Can someone tell me why this doesn't work?

$ \left( \begin{array}{cc}
a  & b \\
c & d \end{array}\right)
$

[quote="TripleM"]Can someone tell me why this doesn't work?

[tex]
\left( \begin{array}{cc}
a & b \\
c & d \end{array}\right)
[/tex][/quote]

It is the fault of the word processor you are using. It is converting the ampersand, "&", to HTML code, "amp;", before passing the information on to this forum's TeX processer. As you can see above, my web browser (Mozilla) gives an error-free result with your unaltered code. Check whether your word processor has a mode for plain text without HTML.

Erin Schram

[quote="Erin J. Schram"]

It is the fault of the word processor you are using. It is converting the ampersand, "&", to HTML code, "amp;", before passing the information on to this forum's TeX processer. As you can see above, my web browser (Mozilla) gives an error-free result with your unaltered code. Check whether your word processor has a mode for plain text without HTML.

Erin Schram[/quote]

Thats weird. I'm using firefox.. and the option here at AoPS for disabling html is checked.

$
\left(\begin{array}{cc}

a & b\\
c & d

\end{array}\right)
$

Try unchecking the disable HTML button.  I tried it with that box checked but it did the same thing that your's did.  I have it unchecked now and it is working. Btw, I am using IE.


modular arithmetic
trigonometry
number theory unsolved
number theory

a) Show that, for prime $p \equiv 1,3$ or $5\pmod  8$ the following equation holds:

$ \displaystyle \sum^{\frac {p-1}2}_{k=1} \tan\left(\frac {k^2\cdot\pi}p\right) = 3  \sum^{\frac{p-1}2}_{k=1} \cot\left(\frac {k^2 \cdot \pi}p\right)$.

b) Prove that the equation in a.) fails for all primes $p \equiv 7 \pmod 8)$.

In order to specify what's difficult and what's easy about this:

$ \displaystyle p=1,5\ (mod\ 8)\Rightarrow p=1\ (mod\ 4) \Rightarrow (\frac{-1}p)=1 \Rightarrow \sum^{\frac{p-1}2}_{k=1} tan(\frac{k^2\pi}p)= \nolinebreak 3 \sum^{\frac{p-1}2}_{k=1} cot(\frac{k^2\pi}p) = \nolinebreak 0$, so we're done.

I guess this leaves the cases $p=3,7\ (mod\ 8)$.

As to the case $p \equiv 3 (mod8)$,
  $\newline$
  We have $\big( \frac{-2}{p}\big)=1$. So
  \[\sum\limits_{k=1}^{\frac{p-1}{2}} \cot\frac{k^2\pi}{p}
  =\sum\limits_{k=1}^{\frac{p-1}{2}} \cot\frac{-2k^2\pi}{p}\]
  and with $\cot2\alpha=\frac{1-\tan^2\alpha}{2\tan\alpha}
  =\frac{1}{2}(\cot\alpha-\tan\alpha).$
  $\newline$
  We get \[\sum\limits_{k=1}^{\frac{p-1}{2}} \cot\frac{k^2\pi}{p}
  =-\frac{1}{2}\Big(\sum\limits_{k=1}^{\frac{p-1}{2}} \cot\frac{k^2\pi}{p}
  -\sum\limits_{k=1}^{\frac{p-1}{2}} \tan\frac{k^2\pi}{p}\Big).\]
  That is what we want.


logarithms
trigonometry
function
calculus
calculus computations

Does anyone know if either of these equations have an explicit elementary solution?

$ f'(t)=\log \left( \frac{1-f(t)}{f(t)}\right)$

$ f'(t)=\tan^{-1}\left( \log \left( \frac{1-f(t)}{f(t)}\right) \right), \; f(0)=1$

If not (likely), does anyone know where/how I can graph this?

I realize that both have a tendency towards zero, and what their [i]general [/i]shapes look like...

Also, does anyone know of any program that will draw out a curve based on a differential equation without necessarily solving for the function?

Thanks!

~ tZf

Both equations are autonomous, of first order. Look for the inverse function $ t=g(s)$, which exists locally, apart from exceptional initial values. In the first case it satisfies the equation $ g'(s)=1/\log((1-s)/s)$.


function
real analysis
real analysis unsolved

I know this theorem for $n=1$, but I need some help to prove it for general $n$:

If $f: \mathbb{R}^{n}\rightarrow\mathbb{R}$ is differentiable, for each $a,b\in\mathbb{R}^{n}$ show that there exist $c=tb+(1-t)a, t\in(0,1)$ such that $f(b)-f(a)=\nabla f(c)\cdot(b-a)$.

This is a consequence of the one-dimensional mean value theorem. Consider the function $g(t)=f(tb+(1-t)a)$ on $[0,1].$

We have $g(0)=f(a),g(1)=f(b),$ and, by the chain rule, $g'(t)=\nabla f(tb+(1-t)a)\cdot (b-a).$ Just apply the one-dimensional theorem to that.


Let $ABC$ be a triangle with orthocenter $H$. Let $H'$ be the symmetric of $H$ w.r.t. the midpoint of $AC$. Prove that $\measuredangle BAH' = 90^{\circ}$.

Let $O$ be the circumcentre and $M$ be the midpoint of $AC$. It is well-known that $BH=2OM$. Since $\angle{H'MO}=\angle{H'HB}$, and $H'H=2H'M$, so $\triangle{H'OM}\sim \triangle{H'BH}$. Therefore $B, O, H'$ are collinear, thus $\angle{BAH'}=90^o$.


function
calculus
integration
derivative
calculus computations

if f(x) is an odd function and g(x) is an even function then which of the following is true?

I. the integral from -a to a of f(x) dx = 0
II. the integral from -a to a of g(x)= the integral from 0 to a of 2 * g(x)dx
III. the integral from -a to a of f(g(x))dx=0

 :maybe:

other one:

if f(x) is an odd function then which of the following is necessarily true?

I. the integral from -2 to 2 of f(x) dx=0
II. the integral from 0 to 2 of f(x) dx is greater than 0 if a is greater than 0
III. the integral from 2 to -2 of f(-x)dx=the integral from 2 to -2 of f(x) dx

please see if you can answer this thank you...

[quote="sneezywalrus"]if f(x) is an odd function and g(x) is an even function then which of the following is true?

I. the integral from -a to a of f(x) dx = 0
II. the integral from -a to a of g(x)= the integral from 0 to a of 2 * g(x)dx
[/quote]

They are true  :)
[hide]$\int_{-a}^{a}f(x)dx=\int_{-a}^{0}f(x)dx+\int_{0}^{a}f(x)dx=I_{1}+I_{2}$

In $I_{2}$ you need let $t=-x$  :wink: [/hide]

[quote="N.T.TUAN"][quote="sneezywalrus"]is there a way you can help me with the second one as well? that would be much appreciated thank you.[/quote]

[hide]$\int_{-a}^{a}g(x)dx=\int_{-a}^{0}g(x)dx+\int_{0}^{a}g(x)dx=I_{1}+I_{2}$

In $I_{1}$ you need let $t=-x$  :wink: [/hide]  :huh:[/quote]

sorry but....what t? you say t should =x but where does t come up? so lost...  :huh:

[quote="N.T.TUAN"]We used change varible  :wink:[/quote]

...still lost...ugh

can you explain it once more sorry....please?

thank you once again in advanced....  :(

[quote="N.T.TUAN"][quote="sneezywalrus"]

...still lost...ugh

can you explain it once more sorry....please?

thank you once again in advanced....  :([/quote]

Ok, you see here http://mathworld.wolfram.com/DefiniteIntegral.html  :wink:[/quote]

but....why?

[quote="N.T.TUAN"]I am sorry. I cann't help you.  :(  Now, you need a text book on Real Analysis. And read it, Please  :)[/quote]

sorry to bother you but one last time can you jus fully answer the question? without an explanation....please? sorry.

Use the fact that if f(x) is an odd function, f(-x) = -f(x).  If g(x) is an even function g(-x) = g(x).

If F'(x) = f(x) (which means that F(x) is the antiderivative/integral of f(x)) and f(x) is odd, then F(x) is even:
f(-x) = -f(x)
integral of f(-x) dx = integral of -f(x) dx
-F(-x) = -F(x)
F(-x) = F(x)

This can be confirmed by taking the derivative of both sides.

Thus, F(a) - F(-a) = 0, which is equivalent to the integral from -a to a of f(x) dx = 0.

Similarly, if G'(x) = g(x) and g(x) is even, then G(x) is odd:
g(-x) = g(x)
integral of g(-x) dx = integral of g(x) dx
-G(-x) = G(x)
G(-x) = -G(x)

This can be confirmed by taking the derivative of both sides.


trigonometry
inequalities
limit
real analysis
real analysis unsolved

does the following series converge:

Sum[Abs[Tan[n]]/n^2,{n,1,Infinity}]

$ \sum _{n\equal{}1}^{\infty } \frac{|\tan
   (n)|}{n^2}$

$ \infty \equal{}\sum _{n\equal{}1}^{\infty } \frac{1}{n}\equal{} \sum _{n\equal{}1}^{\infty } \frac{n}{n^2}\leq\sum _{n\equal{}1}^{\infty } \frac{|\tan (n)|}{n^2}$

I'm not sure how you justify that last step; that inequality doesn't hold outside of the first period.

In any case, I doubt it.  It should be possible to show that this sequence fails the $ n^{th}$ term test by some equidistribution argument.

$ n\le|\tan n|$ is not true.

$ \tan n$ has an irregular structure; it's not too big most of the time but occasionally $ n$ will be a good approximation of $ \left(k\plus{}\frac12\right)\pi$ resulting in sporadic values of $ |\tan n|$ which are large. We'll have to investigate how often $ n$ is close to $ \left(k\plus{}\frac12\right)\pi$ and how close.

[quote="t0rajir0u"]It should be possible to show that this sequence fails the $ n^{th}$ term test by some equidistribution argument.[/quote]If I'm not mistaken, showing that $ \lim_{n\to \infty}\frac {|\tan n|}{n^2}\neq 0$ would prove that the irrationality measure of $ \frac {\pi}{2}$ is no less than $ 3$, which would be a very surprising result.


ARML
geometry
perimeter
ratio
geometric sequence

You may recognize this problem from the ARML problems of the month.  I worked it around 11 P.M. last night, so I'd like to see some other solutions to compare to.  In an effort not to be taken down for plagiarism or something, I made my own diagram to go along with it (which is why it doesn't look very good).

A spiral is formed by connecting the midpoints of adjacent sides of a regular polygon, shading a triangle, and continuing the process as shown. What is the perimeter of such a spiral constructed in a regular hexagon of side length 1?

[img]http://i19.photobucket.com/albums/b158/japanman2012/hexspiral.jpg[/img]

Sorry if my solution is wrong, I've not really bothered checking it, and like I said, I did it past my daily expiration date.

[hide]I divided the perimeter into two halves.  Both of them begin with a half side.  From there, you can tell easily that each corresponding sidelength from thereon will be congruent to it's counterpart on the other half of the perimeter.

From there it's a simple matter of geometry and infinite series (who would've guessed? :)).

So as we all know, in a 30-60 right triangle (formed by cutting one of the individual triangles in half), the ratio of the 60 side to the hypotenuse is $\frac{\sqrt{3}}{2}$, so we can make this the ratio in the infinite series:

$\frac{1}{2} + \frac{\sqrt{3}}{4} + \frac{3}{8} ...$

In other words, the sum for half of the perimeter is:

$\frac{\frac{1}{2}}{1-\frac{\sqrt{3}}{2}}$ or simply put:
$\frac{1}{2-\sqrt{3}}$  Multiply the numerator and the denominator by $2 + \sqrt{3}$ and you have:

$\frac{2 + \sqrt{3}}{1}$  Nice enough, eh?

So the whole perimeter is $4 + 2\sqrt{3}$[/hide]

Is that right?

[hide]I'm not sure, but I got $\frac{7+3\sqrt{3}}{2}$. Basically I used an infinite geometric sequence on two sides of the spirals and then added in the extra .5 length of the outer hexagon. I found the ratio to be $\frac{\sqrt{3}}{2}$ and then the first length of the first side as .5, then the second as $.25\sqrt{3}$. So then $\frac{.5}{1-.5\sqrt{3}}+\frac{.25\sqrt{3}}{1-.5\sqrt{3}}+.5=\frac{7+3\sqrt{3}}{2}$...[/hide]

are you sure you did your addition right at the end?  I put your 3-term string through and it came out to be $4 + 2\sqrt{3}$ at the end, as opposed to your final answer.  I'll punch it through my calculator to make sure I'm doing mine right though :).

Oops. Sorry, I did my calculation thingy by hand and didn't check afterwards... yeah, it's $4+2\sqrt{3}$


is anybody solving h.c verma problems here,if anybody then
can you please solve h.c verma ,newtons laws of motion,28th problem

and also ,the book, fundamentals of mechanics by irodov a good book
for understanding the concepts of mechanics?

does any body know ,is there any site for theory of constraint relation

It may be helpful if you post the problem here. Not every one has seen the text book you are talking about.

I dont know about the H.C.Verma problem, but that book by Irodov is a pretty good option.....the sums are somewhat closer to olympiad sums though.


Help me !!! :(  This's my exercise. Could you help me as soon as pssible?...Thanks

Find X so that:
2*X^4 -X^3 -3*X^2 -3*X +18=0

Oh it's very easy. Because:
$ 2x^4\minus{}x^3\minus{}3x^2\minus{}3x\plus{}18\equal{}0$ $ \Longleftrightarrow$ $ (x^2\minus{}3x\plus{}3)(2x^2\plus{}5x\plus{}6)\equal{}0$

OK!

[quote="thanhnam2902"]Oh it's very easy. Because:
$ 2x^4 \minus{} x^3 \minus{} 3x^2 \minus{} 3x \plus{} 18 \equal{} 0$ $ \Longleftrightarrow$ $ (x^2 \minus{} 3x \plus{} 3)(2x^2 \plus{} 5x \plus{} 6) \equal{} 0$

OK![/quote]

How did you get that?

You can find it by some way.
Example: Let $ 2x^4 \minus{} x^3 \minus{} 3x^2 \minus{} 3x \plus{} 18 \equal{} (ax^2 \plus{} bx \plus{} c)(dx^2 \plus{} ex \plus{} f)$
$ \Longrightarrow a, b, c, d, e, f \Longrightarrow 2x^4 \minus{} x^3 \minus{} 3x^2 \minus{} 3x \plus{} 18 \equal{} (x^2 \minus{} 3x \plus{} 3)(2x^2 \plus{} 5x \plus{} 6)$
Or you use maple and get so.

:o  :o woow..i written in my notebook  :) Thank you!

\[ \text{solve the system of equations}\] \begin{eqnarray*}
&a+2b=-1&\\
&\alpha+2\beta+ab=-3& \\
&\alpha b+\beta a=-3&\\
&\alpha \beta = 18&
\end{eqnarray*}
[hide] \begin{eqnarray*}
&\alpha b+\beta a=-3&\\
&\Leftrightarrow b\alpha^2+3\alpha+18a=0&\\
&\Leftrightarrow b\alpha^2+3\alpha-18(1+2b)=0&\\
&\boxed{\alpha,\beta=6,3}&\\
&\Leftrightarrow 2b^2+b-15=0&\\
&\boxed{a,b=-5,-3}&\end{eqnarray*}[/hide]  :D

[quote="igiul"]
\begin{eqnarray*} & a + 2b = - 1 & \\
& \alpha + 2\beta + ab = - 3 & \\
& \alpha b + \beta a = - 3 & \\
& \alpha \beta = 18 & \end{eqnarray*}
[/quote]

But your equations (the second and third) assume you already know where the factors of the coefficient of $ x^4$ are, specifically $ 2=2\cdot1$.

Wasn't it only that or $ 2^{\frac{1}{2n}}, 2^{1\minus{}\frac{1}{2n}}$?


Support
topology
real analysis
real analysis solved

Given a regular Borel measure $\mu$ on the compact Hausdorff topological space $X$ s.t. $\mu(X)=1$, show that there is a compact subset $K$ of $X$ s.t. $\mu(K)=1$, and $\mu(K')<1$ for all proper compact subsets $K'$ of $K$.

(a Borel measure is a measure defined on the sigma algebra generated by the open sets, and a Borel measure $\mu$ is called regular if for every measurable set $S,\ \mu(S)$ is $\sup\mu(K)$ after all the subsets $K\subseteq S$ which are compact in $X$, and also $\inf\mu(U)$ after all the supersets $U\supseteq S$ which are open in $X$)

Let $\mathcal{F}$ be the family of all open sets of measure zero, and let $S=\cup\mathcal{F}$. We want to show that $S\in\mathcal{F}$. $S$ is trivially open. For every closed $Y\subseteq S$, $Y$ is compact, and can therefore be covered by finitely many elements of $\mathcal{F}$. By the additivity of the measure, $\mu(Y)=0$. By regularity, $\mu(S)=0$. Now define $K=X/S$. Clearly, $\mu(K)=1$. For every closed $Y\subset K$, $X/Y$ is an open set not contained in $S$ and hence not a member of $\mathcal{F}$, so $\mu(X/Y)>0$ and $\mu(Y)<1$.

This did not use the full strength of either compactness or countable additivity, so we could weaken the conditions in two places. We could keep compactness and only require finite additivity in our measure, or we could keep countable additivity and weaken every instance of compactness to "closed, and every open cover has a countable subcover". This would allow $X$ to be a $\sigma$-compact space.


superior algebra
superior algebra theorems

Determine if the operation * gives a group.
Let $ *$ be defined on $ 2Z \equal{} {2n|n \in Z}$ by $ a*b \equal{} a \plus{} b$

Yes, it does.

0 is the identity.

Also, if n belongs to the set then -n belongs too.

Associativity follows immediately from the fact that $ (\mathbb{Z}, \plus{} )$ is a group in its own right.


analytic geometry
symmetry
ratio
geometry
trigonometry
probability
probability and stats

I was trying to find a proof of the following:

Prove the distribution of the quotient of two independent standard normal is a cauchy distribution.

Let $ X$ and $ Y$ be independent standard normal. Then the joint distribution of $ (X,Y)$ is a radially symmetric normal. In particular, it has density $ f(x,y)\equal{}\frac1{2\pi}e^{\minus{}(x^2\plus{}y^2)/2}.$

Now, convert $ (X,Y)$ into polar coordinate $ (R,\Theta).$

Note that because of the radial symmetry, $ \Theta$ is uniformly distributed on $ [0,2\pi],$

You asked about the ratio $ \frac{Y}{X}.$ Note (from the usual polar coordinate geometry) that $ \frac{Y}{X}\equal{}\tan\Theta.$

Your task is now to show that that tangent of an angle uniformly distributed on $ [0,2\pi]$ has a Cauchy distribution. This isn't too hard; I'll leave the details to you.

(Could $ \frac{Y}{X}$ be undefined? Well, yes, but that has probability zero, and any event with probability zero can safely be ignored.)


$ x$ is less than a million. If we put a $ 1$ after $ x$ we get a number which is three times as big as if we had put a $ 1$ before x. What is $ x$?

Let the number without 1 be a

$ 10a\plus{}1\equal{}3(10^n\plus{}a)$, where $ n$ is number of digits of $ a$

$ 7a\equal{}3\cdot 10^n\minus{}1$

Only such number on RHS which is divisible by 7 is 299999

$ a\equal{}42857$


geometry
3D geometry
prism
Pythagorean Theorem

How would you do this problem?

What is the farthest distance between vertices in a rectangular prism with lengths of 3,4, and 12?

[hide="Step 1"] What are the two vertices farthest apart? [/hide] [hide="Step 2"] What triangles have the length between those vertices as their hypotenuse? [/hide]

ive seen those ramdoozles of times  :oops:

the answer is 13. thanks. i got the formula. its the square root of (length^2+width^2+height^2).

[hide]You want the distance between two opposite corners. First, imagine that the 3 by 4 face is touching the ground. Using the Pythagorean theorem, this diagonal is 5. Now, the one we are really solving for is the hypotenuse of the 5-12-? triangle. Using the Pythagorean theorem, this distance is 13. The answer [i]is[/i] 13.[/hide]

i edited my post. i looked up the answer wrong :(


There are $ 9*10^4\equal{}90000$ five-digit integers.

Of these, $ 9*10*10*10*5$ integers have units digits that are less than 5.

So the probability is $ \frac{9*10*10*10*5}{9*10*10*10*10}\equal{}\frac{5}{10}\equal{}\boxed{\frac{1}{2}}$.

[quote="GameBot"]A five-digit integer will be chosen at random from all possible positive five-digit integers. What is the probability that the number's units digit will be less than 5? Express your answer as a common fraction.[/quote]

Actually, you don't need to do all that.
There are 10 possible digits - 0 1 2 3 4 5 6 7 8 9
and out of those 5 are less than 5, so

5/10
=
1/2


This problem took me sometime till I recalled a "quick" way. Anyway, it's a nice problem for this forum.

Suppose that $a,b,c,d$ are numbers which satisfy $a+b+c+d = 10, (a+b)(c+d) = 16, (a+c)(b+d) = 21,$ and $(a+d)(b+c) = 24$. Compute the value of $a^2+b^2+c^2+d^2$.

No calculator allowed.  :)

No... that is the quick way. I wrote it all out, but you can practically assume some parts of it.

[quote="236factorial"]So, what's $a^2+b^2+c^2+d^2$?[/quote]

39

i did  it differently , i found a , b, c, d individually , unlike jesus.

According to the official answer key  :D , they did it the same way as Jesus (well, his was less formula memorizing). The key is the realize that

$a^2+b^2+c^2+d^2=(a+b+c+d)^2-(a+b)(c+d)-(a+c)(b+d)-(a+d)(b+c)$.

Of course, I don't know who would spend their time memorizing this..

Well, by what I was given, I was able to assume that that was the case, and then just proved that it was. ;) That's probably what they did, too, but didn't bother proving anything...

[quote="tram"]Hi, there is another way, which goes through calculating all: a,b,c,d. Namely let (for example): $a+b=x,\ c+d=y$. Then:
$x+y=10$ and $xy=16$, so you can guess, that $x=8,\ y=2$ or reverse. With such tricks, you can probably (I didn't calculate it to the spot) have all: a,b,c,d.[/quote]you can do it by this methiod but u'll have to deal with too many equations...what was ur method eminem?


1-A glass beaker of water is on the balance. A metal ball is hung from the edge of the beaker but outside the beaker. The weight of everything is recorded. The ball is now put inside the beaker hanging from the string, not touching the bottom.

 Does the weight registered on the balance change?

2-You are riding in a train and have just opened a bottle of beer. How much of the beer should you drink so that after putting your partially empty bottle on a table (which is shaking) it would be as stable as possible?

1. The mass/weight measured will be definitly the same, maybe except for a few moments while manipulating with the apparatus and changing the position of the ball.

2. You should drink just enough beer to put the common center of mass of bottle and beer into the lowest possible position. If the bottom of the bottle is cylinder-shaped, its mass $M$, cross-section $S$, density of beer $\rho$, height of the center of mass of the bottle $H$, then height of common center of mass is

$x = \frac{MH + \rho S h^2}{M + \rho S h}$

now when dx/dh = 0 it is in its minimum. It's just messy math from this point.

[hide][quote="shadysaysurspammed"]Does the weight registered on the balance change?[/quote]
Yes, weight = $mgh(m_c)$, where $m_c$ is the center of mass. The height is taken from the center of mass. When you place the ball into the water, it displaces water, causing the center of mass for the whole beaker system to rise, and thus $h$ increases, increasing its weight.[/hide]

ummmmm spoon,isnt weight equal to mg?mgh i for potential energy,right?
what i think is,when the ball with the strin is immersed in the water.due to buoyant force,the tension in string decreases,but,see,when we jump,dont we exert some force on the ground and then jump?in the same way,the molecules of water,to push the ball up(buoyant force) have to push molecules down,action reaction.so those molecules exert force on the bottom of the beaker,hence contributing to weight.but i cant get what to derive from all this :?:

Fair enough... I will change my argument, because what I have is not infact what I meant! This is what I meant...

Ok.. what I meant to say before is weight = ${mg(h_{m_c})}$, where g is a function of h, which we all know it is.

The key is to look at the system as a whole, and not the individual forces within the system.

Weight of system with ball outside:

$F_0 = \frac{GmM}{r^2}$ (where r is taken from the center of mass of the system.)

Weight of system with ball inside:

$F = \frac{GmM}{(r+h)^2}$ (where h is the change in height of the center of mass)

It follows directionaly that

\begin{eqnarray*}
\Delta F &=&  \frac{GmM}{r^2}\left(1-\frac{1}{(1+\frac{h}{r})^2}\right) \\
&=& \frac{GmM}{r^2}\left(1-\left(1+\frac{h}{r}\right)^{-2}\right) \\
&=& \frac{GmM}{r^2}\left(1-1+\frac{2h}{r}\right) \\
&=& \frac{GmM}{r^2}\left(\frac{2h}{r}\right) \\
&=& \frac{2mgh}{r} > 0
\end{eqnarray*}

It is clear that the weight will increase by $\frac{2mgh}{r}$.

Note: I used the assumption that h << r.

[quote="shadysaysurspammed"]u looked at the sum in simpler way spoon,thats nice.
u said that weight would increase,wouldnt it decrease?(cause r+h the whole square is in the denominatotr when the ball is inside  the beaker)

i'm still not too convinced on the assumption of h<<r,is this right?isnt there another way becuse its not necessary that h<<r :?:[/quote]
Yes, hehe, because I did $F_0 - F$, which means I subtracted the previous weight from the new weight and got the positive difference, but you are right, the weight decreases by this amount.

Well, if h is about 1/2 cm and r is the radius of the earth, I would say that h << r.

Kubus is right in the fact that the change would be small. For my purposes I assumed the earth was spherical (which I think has to be assumed here), thus the horizontal movement would not affect the weight.

Basically, if we neglect the change in h, then the mass doesn't change. If we don't neglect the change in h, then the weight decreases. The question wasn't specific, so I figured they were looking for the solution based on the change in the center of mass. Really, the question would have had to be more specific for us to really know.

The total weight registered on the balance remains the same. The water pressure on the bottom of the beaker increases because of the rise of water level. The tension on the string changes size. Consider all of the forces on the beaker.

the above stuff is the solution from where i saw this sum.i knew somethin was wrong,so i asked u guys.
and yeah,h<<R(i thought r was something of the beaker :blush: )

read my post(the one in the beginning),please tell me if what i think there is right,please. :lol:

If the question was asking without reference to the change in g, then for all intensive purposes the internal forces balance out (as they must) and the weight remains the same.

If the mass was 1 kg, and the center of mass changed 1cm, the change in weight would be about 3x10^-8 as Kubus suggested.

However, if for example this was a large tank, like a dunking tank, and a human being is the displacement "ball", the effect would be quite significant. Well... at least measureable ;)

[quote="kubus"]Spoon, now I see what you were trying to say, but those weight changes are going to be of order $10^{-8}$ or so, which is quite unmeasurable in normal circumstances. Wouldn't other things affect the measured weight more? For instance, the center of mass also moves in horisontal direction, and the scales would surely react to this also ...[/quote]

Kubus, I was thinking about what you said some more, and I wanted to ask/suggest some things.

Are you thinking the balance is a lever balance? i.e. one that is on an axle, like a triple beam balance. Because if you were, then what you said makes more sense. It would seem that clearly a change in the horizontal component of the center of mass would affect the balance.

This is true! Unfortunately though, such a balance does not measure weight. It measures torque. If the horizontal component changes (for instance, the system moves farther from the pivot point) then the scale would register a heavier weight. However, the "weight" which is exclusively the vertical component (by definition) of force would not change. The torque exerted on the balance would increase, and the balance would register more weight. This would be a false measurement though, as such a balance doesn't work correctly.

Is that what you were referring to? If so... it was a nice observation.


[quote="Slizzel"]Fint the maximum and the minimum for the folowing term $x_{n}=\dfrac{2^{n}-3^{n}}{6^{n}}$.[/quote]

$n$ is a real number?

[hide]$x_n=\dfrac{1}{3^n} + (-\dfrac{1}{2^n})$
$x_n \ge 2\sqrt{-\dfrac{1}{6^n}}$ is the result of applying AM-GM. What are the restrictions on $n$? :huh:[/hide]

[quote="chess64"]$x_n=\dfrac{1}{3^n} + (-\dfrac{1}{2^n})$
$x_n \ge 2\sqrt{-\dfrac{1}{6^n}}$ is the result of applying AM-GM. What are the restrictions on $n$? :huh:[/quote]
You can't apply AM-GM for negative numbers. :!:
Obviously from the question itself, $n$ must a positive integer.

For the maximum:
Because $n$ is a positive integer. The $2^n-3^n$ is max when $n=1$ so $max=-\dfrac{1}{6}$ I think

i would say that the max is $x_0$ since it is 0, and all the others positive integers give negative numbers, and for min, looking at the graph $x_1$ is the lowest

well consider that
$x_{n}=\dfrac{2^{n}-3^{n}}{6^{n}}=\dfrac{2^{n}-3^{n}}{2^{n} * 3^{n}}=\frac 1{3^n}-\frac 1{2^n}=3^{-n}-2^{-n}$

Finding the max and min values now are a bit strange... (well, we know that for $n>0$ $x_n<0$ - by graphing, and for $n<0$ $x_n>0$

well with (very little) calculus one can proof that if $n\in\mathbb R$ that $x_n$ is minimal if $n=\log_{\frac32}(\log_2 3)$. plugging in u get the minimal $x_n$ but i couldnt simplify that yet ...


[quote="thanhnam2902"]$ 2(sin^4x \plus{} cos^4x) \plus{} \frac {1}{sin^8x} \plus{} \frac {1}{cos^8x} \equal{} 33$[/quote]

$ Equal \leftrightarrow 2(1\minus{}2sin^2xcos^2x)\plus{}\frac{(1\minus{}2sin^2xcos^2x)^2\minus{}2sin^4xcos^4x}{sin^8xcos^8x}\equal{}33 (1)$

Let $ sin^2xcos^2x\equal{}t \in\ [0;\frac{1}{4}]$

$ (1) \leftrightarrow 2(1\minus{}2t^2)\plus{}\frac{(1\minus{}2t)^2\minus{}2t^2}{t^4}\equal{}33$

It 's easy  :wink:

hello, your equation is equivalent with
$ {\displaystyle (35619\plus{}4024\cos(4x)\plus{}1308\cos(8x)\minus{}136\cos(12x)\plus{}\cos(16x))\csc(x)^8\sec(x)^8\sin\left(\frac{\pi}{4}\minus{}x\right)^2\sin\left(\frac{\pi}{4}\minus{}x\right)^2\equal{}0}$.
Sonnhard.


So I'm taking a Mathematical Reasoning class and I've been having problems doing this one problem near the end, it's very hard. Also, not sure if this is the right forum to post this in and if it isn't, feel free to move it.

So here it is:
[b]
In the village of Perfect Reasoning, each employer has an apprentice. At least one apprentice is a thief. To remedy this without embarrassment, the mayor proclaims the following statements:At least one apprentice in this town is a thief. Every thief is known to be a thief by everyone except his or her employer, and all employers reason perfectly. If n days from now you have concluded that your apprentice is a thief, you will come to the village square at noon that day to denounce your apprentice." The villagers gather at noon every day thereafter. If in fact k>=1 of the apprentices are thieves, when will they be denounced, and how do their employers reason? (hint: Study small values of k, and use induction to prove the pattern for all k.)[/b]

Not really sure where to even start with this, or how or what to use induction on. So any help would be greatly appreciated!

[hide="base case and partial"]
1 employer, on day 1 immediately deduces that his apprentice is a thief

2 employers, Each assumes the other employer's apprentice is the thief on day 1. On day two, since no one has been denounced yet, they realize they both made that assumption, so now they both conclude that their own apprentice is a thief.

3 employers, it will take 3 days.

And so on.
[/hide]


For those people who took the SAT on an alternate date because of ARML, did you get your paper report in the mail yet?
My scores have appeared on the collegeboard website but I haven't received my report yet.


Can anyone identify the differences between James Stewart 5e Single Variable book and the normal James Stewart 5e?

The single variable book cuts out the multivariable chapters, from parametric equations to full-blown vector calculus. The single-variable parts are identical in both books.
If you are likely to take a multivariable calculus class using Stewart, get the full length version. If you know you'll stop before that, get the shorter and lighter version.


I am mathematically challenged so I hope that my questionare are not too below the abilities of anyone on this forum.

Is it possible to solve for four unknown variables? Namely, in the equation below, is it possible to determine the real number solution for x, y, z, and a? If so, what are the real number solutions, and how does one get to the solution?

Here is the equation:

17x + 15y + 11z + 11a = 4822

Thanks in advance for your help.

Well, for a problem like this there are an infinate number of solutions.  For instance, you can choose $x$, $y$ and $z$ in any way that you want, and then solve the equation for $a$...

In general, you need the same number of equations as you have unknowns for linear systems of equations.

For instance, $x + y = 20$ has an infinite number of solutions, just consider these: $(x=10,y=10), (9,11), (8,12), (-1,21), (-5,25),...$.  But if you also know that $x - y = 8$, then you can solve the 2nd equation for $y$, and you have $y = x - 8$, then you can substitute this into the first equation, and you get $x + (x - 8) = 20$, and solving for $x$ you get $x = 14$.  Then you can go back to see that $y = 14 - 8 = 6$.

We may want to move this to the getting started forum.


symmetry
calculus
algebra proposed
algebra

[quote="hoangvn.fix"]1, $ x^4 \minus{} 3x^2 \minus{} 2x \plus{} 1 \equal{} (2x \minus{} 1)\sqrt {2x^3 \minus{} 1}$[/quote]

$ \plus{}2.71151017...$

[quote="hoangvn.fix"]2, $ \frac {x^3 \plus{} x^2 \minus{} 9x \plus{} 2}{x \minus{} 2} \equal{} \sqrt {2x^2 \minus{} 6x \plus{} 3}$[/quote]
$ 1 \plus{} \sqrt 2$
$ \minus{}5,267094012...$

[i]Please, could you indicate us in what contest you found these crazy problems ... ?[/i]

dear pco,
thank you for solving my own problems. here is my solution:
1,
$ \Leftrightarrow (x^2-1)^2-x^2-2x=(2x-1)\sqrt{(2x-1)(x^2-1)+(x^2+2x)}$
Setting
$ a=x^2-1$
$ b=\sqrt{(2x-1)(x^2-1)+(x^2+2x)}$
We obtain:
$ \{\begin{array}{c}{a^2-x^2-2=(2x-1)b}\\{b^2-x^2-2=(2x-1)a}\end{array}$

This is a symmetry-system.
2,
$ \Leftrightarrow \frac{x^3-5x-2}{x-2}=\sqrt{2x^2-6x+3}-(x-2)$

If $ \sqrt{2x^2-6x+3}+(x-2)=0$
then $ x=1-\sqrt{2}$ not sastify the equation.
if not then
(2)$ \Leftrightarrow \frac{(x+2)(x^2-2x-1)}{x-2}=\frac{x^2-2x-1}{\sqrt{2x^2-6x+3}+x-2}$
Hence this is easy for you.

[quote="hoangvn.fix"]dear pco,
thank you for solving my own problems. here is my solution:
1,
$ \Leftrightarrow (x^2 - 1)^2 - x^2 - 2x = (2x - 1)\sqrt {(2x - 1)(x^2 - 1) + (x^2 + 2x)}$
Setting
$ a = x^2 - 1$
$ b = \sqrt {(2x - 1)(x^2 - 1) + (x^2 + 2x)}$
We obtain:
$ \{\begin{array}{c}{a^2 - x^2 - 2 = (2x - 1)b} \\
{b^2 - x^2 - 2 = (2x - 1)a}\end{array}$

This is a symmetry-system.
2,
$ \Leftrightarrow \frac {x^3 - 5x - 2}{x - 2} = \sqrt {2x^2 - 6x + 3} - (x - 2)$

If $ \sqrt {2x^2 - 6x + 3} + (x - 2) = 0$
then $ x = 1 - \sqrt {2}$ not sastify the equation.
if not then
(2)$ \Leftrightarrow \frac {(x + 2)(x^2 - 2x - 1)}{x - 2} = \frac {x^2 - 2x - 1}{\sqrt {2x^2 - 6x + 3} + x - 2}$
Hence this is easy for you.[/quote]

Please, hoangvn.fix, could you give us your own clever solution up to the end for these two equations, instead of my ugly numeric calculus one ?

Thanks

for the first i think that i have already give you full solution because that is a symmetry system you can subtract 2 sides of the equation for the second after changing the equation to
(2)$ \Leftrightarrow \frac{(x\plus{}2)(x^2\minus{}2x\minus{}1)}{x\minus{}2}\equal{}\frac{x^2\minus{}2x\minus{}1}{\sqrt{2x^2\minus{}6x\plus{}3}\plus{}x\minus{}2}$
then we have
$ \frac{x\plus{}2}{x\minus{}2}\equal{}\frac{1}{\sqrt{2x^2\minus{}6x\plus{}3}\plus{}x\minus{}2}$
$ x\equal{}\minus{}2$ notsatisfy the equation so we have
$ \frac{\minus{}x^2\plus{}x\plus{}2}{x\plus{}2}\equal{}\sqrt{2x^2\minus{}6x\plus{}3}$
which can be reduce to the quartic equation so you can use ferrary method to solve
hoping this could help you

[quote="hoangvn.fix"] [...]
hoping this could help you[/quote]

 :rotfl:


MATHCOUNTS
big brother
FTW
AMC
AIME
percent
Mafia

[i]Eleven Houseguests enter…They don’t know each others’ names. They don’t know each others’ personalities. They know nothing about what will happen over the next few months…[/i]

[b]Players
CatalystOfNostalgia
chenhsi
Brut3Forc3
erabel
pacman2812
BOGTRO

[color=blue]Jury Members
cf249
AIMEman
crazychinesemaniac
[/color]
[color=red]Evicted
jhredsox93
mz94[/color][/b]

Welcome to this second installment of the reality show, [b]Big Brother[/b]! Why the subtitle you may ask? Well, this season, there will be more power available than ever before. And you eleven will be the ones fighting for it. It can change at the drop of a pin, and you'll never predict the consequences of using it.

This is a game, and we do have rules and structure though:

[i]Big Brother is a power struggle. Each week, the power in the house will shift, via several competitions in which the Houseguests will participate. Here are these competitions:

Head of Household (HoH): This position is the most powerful in the house. It takes place at the start of every week, and everyone competes except for the previous week's HoH. The HoH has the responsibility to nominate two Houseguests for eviction, the worst place to be in the house. If you are still nominated (see Power of Veto) by the Eviction vote, the rest of the house will vote, and potentially send you home.

Luxury Competition: These will be scarce during the game, but there will be at least one. The winner(s) will receive some sort of advantage in the rest of the game. For example, the winner may get an extra vote in an Eviction vote.

Power of Veto (PoV): This competition takes place after the HoH has made nominations. There are (at the beginning of the game) 6 competitors: The HoH, the two nominees, and three HGs selected by random draw. The winner of this competition has the ability to Veto one of the HoH's nominees. They do not have to, however. If a nominee is Vetoed, then the HoH is required to put up a replacement nominee.

The Big Brother game proceeds in Weeks: on the real show, they do indeed last one week. Here, though, their length will be variable. However, for my own purposes, and to keep the game realistic, we'll keep teack of the days as if each week lasted seven. On Day 2, the first HoH competition takes place. Then, on Day 5, the HoH will nominate two other Houseguests for Eviction. On Day 6, the Veto Competition will take place. Then, Day 7, the PoV winner will be able to Veto one of the HoH's nominations, which, if done, will require the HoH replace the nominee. Then, on Day 9, the HGs, excluding the current HoH and nominees, will vote on which of the nominees will be Evicted. Then, after the eviction, we start all over again.

Once we are down to the final three, the game will change. There will be a three-part HoH competition. The first part will feature all three remaining HGs. The winner of Part 1 automatically goes on to Part 3. Then, the other two fight it out in Part 2, where the winner also moves to Part 3. Then, the winner of Part 3 is the final HoH, and gets to single-handedly make the final Eviction, choosing their Final 2 partner.

Then, the last 7 Evicted HGs (the Jury) vote, deciding which of the Final 2 will be crowned the winner of Big Brother 2, and walk away with $ \$500,000$!

There are also several ground rules, which must be followed at all times by everyone:

1. Communication, both inside the thread and out, is allowed at all times, except the following:
a. HoH Competitions
b. PoV Competitions, unless otherwise specified
c. Eviction Votes
d. Any other time I tell you so
2. I expect that most of the game should take place outside this thread: I also require that any discussion groups be made known to me.
3. You are NOT allowed to discuss the game with anyone not in the game, unless you are told otherwise.
4. Once you are evicted, you will be sent special instructions about your future role in the game. As a rule, though, don't talk to anyone still in the game once evicted.
5. Use your good judgment: don't do anything you think I would disallow.
6. This is a game: have fun, and make it fun for others. No flaming, etc.

If you ever have any questions about what you are and aren't allowed to do, send me a pm, or ask in-thread if you don't care that others know. I will never reveal anything someone pms me that isn't a competition.

Finally, the above rules are just the basics. As promised, there will be twists, and therefore additions/changes to the rules. Anything can happen: Expect the Unexpected.
[/i]

Well, that about sums it up. We'll start our very first HoH competition soon, on (BB) Day 2. The battle for power begins...

Good luck...Do Not Assume anything...

I'm here.  Wait, where is everybody?  AAAAAAAH!!!!!!
Wow, I don't know where that came from.  :huh:

Well, it's time to reveal the first HoH competition.

[b]Head of Household One[/b]

There will be five teams of 2 in this competition, chosen by order of in-thread confirmation. So, the first team is cf249 and crazychinesemaniac.

Once you have posted in-thread, you may not talk to anyone outside the thread about the game. Or in-thread. And no discussion of this comp in your confirmation.

So, anyway, one person will be left out...and therefore ineligible to win the first HoH. Sorry...

Each team will now answer a ten question quiz about who The Mole is...err…sorry. Each team will answer a 5 question quiz about personal preferences. Whichever team matches each other on the most questions wins, with the TB being smallest difference in your answers. No talking during this HoH.
1. Last Place, or Second Place?
2. Win no Competitions, or every one possible?
3. Loyalty or Betrayal?
4. “Do Not Assume” or “Expect the Unexpected”?
5. Lie or Cheat?
TB: On a scale of 1-100, how much do you like your partner?

Good luck.

[quote="CatalystOfNostalgia"]I guess I'll confirm.

SamL you have quite a few typos. In the cast list, you forgot the 'c' in pacman2812. In addition, I believe you may have been going for "5 teams of 2." That's the first thing in this thread I read so I was like "whoa 25 players that is quite a showing"

Wait so like, how is the HOH going to be decided between the winning pair? Or is it a joint nomination like in the first week of All-Stars?[/quote]

The irony....I think you meant the "m" in pacman2812.  :P. Also please explain task clearer. For instance, are we allowed to post in this thread till the competition is done, even if its irrelevant to the competition. Yea...how are we going to decide who gets HOH.

Just a note: you may send in your answers once your team is formed.

and will typos ever be a part of the game?

Teams?
cf249
crazychinesemaniac
--
CatalystOfNostalgia
jhredsox93
--
chenhsi
erabel
--
Brut3Forc3
??
--
??
??
--
?? (lonely person)

Who wants to be with me?

:rotfl: I forgot to post already. I lose.

You may or may not want to disregard that...

If you can beat 2 minutes, you've beaten me.

Umm OK?

We have results. In [hide="third place,"]CatalystOfNostalgia, with a time of 0:00. He did not feel a need to post...Arrogance, or savvy play?

Now, first place beat second by a surprisingly low margin...3 minutes! With a time of [hide="2:21,"]pacman2812. And as for [hide="Brut3Forc3..."]2:18! That means pacman2812 moves onto Round Three automatically. Congratulations!

Now, as for Round Two. This round requires AIM which, luckily, both competitors have it. I'm not quite done with it though, so tomorrow should begin the window in which you may take it.[/hide][/hide][/hide]

WOOHOO!!! :D
I'm pretty lucky, because I don't have AIM...
:D

[quote="perfect628"]Now, first place beat second by a surprisingly low margin...3 minutes![/quote]

um...

bump?

um yeah perfect628 come on let's get it moving?

last post was on november 10th. I would declare this abandoned.


linear algebra
matrix
linear algebra unsolved

I am solving an engineering problem. Any suggestion to solve the following problem is highly appreciated. Many thanks in advance.

Problem: There are two groups of square matrices $G_{1}$ and $G_{2}$ such that if $A \in G_{1}$ and $B \in G_{2}$, then

1. $A^{H}B+B^{H}A = 0$,
2. $A^{H}A = B^{H}B = I$

where $^{H}$ denotes transpose conjugate opereation, $0$ and $I$ are all-zero and identity matrices, respectively.

Question: What are the maximum numbers of matrices of $G_{1}$ and $G_{2}$.

Erm... You should be a bit more specific in describing your question:

0) Is the size of matrices fixed?

1) By "groups" do you mean just sets without any structure or actual groups? (I assume the first because you, clearly, cannot have the identity matrix in both $G_{1}$ and $G_{2}$)

2) What exactly do you want to maximize? If, for instance, $G_{2}$ consists of just one element $I$, then $G_{1}$ can be taken the set of all matrices such that $A^{H}=-A$, $A^{2}=-I$, which is an infinite set if the size of your matrices is $2$: you can take $A=\begin{matrix}0 & x\\-\bar x & 0\end{matrix}$ with any complex $x$ satisfying $|x|=1$. If you take $4\times 4$ matrices, you can make both $G_{1}$ and $G_{2}$ infinite.

Many thanks for your attention.

My problem is actually a relaxation of orthogonal designs (proposed by Hurwitz-Radon). Please find answers below.

0) Is the size of matrices fixed?
The size of matrix is not fixed. But you may consider the small sizes first, $2 \times 2$ or $4\times 4$ complex matrices.

1) By "groups" do you mean just sets without any structure or actual groups? (I assume the first because you, clearly, cannot have the identity matrix in both G_{1} and G_{2})

Yes, I mean the two sets.

2) What exactly do you want to maximize? If, for instance, G_{2} consists of just one element I, then G_{1} can be taken the set of all matrices such that A^{H}=-A, A^{2}=-I, which is an infinite set if the size of your matrices is 2: you can take A=\begin{matrix}0 & x\\-\bar x & 0\end{matrix} with any complex x satisfying |x|=1. If you take 4\times 4 matrices, you can make both G_{1} and G_{2} infinite.

I am considering the case such that  the two sets are of the same size. Furthermore, the entries of matrices are restricted in {$0, \pm 1, \pm j$} ($j^{2}=-1$). What is the maximal size of the two sets?

I really appreciate your help.


Solve in natural numbers the equation:

6^y+2^x+3^y*2^(x+y)=17*37-1.

Solve the equation: sqrt(-x^2+4x-3)+2^(x+1)+4^(x-1)-5=0, x in R.

bye bye

lets say that:
Solve in natural numbers the equation:
6^y+2^x+3^y*2^(x+y)=17*37-1.  is the 1st problem
and

Solve the equation: sqrt(-x^2+4x-3)+2^(x+1)+4^(x-1)-5=0, x in R.  is the 2nd problem

for the 1st one:

we write the eqs like so:
17*37=(6^y+1)(2^x+1) from where we get that: 2^x=16, 6^y=36 (because we are looking only for the natural solutions) so we get that x=4 and y=2.

cheers!!!!!!!!! :D :D :D

for the 2nd we have that:

we must have x in [1,3] because of sqrt(-x^2+4x-3) and for x>1 we get that:

sqrt(-x^2+4x-3)+2^(x+1)+4^(x-1)>0+4+1=5. so the only solution is 1.

hehe!!!  it's a nice problem! :D :D


Given a square $ABCD$ with side length $2$ and triangle $CBD$ has a square $GEOM$ inscribed in it, find the area of square $GEOM$.

[quote="Go Around the Tree"]given a square ABCD with side length 2 and triangle CBD has a square GEOM inscribed in it, find the area of square GEOM[/quote]

Hint[hide] Use similar triangles[/hide]

[hide="A picture always helps."]By similar triangles, $\Delta OMC$, assuming that O and M touch BC and DC, respectively, is a 45-45-90 triangle, so MC, also assuming side $s$ is the side of the square, is $MC=\frac{s\sqrt{2}}{2}$. Also, $DM=s\sqrt{2}$ because $\Delta DMG$ is also a 45-45-90 triangle by similar triangles. $DM+MC=2$, so $s\sqrt{2}+\frac{s\sqrt{2}}{2}$. So $\frac{2s\sqrt{2}+s\sqrt{2}}{2}\\=\frac{3s\sqrt{2}}{2}$. So $\frac{3s\sqrt{2}}{2}=2\\3s\sqrt{2}=4\\s\sqrt{2}=\frac{4}{3}\\s=\frac{2\sqrt{2}}{3}$. So the area is $s^{2}=\frac{8}{9}$.[/hide]

Just a little latex clarification, put a space between \Delta and OMC

I notice that people here use \Delta for \triangle and \triangle for \Delta. This may be nitpicking, but those are not interchangeable - \Delta is a letter, \triangle is a symbol.

$\text{awesome, thanks for introducing Latex and getting me into it lol, i thought it was useless before }$

another thing I noticed is that there are 4.5 of the inscribed squares that make up the big square with the triangle made by the diagonal. I think this is true for all problems with a square and a triangle formed by the diagonal with a square inscribed lying on the hypotenuse of the triangle.

Therefore you can use the equation $4.5x=4$

$4$ is the area of the big square with side length 2. $4.5$ is the number of small squares needed to make the big square.

Then you can get $4/4.5$=$8/9$

which is the same.


[quote="vinny919"]Does there exist a function such that $f(x - 1) - f(1-x) = x$, for all real x?[/quote]

No: f(1-1)-f(1-1) =1 --> f(0)-f(0)=1, 0=1, contradiction.

How about for all real x except x=1?  PS How do you write does not equal in tex?

This problem then changes to f(y)=f(-y)+y+1 for all real y except 0.
Then you define a piecewise funtion such that f(y)=1 for y>0 and f(y)=y for y<0.  Would it be possible w/o breaking it up into parts?

[quote="Wumbate"]Then you define a piecewise funtion such that f(y)=1 for y>0 and f(y)=y for y<0.[/quote]

Why?

Uh, I guess that wouldn't work, either.
See when y is positive it works, for example if y=2 then you get 1=-2+2+1.  However if y=-2 then you get -2=1-2+1 which isn't true.

Can it possibly be true?

[quote="Wumbate"]How do you write does not equal in tex?
[/quote]

\neq

$\neq$

sticking an n after the slash or a \not before the slash will negate a lot of relation symbols

There cannot possibly be such a function.
Following from f(y)=f(-y)+y+1, you let y=-a and get f(-a)=f(a)-a+1.  Since a and y are interchangeable, that means f(y)=(f(y)-y+1)+y+1, or 0=2, a contradiction.

Just do x := 2 - x, add the two equations and you get 0 = 2, haha.


Hey. I am wondering if anyone knows the state cutoffs for Massachusetts from 2000-2005.

Any help is appreciated.
max.

[url]http://www.artofproblemsolving.com/Forum/index.php?f=232[/url]post your question here and you might get better answers

[quote="frost13"]I'm pretty sure it was 37 or 38 last year.  At least 4 kids scored 41 or better at chapter this year and all 4 of them were at Nationals last year.  Tough year to be competing in MA!![/quote]

Err... since you seem to know MA, would you happen to know Ohio's cut-off at the state. (the score I would need if I was participating last year to go to nationals- get in the top four written)

Thanks.

I'm just about to go do last year's round :) (I didn't take part last year)

I have contacted my source in Ohio, so I'll let you know if I find out.  I do know Sycamore had 1, 2, 4, 5, 8, & 10 last year.

I might be wrong, but I think the Ohio cuttoff last year was 37

frost do u know if only 4 ppl got 41 or higher on chapter?

I don't KNOW.  I do know 4 people got at least 41 but that is only from 2 of the chapters.  Where are you from in MA?

Kentucky's Cutoff last year was 35
I got 30
Uh...
If i only didn't miss that one target and sprint
I would've destroyed anybody on tiebreakers
and
I would at least get a scholarship
I'm really studying hard right now because
Only 2 7th graders were in the top 10 last year
I was 11th
so I hopefully might make it this year

Well, I just got a 36 on the 2006 State...

But it's kinda late (I think better in the morning), and I had a few interruptions.

I think I'm okay... lets hope I have a good day on Saturday :lol:


Only one CT school on the AMC 8 school honor roll so far...
Greenwich Central Middle School

http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2006-8a/2006-amc8schHonorRoll.shtml

Did anybody get the official results yet?

there's no official results...some schools get them later, and they have to triple check the scores

don't worry

if u know u got a good score as well as 2 or more people, ur fine

-jorian

[quote="dkushnir"]Any perfect scorers in CT?
How many 24s and 23s?[/quote]
One perfect score.The results are up now.

o, btw there were 4 24's and 4 23's
-jorian

I got a 24 on AMC 8. Oh, and if anyone sees my other score that I put up on another forum...That was my MATHCOUNTS score, so my bad!


$ A$ is a fixed point at a height $ H$ above a perfectly inelastic smooth horizontal plane. A light inextensible string of length $ L(> H)$ has one end attached to $ A$ and other is to a heavy particle. The particle is held at the level of $ A$ with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest.

I"m not sure I'd understood "next instantaneously at rest", but seems to be the classical conservation of energy in a pendulum, and I don't think this is the place for that so simple kind of problem.

I'm guessing "instantaneously at rest" means that the instantaneous velocity is 0.

It was there in an exercise of my collisions text. Where does the collision happen?

This is what I thought:

The particle will come down along an arc (taut string) and hit the floor.There,it looses its vertical component of velocity as the collision is inelastic.Then it travels horizontally along the floor (no friction).Here the string is not taut now.Then when the string gets taut again,it experiences a jerk, which cancels it velocity along the direction of string at that moment.With the other component of velocity (perpendicular to string), it moves up with taut string.Then,it instantaneously comes to rest at the max height of its trajectory, which is what we want.

Let the mass of particle be $ m$.
Let $ v$ be the velocity with which it hits the floor.
therefore, $ mgH \equal{} \frac{1}{2} m v^2$
$ v\equal{}\sqrt{2gH}$
let $ \theta$ be the angle the string makes with the horizontal the instant the particle hits the floor.
$ sin(\theta)\equal{}\frac{H}{l}$

The horizontal component of velocity = $ v' \equal{} vsin(\theta) \equal{} \sqrt{2gH} sin(\theta)$

When the string becomes taut again,the angle the string makes with the horizontal will again be $ \theta$.
The component of velocity perpendicular to direction of string =
         $ \sqrt{2gH}*\frac{H^2}{l^2}$

$ \frac{1}{2}m (2gH*\frac{H^4}{l^4}) \equal{} mgh'$

therefore, $ h'\equal{} \frac{H^5}{l^4}$

On colliding with floor, $ v_{_y} \equal{} 0$.

$ v_{_x} \equal{} \sqrt {2gh} \sin\theta \equal{} \sqrt {2gh} \frac {h}{l}$

Since string is inextensible, $ v_{_\perp} \equal{} v_{_x} \sin\theta$ and $ v_{_\parallel} \equal{} 0$ after which the particle undergoes circular motion.

$ \therefore h_{_2} \equal{} \frac {v_{_\perp}^2}{2g} \equal{} \frac {(\not{2}\not{g}h)\frac {h^4}{l^4}}{\not{2}\not{g}} \equal{} \frac {h^5}{l^4}$

One more question:
ii) How many "bumps" does the particle encounter if the collisions are elastic?

[quote="Rajiv"]One more question:
ii) How many "bumps" does the particle encounter if the collisions are elastic?[/quote]

Infinite times??It will never stop.
e will only decrease it by dividing.So,it will become smaller and smaller,but never 0. :?:

And what about losses due to jerking of string?

Jerking will only remove a component of velocity each time.
velocity $ v$ becomes $ v\cos\theta$, a fraction of itself, so, again velocity will tend to zero only with an infinite no. of collisions.
Did you mean to ask, "Find the no. of collisions between two consecutive instances of the string becoming taut." ?

[b]@Arjun:[/b] Rajiv said the collisions are elastic, so velocity will not decrease because of that.


logarithms
real analysis
real analysis unsolved

Let $ A$ be the following condition describing sequence $ (x_{n})$:

$ \forall c>0\quad \quad \exists N\quad \quad \forall n>N\quad \quad  \forall n<m<2n\quad |x_{n}\minus{}x_{m}|<c$

Show that from condition $ A$ does not follow that sequence is convergent - in other words - show that there exists divergent sequence which satisfies the condition $ A$.

The sequence $ x_n\equal{}\log n$ satisfies $ |x_n\minus{}x_m|<\log 2$ for $ n<m<2n$. This is not a counterexample...yet.

I think the way to do this is to use a series with strings of 0's and something like a 1/n term inserted strategically.

This says that it doesn't change much over logarithmically constant intervals. OK, we need something that changes slower than the logarithm.

Counterexample: $ x_n\equal{}\log(\log n)$ (and something at 1).

The estimate is a little easier with $ x_n\equal{}\sqrt{\log n}$:
\[ |x_n\minus{}x_m|\equal{}\frac{|\log n\minus{}\log m|}{\log n\plus{}\log m}\le \frac{\log 2}{\log n}\to 0\]

Except the square roots in the denominator are missing :P.

Set $ y_{3^n }  \equal{} \frac{1}{n}$, $ y_m  \equal{} 0$ otherwise, and set $ x_n  \equal{} \sum\limits_{m \equal{} 1}^n {y_m }$


MIT college Harvard ARML AMC USA(J)MO USAMO

Fortunately, I got accepted to MIT (sorry if you already knew this and it's getting repetitive). But did any other Georgians get in?

Philip
Sean
Linda
etc.

Linda was going for Harvard, and as of Lassiter had no news.

I thought Linda wanted to go to MIT, but her parents wanted her to go to Harvard

Vibin Kundukulam got in. Any tips for us younger juniors?  :D

who's Vibin Kundukulam?

just a few tips: make sure you do the interview. be yourself, and don't be afraid to highlight your accomplishments, but at the same time don't come off as cocky or arrogant.
also, make sure senior year coursework is not significantly easier than what you can handle. there are a lot of talented people vying for acceptance and you don't want a slack schedule to hurt your chances.

Um, nope, drunner isn't Brayden, he's Mark D., 3rd @ Lassiter, iirc. I was his roommate at ARML.

[quote="cleansmyheart"]linda, myself, and everyone else that applied early to harvard should know on friday, at 5pm by email.

i'm betting she'll be accepted, i'll be deferred. i'm shooting for not being rejected, woo!

drunner, did i miss your introduction? the only person that i've seen who's been accepted to MIT is brayden..so i'll assume you're brayden.

semi-finalist for tech's presidential scholarship?

what are your top choices?[/quote]

I'm Mark Doss. I go to Columbus High School. Remember I was on ARML?
did they nominate the GA tech semifinalists yet? hopefully i was nominated  :maybe:
caltech, mit, and gatech are my three top schools.

hey, mark. i never talked to you much..

yeah, i got my semi-finalist letter last week. but i've heard some people haven't even gotten their acceptance letters yet...gatech is odd.

my top choices are harvard, uc berkeley, and mit.

i know carol and brayden are into mit, i haven't heard from anyone else though.

Aah... I hope I get in next year :(

What sort of extracurricular activities look good?

Does ARML look good? Wut about USAMO? (I'm preparing like a monkey for that this year...)

saving the world from global warming or finding a cure for cancer will guarantee a acceptance into mit karth  :P

Science fair? How would going to ISEF look?

( :whistling:  :whistling:  :whistling: :blowingmyownhorn: )

[quote="jhcreinhardt"]high school students still dont understand what success is.  americans are blind to the infinite opportunities around them.  for some reason some really smart people that that they can only succeed if they go to ________ college.  but then again, these smart kids arent known for having any sense, are they.[/quote]

Say that to Mr. Kedlaya or Bill Gates (well, he did drop out of college...)  :wink:

But sometimes, it's true that many successful people (like actors and other top people of businesses) go to 'not so ivy-league' colleges and pwn other people financially.

I doubt if jhcreinhardt was talking about only financial success. I also doubt if Mr. Kedlaya thought, "Oh wow, I'd better go for MIT cause that'd really look good on my job app. and then I could pwn some people financially."

True. I guess it all depends on what one's definition of success is.

[quote="akkaren"]Why would one define success? It just creates an opportunity for your set of beliefs/purpose in life to fail. Note that I don't object to the possibility of the person failing--it's the ideology that shouldn't be given the chance.[/quote]

i dunno seems boring to live a life without purpose/goals.

one of you should make one called "Georgia definition of success"
I am sure that it would be on topic or perhaps "Georgia random thread" or ""
or we could run away from here :o

I for one like doing math for fun when I'm bored. I mean, it's way better than playing video games. Math actually is more interesting, and you don't get bored easily, especially when you're horrible. I don't think you have to be good at math to like it.

Who cares about college apps? There's graduate school also, and if you are truly dedicated enough in school, you won't have to worry about getting into a good school.

USAMO= it's just used to see what good students there are in the US

it's also a goal people who are good at math, like math, or just do it for their college apps like to go for.  :P

that is so right !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
do you think people will stop locking my math books away now!?!!?????

[quote="drunner2007"][quote="Phelpedo"]Most people get bored more easily with math than video games.[/quote]

true, but in the end math is infinitely deeper (unless you're actually programming games and playing them), so it can keep you preoccupied with fascinating ideas.[/quote]

Yeah there are like quintillions of more math ideas than ideas that you can get off of, say Halo II or Runescape...so overall math would keep you more occupied than video games.

[quote="akkaren"][quote="jhcreinhardt"]
i dunno seems boring to live a life without purpose/goals.[/quote]
But do you need to define whether you succeed or not at your purpose? For goals, sometimes it's clear-cut, but they're minor enough that it's not really what I'm talking about.

CAN you tell whether or not you succeed at your purpose in life? Minor purposes... but not anything major enough to be worth reflecting upon or worrying about, or even really measuring except in a casual aside thought during the process of figuring out what to do next.[/quote]

Does it matter if one succeeds as his or her own purpose in life? Life isn't a failure because you didn't fit your definition of a purpose. It's enough just to live with some direction, regardless if you follow it or not..


Players A and B play a series of games with A winning each game with probability 0.6. The overall winner is the first player to have won 2 more games than the other.

Find the probability that A is the overall winner.

After two games, two things can have happened:
- One player has one two games, thus the series ends.
- Each player has won the same number, so we basically start over.

So let's consider strings of two games.
Case 1: $ A$ wins both, thus the series. Probability $ 0.6^2 \equal{} 0.36$.
Case 2: $ B$ wins both, thus the series. Probability $ 0.4^2 \equal{} 0.16$.
Case 3: Each player wins one. We essentially return to the same situation.

This means the probability $ A$ wins the series is $ \frac{0.36}{0.36 \plus{} 0.16} \equal{} \frac{9}{13}$.


Hi,

I work in a retail shop and we have just started processing visa.

For each transaction we get charged a rate of 3% and this % is passed on to the customer.

My problem is finding out the customers final cost to charge them.

Say the items retail cost was 1500 i want to find the number which when taking 3% off brings me back to the original retail price.

I tried multiplying the 1500 by 3% which gave me 1545 but when subtracting it back off i am down to 1498.65

Hope this makes sense, and in advance thankyou all so much for your time and help!

Tim

Following your notation, $ x=y-z$, where $ z=\frac{3}{100}y$
$ x=y-\frac{3}{100}y=\frac{97}{100}y$, then $ y=\frac{100}{97}x$


function
logarithms
algebra unsolved
algebra

find all function wich verify $f(f(x))=x$ and $yf(\frac{f(x)}{y})=xf(\frac{f(y)}{x})$ for all x, y where the expression is defined

i agree with rust

you can notice that $f(x)=1-x$ and $f(x)=(1-x^{n})^\frac{1}{n}$
are solutions

How prove that ?

thank's for your ideas

If $\lambda: \mathbb{R}\to\mathbb{R}$ is a bijective solution of the Cauchy equation $\lambda(x+y)=\lambda(x)+\lambda(y)$,
then $\mu : \mathbb{R}\to\mathbb{R},\ \mu(x)=\{\begin{array}{rl}0&\text{for }x=0\\ e^{\lambda(\ln |x|)}&\text{for }x>0\\-e^{\lambda(\ln |x|)}&\text{for }x<0 \end{array}$ is a bijective solution of the Cauchy equation $\mu(xy)=\mu(x)\mu(y)$.
And then $f(x)=\mu^{-1}(1-\mu(x))$ solves your problem.

Now there are more than $|\mathbb{R}|$ such $\lambda$.
(every bijection between any pair $B,B'$ of $\mathbb{Q}$-bases of $\mathbb{R}$ extends to one)

what the proof or references olorin ?
thx a lot


What do you do when you don't understand the solution to a problem that you gave up on?

I read a solution to a problem here: http://www.artofproblemsolving.com/Wiki/index.php/2007_USAMO_Problems/Problem_5

the solution looked short, there weren't any symbols that I did not recognize, but I don't understand any step or the motivation or inspiration for anything.

[quote="Poincare"]i have felt that many times especially with problems i don't really understand. Think of it like this, just break every step in the solution into small steps and then fit it all together and you will feel immensely proud of your self for understanding the solution  :P[/quote]

I don't think that's the right way to learn math. Just because you can comprehend a solution doesn't mean that there's nothing left to learn from the problem. Understanding the solution and getting something out of the problem are totally different. If you come up with the solution yourself, you'll understand that solution much better than if someone just told you a solution. And if someone does tell you the solution, try to solve the problem another way (especially if you were trying the problem in one method and the solution method is completely different, you should make an honest attempt at trying to solve the problem in your method).

If you "break the solution into small steps", the whole solution will feel somewhat foreign. You might get all of the small steps, but it won't be intuitively clear why we did those steps (especially on solutions that refer to an equation that was proven earlier, I tend to get lost).

@aufha: try to solve similar problems (problems like the USAMO problem might be hard to find, though) or try to find the motivation youself. Try to figure out "what do we need to prove?" and "how can I prove that?"

I'm not very good at solving olympiad problems, either, but when I see a foreign solution, I try to understand that solution (this means reading it multiple times if necessary), so that I learn something from it.

Just so you know, the example link you gave is a solution that virtually [i]no one[/i] thought was motivated, and on that USAMO almost no one got that problem.

yes i know, but i meant that just understand the solution for then later you can go into the details.


Prove that there are infinitely many primes $ p$ such that the total number of solutions mod $ p$ to the equation $ 3x^{3}\plus{}4y^{4}\plus{}5z^{3}\minus{}y^{4}z \equiv 0$ is $ p^2$

[quote="brianchung11"]Prove that there are infinitely many primes $ p$ such that the total number of solutions mod $ p$ to the equation $ 3x^{3} \plus{} 4y^{4} \plus{} 5z^{3} \minus{} y^{4}z \equiv 0$ is $ p^2$[/quote]
If  p is a prime in the form 6k+5 then $ \{x^3|x \equal{} 0,...,p \minus{} 1\}$ is a complete reside mod p . It is obvious because the equation $ x^3\equiv 1 (\mod p)$ has only solution $ x\equiv 1 (\mod p)$ .Because $ \gcd(p,3) \equal{} 1$ ,so the set $ \{3x^3|x \equal{} 0,...,p \minus{} 1\}$ is also a complete reside mod p .
Therefore for each pairs of $ (z_0,y_0)$ there exist only $ x_0$ such that $ p|3x^3 \plus{} 4y_{0}^4 \plus{} 5z_{0}^3 \minus{} y_{0}^4z_{0}$
There are exactly $ p^2$ pairs of $ (z,y)$ chose from the set $ \{0,..,p \minus{} 1\}$ , so the equation mod p also has $ p^2$ solution . This proof completes .


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero such that $ a \ge b \ge c \ge k\sqrt {ab}.$ Determine the least constant $ k > 0$ such that the following inequality holds:
$ \frac {1}{\sqrt {b^2 \plus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt {c^2 \plus{} ca \plus{} a^2}} \plus{} \frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}} \ge \frac {3\sqrt {3}}{a \plus{} b \plus{} c}.$

P/s: This problem is not mine. I've just seen it on Mathscope forum, and after solving, I found it is very interesting so I  post it here.  :blush: :)

This is the link of this problem: http://forum.mathscope.org/showthread.php?t=7046, but I think the solution of the author is not correct!

My way is to denote that
$ F(a,b,c) \equal{} (a \plus{} b \plus{} c)\sum \frac {1}{\sqrt {a^2 \plus{} ab \plus{} b^2}},$
then we can show that $ F(a,b,c) \ge F(\sqrt {ab},\sqrt {ab},c)$ with $ a \ge b \ge c.$
From now, we can find that $ k_{\min} \equal{} \frac {11 \minus{} 3\sqrt {13}}{2}.$
:)


probability
number theory
least common multiple
relatively prime

Two fair dice, each with faces numbered 1 through 6, are rolled at the same time. Each die has five exposed faces, which are summed. The probability that the LCM of two sums is multiples of 4 can be expressed in terms of $\frac {p}{q}$ where p and q are relatively prime integers. Calculate p+q.

Please show your work.

P.S. Last part on p+q has been created by myself. :)

Edit: Thanks for pointing out.

[quote="Silverfalcon"][b]This probability[/b] can be expressed in terms of $\frac {p}{q}$ where p and q are relatively prime integers.[/quote]What probability are you talking about, exactly?  :?:


inequalities
geometry
incenter
circumcircle
inradius
geometry proposed

Given $\triangle ABC$ incenter $I$ and circumcenter $O$ and circumradius $R$ side $a,b,c$ prove that:
$\frac{OI}{R}\ge \frac{|b-c|+|c-a|+|a-b|}{a+b+c}$

This inequality is equivalent to

Given $x \geq y \geq z > 0$. Then $8xyz(x+y+z)^{2}\leq (x+y)(y+z)(z+x)(y+2x)(y+2z)$. But I still don't know how to prove it. :(

ok treegoner, I haven't prove it too, because I found it with maple and cabri,I see it true but I can't give any proof! :) but the following inequality I can prove it:
$5OI \ge |b-c|+|c-a|+|a-b|$

I can prove weaker inequality
$\frac{OI}{2r}\ge \frac{|b-c|+|c-a|+|a-b|}{a+b+c}$ with $r$ is inradius,are anybody interested in ?

In my last message, I have already proved that $2OI \geq max \{|a-b|, |b-c|, |c-a| \}$. Thus if $a \geq b \geq c$, we obtain that $|a-b|+|b-c|+|c-a| = 2(a-c) \leq 4OI$.

Definitely, it is true that $a+b+c \geq 6\sqrt{3}r > 8r$. Therefore, your weaker inequality is true.

Your initial inequality is more tricky. We will show that $a+b+c > 4R$ if $ABC$ is not an obtuse triangle. Indeed

$a+b+c = 2R(sinA+sinB+sinC) > 2R(\sum sin^{2}{A})$

$= 2R(2+2cosAcosBcosC) \geq 4R$.

However, I still don't know how to prove the inequality when $ABC$ is obtuse. Probably it is not true  :roll:


We are given $n$ marbles. Show that there cannot exist 3 sets of colors, so that all colorings of all the marbles with colors from set $A$ + all coloring of all marbles from set $B$ will be equal to the number of colorings of all marbles from set $C$ :)
Hmmmm, can anyone prove this?? Except Andres Wiles :) :D

I didn't see this problem. How could it happen?  :P  :P  :P

Dust, do you mean I should prove Fermat Last Theorem?

Just consider Wiles proof as the official solution.
So, please do not post official solution. And try to find another one  :D

Pierre.

I posted this problem because, when I entered the combinatorics unsolved problems forum, I saw that the last 20-25 posts belonged to Myth, so I thought that maybe he will not notice :lol:  :lol: and will solve this problem :D . Too bad he noticed the equivalence  :wacko:  :roll:
Ok, it was just a joke....

[quote="DusT"]I posted this problem because, when I entered the combinatorics unsolved problems forum, I saw that the last 20-25 posts belonged to Myth[/quote]
I didn't want to increase number of my post, but I want eliminate very bad situation, when many not hard problems lie in "unsolved combinatorics" for years!
So you can help me to do this work. Do you want?

Yes, actually I understood that
[quote] I want eliminate very bad situation, when many not hard problems lie in "unsolved combinatorics" for years! [/quote]
Ok, I will help you, as much as possible, considering my dial-up, only in the evenings connection :(

[quote="Valentin Vornicu"][quote="Myth"][quote="DusT"]Ok, I will help you, as much as possible, considering my dial-up, only in the evenings connection :( [/quote]
It is our common BIG problem :([/quote]Don't you have cable available in your city? Or is it too expensive compared with dial-up costs?

PS I am working on the jumpbox, and should be available soon. Anyway the page size has been decreased massively since the upgrade ;)[/quote]
Actually, I have cable connection now. And I worked many time in Internet during December. Now I just [b]fear[/b] how much money I will need to pay to my internet operator :(

[quote="Valentin Vornicu"][quote="Myth"]Actually, I have cable connection now. And I worked many time in Internet during December. Now I just [b]fear[/b] how much money I will need to pay to my internet operator :([/quote]Wait a minute, you have to pay per-traffic? Couldn't you get the unlimited version? You may download traffic monitor programs. Also make sure you are free of spyware that can increase your traffic. Furthermore find out if there is difference of price between local traffic and international traffic (the aops-mathlinks servers are located in USA).  If there is such a difference, maybe you can find proxies that help you contorize external traffic as local traffic ..[/quote]
Thank you for your advice.
We don't play games here ;) (take into consideration that my first specialization is computer science and applied mathematics).
Of course, I use various methods to decrease traffic. For example, all traffic come to me from specialized intenet service, which compress data amount for 70%. Though, after upgrade you enabled gzip compression, so I don't get benefits from loading pages itself. But this service is able to decrease image sizes by reducing their quality, and I use it for latex-related images.
BTW, one time you used png instead of gif. Why did you return to gif?

All "unlimited traffic" offers are TOO expensive, so only corporate clients may use it :(

[quote="Valentin Vornicu"][quote="Myth"]BTW, one time you used png instead of gif. Why did you return to gif?

All "unlimited traffic" offers are TOO expensive, so only corporate clients may use it :([/quote]Gif renders more nicely, and is easy to make it transparent :)

In Romania unlimited traffic for the major networks is around 40-45 $\$ per month, but with a bandwidth limit of 256kbps (32 kB/s) so actually it's not that unlimited
But the most used solution now is creating of large networks between block apartments which are connected through fiber-optics connection. The price thus drops to almost 9-10 $\$ per month with 1Mbit medium transfer. Of course this method has drawbacks, as many viruses run free on such newtorks, but it solves the price issue.[/quote]

I even didn't hear about such prices!
I pay 1 euro for 10Mb. Take into consideration that my month salary in University is about 25 euro.

[quote="Myth"]I even didn't hear about such prices!
I pay 1 euro for 10Mb. Take into consideration that my month salary in University is about 25 euro.[/quote]The prices above are the ones in Bucharest (where the internet and localnet networks are very expanded, so many people have internet -> thus is cheaper. I assume it's more expensive in the rest of the country  :?

1 euro for 10 Mb that is too much.
I hope that closer too, or after New Year, I will have an optic fiber connection, wich by the way seem quite cheap. Less than 20$ for 500 MB traffic each month plus extra traffic at about the same price...
Btw, I know that in students's dormitories, in Romania, there are very fast connections. My friends told me they sometimes watch movies without downloading them :)

[quote="Myth"]We also have local network at very high speed.
I download films in 2 minutes![/quote]Ah well there's the catch :) Do you pay local traffic as much as external traffic? If you don't look for people selling local proxies (there are tons of them here in Romania).

@dust: ask grobber about internet in campus :) most of the time indeed it's very fast ..

[quote="Myth"]We have free traffic inside our city network. Unfortunatelly, no one provides such proxies.[/quote]I can bet you 10$ that there is someone providing such proxies (especially in Russia :) ).  You just haven't asked about it. Enter mirc / dc++ channels in your city and ask for it. You will be amazed :)

[quote="Myth"]You don't realize all my problems. Local network is bounded to our city only. It contains 1.5-2 thousands computers. We pay for all outgoing traffic to WAN, even if destination server stays near to your home.[/quote]Mikhail I realize what you mean. What happens is that there is a person in your network (for sure) that has proxies for outside communication. This way your traffic is made through his computer and it's considered as local. His traffic is not considered local (but usually you pay him a small amount for this. Having more users they earn some money this way).

[quote="Myth"]Of course, I realized what you offered. But our network doesn't have such  philanthropists :)[/quote]They are not philanthropists. Not in Romania anyway :) They are just trying to make honest bucks. Unfortunately for them this "job" is not quite honest :D:D ...


Let $ \alpha\in\mathbb{R},\,\, T\in\mathbb{R}\setminus \{ 0\}$, and $ f: \mathbb{R}\to \mathbb{R}$ given by $ f(x) \equal{} \sin x \plus{} \cos \alpha x$. Prove that $ f(x \plus{} T) \equal{} f(x)$ if and only if $ \alpha\in\mathbb{Q}$.

I don't think that will ever work for an arbitrary value of T.

I think the proposer of the problem made a mistake in writing the condition. Maybe the equality $ f(x \plus{} T) \equal{} f(x)$ must hold for some $ T$, not for arbitrary $ T$.

There exists $ T\not\equal{}0$ s.t., for all $ x$, $ f(x\plus{}T)\equal{}f(x)$ iff  $ \alpha$ is a rational number.


inequalities
geometry
inequalities unsolved

[quote="andre.l"]Let O and H be the orthocenter and the circumncenter of the triangle ABC, respectively. Prove that $\frac{[AHB]}{[AOB]}+\frac{[AHC]}{[AOC]}+\frac{[BHC]}{[BOC]} \ge 3$.[/quote]

$\frac{[AHB]}{[AOB]}+\frac{[AHC]}{[AOC]}+\frac{[BHC]}{[BOC]}$
$=\frac{[ABC]}{[AOB]}+\frac{[ABC]}{[AOC]}+\frac{[ABC]}{[BOC]} - 6$
$=(a+b+c)\Big( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \Big) - 6$
$\ge 9 - 6 = 3$


calculus
integration
trigonometry
function
complex analysis
complex analysis unsolved

Prove that for $ 0 < \alpha < 1, R > 0$ and complex $ z$ such that $ 0 < |z| < R$, $ z \notin (0,R)$ the following is true:
$ z^\alpha \equal{} \minus{} \frac {\sin{\pi \alpha}}{\pi} \int_R^0 \frac {|t|^\alpha dt}{t \minus{} z} \plus{} \frac {1}{2\pi i} \int_{|z| \equal{} R} \frac {{\xi}^\alpha d\xi}{\xi \minus{} z}$

I've tried to apply integral Cauchy theorem to the analytic branch of the function $ f(\xi) \equal{} \xi^\alpha$, in the complex plane with the slit along the line $ (0, \infty)$, as an integration region i took one small disk around $ 0$ and the one with the radius $ R$ with upper and lower slits along the line $ (0,R)$. The problem is that the expression i got is the following:

$ z^\alpha \equal{} \minus{} \frac {1 \minus{} \exp(2\pi \alpha)}{2\pi i} \int_R^0 \frac {t^\alpha dt}{t \minus{} z} \plus{} \frac {1}{2\pi i} \int_{|z| \equal{} R} \frac {{\xi}^\alpha d\xi}{\xi \minus{} z}$

which is not quite what i need to prove. Would be grateful if someone could share ideas!

Apparently their idea was to write $ {\color{red}\minus{}}R$ as the lower limit in the first integral, which calls for the cut along the negative semi-axis and a different branch of $ \xi^\alpha$ in the circle.

thanks, fedja, indeed, i guess the point was that they cut along the negative semi-axis and one branch had the multiplier $ e^{\pi i \alpha}$, the other one  $ e^{ \minus{} \pi i \alpha}$, and the factor $ \sin{\pi \alpha}$ comes from the difference.
Thanks again :)


[quote="Nekruzjon_eko"]Prove that, for all $n,k$ $\in\N}$$N$ and $(k>1)$, the number $3^{n^{k}}+1$ is not divisible by $5$[/quote]
If m is odd $3^{m}=\pm 2 (mod \ 5).$
If 4|m then $3^{m}=1(mod \ 5)$. It give needed result.

How did you come up with those congruences?


simple nice inequalities
prove that $ a^{2} \plus{} b^{2} \plus{} c^{2} \ge 3$ $ For a \plus{} b \plus{} c \ge 3$ and a, b , c , are non negative

with caushy-shwarz:
$ a^2\plus{}b^2\plus{}c^2\geq\frac{1}{3}(a\plus{}b\plus{}c)^2\geq3$

[quote="anurags92"]thanks..... i am such a fool......
isnt there  ny other way coz before this exercise cauchy is not introduced  :oops:[/quote]
$ ab\plus{}bc\plus{}ac\le a^2\plus{}b^2\plus{}c^2$ because $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\ge 0$ so
$ 3\equal{}\frac{1}{3}(a\plus{}b\plus{}c)^2\equal{}\frac{a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ac)}{3}\le \frac{a^2\plus{}b^2\plus{}c^2\plus{}2(a^2\plus{}b^2\plus{}c^2)}{3}\equal{}a^2\plus{}b^2\plus{}c^2$

$ 3(a^2\plus{}b^2\plus{}c^2)\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(a^2\plus{}b^2\plus{}c^2) \ge a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca) \equal{} (a\plus{}b\plus{}c)^2 \ge 9$.


geometry
circumcircle
ratio
trigonometry
similar triangles
projective geometry
cyclic quadrilateral

Let a triangle $ABC$ and its circumcircle .The tangent to the circle  at $A$ meets $BC$ at $Z$ .Let the second tangent from $Z$ to the circle at $L$ and $AL$ meets $BC$ at $K$ .
Prove that the points $B,C$ , $Z,K$ make a hormonic division and  $AK$ is the symmedian of the triangle $ABC$ .

Let (O) be the circumcircle of $\triangle ABC$ and (P) the circumcircle of the quadrilateral AOLZ with right angles $\angle OAZ,\ \angle OLZ.$ Let (S), (T) be 2 congruent circles with diameters OB, OC, i.e., tangent to (O) at B, C.  Let (S) meet BC again at M. Since OB is a diameter of (S), the angle $\angle OMB = 90^\circ$ is right and M is the midpoint of BC. Similarly, (T) meets BC again at M. Let (P) meet BC again at M'. Since OZ is a diameter of (P), the angle $\angle OM'Z = 90^\circ$ is right and $M' \equiv M$ is the midpoint of BC. Thus (S), (T), (P) meet at O and again at M. Inversion in (O) carries the circle (P), passing through $A, L \in (O)$ and through the inversion center O, into the line AL. The congruent circles (S), (T) tangent to (O) at B, C and passing through the inversion center O go into tangents s, t of (O) at B, C. Consequently, the inversions s, t, AL of (S), (T), (P) in (O) meet at the inversion N of M, hence AL is the A-symmedian of $\triangle ABC.$

Using a central projection, project the circle (O) with the interior point K into a circle (O') and a point K', such that K' is the center of (O'). In this, the $\triangle ABC$ inscribed in (O) is projected into a $\triangle A'B'C'$ inscribed in (O') and the line AK into the line A'K'. Since $K' \in B'C',$ B'C' is a diametr of (O'), K' the midpoint of B'C', and the angle $\angle C'A'B' = 90^\circ$ is right. Thus A'K' is the A'-median of the right $\triangle A'B'C'.$ Tangents to (O) at B, C, intersecting at N on the A-symmedian AK of $\triangle ABC,$ go into parallel tangents to (O') at B', C', both perpenddicular to B'C', hence their intersection N' is at infinity and A'K' is also perpendicular to B'C'. Thus A'K' is the A'-altitude of $\triangle A'B'C'.$ As a result, the $\triangle A'B'C'$ is isosceles, which means that the tangent to (O') at A' is parallel to B'C' and its intersection Z' with B'C' is at infinity. The cross-ratio $\frac{KB}{KC}\cdot \frac{ZC}{ZB}= \frac{K'B'}{K'C'}\cdot \frac{Z'C'}{Z'B'}=-1$ is preserved in this central projection.

Silouan,this lemma is well-known!
Very nice yetti!

[b]Let' see another approach to this lemma:[/b]
Firstly with similar triangles,we prove that $\frac{ZB}{ZC}=\frac{c^{2}}{b^{2}}$.We also know that $AL$ is the polar of the point $Z$,which means that $Z,B,K,C$ make a harmonic division.That means,that$\frac{KB}{KC}=\frac{c^{2}}{b^{2}}$ and so we proved that $AK$ is symmedian for the triangle $ABC$ and particularly for the vertex $A$.
[i]
P.s.In this proof, I use some well known theorems.If you want me to prove any of them-just tell me![/i]

We have triangle $\ ZAC$ similar to triangle $\ ZBA$
$\Longrightarrow\frac{ZA}{ZB}=\frac{ZC}{ZA}=\frac{AC}{BA}$
$\Longrightarrow\frac{ZC}{ZB}=\frac{AC^{2}}{AB^{2}}$
Similarly: $\Longrightarrow\frac{ZC}{ZB}=\frac{LC^{2}}{LB^{2}}$
It follows that: $\frac{ZC}{ZB}=\frac{AC.LC}{AB.LB}=\frac{S(ALC)}{S(BLC)}=\frac{KC}{KB}$
$\Longrightarrow\frac{\overline{ZC}}{\overline{ZB}}=-\frac{\overline{KC}}{\overline{KB}}$
$\Longrightarrow (BCZK)=-1$

[quote="Nick Rapanos"]Silouan,this lemma is well-known!
[/quote]
I know this Nick .It is the lemma you used to prove my exercise .
But I posted here to see onother nice solutions .
And I think that only to read yetti's solution is all the money .
Thank you yetti .

[quote="silouan"]
And I think that only to read yetti's solution [u]is all the money .[/u]
Thank you yetti .[/quote]

Greek expression!!

[size=117][color=darkblue][b][u]The most elementary and short proof[/u][/b] (I like to gambol with ... LaTeX).[/color][/size]

[color=darkred][b][u]Lemma[/u][/b] (simple and well-known) $\Lambda\ (X,Y)\ : \ d=BC\ ,\ X\in d\ ,\ Y\not\in d\Longrightarrow$ $\frac{XB}{XC}=\frac{YB}{YC}\cdot \frac{\sin\widehat{XYB}}{\sin \widehat{XYC}}\ .$[/color]
[hide="Proof of the lemma."]
[color=darkblue]Apply the Sinus' theorem in the triangles $XYB$ and $XYC\ :$

$\{\begin{array}{c}\frac{XB}{\sin \widehat{XYB}}=\frac{YB}{\sin \widehat{YXB}}\\\\ \frac{XC}{\sin \widehat{XYC}}=\frac{YC}{\sin\widehat{YXC}}\\\\ \widehat{YXB}\equiv\widehat{YXC}\mathrm{\ or\ }m(\widehat{YXB})+m(\widehat{YXC})=180^{\circ}\end{array}$ $\Longrightarrow$ $\frac{XB}{XC}=\frac{YB}{YC}{\cdot\frac{\sin\widehat{XYB}}{\sin\widehat{XYC}}\ .}$[/hide][/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote $m(\widehat{KAB})=x$ , $m(\widehat{KAC})=y\ .$

$\{\begin{array}{c}LB=2R\sin x\ ,\ LC=2R\sin y\Longrightarrow\frac{LB}{LC}=\frac{\sin x}{\sin y}\\\\ \Lambda\ (Z,A)\Longrightarrow\frac{ZB}{ZC}=\frac{AB}{AC}\cdot \frac{\sin C}{\sin (A+C)}\Longrightarrow\frac{ZB}{ZC}=(\frac{c}{b})^{2}\\\\ \Lambda\ (Z,L)\Longrightarrow\frac{ZB}{ZC}=\frac{LB}{LC}\cdot\frac{\sin x}{\sin y}\Longrightarrow\frac{ZB}{ZC}=(\frac{\sin x}{\sin y})^{2}\\\\ \Lambda\ (K,A)\Longrightarrow\frac{KB}{KC}=\frac{AB}{AC}\cdot\frac{sin x}{sin y}\Longrightarrow\frac{KB}{KC}=\frac{c}{b}\cdot \frac{\sin x}{\sin y}\\\\ \end{array}\ \ \ \ \Longrightarrow\ \ \ \ \{\begin{array}{c}\frac{LB}{LC}=\frac{c}{b}=\frac{\sin x}{\sin y}\\\\ \frac{ZB}{ZC}=\frac{KB}{KC}=(\frac{c}{b})^{2}\end{array}\ \ \ \ \Longrightarrow$

$\{\begin{array}{cc}1.\blacktriangleright\ & AB\cdot CL=AC\cdot BL\ ,\ \mathrm{i.e.\ the\ quadrilateral\ ABLC\ is\ harmonically\ .}\\\\ 2.\blacktriangleright\ & \mathrm{\ the\ points\ K\ ,\ Z\ are\ conjugate\ w.r.t.\ the\ pair\ of \ points\ \{B,C\}\ .}\\\\ 3.\blacktriangleright\ & \mathrm{\ the\ ray\ [AK\ is\ A\--\ symedian\ in\ the\ triangle\ ABC\ .}\end{array}$

[b]Remark.[/b] $S\in ZO\cap AK\ ,$ where the point $O$ is the circumcenter of the triangle $ABC\ ,$ then $\widehat{KSB}\equiv\widehat{KSC}\ .$
[/color]

Another elementary proof of this problem is as follows:

Let O be the circumcenter of the triangle ABC and we denote as D, E, the intersection points of the circle (O), from the line ZO (D, between Z, O).

We denote also as F, the intersection point of BD, CE and as H, the one of BE, CD. So, the point H is the orthocenter of the triangle DEF (DC, EB, are perpendicular to EF, DF, respectively).

Hence, the line FH, is perpendicular to DE and we wiil prove that passes through the point L.

We denote as N, the intersection point of the lines DE, AL. It is well known that the points Z, N, are harmonic conjugates with respect to the points D, E (Proof. – We denote as D', E', the intersection points of the line ZL, from the lines through D, E respectively and tangent to the circle (O). So, we have that ZD / ZE = DD' / EE' = LD' / LE' = ND / NE).

From the complete quadrilateral FBHCDE, we conclude (easy to prove by Menelaus's and Ceva's theorems) that the line FH, passes through the point N (as the harmonic conjugate of Z, wrt D,E) and so, it is coincided with the line AL, as perpendicular to DE.
We conclude also  that the points Z, K, are harmonic conjugates with respect to B, C.

Let A' be the orthogonal projection of A on BC and M the midpoint of the segment BC.

Because of the perpendicularities of the lines AA', BC and AL, DE, => <A'AK = <MZO, (1)

From the cyclic quadrilateral AZMO => MZO = <MAO, (2).

From (1), (2), => <A'AK = <MAO and the proof is completed (because of AA', is isogonal of AO with respect to the angle <A).

Kostas Vittas.

[quote="vittasko"][color=darkred]An elementary proof of this problem is as follows:
... Hence, the line FH, is perpendicular to DE and passes through the point L ([b][u]easy[/u][/b] to prove if we consider the line ZBC, turn around the point Z.
The intersection point H, of DC, EB, lies always on the line through F and perpendicular to DE). We conclude now, that the lines AL, FH, are coincided ... [/color][/quote]
[color=darkblue][b]Why[/b] $L\in FH\ ?$ The point $F$ is mobile once with the ray $[\overline{ZBC}$from $Z$ and, in the my opimion, you can infer only $FH\parallel AL$ .... [u]Please, detail you ![/u]

I can continue thus : denote $V\in AL\cap ZO\ ,\ S\in HF\cap ZO\ .$ Therefore, $AL\perp ZO\ ,\ HF\perp ZO\Longrightarrow$ $AL\parallel HF\ .$

$\{\begin{array}{c}OD^{2}=OA^{2}=OV\cdot OZ\Longrightarrow OD^{2}=OV\cdot OZ\Longrightarrow\mathrm{V\ is\ conjugate\ of\ Z\ w.r.t.\ the\ points\ D\ ,\ E\ .}\\\\ H\in DC\cap EB\cap FS\ ,\ Z\in BC\cap DE\Longrightarrow\mathrm{S\ is\ conjugate\ of\ Z\ w.r.t.\ the\ points\ D\ ,\ E\ .}\end{array}$ $\Longrightarrow$

$V\equiv S\mathrm{\ and\ }\{\begin{array}{c}V\in AL\\\ S\in HF\end{array}\ ,\ AL\parallel HF\Longrightarrow AL\equiv HF\ .$[/color]

Thank you very nuch dear Virgil, for your remarks and advices.

I tried to solve the problem, with basic geometry tools and I am sorry because of some failing was in the proof.

Now, I have been corrected and completed the notations, in the solution I presented and I hope that there is not any indefiniteness again.

Kostas Vittas.


Suppose you had two algorithms, one the speed of 8n^2, and the other of the speed of 64 * n *(log base 2 of n). At what point would the 64n log2 n become more efficient?

do you mean
a) at what point would te second expressing be smaller (have a faster execution time) no matter what n is chosen as long as $ n > k$

or

b) at what point would the second expression be larger (have a faster speed) no matter what n is chosen as long as $ n > k$\

assuming you mean the first one (because the second option gives an upper bound, not a lower bound)
[hide]
you want
$ 8n^2 > 64 n \log_2 n$
$ n^2 > 8 n \log_2 n$
$ n>8 \log_2 n$ (n is nonzero because you cannot take the logarithm of 0 for any base)
$ 2^n>2^{8 \log_2 n}$
$ 2^n>2^8 \cdot 2^{\log_2 n}$
$ 2^n > 256 n$
i am stuck here
:(
[/hide]

Yes, I meant the first one, and I did get stuck in the same spot, quite ironically.

I'm certain there is a solution, since graphing it led to an intersection point (somewhere around 1.1, if memory serves correct...); however, I have no idea how this problem can be solved algebraically, and since a graph yields a solution, then there has to be an algebraic solution to it.

On the other hand, I am abysmal at all forms of mathematics.

[quote="math154"]There's no elementary solution to this...[/quote]

Sorry for any ambiguity.

I did graph it and the intersection is around $ 1.1$...I meant there wasn't any easy way to FIND the solution (naturally, without a calculator).

This is a [url=http://en.wikipedia.org/wiki/Transcendental_equation]transcendental equation[/url], which doesn't look so easy. :wink:

[quote="stevenmeow"]
[hide]
you want
$ 8n^2 > 64 n \log_2 n$
$ n^2 > 8 n \log_2 n$
$ n > 8 \log_2 n$ (n is nonzero because you cannot take the logarithm of 0 for any base)
$ 2^n > 2^{8 \log_2 n}$
$ 2^n > 2^8 \cdot 2^{\log_2 n}$
$ 2^n > 256 n$
i am stuck here
:(
[/hide][/quote]

Mixed some exponentiation properties there...

$ 2^{8 \plus{} \log_2 n} \equal{} 2^8 \cdot 2^{\log_2 n}$
$ 2^{8 \log_2 n} \equal{} (2^{\log_2 n})^8 \equal{} n^8$

EDIT: I've used this problem as an excuse to play with Newton's method (I don't like graphing :P) -- indeed, the value of $ n$ is [i]very[/i] close to $ 1.1$ (about $ 1.09999...$)

There are two solutions to the equation you all have been playing with.

We want to know when the quasilinear-time algorithm will become [i]more[/i] efficient than the quadratic-time algorithm. Therefore, we are interested in the [b]larger solution[/b].

[hide="Computationally Light Approximation"]$ 8n^2 \equal{} 64 n \log _2 n$
$ n \equal{} 8 \log _2 n$
$ 2^m \equal{} 8m \; ; \qquad 2^m \equal{} n$

If $ m \equal{} 5$, we have $ 32 < 40$ and if $ m \equal{} 6$ we have $ 64 > 48$.

Looking somewhere in between, we find that $ m \equal{} 5.5$ results in $ 2^m \equal{} 32 \cdot 1.414 \approx 45.25$ and $ 8m \equal{} 44$. So $ m$ is less than $ 5.5$, and $ n$ is less than $ 8m \equal{} 44$
$ \textbf{Note: }$ [i]At $ m \equal{} 5.5$, both $ 2^m$ and $ 8m$ have increased from the equality point, so value at the equality point must be less than the lower of $ 2^{5.5}$ and $ 8 \cdot 5.5$.[/i]

Now we shift down a bit and look for a lower bound. Note that$ 2^{1/3} < 1.3$. Since $ 32 \cdot 2^{1/3} < 32 \cdot 1.3 < 8 \cdot 5.\overline{3}$ (ie, $ m \equal{} 5.\overline{3}$ is too low), we can use $ 8 \cdot 5. \overline{3} \equal{} 42.\overline{6}$ as an upper bound.
$ \textbf{Note: }$ [i]Here, $ 8m > 2^m$, so again, $ 8m$ provides the tighter bound.[/i]

This gives you $ \boxed{n \in (42.\overline{6} , 44)}$, which is reasonably narrow for light computation by hand.

Since $ 2^{1/3}$ is significantly less than $ 1.3$, the margin between $ 8m$ and $ 2^m$ is quite a bit larger at $ m \equal{} 5.\overline{3}$ than at $ 5.5$. This, added with the fact that $ 2^m$ is convex in $ m$, suggests that the actual value is much closer to $ 44$ than to $ 42.\overline{6}$. Indeed, the actual value is somewhere in $ (43.5,43.6)$.[/hide]

On a practical note, an efficiency comparision that states that one algorithm is more efficient for all $ n > 1.1$ is making a very strong statement. It means that the less efficient algorithm is pretty much useless compared to the more efficient one, since the less efficient one is only favorable for one particular input, $ n \equal{} 1$, and that too, minimally so!

On the other hand, in this case, the quasilinear is only more efficient for $ n > \sim 43.5$, so if we had to pick one algorithm to use consistently, we would have to decide based on the expected distribution of inputs.


could you post previos mathcounts tests?

oh, well.
it would have been cool to have mathcounts practice.
does anyone know a website that has mathcounts tests?

I wouldn't practice with Mathcounts Tests now, save them for closer to the actual exams, there are very few years of tests, so it's a limited resource you shouldn't use up so quickly.

Will tell Mathcounts team about web page. :cool:

I'm confused: If full reproduction of the Mathcounts tests is illegal and a violation of copyright, are the sites that have old Mathcounts stuff (as mentioned in the sticky thread "Places to get old Mathcounts stuff") illegal? :o Or how would they do it?

iF you pay them money, it's not illegal  :(

Yes, and if you post them without selling them...right?

[quote="Treething"]It can't be that hard. Just modify the questions a little bit, a misspelling here and there, a little change of wording...[/quote] I think that would still be plagiarism... :?

Paraphaseation is the way to go, but you must put something like "originated from Mathcounts Intergalactic competition, 30101" underneath the problem.

[quote="MCrawford"]No, they are copyrighted by MATHCOUNTS and their full reproduction would be a violation of those rights.[/quote]

What if we gave credit to MATHCOUNTS?

[quote="MCrawford"]We have built a new [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_MATHCOUNTS.php]MATHCOUNTS page[/url] explaining both the MATHCOUNTS competition as well as the opportunities that we provide for MATHCOUNTS students.

Let us know if you think there is anything that we should add.

Thanks,
Mathew[/quote]

What about the problems that MATHCOUNTS posts as the problem(s) of the week? :?:

Or we could add the MATHCOUNTS toolbox, which is basically a packet of things that you should understand/memorize that would help. It's at:

http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf

It's a good idea to edit your previous post instead of posting a new one in a few minutes, so that nobody can accuse you of double-posting and spamming. I realize you're new. :welcome:

There's a "MATHCOUNTS bible" out there (somewhere) that tells everything you should know for MathCounts and the rest of life.

Intergalactic competition?! O_O


Function OMK
HOw to solve these type of function questions
2005.5
Find all functions f:R->R such that f(x+y)=f(x)f(y)f(xy) for all real x and y.

2004.4
Find all real f:R->R f(x^2 + y) = 2 f(x) + f(y^2) for all real x,y

2003.5
f is set of postive number, find f which satisfied n:->m
(i) f(xy)= f(x)f(m/y)+f(y)f(m/x)
(ii) f(1)=1/2

thanks for helping me

[quote="hotheadhacker_mathslover"]
2005.5
Find all functions f:R->R such that f(x+y)=f(x)f(y)f(xy) for all real x and y.[/quote]
If $ f(u)\equal{}0$ for some $ u$, then $ f(x\plus{}u)\equal{}0$ $ \forall u$ and so $ f(x)\equal{}0$ $ \forall x$
If $ f(x)\neq 0$ $ \forall x$ then :
$ f(x\plus{}y\plus{}z)\equal{}f(x)f(y\plus{}z)f(xy\plus{}xz)$ $ \equal{}f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x^2yz)$
And so $ f(x^2yz)\equal{}f(xy^2z)\equal{}f(xyz^2)$ $ \forall x,y,z$ and so $ f(u)\equal{}f(v)$ $ \forall u,v\neq 0$ and so $ f(x)\equal{}c$ $ \forall x$ and so three solutions :

$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}\minus{}1$ $ \forall x$
$ f(x)\equal{}\plus{}1$ $ \forall x$
[quote="hotheadhacker_mathslover"]2004.4
Find all real f:R->R f(x^2 + y) = 2 f(x) + f(y^2) for all real x,y[/quote]
$ x\equal{}y\equal{}0$ $ \implies$ $ f(0)\equal{}3f(0)$ and so $ f(0)\equal{}0$
$ x\equal{}0$ $ \implies$ $ f(y)\equal{}f(y^2)$
$ y\equal{}0$ $ \implies$ $ f(x^2)\equal{}2f(x)$
So $ f(x)\equal{}2f(x)$ and a unique solution $ f(x)\equal{}0$ $ \forall x$

[quote="hotheadhacker_mathslover"]2003.5
f is set of postive number, find f which satisfied n:->m
(i) f(xy)= f(x)f(m/y)+f(y)f(m/x)
(ii) f(1)=1/2

thanks for helping me[/quote]
I dont understand "f is set of postive number, find f which satisfied n:->m"
Sorry

thx alot guys~and i noe whr to post them now~


I have the option of doing both this summer. I want to go to the USAMO this year, but I also want to go to the RSI in the summer of 2008. Should I spend more time doing research or doing AoPS 1/2, ACoPS, etc.? I finished my sophomore year, and I'll become a junior this fall. I want to become a mathematician when I grow up.

As far as I know, you don't need any research experience to go to RSI.

If you have a mentor and a project in mind for research, by all means go for it. Personally, I think it's a much more fulfilling experience than training for the USAMO. That said, if you think you're just planning on fiddling around with a problem aimlessly as "research," then perhaps you can get more out of the problem solving and proof writing skills from USAMO practice. The above poster is correct in saying that you don't need any research experience to go to RSI, but it certainly helps, and accomplishing something in your research shows that you have the capability to utilize what RSI has to offer which, in my opinion, is what they are looking for.