limit
function
calculus
trigonometry
Asymptote
algebra
rational function

what is limit notation and how is it used. ive tried google and AoPS wiki. they dont help much because theres terminology i dont get. can someone help me?

$ \lim_{x\rightarrow a}f(x) \equal{} L$ is read as "the limit of f(x) as x approaches a is L".

If that's not what you wanted, then I'm not sure how to help you.

If you're just starting to learn limits, try finding the one-sided limits of a rational function $ \frac{1}{x}$.

[hide="answer to rational function question"]
So, $ \lim_{x\rightarrow 0^{\minus{}}}\frac{1}{x}\equal{}\minus{}\infty$ and $ \lim_{x\rightarrow 0^{\plus{}}}\frac{1}{x}\equal{}\infty$

Note that the superscript - and + indicate whether you're approaching x from the left or right of the graph, respectively.

[/hide]

If you can understand that, you will soon come to understand limits.

I only understand a bit myself but I'll try to reword it based on the above

1. x approaches a but can never equal a
2.the superscript tells you ,if x approaches 0, if the majority of the graph is negative or positive
3. L is as small as it can get

If I am in anyway wrong, feel free to harshly criticize me

[quote="t0rajir0u"][quote="Prime factorization"]$ \lim_{x\rightarrow a}f(x) \equal{} L$ is read as "the limit of f(x) as x approaches a is L".[/quote]

This is how the notation is read, but I think the poster wants an explanation of what it [i]means.[/i]

Intuitively, $ \lim_{x\to a}f(x) \equal{} L$ means "if we pick $ x$ arbitrarily close to $ a$, we can make $ f(x)$ arbitrarily close to $ L$."

[b]Example 1:[/b]  $ \lim_{x\to 1}\frac{x^{2}\minus{}1}{x\minus{}1}\equal{} 2$.  This means that if we pick $ x$ arbitrarily close to $ 1$, we can make $ \frac{x^{2}\minus{}1}{x\minus{}1}$ arbitrarily close to $ 2$.  Indeed,

$ f(1.01) \equal{} 2.01$
$ f(1.001) \equal{} 2.001$
$ f(1.0001) \equal{} 2.0001$

and so forth.  Note that this example is trivial because $ \frac{x^{2}\minus{}1}{x\minus{}1}\equal{} x\plus{}1$, so it is easy to see the result in this case.

[b]Example 2:[/b]  $ \lim_{x\to\infty}\frac{1}{x}\equal{} 0$.  This means that if we pick $ x$ arbitrarily close to $ \infty$ (that is, arbitrarily large), we can make $ \frac{1}{x}$ arbitrarily close to $ 0$ (that is, arbitrarily small).  Indeed,

$ f(1000) \equal{} 0.001$
$ f(10000) \equal{} 0.0001$
$ f(100000) \equal{} 0.00001$

and so forth.

This intuitive notion of arbitrary closeness is the basis for what is called the epsilon-delta definition of a limit.  There are other subtleties involved, but those can wait until a proper calculus course.  :)[/quote]

so would it be something like $ \lim_{x\to 90}\sin (x)\equal{}1$

Yeah, but it's almost always better to use radians in calculus, since it simplifies things hugely. So...
$ \lim_{\theta\rightarrow\pi}\sin{\theta}\equal{}1$
The answers are essentially the same though.

[quote="Icy"]Yeah, but it's almost always better to use radians in calculus, since it simplifies things hugely. So...
$ \lim_{\theta\rightarrow\pi}\sin{\theta}\equal{} 1$
The answers are essentially the same though.[/quote]

$ \frac{\pi}{2}$ ;)

Also, it's better to use \to instead of \rightarrow in limits as in:
$ \lim_{\theta\to\frac{\pi}{2}}\sin\theta\equal{}1$

Oops. :blush:

That's what I meant.

Is there any reason $ \to$ is better than $ \rightarrow$?

[quote="Icy"]Is there any reason $ \to$ is better than $ \rightarrow$?[/quote]
I'm not certain, but I suspect that the reason is for if someone has to read the code as opposed to viewing the image. Otherwise, they appear identical to me.

A better example might be
$ \lim_{\theta\to\frac{\pi}{2}}\tan\theta \equal{}\infty$, an example of an limit equal to infinity, where the approached y value does not really exist.

Infinite limits are also possible, i.e. $ \lim_{x\to\infty}\frac{1}{\theta}\equal{} 0$

EDIT: approaching from the the left, I forgot what that's called. I'm sure there is a term, I read it before. Thanks, azjps.

Yeah, a useful fact is that is that
$ \lim_{x\to\infty}\frac{1}{x^{k}}\equal{} 0$.
Edit: Yes, I meant $ k>0$ and $ x>1$.

No, try $ x\le 1$. You have to have the condition $ x > 1$

EDIT: t0rajir0u is right. my mistake.

temperal what do you mean by your last post?

I believe he means $ k > 0$.  xpmath's statement is false when $ k\le 0$.

oh ok

The message is too small. Please make the message longer before submitting.

[quote="Temperal"]A better example might be
$ \lim_{\theta\to\frac{\pi}{2}}\tan\theta =\infty$, an example of an limit equal to infinity, where the approached y value does not really exist.

Infinite limits are also possible, i.e. $ \lim_{x\to\infty}\frac{1}{x}= 0$[/quote]

Wait, $ \lim_{\theta\to\frac{\pi}{2}}\tan\theta$ does not exist. That is because

\begin{eqnarray*}\lim_{\theta\to\frac{\pi}{2}-}\tan\theta &=&\infty\\ \lim_{\theta\to\frac{\pi}{2}+}\tan\theta &=&-\infty\end{eqnarray*}

In other words, as you approach $ \frac{\pi}{2}$ from the left side on the graph of $ \tan\theta$, we find that due to the vertical asymptote, the graph shoots up to $ \infty$. However, from the right side of the graph, as we approach $ \frac{\pi}{2}$, the graph shoots down to $ -\infty$. So there is no limit here.

An example where the limit approaches infinity is

\begin{eqnarray*}\lim_{x\to 0}\frac{1}{|x|}=\infty\end{eqnarray*}

As we pick arbitrarily small values of $ x$, our answer can become increasingly large. Hence the limit tends towards $ \infty$.


farz konid $b,m,n$ adade tabeii bashand ke $b>1$ va $m$ namosavi ba $n$..sabet konid agar $b^{m}-1$va $b^{n}-1$ avamele yeksani dashtee bashand angah $b+1$ tavani az $2$ hast. ((((shortlist1997))))))

fek konam oon bayad [b]avamele [u]avale[/u] yeksan[/b] bashe...

khob inke vazhehe baiad avamele avale yeksan bashe ..!!!! [/u]

agar baraye $b>1,k>1$ va $b^{k}-1,b-1$, avameleh aval e yeksani dashteh bashand, angah $k$ tavani as 2 ast, va hamchenin $b+1$ tavani as 2 ast.

proof; by contradiction.
farz konid $p$ yek  addad e avale e fard hast ke p|k.  agnah baraye  har amaleh aval e $1+b+b^{2}\cdots+b^{p-1}$, q, khahim dasht $q|1+b+b^{2}+\cdots+b^{k-1}|b^{k}-1$. pas
$1+b+\cdots+b^{p-1}\equiv p\pmod q$ ama ma darim $1+b+\cdots+b^{p-1}\equiv 0\pmod q$.
pas p=q. pas $1+b+\ldots+b^{p-1}=p^{u},\ u>1$. agar gharar bedid $b=vp+1$ tanaghoz ast.
pas $k=2^{t}$. farz kon $r$ amaleh aval e $b+1$ hast, angah $r|b^{k}-1$ pas $r=2$.
va in hamoon chiziye ast ke ma mikhahim.


Find all positive integers $ x,y,z,n$ such that $ x!\plus{}y!\plus{}z!\equal{}5.n!$ where $ k!\equal{}1.2...k$

[quote="every"]Find all positive integers $ x,y,z,n$ such that $ x! \plus{} y! \plus{} z! \equal{} 5.n!$ where $ k! \equal{} 1.2...k$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 3,,2008, and we can't discuss it now

This is problem 9 from Correspondence Problem solving Copetition on Mathematics & Youth Magazine  on Mar, 2008


trigonometry
integration
limit
algebra
polynomial
function
linear algebra

Nordic university-level mathematics team competition was held on 19th March- 26th Marct 2007. Problems were:

1. Does there exist a planar set $X$ with at least two point and the following property: for every $x,y\in X, x\ne y$, there are exactly two points in $X$ on the midpoint normal of the segment $[x,y]$. ("Midpoint normal" means the normal, which goes through the midpoint. )

2. Show that for any $n$ there exists a non-degenerate convex $n$-gon whose sides and diagonals have integer lengths.

3. Show that
\[\sqrt{2}e^{\pi/4}-1\leq \frac{e^{y}(y\cos x+x\sin x-y)}{x-e^{y}(x\cos x-y\sin y)}\]
for any $x,y\in ]0,\pi/4]$.

4. Let $f: \mathbb{R}\to \mathbb{R}$ be a continuous non-negative function. Assume that $\int_{-\infty}^\infty f(x)\,dx=1$ and $r>0$ is fixed. Let
\[I_{n}(r)=\int\cdots \int_{x_{1}^{2}+\cdots+x_{n}^{2}\leq r^{2}}f(x_{1})f(x_{2})\cdots f(x_{n})\,dx_{1}\cdots dx_{n}. \]
Find $\lim_{n\to \infty}I_{n}(r)$.

5. Let $n$ and $k$ be positive integers with $k\leq n/2$. Prove that
\[\sum_{\frac{n}{2}-k<i<\frac{n}{2}+k}\binom{n}{i}\geq \left (1-\frac{n}{4k^{2}}\right )2^{n}. \]
6. Prove that for any natural number $k$ there are unique polynomials $f_{k}(x)$ and $g_{k}(x)$ both having degree $k-1$ such that
\[x^{k}f_{k}(x)=(x-1)^{k}g_{k}(x)+1. \]
Prove also that $|g_{k}(x)|,|f_{k}(x)|\leq \frac{1}{2}\binom{2k}{k}$, whenever $x\in [0,1]$.

7. Let $q\in \mathbb{C}\setminus \{0\}$. Define the polynomial $p_{n}(x)$ by setting $p_{0}(x)=1$ and
\[p_{n}(x)=\prod_{h=1}^{n}(1-xq^{h-1})=\sum_{k=0}^{n}s(n,k)x^{k},\quad x\in \mathbb{Z}^{+}. \]
Show that the polynomial identity
\[\sum_{k=0}^{n}s(n,k)p_{n-k}(-x)=q^{\binom{n}{2}}x^{n}\]
is valid for any $n\in \mathbb{N}$.

8. Let $P(x)$ be a polynomial of degree $n$ with real zeros $a_{1},\cdots,a_{n}$, and let $P^{\prime}(x)$ have zeros $b_{1},\cdots,b_{n-1}$. Let $f$ be a convex function defined on the real line. Prove that
\[\frac{f(a_{1})+\cdots+f(a_{n})}{n}\geq \frac{f(b_{1})+\cdots+f(b_{n-1})}{n-1}. \]
9. Suppose that the square matrices $A$ and $S$ are non-singular and have integer entries and that $\det A=1$. Define $B=S^{-1}AS$. Prove that for some positive integer $m$ all entries of the matrix $B^{m}$ are also integers.

10. Denote by $\mathbb{F}_{2^{n}}$ the finite field of $2^{n}$ elements. The [i]trace onto the prime field[/i] is a mapping $\mathbb{F}_{2^{n}}\to \mathbb{F}_{2}$ defined as $\text{Tr}(x)=x+x^{2}+x^{2^{2}}+\cdots+x^{2^{n-1}}$. As $\mathbb{F}_{2^{n}}$ is an $n$-dimensional vector space over $\mathbb{F}_{2}$, it has a basis $\{\beta_{1},\cdots,\beta_{n}\}$. A basis is [i]self-dual[/i], if $\text{Tr}(\beta_{i}\beta_{j})=0$, when $i\ne j$, and $\text{Tr}(\beta_{i}\beta_{j})=1$, when $i=j$.
  Let $\{\beta_{1},\cdots,\beta_{n}\}$ be a self-dual space and define $n\times n$-matrices $C^{(1)},\cdots,C^{(n)}$ over $\mathbb{F}_{2}$ by the following equation:
\[\beta_{i}\beta_{j}=\sum_{i=1}^{n}C_{ij}^{(l)}\beta_{l}. \]
It is clear that each $C^{(k)}$ is symmetric. Prove that $C_{ij}^{(k)}=C_{kj}^{(i)}$ also holds for each $i,j,$ and $k$.

1. Yes. Take $X$ as a boundary of any convex set.

4. If $M=\sup_{x\in [-r,r]}|f(x)|$ we get $|I_{n}(r)|\leq M^{n}V(r,n)$ where $V$ denotes the volume of $n$-dimensional sphere of radius $r$. Now $V(r,n)=\frac{\pi^{n/2}r^{n}}{\Gamma \left (1+\frac{n}{2}\right )}\leq \frac{c^{k}}{\lfloor \frac{n}{2}\rfloor ! }r^{n}$ where $c$ is a constant. This converges to $0$ as $n\to \infty$.

5. Let $X$ be a random variable, binomially distributed with parameters $n$ and $1/2$. Then
\[P(|X-n/2|<k)=2^{-n}\sum_{n/2-k<i<n/2+k}\binom{n}{i}. \]
The distribution has mean $\mu=n/2$ and standard deviation $\sigma=\sqrt{n}/2$ so Chebyshev's inequality gives
\[P(|X-\mu|<k)=1-P(|X-\mu|\geq k)\geq 1-\sigma^{2}/k^{2}=1-n/4k^{2}. \]

2. We show that for every $n$ there exists a set $X_{n}\subset S^{1}$, consisting of $n$ points on the unit circle $S^{1}$ such that for any $x,y\in X_{n}$ it holds $|x-y|\in\mathbb{Q}$. Then we would obtain the point set with integer distances just by scaling the construction by the LCM of the denominators, right?

[i]Lemma.[/i] The distance between two points on the unit circle seperated by angle $2\alpha$ is given by $2 | \sin(\alpha)|$.

Proof: trivial school trigonometry (sine theorem?).

Now define $\beta$ to be the smallest angle in a Pythagorean triangle, e.g., $\beta=\mathrm{arcsin}\frac{3}{5}$.

Then we claim that the set
$X_{n}: =\{P_{k}=(\cos(2(k-1)\beta), \sin(2(k-1)\beta)): k \in \overline{1,n}\}$
satisfies:
1. $|X_{n}|=n$;
2. For any $x,y\in X_{n}$, it holds $|x-y|\in \mathbb{Q}$.

Indeed:
1) the points $P_{k}$ are distinct since $\frac{\beta}{2\pi}$ is irrational.
2) the distance between $P_{k}$ and $P_{l}$ is $2|\sin((k-l)\beta)|$ by Lemma. Since $\sin(k\beta)$ is a  rational polynomial in $\sin(\beta)$ and $\cos(\beta)$, which themselves are rational by construction, we are done.

Am i right?  :maybe:

In the first problem it was meant to be: "Does there exist a FINITE planer set"..

I did compete this year, our team got 40/60 points.. Couldn't do 3,7,8 :(

I'm trying the problem 8 for hours!

I had a lot of ideas, but i can't find a solution.

Somebody can help me?

Taking the vector of the $ a_i$ taken each $ n\minus{}1$ times majorizes the vector with the $ b_i$ taken each $ n$ times.
This is a nice result that I was not aware of... wowt!!

[quote="Albanian Eagle"]Taking the vector of the $ a_i$ taken each $ n \minus{} 1$ times majorizes the vector with the $ b_i$ taken each $ n$ times.
This is a nice result that I was not aware of... wowt!![/quote]

A quick clarification if you may...(since I dont understand your insight completely) So after you established that the $ a_i$ vector taken $ n \minus{} 1$ times majorizes the $ b_i$ vector taken $ n$ times, did you then use something like Schur convexity? or did you use something else? If you could expand upon that, i'd really appriciate that.

Thanks

[quote="RealZeta"]

A quick clarification if you may...(since I dont understand your insight completely) So after you established that the $ a_i$ vector taken $ n \minus{} 1$ times majorizes the $ b_i$ vector taken $ n$ times, did you then use something like Schur convexity? or did you use something else? If you could expand upon that, i'd really appriciate that.

Thanks[/quote]
Proceed by [url=http://en.wikipedia.org/wiki/Karamata%27s_inequality]Karamata's inequality.[/url]

9. Let $ R \equal{} \mathbb{Z}/k\mathbb{Z}$, where $ k \equal{} \det S$. Matrices over $ \mathbb{R}$ with determinant $ 1$ form a finite group. Then by the Lagrange theorem, we get $ A^N \equiv I\pmod k$ for some $ N$ . Thus $ A^N \equal{} I \plus{} kC$, where $ C$ is a matrix with integer coefficients.  We have $ B^N \equal{} S^{\minus{}1} A^N S \equal{} S^{\minus{}1} (I \plus{} kC)S \equal{} I \plus{} kS^{\minus{}1} CS$ and $ B^N$ has integer coefficients.

Q6. The existence of $ f_k(x)$ and $ g_k(x)$ is guaranteed by the Eucleidean algorithm, since $ \gcd (x^k , (x \minus{} 1)^k ) \equal{} 1$. The Eucleidean algorithm also shows that the restriction on the degrees holds. To prove the uniqueness, substitution $ x\equal{}0$ at the above equation first reveals the constant coefficient of $ g_k$ . Then consecutive differentiation of the above equation and substitution $ x \equal{} 0$ shows that the coefficients of $ g(x)$ are determined up to degree $ k \minus{} 1$. Hence $ g_k$ is unique and clearly the same holds for $ f_k$. For the estimation of absolute value, we can use induction to find $ g_k$ efficiently:
\[ g_k(x)\equal{}(\minus{}1)^{k\plus{}1}\sum_{i\equal{}0}^{k\minus{}1}\binom{k\plus{}i\minus{}1}{i}x^i.\]
To verify that $ (x \minus{} 1)^kg_k (x) \plus{} 1$ is a polynomial divisible by $ x^k$ is straightforward. Thus, if $ x\in [0, 1]$, then
\[ |g(x)\equal{}\sum_{i\equal{}0}^{k\minus{}1}\binom{k\plus{}i\minus{}1}{i}x^i\leq \frac{1}{2}\binom{2k}{k}.\]
Substituting $ x\equal{}1\minus{}y$ shows that
\[ y^k(\minus{}1)^{k\plus{}1}g_k(1\minus{}y)\equal{}(y\minus{}1)^k(\minus{}1)^{k\plus{}1}f_k(1\minus{}y)\plus{}1,\]
which implies, by the uniqueness of $ f_k$ and $ g_k$ , that $ f_k (x) \equal{} (\minus{}1)^{k\plus{}1} g_k (1\minus{}x)$.
The estimation for $ |f(x)|$ follows immediately.


Given n objects arranged in a row. A subset of these subjects is called unfriendly if no two of its elements are consecutive.Show that the number of unfriendly subsets each having k elements is $ {n\minus{}k\plus{}1 \choose k}$

Hint: Reword this question as follows: How many ways are there to color $ k$ squares out of $ n$ black so no two consecutive squares are colored.

In this formulation, we know that every black square except the last one [b]must[/b] be followed by a white square.  What happens when you delete those $ k\minus{}1$ white squares that we know exist?

[b]Equavalent problem.[/b] Let $ S\equal{}\{1,2,...,n\}$. How many subsets $ A$ of $ S$ such that $ A$ does not contain two consecutive integers?

[b]Solution.[/b]
Let $ A\equal{}\{a_1,a_2,...,a_k\} \subset S$ satisfies the problem such that $ a_1 < a_2 < ... < a_k$ and none of $ a_i, a_{i\plus{}1}$ are consecutive. We have $ a_{i\plus{}1} \geq a_{i} \plus{}2$, or $ a_{i\plus{}1} \minus{} (i\plus{}1) > a_i \minus{} i$ and hence, $ \{a_{i} \minus{} i\}_{i\equal{}1}^{k}$ is an increasing sub-sequence of $ \{0,1,...,n\minus{}k\}$. On the other hand, each of those sub-sequence $ \{b_1,b_2,...,b_k\} \subset \{0,1,...,n\minus{}k\}$ yields exactly one set $ A$, defined by $ a_i \equal{} b_i \plus{} i$. Thus we have a bijective. Therefore, the number of subsets $ A$ is equal to the number of subsets with $ k$ elements of $ \{0,1,...,n\minus{}k\}$, which is $ {{n\minus{}k\plus{}1} \choose k}$, QED.

We also can prove that:
The number of subsets with $ k$ elements having exactly $ r\minus{}1$  non-consecutive elements is $ {k \minus{} 1 \choose r \minus{} 1}{n \minus{} k \plus{} 1 \choose r}$
(When $ r \equal{} k$ we have this problem)


eh the last problem was cool, but I was wondering: Which is bigger?

$ 98^{50}\plus{}101^{50}$

$ 99^{50}\plus{}100^{50}$

If $ 101^{50} > 99^{50} \plus{} 100^{50}$, then I strongly suspect that $ 98^{50} \plus{} 101^{50} > 99^{50} \plus{} 100^{50}$...

EDIT:  see [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1181819#1181819[/url] for the proof that $ 101^{50}>99^{50}\plus{}100^{50}$; it was just posted yesterday.


integration
trigonometry
function
calculus
calculus computations

Show that there exists $ c\in (0,\frac{\pi}{2})$ such that $ \int_{0}^{c}\sqrt{\sin x}dx\plus{}c\sqrt{\cos c}\equal{}\int_{c}^{\frac{\pi}{2}}\sqrt{\cos x}dx\plus{}(\frac{\pi}{2}\minus{}c)\sqrt{\sin c}$.

[quote="projektnemesis"]Show that there exists $ c\in (0,\frac {\pi}{2})$ such that $ \int_{0}^{c}\sqrt {\sin x}dx \plus{} c\sqrt {\cos c} \equal{} \int_{c}^{\frac {\pi}{2}}\sqrt {\cos x}dx \plus{} (\frac {\pi}{2} \minus{} c)\sqrt {\sin c}$.[/quote]
[hide="Hint"]
I will use intermediate value theorem. Take $ g(u) \equal{} \int_{0}^{u}\sqrt {\sin x}dx \plus{} u\sqrt {\cos u} \minus{} \int_{u}^{\frac {\pi}{2}}\sqrt {\cos x}dx \minus{} (\frac {\pi}{2} \minus{} u)\sqrt {\sin u}$. We obtain that $ g(u)$ is obviously a continuous function. $ g(\frac {\pi}{2}) \equal{} \int_{0}^{\frac {\pi}{2}}\sqrt {\sin x}dx > 0$ and $ g(0) \equal{} \minus{} \int_{0}^{\frac {\pi}{2}}\sqrt {\cos x}dx < 0$, (because $ f(x) \equal{} \sqrt {\cos(x)}$ is a nonnegative function in ($ 0,\frac {\pi}{2}$)). Then there exist $ c\in (0,\frac {\pi}{2})$ such that $ g(c) \equal{} 0$ and we are done. I am sorry if I am wrong..
[/hide]


The categories and percentages for 536 million juvenile books are given in the table. Of these juvenile books, how many books were Nonfiction?

$ \begin{tabular}{|l r|}
\hline
Fiction& 55\% \\
Nonfiction & 25\% \\
Educational & 10\% \\
Other & 10\% \\
\hline

\end{tabular}$

$ 25\%$, or $ \frac{1}{4}$, of the books are nonfiction.  There are 536 million books, so the answer is \[ \frac{536000000}{4}\equal{}\boxed{134000000}\]

EDIT: beaten, hmm...

not the reason i asked on forum

i asked because 1/4 times 536000000 is 134000000, not 136000000 as it said on the review

Most likely it was a typo when the person put in the question and answer.

unfair!!!
i typed in $ 134000000$
$ 536$ million divided by $ 4$
i got $ 134$ million
so it is probably a glitch

I reported the same problem because when I divided 536000000 by 4 I got 134000000. I think they made an error.


complex analysis
trigonometry
complex analysis unsolved

does there exist a holomorphic surjection from the unit disk onto the whole of the complex plane?

Lots of them, like $ z\mapsto ((z\plus{}1)^{\minus{}1}\minus{}1)^2$

What about,
$ \frac {1}{\cos \left( z \plus{} (1 \plus{} \pi/2) \right)}$
There is an isolated essential singularity at $ \minus{} 1$ now, on the boundary of the disk. So by Picard's theorem and neighborhod of $ \minus{} 1$ inside the disk will get mapped into $ \mathbb{C}$ with an exception of a single point. But this map doesn't miss any points.

EDIT:  :oops:

That expression is never zero. And the singularity at $ z\equal{}\minus{}1$ is a simple pole.


Hmm, the equation should be:
$P = (\frac{dx}{dt}) * (m \frac{d^{2}x}{dt^{2}}) = k$.

Letting $y = \frac{dx}{dt}$, the equation then becomes:
$my* (\frac{dy}{dt}) = k$, so multiplying by $dt$ and integrating from $v$ (initial velocity) to $y$ (current velocity) we get:

$\frac{m}{2}* (y^{2}-v^{2}) = kt$, so then
$y = \sqrt( \frac{2kt}{m}+v^{2})$

Then, $a = \frac{dy}{dt}= \frac{\frac{k}{m}}{\sqrt(\frac{2kt}{m}+v^{2})}= \frac{k}{\sqrt(2kmt+(mv)^{2})}$.

[quote="mli"]Suppose an object produces constant power k.  Find acceleration function in terms of k if velocity at t=0 is v and position at t=0 is p.  I've set (sorry i can't do latex)

k = ( x / t) * d^2x/(dt)^2 but I don't know if this is right, and I can't solve this differential equation because I haven't done real math in a year and I'm getting slower.[/quote]

$\text{Power}= \text{Force}\times\text{velocity}$

$\Rightarrow k = ma\cdot v$

$\therefore a = \frac{\frac{k}{m}}{v}$

Also $v\frac{dv}{dt}= \frac{k}{m}$

$\Rightarrow \int v dv = \int \frac{k}{m}dt$

$\Rightarrow \frac{1}{2}(v^{2}-v_{0}^{2}) = \frac{k}{m}t$

$\Rightarrow v = \sqrt{2\frac{k}{m}t+v_{0}^{2}}$

$\therefore a = \frac{\frac{k}{m}}{\sqrt{2\frac{k}{m}t+v_{0}^{2}}}$


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let $ A$ be a ring (commutative and with 1) in which every element $ x$ satisfies $ x^{n}\equal{} x$ for some $ n > 1$ (depending on $ x$). Show that every prime ideal in $ A$ is maximal.

the property still holds in factor rings. integral domains with the property are obviously fields, just cancel $ x$. thus, every prime ideal is maximal.

-oo-, I don't understand your reply. Please elaborate.

What he is trying to say:

Take a Primideal I and consider $ K: \equal{}R/I$. K is a integral domain because I is prime. The equation $ x^{n}\equal{}x$ carres over to K. An integral domain with this property is a field as one can see by canceling to $ x^{n\minus{}1}\equal{}1\Rightarrow x^{n\minus{}2}\equal{}x^{\minus{}1}$. So I is not only prime, P is maximal.


A circle $ \Gamma$ and a line $ \ell$ is given in a plane such that $ \ell$ doesn't cut $ \Gamma$.Determine the intersection set of the circles has $ [AB]$ as diameter for all pairs of $ \left\{A,B\right\}$ (lie on $ \ell$) and satisfy $ P,Q,R,S \in \Gamma$ such that $ PQ \cap RS\equal{}\left\{A\right\}$ and $ PS \cap QR\equal{}\left\{B\right\}$

Let $ O$ be the center of $ \Gamma$ and let the diagonals $ PR, SQ$ of $ PQRS$ intersect at $ C.$ $ l \equiv AB$ is a polar of $ C$ WRT the circle $ \Gamma$ $ \Longrightarrow$ $ C$ is fixed. Likewise, $ BC, CA$ are polars of $ A, B.$ $ AO \perp BC, BO \perp CA, CO \perp AB$ $ \Longrightarrow$ $ O$ is orthocenter of $ \triangle ABC.$ Let $ A' \in BC, B' \in CA, C' \in AB$ be feet of the altitudes $ AO, BO, CO.$ Inversion in $ \Gamma$ takes $ A, B$ into $ A', B'.$ The angles $ \angle AA'B, \angle AB'B$ are right, $ A', B'$ are on a circle $ \Delta$ with diameter $ AB.$ Therefore, inversion in $ \Gamma$ takes the circle $ \Delta$ into itself, which means that $ \Delta$ is perpendicular to the inversion circle. The circle $ \Delta$ centered on the line $ \l$ is also perpendicular to this line. The non-intersecting line $ \l$ and circle $ \Gamma$ can be transformed by some other inversion into 2 concentric circles $ \Lambda^*, \Gamma^*.$ Image of $ \Delta$ in this inversion is a circle or line perpendicular to these 2 concentric circles, therefore it is a line $ d^*$ through their common center $ X^*.$ Set of all circles $ \Delta$ goes into set of all lines $ d^*$ through $ X^*,$ therefore, set of all circles $ \Delta$ is a pencil circles of intersecting at $ X, Y$ (and centered on the line $ l$). $ XY \perp l$ and by symmetry, $ O \in XY.$

[quote="sinankaral53"]A circle $ \Gamma$ and a line $ \ell$ is given in a plane such that $ \ell$ doesn't cut $ \Gamma$.Determine the intersection set of the circles has $ [AB]$ as diameter for all pairs of $ \left\{A,B\right\}$ (lie on $ \ell$) and satisfy $ P,Q,R,S \in \Gamma$ such that $ PQ \cap RS \equal{} \left\{A\right\}$ and $ PS \cap QR \equal{} \left\{B\right\}$[/quote]
turkish solution


function
real analysis
calculus
calculus computations

Consider the equation $ x'' \equal{}\minus{}kx$ with $ k > 0$ (the spring equation)
Prove the law of conservation of energy, which in this case says that for any solution $ x(t)$ the quantity $ \frac{x'^{2}}{2}\plus{}\frac{kx^{2}}{2}$ is constant.
(in this case, the first summand is the kinetic energy of the spring and the second summand is the potential energy of the spring.

Total energy should be $ E\equal{}\frac{1}2(x')^{2}\plus{}\frac{k}{2}\,x^{2}.$ The first term is kinetic energy which does depend on motion! Kinetic energy is $ \frac{1}2\,mv^{2},$ and we are tacitly assuming that $ m\equal{}1.$

The proof is simple. Just assume that $ x(t)$ is a smooth function that satisfies the differential equation. Compute $ \frac{dE}{dt},$ using the chain rule as appropriate, and using the the fact that $ x$ is a solution to the differential equation.

o, my mistake there.. and thanks for the help


algebra
polynomial
binomial theorem
number theory unsolved
number theory

Consider $x_n$ the coefficient of $ x^n$ in $ (x^2+x+1)^n$. Prove that for prime $p$ we have: $p^2$ divides $x_p-1$.

We write $(x^2+x+1)^p$ as $(x-\varepsilon)^p(x-\bar\varepsilon)^p$, develop the two factors according to the binomial theorem, and the desired result follows immediately by using the fact that $p|\binom pk,\ \forall k\in\overline{1,p-1}$ (actually, this works for primes larger than $3$, and it's not true for $p=3$, as you can easily check: $x_3=7$).

For $p=2$ and $p=3$ the statement is not true. Assume  $p>3$. If $\zeta=e^{\frac{2\pi i}{3}}$, then $x^2+x+1=(x-\zeta)(x-\zeta^2)$ and thus the coefficient of $x^n$ in  $(x^2+x+1)^n$ is equal to $$a_n=\sum_{k=0}^n \binom{n}{k} (-\zeta)^{n-k} \binom{n}{n-k} (-\zeta^2)^k=(-\zeta)^n\sum_{k=0}^n \binom{n}{k}^2 \zeta^k$$ Hence, $a_p\equiv -\zeta^p(1+\zeta^p)=1(\text{mod} \ p^2)$ in the ring $\mathbb{Z}[\zeta]$ and therefore in the ring $\mathbb{Z}$.


function
integration
logarithms
calculus
calculus computations

Hello,

Who can you integrate this $y=2^x$ and derferitate it?

Greets.

hint: Suppose that $u$ is the function of $x$ then $\int a^u du=\frac{a^u}{Ln a}+c$;)

first I will to calculate the derferitate so $y'=(2^x)'$ thus $lny=xln(2)$ then $\frac{1}{y}y'=ln(2)$ and $y'=ln(2)2^x$ is this oké?

Then integrate $Y=\int2^x$ again taking a ln impossible I can 't see  Who can help? Greets

You can write $2^x = e^{x \ln 2}$ and i think you can integrate $\int e^{(\ln 2)x} \, dx$ by yourself.

oké thank you very much You need to see it els you look and look and you have now idea.

So we write it $\int 2^x=\int e^{xln(2)}$ then $u=xln(2) \rightarrow \ u'=ln(2) dx$ and this is easy $\frac{1}{ln(2)}\int e^{u}du$

then $\int2^x=\frac{e^{xln(2)}}{ln(2)}=\frac{2^x}{ln(2)}$

Greets thank you very much.


A board game spinner is divided into three parts labeled $ A$, $ B$
and $ C$. The probability of the spinner landing on $ A$ is 1 and
the probability of the spinner landing on $ B$ is $ \frac{5}{12}$.  What is the
probability of the spinner landing on $ C$? Express your answer as a
common fraction.

WTF?

If the probability of the spinner landing on red is 1, then it will always hit red.

This is a mistake. It says $ 1$, but it's really $ \frac{1}{3}$.

So $ \frac{1}{3} \plus{} \frac{5}{12} \equal{} \frac{3}{4}$, and we subtract this from $ 1$, getting $ \frac{1}{4}$

[hide="this is my answer i think ..."]landing on A =1/3
landing on B=5/12

adding these together we get 1/3+5/12=3/4

we subtract this from 1-3/4=1/4

so the answer is 1/4 or $\boxed{1/4}$[/hide]

[b][i][u]please rate my post thank you!![/u][/i][/b]

please be fair though thanks [=

 :coool:  :coool:  :coool:  :coool:

may the force of coolness be with you

PS i know coolness isn't a word

[hide]Okay, 1\3+5\12=3\4, and you need to subtract 1 so the answer would be 1\4[/hide]
Problem and solution above.


Let $\displaystyle m,n \in \mathbb N$, $\displaystyle m,n > 3$. Prove that the roots $\displaystyle x_1,x_2$ of the polynomial $x^2 - mx + n = 0$ are natural numbers if and only if $\displaystyle \left\lfloor m x_1 \right\rfloor + \left\lfloor m x_2 \right\rfloor$ is a perfect square.

$\displaystyle \star \star \star$

[hide]If $x_1, x_2\in\mathbb N$, then $\lfloor mx_1\rfloor+\lfloor mx_2\rfloor=(x_1+x_2)^2$.
If $x_1x_2\leq0$, then $m\leq0$ which it can't be.
For all nonatural positive real number solutions, we have $x_1+x_2, x_1x_2\in\mathbb N$.
Let $m=x_1+x_2$ and then $\lfloor mx_1\rfloor+\lfloor mx_2\rfloor=\lfloor x_1^2+x_1x_2\rfloor+\lfloor x_2^2+x_1x_2\rfloor=\lfloor x_1^2\rfloor+\lfloor x_2^2\rfloor+2x_1x_2$
Let $\left\{x\right\}$ denote the fractional part of x. We have $\left\{x_1^2\right\}+\left\{x_2^2\right\}=1$ so $\lfloor mx_1\rfloor+\lfloor mx_2\rfloor=(x_z+x_2)^2-(\left\{x_1^2\right\}+\left\{x_2^2\right\})=m^2-1$[/hide]


can anyone help me with the limit that is in the attachment? thank you

Lim (x->3)
thank you :D

$ \lim_{x\rightarrow 3} A^{B}$,

Rationarization of the both denominator and numerator of $ A$ at a time, then rationalize of the denominator of $ B$ gives the answer in the form:

$ \frac {(n \plus{} 1)\sqrt {n}}{n \plus{} 2}$ for some positive integer $ n$.

[quote="kunny"]$ \lim_{x\rightarrow 3} A^{B}$,

Rationarization of the both denominator and numerator of $ A$, then rationalize of the denominator gives the answer in the form:

$ \frac {(n \plus{} 1)\sqrt {n}}{n \plus{} 2}$ for some positive integer $ n$.[/quote]
$ n \equal{} 2$  :wink:

The limit is equal to $ \left( \frac{8}{9} \right)^\frac{\minus{}1}{2}\equal{}\frac {3\sqrt {2}}{4}$


In the triangle $ ABC$ we have $ AB \equal{} 5$ and $ AC \equal{} 6$. The area of the triangle when the $ \angle ACB$ is as large as possible is

$ \text{(A)}\ 15 \qquad \text{(B)}\ 5 \sqrt{7} \qquad \text{(C)}\ \frac{7}{2} \sqrt{7} \qquad \text{(D)}\ 3 \sqrt{11} \qquad \text{(E)}\ \frac{5}{2} \sqrt{11}$

[hide="Solution"]
Consider a circle of radius $ 5$ centered at $ A$, which represents the locus of all possible points $ B$.
[asy]size(150);
pair A=(0,0), C=(-6,0), B=(-25/6,5*11^.5/6);
draw(Circle(A,5));
draw(A--B--C--cycle);
label("\(6\)",(A+C)/2,S);
label("\(5\)",(A+B)/2,NE);
picture p = new picture;
draw(p,Circle(C,0.4));
clip(p,A--C--B--cycle);
add(p);
label("\(\theta\)",C+(.3,0),NE);
label("\(A\)",A,SE);
label("\(B\)",B,NW);
label("\(C\)",C,SW);[/asy]
Clearly $ \theta \equal{} \angle ACB$ is maximized when $ \overline{BC}$ is a tangent to the circle, so $ \angle B \equal{} 90^{\circ}$. Then the area is $ \frac{1}{2}bh \equal{} \frac{1}{2} \cdot 5 \cdot \sqrt{6^2 \minus{} 5^2} \equal{} \frac{5}{2}\sqrt{11}\ \mathrm{(E)}$.
[/hide]

[quote="NogarD"][quote="azjps"][hide="Solution"]
Consider a circle of radius $ 5$ centered at $ A$, which represents the locus of all possible points $ B$.
[asy]size(150);
pair A=(0,0), C=(-6,0), B=(-25/6,5*11^.5/6);
draw(Circle(A,5));
draw(A--B--C--cycle);
label("\(6\)",(A+C)/2,S);
label("\(5\)",(A+B)/2,NE);
picture p = new picture;
draw(p,Circle(C,0.4));
clip(p,A--C--B--cycle);
add(p);
label("\(\theta\)",C+(.3,0),NE);
label("\(A\)",A,SE);
label("\(B\)",B,NW);
label("\(C\)",C,SW);[/asy]
Clearly $ \theta \equal{} \angle ACB$ is maximized when $ \overline{BC}$ is a tangent to the circle, so $ \angle B \equal{} 90^{\circ}$. Then the area is $ \frac {1}{2}bh \equal{} \frac {1}{2} \cdot 5 \cdot \sqrt {6^2 \minus{} 5^2} \equal{} \frac {5}{2}\sqrt {11}\ \mathrm{(E)}$.
[/hide][/quote]

What if $ \overline{BC}$ is 5 units long. Then 6 would be the base line, and the height would be increased compared to your case? Or am I just wrong :P ?[/quote]

In fact, if we wanted to maximize the area of the triangle, we would have $ \frac{1}{2} bc \sin (\angle A) \le \frac{1}{2}bc \equal{} 15$. But as Zellex pointed out the question asks for the case when $ \angle ACB$ is maximized.


calculus
integration
calculus computations

okay, so i got hel on one of the problems, but i am still confused on all of these, they keep getting harder, please help me  with these problems

1. antiderivative of (5e^(-3g)) dg

2.antiderivative of r(e^-r^2) dr

3.the antiderivative of r(radical r^2+2) dr

4. antiderivative of (t^2 + 2)/(t^3 + 6t +3) dt

help please, very confused

These all belong in "Calculus Computations and Tutorials." Please post there, not in "Unsolved and Proposed Problems."

And if you're that confused, you need more help than you're going to get from us. Talk to your teacher or professor. Get tutoring.

What I called the "natural substitution" - that is, substitute for the "thing inside" - does work in each one of these cases. They [i]don't[/i] keep getting harder - they're all at almost exactly the same level of difficulty.


function
limit
trigonometry
parameterization
complex analysis
calculus
calculus computations

hi
here's a question...the general term of a series is t(n)=1/(a*n^4 + 1).....find the sum of this infinite series...
Shyam

We will use the following

[b]Theorem[/b]. If $ f$ is meromorphic function on $ \mathbb{C}$ with poles $ z_{i}, i = 1,\ldots, k$, $ z_{i}\notin\mathbb{Z}$, and $ \lim_{z\to\infty}zf(z) = 0$ then
\[ \lim_{n\to\infty}\sum_{m=-n}^{n}f(m)=-\sum_{i=1}^{k}\mathrm{res}_{z_{i}}\pi f(z)\cot\pi z.\blacksquare\]
Let $ f(z) =\frac{1}{z^{4}+\zeta^{4}}$. $ f$ is meromorphic with poles $ z_{1}=-\zeta\frac{1+i}{\sqrt{2}}, z_{2}=-\zeta\frac{1-i}{\sqrt{2}}, z_{3}=\zeta\frac{1-i}{\sqrt{2}}, z_{4}=\zeta\frac{1+i}{\sqrt{2}}$. We also have (easy to check) $ \mathrm{res}_{z_{i}}\pi f(z)\cot\pi z =\pi\frac{-z_{i}}{4\zeta^{3}}\cot\pi z_{i}$. By the theorem
\[ \lim_{n\to\infty}\sum_{m=-n}^{n}f(m)=\sum_{n=-\infty}^{\infty}\frac{1}{n^{4}+\zeta^{4}}=\frac{\pi}{4\zeta^{3}}\sum_{1}^{4}z_{i}\cot\pi z_{i}.\]
Few more calculations gives
\[ \sum_{n=-\infty}^{\infty}\frac{1}{n^{4}+\zeta^{4}}=\frac{\pi}{2\sqrt{2}\zeta^{3}}\left( (1+i)\cot{\frac{\pi\zeta(1+i)}{\sqrt{2}}+(1-i)\cot{\frac{\pi\zeta(1-i)}{\sqrt{2}}\right)=\frac{\pi}{2\sqrt{2}\zeta^{3}}\frac{\sin{\sqrt{2}\pi\zeta}+\sinh{\sqrt{2}\pi\zeta}}{\cosh{\sqrt{2}\pi\zeta}-\cos{\sqrt{2}\pi\zeta}}.}}\]
Assume that $ a > 0, b: =\sqrt [4]{1/a}$. Then $ \sum_{1}^{\infty}\frac{1}{an^{4}+1}=\frac{1}{a}\sum_{1}^{\infty}\frac{1}{n^{4}+1/a}=\frac{1}{2}\left(\frac{\pi}{2\sqrt{2}b^{3}}\frac{\sin{\sqrt{2}\pi b}+\sinh{\sqrt{2}\pi b}}{\cosh{\sqrt{2}\pi b}-\cos{\sqrt{2}\pi b}}-\frac{1}{b^{4}}\right)$.

You can use the same method when $ a < 0$.

thanks for the soln adam....but i'm still in grade 12 (which means i'm still in school...not in college!!)...so i've not really learnt abt the zeta function u used....jus heard abt it...isn't there a simpler solution??anybody please....

I don't think there is a simpler solution without using complex analysis concepts. Also, the zeta used by Adam. is a parameter, not the zeta function.


Especially those who went there for the first time. Pros and cons and would you go to the same camp next year. Thank you.

Hmm, this doesn't seem to have gotten you very far.  If you're still interested in people's opinions, you might try asking more specific questions.  Or you might troll around in the older threads in this forum for answers to your questions.

Over the summer, I attended a math camp at the University of Michigan. It ws the first summer I did anythng like that. I really enjoyed it. There really weren't many downsides to it, but here's something to consider: If you do a program that you have to stay there for the whole time it runs, you might get a little bit homesick. I did, but I concentrated on making friends and enjoying myself, that it went away pretty quickly.
I hope I helped!
Hikari

i stayed on the comp this summmar but ill go to a camp next summar. but its usually best to take someone you know to camp, cause then you have at least one person to talk to. and its not as hard to make friends then. oh oh booyah!

ah jes. ill make sure to bring the ddr pads! oh oh booyah!


For which $ n\equal{}2^p\cdot 5^r$, $ p,r\ge 1$, the number

$ T\equal{}n\plus{}25$ is a perfect square?

[quote="Inequalities Master"]For which $ n \equal{} 2^p\cdot 5^r$, $ p,r\ge 1$, the number

$ T \equal{} n \plus{} 25$ is a perfect square?[/quote]
$ T \equal{} k^2 \iff (k\minus{}5)(k\plus{}5) \equal{} n$ So $ k\minus{}5 \equal{} 2^x5^y$ $ k\plus{}5\equal{}2^z5^w$ $ k\plus{}5\minus{}(k\minus{}5)\equal{}10\equal{}2^x5^y\minus{}2^z5^w$. We get some cases:
$ x \equal{} 0$: Then $ 2^x5^y$ is odd, but then $ 2^z5^w$ is odd so $ z\equal{}0$ giving $ 5^y\minus{}5^w \equal{} 10$ which has no solution. (In the same way $ z \neq 0$)
$ y \equal{} 0$: Then $ 2^x5^y$ is not divisibe by $ 5$, but then $ 2^z5^w$ is not divisble by $ 5$ so $ w\equal{}0$ giving $ 2^x\minus{}2^z \equal{} 10$ which has no solution. (In the same way $ w \neq 0$)
So we have $ x,y,z,w \neq 0$.
But $ x,z \ge 2$ gives $ 2^x5^y\minus{}2^z5^w$ is divisble by $ 4$, but $ 4 \nmid 10$ so either $ x\equal{}1$ or $ z\equal{}1$.
In the same way: Either $ y\equal{}1$ or $ w\equal{}1$.
$ x\equal{}1,y\equal{}1$: $ 10\minus{}2^z5^w\equal{}10$. No solution.
$ x\equal{}1,w\equal{}1$: $ 2 \cdot 5^y\minus{} 5 \cdot 2^w \equal{} 10(5^{y\minus{}1} \minus{} 2^{w\minus{}1}) \equal{} 10$. And then $ 5^{y\minus{}1} \minus{} 1 \equal{} 2^{w\minus{}1}$. Solution: $ y\equal{}2$, $ z\equal{}3$. Giving $ n \equal{} 5^32^4$ And indeed $ 25 \plus{} 2^45^3 \equal{} 45^2$
$ z\equal{}1,y\equal{}1$: $ 2^x \cdot 5 \minus{} 2 \cdot 5^w \equal{} 10(2^{x\minus{}1} \minus{} 5^{w\minus{}1}) \equal{} 10$ giving $ 2^{x\minus{}1} \equal{} 1 \plus{} 5^{w\minus{}1}$ But $ 1 \plus{} 5^{w\minus{}1} \equiv 2 \bmod 4$ so $ w\equal{}1$ and $ x\equal{}2$ Giving $ n \equal{} 2^35^2$. $ 25 \plus{}2^35^2 \equal{} 15^2$
$ z\equal{}1,w\equal{}1$: $ 2^x \cdot 5^y \minus{} 2 \cdot 5 \equal{} 10$ $ \iff$ $ 2^x5^y \equal{} 20$ giving $ x\equal{}2$ $ y\equal{}1$ and $ n \equal{} 2^35^2$, which we have already found.

So $ n \equal{} 200$ or $ n \equal{} 2000$!


Let A and B be opposite vertices of a cube.  A round-trip path is defined to be a path on the cube from A to B and then back to A, using the least possible number of edges, and without using the same edge more than once.  How many round-trip paths are there?

Can you generalize the problem to a 4-dimensional (hyper)cube?  How about an n-dimensional cube?

[hide]For a cube, I got 24 possibilities.
For a hyper-cube I got 216 possibilities.
For a n-dimensional-cube, I got ${(n-1)^{(n-1)}}\cdot n\cdot 2$ possibilites[/hide]

Could you explain your answer for the (3-dimensional) cube?  That's not the answer I got.

Yeah, I asked someone and they said my answer is wrong, but here's the explanation of the wrong answer if you need it.

[hide]I thought that for the cube, there would be 3 edges to go on for the first edge, and 2 for each one after that, making the possibilites for going from one corner to another 3*2*2 = 12.  Then I thought you would multiply the answer by 2 to get the number of ways to go from one corner, to the other and back.

But I know now that it's really 3*2*1, and that you have to square the number.[/hide]

I agree that there are 3*2*1 = 6 (shortest) paths from A to B, and similarly there are 6 paths from B to A.  But 6*6 will count paths where we used the same edge going forward and back, and that is not allowed.  In other words, 6*6 is an overcount.


calculus
derivative
geometry
calculus computations

I am having trouble with this problem.  Any help how to start it would be appreciated?
[url=http://imageshack.us][img]http://img297.imageshack.us/img297/1382/untitledsk9.jpg[/img][/url]

It can be argued that the rate of change is equal to the area of the surface of oil. The argument is based on the fact that the volume of a cylinder is (area of base)(height).


Looks to me very similar to Tetris. I don't like Tetris, is it worth a try?

(some Wiki pirating)
Blokus is an abstract strategy board game for two to four players, invented by Bernard Tavitian and first released in 2000 by Sekkoïa, a French company. It has won several awards, including the Mensa Select award and the 2004 Teacher's Choice Award.

[img]http://upload.wikimedia.org/wikipedia/en/thumb/9/91/Blokus.jpg/200px-Blokus.jpg[/img]

It's not really that similar to tetris.  Unlike tetris, in which you play against yourself, in Blokus, you have to make your moves such that not only you are in a good position, but so that your opponent has trouble.  So you have to out-think your opponent, and think several moves ahead.  More similar to chess.

how about [url=http://www.boardgamegeek.com/game/24568]Rumis+[/url], the expansion for 3D Blokus aka Rumis?

I've never heard of that game but you should certainly play Blokus.  It had a cult-like following in my HS.

Blokus FTW :D

Rumis+ looks really cool, but only problem is that if you have empty space in the middle, it would be a mess to put a piece in there (if you can)

If you go to official Blokus website, you can play online against others.


Given polynomials $ P,Q \in \mathbb{C}[x]$ with no common factors. Prove that $ R(x,y)\equal{}P(x)Q(y)\minus{}P(y)Q(x)$ is squarefree (has no factors of the form $ T(x,y)^2$ where T is a nonconstant polynomial).

Assume the contrary. Then $$T(x,y)\mid P(x)Q(y)-P(y)Q(x)$$ we would also have $$T(x,y)\mid P'(x)Q(y)- P(y)Q'(x),$$ after taking the $x$-derivative.
Multiplying the first relation by $P'(x)$ and the second by $P(x)$ and taking the difference we get
$$T(x,y) \mid (P'(x)Q(x)- P(x)Q'(x))P(y).$$
Let's assume $P'Q - PQ'$ is non-zero for now. Then $T(x,y)$ is a non-trivial factor of $P(y)$ as a polynomial of $y$ (upon choosing a proper $x$) and similarly a factor of $Q(y)$. If $P'Q-PQ'$ is identical to zero we have $$\frac{P'(x)}{P(x)} = \frac{Q'(x)}{Q(x)}\implies \int \frac{P'(x)}{P(x)}dx = \frac{Q'(x)}{Q(x)}dx \implies \int \frac{dP(x)}{P(x)} = \frac{dQ(x)}{Q(x)}\implies \ln |P(x)| = \ln |Q(x) | + c.$$
The latter happens only if $P(x) = dQ(x)$ for some constant $d$ which is impossible since $P$ and $Q$ have no common factor.


How many subsets of the set $ \{1,2,3,4,...,30\}$ have the property that the sum of the elements of the subset is greater than 232?

My attempt:
The sum of $ 1,2,3,...30$ is $ 465$, and I think it might be related to $ 232$, but I don't know what's next.

Let A be a set such that its sum is greater than 232 then A' will have sum $ \le$ 232, Hence for every set A' exists, means out of  $ 2^{30}$ sets only half have sum greater than 232 hence $ 2^{29}$.


Could you find all powers of two such that if you re-arrange the digits of this number you get another power of two. :blush:  :oops:

A number and a rearrangement of its digits are congruent $ \bmod 9$, but the order of $ 2 \bmod 9$ is $ 6$, which means that the greater of any such powers of two must be $ 2^6$ times greater or more, so they can't have the same number of digits.  The only difficulty is in eliminating the possibility of rearranging zeroes as the leading digit; is this given as valid or invalid in the problem?


if p is prime, prove that p-1 1s (like for 7, you get 1111111) is diviisble by p given that p>5

[hide]
So you have $10^{p-1}+10^{p-2}+...+1$. Notice that these are all different $\mod p$ and none of them is divisible by $p$ (because $p>5$ this works). Thus they must be $1,2,...,p-1 \mod p$. Summing them up we get $\frac{p(p-1)}{2} \mod p$. But since $p$ is odd, the $p-1$ cancels with the $2$ so $\frac{p(p-1)}{2} \equiv 0 \mod p$.[/hide]


Posted before. Here is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14296]link[/url].


This one is very nice.

A 6 x 6 "chessboard" is tiled with 1 x 2 dominos. Prove that there exists a straight line, connecting two opposite sides of the board, which is not cut by any domino.

We can see that each of the 10 lines which connect 2 opposute sides (we don't count the edges of the square) must intersect an even number of pieces: a line like that sepparates a rectangle of dims 6 and k (1<=k<=5), which contains an even number of squares 1X1. The rectangle contains an integer number of pieces 1X2 (covering an even number of little squares) and some squares which have to be near the line we considered, so these last little squares must come in even numbers.

This means that we have >=20=10*2 pieces, but the square is only covered by 18 pieces, so we got a contradiction.

I also had another proof, based on the fact that there must be an even number of horizontal pieces and an even number of vertical ones (we prove this with the help of a colouring of the 6X6 square), but it's a bit longer, so I won't post it.


In the "Add an Attachment" section, press "Browse" to select a graphics file; then click "Add Attachment" to add that attachment to your post. Submit the post as usual.

It has to be of certain types, however.  No .bmp files, for example.

I think zabelman meant in [tex]\LaTeX[/tex] files (since it is in this board). I haven't managed to figure out how to make them work properly yet, so I'm also awaiting enlightenment from someone who is more of a [tex]\TeX[/tex]nician than I.

Yes, I meant importing pictures into [tex]\LaTeX[/tex] files... sorry for the ambiguity.

[quote="Osiris"]Ah.

1. Convert the graphics to encapsulated postscript (.eps file); I use photoshop to do this.

2. In your [tex]\TeX[/tex] file, make sure you

\addpackage{graphicx}

3. Use

\includegraphics[width=width,height=height,angle=angle,scale=scale]{filename}

to import graphics. angle rotates a picture by a given number of degrees counterclockwise.[/quote]
Have you actually gotten this to work? I was unable to the last time I tried (some months ago).

It worked for me. There are a myriad of options when you convert graphics into EPS through Photoshop - from what I remember you have to choose "ASCII". Also, place them in the same folder as the TeX files.

what unit of measurement should I use for width, height, angle?

It's up to you. If you leave the measurements blank, the default measurements are used (I don't know the ratio of pixel/inch). If you specify the length, you can specify them in px (pixel), in (inch), or cm (centimetre).

Angle is in degrees.

I don't have photoshop, but I can create pdfs, and pdfs can be imported into [tex]\LaTeX[/tex] files. But when I make a picture in Paint and print to a pdf, the image takes up just a little corner of the page, and the image import commands in [tex]\LaTeX[/tex] seem to want to stretch the entire page down, so as a result, the image comes out very small. Does anyone know what I can do to fix that problem?

If you have [tex]PDF\LaTeX[/tex], then you can import PDF files.

Don't know if you've already solved this problem, but the pictures tutorial in the LaTeX page should solve it.

Yeah, I used that tutorial, and it worked well. Thanks!


Hello, Mathcounts competitions are coming up, and I'm really excited. I think both myself and my school can do well (in our state). I've done some practicing (and more and more closer to competition). My question- In countdown round, I understand speed is very important. However, when I try to skim the way I do in FTW, it's hard. I get distracted my the moderator reading the question very slowly out loud. I cannot skim when he is speaking. How should I overcome this hurdle?

Thank you in advance, and good luck on your competitions!

Hmm... Are you allowed to bring earplugs or something?

I do FTW a lot, and more than half the time my brothers are fighting about <insert topic here> right behind me, and I've gotten pretty good at shutting them out.

If you do it a lot I think you'll do fine.

There's no way to prevent that. You can't ask them to stop. Closing you ears is pointless with hands; you have to do some work to get the answer!

Instead, ask your parents/coaches/siblings to talk while your studying to prepare. However, don't do this too much as it could affect your performance on the actual test. You'll lose study time if you have terrible concentration.

i'd just like to point out that i find it funny that both countdown related threads have been started by members whose username is pink<something>. anyways, in order to make this not spam:

i'd just focus on reading the problem and ignoring the guy. just tune his (in my case, monotone) voice out, and work. thats the best advice i can give you-sorry.

I've had that problem a lot but there's not much you can do
For me it comes naturally
For others... not so much
You'll have to get used to it before the competition, though

[quote="pinkmuskrat"]Thanks. Hmm. I was thinking, maybe I could plug my ears while skimming the question, then unplug them and start working on the problem (while the moderator is still reading).[/quote]

This probably will make your timings higher..I remember I had the same problem, but finlly I am able to read it while ignoring the guy.  IDK how i could do that but I think if you get used to a guy peaking while reading, then that should help (ie. put TV on and read a book)

The plugging ears idea is pretty bad, I'm sorry to say. If it's a fairly easy problem that can be twitched easily, the time that you take to unplug your ears may be enough for the other person to get a point. I guess you should just get used to distractions and learn how to block them out.

i don't know if this is a good idea, but maybe you could practice at school. If the teacher(orclass) is reading a passage, read a little bit faster than the class and try to block the teacher out.

Attempt this at your own risk :P


[quote="Labiliau"]we use the generalised moyenne :

$(\frac{(4x-x^3)^{1/3}+(x+x^3)^{1/3}}{2})^3 =< (\frac{(4x-x^3)^{1}+(x+x^3)^{1}}{2})^1 = 5x/2 =< 5$

Thus the maximum is $2 \sqrt[3]{5}$[/quote]

I don't believe that your maximum actually occurs though.  You have only found an upper bound.

[quote="siuhochung"][quote="asjsd"]what is that theorem?
thanks![/quote]
power mean inequality.
Equivalent to a special case of holder in this case.[/quote]

o thanks siuhochung !!!


So, I would like to know the definites/factors of your state. For example, in my state...well, nvm (I don't think Tobias counts :P ...). So for example in Michigan there's the Yuan factor...

$\$$10 leet will say something about the 3 factors in MA...

-jorian

I'm the only one returning who was in the top 10 at state last year so am I a factor? This makes no sense whatsoever.

umm...well neal deleted some stuff in another topic so this is kinda hard to explain. it all started when someone had a goal of first at states for michigan. then someone else noted that was hard to do since there was "the yuan" factor. then leet had a comment about the lian/synge/(perfect628) factor for MA. of course texas has kevin chen, we have dan tobias :P , and florida has preet patel.

-jorian

are there any other definates for WI?
-jorian

no. Like I said, I am the only one returning who made top 10 states.

oh, everyone else was in 8th grade?
-jorian

[quote="bpms"]Hmm. Let's see, whos guaranteed to make nationals:

:crickets chirping: .

That's right, no one is guaranteed to make nationals no matter how good they are. Kevin Chen has a very very high chance, but everyone has off days. You also don't know about someone who is unknown, who may become known.
PS: You spelled definite wrong ;)[/quote]

i agree with preet completely there.  no one is guaranteed, i dont care how good you are.  you can have an off day, and there are alot of good competitiors out there who could randomly to really well.

but there are "favorites".  :P   i vote for preet from florida.

I vote for:

No one since Chapters hasn't even begun.

Darnit, I was just about to post "this is where someone takes 'definite' literally" and posts "no one", but then two people beat me to it and posted "no one".

/goes back to lurking

umm...wasn't the point of this thread...

more like, "who has a high chance of returning or going"...something like that

My last name is trabucco, see if you can find my first 1 somewhere on this forum. I agree, I had an off day at chapters last year. Luckilly, it was easy, though.

This is too early to speculate. Half these people might not have even made their school teams for all we know.

But if I were to say anything, I think Massachusetts will be very, very competitive. We have 3 returning 8th graders, then assuming all three of us make it back (which might not neccessarily happen), there are 5-6 people who definitely would get like 35+ on a 2006-like state test, and would certainly deserve to go.

Okay, I have no idea what is going on, but this topic shouldn't even be here.

I believe that we should just do well at Chapters and States in order to realize that we all are capable to make it to Nationals.

And yeah, last year 43 was the cutoff score for our chapter,

I sort of didn't try hard last year, since I got

3rd Place in 2005 and 4th Place in 2006 at Chapters in Written.

But, maybe I was on an off day, since then I got

3rd Place in 2005 and 1st Place in 2006 at States in Written

So, I'm just saying don't let the Chapter Scores Fool Ya because they have like no effect on State Scores.

And I am not "exaggerating" but our chapter has 200 students participate, I think, and the top 4 teams and top 6 individuals make it I believe.

[quote="davidlizeng"]i'm in the same chapter as Ubemaya..
last year there was around 250-300 people or something
and Ubemaya is lying........
44 was the "cutoff" score
a [b]43[/b] would not make it, but a 44 would
the proctor guy said Santa Clara was the hardest
chapter in the U.S.
he might be exxagerating (i can't spell this)[/quote]
Mr. Boyd's county had a 45 cutoff for countdown. There were four perfects last year. There are usually 4+ teams in the 60s. Teams have gotten scores of 65 before. Only a few with scores of 44 made it to states by individual strength. It depended on tiebreakers for 44s to make it. Kevin Chen had a higher written rank at nationals than at county. Though your county is probably about as hard.
About what Mr. Frost said, one kid got 4th last year and 12th this year at states. No one is guaranteed to make nationals. Our written ranks at states in our order ranks at nats were
4, 3, 1, 2, Almost the opposite written order.

[quote="bpms"][quote="davidlizeng"]i'm in the same chapter as Ubemaya..
last year there was around 250-300 people or something
and Ubemaya is lying........
44 was the "cutoff" score
a [b]43[/b] would not make it, but a 44 would
the proctor guy said Santa Clara was the hardest
chapter in the U.S.
he might be exxagerating (i can't spell this)[/quote]
Mr. Boyd's county had a 45 cutoff for countdown. There were four perfects last year. There are usually 4+ teams in the 60s. Teams have gotten scores of 65 before. Only a few with scores of 44 made it to states by individual strength. It depended on tiebreakers for 44s to make it. Kevin Chen had a higher written rank at nationals than at county. Though your county is probably about as hard.
About what Mr. Boyd said, one kid got 4th last year and 12th this year at states. No one is guaranteed to make nationals. Our written ranks at states in our order ranks at nats were
4, 3, 1, 2, Almost the opposite written order.[/quote]

How many people were in countdown in that chapter?
Our chapter had 3 46's, 5 45's, and I think 10 44's. When I said "if you miss 2 sprints you wouldn't make it to states" I wasn't lying, because the lowest individual was 10th place (5th best individual after the top 4 teams) and he got a 44, and by tiebreaker rules he obviously missed 1 target not two sprints.
The first place team last year got 64.5 and the second got 64. I think 4th (Miller) got like a 62.5 or something, I dunno about the rest.
Our team sucks this year (like every year) so I doubt we'll make it to state. :( :( :( :( :(

EDIT: I was looking up last year's chapter information for fun. The website says that the top 4 teams were off by 1 1/4 points, which means that Miller had 63.25 points. Also, 13th place was the last individual who made it to States (which was also 44 though, but he might have missed 1 target not 2 Sprints). And looking at his name, I think he placed 8th at States. I know he placed top 10. And he was a 7th grader last year... :( :( :(

EDIT again: According to some article, last year's team scores were 64.5, 64, 64, 63.25, and 60.

To ubemaya:
In Mr. Boyd's county countdown is official. So 10 people.

[quote="13375P34K43V312"]Hahahah our teams suck...but our individuals are good so it's all good
We got like 56 at chapter last year, that was 5th for our chapter. And other than Mr. Fischer's chapter in Indiana and a couple in New York, mine was probably one of the harder in the east. Like in proportion to population, I think it's one of the harder in the country. Our top ten individual scores were
45 44(me) 44 44 44 43 43 43 43 43 I think.

As to state vs. nationals, our national ranks went 4, 3, 1, 2.[/quote]

is george silvis in your county?

who got all of those scores!!!???

-jorian

Silvis WAS in mine, and he was in 8th grade. Carl, at nats, it was 3,4,2,1.

I did do better than him at nats. At states, he was 1 and I was 2. And I just realized that he was listing the places differently than I was. Disregard what I've said about the subject.

[quote="mathgeniuse^ln(x)"][quote="bpms"][quote="Fanatic"]I don't know how hard the chapter actually is, but the Illinois chapter my brother went to produced 3 National Champions in a span of 5 years... all from different schools, too![/quote]
Um wow. My chapter last year had 1st 3rd 5th and 6th at states but only a 32 was need for countdown at chapter (top ten). Talk about strange trends...[/quote]

32, wow that is low.

We have weird trends in our chapter too.

LIke in 2004 9 41s and 1 40
2005 Like 6 42s 3 43s and 1 44
2006 Like 6 43s 3 44s and 1 45

Yeha, that is totally weird.[/quote]
But didn't 04 have a really hard sprint comparitively?
05 was harder than 06...

I don't know, but I remember that I got 36 that year.  20 sprint 16 target.

The sprint was hard that year, especially #29.

05 wasn't harder than 06, in my opinion, but the results don't show that.


trigonometry
calculus
integration
function
algebra unsolved
algebra

Let $\alpha , \beta , \gamma , \delta$ be positive numbers such that for all $x$, $\sin{\alpha x}+\sin {\beta x}=\sin {\gamma x}+\sin {\delta x}$. Prove that $\alpha =\gamma$ or $\alpha=\delta$.

The keywords are "almost periodic function".
Let $f(x)$ be any finite sum of complex exponentials $\sum c_i e^{{\alpha_i}x}$.
The $c_i$ and $\alpha_i$ can be reconstructed from the set of Fourier integrals $I_{\theta} = $ average value (on $\mathbb{R}$) of $exp (-i \theta t) f(t)$.   $I_\theta$ is zero except when $\theta = \alpha_i$ for some $i$, and the value of the integral is then the coefficient $c_i$.

Using Maclaurin series for $f(x)=\sin{\alpha x}-\sin {\gamma x}$ and $g(x)=\sin {\beta x}-\sin {\delta x}$ and comparing serieses of these two "equal" functions (if the functions are the same then the coefficients in their Maclaurin serieses are the same)$ we receive the answer automatically.

Hint: Take the derivative of the difference of the two sides $4n$ times.

what a funny problem
[hide = solution]
Differentiating once, and setting $x = 0$ we obtain \[\alpha + \beta = \gamma + \delta\] Differentiating two times more and setting \[\alpha^3+\beta^3 = \gamma^3+\delta^3\]
Dividing and doing some algebra we deduce $\alpha \beta = \gamma \delta$ from where it follows that $(\alpha, \beta)$ and $(\gamma, \delta)$ are roots of the same quadratic. Thus $\alpha = \gamma$ or $\alpha = \delta$.
[/hide]


[hide="Solution"]The original equation is equivalent to$\sqrt{x+4}=3-\sqrt{1-x}$
Square both sides:
$x+4=9-6\sqrt{1-x}+1-x$
Thus
$x-3=3\sqrt{1-x}$
Sqrare again:
$x^2-6x+9=9-9x$

$x^2+3x=0$

so $x=0$ or $x=-3$[/hide]

[quote="symmetry"]$\sqrt{x + 4} + \sqrt{1-x} = 3$

Silverfalcon note: I edited Latex for you.[/quote]
I guess it asks for the values of x for which the equation is suttisfied and if this is the case it is very simple.Just square it and cancel the common terms on both sides, what you`ll have left is ((x+4)(1-x))=4
 Square again and solve the optained quadratic equation, you`ll have x=0 or x=-3..Clearly' they are both solutions.
Asking this problem wasn`t a joke, was it?

yeah. i know. i wrote it down on paper and got the answers 0, -3.

Thank you all for your help with my question.


trigonometry
limit
modular arithmetic
function
integration
calculus
algebra

Does the sequence $ \sin (n^{2})\plus{}\sin (n^{3})$ converge?

If I remember it right, Harazi used to post such problems in bundles. :lol: The answer is "No" because the sequence of points $ (n^{2},n^{3})\mod 2\pi$ is dense (actually, uniformly distributed) in $ [0,2\pi]^{2}$ but it would be interesting to see a more elementary approach. Also, the problem is a bit too hard for Computations and Tutorials, so I'm moving it to Unsolved and Proposed problems.

At first when I saw this problem it looked really easy but I am stumped.

How about an easier problem...
$ \sin (n)$

Does this converge?

Unlike (the classic),
$ \lim_{x\to\infty}\sin x$

This one is harder because we can only work on discrete integer points.

The following argument is overkill - it's much, much more than you need to answer sylow_theory's question. But I've always liked it.

Claim: suppose $ \alpha$ is an irrational real number. Define $ x_{n}\equiv2\pi\alpha n\pmod{2\pi}.$ That is, $ e^{ix_{n}}\equal{} e^{2\pi i\alpha n}.$ Suppose $ f$ is a continuous function periodic of period $ 2\pi.$ (For short, $ f\in C(\mathbb{T}).$) Then

$ \lim_{N\to\infty}\frac{1}{N}\sum_{n \equal{} 1}^{n}f(x_{n}) \equal{}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\,dx.$

Corollary: The range of the sequence $ \{x_{n}\}$ is dense in $ \mathbb{T}\equal{} [0,2\pi]$ (with the ends identified.)

Simple consequence of the corollary: the sequence $ \sin n \equal{}\sin\left(\frac{2\pi}{2\pi}n\right)$ takes on values dense in $ [\minus{}1,1]$ and hence diverges. Of course, this uses the fact that $ \pi$ is irrational.

To prove the corollary given the claim, note that if the open interval $ I$ were omitted from the range of the sequence $ \{x_{n}\},$ then we could construct a continuous function $ f$ which is zero on the complement of $ I$ but positive on $ I.$ Then the sum on the left of the claim would be zero while the integral on the right would be positive - a contradiction.

Now, to prove the claim itself:

Step 1: The claim is true when $ f(x) \equal{} e^{ikx}$ for $ k\in\mathbb{Z}$.

There are two cases: either $ k \equal{} 0$ or $ k\ne0.$

If $ k \equal{} 0,$ then $ f(x) \equal{} 1$ everywhere and both the sum on the left and the integral on the right are $ 1.$

If $ k\ne0,$ then the integral on the right is zero. As for the sum on the left:

$ \frac{1}{N}\sum_{n \equal{} 1}^{N}e^{ikx_{n}}\equal{}\frac{1}{N}\sum_{n \equal{} 1}^{\infty}e^{2\pi ik\alpha n}\equal{}\frac{1}{N}\cdot\frac{e^{2\pi ik\alpha}\minus{}e^{2\pi ik\alpha(N\plus{}1)}}{1\minus{}e^{2\pi ik\alpha}}$

Now we use the fact that $ \alpha$ is irrational: $ k\alpha$ cannot be an integer, so $ 1\minus{}e^{2\pi ik\alpha}\ne0$ and the above geometric series computation is correct. Then we have $ \frac{1}{N}$ times a bounded quantity, which clearly tends to zero as $ N\to\infty.$

Step 2: the claim is correct whenever $ f$ is a trigonometric polynomial.

Since both sides are linear expressions in $ f$ and since a trigonometric polynomial is a linear combination of the functions in step 1, this follows immediately.

Step 3: Given $ \epsilon > 0,$ write $ f(x) \equal{} g(x)\plus{}h(x)$ for $ g$ a trigonometric polynomial and $ \|h\|_{\infty}<\epsilon.$ This is a quote of the trigonometric version of the Weierstrass approximation theorem: the trig polynomials are dense in $ C(\mathbb{T}).$

But $ \left|\frac{1}{N}\sum_{n \equal{} 1}^{N}h(x_{n})\right|\le\|h\|_{\infty}$ and $ \left|\int_{0}^{2\pi}h(x)\,dx\right|\le\|h\|_{\infty}.$

With that, and the fact that the claim is true for the trig polynomial $ g,$ we may pass to the limit and prove the claim for arbitrary continuous functions.

---

As I said, extreme overkill given what sylow_theory asked. There are easier proofs. But I do like the argument.


function
trigonometry
algebra
domain
parameterization
algebra unsolved

Find all functions $f: \mathbb R \to \mathbb R$ such that $f(0)\equal{}0$ and
\[f(x \cos B)\minus{}f(x(\cos B)^2)\equal{}x\minus{}x^2 \cos B.\]
For each real number $x$ and each $B  \in \mathbb R \setminus \left\{B : B=k\pi \text{ or } B=k \pi+ \frac{\pi}{2}, k \in \mathbb Z\right\}.$

[quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote]
If $ B \equal{} 0$, we have: $ f(0)\minus{}f(0)\equal{}x \minus{} x^2 \cdot 0 \iff x \equal{}0$, which cannot hold for all $ x$, so there is nu such function... Maybe you missed something?

[quote="Mathias_DK"][quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote]
If $ B \equal{} 0$, we have: $ f(0) \minus{} f(0) \equal{} x \minus{} x^2 \cdot 0 \iff x \equal{} 0$, which cannot hold for all $ x$, so there is nu such function... Maybe you missed something?[/quote]

You should say $ \cos B\equal{}0$ but if I understood well, $ \cos B \equal{} 0$ and  $ \cos B\equal{}\pm 1$ are not included in the problem's domain (see the statement).

[quote="prester"][quote="Mathias_DK"][quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote]
If $ B \equal{} 0$, we have: $ f(0) \minus{} f(0) \equal{} x \minus{} x^2 \cdot 0 \iff x \equal{} 0$, which cannot hold for all $ x$, so there is nu such function... Maybe you missed something?[/quote]

You should say $ \cos B \equal{} 0$ but if I understood well, $ \cos B \equal{} 0$ and  $ \cos B \equal{} \pm 1$ are not included in the problem's domain (see the statement).[/quote]
$ B\equal{}0 \Rightarrow \cos B \equal{} 0$ ;) I didn't really understand what he have written, but the result is the same when $ \cos B \notin \{\minus{}1,0,1\}$..
Then we have:
$ x \equal{} \frac{1}{\cos B}$ gives
$ f(\cos B) \equal{} f(1)$
Plugging in $ x\equal{}1$ gives $ f(\cos B) \minus{} f( (\cos B)^2) \equal{} 1\minus{}\cos B$. But $ f(\cos B) \equal{} f(1) \equal{} f( (\cos B)^2 )$, so $ \cos B \equal{} 1$, which cannot hold for all $ B$. So there is no solution.
I still think there may be some mistake in the problem statement..

[quote="Mathias_DK"]
$ B \equal{} 0 \Rightarrow \cos B \equal{} 0$....
[/quote]
Sorry but I believed that $ B \equal{} 0 \Rightarrow \cos B \equal{} 1$...

[quote="Mathias_DK"]
$ ...f(1) \equal{} f( (\cos B)^2 )...$
[/quote]
Sorry but I don't see that.

Instead I derived the property $ f(x) \minus{} f(\minus{}x)\equal{}4x^2$ putting $ \cos B \equal{} \pm \frac12$ in the original equation and summating the results...and this result is the best I have done until now...

[quote="Mathias_DK"][quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote]
If $ B \equal{} 0$, we have: $ f(0) \minus{} f(0) \equal{} x \minus{} x^2 \cdot 0 \iff x \equal{} 0$, which cannot hold for all $ x$, so there is nu such function... Maybe you missed something?[/quote] [b][u]$ B \in \mathbb{R} \minus{} \{ B: B\equal{}k \pi ,,or,, B\equal{}k \pi \plus{}\frac{ \pi}{2} \ , k \in \mathbb{Z}\}$[/u][/b]

[quote="prester"]
[quote="Mathias_DK"]
$ ...f(1) \equal{} f( (\cos B)^2 )...$
[/quote]
Sorry but I don't see that.[/quote]
$ \cos B$ can be any number in the interval $ (\minus{}1;1) \backslash \{0\}$, so $ f(x) \equal{} f(1) \forall x \in (\minus{}1;1) \backslash \{0\}$ and therefore $ f ( \cos^2 B ) \equal{} f(1)$

[quote="Mathias_DK"][quote="KapioPulsar"][quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote] Sorry i forgot to tell that f is continious  :blush:  :oops:[/quote]
I have already proved that no such $ f$ exists, even without the additional constraint, haven't I?[/quote] f exist i solve it but litle complicated

[quote="Mathias_DK"][quote="prester"]
[quote="Mathias_DK"]
$ ...f(1) \equal{} f( (\cos B)^2 )...$
[/quote]
Sorry but I don't see that.[/quote]
$ \cos B$ can be any number in the interval $ ( \minus{} 1;1) \backslash \{0\}$, so $ f(x) \equal{} f(1) \forall x \in ( \minus{} 1;1) \backslash \{0\}$ and therefore $ f ( \cos^2 B ) \equal{} f(1)$[/quote]

Yes I see it now. Thanks...You're right, no solutions

[quote="prester"][/quote][quote="Mathias_DK"][quote="prester"]
[quote="Mathias_DK"]
$ ...f(1) = f( (\cos B)^2 )...$
[/quote]
Sorry but I don't see that.[/quote]
$ \cos B$ can be any number in the interval $ ( - 1;1) \backslash \{0\}$, so $ f(x) = f(1) \forall x \in ( - 1;1) \backslash \{0\}$ and therefore $ f ( \cos^2 B ) = f(1)$[/quote][quote="prester"]

Yes I see it now. Thanks...You're right, no solutions[/quote] Solutions exist !! I solve it !try think it again!

[quote="KapioPulsar"][quote="Mathias_DK"][quote="KapioPulsar"] Solutions exist !! I solve it !try think it again![/quote]
The way Prester and I interpreted the problem makes it obvious that there are no solution. Can you try to write it down again? Is $ B$ perhaps a given constant, such that the equation only holds for one value of $ B$?[/quote] i say $ B \in \mathbb{R} \minus{} \{ B: B \equal{} k \pi ,,or,, B \equal{} k \pi \plus{} \frac {\pi}{2} \ , k \in \mathbb{Z}\}$ this problem is from math olumpiad in greece 1997-1998 and its as exactly as i write it[/quote]

@mathias_dk: I think that we must apologize to KapioPulsar, because $ f (\cos B) \ne f( \cos^2B)$...I did not understand the statement well since $ B$ and so $ \cos B$ is a parameter and not a variable... :oops:  :?:

[quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote]

I dont like at all this $ \cos(B)$ thing and I suggest the following transformation (equivalent, IMHO) :

Find all functions from $ \mathbb R\to\mathbb R$ such that $ f(0)\equal{}0$ and $ f(xy)\minus{}f(xy^2)\equal{}x\minus{}x^2y$ $ \forall x\in\mathbb R$, $ \forall y\in(\minus{}1,0)\cup(0,1)$

Let $ P(x,y)$ be the assertion $ f(xy)\minus{}f(xy^2)\equal{}x\minus{}x^2y$

Let $ x\in(0,\frac 14)$ :

$ x\in(0,\frac 14)$ $ \implies$ $ 2x\in(0,1)$ and so $ P(\frac 1{4x},2x)$ $ \implies$ $ f(x)\equal{}f(\frac 12)\minus{}\frac 1{8x}$

$ x\in(0,\frac 14)$ $ \implies$ $ x\in(0,1)$ and so $ P(\frac 1x,x)$ $ \implies$ $ f(x)\equal{}f(1)$

And so $ f(\frac 12)\minus{}\frac 1{8x}\equal{}f(1)$ $ \forall x\in(0,\frac 14)$ : contradiction.

So no solution.
Sorry  :roll: .

[quote="Mathias_DK"][quote="KapioPulsar"]find all fruction R->R ,$ f(0) \equal{} 0$, $ f(xcosB) \minus{} f(x(cosB)^2) \equal{} x \minus{} x^2cosB$,  x->R and for every B belongs R[u][i][b]-{B:B=kπ or B=kπ+π/2, k belongs to Z}[/b][/i][/u][/quote][/quote]
If $ B \equal{} 0$, we have: $ f(0) \minus{} f(0) \equal{} x \minus{} x^2 \cdot 0 \iff x \equal{} 0$, which cannot hold for all $ x$, so there is nu such function... Maybe you missed something?[/q
b is never 0 because b belongs to R-{b=kπ or b=kπ+π/2, κεΖ} so for k =0 we have b=0 that's not acceptable...we cannot pace b=0

[quote="kyros"]
b is never 0 because b belongs to R-{b=kπ or b=kπ+π/2, κεΖ} so for k =0 we have b=0 that's not acceptable...we cannot pace b=0[/quote]
Don't be such a fool! Read the rest of the posts ;)


The length of a diagonal of a square is $ \sqrt{2} \plus{} \sqrt{3}$ units. What is the area of the square? Express your answer in simplest form as $ \frac{a}{b} \plus{}\sqrt{c}$, where $ \frac{a}{b}$ is a common fraction and $ c$ has no
perfect square factors other than 1 .

well, we know in a square of side length s, the diagonal is $ s\sqrt {2}$

in this case, $ s\sqrt {2} \equal{} \sqrt {2} \plus{} \sqrt {3}$; $ s \equal{} \frac {\sqrt {2} \plus{} \sqrt {3}}{\sqrt {2}} \equal{} 1 \plus{} \frac {\sqrt {3}}{\sqrt {2}} \equal{}1 \plus{} \frac {\sqrt {6}}{2}$

the area is $ (1\plus{}\frac{\sqrt{6}}{2})^2\equal{}1\plus{}\frac{6}{4}\plus{}2\cdot\frac{\sqrt{6}}{2}\equal{}\frac{5}{2}\plus{}\sqrt{6}$

You can also note that the area of a paralellogram is $ d_1d_2/2$, where $ d_1$ and $ d_2$ are the diagonals.  So instead you'd do $ (\sqrt2\plus{}\sqrt3)^2/2\equal{}(5\plus{}2\sqrt6)/2\equal{}5/2\plus{}\sqrt6$.

[quote="dysfunctionalequations"][quote="iin77"]You can also note that the area of a paralellogram is $ d_1d_2/2$, where $ d_1$ and $ d_2$ are the diagonals.  So instead you'd do $ (\sqrt2 \plus{} \sqrt3)^2/2 \equal{} (5 \plus{} 2\sqrt6)/2 \equal{} 5/2 \plus{} \sqrt6$.[/quote]

Not just any parallelogram... only rhombi.[/quote]
A square is a rhombus is it doesn't matter.  Phew!  :)


What is the sum of the three smallest positive integers that have an odd number of positive factors?

Perfect squares all have an odd number of factors, and they are also the only numbers with an odd number of factors.

Think about it. Factors come in pairs, so they are always an even number of them, unless one of them is used twice, ie. 3*3.

The sum of the three smallest is $ 1 \plus{} 4 \plus{} 9 \equal{} 14$.

Every positive integer except for the perfect squares has an even number of positive factors. So the first three smallest positive integers with an odd number of positive factors, or in other words, the first three perfect square integers, are 1, 4, and 9. Their sum is $ 1\plus{}4\plus{}9\equal{}\boxed{14}$.

EDIT: Beaten.


Two circles are externally tangent at point $ A$. Let $ BC$ a common tangent. A line through $ A$ and parallel to $ BC$ meets the circles at $ D,E$. If $ M$ is the midpoint of $ DE$ , prove that $ MB \perp MC$.

Consider two circles $ (O_1)$ and $ (O_2)$ are externally tangent at $ A$ and $ BC$ a common tangent, where $ B$ lies on the circle $ (O_1)$ and $ C$ lies on the circle $ (O_2)$. A line through $ A$ and parallel to $ BC$ intersects $ (O_1)$ at $ D$ and $ (O_2)$ at $ E$. We have triangles $ ABD$, $ ACE$ are two isosceles triangles. So if we call $ G$, $ H$ are the intersection of $ BO_1$, $ CO_2$ with the segment $ DE$, respectively, then we have $ G$ is the middle point of $ AD$ and $ H$ is the middle point of $ AE$. It follows that $ DE\equal{}2\cdot GH$. On the other hand, we can easily to see that the quadrilateral $ BGHC$ is a rectangle, hence $ BC\equal{}GH$, i.e. $ BC\equal{}\frac{1}{2}DE$. So we conclude that $ BCMD$ and $ BCEM$ are two parallelogram.
Hence $ \angle BCM\plus{}\angle CBM\equal{}\angle BDM\plus{}\angle CEM\equal{}\angle ABC\plus{}\angle ACB\equal{}90^\circ$. Our solution is completed.

[quote="April"]Consider two circles $ (O_1)$ and $ (O_2)$ are externally tangent at $ A$ and $ BC$ a common tangent, where $ B$ lies on the circle $ (O_1)$ and $ C$ lies on the circle $ (O_2)$. A line through $ A$ and parallel to $ BC$ intersects $ (O_1)$ at $ D$ and $ (O_2)$ at $ E$. We have triangles $ ABD$, $ ACE$ are two isosceles triangles. So if we call $ G$, $ H$ are the intersection of $ BO_1$, $ CO_2$ with the segment $ DE$, respectively, then we have $ G$ is the middle point of $ AD$ and $ H$ is the middle point of $ AE$. It follows that $ DE \equal{} 2\cdot GH$. On the other hand, we can easily to see that the quadrilateral $ BGHC$ is a rectangle, hence $ BC \equal{} GH$, i.e. $ BC \equal{} \frac {1}{2}DE$. So we conclude that $ BCMD$ and $ BCEM$ are two parallelogram.
Hence $ \angle BCM \plus{} \angle CBM \equal{} \angle BDM \plus{} \angle CEM \equal{} \angle ABC \plus{} \angle ACB \equal{} 90^\circ$. Our solution is completed.[/quote]

 Dear April,
 after your nice solution I can state  a variation of the original problem:

  [color=red][b]If $ N$ is the midpoint of $ BC$ , prove that $ NA\equal{} NM$ .[/b][/color]

 A little different solution can be done , if we note that lines $ DB,EC,MN$ concur and $ DB,EC$ are perpendicular. Of course I  can not skip the fact that $ DE\equal{} 2BC$.

   Babis


linear algebra
matrix
vector
inequalities
linear algebra unsolved

Assume $ A \in M_{s \times n}(\mathbb{K})$,  $ B \in M_{n \times m}(\mathbb{K})$

show that: If $ rank(AB)\equal{}rank(B)$

then $ rank(ABC)\equal{}rank(BC)$

I can only do it when $ A,B,C$ are square matrix of same size, so I only give a hint
[hide="Hint"]
We will prove by contradiction by the assumption I have stated above : Note that if the consequent is false, then $ rank(ABC)<rank(BC)$, then $ null(ABC)>null(BC)$, it means that there exist a vector over $ K$, name $ x$ such that $ BCx$ is not equal to $ 0$ but $ ABCx\equal{}0$. Now write $ y\equal{}Cx$, then $ By$ is not equal to $ 0$ but $ ABy\equal{}0$, (and we know that $ Ker(B)$ is a subset of $ Ker(AB)$), so that $ null(B)\leq null(AB)$ but with the description above we have $ null(B)<null(AB)$ which says that $ rank(B)>rank(AB)$, the proof is finished.
[/hide]

[quote="Qruni"]I can only do it when $ A,B,C$ are square matrix of same size, so I only give a hint
[/quote]

If $ rank(AB)\equal{}rank(B)$ $ \Rightarrow$ $ \begin{pmatrix}
AB &0 \\
 B&BC
\end{pmatrix} \sim \begin{pmatrix}
AB &0 \\
 0&BC
\end{pmatrix}$

becasue: $ \begin{pmatrix}
AB &0 \\
 B&BC
\end{pmatrix} \rightarrow \begin{pmatrix}
0 &\minus{}ABC \\
 B&BC
\end{pmatrix}\rightarrow \begin{pmatrix}
0 &ABC \\
 B&0
\end{pmatrix}$

So $ rank \begin{pmatrix}
0 &ABC \\
 B&0
\end{pmatrix} \equal{} rank \begin{pmatrix}
AB &0 \\
 0&BC
\end{pmatrix}$

then we get the conclusion.

now I only need to proof : $ rank(AB)\equal{}rank(B) \Rightarrow \begin{pmatrix}
AB &0 \\
 B&BC
\end{pmatrix} \sim \begin{pmatrix}
AB &0 \\
 0&BC
\end{pmatrix}$

we have $ Q \in GL_{m\times m}(\mathbb{K})$ such that: $ BQ\equal{}\begin{pmatrix}
B_1 &0 \\
 B_2&0
\end{pmatrix}$
and $ ABQ\equal{}\begin{pmatrix}
A_1 &A_2 \\
 A_3&A_4
\end{pmatrix}\begin{pmatrix}
B_1 &0 \\
 B_2&0
\end{pmatrix}\equal{}\begin{pmatrix}
A_1B_1 \plus{}A_2B_2&0 \\
 A_3B_1\plus{}A_4B_2&0
\end{pmatrix}$

then $ \begin{pmatrix}
AB &0 \\
 B&BC
\end{pmatrix}\begin{pmatrix}
Q &0 \\
 0&I
\end{pmatrix}\equal{}\begin{pmatrix}
\begin{pmatrix}
A_1B_1\plus{}A_2B_2 &0 \\
 A_3B_1\plus{}A_4B_2&0
\end{pmatrix} &0 \\
 \begin{pmatrix}
B_1 &0 \\
 B_2&0
\end{pmatrix}&BC
\end{pmatrix}$

then exist $ P_1 \in GL_{s\times s}(\mathbb{K}), P_2 \in GL_{n\times n}(\mathbb{K})$ such that $ \begin{pmatrix}P_1&0\\0&P_2\end{pmatrix}\begin{pmatrix}
\begin{pmatrix}
A_1B_1\plus{}A_2B_2 &0 \\
 A_3B_1\plus{}A_4B_2&0
\end{pmatrix} &0 \\
 \begin{pmatrix}
B_1 &0 \\
 B_2&0
\end{pmatrix}&BC
\end{pmatrix}\equal{}\begin{pmatrix}
\begin{pmatrix}
I_r &0 \\
 0&0
\end{pmatrix} &0 \\
 \begin{pmatrix}I_r &0 \\
 0&0
\end{pmatrix}&P_2BC
\end{pmatrix}$ (Assume $ rank(AB)\equal{}rank(B)\equal{}r$)

Obviously

 $ rank\begin{pmatrix}AB &0\\ B&BC\end{pmatrix}\equal{}rank\begin{pmatrix}
\begin{pmatrix}
I_r &0 \\
 0&0
\end{pmatrix} &0 \\
 \begin{pmatrix}I_r &0 \\
 0&0
\end{pmatrix}&P_2BC
\end{pmatrix}\equal{}r\plus{}rank(P_2BC)\equal{}rank(AB)\plus{}rank(BC)\equal{}rank\begin{pmatrix}AB&0\\0&BC\end{pmatrix}$

so $ \begin{pmatrix}AB &0\\ B&BC\end{pmatrix} \sim \begin{pmatrix}AB&0\\0&BC\end{pmatrix}$

now work out the problem :roll:

By Frobenius inequality, $ rank(AB) \plus{} rank(BC) \leq rank(ABC) \plus{} rank(B)$. Since $ rank(AB) \equal{} rank(B)$, we have $ rank(BC) \leq rank(ABC)$. But clearly we also have $ rank(ABC) \leq rank(BC)$. Therefore, $ rank(ABC) \equal{} rank(BC)$. $ \Diamond$


ratio
number theory proposed
number theory

A calculator can perform two different operations:

a) transform a given number $x$ into $2x - 1$,
b) multiply a given number by 2.

Is it possible to transform 2005 into the fifth power of an integer using this calculator?

Let $x_1=2005$ and let $x_1, x_2, ...$ be a sequence with $x_{i+1}=2x_i$ or $x_{i+1}=2x_i-1$ for each $i$. Let $y_i=x_i-1$. Then the sequence $y_i$ satisfies $y_1=2004, y_{i+1}=2y_i$ or $y_{i+1}=2y_i+1$ for each $i$. We want $y_n=k^5-1$ for some $n, k$.

Now looking at $y_i$ in base 2, we see that in the sequence we can get any number whose base 2 representation begins with $11111010100$ (that is, $2004$) and only these numbers. And we want one of these numbers to be of the form $k^5-1$. Obviously we can find one: since $\frac{(k+1)^5-1}{k^5-1}$ gets arbitrarily close to $1$ for large enough $k$, and since $\frac{11111010101d_1d_2...d_m}{11111010011e_1e_2...e_m} \geq \frac{11111010101}{1111101100}=r$ which is fixed, eventually the ratio $\frac{(k+1)^5-1}{k^5-1}$ will be smaller than $r$ and then a number of the form $k^5-1$ will fall in the required interval.


Let be $ X$ a circunference and $ A$, $ B$ and $ C$ three points on $ X$. Let $ Y$ be another circunference tangent to $ AB$, $ AC$ and $ X$ with center inside $ X$. $ L$ and $ N$ are the intersecting points of $ AB$ and $ AC$ with the circunference $ Y$ respectively. Prove that the midpoint of $ LN$ is the incenter of $ ABC$

it has been posted before
Y is called mixtilinear of triangleABC

Dear Mathlinkers,
did somebody know whom introduce this kind of mixtilinear circle?
Thank in advance
Sincerely
Jean-Louis

Dear Jean-Louis,

These circles were introduced by Leon Bankoff in [i]L. Bankoff, A Mixtilinear Adventure, Crux Math. with Math. Mayhem 9 (1983), 2-7 [/i].

Best regards,
Sincerely,
Cosmin


ratio trigonometry Euler geometry calculus derivative function

heres d 1st question...and the second,....both from RMO 2003(and i dunno answers to both) :(
1. Let ABC be a triangle in which AB = AC and angle CAB = 90 degree. Suppose M and N are points on the hypotenuse BC such that BM^2 + CN^2 = MN^2. Prove that 6 MAN = 45 degree.
2.Suppose P is an interior point of a triangle ABC such that the ratios
[d(A,BC)/d(P,BC)] ,[ d(B,CA)/d(P,CA) ],[d(C,AB)/d(P,AB)]
are all equal. Find the common value of these ratios. [ Here d(X, Y Z) denotes the perpendicular distance from a point X to the line Y Z.]

I probably got the first one
here's the hint
[hide]
1.Draw perpendicular on BC from A,let the foot of the perpendicular be X,it bisects BC.
2.let angMAX =x,angNAX = y
3.prove BM x CN = 2 x XN x MX
4.Use compound angles(trig),prove tan(x + y)= 1.
[/hide]

Some hints for the first question-
[hide]1. Manipulate the given condition to your advantage.
2. Use standard trigonometric formulae. eg-sine rule, tan 45=1,etc.
3. Use the fact that ABC is a right angled isosceles triangle and that its angles measure 45, 45 and 90.[/hide]

Hope this helps!

[quote]1.Draw perpendicular on BC from A,let the foot of the perpendicular be X,it bisects BC.
2.let angMAX =x,angNAX = y
3.prove BM x CN = 2 x XN x MX
[/quote]

@ scary math i didnt understand what you have done in the third step or rather how you have proved it...plz expand..
@d1994 do you think i didnt use sine , cosine rule..i did..i evn tried to construct a rt triangle at MN but i cant get the damn question!!!!...n scary math I think you should try to use BM^2(or BM square) + NC^2 = MN^2...RMO ppl would not have ggiven it if it wasnt to be used...

Ofc I used dat,here's the expansion
[hide]
BM + XM = XN + CN
BM - CN = XN - CN
BM^2 + CN^2 -2BMCN = XN^2 + MX^2 -2XNMX
MN^2 = 2BMCN + XN^2 + MX^2 -2XNMX
MN^2 = 2BMCN + (XN + MX)^2  - 4XNMX
MN^2 = 2BMCN + MN^2 -4XNMX
BMCN = 2XNMX (Proved)
Now take
tanA = XN/(XN+ NC)
tanB = MX/(MX + BM)
This should be sufficient :) [/hide]

Let the feet of the perpendiculars from Point $ A,B,C$ to $ BC,CA,AB$ respectively be $ D,E,F$.
Let the feet of the perpendiculars from Point $ P$ to $ BC,CA,AB$ respectively be $ X,Y,Z$.
Let $ AP$ meet $ BC$ at $ A'$ , $ BP$ meet $ CA$ at $ B'$ and $ CP$ meet $ AB$ at $ C'$.

WE have,
$ \frac {AD}{PX} \equal{} \frac {AP}{PA'} \equal{} \frac {BE}{PY} \equal{} \frac {AQ}{QB'} \equal{} \frac {CF}{PZ} \equal{} \frac {CR}{RC'}$
Therefore, By gergonne Euler theorem(statement and proof in attachment), we have
$ \frac {AP}{PA'} \plus{} \frac {AQ}{QB'} \plus{} \frac {CR}{RC'} \equal{} 2$
Since the are all equal all of them are equal to $ 2/3$ thus proving that the given point $ P \equal{} G$(i.e  Centroid of the triangle........
WE ARE DONE
$ Q.E.D.$

[quote="sahilsharma94"][quote]BM + XM = XN + CN
[i]BM - CN = XN - CN [/i]

[/quote]
quite a brillant proof...but how can you use BM-CN = XN-CN unless u prove that BM = XN or CN= XM[/quote]

It is actually a typo
instead of $ BM\minus{}CN \equal{} XN\minus{}CN$ it is $ BM\minus{}CN\equal{}XN\minus{}MX$

This problem deos have pretty many solutions I see.
 :lol: Sorry for the typing error :P

@ sahil- Upon introspection, constructing a perpendicular to the hypotenuse is a much better option than using standard trigonometric formulae. From experience, I have learnt that it is better NOT to use trigonometry for standard geometry questions, except as a last resort.
Scary math's proof is pretty much on the same lines to what I did. Sometimes it helps to backtrack a bit from what you need to prove to arrive at something that you already know, in this case AM^2+CN^2=MN^2.

@d1994
Do they deduct marks in RMO for using trig in Geometry?Coz I never heard anything like dat before :(

No, I dont think they deduct marks....But trigonometric proofs often get messy. Pure geometry proofs on the other hand look far more elegant.

I am pretty messy :P Therefore,thanx for the advice.
Anyway,I am stuck with divisibility and FTA :(
1.Prove
x^4 + y^4 + z^4 - w^4 = 1995
Find all integer solutions of x,y,z,w
2.If each person in a group of n people is a friend of atleast 1/2 the people in the group,then prove that it is possible to seat them in a circle so that everyone sits next to a friend of his/hers.
3.Plz tell what this ques mean
x^2 + y^2 = z^n has a solution in N,for all n є N

[quote="Scary math"]Anyway,I am stuck with divisibility and FTA :(
1.Prove
x^4 + y^4 + z^4 - w^4 = 1995
Find all integer solutions of x,y,z,w[/quote]

[hide="HINT"]
[b]take modulo 16[/b]
[/hide]

[quote]
1.Prove
x^4 + y^4 + z^4 - w^4 = 1995
Find all integer solutions of x,y,z,w
[/quote]
If i'm not mistaken, i Think fermat's last theorem can be useful(still working out how to apply it though)..guys u know what...although its called a theorem...its really not  theorem...it has not actually been proved apart from for some special cases.....

Can any1 plz find the third derivative of tanx(wid explaination)..i wanna check my answer

Here is general version of the  2006 RMO problem.(Easier)
$ X$ is a set of positive integers greater than or equal to $ 8$ and let $ f: X \rightarrow X$ be a function such that $ f(x\plus{}y)\equal{}f(xy)$ for all $ x\geq 4,y\geq 4$.Prove that $ f$ is a constant function.

[quote]
@d1994
Do they deduct marks in RMO for using trig in Geometry?Coz I never heard anything ...[/quote]
[quote]

Re: queryThursday, August 13, 2009 10:59 PM
From: This sender is DomainKeys verified"Amitabha Tripathi" <at1089@gmail.com>Add sender to Contacts

You may use Calculus to solve questions in RMO/INMO. However, these questions will not need the use of Calculus, and often it may not be possible to devise a method to use it. Use of Calculus (or any other branch of Mathematical knowledge) wil certainy not result in deduction of marks.

[/quote]

so my regional coordinator says use calculus...so it implies u deff can use triggo in geometry

@ plz solve dis question..its dere in challenge n thrills  ........solved one but i dont get a word of wat dey say..cn some1 plz post a detailed solution....

On a rainy day , 5 gentlemen A,B,C,D,E attend a party. Dey leave deir umbrellas in a room..When dey return...the umbrellas get mixed up and the umbrella are returned to them in such a way that no1 gets his own umbrella. in How many ways can this happen?

the thing that is being referred here is derangement of 4 things,and is quite famous.
the formula for derangement of $ x$ things is  $ x! \left(1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \cdot \ \cdot\right)$
Having said this i don't think it'll come in RMO as actual proof is little bit complex(means long,it uses PIE)
Combinatorics in RMO has been easy for past 2-3 yrs or so.

abt using trig or anything,it is not much welcomed but they don't deduce marks if ur soln is completely right.but be ready for strict marking even for a little flaw  :)( though that happens with any method)

[quote]
the thing that is being referred here is derangement of 4 things,and is quite famous.
the formula for derangement of  things is
Having said this i don't think it'll come in RMO as actual proof is little bit complex(means long,it uses PIE)
Combinatorics in RMO has been easy for past 2-3 yrs or so.
[/quote]

@ skand evn i know its derangement formula...actually i also did it by derangement but dey did it by inclusion exclusion principle but i was not able to understand...so plz can you or any1 else ACTUALLY EXPLAIN using an example that HOW does IEP work n how we know which permutations of which properties are to be  taken...how dey are to be summed n which ones are to be subtracted n added etc etc lyk dey say |A`B`C`|=N - (summation of ) A + (summation of) AB - ABC ..now how do i know which properties are to be taken in AB summation etc etc(PS- refer to challenge n thrills of pre-college PG no.-558-559

4. Suppose N is an n-digit positive integer such that
(a) all the n-digits are distinct; and
(b) the sum of any three consecutive digits is divisible by 5.
Prove that n is at most 6. Further, show that starting with any digit one can find a six-digit
number with these properties.

PS..any1 who does challenges and thrills of pre-college maths...can you plz tell me the worked out solutions of chap 9 end chap questions :-Q-8 to Q-20

hello,welcome here :)

What in these pages being referred to is something simple:
if you have to count ways in which properties say A and B do not hold true,first we count no of ways such that A doesn't hold true,then we count no of ways in which B doesn't hold true and then we subtract them(A and B) from total no of ways to find no of ways in which A and B do not hold true,but in this process we have subtracted no of elements in which both A and B don't hold true two times(1st while sub A,then by sub B),so we add that once to ensure that we have done it all right.

It helps(in my opinion) to know about set theory.Similar formula holds for no of elements in a set.

Or another way is to draw a Venn Diagram(it helped me to comprehend)

[quote="sahilsharma94"].
PS..any1 who does challenges and thrills of pre-college maths...can you plz tell me the worked out solutions of chap 9 end chap questions :-Q-8 to Q-20[/quote]
Wait for 1 more week,hopefully I 'll done with that chapter by then. :)

Q 4
Let the number be
a1 a2 a3 a4 a5 a6 a7
a1+a2+a3=0mod5
a2+a3+a4=0mod5
a3+a4+a5=0mod5
a4+a5+a6=0mod5
a5+a6+a7=0mod5

if a1=x mod 5
   a4=xmod5
   a7=xmod5
contradiction. (we can find pairs like (1,6),(2,7),(3,8)etc but not triplets).

The mod part in CTPC is kinda confusing.Can any1 suggest me a good book with proper explanations? :(

[quote="sahilsharma94"][quote]
Q 4
if a1=x mod 5
a4=xmod5
a7=xmod5
contradiction. (we can find pairs like (1,6),(2,7),(3,8)etc but not triplets).
[/quote]
can you plz expand on the above mentioned steps!!!!like if a1,a4,a7=xmod5,what is the contradiction???

@skand plz appy derangement formula/IEP to the following principle..its precisly this wot I cudnt understand..:- Find the no. of ways of dealing a 5-card hand from a regular 52-card deck such dat d hand contains one card in each suit...i did it by P n C..but cudnt see how to use IEP :blush:[/quote]

From the first and second eqn, take a2, a3 to the other side.
from the fourth and fifth take a5,a6 to the other side.
so now,
a1=a4=a7=x mod 7 (a1,a4,a7<10)
this is a contradiction. For three numbers congruent same modulo 5 ,one is greater than 10(for eg. 1,6,11 ; 2,7,12 etc). digits of a  number cannot be greater than ten.
@ scary math-What is CTPC?  :oops:
Niven zuckerman or Engel is good for number theory.

@grafitti
Thanx
Challenge and thrill of pre-college mathematics.

[quote="sahilsharma94"]
If i'm not mistaken, i Think fermat's last theorem can be useful(still working out how to apply it though)..guys u know what...although its called a theorem...its really not  theorem...it has not actually been proved apart from for some special cases.....
[/quote]
You're deeply mistaken
it has already been proved
[url=http://en.wikipedia.org/wiki/Fermat%27s_last_theorem]see this[/url]

[quote]The ultimately successful strategy for proving Fermat's Last Theorem was first described by Gerhard Frey in 1984.[101] Frey noted that if Fermat's equation had a solution (a, b, c) for exponent p>2, the corresponding elliptic curve[note 1]

y2 = x (x − ap)(x + bp)
would have such unusual properties that the curve would likely violate the Taniyama–Shimura conjecture.[102] This conjecture, first posed in the mid-1950s and gradually refined through the 1960s, states that every elliptic curve is modular, meaning that it can be associated with a unique modular form.

Following this strategy, the proof of Fermat's Last Theorem required two steps. First, it was necessary to show that Frey's intuition was correct, that the above elliptic curve is always non-modular. Frey did not succeed in proving this rigorously; the missing piece was identified by Jean-Pierre Serre. This missing piece, the so-called "epsilon conjecture", was proven by Ken Ribet in 1986. Second, it was necessary to prove a special case of the Taniyama-Shimura conjecture, raising it from a mere conjecture to a theorem. This special case (for semistable elliptic curves) was proven by Andrew Wiles in 1995. Summarizing, the epsilon conjecture showed that any solution to Fermat's equation could be used to generate a non-modular semistable elliptic curve, whereas Wiles' proof showed that all such elliptic curves must be modular. This contradiction implies that there can be no solutions to Fermat's equation, thus proving Fermat's Last Theorem.

[quote]
earlier i had read only this[/quote]Many proofs for specific exponents use Fermat's technique of infinite descent, which Fermat used to prove the case n = 4, but many do not. However, the details and auxiliary arguments are often ad hoc and tied to the individual exponent under consideration.[88] Since they became ever more complicated as p increased, it seemed unlikely that the general case of Fermat's Last Theorem could be proven by building upon the proofs for individual exponents.[quote][/quote][/quote]

Fermats last theorem might be famous but in number theory especially in maths olympiads, fermats little theorem and eulers totient are more useful.
Question :
prove that four squares cannot be in AP.

Jean-pierre serre in on facebook,so is Swarnendu Datta. :)

[quote]
Jean-pierre serre in on facebook,so is Swarnendu Datta. [/quote]

and who are dey??!???

[quote="sahilsharma94"]Following this strategy, the proof of Fermat's Last Theorem required two steps. First, it was necessary to show that Frey's intuition was correct, that the above elliptic curve is always non-modular. Frey did not succeed in proving this rigorously; the missing piece was identified by [color=red]Jean-Pierre Serre[/color]. [/quote]
Eh u urself spoke of him without evn knowing who he is............... :rotfl:
BTW Swarnendu Datta represented India at IMO 4 times,brought 3 silvers 1 bronze :D


geometry
3D geometry
percent
percent increase

If each edge of a cube is increased by $ 50\%$, what is the percent increase in the surface area of the cube?

Let the original side length be 2.

So the surface area is 24.

Now increase 2 by 50%, which is the same as multiplying by 1.5, to get 3.

So the new surface area is 54.

We want to find the percent increase in the surface area. 54-24 = 30. So it increased by 30 square units which is 30/24 = 1.25 or 125%.

So, let's say that the original length of each side of a cube is 1

The formula for the surface area of a cube is x^2*6, where x is the length of one side, so therefore the surface area of the cube is 6

So, now we have to increase each side of a cube by 50%, which is the same as 150% of the original side

Now, to find 150% of 1, we can multiply 1 by 1.5 to get 1.5

Now, that we know the length of one side of the cube we can plug this in to the surface area formula

So, the new surface area of the cube is 13.5

We want to know the percent INCREASE in the surface area of the cube, so we have to subtract 6 from 13.5 which is 7.5

Now, we have 7.5 as the numerator and 6 as the denominator, so 7.5/6= 1.25

The question asks us to put our answer in percent, so 1.25 is equal to our answer of 125%

[hide="Alternative Solution"]The surface area of a cube with side length $s$ is $6s^2$.  Thus, the surface area of a cube with side length $\frac32s$ is simply $6\left(\frac32s\right)^2 = \frac{27}2s^2$.  Thus, the percent increase is equal to $\frac{\frac{27}2s^2 - 6s^2}{6s^2} \times 100\% = \frac{\frac{15}2}{6} \times 100\% = \boxed{125\%}$.[/hide]


There is a rectangle which is divided up into 8 squares across, 5 squares down.
If we start at the left bottom corner and end up at the right top corner, how many ways is there of travelling? (i.e. 13squares distances)

Is there a formula?

Well you need to travel thirteen squares.  You can imagine it like a sequence of ups and rights (I assume we can only go up and right).

We can think about how to get 5 ups.  After choosing 5 out of thirteen spaces to be up, the rest have to be right.  So our answer would be $ \binom{13}{5}$

We could also do it the other way.  If we thought about how to choose 8 rights we would get $ \binom{13}{8}$

Note that $ \binom{13}{5} \equal{} \binom{13}{8}$

By the way the answer is 1287.

is there a way to do it for:
i)3D? e.g. 3x5x8
ii)additional paths created inbetween?

I think ii) is a case bash, but i) seems to be doable (maybe both have formulae?!)

Well for i I think it's about the same thing if you're going on the surface of the box.  Actually I think it depends on what the question is asking for exactly.  What do you have in mind?

Well for ii I'm going to assume you mean that I choose some point in the middle of the rectangle and I'm trying to travel from that point to one of the corners (or a different point).  This is the exact same thing because we just create another smaller rectangle by doing this.

For i), the question has a similar solution if you are working on an $ a$ x $ b$ x $ c$ 3D grid. In this case, the answer for the number of paths (take the 3x5x8 example).  There are a total of $ \binom{16}{3}$ ways to pick when to move over in the 3 direction.  And after that there are $ \binom{13}{5}$ ways to pick which of the remaining moves are in the 5 direction. So the answer of the total number of paths is $ \binom{16}{3} \binom{13}{5}$.

This generalizes to an a x b x c grid using the trinomial formula $ \binom{a\plus{}b\plus{}c}{a,b,c}$