Hey guys,

I decided to self-study AP Phyiscs C ,since my school only offers Phyiscs B...and I took AP Phy. B this year.

Does anyone have any recommendations on what textbooks I have to use to study for C?

Thanks in advance,

I believe than these two books are very good for self-study:

http://www.amazon.com/gp/product/0471320579/102-8528060-5584141?v=glance&n=283155

and

http://www.amazon.com/gp/product/0471401943/qid=1150185947/sr=1-9/ref=sr_1_9/102-8528060-5584141?s=books&v=glance&n=283155 (this is for C)

A greeting!

 :lol:

Should I buy Fundamentals of Physics or Physics 1 & 2 ?

Thanks,

I don't know Fundamentals of Physics, but I think that Physics vol. 2 is good for C. In this book is necessary a little of calculus (no indispensable), is rigorous, has many exercises and don't indispensable to have Physics vol. 1.

 :bye:

[quote="drunner2007"]i don't know. i did the same thing your doing my 11th grade year. I had fundamentals of physics by halladay, resnick, ...
i didn't like something about the book. on the contrary, i didn't think it was that rigorous. some of the exercises seemed like a waste of time (they weren't designed well).
but I have only used that book for a calculus based physics course, so I havn't seen every book out there.
I have done a little bit of work in a more advanced text, physics for scientists and engineers by giancolli. that book was much nicer. my physics b text was better as well.[/quote]
so drunner2007:
you mainly used the Fumdamentals of Physics to study for AP Physics C? i see that you don't think the book is rigorous..but do u think generally it was a good book for prep. ap C test?
also i'd greatly appreciate if u can suggest me how to study for it (since you've already done it).
Thanks in advance..

Fundamentals of physics worked for me.  It is certainly rigorous enough, because it what most people used who made US physics team.  Also, more than half of the people who take AP Physics at my school get 5's on both parts of C, and we use that book.
Many of the problems in it are harder than AP test questions, so you will be well prepared for the AP test.

some of the problems are harder than the ap exam of course, but I saw lots of problems that were really simple one-step or two-step problems with ugly numbers. like i said before, i havn't seen all the introductory calc based physics books out there so maybe it is the best. overall, you can make it through the course with the text but i wasn't that impressed with it somehow.

kimby, if you want to self-study the course, make sure you stay disciplined. you should probably have a physics teacher be a mentor so that he can give you a good schedule to follow and help you out if you ever get stuck on a hard section.

most of the calculus used is fairly basic, because they design these texts for college freshman that are taking calculus 1 concurrently. however, there are some problems in the book that will require calculus 2 or even 3. fortunately, those won't be on the ap exam. the ap exam is more focused on understanding the pure physical principles, then applying some fairly straight forward math (about half calculus). mainly worry about getting lots of practice with exercises, and make sure to focus on problems that will really put your understanding of principles.

if you took physics b last year you will have a nice headstart. but you also need to keep in mind that since lots of other people are doing the same (taking c after b) that will make it less of an advantage since the ap score you get depends on how you did compared to the average. so review those sections that are very close to what you studied in b, but focus a little more time on the material that is new.

i think that the hardest part of the course is the rotational stuff and the magnetism stuff. also don't forget to study waves! don't be suprised if you see a lot of the same stuff you saw in b (on the ap exam), but make sure you still take it seriously.

good luck  :)


Way too overposted.. please don't post problems from well known olympiads.. check the resources section please :)
http://www.mathlinks.ro/Forum/viewtopic.php?p=436492#436492


parameterization
LaTeX
linear algebra
matrix
geometry
calculus
integration

Is there any command I can alter the space between two lines in latex. For example I have two equations

\begin{equation}    E=mc^2          (1)  \end{equation}

\begin{equation}   PV=nRT           (2)   \end{equation}

and I have always found the space between the two always too wide. how can I narrow down the space between the two equations in displaymath mode  :?:

Try using
[code]\begin{eqnarray}
E=mc^2 \\
PV=nRT
\end{eqnarray}[/code]
so
$\begin{equation}E=mc^{2}\end{equation}\begin{equation}PV=nRT \end{equation}\text{becomes}\begin{eqnarray}E=mc^{2}\\ PV=nRT \end{eqnarray}$

Yes, but how to push all the equations and their numbers aling from the left neatly, i.e. I want to objects within {equarray} to comes out like:
(1)  E=mc^2
(2)  PV=nRT

Assuming you want this for the whole document then use
[quote]\documentclass[12pt,fleqn,leqno]{article}
\begin{document}
\begin{eqnarray}
E=mc^2 \\
PV=nRT
\end{eqnarray}
\end{document}[/quote]
Note the parameters in \documentclass. \fleqn will show displayed formulae flush left and \leqno puts equation numbers on the left

If you want different settings for different euqations then have a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110627]numerate equations in an \itemize environment[/url] topic.

Well, the "E=mc^2" have more spacing than "PV=nRT" from the equation numbers. One more thing how can we put all the following steps with equality signs all align, i.e. I want to have something like this:

(1)   E = mc^2
          = m_0\gamma c^2
          = m_0 ((v-at)/\sqrt{1-(v/c)^2) c^2

(2)   PV = nRT
            = .......
            = .......
            :
            :
with all "=" signs in the same psoitions

Try $\AmS-\text{\LaTeX}$'s align environment. The \notag command is to prevent that line being numbered.
[quote]\documentclass[12pt]{article}
\usepackage[fleqn,leqno]{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}
\begin{align}
E & = mc^2 \\
& = m_0\gamma c^2 \notag \\
& = m_0 ((v-at)/\sqrt{1-(v/c)^2}) c^2 \notag \\
PV & = nRT \\
& = ....... \notag \\
& = ....... \notag
\end{align}
\end{document}[/quote]

It seems the above is inconsistent with my setting at the beginning opf the file:

\documentclass[10pt,twoside]{siamltex}
\usepackage{amsfonts,epsfig}    :blush:

siamltex.cls redefines 'proof' so if you remove the line
\usepackage{amsthm}
everything should work.

Do note that siamltex uses an equation numbering system that incorporates the section number so you get
(1.1)
(1.2)
(1.3)
etc
instead of
(1)
(2)
(3)

To be consistent you would probably be best to stick to this style of numbering but if you must change it then use the line
\renewcommand{\theequation}{\arabic{equation}}

Well, I still can't able to reduce the distance between the lines between the words and the \begin{equation} . For example, If I want to shorten the distance between the words and the two equations in the following

     This is the equation I want to use:

                E = m c^2
                PV = nRT

     The next one is.......

what should I do?

It looks fine to me as it is but if you want to change the spacing after any particular line use
\vspace{-1ex}
1ex is the height of the letter x in the current font and you can use decimals like -1.2ex.
Do make sure that the printout looks good and not squashed.

Another idea is to include text inside an equation or align environment using
\text{the next equation is} \notag \\

Yes, that is good and really helpful
 :)
How about spacing between equations inside table, for example,
I have found the spacing between lines in the following too narrow, how can we increase or reduce the height of the table below? :?:

[code]\begin{tabular}{|l|l|}
\hline aa & a=\frac{1}{2}  \\ \hline n=0 & \frac{a}{b}
\\ \hline n=1 & \frac{1+B}{C} \\ \hline
n=2 & {\frac{a}{b}(1+{\rm Tr}(A)-\frac{\det(A)}{2})} \\
\hline n=3 &
{\frac{2}{3} ( 1+{\rm Tr}(A)-\frac{ 2 \det (A)}{2} } \\ \hline
\end{tabular}[/code]

Yes that's difficult to do well. One way is to follow the idea in the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=struttab]TeX FAQ's Spacing lines in tables[/url], but I have used a different way.

Firstly, replace tabular by array. They do the same things but array is used in math mode so you won't need any dollar signs in the array, just a pair outside it.

Secondly, extra line spacing works but creates gaps in the vertical lines. To avoid this create an array in the cells on the right-hand side - these can have spacing as you wish but as they are seen as a single entity so keeping the vertical lines.
[code]$\$$\begin{array}{|l|l|}
\hline aa & \begin{array}{c} \\ a=\dfrac{1}{2} \\\\ \end{array}
\\ \hline n=0 & \begin{array}{c} \\ \dfrac{a}{b} \\\\ \end{array}
\\ \hline n=1 & \begin{array}{c} \\ \dfrac{1+B}{C} \\\\ \end{array}
\\ \hline n=2 & \begin{array}{c} \\ \dfrac{a}{b}\left(1+\text{Tr}(A)-\dfrac{\det(A)}{2}\right) \\\\ \end{array}
\\ \hline n=3 & \begin{array}{c} \\ \dfrac{2}{3} \left( 1+{\rm Tr}(A)-\dfrac{ 2 \det (A)}{2}\right) \\\\ \end{array}
\\ \hline
\end{array}$\$$[/code]
gives \[\begin{array}{|l|l|}\hline aa & \begin{array}{c}\\ a=\frac{1}{2}\\\\ \end{array}\\ \hline n=0 & \begin{array}{c}\\ \frac{a}{b}\\\\ \end{array}\\ \hline n=1 & \begin{array}{c}\\ \frac{1+B}{C}\\\\ \end{array}\\ \hline n=2 & \begin{array}{c}\\ \frac{a}{b}\left(1+\text{Tr}(A)-\frac{\det(A)}{2}\right) \\\\ \end{array}\\ \hline n=3 & \begin{array}{c}\\ \frac{2}{3}\left( 1+{\rm Tr}(A)-\frac{ 2 \det (A)}{2}\right) \\\\ \end{array}\\ \hline \end{array}\] A few things to note:
1. \dfrac gives a display fraction which looks better (though the you won't see the difference in this forum)
2. {\rm Tr} is for compatibility with old versions of LaTeX and so I have used \text{Tr} instead.
3. I have used \left( and \right) to get the right size brackets
4. The table now is quite large but you can play with the line spacing as you would like. For example, \\*[2ex] could replace \\\\

As is the way with these things I have just realised that there is a simpler way which is to use \arraystretch[code]\renewcommand\arraystretch{2.5}
$\$$\begin{array}{|l|l|}
\hline aa & a=\dfrac{1}{2}
\\ \hline n=0 & \dfrac{a}{b}
\\ \hline n=1 & \dfrac{1+B}{C}
\\ \hline n=2 & \dfrac{a}{b}(1+\text{Tr}(A)-\dfrac{\det(A)}{2})
\\ \hline n=3 & \dfrac{2}{3} \left( 1+{\rm Tr}(A)-\dfrac{ 2 \det (A)}{2}\right)
\\ \hline
\end{array}$\$$[/code]This also leads to a smaller table.

Yes, that's the most helpful :lol:
Just one more query:
If, say in one of the entries in the table, I have to have it in two lines because the equation is too long. Say I want to put the entry  E = mc^2+PVnRT+Q = -dE/dt as in the following

n=0     aa     {a\b}(1+Tr(B)

n=1     ab     E = mc^2+PVnRT
                     +Q = -dE/dt

n=2     ac      Z=Z_0e^{--ih/t)

what should I write for the entry "E=mc^2..." ?  :?:

In that case use the first method (not using \arraystretch) so then you can use \\ inside the inner array. So the n=1 row has for the right-hand cell
 [code]\begin{array}{c} \\ E = mc^2+PVnRT \\ +Q = -dE/dt  \\\\\end{array}[/code] which gives
$\begin{array}{c}\\ E = mc^{2}+PVnRT \\+Q =-dE/dt \\\\\end{array}$
You could even line up the equals signs as in [code]\begin{array}{ll} \\ E & = mc^2+PVnRT +Q \\ & = -dE/dt  \\\\\end{array}[/code] to get
$\begin{array}{ll}\\ E & = mc^{2}+PVnRT+Q \\ & =-dE/dt \\\\\end{array}$

So, I have tried the following but I found the $\frac{2\sqrt{2}}{e\pi}$ on the right hand side moved up a bit. How to push it down a little and to be aligned with those on the left hand sides?

[quote]\begin{array}{|l|l|l|l|} \hline {~} & \begin{array}{c} \\
s=\dfrac{1}{3} \\\\ \end{array} & \begin{array}{c} \\ s=\dfrac{1}{2}
\\\\ \end{array} & \begin{array}{c} \\ s=\dfrac{2}{3} \\\\
\end{array}
\\
\hline n=0 & \begin{array}{ll} \\ \dfrac{3\sqrt{3}}{e\pi} \\\\
\end{array} & \begin{array}{ll} \\ \dfrac{2}{\pi} \\\\ \end{array}
& \begin{array}{ll} \dfrac{2\sqrt{2}}{e\pi}  \\\\
\end{array}
\\
 \hline
\end{array}[/quote]

$\begin{array}{|l|l|l|l|}\hline{~}& \begin{array}{c}\\ s=\frac{1}{3}\\\\ \end{array}& \begin{array}{c}\\ s=\frac{1}{2}\\\\ \end{array}& \begin{array}{c}\\ s=\frac{2}{3}\\\\ \end{array}\\ \hline n=0 & \begin{array}{ll}\\ \frac{3\sqrt{3}}{e\pi}\\\\ \end{array}& \begin{array}{ll}\\ \frac{2}{\pi}\\\\ \end{array}& \begin{array}{ll}\frac{2\sqrt{2}}{e\pi}\\\\ \end{array}\\ \hline \end{array}$

Plus:  should we use \begin{array}{lll} ... \end{array} instead of \begin{array}{lll}...\end{array} if the equation is to be broken into three lines?

You forgot the \\ before \dfrac{2\sqrt{2}}{e\pi} in the last row.
[quote="bchui"]
Plus:  should we use \begin{array}{lll} ... \end{array} instead of \begin{array}{lll}...\end{array} if the equation is to be broken into three lines?[/quote]"instead"?? They are the same.

There is no need to have 3 columns, which is what \begin{array}{lll} does. You can use \\ to write 3 rows.
I did use \begin{array}{ll} for 2 columns at one point above because I wanted to line up the equals sign so the left of the equals sign went in the first column then the = plus the rest of the equation went into the next column.

So you can use \begin{array}{l} or \begin{array}{c} in all your lines above. Really you only need one column in those cases.

Yes, you are right! :lol:

There is a further probelm I encountered. When I typed the following lines, I was trying to make spacing between the section heading and the words "Next, we..." but I found that the wordings always moved up a line. what should I put instead of "\vspace{3ex}? I tried \medskip but it didn;t work

[quote]\section{Generalized Hamiltonian} \label{Ham-sec}
\vspace{3ex}
Next, we consider

\being{equation}
\det(\lambda I_n-A)=0
\end{equation}

which is the characteristic equation ....
[/quote]

How can we flush the last sentence "which is ..." to the left? I always found some spacings for the sentences after an equation ?

There's nothing wrong with \vspace{3ex} if that gives the effect you want. \medskip doesn't work as there is no paragraph there. If you want the same line spacing after [i]every[/i] section then use [code]\newcommand{\mysection}[1]{\section{#1} \vspace{3ex}}[/code] and write for example,
[code]\mysection{Generalized Hamiltonian} \label{Ham-sec}[/code]
To remove an indent on a line use \noindent. Thus [code]\noindent which is the characteristic equation .... [/code]

If the sentence is too long, I always found that there is black box at the right end of the line. How could we remove the nasty balck box?

I can't reproduce the black box even if I used some of the constructions that you use. I tried [code]\documentclass[10pt,twoside]{siamltex}
\usepackage{amsfonts,epsfig}
\usepackage[fleqn,leqno]{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcommand{\mysection}[1]{\section{#1} \vspace{3ex}}

\begin{document}
\mysection{Generalized Hamiltonian} \label{Ham-sec}

Next, we consider

\begin{equation}
\det(\lambda I_n-A)=0
\end{equation}

\noindent This sentence is far too long, in fact it goes on and on and there's no way it can fit on one line and I wonder what will happen
\end{document}[/code]and it was fine.

Perhaps you could give an example like this where you get the black box?

When I run the following formula, I have found that the equations all flush to the right. How can we moved it back to the left hand side with equation number in the first line only and no equation numbers in all the following lines?

[quote]\begin{align}
{\rm Tr}^k(BB^T)  = \det\left[\begin{array}{cccc}\eta_1& { }&{ }& B_1\\
0&\eta_2&{ }& { }\\
\vdots&\vdots&\ddots& { }\\
0&0&\ldots&\eta_k\end{array}\right]\\
 + \ldots +\det (\left[\begin{array}{cccc}\eta_{n-k+1}& { }&{ }& B_k\\
0&\eta_{n-k+2}&{ }& { }\\
\vdots&\vdots&\ddots& { }\\
0&0&\ldots&\eta_n\end{array}\right] \notag \\
aaaa \notag\\
nnnn \notag
\end{align}[/quote]

The same problem appears when I tried the following:

[quote]\begin{eqnarray}
{\rm Tr}^k(BB^T) &= {\rm Tr}^k((P^{-1}AP)(P^{-1}AP)^T)\\
&= {\rm Tr}^k
(P^{-1}APP^TA^T(P^{-1})^T) = {\rm Tr}^k(P^{-1}APP^{-1}A^TP) \notag \\
&= {\rm Tr}^k(P^{-1}AA^TP)= {\rm Tr}^k(AA^TPP^{-1}) = {\rm Tr}^k(AA^T) \notag
\end{eqnarray}[/quote]

What I want is to have the first line an equation number on the left and all the following lines with no equations numbers. could this be done together with all the equality signs aligned?

We've covered some of this before. To get equations on the left and numbering on the left use
\usepackage[fleqn,leqno]{amsmath}
in the preamble.
In your example you only get one equation number as you wanted. On the following lines \notag stops further equation numbers.
Finally to get the equals signs nicely lined up just replace eqnarray by align. In fact eqnarray shouldn't ever be used - see [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=eqnarray]Why not use eqnarray?[/url]

One more thing, I found the \vspace{3ex} doesn;t work to skip a line for the sentence "Before we..." after the \section heading. What should I do?

[quote]\section{Generalized Function} \label{Gen-sec}

\vspace{3ex}

Before we move on to the next section, we need to quote the following results...[/quote]

Yes, well, I simply want to have the following lines all flushed left, what should I do?

[quote]\begin{align}
{\rm Tr}^k(BB^T)  = \det(\left[\begin{array}{cccc}\eta_1& { }&{ }& B_1\\
0&\eta_2&{ }& { }\\
\vdots&\vdots&\ddots& { }\\
0&0&\ldots&\eta_k\end{array}\right]
\left[\begin{array}{cccc}\eta_1& 0&\ldots & 0\\
{ }&\eta_2&\ldots & 0\\
{ }& { }&\ddots & \vdots \\
B_1& { }& { }& \eta_k\end{array}\right])  \\
 + \ldots +\det (\left[\begin{array}{cccc}\eta_{n-k+1}& { }&{ }& B_k\\
0&\eta_{n-k+2}&{ }& { }\\
\vdots&\vdots&\ddots& { }\\
0&0&\ldots&\eta_n\end{array}\right]
\left[\begin{array}{cccc}\eta_{n-k+1}& 0&\ldots & 0\\
{ }&\eta_{n-k+2}&\ldots & 0\\ { }& { }&\ddots & \vdots \\
B_k& { }& { }& \eta_n\end{array}\right])  \notag  \\
= (\eta_1\eta_2\cdot\cdot\eta_k)^2+\ldots
(\eta_{n-k+1}\eta_{n-k+2}\ldots \eta_n)^2 \notag
\end{align}[/quote]

[quote="stevem"][quote="bchui"]One more thing, I found the \vspace{3ex} doesn;t work to skip a line for the sentence "Before we..." after the \section heading. What should I do?[/quote]
That's how siamltex works. If you look at its documentation (compile docultex.tex) you'll see it starts its sections like this:
[quote]  [b] 2. Headings.[/b] The top matter of a journal paper falls into a standard format.[/quote]
So the first line of a section starts on the same line as its heading. I would stick with this as anyone requiring you to use this style expects it to have sections like this.

But if you [i]must[/i] break the convention, then you need a line to add spacing to (see the error you get if you replace \vspace{3ex}  by \\) so make a fake line with a space on it with
\ \vspace{3ex}[/quote]

I just tried that but it didn;t work:

[quote]LaTex Error:  Thre is no line here to break[/quote]

However, I found the following trick does work:

[quote]\section{Generalized Function} \label{Gen-sec}
${ }$
\vspace{3ex}
The following facts are required[/quote]

 :D

[quote="bchui"]Yes, well, I simply want to have the following lines all flushed left, what should I do?[/quote]Use the flalign environment with an extra & on the end of each line and \nonumber instead of \notag
Have look at page 54 of [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Herbert Voß Math Mode[/url] to see what I mean.
[code]\begin{flalign}
	\text{Tr}^k(BB^T) & =	\det\left(
		\begin{bmatrix}
			\eta_1& { }&{ }& B_1\\
			0&\eta_2&{ }& { }\\
			\vdots&\vdots&\ddots& { }\\
			0&0&\ldots&\eta_k
		\end{bmatrix}
		\begin{bmatrix}
			\eta_1& 0&\ldots & 0\\
			{ }&\eta_2&\ldots & 0\\
			{ }& { }&\ddots & \vdots \\
			B_1& { }& { }& \eta_k
		\end{bmatrix}
		\right) &\\
	& + \ldots +\det\left(
		\begin{bmatrix}
			\eta_{n-k+1}& { }&{ }& B_k\\
			0&\eta_{n-k+2}&{ }& { }\\
			\vdots&\vdots&\ddots& { }\\
			0&0&\ldots&\eta_n
		\end{bmatrix}
		\begin{bmatrix}
			\eta_{n-k+1}& 0&\ldots & 0\\
			{ }&\eta_{n-k+2}&\ldots & 0\\ { }& { }&\ddots & \vdots \\
			B_k& { }& { }& \eta_n
		\end{bmatrix}
		\right) & \nonumber \\
	&= (\eta_1\eta_2\cdot\cdot\eta_k)^2+\ldots
	(\eta_{n-k+1}\eta_{n-k+2}\ldots \eta_n)^2  & \nonumber
\end{flalign}[/code]I've taken the opportunity to replace \left[\begin{array}{cccc} by the AMSMath version \begin{bmatrix} which is a bit simpler.

One other thing - the default margins aren't wide enough (hence the black box you were complaining about earlier). Reduce them. You could for example use
\usepackage[margin=1.5in]{geometry}

How about enlarging the height of the bracket symbol $($ and the vertical line in the following integration sign:

[quote]\int_{x=0}^1 \frac{f(x)}{x} dx =
 \big( \frac{g(x)}{x} \big)\vert_{x=0}^1 [/quote]

 :?:

Use \left( and \right) for (...). For the vertical line, treat it as the right side of |...| but without the first |. You do this with \left. and \right| (note [b].[/b] after \left) to get
[code]\int_{x=0}^1 \frac{f(x)}{x} dx = \left.\left( \frac{g(x)}{x} \right)\right|_{x=0}^1[/code] which looks like
$\textstyle\int_{x=0}^{1}\frac{f(x)}{x}dx = \left.\left( \frac{g(x)}{x}\right)\right|_{x=0}^{1}$

Mind you, the following looks even better using display mode and \, before dx
[code]\displaystyle\int_{x=0}^1 \frac{f(x)}{x}\, dx = \left.\left( \frac{g(x)}{x} \right)\right|_{x=0}^1[/code]
$\int_{x=0}^{1}\frac{f(x)}{x}\, dx = \left.\left( \frac{g(x)}{x}\right)\right|_{x=0}^{1}$

(Note: because of the way the forum operates I had to add \textstyle to show the first image. You won't need it in your document)

Yes, there very helpful! :lol:

There is another problem which looks pretty hard :

(1)  I want to draw a table of $f_{s}(n)$ as follow

       s | n      1          2          3          4
       s=1       f_1(1)    f_1(2)   f_1(3)    f_1(4)
       s=2        :             :          :            :

How could we draw a diagonal line which separates s and n at the top left corner of the table?

(2)  In the middle of a section of the paper, I want to have something emphasised, like

       The physical meaning of $E=mc^{2}$ could be stated as:

                The total energy stored in any object is
                directional proportional to its mass

       Therefore,......

How can we have the emphasised sentence "The total energy...." appeared as italic and aligned in the middle of the page?

1. The diagonal line in the table is best done using the [url=http://texcatalogue.sarovar.org/entries/slashbox.html]slashbox package[/url] (compiling slashbox.tex gives the documentation).

2. Use the center environment along with \emph for italics.

Both questions are illustrated in the following code:
[code]\documentclass{article}
\usepackage{slashbox}
\begin{document}
\noindent
$\begin{array}{|c|c|c|c|c|}
\hline
$\backslashbox{$s$}{$n$}$ & 1 & 2 & 3 & 4 \\ \hline
s=1 & f_1(1) & f_1(2) & f_1(3) & f_1(4) \\ \hline
s=2 & : & : & : & : \\ \hline
\end{array}$
\bigskip \\
The physical meaning of $E=mc^2$  could be stated as:
\begin{center}
\emph{The total energy stored in any object is \\
directional proportional to its mass }
\end{center}
Therefore ...
\end{document}[/code]

Well, how can we have the entences in the \begin{center} all aligned from the left hand side?

[quote]\begin{center}
\emph{The total energy stored in any object is \\
directional proportional to its mass }
\end{center}[/quote]

and, how to produce the inverted " sign, for I want to have a quoted sentence:

[quote]this is a more general version of "Stokes Theorem"[/quote]

but the first " sign always appeared as inverted

[quote="bchui"]Well, how can we have the entences in the \begin{center} all aligned from the left hand side?[/quote]How can you have both centered and left-aligned? It has to be one or the other.

[quote="bchui"]but the first " sign always appeared as inverted[/quote]That's how it should be and is the professional way to typeset text.

Could we possibly put the sentence within \center inside a box so that the sentence would be aligned to the left?

I I think you mean that the first line is centered and the remaining lines start at the same indentation as the first line.

Does this do what you want?
[code]\begin{center}
\parbox{7cm}{\emph{The total energy stored in any object is \\
directional proportional to its mass}}
\end{center}[/code]Adjust the 7cm (=the width of the box) to suit the sentence. To ensure it is centered, repeat the sentence in the line before the parbox and change the 7cm until it lines up, then remove that extra sentence.

I am trying to include a figure in the latex file. There is a special command for the symbol "FIG" so that it looks smaller than the usual size. Can you recall what that is?

How about \renewcommand\figurename{\footnotesize Fig.}?
You can of course replace \footnotesize and Fig. with whatever you want. This will change the size of the number and caption as well which I imagine you want otherwise it will look odd.

[quote="bchui"]:blush: That doesn;t work at all and the size of the words appeared as just the same as 10pt.

I wonder how could we alter the size of the characters without changing the setting at the top of the article.

For example, if I want to reduce the size of $i_{1},\ldots i_{k}$ in the following, what should I do?

[quote]\sum_{\begin{array}{c}1\leq i_1\ldots i_k\leq
n;\\ 1\leq l_1\leq k \end{array}} \lambda [/quote][/quote]
You can use \substack command. With the packages
[code]\usepackage{amsmath}
\usepackage{amsintx}[/code]
type
[code]\sum_{\substack{1\leq i_1\ldots i_k\leq
n;\\ 1\leq l_1\leq k }} \lambda[/code]
This gives
$\sum_{\substack{1\leq i_{1}\ldots i_{k}\leq n;\\ 1\leq l_{1}\leq k }}\lambda$.

[quote="bchui"]:blush: That doesn;t work at all and the size of the words appeared as just the same as 10pt.[/quote]
Try \renewcommand\figurename{\scriptsize Fig.} or even \renewcommand\figurename{\tiny Fig.}

[quote="bchui"]Yes, but where can I download  "amsintx.sty". When I run the latex, it cannot find the file[/quote]
Both packages contain the command \substack and if you run only amsmath, it will work. If you want the expression $\sum_{\substack{1\leq i_{1}\ldots i_{k}\leq n;\\ 1\leq l_{1}\leq k }}\lambda$ to look fine in an inline equation in a $\LaTeX$ document (i.e. not in the forum) use \displaystyle command before the whole expression:
[code]\displaystyle\sum_{\substack{1\leq i_1\ldots i_k\leq n;\\ 1\leq l_1\leq k }} \lambda[/code]
In the forum it isn't necessary, beacuse \displaystyle is automatically enabled.


For this problem, just do the math directly; no twists.

$ \frac {8!}{(2^6)}$

$ \equal{} \frac {8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{64}$

$ \equal{} \frac {7\cdot6\cdot5\cdot4\cdot3\cdot2}{8}$

$ \equal{} 7\cdot6\cdot5\cdot3$
$ \equal{} 42\cdot15$
$ \equal{} 420 \plus{} 210$
$ \equal{} \fbox{630}$


Find all integers x and y such that
x(x+1)=2y(y+1).

It give Pelle equation $u^2-2v^2=-1,u=2x+1,v=2y+1$ Elementary solution u_0=v_0=1, a=3,b=2. All solution given by
$u_n+v_n\sqrt 2 =\pm(1+\sqrt 2)(3+2\sqrt 2)^n,n\in Z$ It is easy to prove that $u_{n+1}=3u_n+4v_n,v_{n+1}=3v_n+2u_n$
Therefore all $u_n,v_n$  are odd, and it give $x_{n+1}=3x_n+4y_n+3,y_{n+1}=3y_n+2y_n+2$.Let ${x_0=0,y_0=0}$, then all solution give by sequences $(x_n,y_n),(-1-x_n,y_n),(-1-x_n,-1-y_n),(x_n,-1-y_n)$.

Thanks for your prompt answer, Rust!
I was wondering if anyone could find a more elementary approach, I mean without reducing this to Pell's equation.
Any ideas out there?

Well, you could formulate it without using the term 'Pell' and noone would even think of it as nonelementary  :P
Pell doesn't use anything special or any nonelementary technique, and there should be no solution basically different from the Pell one.

All right, thank you.
I just thought Pell's proof could be simplified for this special equation, in some smart way. Never mind, then.


trigonometry
real analysis
real analysis unsolved

Find the [url=http://en.wikipedia.org/wiki/Natural_density]density[/url] of the set $ A: \equal{}\left\{n\in\mathbb N\colon \sum_{k\equal{}1}^n \sin(k)<0\right\}$ in $ \mathbb N$.

We have: $ \sum_{k \equal{} 1}^n \sin(k) \equal{} \frac {1}{2\sin\frac {1}{2}}\left(\cos\frac {1}{2} \minus{} \cos\left(n \plus{} \frac {1}{2}\right)\right)$

$ \sin\frac {1}{2} > 0$, so here should be $ \cos\left(n \plus{} \frac {1}{2}\right) > \cos\frac {1}{2}$.

This gives: $ n\in\left( \minus{} \frac {\pi}{3} \minus{} \frac {1}{2} \plus{} 2\pi m,\; \frac {\pi}{3} \minus{} \frac {1}{2} \plus{} 2\pi m\right)$ for $ m\geq 0$ and we should find how many numbers $ \leq n$ could be in this intervals.

I think the answer is number of integers from interval $ \left( \minus{} \frac {\pi}{3} \minus{} \frac {1}{2},\; \frac {\pi}{3} \minus{} \frac {1}{2}\right)$ divide number of integers from the interval $ (0, 2\pi)$. And it will be: $ \frac {2}{6} \equal{} \frac {1}{3}$

All is fine until the very end. The density of integers such that $ n\in E\mod 2\pi$ where $ E\subset[0,2\pi]$ is a union of intervals is $ \frac{|E|}{2\pi}$ (integers are "uniformly distributed" modulo $ 2\pi$) :)

yes fedia, it is not so easy....(I studied high math only one year in the institute and don't know a lot of theory that you use... not only in this post... :( ). I made a mistake, it should be

$ n\in\left(2\pi m \minus{} 1,\;2\pi m\right)\; m\geq 1$

So, for evefy $ m$ there exist such $ n$. Period of this intervals is $ 2\pi$. So, for big $ n$ number of intervals will be approximatly $ \frac {n}{2\pi}$ and the answer for the problem is $ \frac {1}{2\pi}$.


geometry
quadratics
cyclic quadrilateral

The TRML 2005 team round (excluding the geometry ones)!

1. Take out all numbers containing a nine in the set of 3 digit numbers.  Multiply the elements.  How many zeros are at the end?

5. Let a be a real number.  Find the interval for a if the following conditions are satisfied:
$ x^2 + 4 a x -4 a + 3=0$
$x^2+(a-1)x+a^2=0$
$2 x^2 - (4 a+1)x+2a^2-1=0$

6. Let a,b,c be integers.  a>b and they satisfy:
$a+b=60$
$ab=c^2-72c+1920$
Find a.

8. A cyclic quadrilateral has sides 1,4,8,7. Find the radius.

9. If x and y satisfy:
$log({x^3 +\frac{1}{3} y^3 + \frac{1}{9}})=log(x)+log(y)$
Find (x,y)

10.  If x and y are integers, $x \ge y$ and
$\frac{x+y}{x^2-xy+y^2}=\frac{3}{7}$
Find (x,y).

Why excluding the geometry ones?

By the way, problem 8 can be solved by using [url=http://mathworld.wolfram.com/CyclicQuadrilateral.html]the special case of Bretschneider's formula.[/url]

They seem not so difficult....
however they are interesting :lol:

[hide="Problem 9"]We have the restriction $x,y>0$.
The equation is equivalent to
$x^3+\frac{1}{3} \cdot y^3+\frac{1}{9}=xy$
However since $x,y>0$
from AM-GM we get
$x^3+\frac{1}{3} \cdot y^3+\frac{1}{9}\geq3\sqrt[3]{x^3 \cdot \frac{1}{3} \cdot y^3 \cdot \frac{1}{9}}=xy$
so the equality holds
that is $x^3=\frac{1}{3} \cdot y^3=\frac{1}{9}$.....[/hide]

Does #6 has solution ? Here is how I work out ...

Suppose there are solution for $a,b,c$ then from those equations

$a(60-a)=(c-36)^2+624$  :arrow:  $(c-36)^2+(a-30)^2=276$

Since $276\equiv 0 \mod4$ It follows both $c-36$ and $a-30$ are even .

But even perfect square only end with digit $0,4,6$ and the only one add up to $6$ is $0+6$ . So from here one must be $0$ or $10^2$ , but both these wont give us a perfect square . Hence no solution for this system of equation .

Is there anything wrong  :?

Problem 10
[hide]
We know
$x+y = 3n$
$x^2 - xy + y^2 = 7n$

So
$x^2 + 2xy + y^2 - 3xy = (x + y)^2 - 3xy = (3n)^2 - 3xy = 9n^2 - 3xy = 7n$
$\Rightarrow 9n^2 - 7n = n(9n - 7) = 3xy$

We know n must be some multiple of 3, so trying to solve when $n = 3$ gives
$3(20) = 3xy \Rightarrow xy = 20$
$x + y = 9$

Placing these values into a quadratic,
$x^2 - 9x + 20 = (x - 5)(x-4)= 0$

So, the answer is (5, 4).[/hide]

[quote="fourierseries"]Sorry, I'm not sure how to input the diagrams for the geometry problems.  :P[/quote]
Add attachments.
Actually, we are never given diagrams in Math contests so you can just describe the problems.

[quote="fourierseries"]Okay. I shall post the geometry problems shortly after I make the attachments :D[/quote]
Are they available now?  :)


trigonometry
quadratics
algebra
quadratic formula

If 0<a<30 and $ sin^{6}a\plus{}cos^{6}a\equal{}7/12$, find cos a. Well actually, I solved this problem, but I want to see if there is an easier method, so if you post, can you post your method? Thanks! (My answer is$ \sqrt{30}/6$)

[quote="rob22"]For simplicity of notation, let $ x = \sin{a}$ and $ y = \cos{a}$. It is known that $ x^2 + y^2 = 1$. Thus, we are trying to solve:

$ x^6 + y^6 = \frac {7}{12}$
Expanding the LHS:
$ (x^2 + y^2)(x^4 - x^2y^2 + y^4) \\
x^4 - x^2y^2 + y^4 \\
x^4 - x^2(1 - x^2) + (1 - x^2)^2 \\
3x^4 - 3x^2 + 1$

Now, just solve $ 3x^4 - 3x^2 + 1 = \frac {7}{12}$, making sure $ 0 \leq x \leq \frac {1}{2}$ Quadratic Formula Yields:

${ x = \frac {\sqrt {6}}{6}}$
Going back to $ x^2 + y^2 = 1$, and that ${ \frac {\sqrt {3}}{2}} \leq y \leq 1$ it is easy to find that: ${ y = \frac {\sqrt {30}}{6}}$ <-------------- answer[/quote]

IMHO, that was brilliant!

Yes, thank you, I can see how that works.

My Way: $ (sin^{2}a\plus{}cos^{2}a)^3\equal{}1$
$ sin^6a\plus{}cos^6a\plus{}3sin^2cos^2(sin^2a\plus{}cos^2a)\equal{}1$
$ 7/12\plus{}3sin^2acos^2a(1)\equal{}1$
$ sin^2acos^2a\equal{}5/36$. From here, you just substitute $ x\equal{}cos^2a$ so $ x(1\minus{}x)\equal{}5/36$, and you solve than and take the square root of x to get cos a. I just wanted to ask if this way is as practical as rob22's (You guys probably know as you are more experienced). Thanks!

No I'm just asking which solution would be quickest. Thanks!

Your solution is quicker, but I think Rob's is a little better.

Agreed.

There is always more then one way to solve a question.  Both solutions are perfectly good, one was just quicker.  Perhaps on a different question Bengal's approach makes more sense.


Yes, do you guys have a date planned yet?
Because we are choosing a date for our school's competition, and are trying to avoid Princeton.

I don't know the exact date, but when I asked, I was told toward the end of January, which is a change from the previous November competitions.

That would be an interesting change of pace. Isn't HMMT supposed to be like Feb. 7 or something crazy like that?

Doing it at the end of January would make it really hard for teams to do both.

....things to consider. Normally I wouldn't have a problem with it if HMMT was at its usual date at the end of February. Aside from that Nov. date, or an earlier January date, there aren't a whole lot of weekends you can do it on.

HMMT should be on the 21st, and Standford MC is also on the 21st, but none of them are close to pumac which is the 31st of January.

[quote="spse"]When is Princeton Math Competition 2008?  Anyone in NJ wants to do it?  We can start a team and prepare in the summer.[/quote]

crap not in NJ


calculus
integration
algebra
polynomial
limit
function
real analysis

[quote="kunny"]Let $f(x)$ be the polynomial with respect to $x,$ and $g_{n}(x)=n-2n^{2}\left|x-\frac{1}{2}\right|+\left|n-2n^{2}\left|x-\frac{1}{2}\right|\right|.$
Find $\lim_{n\to\infty}\int_{0}^{1}f(x)g_{n}(x)\ dx.$[/quote]

The sequence of functions $g_{n}(x)\rightarrow\delta\left( x-\frac{1}{2}\right)$ as $n\rightarrow\infty ,$ where $\delta$ is the Dirac delta function(al). According to the so-called snifting property of the delta function, we have

$\lim_{n\to\infty}\int_{0}^{1}f(x)g_{n}(x)\ dx = f \left(\frac{1}{2}\right)$

where I have hand-waved the rigorous application of the Lebesgue Dominated Convergence Theorem (LDCT). --Ben

You can think of the Dirac distribution at $x$ as the limit of a sequence of functions $d_{n}$  with $\int_{\mathbb{R}}d_{n}(t) dt= 1$ and $d_{n}(x) \to \infty$ as $n \to \infty$ but for any $y \ne x$, $d_{n}(y) \to 0$. It's like a sequence of hills which each are taller and more narrow than the previous.

It's not really a function because it's value at $x$ isn't defined (everywhere else it's 0). But it exists as a distribution (a tempered distribution in fact), and it's easy to think of as the limit of a sequence of functions.

One of the most famous properties of the Dirac distribution is that for any function $f$ and real $t \in [a, b]$, we have
$\int_{a}^{b}f(t)\delta(x-t) dt = f(x)$

You can see why this is true by thinking of the limit. Each successive function $d_{n}$ is more and more concentrated around $x$, so that's the only part of $f$ that matters. And since $\int d_{n}(t) dt = 1$, we're basically averaging $f$ over smaller and smaller intervals around $x$, until it's basically $f(x) \cdot 1$ in the limit. (Note: this argument was NOT meant to be rigorous!)

In other words, the Dirac distribution is the identity for the convolution operation:
$f * g(x) = \int_{\mathbb{R}}f(y)g(x-y)dy$
$f * \delta(x) = \delta * f(x) = f(x)$

Anyway, [b]benorin[/b] first notes that $g_{n}(x) \to \delta_{1/2}$ and then uses the property I mentioned.

This is a plot of the sequence of functions $g_{2n+1}(x),n=0,1,\ldots ,12$

[url=http://www.uploadfile.info][img]http://www.uploadfile.info/uploads/22e0ff79d4.jpg[/img][/url]

[quote="benorin"]The sequence of functions $g_{n}(x)\rightarrow\delta\left( x-\frac{1}{2}\right)$ as $n\rightarrow\infty$ , where $\delta$ is the Dirac delta[/quote]
Converges in what sense? How can you say that anything converges to the Dirac delta?

I'll answer my own question: the sequence converges in the only sense available to us for distributions, which is [i]weakly[/i].

We say that $f_{n}\to f$ weakly (in the sense of distributions) iff for all test functions $\varphi$ (test functions being $C^{\infty}$ compactly supported or something like that).

$\int f_{n}\varphi \to \int f\varphi$

In particular, $f_{n}$ converges to $\delta(x-a)$ weakly iff for all test functions $\varphi,$

$\int f_{n}\varphi \to \varphi(a).$

So when benorin says that $g_{n}\to\delta(x-1/2),$ he's actually stating the conclusion he wants. A little bit of argument about why $g_{n}$ does actually converge (weakly) to the appropriate shifted delta might be in order.

--

Xevarion was being informal, but basically correct. Note that he also was talking about weak convergence in the sense of distributions.

[quote="Kent Merryfield"][quote="benorin"]The sequence of functions $g_{n}(x)\rightarrow\delta\left( x-\frac{1}{2}\right)$ as $n\rightarrow\infty$ , where $\delta$ is the Dirac delta[/quote]

$\vdots$

So when benorin says that $g_{n}\to\delta(x-1/2),$ he's actually stating the conclusion he wants. A little bit of argument about why $g_{n}$ does actually converge (weakly) to the appropriate shifted delta might be in order.

--

Xevarion was being informal, but basically correct. Note that he also was talking about weak convergence in the sense of distributions.[/quote]

Indeed, I entirely hand-waved the issue of convergence. Since the delta functional is not a function, and there are infinitely many sequences of functions that would converge to the delta function, I followed simple criteria for declaring such "convergence": any sequence of functions such that $f_{n}(x)\rightarrow 0, \forall x\neq 0$ where the convergence is, say pointwise, and such that $f_{n}(0)\rightarrow\infty$ is a delta sequence.

[quote="benorin"]I followed simple criteria for declaring such "convergence": any sequence of functions such that $f_{n}(x)\rightarrow 0, \forall x\neq 0$ where the convergence is, say pointwise, and such that $f_{n}(0)\rightarrow\infty$ is a delta sequence.[/quote]

Whoa! Let's be careful here! That's not a very good criterion.

Let me give you several sequences of functions. In each case, the function is zero outside the interval $(-1/n,1/n).$

$f_{n}(x)=\sqrt{n}$ on $(-1/n,1/n)$
$g_{n}(x)=n$ on $(-1/n,1/n)$
$h_{n}(x)=5n$ on $(-1/n,1/n)$
$k_{n}(x)=n^{2}$ on $(-1/n,1/n)$

Note that all of these functions tend pointwise to zero except at zero and tend to infinity at zero. Consider them all as sequences of distributions. What are their weak limits?

$f_{n}\to 0$
$g_{n}\to\delta$
$h_{n}\to5\delta$ (Think about that - how have you distinguished $\delta$ from $5\delta?$)

And finally, $k_{n}$ does not converge to any distribution.

Here's another example: $f_{n}(x)=\frac{\sin{n\pi x}}{\pi x},$ extended to be continuous. Then $f_{n}$ does not come close to meeting the conditions you have laid out; nevertheless, considered as a distribution (meaning smooth test functions), $f_{n}\to \delta.$

Here's a theorem. This moves a little away from distributions, since I'm not using smooth test functions here, merely continuous ones. But then, the Dirac delta can also be considered a measure as well as a distribution. (And you don't have to understand a word of this introductory paragraph to understand the proof.)

Suppose the $f_{n}$ is a sequence of nonnegative integrable functions (you can use Riemann integrability if you'd like) on $[-M,M]$ such that:

1. $\forall\,\eta>0,\,\exists\,N$ such that if $N>n,\ f_{n}(x)=0$ for $|x-a|\ge\eta.$

2. $\lim_{n\to\infty}\int_{-M}^{M}f_{n}(x)\,dx=1.$

Then if $g$ is an integrable function on $[-M,M]$ that is continuous at $a,$ then

$\lim_{n\to\infty}\int_{-M}^{M}f_{n}(x)g(x)\,dx=g(a).$

Proof: Write

$\int_{-M}^{M}f_{n}(x)g(x)\,dx= \int_{-M}^{M}f_{n}(x)g(a)\,dx+\int_{-M}^{M}f_{n}(x)(g(x)-g(a))\,dx$

Let's look at the first integral:

$\int_{-M}^{M}f_{n}(x)g(a)\,dx= g(a)\int_{-M}^{M}f_{n}(x)\,dx \to g(a)$ by hypothesis.

Now we estimate the second integral:

$\left|\int_{-M}^{M}f_{n}(x)(g(x)-g(a))\,dx\right|\le \int_{-M}^{M}f_{n}(x)|g(x)-g(a)|\,dx$

For $\epsilon>0,$ find $\delta > 0$ such that $|g(x)-g(a)|<\frac{\epsilon}2$ if $|x-a|<\delta$

Now find $N_{1}$ such that for $n>N_{1},\ f_{n}(x)=0$ for $|x-a|\ge\delta.$

Find $N_{n}$ sucn that for $n>N_{2},\ \int_{-M}^{M}f_{n}(x)\,dx<2.$

Let $n>\max\{N_{1},N_{2}\}.$ Then

$\int_{-M}^{M}f_{n}(x)|g(x)-g(a)|\,dx= \int_{a-\delta}^{a+\delta}f_{n}(x)|g(x)-g(a)|\,dx$

$\le \int_{a-\delta}^{a+\delta}f_{n}(x)\cdot\frac{\epsilon}2\,dx$

$\le \frac{\epsilon}2\int_{-M}^{M}f_{n}(x)\,dx\le\epsilon.$

--

This is not a strongest possible theorem - you could tweak some of the hypotheses to make them weaker than I've stated. But it's strong enough to cover this particular problem.


ARML
Putnam
MATHCOUNTS
geometry
power of a point

we were really strong in everything... especially Individual.

We had 3 7's, 1 6's (that was me), many many 5s and those were just off of many mistakes.

Also, we did pretty well on Power Round and Team Round if I remember.

But we only scored 6 off of relay =(

this seems like a sort of strange topic to have

yea this is a strange topic... let's see um... we're really weak at nothing. We're extremely strong at Power and Individual... we're also good at Team and Relay. So yea.

Haha, Ohio went from 0 for relay in 2005 to 18 in 2006.

I attempted to crunch some numbers on this a couple of years ago.  I used a funny formula looking at what percentage of a team's total score came from any given round, and comparing that percentage to the average of ten top teams.  Here's what I got, using data through 2003:

Minnesota is strong on individual and weak on team and relay.
Minnesota Maroon (and, I suspect, most B teams) is strong on team and power and weak on individual and relay.
Chicago is strong on relays and weak on power.
NYC is strong on individual and not particularly weak on anything.
TJ has no big strengths, but a weakness for power.
Montgomery and Texas are balanced--no big strengths or weaknesses.
Massachusetts and SFBA are both strong on relays and weak on team.
AAST is strong on individual and weak on relays.
Exeter is strong on team and weak on individual.

I've long believed that Chicago was far and away the best team in the country, and would win another championship as soon as the other teams at the top falter on relays.  After last year, we'll see if Montgomery can continue to challenge them for that.

In terms probably pretty similar to Sly Si's system, we (Southern California) jumped into Division A as a first-year team in 2002 by being strong on team and power, middling on individual, and weak on relays. Since then, we've become fairly balanced among team, power, and individual, but relays remain our weak point.

That 2002 team and power performance depended quite heavily on a single individual acting as the leader - and he personally got 8 on the individual round that year. (You San Diego people might have met him - he went to UCSD as an undergrad.) Our load is now much more widely distributed.

Keep in mind the numbers are only through 2003.

I think most teams would say that they tend to do the worst on the Super Relay (if you go by percentage of answers that are correct).

But about the relay: I think teams sometimes think they stink at relays because they're not scoring nearly as many points as on team or power (all three events are out of 40).

That's kind of misleading, because the relay event is not so much a [i]team[/i] event as it is an individual event where you only get credit on your question if 2 other people get their questions correct as well. (Granted, there is a slim possibility of getting the right answer if one or more relayers makes a mistake.)

Consider that the best teams in the country get maybe 75% of all possible points on the individual event. Now ask yourself how low that number would be if you graded the individual round with the rule that a student's answer gets credit only if their entire relay team got it correct.

Now I realize something I have in my records: our rank in each component in each of the years we've competed. And now that I go back and look, our team rank in that first year was equal in individual and relay - and we were carried best by our team round performance that year.

To break into the very top ranks, you need points from everyone on the individual round. It's fair to say that the team an power round most effectively measure the "front half" of your team, the people who do more than their share of the work in the room. But the individual round is really about the "back half" of your team. What does that make relays? I've never figured that out.

Here's a table of our team's performance over the last five years. (We're Southern California, and this is just our A team.) The ranks are given in a Putnam-like "average the ties" style.

$\begin{array}{cccccc}&2002&2003&2004&2005&2006\\ \text{Team}&9&14&27.5&19.5&12\\ \text{Power}&16.5&13.5&12&21&18\\ \text{Indiv.}&28&29&16.5&12&10.5\\ \text{Relay}&28&32.5&32&14.5&25\\ \text{Overall}&19&19&17&13&14\end{array}$

[quote="Altheman"][quote="jackb1115"][quote="probability1.01"]Dude if I were on Chicago ARML I suspect we wouldn't have as much trouble w/ power.[/quote]

A couple of other team members and myself were considering buying you a plane ticket and carting you out here.  Fell through though =)[/quote]

is that really the only problem? i thought it was that the coaches were like: he missed the tryout [same situation as tony liu][/quote]

Yeah uhhh that was pretty random as I emailed one of your coaches long before the tryouts but he never responded... eh although I probably was not going to try out anyway considering how far away it is.

[quote="Sly Si"]Does anyone know where to get old records of ARML standings and scores?  When I was getting my data, I used the old ARML site, but that got taken down a while ago.  Consequently, I won't ever be able to add teams like North Carolina or Lehigh to my stats.[/quote]

But the results, scores, etc. are still up on [url]http://www.arml.com[/url] for the past 6 years... Is that not what you want?

By the way, it appears we're extremely good all around (we won the team in a six-way tie, won the individual in a two-way tie, won the power in a two-way tie, and came out average on the relay).

well, that would be over the span of one year, which is admittedly a fairly small sample size

[quote="probability1.01"][/quote][quote="Altheman"][quote="jackb1115"][quote="probability1.01"]Dude if I were on Chicago ARML I suspect we wouldn't have as much trouble w/ power.[/quote]

A couple of other team members and myself were considering buying you a plane ticket and carting you out here.  Fell through though =)[/quote]

is that really the only problem? i thought it was that the coaches were like: he missed the tryout [same situation as tony liu][/quote][quote="probability1.01"]

Yeah uhhh that was pretty random as I emailed one of your coaches long before the tryouts but he never responded... eh although I probably was not going to try out anyway considering how far away it is.[/quote]

Alex A -- Tony isn't on the team b/c IMSA's graduation ceremony is that weekend.

Out plan was to fly you out for tryouts, and most likely then again for the practices =)

oh ok i heard something vague and i filled the details incorrectly...

To get really old ARML results, you should use what any time traveler would use: the Wayback machine (found at http://www.archive.org)    :lol:

Here is what the Archive has for ARML.

http://web.archive.org/web/*/http://www.arml.com

This page has PDFs of some of the "ARMLogs" back to 1994:

http://web.archive.org/web/20010401175626/http://arml.com/

Here is the link to the Spring 1994 (result from 1993):

//web.archive.org/web/20010603165718/arml.com/ARMLLOGS/1994.pdf

[quote="Karth"]Um.... TJ bad at power round? Is there something wrong  :P

*cough Jacob Steinhardt*[/quote]

Eh Zhai and Tony and Arnav all own me easily, so not much of an argument. The only argument I can see is that most of our A team can now solve power round problems, mainly because we have focused a lot on it this year due to not doing well in the past.

[quote="worthawholebean"]Everywhere.[/quote]

I agree, when you're divided up into groups of 9 to do a team round and 3 people supply all the answers while the 6 other people in total come up with 1 wrong answer, you know you have a problem.

Or when it takes 4 minutes for the first person to pass back a simple Power of a Point question.

Or when your state's practices are giving you MathCounts State level problems which aren't even close to ARML level.

[quote="Karth"]Um.... TJ bad at power round? Is there something wrong  :P

*cough Jacob Steinhardt*[/quote]

Um, isn't he also really good at USACO and not going to ARML this year?

yeah, stop making excuses.  john pardon went to some comp sci competition and look where NC ended up: dead first.  but yeah pardon probably wouldn't have liked the pink anyway... :P

isn't it also awesome when you have like 6 former/present MOPpers on your team?  (4 if you don't count red, 5 if you don't count former)

but yeah like we have to match that this year (somehow)


If $ a,b,c \geq 0$ prove that $ \sum a^2 \sqrt{(a\plus{}b)} \sqrt{(a\plus{}c)} \geq \sum 2a^2b.$  (The sums are cyclic.)
I don’t know whether this problem is easy or hard—all I can find is a complicated SOS proof.  Can you find an easy proof?

Can u  post your SOS proof (However complicated) here?

[quote="dabug"]If $ a,b,c \geq 0$ prove that $ \sum a^2 \sqrt {(a \plus{} b)} \sqrt {(a \plus{} c)} \geq \sum 2a^2b.$  (The sums are cyclic.)
I don’t know whether this problem is easy or hard—all I can find is a complicated SOS proof.  Can you find an easy proof?[/quote]
My solution :)

can_hang2007, your solution is great!  I thought I had made up this inequality the other day, but I guess not.

Here are the details of my solution.  All the summations are cyclic.

First some preliminaries we’ll need later:

\[ \sum (a\minus{}b)^3\equal{} \sum (a^3\minus{}3a^2b\plus{}3ab^2\minus{}b^3) \equal{} \sum (\minus{}3a^2b\plus{}3ab^2),\]
and
\[ \sum(2a\plus{}b)(a\minus{}b)^2\equal{}\sum(2a^3\minus{}3a^2b\plus{}b^3)\equal{}\sum(3a^3\minus{}3a^2b).\]

Now multiply the original inequality by 2 and add $ \sum(\minus{}a^3\minus{}a^2b\minus{}a^3\minus{}a^2c)$ to both sides to get the equivalent form

\[ \sum (a^2(\minus{}(a\plus{}b)\plus{}2\sqrt{a\plus{}b}\sqrt{a\plus{}c}\minus{}(a\plus{}c)) )\geq \sum(4a^2b\minus{}2a^3\minus{}a^2b\minus{}a^2c).\]
The left side is $ \sum(\minus{}a^2(\sqrt{a\plus{}b}\minus{}\sqrt{a\plus{}c})^2),$ and multiplying by $ (\sqrt{a\plus{}b}\plus{}\sqrt{a\plus{}c})^2)$ over itself we rewrite the left side as

\[ \sum(\minus{}a^2 \frac{(b\minus{}c)^2}{(\sqrt{a\plus{}b}\plus{}\sqrt{a\plus{}c})^2})\equal{}\sum(\minus{}c^2\frac{(a\minus{}b)^2}{(\sqrt{a\plus{}c}\plus{}\sqrt{b\plus{}c})^2}).\]

The right side is \[ \sum(3a^2b\minus{}2a^3\minus{}a^2c)\equal{}\sum((3a^2b\minus{}3ab^2)\plus{}(\minus{}2a^3\plus{}2ab^2))\equal{}\sum(\minus{}\frac 13 (a\minus{}b)^3\minus{}\frac 23 (2a\plus{}b)(a\minus{}b)^2)\equal{}\sum (\minus{}\frac{5a\plus{}b}{3}(a\minus{}b)^2).\]

Subtracting the right side from both sides, our original inequality is now equivalent to
\[ \sum(\frac{5a\plus{}b}{3}\minus{}\frac{c^2}{(\sqrt{a\plus{}c}\plus{}\sqrt{b\plus{}c})^2})(a\minus{}b)^2 \geq 0.\]

Now $ (\sqrt{a\plus{}c}\plus{}\sqrt{b\plus{}c}) ^2> (2\sqrt{c})^2>3c,$ so it would be sufficient to prove that

\[ \sum((\frac{5a\plus{}b}{3}\minus{}\frac{c}{3})(a\minus{}b)^2) \geq 0.\]

Now we use the Squaring Analysis Theorem that has been discussed often on these pages.
Letting $ S_c\equal{}\frac{5a\plus{}b\minus{}c}{3}$ , with similar definitions for $ S_a$ and $ S_b,$ we must show that $ \sum S_c(a\minus{}b)^2 \geq 0.$  There are two cases to consider: $ a \geq b \geq c$ and $ a \geq c \geq b.$  We prove the first case—the second is similar.
We clearly have $ S_b\equal{}\frac{a\minus{}b\plus{}5c}{3} >0.$  Also, $ S_b\plus{}S_a\equal{}\frac{4b\plus{}6c}{3} >0$ and $ S_b\plus{}S_c\equal{}\frac{6a\plus{}4c}{3} >0.$  By the Squaring Analysis Theorem ,  our inequality is established.

There probably is a much easier way that I can’t think of!

[quote="dabug"]can_hang2007, your solution is great!  I thought I had made up this inequality the other day, but I guess not.

Here are the details of my solution.  All the summations are cyclic.

First some preliminaries we’ll need later:
\[ \sum (a \minus{} b)^3 \equal{} \sum (a^3 \minus{} 3a^2b \plus{} 3ab^2 \minus{} b^3) \equal{} \sum ( \minus{} 3a^2b \plus{} 3ab^2),
\]
and
\[ \sum(2a \plus{} b)(a \minus{} b)^2 \equal{} \sum(2a^3 \minus{} 3a^2b \plus{} b^3) \equal{} \sum(3a^3 \minus{} 3a^2b).
\]
Now multiply the original inequality by 2 and add $ \sum( \minus{} a^3 \minus{} a^2b \minus{} a^3 \minus{} a^2c)$ to both sides to get the equivalent form
\[ \sum (a^2( \minus{} (a \plus{} b) \plus{} 2\sqrt {a \plus{} b}\sqrt {a \plus{} c} \minus{} (a \plus{} c)) )\geq \sum(4a^2b \minus{} 2a^3 \minus{} a^2b \minus{} a^2c).
\]
The left side is $ \sum( \minus{} a^2(\sqrt {a \plus{} b} \minus{} \sqrt {a \plus{} c})^2),$ and multiplying by $ (\sqrt {a \plus{} b} \plus{} \sqrt {a \plus{} c})^2)$ over itself we rewrite the left side as
\[ \sum( \minus{} a^2 \frac {(b \minus{} c)^2}{(\sqrt {a \plus{} b} \plus{} \sqrt {a \plus{} c})^2}) \equal{} \sum( \minus{} c^2\frac {(a \minus{} b)^2}{(\sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} c})^2}).
\]
The right side is
\[ \sum(3a^2b \minus{} 2a^3 \minus{} a^2c) \equal{} \sum((3a^2b \minus{} 3ab^2) \plus{} ( \minus{} 2a^3 \plus{} 2ab^2)) \equal{} \sum( \minus{} \frac 13 (a \minus{} b)^3 \minus{} \frac 23 (2a \plus{} b)(a \minus{} b)^2) \equal{} \sum ( \minus{} \frac {5a \plus{} b}{3}(a \minus{} b)^2).
\]
Subtracting the right side from both sides, our original inequality is now equivalent to
\[ \sum(\frac {5a \plus{} b}{3} \minus{} \frac {c^2}{(\sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} c})^2})(a \minus{} b)^2 \geq 0.
\]
Now $ (\sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} c}) ^2 > (2\sqrt {c})^2 > 3c,$ so it would be sufficient to prove that
\[ \sum((\frac {5a \plus{} b}{3} \minus{} \frac {c}{3})(a \minus{} b)^2) \geq 0.
\]
Now we use the Squaring Analysis Theorem that has been discussed often on these pages.
Letting $ S_c \equal{} \frac {5a \plus{} b \minus{} c}{3}$ , with similar definitions for $ S_a$ and $ S_b,$ we must show that $ \sum S_c(a \minus{} b)^2 \geq 0.$  There are two cases to consider: $ a \geq b \geq c$ and $ a \geq c \geq b.$  We prove the first case—the second is similar.
We clearly have $ S_b \equal{} \frac {a \minus{} b \plus{} 5c}{3} > 0.$  Also, $ S_b \plus{} S_a \equal{} \frac {4b \plus{} 6c}{3} > 0$ and $ S_b \plus{} S_c \equal{} \frac {6a \plus{} 4c}{3} > 0.$  By the Squaring Analysis Theorem ,  our inequality is established.

There probably is a much easier way that I can’t think of![/quote]

I think this solution is not correct as

\[ \sum(3a^2b \minus{} 2a^3 \minus{} a^2c) \equal{} \sum((3a^2b \minus{} 3ab^2) \plus{} ( \minus{} 2a^3 \plus{} 2ab^2)) \equal{} \sum( \minus{} \frac 13 (a \minus{} b)^3 \minus{} \frac 23 (2a \plus{} b)(a \minus{} b)^2) \equal{} \sum ( \minus{} \frac {5a \plus{} b}{3}(a \minus{} b)^2).
\]

should be
\[ \sum(3a^2b \minus{} 2a^3 \minus{} a^2c) \equal{} \sum((3a^2b \minus{} 3ab^2) \plus{} ( \minus{} 2a^3 \plus{} 2ab^2)) \equal{} \sum( \minus{} (a \minus{} b)^3 \minus{} \frac 23 (2a \plus{} b)(a \minus{} b)^2) \equal{} \sum ( \minus{} \frac {7a \minus{} b}{3}(a \minus{} b)^2).
\]
Am I right? :wink:


geometry
3D geometry
tetrahedron
prism
pyramid
analytic geometry
function

What'd you get?

150. (30, 0, 0) (Right, wrong, blank)

Oh snap, nice job.

I got 129 (21 right, 2 skipped, 2 wrong :oops:).

<edit> It seemed harder than last year... (i got 119.5 last year)

Congratulations.

I wish I could get 30 questions right on the AMC 12.  I can't even do 25.

I thought there were only 25 questions on the AMC...

Are we allowed to discuss specific questions now?

I got a 132!!! :o Hurrah! 22 right, 3 wrong (one/two stupid mistakes, one guess).

I studied 2006AMC12A#25 the morning before and lo and behold I opened the '07AMC12A booklet and #25 is similar to the previous #25! :) So the first one I solved was #25 :D.

Eh Saiem, you might have gotten a perfect, but you can't count.

129!

Got 16 and 24 wrong, and 17 and 19 omitted.

So pretty good, too bad I didn't see how to do 17 and got 24 wrong and guessed on 16.

Hey, 25 was the first one I solved as well  :D

Other people at my school didn't do too well, one kid thinks he got a 130.5, another got like a 118.

Hey, what was #17 again? I don't have the problems anymore (strange circumstances :roll:) and the only record of it on my scratch paper is "17. $\boxed{B}$" so I think I solved it in my head or something...

What did you guys get for the question about cutting tetrahedrons from the corners of a square?

I think it was around 20 or something...

Oh, you just go like $s+\frac{s}{\sqrt{2}}+\frac{s}{\sqrt{2}}= 1\implies s=\sqrt{2}-1$ and the answer is $8\cdot \frac{1}{3}\left(\frac{s}{\sqrt{2}}\right)^{3}$ which is B. I accidentally did $\frac{1}{3}\left(s\right)^{3}$ instead, which is A :oops:

Why $\left(\frac{s}{\sqrt{2}}\right)^{3}$?
Don't you have to find the heights of the prisms?
I got something with a $10$ and a $7$ and maybe a $\sqrt{2}$?

The answer with 10, 7, 3, and $\sqrt{2}$ was correct.

woot. then maybe i got it right.  :)
I was thinking I calculated the heights wrong.  :(

What prisms? They're pyramids. The area of a pyramid is $\frac{1}{3}\cdot B\cdot h$ where $B$ is the area of the base and $h$ is the height. $s=\sqrt{2}-1$ is the length of a side of the octagon. $\frac{s}{\sqrt{2}}$ is the length of a side of the base of the pyramid. The area of the base is $\frac{1}{2}\cdot \left(\frac{s}{\sqrt{2}}\right)^{2}$ and the height is $\frac{s}{\sqrt{2}}$ so the volume of each pyramid is $\frac{1}{6}\cdot \left(\frac{s}{\sqrt{2}}\right)^{3}$ and there's 8 of them (i dropped the 1/2 in my last post)

I am very disappointed with my performance.  I was expecting close to 140 but instead got only 130.5. 21 correct, 3 blank.

I made a stupid mistake on #9!!!!

In addition, I solved 1-18 in 30 minutes, then spent 20 minutes solving #19.  :mad:  shoelace thm would have been so useful, but I didn't fully remember it.

123.  (20,3,2)  :(

I made a bunch of absolutely stupid mistakes.  I missed #7 and #21 for some inexplicable reason, probably because I was gunning through the easy ones to save time and I must have had a brain fart.  I had the correct expression for #24 on my scratch paper, but I mistyped it into my calculator and thus got the wrong answer.   I left 19 and 25 blank.  I didn't see the bijection on 25, which I guess is kind of acceptable, but I shouldn't have left 19 blank. I just saw ugly numbers and figured that it wouldn't be worth the time or 3.5 points to do the arithmetic, even though afterwards I saw the answer falls out nicely.

Hopefully, I'll do better on the B exam.

[quote="Danbert"][quote="stupidkid"]
Sure, but if you're going to spend time programming a calculator during the 75 minutes, then you're probably not going to do that well on the test.[/quote]
Not necessarily true!  I spent three minutes writing a program, it ran for less than a minute, and I worked on other problems while it was running.  I got the right answer for that problem.  Score: 144[/quote]
Which problem was that?

[quote="chess64"]You could have written it on a TI-83 too.[/quote]

What would the program be then?

I can't find sumdig on my TI-83.

For a 'sumdigit' program, you could do something like:

[code]
0 store s
for(a, 1, int(log(n))+1)
n/10 store b
iPart(b) store n
fPart(b)*10+s store s
end
disp s
[/code]

Ironically, I don't own any sort of graphing calculator.  I've never found the need for one; I just use the ones in class for programming "interesting" things.

[quote="stupidkid"]
Sure, but if you're going to spend time programming a calculator during the 75 minutes, then you're probably not going to do that well on the test.
[/quote]

Actually, it could be the case that some really smart people are slow workers, and writing a program could actually be advantageous.  When it comes to the AIME then, the effects of slow-working are less noticeable.

[quote="1234567890"]I managed to find another school to take the AMC 12 at after a "cold day".

Unfortunately, I did really poorly: 19 right, 3 wrong, 3 blank, for 118.5. All of my mistakes were really stupid.  :(

Too depressed to talk about them now......[/quote]

This isn't my score, actually. Apparently, I misbubbled on a question and got a 112.5.   :maybe:

Oh, well. I'll just do better the B test.  :)

how do u know that u misbubbled a question?

Our proctor hand-graded our booklets for us (I circled the answers on the booklet, and I then transferred my circles onto the scantron). So I circled the wrong answer for one of them.  :(

[quote="1=2"][quote="Potato Theory"]For a 'sumdigit' program, you could do something like:

[code]
0 store s
for(a, 1, int(log(n))+1)
n/10 store b
iPart(b) store n
fPart(b)*10+s store s
end
disp s
[/code][/quote]

I tried it, but it gave a non-real answer. Did you define the variables?????[/quote]
Just stick an "Input n" at the beginning of it
[code]
input n
0 store s
while n>0
10fPart(n/10)+s store s
iPart(n/10) store n
end
disp s
[/code]

[quote="stupidkid"]
What do you mean? Everything that I did on the 89 could've been done on the 83. Mind giving an example of how the 89 gives an unfair advantage?[/quote]

CAS

[quote="JSteinhardt"]Okay, so Shoelace theorem is just equivalent to cross products. That doesn't seem particularly nice though, you have to break it into four cases and add up the $x$-coordinates. I was looking for a solution by symmetry.

EDIT: Speaking of which, are we going to start all the solution threads for stuff?[/quote]

[b]cross[/b] products?
don't you mean dot products? (don't want to yield a whole new vector)
(Consider $Y=\begin{bmatrix}y_{1}\\ y_{2}\\ \\ \\ \cdots \\ \\ y_{n}\end{bmatrix},X=\begin{bmatrix}x_{1}\\ x_{2}\\ \\ \\ \cdots \\ \\ x_{n}\end{bmatrix}$ with modular indices then $A=\left|\frac{Y_{n+1\to n}\cdot X-X_{n+1 \to n}\cdot Y}{2}\right|$

That is equivalent to half the magnitude of the sum of the cross products of every two adjacent vertices with respect to an arbitrary point on the interior of the polygon.

Basically, but when you do it just by lining the x and y coordinates up in columns, it makes it a ton faster to find the area of some irregular polygons...

[quote="mathgeniuse^ln(x)"]I was wondering, but where is store on a TI-84.[/quote]
Above the [ON] button.


trigonometry
inequalities
inequalities unsolved

Let $ABCD$ be an inscribable quadrilateral of sides $a,b,c,d$. Then

$ \frac {8\sqrt{abcd}}{(a+b)^2+(c+d)^2} \leq \frac{\sin \hat A}{\sin \hat B} \leq \frac{(a+d)^2+(b+c)^2}{8\sqrt{abcd}} $

I'll just solve the first inequality as the second is pretty much the same:

It is easy to find that: $\displaystyle
\frac{\sin B}{\sin A}=\frac{ad+bc}{ab+cd}.$ We have that $\displaystyle \frac{1}{ab+cd} \geq \frac{4}{(a+b)^{2}+(c+d)^{2}}$ and $ad+bc \geq
2\sqrt{abcd}$. We multiply the last two relations which gives: $\displaystyle
\frac{ad+bc}{ab+cd} \geq \frac{8\sqrt{abcd}}{(a+b)^{2}+(c+d)^{2}}.$


I'm working on a modeling problem and stumbled on this equation (with $x$ being a variable and everything else are constants):

$x+ax^r+b = 0$

Does anyone know how to solve this equation?

Thanks a bunch,.

I am sure there is no general solution in terms of the constants. For some few $r$ there would be an explicit solution. You can of course compute it numerically if you know the constants.

PS I just checked in Maple because I doubted myself (as no good mathematician should in this kind of situation). Maple agrees with me.

[quote="Xevarion"]I am sure there is no general solution in terms of the constants. For some few $r$ there would be an explicit solution. You can of course compute it numerically if you know the constants. [/quote]

I just talked to my mentor and he agrees too. It seems like I hit a brick wall already :|. Thanks though.

The equation is a so-called trinomial equation, and can be solved using hypergeometric series.

See http://arxiv.org/PS_cache/math-ph/pdf/0303/0303016.pdf


function
algebra
polynomial
algebra proposed

Let be $ f: \mathbb{R}\rightarrow \mathbb{R}\ ,\ f(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d\ ,\ (a,b,c,d\in \mathbb{R})$ such that $ f(2)\plus{}f(5)<7<f(3)\plus{}f(4)$ . Prove that $ (\exists)u,v\in \mathbb{R}$ such that $ u\plus{}v\equal{}7$ and $ f(u)\plus{}f(v)\equal{}7$ .

[hide]
Consider the function $ P(x) \equal{} f(x) \plus{} f(7\minus{}x)$.

Note that $ P(x)$ is a polynomial (and thus continuous), and that $ P(2) < 7 < P(3)$.

Thus, for some $ u \in (2,3)$ we have $ 7 \equal{} P(u) \equal{} f(u) \plus{} f(7\minus{}u) \equal{} f(u) \plus{} f(v)$ as desired.
[/hide]


If a side of a regular pentagon is n+2, then its perimeter, in terms of n, would be

[quote="236factorial"]If a side of a pentagon is n+2, then its perimeter, in terms of n, would be[/quote]

Assuming it's a regular pentagon...

[hide]The perimeter would be $5(n+2)=5n+10$[/hide]


If the inradius of an equilateral triangle is 5, find the triangle's height.

Nice simple geometry problem. Happy Thursday!  :)

[hide]Since in the equilateral triangle the bisectrix, the median and height are coincidents, 5 is 1/3 of the height. Then triangle's height = 15[/hide]


geometry
3D geometry
sphere
calculus
calculus computations

I need to know the area of the intersection between a sperical shell and a plane in spherical coordinates. By "shell" I mean a sphere with some differential thickness dR. Basically, I know that the intersection of a sphere and a plane is a circle. But I want to consider this sphere having some differential thickness and determine the subsequent differential area of intersection with the plane.

Can someone please help me out with formulating this problem? Thanks!

I'm not quite sure how to explain it yet, but you should come up with the nice result that the area is $2\pi R\,dR$ independent of where on the sphere it intersects the plane; the tilt of the sphere away from the equator balances the smaller cross-section exactly.


induction
combinatorics proposed
combinatorics

This one is quite trivial. One just substitutes the expression of $F_{i}$ in terms of powers of Golden section and its conjungate using Binet's identity and applies Newton's binom expansion formula twice (for both sums of the powers of the Golden section and its conjungate). It is straightforward to check that the resulting number is indeed $F_{2n}$ - apply Binet's identity once more.

Well I didn't really use Lucas numbers and Fibonacci sequence to obtained....

As $F_{2n}= F_{2n-1}+F_{2n-2}$ I wondered what was $F_{2n-1}$ and I combined them...

So for $F_{2n}$ and $F_{2n+1}$ I used induction over n...

And if we have that for $every$ $k \leq  n$ is true then $F_{2n}$ can be proved...

I think you can understand me....


In a PDF on inequalities there is an exercise by Darij to show $a^{3}+b^{3}+c^{3}+3abc-2b^{2}a-2c^{2}b-2a^{2}c \geq 0$ where a,b,c are sides of a triangle. I think I have solved it, but the proof seemed way too simple and the inequality really blunt. Can you spot anything wrong with the below.

Rearrangement gives $\sum_{cyclic}ac(b-a)+a^{3}\geq 0$, and applying Ravi's substitution $\sum_{cyclic}(y+z)(x+y)(x-y)+(y+z)^{3}= \sum_{cyclic}(y+z)[(x+y)(x-y)+(y+z)^{2}]=\sum_{cyclic}(y+z)[x^{2}+z^{2}+2yz] \geq 0$.
That it is positive is trivial given x,y,z>0.

Does this proof work? Sorry if I'm being stupid one way or another.

Take substituitons $a=x+y,b=y+z,c=z+x$ and if you make calculations right you must reach $\sum_{cyc}x^{2}y\geq 3xyz$.  :)

[quote="Ilthigore"]
Rearrangement gives $\sum_{cyclic}ac(b-a)+a^{3}\geq 0$[/quote]

Isn't it $\sum_{cyclic}ac(b-\boxed{2}a)+a^{3}$?


Can anybody post some Masters Round problems from the past so everyone can solve them just for fun?

they used to have some in  the previous competitions...

i think they are only pubished for national questions... :play_ball:

[quote="mathgeniuse^ln(x)"]I don't think MathCounts is giving anymore in their competition books, so yeah.

But I will try.[/quote]

Mathcounts put one problem from 1991 or something in their All-Time Best Mathcounts Problems book. It was about putting lightbulbs in a circle and removing every other one--kinda like the problem where King Arthur goes crazy and kills off all his knights except one. And it asks you to find which lightbulb will "survive" for a given number of lightbulbs.

i think u mean this:

[img]http://www.artofproblemsolving.com/Classes/MATHCOUNTS/Images/lightbulbs.gif[/img]

If bulb number 1 is removed, bulb number 2 is skipped, bulb number 3 is removed, bulb number 4 is skipped, bulb number 5 is removed, bulb number 6 is skipped, bulb number 2 is removed, bulb number 4 is skipped, bulb number 6 is removed, then clearly bulb number 4 is the last to be removed. It is called "the survivor."

1. Using this progression, determine "the survivor" if the number of bulbs initially is:

(a) 4

(b) 5

(c) 7

(d) 8

(e) 9

2. Based on your observations,

(a) predict "the survivor" if there are 22 bulbs.

(b) explain how you arrived at the solution of "the survivor" for 22 bulbs.

3. Explain how you would predict "the survivor" for n bulbs.

4. Suppose "the survivor" is bulb number 4. What can you say about the number of bulbs initially?

[hide="3"] Define $f(x)$ to be the "survivor" of $x$ lightbulbs. For $k, n \in\mathbb{Z^{+}}$ and $n<2^{k}$, $f(2^{k})=2^{k}$ and $f(2^{k}+n)=2n$.

Not sure how to write an explicit formula.  :maybe: [/hide]

[quote="anirudh"]there is one in the All Time Greatest MATHCOUNTS Problems[/quote]

the one i posted is from teh all time greatest

[quote="anirudh"]there is one in the All Time Greatest MATHCOUNTS Problems[/quote]
We know. About 3 posts before you, mathnerd314 told us about the lightbulb problem.  :wink:


Hypotenuse of the 9-12-x right triangle is 15 (scaled 3-4-5) so AB is the hypotenuse of a 15-20-x triangle, which is 25 (again, scaled 3-4-5).


prove by contradiction that if X is a finite set and f : X->Xis onto,then f is one-one.

[quote="madhuri"]prove by contradiction that if X is a finite set and f : X->X is onto, then f is one-one.[/quote]

Have you considered instead declaring yourself to be a Constructivist and therefore forever eschewing the use of Proof by Contradiction? ;)

The below is quoted from a National University of Singapore web site, for a class [url=http://www.thecore.nus.edu.sg/spt/ccsp03/tutorial2.html]Mathematical Ideas[/url] taught by Peter Pang. [quote] The Constructivists are sort of half way in between. They believe that the natural numbers are intuitive (god-given) and the rest is to be constructed (by man) from the natural numbers. While there are doubts as to whether the natural numbers are really intuitive, the most serious blow to the constructivists is that their philosophy forces upon them a more rigorous logic than that practiced by most mathematicians. More specifically, as they insist upon constructivistic methods, the constructivists cannot accept the law of excluded middle. As a result, they are not able to accept proof by contradiction, which is one of the most powerful tools for non-constructivist mathematicians.[/quote]

Here at this forum, we like to celebrate the elegance of mathematical proof, and proofs by contradiction are usually the least elegant of proofs. Proofs by contradiction tend to be short but boring and unenlightening.

For a proof by contradiction, just assume the opposite of what you want to prove, that is: "Assume that the function f is not one-to-one," and start deriving new properties until you derive something that contradicts an earlier statement, such as X has no fixed size or f is not onto. Then say, "Therefore, by contradiction, f is one-to-one."

Erin Schram
Mathematician


calculus
integration
number theory unsolved
number theory

Q. Find all integral solutions of the equation:--

      $1 + 3 + .... + 3^{m} = n^2$

where $m \geq 0$.

Your equation$\leftrightarrow 3^{m+1}=2n^2+1$
Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=411125&highlight=#411125]here[/url]
A nice solution of [b]nthd[/b]


email
AMC
AMC 12
USA(J)MO
USAMO
AIME
MIT

I applied...the only thing i have going for me is that i'm very well rounded (excel at other nontypical math things) that I will have taken abstract algebra by RSI and that I have a rec from a brown professor. Also my essay is written very well. My SAT is good. Do I have a shot?
What about you guys?
When do we find out?

I find it weird nobody has replied to this...I'm sure many people on the forum are applying.

a chance thread? go to CC -______-

but idk you sound ok; just wait?

haha yea thats pretty much what I'll have to do [wait that is]
U went right?

I find it really funny how much of an overlap AOPS has with CC. You say CC anywhere else on the internet and nobody will know what you're talking about.

Good luck! What are you planning to research?

do they notify you buy email earlier than March 26, or only by mail?

I'd really like to do RSI next year... I'm a sophomore, and I've got pretty good SAT scores, and I'm also pretty well-rounded... My brother did TASP when he was in high school, but I am trying to spend my summers in math and science.

@mathemonster
where did you get that date? (march 26th)

i think the website.
EDIT: buzzer how'd u do on the AMC12?

yea i got a rec from a tenured ivy leauge prof. which is im sure why im an alternate at all

sooo...did you guys get in in the end?:)

[quote="timwu"]By the way, will RSI rejection affect your future application to MIT? (I hope not...)[/quote]
I doubt it. (This was not the case for me, at the very least.)

No, so far we haven't got in..I don't have much hope. But I am doing math research anyways so idrc

Doing research with a prof ^.^ That is so nice, Bijection!

By the way...Does any of you know if Romanian students can participate? i sent an email to the people at rsi the other day asking this, and i got this answer whose meaning i can't honestly comprehend :)

"We don't have any sponsors for Romanian students".

Can someone please help me with this?:)
*i would like to apply for RSI 2011*

also..what is the ideal preparation for RSI?

RSI does not accept applications directly from non-US students.  They only accept students from countries that have a sponsoring governmental or other organization, and you then would apply to that organization rather than to RSI directly.

However, based on the response that you got, it looks like Romania does not have such a sponsoring organization for 2011, so they will not be taking any Romanian students.

Thanks!
That makes sense now :)

Just one more thing: if someone is studying in some other country than the one he is a citizen of, can he apply to the RSI sponsors in the country he is studying in? :)

@Kewi
yea research with a prof..good luck getting in for 2011

@ Bijection

What are you going to research? :)

combinatorics i hope..i doubt ill get a result though

[quote="kewi"]Do you guys think that if I go talk to some sort of company and ask them to be a sponsors for romanian students willing to attend rsi and they will accept, romanian students will be eligible for rsi 2011?
I mean..anyone wanting to be a sponsor is welcome to be one, isn't he?

:D[/quote]

do try! good luck!