Is the HP 49 G+ graphing calculator good for AP Statistics?  My teacher wants me to get a TI 83 or 84 but I would rather get this(or an 89) since they are more powerful.  Does anyone have any suggestions?

All advice I read says the TI-83+ (or its recent replacement, the 84+) is the way to go for studying statistics. The feature set includes many statistics-friendly features, and just about all introductory statistics books these days include keystroke guides for that model.

Hmm... I'm not sure, but I would recommend getting the TI... They are more popular and a lot of math books depend on the specific calculator.  Another good reason is that there are a lot of games for the TI's  :lol: .  Haven't seen any HP games, but who knows.

I hate TI-89's for statistics stuff.  I use it for everything that does not involve lists.  I've just always felt that it's list capabilities are harder to use than the TI-83's.

I have a TI-89 Titanium, and it's pretty cool, with lots of memory (like 5 times more than 84+ I think). It also has better graphics and tons of programs, such as cellsheet, eepro,contacts, finance, poly root finder,simul equ solver, and a statistics program that comes in like french and dutch. Overall, it's pretty good and there are good games for it for free. :D  :D

The only thing is that the 89 is more expensive than the 84: mine costs $130 or something like that.


college
real analysis
real analysis solved

this cake is dedicated to your afternoon :lol:
assume that $S$ is uncountable in $\mathbb{R}$ prove that there exists $t\in \mathbb{R}$ such that both $S\bigcap (-\infty,t),S\bigcap (t,+\infty)$ are uncountable .what is the cardinal of t? :blush:

To prove the existence, consider the set $m$ (assume it's non-void; if it's void, it's even easier for us to solve the problem) containing $x$ s.t. $S\cap (x,\infty)$ is countable, and take $t=\inf m$. It's easy to show that $S\cap(t,\infty)$ is also countable. Consider now the set $M$ of $x$ (also assumed non-void) s.t. $S\cap(-\infty,x)$ is countable, and take $T=\sup M$. We also get $S\cap(-\infty,T)$ to be countable. If $T\ge t$, then $S$ is countable, so $T<t$. Now take any number between $T$ and $t$.

In this case, the set seems to have continuum cardinality (the interval $(T,t)$). I think it's the same in the other cases, when either $m$, or $M$, or both are void.

[Edit:

I assumed it obvious that if $m$, for instance, is not void, then it's bounded from below, because otherwise we can write $S$ as a countable union of countable sets; something similar holds for $M$]

I remember it is one of the problems of Berkeley Examination

@Grobber:
 If $m$ is void then we can define $\inf{m} = +\infty$
 If $M$ is void then we can define $\sup{M} = -\infty$
 Just to make it easier to present!! Isn't it? :D

 @sam-n : mods say about naming topics carefully all the times!! You should be careful :D

 Why Myth's so silent??


LaTeX
function
calculus
derivative
integration
calculus computations

i have this

Find f.
f ''(t) = 4e^t + 2sin(t)
f(0) = 0
f(pi) = 0
f(t)=__________

but i dont know how to do this

Let me put your question in $ \text{\LaTeX}$ to make it more readable.

We have for some function $ f$ that:

$ f''(t) = 4 e^t + 2 \sin(t)$
$ f(0) = 0$
$ f(\pi) = 0$

Find $ f(t)$.

Now, we know the second derivative of $ f$. So if we would integrate the given expression on both sides we would regain $ f$. Do you know how to do this? (Be mindful of the constants you get when doing indefinite integration!) After you have done this, use the other two facts to solve for the constants you got while integrating.

In case something is still unclear, please show us what you have done and where exactly the difficulties are.

well i am having difficulties with getting the antiderivative of e

if i have 4e^t, what is the antidirivative..?

also , it gives a values for the normal function, so how can i solve for the constant for the first derivative

Let's denote a function by small letters $ f, g$ etc, and their antiderivatives by capitals $ F, G$ etc.

We have that for $ f(t) \equal{} e^t$, $ F(t) \equal{} e^t \plus{} C$, where $ C$ is some constant.

Furthermore, if we take $ g(t) \equal{} a \, f(t)$, we have $ G(t) \equal{} a \, F(t) \plus{} C$, if $ a$ is some number.

This means that the antiderivative of $ 4 e^t$ is again $ 4 e^t$ (plus some constant, to be precise).

I will do a similar (easier) problem, to show how it goes.

----------------------------------------------
If we know for some function $ f$ that:

$ f''(t) \equal{} e^t$
$ f(0) \equal{} 0$
$ f(1) \equal{} 0$

Find $ f$.
----------------------------------------------

We have $ f'(t) \equal{} \int f''(t) \, dt \equal{} \int e^t \, dt \equal{} e^t \plus{} C$, where $ C \in \mathbb{R}$.

This means that $ f(t) \equal{} \int f'(t) \, dt \equal{} \int \left[ e^t \plus{} C \right] \, dt \equal{} e^t \plus{} C t \plus{} D$, where $ D \in \mathbb{R}$.

Now, we now have that $ f(t) \equal{} e^t \plus{} C t \plus{} D$. We are ready to use the other two equations to find $ C$ and $ D$.

If we use the first one, we get $ 0 \equal{} f(0) \equal{} e^0 \plus{} C 0 \plus{} D \equal{} 1 \plus{} D$, so $ D \equal{} \minus{}1$.

If we then use the second one, we get $ 0 \equal{} f(1) \equal{} e^1 \plus{} C 1 \minus{} 1$, so $ C \equal{} 1 \minus{} e$.

We then get that $ \boxed{f(t) \equal{} e^t \plus{} (1\minus{}e)t \minus{} 1}$.

Are you able to do your problem now?

i get it now, but what if it were e^(t^2 -t)

would the anti D. be the same thing?

Definitely not.

"Finding antiderivatives" is also known as indefinite integration. The most important thing you need to know about finding derivatives is the chain rule. It follows that the most important thing you need to know about finding antiderivatives is the chain rule read backwards, which is to say, substitution.

Here's a derivative: $ \frac{d}{dt}e^{t^2\minus{}t}\equal{}(2t\minus{}1)e^{t^2\minus{}t}.$

The corresponding antiderivative would be that the antiderivative of $ (2t\minus{}1)e^{t^2\minus{}t},$ also written as $ \int (2t\minus{}1)e^{t^2\minus{}t}\,dt,$ is $ e^{t^2\minus{}t}\plus{}C.$

But note that that does [i]not[/i] tell us anything at all about $ \int e^{t^2\minus{}t}\,dt$ (the antiderivative of $ e^{t^2\minus{}t}.$) And in fact, while that antiderivative exists, it cannot be represented in a finite number of steps using functions whose names you know.

That happens some of the time with antiderivatives. In fact, I've gone so far as to call it the normal state of affairs; the ones we can represent in closed form are the exceptions.

for example for this question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
(3 being on the bottum)
g(x) = integral sign (from 3 to x) e^(t^2-t) dt

i am getting the anti D. to be (2t-1)e^(t^2-t)

so that would mean the answer is

(2x-1)e^(x^2-x)   -    5e^6

i am getting this wrong, can u explain why it is

Whoa, whoa, whoa!

$ \frac {d}{dx}\int_3^xe^{t^2 \minus{} t}\,dt.$

Step one. Stop trying to answer this. Proceed no further. And ignore what you've already written, since it is total nonsense.

Step two. Look in your textbook, or online, or wherever you look to find the definition of the Fundamental Theorem of Calculus. There are two versions, or two aspects of the FTC; what you want is the "differentiation of the integral" version. It's there somewhere - it's in every calculus book that I know about. Find it. Read it carefully. Read it again. Look at the examples around it.

And don't come back until you've done that.

... and given what you posted, I suspect that the relevant part is labeled "part 1" in the textbook you took this problem from.

i get why that what i wrote was wrong

but that leaves me with getting the antiderivative of e^(t^2-t), there is no similar question like this, nor could i find on the internet...

if i could get how to do this, i can figure this question out...

[quote="Kent Merryfield"]Look in your textbook, or online, or wherever you look to [b]find the definition of the Fundamental Theorem of Calculus[/b]. There are two versions, or two aspects of the FTC; what you want is the "differentiation of the integral" version. It's there somewhere - it's in every calculus book that I know about. Find it. [b]Read it carefully.[/b] [i][b]Read it again.[/b][/i] [u][i][b]Look at the examples around it.[/b][/i][/u]

And don't come back until you've done that.[/quote]

ok i figured that out (finally...)

but what would be the Anti D of 6^x

doing (6^(x+1))(1/(x+1)) would be wrong?

You would really benefit from some quality time spent reading your textbook and notes.  Yes, what you've written is wrong: you're applying a formula in a completely invalid context.  As [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1679546#1679546]someone recently noted[/url], when you remember a theorem (one special case is a computational rule of this sort), you must remember not only the [i]conclusion[/i] of the theorem but also the [i]conditions under which the conclusion is valid[/i].  The function $ 6^x$ is an exponential function (the variable appears in the exponent) -- you would be well-served to spend a few minutes reading the sections of your text and notes that deal with derivatives and integrals of exponential functions.

k i figured it out
but how would u do the Anti D of

4
--------
t^2 + 1


function
real analysis
real analysis unsolved

Hi everyone,
I guess this is the right place to post this question. I can't solve the following diff. eq. Can anyone help me with it? Of course the solution to modified Bessel is known but I don't know what to do when there are other functions. In short, I can't seem to find the particular solution.
Thanks a lot,

Maple says a particular solution to
\[ {\frac {d^{2}}{d{r}^{2}}}X \left( r \right) \plus{}{\frac {{\frac {d}{dr}}X
 \left( r \right) }{r}}\minus{}{A}^{2}X \left( r \right) \equal{}B\plus{}C
{{\rm J}_0\left(D  r\right)}\]
is
\[ X \left( r \right) \equal{}\minus{}{\frac {r{{\rm I}_0\left(Ar\right)}C{A}^{3}
{{\rm K}_1\left(Ar\right)}{{\rm J}_0\left(D\,r\right)}\plus{}r
{{\rm K}_0\left(Ar\right)}C{A}^{3}{{\rm I}_1\left(Ar\right)}
{{\rm J}_0\left(D\,r\right)}\plus{}B{D}^{2}\plus{}B{A}^{2}}{{A}^{2
} \left( {D}^{2}\plus{}{A}^{2} \right) }}\]

[quote="blueeverest"]I will see if this [particular] solution works and let you know. I tried to find it in Mathematica but the answer apparently didn't make any sense. So there is NO analytical way to figure out the solution here, correct?
Again, thanks for the help.[/quote]

Of course there is an analytical way.  It is called "variation of parameters".  If you know the general solution to the homogeneous equation (as we do in this case, in terms of Bessel functions) it amounts to doing some integrals.


superior algebra
superior algebra unsolved

Let $ A$ be the ring with the identity, and $ (ab)^2\equal{}a^2b^2 \forall a,b\in A$. Prove that $ aba\equal{}a^2b$ and $ A$ is commutative.

$ (ab)^2 \equal{} a^2b^2$ and $ (a(b \plus{} 1))^2 \equal{} a^2(b \plus{} 1)^2$ give $ aba \equal{} a^2b$.
$ aba \equal{} a^2b$ and $ (a \plus{} 1)b(a \plus{} 1) \equal{} (a \plus{} 1)^2b$ give $ ab \equal{} ba$.


calculus
integration
trigonometry
calculus computations

[hide="3"]
$ \int{\left({x^{7m} \plus{} x^{2m} \plus{} x^{m}}\right)}\left({2x^{6m} \plus{} 7x^{m} \plus{} 14}\right)^{1/m}dx$
multiply and divide by $ x$
$ \int (x^{7m \minus{} 1} \plus{} x^{2m \minus{} 1} \plus{} x^{m \minus{} 1})(2x^{7m} \plus{} 7x^{2m} \plus{} 14x^{m})^{\frac {1}{m}} dx$
put $ 2x^{7m} \plus{} 7x^{2m} \plus{} 14x^{m} \equal{} y$
$ \equal{}\int \frac {y^{\frac {1}{m}}}{14} dy$
now u can proceed .
[/hide]

[hide="4"]
put $ x \equal{} 2a cos (2 \theta)$
$ \int (\minus{}2a^2 sin(4 \theta) d(\theta))$

[/hide]

hello, for 2) rewriting your integral in the form
$ \int\frac{t\sqrt{t\plus{}5}}{(1\minus{}t^2)^2}\minus{}\frac{t\sqrt{t\plus{}4}}{(1\minus{}t^2)^2}\,dt$.
Sonnhard.


Is there any example of subset of $ \mathbb{R}^3$ which satisfy the scalar multiplication properties by does not satisfy the addition property?

I can only think of subset which satisfy addition property but not scalar multiplication property like $ \{(x,y,z)\in\mathbb{R}^3|x\geq 0\}$

Anyone ?

Take for example $ \left\{\left(x,y,z\right)\in\mathbb{R}^{3}\mid x\equal{}y\equal{}0\right\}\cup \left\{\left(x,y,z\right)\in\mathbb{R}^{3}\mid y\equal{}z\equal{}0\right\}$.


Solve in real number the system equation :

\[ \begin{array}{*{20}{c}}
   {({x^2} \minus{} 6x \plus{} 13)y \equal{} 20}  \\
   {({y^2} \minus{} 6y \plus{} 13)z \equal{} 20}  \\
   {({z^2} \minus{} 6z \plus{} 13)x \equal{} 20}  \\
\end{array}\]

 :arrow:

If $ x\equal{}y$, then $ (x\minus{}4)(x^2\minus{}2x\plus{}5)\equal{}0$ or $ x\equal{}y\equal{}4\equal{}z$.
Because $ x^2\minus{}6x\plus{}13\equal{}(x\minus{}3)^2\plus{}4$  we get $ 0<x,y,z\le 5 \to \frac{20}{13}<x,y,z\le 5\to 2.5\le x,y,z\le 5.$
If $ x>4$, then $ 2.5\le z<4$ and $ 2.5\le y<4$ and $ (y^2\minus{}6e\plus{}13)z<20$.
If $ x<4$, then $ (z\minus{}3)^2>1\to z>4$. In these same $ y<4$ and $ (x^2\minus{}6x\plus{}13)y<20$.
Therefore $ x\equal{}4\to y\equal{}z\equal{}4$ unique solution,

[quote="Rust"]
$ 0 < x,y,z\le 5 \to \frac {20}{13} < x,y,z\le 5\to 2.5\le x,y,z\le 5.$[/quote]

Can you Explain for me :?:

[quote="Rust"]$ f(x) \equal{} x^2 \minus{} 6x \plus{} 13 \equal{} (x \minus{} 3)^2 \plus{} 4$. Therefore
\[ 0 < x,y,z\le 5\to 4\le x^2 \minus{} 6x \plus{} 13\le 13\to \frac {20}{13}\le x,y,z\le 5.\]
$ f(\frac {20}{13} \equal{} \frac {19^2}{13^2} \plus{} 4 \equal{} \frac {1037}{169}$. It give $ 3 < \frac {169*20}{1037}\le x,y,z\le 5.$[/quote]

Thankyou, [b]Rust[/b]. I have some idea for this : :)
Denote $ f(t) \equal{} \frac {20}{t^2 \minus{} 6t \plus{} 13}$, then we have : $ \begin{array}{*{20}{c}} {y \equal{} f(x)} \\
{z \equal{} f(y)} \\
{x \equal{} f(z)} \\
\end{array}$
or $ \begin{array}{*{20}{c}} {y \equal{} f(x)} \\
{z \equal{} f(y)} \\
{x \equal{} f(f(x))} \\
\end{array}$

then we examine $ f(x)$  :wink:

What do you think? dear mathlinker:?:

[quote="Rust"]
\]
$ f(\frac {20}{13} \equal{} \frac {19^2}{13^2} \plus{} 4 \equal{} \frac {1037}{169}$. It give $ 3 < \frac {169*20}{1037}\le x,y,z\le 5.$[/quote]

Dear, I think it is not right! because if $ x \ge \frac {20}{3}$  then not sure that $ f(x) \le f(\frac {20}{13})$.  Because $ f(x)$  decrease on $ [2,5;3]$ and increase on $ [3;5]$

 :wink:  :!:

Will you explain clearly for me that :$ x,y,z \in [3;5]$  :huh:


function
induction
algebra unsolved
algebra

Prove that:Exist exactly 1 function $ f$:$ \mathbb{N*}\rightarrow\mathbb{N*}$ such that:
 1,$ f(2)\equal{}2$
 2,$ f(m.n)\equal{}f(m).f(n)$ $ \forall m,n\in\mathbb{N*}$.
 3,$ f(m)<f(n)$ with $ m<n$.

$ f(1)\equal{}1$
We prove that
$ f(n)\equal{}n$
Check $ n\equal{}2,3,4$
Suppose chose for k
We will prove for $ k\plus{}1$
If $ k\plus{}1$ is compose number then
$ k\plus{}1\equal{}pq$ where $ p,q>1$
So
$ f(k\plus{}1)\equal{}f(p)f(q)\equal{}k\plus{}1$
If $ k\plus{}1$ is a prime then
$ f(k\plus{}2)\equal{}k\plus{}2$
From 2 then
$ f(k\plus{}1)\equal{}k\plus{}1$
So
$ f(n)\equal{}n ,\forall n$

I'm sorry,but I want you prove exist exactly one function $ f$ satisfing of the condition!!

[hide]
$ f(m)<f(n)\Rightarrow f(m)\plus{}1\leq f(n)$ when $ m<n$. It follows that $ f(a)\plus{}b\leq f(a\plus{}b)$ for all positive integers $ a,b$. As a result, $ f(1)\leq 1$ so $ f(1)\equal{}1$. For any $ N>2$, there exists $ M$ such that $ 2^{M}>N$. We then have
\[ N\equal{}f(2)\plus{}(N\minus{}2)\leq f(N)\leq f(2^{M})\minus{}(2^{M}\minus{}N)\equal{}N\]
(because $ f(2^{M})\equal{}2^{M}$ for all positive integers $ M$) so $ f(N)\equal{}N$. So $ f$ must be the identity function.
[/hide]

Yes unique $ f(n)$
What is your question?
My solution prove that exist unique solution of the function.

It is problem 8 from Canadian Math Olympiad 1969.
All solutions proves that the only function that satisfies these conditions is $ f(n) \equal{} n$. See the attached file.

[quote="duy_nguyen"]@TTsphn:You trying to show that the function $ f(n) \equal{} n$ satisfying of the condition by induction,but not unique.[/quote]
I think that from TTsphn's solution you could see that $ f(n)\equal{}n$ is the unique function. :maybe: He showed that every value of $ f(n)$ is unique and equal to $ n$.


A magic square is an array of numbers in which the sum of the numbers in each row, in each column, and along the two main diagonals are equal. The numbers in the magic square shown are not written in base 10. For what base will this be a magic square?

[asy]unitsize(0.75cm);
for (int i=0; i<4; ++i) {
draw((0,i)--(3,i),linewidth(0.7));
draw((i,0)--(i,3),linewidth(0.7));
}
label("1",(1.5,2),N);
label("2",(2.5,0),N);
label("3",(0.5,1),N);
label("4",(0.5,0),N);
label("10",(1.5,1),N);
label("11",(2.5,2),N);
label("12",(2.5,1),N);
label("13",(0.5,2),N);
label("14",(1.5,0),N);[/asy]

Let the numbers be in base $ n$.  Then we can set an equality between the first column and first row in terms of $ n$.

$ 3\plus{}3\plus{}4\plus{}n\equal{} 3\plus{}1\plus{}1\plus{}n\plus{}n$

$ 10\plus{}n\equal{} 5\plus{}2n$

$ 5\equal{}n$

We can check to ensure that $ \boxed{5}$ is the correct base.


logarithms
number theory
relatively prime
algebra proposed
algebra

Given $ m$ and $ n$ as relatively prime positive integers greater than one, show that
\[ \frac{\log_{10} m}{\log_{10} n}\]
is not a rational number.

[quote="Leon"]Given $ m$ and $ n$ as relatively prime positive integers greater than one, show that
\[ \frac {\log_{10} m}{\log_{10} n}\]
is not a rational number.[/quote]

$ \frac {\log_{10} m}{\log_{10} n}\equal{}\frac pq$ $ \implies$ $ m^q\equal{}n^p$ which is obviously impossible if $ m$ and $ n$ as relatively prime positive integers greater than one

[quote=ATGY]Note that $\[ \frac{\log_{10} m}{\log_{10} n}\] = \log_n(m)$. Clearly, since both of them area greater than $1$ and they are relatively prime, $\log_n(m)$ is irrational. Hence proved.[/quote]

This is not evident, must be reduced to the #2.

Assuming that $\frac{\log_{10} m}{\log_{10} n} = \frac{p}{q}$, where $p,q \in \mathbb{Z}$ and $q\neq0$. $\Rightarrow \log_n m= \frac{p}{q} \Rightarrow \log_n m^q= p \Rightarrow n^p=m^q.$

But, we know that $m$ and $n$ are relatively prime positive integers $\Rightarrow gcd(m,n)=1$. But from the above equation, we get that $m\vert n^p$ and $n\vert m^q$, which is only possible when $p=q=0$, as they are coprime, but as $q\neq0$, we have a contradiction. Q.E.D.[b][/b]


induction
inequalities unsolved
inequalities

If my computations are right, then $a_1+...+a_n=1-(2n-1)a_n$.

Thanks harazi but can you show me how you can get it? :D

By simple induction!

Your sequence is defined by $a_1=\frac12$ and by the recursion equation $a_{n+1}=\frac{2n-1}{2n+2}\cdot a_n$. Now, you can prove the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ by mathematical induction as follows:

For n = 1, the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ is trivial, since it then simplifies to $a_1=1-a_1$, what is clearly true because of $a_1=\frac12$.

Now assume that the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ holds for some n = k. We must show that it also holds for n = k + 1.

In fact, since the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ holds for n = k, we have

$a_1+a_2+...+a_k=1-\left(2k-1\right)a_k$;

thus,

$a_1+a_2+...+a_{k+1}=\underbrace{\left(a_1+a_2+...+a_k\right)}_{=1-\left(2k-1\right)a_k}+a_{k+1}=\left(1-\left(2k-1\right)a_k\right)+a_{k+1}$
$=1-\left(\left(2k-1\right)a_k-a_{k+1}\right)$.

On the other hand, by the recursive definition of the sequence $\left(a_n\right)$, we have $a_{k+1}=\frac{2k-1}{2k+2}\cdot a_k$; hence, $\left(2k+2\right)a_{k+1}=\left(2k-1\right)a_k$, and thus, $\left(2k-1\right)a_k=\left(2k+2\right)a_{k+1}=2\left(k+1\right)a_{k+1}$. Thus,

$\left(2k-1\right)a_k-a_{k+1}=2\left(k+1\right)a_{k+1}-a_{k+1}=\left(2\left(k+1\right)-1\right)a_{k+1}$.

Hence,

$a_1+a_2+...+a_{k+1}=1-\underbrace{\left(\left(2k-1\right)a_k-a_{k+1}\right)}_{=\left(2\left(k+1\right)-1\right)a_{k+1}}=1-\left(2\left(k+1\right)-1\right)a_{k+1}$.

Thus, the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ is proven for n = k + 1. Hence, our induction proof of the formula $a_1+a_2+...+a_n=1-\left(2n-1\right)a_n$ is complete.

  Darij


Prove that from $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} ab \plus{} bc \plus{} ac$ come up that $ a \equal{} b \equal{} c$  :)

Not really Pre-Olympiad.  And you probably want the condition $ a, b, c \in \mathbb{R}$ because there is a well-known related problem given below.

[hide="Note"] We can assume WLOG that $ a, b, c \ge 0$ since the terms on the LHS are all positive. [/hide]
[hide="Method 1"] $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca \Leftrightarrow (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 0$ [/hide]
[hide="Method 2"] $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$ by (three applications of) AM-GM, with equality if and only if $ a \equal{} b, b \equal{} c, c \equal{} a$. [/hide]
[hide="Method 3"] $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$ by Rearrangement, with equality if and only if the sequences $ (a, b, c), (b, c, a)$ are similarly sorted (which requires $ a \equal{} b \equal{} c$). [/hide]
[b]Follow-up:[/b]  Prove that if $ a, b, c \in \mathbb{C}$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca$, then $ a, b, c$ form an equilateral triangle in the complex plane.

[quote="Elio (n)"]Oh.... man . Three methods of solutions in only 3 minutes. :o

Can I quest torajrou what grade is he? :)[/quote]

Well , anyone who has studied inequalities enough would know those three methods. Your problem is used as an example in many books for these methods and is very well known.

:)

12th.  :)  Anaxerzia is correct, though.  The follow-up question, while also standard, is a little more interesting (it is not, for one thing, an inequality question anymore) and implies the result for real numbers (an equilateral triangle on the real line is a point).  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1508094362&t=184462]another thread[/url] where the follow-up was discussed.

[hide="Method 4."]

For $ a,\ b,\ c\in\mathbb{R},$

$ a^2\plus{}b^2\plus{}c^2\equal{}ab\plus{}bc\plus{}ca$

$ \Longleftrightarrow a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}\left(a\minus{}\frac{b\plus{}c}{2}\right)^2\plus{}\frac{3}{4}(b\minus{}c)^2\equal{}0$

$ \Longleftrightarrow a\minus{}\frac{b\plus{}c}{2}\equal{}0$ and $ b\minus{}c\equal{}0$, yielding $ a\equal{}b\equal{}c$.

[/hide]

[hide="Method 5"]
I think every possible proof must be a bit similar.
As [b]t0rajir0u[/b] sadi we can assume $ a,b,c \geq 0$
thus we have $ (a \plus{} b \plus{} c) \equal{} 0 \iff a \equal{} b \equal{} c \equal{} 0$
We also have
$ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca \equal{} 0 \iff (a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} ab \minus{} bc \minus{} ca) \equal{} a^{3} \plus{} b^{3} \plus{} c^{3} \minus{} 3abc \equal{} 0$
thus $ \frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{3} \equal{} abc$ which yields from AM-GM $ a \equal{} b \equal{} c$ and the conclusion follows :wink: [/hide]


geometry
geometric transformation
blogs
articles

well... actually, I was translating it manually. you know, automatic translation isn't good at all.

but, if you feel you can read it with that translation, feel free and enjoy it. welcome to our magical blog  :)

jackdillon

lets learn japanese and translate this blog to japanese?!?!

or latin!!!!

wooooo

thats goood

i will learn french

I've created a good article there, for you AoPSer's

enjoy it and have fun!

if you also want to put articles there, the pass of the blog is "biscoitodeleite"

enjoy it and have fun! let's do a mega blog!


inequalities
complex numbers
absolute value
advanced fields
advanced fields unsolved

I'm trying to understand a proof of Selberg's Identity. That is let $ y_j$ $ (1\leq j\leq N)$ be arbitrary element of an inner product space H and let $ x\in H$ Then prove that:

$ \sum_{j = 1}^N(\frac {|\langle x,y_j\rangle|^2}{\sum_{k = 1}^N|\langle y_j, y_k\rangle|})\leq\parallel{}x\parallel{}^2$.

I'm working through the proof on p56 of Inner product structures by Vasile I. Istrățescu.

There is a part where it says if x and y are complex numbers then: $ |xy|\leq\frac {1}{2}(|x|^2 + |y|^2)$ how do you prove this?

It then says that: Hence if $ \lambda_j$ are complex unmbers then:

$ \sum_{j,k = 1}^N\lambda_j\lambda_k\langle y_j,y_k\rangle = \frac {1}{2}\sum_{j,k = 1}^N(|\lambda_j|^2 + |\lambda_k|^2)|\langle y_j,y_k\rangle|\leq\sum_{j,k = 1}^N|\lambda_j|^2|\langle y_j,y_k\rangle|.$

Can someone explain how this follows please

Or if its easier can someone explain how to prove the identit in another way? Ta

[quote="zelda_leonhart"]There is a part where it says if x and y are complex numbers then: $ |xy|\leq\frac {1}{2}(|x|^2 \plus{} |y|^2)$ how do you prove this?[/quote]
Note $ |xy| \equal{} |x| \ |y|$, so the statement is equivalent to $ 0 \leq \frac{|x|^2 \minus{} 2 |x| \ |y| \plus{} |y|^2}{2}$, or $ 0 \leq \frac{(|x| \minus{} |y|)^2}{2}$ which is clear.

[quote="zelda_leonhart"]It then says that: Hence if $ \lambda_j$ are complex unmbers then:

$ \sum_{j,k \equal{} 1}^N\lambda_j\lambda_k\langle y_j,y_k\rangle \equal{} \frac {1}{2}\sum_{j,k \equal{} 1}^N(|\lambda_j|^2 \plus{} |\lambda_k|^2)|\langle y_j,y_k\rangle|\leq\sum_{j,k \equal{} 1}^N|\lambda_j|^2|\langle y_j,y_k\rangle|.$

Can someone explain how this follows please[/quote]
I think the $ \equal{}$ and $ \leq$ should be switched. The inequality follows by applying $ |xy|\leq\frac {1}{2}(|x|^2 \plus{} |y|^2)$ to $ \lambda_j \lambda_k$ in every term of the summation. The equality comes from grouping terms differently. The term for $ j,k$ and the term for $ k,j$ is $ \frac{1}{2}\left( (|\lambda_j|^2 \plus{} |\lambda_k|^2)|\langle y_j,y_k\rangle| \plus{} (|\lambda_k|^2 \plus{} |\lambda_j|^2)|\langle y_k,y_j\rangle| \right)$, which groups into $ |\lambda_j|^2 |\langle y_j,y_k\rangle| \plus{} |\lambda_k|^2 |\langle y_j,y_k\rangle|$, which are two terms in the second summation. (We're using the inner product being symmetric here)

[quote="MellowMelon"](We're using the inner product being symmetric here)[/quote]

I thought the inner product was conjugate symmetric? Am I wrong?

Also from the first expression to the second where does the absolute value signs around the inner product come from?

[quote="MellowMelon"]Right, conjugate symmetric. But $ |z| \equal{} |\overline{z}|$ and we're only dealing with the magnitudes of the inner products, so that's not an issue. Sorry for skipping ove rthat.

The absolute value signs themselves I'm guessing are the triangle inequality... not positive about that. Though to say
$ \sum_{j,k \equal{} 1}^N\lambda_j\lambda_k\langle y_j,y_k\rangle \leq \frac {1}{2}\sum_{j,k \equal{} 1}^N(|\lambda_j|^2 \plus{} |\lambda_k|^2)|\langle y_j,y_k\rangle|$
is comparing a complex number (LHS) to a real (RHS), which doesn't make sense. Maybe it's supposed to be
$ \left| \sum_{j,k \equal{} 1}^N\lambda_j\lambda_k\langle y_j,y_k\rangle \right| \leq \frac {1}{2}\sum_{j,k \equal{} 1}^N(|\lambda_j|^2 \plus{} |\lambda_k|^2)|\langle y_j,y_k\rangle|$
in which case it's definitely the triangle inequality.

At this point I'm making too many corrections to their statement, so I think someone else is going to have to give a more definitive answer.[/quote]

Well basically I am worried about the text. I found typos elsewhere in the book and was questioning its worthyness.

Most of the possible corrections I thought should be made were the ones you mentioned. Its just that part about the absolute value signs thats bothering me. I too thought it should be the absolute value of the whole thing on the LHS, but I'm not sure that then fits in with the rest of the proof. I'll have a look tomorrow and see what I can see.


What is the sum of all possible values of $ k$ for which $ x^2 \plus{} kx \minus{} 12x \plus{} 16$ is the square of a binomial?

This will be a square of a binomial if it's in the form:

$ x^2 \plus{} 8x \plus{} 16$

or

$ x^2 \minus{} 8x \plus{} 16$

$ k \minus{} 12 \equal{} 8$

$ k \equal{} 20$

and

$ k \minus{} 12 \equal{} \minus{}8$

$ k \equal{} 4$

$ 4 \plus{} 20 \equal{} \boxed{24}$

The middle term in a binomial square will always be $ \frac{1}{2}$ the value of the multiplied first and last terms.

In other word(or should I say, symbols):

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

$ (a \minus{} b)^2 \equal{} a^2 \minus{} 2ab \plus{} b^2$

Find out more about special cases and factoring in my blog, Encyclopedia Mathematica!  :D


geometry
3D geometry
sphere
calculus
integration
real analysis
real analysis unsolved

The surface area of the part of the sphere $x^2+y^2+z^2=4$ that lies within the cylinder $x^2+y^2=2y$.

I looked at and thought: Easy convert to polar! But after putting it all together in polar and crunching it I ended up getting stuck with a nasty integral that involved integrating: $(sin^2\theta+1)^{\frac{3}{2}}$ if that was a minus sign, no problem. Even my TI-89 refuses to integrate that.

lol! My professor gave us a very similar problem but also gave us a hint. I am also stumped but maybe you can get it with his hint! He gave us: $x^{2}+y^{2}+z^{2}=9$ and $x^{2}+y^{2}=3y$ and the hint: $\int\frac{x}{\sqrt{a^{2}-x^{2}}}\,dx=-\sqrt{a^{2}-x^{2}}+c$.

$A(S)=\int\int_{E}\sqrt{1+\frac{x^{2}}{4-x^{2}-y^{2}}+\frac{y^{2}}{4-x^{2}-y^{2}}} = \int\int_{E}\sqrt{1+\frac{x^{2}+y^{2}}{4-x^{2}-y^{2}}}$ =
$\int\int_{E}\sqrt{\frac{4-x^{2}-y^{2}+x^{2}+y^{2}}{4-x^{2}-y^{2}}}$ = $\int\int_{E}\sqrt{\frac{4}{4-x^{2}-y^{2}}}$ = $\int\int_{E}\frac{2}{\sqrt{4-x^{2}-y^{2}}}$

Not sure where the hint kicks in...but I am stuck there cause I don't know what $E$ is....

That integral (in the original post) is a variation on an elliptic integral, and it doesn't have an elementary antiderivative.

There are several possibilities:
1: The problem is evil.
2: You made a mistake somewhere.
3: You should have integrated in a different order.

I think #3 is most likely.

Now I feel stupid....I already did this problem here on the boards for polar at :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61269]Polar Conversion[/url]

@lajekahr:
I have no idea how you would do this without a polar conversion sorry can't help there!

Wow, that is a fairly easy Polar Conversion....Why on Earth would my professor give us a [b]hint[/b] using Cartesian???

Thanks for that link sangre. I think I will just answer using the polar conversion cause...well...cause it makes more sense. Unless someone here can show me how that [i]hint[/i] could possibly be used....


prove if a,b,c>0 and abc=1 then
$ a^3\plus{}b^3\plus{}c^3\minus{}3>\equal{}\sum\frac{a}{b\plus{}c}\minus{}\frac{3}{2}$

[quote="tranquoc"]prove if a,b,c>0 and abc=1 then
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3 > \equal{} \sum\frac {a}{b \plus{} c} \minus{} \frac {3}{2}$[/quote]
it's easy to prove by use SOS method
the inequality is equivalent to
$ \sum {S_c(a\minus{}b)^2}\geq 0$ with
$ S_a\equal{}a\plus{}b\plus{}c\minus{}\frac{1}{(a\plus{}b)(a\plus{}c)}>0$
similar with $ S_b,S_c$ we get the same result.so the inequality has been proved.!

$ a^3\plus{}b^3\plus{}c^3 \ge a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c}\equal{}\sum \frac{a}{\sqrt{bc}} \ge \sum \frac{2a}{b\plus{}c} \ge \\\sum \frac{a}{b\plus{}c} \plus{}\frac{3}{2}$
Done :D

[quote="tranquoc"]prove if a,b,c>0 and abc=1 then
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3 > \equal{} \sum\frac {a}{b \plus{} c} \minus{} \frac {3}{2}$[/quote]
By the hypothesis, we have
$ a^3\plus{}b^3\plus{}c^3\equal{}\sum\frac{a^2}{bc}\geq 4.\sum (\frac{a}{b\plus{}c})^2\geq\frac{4}{3}.(\sum\frac{a}{b\plus{}c})^2\geq\sum\frac{a}{b\plus{}c}\plus{}\frac{3}{2}$

This inequality is true ,but very easy:
$ a^3\plus{}b^3\plus{}c^3\plus{}\frac{3}{2} \geq 3\sum\frac{a}{b\plus{}c}$
AND you???????????????????/

;).
$ 2LHS  \ge 3\sum a^2 \ge  3\sum a\sqrt{a} \equal{} 3\sum   \frac{a}{\sqrt{bc}} \ge 2RHS$
=>done

It's equivalent to
$ \sum (a\minus{}b)^2[\frac{1}{2}(a\plus{}b\plus{}c)\minus{}\frac{3}{2(a\plus{}c)(b\plus{}c)}]\geq 0$ (by hypothesis $ abc\equal{}1$)

By schur $ a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq \sum ab(a\plus{}b)$

abc=1-->  $ a^3\plus{}b^3\plus{}c^3\plus{}3 \geq \sum\frac{a\plus{}b}{c} \geq 4\sum\frac{a}{b\plus{}c}$

Done :rotfl:


Let's call three points $ A, B, C$. They represent three cities on a
map. There are several roads connecting any given pair of cities. To
go from one city to another, you can either take one of the direct
roads connecting the cities, or take one of the numerous possible
routes via the third city.

There are $ 106$ ways to go from $ A$ to $ B$, and $ 74$ ways to go from $ A$ to $ C$.
You have an even wider choice to go from $ B$ to $ C$.

In how many ways can you go from $ B$ to $ C$?

[hide="Hint"] Let $ a$ be the number of roads connecting $ B$ and $ C$, and so forth.  How many total ways are there to go from $ A$ to $ B$ (or whatever) in terms of $ a, b, c$? [/hide]

Okay, I haven't been online for quite some time now, so I'm rusty...

[hide="Solution"]
Let $ x$ be the number of paths from $ A$ to $ B$.
Let $ y$ be the number of paths from $ A$ to $ C$.
Let $ z$ be the number of paths from $ B$ to $ C$.

So, the total number of ways from $ A$ to $ B$ is $ x\plus{}yz$ because there are only $ x$ ways to get directly from $ A$ to $ B$ and $ yz$ ways to get from $ A$ to $ C$ to $ B$.

So, similarly, the total number of ways from $ A$ to $ C$ is $ y\plus{}xz$.

So, we have two equations :
$ y\plus{}xz \equal{} 74$ and $ x\plus{}yz\equal{}106$. Subtracting them and factoring gives us $ (y\minus{}x)(z\minus{}1)\equal{}32$. I believe the only factors that work are $ 4,8$(that satisfy the condition that there are more choices to go from $ B$ to $ C$), so solving gives us $ z\equal{}5, x\equal{}11, y\equal{}19$.

The answer is $ \boxed{214}$
[/hide]

[quote="Zellex"]Okay, I haven't been online for quite some time now, so I'm rusty...

[hide="Solution"]
Let $ x$ be the number of paths from $ A$ to $ B$.
Let $ y$ be the number of paths from $ A$ to $ C$.
Let $ z$ be the number of paths from $ B$ to $ C$.

So, the total number of ways from $ A$ to $ B$ is $ x\plus{}yz$ because there are only $ x$ ways to get directly from $ A$ to $ B$ and $ yz$ ways to get from $ A$ to $ C$ to $ B$.

So, similarly, the total number of ways from $ A$ to $ C$ is $ y\plus{}xz$.

So, we have two equations :
$ y\plus{}xz \equal{} 74$ and $ x\plus{}yz \equal{} 106$. Subtracting them and factoring gives us $ (y\minus{}x)(z\minus{}1) \equal{} 32$. I believe the only factors that work are $ 4,8$(that satisfy the condition that there are more choices to go from $ B$ to $ C$), so solving gives us $ z \equal{} 5, x \equal{} 11, y \equal{} 19$.

The answer is $ \boxed{214}$
[/hide][/quote]

How do you know that the alternate routes are a product of two numbers?

Say you wanted to get from $ A$ to $ C$ to $ B$.  There are $ y$ ways to get from $ A$ to $ C$ and $ z$ ways to get from $ C$ to $ B$, so the total number of possible combinations you can choose is $ yz$.  This is the basic [url=http://en.wikipedia.org/wiki/Rule_of_product]Rule of Product[/url].

That makes sense, I was thinking about it the wrong way. Thanks :D


Me and a friend are starting a club at our school and i was wondering if any of you would take some time and make a design on like photoshop for us. It's for an 8 by 11.5 sheet of paper, the club we are starting is IFU (Interscholastic Film Union). Thanks

What do you want on it (ie text, graphics, logo(?)...).  And, like ToBe said, what exactly is it? An ad, sign, ...

I would help but I have to leave for camp in a few days...


Let $ F(X_0)\equal{}\sum_{i\equal{}1}^N |X_i\minus{}X_0|^\beta$, Where $ N$ is a fixed positive integer. $ \beta > 1$ and $ X_i < X_{i\plus{}1}$ for all $ i$.
 Is $ F(X_0)$  convex in the region $ [X_1, X_N]$.
Please give outline of proof or reference.

Well, $ x\rightarrow |x|^p$ is convex on $ R$ for $ p > 1$, so yes.

Thank you. Please suggest some good references where I can study properties
of product of convex function which in general need not be convex.


vector
superior algebra
superior algebra unsolved

Let $K\subseteq \mathbb{C}$ a field with the usual operations such that:
1. The field K has only 2 endomorphisms f and g
2.$f(x)=g(x)\Rightarrow x\in \mathbb{Q}$
Prove that we can find a positive integer $d$, squarefree and $d\neq1$, such that $K=\mathbb{Q}(\sqrt{d})$

I had a "crazy" idea at first and I would like to know if it is nonsense
You easily get that $\mathbb{Q}\subset K$ and that one of the endomorphisms is $1_{K}$. Can I take now a base of K over Q such that 1 is in the base and prove that it has only 2 elements? (I thought that if more than 2 elements are in the base than I can find more than 2 endomorphisms, but I can't figure out if this is true :maybe: ). Maybe, as I said before, it is all nonsense :oops: But I would appreciate your opinion and your solutions :)

[quote="ciprian"] (I thought that if more than 2 elements are in the base than I can find more than 2 endomorphisms, but I can't figure out if this is true :maybe: )[/quote]

this depends on whether you consider [i]field[/i] or [i]vector space[/i] endomorphisms. i think the assertion deals with the field structure, so that it's not as easy as you thought.

I am totally lost now :maybe:  In the problem the endomorphisms are for fields...Can you please explain further -oo-? What is the difference?  :( Could any solution be given on my idea?

Hi.

Of course f or g is the identity. Say f=id. Now what can g^2 be? If we have $g^{2}=f=id$ than g is an automorphism.
If $g^{2}=g$ then consider the subfield $F: =\{k\in K | g(k)=k\}=im(g)$. We now have $g|F=id_{F}$. One can easily see that $F=K$ or g cannot be injective. In both cases we get a contradiction to $f\neq g$ or to the fact that g is a field endomorphism.

So we have that f and g are in fact automorphisms of K and form the galoisgroup of $K|\mathbb{Q}$. The fixed field of the galoisgroup is $\mathbb{Q}$ because of condition 2.
So the extension $K|\mathbb{Q}$ is galois $\Rightarrow |K: \mathbb{Q}|=2\Rightarrow K=\mathbb{Q}(\sqrt{d})$ with some d which has no squareroot in $\mathbb{Q}$. In fact d can be choosen to be squarefree.

[quote="Gockel"]
If $g^{2}=g$ then consider the subfield $F: =\{k\in K | g(k)=k\}=im(g)$. We now have $g|F=id_{F}$. One can easily see that $F=K$ or g cannot be injective. In both cases we get a contradiction to $f\neq g$ or to the fact that g is a field endomorphism.[/quote]

it suffices to say that automorphisms form a group :)
as for the rest, you've got it: every automorphism of $K$ is an automorphism on $\mathbb{Q}$, so $[K: \mathbb{Q}]=2$, and you've finished..

Yes, the automorphisms do, but the first posts only says that g is an endomorphism, so we have to proof, that it is surjective.


function
algebra
polynomial
real analysis
real analysis unsolved

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be an aditive function and $P: \mathbb{R} \rightarrow \mathbb{R}$ be a polynomial function (which is not constant).

Prove that:

$P \circ f$ has primitives or $e^f$ has primitives

implies

$\exists c \in \mathbb{R}$ s.t. $f(x)=cx$, $\forall x \in \mathbb{R}$

[i] Indeed, see http://www.mathlinks.ro/Forum/viewtopic.php?highlight=additive&t=32514 [/i]

I've posted this one, and I believe fedja posted a couple of solutions.


ARML
AwesomeMath
summer program
MATHCOUNTS
AMC
USA(J)MO
USAMO

meh did this like, happen randomly?

And to make your computer boot from CD, you need to edit the BIOS boot options by holding down F2 right after you power up the computer. You should then be in the setup.. the boot options are pretty easy to find. Just don't mess anything else up.

actually i found it

it was under boot...

i moved the CD rom up to 1st order and now i think its reading the CD first as i start up the computer...
i think the cd is running...ahh too bad i had to lose all my files but atleast i wont have to buy a new computer

thanks

and btw federer will win US open^

and when are you coming back?  lol so our arml team can win next year.

hmm I hope you didn't have any files that were too important.. I have all mine from last December backed up on CDs, and my current ones on an external drive. XD paranoia.

mm yeah I hope he wins. I felt so bad for him when he lost Wimbledon.

I fail ARML. :(

meh.. you can't just tell someone to "get ubuntu" and expect him/her to love it. I don't think Linux is worth the trouble for the average user. No, get a mac. I swear - zero maintenance. :D

umm yea...but thats only because not many people have macs...once more and more people get macs, they will discover more and more errors...

actually wait donald young will win...

Um why'd you delete my post? Personally insulted =P

^huh?

besides why arent you at awesomemath with those other pesky 8th/9th graders...

lol you couldve seen lyndon face to face...

Wow... Lyndon went?
Heh.  But I think what #H34N1 is right:
[quote="#H34N1"]I could learn more here than what I could at AMP. lol?[/quote]
There are some things that do indeed help more than others.  AMP is uncomparable to something such as AMY, MOSP, or even working through a good math textbook (preferably AoPS :thumbup: ).
If you want to go to a camp purely for recreational activities or for math, then go to such camps.  A combination of those two elements is not necessarily better, as is in this case.

meh aops textbooks don't really cover olympiad math and their concepts are really trivial

[quote="undefined117"]mm yeah I hope he wins. I felt so bad for him when he lost Wimbledon.[/quote]
Federer doesn't deserve to win. :)

hmm weird do you guys take classes here or something?

i just bought a couple of there books...yea i would agree pretty helpful considering i was number 8 on my mathcounts team in 7th grade...but all the books help with computational stuff and formulas/concepts used for Mathcounts/AMC/AIME...not USAMO yet

but in AMP i think you learn more proofs and its more hands on and they will help you a lot more with USAMO and olympiad type problems...besides youll meet a lot of smart people...

umm recreational?  i think it obviously helps...look at the people that went last year and look at where they are now...

btw do the olympics use hardcourts?

the only thing I know about is the red rings of death for xbox360... happened here just yesterday

[quote="the future"]hey since you guys are good with computers, anyone wana help me?

on my windows xp every time i turn the computer on, it goes to the loading screen (where theres that little loading bar at the bottom)...

then it goes to this screen that says "this computer has had a problem and blah blah blah"

and it asks me to start in safe mode, safe mode with networking, last point where it worked, or normally but i have experiemented with all 4 of the options and every time i try to start up with any of the options, i get to that stupid BSOD where it gives me an error message...

i can never get on to the log on screen, no matter how i start it i get the blue screen of death and i cant seem to get pass it...also i try to reinstall windows but everytime i put in the CD it never gives me the option to run that CD...anybody know how to apporach this problem?

btw im writing this on another computer...and anyone know why its causing this problem?[/quote]

[url=http://distrowatch.com/]Click[/url] to fix.

[quote="#H34N1"]Get ubuntu[/quote]
lolz, n00b distro. windoze vista ultimate 64bit is t3h sh1tz.


I received the following error message when trying to 'mark all topics as read' in the F&G forum.

DEBUG MODE

SQL Error : 1064 You have an error in your SQL syntax near ') AND forum_id = 138 AND topic_moved_id = 0' at line 3

SELECT topic_id FROM phpbb_topics WHERE topic_id IN () AND forum_id = 138 AND topic_moved_id = 0

Line : 278
File : /var/www/Forum/viewforum.php

Can you try this a couple more times and see if it always happens?

it happened repeatedly that night, and when trying to do the same thing tonight, it happened sever times again

Very weird.  What browser are you using?

Have you tried it on any other browsers?

Has anyone else seen this same error?

You must have another browser other than Netscape, for example Internet Explorer. Or you could download Firefox 1.0 which is a pretty good one, and it's freeware. Does your error persist only on those forum? Or on all marked read links across the site?

it's only happened on the round table and f&g forums

[quote="Valentin Vornicu"]Are you a moderator of those two sections?[/quote]

she mods F&G...

but I agree, definitely get something else...if you don't like IE, go for firefox...


Anyone mind giving me an explanation to that problem? I am having trouble figuring the one out in the solutions thread.

I dunno... I can't really understand the problem... maybe you should post it :P.

4. The walkway to the front of Euler Elementary School is a  grid. ACME Sidewalk company has been chartered to tile it with 15 non-overlapping  concrete slabs. Compute the number of tilings ACME can choose from for this task.

The answer is $\infty$. Assuming that the grid can be any grid we want, and the concrete slabs can be any size and shape ;).

Sorry, the latex didn't copy. The walkway is a 15 by 6 grid and each tile is a 2 by 3 grid.

Basically, it's the number of ways you can be all like 2a + 3b = 15. Which is like (0, 5) or (3, 3) or (6, 1). And that gives you 1 + 20 + 7 = 028 I think or you can do it the real way which is recursion but that means I can't be lazy :( ok I'll go take a look at what people said in the original thread.


conics
ellipse
AMC
AIME
geometry
analytic geometry
graphing lines

Basically the same question as 2001 AIME #5 but I added last part and changed numbers.  :D

An equilateral triangle is inscribed in the ellipse whose equation is $ x^2 \plus{} 9y^2 \equal{} 9$. One vertex of the triangle is (0,1), one altitude is contained in the $ y$-axis, and the area of the equilateral triangle is in the form $ \frac{m \sqrt{n}}{o}$ where $ n$ is square free and $ m$ and $ o$ are relatively prime positive integers. Find $ m\plus{}n\plus{}o$.

Level: AIME since it's off from actual AIME problem  :rotfl:

solution

[hide]First we draw the ellipse and note that (0,1) is a covertex of the ellipse.  Since the y axis is an altitude, we can conclude that the triangle is symmetric about it.  Draw Triangle ABC, with A being the point (0,1).  Now extend the altitude on the y axis to the ellipse (it is now the minor axis), and call the intersection point D.  Notice how we now have a cyclic quad ABDC, and triangle BCD can be broken into 30-60-90 triangles.  Using Ptolemy's Theorem, we have that $ \dfrac{2s^2}{\sqrt{3}}\equal{}2s$.  Solving yields $ s\equal{}\sqrt{3}$ and area $ \dfrac{3\sqrt{3}}{4}$.  The answer then is $ 10$.[/hide]

Hmm, I did not get that. Maybe someone else will try?

I'm pretty sure my answer is right because I solved it in same manner as the official AIME problem did.  :lol:

Can you post their solution or tell me where I went wrong?

[hide="Solution"]One side of the triangle must be parallel to the x-axis.So, one of the sides of the triangle that goes through (0,1) has slope of sqrt3(As equileteral triangles have angle).Let (x,y) be a point on the line.Then,x/y-1=sqrt3.So,
x=ysqrt3-sqrt3 is the equation of the line.
x^2+9y^2=9
or, 3(y-1)^2+9y^2-9=0
or,12y^2-6y-6=0.
or,y=(6+-sqrt(36+288)/24=1 or -1/2.
x=sqrt3*-3/2.(other gives the point (0,1)).
We get the sidelength sqrt(27/4+9/4)=3.
So, area =sqrt3*9/4.So, the answer is 16.[/hide]

I got the same answer as [b]Altheman[/b].

[b]@ azn6021023:[/b] ABDC is not cyclic, as it is inscribed in an ellipse and not a circle. If you want to use Ptolmey's, do an x-axis direction compression by a factor of 3, first.

[b]@ Cronus:[/b] To make the slope of the line $ \sqrt {3}$, you want $ \frac {y \minus{} 1}{x} \equal{} \sqrt {3}$ instead of $ \frac {x}{y \minus{} 1} \equal{} \sqrt {3}$

*mentally slaps self for thinking ellipses are circles.*

Thanks TZF.I think I should do the same as azn. :D

Wait, how would I do an x direction compression to apply Ptolemy's?

Let $ x\equal{}3z$. Now, instead of graphing in the x-y plane, lets graph in the z-y plane:

The ellipse equation now becomes $ (3z)^2\plus{}9y^2\equal{}9 \implies z^2\plus{}y^2\equal{}1$, which is a circle. Now you can find the area by whatever means, and then multiply by 3 to go back into the x-y plane.

Although, this would also make the triangle non-equilateral, so I don't know if you'd want to use Ptolmey's now anyway.

How'd you get that? See post #5, I think its 442...

Answer is indeed 442.

[hide="Detailed Solution"]

First of all, it is important to know that the slope of equilateral triangle from the top vertex to one of the bottom ones is $ \minus{} \sqrt {3}$ (or $ \sqrt {3}$ if the slope is positive).  So, using the given point (0,1), we know that this line will be $ y \equal{} \minus{} \sqrt {3} x \plus{} 1$. Plug this into the $ x^2 \plus{} 9y^2 \equal{} 9$ to get

$ x(28x \minus{} 18 \sqrt {3}) \equal{} 0$

So, $ x \equal{} \frac {9 \sqrt {13}}{14}$. Now, the key part is that if you multiply this by 2, you'll get the side of the triangle (this is equilateral triangle so the bottom vertices have same $ x$-coordinates, but just opposite in positive/negative.

Therefore, the side length is $ \frac {9 \sqrt {3}}{7}$. Hence, using area formula for an equilateral triangle $ A \equal{} \frac {s^2 \sqrt {3}}{4},$ we get $ A \equal{} \frac {243 \sqrt {3}}{196}$ and the answer is $ 243 \plus{} 3 \plus{} 196 \equal{} \fbox{442}$. [/hide]

It seems like no one likes problems involving conics haha.

Edit: A further comment about this problem

[hide="Regarding Altheman Solution"]

What Altheman did was to find the $ y$ coordinate of the result and using the distance formula with the given vertex. While this results in same answer, above method is easier because it relies less on brute calculation. It is AIME so it really doesn't matter as far as time goes but in the terms of accuracy, my way is better (and more secured). [/hide]

There are a few good conic problems but they are generally computational (you have to solve a pair of quadratics, etc). Also most of the theory is developed with foci/directricies, etc. (sp?) There are still nice problems though... There have been a lot of good construction problems with conics too.

Yeah, I recall some conics problem, and they seem to be great. I'll hopefully make some modifications soon...

I find conics problems tend to be more interesting when the numbers are clean but the axes/asymptotes don't line up with the x and y axes.


Matt will arrange four identical, dotless dominoes (shaded 1 by 2 rectangles) on the 5 by 4 grid to the right so that a path is formed from the upper left-hand corner $ A$ to the lower righthand corner $ B$. In a path, consecutive dominoes
must touch at their sides and not just their corners. No domino may be placed diagonally; each domino covers exactly two of the unit squares shown on the grid. One arrangement is shown. How many distinct arrangements are possible, including the one shown?

[asy]size(101);
real w = 1; picture q;
filldraw(q,(1/10,0)--(19/10,0)..(2,1/10)--(2,9/10)..(19/10,1)--(1/10,1)..(0,9/10)--(0,1/10)..cycle,gray(.6),linewidth(.6));
add(shift(4*up)*q); add(shift(3*up)*shift(3*right)*rotate(90)*q); add(shift(1*up)*shift(3*right)*rotate(90)*q); add(shift(4*right)*rotate(90)*q);
pair A = (0,5); pair B = (4,0);
for(int i = 0; i<5; ++i)
{draw((i,0)--(A+(i,0))); draw((0,i)--(B+(0,i)));}
draw(A--(A+B));
label("$A$",A,NW,fontsize(8pt)); label("$B$",B,SE,fontsize(8pt));[/asy]

Note that this problem is exactly the same as if you would trace the path travelled by the centre of each domino's two squares.  Thus since it is moving from the centre of the top left square to the centre of the bottom right square, you are actually tracing the path of the centre of each square from point (0,0) to point (3,4), where the y axis is directed downwards.

Thus the centre has to make 3+4 = 7 moves to the right or downwards, of which 3 are to the right, and 4 are downwards.  We see that there are thus 7 choose 3 = 7 choose 4 = 35 ways to do this.  (This is a normal path-counting problem; out of the 7 total moves, 3 moves need to be chosen to go to the right, and 4 chosen to go downwards thus 7choose3.)


Prove that there are no triangle satisfying the following condition:

CosA + CosB + CosC = 1

here,A,B,C are the angles of the triangle.

Because:
$\cos A+\cos B+\cos C=1\Rightarrow \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=0$. From that one of them $A, B$ or C equal 0, that is impossible because $A,B,C$ are angles in triagle $ABC$

could you explain how you reached your second equation?

Okay
$\cos A+\cos B+\cos C=1\Leftrightarrow\cos A+\cos B+\cos C-1=0$
$\Leftrightarrow 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)-2\sin^{2}\frac{C}{2}=0$
$\Leftrightarrow \sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)=0$
$\Leftrightarrow \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=0$

done :o

Is there any particular way you figured out that you had to do it that way or did it just come to you or have you seen it before?  Just curious to know what to look for to change it to that.


I'm stuck on this problem, I've tried repeatedly to get the answer but I feel that I'm doing something wrong.  It's been a while since I've done advanced algebra/clac, if someone could post a step by step response I'd really appreciate it.  Thanks.

  $ 1$
________    $ \minus{}1$
$ sqrt(x)$
___________
     $ x\minus{}1$

You're trying to rationalize? :maybe:

Multiply top and bottom by $ x$... and it's rationalized.... :huh:


The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked $x$?

[asy]
defaultpen(linewidth(0.7));
path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3);
draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin);
draw(shift(1,0)*p, dashed);
label("$x$", (0.3,0.5), E);
label("$A$", (1.3,0.5), E);
label("$B$", (1.3,1.5), E);
label("$C$", (2.3,1.5), E);
label("$D$", (2.3,2.5), E);
label("$E$", (3.3,2.5), E);[/asy]

$
\mathbf{(A)}\;  A\qquad
\mathbf{(B)}\;  B\qquad
\mathbf{(C)}\;  C\qquad
\mathbf{(D)}\;  D\qquad
\mathbf{(E)}\;  E$

Just from mentally folding it, I get [hide]C. C[/hide]

[hide]C i realized that a and b could easily touch. I figured basically everything else except c could fold over and touch x[/hide]

[hide]
I mentally fitted the cube together and got C[/hide]

[hide]I folded it mentally and got C.[/hide]

Guys, I don't think we need 50 people saying the exact same thing about how they folded it mentally.... if you have a new solution, great, but I dont't think we need 50 of the exact same one.

/requestlock don't revive 12 year old threads.

[quote=Math1331Math]Wait actually  if ur desperate tear out piece of paper label and fold  :-D[/quote]

hey that's me

You don't even need to go more than 3 folds, x up, b up, c to the side, and you are done. D and e are irrelevant.

Or, in general whenever it is one of these questions, the answer is $C$ like 90% of the time.  So if you don't know, guess $C$ :P

But this thread is 12 years old. It should be locked.

[hide]It is definitelt gonna be C.C is the base, D is in front C, E is at right side of C, B is at the left, A is behind C, and x is on the top.[/hide]

[quote=FieryEntei]Thanks for adding a lot to this thread. With your post, I finally understand the concept that like 40 people have already posted above.[/quote]

Don't be so mean the user's new  :)


trigonometry
email
real analysis
real analysis theorems

To protect his food from hungry bears, a boy scout raises his food pack with a rope that is thrown over a tree limb at height $h$ above his hands. He walks away from the vertical rope with constant velocity $v_{boy}$, holding the free end of the rope in his hands (Fig. P2.71).

Show that the speed $v$ of the food pack is given by $\frac{x. v_{boy}}{\sqrt{x^{2}+h^{2}}}$ where $x$ is the distance he has walked away from the vertical rope.

Now in the manual of this book, here is the solution:
[quote]In a walking distance $\Delta x$, in a time $\Delta t$, the length of rope $l$ is only increased by $\Delta x \sin \theta$
$\therefore$ The pack lifts at a rate $\frac{\Delta x}{ \Delta t}\sin \theta = v_{boy}. \frac{x}{l}= \boxed{\frac{x. v_{boy}}{\sqrt{x^{2}+h^{2}}}}$[/quote]
[size=117][color=blue]I cannot figure out how he found[/color][/size] $\Delta x \sin \theta$  :?:

I would not use the $\sin$ at all: since the hypotenuse has length $\sqrt{x^{2}+h^{2}}$, the rate at which its length increases is $\frac{x\.x}{\sqrt{x^{2}+h^{2}}}$

NaZAf

I have a complete solution manual of all Serway's chapters. If you want it :) pm me ur emailadress

OK. I have the instructor's manual book. Anyway, my email address is this: Nadeem.af@gmail.com

Are the solutions well explained??


To clear up any confusion about what may or may not be discussed about the school and chapter competitions, here are the official rules.

You [b]may[/b] discuss your and/or your school's [b]ranking[/b] only.

You [b]may not[/b] discuss:

- anybody's [b]scores[/b], either your own or anybody else's, either individual or school, whether in the school or chapter round;

- anybody's else's rankings besides your own or your own school's;

- anything at all about the problems, including whether your thought they were "easy" or "hard", whether you felt that you did better or worse than you were expecting, or anything at all about the subject matter.

We will tell you when it is OK to freely discuss the school and chapter contests; this will not be until all chapter contests in all states are concluded in late February.

[quote="jhredsox"]what about your arch rival's score from your school (aka dan :D )?

Cause I already violated this.

jorian[/quote]
You better edit it then :mad:

[quote="jhredsox"]ahh, darnet.

it's posted all over the place (not really)

it won't hurt

sry again

jorian[/quote]
Are you that lazy?!?!?!?
When the admins say something, they mean it!
AoPS could get in trouble.
Use the search feature

ok, fine. i think i only publically put in my blog, but no one goes there.

it ruins the fact that i beat dan (which u can kind of figure out cause i got 1st)

can i say i beat him :rotfl:

then why'd u already say u got 6th, and make a topic about it?

i made a topic about it???

you mean in my blog?

jorian

i'm pretty sure you can

if not, sorry for violating this rule :oops:  :(

jorian

[quote="13375P34K43V312"]Are we allowed to discuss scores with people whom you are sure that they have already taken the chapter test (because you looked on the Mathcounts webpage and saw their chapter date) privately, via PM and/or AIM and/or e-mail?[/quote]

yeah I think the Admins wouldn't really mind if you pmed people who have already completed the chapters and discuss the problems, etc.

Just keep in mind that sometimes a competition will be delayed by bad weather, so just checking that the scheduled date has gone by is not really enough.  I don't know if there's an easy way of checking whether a certain chapter competition has actually happened or not.

ya, notice why we can't talk about the test til the 26th...

-jorian

A possibility would be to ask them the genre of like, the first problem (algebra, geometry, number theory, counting, etc.)

[quote="DPatrick"]These are not our (AoPS's) rules, there are MATHCOUNTS's rules.  If you have a question about the specifics of the rules, then you should take it up with your coach and/or chapter coordinator.

Don't lose sight of the forest for the trees: the purpose of the rules prohibiting discussion is to protect the integrity of the chapter contests, by not disclosing any aspect of them to students (or coaches) who haven't taken them yet.  Keep this in mind and act appropriately.

I personally have no knowledge of the chapter competition myself, and I want to keep it that way, as I will be helping out at our local chapter (San Diego) next weekend.[/quote]
So wait, can we discuss the problems via AIM if we are 100% sure the other person has taken chapters?

[quote="bpms"]So wait, can we discuss the problems via AIM if we are 100% sure the other person has taken chapters?[/quote]

How can DPatrick be any clearer?  AoPS doesn't establish the rules for MATHCOUNTS.  MATHCOUNTS establishes the rules for MATHCOUNTS.

You absolutely and positively may not use any of the facilities provided by AoPS to jeopardize the integrity of the contest!  Period!  No exceptions!  DPatrick previously stated AoPS's interpretation of what you [b]may[/b] do.  You are expected to strictly follow his posted rules.  Below, I will emphasize some of what you [b]may not[/b] do, based on my interpretation of the rules and the risks to contest integrity...

Do not discuss the problems with anyone outside your own school!  Do not use AIM to discuss the problems.  Do not use PM to discuss the problems.  Do not use e-mail to discuss the problems.  Do not use the telephone to discuss the problems.  Do not use the Forum to discuss the problems.  DO NOT DISCUSS THE PROBLEMS WITH ANYONE OUTSIDE YOUR OWN SCHOOL!  PERIOD!

DPatrick has specifically stated he doesn't want to know anything about the contest.  But in his role as an AoPS administrator, he may have to access logs or backup files.  Anything that passes through AoPS is accessible to him.  Don't put him in the awkward ethical position of seeing material he mustn't yet see.

Do you think anyone at your ISP or your friend's ISP or at any of the dozen or so ISP's between you might be a coach or a sibling or a parent of a MATHCOUNTS student?  Hundreds of people have access to every message you send.  Don't put them in the awkward position DPatrick wants to avoid.

Has your computer or your friend's computer or any of the dozens of computers between you ever been infected by a virus or spyware?  Some viruses and spyware perform "data mining", whereby they attempt to capture e-mails, IM's, and other personal information.  Don't let a data mining virus send the MATHCOUNTS questions to an unethical computer hacker, who may share them with his unethical little brother.

MATHCOUNTS takes extraordinary steps to protect the integrity of the contest, even when they are communicating between the National office and their State and Chapter coordinators.  For good reasons that they do not have to explain to you or me or DPatrick, MATHCOUNTS doesn't even permit discussion of the questions on their own private Forum, accessible only to their own coordinators.

Just so you know how strongly I feel about this... if I ever find out an AoPS student has leaked information that has jeopardized a competition, I will personally rat you out to the MATHCOUNTS Foundation, and I will make it my personal mission to get you banned for life from all future mathematics competitions and all summer math programs.

If you feel compelled to discuss the questions, discuss them only face-to-face with members of your own school and with your own coach.  You can wait until MATHCOUNTS releases the questions on their own Web Site (usually near the end of the month, depending on weather and other factors).

I'm sorry, now I feel really embarrassed that I didn't consider those scenarios :oops:

Ok, guys, this has gotten to spam material (aka the last post). This was an [b]announcement[/b] in the first place, and not something to ask what if questions. I know I contributed a little bit at the beginning, but that was just a question regarding what I had previously posted. I think rcv went a little far in what he said, but his general principle is right. And anyways, don't you have better things then talk about talking about a math competition? Shouldn't you be studying for state if you made it and not worrying about the previous chapter round?

I hope you guys have a nice Presidents' Week, as you can talk the Monday afterwards.

-Jorian Hoover

[quote="BFG"][quote="rcv"]

I will make it my personal mission to get you banned for life from all future mathematics competitions and all summer math programs.

[/quote]

 [b]banned for life from all future mathematics competitions and all summer math programs???[/b]

What a pity that you are only a commom person who has a very limited power (nothing more than a small forum's moderator). You should be the President of the World Mathematics Association in order to carry on your great mission, if there is one.[/quote]
Tell me, would you want someone who has cheated in past competitions to compete in yours or in summer camp. Also, do you know who rcv is? Trust me, if rcv tells Mr. Rusczyk that you shouldn't compete, he can tell all of the well known competition heads that one shouldn't compete.

Besides, cheating is wrong.

Well Mr. Verser didn't say that he *would* succeed, he said he'd make it his personal mission to get you banned. Further, he isn't a common person with a very limited power; he may not actually be in charge of any of the summer programs or competitions, but I don't doubt if he requested one of the heads to ban someone from their competition they'd comply.

[quote="7h3.D3m0n.117"]yeah I think someone should lock this topic

@rcv: that was a little  bit harsh and i don't think bpms would ever tell another person specific questions and help for MC competitions....[/quote]
Um, no I don't think that was harsh, I was being stupid.

[quote="7h3.D3m0n.117"][quote="rcv"]I will make it my personal mission to get you banned for life from all future mathematics competitions and all summer math programs.[/quote]

i think that is harsh  :lol:[/quote]
Cheating on competitions deserves banning from them.

This is an announcement guys!

Not a debate topic!

Who cares if it's harsh? It's his opinion!

-jorian

yeah someone lock this before it gets worse  :lol:

Can we discuss chapter questions tomorrow?

No, weather may delay some competitions.

I think everyone chapters are done taking their tests. It ends on Feb. 25th, and the time is 7:35 p.m. Feb. 25th. So everyone should be done by now.

They said they would [i]tell us[/i] when we could discuss....I wouldn't chance it. Refer to perfect's post

[color=red][b]Do not discuss Chapter Competitions until we tell you it's OK.[/b][/color]

There are a few Chapter Competitions [b]scheduled[/b] for the week of February 26, and weather may have delayed others.

We will NOT give the OK until MATHCOUNTS tells us it's OK, or until MATHCOUNTS posts the 2007 Chapter Competitions on their Web Site.  Usually this is a couple of days (sometimes longer) after the very last Chapter Competition has concluded.

[color=red][b]Do not discuss Chapter Competitions until we tell you it's OK.[/b][/color]

Update: we have heard from the national MATHCOUNTS office, and there are still chapters that have not had their competition yet.

So please still do [b]not[/b] discuss the chapter competitions (except that it's ok so say your and/or your school's [b]rank[/b], but not your [b]score[/b]).

[color=red][b]Do not discuss Chapter Competitions until we tell you it's OK.[/b][/color]

[b]Another update:[/b]  The MATHCOUNTS National office has informed us that they do not expect to release the Chapter Competitions until Monday, March 5.

The National Office says there were "some delayed competitions" and we "need to make sure all [Chapters] are complete before posting the tests".

If your State Competition is on March 3, I know this stinks.  But it's not my decision or AoPS's decision.

We all know the integrity of every last Chapter Competition is very important.  So, if the National MATHCOUNTS office isn't releasing the tests until next week, we can't talk about the tests until next week.

[color=red][b]Do not discuss Chapter Competitions until we tell you it's OK.[/b][/color]


induction
algorithm
number theory solved
number theory

A number written in decimal expansion can be transformed by the following operations:

(1) put a 4 at the end of the number (e.g. 12 -> 124)
(2) put a 0 at the end of the number
(3) divide by 2 if the number is even

Your starting number is 4. Prove that by a sequence of operations above every n \in N can be generated !

mr: estimate
hr: working backwards, partial problems, distinguishing cases,
    analysis of the situation

Let's call a number reachable if it can be reached by the operations, 10*n+4,10*n,n/2.

Is 17 reachable ?
Yes and the following sequence (to be read from right to left) will make the trip : 17 34 3 6 12 24 2 4

How to get it ? Start we 17 and double the number until we reach a number N ending with 4, then take (N-4)/10 and repeat.

I'll try to prove by induction that the algorithm always works (and ends).

Let N, and let suppose for every n < N, n is reachable.

. If N ends with 1,2,3,6,7,8 it's easy to see that at most 3 'x 2' (multiply by 2) are necessary to get a number N' ending with 4. Since 2^3 < 10, (N'-4)/10 < N so induction works.

. If N ends with 5, 2*N ends with 10, 2*N/10 < N + induction

. If N ends with 9, 4 doubling are necessary to get a number N' ending with a4. But a is always even, (N'-4)/4 < N + induction

The induction is easy to start (especially with a computer at hand  :D ).

Hope it's correct

PS forgive my ignorance (and my curiosity) but what does BWM mean ?

I think BWM = Bundes Wettbewerb Mathematik or so.
(German Mathematical Olympiad)

Hi i/3 and Arne,

BWM means Bundeswettbewerb Mathematik which might be translated by Federal Mathematical Competition.  It was established as a homework competition in 1970 in Western Germany (Federal Republic of Germany).
There are two stages with four problems each. Required mathematical skills are limited but heuristical  skills are needed. Though the heuristical level is below the IMOs.
There is another mathematical olympiad being established in Eastern Germany (German Democratic Republic). It is called Deutsche Mathematik-Olympiade (German Mathematical Olympiad). These two competitions and a further possibility are used to select people for the FRG IMO selection tests.
Anybody wanting more detailed information according Germany might contact me. I could send a short file providing a description in English !


For what ordered triple of integers ($a$,$b$,$c$,) is $x = \frac{3}{\sqrt[3]{7} - 2}$ a solution of $x^3 + ax^2 + bx + c = 0$?


1.Suppose that $ S_1,S_2,S_3,...$is a sequence of connected,compact subsets of plane and
$ S_1\supset S_2 \supset S_3...$Is $ S\equal{} \cap S_n$ connected?Give a proof or counter-example.
2.Consider the situation for a nested decreasing sequence of compact path-connected sets.

1. Suppose $ S\equal{}A\cup B$, where $ A$ and $ B$ are disjoint, nonempty, and closed in $ S$ (hence compact). We have $ d\equal{}\mathrm{dist}(A,B)>0$. Let $ C\equal{}\{x\in\mathbb{R}^2\colon d/3\le \mathrm{dist}(x,A)\le 2d/3\}$. Then $ S_n\cap C$ is a nested sequence of compact nonempty sets with empty intersection, a contradiction.

This seems to work in any proper metric space. ("Proper" = all bounded closed sets are compact.)

2. Take your favorite compact connected, but non-path-connected set. Approximate it by a sequence of closed neighborhoods.


Starting from the numbers $1, \frac{1}{2}, \frac{1}{3}, ...,\frac{1}{100}$, the following operation is  performed until only one number remains: Choose two numbers, say $a$ and $b$, and replace them with $a+b+ab$. Determine the remaining number.

I'm quite sure it appeared before:

The trick is to consider $(a+1)(b+1)=a+b+ab+1$ so we conclude that defined operation is commutative. Hence we want to count:

$(1+1)(\frac{1}{2}+1)...(\frac{1}{100}+1)-1=2 \frac{3}{2} \frac{4}{3}....\frac{101}{100}-1=100$

[quote="Megus"]we conclude that defined operation is commutative[/quote]
And also associative (we need this).

A beautiful problem :!:  :)


If $\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0,$ prove that $[\alpha^{6}] = 3$ (Here $[.]$ is the greatest integer function.)

[hide="idea"]
Consider function $f(x)=x^{5}-x^{3}+x-2$.
$f'(x)=5x^{4}-3x^{2}+1 > 0$    $\forall x \in R$
Function $f(x)$ is monotonically increasing over $R$, so there is only one root.
$\alpha^{6}\in [3, 4) \Rightarrow \alpha \in [\sqrt[6]{3}, \sqrt[6]{4})$
All we have to show is $f(\sqrt[6]{3}) \le 0 < f(\sqrt[3]{2})$
[/hide]

i think olympiad probelms can be solved without calculus. :)

Again an INMO problem already posted.
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=344673#p344673[/url].


http://www.wheatoncollege.edu/Academic/academicdept/MathCS/Faculty/tratliff/writing/calculus-I/rous.html

i need to figure ou what the population of the 2 groups will be in 20 years. the problem is that i have no clue how to figure out the rate since it varies every year.

Well, yes- that's the point. You need a model with those properties- both oscillation and decay, centered around some point other than zero.

so there is there a way to figure it out? my teacher never taught us this, she wants us to go and research it ourselves, but i cant seem to find it anywhere.


As the diagram shows:
Given a triangle ABC with area 1, points D,E,F, and G trisect BC and AC. Find the area of quadrilateral MDEN.

[hide="another hint"]You should find the value of $\frac{BN}{BF}$ and $\frac{BM}{BF}$ to apply this formula:
Given a triangle $ABC$. $M,N \in AB,AC$. Then $\frac{S(AMN)}{S(ABC)}= \frac{AM}{AB}. \frac{AN}{AC}$
Thus, you can find the area of the triangle $BCF$ and $BMD$ and subtract them, you will get the area of the quadrilateral $MDEN$ :)
For example:
$\frac{BN}{NF}= \frac{S(ABE)}{S(AEF)}=3$ then $\frac{BN}{BF}= \frac{3}{4}$
$\Rightarrow \frac{S(BNE)}{S(BCF)}= \frac{BN}{BF}. \frac{BE}{BC}= \frac{3}{4}. \frac{2}{3}= \frac{1}{2}$[/hide]

[quote="JavaMan"][hide]
I got 1/9, is that right?
[/hide][/quote]
I got $S(MDEN)= \frac{5}{42}S(ABC)$ :D

How'd you get [hide]BN/NF = 3[/hide]

[hide="justification"]
edited;
I'm gonna use triangle areas.

let $\theta = m\angle ANF$
$\frac{1}{2}\sin{\theta}(AN\cdot NF+NF\cdot NE) = \frac{2}{9}$
$\boxed{\frac{1}{2}\sin{\theta}NF \cdot AE = \frac{2}{9}}$ since AE = AN + NE

$\frac{1}{2}\sin{\theta}(AN\cdot BN+BN\cdot NE) = \frac{2}{3}$
$\boxed{\frac{1}{2}\sin{\theta}BN \cdot AE = \frac{2}{3}}$ since AE = AN + NE

Dividing equations gives $BN/NF = 3$
[/hide]

[b]Note: Ignore; I spotted my mistake.[/b]

[quote="JavaMan"]How'd you get [hide]BN/NF = 3[/hide]

[hide="justification"]
I'm gonna use triangle areas.

let $\theta = m\angle ANF$
$\frac{1}{2}\sin{\theta}(AN\cdot NF+NF\cdot NE) = \frac{1}{3}$
$\boxed{\frac{1}{2}\sin{\theta}NF \cdot AE = \frac{1}{3}}$ since AE = AN + NE

$\frac{1}{2}\sin{\theta}(AN\cdot BN+BN\cdot NE) = \frac{2}{3}$
$\boxed{\frac{1}{2}\sin{\theta}BN \cdot AE = \frac{2}{3}}$ since AE = AN + NE

Dividing equations gives $BN/NF = 2$
[/hide][/quote]
Look at the diagram and just measure, you will see that $\frac{BN}{NF}=3$ :)
[hide="reason"]$\frac{BN}{NF}= \frac{S(ABN)}{S(ANF)}= \frac{S(BNE)}{S(ENF)}= \frac{S(ABE)}{S(AEF)}$
But $S(ABE)= \frac{2}{3}S(ABC)$ and $S(AEF)= \frac{2}{3}S(AEC)= \frac{2}{9}S(ABC)$
Thus, $\frac{BN}{NF}= \frac{\frac{2}{3}}{\frac{2}{9}}=3$[/hide]

[quote="JavaMan"]Yup! Should've seen it :lol:[/quote]
Yes, but I just want to show simpler solution to it :D

[quote="JavaMan"]
I got 1/9, is that right?
[/quote]

tunganh has got the right answer.:)

[quote="tunganh"][quote="HTA"][hide="hint"]from A , C , E draw the line parpendicular to the line BF , you can find the value of , $\frac{NE}{NA}$[/hide][/quote]
Or you can use my above method and easily find out the value of $\frac{AM}{AD}$ and $\frac{AN}{AE}$ :D[/quote]

Another way to get the value of $\frac{AM}{AD}$ and $\frac{AN}{AE}$ :
[hide]
From D,E draw lines pararell to AC and intersect with BF at P, Q,respectively. $\frac{AM}{DM}=\frac{FA}{PD}=\frac{FA}{\frac{1}{3}FC}=\frac{FA}{\frac{1}{6}FA}=6$
[/hide]

[quote="HTA"][hide="hint"]from A , C , E draw the line parpendicular to the line BF , you can find the value of , $\frac{NE}{NA}$[/hide][/quote]
i also use this method , tunganh .After we find out $\frac{NE}{NA}$ , we obtain the values of $\frac{AN}{AE}$ and $\frac{AM}{AD}$ , so use this method in triangle ADE   :wink:

a more general method - Menelaus's Theorem.  i'll state it using the original diagram for this problem.

Consider a triangle $CBF$.  Let $E$ be a point on line $CB$, let $N$ be a point on line $BF$, let $A$ be a point on line $FC$.  Then $A,N,E$ are collinear [b]if and only if[/b]

$\frac{CE}{EB}\cdot \frac{BN}{NF}\cdot \frac{FA}{AC}= 1$.

if you are using directed segments (which you almost never do) then the right-hand side is -1.  also, note the pattern of letters in the fractions - (vertex) to (point on line through that vertex) divided by (point on line through that vertex) to (other vertex on the line) and so on cyclically around the vertices.  also note the 'if and only if' - if they are collinear, then the equation holds.

in this case Menelaus gives $\frac{1}{2}\cdot \frac{BN}{NF}\cdot \frac{2}{3}= 1$, hence $\frac{BN}{NF}= 3$


induction
LaTeX
floor function
ceiling function

Let p_0, p_1, p_2, be a sequence of integers defined by

p_1 =0 , p_2 = 2 and

p_n = p_[n/2] + p_[n/2] + 1  for all n in Z+,  where n >= 3.

_ denotes subscript. sorry no latex.

[] denotes floor for first subscript n/2 and then ceiling for the second subscript n/2.

Prove for all non-neg that P_n = n -1

You don't need to have $ \text{\LaTeX}$ installed on your system to use it on AoPS; simply place dollar signs around what you want to be parsed:

[color=blue]Let $ p_0$, $ p_1$, $ p_2$, be a sequence of integers defined as $ p_1 = 0$, $ p_2 = 2$, and $ p_n = p_{\left\lfloor n/2\right\rfloor} + p_{\left\lceil n/2\right\rceil} + 1\;\forall\;n\ge3$.

Prove for all non-negative $ n$ that $ p_n = n - 1$.[/color]

My question is, what happened to $ p_0$'s definition? :)


I don't understand the solution given here:

[url]http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12A_Problems/Problem_9[/url]

It says that when he gets home he is 6/7 of the way. Is that distance, or time or what? And how did you come to that conclusion. Also, how come his current position is 6/7/2?

Identity

You're not alone! The problem is straightforward to work out but I can't follow the solution's logic at all.

I think the units of distance the solver of the problem used is how far Yan can ride his bike in the time that he takes to get to the stadium with either method. Not only is this implicitly defined and unnecessary, I think he used it wrong. The $ \frac {6}{7}/2$ step is incorrect, and the answer came out of nowhere.

I replaced the solution.

MellowMelon.

The solution you posted is incorrect. The ratio turns out to be $ \minus{} \frac {3}{4}$

I believe this may be the correct solution.

$ a$ is the distance to the stadium.
$ b$ is the distance to home.
$ w$ is the speed of walking.

$ t \equal{} \frac {d}{s}$
$ \frac {a}{w} \equal{} \frac {b}{w} \plus{} \frac {a \plus{} b}{7w}$
$ \frac {a}{w} \equal{} \frac {8b \plus{} a}{7w}$
$ 7a \equal{} 8b \plus{} a$
$ 6a \equal{} 8b$
$ \frac {3}{4} \equal{} \frac {b}{a}$

This solution was boring and not very clever. Can anyone provide a really nice and succint solution?
That would be awesome!!!  :diablo:

Woops. Sorry, corrected it. In my defense, your final line needs to have $ b/a$ instead of $ a/b$. :P

[quote="Farenhajt"][hide="No equations"]If Yan were at $ {6\over 7}$ of the [u]whole[/u] way, then it would take him the same amount of time to walk the remaining distance and to ride the whole distance from the home to the stadium. Hence he must be at the midpoint of the first $ {6\over 7}$ of the way, since the time it would took him to get to the last seventh must be the same as the time it would took him to get back home. Therefore, he traveled $ {3\over 7}$ of the way, hence the desired ratio is $ {3\over 4}$.

In general, if he rides $ n$ times faster than he walks, the ratio will be $ {n \minus{} 1\over n \plus{} 1}$[/hide][/quote]

That was beautiful !!