigned two different random positive integers. One says to the other "My number is in the Fibonacci sequence and it's a two digit prime!" the other says "What coincidence! Mine is too.". Find the sum of their numbers.

2. How many more 0's does $4269!$ have at the end than 1 googol ($10^{100}$)?

3. The supplement of an angle is 5 times the complement of the same angle. Find the measure of the angle.

4. Arrange the following numbers from greatest to least: $10^{30},2^{100},3^{70}$

5. I start standing at point A. I go 3 miles directly west, then 5 miles directly north, then 5 miles directly west, then 20 miles directly south to point B. How many miles am I away from point A now?

6. What is the sum of all positive integers less than 100 with an odd amount of divisors?

7. Compute: $25007^2-24993^2$

8. What is the product of the solutions to $x^2-5544x-90119=0$?

9. Compute:

$\displaystyle\sum_{n=1}^{17} (n^3+n^2+n+1)$

10. What is the number of divisors of $2005*2006*2007$?

[hide="Answers - Check when done"]1. 102
2. 964
3. $\frac{270} {4}$ or $67.5$
4. $3^{70},2^{100},10^{30}$
5. 17 miles
6. 285
7. 700,000
8. -90119
9. 25364
10. 192[/hide][/quote]
[hide="1"]

Haha. $13$ is the first one, $89$ is the second. The sum is $102$.
[/hide]
[hide="2..? question"]
Hmm... i got $763.$ I got $4269/5=853$ rounded down. You add 10 for the extra powers of 5. There are 100 zeroes after the googol. 863-100=763.[/hide]
[hide="3"]$180-x=5(90-x)$. $x=67.5$[/hide]
[hide="4"]Ugh. THis one i used a calc. $100^{30}$ ? $2^{100}$. Taking logs of both sides, we see that $2^{100}$ is larger. DOing the same to the power of three, the $3^{70}$ is larger. $3^{70}, 2^{100}, 100^{30}$[/hide]
[hide="5"]THe net length west is 8, and the net length south is 15. the pythagorean theorem makes the distance $17$. [/hide]
[hide="6"]These are just the square numbers. They add up to $285$. [/hide]
[hide="7"]$25007^2-24993^2=(25007-24993)(25007+24993)=14(50000)=\boxed{700000}$[/hide]
[hide="8"]THe product is $\frac{c}{a}=\boxed{-90119}$.[/hide]
[hide="9"]I'm sure there's a better way to do this than use a calculator. Does it have anything to do with bases? How do you sum two numbers in different bases without actually converting? if that has anything to do wiht this....  :roll: [/hide]
[hide="10"]$2005=401\times5, 2006=2\times17\times59, 2007=3^2\times223, 2005\times2006\times2007=401\times5\times2\times17\times59\times3^2\times223.$ BUt the the answer is less than 64...argh. [/hide]

[hide="Hints/comments for ch1n353ch3s54a1l"]#2 would be much easier just to do with the formula, but you should still be able to count the extra 5's

#4 you didn't need a calculator.

$2^{100}=1024^{10}$ and $10^{30}=1000^{10}$ Notice the difference. So $2^{100}>10^{30}$

$3^{70}=9^{35}$ and $2^{100}=8^{33\frac{1} {3}}$

#9 is multiple formulas. You can do it without a calculator, I'm just lazy.

#10 you misread the question I think[/hide]

[quote="ch1n353ch3s54a1l"]Thanks :)

There's a formula for #2? Can you tell me what it is?  :)

[hide="9"]There are 17 ones. the sum of the Ns up to 17 are $\frac{n(n+1)}{2}$ and that is 153. The sum of squares formula makes the sum of the squares 1785, for the cubes, the sum for the first few cubes makes a pattern, and by the pattern, the sum of the cubes to 17 is 23409. 23409+1785+153+17=25364 [/hide]

still don't EXACTLY get #10.. but...yeah i'll figure it out eventually (i hope)  :P[/quote]

The formula is listed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433323#p433323]here in the formula topic.[/url]

#10's meaning is asking how many divisors of that number are there? (for example, 6 has 4 divisors (1,2,3,6)

There's a formula for that one too.

Just in case you didn't want to look through 8 pages of stuff,
[quote="ccy"]# of 0s at the end of $n!$ is $\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{5^2}\right\rfloor+\left\lfloor\frac{n}{5^3}\right\rfloor+...$[/quote]

For number 10, here's a hint:
[hide=":) Click Me"]
$2005=5\cdot401$
$2006=2\cdot17\cdot59$
$2007=3^2\cdot223$
[/hide]

Yeah, I got that far, and then I got this weird formula which really didnt' work.........I got somewhere in the hundred billions I think... :roll:

[hide="1"]The two numbers are 13 and 89, just go through the Fibonnaci numbers, it won't kill you. Their sum is 102.[/hide]
[hide="2"]Too lazy to explain how to do it right now, but the formula's probably been posted under Formulas. 4269! has 1064 zeros, $10^100$ has 100, answer is 964.[/hide]
[hide="3"]Write up the equation 5(90-x)=180-x. Solve for x and you get 45 degrees.[/hide]
[hide="4"]$3^70,2^100, 10^30$?[/hide]
[hide="5"]Make a right triangle with legs 8 and 15, solve for hypotenuse AB and you get 17.[/hide]
[hide="6"]All the positive integers with an odd amount of divisors are perfect squares, so the sum of all the perfect squares less than 100 is 236.[/hide]
[hide="7"]Factor as a difference of squares:(25007-24993)(25007+24993)=14*50000=700000.[/hide]
Don't have time to post the rest right now. I'll add them to this  post later.[/hide]

[quote="ch1n353ch3s54a1l"][quote="JesusFreak197"][hide="2"]$4269!$ has 1064 $0$'s at the end, so it has $\boxed{964}$ more $0$'s than a googol.[/hide]

[hide="10"]$2005\cdot 2006\cdot 2007=2\cdot 3^2\cdot 5\cdot 17\cdot 59\cdot 223\cdot 401$, so we have $2(3)(2)(2)(2)(2)(2)=\boxed{192}$ factors.[/hide][/quote]

Wait, sorry. For #10 how did you get $2(3)(2)(2)(2)(2)(2)$? Someone told me the formula was like $(a_1+1)(a_2+1)...(a_n+1)$... :| :dry:[/quote]
It is, but the a's are the powers of the prime factors. The proof is really easy, I encourage you to try it.

[quote="ch1n353ch3s54a1l"][quote="JesusFreak197"][hide="2"]$4269!$ has 1064 $0$'s at the end, so it has $\boxed{964}$ more $0$'s than a googol.[/hide]

[hide="10"]$2005\cdot 2006\cdot 2007=2\cdot 3^2\cdot 5\cdot 17\cdot 59\cdot 223\cdot 401$, so we have $2(3)(2)(2)(2)(2)(2)=\boxed{192}$ factors.[/hide][/quote]

Wait, sorry. For #10 how did you get $2(3)(2)(2)(2)(2)(2)$? Someone told me the formula was like $(a_1+1)(a_2+1)...(a_n+1)$... :| :dry:[/quote]

We did this at a mathcount practice... i ll try to explain to you the best i can

lets use 24 for example

as you know... its prime factoration is 2^3*3

Now basically you just get the power of the distinct prime factors... and add one, and the total product is how many factors you can get

for example 2 has a power of 3, so make it 4. 3 has a power of one, make it 2. 2*4=8 factors!

This is a probability thing, you can use the primes to create factors. In this example, you can use the "2" 4 times. Because you can use it once, twice, three times, or not use it at all. Same thing applies to the 3, use it or not use it. Then you get all the possible outcome, which is 2*4=8

sorry i m bad at explaining stuff :huh:

Ohh...I guess I just misunderstood the guy who told me... It didn't seem likely anyways  :P Thanks  :)

[hide="5"][size=150]17.[/size][size=100] i drew it and used pythagorean theorem.[[/size][/hide]

I don't get the notation for number nine. what does the 'e'-looking symbol mean? why are there numbers above and below it. It may just be a computation problem but I can't solve it if I don't know what it means can I?

The $\sum$ symbol means sum.

For example: $\sum_{n=1}^{5}n$ means to start at $n=1$ and go to $n=5$, adding the value $n$.

So the example $\sum_{n=1}^{5}n$ would become $\sum_{n=1}^{5}n=1+2+3+4+5$.


I know where Drunken_Math got the error: tan 60 doesn't equal $\frac{\sqrt3}{3}$

[quote="shadysaysurspammed"]This sum is too easy to be in here,this should have been in moderate or gettin started :rotfl:[/quote]

I don't know who moved it here  :?


Hi FishDragon;
  It is not that hard. I gave you my very long sols.
$(1) (x-1)(y^2+6)=y(x^2+1)$
$(2)(y-1)(x^2+6)=x(y^2+1)$
 We have (1)-(2) and found that $x=y$ or $x+y=2xy-7$
1.$x=y$ It is easy. Right
2.$x+y=2xy-7$ . Just (1)+(2) . You will solve $(xy)$ immediately. And it is easy too
Bye


[quote="thanhnam2902"]Let $ ABC$ is a triangle. Prove that: ${ \boxed{p^2\ge 16Rr - 5r^2}}$[/quote]
It's equivalent to $ \sum_{cyc}(a^3 - 2a^2b - 2a^2c + 3abc)\leq0,$
which is Schur for $ a = y + z,$ $ b = x + z$ and $ c = x + y.$ :wink:

it is equivalent to $ IG^2\geq 0$  :wink:


Anyone else that is beastly at the cello playing this song? Can I have a few tips? :D

Make a few attempts at playing the Suite 6 Prelude just to give you an appreciation of how easy Suite 1 is?

(Disclosure: I'm a violist, not a cellist - but we do see transcriptions. And I can't come anywhere close to playing Suite 6.)

Tips?

Practice...

Actually, listening to recordings of it helps.

Try strumming the chords blocked.

If I were you, start on the other movements of the suite. In my opinion, the level of difficultly (from least to greatest) goes as follows:
*Minuet
*Giuge (Or is it Guige?).
*Prelude
*Courante
*Allamande
*Sarabande

[quote="Dojo"]*Giuge (Or is it Guige?).[/quote]
It's "Gigue." As in "jig": a fast dance, usually in 6/8 time. In some other settings, a jig (or "gigue") may have dotted eighth - sixteenth - eight rhythms, but Bach doesn't do that, not in these suites.

Dojo is probably right about starting with the Minuet of Suite 1. Maybe after that, the next easiest movement (easiest as in quickest to a generally pleasing result) of the whole collection might be the Bouree from Suite 3 (the C major suite).

My personal favorite movement to play (and I can only play a few of them) is the Sarabande from Suite 5 (C minor). So very few notes - and each note carries so much musical context with it. Bach as an abstract minimalist.

I'm only playing the prelude for a concert. I have a bit trouble doing the dynamics with the part on the second page with the pattern:

f# a
d  a
e  a
f# a
g  a
...........

My private teacher told me to block them.

That's a good idea. I would also consider dropping off the A string and just practicing the D string notes. (Which should be everything except the repeated a).

Here's my fingering:

3  0
0  0
1  0
3  0
4  0
1  0
3  0
0  0
4  0
1  0
3  0
0  0
4  0
1  0
3  0
0  0
1  0
3  0
4  0
1  0 (Moves into 4th position.)
3  0
0  0
1  0
3  0
2  0 (Moves into 5th position.)
.....

Basically I never put a finger on the A string. It sounds pretty cool that way.

[quote="jonathanchou711"]Basically I never put a finger on the A string. It sounds pretty cool that way.
[/quote]It is absolutely the intended way. It just doesn't work right if you alternate between a fingered A string and an open A string, and the open string is definitely meant to be used.
Crossing over to higher notes on the D string that the A string feels weird at first; you have to override some of your instincts to make it work.

There's a word for this: [url=http://en.wikipedia.org/wiki/Bariolage]bariolage[/url].

(Since jmerry and I are both violists, your fingerings don't mean anything to us.)

Well I didn't really know that. Is Dojo a cellist?

I am a cellist.

That fingering is almost good, although I personally do:

(D string notes only, measure 33)

3
0
1
3
4
1 (4th position)
3
0
1
1 (5th position)
2
0
1 (6th position)
2
3
1
2
3 (4th position)
4
1
3
1
3
2 (2nd position)
4
2
4
1
4
3 (1st position)
4
1

Anyone have any references I can do to make my A strings softer or make my notes sound smoother?  :maybe:

I would practice synchopated rhythms. Make the A string note shorter. Also, try to "flick" your bow hand on your down bow, assuming you use down bow on the D string notes.

Can someone explain to me what makes the 1st Cello Suite demanding on bowed instruments? If find them easy to play on the guitar (there is a Lute transcription BWV 995 of the 5th Cello Suite, which is harder), so its seems the sound generation with a bow is more difficult?

On the contrary, the Violin Sonatas and Partitas are medium difficult up to very difficult on the guitar IMHO.

It may have to do with a technique that is easier for guitarists? The 1st suite is not as demanding as you may hear. Once you learn it, it becomes extremely easy.

The short answer:
[quote="jonathanchou711"]Well I've only been playing for a year.[/quote]
It's not hard. It's the easiest of the suites. You're also less likely to see #1 transcribed for guitar than some others, because it relies so heavily on open strings as self-accompaniment. When the strings are tuned in fourths, that doesn't work so well.

[quote]You're also less likely to see #1 transcribed for guitar than some others, because it relies so heavily on open strings as self-accompaniment. When the strings are tuned in fourths, that doesn't work so well. [/quote]

The lute transcription of the first cello suite suffers to some extent from the same problem for a similar reason, but I think the lute version sounds richer from the resonance of all the additional strings.  I have a recording of it by Nigel North, which sounds amazing.  It seems to be one of the easier Bach suites to play on lute, and so you see it more frequently than some of the others, especially the suites that Bach wrote specifically for lute, for example:

[quote]there is a Lute transcription BWV 995 of the 5th Cello Suite, which is harder[/quote]

As written, Bach's lute transcription of his 5th cello suite is unplayable on the lute of Bach's day, the 13 course Baroque lute.  The range is too low, there is an awkward chord in the Prelude, and an impossible one in the Allemande.  Even people who play Bach's lute suites [i]on the lute[/i] are working off of transcriptions!

So, then, what instrument did Bach intend it for?  A five-string cello?  (I'm sorry, I'm pretty naive when it comes to bowed-string instruments.)  With the lute suites, one possibility is that Bach composed them on/for a lute-harpsichord kind of instrument called the Lautenwerk, not a real lute, which would obviously have completely different technical capabilities.[/quote]

Yes, it was a five-string cello. I don't know if there are any of those still around, but there are a few five-string viola/violins out there; I've seen an electric version.

For another exotic instrument, the other piece on my music stand is a transcript of the Schubert [url=http://en.wikipedia.org/wiki/Arpeggione]Arpeggione[/url] sonata. [Random note: I clicked over to the cello article out of curiosity. It has some recordings as examples- and the very first one one the list is the prelude from suite #1.]

[quote="jmerry"]For that matter, the sixth cello suite was written for an instrument with five strings (CGDAE). It's playable without that string, but it does go very high at times.[/quote]
Not only is the position high, it creates an awkward thumb position. The thumb position is used to "simulate" the missing E string. The use of a E string would eliminate the need for thumb position.

[quote="jmerry"]	Yes, it was a five-string cello. I don't know if there are any of those still around, but there are a few five-string viola/violins out there; I've seen an electric version.[/quote]
There are a few five string electric celli as well. I also believe that, with enough money, you can ask for a custom made 5 stringed instreument.


integration
trigonometry
college contests

Let $f(x),a>0$ such that:$f'(x)\ge a \forall x \in \mathbb R$.Known that:
\[ 0<\int^{\frac{\pi}{2}}_0 f(x) \sin x dx <a\]
Prove that in $[0;\pi /2]$ the equation $f(x)=0$ has the unique root.

[quote="zhaobin"]prove f(0)<0[/quote]
I think that zhaobin wanted to proposed something like this:
$\int_0^{\frac{\pi}{2}}f(x)sinxdx=\int_0^{\frac{\pi}{2}}f(x)d(-cosx)=f(0)+\int_0^{\frac{\pi}{2}}cosxf'(x) \geq a+f(0)$ from where
it follows that $f(0)<0$.


Let a,b,c is real numbers such that a+b+c=0 .Prove that :
$ \frac{ab}{c^2}\plus{}\frac{bc}{a^2}\plus{}\frac{ca}{b^2} \leq\frac{\minus{}15}{4}$

[quote="tin_124"]Let a,b,c is real numbers such that a+b+c=0 .Prove that :
$ \frac {ab}{c^2} \plus{} \frac {bc}{a^2} \plus{} \frac {ca}{b^2} \leq\frac { \minus{} 15}{4}$[/quote]
For $ a\neq0,$ $ b\neq0$ and $ c\neq0$ we obtain:
 $ \frac {ab}{c^2} \plus{} \frac {bc}{a^2} \plus{} \frac {ca}{b^2} \leq\frac { \minus{} 15}{4}\Leftrightarrow(a\minus{}b)^2(2a\plus{}b)^2(a\plus{}2b)^2\geq0.$