That is funny :) I'm not sure why it happens  :maybe:

It's because the LaTeX syntax is a mess. \alpha is a math mode command so shouldn't be in \text. What Latex does is to try and sort out the problem by adding dollar signs and then the renderer gets all confused with the dollar signs.
For an interesting view try it in a LaTeX document and you get the following errors:
[code]! Missing $ inserted.
<inserted text>
                $
l.6 \text{ Why does \alpha repeat?}

I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Try loading font information for U+msa on input line 6.
(C:\texmf\tex\latex\amsfonts\umsa.fd
File: umsa.fd 2002/01/19 v2.2g AMS font definitions
)
LaTeX Font Info:    Try loading font information for U+msb on input line 6.
 (C:\texmf\tex\latex\amsfonts\umsb.fd
File: umsb.fd 2002/01/19 v2.2g AMS font definitions
)
! Extra }, or forgotten $.
\mbox #1->\leavevmode \hbox {#1}

l.6 \text{ Why does \alpha repeat?}

I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

! Missing $ inserted.
<inserted text>
                $
l.7 \end{document}

I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

! Missing } inserted.
<inserted text>
                }
l.7 \end{document}

I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.[/code]You are lucky that anything at all has come out at the end!

The moral of the story is not to use math mode inside text mode and \text should be used very sparingly in AoPS.

Because it is rendered as image, thus causes longer loading times, more work for the server, et cetera (and in my oppinion, often looks awful). Math mode is for math, not for epic texts.

[quote="fedja"]Yes, stevem, as I wrote, I understood perfectly well that $ \text{\LaTeX}$ was garbled but the normal reaction to that should be "unparseable or dangerous LaTeX formula" (or whatever that standard phrase is), not repeating the text 3 extra times in diminishing fonts. The point was not how to write the phrase correctly but why the AoPS parser or whatever reacts in such peculiar way and whether this may be an indication of some bug :).[/quote]
The problem is with LaTeX not AoPS. LaTeX tries to correct the syntax and gets it wrong but does produce an image which AoPS produces. The "Unparseable or potentially dangerous latex formula" error is when LaTeX itself cannot compile the formula.

So what you are asking for is for AoPS to second-guess LaTeX and say when it thinks it got it wrong and I don't think that's practicable. For example, at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217436 there was a \right) missing which LaTeX compiled slightly wrongly and AoPS faithfully reproduced the incorrect result.

In the text-mode case, AoPS could check for math mode commands inside text boxes, but there are likely to be zillions more such syntax errors which I think is best left to LaTeX to deal with imperfectly.

What might be a practical possibility is to enhance the TeXer to show the errors in the LaTeX log which would help to find the problem. If not, then the user could use their own computer or one of the 4 online document compilers listed at [url=http://sixthform.info/steve/wordpress/?p=59]Online LaTeX[/url] in the second list there.

[quote="ZetaX"]That's the same answers I always get if I report some bug caused by wrong code: a couple of people try to tell you that the code is the problem. Well, if the code is always the problem, why do we have error messages at all¿ ;)
There may be some more severe bugs caused by the same, or at least unexpected behaviour, thus the bug should be observed nonetheless. The code doesn't do as it should (either producing something expected or giving an error message), so somethings wrong on the TeXer.

I sometimes felt a bit like I need to defense why I submitted that bug instead of just being silent...[/quote]
I hope I explained this above. It is a problem with the LaTeX compiler - it does produce error messages but these aren't shown - they are rather verbose as you see above so it's not practical to put them in a forum though they could be added to the TeXer.

Basically all AoPS does is put the formula into a LaTeX document, compile it and cut the image out and put it on the forum. It leaves the syntax checking to LaTeX which, as I say, is far from perfect. Compiling it on your own computer or another online service which shows the log may reveal the problem, although sometimes the log obscures rather than clarifies.

[quote="#H34N1"]Wait why sould we use text sparingly?[/quote]
I would add that you should write just as you would in a LaTeX document. So in a document you would say:
[code]This is text but this is maths $x=1$ followed by lots more text[/code]and you wouldn't use \text there so no need to do so in the forum.

Where \text is useful, in both documents and the forum, is in things like
$ |x|\equal{}\begin{cases}
x & \text{if } x \ge 0 \\
\minus{}x & \text{if } x < 0
\end{cases}$
where the 'if' should be left as text and not as i times f.

[quote="stevem"][...] Compiling it on your own computer or another online service which shows the log may reveal the problem, although sometimes the log obscures rather than clarifies.[/quote]
I think you didn't get my point. Fedja and me knew very well what and where the error is. The problem is that always, if one of us reports some non-intended behaviour of the TeXer, most answers only tell us that we have to expect this because our code is wrong at position x. I don't think we have to expect completely unreasonable behaviour. And I send that reports because I cannot exclude that there is correct code that gets affected, too (Valentin has changed a lot of things and added own stuff, etc...), or that one can use this to destroy the forum's formating (yes this happened at I know at least one TeX that does).

Actually I think it's my point that you don't seem to get. I agree that if the forum's formatting is upset that does need to be dealt with, but minor stuff like the code repeating is it's own error report. It isn't difficult to make a complete mess of LaTeX code but I don't think it's AoPS's job to sort it out for you, beyond what LaTeX itself does. If it were possible for AoPS to correct every syntax error then that would allow people to get into bad habits and write code that won't work in documents.

On the other hand, there's a balance to be met; XHTML for example insists on exact code and if there's even a tiny error it completely refuses to parse the code whereas HTML is a bit more forgiving. I wouldn't want the strictness of XHTML but equally I wouldn't want an anything goes system.

Putting it another way, getting strange results is a good way to learn the correct syntax. The repeating example is a good way to learn all about LaTeX's two input modes.

I would totally agree with stevem if the standard $ \text{\LaTeX}$ run in batch mode would produce the same image as the one that appeared on AoPS but it does not: no matter how hard I try, it just gives 42 errors and the empty page as an output. So this particular glitch does not come from the standard way  $ \text{\LaTeX}$ handles the input but rather from handling TeX code on AoPS. I have absolutely no idea whether such a bug is dangerous for anything or not; I'm just in habit to report anything strange I see in the site's handling the code. To figure out whether it really requires any action and what exactly should be done about it is primarily Valentin's business, not mine: computer security is not my cup of tea, so to speak. I completely agree that there are much more ways to set up an erratic $ \text{\LaTeX}$ formula than one can possibly take care of. As long as the results of the errors are predictable and/or agree with what you see on your home computer, there is no point in commenting on them, indeed. But sometimes (like in this case) the outcome is completely unexpected. Then I just hand it over to better experts than I to look at. My point was that stevem's post with full TeX log only explains why [b]something other[/b] than the sentence $ \text{Why does }\alpha\text{ repeat?}$ may appear (which I and ZetaX understand ourselves) but it fails to explain why [b]exactly this[/b] effect occurs. Pehaps, I'm a bit paranoid about strange effects in how $ \text{\LaTeX}$ and Asymptote are handled on the site, but to ignore a stupid report when it is stupid is quite easy and to notice a bad bug without such stupid reports may be much harder. That's basically all I wanted to say :). I apologize for this fancy error report's turning into a long argument :blush:

[quote="fedja"]$ \text{ Why does \alpha repeat?}$
I know that it is erratic $ \text{\LaTeX}$ but why such a strange reaction? :?[/quote]
I have finally nailed this problem. It took me a while to solve as it is only a problem that occurs in \displaystyle (which AoPS uses by default). If tried in text mode you get no output at all. Using [code]$\displaystyle \text{ Why does \alpha repeat?}$ or \[\text{ Why does \alpha repeat?}\][/code] in a document will give the same strange output.

The problem doesn't occur with \mbox where you get: $ \mbox{ Why does \alpha repeat?}$ but only with amsmath's implementation of \text. If you look at the error log there seem to be a number of font changes. This is because \text uses the TeX command \mathchoice. This is where the repeats occur - I'll quote Knuth's TeXbook (page 151) to explain (my bold text):
[quote]TeX has a special feature called \mathchoice that comes to the rescue in such situations: You write

\mathchoice{<math>}{<math>}{<math>}{<math>}

where each <math> specifies a subformula. TeX will choose the first subformula in style D or D′, the second in style T or T′, the third in style S or S′, the fourth in style SS or SS′. ([b]TeX actually typesets all four subformulas, before it chooses the final one[/b], because the actual style is not always known at the time a \mathchoice is encountered; for example, when you type ‘\over’ you often change the style of everything that has occurred earlier in the formula.[/quote]
([i]D,D' = displaystyle, T,T' = textstyle, S,S' = scriptstyle, SS,SS' = scriptscriptstyle; D',T',S',SS' are cramped versions which don't raise exponents so much but are otherwise the same[/i])

It's clear then that TeX never makes a choice due to the errors so leaves the 4 styles there, which explains the different font sizes. Thus it is not the fault of AoPS but the underlying TeX engine.

[quote="stevem"]It's clear then that TeX never makes a choice due to the errors so leaves the 4 styles there, which explains the different font sizes. Thus it is not the fault of AoPS but the underlying TeX engine.[/quote]The defense rests its case, your Honor!


Prove that the square of every odd number leaves a remainder of 1 when divided by 8.

The odd number can be written as $(2n-1)$.

Therefore, $(2n-1)^2=4n^2-4n+1 \equiv 1(\mod{4})$. However, since there are two terms $\equiv 0(\mod{4})$, then they are also $\equiv 0(\mod{8})$. The $1$ left over is obviously $\equiv 1(\mod{8})$, so all odd squares are $\equiv 1(\mod{8})$

yeah. also, to show that the two multiples of $4$ add to give a multiple of $8$ you could factor the sum as $4n(n+1)$ and you know that one of $n$ and $n+1$ is a multiple of two, which tells you that the expression is a multiple of $8$.

[quote="JBL"][quote="Iversonfan2005"]since there are two terms $\equiv 0(\mod{4})$, then they are also $\equiv 0(\mod{8})$.[/quote]

This is wrong -- consider the sum 4 + 8.  You need to make the extra step drunner took (or casework).[/quote]

oh sorry, I should've mentioned that :blush:


VALENTINO:

I dont know that you agree or not but this question is not at the level of the other $IMO$ problems.I know that the first problem should be easier than the other but not this much .Do you agree :?  :?  :? ?

after cumminating with darij

well ,it is was not so easy at all as darij's word so i take back my word :)

Darij, have a good exam tommorow

[quote="lomos_lupin"]VALENTINO:

I dont know that you agree or not but this question is not at the level of the other  problems.I know that the first problem should be easier than the other but not this much .Do you agree    ?

after cumminating with darij

well ,it is was not so easy at all as darij's word so i take back my word

Darij, have a good exam tommorow[/quote]
Hi dear tavakolli.
I don't agree with U I think 2 geometry problem were very very easy.I solved both of them in 10 minutes.
I think this IMO were easy .
The problem 6 most hard problem take 1 minute to solve.


function
linear algebra
matrix
induction
combinatorics proposed
combinatorics

Let $ A$ be a set and $ A_1, \ldots, A_n \subset A$ be sets of size $ n$. Prove the existence of injections $ f_i: \{1, \ldots, n\} \rightarrow A_i$ ($ i \equal{} 1, \ldots, n$) such that the functions $ g_j: \{1, \ldots, n\} \rightarrow A$ ($ j \equal{} 1, \ldots, n$) defined by $ g_j(i) \equal{} f_i(j)$ are also injective.

The requirement is tantamount to building a $ n \times n$ matrix, with the entries of row $ R_i$ made by the (distinct) elements of $ A_i$, such that the columns $ C_j$ are also each made by distinct entries. The base case $ n\equal{}1$ of induction by $ n$ is trivial. Then, pick distinct elements $ a_i \in A_i$ (clearly possible, since after having picked elements $ a_1,a_2,\ldots,a_k$, with $ k<n$, there is always available some $ a_{k\plus{}1} \in A_{k\plus{}1} \setminus \{a_1,a_2,\ldots,a_k\}$).
Now use the induction hypothesis for the sets $ A_i \setminus \{a_i\}$, $ 1 \leq i \leq n\minus{}1$ in order to build a $ (n\minus{}1) \times (n\minus{}1)$ matrix, and build column $ C_n$ with the elements $ a_1,a_2,\ldots,a_n$ (in this order), and row $ R_n$ with the elements of $ A_n$ (in some order). This creates a $ n \times n$ matrix where the only way a column may contain duplicate entries ([i]bad[/i]} column), is for its entry on row $ R_n$ to be duplicated on some other row.

Moreover, from all the $ n \times n$ matrices built this way, let us select one with [b]minimal[/b] number of bad columns. If none, we are done, so assume there is at least one bad column $ C$, with entry $ a$ on row $ R_n$ duplicated on some other row. Then there must exist some other column $ C'$ with no entry equal to $ a$ (since $ a$ can appear at most $ n$ times in the matrix). Let us define $ \varphi : C \to C'$ by $ \varphi(c_i) \equal{} c'_i$ (each entry in $ C$ is mapped onto the corresponding entry in $ C'$, situated on the same row). Define an [i]alternating path[/i] between the two columns as being a sequence of entries $ c_1\varphi(c_1)c_2\varphi(c_2)\ldots c_k\varphi(c_k)$, with $ c_i$ entries in $ C$, and its [i]swap[/i] as being the replacing of entries $ c_i$ with $ \varphi(c_i)$ ande vice-versa. By the above, $ \varphi(a) \neq a$. The swap of the alternating path $ a\varphi(a)$ makes $ C'$ a good column, so $ C'$ must have been good before the swap (otherwise we contradict the minimality of the number of bad columns). Now we must assume $ \varphi(a)$ exists as an entry in $ C$, otherwise $ C$ becomes good. Notice that $ \varphi^2(a) \neq a$, since $ a$ is not an entry in $ C'$, and $ \varphi^2(a) \neq \varphi(a)$, since both are entries in $ C'$, which is a good column. Consider the swap of the alternating path $ a\varphi(a)\varphi(a)\varphi^2(a)$; it also leaves the column $ C'$ good, so we must assume $ \varphi^2(a)$ exists as an entry in $ C$, otherwise $ C$ becomes good.

Iterating this procedure obliges $ \varphi^{k\minus{}1}(a)$ to be an  entry in $ C$, and moreover $ \varphi^{k}(a) \neq \varphi^j(a)$, for all $ k > j \geq 0$, otherwise we can operate the appropriate swap and turn $ C$ into a good column, while leaving the column $ C'$ good. But $ \varphi^{n\minus{}1}(a)$ cannot be an entry in $ C$, since there is  no more room left ($ a$ appears twice as an entry in $ C$), therefore the procedure must stop for some $ k \leq n\minus{}1$, and the corresponding swap turns $ C$ into a good column, while leaving the column $ C'$ good, contradicting the minimality invoked at the start.


Let $ a$, $ b$, $ c$, $ x$, $ y$, and $ z$ be real numbers that satisfy the three equations
\begin{align*}
  13x +  by +  cz &= 0 \\
   ax + 23y +  cz &= 0 \\
   ax +  by + 42z &= 0.
\end{align*}Suppose that $ a \ne 13$ and $ x \ne 0$.  What is the value of
\[ \frac{13}{a - 13} + \frac{23}{b - 23} + \frac{42}{c - 42} \, ?\]

[hide]Label the equations (1), (2), (3).
By (2)-(1), we have $ (a\minus{}13)x\plus{}(23\minus{}b)y\equal{}0 \iff (a\minus{}13)x\equal{}(b\minus{}23)y$, and by (3)-(2), we have $ (b\minus{}23)y\plus{}(42\minus{}c)z\equal{}0 \iff (b\minus{}23)y\equal{}(c\minus{}42)z$. Thus, we have $ (a\minus{}13)x\equal{}(b\minus{}23)y\equal{}(c\minus{}42)z\equal{}k$. Since $ a\neq 13, x\neq 0$, we have $ k\neq 0$.

Now, adding $ (a\minus{}13)x$ to (1) tells us that $ ax\plus{}by\plus{}cz\equal{}k$. From this, we have $ ax\plus{}by\plus{}cz\equal{}(a\minus{}13)x \iff \frac{1}{a\minus{}13}\equal{}\frac{x}{ax\plus{}by\plus{}cz}$, or $ \frac{13}{a\minus{}13}\equal{}\frac{13x}{ax\plus{}by\plus{}cz}$. Similarly, we have $ \frac{23}{b\minus{}23}\equal{}\frac{23y}{ax\plus{}by\plus{}cz}, \frac{42}{c\minus{}42}\equal{}\frac{42z}{ax\plus{}by\plus{}cz}$. Thus, our desired sum is equal to $ \frac{13x\plus{}23y\plus{}42z}{ax\plus{}by\plus{}cz}$.

To compute this, we add (1),(2),(3) to obtain $ 13x\plus{}23y\plus{}42z\plus{}2(ax\plus{}by\plus{}cz)\equal{}0 \iff 13x\plus{}23y\plus{}42z\equal{}\minus{}2(ax\plus{}by\plus{}cz)$. Since $ ax\plus{}by\plus{}cz$ is nonzero, this means that $ \frac{13x\plus{}23y\plus{}42z}{ax\plus{}by\plus{}cz}\equal{}\boxed{\minus{}2}$.[/hide]

Very nice!

Basically the assumption of $ x\ne 0$ says that this system has a nontrivial solution in (x,y,z) so the det. of the coeffecient matrix is zero.

Now the determinant of the following is zero

13 b c
a 23 c
a b 42

we can do operations such as adding rows or dividing out a number from a column without changing the problem...

13-a  0  c-42
0  23-b   c-42
a   b    42

1  0   1
0  1   1
a/(13-a)  b/(23-b) 42/(c-42)

Using $ A\equal{}\frac{13}{a\minus{}13}$, etc

1  0 1
0 1 1
-A-1   -B-1   C

A+B+C+2=0 so the answer is $ \boxed{\minus{}2}$.

Nothing new (storage)
[hide=solution]Notice that if we subtract pairwise equations, we have

$$(13-a)x=(23-b)y=(42-c)z = k$$

Now, plugging in $13-a = \frac{k}{x},$ $23-b = \frac{k}{y}$ and $42-c = \frac{k}{z}$ into our desired expression, we get

$$-\frac{13x+23y+42z}{k}$$

as our desired quantity.

Finally, let $m = 13x+23y+42z$ and $n=ax+by+cz.$  Adding up all of our equations, we get $3k = m-n$ and $m+2n = 0.$  Plugging in $n=-\frac{m}{2}$ back into the other equation and solving, we get $m=\frac{k}{2}$ to get $\boxed{-2}$ as our answer[/hide]


1.|A|=|B|=n  how bijection f:A->B are there ?

2.show that (a,b)~(0,1) where a<b

[hide="1"]

For the first element of A, you have n possibilities for its image, next n-1,...

n! is the answer [/hide]


quadratics
modular arithmetic
number theory

[quote="DaMasterbomb"]Here are two trivial number theory proofs. I'm not sure whether this should go in HSB or here.

1. Prove that for any positive integers $a, b, c$ such that $a^{2}+b^{2}=c^{2}$, at least one of the numbers ($a, b, c$) must be divisible by 3.

2. Compute, with proof, the minimum value of the expression $\frac{a}{b}+\frac{b}{a}$ for positive integers $a$ and $b$.[/quote]
[hide="1"] Let's consider this in $\mod3$. The quadratic residues are $0$ and $1$. So, the only possibilities are $0+1\equiv1$ or $0+0\equiv0$. Thus, at least one of the integers is divisible by $3$. [/hide]
[hide="2"] AM-GM: :D
$\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ba}}=\boxed{2}$ [/hide]

1. [b]Proof by contradiction:[/b] Assume that neither a, b or c are divisible by 3. Then $a^{2}\equiv b^{2}\equiv c^{2}\equiv 1 \pmod{3}$, which implies that $2 \equiv 1 \pmod{3}$ because $a^{2}\:+\: b^{2}\: =\: c^{2}$. This contradiction (i.e. $3|1$) completes the proof.

2. Set $x \: =\: a/b \: >\: 0$. Then

${\textstyle \frac{a}{b}\:+\: \frac{b}{a}\;=\; x \:+\: \frac{1}{x}\;=\; (\sqrt{x}\:-\: \frac{1}{\sqrt{x}})^{2}\:+\: 2.}$

Hence the minimum value of ${\textstyle \frac{a}{b}\:+\: \frac{b}{a}}$ is 2 (occurs when a = b since $\sqrt{x}= 1/\sqrt{x}$ gives $x = 1 = a/b$).

[hide="1"]Using contradiction, assume that none of the numbers are divisible by 3. Then $n\equiv 1,2 \pmod{3}\implies n^{2}\equiv 1 \pmod{3}$, so that $a^{2}+b^{2}\equiv 2\pmod{3}$, but $c^{2}\equiv1\pmod{3}$. Thus, at least one is divisible by 3.[/hide]
[hide="2"]From AM-GM, $\frac{a}{b}+\frac{b}{a}\ge 2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=\boxed{2}$.[/hide]


let a,b,c>0
Prove that
$ \left( 1\plus{}\frac{a}{b} \right) \left( 1\plus{}\frac{b}{c} \right) \left( 1\plus{}\frac{c}{a} \right) \ge 2 \left( 1 \plus{}\frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}} \right)$

[quote="aktyw19"]let a,b,c>0
Prove that
$ \left( 1 \plus{} \frac {a}{b} \right) \left( 1 \plus{} \frac {b}{c} \right) \left( 1 \plus{} \frac {c}{a} \right) \ge 2 \left( 1 \plus{} \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}} \right)$[/quote]
$ [2,1,0]\ge[\frac53,\frac23,\frac23]$ by Muirhead  :)

Or if spanferkel's solution isnt clear enough:
by setting $ \sqrt[3]{a}\equal{}x,\sqrt[3]{b}\equal{}y,\sqrt[3]{c}\equal{}z$ the inequality after some ''work'' can be rewritten as:

$ \sum_{sym}x^6y^3 \ge 2 \sum x^5y^2z^2.$

which is true by muirhead.....

We only need to prove that

$ \sum_{cyc}\left(\frac{a}{b}\plus{}\frac{a}{c}\right) \ge \frac{2(a\plus{}b\plus{}c)}{\sqrt[3]{abc}}$

Acording to Am-Gm inequality we have:

$ \frac{a}{b}\plus{}\frac{a}{c} \plus{}1\ge \frac{3\sqrt[3]{a^2}}{\sqrt[3]{bc}}\equal{}\frac{3a}{\sqrt[3]{abc}}$

$ \Rightarrow  \sum_{cyc}\left(\frac{a}{b}\plus{}\frac{a}{c}\right) \ge \frac{2(a\plus{}b\plus{}c)}{\sqrt[3]{abc}}\plus{}\frac{(a\plus{}b\plus{}c)}{\sqrt[3]{abc}}\minus{}3 \ge RHS$


[quote="dhernandez"][quote="rohitsingh0812"]Q. Let $A_1,A_2,....,A_n$ be arithmetic progressions of positive integers, each of k terms, such that any two of the have at least two common elements. Suppose b of the APs have common difference $d_1$ and others have common difference $d_2$, where $0<b<n$. Prove that:--

     $b \leq 2 ( k - \frac{d_2}{gcd(d_1,d_2)}) -1$[/quote]

ALL the others or just some others?

Daniel[/quote]
All of them.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ \frac{b\plus{}c}{2b^2\plus{}3bc\plus{}2c^2}\plus{}\frac{c\plus{}a}{2c^2\plus{}3ca\plus{}2a^2} \plus{}\frac{a\plus{}b}{2a^2\plus{}3ab\plus{}2b^2} \ge \frac{18}{7(a\plus{}b\plus{}c)}.$
Can we generalize this inequality? ;) :)

In the attachment you'll find my solution. It's not nice anyway, but solution is solution^^

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ \frac {b \plus{} c}{2b^2 \plus{} 3bc \plus{} 2c^2} \plus{} \frac {c \plus{} a}{2c^2 \plus{} 3ca \plus{} 2a^2} \plus{} \frac {a \plus{} b}{2a^2 \plus{} 3ab \plus{} 2b^2} \ge \frac {18}{7(a \plus{} b \plus{} c)}.$
Can we generalize this inequality? ;) :)[/quote]
For more information about this inequality, we have
$ \frac{1}{4b^2\minus{}bc\plus{}4c^2} \ge \frac{1}{4b^2\minus{}\frac{2b^2c^2}{b^2\plus{}c^2}\plus{}4c^2} \equal{}\frac{b^2\plus{}c^2}{2(2b^4\plus{}3b^2c^2\plus{}2c^2)},$
and from the above inequality:
$ \sum \frac{b^2\plus{}c^2}{2b^4\plus{}3b^2c^2\plus{}2c^4} \ge \frac{18}{7(a^2\plus{}b^2\plus{}c^2)},$
we can obtain the nice and interesting result of Vasc is:
$ \sum \frac{1}{4b^2\minus{}bc\plus{}4c^2} \ge \frac{9}{7(a^2\plus{}b^2\plus{}c^2)}.$
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ \frac {b \plus{} c}{2b^2 \plus{} 3bc \plus{} 2c^2} \plus{} \frac {c \plus{} a}{2c^2 \plus{} 3ca \plus{} 2a^2} \plus{} \frac {a \plus{} b}{2a^2 \plus{} 3ab \plus{} 2b^2} \ge \frac {18}{7(a \plus{} b \plus{} c)}.$
Can we generalize this inequality? ;) :)[/quote]
The following generalization holds:

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. If $ \minus{} 2 < k < 2$, then
$ \sum\frac {(b \plus{} c)[(2 \minus{} k)a \plus{} (k \minus{} 1)(b \plus{} c)]}{b^2 \plus{} kbc \plus{} c^2} \ge \frac {6k}{k \plus{} 2}.$

This inequality can be rewritten in the known form

$ \sum\frac {a(b \plus{} c) \plus{} (k \minus{} 1)bc}{b^2 \plus{} kbc \plus{} c^2} \ge \frac {3(k \plus{} 1)}{k \plus{} 2},$

and is true for $ k > \minus{} 2$.

@VasC: how do you find these generalisations?  :)


I'm completely new to the topic. Could someone explain a few things to me about going about problems that involve this.  I heard of it in my Advanced Mathematics class. It seemed difficult, and I believe it will help with other topics in math as well. I also want to be able to dazzle all the others in my advanced class and my honors class. Thanks!

"Number theory" is familiar enough, but what exactly is "number theory sequencing"?


number theory
greatest common divisor
superior algebra
superior algebra solved

is easy to see  that if  $m|n$ canonical morphism $\mathbb{Z}_n\longrightarrow\mathbb{Z}_m$ is suryective, and sends invertible clases in invertible clases. My question is:
the restriction of this morfism $\mathbb{Z}_n*$ is suryective over $\mathbb{Z}_m*$?

I was wondering me about it because I was wanting to prove that if $m|n$ a primitive root mod $n$
is also a primitive root mod $m$.

Yes it's surjective. It's enough to prove it for $p=\frac{n}{m}$ being prime and that's trivial since for every $a$ with $gcd(a,m)=1$ one of the numbers $a$ and $a+m$ is coprime to $n$ and so represents a invertible residue class modulo $n$. (prove works by simple calculating the $gcd$)


A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:

$\textbf{(A)}\ 33\text{ mph} \qquad
\textbf{(B)}\ 34\text{ mph} \qquad
\textbf{(C)}\ 35\text{ mph} \qquad
\textbf{(D)}\ 36\text{ mph} \qquad
\textbf{(E)}\ 37\text{ mph}$

$ t\equal{}\frac{d}{v}$.  $ t_{1}\equal{}4$ and $ t_{2}\equal{}3$.  Then $ v\equal{}\frac{d}{t}\equal{}\frac{240}{7}\approx 34$

[hide]So it takes 4 hrs to go and 3 back so the time is 7 hrs, and the total distance is 120x2 cause he goes twice or 240

so its distance over time equals rate so (d=rt so r=d/t) 240/7 is closest to 34

B[/hide]

Myself and Urc already solved so I don't see the point of the last post.


geometry
perimeter
3D geometry
sphere
IMO Shortlist
combinatorial geometry
IMO Longlist

Let $m$ be a convex polygon in a plane, $l$ its perimeter and $S$ its area. Let $M\left( R\right) $ be the locus of all points in the space whose distance to $m$ is $\leq R,$ and $V\left(R\right) $ is the volume of the solid $M\left( R\right) .$

[i]a.)[/i] Prove that \[V (R) = \frac 43 \pi R^3 +\frac{\pi}{2} lR^2 +2SR.\]

Hereby, we say that the distance of a point $C$ to a figure $m$ is $\leq R$ if there exists a point $D$ of the figure $m$ such that the distance $CD$ is $\leq R.$ (This point $D$ may lie on the boundary of the figure $m$ and inside the figure.)

additional question:

[i]b.)[/i] Find the area of the planar $R$-neighborhood of a convex or non-convex polygon $m.$

[i]c.)[/i] Find the volume of the $R$-neighborhood of a convex polyhedron, e. g. of a cube or of a tetrahedron.

[b]Note by Darij:[/b] I guess that the ''$R$-neighborhood'' of a figure is defined as the locus of all points whose distance to the figure is $\leq R.$

God! This is going to be awfully hard to explain in a rigorous manner, but I'll take a whack at it. :)

First of all, let's notice what $M(R)$ "looks like". If we place spheres of radius $R$ centered at each of the vertices of the polygon, then $M(R)$ is precisely the convex hull of the spheres. Alternatively, you may think that we place a sheet of rubber around the whole thing (polygon and spheres), and the set is what we obtain after the sheet has shrinked as much as possible.

a) In order to compute the volume, we need to compute the volume of three parts: the prism having the polygon as the median plane and having a height of $2R$ (this has volume $S\cdot R+S\cdot R=(2\cdot S)R$), the half-cylinders around the edges (this part has volume $\frac 12\sum a_i\cdot \pi R^2=(\frac \pi 2\cdot l)R^2$, where $a_i$ are the sides of the polygon), and whatever remains around the vertices. This last part is made up of sphere slices, which we can slide around on the sides of the polygon so that they form a sphere of radius $R$, so the volume is $(\frac 43\cdot \pi)R^3$.

b) The area is, again the sum of the areas of the three regions we have defined above: the area of the bases of the prism ($2S$), the areas of the half-cylinders, which is $l\cdot \pi R$, and the area of a sphere of radius $R$, which is $4\pi R^2$. The total area is thus $4\pi R^2+(\pi\cdot l)R+2\cdot S$.

c) Something similar to a). I'll be back later to edit this.


Hey guys I am not sure if it belongs here, but I was curious since for my AP stats project I asked this question I was wondering what AOPS people thought about the same topic.

That's a tough question to answer. Math doesn't have definite boundaries between different branches, so it is all mixed together.

Also, I think that it really depends on the person. How I see it, algebra is more definite while geometry requires creativity.

Personally, I am taking both Algebra 1 and Geometry this year. Material covered in geometry is more broad while algebra has less material, but it requires practice.

Depends on what level, and to be fair, both can be extremely challenging in their own respects, so I don't think this question has a fair answer.

In any case, I'm not particularly great at either :|

The question was written to actually feel kind of vague. At school it seems that the feelings about them are mixed quite well, but there seems to be trend where people who are taking higher level math (AP calc for example) feel that Geometry is harder. I thought it might have been because people have been exposed to more algebra (like 3 year compared to 1 year for geometry). So I was wondering about the people who actually know math beyond what is taught at school and know that Geometry is more than just a memorization of formulaes.

Well the previous posts only talk about curriculum level math. I'd say that at competitions, geometry tends to be harder than algebra. Both topics are a lot more than memorizing formulas though.

I believe Geometry is harder because of all the memorization of Theorems and Postulates.

I also agree that many higher level mathematics students believe that Geometry is harder, since all the "smart" people in the Math club at my school say that Geometry is harder for them than Algebra.

 :arrow: I didn't really like Geometry because I detested the proofs and all (main reason). :wink:

Looking at the poll, many people think Geometry is harder. But, to compare them is like comparing apples to oranges. Which is better? Both have positives (easy) and negatives (hard) in their own aspects. (also like comparing Physics with Biology...which is harder :?: )

So you might be comparing the wrong things; it really does depend on opinion and experience.

well mathskills, although i agree with you as far as the theorems, postulates, conjectures, etc., i have to disagree with you on the hatred of proofs. I think that geometry is harder than algebra, but both rely on each other, however i love geometry for it's application of deductive/inductive reasoning. In algebra, that is almost never used.. :) but that's just my opinion

$ The Origami Guy$

HMMT geometry killed me so badly I'm gonna have to say Geo D:

[quote="Ihatepie"][quote="Mathskills"][quote="origamiguy777"]I think that geometry is harder than algebra, but both rely on each other, however i love geometry for it's application of deductive/inductive reasoning. In algebra, that is almost never used.. :) but that's just my opinion

$ The Origami Guy$[/quote]

Algebra is more definite, while geometry is more "perfect" but open to increased possibilities. Such as your example, deductive/inductive reasoning...that includes actual "creative" thinking...in algebra, every "x" and every "y" has a definite answer, unlike geometry, whose concepts can be altered and changed to fit a variety of different possibilities, if you follow me...

 :idea: Well, anyway, different people have different opinions.  :idea:[/quote] No? There can be different possibilities for answers in both geometry and algebra.[/quote]

Well, I just find the change more often in Geometry than Algebra. I'm not pointing to the fact that when you factor a polynomial of [i]ax^2 + bx + c[/i] you can get 2 answers, but I'm pointing to the fact that in Geometry, you are more "creative" when selecting answers--Geometry is 3D and...well...anyway, Algebra has more of "definite" feeling to it. If you don't follow me, that's OK, not many people do... :wink: I'm not very good at explaining my opinions, unfortunately.


Hello!

$ x,y,z$ are non-negative real numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}1$. Prove that
\[ \sum_{cyclic} (1\minus{}2x^2)(y\minus{}z)^2\geq 0.\]

$ \sum_{cyc} (1 \minus{} 2x^2)(y \minus{} z)^2$

$ \equal{} \sum_{cyc} (y^2 \plus{} z^2 \minus{} x^2)(y \minus{} z)^2$

$ \equal{} \sum_{cyc} x^2(x\minus{}y)(x\minus{}z)$

$ \geq 0$


What was the top 4 scores for New York last year? and what are some good scores to make nats?

thanks!!!'
 :ddr:  :spider:  :starwars:

this should probably go in the new york forum.

well, michigan is comparable (i guess) to ny, so i think the cut off for nats was a ... 38. not sure...

darn if i was in michigan last year, i would have made nats
oh well
7th place is ok

Yea well Texas and California and some other states are in another world so 7th place in Texas is like 1st or 2nd in any other state. Does anyone know the cutoff for the Department of Defense (they have a team i think)?

[quote="jonathanchou711"]This is my first year... (Kind of cause 6th grade didn't count) I got 1st at chapter and I'm not sure what to get on State.... Well, the good thing is I'm in Oklahoma. OK is retarded.... It is stupido... I don't know. Maybe a 35 is a good score in NY...   :wink:[/quote]

Aim for the 40's, if possible.

Even though I live in a small state, it's really hard to make nats, which sucks. :( Maybe i'll make it this year? I hope!!! :maybe:

Wow.... New York isn't as good as I thought. (sorry to anyone who live there) Those scores are.... low...

New York was yet another example of why state scores don't matter at all. If I remember correctly, they had three competitors in the top 57 and another was quite close.

Seriously though; how about you stop worrying about the cutoff and work so you can actually reach it?

Okay, here's probably the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=528556888&t=196445]best thread[/url] that I've seen in the Mathcounts forum, because of just a couple of posts (12 and 14). I probably can't remember all the threads I've seen, but that was pretty good.

The same thing applies to state.

[hide="quote 1"][quote="CatalystOfNostalgia"]If I were going back to Nationals I would probably not waste my time worrying how other people are preparing, and preparing yourself.

You should probably try it.[/quote]

Replace Nationals with State though.[/hide]

[hide="quote 2"][quote="frost13"]Gotta agree with Catalyst on this one.  If you knew you only needed a 38 wouldyou leave 6 blank since you didn't NEED them?  Areyou going to try HARDER if you think you need a 39?  Go do some problems, write some problems, prove some solutions, understand what you did wrong on the last one.

Assume you need to get them all right and go out and try to get them all right.  Or, assume you have no chance and just enjoy the challenge and what ever will be will be.

Good luck to everyone, and relax!!!!![/quote][/hide]

[hide="quote 3"][quote="CatalystOfNostalgia"]I would assure you, all these numbers would probably be higher if you stopped coming on this forum (by that I mean the MATHCOUNTS forum).

Also, don't worry about the scores you get in practices, they're just numbers. Learn how to solve the problems. Then the MATHCOUNTS people will tell you your score when you get to Nationals. You do your job, they do theirs. They don't do yours, and you don't have to do theirs.[/quote]

Replace Nationals with State again. Bleh.[/hide]

Of course, all this really shows is that I have no life and am a hypocrite. Good for you.

I would imagine that a [color=red][size=75]<deleted by moderator>[/size][/color] would make you CD, but here idk, since I got [color=red][size=75]<deleted by moderator>[/size][/color] and got 4th at my chapter (1,2,3 WERE on my team but we got disqualified because we were too strong)

Also, don't forget to view my topic regarding protests. Read and protest with us!

[quote="Ravi B"]Last year, in New York State, the top 4 scores at the state competition were 36, 35, 33, and 33.[/quote]

It was 36, 35, 32, and 31 I think...

The scores I listed are from the official sheet given to NY coaches last year, which I am looking at now.  To be precise, here are last year's top 20 NY scores:  36, 35, 33, 33, 32, 31, 30, 30, 30, 29, 29, 29, 28, 28, 28, 28, 27, 27, 27, 27.


[quote="m4467"]given a real number $ a$ such that $ |a| < 1$. Prove $ \lim_{n\rightarrow\infty} na^n \equal{} 0$. (note that $ a$ could be negative.)[/quote]

Notice that
we might assume $ a > 0$ because

$ lim |Un| \equal{} 0 \implies lim Un\equal{}0$

For $ n> \frac 1{1\minus{}\sqrt{|a|}}$ there is  $ n|a|^n > (n\plus{}1)|a|^{n\plus{}1}\geq 0$ so $ n^2|a|^n$ is bounded : $ 0\leq n^2|a|^n < b$,  so $ 0\leq n|a|^n  < \frac bn$ .. and so on ..
 :cool:

It is an result :
$ a > 1$ then $ \lim_{n\to\infty}\frac {a_n}{P(n)} \equal{} \infty$
To solve you problem you can use :
If $ \lim_{n\to\infty}\frac {x_{n \plus{} 1}}{x_n} < 1$ then $ \lim_{n\to\infty}x_n \equal{} 0$


Does there exists a function $ f: R\rightarrow R$ such that $ \forall x,y \in R$,
$ f(x^2y \plus{} f(x \plus{} y^2)) \equal{} x^3 \plus{} y^3 \plus{} f(xy)$

No there exist function $ f$.
Because:
+)$ f$ is bijection
+)$ f(0)\equal{}0$
+)$ f(f(x))\equal{}x^3$
+)$ f(f(y^2))\equal{}y^3$


Solve the equation system:

$ \left\{ \begin{array}{l}
 x \plus{} y \plus{} (z2 \minus{} 8z \plus{} 14)\sqrt {x \plus{} y \minus{} 2}  \equal{} 1 \\
 2x \plus{} 5y \plus{} \sqrt {xy \plus{} z}  \equal{} 3 \\
 \end{array} \right.$

hello, do you mean $ \{ \begin{array}{l}
x + y + (z^2 - 8z + 14)\sqrt {x + y - 2} = 1 \\
2x + 5y + \sqrt {xy + z} = 3 \\
\end{array} $.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \{\begin{array}{l} x + y + (z^2 - 8z + 14)\sqrt {x + y - 2} = 1 \\
2x + 5y + \sqrt {xy + z} = 3 \\
\end{array} $.
Sonnhard.[/quote]

Yes, there are 2 equations.

[quote="borislav_mirchev"]http://www.mathlinks.ro/viewtopic.php?t=182073[/quote]

Oh! I'm sorry, I don't know this equation system was posted. Moderator can lock os delete this Topic!! :wink:


geometry
geometric transformation
reflection
MathPath
AIME
symmetry

A beam of light strikes $\overline{BC}$ at point $C$ with angle of incidence $\alpha=19.94^\circ$ and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments $\overline{AB}$ and $\overline{BC}$ according to the rule: angle of incidence equals angle of reflection.  Given that $\beta=\alpha/10=1.994^\circ$ and $AB=AC,$ determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at $C$ in your count.

[asy]
size(250);defaultpen(linewidth(0.7));
real alpha=24, beta=32;
pair B=origin, C=(1,0), A=dir(beta), D=C+0.5*dir(alpha);
pair EE=2*dir(180-alpha), E=intersectionpoint(C--EE, A--B);
pair EEE=reflect(B,A)*EE, EEEE=reflect(C,B)*EEE, F=intersectionpoint(E--EEE, B--C), G=intersectionpoint(F--EEEE, A--B);
draw((1.4,0)--B--1.4*dir(beta));
draw(D--C, linetype("4 4"),EndArrow(5));
draw(C--E, linetype("4 4"),EndArrow(5));
draw(E--F, linetype("4 4"),EndArrow(5));
draw(F--G, linetype("4 4"),EndArrow(5));
markscalefactor=0.01;
draw(anglemark(C,B,A));
draw(anglemark((1.4,0), C,D));
label("$\beta$", 0.07*dir(beta/2), dir(beta/2), fontsize(10));
label("$\alpha$", C+0.07*dir(alpha/2), dir(alpha/2), fontsize(10));
label("$A$", A, dir(90)*dir(A));
label("$B$", B, dir(beta/2+180));
label("$C$", C, S);[/asy]

The way i solved this was to copy the angle out around itself so it froms a kind of circle with $B$ at he center. Then its just the number of times the incidence beam (extended) crosses these lines.

I believe paladin8 solved a problem like this in a similar fashion not to long ago, so you can do a search for that. Or it might have been someone else. Im not sure.

He/She was Faranhajt, if that's the correct spelling (I don't feel like looking for the correct spelling of his/her name).

[quote="seamusoboyle"]I believe paladin8 solved a problem like this in a similar fashion not to long ago, so you can do a search for that. Or it might have been someone else. Im not sure.[/quote]

Heh, seems like I'm getting credit for something I never did.

This is a different answer from the one I obtained in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1771865547&t=124827]this topic[/url]. Why the discrepancy?

I think [b]knexpert[/b] looked for how long it took to reverse direction, and then doubled it, which I don't think gives an accurate answer, since when it reverses direction it does not necessarily do so symmetrically.

You looked for how many bounces in total, which does not assume symmetry between going in and going out.

[quote="seamusoboyle"]The way i solved this was to copy the angle out around itself so it froms a kind of circle with $ B$ at he center. Then its just the number of times the incidence beam (extended) crosses these lines.

So the answer is 71.  $ \frac{180\minus{}(2 \times 19.94)}{1.994}\plus{}1$.

Um, is there a correct solution with an explanation?  Thanks!

zeitnot is correct. Reflect the entire thing across the surface of the first impact and keep replicating it, since the reflection is exactly the path that would occur in the non-reflected side. If there are $ n$ reflections, then the path can be described by reflecting it $ n$ times across each of its reflection surfaces.

[img]http://www.artofproblemsolving.com/Classes/AIME/Images/1994aime14bbe.gif[/img]

Hope that picture makes it clear.

I think it should be 81. Each time the beam of light hits a mirror, the angle of incidence increases 1.994 degrees until it exceeds (180-1.994) degrees. Since another reflection would make the angle of incidence over 180 which is impossible, it will never be able to hit the other side. Therefore, the answer is $\lfloor\frac{180-19.94}{1.994}\rfloor+1=81$

darkprince and zeitnot

it should be 180-a instead 180-2a. However the official answer is 180-2a. Is there an explaination?

Thanks

Another solution: Note that each time the beam hits a side, the angle of incidence is $\frac{2}{10}\alpha$ more than the previous angle of incidence on the same side. $\frac{n}{10}\alpha\le{90}$ so $n=45$ and this last hit is on side $AB$. Since, the initial angle was $\frac{10}{10}\alpha$, the beam makes $36$ reflections. Hence, there are $72-1=71$ total reflections since the last one doesn't count.

@Diehard, how can you assume that the reflections out are symmetrical to the reflections in (i.e. why can you just multiply 36 by 2?)

Also, I followed Phelphedo's thread and I'm still not convinced why the formula is not  $ \left \lceil \frac{180-\alpha}{\beta} \right \rceil $

[quote=muti66][quote=knexpert]This is really similar to a problem I did at MathPath, except it was with pool tables.

[hide="Solution"] We can generalize this problem for any incedent angle $a$ and angle between $AB$ and $BC$ $b$. Draw a perpendicular to $BC$ through $C$. Using the angles of the triangle, we find that the second incident angle is $a+b$. Using a similar process, we find that the nth incident angle is $a+(n-1)b$. To find the number of bounces before it starts reflecting back, we can set the expression equal to 90 and take the floor of $n-1$, which means that $n=\left ceiling \frac{90-a}{b}\right ceiling$, so $n=36$. Double this to get the total number of bounces, which is $\boxed{72}$.[/hide]

This probably isn't completely correct. Please feel free to comment.[/quote]

Wait why did you double it at the end, also, what is the real answer, anyone care to post a solution/confirm an answer?[/quote]

I don't believe that answer is correct, when I attempted this problem, I got $71$ since $\left\lfloor 70.3 \right\rfloor + 1 = \boxed{71}$.

The solution is in theory except for the ceiling function part, correct, the user just did the computation slightly wrong; he/she took the ceiling function and forgot to add 1 at the end for $n-1$. I don't understand why you, take the floor, or double it though...

The answer is not $81$ because the problem is not asking for the number of times the light beam bounces off the infinitely long rays; it's asking for the number of times it bounces off the specific segments, so the last $10$ bounces don't actually work because of their position.

I attained $081$ and see why this is wrong, as per u/Hamstpan38825's explanation. What I don't see, however, is why $\lceil\tfrac{180-2\alpha}{\beta}\rceil$ indeed works: how might one measure when the beam surpasses the [i]segments[/i] $AB$ and $AC$?

Searching "$\beta=\alpha/10$" yields the following.
[list]
[*] As for [url=https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_14]the Wiki solution[/url], it is evident whence the equation $k\beta\le180(^\circ)-2\alpha$ (or $\alpha+k\beta\le180^\circ-\alpha$) was derived, but I fail to see why all these intersections, when reflected back into initial "frame of reference", lie within $ABC$. [/*]
[*] As for [url=https://artofproblemsolving.com/community/c6h66935]the other thread[/url], where is $Z$ located, and how can one infer that $ABZ$ is even nondegenerate? By inspection, it seems as though the solution assumes that $Z$ is the reflection of $A$ across $BC$, which I don't see why is the case... [/*]
[*] [url=https://artofproblemsolving.com/community/c5h467446]c5h467446[/url] and other solutions above seem to assume that the reflections toward and away from $A$ are "symmetric", which certainly cannot be assumed without justification. [url=https://artofproblemsolving.com/community/c5h509006]c5h509006[/url] and [url=https://artofproblemsolving.com/community/c5h2367976]c5h2367976[/url] express similar confusions. [/*]
[*] ([url=https://artofproblemsolving.com/community/c4h2352501]c4h2352501[/url] bespeaks that somehow, hamstpan was once upon a time indeed immature. :')) [/*]
[/list]
I'd appreciate any help on this. Thank you!

Let $N$ be the answer and define points $P_i$ for $1\le i\le N$ as the $i$th point the beam hits, where $C = P_1$. Also, let $\theta_i$ be the angle made by the beam with the line segment it intersects on the $i$th intersection.

It is easy to see that $\theta_n = \theta_{n-1} + \beta$, and since $\theta_1 = \alpha$, we have $\theta_n=\alpha + \beta(n-1) = \frac{9\alpha}{10} + \frac{\alpha}{10}n$. Now, by the law of sines in $\triangle P_{n+1}P_nB$, we see that
\[ BP_{n+1} = \frac{\sin \theta_n}{\sin (180^\circ - \theta_{n+1})}\cdot BP_n = \frac{\sin\theta_n}{\sin\theta_{n+1}}BP_n.\]
Taking a product and telescoping suffices to obtain $BP_n = \frac{\sin\theta_1}{\sin\theta_n}BP_1 = \frac{BC\cdot \sin\alpha}{\sin\theta_n}$ in general. Hence, we want the largest value of $n$ such that
\[ \frac{BC\cdot \sin\alpha}{\sin\theta_n}\le BC\iff \sin\theta_n\ge \sin\alpha.\]
It suffices to find the smallest $n$ such that this inequality fails, which is the smallest value of $n$ satisfying $\frac{9\alpha}{10} + \frac{\alpha}{10}n> 180^\circ - \alpha$. It is easy to see that this value of $n$ is $\left\lceil 10\cdot \left(\frac{180^\circ}{\alpha} - \frac{19}{10}\right)\right\rceil = 72$. Hence, all $n\le 71$ satisfy the inequality, so $N = \boxed{071}$.

Indeed, I'd suspected there would have to be some sort of angle to length relation such as law of sines. Thank you.


If I am not wrong, this problem is in Old and New.


algebra
polynomial
induction
number theory unsolved
number theory

Prove that:
If a polynomial has rational value at all irrational points, it must be a constant.
 :)  :)

No one is interesting? It is a very beautiful problem. :)

[quote="puuhikki"]I think the following works. Assume contrary that polynomial $f$ is non-constant. Then for some $a,b\in\mathbb{R}$ there exists $c,d\in\mathbb{R},c<d$ such that $f(a)=c$ and $f(b)=d$. (you can assume $c<d$, otherwise consider $-f$). In $[c,d]$ there are uncountable many reals and mapping $f|[a,b]\to [c,d]$ is surjective. But in $[a,b]$ $f$ gets only countably many values in rational points and in irrational points, so that $[c,d]$ is countable. A contradiction.[/quote]

  Thank you for your sols. Unfortunately, I knew nothing about map nor countable... so I can't understand it. May you explain more detail please? At the same time, I will read more document about countable, uncountable...

A mapping is just a synonym for a function. A set $A$ is said to be countable if it is finite or $\text{card }A=\text{card }\mathbb{N}$. $A$ is uncountable if it is not countable. Now every non-empty interval of reals is uncountable, rationals are countable, a subset of a countable set is countable, and a countable union of countable sets is countable.

I have an other solution

we'll use induction . we suppose that for all polynomials of degree less than $r$ we have the theoreme .
we considere a polynimial $P \equiv{ a_nX^n+a_{n-1}X^{n-1}+....+a_1X+a_0}$ such that $\forall r$ irrational we have $P(r)$ is rational
the ploynimial  $\Delta P\equiv{P(X+1)-P(X)}$ has the same proprity as $P$ ( we have also $\deg \Delta P\le{r-1}$) so by the induction hypothese we can say that  $\Delta P\equiv\lambda$
 than we can proof easely that $P\equiv{aX^2+bX+c}$
 the problem becomes easy we have only to proof that $a=b=0$ :D


Here's the question that I think all of the given answers are wrong(multiple choice):
Here are seven numbers: –9 ; 0 ; –5 ; 5 ; –4 ; –1 ; –3. We arranged six of them in groups of two so that the sum in each group is the same. Which number remains?
A) 5 B) 0 C) –3 D) –4 E) –5
I think the right answer must be -1. In addition, I couldn't find the official solution for this contest, so I need to discuss about this question!

Well, you are wrong -- if you'd posted your proposed solution, I could have explained why, but instead, I'll give you a hint:

[hide]Consider everything modulo 3.  (Why 3?  Well, because there are 3 pairs with the same sum is one reason.)[/hide]

Thank you very much,JBL! I misread the problem and that's why I find it wrong! :blush:(It said divided into three pairs, not two pairs as I thought at first)


Let $ a, b, c$ be [b]real[/b] numbes satisfying $ a\plus{}b\plus{}c\equal{}3$. Prove that  :) :
\[ \frac{(a\plus{}b)^2}{4a\plus{}b^2\plus{}c^2}\plus{}\frac{(b\plus{}c)^2}{4b\plus{}c^2\plus{}a^2}\plus{}\frac{(c\plus{}a)^2}{4c\plus{}a^2\plus{}b^2} \le 2\]

[quote="nguoivn"]Let $ a, b, c$ be [b]real[/b] numbes satisfying $ a \plus{} b \plus{} c \equal{} 3$. Prove that  :) :
\[ \frac {(a \plus{} b)^2}{4a \plus{} b^2 \plus{} c^2} \plus{} \frac {(b \plus{} c)^2}{4b \plus{} c^2 \plus{} a^2} \plus{} \frac {(c \plus{} a)^2}{4c \plus{} a^2 \plus{} b^2} \le 2\]
[/quote]

only prove

$ \frac {(a \plus{} b)^2}{\frac{4}{3}a(a\plus{}b\plus{}c) \plus{} b^2 \plus{} c^2} \plus{} \frac {(b \plus{} c)^2}{\frac{4}{3}b(a\plus{}b\plus{}c) \plus{} c^2 \plus{} a^2} \plus{} \frac {(c \plus{} a)^2}{\frac{4}{3}c(a\plus{}b\plus{}c) \plus{} a^2 \plus{} b^2} \le 2$

[quote="nguoivn"]Let $ a, b, c$ be [b]real[/b] numbes satisfying $ a \plus{} b \plus{} c \equal{} 3$. Prove that  :) :
\[ \frac {(a \plus{} b)^2}{4a \plus{} b^2 \plus{} c^2} \plus{} \frac {(b \plus{} c)^2}{4b \plus{} c^2 \plus{} a^2} \plus{} \frac {(c \plus{} a)^2}{4c \plus{} a^2 \plus{} b^2} \le 2\]
[/quote]
Actually, this inequality follows from the known result:

If $ x,$ $ y,$ $ z$ are nonnegative real numbers such that $ x\plus{}y\plus{}z>0,$ then
$ \frac{x}{4x\plus{}4y\plus{}z}\plus{}\frac{y}{4y\plus{}4z\plus{}x}\plus{}\frac{z}{4z\plus{}4x\plus{}y} \le \frac{1}{3}.$

[quote="hedeng123"]
only prove

$ \frac {(a \plus{} b)^2}{\frac {4}{3}a(a \plus{} b \plus{} c) \plus{} b^2 \plus{} c^2} \plus{} \frac {(b \plus{} c)^2}{\frac {4}{3}b(a \plus{} b \plus{} c) \plus{} c^2 \plus{} a^2} \plus{} \frac {(c \plus{} a)^2}{\frac {4}{3}c(a \plus{} b \plus{} c) \plus{} a^2 \plus{} b^2} \le 2$[/quote]
How did you prove it with $ a, b, c$ are the real numbers?  :maybe:
[quote="can_hang2007"]
Actually, this inequality follows from the known result:

If $ x,$ $ y,$ $ z$ are nonnegative real numbers such that $ x \plus{} y \plus{} z > 0,$ then
$ \frac {x}{4x \plus{} 4y \plus{} z} \plus{} \frac {y}{4y \plus{} 4z \plus{} x} \plus{} \frac {z}{4z \plus{} 4x \plus{} y} \le \frac {1}{3}.$[/quote]
Yes, you are right  :oops:


Solve the following equation in positive integers for $a \leq b \leq c$

\[ a + b + c +ab +bc +ca = abc +1 \]


rotation
limit
geometry proposed
geometry

An airplane wants to go from a point on the equator, and at each moment it will go to the northeast with speed $v$. Suppose the radius of earth is $R$.
a) Will the airplane reach to the north pole? If yes how long it will take to reach the north pole?
b) Will the airplne rotate finitely many times around the north pole? If yes how many times?

Let $l$ be distance of the point with the north pole and $x$ the distance it has gone at each time. Suppose it moves $\Delta x$ meters towards the northeast. if $\Delta x$ is small enough we can assume that it has moved in a plane. Then:\[ \displaystyle \Delta l=\sqrt{l^2+\Delta x^2-2l\Delta xcos(45)}-l \]. So $\lim_{\Delta x\rightarrow0}{\frac{\Delta l}{\Delta x}}=cos(45)=\frac{\sqrt{2}}{2}$. So after $\sqrt{2}l$ meters it reaches the north pole.It rotates infinate times around the north pole because we can compute$\lim_{\Delta x\rightarrow 0}{\frac{sin(\Delta \theta)}{\Delta x}=\frac{1}{\sqrt{2}l}}$.

Is it assumed that the earth is perfectly circular?


why do i get thsi G.P. of words [url=http://www.mathlinks.ro/viewtopic.php?p=929566#929566]here[/url]

Because you put _ inside \text where it causes problems. _ is not allowed in text mode only in math mode. LaTeX puts in extra dollar signs to try to cope so causing the new symbols.

It is only the letters that you want in text mode but _ in math mode so move a couple of } and instead of \text{XeF_{4}},\text{XeF_{6}} write \text{XeF}_{4},\text{XeF}_{6} which gives $ \text{XeF}_{4},\text{XeF}_{6}$

In documents there is a \textsuperscript but not a \textsubscript, but anyone who wants to type chemistry should use the excellent [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=mhchem]mhchem package[/url] where \ce{XeF4},\ce{XeF6} gives [img]http://sixthform.info/gblog/pictures/bc6315a384f5adaf85227d4e29a7d692.gif[/img]

thanks stevem  :) well isn't the package nebled in this forum i tested in the test why don't i get it?

mhchem is not enabled in the forum, so you can only use it in documents. You would have to ask for it on the forum.


inequalities
geometry
inradius
circumcircle
cyclic quadrilateral
area of a triangle
inequalities unsolved

Let $ABCD$ be a quadrilateral of sides $a,b,c,d$ inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  We have

$ 8Rr\sqrt{2} \leq (a+c)^2 $

We have to prove the inequality $8\sqrt2\cdot Rr\leq\left(a+c\right)^2$.

Well, let $s=\frac{a+b+c+d}{2}$ be the semiperimeter of the quadrilateral ABCD. Since the quadrilateral ABCD is circumscribed, the Pitot theorem yields a + c = b + d. Thus, $s=\frac{a+b+c+d}{2}=\frac{\left(a+c\right)+\left(b+d\right)}{2}=\frac{\left(a+c\right)+\left(a+c\right)}{2}=a+c$. Consequently, our inequality $8\sqrt2\cdot Rr\leq\left(a+c\right)^2$ is equivalent to $8\sqrt2\cdot Rr\leq s^2$.

The area of an arbitrary polygon which has an incircle equals the semiperimeter of the polygon times the inradius. Thus, since s is the semiperimeter and r is the inradius of the circumscribed quadrilateral ABCD, the area F of the quadrilateral ABCD equals F = sr. Thus, $r=\frac{F}{s}$.

By a well-known formula for the circumradius of a cyclic quadrilateral, the circumradius R of the quadrilateral ABCD can be found from the sides a, b, c, d and the area F by the formula

$R=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4F}$

(for the proof of this formula, see the note below). Thus,

$Rr=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4F}\cdot \frac{F}{s}=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4s}$.

Hence, the inequality in question, $8\sqrt2\cdot Rr\leq s^2$, becomes

$8\sqrt2\cdot \frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4s}\leq s^2$.

This simplifies to

$2\sqrt2\cdot \frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{s}\leq s^2$.

Multiplying this with s, we get

$2\sqrt2\cdot\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}\leq s^3$.

But this inequality is rather easy to show: By AM-GM, we have

$\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)\leq\left(\frac{\left(ab+cd\right)+\left(bc+da\right)+\left(ac+bd\right)}{3}\right)^3$;

in other words,

$\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)\leq\left(\frac{ab+cd+bc+da+ac+bd}{3}\right)^3=\left(2\cdot\frac{ab+cd+bc+da+ac+bd}{6}\right)^3$
$=8\left(\frac{ab+cd+bc+da+ac+bd}{6}\right)^3$,

so that

$\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}\leq \sqrt{8\left(\frac{ab+cd+bc+da+ac+bd}{6}\right)^3}=2\sqrt2\left(\sqrt{\frac{ab+cd+bc+da+ac+bd}{6}}\right)^3$.

By the symmetric mean inequality,

$\sqrt{\frac{ab+cd+bc+da+ac+bd}{6}}\leq\frac{a+b+c+d}{4}$,

so that $\sqrt{\frac{ab+cd+bc+da+ac+bd}{6}}\leq\frac12\cdot \frac{a+b+c+d}{2}=\frac12\cdot s$;

thus,

$\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}\leq 2\sqrt2\left(\sqrt{\frac{ab+cd+bc+da+ac+bd}{6}}\right)^3$
$\leq 2\sqrt2\cdot \left(\frac12\cdot s\right)^3=2\sqrt2\cdot \frac18\cdot s^3=\frac{1}{2\sqrt2}s^3$.

Hence,

$2\sqrt2\cdot\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}\leq s^3$.

And the inequality is proven. $\blacksquare$

I promised in the above to prove the formula

$R=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4F}$.

Here is the promised proof:

Consider not only the sidelengths a = AB, b = BC, c = CD and d = DA, but also the diagonals e = AC and f = BD of the quadrilateral ABCD. By a really well-known formula, the area of a triangle with sidelengths $a_1$, $b_1$, $c_1$ and circumradius $R_1$ is equal to $\frac{a_1b_1c_1}{4R_1}$. Applying this to the triangles DAB and BCD, which both have the same circumradius R (since the circumcircles of triangles DAB and BCD coincide with the circumcircle of the cyclic quadrilateral ABCD), we see that the area of the triangle DAB equals $\frac{DA\cdot AB\cdot BD}{4R}=\frac{daf}{4R}$, and the area of the triangle BCD equals $\frac{BC\cdot CD\cdot BD}{4R}=\frac{bcf}{4R}$. But the quadrilateral ABCD is formed by the triangles DAB and BCD, and thus, the area F of the quadrilateral ABCD is the sum of the areas of these triangles DAB and BCD. Hence,

$F=\frac{daf}{4R}+\frac{bcf}{4R}$.

In other words, $F=\frac{daf+bcf}{4R}=\frac{\left(bc+da\right)f}{4R}$, so that $4R=\frac{\left(bc+da\right)f}{F}$. Similarly, $4R=\frac{\left(ab+cd\right)e}{F}$. Multiplying these two equations together, we get

$\left(4R\right)^2=\frac{\left(bc+da\right)f}{F}\cdot\frac{\left(ab+cd\right)e}{F}$.

In other words,

$\left(4R\right)^2=\frac{\left(ab+cd\right)\left(bc+da\right)ef}{F^2}$.

By the Ptolemy theorem, ef = ac + bd; thus,

$\left(4R\right)^2=\frac{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}{F^2}$.

Taking the square root, we get

$4R=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{F}$,

and thus

$R=\frac{\sqrt{\left(ab+cd\right)\left(bc+da\right)\left(ac+bd\right)}}{4F}$,

and the proof is complete.

  Darij


number theory
least common multiple
relatively prime

We just have to show that $n$ and $n+1$ are relatively prime for any positive integer $n$ ;)

if you divide n +1 by n, there is always a remainder of 1 therefore they are relatively prime, and their lcm is therefore $\frac{n*(n+1)}{1} = n*(n+1)$

[quote="a_vakilian"][quote="amirhtlusa"]gcd(n,n+1) is 1, because one of them is even, the other one odd....with no common primes.[/quote]
you are wrong because maybe both of them are divisable by ddd number like $3$.
the proof is it:consider $(n,n+1)=d$ so $d|n+1-n$ so $d|1$ so $d=1$!
I think this proof is OK ;)[/quote]I think it is.  I did (almost) the exact same proof at a summer camp this year.

Well, thanks to you all guys :)
(I think we can also prove n and n+1 are relatively primes easily using Bezout..)

[quote="a_vakilian"]Exuse me :blush:
but I think you should explain more :P
Again Exuse me :oops:[/quote]

its ok :)

andyes, there are many ways to prove that, but the best way is simply what a_vakilian said...

[quote="amirhtlusa"][quote="a_vakilian"][quote="amirhtlusa"]gcd(n,n+1) is 1, because one of them is even, the other one odd....with no common primes.[/quote]
you are wrong because maybe both of them are divisable by ddd number like $3$.
 [/quote]
REMEMBER WHAT I SAID AT THE END:

[b]WITH NO COMMON PRIMES[/b][/quote]Shouting something doesn't make it true. You didn't prove it, just said it.


can anyone help me with this quesion? http://myweb.tiscali.co.uk/myticalipage/question.jpg

What is the inequality I need to know here? I've tried to look it up in so many books, and couldn't find it. i'm really desperate!

please help! thank you

I guess the inequality between the AM-HM (Arithmetic-Harmonic Mean) works: \[\frac{\sum_{i=1}^n a_i}{n} \geq \frac{n}{\sum_{i=1}^n \frac{1}{a_i}}\]

thanx, no wonder i don't know, i've never heard of this inequality?

You can find proofs and description here:

http://planetmath.org/encyclopedia/ArithmeticGeometricMeansInequality.html


I wonder how many inequalities has this book and especially what will happen when they will end. And now the solution:
      Prod (1+1/x)>=(1+1/(xyz)^1/3)^3>=64. I used Huygens inequality. Anyway, the problem can be done allmost in any way.

I don't know Huygens inequality.
Can you please explain it?

It's problem C75. Maybe you didn't know its name, but I'm sure you knew it.

If we put  \sum x instead of 1 we shall prove that \pi (2+x\y+z\y) \geq 64 which is obios using AM-GM We have equality if x=y=z=1\3


[size=200][b]Distinguishabililty[/size][/b]

The concept of being able to differentiate between objects is called [b]distinguishability[/b].
If objects are distinguishable, then we are able to tell them apart from each other. If they are not distinguishable, or [b]indistinguishable[/b], then they are exactly alike. For example, we are able to distinguish between
\[ \triangleleft \ \Diamond \ {\bigtriangleup}\]
but not between
\[ \odot \ \odot \ \odot\]
[b][size=150]Problems:[/size][/b]

[b]Problem 1:[/b] How many ways are there to distribute $ 3$ peppermints amongst $ 4$ children?
(a) Both are distinguishable
(b) The peppermints are distinguishable, but the children are not
(c) The childred are distinguishable, but the peppermints are not
(d) Both are indistinguishable

[b]Problem 2:[/b] There are $ 5$ blue pencils, $ 3$ black pencils, and $ 2$ red pencils. If the pencils are to be placed in $ 10$ slots with no more than $ 1$ pencil per slot, and pencils of the same color must all be adjacent to each other, how many ways are there to arrange the pencils?

[b]Problem 3:[/b] How many ways are there to distribute $ 4$ distinguishable balls amongst $ 3$ indistinguishable boxes?

[b]Problem 4:[/b] Find all nonnegative integers $ a,b,$ and $ c$ is $ a \plus{} b \plus{} c \equal{} 7$.

[b][size=150]Solutions:[/size][/b]

[b]Problem 1:[/b] How many ways are there to distribute $ 3$ peppermints amongst $ 4$ children?
(a) Both are distinguishable
(b) The peppermints are distinguishable, but the children are not
(c) The childred are distinguishable, but the peppermints are not
(d) Both are indistinguishable

[i]Solution to Problem 1:[/i]

(a) We can choose $ 4$ people for each peppermint to go to resulting in $ 4^3 \equal{} \boxed{64}$ ways.

(b)
[i]Case 1:[/i] We give all three peppermints to $ 1$ child. Since the children are distinguishable, there are $ \dbinom{4}{1} \equal{} 4$ ways to do that.

[i]Case 2:[/i] We give $ 1$ peppermint to one child and $ 2$ peppermints to another. Since the peppermints are indistinguishable, it does not matter which ones to we choose to go to which child. Now, there are $ \dbinom{4}{2} \equal{} 6$ ways to do that.

[i]Case 3:[/i] We give $ 1$ peppermint each to $ 3$ children. There is $ \dbinom{4}{3} \equal{} 4$ ways to do that. Seeing as we've covered all the cases, the answer is $ 4 \plus{} 6 \plus{} 4 \equal{} \boxed{14}$.

(c)
[i]Case 1:[/i] We can give $ 3$ peppermints to any child. Since the children are indistinguishable, it does not matter which. Thus, there is $ \dbinom{3}{3} \equal{} 1$ way to do that.

[i]Case 2:[/i] We can give $ 1$ peppermint to any child and $ 2$ peppermints to another. There is $ \dbinom{3}{2} \equal{} 3$ ways to do that. Once again, which children recieve the candy doesn't matter.

[i]Case 3:[/i] We can give $ 1$ peppermint each to $ 3$ children. We have $ 3$ choices for the first peppermint, $ 2$ choices for the second, and $ 1$ choice for the last which is $ 3\cdot2\cdot1 \equal{} 6$. Now, the answer is $ 1 \plus{} 3 \plus{} 6 \equal{} \boxed{10}$.

(d) Since both are indistinguishable, it doesn't matter which peppermints which child gets or what peppermints are given to a certain child. The only ways are
\[ 3,0,0,0\]

\[ 1,2,0,0\]

\[ 1,1,1,0\]
Hence there are $ \boxed{3}$ ways. $ \Box$

[b]Problem 2:[/b] There are $ 5$ blue pencils, $ 3$ black pencils, and $ 2$ red pencils. If the pencils are to be placed in $ 10$ slots with no more than $ 1$ pencil per slot, and pencils of the same color must all be adjacent to each other, how many ways are there to arrange the pencils?

[i]Solution for Problem 2:[/i] Since pencils of the same color are indistinguishable, all that matters is the ordering of the color groups. There are $ 3$ color groups so the answer is $ 3! \equal{} \boxed{6}$. $ \Box$

[b]Problem 3:[/b] How many ways are there to distribute $ 4$ distinguishable balls amongst $ 3$ indistinguishable boxes?

[i]Solution for Problem 3:[/i]

[i]Case 1:[/i] We put all $ 4$ in one box. There's only $ \dbinom{4}{4} \equal{} 1$ way to do that as the boxes are indistinguishable.

[i]Case 2:[/i] We put $ 1$ in one box and $ 3$ in another box. We need to choose which three goes to the second box so this is $ \dbinom{4}{3} \equal{} 4$.

[i]Case 3:[/i] We put $ 2$ marbles each of two boxes. There are $ \dbinom{4}{2} \equal{} 6$ ways to choose which $ 2$ go to the first box and remaining ones go in the second. Remember, it does not matter which box the marbles go into as the boxes are indistinguishable.

[i]Case 4:[/i] We put $ 2$ marbles in the first box, $ 1$ in the second, and $ 1$ in the last. There are $ \dbinom{4}{2} \equal{} 6$ ways to choose which ones go into the first box. We have $ \dbinom{2}{1} \equal{} 2$ ways to choose which one goes into the second box and the remaining one goes into the last box. Thus, there are $ 6\cdot2 \equal{} 12$ ways. Now that we're done with all cases, $ 1 \plus{} 4 \plus{} 6 \plus{} 12 \equal{} \boxed{23}$. $ \Box$

[b]Problem 4:[/b] Find all nonnegative integers $ a,b,$ and $ c$ is $ a \plus{} b \plus{} c \equal{} 7$.

[i]Solution for Problem 4:[/i] Instead of tackling this with brute casework, let's work for a more elegant solution. Let us have seven balls and $ 2$ dividers. With two dividers, we distribute the seven balls into $ 3$ seperate groups corresponding to $ a,b,$ and $ c$. We also note that as there are exactly $ 7$ balls, the sum is always $ 7$! Now all we have to do is find the number of ways we can choose the dividers. This is simply $ \dbinom{7 \plus{} 2}{2} \equal{} \boxed{36}$. $ \Box$

[b]Important![/b] This is an example of a  formula called [b]Balls and Urns[/b]. This formula states that if we are distributing $ n$ indistinguishable objects to $ k$ distinguishable objects, the number of ways to do so is
\[ \dbinom{n \plus{} k \minus{} 1}{k \minus{} 1}\]
[b][size=150]Review Problems:[/size][/b]

[b]1[/b] Find the number of ways to distribute $ 7$ identical pencils to $ 5$ different children.

[b]2[/b] How many ways are there to distribute $ 8$ indistinguishable balls to $ 5$ distinguishable boxes?

[b]3[/b] For how many nonnegative $ a,b,c$ is $ a \plus{} b \plus{} c \equal{} 11$?

[b]4[/b] Ice-cream-o-rama is eager to advertise how many flavors it has. But it really only has three basic flavors: chocolate, vanilla, and strawberry. However, they can make "new" flavors by taking four scoops of ice cream of those basic flavors and blending them together. Different proportions of the basic flavors give different new flavors. How many flavors can Ice-cream-o-rama now advertise? [i](Source: ALCUMUS)[/i]

[b]5[/b] There are an infinite amount of chocolate chip cookies, Oreo's, crackers, and biscuits. In how many ways can Bobby choose $ 5$ of these?

[b]6[/b]$ \star$ Compute the number of ways to distribute $ 6$ distinguishable balls to $ 4$ indistinguishable urns.

[b]7[/b]$ \star$ For how many positive $ a,b,c$ is $ a \plus{} b \plus{} c \equal{} 10$?

Was this supposed to be posted in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282647&start=40]Tip of the Week?[/url]


algebra
polynomial
superior algebra
superior algebra unsolved

I know how to prove the following:

Let $ F$ be a field and $ F_0$ a subfield of $ F$ with $ q$ elements. Let $ n: \equal{} [F: F_0]$. Then $ F$ has $ q^n$ elements.

Can this be used to prove that if there exists a field with $ k$ elements and $ n$ is a fixed positive integer, then there exists a field with $ k^n$ elements? I can't choose $ F \equal{} \mathbb{F}_{k^n}$ and $ F_0 \equal{} \mathbb{F}_k$ since I don't know whether a field with $ k^n$ elements exists.

You are not given anything about the values $ n$ can take, so no.  That conclusion requires extra work.

It would suffice to show there is a degree $ n$ irreducible polynomial over $ F$.


inequalities
geometry
trigonometry
inequalities unsolved

I found an inequality in my book, that is:

If the lengths of sides of a quadrilateral $ABCD$ with area $S$ are $a,b,c,d$, then $S \le (abcd)^{\frac{1}{2}}$, equality holds if and only if $ABCD$ are cyclic.

But when I checked the equality case, it seems that the equality case it said is wrong. So I want to ask, despite of the equality case, is this inequality true?

[quote="yptsoi"]I found an inequality in my book, that is:

If the lengths of sides of a quadrilateral $ABCD$ with area $S$ are $a,b,c,d$, then $S \le (abcd)^{\frac{1}{2}}$, equality holds if and only if $ABCD$ are cyclic.[/quote]

This inequality is wrong. Actually, I guess the author of the book had meant to say $S\leq \sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$, where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral ABCD. This inequality follows from the general formula

$S=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right) -abcd\cos ^{2}\frac{B+D}{2}}$

proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=15973 , and indeed, equality holds in $S\leq \sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}$ if and only if the quadrilateral ABCD is cyclic.

  Darij

I am not sure since my book said that the inequality was proven by a chinese in 1987, so his inequality is probably not the classical inequality you mentioned.

What i actually find  [this is expansion of a problem 10.55 a) from Prasolov "Zadaczi po planimetry"] is:

$S \leq \frac{\sqrt{3}}{8} (abcd)^{\frac{1}{2}}(\sum_{cyc} \frac{(abc)^{\frac{1}{3}}}{d^{\frac{1}{2}}})$ so it suffices to prove:

$\frac{\sqrt{3}}{8}(\sum_{cyc} \frac{(abc)^{\frac{1}{3}}}{d^{\frac{1}{2}}}) \leq 1$


If xyz=1, then prove that (xy-y+1)(yz-z+1)(xz-x+1)<=1

It's just a classical Schur problem :) (Proposed by T. Andreescu for something I think) I remember seeing a proof in the really famous chinese math magazine (in english)

I'm not sure it's the same problem. I think you are thinking of IMO 2000 #2?

[hide="Well same idea (and the problem given is wrong, inequality sign flipped)"]
Since $xy-y+1=xy-y+xyz=y(x-1+yz)$, we have $LHS=\prod y(x-1+yz) = \prod (x-1+\frac{1}{x})=\prod (x+\frac{1}{x}-1)$
Now by am gm, $x+\frac{1}{x}\geq 2$ proves it.[/hide]

[quote="me@home"]Well same idea (and the problem given is wrong, inequality sign flipped)]
[/quote]

um. let x=1/2 , y=2, z=1 , lhs=0


Solve the equation:
$ (x^2\plus{}1)|x^2\plus{}2x\minus{}1|\plus{}6x(1\minus{}x^2)\equal{}(x^2\plus{}1)^2$

[quote="tnhh"]Solve the equation:
$ (x^2 \plus{} 1)|x^2 \plus{} 2x \minus{} 1| \plus{} 6x(1 \minus{} x^2) \equal{} (x^2 \plus{} 1)^2$[/quote]
1) $ x^2\plus{}2x\plus{}1\ge 0$
Then the equation becomes $ (x^2\plus{}1)(x^2\plus{}2x\minus{}1)\plus{}6x(1\minus{}x^2)\equal{}(x^2\plus{}1)^2$ $ \iff$ $ (x\minus{}1)(2x^2\plus{}3x\minus{}1)\equal{}0$

$ \iff$ $ x\in\{\frac{\minus{}3\minus{}\sqrt{17}}4,\frac{\minus{}3\plus{}\sqrt{17}}4,1\}$ and so $ x\equal{}1$ since $ x^2\plus{}2x\plus{}1<0$ for the two other values

2) $ x^2\plus{}2x\plus{}1\le 0$
Then the equation becomes $ \minus{}(x^2\plus{}1)(x^2\plus{}2x\minus{}1)\plus{}6x(1\minus{}x^2)\equal{}(x^2\plus{}1)^2$ $ \iff$ $ 2x(x\plus{}1)(x^2\plus{}3x\minus{}2)\equal{}0$

$ \iff$ $ x\in\{\frac{\minus{}3\minus{}\sqrt{17}}2,\minus{}1,0,\frac{\minus{}3\plus{}\sqrt{17}}2\}$ and so $ x\in\{\minus{}1,0\}$ since $ x^2\plus{}2x\plus{}1>0$ for the two other values

Hence the three solutions : $ \boxed{x\in\{\minus{}1,0,\plus{}1\}}$


$ y\equal{}x\plus{}3$

the equation is equivalent with $ \sqrt[3]{(\frac{y}{y\minus{}3})^4}\equal{}\frac{3}{2}$


How many different four-digit numbers can be formed by rearranging the four digits in 2004?

You used permutations...nPr. The n number of things you take r at a time.

nPr= n!/(n-r)! = 4!/(4-4)! = 4! = 4*3*2*1 = 24

Then because it needs to be a four-digit number, you subtract two because of the possibilities, 0024 and 0042, giving you 22 different 4-digit numbers.

I got something different.
First, we pick the thousands digit.  Since the number must be a four digit number, it can only be $ 2$ or $ 4$.  So we have $ 2$ choices.  For the other three digits, we have two zeros and either $ 2$ or $ 4$ (whichever one that wasn't used for the thousands digit; let's call it $ x$).  We want to arrange the two zeros and $ x$ in the other three digits.  If we choose which digit we want $ x$ to be, then the other two must be $ 0$.  We can choose the digit which we want $ x$ to be in $ 3$ ways (hundreds digit, tens digit, or units digit).  Therefore, our answer is $ 2\cdot 3\equal{}\boxed{6}$.


Hello,
I need help with the following problem:
[img]http://i48.tinypic.com/10xfqzq.png[/img]

I don't know what to try, could someone please help out?

Thank you!

Each coordinate $ n_i$ is a continuous function of the $ n$'s and $ f$'s, since it's a projection

$ (n_1,\dots,n_d,f_1,\dots,f_k) \mapsto n_i$.

So you now need to check that, for each $ r \ge 0$, we have that

$ (n_1,\dots,n_d,f_1,\dots,f_k) \mapsto f_i(r)$

is continuous. Note it suffices to show that $ f_i \mapsto f_i(r)$ is continuous. But

$ f_i \equal{} (f_i(0),f_i(1),\dots)$, so this is a projection onto the $ r$th coordinate, and is therefore continuous.

Thank you so much!
I will look into it more carefully and I hope it won't be a problem if I happen to have additional questions.

I have another problem, similar to this one.
I'm really confused and insecure when it comes to product topology. We haven't done any examples in class, and I can't seem to understand it on my own. :(

We have to prove that $ B$ is homeomorphic to $ B^k$, where everything is defined as above,
that is, $ B$ is the Polish set $ \mathbb{N}^{\mathbb{N}}$ with product topology, and every $ \mathbb{N}$ has discrete topology.

The suggested homeomorphism is:
$ \alpha\equal{}(f_0, f_1, \ldots)$, $ \psi(\alpha)\equal{}(f_0(0), f_0(1), f_1(0), f_0(2), \ldots)$
How do I prove that this is a homeomorphism?
I can write what I've done so far, but it's all messy and I'm pretty sure I did it all wrong...

Thank you.


Let $ABC$ be a triangle with $AB = AC$ and let $D$ be a point on $BC$ such that the incircle of $ABD$ and the excircle of $ADC$ corresponding to $A$ have the same radius. Prove that this radius is equal to one quarter of the altitude from $B$ of triangle $ABC$.

Solution of Geece 95/2:

Let $AD=d,BC=a,AB=b$ and let the given incircle and excercle touch $BC$ at $P,Q.$

Then with the basic properties we find $BD=\frac{a}{2}+b-d,DC=\frac{a}{2}-b+d.$

We can lso caalculate that $d=b-\frac{a^2}{8b}.$

Then the asked radius in square is $\frac{a^2}{16}-\frac{a^4}{64b^2}.$

So the quesion is equivalent with proving that
$ h_b^2=a^2-\frac{a^4}{4b^2}.$

Let the altitude of $B$ interset $AC$ in $E$ wehave to prove$CE=\frac{a^2}{2b}$
or in $\triangle CEB$ that $cos \gamma =  \frac{a}{2b}$ which is trivial to prove by the sine law in $ABC.$

This geometry is just bashing and nothing else.No food of fun in it.
[u][b]Try this one[/b][/u]
In a scalene triangle ABC the altitudes AP and CQ are dropped from the vertices A and C to the sides BC and AB.The area of the triangle ABC is equal to $18$, the area of triangle BPQ is $2$ and the length of the segment PQ is $2\sqrt{2}$.Prove that the distance between the centroid of triangle ABC and vertex B is $2\sqrt{2+\sqrt{2}}$


geometry
trapezoid
similar triangles
geometry proposed

$X$ is a point on a circle with diameter $AB$. $t_A$, $t_B$, $t_X$ are the tangent to the circle from these points. Let $Z = AX \cap t_B$, $Y = BX \cap t_A$.

Prove that $YZ$, $t_X$, $AB$ are concurrent.

$P \equiv AB \cap t_X$ and $Q$ is the orthogonal projection of $ X$ on $AB.$ Line $XQ$ is the polar of $ P$ with respect to the circle with diameter $ AB$ $ \Longrightarrow$ $(A,B,Q,P) \equal{} \minus{} 1.$  Since $ t_A \parallel t_B \parallel XQ,$ then denote $ T_{\infty}$ the infinity point of $ t_A,t_B,XQ.$  Line $ YZ$ cuts the sidelines $t_A,t_B,AB$ of the degenerate triangle $ ABT_{\infty}$ at $ Y,Z,P'.$ Thus, the pencil $ T_{\infty}(A,B,X,P')$ is harmonic, i.e $ (A,B,Q,P') \equal{} \minus{} 1$ $ \Longrightarrow$ $ (A,B,Q,P) \equal{} (A,B,Q,P') \equal{} \minus{} 1,$ i.e. $ P \equiv P'$ $ \Longrightarrow$ $ P \equiv t_X \cap YZ \cap AB.$

This is a (degenerate) case of Pascal's Theorem.

Let $ YZ$ and $ AB$ meet at $ C$. Let $ CX$ meet $ BZ$ at $ D$.
Note that $ ABZY$ is a trapezoid. Thus, by Ceva and similar triangles, $ D$ is the midpoint of $ BZ$ (this is in fact true for any trapezoid when the nonparallel sides are extended to meet). But since $ \triangle ZXB$ is a right triangle, $ DX \equal{} DB \equal{} DZ$, so $ \angle DXB\equal{}\angle DBX\equal{}\angle XAB$, so $ DX$ is tangent to the circle.

EDIT: I did it in the wrong direction before, fixed it.


The Dynamite game show has two types of questions: bang questions, worth 6 points each; and snap questions, worth 5 points each. Assuming there is no limit to the number of questions that can be answered, what is the greatest number of points that cannot be attained?

We have $ 6(5)\minus{}(6\plus{}5)\equal{}\boxed{19}$.


1.$ (xy\plus{}yz\plus{}xz)^2>(x\plus{}y\plus{}z)(x^3\plus{}y^3\plus{}z^3)$
Please, dont just say HOLDER

2.$ 9(x\plus{}y)(y\plus{}z)(z\plus{}x)\geq 8(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)$

[hide="2"]
Ok so (2,1,0) majorizes (1,1,1) so $ \Sigma{x^2y} \geq \Sigma{xyz}$

Since I'm to lazy to type let the RHS be S.

So we have $ S \geq 6xyz$

Thus, $ 9(2xyz\plus{}S) \geq 8(S\plus{}3xyz)$ and we win.[/hide]

Problem 1 doesn't seem to hold. $ x \equal{} y \equal{} z \equal{} 1$ yields $ 3^2 > 9$ which is clearly false.

EDIT: If you reverse the inequality sign, it follows from Cauchy-Schwarz (a special case of Holder's).


abstract algebra
topology
group theory
superior algebra
superior algebra theorems

Suppose A is an abelian profinite group, R/Z is euiped with quotient topology induced from real line.
then Hom(A, R/Z)  is an discrete abelian torsion group. Hom means all continuous map from A to R/Z.
Why the  map be torsion?

Does it help you to know which closed subgroups of $ \mathbb R/\mathbb Z\equal{}S^1$ there are?


We use the path counting method, since casework is a haphazard approach.

 [geogebra]a4d97e7716fe30920c7ebb576012171ca21b33c7[/geogebra]

Unless I made a stupid addition error, the answer's $ \boxed{91}$.

Whoa, what is that method and why does it work?

It's PIE.

$ \dbinom{11}3$ counts paths with no restriction.

However, we are obviously overcounting. By PIE,

$ \dbinom42\dbinom71$ counts the paths that go to the point $ (2,2)$ (which are all invalid; assume $ A$ is $ (0,0)$ and $ B$ is $ (8,3)$) and then go to $ B$ with no other restrictions,

$ \dbinom83\dbinom30$ counts the paths that go to the point $ (5,3)$ and then to $ B$ with no other restrictions (which are also all invalid), and

$ \dbinom42\dbinom41\dbinom30$ corrects for the overcounting (i.e. some cases go to both $ (2,2)$, then $ (5,3)$, then $ B$).

Thus, our answer is $ \dbinom{11}3\minus{}\left(\dbinom42\dbinom71\plus{}\dbinom83\dbinom30\minus{}\dbinom42\dbinom41\dbinom30\right)\equal{}\boxed{91}$.

uh math154, the first method you used, how did you get how many ways there are for each point. Also why does it seem like the possibilites add up?

The concept's actually pretty simple: if we let the number of ways to go to the point $ (x,y)$ be $ A_{(x,y)}$, then $ A_{(x,y)} \equal{} A_{(x \minus{} 1,y)} \plus{} A_{(x,y \minus{} 1)}$, where $ A_{(m,n)} \equal{} 0$ for all nonexistent points $ (m,n)$.

I kinda saw that from the diagram, but I can't intuitively see why.

Is it because if you have two sides that have a certain number of paths, then it will be the sum in this case because the new pathways must come from the old and be continued by one of the two choices thus leaving it to be added.

Yes, that is precisely why. :) It's because the only way to go to some point $ (x,y)$ is to either have first been at $ (x\minus{}1,y)$ or $ (x,y\minus{}1)$.

PIE - $\textstyle{Principle \ of \ Inclusion-Exclusion}$.

PIE is generally used to find the desired quantity of (whatever criteria) by complementing it from a complete set of (whatever original set of things).


[quote="bthings_2000"]One morning, each member of a family drinks an 8 oz combination of milk and coffee. The amount of each ingredient varies from cup to cup but is never zero. If one girl drinks 1/4 of the milk and 1/6 of the coffee in her combination, how many people are in the family?[/quote]

is this multiple choice (it's an AMC-12)? if so, can you send the answer choices?

I believe that was AMC-10 2000 #22.  I didn't really understand it...

The answer choices were:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

theres going to be between 4 and 6 people non-inclusive, since she has between 1/4 and 1/6 of the stuff.
ergo, there are 5 people


algebra
function
domain
geometry
rectangle
analytic geometry
real analysis

I have a question:

$ D$ is a regular domain $ \Rightarrow D$ is simply connected???

I have a question, too:

$ D$ is a regular domain $ \implies$ what is a regular domain???

The solution of this question is just a consequence of the more general problem,

"What do the words 'regular' and 'normal' mean, and why can't mathematicians find better names for these properties?"

For example $ D\subset{R^{2}}$ is a normal regular domain relative to x:
$ D=\left\{(x,y)\in{R^{2}}: x\in[a,b],\alpha(x)\leq y\leq \beta(x)\right\}\alpha: [a,b]\to R, \beta: [a,b]\to R$
$ \alpha\in{C^{1}},\beta\in{C^{1}}$in [a,b]

D is a reular domain $ \Leftrightarrow D=\bigcup_{k=1}^{n}D_{k}: D_{i},D_{j}$ are normal regular domain  with no internal poin in common $ \forall{i=1,2,...n},\forall{j=1,2..n},i\neq{j}$
Sorry but It's written in my book

Then no, a regular domain is not necessarily simply-connected. A rectangle oriented along coordinate axes is a normal regular domain, and you can make a non-simply-connected domain out of four such rectangles.

Normal regular domains are simply-connected. For example, you can continuously shrink such a domain onto the curve $ y=(\alpha(x)+\beta(x))/2$, which itself can be shrunk into a point.


What is the smallest possible cardinality of the set of limit points of an uncountable subset of $ \mathbb{R}$?  Is it possible for the measure of the closure of an uncountable subset of $ \mathbb{R}$ to be zero?

[quote="t0rajir0u"]What is the smallest possible cardinality of the set of limit points of an uncountable subset of $ \mathbb{R}$? [/quote]
I think $ \mathfrak{c}$. See also [url=http://en.wikipedia.org/wiki/Cantor-Bendixson_theorem]Cantor-Bendixson Theorem[/url] (especially [url=http://planetmath.org/encyclopedia/CantorBendixsonTheorem.html]this[/url] special case).
[quote="t0rajir0u"]Is it possible for the measure of the closure of an uncountable subset of $ \mathbb{R}$ to be zero?[/quote]
Yes, for example Cantor Set is uncountable, closed and has measure zero.


calculus
integration
trigonometry
search
complex analysis

How would one go about integrating $ \int_0^{\infty} \frac{\cos z}{\sqrt{z}} dz$? This should be the Fresnel integral after a substitution, so I'd like to know if there are methods for its evaluation without passing back to the Fresnel formulation, i.e. just working with this integral in its present form. Thanks.

hsiljak's post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=209936939&t=213491]this thread[/url] has a promising approach.


[quote="TwinDracula"][b]square and circle share the same center ANSWER(i think)[/b]
finding the radius
[b]#7266 A square with side length 2 and a circle share the same cent [...] [/b]
[416] USA - 2005 - AMC8

A square with side length 2 and a circle share the same center.
The total area of the regions that are inside the circle and
outside the square is equal to the total area of the regions that
are outside the circle and inside the square. What is the radius
of the circle?

so....
we have a circle and a square

area of circle-area of square = 0 because the total area of the regions that are inside the circle and
outside the square is equal to the total area of the regions that
are outside the circle and inside the square.

Therefore, area of circle=area of square

The area of the square is 2*2 or 4.

so... pi*r^2=4
r^2=4/pi (divided both sides by pi)
r=sqrt(4/pi) (square root of both sides)
r=2/sqrtpi

[b]2/sqrtpi[/b] is the answer... i think ... to the non-answered problem

plzzzzz rate this post excellent with 6 cubes[/quote]

1. Post via the Ask on forum button
2. Do not post a solution. Why are you posting a solution to your own problem.
3. Why do you ask people to rate the post 6 cubes??
4. When writing a solution, use $ \text{\LaTeX}$ make it easy to read.
5. Since this is completely unreadable and not understandable, I rate this post a 1.


Hi everyone,

Does anyone know how to convert degrees into radians using a TI-83?

Thank you very much

Multiply by $ \pi$ and divide by $ 180$  :P

Or if you want to make it so your calculator gives answers in radians (which i don't suggest), just go to mode, and on the third row select "radian" instead of "degree"


topology
advanced fields
advanced fields unsolved

Show that every smooth map from $ S^1 \times S^3$ to $ S^2 \times S^2$ has degree 0.

very nice!
[hide]for a closed oriented 4-dimensional manifold, call $ [M]$ and $ \phi_M$ the fundamental class and its dual in $ H_4(M)$ and $ H^4(M)$ respectively (integer coefficients to be understood), and take a map $ f: X\to Y$ between the two spaces. define $ \alpha, \beta \in H^2(Y)$ to be generators of the cohomology ring $ H^*(Y)$, so that $ \alpha\cup\beta \equal{} \pm \phi_Y$ (e.g., $ \alpha$ and $ \beta$ can be chosen as poincarè dual to the fibers $ S^2\times *$ and $ *\times S^2$ in $ Y$).
$ \deg f \equal{} \langle \phi_Y, f_*[X]\rangle \equal{} \langle f^*\phi_Y, [X]\rangle \equal{} \langle f^*(\alpha \cup \beta),[X]\rangle \equal{} \langle f^*\alpha \cup f^*\beta,[X]\rangle \equal{} 0$, because $ H^2(X)\equal{}0$[/hide]
the next natural question is whether the set of homotopy classes of maps $ S^1\times S^3\to S^2$ is trivial or not (e.g. if every map is homotopic to a constant): my guess would be "no, it's not trivial", but i don't see any straightforward way to prove it from $ \pi_3(S^2)\equal{}\mathbb{Z}$.

The important part of any proof is using the fact $ H^2 (S^1 \times S^3) \equal{} 0$. Then you can use your favorite argument to finish the proof. Mine is using Künneth formula.


Prove that

$ 0\leq yz\plus{}zx\plus{}xy\minus{}2xyz\leq 7/27,$

where $ x,y$ and $ z$ are non-negative real numbers for which $ x\plus{}y\plus{}z\equal{}1$.

Follows Schur Inequality ,$ (x \plus{} y \plus{} z)^3\ge 4(xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z) \minus{} 9xyz$ we have
\[ \sum {xy}\le \frac {1 \plus{} 9xyz}{4}\le 2xyz \plus{} \frac {7}{27}
\]
because from $ AM\ge GM$, $ xyz\le \frac {1}{27}$

 Also \[ xy\plus{}yx\plus{}zx\ge \sqrt{3xyz(x\plus{}y\plus{}z)}\equal{}\sqrt{3xyz}\ge 2xgz\]

This problem was posted several times. Here is a link that has a number of different solutions: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33007


The complex number Z=X+iY is such that IzI<1 and Y>0

Prove 0 <arg(Z+2)< $\frac{\pi}{6}$

[hide="Solution"] The locus of $z+2$ in the complex plane is the upper half of a semicircle of radius $1$ centered at $(2, 0)$.  The tangent to this semicircle passing through the origin is the line with the greatest angle to the $x$-axis, so its intersection with the semicircle is the point of greatest argument.  (The points of smallest argument are the points approaching the $x$-axis.  They do not touch there because their imaginary parts are strictly positive.)

Let's calculate that angle.  Let's call the points in question $A(0, 0), B(2, 0), C(x+2, y)$, so that $\angle C = \frac{\pi}{2}$.

Clearly $AB = 2$.  Because $BC$ is a radius, $BC = 1$.  Then of course $\angle BAC = \arcsin \frac{1}{2}= \frac{\pi}{6}$.

Because we specified that $|z| < 1$, our locus does not contain the perimeter of the semicircle, so this is a strict upper bound. [/hide]


search
combinatorics proposed
combinatorics

We are to colour all the edges and diagonals of an n-gon such that no two lines of the same colour meet at any vertex. Find the minimum number of colours required

You are asking about the edge-chromatic index of the complete graph.  For background on edge-coloring, see http://mathworld.wolfram.com/EdgeChromaticNumber.html or http://en.wikipedia.org/wiki/Edge_coloring -- for a discussion of this particular problem, see http://scienceblogs.com/goodmath/2007/07/edge_coloring_and_graph_turnin_1.php (though I bet that this has been discussed on the forum at some point -- try a search!).

Thankyou JBL. That was very much helpful


i don't think they are stupid and embarissing.  I think mine are.

[quote="juicybooty911"]man, i was reading some of my old posts, and they are like so stupid and embarrassing. :ewpu:[/quote]
haha that is just like math proofs. every time i go back and read my old proofs, there is always something that i can correct, never perfect!!! but i guess it is unlike stupidity and embarrassment... :D


Does anyone know how this year's quarterfinal exam is going to look like? Is it going to be AIME style format or like the preliminaries from previous years? Also is it just mechanics or should I also go over Electricity and Magnetism?

http://www.aapt.org/olympiad2008/

The exam is 1 hour in length, and will test mechanics and electrostatics. I assume it will be like previous years' open response exams, or at least I guessed based on the description.

Does electrostatics include capacitors? What specifically does electrostatics include?


A rectangular lawn measures 60 feet by 80 feet. The edges of the lawn are torn up to install a sidewalk of uniform width around it. The area of the new lawn is one-sixth the area of the old. How many feet wide is the sidewalk?


Does anyone know example of set $A \in \mathbb{R}$ that is not Borel?

For example, we can take the well-know set which is not Lebesgue measurable exactly $A=A' \bigcap (0,1)$, where in $A'$ is the set of representatives of $\mathbb{R} {/ \mathbb{Q}}$.
If it was Borel, it would be Lebesgue measurable

Or if you just want an existence proof, without the actual construction, use the fact that the cardinality of the collection of Borel sets is $c$ (the concinuum cardinality), whereas the cardinality of the collection of all real sets is $2^c>c$.


Eight percent of what number is eight less than eighty percent of eight hundred?

[hide] We have $ x(80\%)\equal{}800(80\%)\minus{}8 \implies \frac{4x}{5}\equal{}56 \implies x\equal{}\boxed{70}$.[/hide]


is there some theoreom that says that if $ a_{1} \equiv a_{2} \pmod{n}$ and $ b_{1} \equiv b_{2} \pmod{n}$ then $ a_{1}^{b_{1}} \equiv a_{2}^{b_{2}}\pmod{n}$ or something like that? with those exponent thingies? thx. :D

No; for example, $ 2\equiv5\pmod3$, but obviously, $ 2^2\equiv1\not\equiv5^5\equiv2\pmod3$


Let $ M,N$ be tow points outside of $ \triangle ABC$ such that: $ AM\equal{}AN$ and $ \angle AMB\equal{}\angle ANC\equal{}90^0$ .Let $ F$ be the points on the side BC such that: $ \frac{BF}{FC}\equal{}\frac{BM}{CN}$ .Prove that: $ \angle FMN\equal{}\angle CAN$
I think that this problem not diffcult but I can't slove it

[quote="livetolove212"]http://www.mathlinks.ro/viewtopic.php?t=291093[/quote]
thank you very much


On Friday, a snowboard originally priced at $100 was
discounted 50%. On Monday, that sale price was reduced by
30%. In dollars, what is the price of the snowboard after the
Monday reduction?


The number of elements of the solution set of inequality $ \frac{2x}{x\minus{}2}\leq1$, in $ \mathbb{Z}$ , is:

a) 0
b) 1
c) 4
d) 3
e) 2

If $ x > 2$ then $ 2x \le x-2$ and so $ x \le -2$ which cannot hold.

If $ x < 2$ then $ 2x \ge x-2$ and so $ \4x \ge -2$.

Therefore, for integer values of $ x$, we have $ x= -2, -1, 0$ or $ 1$, that is, the number of elements in the solution set is $ 4$.

Thanks Andrew. Very simple method, very good.

[geogebra]26b266bdba4113f602b45bdcd98157b1cba5abcb[/geogebra]
$ y \equal{} \frac {2x}{x \minus{} 2} \equal{} \frac {4}{x \minus{} 2} \plus{} 2$

So, $ \frac {2x}{x \minus{} 2} \le 1 \iff \minus{} 2 \le x < 2$.
Therefore $ x \equal{} \minus{} 2, \minus{} 1, 0, 1$.
Thus, $ \boxed{C}$.


induction
superior algebra
superior algebra solved

Let $ \left(a_n\right)_{n\geq 1}$ be a sequence such that
$ a_{n\plus{}1}\equal{}\frac{n}{n^2\minus{}1}\left(a_1\cdot a_2\plus{}a_2\cdot a_3\plus{}...\plus{}a_{n\minus{}1}\cdot a_n\right)$ for any $ n\geq 2$ and such that $ a_1$ and $ a_2$ lie in $ \left(0,\frac12\right)$.
Prove that the sequence $ \left(a_n\right)_{n\geq 1}$ converges.

The general term can be write such that :
a(n+1) = (n/(n^2-1))*(a(n)*((n-1)^-1)/(n-1) + a(n-1)*a(n)) = (n/(n+1))*a(n)*(1+a(n-1)/(n-1))

Now we  prove by using  induction : 0 =< a(n) < n/(n+1 for n >= 1, and  a(n+1) =< a(n) for n > 1
The initial step is manifestly thru, because of 0 =< a(1) =< 1/2 , 0 =< a(2) =< 2/3
After some calculation we have :
0=< a(n-1) =< (n-1)/n   ===>   0 < 1+a(n)/(n-1) =< (n+1)/n   ===>  0 =< a(n+1) =< a(n)
(0=< a(n-1) =< (n-1)/n) & (a(n) =< n/(n+1))  ===> 0 =< a(n+1) =< (n/(n+1))^2* (1+1/n) = n/(n+1) < (n+1)/(n+2)

Manifestly the sequence {a(n)} converges because it decreases and is bounded.

I am very sorry. I made a mistake in my last post. I hope this following one is correct.

The general term can be write such that: a(n+1) = (n/(n^2-1))*(a(n)*((n-1)^2-1)/(n-1) + a(n-1)*a(n))

Now we prove by using induction: 0 =< a(n) < 1 for n >= 1, and  a(n+1) =< a(n) for n > 1
The initial step is manifestly thru, because of 0 =< a1 =< 1 , 0 =< a2 =< 1 , 0 =< a(3) = (2/3)*a1*a2 =< 2/3
After some calculation we have:
0=< a(n-1) =< 1   ===>   0 < (n-1)^2-1)/(n-1)+a(n-1)  =< (n^2-n-1)/(n-1) < (n^2-1)/n ===> 0 =< a(n+1) =< a(n)

Manifestly the sequence {a(n)} converges because it decreases and is bounded.


It's either that I'm lucky, dumb, or smart on this problem:

It was from AoPS book one.  Find [tex]a^3+b^3[/tex]        if [tex]a^2+b^2=ab=4[/tex]

I got correct answer which was zero but my way was totally different compared to answer in art of problem solving book.

Here it goes:

[tex]a^3+b^3 = (a+b)(a^2-ab+b^2)[/tex]
[tex]a^3+b^3 = (a+b)(a^2+b^2-ab )[/tex]
[tex]a^3+b^3 = (a+b)((4)-(4)[/tex]
[tex]a^3+b^3 = (a+b)(0)[/tex]
[tex]a^3+b^3 = 0[/tex]

So, we have 0 for our answer.  For the ppl who knows math better than I am, is my method just a dumb method or a method that is out of really good thinking. (I didn't really think, though)

That method is probably just as good as the solution in the book.  I think it is good that you got it using a different method than the book taught.  Just be sure to understand and be able to use the book's method because your way wouldn't help much if the question asked for [tex]a^5+b^5[/tex] or something else like that.

This is how I would do it:

(a^2+b^2)^2=4^2
a^4+2a^2b^2+b^4=16
(a^4+b^4)=16-2a^2b^2

ab=4 so a^2b^2=16

2*a^2*b^2=32

16-32=-16

However, I don't see why your method is wrong...

EDIT: Whoops, they are looking for a^3+b^3, not a^4+b^4, my bad!

Both methods work, but yours seems easier than the one in the answer book.  (for those without AoPS volume 1, the answer book's solution was to square a+b to find a numerical value for a+b, then cube a+b and use substitution).

I agree with nr1337. The book's solution was harder to understand than yours.