[b]1/[/b]Let $ x \plus{} y \equal{} 2008$ and $ x, y$ are integers.
Find the max and min value of
a) $ A \equal{} x!\cdot y!$
b) $ B \equal{} x! \plus{} y!$
[b]2/[/b]Do the same problem with $ x \plus{} y \equal{} 2009$

[quote="Yongyi781"]It is assumed that $ x$ and $ y$ are integers, right? :maybe:[/quote]
Yeah, right! :)

[hide="Part A"]$ 2008! \equal{} \binom{2008}{x} \cdot A$

To maximize $ A$, we minimize $ \binom{2008}{x}$

Assuming you want $ x,y \geq 0$, the minimum occurs $ x\equal{}0$ or $ x\equal{}2008$, in which case $ \binom{2008}{x} \equal{} 1 \implies A \equal{} 2008!$

The minimum value occurs when we maximize $ \binom{2008}{x}$, which happens when $ x \equal{} y \equal{} 1004 \implies A \equal{} (1004!)^2$[/hide]

[quote="TZF"][hide="Part A"]$ 2008! \equal{} \binom{2008}{x} \cdot A$

To maximize $ A$, we minimize $ \binom{2008}{x}$

Assuming you want $ x,y \geq 0$, the minimum occurs $ x \equal{} 0$ or $ x \equal{} 2008$, in which case $ \binom{2008}{x} \equal{} 1 \implies A \equal{} 2008!$

The minimum value occurs when we maximize $ \binom{2008}{x}$, which happens when $ x \equal{} y \equal{} 1004 \implies A \equal{} (1004!)^2$[/hide][/quote]
Nice solution! :coolspeak:
Can anyone prove part b :?:

For part B, just use Jensen's Inequality.

For #2, everything is the same, except that since 2009 is an odd number, 2a. gets minimized when $ (x,y)\equal{}(1004,1005)$ or $ (1005,1004)$

[quote="TZF"]For #2, everything is the same, except that since 2009 is an odd number, 2a. gets minimized when $ (x,y) \equal{} (1004,1005)$ or $ (1005,1004)$[/quote]
I think you mean "maximized."

Nah

Recall that $ x!y! \equal{} \frac{2009!}{\binom{2009}{x}}$. Then, $ x\equal{}1004$ or $ x\equal{}1005$ will maximize the denominator, and therefore minimize $ x!y!$

Oh, sorry, I thought it was $ \binom{2009}{x}$ being maximize.

[quote="TZF"]For part B, just use Jensen's Inequality.

For #2, everything is the same, except that since 2009 is an odd number, 2a. gets minimized when $ (x,y) \equal{} (1004,1005)$ or $ (1005,1004)$[/quote]
Can you prove with another method? :huh:
I haven't learned about Jensen's Inequality! :P

More problems:
Let $ x\plus{}y\plus{}z\equal{}2008$. Find the min and max values of:
$ A\equal{}x!\plus{}y!\plus{}z!$
$ B\equal{}x!\cdot y!\cdot z!$

basically for x+y = 2009
assume x> y
if x-y > 1 we will prove that x!+y! > (x-1)!+(y+1)! which is equivalent to (x-1)!(x-1)>y!y which is true
so obviously , the minimum occur if and only if x-y =1

we can use the same method for the last problem


[quote="rem"]Most contests on AOPS have an option of "downloading a PDF version". However, when I click on the option, the same page comes up. Can someone please help me?

Thank you.[/quote]
This has been reported before, it isn't a bug it's intentional.

[quote="#H34N1"]Why is it intentional :([/quote]
Because it's against copyright laws...
How would you feel if you were selling a book of problems, but the book was online and easily available at the place where most who would buy your book are members?

[quote="ccy"]I really don't think it is intentional; the administrators have a lot of more important things on their hand than debugging pdftex issues.[/quote]I'm moving so it'll be another 2 weeks until I can sort all that stuff out. Keep the reports coming though I will certainly deal with all of them sooner or later.

[quote="bpms"][quote="#H34N1"]Then why HTML format? Its still available online[/quote]
It isn't easily printable[/quote]

Copy&Paste into MS Word. :)

[quote="Valentin Vornicu"]I'm moving so it'll be another 2 weeks until I can sort all that stuff out. Keep the reports coming though I will certainly deal with all of them sooner or later.
[/quote]

That'd be great. :)

I actually wrote a [url=greasemonkey.mozdev.org]Greasemonkey[/url] script to make it well formatted for printing.


trigonometry
geometry
circumcircle
similar triangles
geometry unsolved

in a triangle $ ABC$  let $ M$  a point in $ [BC]$ such that $ AB\equal{}MC\equal{}1$ and $ \widehat{BAM}\equal{}90$° and $ \widehat{MAC}\equal{}30$°
calcule $ BM$

The equivalent trigonometry problem:

Let $ \alpha$ be an angle in $ (\frac{\pi}{6},\frac{\pi}{2})$ such that $ 2\sin{(\alpha \minus{} \frac{\pi}{6})}\tan{\alpha} \equal{} 1$
Calculate $ \frac{1}{\sin{\alpha}}$

$ \angle CAB \equal{} \angle CAM \plus{} \angle BAM \equal{} 120^\circ\ \Longleftrightarrow\ a^2 \equal{} b^2 \plus{} bc \plus{} c^2.$ Let AK bisect $ \angle CAB$ internally, $ AK \equal{} \frac {\sqrt {bc [(b \plus{} c)^2 \minus{} a^2}]}{b \plus{} c} \equal{} \frac {bc}{b \plus{} c},$ $ BK \equal{} \frac {ca}{a \plus{} b},$ $ KC \equal{} \frac {bc}{a \plus{} b}.$ AM now bisects the $ \angle CAK$ of the $ \triangle AKC$ with sides $ a' \equal{} KC \equal{} \frac {bc}{a \plus{} b},\ b' \equal{} b,\ c' \equal{} AK \equal{} \frac {bc}{b \plus{} c}.$ $ MC \equal{} \frac {a'b'}{b' \plus{} c'} \equal{} \frac {ab}{b \plus{} 2c}.$ Given is $ MC \equal{} AB \equal{} c,$ hence $ \frac {ab}{b \plus{} 2c} \equal{} c.$ Then $ KM \equal{} \frac {c'a'}{b' \plus{} c'} \equal{} \frac {abc}{(b \plus{} c)(b \plus{} 2c)} \equal{} \frac {c^2}{b \plus{} c}$ and $ BM \equal{} BK \plus{} KM \equal{} \frac {c(c \plus{} a)}{b \plus{} c}.$

Substituting now $ c \equal{} AB \equal{} 1,$ $ d \equal{} BM \equal{} \frac {a \plus{} 1}{b \plus{} 1} \equal{} \frac {x}{y},$ which we need to calculate using 2 equations:

$ a^2 \equal{} b^2 \plus{} bc \plus{} c^2,\ \ \ ab \equal{} c(b \plus{} 2c)$

$ a^2 \equal{} b^2 \plus{} b \plus{} 1,\ \ \ ab \equal{} b \plus{} 2$

Substituting $ x \equal{} a \plus{} 1,\ y \equal{} b \plus{} 1,$ where $ x > 1,\ y > 1,$

$ (x \minus{} 1)^2 \equal{} (y \minus{} 1)^2 \plus{} y,\ \ \ (x \minus{} 1)(y \minus{} 1) \equal{} y \plus{} 1$

$ x(x \minus{} 2) \equal{} y(y \minus{} 1),\ \ \ xy \minus{} (x \plus{} y) \equal{} y$

From the 2nd equation, $ x \equal{} \frac {2y}{y \minus{} 1}.$ Substituting to the 1st one, we get $ 4y \equal{} y(y \minus{} 1)^3.$ Since y > 1, we can reduce, yielding $ y \equal{} \sqrt[3]{4} \plus{} 1.$ Then $ x \equal{} \frac {2y}{y \minus{} 1} \equal{} \sqrt[3]{2} \left (\sqrt[3]{4} \plus{} 1 \right)$ and $ BM \equal{} d \equal{} \frac {x}{y} \equal{} \sqrt[3]{2}.$ Not constructible.

[quote="greatestmaths"][color=darkred]In a triangle $ ABC$  let $ M$  a point in $ [BC]$ such that $ AB = MC = 1$ and $ \widehat{BAM} = 90$° , $ \widehat{MAC} = 30^{\circ}$. Calculate $ BM$.[/color] [/quote]
[color=darkblue]Denote $ MB = x$. Observe that $ \|\begin{array}{c} m(\widehat {BAC}) = 120^{\circ} \\
 \\
a = x + 1\ ;\ c = 1 \\
 \\
AM^2 = x^2 - 1\end{array}$.

[b][u] Proof I.[/u][/b] $ \frac {MB}{MC} = \frac {AB}{AC}\cdot\frac {\sin\widehat {MAB}}{\sin\widehat {MAC}}$ $ \implies$ $ \frac x1 = \frac 1b\cdot \frac {1}{\frac 12}$ $ \implies$ $ \boxed {\ xb = 2\ }\ .$

Aplly the [b]generalized Pythagoras' theorem[/b] in $ \triangle\ ABC$ : $ (x + 1)^2 = b^2 + b + 1$ $ \implies$

$ (x + 1)^2 = \frac {4}{x^2} + \frac 2x + 1$ $ \implies$ $ x^3(x + 2) = 2(x + 2)$ $ \implies$ $ \boxed {\ x = \sqrt [3]2\ }\ .$

[b][u]Proof II.[/u][/b] Apply the [b]generalized Pythagoras' theorem[/b] to the side $ [BC]$ in $ \triangle\ ABC$ :

$ (x + 1)^2 = b^2 + 1 + b$ $ \implies$ $ \boxed {x(x + 2) = b^2 + b}\ \ (1).$

Aplly the [b]Stewart's theorem[/b] to the cevian $ [AM]$ in $ \triangle\ ABC$ :

$ (x^2 - 1)(x + 1) + x(x + 1) = 1 + b^2x$ $ \implies$ $ \boxed {x^2(x + 2) = 2 + b^2x}\ \ (2).$

From the relations $ (1)$ and $ (2)$ obtain $ x(b^2 + b) = 2 + b^2x$ $ \implies$ $ xb = 2$ $ \implies$

$ x^2(x + 2) = 2 + 2b$ $ \implies$ $ x^3(x + 2) = 2(x + 2)$ $ \implies$ $ \boxed {\ x = \sqrt [3] 2\ }\ .$  [/color]

Construct a perpendicular to BC at C. It will intersect AM on the
  circumcircle of ABC. From the similar triangles and maybe
  with the help from Ptolemy theorem (and without the  "sophisticated"
  rigamarole ) calculate BM to what it is  supposed to be.

  Thank you.
  M.T.

[quote="greatestmaths"][color=darkred]In a triangle $ ABC$  let $ M$  a point in $ [BC]$ such that $ AB \equal{} MC \equal{} 1$ and $ \widehat{BAM} \equal{} 90$° , $ \widehat{MAC} \equal{} 30^{\circ}$. Calculate $ BM$.[/color] [/quote]
[color=darkblue][b][u]Proof III[/u][/b] ([u]without the "sophisticated" rigamarole - the Ptolemy's theorem[/u]).

Denote $ N\in (AC)$ for which $ MN\perp MA$ and $ BM \equal{} x$ , $ MN \equal{} y$.  Observe that

$ MN\parallel AB$ , $ AN \equal{} 2y$ and $ AM \equal{} y\sqrt 3$. Thus, $ 3y^2 \plus{} 1 \equal{} x^2$ and $ \frac {1}{x \plus{} 1} \equal{} \frac y1$.

Therefore, $ (x \plus{} 1)^2(x^2 \minus{} 1) \equal{} 3$ $ \implies$ $ (x \plus{} 2)(x^3 \minus{} 2) \equal{} 0$ $ \implies$ $ x \equal{} \sqrt [3]2\ .$[/color]


calculus
integration
function
inequalities
trigonometry
calculus computations

Prove that:

\[ \int_0^\infty\frac x{x^{\pi+2}+1}dx\le \frac \pi {12} \]

impossible see it by numerical integration

It is true. The integral turns out to be about $0.650$, but $\frac{\pi}{12} \approx 0.262$.

By the way, thanks for introducing me to a [url=http://mathworld.wolfram.com/LerchTranscendent.html]new special function[/url] that I hadn't heard of before. :D

edit: It would be cool to see a proof without a computer that evaluates the integral, or gives a decently good upper bound. I'll try to do it later today if I get a chance.

Of course it has a solution without using a calculator, else it haven't been proposed for a math contest. Here it is:

$\int_0^\infty\frac x{x^{\pi+2}+1}dx\le \int_0^\infty\frac x{x^{5}+1}dx=$
$\int_0^1 \frac{\frac 1 y}{(\frac 1 y)^5+1} \frac 1 {y^2}dy=\int_0^1 \frac{y^2}{1+y^5}dy\le$
$\int_0^1 \frac{y^2}{y^6+1}dy=\frac{\pi} {12}$.

A couple of questions:

Are you sure $\frac{x}{x^{\pi+2} + 1} \le \frac{x}{x^5 + 1}$ for all positive $x$? I don't think it is. What about $x \in (0,1)$? How do you justify that first inequality, then?

When you substitute $x = \frac{1}{y}$, shouldn't your new bounds of integration still be $0$ and $\infty$? (That'll give you $\frac{\pi}{6}$, but it's still too small.)

But the first one is the one that matters. Try evaluating the integral numerically -- you'll see that the inequality doesn't hold.

edit: Maple tells me the actual value is $\frac{\pi}{(\pi + 2)\sin(\frac{2\pi}{\pi+2})} \approx 0.650158 > \frac{\pi}{12}$.

[quote="Ramanujan"]I think the problem should be $\int_1^\infty\frac x{x^{\pi+2}+1}dx\le \frac \pi {12}$     :?[/quote]

But then it would be easy since $\frac x{x^{\pi+2}+1} \leq \frac{x^2}{x^6 + 1}$ and $\int_1^\infty \frac{x^2}{x^6 + 1}dx =\frac \pi {12}$ .  :?


[quote="shfdfzhjj"]$ x_1,x_2,\ldots,x_n\ge0,x_1 \plus{} x_2 \plus{} \cdots \plus{} x _n \equal{} 1$,then
$ \sum_{cyc}\sqrt {1 \minus{} x_1x_2}\ge\sqrt {n^2 \minus{} 1}$[/quote]
The inequality holds only for $ n\equal{}2$ and $ n\equal{}3.$ For $ n \ge 4$, you can try $ x_1\equal{}x_1\equal{}1/2,x_3\equal{}\cdots \equal{}x_n \equal{}0.$ And in fact, for all $ n \ge 4$, we have
$ n \ge \sum^n_{i\equal{}1} \sqrt{1\minus{}x_ix_{i\plus{}1}} \ge n\minus{}1\plus{}\frac{\sqrt{3}}{2}.$ (In here, we denote $ x_{n\plus{}1} \equiv x_1$).

What about $ \sum_{i\equal{}1}^n{x_i^2}\equal{}1,x_i\ge0$
$ \sum_{i\equal{}1}^n{\sqrt{1\minus{}x_ix_{i\plus{}1}}}\ge\sqrt{n(n\minus{}1)}$ :maybe:


Let $a$ be the largest positive root of $x^{3}-3x^{2}+1=0$.Prove that $[a^{1788}]$ and $[a^{1988}]$ are divisible by $17$.

thanks rust
wow
thats a nice solution. cud u pleeees tell me how u thot of using $x_{n}$?
i mean, how does it strike wat to use in such conditions?
actually, in my country, there isnt much scope for lrning all these coool and nice stuff. its only the bookish stuff that counts here. but i like these. however, i dont get adequate stuff.
pls help me.
thanks again


integration
calculus
calculus computations

Calculate :
 1)$ \int \int \int_{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \equal{} 1}{(x_1x_2 \plus{} x_2x_3 \plus{} x_1x_3)dx_1dx_2dx_3}$
  2)$ \int \int ....\int_{x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} ... \plus{} x_n^2 \equal{} 1}{(\sum_{1 \leq i < j \leq n}{2x_ix_j})dx_1dx_2...dx_n}$
 It hope that you are interested in it ... :play_ball:

I hope that you are interested in it  ....  :play_ball:  thanks you very much


i'm officially amused.   :lol: so why is that "required for canadian residents only"?

I am missing something. What is this sweepstake?

did you click on the link and look at the 9th line of the entry form?  like i said, i'm amused  :lol:

oops. I had gone to the tenth line! I am getting some diead aboutthe the standard of the USAMO though!

Perfect, Penguin!
You would cruise through the USAMO!

Yay!
Then again, I would have to qualify first.

[quote="Ignite168"]http://www.buildabear.com/lp/CelebearateMom/celebearateMom_entryForm.aspx

Look at the 9th line ;)[/quote]
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
 :rotfl:  :rotfl:    :rotfl: They are all laughing

why is this only for Canadian residents? cuz they are smart?

minor side note: does anyone else think "cele-bear-ate" is extremely corny/odd-sounding?

[quote="kheg_k"]i think it s one of those things, like those validation codes u type in when signing up for an online service, game, ... ect, ect

except for some reason maybe build a bear has protection from US computers?

i dunno[/quote]

Umm, why would you need to prevent automated bots from mass mailing a contest form?

I think it combines celebrity with bear and ate (for nice applications of 'bear' 'ate' see the second post of the never ending story!)


trigonometry
calculus
derivative
function
integration
limit
modular arithmetic

[b]Summary:[/b]

Can anyone give me any examples of situations in which the "closed" form of a series is a useful indicator of its behavior outside of the series' interval of convergence?

[b]Motivation:[/b]

Someone I know tells me she "figured out" the shape of a sawtooth wave from it's Fourier Series expansion, without knowing a thing about Fourier Series. She was using the sawtooth defined by $ \sum_{k \equal{} 1}^{\infty} \frac {\sin k t}{k}$. She told me she first found the roots, and then took the derivative. For the derivative, she got $ \sum_{k \equal{} 1}^{\infty} \cos kt$, which she expressed using polar complex numbers, and then used the geometric series formula to get a derivative of $ \minus{} 1/2$, except at points of discontinuity.

It seems to me that her result describes this particular sawtooth correctly. However, the use of the geometric series formula was invalid, since the series does not converge in the first place.

Is this by coincidence? If not, are there other cases in which switching between a function and it's series outside the interval of convergence is actually meaningful?

There's the whole thing about formal power series which is quite useful. I've only seen them put to use in combinatorial environment, however. Perhaps this is the sort of thing you mean?

The use of the geometric series formula can be explained by various methods for [url=http://en.wikipedia.org/wiki/Divergent_series]summing divergent series[/url].  Cesaro summation is a popular one in Fourier analysis, so that's probably the appropriately "rigorous" argument here.

A general situation in which this is true goes by the name of [url=http://en.wikipedia.org/wiki/Analytic_continuation]analytic continuation[/url], where it's often still possible to assign meaningful values outside of a series' radius of convergence unless the circle of convergence forms a "natural barrier."  For example, this is true of $ f(x) \equal{} \frac {1}{1 \minus{} x}$ but not true of

$ f(x) \equal{} \prod_{k \equal{} 1}^{\infty} \frac {1}{(1 \minus{} x^k)}$.

(Minor typo: I'm sure TZF meant those series to start at $ k=1.$

For one thing, in my mind, looking at a Fourier series like this from a power series perspective is not really looking "outside" the radius of convergence but rather looking on the circle of convergence.

One thing which I exploit as much as I can: the language of distributions and weak convergence justifies a considerable number of things that lie outside of classical convergence. From that point of view, this result is no coincidence at all.

We start with $ f(t)=\sum_{k=1}^{\infty}\frac{\sin kt}t.$ I won't worry too much about classical convergence and just take for granted that it most likely converges weakly.

As long as we're talking both weak limits and weak derivatives, we may interchange limit and integral without restriction, so $ f'(t)=\sum_{k=1}^{\infty}\cos kt$

As you've noted, that series doesn't converge classically anywhere - but it should still converge weakly. We try using the language of geometric series. One way to get at this is to introduce an $ r<1.$
\begin{align*}f'(t)&=\sum_{k=1}^{\infty}\cos kt\\ &=\lim_{r\to 1^-}\sum_{k=1}^{\infty}r^k\cos kt\\ &=\lim_{r\to1^-}\text{Re }\left(\sum_{k=1}^{\infty}r^ke^{ikt}\right)\\ &=\lim_{r\to 1^-}\frac{r\cos t-r^2}{1-2r\cos t+r^2}\end{align*}
(I left out the intermediate computational steps involving the geometric series, but you can reproduce those for yourself easily enough.)

Now, look at that last limit. In classical terms, it diverges where $ t\equiv 0\pmod{2\pi}$ but converges pointwise everywhere else to $ -\frac12.$ That "everywhere else" convergence is reasonably nice - uniform on compact subsets of $ (0,2\pi),$ for instance. But there's clearly also a portion of this supported on $ \{0\}$ that can't be described as a function.

Since we're looking at the derivative of a periodic function, we should have that $ f'(t)$ integrates to zero over any full period. Since $ \int_0^{2\pi}-\frac12\,dt=-\pi,$ we must actually have that
\[ f'(t)=\pi\delta(t)-\frac12\]
where that's the Dirac delta.

Integrating this back gives the sawtooth wave, including correctly giving the location and size of the jump discontinuity.

I find this kind of thinking - using weak derivatives and weak limits - to be very attractive when working with Fourier series and Fourier transforms computationally. It has an Eulerian (calculate first and justify later) flavor.


n is natural number, p is prime, prove:

p^2 divides ((((np)!)/((np-p)!)(p!)) - n)

 :ninja:

Newton¿

In general we have $ \binom{pa}{pb} \equiv \binom{a}{b} \mod p$.
This still holds $ \mod p^2$ and if $ p>3$ we get it $ \mod p^3$. It can be sharpened even more (but the gets a bit messy to prove).


abstract algebra
superior algebra
superior algebra unsolved

If $ f: A\rightarrow B, g: C\rightarrow D$ are two module isomorphisms of R-modules, then show that $ f\otimes g$ is an isomorphism.

i'm having trouble with this because $ f(a)\otimes g(c)\equal{}f(a')\otimes g(c')$ does not necessarily imply $ f(a)\equal{}f(a'),g(c)\equal{}g(c')$

You have inverse isomorphisms $ f^{\minus{}1}$ and $ g^{\minus{}1}$. Now try to prove that $ f^{\minus{}1} \otimes g^{\minus{}1}$ is the inverse of $ f \otimes g$.


[hide]Foiling the right side,

$4x^{2}+20x+a = 4x^{2}+4bx+b^{2}$

Equating like terms yields the system:

$20x = 4bx$ and $a = b^{2}$, which can easily be solved to find that $(a, b) = (25, 5)$, so $a+b = 30$.[/hide]

as part of the problem statement, you should say that that if that expression is true for all x, then find a+b

[quote="xxxxx"]If $4x^{2}+20x+a= (2x+b)^{2}$, find $a+b$[/quote]
It depends ...
If the author have meant "If for all $x$..." then [b]I Am Me[/b] is right and we can find not only $a+b$ but each of $a$ and $b$.
If the author have meant "If for any ... " then we can take $x=0$ and have $a=b^{2}$ with infinitly many possibilities for $a+b$.

[hide="solution"]We can subtract equal terms 4x^2 and get

$20x+a=4xb+b^{2}$

$b=5, a=25, a+b=30$[/hide]


Does anyone know if there is a way to use Latex commands in a ppt (powerpoint presentation)?  Thanks for any help.

If you use the "prosper" setting, it is just like powerpoint.

You might have to download some packages though. It works great.

Hi,
Try [url=http://texpoint.necula.org/]TexPoint[/url].
SSN.

Sorry to bump an old thread, but I was wondering if there are other free programs that can do this?

[url=http://users.ecs.soton.ac.uk/~srg/softwaretools/presentation/TeX4PPT/]TeX4PPT[/url] seems to be free, but you really ought to be using Latex's [url=http://latex-beamer.sourceforge.net/]Beamer[/url]. The results are stunning and put PowerPoint to shame. You won't need Office at all because the output is pdf. MiKTeX will install beamer for you if it isn't already installed.

Try looking at [url=http://latex-beamer.sourceforge.net/beamerexample1.pdf]this example[/url] on the beamer page. When it has downloaded, press CTRL-L and then the arrow buttons (or mouse) to go through it. ESC goes back to the normal screen. I had great fun producing [url=http://www.sixthform.info/steve/Slideb.pdf]How Was I Made?[/url] as an introduction to LaTeX presentations for a non-technical audience. (You may have to press Esc then Ctrl-L at the start).

Once you are hooked by beamer there is no way you'd ever want to go back to PowerPoint.

How do I get MikTex to automatically install Beamer?

Which version do you have?
Version 2.5: You can set MiKTeX to automatically install new packages as needed. Start->All Programs->MiKTeX 2.5->Settings->Install missing packages on-the-fly. I suggest you set this to Ask me first. Then when you compile a document with the beamer documentclass MiKTeX will install beamer.

Otherwise:
Version 2.4: Start->All Programs->MiKTeX->MiKTeX Package Manager
Version 2.5: Start->All Programs->MiKTeX->Browse Packages
Wait a while until MiKTeX loads all the package names, then right-click on beamer and select install. The package is quite large so may take a little while to download if it's not already on your computer.

Wow wow wow wow wow wow

Steve, would you post the tex file for your document so we can learn how to do things like that?  Thanks!

That's because you are starting at the wrong end. Start with simple examples by reading [url=http://www.tug.org/pracjourn/2005-4/mertz/mertz.pdf]Beamer by Example[/url] which is meant for beginners. This will help you to build up slowly.

[quote="stevem"]OK, but I don't know how helpful it will be because it contains lots of customisations. Anyway, here it is along with the images and the customised theme I renamed as Poole, based, if I remember correctly, on the Warsaw theme supplied with Beamer.[/quote] Wow Steve! How long did it take you to make that? :o

It's cool, but I have to agree with Phoenix257! :ninja:

[quote="beans"]Wow Steve! How long did it take you to make that? :o [/quote]Hard to say as I played with it for some weeks as an exercise in learning to use Beamer; I was upgrading from Prosper, which is another LaTeX presentation package, but nowhere as near as good as Beamer.

[quote="stevem"]
Version 2.5: Start->All Programs->MiKTeX->Browse Packages
Wait a while until MiKTeX loads all the package names, then right-click on beamer and select install. The package is quite large so may take a little while to download if it's not already on your computer.[/quote]

Hi all,

I'm using Version 2.6 but I assume it would be similar.
Normally I would let TexnicCentre download the package automatically but I'm capped at home, so had to download the .tar.gz from elsewhere and bring it home.

Now that I've got it, how do I get MiKTeX to install from this file? What do I need to name the file? When I pick the directory this file is in as the "local repository" to download from, I am informed that this "directory does not seem to be a local package repository."

You can't get MiKTeX to install from this file as it uses its own file format for installation. You will have to uncompress the files and read the instructions (probably in a readme file) to install manually.

Edit: I think I've found the solution.

Thanks stevem!

I'm unsure if anyone's suggested [url=http://www.elevatorlady.ca]Aurora[/url] yet. But it's a terrific plug in for Office apps that supports 2007 versions and uses MiKTeX to generate transparent PNGs or its own native vector format.

So in fact you can now use PowerPoint and its brilliant Presenter View, with LaTeX without being stuck with Adobe Reader's primitive presentation features.

Can I make Beamer work on TeXShop?

Sorry to revive.

Also, TexPoint doesn't work on Office 2008, and Aurora is Windows only.


[quote="Mathsniper"]我覺得港澳地區人群中質素存在參差, 但人才是有的, 我喜歡研究數學, 但同學們都不甚喜歡, 全校幾千人喜歡數學的十只手指頭數得到, 以前因為課都沒有聽做數學比老師沒收數學書, 還降操行... 但我不介意, 只要有機會比我學數學就行了, 我不介意這一切的外來壓力. 現在, 學校把我從初二帶到高二這階段, 另我了解到數學的真諦, 但現在... 看到多麼的數學問題, 概念, 以前數學家(好像愛多士, 歐拉, 柯西, 格林, 古隡等)的數學創造能力, 雖然我沒有這數學大喜中的聖人的境界, 但我知道這一真理, 就是, 不論在何時何地, 只要我對數學持著一種執著, 堅持, 才是我生存的意義! 我知, 我不人才, 但只要有你們一群喜歡數學的研究者, 我堅信智慧肯定有所提高.
至此, 一起討論數學吧!  :lol:[/quote]

 :) 你要继续努力哦，数学是非常beautiful的，丘成桐先生就是你的好榜样呀。
爱好＋努力＋正确的方法＝成功。
With Best Wishes!

yaaaaaaa~! 我在查看中國百科全書(數學卷)時, 都有留意到丘成桐先生的研究, 他在微分幾何上表現凸出.

:arrow: 是的，他和他的老师陈省身先生都在微分几何上作出了举世瞩目的成就。是华人的楷模。

[quote="Mathsniper"]
看到多麼的數學問題, 概念, 以前數學家(好像愛多士, 歐拉, 柯西, 格林, 古隡等)的數學創造能力, 雖然我沒有這數學大喜中的聖人的境界, 但我知道這一真理, 就是, 不論在何時何地, 只要我對數學持著一種執著, 堅持, 才是我生存的意義! 我知, 我不人才, 但只要有你們一群喜歡數學的研究者, 我堅信智慧肯定有所提高.

我想跟你说的是这个时代真正爱数学的人已经不多了，可能琳琅满目的高科技产品引去了大家的视线。
两年前我还在苦苦挣扎时，一位教授写信给我说：Mathematics is beautiful and we do it for love。
把这句话送给你，希望你开心地学习。
看得出你对数学家还瞒了解的，多读些数学家的故事会有所启迪的。

With Best Wishes！

其實我初一時是學Computer Science的, 從Visual Basic, Pascal, C, C++, Java, 到Algorithm, Data Structure的研究, 越來越明白到數學的重要性, 慢慢我開始學數學了(研讀), 開頭是跟學校數學培訓學會的, 但隨後感到知識十分缺乏, 便開始投開大量時間學習, 包括上課時間....(是否很壞呢? :P 不過已經比老師話得多... 不理會了), 到現在, 真正用了2年半時間學習(初三頭到高二, 即現在), 現在回頭一看, 自己的數學能力已經提升了, 我亦為此而喜悅...

Well, maybe guys in North America loves Computer science far more than asians.
I know that at least 5% of a british middle school students can program VB
And btw, anybody knows 沈尹默老先生, he's my teacher's teacher, and master in the field of chinese poetry and 书法...

Never heard of him...... I think the master of 书法 was 启功.

[quote="Dreamer"]其實我初一時是學Computer Science的, 從Visual Basic, Pascal, C, C++, Java, 到Algorithm, Data Structure的

了不得呀，  :o 这些东西大陆一般要到大学才学习。
Data Structure 应该比较难。
中学就学到这么多东西，将来一定不简单。 :lol:[/quote]

如果參加電腦競賽(IOI那一類比賽)，那也不算奇怪..

[quote="ifai"][quote="Mathsniper"]以前因為課都沒有聽做數學比老師沒收數學書, 還降操行... 但我不介意, 只要有機會比我學數學就行了, 我不介意這一切的外來壓力[/quote]
那其它課學得怎樣? 可知"專才"是沒用的, "通才"才能更深入的了解數學的真諦!![/quote]
why do you think so?
我覺得通才不可能存在

[quote="ifai"]Mathsniper, which school do you study in?[/quote]
我就讀濠江中學的 :)

我觉得专才和通才都挺有用，

专才比如阿兰.图灵 ,通才比如冯.诺依曼。他们都挺伟大。 :roll:

[quote="Dreamer"]各位放假否？ ;)[/quote]
明天开考, 下下周放假.

這星期正在考試, 下星期大放假  :lol:

考完了，还得补课，补到25号。

從現在放到年約初十 :D hahaha~

[quote="dingdongdog"]考完了，还得补课，补到25号。[/quote]
erk, toooooo bad. I'm sorry for u guys
Well, though we canadians don't have a lot of activities like that, but I hafta do some SAT practise indeed.
However, i don't have a spring festival break! :(

在加拿大假期时间是怎样安排的？不过春节吗， :?:

Well, Canadians are now really likely to be called a recessive coutry to America, or just call it "The Country of 'Jack's" so, we pretty much have the same plan as americans, Christmas Break from Dec. 17 to Jan 2 (Eastern part, like Ontario have a plan of Dec 23 to Dec 8). then no more breaks during the first two months of the year.
:(
But anyways, I can have a great time in learning whatever I want these days. Plus i just joined NOML, anyone here does so???(NOML.org)

anyway,studying is a good thing.
I shall have a holiday from Jan. 22 to Feb.26 :rotfl:

I know nothing about NOML :blush: .
Maybe someone else knows it.

請問這裡有否港澳應屆高中準畢業生想報考國立臺灣大學或國立成功大學數學系?
PS: 台灣還有哪些大學的數學系較好?

[quote="nirvanatear"]eR...
我也在加拿大...
有兴趣就加我MSN:
nirvanatear@hotmail.com
另外15楼的...
TURING也不算专才.
他研究的数学其实是逻辑.
但其实归结起来.
数学是人类一切知识的基础.
it's the only lore whom all other lores could based on[/quote]
eh really?
在哪?Ontario? Alberta? BC?

Newfoundland~
just kidding.
BC la
Burnaby North
听说过吗?

[quote="Mathsniper"]我覺得港澳地區人群中質素存在參差, 但人才是有的, 我喜歡研究數學, 但同學們都不甚喜歡, 全校幾千人喜歡數學的十只手指頭數得到, 以前因為課都沒有聽做數學比老師沒收數學書, 還降操行... 但我不介意, 只要有機會比我學數學就行了, 我不介意這一切的外來壓力. 現在, 學校把我從初二帶到高二這階段, 另我了解到數學的真諦, 但現在... 看到多麼的數學問題, 概念, 以前數學家(好像愛多士, 歐拉, 柯西, 格林, 古隡等)的數學創造能力, 雖然我沒有這數學大喜中的聖人的境界, 但我知道這一真理, 就是, 不論在何時何地, 只要我對數學持著一種執著, 堅持, 才是我生存的意義! 我知, 我不人才, 但只要有你們一群喜歡數學的研究者, 我堅信智慧肯定有所提高.
至此, 一起討論數學吧!  :lol:[/quote]

也許我以前也有你的經驗，那時的我和現在的我都認為要放開懷抱去見識這個世界，多認識不同的人 (不一定數學強的人)，話不定對數學有新的體會和突破！

BTW, 你有冇參加港澳數學奧林匹克？成績如何？


If $x,y,z$ are sides of a triangle, prove that \[ | x^2(y-z) + y^2(z-x) + z^2(x-y) | < xyz. \]


I thought this was interesting.  My first post here so I hope it's appropriate.  I believe the plot below is correct.  If not, then by all means prove me wrong.

Derive an expression for explicitly determining all the points satisfying:
\[ \sin(x^2 y) \equal{} \ln(x/y)
\]
Here's a plot of the entire set and a specific set I calculated directly(the blue and red dots).

This is my approach:

Set $ t \equal{} x^2y\neq 0$, because $ x,y\neq 0$

Then $ y \equal{} \frac {t}{x^2}$ and $ \frac {x}{y} \equal{} \frac {x^3}{t}$

$ \sin(x^2y) \equal{} ln\left(\frac x y\right) \Rightarrow$

$ \sin t \equal{} ln\left(\frac {x^3}{t}\right) \Rightarrow$

$ e^{\sin t} \equal{} \frac {x^3}{t} \Rightarrow x^3 \equal{} te^{\sin t} \Rightarrow$

$ x \equal{} (te^{\sin t})^{1/3}$, and

$ y \equal{} \frac {t}{x^2} \equal{} \frac {t}{(te^{\sin t})^{2/3}} \equal{} \frac {t^{1/3}}{(e^{\sin t})^{2/3}} \equal{} \left(\frac {t}{e^{2\sin t}}\right)^{1/3}$

So we have the system of the equations:

$ x(t) \equal{} (te^{\sin t})^{1/3}$

$ y(t) \equal{} \left(\frac {t}{e^{2\sin t}}\right)^{1/3}$

Hi Pontios, that's much nicer than what I came up with:  I allowed the range $ \ln(x/y)=(-1,1)$ or $ \frac{1}{e}\leq \frac{x}{y}\leq e$.  That gives the two equations $ y=\frac{1}{e} x$ and $ y=ex$ which gives the bounds on the plot.  Then I substitute $ y=ax$ for $ \frac{1}{e}\leq a\leq e$ into the equations and obtain:
\begin{align*}
x_1(a)&=\sqrt[3]{\frac{\arcsin(-ln(a))+2n\pi}{a}}\\
x_2(a)&=\sqrt[3]{\frac{\arcsin(-ln(a))+(2n+1)\pi}{a}}\\
y_1(a)&=ax_1(a)\\
y_2(a)&=ax_2(a)
\end{align*}
I wonder how we might come up with expressions for complex variables $ x$, $ y$, and $ t$?  In your derivation above, we'd have to substitute:

\begin{align*}
\sin(t)&=\log(x^3/t)\\
e^{\sin(t)}&=\exp(ln|x^3/t|+i\arg(x^3/t))
\end{align*}
where I'm using the standard notation ``$ \log$'' for complex logarithm (to base e).  This would not allow isolation of the $ x^3/t$ term. Perhaps though this is not appropriate for this forum.


geometry
rectangle
trigonometry
inequalities
geometry unsolved

hi!!

Given $ ABC$ a triangle rectangle in $ B$ and $ \alpha$ measure  of
the angle formed by the median descended from $ A$ and the hypotenuse

prove that:

$ \sin{\alpha}\le \frac{1}{3}$

[url=http://www.casimages.com][img]http://nsa11.casimages.com/img/2009/12/10/091210123334515101.png[/img][/url][/img]

thanx.

[quote="reason"]hi!!

Given $ ABC$ a triangle rectangle in $ B$ and $ \alpha$ measure  of
the angle formed by the median descended from $ A$ and the hypotenuse

prove that:

$ \sin{\alpha}\le \frac {1}{3}$
[/quote]
Applying the sinus rule in $ \triangle ACM$,we easily get : $ sin{\alpha} \equal{} \frac {AB.BC}{2AC.AM}$ . Hence,It suffices to prove that : $ 4AC^2.AM^2 \geq 9AB^2.BC^2$,using $ AM^2 \equal{} AB^2 \plus{} \frac {AC^2}{4}$ and $ AC^2 \equal{} AB^2 \plus{} BC^2$,the inequality may be reduced to $ (AC \minus{} 2AB)^2 \geq 0$ which is true.

[quote="rachid"][quote="reason"]hi!!

Given $ ABC$ a triangle rectangle in $ B$ and $ \alpha$ measure  of
the angle formed by the median descended from $ A$ and the hypotenuse

prove that:

$ \sin{\alpha}\le \frac {1}{3}$
[/quote]
Applying the sinus rule in $ \triangle ACM$,we easily get : $ sin{\alpha} \equal{} \frac {AB.BC}{AC.AM}$ . Hence,It suffices to prove that : $ 4AC^2.AM^2 \geq 9AB^2.BC^2$,using $ AM^2 \equal{} AB^2 \plus{} \frac {AC^2}{4}$ and $ AC^2 \equal{} AB^2 \plus{} BC^2$,the inequality may be reduced to $ (AC \minus{} 2AB)^2 \geq 0$ which is true.[/quote]

yes  :wink:  ; but we have : $ sin{\alpha} \equal{} \frac {AB.BC}{2.AC.AM}$ .

Applying sine rule in $ \triangle ACC_1$ we get $ sin \alpha\equal{}\frac{1}{\sqrt{10}} < \frac{1}{\sqrt{9}}$, done.

Best regards,
sunken rock


function
integration
real analysis
real analysis unsolved

The sum $ \sum_{n\equal{}1}^\infty \frac{n^k}{{2n\choose n}}$ always has the form $ p_k\plus{}q_k\frac{\pi}{\sqrt{3}}$ with $ p_k,q_k\in\Bbb{Q}$ .

Can you give an explicit formula for $ p_k$ and $ q_k$ ?

The formula (51) on  [url=http://mathworld.wolfram.com/BinomialSums.html]mathworld[/url] seems to be wrong.

[quote="Doe John"]The formula (51) on  [url=http://mathworld.wolfram.com/BinomialSums.html]mathworld[/url] seems to be wrong.[/quote]
Why it is wrong?

I computed both expressions with Mathematica and got different results.

k = 1;

FullSimplify[Sum[n^k/Binomial[2*n, n], {n, 1, \[Infinity]}]]

$ \frac{2}{27}\left(9 \plus{} \sqrt{3}\, \pi\right)$

Simplify[1/2*(-1)^(k - 1)*Sum[StirlingS2[k + 1, j]*Binomial[2*j, j]/3^j*(2*Pi/3/Sqrt[3] +Sum[3^i/(2 i + 1)/Binomial[2 i, i], {i, 0, j - 1}]), {j, 1, k + 1}]]

$ \frac56\plus{}\frac{4 \pi}{9\sqrt{3}}$

I solved the k = 0 case, i feel like there is a better way to go about this, but here's my solution

Denote
\[ S = \sum_{k = 1}^{\infty}\frac {1}{\binom{2k}{k}}
\]
Let's note a property of the Beta Function, namely
\[ (2k + 1)B(k + 1,k + 1) = \frac {1}{\binom{2k}{k}}
\]
Recall
\[ B(k + 1,k + 1) = 2\int_0^{\frac {\pi}{2}} \left(sin(\theta)cos(\theta)\right)^{2k + 1}d\theta
\]
$ \therefore S = 2\sum_{k = 1}^{\infty}(2k + 1) \int_0^{\frac {\pi}{2}} \left(sin(\theta)cos(\theta)\right)^{2k + 1}d\theta$ $ = \2 \int_0^{\frac {\pi}{2}} 2\sum_{k = 1}^{\infty}(2k + 1)\left(sin(\theta)cos(\theta)\right)^{2k + 1}d\theta (*)$
Let's exploit some properties of the geometric series.
First, let's note that $ (with |x| \leqslant 1)$
\[ \sum_{k = 0}^{\infty} x^k = \sum_{k = 0}^{\infty} x^{2k} + \sum_{k = 0}^{\infty} x^{2k + 1}
\]
But we already know that
\[ \sum_{k = 0}^{\infty} x^k = \frac {1}{1 - x} \implies \sum_{k = 0}^{\infty} (x^2)^k = \frac {1}{1 - x^2}
\]

\[ \therefore \sum_{k = 0}^{\infty} x^{2k + 1} = \frac {1}{1 - x} - \frac {1}{1 - x^2} = \frac {x}{1 - x^2}
\]
From (*) we know we interested in sums of the form
\[ \sum_{k = 1}^{\infty} (2k + 1)x^{2k + 1}
\]
Which we can find.
\[ \sum_{k = 0}^{\infty} (2k + 1)x^{2k + 1} = \sum_{k = 0}^{\infty} x\frac {d}{dx}x^{2k + 1} = x\frac {d}{dx} \sum_{k = 0}^{\infty} x^{2k + 1} = \frac {x(1 + x^2)}{(1 - x^2)^2}
\]
Which means that
\[ \sum_{k = 1}^{\infty} (2k + 1)x^{2k + 1} = \frac {x(1 + x^2)}{(1 - x^2)^2} - x
\]
So we have found that
\[ S = 2 \int_0^{\frac {\pi}{2}} (\frac {x(1 + x^2)}{(1 - x^2)^2} - x) d\theta
\]
where $ x = sin(\theta)cos(\theta)$.

I put this into the wolfram online integrator and found that
\[ S = \frac {2}{27}(\frac{9}{2} + \sqrt {3}\pi)
\]


vector
Functional Analysis
linear algebra

If $ W$ is a subset of vector space $ V$ , we can show that

$ W\subset (W^{\perp})^{\perp}$

using argument like

letting $ w\in W$ then $ w\cdot y\equal{}0$ $ \forall y\in W^{\perp}$. Hence $ w\in (W^{\perp})^{\perp}$

But why when we want to prove $ W\equal{}(W^{\perp})^{\perp}$ , we need $ V$ to be finite and then the dimension theorem? Can't we just prove that $ (W^{\perp})^{\perp}\subset W$ using similar argument like

letting $ x\in (W^{\perp})^{\perp}$ then $ x\cdot w\equal{}0$ $ \forall w\in W^{\perp}$ .Hence $ x\in W$ ?? (without assuming it to be finite?)

Can someone please explain ?

Here's a Hilbert space answer. Let our vector space be the real inner product space $ \ell^2$ of square summable sequences, with the usual inner product $ \langle x,y\rangle\equal{}\sum_{j\equal{}1}^{\infty}x_jy_j.$

Now, let $ W$ be the ordinary span of $ \{e_2,e_3,e_4,\dots\}.$ That is $ x\in W$ iff $ x_1\equal{}0$ and only finitely many $ x_j$ are anything other than zero. $ W$ is a subspace of $ \ell^2;$ it is not a closed subspace.

$ W^{\perp}\equal{}\text{span}(\{e_1\}).$ But then $ (W^{\perp})^{\perp}$ consists of all $ x\in\ell^2$ such that $ x_1\equal{}0.$ That is not the same as $ W;$ rather, it is the (nontrivial) closure of $ W.$

I'll make it even shaper. $ (W^{\perp})^{\perp}$ is [b]much[/b] larger than $ W$ in the following sense: $ W$ has a countable Hamel basis; $ (W^{\perp})^{\perp}$ does not.


Ms. Hamilton's eighth-grade class wants to participate in the annual
three-person-team basketball tournament.  The losing team of each game
is eliminated from the tournament.  If sixteen teams compete, how many
games will be played to determine the winner?

Since there are 16 teams competing, and we want one winner, we play $ \boxed{15}$ games.

To expand, since it is a single-elimination tournament, everyone except for one team must be eliminated (the winning team) and because one team is eliminated every game, the number of games in a single-elimination tournament with $ n$ teams is $ n\minus{}1$


geometry
3D geometry
linear algebra
matrix
tetrahedron
rectangle
invariant

At the vertices $A, B, C, D, E, F, G, H$ of a cube, $2001, 2002, 2003, 2004, 2005, 2008, 2007$ and $2006$ stones respectively are placed. It is allowed to move a stone from a vertex to each of its three neighbours, or to move a stone to a vertex from each of its three neighbours. Which of the following arrangements of stones at $A, B, \ldots , H$ can be obtained?
$(\text{a})\quad 2001, 2002, 2003, 2004, 2006, 2007, 2008, 2005;$
$(\text{b})\quad 2002, 2003, 2004, 2001, 2006, 2005, 2008, 2007;$
$(\text{c})\quad 2004, 2002, 2003, 2001, 2005, 2008, 2007, 2006.$

Is it by any chance true that all three of them are attainable? That's what seems to be true, but I there's something missing, especially since I didn't devise the sequence of moves. I used a matrix ($8\times 8$) and showed that the linear system of equations represented by it is compatible. Basicly, I denoted by $a,b,c,d,e,f,g,h$ the number of moves of the first kind operated on the vertices (labeled in your first picture, in the same order, $2001,2002,2003,2004,2005,2008,2007,2006$). Then we get a system of eqns of the type $3a=b+d+e$, etc. The matrix formed by the coefficients ($3$ and $-1$'s) has rank $7$ and the sum of the elements on each column is $0$. For any result column, the sum of the elements is also $0$ because the total sum of the numbers is constant. This means that the new column can't form any $8\times 8$ non-zero determinant with any other $7$ columns of the original matrix, thus making the system compatible. This means that we can find $a,b,c,d,e,f,g,h$.

What's not clear is whether these solutions are natural for all three separate cases. Maybe in some cases they're not natural numbers and then we're toast :). Anyway, I tried.. :).

I suggest you to try a more elementary approach. It's problem 1, after all!   ;)

Ok, I know for sure that the first one isn't attainable. Consider the two tetrahedrons $2005,2004,2002,2007$ and $2001,2006,2003,2008$. The difference between the sums of the vertices of these two tetrahedrons is $0$ at first and it can only decrease/increase by $6$ at a time (after a move), so it can't be $4$, as in the first picture.

In the other two pictures the differences are $0,-6$, so I'll have to find other arguments (I'm sure at least one of them is attainable, though :)).

Sorry for the repeating nagging (:D), but I've just realized that the third config isn't attainable either: Consider the two rectangles $2001,2006,2007,2002$ and $2005,2004,2003,2008$. The difference between the sums of their vertices is $-4$, and it can increase/decrease by $4$, so it can't be $2$, like in the third picture.

As for the second one, it's obvious: just perform the first kind of move on the vertices labeled $2008,2004$ at the beginning.

So, the answer is: We can obtain only (b).

Yesssss!   :P
Note that a single invariant could prove both (a) and (c) are not attainable.
Consider the vertices labelled 2005, 2001, 2008, 2006 in the first picture: their sum has constant parity, and in the first picture it's even, while in (a) and (c) it's odd.


Here is the Indian team for IMO 2004.

     1.  Vipul Naik ( IMO 2003 , silver medallist )
     2.  Anand Deopurkar ( IMO 2003 , bronze medallist )
     3.  Kshipra Bhawalkar
     4.  Anupam Prakash
     5.  Rohit Joshi
     6.  Abhishek Dang

 Note : From 1 to 5 are 12th graders and Abhishek is a 9th grader , a brilliant fellow.

Hi Myth.
            I suppose that vipul ( a new member of mathlinks ) is the same Vipul Naik who is on the team . That's just my guess , I may be wrong.

PS : The leaders are R.B Bapat and V.V Acharya


I'll make it look nicer :D:

x13 - 3x1 + 2 = 2x2
x23 - 3x2 + 2 = 2x3
...
x993 - 3x99 + 2 = 2x100
x1003 - 3x100 + 2 = 2x1.

Try to use the   and   commands (without the spaces ;)).


Check the following code:
[code]#include<stdio.h>
#include<string.h>
char f[10][x]; /*replace 'x' by a suitable number*/

main()
{

f[0][0]='1';
int c=strlen(f[0]);
printf("%d",c);
}
[/code]
I am using a gcc C compiler...

When (just like above) I declare the 2-dimensional array globally, the program gives me the output (for any 'x') as 1...

Now look at this (array declared inside main()):
[code]#include<stdio.h>
#include<string.h>

main()
{
char f[10][x]; /*replace 'x' by a suitable number */
f[0][0]='1';
int c=strlen(f[0]);
printf("%d",c);
}[/code]
This piece of code gives me various outputs depending on 'x'...
EX:
x=1, output=30
x=4, ouput=60
x=10,output=1
x=100, output=5

Now I have two doubts:
1) Why is this funny thing occurring?
2) As far as I know ( and I know very little),
the strlen() function needs a '\0' character at the end of character array to be able to recognise it as a string and calculate its length....

I wrote the above code for 1-dimensional array and got the [b]size of array [/b]as output for strlen() and I expected it as there is no '\0' in the code...

But in the above code, I have not used '\0' character...
So both pieces of codes should not work (i.e. they should show the total number of columns of the 2-D array and output=x)....
But I see that in the 1st case, it works properly....

And even if I am to assume that the comp. did sth to put a '\0', then why is it differentiating between the 'local' and 'global' declaration cases...

Thanks...

Variables declared globally are initialised to 0 (= '\0'), whereas local ones aren't. Calling strlen on something which doesn't end in a 0 byte will cause unexpected results as you've seen.


Prove that if $a,b,c >0$ then      $\sum\frac{\sqrt a}{b+c}\geq\frac{3\sqrt 3}{2\sqrt{a+b+c}}$

Hölder's inequality yields

$\left(a+b+c\right)\left(\frac{\sqrt{a}}{b+c}+\frac{\sqrt{b}}{c+a}+\frac{\sqrt{c}}{a+b}\right)^2\geq\left(\sqrt[3]{a\cdot\left(\frac{\sqrt{a}}{b+c}\right)^2}+\sqrt[3]{b\cdot\left(\frac{\sqrt{b}}{c+a}\right)^2}+\sqrt[3]{c\cdot\left(\frac{\sqrt{c}}{a+b}\right)^2}\right)^3$
$=\left(\left(\frac{a}{b+c}\right)^{\frac23}+\left(\frac{b}{c+a}\right)^{\frac23}+\left(\frac{c}{a+b}\right)^{\frac23}\right)^3$.

Now, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=5520 problem 1 for $k=\frac23$, we have

$\left(\frac{a}{b+c}\right)^{\frac23}+\left(\frac{b}{c+a}\right)^{\frac23}+\left(\frac{c}{a+b}\right)^{\frac23}\geq 3\cdot\left(\frac12\right)^{\frac23}$.

Thus,

$\left(a+b+c\right)\left(\frac{\sqrt{a}}{b+c}+\frac{\sqrt{b}}{c+a}+\frac{\sqrt{c}}{a+b}\right)^2$
$\geq\left(3\cdot\left(\frac12\right)^{\frac23}\right)^3=3^3\cdot\left(\frac12\right)^2=\frac{27}{4}=\left(\frac{3\sqrt3}{2}\right)^2$.

Now, taking the square root, we get

$\sqrt{a+b+c}\left(\frac{\sqrt{a}}{b+c}+\frac{\sqrt{b}}{c+a}+\frac{\sqrt{c}}{a+b}\right)\geq\frac{3\sqrt3}{2}$,

so that

$\frac{\sqrt{a}}{b+c}+\frac{\sqrt{b}}{c+a}+\frac{\sqrt{c}}{a+b}\geq\frac{3\sqrt3}{2\sqrt{a+b+c}}$.

This proves your inequality.

  darij


Find the integer solutions of the equation:
(x^3)+(y^3)-2xy+x+y+2=0
I wait your solutions

put x+y=a
and xy=b Then the solution is easy! :)  :D

Actually, Socrates is right (isn't he always ;))
Writing in terms of $s=x+y$ and $p=xy$ the equation becomes
\[s^3-3sp-2p+s+2=0
\]
or, equivalently,
\[27p=9s^2-6s+13+\frac {28}{3s+2}\]
We deduce that $\frac {28}{3s+2}$ is an integer, hence $s\in  \{-10,-3,-2,-1,0,4\}$.
By inspection, we find that the only value yielding integer values for $x,y$ is $s=-1$ and then $(x,y)=(-1,0)$ or $(0,-1)$.


If the sum of two numbers is 5, and the sum of the reciprocals of those two numbers is $\frac{2}{3}$, find the sum of the squares of the two numbers.

[quote="now a ranger"]x+y=5 since both numbers add up to 5

therefore x squared plus y squared equals 25.[/quote]
$(x+y)^{2}=x^{2}+2xy+y^{2}$  :wink:

[quote="now a ranger"]x+y=5 since both numbers add up to 5

therefore x squared plus y squared equals 25.[/quote]
You may want to check what your ideas are when you have them and they may not work. If $(3+4)^{2}=3^{2}+4^{2}$ then $7^{2}=5^{2}$ and 7 would equal 5. Also, hide your solutions when you post them.


calculus
derivative
logarithms
function
limit
analytic geometry
graphing lines

[quote="Icosahedroid"]What is the derivative of x^x[/quote]
We have $ y\equal{}x^{x}$. Taking the natural log of both sides results in
$ \ln y\equal{}x\ln x$. Differentiating both sides with respect to $ x$ results in
$ \frac{y'}{y}\equal{}\ln x\plus{}1$
$ y'\equal{}(\ln x\plus{}1)y$
Recall that $ y\equal{}x^{x}$. Substituting this back in gets
$ y'\equal{}x^{x}(\ln x\plus{}1)$

In general, when you have the variable in the power, taking the natural log is the first step in solving these problems.

Rewrite $ x^{x}$ as $ x^{x}\equal{}e^{\ln x^{x}}\equal{}e^{x\ln x}$ to differentiate.

What ckck said and what kunny said are different in tactics - the steps are performed in a slightly different order, the intermediate results look a little different - but in substance, they are the same method.

Since ckck provided the answer, here is a consequence of that answer:

The function $ f(x)$ is decreasing on $ (0,e^{-1}]$ and increasing on $ [e^{-1},\infty).$ So it has a minimum at $ x=e^{-1}$ and that minimum value, $ e^{-e^{-1}},$ is positive and less than $ 1.$ It follows that $ \lim_{x\to0^+)}$ must exist and must be positive. (We can do a few more things to show that the limit is $ 1.$) But we also have that $ \lim_{x\to0^+}f'(x)=-\infty.$ The graph gets infinitely steep there.

How did you determine that the function is decreasing on (0,1/e]? You can't find the minimum by setting the derivative equal to zero because the graph of f'(x) is discontinuous at x=0.

It's decreasing on that interval because the derivative is negative on that interval. That's it - don't look for monsters under the bed that aren't there.

Uh - that's kind of a useless thing to say. [i]This[/i] particular function happens to have a limit of $ 1,$ but in a calculus limit context (which this is), $ ``0^0"$ is an indeterminate form, which means that you could find instances in which its limit is anything at all or even nonexistent.

[quote="Kent Merryfield"]

The function $ f(x)$ is decreasing on $ (0,e^{ \minus{} 1}]$ and increasing on $ [e^{ \minus{} 1},\infty).$ So it has a minimum at $ x \equal{} e^{ \minus{} 1}$ [/quote]

I say that $ f(x)$ is decreasing on $ (0, e^{\minus{}1})$ and increasing on $ (e^{\minus{}1}, \infty)$,
because at $ x \equal{} e^{\minus{}1}$, it has a zero slope.  You would not include the $ x$-value
in either intervals because that is where the slope is zero.  At that $ x$-value, it does
not lie on the curve where there is a negative slope or a positive slope.

That's a technical point, but I did say it correctly.

The statement $ ``f(x)$ is (strictly) increasing on the set $ S"$ means this: for all $ x,y\in S$ such that $ x<y,$ then $ f(x)<f(y).$

By that standard, the function $ f(x)\equal{}x^2$ is increasing on $ [0,\infty).$ If you choose any $ 0\le x<y<\infty,$ then $ x^2<y^2.$ Do note that that is still true if you let $ x\equal{}0.$

By exactly the same standard, $ f(x)\equal{}x^2$ is decreasing on $ (\minus{}\infty,0].$ There's nothing at all wrong with the overlap there, the fact that $ 0$ belongs to both sets.

The reason there's nothing wrong with that is that the the statement $ ``f(x)$ is increasing at the point $ a"$ is meaningless. A function cannot be increasing at a single point because the very definition of increasing requires at least two points. Hence for the example of $ f(x)\equal{}x^2,$ it would not be correct to say that that function is increasing at $ 0,$ even though it would be correct to say that the function is increasing on $ [0,\infty).$

Granted, one would often hear someone say that the function $ f(x)\equal{}x^2$ is increasing at $ x\equal{}1,$ but that's not really the correct statement - it's a sloppy contraction of a longer statement, "There exists a neighborhood of $ 1$ such that $ f(x)$ is increasing on that neighborhood." A sufficient (but not necessary) condition for being able to say that is that $ f$ has a continuous derivative and $ f'(1)>0.$ (Note that $ f'(1)>0$ isn't even sufficient for that; you also have to invoke something like the continuity of the derivative to be sure.)


Hey dudes!

This is my first post EVER on these boards...

I consider myself a Chronicles of Narnia FANATIC. My homepage is http://narniaresources.com (the best site EVER! Help take part in the worlds greatest movie!). If you want to talk narnia with me, you will become my best friend.

See ya!
-OneFtFwd

i read all of the series in 4th grade.

did you know that this series is the only childrens book the author wrote? :huh:

wait, you said that's [i]your[/i] home page as in you made that? Wow, thats really nice.

Neat homepage! I just reread the chronicles of narnia. I can't wait for the movie to come out.

He was also friend with J.K. Tolkin
Isn't the trailer copywrited

Did I miss something? I thougt it was J.R.R. Tolkien, or am I mistaken?

i think he mixed up JK Rowling and JRR Tolkien.


24. Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio
randomly selects a number from the set {1, 2, . . . , 10}. The probability that
Sergio’s number is larger than the sum of the two numbers chosen by Tina is
(A) 2/5 (B) 9/20 (C) 1/2 (D) 11/20 (E) 24/25

Would someone be so kind as to explain the solution to this problem very slowly and step-by-step to me?
I couldn't understand the solution given on the AMC website (I think that is where I found the solution)

Try listing out the possible values for Tinas sums.

There are 10 choices

1+2=3
1+3=4
1+4=5
1+5=6
2+3=5
2+4=6
2+5=5
3+4=7
3+5=8
4+5=9

Then compare them with 1,2,3,4,5,6,7,8,9,10

to get the probablility.

We can use a [url=http://mathworld.wolfram.com/BinomialCoefficient.html]binomial coefficient[/url] to determine how many possible choices Tina has.

[hide="my solution"]
$ \binom{5}{2}\equal{}\frac{5!}{2!\cdot3!}\equal{}\frac{10\cdot12}{12}\equal{}10$

Listing Tina's sums, we'll subscript to show their frequency of occurrence.

$ 3_{1/10}$
$ 4_{1/10}$
$ 5_{2/10}$
$ 6_{2/10}$
$ 7_{2/10}$
$ 8_{1/10}$
$ 9_{1/10}$

Now remember that the chance of two events whose measurements are $ p$ and $ q$ occurring at once is $ pq$. In this case, $ p$ is always $ 1/10$. Note that $ q$ is the sum of the subscripts  on all of Tina's sums that are less than Sergio's number, which corresponds to $ p$. I summed up $ q$ and multiplied by $ p$ to find the net probability.

$ \displaystyle\frac{1}{10}\left(\frac{10\plus{}9\plus{}8\plus{}6\plus{}4\plus{}2\plus{}1\plus{}0\plus{}0\plus{}0}{10}\right)\equal{}\boxed{\frac{2}{5}}$

Thus, the answer is [b]A[/b][/hide]
 :lol:

iguil, could you slowly explain where the final sum above the ten in the fraction comes from to me? I followed your solution until the last two sentences of your final paragraph but had difficulty seeing how they translated into the bottom fraction.

[hide="my reasoning"]Your total grade in a class is determined by a sum of your completion of a section of assignments, be it tests, quizzes, homework, participation.... Lets say tests are worth 50%, quizzes are 30%, and homework is 20%. Assume you got 93% of the total possible homework points, 82% of the possible tests points, and 73% of the possible quiz points.

Your grade is NOT found by $ \frac {93 \plus{} 82 \plus{} 73}{3} \equal{} 82.67$
This is true if and only if tests, homework, and quizzes are worth the same possible amount.  Instead, consider the objective equation $ 50T \plus{} 30Q \plus{} 20H \equal{} G$. Note that this retains the teacher's grading rubric.  If tests make up 50% of your grade, you must treat them as such. Using this equation, we find the grade to be

$ G \equal{} 50(.82) \plus{} 30(.73) \plus{} 20(.93)$
$ G \equal{} \frac {50(82) \plus{} 30(73) \plus{} 20(93)}{100}$
$ G \equal{} \frac {4100 \plus{} 2190 \plus{} 1860}{100}$
$ G \equal{} 81.50$

Lets verify that. Assume that there are 100 possible points in the class. Thus, the sum of your points is equivalent to your grade....

$ G \equal{} 50(.82) \plus{} 30(.72) \plus{} 20(.93)$
$ G \equal{} 81.50$  :) [/hide]Look again at the problem. Do you see how this was implemented?
[b]
EDIT:[/b] Note that the rates in the fraction in question correspond to $ \{10, 9, 8,\ldots,1\}$ respectively.

You are right. I don't understand how it was implemented.

"Now remember that the chance of two events whose measurements are p and q occurring at once is pq. "
What are the measurements of events?

"In this case, p is always 1/10. "
Where p is what?

"Note that q is the sum of the subscripts on all of Tina's sums that are less than Sergio's number, which corresponds to p. I summed up q and multiplied by p to find the net probability.  "
Since I don't understand what p is, it follows that I don't understand that the rest of the solution.

"EDIT: Note that the rates in the fraction in question correspond to $ \{10, 9, 8,\ldots,1\}$ respectively."
What are the rates?

Is there vocabulary that I don't know? I'm sorry that I'm taking so long to understand this.

It's not you, it's me. Sometimes I forget I jump the gun. A lot of my math is intuitive.  When I said 'measurements' I meant probability.  As for rates... disregard that. And try not to get discouraged, bear with me.

[hide]$ p$ is the probability of Sergio getting the number $ k$. This is always 1/10, as there are no multiplicities in the set $ \{10,9,8,7,6,5,4,3,2,1\}$, which has 10 distinct possibilities.

$ q$ is the probability that Sergio's value $ k$ is greater than the sum of Tina's two numbers (call this sum $ s$). $ q$, unalike $ p$, varies as $ k$ changes. This is sort of where that 'grade rubric' came in.  For each $ k$, there exists $ n$ instances of $ s$ such that $ k>s$. If we allow $ q\equal{}\frac{1}{10}\sum n$ we can quickly compute $ q$.

$ k\equal{}10\Longrightarrow n\equal{}10$
$ k\equal{}9\Longrightarrow n\equal{}9$
$ k\equal{}8\Longrightarrow n\equal{}8$
$ k\equal{}7\Longrightarrow n\equal{}6$
$ k\equal{}6\Longrightarrow n\equal{}4$
$ k\equal{}5\Longrightarrow n\equal{}2$
$ k\equal{}4\Longrightarrow n\equal{}1$
$ k\equal{}3,2,1\Longrightarrow n\equal{}0$

Thus, $ q\equal{}4\Longrightarrow pq\equal{}\frac{2}{5}$[/hide]
 :maybe: I hope that helps. I don't know if I can do much better.  If you're still completely lost I would suggest asking your math teacher about probability and he/she can probably tie off loose ends.

Yes, that helps. For each value of k do we simply look at the possible pairs of numbers from Tina's list to figure out n? It wouldn't be too difficult for us to list them out... or at least see where our list would go for each value of k, right?

Thank you very much for your time, igiul and the future.

[quote="aufha"]Yes, that helps. For each value of k do we simply look at the possible pairs of numbers from Tina's list to figure out n? It wouldn't be too difficult for us to list them out... or at least see where our list would go for each value of k, right?
[/quote]

You might say that.  The only really important thing about Tina's two numbers is that there are 10 possible combinations even though some of those combinations have the same sum. For instance $ \{2,4\}, \{1,5\}$ and $ \{1,4\}, \{2,3\}$.  This means that the important list isn't the list of Tina's numbers. Instead, it's the list of the sums of those pairs so we can see how many $ s$ are less than any given $ k$. That number is in fact $ n$.

So... yeah. One down, many to go. :P

And don't worry about my time man, I'm glad to help.

$ S\equal{}\{1, \; 2, \; 3, \; 4, \; 5 \}$ has the property that $ n \in S \implies 6\minus{}n \in S$.

Thus, if $ (a, \; b)$ is a possible pair of numbers for Tina, then $ (6\minus{}a, \; 6\minus{}b)$ is an equally possible pair.

Averaging complementary pairs gives the average pair $ \{3, \; 3 \}$.

[i]This was not a possible pair in the first place, so we don't have to account for this being its own complement.[/i]

So, the average sum to beat is 6, and the probability is $ \frac{10\minus{}6}{10} \equal{} \frac{2}{5}$

This should have been posted under American Mathematics Competitions. This problem is too hard for High School Basics level.


Yep,it's simple but i'd like to know

There are 3 coloured balls that can be spun in a lotto motion with a funnel for the ball to dropout
The red ball comes out which is a one third probabilty.
What's the formulae for the red ball to come out a 2nd time and a 3rd and 4th and so on.
Is it correct that 2 times would be 1/2 x1/3 equals 1/6 or 0.16 ththe
3 times 1/3 X 1/6 equals 1/18 or .05
4 times 1/4 x 1/18 equals 1/72 0.013 etc etc
Whats the formalue? wcosmos@dodo.com.au

Please don't post math problems in this forum.  It belongs in High School Basics.  Could an admin please move it?

Hi ozzie. You post your problem at: http://www.mathlinks.ro/index.php?f=150


trigonometry
geometry
rectangle
ratio
geometry unsolved

[quote="ashrafmod"]If a diameter AB of a circle meet any chord CD, not a diameter, in E, and if AF, BG be drawn perpendicular to CD, then CG = DF.[/quote]
[color=darkblue][b]Proof 1.[/b] Denote the circle $w=C(O,R)$, $m(\widehat{BCD})=m(\widehat{BAD})=x$, $m(\widehat{ABC})=m(\widehat{ADC})=y$. Therefore,
$BC=2R\cos y\ ,\ CG=BC\cos x\Longrightarrow CG=2R\cos x\cos y\ \ (1)\ .$
$AD=2R\cos x\ ,\ DF=AD\cos y\Longrightarrow DF=2R\cos x\cos y\ \ (2)\ .$
From the relations $(1)$ and $(2)$ obtain $\boxed{\ CG=DF\ }\ .$

[b]Proof 2.[/b] Denote the projection $P$ of the center $O$ to the chord $CD$ and $m(\widehat{OEP})=\phi$.
From the relations $PC=PD$ and $PG=PF=R\cos \phi$ obtain $\boxed{\ CG=DF\ }\ .$
[/color]

See the attached diagram.

Extend $AF$ and $BG$ to get a rectangle $AA'BB'$ (it's a rectangle since $\angle AA'B=\angle BB'A=90^\circ$ (why?) and $AA'\parallel BB'$ (why?) and $AB'\parallel A'B$ (why?)). Also draw $OM$ where $M$ is the midpoint of $CD$. Then $OM\perp CD$ (why?) and after that it's quite obvious that the whole picture is symmetric in respect to $OM$, hence $CG=DF$

[quote="ashrafmod"]If a diameter AB of a circle meet any chord CD, not a diameter, in E, and if AF, BG be drawn perpendicular to CD, then CG = DF.[/quote]

 [color=blue]I have not yet read the solutions , but :[/color]
  If we take the midpoint $M$ of $CD$ , then $OM // AF // BG$. Since $O$ is the midpoint of $AF$, point $M$ is also the midpoint of $FG$ ( $AFBG$ is a trapez, $O$ is the pidpoint of a diagonal, $OM // AF$ ect).
Thus
 $GD= MD-MG = MC-MF=FC$

  [u]babis[/u]

[color=red] EDIT[/color]

 [i]Now I have seen the figure of Fahrenajt's solution. My solution in not ''too'' different than his one.[/i]

I think that we can see a useful idea in this very simple problem.

The orthogonal ( or parallel ) projections on a line, of three given collinear points, form two segments with the same ratio, as the given points ( easy to proof by Thales theorem ).

So ( see Farenhajt's configuration ) because of OA = OB, =>  MG = MF and ( here ) the proof is completed.

Kostas Vittas.

PS - In other words, if we draw three arbitrary parallel lines, pass through three given collinear points, then the intersection points of these parallel lines, from another arbitrary also line, form two segments with the same ratio as the given points.


superior algebra
superior algebra unsolved

Please help me undertsand maximal ideals as well as

prove that if R is a commutative ring with unity, and i is an ideal of R, then I is a miaximal ideal if and only if R/I is a field.

Thank you

hi,
It will be nice if u could tell where exactly u got lost while reading the proof as there is no point in typing the whole proof here,u could find it in all standard books on ring theory.
regards
Rohith


Hi, I'm wondering if anyone took the UMD math test today, and how do you think you did on it ?

Probably about 18 right and 3 wrong (4 skipped).

Most likely 18 right and 1 wrong, 6 skipped.  It was harder than previous tests, I thought.

Could someone post the questions?

Thanks

They will be posted at  http://www.math.umd.edu/highschool/mathcomp/2006.shtml next week.

For some reason I had a lot of trouble with the recursive function one . . . I tried to find the elegant solution, didn't, started brute forcing it, realized that would take way too much time, and ended up leaving it blank.

Thank You very much for the link.

These questions seem hard compared to the AMC 12...the 2005 I mean...

[quote="chess64"]I can't believe the last one was 1... and for that $10^{2006}$ I switched it to base 3 and tried a smaller exponent, and I counted the same amount of 1's and 2's. Then after the competition I counted again and I realized the difference was 1 :([/quote]

I realized that the last one was 1, but only after the competition, with the aid of a calculator. How do you do it without a calc?

It will be interesting to see what happens with the Round Two.  Last year, the cutoff for Part II was 54 (the 52 I said in class was off by 2 points).  About 2300 students took the test and 220 qualified for the final round.  Sounds like you all have a good shot at it.  Good luck!

Best,
Dr. A.

[quote="mathgeniuse^ln(x)"][quote="mdk"]the last one states:

The value of

1/(2sin(10deg))- 2sin70 deg

is which of the following:
a. 1  b. sin80 deg  c. sqrt(2)  d. sin 50 deg[/quote]

Here is the solution

[hide="How did I do this"]
$\frac{1}{2\sin 10^\circ}-2\sin 70^\circ$

$\frac{\cos 10^\circ}{2\sin 10^\circ \cos 10^\circ}-2\sin 70^\circ$

$\frac{\cos 10^\circ}{\sin 20^\circ}-2\sin 70^\circ$

$\frac{\cos10^\circ-2\sin 70^\circ \sin 20^\circ}{\sin 20^\circ}$

$\frac{\cos 10^\circ+\cos 90^\circ-\cos 50^\circ}{\sin 20^\circ}$

$\frac{\cos 10^\circ-\cos 50^\circ}{\sin 20^\circ}$

$\frac{\cos 30^\circ \cos 20^\circ+\sin 30^\circ \sin 20^\circ-\cos 30^\circ \cos 20^\circ+\sin 30^\circ \sin 20^\circ}{\sin 20^\circ}$

$\frac{2\sin 30^\circ \sin 20^\circ}{\sin 20^\circ}=2\sin 30^\circ=1$

$A. 1$[/hide][/quote]

I don't understand what you did from here                   cos10deg -2sin(70deg)sin(20deg) note that sin(70deg)=        cos(20deg), so you get (cos(10deg)- sin(40deg))=                 sin(20)/sin(20)=1 and voila!

Thanks for the inspiration.

Hmmm...that is true...I just elaborated on it...

But yeah, you could do that...but still my method works...

Results are in for the 2006 UMD.  Homeschool Math Clubbers, I posted the results over in the Halo topic.

Congratulations, everyone!  :-)

[quote="mathgeniuse^ln(x)"]I took the contest when it was posted...what would a 94 put you...24 correct 1 blank...[/quote]

I think top 5, at least.

Oh, well thanks.

94 is the 3rd Highest Score Possible right.

100, 96, 94....

24 correct and 1 blank would give you a 96, the second-highest possible score.  I think only Brian Lawrence and maybe Richard McCutchen would get a 100.

[quote="mathgeniuse^ln(x)"]But I got 24 right and 1 wrong...=94

Sorry if I made a typo in my post...[/quote]

Great job. Which one did you get wrong?

Number 22, the one with the greatest length of chord, or least length, with answer of 20.

#25, was cool.

#24, just notice a telescoping sequence.

#23, 1/2absinC

#22, I just want to kick my self.

#21, I liked it...

Oh right...I was confusing it with the other question...but still I want to kick myself...


Slashdot today links to this story:

[url]http://mfeldstein.com/index.php/weblog/permalink/blackboard_patents_the_lms[/url]

Here's a quote from the blog entry:

[quote]I’m surprised there hasn’t been more uproar about this yet. The ever-brilliant US Patent and Trademark Office has apparently granted Blackboard a patent for...well...pretty much anything remotely related to learning management systems. As I read it, Blackboard basically owns the patent on any sort of groupware at all that is used for teaching purposes. This could have very serious consequences for both proprietary and Open Source competitors--and I define “competitors” as loosely as possible.[/quote]

Could this be a problem for AoPS?

check the bottom: it says
"Powered by phpBB © 2001, 2002 phpBB Group"

if its copyrighted by them, then its okay


find all continuous functions $ f: R \rightarrow R$ such that for all $ x,y \in R$ :
$ (x\minus{}y).f(x\plus{}y)\minus{}(x\plus{}y).f(x\minus{}y)\equal{}4xy(x^2 \minus{} y^2)$

This task can be solved without condition "all continous...".

Let $ f(x) \equal{} g(x) \plus{} x^3$. So:
$ (x \minus{} y)((x \plus{} y)^3 \plus{} g(x \plus{} y)) \minus{} (x \plus{} y)((x \minus{} y)^3 \plus{} g(x \minus{} y)) \equal{} 4xy(x^2 \minus{} y^2)$;
$ (x \minus{} y)g(x \plus{} y) \equal{} (x \plus{} y)g(x \minus{} y)$.

For $ x \equal{} y \plus{} 1$ we will have:
$ g(2y \plus{} 1) \equal{} (2y \plus{} 1)g(1)$.

So, $ g$ is linear and $ g(x) \equal{} kx$ for some $ k\in R$.
It's obviously, that all $ f(x) \equal{} x^3 \plus{} kx$ is solution.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
search
AMC 10

Could someone post previous years required AIME score for USAMO?  Thank you!

just on the same topic is AMC 10 harder or Mathcounts?

I never did MATHCOUNTS, but since it is geared mainly to middle school students, I would say AMC 10 is harder.  The AMC 10 typically contains 2-3 tricky questions at the end.

There's also a floor value. Maybe you should search old threads or perhaps look at the FAQ.

On the topic of Mathcounts vs. AMC 10, I personally thought Nats was way harder than AMC 10. A friend of mine used AMC 10s to prepare for State though (after finishing old state tests).

[quote="FantasyLover"][quote="jmerry"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1221302#1221302]Link.[/url][/quote]

uh so you needed 181 to qualify for usamo in 2007?[/quote]

If you were in 10th grade or lower.

[quote="JRav"]It's based on an index that combines AIME and AMC.  If you score in the 140-150 range on the AMC, you may need only a 3-5 on the AIME.  If you score closer to 100 on the AMC, you'll need somewhere around 8-11 to qualify.[/quote]

There is a floor value you must score to qualify. usually its 7 or 8.

Post-expansion the floor has been 6 both years. (since you could not have qualified with a 5 last year)

[quote="FantasyLover"]if you get 140-150 in AMC, you only need to get 3-5 on the AIME??
---

UGH I SHOULDVE TAKEN IT IN 2007[/quote]
I think FantasyLover is living up to his screen name. I'm not aware of anyone in recent memory qualifying with a 5 or lower on the AIME, and the reason the qualifying scores were lower in 2007 is that the tests were more difficult - especially the AIME. Had you taken it in 2007, you would probably have gotten a lower score than you did in 2008.

This isn't the SAT where the test makers try out test items beforehand and do extensive statistical norming so that the same score is supposed to mean the same thing. The AMC doesn't - in practice, can't - pre-screen its questions, and the the meaning of a particular score will vary from year to year.[/quote]

[quote="FantasyLover"][quote="JRav"]It's based on an index that combines AIME and AMC.  If you score in the 140-150 range on the AMC, you may need only a 3-5 on the AIME.  If you score closer to 100 on the AMC, you'll need somewhere around 8-11 to qualify.[/quote]

if you get 140-150 in AMC, you only need to get 3-5 on the AIME??

i thought you go to USAMO when AMC+10*AIME=200 something...[/quote]

Well, if you quoted the whole thing, you'd see he's quoting JRav.  I'd say if you scored 140-150 on AMC 12, a 7-8 shoudl make it, 6 if you're lucky.  FOr AMC 12, probably 6-8.

If you're scoring in the 140-150 range, chances are, you're going to do pretty well on the AIME anyway.

so if i score not extremely high on the AMC's but i do very well on the AIME is there a chance for me to make USAMO?

Umm define "not extremely high" and "very well." Also what grade are you in?


FTW
MATHCOUNTS
number theory
prime numbers

Firefighter Steve was born in 1950. Steve had a grandmother who was born in a year which is the product of two prime numbers, one of which is one less than twice the other. In what year did Steve's grandmother celebrate her 90th birthday?

[quote="Jongao"]guess and check?
(come on, the grandmother cant be born in 2*3=6)
I got 31*61=1891,so the answer is 1891+90=1981[/quote]
That's kinda like what I did-I have no number theory knowledge :P
@ GameBot, can you please stop posting every single FTW problem? it's like spam

lol its 'bot'.

when you review the problem and if you don't get the problem, click 'Ask on Forum' button.

the problem comes here. :D

I don't get bots, are they computers or humans?.......

Why should it be $1981$ only? How do we prove there is no other solution?

Solution to this? (except for guess and check.)

@above I think it's supposed to be a guess and check question... (plus I recognize it from mathcounts sprint)


Give $x,y,z$ is positive integer such that $xy=z^{2}+1$
Prove that there exists $a,b,c,d \in Z$ such that $x=a^{2}+b^{2},y=c^{2}+d^{2}$

For all divisors p $z^{2}+1$ we have $(\frac{-1}{p})=1$. It give, that all prime divisors are $p=2$ or $p=1(mod 4)$. We have $2=1^{2}+1^{2}$, $p=a^{2}+b^{2}$. If $x=a^{2}+b^{2}=Norm(a+bi),y=c^{2}+d^{2}=Norm(c+di)$, then $xy=Norm((a+bi)(c+di))=(ac-bd)^{2}+(ad+bc)^{2}$. It give, that for all divisors $x|(z^{2}+1)$ exist a,b, suth that $x=a^{2}+b^{2}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=60555
and more general
http://www.mathlinks.ro/Forum/viewtopic.php?t=45329 .


Aflati solutiile intregi ale ecuatiei: $y^{3}=6x^{2}+2$.

[hide="Solutie"]Ecuatia se reduce la $(36x)^{2}= (6y)^{3}-432$. Substituim $y_{1}=36x$, $x_{1}=6y$ si se obtine $y_{1}^{2}= x_{1}^{3}-432$, care este chiar [url=http://en.wikipedia.org/wiki/Mordell_equation]ecuatia lui Mordell[/url] pentru $k=-432$. Solutiile sunt date în acest caz de $(x_{1},y_{1}) = (12, \pm 36)$ (vezi [url=http://tnt.math.metro-u.ac.jp/simath/MORDELL/]aici[/url], de exemplu), adica $x=\pm 1, y=2$.[/hide][hide="P.S."]De unde e problema?[/hide]

Problema este facuta de mine.  :D
Este o aplicatie a Marii teoreme a lui Fermat pentru cazul $n=3$. Ecuatia se mai poate scrie sub forma $(x+1)^{3}=y^{3}+(x-1)^{3}$. Rezulta ca $(x-1)y(x+1)=0$. Se obtin solutiile $(x,y)=(\pm1,2)$.


search
number theory
number theory unsolved

Solve the diophantine equation $3a^{3}+4b^{3}+5c^{3}=0$.

:blush: Sorry for my typo. I've edited the problem.

ZeTaX, it's amazing how well you remember all these threads (or do you search each time?).

I don't remember all those links etc. directly.
But if I remember this or a similar problem, I take a search. And then I normally combine what I remember from the thread with searching.

For this one, I remembered:
a) I posted there
b) it is in Number Theory (which is also clear if I wouldn't remember)
c) the global-to-local-pribciple was mentioned

=> I searched this subforum for "local" and threads I posted in.


Okay, here is a bet - there will be at least one topic in the G&FF (other than this) concerning Christmas (and i won't be the one starting it).

Put your bets for or against this, people! :)

the engineer one...i place a bet of 2 cents :lol:

-jorian

I bet a million dollars as I will make one if the engineering one does not count

I bet 50 dollars that Ars will make one if the engineering one does not count

All the money I have, considering, even if you change it so no one who bet can make it, i'll have one of my friends do it.


[quote="kstan013"]So should we all like meet together at nats...or is that just uncool?[/quote]
Hmm, it's best not to plan these things.
Just let them happen on their own.
Planning them usually ends up oddly.
Just my experience.

I agree with diophantient partially. With so many people and so much commotion, it won't even be possible to try to get everyone to meet? Some people might not know..and u no the rest.

But, if u can get it to happen. And still go through all that pain. Hey, i'm for it.

GO MICHIGAN!!!![/quote]

[quote="kstan013"]So should we all like meet together at nats...or is that just uncool?[/quote]

My gosh, That is the coolest thing I have ever heard of.

Where will it be?! Tell me! Tell me! Tell me!!!

Sorry, I'm always sugar high when I wake up. :P

Sounds cool, but I need more info, where and when will it be...

I won't be able to come, because I live in NY, and I will be leaving for Texas in 9 minutes and 40 seconds.

Nats is over. No need to wake this up again.


geometry
geometric transformation
rotation
trigonometry
circumcircle
trig identities
Law of Cosines

Three sides of a triangle determine a unique triangle up to rotation and symmetry.

If im given three sides of a triangle, how much information can I obtain about the triangle?

Is this enough information to draw a triangle?

With what?  Straightedge and compass?  Given the lengths of the three sides of a triangle, you can deduce everything.  That's what "unique" means.

if I was given three sides of a triangle, if i wanted to draw it by hand it would be take some trial and error.

unless I knew an angle.

I believe my real question is how do I determine the angles if given all three sides.

if you have all three lengths of the triangle you can quite easily get the angles with the cosine law:

c^2 = a^2 + b^2 - 2ab*cos(y)

y is angle  ACB

[quote="Fraenkel"]If im given three sides of a triangle, how much information can I obtain about the triangle?[/quote]
You can find the angles using the Law of Cosines and then the Law of Sines, and then you can find everything else if you need to (median lengths, circumradius, etc.).

[quote="Fraenkel"]Is this enough information to draw a triangle?[/quote]
Yes, and this is how:

[list][*]Start by constructing the first side of the triangle.
[*]Now, construct two circles with radii equal to the lengths of the other two sides, one at each endpoint of the first side.
[*]Finally, connect the endpoints of the first side to the intersection of the circles.[/list]


Two people (labeled A and B) play a game where they take turns counting positive integers. For example, A may start by saying "1," then B will say "2," A will say "3," and so on.
Let $m$ and $n$ be relatively prime integers. If a person comes says an integer that is a multiple of $m$, they write down $m$, and similarly for $n$. If an integer is a multiple of $m$ and $n$, then they write down $mn$.
So while playing the game a string of $m$'s and $n$'s is created.
It turns out that the pattern of $m$'s and $n$'s is periodic. Find the period of this pattern (i.e., the number of characters in the repeating segment of the string).

[hide]A period should clearly occur every $mn$ numbers.  In that, $m$ is written $n$ times, and $n$ is written $m$ times.  So the period should be $\boxed{m+n}$.[/hide]


geometry AMC AIME probability USA(J)MO USAMO Alcumus

I am debating as to whether I should get Introduction to Geometry, Introduction to Number Theory and Introduction to Counting and Probability. Last year, I made AIME (although I did really badly) and over the summer I did a lot of math problems. Do you think that I need to get the above books, or should I just move on to AOPS II. Also, I plan to do Intermediate Algebra and Intermediate Counting and Probability. Should I go to these without doing the introduction books after only having done AOPS I.

My goal is to this year make USAMO. I find that my ideas are not very organized in my head and even though I know most of the concepts in some of these introductory books, I am only considering getting them primarily to help me review and organize my ideas. My main dilemma is whether I will have ample opportunity to review in earlier sections of intermediate and olympiad books or I will be stuck without review.

I would suggest starting the Intermediates and see where you're at. Chances are, if you can do the Intermediates, you know everything from the intros. Also do some AIME's or other competitions and find your weak points.

Hope this helps  :lol:

Go to the classes page, and try the diagnostics tests ("do you need this course?") for the intro books you mentioned, and then try the "are you ready for this course?" for the interm books. Judging by how well you do on those, you can then make your decision. :)

Thanks! I think that I will buy introduction and intermediate counting and probability.

introduction as in the whole intro series? Oh wait...do u mean intro c+p and interm c+p?

You should also consider working through ACoPS. You don't necessarily have to buy it. There are tons of free copies online.

could you post a link of where i might find a free copy?

yeah i think so........

the sites didnt really seem that trustworthy anyways.

Thanks for the advice. I have ACOPS already. While I can do problems in the geometry section, my algebra, number theory and combinatorics are relatively bad. Do you think that this is a good plan to work through books?

[b] Plan: [/b]
Review AOPS V1 & Intro to Geometry and Number Theory (quick review of concepts and the problems marked as difficult)
Intro to Counting and Probability
AOPS V2
Interm. Algebra
Interm. Counting and Probability
ACOPS
PSS
Other books (e.g. IMO compendium, Challenging Problems in Geometry, Geometry Revisited, 104 problems in NT and Math Olympiad Challenges)

Nice plan! What happened to AOPS V2? Just curious.

I also found that I could not solve one problem in the introduction to C + P post-test (one as in a single one not all of them; I think it was the second last one). Should I do intro or will AOPS V1 and V2 combo chapters and interm. C + P have enough review?

Thanks. These tips have been really helpful. Just out of curiosity, could you guys solve any USAMO or IMO problems before beginning to study these books?

[quote="AIME15USAMO"]Hmm, I forgot about V2...even though I started working on it yesterday  :oops:

Anyway, volumes 1 and 2 should be fine for ur C+P. Also, try alcumus. It's good review to see how well you do at C+P.[/quote]

Alcumus is awesome! Their learning algorithm is getting pretty good...

Yeah, but Alcumus is more Intro to C/P level, so you can also use the Do you know? tests on the Online Classes page.

oh here's a rather "used" studying schedule for USAMO:

-ACoPS
-First steps for math olympians(good to review concepts)
-AIME/AMC12 problems
-PSS

= pwnage

Even if can't get much in the books done, make sure you get some olympiad problems done. Don't start with the USA olympiads or China's olympiads. More appropriate would be the canada, ireland or maybe some of the earlier Russian.

[quote="Paiev"]^That's a rather poor schedule... First Steps is for like beginners to AMC while ACoPS is for like beginners to USAMO.

[quote]Review AOPS V1 & Intro to Geometry and Number Theory (quick review of concepts and the problems marked as difficult)
Intro to Counting and Probability
AOPS V2
Interm. Algebra
Interm. Counting and Probability
ACOPS
PSS
Other books (e.g. IMO compendium, Challenging Problems in Geometry, Geometry Revisited, 104 problems in NT and Math Olympiad Challenges)[/quote]

I hope you don't plan on doing all this within any reasonable time frame. Like before USAMO 2010. Or USAMO 2011. You could spend from now until USAMO 2011 just with Engel and still not finish it.[/quote]

first steps is to review for the AMCs. If you get to into taking lots of time to solve problems, like USAMO level problems, it's good to take a step back and do lower level problems. A black MOPer told me that. As far as Engel and Zeitz are concerned, their books are great for making the leap to USAMO. after that books such as geo revisited are essential.

[quote="limac"][quote="AIME15USAMO"]That is pretty much what my plan is. Here is mine exactly.

Finish Intro to Geo (I'm on chapter 17)
Precalculus
ACoPS
AIME's for practice
Intermediate Algebra
Intermediate C+P
WOOT
PSS
Practice USAMO's

I am planning on finishing these before 2011.[/quote]

I am pretty sure Intermediate Algebra should be done before starting Precalc.[/quote]

I took the "Are you ready for this?" precalc diagnostics test and I was ready  :P I also asked Mr. Rusczyk and he said it was fine. Thanks for your suggestion though  :lol:

Also: [quote="rrusczyk"]If you get all the "Do you Know" questions right, you are good for that part of AIME.[/quote]

That might help too.

[quote="batteredbutnotdefeated"]first steps is to review for the AMCs. If you get to into taking lots of time to solve problems, like USAMO level problems, it's good to take a step back and do lower level problems. A black MOPer told me that. As far as Engel and Zeitz are concerned, their books are great for making the leap to USAMO. after that books such as geo revisited are essential.[/quote]

That's fine and all, but if you're doing ACoPS you won't get anything out of First Steps that you wouldn't get by doing some AMC/AIME problems (which are already in your list).

Well, I've been hearing that AcOPS is easier than Engel PSS, so you might try AcOPS first.

Also, I don't know if this is true...
but I also heard that no black-mopper has ever finished PSS.

[quote="iYOA"]in an older thread i asked how long it takes to finish PSS and i think pythag101 said that it took some black-moppers 2 months[/quote]

They probably didn't do all the problems. Olympiad problems can take 2 hours, and Engel and over 1200 problems. Do da math.

2400 hours...in 61 days = more hours than there are in a day. But then again, black MOP-ers can probably do olympiad problems more than the average AoPSer.


trigonometry
summer program
MathPath
geometry
geometric transformation
rotation
angle bisector

Do you use the calculator or do it by hand?

The handy chart taped in my calculator's battery case with the radical format of pretty much all major sines, cosines, and tangents works nice.

[quote="mustafa"]I know the sine, cosine, tangent for 30, 45, 60, 90 by memory, and then I use the adding and double angles, and half angles and stuff for the rest. Or my calculator. I like that best.[/quote]

Yeah, my teacher made us memorize sixteen points on the unit circle, complete with degrees, radians, and coordinates. And we have to label a blank unit circle in 90 seconds. :o

At MathPath 2005 we "learned" how to calculate $\sin 18^{\circ}$ by hand. At least, I wrote it down in my notes and promptly forgot about it...

We had to memorize all multiples of 30, 45 degrees around the unit circle in degrees and for radians (which really aren't that bad), and I wrote a few contest problems concerning sin15, so I have commited that to memory, and to remember sin75=cos15, you just change the minus sign to a plus sign - fairly simple.

But before I learned double/half angle, I used Angle Bisector theorem...very fun

I memorized 30, 45, 60 and 90. Then, by using hte half-agnle formula sooooooooo much, I sorta know 15, and that stuff...

[quote="mathgeniuse^ln(x)"][quote="mathnerd314"]I would rather use the calculator, but noooooo, we have to do most of our calculations by hand in my class. And I don't have a TI-89, so using a calculator wouldn't give the pretty radical anyway.[/quote]

Wait they give radicals...dude that means anyone with it can solve any equation doesn't it...?[/quote]

No, as someone (Prof. D?) said, you can't express $\sin 1^{\circ}$ with radicals. Or something like that. I just remember that he then proceeded to define a rotation as $480^{\circ}$ and consequently show how it was possible :P.

Edit: I mean, a full circle rotation...

I can evaluate any angle in the form $(15k)^\circ$ or $\frac{k\pi}{12}$, where $k\in\mathbb{Z}$.  I memorized the radical forms.

[quote="13375P34K43V312"]I know the values, but never remember which goes with which, so I just use the unit circle for multiples of 30 and 45. Then I use the double/addition/half angle formulas.[/quote]
Same, except I use the right triangle :) . I write everything in terms of sines and cosines, and if I need even less variants I change the cosines to sines.

You can easily evaluate any thing half of the previous, by using the Angle Bisector Theorem in a right triangle.

How do you find sin 18? I've never learned it? Maybe a hint would be good, that would allow me to figure it out on myself.

Small hint:

$72^\circ+18^\circ=90^\circ$.

Notice that you can use the double-angle formulas twice.

:wallbash_red:  :wallbash:  :wallbash_red:  I totally should have figured that out!!! That is really cool!!!

Also, you can use the fact that 90+18=108, 108 degrees being the interior angle of a regular pentagon.

[quote="13375P34K43V312"]I remember sin18 as well.[/quote]

Yeah, we did that on the last day of MathPath 2005...

It's still in my blue notebook, which is currently hiding somewhere in my room...

I just recall 30, 45, and 60, and the multiples of 90, and I use addition and subtraction formulas.

For example:

$\sin{15}=\sin{60-45}=\sin{60}\cos{45}-\sin{45}\cos{60}=\frac{\sqrt{2}}{2}\left(\sin{60}-\cos{60}\right)$

$\frac{\sqrt{2}}{2}\left(\sin{60}-\cos{60}\right)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}-1}{2}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$

Then use that to compute bigger angles, like, say, 105.

For sin, I'll use a power series out to 7!
For cos, I memorize some of the easier ones and use formulas.

eww the power series don't generally turn out nice though.

[quote="nutz_for2.718281828"][quote="bpms"][quote="lingomaniac88"]You could figure out $\sin(10^\circ)$ by using the triple-angle formula $\sin3\theta=3\sin\theta-4\sin^{3}\theta$.[/quote]
But that gives a butt ugly answer with the cubic right?[/quote]

What about the 1 third angle formula...[/quote]
That does too... The formulas like the half angle are so ugly that always try to do the problems with stuff like the difference formulas

We can use this formula to find values of cotan
$\boxed{cotana=cotan2a+\sqrt{1+cotan^{2}2a}}$
Ex: $cotan7\&o30' = cotan15^{o}+\sqrt{1+cotan^{2}15^{o}}$
similar for $a=15^{o}$ easy


This is a similar contest to the one that NoSoupFor you held. I am going to hold a contest everyone gets to test their graphical disign work. All programs will be accepted and I want you to PM me If you want to join. After ten users I will post a topic. Probably round this time  in the day. Please also put the graphics in attachments. I shall put in a round for each one put it will just be an ispiring one. The rounds shall take about a week or untill everyone has posted a reply. There will also be ten rounds.

Once again:

Can you submit hand drawn pieces??? I fail at computer work..


algorithm
geometry
geometric transformation
rotation
puzzles

Two men lost in a desert have between them only a single container of water, of arbitrary size.  Wanting to be fair, they decide that they want to split the water evenly between themselves, but have no measuring cups or indeed any other tools at all.

Can they do it?  If not, what restrictions must be placed on the container?

Is it like a regular bottle of water? meaning does it have the twist cap thing? if it does just switch off tiny cups

What happens if they have an undetermined amount of water less than the volume of the cap?

What does the container look like? What is it, specifically? Are we allowed to assume an idealized desert, or must we factor in evaporation, heat, and that they might not survive long enough to complete a given algorithm?

They could pour the water on the ground :rotfl:  If they both get 0, they both get the same amount.

I will be a little more (or less, as the case may be) specific.

- The container is an arbitrary closed surface in three dimensions, except for its opening, which is also arbitrary (but can, I suppose, be assumed to be flat.)
- The container's entire volume is filled with water.
- Both men want exactly half of this volume.
- This is an idealized environment, and nothing else happens.

[hide]Pour a little of the water out and make a mark where the water level is. Turn it upside-down and make another mark where the water level is. Repeat this process until the two marks coincide.[/hide]

Or use the Banach-Tarski Paradox. :D

I considered that, but it fails if the container is horridly non-convex.

[hide]One of the men drink from the cup until they see that the water is level with the upper bottom edge of the cup. He has now drunk exactly half the water. The other man then just drinks the rest.[/hide]
Is that correct?

That would work for a vessel created by the rotation of a line around a vertical axis (or even symmetric across a plane that divides the container in half), and which has a bottom. The problem involves an arbitrary closed surface.

I think we need some sort of requierment of symmetry.


algebra
polynomial
superior algebra
superior algebra unsolved

How do I show that the extension $\mathbb{Q}(i, \sqrt{2}, \sqrt{i+\sqrt{2}})$ is not Galois over $\mathbb{Q}$? It is equivalent to show that $\sqrt{3}$ does not belong to this extension, and even though all of that looks rather obvious, I cannot write it down properly ...

The minimal polynomial of $a=\sqrt{i+\sqrt{2}}$ is
\[ (x^4-1)^2 + 8 \]
It has the roots $\pm \sqrt{\pm i \pm \sqrt{2}}$ which aren't in the extension. Hence, it is not normal.

Yeah, that's what I did. But I couldn't show that $\sqrt{i-\sqrt{2}}$ is not in the extension. How do you do that ?

suppose $\sqrt3 \in \mathbb{Q}(i, \sqrt2, \alpha)$, where $\alpha = \sqrt{i+\sqrt2}$.
observe that the minimal polynomial of $\alpha$ over $\mathbb{Q}(i, \sqrt2)$ is $\mu_\alpha(x) = x^2-(i+\sqrt2)$, while the minimal polynomial of $\sqrt3$ over the same field is still $\mu_{\sqrt3}(x) = x^2-3$ (i guess you don't need the proof of these two "trivial" facts).
so $[\mathbb{Q}(i, \sqrt2, \alpha): \mathbb{Q}(i, \sqrt2)] = [\mathbb{Q}(i, \sqrt2, \sqrt3): \mathbb{Q}(i, \sqrt2)] = 2$, and then, for our hypothesis, $\mathbb{Q}(i, \sqrt2, \alpha) = \mathbb{Q}(i, \sqrt2, \sqrt3)$.
but that would mean that $\mu_{\sqrt3}$ and $\mu_{\alpha}$ are traslated, so that there exist $a, b \in \mathbb{Q}(i, \sqrt2)$ such that $\mu_{\sqrt3}(x) = \mu_{\alpha}(ax+b)$. but then $a^2 = 1, 2ab = 0, b^2-2-i = -3$, which is absurd.

it seems like it's working, isn't it?

[quote="ma_go"]but that would mean that $\mu_{\sqrt3}$ and $\mu_{\alpha}$ are traslated[/quote]

I didn't think about it. Thanks.


geometry
inradius
algebra proposed
algebra

a triangle has sides $ a,b,c \in N$ and inradious $ r \in N$.

1)prove that if $ r$ is fixed, only a finite number of triangles with this proporty exist.

2)prove that if $ r\equal{}1$ so the triangle has a right angle.

bye
paolo


Let $ A(3,3),B(\minus{}2,1),C(1,\minus{}2)$, $ T$ is a point in and on $ \triangle ABC$,

$ f(x,y)\equal{}\max{\{2x\plus{}y,x^2\minus{}y\}}$, find out the range of value of $ f(x,y)$


Let A and B be matrices in Mn(C), s.t. AB-BA has rank=1. Prove that (AB-BA)^2=0n. This was given at some IMC (forgot the year).

I don't know the terminology you're using -- would you mind explaining what everything means?  (I know what a matrix is, but that's about the only part of it that I understand.)

Ok:

The rank is the maximum order of a minor =/=0 from the matrix. A minor of order k is a determinant formed with the elements from the intersections of k rows and k columns of the matrix. On is the matrix which has all elements equal to 0 and has order n (a square matrix nxn). Hope it's clear now... :D

And I think by Mn(C) you mean the set of n by n matrices with complex entries.

Exactly. Aren't these standard notations? I mean I thought Mn(A) is generally recognized as the set of square matrices with elements in the set A. And I also thought In and 0n are known as the unit matrix and the 0 matrix. Maybe you guys have other notations for all these. If so, please enlighten me! :D  :D

Sort of standard.  It would help to use subscripts, like Mn(C), In, and 0n.  The use of I for the identity matrix and 0 for the zero matrix is standard, though we often omit the n (when clear from context).  I've also seen Mn, though not as often; in this case I would just write out the words.


geometry
parallelogram
perimeter
rhombus
Pythagorean Theorem

This problem seems kind of easy, but I'm not sure if I'm assuming anything...

The lengths of the diagonals of a parallelogram are 10 and 24. If the length of one side of the parallelogram is 13, what is the perimeter of the parallelogram?

I think it is supposed to be 52 but I assumed that the diagonals were perpendicular bisectors....  :?:

they are perpendicular bisectors in a rhombus, but not necessarily a general parallelogram.

they would bisect each other though.

[quote="ckck"][quote="Quevvy"][quote="ragnarok23"]divide it into two triangles (with the diagonals) and use stewart's

or use the fact that the sum of the squares of all sides of the parallelogram is equal to the sum of the squares of the diagonals[/quote]
What is Stewart's Theorem (or Formula, or whatever)?[/quote]
Stewarts formula is to find the length of a median.
http://mathworld.wolfram.com/StewartsTheorem.html[/quote]
I can't see how this would apply to a parallelogram :huh:

here is what i got:

[hide]
Actually, since we know one side is 13, and half the diagonals are 5 and 12, the diagonals actually ARE perp. bisectors. So the parallelogram is a rhombus, making the perimeter $4(13)=\boxed{52}$

so his assumption was correct.
[/hide]

[quote="ckck"]Well, you notice that when you drw everything out, there is a triangle with side lengths5 and 13, I assumed it had to be right. I don't know if you can do that, though.[/quote]

The converse of the pythagorean theorem is true:
if $a^{2}+b^{2}=c^{2}$ then the angle between a and b is right.
if $a^{2}+b^{2}>c^{2}$ then the angle between a and b is obtuse.
if $a^{2}+b^{2}<c^{2}$ then the angle between a and b is acute.

Dude, using medians, or even the parallelogram is overkill for the median (at least I think this because the parallelogram requires the weird square sum thing).
My favorite way:
[hide]Find the length of the altitude to the side that the median is hitting. Then use pythag to find the length of the median.[/hide]


Suppose  we throw two unbiased dice.We want to find the probability of having their sum 6?
isn't the answer dependent on our sample space?
I could have my sample space as the set $ S \equal{} [2,3,4 ,\cdots,12]$ that is the possibe sums .So my probability P becomes $ \frac {1}{11}$
if i choose my sample space to be $ S \equal{} [(x,y): x,y \in \mathbb{N}  1\leq x,y \leq 6]$
then the probability becomes $ \frac {5}{36}$.
is the probability sample space dependent?

[quote]So my probability P becomes $ \frac {1}{11}$[/quote]
No, with this sample space not all outcomes are equally likely: say, $ 7$ is more likely than $ 2$.

ok now i think i got it the formula probability of A is $ P(A)\equal{}\frac{n(A)}{n(S)}$
only hold if all events are likely


[hide]We work backwards, and use zero-product property to get 7x+4=0 or x-3=0. Multiplying the binomials we attain the quadratic $7x^{2}-17x-12=0$.[/hide]

[quote="Ktradsb1"]Give an example of a quadratic equation that has two solutions: x=-4/7 or x=3.[/quote]
[hide] Anything in the form $0=a(x+\frac{4}{7})(x-3)$ for any real $a$, but it works well when $a$ is a multiple of $7$. :D [/hide]

[quote="mtms5467"][hide]We work backwards, and use zero-product property to get 7x+4=0 or x-3=0. Multiplying the binomials we attain the quadratic $7x^{2}-17x+12=0$.[/hide][/quote]

isn't it [hide]$7x^{2}-17x-12$ instead of $+12$?  :maybe:[/hide]

[hide="There are a lot of solutions"]There are a family of solutions:
n(x+4/7)(x-3)=0
n(7x+4)(x-3)=0, where n is any real number (Does it also work is n is complex? :maybe: )[/hide]

You backtrack

(7X+4)(X-3)=0

7x     4

x       -3

7x*x- 17x-12=0

P.S How do you show x squared?

Type x^2 and put dollar signs around it.

$x^{2}$=x^2

a better place for that question is the Latex forum  :wink:

[quote="Ktradsb1"]Give an example of a quadratic equation that has two solutions: x=-4/7 or x=3.[/quote]

[hide]Technically...$0x^{2}+0x+0=0$ is a quadratic.

But being more serious

$a(7x+4)(x-3)=0$, works.

I think a should be a real number, because we don't talk about complex numbers in this forum.

And being serious
a should be a non-zero real number.[/hide]

[quote="Ktradsb1"]Give an example of a quadratic equation that has two solutions: x=-4/7 or x=3.[/quote]

If $a$ and $b$ are the solutions to $ax^{2}+bx+c=0$, then $ax^{2}+bx+c$ can be factored as $a(x-a)(x-b)$

[quote="vishalarul"]Does it also work is n is complex? :maybe:[/quote]
Yes, it does. A quadratic has always 2 roots in $\mathbb{C}$, so you can get complex coefficients if you want to. (Because the roots are fixed by the factorisation.)

But I guess that [quote="mathgeniuse^ln(x)"]$a$ should be a real number, because we don't talk about complex numbers in this forum.[/quote]

[quote="Ktradsb1"]Give an example of a quadratic equation that has two solutions: x=-4/7 or x=3.[/quote]

[hide]anything with b^2-4ac>0[/hide]
[hide]such as 10000000x^2+x+100000000000000000000000 :rotfl:  :rotfl: [/hide][/hide]


I deleted it,sorry because I don't know that 's your ineq .
Could moderator delete that topic ?

xin lỗi, không biết bạn là ai nhưng đây là bdt của tôi, ít nhất bạn cũng nên ghi tên tác giả hay nói với tôi chứ, sao tôi mới post bài này trên 3t xong là bạn post ngay sang bên này thế. Như vậy người ta gọi là ăn cắp đấy  :mad:  :mad:  :mad:


calculus
integration
trigonometry
calculus computations

integral sign of 0 to t

(21e^8)sin(t-s)ds

i keep getting 0 as final, but it is wrong

The way I read it, you wrote $ \int_0^t21e^8\sin(t\minus{}s)\,ds$

I suspect that's not what you meant, including that I suspect that's not supposed to be $ e^8,$ but I'm not that good a mind reader.

Click on what I wrote (or what anyone else here writes) to see the codes needed to make it look like that; put that between dollar signs to have it show up on the forum.

i made one mistake, change the power of 8 to power of s

Let $ t\minus{}s\equal{}u,\ \int_0^t e^t\sin (t\minus{}s)\ ds\equal{}e^s\int_0^t e^u\sin u\ du$.

kunny im pretty sure you shouldnt have an s anywhere since its a dummy variable.

$ \int_0^t e^s \sin (t\minus{}s)ds\equal{} e^t\int_0^t e^{s\minus{}t} \sin(t\minus{}s)ds \equal{} e^t \int_0^t e^{\minus{}w}\sin w dw$

$ \equal{}\frac{e^t \minus{} (\cos t \plus{}\sin t)}{2}$ after two by parts integrals