There are 30 men and 40 women in the Town Library Club.  They wish to form a 7-person steering committee with 3 men and 4 women.  In how many ways can they form the committee?

[quote=forester2015][hide=Post #4]
 [url=aops.com/community/user/221110][b]TLiu[/b][/url] · 4 minutes ago [url=aops.com/community/p20001591](view)[/url]

 Hello, I think this problem is a bit too bashy for Alcumus and should be removed as it requires the calculation of a 9 digit number!

 [rule]
[color=#5b7083][aops]x[/aops] 4[color=transparent]0000[/color] [aops]Y[/aops] 0 [/hide]

Yes but we can use calculators in Alcumus.[/quote]
[quote = LauraZed]The general consensus in this thread is correct – in general, we wouldn't advise using a calculator while solving Alcumus problems. However, if a problem distinctly says you may use one, like the problem JustKeepRunning quoted above, you can definitely use a calculator for the specified step(s).

[/quote]

Yes, but you can use it. Tho it's not encouraged.

Oh yeah I can just use my dottedcaculator.


articles LaTeX integration trigonometry calculus geometry

The latest issue of [url=http://tug.org/pracjourn/]The PracTeX Journal[/url] has a couple of articles that may be of interest to AoPS users:[list][*][url=http://tug.org/pracjourn/2006-4/henderson]A Beginner's Guide to METAPOST for Creating High-Quality Graphics[/url]
[*][url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url][/list]

Thanks for the links to the articles (and for all your wonderful advice on LaTeX).  From now on, I will replace eqnarray with align whenever I write a paper.

Is it possible to use the align environment on this forum?  My attempts to gave me an "Unparseable or potentially dangerous latex formula".

There seems to be a bug with the align environment on this forum. I'll ask Valentin if he would look into it.

Until the align environment bug is fixed, a workaround is to add \par before \begin{align} or \begin{align*} to get:
$\par\begin{align}f(x) &= \int\sin x\: \text{d}x\\ g(x) &= \int \frac{1}{x}\: \text{d}x \end{align}$
or
$\par\begin{align*}f(x) &= \int\sin x\: \text{d}x\\ g(x) &= \int \frac{1}{x}\: \text{d}x \end{align*}$

Why does [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php]the "Math in $\LaTeX$" tutorial[/url] still list eqnarray rather than align? :huh:

I don't really see what's wrong with eqnarray. I mean it does what you want it to do!

Did you read the article that the first post links to?  In eqnarray, the spacing is off, and the equation numbers are buggy.

As [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=eqnarray]Why not use eqnarray?[/url] says "[i]The environment eqnarray is attractive for the occasional user of mathematics in LaTeX documents[/i]" but it then goes on to say "[i]it makes a serious mess of spacing[/i]" as is shown in 2.1 of the Avoid eqnarray! article. That article shows that there are other problems too.

So why not use the grown-up method viz align which always works and which is "[i]which is designed with the needs of mathematicians in mind (as opposed to the convenience of LaTeX programmers)[/i]"?

I skimmed through it :oops:  But I don't need to number my equations. (This is my first time using eqnarray you see, so I suppose it's best that I stop right there!)

I have a question for align though. I'm confused with the {} symbols that were used. What do they do and are they really necessary?

EDIT: Can someone please tell me what's wrong with the following:

\begin{align*}
v\dfrac{dv}{dx}\underline{k} &= -g\underline{k} -\gamma v^2 \underline{k}
\end{align*}

It doesn't appear on the pdf file! :(  Thanks.

[quote="beans"]I have a question for align though. I'm confused with the {} symbols that were used. What do they do and are they really necessary?[/quote]You can see the difference here:
With {}
\begin{align*}x ={}& x \\ &+x \\ ={}& x \end{align*}
without {}
\begin{align*}x = & x \\ &+x \\ = & x \end{align*}
[quote="beans"]
EDIT: Can someone please tell me what's wrong with the following:

\begin{align*}
v\dfrac{dv}{dx}\underline{k} &= -g\underline{k} -\gamma v^2 \underline{k}
\end{align*}

It doesn't appear on the pdf file! :(  Thanks.[/quote]Have you looked at the error messages? Click on the ->! on the toolbar on the top right (keep pressing for more irrelevant messages) and it will tell you about a paragraph ending before \align* is complete which is LaTeX-speak for 'remove the blank line just before \end{align}'

Best go and stick some {} into my document then!

[quote="stevem"] Have you looked at the error messages? Click on the ->! on the toolbar on the top right (keep pressing for more irrelevant messages) and it will tell you about a paragraph ending before \align* is complete which is LaTeX-speak for 'remove the blank line just before \end{align}'[/quote]

Thanks a lot! You see I have loads of error messages, warnings and bad boxes, so I just ignore them :oops:  (they don't make much sense you see) but thanks a lot :D

[quote="beans"]Best go and stick some {} into my document then![/quote]You won't need to do that unless you have something like that second line. Normally you get
\begin{align*}x = x \\ = x \end{align*}
which doesn't need {} and is fine.

[quote="beans"]Thanks a lot! You see I have loads of error messages, warnings and bad boxes, so I just ignore them :oops:  (they don't make much sense you see) but thanks a lot :D[/quote]Trying to get rid of those messages is the way to learn more about LaTeX, though I admit it takes time and I know you are short of that.

[quote="stevem"]You won't need to do that unless you have something like that second line. Normally you get
\begin{align*}x = x \\ = x \end{align*}
which doesn't need {} and is fine.[/quote] Oh I see now. I was also(!) wondering about the spacing between the lines. You see I've got quite a few integral signs about and the document doesn't look neat because it seems that the line above is sitting on the line below! Is there anything that can be done about this?

[quote="stevem"]Trying to get rid of those messages is the way to learn more about LaTeX, though I admit it takes time and I know you are short of that.[/quote] You got the time bit right. I did try to work through the errors on my first coursework, that's why it's still not complete! (and the fact that I need matlab :D ). I just want this done for tomorrow, and maybe later will try to get rid of the error messages. :)

[quote="stevem"][quote="beans"]I was also(!) wondering about the spacing between the lines. You see I've got quite a few integral signs about and the document doesn't look neat because it seems that the line above is sitting on the line below! Is there anything that can be done about this?[/quote]That shouldn't happen. Could you post a minimal document showing that problem? By that I mean everything including the preamble (ie before \begin{document}) but just including a few lines where the integrals look wrong.

Oh and if you include it on the forum inside [code ] ... [/code ] it won't misinterpret the dollar signs.[/quote]Well I might have slightly exageratted the thing- you see I like a lot of space it seems! :oops:

It's readable, but anyway heres what you asked for:
[code] \documentclass[titlepage, a4paper]{article}
\oddsidemargin=0in
\usepackage{fancyhdr}
\usepackage{amssymb}
\usepackage{upquote}
\usepackage{amsmath}
\usepackage[top=3.5cm,bottom=2.7cm,text={6.5in,9in},centering]{geometry}
\usepackage[pdftex]{graphicx}
\pagestyle{fancy}
\rhead{}   % My name
\lhead{Coursework}
\cfoot{\thepage}
\parindent=0in
\title{{\Huge } \\ {\Huge Coursework}}
\author{{\Large }} % Author
\date{}
\begin{document}

\begin{align*}
-\displaystyle\int \dfrac{v}{g+ \gamma v^2} \:dv \:\underline{k} \quad  = & \quad \displaystyle\int \; dx :\underline{k}\\
\frac{-1}{2\gamma }\ln(g+\gamma v^2) \underline{k} \quad = & \quad x\underline{k} + C
\end{align*}\\ [/code]

It's erm.. messy! The -1 seems to hang about the integral, but is that just me being fussy? :-0

[quote="stevem"]You can change the vertical spacing by replacing \\ between the 2 equations with \\*[2ex] which I agree looks better (2ex means twice the height of an x in the current font, of course you can change it to give as much space as you like).

By the way, you don't need \displaystyle here as the align environment does that for you automatically.[/quote] Thanks once again Steve. :)  It worked a treat- looks pretty darn neat! And you saved me from a lot of typing as well. (I've been going through the errors, and it seems I have lots of extra { flying about, although I don't understand what bad boxes are!)

[quote="beans"]although I don't understand what bad boxes are!)[/quote]They are the least important things to worry about which is why TeXniCenter relegates them to below Errors and Warnings.

In the example you gave you had
\end{align*}\\
The \\ isn't needed on the end and gives "underfull \hbox (badness 1000)". That means that LaTeX couldn't justify the line left and right properly, which isn't surprising because there isn't anything there. Badness is an estimate of how much stretching LaTeX had to do to fill the line, so the lower the better. Remove the \\ and the warning goes.

Sometimes LaTeX can't break a line in the right place eg due to a long formula or poor hyphenation so it will give a overfull \hbox warning. This will be much more obvious as you'll see text go into the margin - though if it's only by a tiny amount you might ignore it rather than having to make things look awkward.

Thus these bad boxes* are to alert you to possible problems, but if the page looks right then you can ignore them.

*LaTeX puts text into boxes behind the scenes so it can get the spacing right. Justification, for example, means tweaking things slightly with a bit of spacing. The methods it uses are complicated which is why it produces much better results than anything a word processor can achieve.


probability
calculus
integration
inequalities
function
limit
real analysis

In the evaluation of P(X > Y) over some given probability space, can one assume that X and Y takes on finitely many values over a \Omega?  I think this is probably a stupid question.  However, I was speaking to my friend regarding this question and in the context of an integration problem, and he showed me to the following website:

http://www.artofproblemsolving.com/Forum/topic-35665.html

Here fedja shows this to be true in the context of a problem for the minimum of measures for three different random variables. However, i cannot understand how one can simply say that a random variable takes on finitely many values WLOG in the evaluation of P(X > Y) simply because it is the limit of an infinite sequence of simple Borel measurable functions.  I believe that I am overlooking something simple.  Please clarify if possible, I would much appreciate it.

I should let fedja answer this himself, but off the top of my head ...

We're trying to prove a sharp inequality for random variables. Suppose there is a set of random variables which violates this inequality. Then those random variables (that is measurable functions) can be arbitrarily closely approximated in some reasonable sense (in measure? in distribution?) by random variables that take on finitely many values (aka simple functions). Then the approximating random variables would also violate the inequality.

One would have to supply details for that to make any sense, but it must be something like that.

Yes, this is all about approximation, as Kent said. Let $X$ and $Y$ be two arbitrary real valued random variables. Let $\widetilde X=\frac kn$ if $\frac kn\leqslant X<\frac {k+1}n$ ($-n^2\leqslant k \leqslant n^2$), $\widetilde X=-n$ if $X<-n$, $\widetilde X=n$ if $X\geqslant n$ and similarly for $Y$. It is easy to see that $\widetilde X>\widetilde Y$ implies $X>Y$. On the other hand, $X>Y+\frac 1n$, $|X|,|Y|<n$ imply $\widetilde X>\widetilde Y$. It remains to note that $\lim_{n\to\infty}P\{X>Y+\frac 1n, |X|,|Y|<n\}=P\{X>Y\}$ (the events on the left form an increasing sequence whose union is the event on the right).


calculus integration trigonometry logarithms function limit algebra

I asked this question on a quiz yesterday in a second-semester calculus class. The results were, shall we say, not encouraging.

So what is it? A totally evil question, or should they have gotten it?

Compute $\int_{0}^{\infty}\frac{x}{x^{3}+x^{2}+x+1}\,dx$

It's not that evil i think but i can't judge them  :D

I found $\frac{ \pi }{4}$ .

[quote="Ramanujan"]We can also find an antiderivative or use the substitution $x= \tan \theta$ .[/quote]
You are right, Ramanujan! Now, I think the question isn't evil at all. I take back my words. They should have got it! ;)

We make the substitution $x = \tan t$ to get

$I = \int_{0}^{\frac{\pi}{2}}\frac{\tan t \cdot \sec^{2}t}{(\tan t+1)\sec^{2}t}\,dt = \int_{0}^{\frac{\pi}{2}}\frac{\tan t }{(\tan t+1)}\,dt = \int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\sin t+\cos t}\,dt$

$\Rightarrow I = \int_{0}^{\frac{\pi}{2}}\frac{\cos t}{\sin t+\cos t}\,dt$

$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}\frac{\sin t+\cos t}{\sin t+\cos t}\,dt = \frac{\pi}{2}$

$\Rightarrow \boxed{I = \frac{\pi}{4}}.$

The methods presented in this thread are very nice, but you can do the integral via the usual fundamental theorem of calculus way by evaluating \[\int^{\infty}_{0}\frac{xdx}{x^{3}+x^{2}+x+1}=\int^{\infty}_{0}\left(-\frac{1}{2}\cdot\frac{1}{x+1}+\frac{1}{2}\cdot\frac{x+1}{x^{2}+1}\right)dx\] The problem is that you get into a $\infty-\infty$ situation if you immediately try to evaluate at infinity.  However, the contribution from the \[-\frac{1}{2}\cdot\frac{1}{x+1}\] term cancels out the contribution from the \[\frac{1}{2}\cdot\frac{x}{x^{2}+1}\] term as $x\to\infty$ (this is just a straightforward calculus $1$ limit)
and you are left with \[\frac{1}{2}\int^\infty_{0}\frac{dx}{1+x^{2}}=\frac{1}{2}\cdot \frac{\pi}{2}=\frac{\pi}{4}\]

What I was expecting was essentially blahblahblah's straightforward solution, helped by an observation you can make in the beginning: the integrand behaves like $x^{-2}$ as $x\to\infty,$ hence we know that the integral converges. We'll bundle the logarithms together to make that limit happen.

$\left.\frac12\left(-\ln(x+1)+\frac12\ln(x^{2}+1)+\arctan x\right)\right|_{0}^{\infty}$

$=\left.\frac12\left(\ln\left(\frac{\sqrt{x^{2}+1}}{x+1}\right)+\arctan x\right)\right|_{0}^{\infty}$

The problem is that I left too many possible ways to get the problem wrong - not factoring right, doing the partial fractions wrong, not knowing what to do with the integral of $\frac{x+1}{x^{2}+1},$ mishandling the evaluation of the logarithms ...

blahblahblah, you method is indeed correct, but as you also mentioned, we need to be very careful while computing the limits, and that is precisely why I think the students might have got stumped on that question!

I"ll find as a rule, identically to Blahblah ... the function $F\in \int\frac{x}{(x+1)(x^{2}+1)}\ dx$ on the interval  $[0,\infty )$ for which $F(0)=0\ .$
Obtain $F(x)=\ldots\ldots=\frac{1}{2}\cdot\left[\frac{1}{2}\cdot\ln (x^{2}+1)+\arctan x-\ln (x+1)\right]$, i.e. $F(x)=\frac{1}{2}\cdot\left[\arctan x+\ln\frac{\sqrt{x^{2}+1}}{x+1}\right]\ .$
Therefore, $\int_{0}^{\infty}\frac{x}{(x+1)(x^{2}+1)}\ dx=\lim_{x\to\infty}F(x)=\frac{\pi}{4}\ .$ I didn't understand where is the difficulty to solve this problem by the your students.

I think that he already explained it... But what was the most common [i]mistake[/i]?


[quote="Magnara"]Given that $a,b\in{R}$, $a>0, b>0$, show that $\sqrt{2}$ lies between $\frac{a}{b}$ and $\frac{a+b}{a+2b}$.[/quote]

Let a = b = 1.

Then sqrt (2) is clearly NOT between 1 and 3/2.
Now, if we flip the other fraction, we get that (a+2b)/(a+b) = 1+b/(a+b).  So let's see if this works.

Assume a/b < sqrt (2), so a < bsqrt (2).

Then substituting this into the other fraction, we get:

1+b/(a+b) > 1+1/(sqrt (2)+1) = sqrt 2

Now, assume a/b > sqrt (2), so a > bsqrt 2.

Now we get 1+b/(a+b) < 1+1/(sqrt (2)+1) = sqrt 2

Now, assume a/b = sqrt(2), so a/b = 1+b/(a+b) = sqrt 2 and we can say that it is trivially between them.

By trichotomy, these are all possibilities, and hence the result follows.

[quote="Magnara"]Woops.  I put down the problem wrong.... Sorry :(

Fixed.  Though you still managed to see that and fix it[/quote]

I took a guess.  Interesting problem, though not especially difficult - the result is somewhat surprising, though.

[quote="Magnara"]Given that $a,b\in{R}$, $a>0, b>0$, show that $\sqrt{2}$ lies between $\frac{a}{b}$ and $\frac{a+2b}{a+b}$.[/quote]\
sqrt 2=1+1/(2+1/(2+1/(2+1...))).  By taking it 1 level, we get 1. Another level gives 3/2
it follows p+2q/p+q exactly. We always stop at a two.
lets try the first few
1+1/2>1+1/(2+1/(2+1/...))>1+1/(2+1/2)
subtract 1 from all sides
well the link between the first two are obvious
the link between the second two is not so obvious but the denominator in the 1/ part must be larger than the other since their numerator part is the same.  And 1/(2+1/(2+... is larger than 1/2.  Conclusively, we can do this for all cases

[quote="Soroban"]Hello, Magnara!

[quote]Given that $a,b\in{R}$, $a>0, b>0$,
show that $\sqrt{2}$ lies between $\frac{a}{b}$ and $\frac{a+2b}{a+b}$.[/quote]
Clearly, $\frac{a}{b} \neq \frac{a + 2b}{a + b}$

If $\frac{a}{b} = \frac{a + 2b}{a + b}$, then $\sqrt{2} = \frac{a}{b}$ which makes $\sqrt{2}$ a rational number.
[/quote]

This is not a problem because we have $a,b\in\mathbb{R}$, not $a,b\in\mathbb{Q}$.

[quote="blahblahblah"][quote="Magnara"]Interesting problem, though not especially difficult - the result is somewhat surprising, though.[/quote][/quote]

Exactly my thoughts.

[quote="nolachrymose"]What is $R$? The set of reals?[/quote]

Yeah.  R is reals, Q is rationals (those reals numbers that can be expressed as the ratio of two integers), Z is the integers.  Oh, and C is the complex numbers.

W is the set of Whole numbers {0,1,2,...} and N is the Natural Numbers {1, 2, 3, ...}

There's not a general consensus on whether $\mathbb{N}$ includes 0 or not.

Frankly, if confusion may arise, writing $\mathbb{Z}^{+}$ for positive integers and $\mathbb{Z}^{*}$ for nonnegative integers is often the best way.


modular arithmetic
number theory unsolved
number theory

Solve the natural equation:
$ 3^{x}\plus{}4 \equal{} 7^{y}$

The source problem is :

Solve the integer equation :

$ 3^{x}\plus{}4^{y}\equal{}7^{z}$.

And your problem can be solved by congruence. :P

I am very busy now so i can't post my solution . If no any body post they solution , i will post it later. :wink:

Oh! I have read the solution of the problem: $ 3^{x}\plus{}4^{y}\equal{} 7^{z}$. But it is not enough! All work we have to do is this problem

[hide]
Consider it modulo 8; then we have $ 4^{x}\plus{}3\equiv (\minus{}1)^{y}\pmod{8}$. For $ x\ge2$, we have $ 4^{x}\equiv 0\pmod{8}$, which means $ 3\equiv\pm 1\pmod{8}$, a contradiction. So, since $ x\equal{}1$ works, ($ x\equal{}y\equal{}1$), it is the only solution.
[/hide]

Oh sorry! I posted a mistake problem. I  repaired it below:
Solve the natural equation: $ 3^{x}\plus{}4\equal{}7^{y}$

You can find it on Number Theory forum, i have solved it before

[hide]Did you see this solution on THTT nktp :P [/hide]

this case is harder but stil solvable by congruence. like this:

It's easy to find the solution $ (1,1)$. now lets assume $ x,y>1$. now checking the equasion by $ mod8$ we get $ x\equal{}y\equal{}1\mod2$
Now by $ mod7$ we get that $ 3^{x}\equal{}3\mod7$ meaning $ x\equal{}1\mod6$. and by $ mod9$ we get $ 7^{y}\equal{}4\mod9$ meaning $ y\equal{}2\mod3$ and consequently $ y\equal{}5\mod6$ since $ y\equal{}1\mod2$

now by $ mod10$ and with knowledge from above we can see that $ 3^{x}\equal{}3\mod10$ and $ 7^{y}\equal{}7\mod10$ which means $ x\equal{}y\equal{}1\mod4$. knowing $ y\equal{}5\mod6$ and $ x\equal{}1\mod6$ we can conclude that $ y\equal{}5\mod12$ and $ x\equal{}1\mod12$

now $ 3^{12k\plus{}1}\equal{}3^{1}\equal{}3\mod13$ so $ 3^{x}\plus{}4\equal{}7\mod13$ but $ 7^{12l\plus{}5}\equal{}7^{5}\equal{}11\mod13$ which means there are no other solutions for the equasion other than $ (1,1)$


pigeonhole principle
geometry
rectangle
geometric transformation
reflection
group theory

Jerry is trying to fill in positive numbers 1,2,3,...,12 into a $ 3\times 4$ table. Two rules have to be followed:
1) For each two adjacent boxs, the number in the right one is larger than that in the left one.
2) For each two adjacent boxs, the number in the upper one is larger than that in the lower one.
An example has been shown in the attached figure. How many ways can Jerry fill in all the numbers?

[hide]
Is 36 correct?
1 needs to be placed on the bottom left.
But the next two that compose the next diagonal, 2 and 3 can be switched within the diagonal but not moved to any other square or a square with greater magnitude would have to be placed below or to the left of it by a pigeonhole argument.
So we can switch up the diagonals but not anything else.
We have 2 diagonals of 1, 2 diagonals of 2 , and 2 diagonals of 3.
$ 1^2 * 2^2 * 3^2 \equal{} 36$.
[/hide]

In fact one could put both $ 2$ and $ 3$ on the left column or the bottom row.  :wink:

The general result is given by the [url=http://en.wikipedia.org/wiki/Young_tableau#Dimension_of_a_representation]hook-length formula[/url], which shows the answer is $ \frac {12!}{6 \cdot 5^2 \cdot 4^3 \cdot 3^3 \cdot 2^2 \cdot 1} \equal{} 462$. (there's a [url=http://www.research.att.com/~njas/sequences/A151334]combinatorial interpretation[/url] that is analogous to that of the catalan sequence)

azjps, where do you pull these random, obscure formulas/theorems from?

Really it would have made more sense to link to [url=http://www.research.att.com/~njas/sequences/A005789]A005789[/url] (though in fact the two sequences are the same -- exercise for the reader ;) ).

The hook-length formula is a powerful result from algebraic/enumerative combinatorics, and shows up in a great many college-level textbooks in that area (e.g., Sagan's [i]The Symmetric Group[/i] or Stanley's [i]Enumerative Combinatorics[/i]).  This is a strange question to ask not in that context.

A friend and I had tried to come up with a direct combinatorial proof for the case of $ 3 \times n$ rectangles (analogous to the reflection principle by which one can prove that Catalan numbers have the formula that they do), but we had no luck, if I remember correctly.

123456789 was kidding  :)  but agreed about the linked sequence.

[quote="cipos2"]
Is 36 correct?
1 needs to be placed on the bottom left.
But the next two that compose the next diagonal, 2 and 3 can be switched within the diagonal but not moved to any other square or a square with greater magnitude would have to be placed below or to the left of it by a pigeonhole argument.
So we can switch up the diagonals but not anything else.
We have 2 diagonals of 1, 2 diagonals of 2 , and 2 diagonals of 3.
$ 1^2 * 2^2 * 3^2 \equal{} 36$.
[/quote]

Well...that's not correct.


$f:Q^{+}\rightarrow R$ such that : $f\left( 1997\right) =1$ and
$\left( \forall x,y\in Q^{+}\right) :f\left( x+y\right) \leq f\left( x\right) +f\left( y\right) $ and $f\left( xy\right) =f\left( x\right) f\left( y\right) $
prove that : $f\left( 2005\right) \leq 1$

And this one was solved by me here... Search...

This problem is used on the french correspondance program and has to be solved by 29th March.

Please don't discuss this problem before....

Can you delete the link please ?

I'v deleted it, even if it is easy to find it back...
More generally, Erdös, please pay attention to the dateline in the correspondance program before posting it here (I like the situation to give advices to a guy named Erdös  :)  ).

Pierre.


If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then the sum of first 110 terms is:

$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$

[hide]Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.

Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$
Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$

Solving the system, we get $d=-\frac{11}{50}$, $a=\frac{1099}{100}$. The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$

$\therefore \boxed{D}$[/hide]

Which is the toughest competition in high school,AMC12?

In high school you can go all the way up to IMO  :D


We will play a game where we decide which events of this decade are the most important. Now is the "buildup phase". Just put another event that occurred this decade on the list. I will post the rest of the rules later. Copy-paste the list each time.

1. Death of Michael Jackson (2009)

1. Death of Michael Jackson
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau.
3. Barack Obama wins Nobel Peace Prize(w00t)

Put the years after.

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau.  (2009)
3. Barack Obama wins Nobel Peace Prize (2009)
4. 9/11 (2001)

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. Barack Obama wins Nobel Peace Prize (2009)
4. 9/11 (2001)
5. Bush elected  :( (2000)

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. Barack Obama wins Nobel Peace Prize (2009)
4. 9/11 (2001)
5. Bush elected (2000)
6. Red Sox win WS (2004)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009) (boy I'm being to nice to this guy)
_________________

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. Barack Obama wins Nobel Peace Prize (2009)
4. 9/11 (2001)
5. Bush elected (2000)
6. Red Sox win WS (2004)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009) :P

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. Barack Obama wins Nobel Peace Prize (2009)
4. 9/11 (2001)
5. Bush elected (2000)
6. Red Sox win WS (2004)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald (2007)

OK that's enough for now. So every turn, you have two options:

1. Add a new event at the bottom.
2. Switch two adjacent events.

The goal is to put the most significant events at the top. mtfbwy

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. 9/11 (2001)
4. Barack Obama wins Nobel Peace Prize (2009)
5. Bush elected (2000)
6. Red Sox win WS (2004)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald (2007)

I want Obama at the BOTTOM.

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. 9/11 (2001)
4. Barack Obama wins Nobel Peace Prize (2009)
5. Red Sox win WS (2004)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald (2007)

I want Bush at the BOTTOM.

1. Death of Michael Jackson (2009)
2. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
3. 9/11 (2001)
4. Barack Obama wins Nobel Peace Prize (2009)
5. Red Sox win WS (2004)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald (2007)
14. The first known instance of a rickroll occurs (2007)
15. The iPHONE debuted. (2007)

1. Death of Michael Jackson (2009)
2. 9/11 (2001)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Barack Obama wins Nobel Peace Prize (2009)
5. Red Sox win WS (2004)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)

1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Barack Obama wins Nobel Peace Prize (2009)
5. Red Sox win WS (2004)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)
_________________

1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Red Sox win WS (2004)
5. Barack Obama wins Nobel Peace Prize (2009)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)

I switched 4 and 5.

But...the decade isn't over yet...not until 2011...

1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Red Sox win WS (2004)
5. Barack Obama wins Nobel Peace Prize (2009)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)
19. The economy collapses(2008)
20. 007math joins AoPS and fails (2007)
21. Death of Billy Mays (2009)
22. AMC 8 banned calculators(2009)
_________________

1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Red Sox win WS (2004)
5. Barack Obama wins Nobel Peace Prize (2009)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)
19. The economy collapses(2008)
20. 007math joins AoPS and fails (2007)
21. Death of Billy Mays (2009)
22. AMC 8 banned calculators(2009)
23. Darryl Wu becomes theyoungest person to become MATHCOUNTS National Champion, at age 11, 6th grade. (2008)

1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Red Sox win WS (2004)
5. Barack Obama wins Nobel Peace Prize (2009)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)
19. The economy collapses(2008)
20. 007math joins AoPS and fails (2007)
21. Death of Billy Mays (2009)
22. AMC 8 banned calculators(2009)
23. Darryl Wu becomes the youngest person to become MATHCOUNTS National Champion, at age 11, 6th grade. (2008)
24. ernie (the AoPS user) nabs his 1337th post. (2008)
25. AoPS debuts(2003)

[hide=Switch 20 and 21]
1. 9/11 (2001)
2. Death of Michael Jackson (2009)
3. The President of Guinea-Bissau, João Bernardo Vieira, is assassinated during an armed attack on his residence in Bissau. (2009)
4. Red Sox win WS (2004)
5. Barack Obama wins Nobel Peace Prize (2009)
6. Bush elected (2000)
7. 'Mission Accomplished" Sign for Iraq War(2003)
8. Saddam Hussein executed (2006)
9. John Kerry loses. (2004)
10. Tiger Woods cheats on wife.(2009)
11. Mark Sanford had an extramarital affair with a mistress in Argentina(2009)
12. Panic! At the Disco splits up, inserts ! back into their name. (2009)
13. Britney Spears goes bald. (2007)
14. The first known instance of a rickroll occurs. (2007)
15. The iPhone debuted. (2007)
16. Launch of Wikipedia (2001)
17. Mathcounts makes sprint round a lot harder(2004)
18. Jibjab starts its "Year in Review"(2005)
19. The economy collapses(2008)
20. Death of Billy Mays (2009)
21. 007math joins AoPS and fails (2007)
22. AMC 8 banned calculators(2009)
23. Darryl Wu becomes the youngest person to become MATHCOUNTS National Champion, at age 11, 6th grade. (2008)
24. ernie (the AoPS user) nabs his 1337th post. (2008)
25. AoPS debuts(2003)


a,b,c in N.

Find a,b,c such that 2(ab+bc+ca)=a+b+c+abc.

I will wait for your nice answers. :blush:

Thank you.

I am getting $ c \equal{} \frac{2ab \minus{}a \minus{} b}{ab \minus{} 2a \minus{} 2b \plus{} 1}$


What's the condition of positive real numbers $a,b,c$ such that:$\forall\ n\in\mathbb{N},n\neq 0\ [na].[nb]=[n^2c]$
[$x$] is the greatest integer $\le x$

[quote="Rust"]c=ab,a,b - integer.[/quote]
Thanks for your help!But why,Rust?

Consider 1)c>ab: $[cn^2]=[abn^2+(c-ab)n^2]>[an][bn]$ if n is large.
2) If c<ab: $[cn^2]=[abn^2-(ab-c)n^2]\le [anbn]-[(ab-c)n^2]<[an][bn]+an+bn-[(ab-c)n^2]<[an][bn]$ if n is large.
3)c=ab, $[abn^2]=[([an]+\{an\})([bn]+\{bn\})]\ge [an][bn]+[[an]\{bn\}+[bn]\{an\}]>[an][bn]$ if n is large and a or b is not integer.


inequalities
algebra
difference of squares
special factorizations
number theory
prime factorization

Hello everyone,

Could anyone please offer a hint as to what I should do with the absolute values? Do I set up two forms of the equation; one for the positive value and one for the negative value?

Thanks.

---

1. $ \frac{|4|}{m^{2}\minus{}16}\ \plus{}\frac{m\minus{}3}{m\plus{}4}\ \equal{}\frac{1}{m\minus{}4}$

Answer in book: [hide]$ m \equal{}{2,6}$[/hide]

2. $ \frac{1}{|3x\minus{}1|}\ <\frac{2}{x}$

Answer: [hide]$ 0 < x <\frac{2}{7}$ or $ x > 0.4$[/hide]

First isolate the absolute value. Then you can take + or - values.

Be careful in the second one, taking the reciprocol of something always flips the inequality sign.

[hide="hint for number 1"]

$ |4| \equal{} 4$ then use difference of squares, then find a common denominator.
[/hide]

[hide="hint for number 2"]

$ |3x\minus{}1| >\frac{x}{2}$

$ 3x\minus{}1 >\frac{x}{2}\;\cup\; 3x\minus{}1 <\minus{}\frac{x}{2}$

[/hide]

Hello,

Thank you for your replies, buzzer11 and Prime factorization.

Just one question though: What did you do to the inequality to get the reciprocal of $ \frac{1}{|3x\minus{}1|}?$

[quote="Ravi B"]Prime Factorization, your answer to #2 is incorrect.  You first need to state that $ x > 0$ before taking the reciprocal of both sides.  Then you need to incorporate that condition into the final answer.[/quote]

You need two cases: one for the postive and another for the negative.

Well, you can deduce that $ x$ is positive just by looking at the original inequality.

Sorry for my last post  :blush:

[hide=" this should be right now"]

Okay, I believe the actual solution would be attained by transposing $ \frac{2}{x}$, then finding a common denominator. Construct a table of signs and you have your answer.

[/hide]


[quote="MTM"]Is it possible to express sin 2x as a polynomial of sin x?[/quote]

Suppose you have $ \sin(2x)\equal{}P(\sin(x))$. This would imply $ P(x)^2\plus{}(1\minus{}2x^2)^2\equal{}1$ ($ \forall x\in[\minus{}1,1]$, so for all $ x$) and so $ P(x)^2\equal{}4x^2(1\minus{}x^2)$, which is impossible since $ \plus{}1$ and $ \minus{}1$ are simple (not multiple) roots of RHS and would need to be multiple roots of LHS.


Pat and Lee play a game where they take turns removing anywhere between $ 1$ to $ 100$ beads from a pile of $ 2008$ beads. The person to remove the last bead wins. If Pat goes first and both players play perfectly, how many beads must Pat remove to ensure that he wins the game?

[hide="Solution"]Given: Pat will win if he takes the 2008th bead.

To guarantee taking the $ 2008^{\text{th}}$ bead, he must also take the $ 1907^{\text{th}}$ bead. Then, Lee takes $ n$ beads, and Pat wins by taking $ 101 \minus{} n$ beads.

To guarantee taking the $ 1907^{\text{th}}$ bead, Pat must also take the $ 1806^{\text{th}}$ bead, etc.

If we can continue this line of reasoning, we see that the problem is logically equivalent to: Determine $ 2008\pmod{101}$.

$ 2008\equiv\boxed{89}\pmod{101}$[/hide]

[hide="Answer"]This is a common game structure.

For each number from 1 to 100, there is another number such that the two add to 101.

After Pat makes the first move, Pat can react to Lee's moves such that the total beads removed each round is 101. The goal is to leave Lee with 1 bead.

The remainder when 2008 is divided by 101 is 89. Thus, on the first round, Pat should take 88, and on each subsequent round, Pat should take 101 minus what Lee had just taken.[/hide]

TZF, why do you subtract $ 1$ from $ 2008\pmod{101}$?

Oh, I thought the player to remove the last bead lost. So I wanted Lee to be left with 1. But I guess not, so Lee should be left with nothing and it's just 89 -- thanks


Find all x such that:
$ \frac{cos^3x\minus{}sin^3x}{\sqrt{sinx}\plus{}\sqrt{cosx}}\equal{}2cos2x.$

[quote="mrmohamed"][quote="MR.129"]Find all x such that:
$ \frac {cos^3x \minus{} sin^3x}{\sqrt {sinx} \plus{} \sqrt {cosx}} \equal{} 2cos2x.$[/quote]
[color=blue][b]Hi all
I get that
[size=150]X = (pi/4) + 2 n pi             n = 0 , 1 ,2,3,...........[/size]
mrmohamed[/b][/color][/quote]
Can you post full solution ? Thank you.

[quote="MR.129"]Find all x such that:
$ \frac {cos^3x \minus{} sin^3x}{\sqrt {sinx} \plus{} \sqrt {cosx}} \equal{} 2cos2x.$[/quote]

Hello
my solution
[url=http://www.servimg.com/image_preview.php?i=459&u=12899977][img]http://i19.servimg.com/u/f19/12/89/99/77/almonk13.gif[/img][/url]


modular arithmetic
algebra unsolved
algebra

This problem from VietNam Mathematical Olympiad 1999. You need post it in Olympiad box, i think so.

Oh, sorry. I don't know. It's just a hard problem my teacher gave me but I haven't solved yet.
@moderators: please move this topic to Olympiad box! Thanks!

${x_{n}+y_{n}\sqrt{5}\over 2}=\left({1+\sqrt{5}\over 2}\right)^{2n+1}$
Also $x_{n}=\left({1+\sqrt{5}\over 2}\right)^{2n+1}+\left({1-\sqrt{5}\over 2}\right)^{2n+1}=\left[ \left({1+\sqrt{5}\over 2}\right)^{2n+1}\right]$

and $x_{n}=F_{2n+2}+F_{2n}$,$y_{n}=F_{2n+1}$, where
$F_{0}=0,F_{1}=1,F_{n+1}=F_{n}+F_{n-1}$ are the Fibonacci Numbers.

[quote="hikaru123"]b. Suppose $a,b\in Z$ and $a^{2}-5b^{2}+4=0$
Prove $\exists k\in N$ that $x_{k}=a$ and $y_{k}=b$[/quote]
It's $a,b\in\mathbb{N}_{0}$! You can always change signs of $a,b$.
Also $a\not =0$ and $(a,b)\not =(1,0)$, as $a^{2}-5b^{2}=-4$.
So ${a+b\sqrt{5}\over 2}\geq 1$ and $\left({1+\sqrt{5}\over 2}\right)^{k}\leq{a+b\sqrt{5}\over 2}<\left({1+\sqrt{5}\over 2}\right)^{k+1}$ for some $k\in\mathbb{N}_{0}$.
Then $1\leq{a+b\sqrt{5}\over 2}\left({-1+\sqrt{5}\over 2}\right)^{k}<{1+\sqrt{5}\over 2}$.  (here ${1+\sqrt{5}\over 2}\cdot{-1+\sqrt{5}\over 2}=1$)

Now the set $R: =\{{x+y\sqrt{5}\over 2}: \ x,y\in\mathbb{Z},x\equiv y\pmod{2}\}$ is closed under summation and multiplikation
and there is a well-defined and multiplicative map $N: R\rightarrow \mathbb{Z}$
with $N\left({x+y\sqrt{5}\over 2}\right)={x^{2}-5y^{2}\over 4}$. ($N(zw)=N(z)N(w)$ for all $z,w\in R$)

Also $a^{2}-5b^{2}=-4$, so $a\equiv b\pmod{2}$, and therefore ${a+b\sqrt{5}\over 2}\in R$ with $N\left({a+b\sqrt{5}\over 2}\right)=-1$.

We can write ${a+b\sqrt{5}\over 2}\left({-1+\sqrt{5}\over 2}\right)^{k}={u+v\sqrt{5}\over 2}$ with $u,v\in\mathbb{Z}$ and $1\leq{u+v\sqrt{5}\over 2}<{1+\sqrt{5}\over 2}$
and ${u^{2}-5v^{2}\over 4}=N\left({u+v\sqrt{5}\over 2}\right)=N\left({a+b\sqrt{5}\over 2}\right)N\left({-1+\sqrt{5}\over 2}\right)^{k}=(-1)^{k+1}$.
Then ${u-v\sqrt{5}\over 2}=\pm{2\over u+v\sqrt{5}}\in [-1,1]$.
So $u={u+v\sqrt{5}\over 2}+{u-v\sqrt{5}\over 2}\geq 0$ and $v\sqrt{5}={u+v\sqrt{5}\over 2}-{u-v\sqrt{5}\over 2}\geq 0$.
$u^{2}-5v^{2}=\pm 4$ gives $u\not =0$ and $u\geq 1$.
${1+v\sqrt{5}\over 2}\leq{u+v\sqrt{5}\over 2}<{1+\sqrt{5}\over 2}$ gives $v=0$.
And $u^{2}=u^{2}-5v^{2}=4(-1)^{k+1}$ gives $u=2$ and $k=2n+1$ odd.
And now ${a+b\sqrt{5}\over 2}={u+v\sqrt{5}\over 2}\left({1+\sqrt{5}\over 2}\right)^{k}=\left({1+\sqrt{5}\over 2}\right)^{2n+1}={x_{n}+y_{n}\sqrt{5}\over 2}$
and $(a,b)=(x_{n},y_{n})$, as $\sqrt{5}$ is irrational.


trigonometry
calculus
calculus computations

easy
Or
$(\sin+\cos)(1-\sin\cos)=1$
denote
$\sin+\cos=y$
then
$(1-y^{2})/2=\sin\cos$
aNd I think you can finished yourself :wink:

This isn't calculus: it's equivalent to
\[\sin^{2}\phi (1-\sin \phi)+\cos^{2}\phi (1-\cos \phi )=0\]
and the left hand side is a sum of two nonnegative terms.  :wink:

[quote="scorpius119"]This isn't calculus.[/quote]

You are right. This problem can be solved in 10 seconds. :D

[quote="kunny"]You are wrong.[/quote]
Why? I only show idea. And it was correct
$y(\frac{1+y^{2}}{2})=2$
then $(y-1)(y^{2}+y+2)=0$ or
$y=1$
$\cos(\alpha-\frac{\pi}{4})=\frac{1}{\sqrt{2}}$ :lol:

Yes, your idea was right, but $\sin \cos =$ was wrong. Now you seem to have edited. :)


Let $ ABCDEF$ be a regular hexagon and let $ G$ be the midpoint of $ BF$.Choose a point $ I$ on $ BC$ such that $ BI\equal{}BG$.Let $ H$ be a point on $ IG$ such that $ \angle CDH\equal{}45^o$,and $ K$ is a point on $ EF$ such that $ \angle DKE\equal{}45^o$.Prove that $ DHK$ is an equilateral triangle.

[quote="whywhywhy"]Let $ ABCDEF$ be a regular hexagon and let $ G$ be the midpoint of $ BF$.Choose a point $ I$ on $ BC$ such that $ BI \equal{} BG$.Let $ H$ be a point on $ IG$ such that $ \angle CDH \equal{} 45^o$,and $ K$ is a point on $ EF$ such that $ \angle DKE \equal{} 45^o$.Prove that $ DHK$ is an equilateral triangle.[/quote]
Please close this topic. It is a problem in the [b]MY magazine[/b], and it is not still out- dated


conics
hyperbola
geometry
incenter
geometry unsolved

If $\ H$ is a hyperbola, $\ F_1, F_2$ it's foci and $\ M\in H$ find the locus of the triangle's $\ F_1MF_2$ incenter.
[i]Proposed by Virgil Nicula [/i]

[hide][url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hyperbola&t=65102]Here[/url] we found that, if the hyperbola has equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then the abscissa of the incenter is equal to $a$.

Actually, my solution there, was incomplete, because I checked only the case $MF_1-MF_2=2a$, in which $M$ is on the right branch.

Accurately, the hyperbola has one more branch with equation $MF_2-MF_1=2a$

Then the abscissa of the incenter is equal to $-a$.

In other words, if $A$ and $B$ are the intersections of the hyperbola with the segment $F_1F_2$ and $AA',BB'$ are perpendicular to $F_1F_2$ then the incenter of $\triangle F_1MF_2$ belongs to the line $AA'$ or $BB'$
(it depends on the branch of $M$)

[hide="Note"]

The incenter $I$ cannot be anywhere on these lines.

Say for example that $I \in AA'$. Draw the circle $(I,IA)$ and bring tangent rays $t_1,t_2$ from $F_1$ and $F_2$ respectively.
These rays must intersect each other.

Let $K \in AA'$ such that $KA^2 = F_1A \cdot F_2A$ and $K'$ the symmetric point of $K$ w.r.t. $A$.

If you choose the point $I$ such that $IA=KA$ then the rays are parallel, so you have not a $M$.

If $IA > KA$ , then the rays $-t_1, -t_2$ intersect each other

(with $-t_1$ I mean the opposite ray of $t_1$)

In other words, the line of $t_1$ intersects the line of $t_2$ at the opposite direction and then $I$ is excenter and we don't that.

Finally the locus of $I$ is the (open) segment $(KK')$

Similarly for the line $BB'$, it's a segment $(LL')$, not all the line $BB'$[/hide]

[/hide]


consider the number 2^9876543. Determine
its last three (rightmost) digits
its first three (leftmost) digits

This question is from the OUSMI application quiz. I ask that no one answers it before the May 15 deadline.

Oakland U's summer math program. Apply [url=http://www.oakland.edu/links/math/ousmi04/ousmi04.html]here[/url].


inequalities
search
geometry
inequalities unsolved

It's has been posted before, please look here: http://www.mathlinks.ro/viewtopic.php?search_id=648048130&t=3171

Thank you very much. But I don't  find solution for the problem . Can't you give me the solution

[color=darkblue]Hi thanhak5. Now, I give you a way to done your problem. But it's long. I use some inequalities in triangle. I will don't prove some inequalities in triangle. You can look it at some books.

Note some inequalities:
$  \| \begin {array} {c} m_a.m_b.m_c\ge p.S , (*) \\
\frac {a}{m_a} + \frac {b}{m_b} + \frac {c}{m_c} \ge 2\sqrt {3} , (**) \\
p.S \ge \frac {3\sqrt {3}abc}{8}, (***) \end {array}$

$ \bullet$ $ \frac { m_a.m_b }{ab} + \frac { m_b.m_c }{bc} + \frac { m_c.m_a }{ac} \ge \frac {9}{4} \Longleftrightarrow m_a.m_b .c + m_b.m_c.a + m_c.m_a .b \ge \frac {9abc}{4}$

$ \bullet$ Apply $ (*), (**), (***)$  we have:

$ m_a.m_b .c + m_b.m_c.a + m_c.m_a .b = m_a.m_b.m_c .\frac {c}{m_c} + m_b.m_c.m_a.\frac {a}{m_a} + m_c.m_a.m_b .\frac {b}{m_b}$
 $ \ge p.S(\frac {a}{m_a} + \frac {b}{m_b} + \frac {c}{m_c})\ge \frac {3\sqrt {3}abc}{8}.2\sqrt {3} = \frac {9abc}{4}$. Then OK![/color]

[i][color=darkblue][u]Note:[/u] $ p = \frac {a + b + c}{2}$, $ S$ is area of $ \triangle ABC$. [/color][/i]

Thank you Thanhnam2902 very much . I can prove (**) . I will try slove  the problem (*),(***) . I think my problem is very strong and we can't use other Ineq... Because from my problem We can slove the problem (*)
  $ \frac {m_a}{a} \plus{} \frac {m_b}{b} \plus{} \frac {m_c}{c} \ge \frac {3.\sqrt {3}}{2}$
  The problem (***) is not right
  (***) :  $ p.S \ge \frac {3.\sqrt {3}abc}{8}$
             <=> $ p.\frac {abc}{4R}  \ge  \frac {3.\sqrt {3}abc}{8}$
             <=> $ sinA \plus{} sinB \plus{} sinC   \ge  \frac {3.\sqrt {3}}{2}$

[quote="thanhak5"]For the traingle . Prove that

  $ \frac { m_a.m_b }{ab} \plus{} \frac { m_b.m_c }{bc} \plus{} \frac { m_c.m_a }{ac} \ge \frac {9}{4}$
Could you help me[/quote]
Let $ a \equal{} y \plus{} z,$ $ b \equal{} x \plus{} z$ and $ c \equal{} x \plus{} y.$
Then $ x,$ $ y$ and $ z$ are positive numbers and $ 2a^2 \plus{} 2b^2 \minus{} c^2 \equal{} 4z(x \plus{} y \plus{} z) \plus{} (x \minus{} y)^2.$
Hence, $ \sum_{cyc}\frac { m_b m_c }{bc} \equal{} \sum_{cyc}\frac {\sqrt {\left(4z(x \plus{} y \plus{} z) \plus{} (x \minus{} y)^2\right)\left(4y(x \plus{} y \plus{} z) \plus{} (x \minus{} z)^2\right)}}{4(x \plus{} y)(x \plus{} z)}\geq$
$ \geq\sum_{cyc}\frac {4\sqrt {yz}(x \plus{} y \plus{} z) \plus{} |(x \minus{} y)(x \minus{} z)|}{4(x \plus{} y)(x \plus{} z)}\geq$
$ \geq\sum_{cyc}\frac {4\sqrt {yz}(x \plus{} y \plus{} z) \plus{} (x \minus{} y)(x \minus{} z)}{4(x \plus{} y)(x \plus{} z)}.$
Id est, it remains to prove that $ \sum_{cyc}\left(4\sqrt {yz}(x \plus{} y \plus{} z)(y \plus{} z) \plus{} (x \minus{} y)(x \minus{} z)(y \plus{} z)\right)\geq9(x \plus{} y)(x \plus{} z)(y \plus{} z),$
which equivalent to $ \sum_{cyc}(x \plus{} y \plus{} z)\sqrt {xy}\left(\sqrt x \minus{} \sqrt y\right)^2\geq0.$
The first inequality is doubly to the second.

Thank arqady very much . You used to S.O.S , I think my problem is diffcult . Can you give me a short solution  for problem

[quote="thanhak5"]For the traingle . Prove that
 $ \frac {ab}{m_a.m_b} \plus{} \frac {bc }{m_b.m_c} \plus{} \frac {ca}{m_c.m_a} \ge 4$
 and
  $ \frac { m_a.m_b }{ab} \plus{} \frac { m_b.m_c }{bc} \plus{} \frac { m_c.m_a }{ac} \ge \frac {9}{4}$
Could you help me[/quote]
$ 1)$
$ m_am_b \le \frac {1}{2}c^2 \plus{} \frac {1}{4}ab$
$ m_bm_c \le \frac {1}{2}a^2 \plus{} \frac {1}{4}bc$
$ m_cm_a \le \frac {1}{2}b^2 \plus{} \frac {1}{4}ca$ $ \equal{} > 3m_am_bm_c\le \frac {1}{2}(m_aa^2 \plus{} m_bb^2 \plus{} m_cc^2) \plus{} \frac {1}{4}( m_cab \plus{} m_abc \plus{} m_bca)$;
$ 2m_aa \le m_cb \plus{} m_bc$
$ 2m_bb \le m_ac \plus{} m_ca$
$ 2m_cc \le m_ba \plus{} m_ab \equal{} > m_aa^2 \plus{} m_bb^2 \plus{} m_cc^2 \le m_abc \plus{} m_bca \plus{} m_cab$;
So we have : $ 3m_am_bm_c \le \frac {3}{4}( m_cab \plus{} m_abc \plus{} m_bca) < \equal{} > \frac {ab}{m_am_b} \plus{} \frac {bc }{m_bm_c} \plus{} \frac {ca}{m_cm_a} \ge 4$; :)
$ 2)$
For triangle with sides ($ a,b,c$) and medians ($ m_a,m_b,m_c$) we have triangle with sides ($ m_a,m_b,m_c$) and medians ( $ \frac {3}{4}a,\frac {3}{4}b,\frac {3}{4}c$ );
So we have :  $ \frac {m_am_b}{\frac {3}{4}a \frac {3}{4}b} \plus{} \frac {m_bm_c }{\frac {3}{4}b \frac{3}{4}c} \plus{} \frac {m_cm_a}{\frac {3}{4}c \frac {3}{4}a} \ge 4$; :)

We have
   $ m_am_b \le \frac {1}{2}c^2 \plus{} \frac {1}{4}ab$
  $ < \equal{} > (2b^2 \plus{} 2c^2 \minus{} a^2)(2a^2 \plus{} 2c^2 \minus{} b^2) \le (2c^2 \plus{} ab)^2$
          .........................
  .....................................
  $ < \equal{} > (a \minus{} b)^2[(a \plus{} b)^2 \minus{} c^2] \ge 0$
 Thank  Sergic Primazon very much . Your solution is shorter than Arqady's solution
 But
    $ 2m_c.c \le m_b.a \plus{} m_a.b$
 I can't slove it


email Alcumus geometry rectangle \/closed

This is a random idea, but beside where it says the number of posts, it could also say "Number of hours online". You could also have "Hours online per day".

Or you could also have "last online"
But how could the site keep track of how long you've been on the site each day?

The website could use something like cookies?

It can detect when you are online, I think, because it says so under your avatar. The site could just time this.

Seems like a good idea, I'd like it a lot. Not sure how it would be implemented, and also uncertain about if the admins like it. Kind of doubt it.

It wouldn't really show who is productive, because a member can sign in to AOPS and then leave and do anything they like without doing anything on
AOPS.

Ya that would be a good idea, though I dont think the administrators would implement it with all the stuff they need to do already (add to alcumus, etc.)

[quote="yunhua98"]I don't think this will happen, as it will enlarge the spamming problem, spammers could look for people who are always online, then spam them, knowing most people have their email on to alert them of replies. (Or at least I do that. :P)[/quote]

lol, wow :lol:

Because clearly it's good to encourage compulsive roaming around G&FF all day.

[quote="r15s11z55y89w21"][quote="Dojo"][quote="myyellowducky82"]Or you could also have "last online"[/quote]

There already is one. Look in the left corner.[/quote]

I don't see it. Could you please post a screenshot?[/quote]

At the top, under the blue rectangle containing all the buttons like "Profile", etc., it says:

You last visited on Yesterday, at 5:46 pm
The time now is Mon Jun 29, 2009 7:17 am
All times are UTC - 8

or whatever.

I think what myyellowducky82 means that there should be one for every user, like beside their avatar.

[quote="myyellowducky82"]Or you could also have "last online"[/quote]I do have this in the code already, but we need to put up an option for users to allow other people to see when their last online date was.

As for the total time spent online, some people (and parents) might not like that information :D


geometry
AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 12 B

Feel free to post.

Prediction:

AMC 10B Cutoff 2007: 111
AMC 12B Cutoff 2007: 94.5

Prediction:

AMC 10B Cutoff 2007: 115.5
AMC 12B Cutoff 2007: 97.5

About the same as last time.. so 97 and 117.  It could dip lower or be a bit higher depending on how many people took it, but I don't see a reason for it being drastically different.

[quote="xingster14"]i want to be optimistic, so im going to say at least ten or so points off the actual target .[/quote]

AMC 12B: 110
AMC 10B: 130

 :D

any chance it could actually bet 94.5? (hoping)

My [biased] view of the AMC 10B [no experience with the 12B] will be that the cutoff will be close to 100.

[quote="neothe0ne"]I hope 12B's cutoff will be 93...... >_>[/quote]
hey man im with you on that

wasn't last year(2006)'s amc 10b cut off 120?

I personally thougt this year's test was harder. My friends thought so too. (maybe because of more geometry problems...idk)

So.. i'm guessing something like 115.

solafidefarms, didn't you get a perfect score? (prior post!) If you clearly found it so "easy", why would you think the cutoff would be lower and not higher?

Personally, can't comment on 12 B, but guessing 10 B will be either 117 or remain the 120 it starts off as...

Why do most on here think AMC B cutoff will be lower than AMC A - have you taken Both? You thought B was so much harder?

if it is at least 7.5 points below, I will be the happiest person alive.

The cutoff might still be 100, because this was a harder test than amc a, BUT it was easier to pass (as for the A's, the whole test as a whole was about average, but like for 12b, the first 15 were way easier than the first 15 for amc 12b, but the last 10 was impossible compared to the last 10 for amc 12a-and you only need 14 right to pass)-so the average score will proablaby be lower than the amc a's, but the cutoff may be higher

[quote="solafidefarms"]Eh, getting a perfect doesn't mean I thought it was an easy test :P... I hit problem 14 and said "what happened to this test? this is a hard problem, and I'm only to problem 14 ! uh-oh..." So if someone hit 14 and stopped, they'd get a score of about 96, and that would represent someone who got all of the really easy ones right. As more than just the non-hard questions have to be answered to qualify, I'd expect the cutoff to be about 100.[/quote]

ya, I hit 13 after doing the first 12 in 20 minutes and was like, wow, this is way harder... (then I failed the rest of the test)

Yeah, I did those first 13 in 6 minutes [good old watches] so I had lots of time to finish the others. :)

same here, I did like the first 18 in 30 mins, and i just wasted the other 45 minutes figuring out 3 problems(guessed on th rest-1 right).. when i shouldve used the time drawing on my paper going back and checking #6 and #17 for reading errors..

Would the lowered cutoff for AIME affect the index score for USAMO?

[quote="paladin8"][quote="vishalarul"]AMC 12B: 98.5
AMC 10B: 118.5

Just a guess.[/quote]

Maybe if 98.5 was a possible score :wink:.[/quote]

It's interesting how the 1.5 points per blank rule drastically decreases the number of possible scores.
For this reason i predict:
10B: 117
12B: 100

amc12B cutoff

99 will be perfect for me

[quote="randomdragoon"]It would be sort of hilarious (in a cynical, sarcastic way) if they took the cutoff score and then decreased it by 1 point before publishing it, to make the people who just missed it look like they just missed it by .5 point.[/quote]
It would be genius though.
Pure genius.
besides, the hillarity would be endless.

[quote="randomdragoon"]It would be sort of hilarious (in a cynical, sarcastic way) if they took the cutoff score and then decreased it by 1 point before publishing it, to make the people who just missed it look like they just missed it by .5 point.[/quote]

How about a 99.1 cutoff? :P

At least the 99.5 cutoff looks realistic to people who don't think about these things.


The radius of the inscribed circle is $6cm$. What is the number of centimeters in the length of $\overline{AB}$? Express your answer in simplest radical form.

[img]7230[/img]

It is really a quarter circular arc, but w/e

And yeah, your equation is right.

[quote="indianamath"][hide]6+(r*root2/2)=r

solving for r gives us

r=12+6*root2[/hide][/quote]
idont get why that equation is used

hint: draw the tangents and you have special triangles. (I'm pretty sure)

Okay, to solve this problem [hide="hint"]note the 60 degree angle, that means it's a 30-60-90. Also note the formula for the inradius. [/hide]

[hide]$30-60-90=x, x\sqrt{3}, 2x$

so $x+x\sqrt{3}-2x=x\sqrt{3}-x=6$

$\frac{6}{\sqrt{3}-1}$

Rationalize denominator to get $\frac{6\sqrt{3}+6}{2}=x$

So $2x=\boxed{6\sqrt{3}+6}$[/hide]

The correct answer is
[hide]
$12+12\sqrt{3}$
[/hide]
Don't bother arguing, because I know that for a fact

[hide]
First, call the right angle something, say C, and call the center of the circle O.
Draw the radii of the circle to the sides of the triangle. If you call the point where the circle meets AB E and the intersection of the circle and AC F, draw quadrilateral OEBF. Then draw BO. This splits the quadrilateral into two congruent 30-60-90 triangles. BE=BF=$6\sqrt{3}$. FC = 6. Therefore, BC = $6+6\sqrt{3}$ AB is double that, or $12+12\sqrt{3}$
[/hide]


search
geometry
geometric transformation
rotation

Given the midpoints of a 17-gon, find its vertices (rule and compass only, of course  :) ).

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1736793246&t=191442

(This was searching for "midpoints vertices."  A smarter search, e.g. using the word "Engel," might turn up threads with more details.)

[hide="Hint"]Consider the $ 180^{\circ}$ rotation operation about a midpoint.[/hide]

[hide="Outline of solution"]If we consider the composition of all such rotation operations, starting at one midpoint $ M$ that is the midpoint of $ AB$, it is also a $ 180^{\circ}$ rotation. Since $ A$ is a fixed point, $ A$ is the center of rotation. If we take an arbitrary point $ P$ and perform each individual rotation on it, $ A$ will be the midpoint of $ PP'$.[/hide]


What is the value of $ (4^3 \plus{} 2^3) \minus{} (3^3 \plus{} 1^3)$?

Evaluate to get
\[ 64\plus{}8\minus{}27\minus{}1\equal{}\boxed{44}
\]


Also in (like Jeffrey: deferred from MIT).

It looks like drunner has a tough choice to make.  :)

[quote="zanttrang"]I'm in - deciding between Caltech and MIT is a difficult (but not unpleasant, certainly) decision![/quote]

Definitely come to MIT's CPW (college preview weekend).  I wasn't able to make it last year but I've heard that it's quite helpful in deciding where to go ;)

[quote="PTynan89"]Wow, mine came today.

I got deferred... (my second deferral)  :([/quote]

Don't worry, you'll get in.
Congratulations to everyone.

I got in as well...just got my letter and screamed my head off, lol.  Caltech was my #1 choice so I'm definitely happy about this.   :)

[quote="mdk"][quote="drunner2007"][quote="Go Around the Tree"]wow drunner.

you must be quite well rounded  :P

How many schools have you applied to? and I suppose you got into all of them[/quote]

haha i'm somewhat well rounded. so far i've applied to ga tech, mit, and caltech, and yes i have gotten into all of them. but i'm sure there are others on this forum that have gotten into these schools, and lots of the ivies.[/quote]

It is still a very good job. Keep up the good work.[/quote]

$\text{maybe others on the forum got in because AoPS people are S-M-A-R-T! }$


Let $ abc\equal{}1$ and $ a,b,c>0$. Prove that
$ \frac{1}{a\plus{}2b}\plus{}\frac{1}{b\plus{}2c}\plus{}\frac{1}{c\plus{}2a}\leq 1$

it's wrong
    when $ b$  and $ c  \to  0$   and  $ a\equal{}\frac{1}{bc}  \to  \infty$
i think   this  problem's made   from  ur other problem


Any proof for

Given $\triangle ABC$, let $P$ and $P'$ such that:
$\frac{AP}{AP'}=\frac{BP}{BP'}=\frac{CP}{CP'}$, then $P$ and $P'$ are inverses wrt the circumcircle of $\triangle ABC$.

[quote="hucht"][quote="Amir.S"]points P and P' are isodynamic points wrt ABC and cause appolonian circles are perpendicular to circumcircle of ABC therfore:
$OP.OP'=OA^{2}$[/quote]

Unfortunately not, those points are any points, It's not necessary to $P$ and $P'$ are isodynamic.[/quote]

yes you are right I took some mistakes I edit it now.

[quote="hucht"][quote="mecrazywong"]Simply draw the Apollonian Circles.[/quote]

and next??[/quote]
Well...it follows from the fact that two circles can at most intersect at 2 points.

[quote="Amir.S"][quote="hucht"][quote="Amir.S"]let angle bisectors of $BPC$  intersect with $BC$ at $X,Y$ hence $(XY,BC)=-1$ let midpoin of $XY$ be $M$ therefore $MX^{2}=MB.MC=MT^{2}$($T$ the intersection point of circumcircle of $\triangle PXY$ and circumcircle of $ABC$) therefore  these two circle are perpendicular.you can prove that $P'$ is on circumcircle of $\triangle PXY$.

therefor $OP.OP'=OA^{2}$[/quote]

I haven't understood your soution at all. I have got that these circles are orthogonal and it's obvious that $P'$ is on it's crcumcircle but I don't understand the last part "therefore $OP.OP'=OA^{2}$"[/quote]

ok,notice that circumcircle of $ABC$ is orthogonal to circumcircles of $\triangle PXY,\triangle PX'Y',\triangle PX''Y''$ (X',X'",Y',Y'' are define like X,Y but X',Y' on AC and X'',Y'' are on AB)

so it means that the power of $O$ wrt these circles are equal.hence P,P',O are collinear . therefore $OP.OP'=OT^{2}=R^{2}$

Is it clear now?[/quote]

Yeah!


For $ 1)$
 $ \begin{bmatrix}1 & cos^{2}(a \minus{} b) & cos^{2}(a \minus{} c) \\
cos^{2}(b \minus{} a) & 1 & cos^{2}(b \minus{} c) \\
cos^{2}(c \minus{} a) & cos^{2}(c \minus{} b) & 1\end{bmatrix} \equal{} \begin{bmatrix}cos^2(a) & sin^{2}(a) & 2sin(a)cos(a) \\
cos^{2}(b) & sin^2(b) & 2sin(b)cos(b) \\
cos^{2}(c) & sin^{2}(c) & 2sin(c)cos(c)\end{bmatrix}\begin{bmatrix}cos^2(a) & cos^{2}(b) & cos^{2}(c) \\
sin^{2}(a) & sin^2(b) & sin^{2}(c) \\
sin(a)cos(a) & sin(b)cos(b) & sin(c)cos(c)\end{bmatrix}$
  and
   $ \det\begin{bmatrix}cos^2(a) & sin^{2}(a) & sin(a)cos(a) \\
cos^{2}(b) & sin^2(b) & sin(b)cos(b) \\
cos^{2}(c) & sin^{2}(c) & sin(c)cos(c)\end{bmatrix} \equal{} sin(a \minus{} b)sin(b \minus{} c)sin(c \minus{} a)$ ,
  Can  You  calculate it ? It is isn't dificult.
a)[color=red] General short For $ (x_1,x_2,..,x_n) \in \mathbb{R}^n$ put $ M \equal{} \det(cos^2(x_i \minus{} x_j))_{1 \leq i \leq n,1 \leq j \leq n}$ then $ \det(M) \equal{} 0$,if $ n \geq 4$[/color]
b) [color=blue] General large For $ (x_1,x_2,..,x_n) \in \mathbb{R}^n$ put $ M \equal{} \det(cos^m(x_i \minus{} x_j))_{1 \leq i \leq n,1 \leq j \leq n}$ then $ \det(M) \equal{} 0$,if $ n > m \plus{} 1$[/color]
 For $ 2$) Let $ B(x_1,y_1),C(x_2,y_2),A(x_3,y_3)$ then $ AB \equal{} C,BC \equal{} a,AC \equal{} b$ then use formular $ 2S_{ABC} \equal{} \sqrt {\frac {(a \plus{} b \plus{} c ) (b \plus{} c \minus{} a) (c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}{4}}$ .....

awesome sir

b) [color=blue] General large For $ (x_1,x_2,..,x_n) \in \mathbb{R}^n$ put $ M \equal{} \det(cos^m(x_i \minus{} x_j))_{1 \leq i \leq n,1 \leq j \leq n}$ then $ \det(M) \equal{} 0$,if $ n > m \plus{} 1$[/color]
 For $ 2$) Let $ B(x_1,y_1),C(x_2,y_2),A(x_3,y_3)$ then $ AB \equal{} C,BC \equal{} a,AC \equal{} b$ then use formular $ 2S_{ABC} \equal{} \sqrt {\frac {(a \plus{} b \plus{} c ) (b \plus{} c \minus{} a) (c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}{4}}$ .....[/quote]


vector
calculus
conics
ellipse
derivative
trigonometry
analytic geometry

I've seen a messy derivation of Kepler's 1st law from Newton's laws using vector calculus in polar, but I'm wondering whether there is any nicer derivation. In particular, is there a nice way to do it using the locus definition of an ellipse? It seems really surprising that such a simple fact (i.e. that an inverse square law creates a conic) would require such a complicated proof.

There is indeed an elegant and enlightening proof of Kepler's first law that doesn't use tedious calculations. It was originally developed by Newton and improvised by Feynman.

I had seen this proof in a lecture, but unfortunately, it was a long time ago and I've misplaced my notes. I'll be grateful if someone finds it for me. :)

I have a proof of Keplers first law using polar, but its very elegant, and short. Not more than 5-6 lines.  Should I post it?:D
But yeah it would be good to see a purely geometrical proof too. I will also try to find it :D

Yes, please post it.  I remember trying to prove it once, but I ended up with a system of differential equations that I couldn't solve.

Ok here it is :D

We no that gravitational force is central, hence momentum is conserved. So the equation of motion of a particle of mass $ m$ in a gravitational field, in polar coordinates.
\[ m(\ddot r \minus{}r\dot\varphi^2)\equal{}\minus{}\gamma \dfrac{mM}{r^2}\]
Momentum is conserved:
\[ r^2\dot\varphi\equal{}C\equal{}\text{constant}\]
So we want to solve the equation of motion, but we only want to obtain, the trajectory of the particle, that is only $ r\equal{}r(\varphi)$.
Hence we do the usual trick, of writing $ \ddot r$ differently:
\[ \dot r \equal{} \dfrac{dr}{d\varphi}\dot\varphi\]
Now lets plug $ \dot\varphi$ form the conservation of momentum equation in here:
\[ \dot r \equal{} \dfrac{dr}{d\varphi}\dfrac{C}{r^2}\]
Now lets use the following substitution, because we get a simpler expression:
\[ r\equal{}\dfrac{1}{n}\]
Then the above equation:
\[ \dot r \equal{} \minus{}\dfrac{1}{n^2}\dfrac{dn}{d\varphi}Cn^2\equal{}\minus{}C\dfrac{dn}{d\varphi}\]
Hence $ \ddot r$:
\[ \ddot r \equal{}\minus{}c^2n^2\dfrac{d^2n}{d\varphi^2}\]
Now plugging this back into the equation of motion, we get:
\[ \dfrac{d^2n}{d\varphi^2}\plus{}n\equal{}\dfrac{\gamma M}{C^2}\]
Looking at this differential equation, it doesnt look that easy to solve..., but it can be. So here is the big trick with which we can solve it in a second. :D
We know that $ \gamma M/C^2$ is constant, hence its derivative, with respect to $ \varphi$ is zero. Hence we can write the above differential equation as this:
\[ \dfrac{d^2}{d\varphi^2}\left(n\minus{}\dfrac{\gamma M}{C^2}\right)\plus{}\left(n\minus{}\dfrac{\gamma M}{C^2}\right)\equal{}0\]
Hmm. Now what is this? well its the equation of non damped 1D harmonic oscillator, for $ n\minus{}\dfrac{\gamma M}{C^2}$. We know that its solution is $ \cos..$. So we get:
\[ n\minus{}\dfrac{\gamma M}{C^2}\equal{}A\cos(\varphi\plus{}\alpha)\]
$ A$ and $ \alpha$ are the usual constants.
We know that $ n\equal{}1/r$, so we can express $ r$ from this equation:
\[ r\equal{}\dfrac{C^2}{\gamma M}\cfrac{1}{1\plus{}\dfrac{AC^2}{\gamma M}\cos(\varphi\plus{}\alpha)}\]
Now put
\[ p\equal{}\dfrac{C^2}{\gamma M} \; \; \; \; \; e\equal{}\dfrac{AC^2}{\gamma M}\]
Then we get:
\[ r\equal{}\dfrac{p}{1\plus{}e\cos(\varphi\plus{}\alpha)}\]
Which is the equation of a conic section.
As you see its not long and messy at all :D

Some of the notation was a bit confusing, but I got it..
$ r,\phi$ are the polar coördinates and $ \gamma\equal{}G$ is the usual Newton's gravitational constant?

Also,
[quote="Thaakisfox"]
Then the above equation:
$ \dot r \equal{} \minus{}\dfrac{1}{n^2}\dfrac{dn}{d\varphi}Cn^2\equal{}\minus{}C\dfrac{dn}{d\varphi}$
Hence :
$ \ddot r \equal{}\minus{}c^2n^2\dfrac{d^2n}{d\varphi^2}$
[/quote]
Could you explain this step? - I didn't get it

Yes, $ r$ and $ \varphi$ are the coordinates, and $ \gamma$ is $ G$ the gravitational constant...

well we have: $ r \equal{} 1/n$ hence, $ \frac{dr}{d\varphi} \equal{} \minus{} \frac {1}{n^2}\frac {dn}{d\varphi}$. So
\[ \dot r \equal{} \frac {dr}{d\varphi}\frac {C}{r^2} \equal{} \minus{} \frac {1}{n^2}\frac {dn}{d\varphi}Cn^2 \equal{} \minus{} C\frac {dn}{d\varphi}
\]
Now we know that:
\[ \ddot r \equal{} \dfrac{d\dot r}{d\varphi}\dfrac{d\varphi}{dt} \equal{} \dfrac{d\dot r}{d\varphi}\dot\varphi
\]
So we want:
\[ \dfrac{d\dot r}{d\varphi} \equal{} \minus{} \dfrac{d}{d\varphi}C\frac {dn}{d\varphi} \equal{} \minus{} C\frac {d^2n}{d\varphi^2}
\]
So:
\[ \ddot r \equal{} \dfrac{d\dot r}{d\varphi}\dot\varphi \equal{} \minus{} C\frac {d^2n}{d\varphi^2}\dfrac{C}{r^2} \equal{} \minus{} C^2n^2\frac {d^2n}{d\varphi^2}
\]
Its just manipulation with the derivatives... :D

That is brilliant.  Although the $ r \equal{} 1/n$ part seems unmotivated.


Let $ ABC$ be triangle with orthocenter $ H$ let $ G_a,G_b,G_c$ be centroid of $ HBC,HCA,HAB$,$ HA\cap G_bG_c \equal{} A_1,$ $ HB\cap G_cG_a \equal{} B_1,$ $ HC\cap G_aG_b \equal{} C_1,$ $ G_aC_1\cap A_1B_1 \equal{} C_2,$ $ G_bB_1\cap B_1C_1 \equal{} B_2$, let $ A_2$ be projection of $ A_1$ on $ B_2C_2$, let $ B_3,C_3$ be circumcenter of $ (A_1A_2G_b),(A_1A_2G_c)$, the line $ B_3C_3$ intersect $ A_2G_b,A_2G_c$ at $ M,N$ prove that triangle $ A_2MN$ is isosceles triangle.

[quote="encyclopedia"]Hi dear kaka__2004 of course $ B_3C_3\perp A_1A_2$ so you shoud prove $ A_1A_2$ is bisector of $ \angle MA_2N$, goodluck, I will post solution tomorrow if nobody solve.[/quote]
hi,[b]encyclopedia[/b],can you post your idea?thank you[/b]

Nice problem , encyclopedia . However , you have a mistake in typing
The problems must be :
[quote="encyclopedia"]Let $ ABC$ be triangle with orthocenter $ H$ let $ G_a,G_b,G_c$ be centroid of $ HBC,HCA,HAB$,$ HA\cap G_bG_c \equal{} A_1,$ $ HB\cap G_cG_a \equal{} B_1,$ $ HC\cap G_aG_b \equal{} C_1,$ $ G_aC_1\cap A_1B_1 \equal{} C_2,$ $ G_bA_1\cap B_1C_1 \equal{} B_2$, let $ A_2$ be projection of $ A_1$ on $ B_2C_2$, let $ B_3,C_3$ be circumcenter of $ (A_1A_2G_b),(A_1A_2G_c)$, the line $ B_3C_3$ intersect $ A_2G_b,A_2G_c$ at $ M,N$ prove that triangle $ A_2MN$ is isosceles triangle.[/quote]

Here is my solution :
Let $ E,F$ is the midpoints of $ HB,HC$. We have :
$ \frac{AG_c}{AE}\equal{}\frac{AG_b}{AF}\equal{}\frac{1}{3} \Rightarrow G_bG_c // MN$ and $ EF // BC \Rightarrow G_bG_c //BC \Rightarrow G_bG_c \perp AH$. Similarity , $ G_aG_c \perp BH,G_aG_b \perp CH$
In the other side , $ \frac{G_bA_1}{G_bA_1}.\frac{G_bC_1}{G_aC_1}.\frac{G_aB_1}{G_cB_1}\equal{}\frac{HB}{HC}.\frac{HA}{HB}.\frac{HC}{HA}\equal{}1 \Rightarrow G_aA_1,G_bB_1,G_cC_1$ are concurrent.
$ \Rightarrow (G_b,G_c,A_1,B_2)\equal{}\minus{}1 \Rightarrow (A_2G_b,A_2G_c,A_2A_1,A_2B_2)\equal{}\minus{}1$
Let $ MN \cap A_1A_2 \equal{} I$
Because of $ MN // B_2C_2$ that means $ MN//AB_2$. Following the Harmonic Devision , we have $ I$ is midpoint of $ MN$
$ \Rightarrow A_2MN$ is isosceles triangle.
Q.E.D


function
combinatorics unsolved
combinatorics

How can we construct a bijective function from (0,1) to [0,1]?
 :(

[hide="hint for a quick proof"]
The Schroder-Bernstein theorem states that if there are injections $ A\to B$ and $ B \to A$ then there exists a bijection $ A \to B$
[/hide]

[hide="more constructive"]
Let $ \lbrace q_k \rbrace_{k\in \mathbb{N}}$ be an enumeration of the rationals with $ q_1\equal{}0$ and $ q_2\equal{}1$.  Then define the function $ \phi: [0,1]\to (0,1)$,
\[ \phi(x) \equal{} \begin{cases} q_{k\plus{}2} & \text{ if } x\equal{}q_k, k\in\mathbb{N} \\ x & \text{ if } x\notin \mathbb{Q} \end{cases}\]
[/hide]

Those are very beautiful ideas.
I've never seen a construction using rational number like that before.
It's great! Thak you.  :D

a more constructive proof
[url=http://answers.yahoo.com/question/index;_ylt=Ave09BJT.PtoPisSqmZZNNQjzKIX;_ylv=3?qid=20070201104401AAjLGXT]here[/url]
I saw this problem several times, so I assume it's from a textbook
I recommend you to say what was your contribution to the problem before posting it on this forum


geometry
circumcircle
trigonometry
ratio
rectangle
analytic geometry
geometric transformation

Let $ABC$ be a triangle such that $CA \neq CB$,
the points $A_{1}$  and $B_{1}$ are tangency points for the ex-circles relative to sides $CB$ and $CA$,
 respectively, and $I$ the incircle.
 The line $CI$ intersects the cincumcircle of the triangle $ABC$ in the point $P$.
 The line that trough $P$ that is perpendicular to $CP$, intersects the line $AB$ in $Q$.
 Prove that the lines $QI$ and $A_{1}B_{1}$ are parallels.

WLOG, let a = BC < CA = b. Let $A_{1}B_{1}$ meet AB at $C_{1},$ let $K \in AB$ be foot of the angle bisector $CI \equiv CP,$ Let parallel to CK through $A_{1}$ meet AB at L and let M be the midpoint of c = AB. As usual, R is the triangle circumradius, s its semiperimeter, and $\triangle$ its area.

$\frac{IK}{CK}= \frac{c}{a+b+c},$ $\frac{A_{1}L}{CK}= \frac{s-c}{a},$ $\frac{A_{1}L}{IK}= \frac{2s(s-c)}{ca}.$

By Menelaus theorem for the $\triangle ABC$ cut by the line $A_{1}B_{1}C_{1},$ $\frac{C_{1}B}{C_{1}A}= \frac{s-a}{s-b}.$ In addition, $C_{1}B-C_{1}A = c,$ hence $C_{1}B = \frac{c(s-a)}{b-a}.$ Since $\frac{BL}{BK}= \frac{s-c}{a}$ and $BK = \frac{ca}{a+b},$ we have $BL = \frac{c(s-c)}{a+b}$ and $C_{1}L = C_{1}B-BL = \frac{bc^{2}}{b^{2}-a^{2}}.$

The right $\triangle PQK \sim \triangle MPK$ are similar, hence $\frac{PK}{QK}= \frac{MK}{PK},$ $QK = \frac{PK^{2}}{MK}= \frac{MK^{2}+MP^{2}}{MK}.$ $MK = BM-BK = \frac{c(b-a)}{2(a+b)}$ and

$MP = R(1-\cos C) = \frac{4R(s-a)(s-b)}{2ab}= \frac{c(s-a)(s-b)}{2 \triangle}$

$MP^{2}= \frac{c^{2}(s-a)(s-b)}{4s(s-c)}$

$QK = \left((b-a)^{2}+\frac{(s-a)(s-b)}{s(s-c)}(b+a)^{2}\right) \cdot \frac{c^{2}}{4(b+a)^{2}}\cdot \frac{2(b+a)}{c(b-a)}=$

$= \frac{s(s-c)(b-a)^{2}+(s-a)(s-b)(b+a)^{2}}{s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}=$

$= \frac{[(b+a)^{2}-c^{2}](b-a)^{2}+[c^{2}-(b-a)^{2}](b+a)^{2}}{4s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}=$

$= \frac{4abc^{2}}{4s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}= \frac{abc^{3}}{2s(s-c) (b^{2}-a^{2})}$

$\frac{C_{1}L}{QK}= \frac{2s(s-c)}{ca}= \frac{A_{1}L}{IK}$

The $\triangle QKI \sim \triangle C_{1}LA_{1}$ are centrally similar by SAS, hence $QI \parallel C_{1}A_{1}\equiv B_{1}A_{1}.$

[url=http://www.artofproblemsolving.com/community/c1090h1163069][color=red][b]PP[/b][/color][/url][color=darkred]. Let $\triangle ABC$ so that  $a\ne b$ with circumcircle $\alpha =C(O,R)$ and incircle $w=\mathbb C(I,r)$ .

 Let  $\left\{\begin{array}{c}
A_{1}\in BC\cap w_a\\\\
B_{1}\in AC\cap w_b\\\\\
\{C,P\}=CI\cap\alpha\end{array}\right\|$  and $Q\in AB$ so that $PQ\perp PC$ . Prove that $IQ\parallel A_{1}B_{1}$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $a>b$. Denote the diameters $[CC']$, $[PP']$ of $\alpha$ and $\left\{\begin{array}{c}K\in CI\cap AB\ ,\ S\in C'P'\cap AB\\\\ U\in CB\cap IQ\ ,\ V\in CA\cap IQ\end{array}\right\|$.

$1\blacktriangleright$ At first I"ll find the position of $Q\in AB$, i.e. the ratio $\frac{QA}{QB}$. Thus, $\left\{\begin{array}{c}P'C'\parallel CP\implies C'S\perp C'Q\\\\ P'A=P'B\implies \widehat{AC'S}\equiv\widehat{BC'S}\end{array}\right\|$ $\implies$

the division $(A,B,S,Q)$ is harmonically, i.e. $\frac{QA}{QB}=\frac{SA}{SB}=$ $\frac{C'A}{C'B}=\frac{2R\sin\widehat{ACC'}}{2R\sin\widehat{BCC'}}=$ $\frac{\sin(90^{\circ}-B)}{\sin(90^{\circ}-A)}=$ $\frac{\cos B}{\cos A}$ $\implies$

$\boxed{\frac{QA}{\cos B}=\frac{QB}{\cos A}}=\frac{c}{\cos B-\cos A}$.

$2\blacktriangleright$ At second I"ll find the positions of $U\in CB$, $V\in CA$ and I"ll value the ratio $\frac{CV}{CU}$. Observe that $\boxed{\frac{CB_{1}}{CA_{1}}=\frac{p-a}{p-b}}$

and $QK=QB+BK=\frac{c\cdot\cos A}{\cos B-\cos A}+\frac{ca}{a+b}=$ $c\cdot\frac{(a+b)\cos A+a(\cos B-\cos A)}{(a+b)(\cos B-\cos A)}=$

$c\cdot\frac{b\cos A+a\cos B}{(a+b)(\cos B-\cos A)}=$ $\frac{(b^{2}+c^{2}-a^{2})+(a^{2}+c^{2}-b^{2})}{2(a+b)(\cos B-\cos A)}$ $\implies$ $\boxed{QK=\frac{c^{2}}{(a+b)(\cos B-\cos A)}}$ $\implies$

$\left\{\begin{array}{c}\frac{QA}{QK}=\frac{c\cos B}{\cos B-\cos A}\cdot\frac{(a+b)(\cos B-\cos A)}{c^{2}}\implies \frac{QA}{QK}=\frac{(a+b)\cos B}{c}\\\\ \frac{QB}{QK}=\frac{c\cos A}{\cos B-\cos A}\cdot\frac{(a+b)(\cos B-\cos A)}{c^{2}}\implies \frac{QB}{QK}=\frac{(a+b)\cos A}{c}\end{array}\right\|$. Apply the [u][b]Menelaus[/b]' theorem[/u] to the transversal $\overline{QUIV}$

in:  $\left\{\begin{array}{cccccc}
\triangle ACK\ : & \frac{QA}{QK}\cdot\frac{IK}{IC}\cdot\frac{VC}{VA}=1 & \implies & \frac{(a+b)\cos B}{c}\cdot \frac{c}{a+b}\cdot \frac{VC}{VA}=1 & \implies & \frac{VA}{\cos B}=\frac{VC}{1}=\frac{b}{1+\cos B}\\\\
\triangle BCK\ : & \frac{QB}{QK}\cdot\frac{IK}{IC}\cdot\frac{UC}{UB}=1 & \implies & \frac{(a+b)\cos A}{c}\cdot\frac{c}{a+b}\cdot\frac{UC}{UB}=1 & \implies & \frac{UB}{\cos A}=\frac{UC}{1}=\frac{a}{1+\cos A}\end{array}\right\|$ $\implies$

$\frac{CV}{CU}=\frac{b}{a}\cdot\frac{1+\cos A}{1+\cos B}=$ $\frac{b}{a}\cdot\frac{2\cos^{2}\frac{A}{2}}{2\cos^{2}\frac{B}{2}}=$ $\frac{b}{a}\cdot\frac{p(p-a)}{bc}\cdot\frac{ac}{p(p-b)}$ $\implies$ $\boxed{\frac{CV}{CU}=\frac{p-a}{p-b}}$.

In conclusion, $\frac{CV}{CU}=\frac{CB_{1}}{CA_{1}}$ $\implies$ $\triangle CVU\sim\triangle CB_{1}A_{1}$ $\implies$ $UV\parallel A_{1}B_{1}$ $\implies$ $\boxed{IQ\parallel A_{1}B_{1}}$.[/color]

Let $(I)$ touch $BC,AC$ at $F,E$. $PQ$ meets $(O)$ at $K$
Let the perpendicular line from $I$ to $BK,AK$ meet $BK,AK$ at $M,N$
We have $IN//AC$ and $IM//BC$ so $\angle NIM=\angle ACB$
Because $NAEI, MIFB$ are rectangle so $IN=AE=CB1$ and $IM=FB=CA1$
Thus triangle $IMN$ and $CA1B1$ are congruent.
So $\angle MNI=\angle A1B1C$. But $IN//AC$ so $MN//A1B1$ $(1)$
we have $\angle PQB=\angle PAB-\angle KBA=\angle PBA-\angle KBA=\angle PBK=\angle PAK$
so triangle $PAK$ and $POA$ are similar.
Then $PK*PO=PA^{2}=PI^{2}$
Thus triangle $PKI$ and $PIQ$ are similar
then $\angle PKI=\angle PIQ$
we have $K,N,P,M,I$ are cyclic so
$\angle INM=\angle IKM=\angle IKP-\angle PKM=\angle PIQ-\angle BCP=$$\angle PIQ-\angle ACP=\angle PIQ-\angle NIP=\angle NIQ$
Thus $MN//IQ$ $(2)$
From (1) and (2) $QI//A1B1$ $q.e.d$

I'm changing the notations according to my solution.

[quote]Let $ ABC$ be a triangle such that $ AB \neq AC,$ the points $ B_{1}$  and $ C_{1}$ are tangency points for the ex-circles relative to sides $ AC$ and $ AB,$ respectively and $ I$ the incircle. The line $ AI$ intersects the cincumcircle of the triangle $ ABC$ in the point $ P.$ The line that goes trough $ P$ perpendicular to $ AP$ intersects the line $ BC$ in $ Q.$ Prove that $ QI$ and $ B_{1}C_{1}$ are parallel.[/quote]
Let us use barycentric coordinates with respect to $ \triangle ABC.$ Then

$ B_1 (s \minus{} a: 0: s \minus{} c) \ , \ C_1 (s \minus{} a: s \minus{} b: 0)$

Line $ B_1C_1$ has infinite point $ T_{\infty} ((s \minus{} a)(s \minus{} c): b(s \minus{} b): \minus{} c(s \minus{} c))$

Since $P$ is the midpoint of the arc $ BC$ of the circumcircle, then its coordinates are

$ P ( \minus{} a^2: b(b \plus{} c): c(b \plus{} c))$

Line $ \mathcal{L}_a$ passing through $ P$ and orthogonal to $ AP$ is parallel to the external bisector of $ \angle BAC,$ hence they have the same infinite point $ (c \minus{} b: b: \minus{} c).$ Its equation is then

$ \mathcal{L}_a \equiv bc(b \plus{} c)x \minus{} cS_Cy \minus{} bS_Bz \equal{} 0 \Longrightarrow Q \equiv BC \cap \mathcal{L}_a \equiv (0: bS_B: \minus{} cS_C)$

$ \Longrightarrow IQ \equiv abcx \minus{} cS_Cy \minus{} bS_Bz \equal{} 0.$ Coordinates of its infinite point $ R_{\infty}$ are:

$ R_{\infty} (cS_C \minus{} bS_B: b(S_B \plus{} ac): \minus{} c(S_C \plus{} ab))$

Now, using the identities

$ cS_C \minus{} bS_B \equal{} 2s(s \minus{} a)(b \minus{} c) \ , \ S_B \plus{} ac \equal{} 2s(s \minus{} b) \ , \ S_C \plus{} ab \equal{} 2s(s \minus{} c).$

$R_{\infty} ((s \minus{} a)(b \minus{} c): b(s \minus{} b): \minus{} c(s \minus{} c)) \Longrightarrow T_{\infty} \equiv R_{\infty} \Longrightarrow IQ \parallel B_1C_1$

let $ \{L,C \}\equal{}CO \cap (O)$ and obviously we have $ PQ \cap (O)\equal{} \{P,L \}$let $ \{B',B \}\equal{}BI \cap (O)$ and define similarly $ A'$.Let $ S_1\equal{}LA' \cap BC$ and $ S_2\equal{}LB' \cap AC$.by pascal's theorem for $ ACBB'LA'$ and $ BCPLA'A$ we get the collinearity of $ S_1,S_2,I,Q$.it remains to prove that $ S_1S_2 \parallel{} A_1B_1$.so we need to compute $ CS_1,CS_2$ for proving $ \frac{CS_1}{CS_2}\equal{}\frac{p\minus{}b}{p\minus{}a}$.but we have $ \frac{CS_1}{BC}\equal{}\frac{CL}{BL\plus{}LC}\equal{}\frac{1}{1\plus{}\cos \angle A}$ and similarly for $ \frac{CS_2}{AC}\equal{}\frac{1}{1\plus{}\cos \angle B}$.now we must prove $ \frac{1\plus{}\cos \angle B}{1\plus{}\cos \angle A}\equal{}\frac{(p\minus{}b)b}{(p\minus{}a)a}$ which is obvious since $ \cos \angle B\equal{} \frac{a^2\plus{}c^2\minus{}b^2}{2ac}$ and similarly for $ \cos \angle A$.

My solution:

Let $ A_2, B_2 $ be the projection of $ I $ on $ CB, CA $ .
Let $ C'=PQ \cap \odot (ABC) $ and $ A', B' $ be the projection of $ I $ on $ C'A, C'B $, respectively .

Since $ IA'=B_2A=CB_1, IB'=A_2B=CA_1 $ and $ IA' \parallel CA, IB' \parallel CB $ ,
so $ \triangle IA'B' $ and $ \triangle CB_1A_1 $ are homothety (also congruent) .
Since $ I, A', B', C', P $ are concyclic at $ \odot (IC') $ ,
so $ \angle PQA= \angle PAB- \angle APQ =\angle ACP -\angle ACC'=\angle C'AP $ ,
hence we get $ \triangle PAC' \sim \triangle PQA $ and $ PI^2=PA^2=PC' \cdot PQ $ .
i.e. $ \triangle PIC' \sim \triangle PQI $
Since $ \angle B'A'I=\angle B'C'I=\angle PC'I -\angle PC'B'=\angle QIP- \angle A'IP=\angle QIA' $ ,
so we get $ QI \parallel A'B' $ . i.e. $ QI \parallel A_1B_1 $

Q.E.D

Dear Mathlinkers,
see
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=576791942&t=189585

Sincerely
Jean-Louis

I found another simpler proof by using [b]Reim theorem[/b] :)

My solution:

Let $ Z $ be the tangent point of $ \odot (I) $ with $ AB $ .
Let a line passing through $ A $ and parallel to $ A_1B_1 $ intersect $ CP, CB $ at $ X, Y $, respectively .

From $ \angle IPQ=\angle IZQ=90^{\circ} \Longrightarrow I, P, Q, Z $ are concyclic . ... $ (1) $

From $ \frac{AX}{XY}=\frac{CA}{CY}=\frac{CB_1}{CA_1}=\frac{AZ}{ZB} \Longrightarrow XZ \parallel CB \Longrightarrow A, P, X, Z $ are concyclic ([b]Reim theorem[/b]). ... $ (2) $

Combine with $ (1), (2) \Longrightarrow IQ \parallel XA $ ([b]Reim theorem[/b]). i.e. $ IQ \parallel A_1B_1 $

Q.E.D

Dear,
very nice proof with my favorite theorem....which make some proof very elegant in my mind.
Sincerely
Jean-Louis

Let $QI$ meet $CB,CA$ at $A_2,B_2$ respectively, incircle touches $AB$ at $Z$.

WLOG assume $CA>CB$. Note $I,Z,P,Q$ concyclic, so $\angle APZ=\angle APC + \angle IPZ= \angle ABC + \angle IQZ=\angle CA_2B_2$, similarly $\angle BPZ=\angle CB_2A_2$.

Thus $\frac{CA_1}{CB_1}=\frac{BZ}{AZ}=\frac{\sin \angle BPZ}{\sin \angle APZ}=\frac{\sin \angle CB_2A_2}{\sin \angle CA_2B_2}=\frac{CA_2}{CB_2}$, so $A_1B_1\parallel A_2B_2\equiv QI$.

Let $R$ be the intersection point of $PQ$ with the circumcircle of $\triangle ABC$. Let $M$ be the midpoint of $RI$. We want to say that $QI$ and $A_{1}B_{1}$ are parallels. $PB^2=PI^2=PR.PQ$, so using the inversion with the center $P$ and the radius $PB$, we have to prove that $PM$ is perpendicular to $A_{1}B_{1}$. Because the line $QI$, goes to the circumcircle of $\triangle PRI$ and $\angle RPI=90$, so $M$ is the center of the circumcircle of $\triangle PRI$. Let $T$ be the midpoint of the arc $ACB$ of the circumcircle of $\triangle ABC$. And let $O$ be its circumcenter. Let $L$ and $N$ be the reflection points of $I$ and $M$ trough $O$. The reflection of $R$ and $P$ trough $O$ are $C$ and $T$. So it is enough to prove that $TN$ is perpendicular to $A_{1}B_{1}$. $N$ is the midpoint of $CL$. $L$ is the circumcenter of the circumcircle of the excenters of $\triangle ABC$. So $LA_{1}$ and $LB_{1}$ are perpendicular to $AC$ and $BC$. So $LB_{1}CA_{1}$ is concyclic with the center $N$. $AA_{1}=BB_{1}$, so $TB_{1}=TA_{1}$ and $CTB_{1}A_{1}$ is concyclic. So $CTB_{1}LA_{1}$ is concyclic with the center $N$. So $TN$ is the perpendicular bisector of $A_{1}B_{1}$. So $TN$ is perpendicular to $A_{1}B_{1}$.

my solution (L.B.T) : X = CO cut (O)) , (I) contact AB = D , PD cut (O) = Y . X,I,Y collinear (easy) . YA/YB=AD/BD=CB1/CA1 ==> YBA=CA1B1
A1B1 cut AB = Z . A1ZA = B-CA1B1 = B-YBA=CBY=IPD=IQD


linear algebra
matrix
linear algebra unsolved

Prove that $ \exists X,Y,Z\in \mathcal{M}_n(\mathbb{C})$ such that
 a)$ X^2\plus{}Y^2\equal{}A$
 b) $ X^3\plus{}Y^3\plus{}Z^3\equal{}A$  , where $ A\in \mathcal{M}_n(\mathbb{C})$

Something much stronger: for any $ p,q$ and any $ A$ one can find $ B,C$ such that $ A\equal{}B^p\plus{}C^q$. Even more, $ B$ can be taken scalar. Indeed, just take a nonzero $ x$ such that $ A\minus{}x^pI_n$ is invertible and use that the exponential is surjective from $ M_n(C)$ to $ GL(n,C)$.
Of course, I did use some nontrivial facts, but I think it gives more insight than some boring choices of matrices. And it shows that Waring's problem is not so interesting in $ M_n(C)$. Now, let's make the problem more interesting: replace $ C$ by $ Q$ and $ Z$!

Hi svejk,
    write correctly yours questions; I spent time in view of this problem: find $ X,Y,Z$ s.t. a) AND b).
  As Harazi said, the problem reduced to a) or to  b) is straightforward.

Hi [b]loup blanc[/b] ! You are right , the problem might be confusing the way I wrote it.I believe that you considered them simultanioulsy.I'm sorry for this.Thank you for your interest.

[quote="harazi"]...Now, let's make the problem more interesting: replace $ C$ by $ Q$ and $ Z$![/quote]
If we take $ \mathbb{Q}$, we can't say anymore that "for any $ p,q$ and any $ A$ one can find $ B,C$ such that $ A \equal{} B^p \plus{} C^q$.".

Take for instance $ A$ nilpotent and $ p \equal{} 0$.


What mathematical fact is in the upper left corner of the IMO 2004 Greece emblem? Thank you.


linear algebra
matrix
combinatorics unsolved
combinatorics

An increasing sequence of positive integers is said to be "alternating" if the first term starts with an odd number and second term is an even number and third number is odd and so on........
A sequence containing no terms is defined to be alternating . If $ A(n)$ denotes the number of alternating sequences involving the integers from the set $ \{1,2,3........n \minus{} 1,n\}$ alone .
For example ,when $ n \equal{} 1$ ,the set to be considered is $ \{1\}$ the possible alternating sequences are $ \{  \} , \{1 \}$ ,thus $ A(1) \equal{} 2$ ,similarly $ A(2) \equal{} 3$
Find the value of $ A(20)$

Can we establish a recursive relation between the $ A(n)$'s ..?

Track two subsets: $ B(n)$ is the number of alternating sequences ending with something even, while $ C(n)$ is the number ending with something odd. The empty sequence counts as ending with something even, since the next entry must be odd.

Initialize with $ B(0)\equal{}1$, $ C(0)\equal{}0$. When we add $ n\plus{}1$ to the set, we can append it to a sequence or not. If $ n$ is even, $ B(n\plus{}1)\equal{}B(n)$ and $ C(n\plus{}1)\equal{}B(n)\plus{}C(n)$. If $ n$ is odd, $ B(n\plus{}1)\equal{}B(n)\plus{}C(n)$ and $ C(n\plus{}1)\equal{}C(n)$.
Start at something even and compose two steps: $ B(2k\plus{}2)\equal{}B(2k\plus{}1)\plus{}C(2k\plus{}1)\equal{}B(2k)\plus{}B(2k)\plus{}C(2k)$ and $ C(2k\plus{}2)\equal{}C(2k\plus{}1)\equal{}B(2k)\plus{}C(2k)$. In matrix form, $ \begin{pmatrix}B\\C\end{pmatrix}_{2k\plus{}2}\equal{}\begin{pmatrix}2&1\\1&1\end{pmatrix}\cdot \begin{pmatrix}B\\C\end{pmatrix}_{2k}$
To go from zero to $ 20$, raise that matrix $ M\equal{}\begin{pmatrix}2&1\\1&1\end{pmatrix}$ to the 10th power. To make that easier, diagonalize $ M$. Its trace is $ 3$ and its determinant is $ 1$, for eigenvalues of $ \frac{3\pm\sqrt{5}}{2}$. Hmm- that looks Fibonacci-ish; in particular, those are the squares of $ \frac{1\pm\sqrt{5}}{2}$. Can we shortcut the whole thing?

If $ n$ is even, $ C(n\plus{}1)\equal{}B(n)\plus{}C(n)\equal{}B(n\minus{}1)\plus{}C(n\minus{}1)\plus{}C(n)$. That gives $ [B(n\plus{}1)\plus{}C(n\plus{}1)]\equal{}[B(n)\plus{}C(n)]\plus{}[B(n\minus{}1)\plus{}C(n\minus{}1)]$. The same applies to the other parity, with the roles of $ B$ and $ C$ interchanged.
OK, now we start with $ A(0)\equal{}1$, $ A(1)\equal{}2$, and $ A(n\plus{}1)\equal{}A(n)\plus{}A(n\minus{}1)$. This gives $ A(n)\equal{}F_{n\plus{}2}$, so $ A(20)$ is the $ 22$nd Fibonacci number $ 17711$.

you can prove the recursion more directly as follows. (as in jmerry's solution, we assume the empty sequence ends in an even number.)

the alternating sequences of type $ n$ are of two types: those using $ n$, and those of type-$ n\minus{}1$. the latter type obviously number $ A(n\minus{}1)$. so we are left with proving that the first type number $ A(n\minus{}2)$. an alternating sequence using $ n$ must obviously end in $ n$. now, to each alternating sequence $ S$ of type $ n\minus{}2$, associate an alternating sequence ending in $ n$ as follows: if $ S$ ends in a number of the opposite parity as $ n$ (i.e., $ n\minus{}3, n\minus{}5, ...$), append $ n$ to it, if $ S$ ends in a number of the same parity as $ n$ (i.e., $ n\minus{}2, n\minus{}4, ...$), append $ n\minus{}1, n$ to it. we've got a bijection, and that does it.


calculus
integration
analytic geometry
symmetry
calculus computations

Why is $ \int\int_S(x^2\plus{}y^2\minus{}2z^2)dA\equal{}0$ for $ S$ being the unit sphere.

I can explicitly show that the surface integral is indeed zero by spherical coordinates, but what im wondering is what is the symmetry argument?

$ \int_S x^2\,dA\equal{}\int_S y^2\,dA\equal{}\int_S z^2\,dA\equal{}C$, by rotational symmetry of the sphere. The integral is $ C\plus{}C\minus{}2C\equal{}0$.

And the easiest way to find $ C$ (should you want to) is again by symmetry: $ 3C\equal{}\iint_S(x^2\plus{}y^2\plus{}z^2)\,dS\equal{}\iint_S 1\,dS\equal{}4\pi$, hence $ C\equal{}\frac{4\pi}{3}$.

[quote="jmerry"]$ \int_S x^2\,dA \equal{} \int_S y^2\,dA \equal{} \int_S z^2\,dA \equal{} C$, by rotational symmetry of the sphere. The integral is $ C \plus{} C \minus{} 2C \equal{} 0$.[/quote]

more details?

[quote="mlok"][quote="bookworm_vn"]more details?[/quote]

Why?[/quote]

Aha  :blush: just for completely.


FTW AMC USA(J)MO USAMO inequalities MATHCOUNTS AIME

i decided to make a tournament.

Rules:

There will be four teams. 8 people will sign up for each section.

Questions will be posed. They will each be worth a certain amount of points.

Question difficulty will vary.

Obviously, Harder questions will be worth more points.

If anyone wishes to help me make questions rather then participate, feel welcome to say so.

This is essentially a relay. You will send your answer to the next player on your team.

You cannot discuss this with anyone.

Nothing may be sent or communicated in any way, besides the answer of course.

Order will be determined randomly.

The problems will get harder as the go along.

The last person will send his/her answer to me, and I will post their time. After all teams have finished, the 2 slowest teams will be eliminated.

Round 2 will then begin. Same rules.

The slowest team is then eliminated.

The eight members on the top team will then compete. I will send a sprint (with my own questions) to each of them. The top 2 will join me on FTW for a CD (with my questions, not FTW) at a time they designate, in a passworded game. The password will be sent to these two.

The winner will then face me in a bset-of-3 FTW CD. Obviously, I'll lose, but it makes the winner happier.

Signups begin now.

I'll make problems.

Warning: I can make mc problems, but I am much better at making USAMO problems...

The problems will be hard. #6~8 should be just below USAMO level.

Most problems should be proofs. I'm making #1,2 which should be a little harder then MC level. Maybe mid AMC 10

[b]i join[/b]

do we pick our own teams?

if so

request bugi, dafjbomb, nikeballa96, lotsofmath, izzy, xpmath, discrete math.

[quote="ReseesPieces"]i wanna play.....but how the crap do the rules work  :maybe:  i didnt get that[/quote]

What don't you understand?

I'm in.

Request for team in no particular order: Bubbles(AIME15), Nikey(nikeballa96), Snowbunny(mz94), Fanfan(FantasyLover), Peanuts(xpmath), Fifi(58...x), dafjbomb.

I'm in. Yay proofs!
I request Izzy, xpmath, AIME15, dafjbomb, Fanfan

Team names can be changed.

[b] Team A [/b]

[b] Team B [/b]

[b] Team C [/b]

[b] Team D [/b]

Choose your team and post it. partial teams can be posted. You can't join a team that has people on it unless the people on it agree.

[b] SIGNUPS CLOSE ON FEBUARY 5TH [/b].

plenty of time.

I'm going to change the rules a little. You sned your answer or solution to me. I then send the problem to the next person. If I'm not on, and someone submits to me earlier then another, I'll send the problem a few minutes earlier to the faster team.

This allows for proof problems.

Oh my god it is a tournament by BOGTRO

I will join

Well don't just use random.org Again it ca be unfair.
Maybe seeding or something?

Definition of fair:

If a team has at least 2 people who are capable of getting at least 10 on the 2008 USAMO, that team is unfair.

Only applies to mc, does not apply to mop or usamo things.

[quote="isabella2296"]I have a question. What makes a team unfair? If the people are too good?

Oh hey, I think Bobby Shen's MathCounts team is too smart. That's unfair! That team isn't allowed to participate! /sarcasm

The purpose of teams, it seems to me, is a group of people who want to work together and win, not to be "fair".[/quote]

I swear someone's posted that before...

While izzy makes a oint, I am not trying to make this like MC, I am trying to give everyone an equal chance.

So random.org may be used, I might keep teams as they are, or I might give a quick test and determine teams from that.

Anyway, if 20 people don't sign up by the 10th, this is either going to be an individual or a small team event. You guys can choose.

BTW, happy birthday athunder!!

[quote="isabella2296"]I have a question. What makes a team unfair? If the people are too good?

Oh hey, I think Bobby Shen's MathCounts team is too smart. That's unfair! That team isn't allowed to participate! /sarcasm

The purpose of teams, it seems to me, is a group of people who want to work together and win, not to be "fair".[/quote]

I thought this isn't MC, just a online tournament :wink: In real MC, you can't make teams equal, but  here you can make equal teams, and it's better. If you are in the best team, it would be boring, and if you are in worse, you could say immediately that you lost probably.

How was I supposed to know when your birthday was?  :P
Happy late birthday then.

Bleh we should start soon.

yeah, seriously, its not like you told us. happy late birthday.

i agree, lets start. and we choose our teams.

All right.

Lets begin!

[b]FINAL Signups (14):
007math
ReseesPieces
AIME15
Twin Prime Conjecture
isabella2296
Bugi
Fantasylover
nikeballa96
athunder
dafjbomb
infinitypi
mz94
Poincare
xpmath[/b]

----------Round 1---------

This is a round to decide teams. I will post questions one by one. If it takes you n hours, to the nearest hour, you get 36-n points. You have a maximum of 48 hours.

I will then group teams accordingly.

Round 1, Question 1:

Define a number n to be "Awesome" if its digits alternate higher, lower, higher, etc. For example, 2879 is a "Awesome number". Compute the number of 5 digit "awesome" numbers.

Solutions are required. You get bonus points based on your solution. For example, for this sample question:

x and y are 2 digit integers such that the reversed digits of x is equal to y^2. Find all such pairs of numbers.

5 point solution: Let $ x\equal{}10a\plus{}b$. Thus, the number reversed is $ 10b\plus{}a$. The maximum value is $ 90\plus{}9\equal{}99$, since a and b are digits. Thus, the number is always 1 or two digit.

Consider y. If y is 2-digit, then $ y \geq 10 \implies y^2 \geq 100$. Thus, there are no such pairs.

0 point solution: Brute forcing all numbers x, there are no pairs.

partial credit for that particular problem will be unlikely. Generally, if your solution is perfect, you get 5 points. If your solution has a small flaw, 4 points. If there is a flaw that gets you the wrong answer, 2 or 3 points.

Good luck to everyone.

Ok. I've recieved two solutions. This fails.

I'm throwing out that problem. If more people don't submit this, this tournament will have to be canceled.

THIS SHOULDN'T TAKE VERY LONG

Show that for positive a,b,c, the following inequality holds

$ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ac$. Determine the equality condition.

TIME BEGINS NOW.

Sigh... this thing is canceled. Complete lack of activity.

It might help if we got reminder pms. :maybe:

Also, people live in different time zones and have different schools. Someone who had school off or was home schooled on the day you posted the problem and was on would have a much greater advantage than someone who had school that day and thus couldn't come on.