algebra
function
domain
polynomial
calculus
calculus computations

Have no idea how to solve this kind of equation $ 2^x\equal{}x^3$.  :?:

The equation has two roots. That's obvious from the graph of these functions. The question of finding them is tough: it can be found approximately using Taylor's expansion of LHS, or another approximation-iteration process, but I do not believe that exact value of it can be found. Mathematica 6's answer is not closed:

x == -(3 ProductLog[-Log[2]/3])/Log[2] ||
 x == -(3 ProductLog[-1, -Log[2]/3])/Log[2]

Or, numerically

x == 1.37347 || x == 9.93954

That's why I believe that the equation doesn't have a closed form of solution, and that the problem lies in the domain of numerical methods.

There are probably an infinite number of complex roots since the equation has the infinite polynomial representation

1+xln2+(xln2)^2/2!+(xln2)^3/3!+... - x^3 = 0

Yes, you're correct. There are three infinite families of solutions. But for the sake of education  :lol: I closed the domain on reals in my analysis...

Thanks everybody. It has two solutions, so the answer says. Actually I thought a little bit and understood that I have to use the graphs of the functions.

I got another problem now...

Can't solve this one $ (0,2)^{x\plus{}1}\equal{}\sqrt{35\plus{}5x}.$

I know that the answer is -2, but don't know how to prove it mathematicaly.

[quote="@petko"]Try tu use [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert's $ W$ function[/url] in your solution.[/quote]
I don't know this function but I think you are trying to complicate a very simple problem :wink:

Try to consider the function $ f(x)\equal{}(0,2)^{x \plus{} 1} \minus{}\sqrt {35 \plus{} 5x}$.
It is continuous and strictly decreasing in it's domain $ [\minus{}7,\infty)$.
We have $ f(\minus{}7)>0$, $ f(0)<0$.
Hence (Darboux property) there is exactly one $ t$ s.t. $ f(t)\equal{}0$.

And now you may try to guess this number. It's not hard to guess $ t\equal{}\minus{}2$.
The proof above shows that $ \minus{}2$ is the only one such number, so it is the only solution to Elena's equation

[b]knoppix[/b] well... I suggested using W function as rather general solution  :whistling: . You are right - some equations are simple as this one, but... some aren't ;)

Thanks guys! I don't think I have to use this W equation, because this test is just for school. It shouldn't be that complicated.
Anyway thanks a lot!


how do you put an image on ur signarture


function
vector
calculus
derivative
Functional Analysis
real analysis
real analysis unsolved

We can prove that $ u(x) \equal{} |x|^{ \minus{} s}\in W^{1,p}(B^n(0,1))$ for $ (s \plus{} 1)p < n$.

Now, take $ (a_k)_{k \equal{} 1,2,...}$ sequence of all points of $ B^n(0,1)$ which has rational coordinates.

Take $ s\in(0,n \minus{} 1)$, $ (s \plus{} 1)p < n$ and construct a function

                $ \displaystyle u(x) \equal{} \sum_{k \equal{} 1}^\infty{\frac {1}{2^k}|x \minus{} a_k|^{ \minus{} s}}$

And my question:

Why this function is well defined (I assume almoust everywhere) and is in $ W^{1,p}(B^n(0,1))$.

I found remark that these $ |x \minus{} a_k|^{ \minus{} s}$ functions are vectors in Banach space (not sure if important :) ).

It is well-defined both almost everywhere and as an element of $ W^{1,p}$ because the series of the norms converges and that is more than enough to ensure that the sequence of partial sums has a limit in a Banach space. The whole point of this example was just to show that there exist everywhere discontinuous functions that have derivatives in Sobolev's sense.

It is because if $ w(x)\equal{}|x\minus{}a_k|^{\minus{}s}$  then we can for example write:

$ \parallel{}w\parallel{}_{W^{1,p}(B(0,1))}\leq\parallel{}w\parallel{}_{W^{1,p}(B(a_k,2))}\equal{}\parallel{}u\parallel{}_{W^{1,p}(B(0,2))}\equal{}M$

and we have one constant for all k, right ?


What is the least integer whose square is 48 more than its double?

$ x^2\equal{}48\plus{}2x\implies x^2\minus{}2x\minus{}48\equal{}0\implies (x\minus{}8)(x\plus{}6)\equal{}0$, the least value that satisfies this is $ x\equal{}\boxed{\minus{}6}$.


probability
logarithms
graph theory
expected value
probability and stats

$ R_k(m,n)$ is the least number such that for each coloring of $ k$-subsets of $ \{1,2,\dots,R_k(m,n)\}$ with blue and red colors, there is a subset with $ m$ elements such that all of its k-subsets are red or there is a subset with $ n$ elements such that all of its $ k$-subsets are blue.
a) If we give a direction randomly to all edges of a graph $ K_n$ then what is the probability that the resultant graph does not have directed triangles?
b) Prove that there exists a $ c$ such that $ R_3(4,n)\geq2^{cn}$.

Anyone answers to this problem? :) Is a) and b) related?

The answer of a) is $ \frac{n!}{2^{\binom{n}{2}}} $. I don't see anything how to solve b). :(
I'm trying to show that there exist natural number $ a $ such that $ R_{3}(4,an)\geq 3^{n} $ for all $ n $. This problem seems to be difficult. :(

Well, I don't know much about words in mathematics so wait till I understand what your saying. :) I'm sorry.
And actually there are many ways to get no four vertices such that all triples among them are red but the other condition is too hard. :(
p.s. Maybe what your saying can work too. :)

[quote="JBL"]Here's an idea of a possible connection (but I have only thought about this for 5 minutes).  Given a tournament $T$ (i.e., an orientation of a complete graph), color its triples of vertices with the colors red = "forms an oriented cycle" and blue = "forms an oriented transitive triple."  It is easy to see that this coloring has no four vertices such that all triples among them are red.  Can we use this?[/quote]

I think "no". We have to make a good tournament "T" that suffices  the condition in the problem but not containig blue triples seem too hard. :(

Working on that idea - what does a set of $n$ in which every subset of three form an oriented transitive triple look like? It's a completely ordered set, in which each member beat those of higher rank and lost to those of lower rank.

Now, we need to build a tournament of $k$ players in which the largest completely ordered subset is of size comparable to $\log k$. That sounds... doable. Taking the hint of the other question, let's go probabilistic. For convenience, let's call a completely ordered subset a "ladder".

Choose all $\binom{k}{2}$ match results by coinflip. The probability that a given size $n$ subset are completely ordered is $2^{-\binom{n}{2}}\cdot n!$. Over all such subsets, that gives us an expected number of $\binom{k}{n}\cdot 2^{-\binom{n}{2}}\cdot n!$ size-$n$ ladders. If $\left(2^{\frac{n-1}{2}}\right)^n > k(k-1)\cdots (k-n+1)$, this is less than $1$ and there is some size-$k$ tournament with no size-$n$ ladders. Taking $k=2^{\frac{n-1}{2}}$ here works.

By this and previous results, $R_3(4,n)\ge 2^{\frac{n-1}{2}}$.

This is not sharp. In all but the smallest of cases, you won't get sharp bounds for Ramsey numbers without enormous amounts of work. Clever construction could certainly do much better.

[Edit - fixed a missing parenthesis.]

Are there known results similar to this problem? :)


limit
trigonometry
real analysis
real analysis unsolved

Given that $f(x)+f'(x) \rightarrow 0$ as $x \rightarrow 0,$ prove that both $f(x) \rightarrow 0$ and $f'(x) \rightarrow 0$ as $x \rightarrow 0$.

Your problem seems to be flawed.. just consider $f(x)=\exp(-x)$..

[quote="zhaobin"]I think the problem may $f(x)+f'(x) \to 0$ when $x \to \infty$[/quote]
Then we can apply L'Hospital rule:

$\lim_{x\to+\infty}f(x) = \lim_{x\to+\infty}\frac{e^{x}f(x)}{e^{x}}= \lim_{x\to+\infty}\frac{(e^{x}f(x))'}{(e^{x})'}= \lim_{x\to+\infty}\frac{(e^{x}f(x))'}{(e^{x})'}= \lim_{x\to\infty}f(x)+f'(x) = 0$

A similar argument works for $x\to-\infty$.

It follows immediately that $f'(x) = (f(x)+f'(x))-f(x)\to 0$.

Note that L'Hospital rule will work if we have only $g(x)\to \infty$ in $\lim \frac{f(x)}{g(x)}$. We don't need $f(x)\to\infty$.

[quote="liyi"]Note that L'Hospital rule will work if we have only $g(x)\to \infty$ in $\lim \frac{f(x)}{g(x)}$. We don't need $f(x)\to\infty$.[/quote]
It's a nice thought, but it doesn't work yet.  In order to apply l'Hospital, you need the existence of some limits.  If the limits simply didn't exist, things will go wrong.  For example, let $f(x) = \sin x$ and $g(x) = x$.  Since we don't know that $f(x)$ has a limit, we can't apply l'Hospital.  But of course, in this case, $f(x)$ having a limit implies the result of the problem in a much more straight-forward manner.

[quote="JBL"]It's a nice thought, but it doesn't work yet.  In order to apply l'Hospital, you need the existence of some limits.  If the limits simply didn't exist, things will go wrong.  For example, let $f(x) = \sin x$ and $g(x) = x$.  Since we don't know that $f(x)$ has a limit, we can't apply l'Hospital.  But of course, in this case, $f(x)$ having a limit implies the result of the problem in a much more straight-forward manner.[/quote]
Ok, what we need is the existence of limit of $\lim f'(x)/g'(x)$. If this limit exist, we can assert that
$\lim f(x)/g(x) = \lim f'(x)/g'(x)$.

In your example, L'Hospital fails because $\lim_{x\to\infty}f'(x)/g'(x) = \lim_{x\to\infty}\cos x$ does not exist.

However, in the original problem, luckily it does.


[hide="click hint"]
$\sum_{i=1}^{n}\frac{1}{a_{i}a_{i+1}}$ $=$ $(\frac{1}{d})\sum_{i=1}^{n}(\frac{1}{a_{i}}-\frac{1}{a_{i+1}})$ where $d$ is the common difference.
[hide][/hide][/hide]

Er...im sry...i cant understand that...can u pls explain using simple formulas of arithmetic progression?

Thank you

The identity as it is given is wrong.  Rather, it should be

$\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}+...+\frac{1}{a_{n}a_{n+1}}= \frac{n}{a_{1}a_{n+1}}$

Note the change in the subscript in the LHS.

[hide="Solution"] Let $a_{i}= a+di$ where $d$ is the common difference.  pardesi is giving the identity

$\frac{1}{a_{i}a_{i+1}}= \frac{1}{(a+di)(a+di+d)}= \frac{1}{d}\left( \frac{1}{a+di}-\frac{1}{a+di+d}\right) = \frac{1}{d}\left( \frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right)$

This is known as the [b]method of partial fractions[/b], and it is one common way of summing series of this type.  To give a simpler example,

$\frac{1}{12}= \frac{1}{3}-\frac{1}{4}$

These types of identities are useful because we can [b]telescope[/b] them, by which I mean that terms in the middle cancel, like so:

$\sum_{i=1}^{n}\frac{1}{a_{i}a_{i+1}}= \frac{1}{d}\sum_{i=1}^{n}\left( \frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right) = \frac{1}{d}\left( \frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}{a_{3}}+\frac{1}{a_{3}}-\frac{1}{a_{4}}+-...+\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right)$

Notice that every term except $a_{1}, a_{n+1}$ is both added and subtracted, so we can ignore them.  In other words,

$\sum_{i=1}^{n}\frac{1}{a_{i}a_{i+1}}= \frac{1}{d}\left( \frac{1}{a_{1}}-\frac{1}{a_{n+1}}\right) = \frac{1}{d}\left( \frac{1}{a+d}-\frac{1}{a+d+dn}\right) = \frac{n}{(a+d)(a+d+dn)}= \frac{n}{a_{1}a_{n}}$

As desired. [/hide]


I know there is information about this on the classroom page but I don't really get it. If audit includes transcript, classes, and two challlenge problems like whole classes, why the whole classes are there and way expensive than audits?

I'm probably going to enroll on geometry class but before I do, I want to know this.

Thanks.

From our [url=http://www.artofproblemsolving.com/Classes/AoPS_C_FAQ.php]classes FAQ page[/url]:

[b]What is the difference between audited and full enrollment?[/b]

The only differences are the grading of the Challenge sets and the Certificiates of Completion. Students with a full enrollment may send their solutions to the instructors and will receive thorough comments on their work. They are also eligible to receive a Certificate of Completion for their work. Students with an Audited enrollment are not eligible for either feedback on Challenge sets or a Certificate of Completion.

[quote="RC-7th"]Doesn't audited mean something that was examined?
What exactly is a certificate of completion?

Thanks for answering them.[/quote]

Audited means you get to listen in the class but not complete the assignments and stuff. In this case, the only difference is having your midterm and final graded. Keep in mind that everyone receives full solutions for the problems.

A certificate of completion is just what it implies - a certificae of comlpetion. I'm pretty sure it doesn't serve other purposes, but somebody correct me if I'm wrong. Perhaps some colleges/high schools may count it for credit?

[quote="Rep123max"]I am pretty sure if you enroll in whole Independent Study, you can get AoPS to get your school to give you credit (as long as the school isn't really picky).[/quote]
We can [b]not[/b] guarantee this.

Credit at your school for AoPS courses is done only on a case-by-case basis, and involves you negotiating with your school.  We can certainly tell your school that you completed one of our courses (in fact, that's partially the purpose of the Certificate of Completion -- also, it looks really nice on your wall!), but whether or not your school will give you any kind of "credit" for that is up to them.  We can in no way "get" your school to give you credit -- you have to negotiate that individually with your school.

If you want to try to get school credit for an AoPS course, please contact us and we will be happy to talk to someone at your school on your behalf.  But we can't promise anything -- the final decision is up to your school.


geometry
rectangle
calculus
calculus computations

1-
A n ew branch bank is to have a floor area of 3500ft(2). it is to be rectangle with three solid brick walls and decorative glass front.the glass cost 1.8 times as much as the brick wall per linear foot.what dimension of the bulidning will minimize the cost of matirials for the walls and the front ??

plz help

The rectangle has two dimensions, $l$ and $w$.  It has $2w + l$ feet of brick and $l$ feet of glass.  We are given the condition $lw = 3500$ and we wish to minimize $C = 2w + l + 1.8 (l) = 2w + 2.8 l$.

There are a number of ways to do this, and they don't all involve calculus.  Here's a very simple one.  Using AM-GM:

$\frac{ 2w + 2.8 l}{2} \ge \sqrt{ (2w)(2.8 l)} = \sqrt{5.6 \cdot 3500}$
$C \ge 2 \sqrt{ 5.6 \cdot 3500}$

With equality when  $2w = 2.8 l$.


I'm missing something here?

You can plug almost any numbers into the variables--for example, a=42, b=17, s=8888888, t=123454321, and then get a unique d that satisfies all the conditions. In this case, 2472056753.

[quote="Astronum"]Find the integers $a, b, d, s, t$ such that the following hold:

[b](1)[/b] $a>0, b>0$

[b](2)[/b] $d = sa+tb$ and

[b](3)[/b] $d \neq \gcd(a,b)$

:juggle:[/quote]

[hide="Just 1 of the solutions out there."]$a=1$
$b=2$
$s=3$
$t=4$
$d=11$[/hide]
mathnerd314's right. This problem has $\infty$ solutions.

Ok... I was afraid of that. I am studying number theory from a book that as no solutions for their exercises, so I wished your conformation. Thank you! :lol:

:juggle:

Here's another one:

$d = a+b$

:D

Really, this problem is trivial.  Two much more important exercises:

1)  $d = gcd(a, b) \implies d | (ax+by) \forall x, y \in \mathbb{Z}$

2)  $d = gcd(a, b) \implies \exists x,y \in \mathbb{Z}: d = ax+by$


linear algebra
matrix
trigonometry
algebra
system of equations
algebra unsolved

Solve the following system of equations in $\mathbb{R}$ where all square roots are non-negative:

$
\begin{matrix}

a  - \sqrt{1-b^2} + \sqrt{1-c^2} = d \\
b -  \sqrt{1-c^2} + \sqrt{1-d^2} = a \\
c -  \sqrt{1-d^2} + \sqrt{1-a^2} = b \\
d -  \sqrt{1-a^2} + \sqrt{1-b^2} = c \\

\end{matrix}
$

Apply the substitution $a=\sin\alpha$, ..., where $\alpha,...\in[-\pi/2,\pi/2]$.
Summing up (1) and (2) we obtain $\sin\beta-\cos\beta=\sin\delta-\cos\delta$. It follows that $\beta=\delta$ OR $\beta+\delta=-\pi/2$. The same for $\alpha$ and $\gamma$. Now it is routine to analyze 3 variants.

I found that there are two sets of solutions of it...... can someone verify those?
(x,x,x,x) and (x,sqrt(1-(x^2)),x,sqrt(1-(x^2)) as ordered pairs .. where x is non-negative


ceiling function
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer. Let $ A \equal{} \{ \minus{} n, \minus{} n \plus{} 1, \minus{} n \plus{} 2,...,n\}$, and $ S$ be a subset of size $ k \equal{} \lceil{\frac {\sqrt {8n \plus{} 1} \minus{} 1}{2}}\rceil \plus{} n \plus{} 1$. Prove that every element of $ A$ is a sum of distinct elements of $ S$.

Here is my solution:
Redenote the number $k=C+n+1$ where $C$ is the root of $C^2-C=2n$
We see that $k$ also the least number that satisfy this property, because we can construct $k'=C+n$ not every elment can representable by oversized
It is easy to observe that there are more than $C$ negative number because it is atmost $n+1$ nonnegative numbers $(1)$
We use the induction principle, it is obviously true for $n=1,2$.
Assume that it is also true for $n>2$, we choose two number $i,j$ such that $i+j<-3$ (feasible because $(1)$ )
When $n$ increase, $k$ is also increase at least $2$, the worst situation is when one is $0$, WLOG other is positive, denote that is $L$
Because $0<L<=n+1$ thus $n+1-L$ lies in $[0,n]$ so by inductive hypothesis, it can represent by a sum of distinct elements of $S$
Now consider $S'=S/{i,j}$, by induction for $n-2$ we have it can fulfill $[-n+2,n-2]$
So from $i+j<-3$, $-(n+1)-(i+j)>-n+2$ so it lies inside the scope of $S'$ so it also reprenstable
Therefor we choose $i,j$ and $S''$, we have $i+j+S''=-(n+1)$ so $S$ can fulfill $[-n,n]$ as we desired


nobody intricates a possible solution???

bravo,no problem,it's also a trend,who can solve it?


Let $ X$ and $ Y$ be independent random variables with the same geometric distribution $ {q^k p}$. Find $ P(X \geq 2 Y)$.

Notation: q = 1-p and the geometric distribution is defined as the probability that the 1st success in a sequence of Bernoulli trials occurs at the trial number 1+k (where k=0,1,2,...)

I don't see how the answer can be independent of p and q.

[quote="fedja"]Well, just compute it in terms of $ p$ and $ q$ then.[/quote]

That's not too helpful.

Think about the extreme cases. As p approaches 1, what happens to $ P(X \ge 2Y)$? It goes to 1, as both random variables acquire discrete distributions at k=0. Now, what happens when p is smaller? Is this probability zero? Of course not!

A quick calculation would suggest the answer is \[ \frac{1}{1\plus{}q\plus{}q^2},\] which can be gotten by summing over values of $ Y$. See if you can work that out.

Actually, it goes to zero at $ p \equal{} 1$. $ X$ and $ Y$ both become $ 1$, not $ 0$. You're starting that sum in the wrong place, so there's actually a $ p$ in the numerator.

?

The way it's been defined in the OP, X and Y take on values (0,1,2,...), no?

OK, they do. It's a nonstandard definition.

[quote="Nukular"]

Think about the extreme cases. As p approaches 1, what happens to $ P(X \ge 2Y)$? It goes to 1, as both random variables acquire discrete distributions at k=0.
[/quote]

Sorry-what do you mean "acquire discrete distributions at k=0"? Both random variables were already discrete (I thought) since there range is countable.

[quote="Nukular"]
A quick calculation would suggest the answer is
\[ \frac {1}{1 \plus{} q \plus{} q^2},
\]
which can be gotten by summing over values of $ Y$. See if you can work that out.[/quote]

Yes-that's also the answer in the back of my book.


function
real analysis
probability and stats

Let $ X$ be non degenerated random variable. Prove that random variables $ a_{n}X \plus{} b_{n}$ converge in distribution to random variable $ aX \plus{} b$ if and only if $ a_{n} \rightarrow a$ and $ b_{n} \rightarrow b$

[hide=" my idea for <="]
We want to prove that
\[ \mathbb{E}f(a_{n}X \plus{} b_{n}) \rightarrow \mathbb{E}f(aX \plus{} b)\]
for every bounded and continous function, when we know that $ a_{n} \rightarrow a$ and $ b_{n} \rightarrow b$.
Let $ g_{n}(x) \equal{} a_{n}x \plus{} b_{n}$ and $ g(x) \equal{} ax \plus{} b$, then $ g_{n} \Rightarrow g_{n}$ and we can rewrite our thesis as follow
\[ \mathbb{E}f(g_{n}(X)) \rightarrow \mathbb{E}f(g(X))\]
for every bounded and continous function, when we know that $ a_{n} \rightarrow a$ and $ b_{n} \rightarrow b$.

$ f\circ g_{n} \Rightarrow f\circ g$ so by Lebesgue dominated convergence theorem we are done.

Is it good ? Anyone has idea for $ \Rightarrow$?

[/hide]

Recall that convergence in distribution is equivalent to (pointwise) convergence of respective characteristic function. That should make your problem really easy  :lol: .

The "only if" part is not true. Take, e.g., $ X$ symmetric, $ a_n \equal{} ( \minus{} 1)^n$, $ b_n \equal{} 0$.


linear algebra
matrix
linear algebra unsolved

What can we say about the set of matrices $nxn$ with real entries and having the property that the set of matrices that are similar to them and commute with them is finite?

Only a comment :
The similiarity class of a matrix $A\in M_n(\mathbb F)$ is finite ($\mathbb F$ is an infinte field.) iff there exist $c\in\mathbb F$ which $A=cI_n$


Find the integer solution of the system
$ \begin{cases}

  (x\plus{}1)(y\plus{}1)(z\plus{}1)^2\equal{}4(z\plus{}19)^2  \minus{}99 \\

  (y\plus{}1)(z\plus{}1)(x\plus{}1)^2\equal{}4(x\plus{}19)^2 \minus{}99 \\

  (z\plus{}1)(x\plus{}1)(y\plus{}1)^2\equal{}4(y\plus{}19)^2 \minus{}99

\end{cases}$ :?:
__________________________________
$ \textbf{Azerbaijan Land of Fire }$ :ninja:

[hide="Solution"]
The only solution is $ x \equal{} y \equal{} z \equal{} 6$.

First, subtract the first two equations:
\[ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)^2 \minus{} (x \plus{} 1)^2(y \plus{} 1)(z \plus{} 1) \equal{} 4(z \plus{} 19)^2 \minus{} 99 \minus{} 4(x \plus{} 19)^2 \plus{} 99\]

\[ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)(z \minus{} x) \equal{} 4(z \minus{} x)(x \plus{} z \plus{} 38)\]

\[ (z\minus{}x)[(x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \minus{} 4(x \plus{} z \plus{} 38)] \equal{} 0\]

Using the 0 product property, we see that either $ x \equal{} z$, or $ (x\plus{}1)(y\plus{}1)(z\plus{}1) \equal{} 4(x\plus{}z\plus{}38)$.

Doing this for all other combinations of these equations, we see that this system can only hold if $ x \equal{} y \equal{} z$. So, looking at an equation, we see that:
\[ (x \plus{} 1)^4 \equal{} 4(x \plus{} 19)^2 \minus{} 99\]

\[ 99 \equal{} (x^2 \plus{} 4x \plus{} 39)( \minus{} x^2 \plus{} 37)\]
The last step brought the 99 over to the other side and used difference of perfect squares to factor. We know that each of these factors are factors of 99. So we can make a table using the $ ( \minus{} x^2 \plus{} 37)$ factor. The other factor can only be positive, so this factor must be positive. This is only going to be a factor of 99 when $ x \equal{} \pm2$ or $ x \equal{} \pm6$. Trying $ x \equal{} \pm 2$ we see that the possible values are $ 99 \equal{} (33)(43 \pm 8)$, which is not true if you take the positive or negative sign. So we try $ x \equal{} \pm 6$. We see that the equation becomes $ 99 \equal{} (1)(75 \pm 24)$. This is only true if we take the positive sign, so we see that the only solution is $ x \equal{} y \equal{} z \equal{} 6$.
[/hide]

EDIT: Fixed the dividing by 0 issue. Thanks, pluristiq.

you canceled $ x\minus{}z$, which you cant do if $ x\equal{}z$...


[quote="SpaceWeather.com"]LUNAR ECLIPSE:  On Wednesday night, February 20th, the full Moon will turn a delightful shade of red and possibly turquoise, too. It's a total lunar eclipse—the last one until Dec. 2010.  Sky watchers in Europe, the Americas, parts of the Middle East and Africa are favored for good views of the two-hour event.[/quote]
 :)

There is [url=http://www.spaceweather.com]timezone information[/url] and [url=http://shadowandsubstance.com/]animated timezone information[/url].

For $ \text{EST (UT \minus{} 5 or GMT \minus{} 5)}$ it starts at $ 19: 37$, peaks at $ 22: 26$, and ends at $ 1: 16$.
For $ \text{UT ( or GMT)}$ start at $ 0: 37$, peaks at $ 3: 26$, and ends at $ 6: 16$.

By "start" I mean when the moon starts to move into the penumbra (the lighter part of the Earth's shadow), peak is when it is in the umbra (the darker region of the the Earth's shadow), and the end is when it leaves the penumbra.  The link the animated timezones shows a visual of this.
 :wink:
Also, to get more aquianted with night sky, the lunar eclipse happens in the constellation Leo, the Lion.  Above the moon is Leo's brightest star Regulus.  Below the moon a bit further away is Saturn, if you have binocular or even a telescope, then you can see its fantastic rings.

Awesome... I'll be outside looking at the moon today... :lol:

Looks like tonight is gonna be an all nighter. Again.

hunter34 Again?

Is anyone going to try to find and look at Saturn?

I want to...
Don't have a good telescope and cannot find the binoculars.

Edit: Nevermind! Found 2 binoculars and... a mini telescope lol. I am going to look for Saturn

:lol: Cool!  It is below the Moon and a little to the left.  It looks like a golden, non-twinkling "star".

There is going to be a Total SOlar eclipse on August 1.........

I saw it! And I saw Saturn...
Took a few pics, too xD

It was [i]suppose[/i] to be cloudy last night but I decided to try and look anyway.  I took a couple pictures of both the Moon and Saturn.  :) Like I said in my blog, I might post them.  Is there like a slideshow code or something?

[quote="7h3.D3m0n.117"]No  :(  unfortunately cloud and curfew prevented the sight of the lunar eclipse for me.[/quote]

I really wanted to see it, probably the only time I would've even [i]cared[/i] about it...Didn't the news say it would be a red moon too? I thought I'd heard something like that. Curfew messes things up, and there was a blanket of snow and a bunch of clouds, like 7h3.D3m0n.117 said. Oh, well. I'll wait another two years... :(

The moon [i]does[/i] appear to be red during a total lunar eclipse.  My [url=http://www.mathlinks.ro/viewtopic.php?t=161862]post[/url] from the eclipse a few months ago explains it more.

The sky was partly cloudy in my area, but I did manage to get a really good view of the eclipse and Saturn as well. I wish I would have thought to take pictures...You should post yours, AstroPhys. :)


It was my first post  :lol:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=29109[/url]

I wonder why the  German TST problem  have been posted in mathlinks :(

many tst problems are from past imos or national olympiads. So it's nothing strange to see this happens.

I had seen this problem long ago in a chinese book.

I had seen this problem  long ago in a  RUSSIAN book (Mosсow math olympiad book) :roll:  :roll:  :arrow:  :P  :P  :arrow:  :ninja:  :ninja:  :roll:  :oops:  :roll:


If $ a_1,a_2,...,a_n$ are positive integers and distinct, then:

\[ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2 \le \frac{2n\plus{}1}{3} (a_1\plus{}a_2\plus{}...\plus{}a_n)\]

[quote="moldovan"]If $ a_1,a_2,...,a_n$ are positive integers and distinct, then:
\[ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \le \frac {2n \plus{} 1}{3} (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)\]
[/quote]

try $ n \equal{} 3,a _1\equal{} 1,a_2\equal{} 2,a_3\equal{} 10$


limit
integration
real analysis
real analysis unsolved

Let $ f \in L^1[0,1]$. Show that

$ \lim _n \int_{[0,1]}f(t)sin(2\pi n t)d(t) \equal{} 0$

do you know about Riemman Lebesgue lemma ?
This is about : Fourier coefficients tends to $ 0$ as ...


symmetry
ceiling function
floor function
number theory proposed
number theory

Let $ a,b$ be positive integers such that we find an integer $ c$, satisfying,
\[ a^2c \minus{} b^2(c \plus{} 1) \equal{} 1
\]
then prove that $ \mid{a \minus{} b}\mid$ is a square.

Counterexamples: $ (a \equal{} 17,b \equal{} 12,c \equal{} 1)$, $ (a \equal{} 99,b \equal{} 70,c \equal{} 1)$, ...
More generally: $ (a \equal{} b_{k \plus{} 1} \plus{} d_k,b \equal{} b_{k \plus{} 1},c \equal{} 1)$ for $ k\geq 2$ where $ \{b_k\}$ is OEIS A001542 and $ \{d_k\}$ is OEIS A001653.  By symmetry there's also a similar infinite sequence of counterexamples with $ c \equal{} \minus{} 2$.

What [i]is[/i] true, at least for the solutions I've found, is that $ 2|a \minus{} b|^2 \minus{} 1$ is always a perfect square (from the definition of $ \{d_k\}$).

I told him already that the problem seems wrong, but was not being able to site the counterexamples! Thanks, JoeBlow!
 :)
But what is "OEIS A001542" & "OEIS A001653"?

More explicitly,
\[ d_k \equal{} \left \lceil \frac {2\plus{}\sqrt{2}}{4}\left( 3 \plus{} 2\sqrt {2}\right)^{k}\right \rceil \\
 \\
b_k \equal{} \left \lfloor \frac {\sqrt {2}}{4}\left (3 \plus{} 2\sqrt {2}\right )^k \right \rfloor
\]
OEIS is the [i]Online Encyclopedia of Integer Sequences[/i]
http://www.research.att.com/~njas/sequences/A001542
http://www.research.att.com/~njas/sequences/A001653

Hello JoeBlow! My friend has modified his problem. Here it is:
$ a^2c\minus{}b^2(c\plus{}1)\equal{}a\minus{}b$

All other conditions are as it is.
Will u check if this one holds!
 :)

$ \iff (a\minus{}b)(c(a\plus{}b)\minus{}1)\equal{}b^2$
Checking coprimality of the factors is straightforward.

[quote="Zetax"]$ \iff (a \minus{} b)(c(a \plus{} b) \minus{} 1) \equal{} b^2$
Checking coprimality of the factors is straightforward.[/quote]
But it should depend on the choice of $ a,b$, doesn't it? We cannot comment straightforward that $ (a \minus{} b)$ and $ (c(a \plus{} b) \minus{} 1)$ are co-primes,can we?? :maybe:

Moreover, if they are co-primes, then it gives rise to the fact that $ (a\minus{}b)>(a\plus{}b)$.

Any prime divisor of $ a \minus{} b$ divides $ b^2$, thus $ b$, thus $ a$, thus $ c(a \plus{} b)$, thus not $ c(a \plus{} b)\minus{}1$.


geometry
circumcircle
power of a point
radical axis
geometry unsolved

In a triangle $ABC$, let $M$ and $N$ be the midpoints of the sides $AC$ and $AB$, respectively. Let $G$ be the centroid, i. e. the point of intersection of the medians of the triangle $ABC$. Furthermore, let the lines $BM$ and $CN$ intersect the circumcircle of triangle $ABC$ at the points $P$ and $Q$, respectively. The tangents to the circumcircle at the points $P$ and $Q$ intersect the lines $AC$ and $AB$ at $X$ and $Y$. Prove that the line $OG$ is perpendicular to the line $XY$, where $O$ is the circumcenter of triangle $ABC$.

I solved this nice problem. Try to solve it, it's easy!

[color=red][Moderator edit: Clarified the wording of the problem and put all the contents into one post.][/color]

Yes, really very nice :)

Let $OF\perp XY$, where $F\in XY$.
Then it's easy to know the sets of points $\{ P,Q,M,N\},\{ Y,Q,N,O,F\}, \{X,P,M,O,F\}$ are concyclic. Since $G=QN\cap PM$ and the 3 radical axis are concurrent, $G$ lies on $FO$, i.e. $GO\perp XY$.

The problem is not hard at all.  :)

$MN//BC \Rightarrow \angle NMB = \angle MBC = \angle CQP \Rightarrow P,Q,N,M$ are cyclic $\Rightarrow GN \cdot GQ = GM \cdot GP$.

$\angle XPO = \angle XMO = 90^{\circ} \Rightarrow P,M,O,X$ are concyclic.

$\angle YQO = \angle YNO = 90^{\circ} \Rightarrow Q,N,O,X$ are concyclic.

Therefore G is on the radical axis of these two circles. Let $O_{1},O_{2}$ be the circumcenter of these two circles respectively. Since $O_{1},O_{2}$ are the midpoints of $XO,YO$ respectively, by midpoint theorem, $O_{1}O_{2} \perp XY$. Hence $GO \perp O_{1}O_{2} \Rightarrow GO \perp XY$.


Label one disc “1”, two discs “2”, three discs “3”, . . . , fifty discs ‘‘50”. Put these 1+2+3+···+ 50 = 1275 labeled discs in a box. Discs are then drawn from the box at random without replacement. What is the minimum number of discs that must me drawn in order to guarantee drawing at least ten discs with the same label?

[hide="solution"]You can draw all discs labeled 1-9 which gives 45.  We can also draw 9 of all the discs labeled 10-50 which gives 369.  Therefore 45+369+1=[b]415[/b].[/hide]


[hide]First, find the roots of the quadratic function.
$2x^2+1=3x$
$2x^2-3x+1=0$
$(2x-1)(x-1)=0$
$x=\frac{1}{2}$ or $x=1$

Therefore, the sum of the roots is $1\frac{1}{2}$ or $\frac{3}{2}$[/hide]

[hide]The sum of the roots of a quadratic equation is -b/a in standard form.
Standard form: $2x^2 - 3x +1 = 0$
that looks like -(-3)/2, or $\frac{3}{2}$ for this answer. [/hide]

[quote="MCrawford"]What is the sum of the roots of $2x^2 + 1 = 3x$?  Express your answer as a common fraction.[/quote]

[hide]$2x^2-3x+1=0$

$x^2-\frac{3}{2}x+\frac{1}{2}=0$

the sum is the negative coefficient of the x

$\frac{3}{2}$[/hide]

Well done. I completely forgot about that method.

[hide]we can see the negative 2nd term is the sum of the roots by:

(ax+b)(cx+d) = (ac)x^2 + (ad+bc)x + (bd)

The roots are -b/a and -d/c.  The sum of these roots is -b/a + -d/c = -bc-ad/ac = -(bc+ad)/ac = - coefficient of x divided by coefficient of x^2[/hide]

[quote="rcv"]The first solution is at the right level for this forum.

The other solutions are correct, but I think a little too advanced for the "Basics" Forum.  (Nice derivation, Barnacle.)[/quote]

I would've done that, but I was worried that the answer had complex roots, like this one
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35232[/url]

[quote="rcv"]The first solution is at the right level for this forum.

The other solutions are correct, but I think a little too advanced for the "Basics" Forum.  (Nice derivation, Barnacle.)[/quote]

Really? I thought this was perfectly appropriate for this. Anyway, mods are always correct.

$\displaystyle 2x^2-3x+1=0\\ \frac{-b}{a}=\frac{3}{2}$

[quote="math92"]what is a quadratic root sum????[/quote]

It's the sum of the roots of a quadratic equation.

[quote="236factorial"][quote="math92"]what are roots???[/quote]

I think they are the numbers that make an equation true.

In this case, 1 and 1/2 are roots.[/quote]

you are right 236!, they are the solution(s)
and the product of the roots is $\frac ca$


Let $ k, n$ be positive integers.  Show that

$ \frac{(kn)!}{k!^n n!}$

is an integer.

[hide]$ \frac {(kn)!}{k!^{n}n!}$ is the number of ways of dividing a set of $ kn$ elements into $ n$ subsets with $ k$ elements in each subset.[/hide]


LaTeX
Euler
number theory unsolved
number theory

For every natural numer n, Q(n) denote the sum of the digits in the decimal representation of n. Prove that there are infinitely many natural number k with Q(3^K)>Q(3^k+!)

if k is 1~30 then k=11,14,16,18,20,27

You could at least use brackets or even better: $\LaTeX$.
It's not clear if you want $3^{k+1}$ or $3^{k}+1$ for the second term.
And by the way, better spelling and at least one check of what you wrote is not suggested but required, especially if you make that many errors! ["Q(3^K)>Q(3^k+!)"]

Yes,it is $Q(3^{k})>Q(3^{k+1})$
To solve this problem,we can used $Q(n) \equiv n (mod 9)$
And $Q(3^{k+n})>Q(3^{k})+9n,k>N_{0}$
Other $Q(n)<9(lgn+1)\to 9(lg(k+n)+1)>9n+Q(3^{k})$
When $n\to+\infty$ we have contraction

[quote="ZetaX"]Whatever $!$ is ;)

But prove the generalisation:
Let $a,b \geq 2$ and $a$ not a power of $b$. Let $Q_{b}(n)$ be the sum of digits of $n$ if written in base $b$. Show that there are $\infty$ many $k$ with $Q_{b}(a^{k}) < Q_{b}(a^{k+1})$.[/quote]
ZetaX,I think in Leonhard Euler first post it is $Q(3^{k})>Q(3^{k+1})$
but not $Q(3^{k})<Q(3^{k+1})$
I think your problem is easy because
if $a$ not a power of $b$,then for every postive integer $t$ there exist $k$ such that $Q(a^{k})$ begin with $t$ in base $b$
and I think the orginall problem has been posted,but no one give a proof.


Gauss algebra polynomial calculus integration function domain

Let $ A$ be a commutative ring and let $ A[x]$ be the ring of polynomials in an indeterminate $ x$, with coeﬃcients in $ A$. Let $ f \equal{} a_0 \plus{} a_1x \plus{} \cdots \plus{} a_nx^n \in A[x]$. $ f$ is said to be primitive if $ (a_0, a_1, \cdots, a_n)\equal{}(1)$. Prove that if $ f$, $ g \in A[x]$, then $ fg$ is primitive iff $ f$ and $ g$ are primitive.
[This is Atiyah-Macdonald #1.2d]
I am particularly confused with $ \Leftarrow$ which I assume uses Gauss Lemma.

For the $ \Rightarrow$, I have:
Suppose that $ fg$ is primitive. Now suppose that $ f$ were not primitive.
Then $ (a_0, a_1, \ldots, a_n) \equal{} (1)$ so there is a common factor to all of the terms. But then $ fg$ would have a common factor as well. Thus $ f$ is primitive, and so must $ g$ be.

I am sorry to say, but you cannot use Gauss' lemma here - the ring $ A$ is not supposed to be a principal ideal domain. The problem is much harder.

  darij

Is there anything wrong with this proof?

Suppose that the coefficients of $ fg$ do not generate $ A$.  Then they all lie in some prime ideal $ \mathcal{P}$.  Now consider the equation $ \overline{f}\overline{g}\equal{}\overline{fg}\equal{}\overline{0}$ in $ A/\mathcal{P}$.  Since we are in an integral domain, comparing leadings terms we must have $ \overline{f}\equal{}\overline{0}$ or $ \overline{g}\equal{}\overline{0}$, in other words one of $ f$ and $ g$ must be non-primitive.

[quote="xboxlive89128"]Then $ (a_0, a_1, \ldots, a_n) \equal{} (1)$ so there is a common factor to all of the terms.[/quote]

I'm assuming that this is a typo and you meant to say $ (a_0, a_1, \ldots, a_n) \neq (1)$.  But you're still quite mistaken; just because a set of elements does not generate the whole ring does not mean that they have a common factor.  One example is the elements $ 2$ and $ 1\plus{}\sqrt{\minus{}3}$ in $ \mathbb{Z}[\minus{}3]$.  You can still modify your proof to work, but you absolutely cannot talk about common factors.

[quote="xboxlive89128"]I am particularly confused with $ \Leftarrow$ which I assume uses Gauss Lemma.
[/quote]Use that fact that every proper ideal is contained in a maximal ideal and that $ A[X]/\mathfrak m[X]\eqsim (A/\mathfrak m)[X]$ is an integral domain if $ \mathfrak m$ is maximal.
Gauß lemma is not needed here as was already pointed out before - and the premises for it are not assumed, anyway.
[quote]For the $ \Rightarrow$, I have:
Suppose that $ fg$ is primitive. Now suppose that $ f$ were not primitive.
Then $ (a_0, a_1, \ldots, a_n) \equal{} (1)$ so there is a common factor to all of the terms. But then $ fg$ would have a common factor as well. Thus $ f$ is primitive, and so must $ g$ be.[/quote]Well, this is not true in general as was said already before. But use the hint on maximal ideals for the other implication - and it should work out without any problems. ;)


inequalities
logarithms
inequalities proposed

If $a, b, c$ are real numbers greater than $1$, prove that for any real number $r$

\[(\log_{a}bc)^{r}+(\log_{b}ca)^{r}+(\log_{c}ab)^{r}\geq 3 \cdot 2^{r}. \]

Let $ ln(a) \equal{} x$ ,$ ln(b) \equal{} y$ and $ ln(c) \equal{} z$ ,$ x,y,z > 0$ so the inequality to prove is
\[ \sum \left(\frac {x \plus{} y}{z}\right)^{r}\geq 3\cdot 2^{r}\]
Since $ r>0$
By AM-GM we have
\[ \frac {1}{3}\sum \left(\frac {x \plus{} y}{z}\right)^{r}\geq \sqrt [3]{\left (\frac {(x \plus{} y)(y \plus{} z)(x \plus{} z)}{xyz}\right )^{r}}\geq 2^{r}\]
and because $ r > 0$ the last one is equivalent to
\[ (x \plus{} y)(y \plus{} z)(x \plus{} z)\geq 8xyz\]

[quote="Svejk"]Let $r=-t$ with $t>0$ an the inequality to prove is
\[\sum \left ( \frac{x}{y+z}\right)^{t}\geq \frac{3}{2^{t}}\]
which has already been posted on this forum.
[/quote]Actually that inequality is not true. For $t=\frac{1}{2}$ for example, the infimum of the LHS is 2.

Actually you never get to this case, as $a,b,c>1$ implies that $\ln a, \ln b, \ln c > 0$.

We have $LHS \geq \frac{(log_{a}b+log_{a}c+log_{b}c+log_{b}a+log_{c}a+log_{c}b)^{r}}{3^{r-1}}$

So, we have to prove that $(log_{a}b+log_{a}c+log_{b}c+log_{b}a+log_{c}a+log_{c}b) \geq 6$.

See that $log_{a}b+log_{b}a \geq 2\sqrt{log_{a}b.log_{b}a}= 2$, and the problem is solved!

Bonjour Monsieur e.lopez,

canst thou please explaineth to me how it is you have so magnificently ascertained that first line of your most delightful proof?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=611970&p=3640278#p3640278]Turkmenistan National Math Olympiad 2012 Grade 9[/url]

Here is nice solution.
Let $(\log_{a} bc) ^r+(\log_{b} ca) ^r+(\log_{c} ab)^r=X$
From $AM-GM$ inequality we have that $\frac{X}{3} \geq \sqrt[3]{(\log_{a}bc)^r (\log_{b}ca)^r (\log_{c}ab)^r} =[(\log_{a}b+\log_{a}c) (\log_{b}c+\log_{b} a) (\log_{c}a+\log_{c}b)] ^\frac{r} {3}$ and again we can say that this is $\geq [2 \sqrt {\log_{a}b \cdot \log_{a}c} \cdot 2\sqrt {\log_{b}c \cdot \log_{b}a} \cdot 2\sqrt {\log_{c}a \cdot \log_{c}b}] ^\frac{r}{3}$,or
$2^r[\log_{a}b \cdot \log_{b}a \cdot \log_{a}c \cdot \log_{c}a \cdot \log_{b}c \cdot \log_{c}b]^\frac{r} {6} =2^r$
So we now get that $X=(\log_{a} bc) ^r+(\log_{b} ca) ^r+(\log_{c} ab) \geq 3 \cdot 2^r$
$Q.E.D$ :)


induction
real analysis
real analysis unsolved

hi,
Let $ (x_{n})_{n \geq 1}$ be a strict monoton increasing series of natural numbers s.t.
\[ x_{n \plus{} 2} \plus{} x_{n} > 2x_{n \plus{} 1}, \forall n \geq 1.
\]
Prove that the number
\[ \theta \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {1}{10^{x_{n}}}},
\]
is irrational.

note that :
$ x_{n \plus{} 1} \minus{} x_{n} \geq 1 \plus{} x_{n} \minus{} x_{n \minus{} 1} \geq 2 \plus{} x_{n \minus{} 1} \minus{} x_{n \minus{} 2} \geq ... \geq n \minus{} 1 \plus{} x_{2} \minus{} x_{1} \geq n$
since $ x_{n \plus{} 2} \plus{} x_{n} > 2x_{n \plus{} 1}$  and $ x_{n}$ is sequance strict monoton increasing  natural numbers.
thus we see :
\[ \sum_{k \equal{} N \plus{} 1}^{\infty}\frac {1}{10^{x_{k}}} \equal{} \left| \theta \minus{} \sum_{k \equal{} 1}^{N}\frac {1}{10^{x_{k}}} \right| \equal{} \left| \theta \minus{} \frac {P_{N}}{10^{x_{N}}}\right|
\]
where $ P_{N} \in \mathbb{N}$
thus:
\[ \left| \theta \minus{} \frac {P_{N}}{10^{x_{N}}}\right| \equal{} \sum_{k \equal{} N \plus{} 1}^{\infty}\frac {1}{10^{x_{k}}} \leq \frac {C}{10^{x_{N \plus{} 1}}} \equal{} o\left(\frac {1}{10^{x_{N}}} \right)
\]
since
$ 0\leq\frac {C}{10^{x_{N \plus{} 1}}} 10^{x_{N}} \equal{} \frac {C}{10^{x_{N \plus{} 1} \minus{} x_{N}}} \leq \frac {C}{10^{N}} \to 0$ as $ N \to \infty$
show yourself that if
$ \left| \alpha \minus{} \frac {p}{q} \right| \equal{} o\left( \frac {1}{q} \right)$ then $ \alpha$ is no rational.

An other one . :lol:  . You prove by induction that $ x_{n\plus{}1} \geq n \plus{} x_n$
We can write $ \theta \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {1}{10^{x_{n}}}}\equal{} \sum_{k \equal{} 0}^{\infty}{\frac {C_k}{10^k}}$ where $ C_{x_n}\equal{}1$ and $ C_k\equal{}0$ for  $ x_n < k < x_{n\plus{}1}$
Manifestly for all $ m_0$ the sequence of figures $ C_{m_0\plus{}n}$ is not periodically, so $ \theta$ is not rational .
 :cool:


calculus
integration
function
analytic geometry
geometry
vector
linear algebra

B(m,n)=gamma(m).gamma(n)/gamma(m+n)
while changing the variables x to rcosz, y=rsinz and taking their differentials we tactly assume the terms containing (dz)^2 & (dr)^2 terms vanish. we only consider terms containing dzdr in the double integral in place of dxdy can somebody please explain why (dz)^2 & (dr)^2 terms vanish or not considered for the double integral?

well i tried that but the problem still remains. (dz)^2, (dr)^2 terms were not included for the double integral but da.db is not exactly equal to the term containing dzdr only. pls check. i cant find the term theta so i used z in place of theta.
rsinz=b. rcosz=a.
da=-rsinzdz+ coszdr
db=rcoszdz+ sinzdr

da.db=rdzdr???????
pls check the link before answering.

[quote]da.db=rdzdr??????? [/quote]

[url]http://en.wikipedia.org/wiki/Polar_coordinate_system[/url]
you know  the polar coordinate, Jacobian , dont you?

The point is that what matters is area- and the vectors $ (dr,dr)$ enclose zero area, just as $ (dx,dx)$ do.

still dadb=rdrdz dont match. to clarify the question pls check the 1st link provided in the replies.

What do you think the product of two vectors should be? That's not $ (da)(db)$, it's $ da\times db$, the cross product.

whether its daxdb or dadb modulus it should be same for to eveluate the integrals we are supposing the variables in perpendicular direction. besides the whole problem in  the link is not done supposing them as vectors.
even if u do consider them to be vectors then a=rcosz, b=rsinz.
da=-rsinzdz+coszdr
db=rcoszdz+sinzdr
even if we neglect dr^2 & dz^2 terms we are left with
dadb=-rsin^2zdzdr+rcos^2zdrdz not rdrdz.

[url]http://en.wikipedia.org/wiki/Integration_by_substitution[/url]

[url]http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant[/url]

[url]http://en.wikipedia.org/wiki/Polar_coordinates#calculus[/url]

Check, check again, then recheck...


function
calculus
derivative
complex numbers
complex analysis

Let $ f(z)\equal{}u(x,y)\plus{}iv(x,y)$ be real differentialbe on an open set.  Given that the real and imaginary parts of $ f(z)$ and $ zf(z)$ are harmonic, show that $ f$ is analytic

Just take partials:  If they're harmonic then they all satisfy Laplace's equation:
$ u_{xx}\plus{}u_{yy}\equal{}0$ and $ v_{xx}\plus{}v_{yy}\equal{}0$ as well as the $ (x\plus{}iy)f$ form:
\[ \frac{\partial}{\partial x^2}\bigg(xu\minus{}yv\bigg)\plus{}\frac{\partial}{\partial y^2}\bigg(xu\minus{}yv\bigg)\equal{}0\]
\[ \frac{\partial}{\partial x^2}\bigg(xv\plus{}yu\bigg)\plus{}\frac{\partial}{\partial y^2}\bigg(xv\minus{}yu\bigg)\equal{}0\]
Do those last two and simplify, assuming the first two, yields the Cauchy-Riemann equations.  Know what, you can do it manually and maybe you need the practice.  But also you can plug them into Mathematica:

[code]val1 = D[x*u[x, y] - y*v[x, y], x, x]
val2 = D[x*u[x, y] - y*v[x, y], y, y]
FullSimplify[val1 + val2]
rval1 = D[x*v[x, y] + y*u[x, y], x, x]
rval2 = D[x*v[x, y] + y*u[x, y], y, y]
FullSimplify[rval1 + rval2]
[/code]

I'm surprised to learn that I could add "Assumptions" in the FullSimplify commands above.  Note if you enter the code above, you'll get long expressions containing the second partials which you need to cancel as per Laplace's equation.  However, if I specify the Laplace criteria as assumptions, I arrive at the Cauchy Riemann equations, in Mathematica syntax, directly.  Derivative[0, 1] means partial with respect to y and Derivative[1,0] is with respect to x.

[code]In[69]:=
val1 = D[x*u[x, y] - y*v[x, y], x, x];
val2 = D[x*u[x, y] - y*v[x, y], y, y];
FullSimplify[val1 + val2 == 0, Assumptions ->
   {D[u[x, y], x, x] + D[u[x, y], y, y] == 0 &&
     D[v[x, y], x, x] + D[v[x, y], y, y] == 0}]
rval1 = D[x*v[x, y] + y*u[x, y], x, x];
rval2 = D[x*v[x, y] + y*u[x, y], y, y];
FullSimplify[rval1 + rval2 == 0, Assumptions ->
   {D[u[x, y], x, x] + D[u[x, y], y, y] == 0 &&
     D[v[x, y], x, x] + D[v[x, y], y, y] == 0}]

Out[71]=
Derivative[0, 1][v][x, y] == Derivative[1, 0][u][x, y]

Out[74]=
Derivative[0, 1][u][x, y] + Derivative[1, 0][v][x, y] == 0 [/code]


vector
linear algebra
matrix
parameterization

Is the following statement true? If it is, why?

The three vectors
\[ a \equal{} \begin{pmatrix}
                         k\\
                         1\\
                         1
               \end{pmatrix},
         b\equal{} \begin{pmatrix}            1\\ k\\ 1   \end{pmatrix},
         c\equal{} \begin{pmatrix}      1\\ 1\\ k\\        \end{pmatrix}
\]
are linearly independent when $ k\equal{}\minus{}2.$

Linear independence in your case means that:
\[ \alpha a \plus{} \beta b \plus{} \gamma c \equal{} 0 \Rightarrow \alpha \equal{} \beta \equal{} \gamma \equal{} 0.
\]
Therefore just solve the upper system and you will obtain:

1. $ k \equal{} 1: \quad$ $ \alpha \equal{} \minus{} \gamma \minus{} \beta, \gamma, \beta$ arbitrary reals. $ \Rightarrow a, b, c$ lineary dependent

2. $ k \equal{} \minus{} 2: \quad$ $ \alpha \equal{} 2\gamma \minus{} \beta, \gamma, \beta$ arbitrary reals. $ \Rightarrow a, b, c$ lineary dependent

3. $ k\neq { \minus{} 2, 1}: \quad$ $ \alpha \equal{} \beta \equal{} \gamma \equal{} 0$. $ \Rightarrow a, b, c$ lineary independent

More general way is to put the vectors into the $ 3\times 3$ matrix and study its rank in dependence to the real parameter $ k$.

In this case you can also study when the triple product is $ 0$.

1. $ (\vec{a}\times \vec{b})\cdot \vec{c}\neq 0 \Rightarrow \vec{a},\vec{b},\vec{c}$ lineary independent.

2. $ (\vec{a}\times \vec{b})\cdot \vec{c}\equal{} 0 \Rightarrow \vec{a},\vec{b},\vec{c}$ lineary dependent.

\[ (\vec{a}\times \vec{b})\cdot \vec{c}\equal{}k^{3}\minus{}3k\plus{}2\equal{}(k\plus{}2)(k\minus{}1)^{2}\]

$ \mbox{a, b and c linear independent}\Leftrightarrow \left|\begin{array}{ccc} k & 1 & 1 \\
1 & k & 1 \\
1 & 1 & k \end{array}\right|\neq 0\Leftrightarrow k^3 \minus{} 3k \plus{} 2\neq 0 \Leftrightarrow \\
\Leftrightarrow (k \minus{} 1)^2(k \plus{} 2)\neq 0\Leftrightarrow k\notin \{1; \minus{} 2\}.$


Imagine you have a hollow sphere.  Now imagine you fill it with matter such that it takes the shape of a sphere.  Essentially, that's all it is.  Think of the earth.

Is it the same thing as homogeneous mass distribution? and, MUST spherical mass distribution has constant density throughout? How about the homogeneous version?

No, its not the same. Just generally sphcerical distribution means, you "fill" up a shpere with some mass. This need not be homogenous, but can be homogenous.
The only thing homogenous in terms of mass, means, is that the mass density does not vary with position or time. That is its constant throughout the sphere.

The type of distribution tells you nothing about whether or not it is homogeneous.  Any distribution can be homogeneous or non-homogeneous, and homogeneous and non-homogeneous distributions can take on any shape.

For example, if I say that a body has a spherical mass distribution given by $ \rho\equal{}r\alpha$ where $ \alpha$ is some constant, then this has spherical distribution but the amount is dependent on the radius.


For any real number b,$ f(b)$ denote the maximum value of the function$ |sinx\plus{}\frac{2}{3\plus{}sinx}\plus{}b|$
 over all $ x\epsilon{R}$  then the minimum value of f(b) over all $ b\epsilon{R}$ is
 a)$ \frac{1}{4}$         b)$ \frac{1}{2}$

c)$ \frac{3}{4}$          d)$ \frac{1}{8}$

 $ |.|$denotes modulus function

would  u accept  a solution  using calculus  ?

Ya,But solution other than using calculus will be more preferred... :D

then  i  gotta  think  ,  sorry running  out  of  time  now.

will be  back  with  a soln  the  next time i login.  good day :)


Determine all integer solutions of the equation:
\[ x^{2}+15^{a}=2^{b}\]
with $ a, b$ integers.

Obviously, we need $ a, b \ge 0$. Now, if $ b = 0$, then $ a = x = 0$ as well; if $ b = 1$, then $ a = 0$ and $ x = \pm 1$; if $ b \ge 2$, then $ 0 \equiv x^{2}+15^{a}\equiv x^{2}+1 \bmod 4$, which is impossible.

I think $ 15^{a}\equiv (-1)^{a}\mod{4}$ which means that all the last equation tells us is that $ a$ is odd and $ x$ is odd...

You're absolutely right! And I should do better to go to bed... :blush:

[quote="jonysatie"]Determine all integer solutions of the equation: $ x^{2}+15^{a}=2^{b}$ .[/quote]
See no solution posted .. if there  is any interest  - peek:

[hide]1. $ a<0$   Easy to prove there are no solutions.
2. $ a=0$   Only solutions are when  $ b=0 (x=0)$ and $ b=1 (x=\pm1)$  [hide](If $ b>1$ look at mod 4) ($ x^{2}$ can not be $ -1 \bmod 4$) [/hide]
3.  when $ a>0$ (and  obviously then b>1)
    A).  $ x$  has to be odd, and is not divisible by $ 3$ or $ 5$
    B).   $ b$ has to be even (say $ b=2c$   [hide](Look at mod 3, since LHS is 1 mod 3, $ 1 \equiv 2^{b}\equiv (-1)^{b}\bmod 3$ [/hide]
So write  $ 15^{a}= (2^{c}+x)(2^{c}-x)$
[hide]Now  both factors can not be divisible by 3  (Otherwise their difference (=2x) would be divisible by 3)
Similarly only one factor can be divisible by 5
Hence Either : $ 15^{a}= 2^{c}+x , 1=2^{c}-1$  or $ 5^{a}=2^{c}+x,3^{a}=2^{c}-x$
Or adding these two  $ 15^{a}+1 = 2 \cdot 2^{c}$  or $ 5^{a}+3^{a}= 2 \cdot 2^{c}$
In either case, $ a$ has to be $ 1$   (Look at mod 32 in the first case, and  mod 16 in the second case)
(and c is either 3 or 2) [/hide]
From this we get  Only  other solution  is  $ a=1$
and  two  possibilities for $ b$   $ (b=3,x=\pm 1)$   and $ (b=6,x=\pm 7)$  [/hide]

The [b]L[/b]eft [b]H[/b]and [b]S[/b]ide of the equation.

[quote="Gyan"][hide]
From this we get  Only  other solution  is  $ a \equal{} 1$
and  two  possibilities for $ b$   [u]$ (b \equal{} 3,x \equal{}\pm 1)$[/u]   and $ (b \equal{} 6,x \equal{}\pm 7)$  [/hide][/quote]

[hide="You mean"]
$ b \equal{} 4,x\equal{}\pm1$, right?
[/hide]


Geia sas paidia!
Eimai o ilias apo 8essaloniki k molis ekana eggrafh sto forum. Parakolou8w edw k polu kairo tis suzhthseis sas k prepei na paradextw oti polles apo autes me exoun voh8hsei pragmatika.
Elpizw na voh8hsw k egw s auth th prospa8eia pou ginetai.

Filika,
Ilias

Ηλία σε καλωσορίζουμε !

   Περιμένουμε μα χαρά τις απόψεις , τις ασκήσεις , τις λύσεις ή τις προτάσεις σου .

 Μπάμπης

Καλώς ήρθες Ηλία. Είναι ευχάριστο που το Forum αποκτά καινούργια μέλη. Πολλά μυαλά είναι πάντα καλύτερα από ένα.

Sas euxaristw polu gia to kalosorisma sas!Elpizw pragmatika na prosferw sto forum me ton kalutero dunato tropo!

Hlias


If 3 small bags and 4 large bags contain 84 apples, while 5 small bags and 3 large bags contain 85 apples, and each small bag as well as each large bag contains the same number of apples, how many are in 2 small bags and 5 large bags?

[hide=" click to reveal hidden content"]

$ 3s\plus{}4l\equal{}84$
$ 5s\plus{}3l\equal{}85$

$ 9s\plus{}12l\equal{}252$
$ 20s\plus{}12l\equal{}340$

$ 11s\equal{}88$
$ s\equal{}8$
$ l\equal{}15$

$ 2s\plus{}5l\equal{}91$.

[/hide]


http://www.gotoquiz.com/politics/political-spectrum-quiz.html

Here.  I got ranked "left social libertarian".

What did you get?

Apparently I'm a center-left moderate social authoritarian.

Left social libertarian FTW! :D

oh wait it wasn't a contest?   :oops:

atomicwedgie and I must be the most liberal people in AoPS!   :o

I am a left moderate social authoritarian.
Left: 7.04, Authoritarian: 1.46

center-left moderate social authoritarian.
Left: 1.35, Authoritarian: 1.39

You are a left moderate social libertarian.
Left: 5.59, Libertarian: 1.49

remember ernie's sig. "If pro is the oppisite of con, what is the opposite of progress? :P "

I got Left social Moderate :huh:
Left 3.5 libertarian 0.14

You are a centrist moderate social libertarian.
Left: 0.74, Libertarian: 1.68

o.0

Left social moderate; 4.13 left, 0.53 Libertarian

[quote="mathemagician1729"]You are a center-left social moderate.
Left: 2.77, Authoritarian: 0.17

I feel like the indecisive Louis XVI..[/quote]

That's not a good sign...

You are a centrist social moderate.
Left: 0.53, Libertarian: 0.88


[quote="Differ"]Can we post Mandelbrot questions on AoPS?[/quote]

I don't see any reason why you should not be allowed to post them. But please clearly specify the source including year etc.  :)


geometry
geometric transformation
rotation
trigonometry
linear algebra
matrix

This is a question from my school textbook, I have tried it so many times but find it impossible, that said it comes from an exercise which had some very dubious question with even more dubious answers:

f is the transformation which maps (x, y) onto (x',y')
x' = (cosA)x - (sinA)y
y' = (sinA)x + (cosA)y
for some constant A

Write down the image p' of the point p(a, b) and the image of q' of the point q(c, d) under f (i got an answer for this involving lots of sin and cos but I don't know if it's right, i can't prove the second part)

Prove that the distance p'q' = distance pq

The answer should be pretty clear; the linear transformation is a rotation and all rotations are isometries (distance preserving functions).  But here's the algebra:

under $f$,

$(a,b)\to (\cos\phi a - \sin\phi b,\sin\phi a+\cos\phi b)$
$(c,d)\to (\cos\phi c - \sin\phi d,\sin\phi c+\cos\phi d)$

$d(p,p')=\sqrt {(a-c)^2+(b-d)^2}$

$d(f(p),f(p'))=\sqrt {[\cos\phi(a-c)-\sin\phi(b-d)]^2+[\sin\phi(a-c)+\cos\phi(b-d)]^2}$

Collecting and using the identity $\cos^2\phi+\sin^2\phi=1$ a few times, you should find

$d(p,p')=d(f(p),f(p'))$

which is the desired result.

A "more algebraic" proof.

Let \[A=\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}\]
Thus the transformation can be written as
\[ \begin{pmatrix}
x' \\ y'
\end{pmatrix}
=
A
\begin{pmatrix}
x \\ y
\end{pmatrix}
 \]
Hence
\[ \begin{pmatrix}
x_1' \\ y_1'
\end{pmatrix}
=
A
\begin{pmatrix}
x_1 \\ y_1
\end{pmatrix},\quad
\begin{pmatrix}
x_2' \\ y_2'
\end{pmatrix}
=
A
\begin{pmatrix}
x_2 \\ y_2
\end{pmatrix}
 \]
It follows that
\[
\begin{pmatrix}
x_1'-x_2' \\ y_1'-y_2'
\end{pmatrix}
=
A
\begin{pmatrix}
x_1-x_2 \\ y_1-y_2
\end{pmatrix}
\]
We denote it by $\beta=A\alpha$ and we are going to prove that $\|\beta\| = \|\alpha\|$
Thus taking transpose of both sides and we get
$\beta^T = \alpha^TA^T$
So $\beta^T\beta = \alpha^TA^TA\alpha$. Notice that $A^TA=E$, we get $\beta^T\beta = \alpha^T\alpha$, i.e.,  $\|\beta\| = \|\alpha\|$.


Let (sin a-sin b)/sin (a-b) = x
     (cos a-cos b)/sin (a+b) = y

Express sin (a-b) in terms of x and y.

Pls show your workings

$ x\equal{}\sin a \minus{} \sin b \equal{} 2\cos \frac{a\plus{}b}{2} \sin \frac{a\minus{}b}{2}$

$ y\equal{}\cos a \minus{} \cos b \equal{} \minus{}2 \sin \frac{a\plus{}b}{2} \sin \frac{a\minus{}b}{2}$

so $ \displaystyle xy\equal{}\minus{}2 \frac{\sin^2 \frac{a\minus{}b}{2}}{\sin (a\minus{}b)}\equal{}\frac{\cos (a\minus{}b)\minus{}1}{\sin(a\minus{}b)}$, and the rest is basic algebra.

by the way, the algebra section suits this problem a bit better. or maybe even the pre-olympiad.


calculus
integration
function
complex analysis

Doesn't $ \int_{|z|\equal{}2} \frac{dz}{\bar{z}} \equal{} 0$ by Cauchy-Goursat?

What makes you say that?

That function is not holomorphic in |z|=2.

I'd go with the fact that it's $ \int_{|z|\equal{}2}\frac{z}{4}\,dz$, the integral of an analytic function on the disk.

jmerry so it is an integral of an analytic function and so Cauchy Goursat applies right?

But $ \text{Re}(\int f(z)) \equal{} 0$ and $ \text{Im}(\int f(z)) \neq 0$?

The original function is not analytic. There happens to be an analytic function which agrees on that curve- which is all that matters.

Yes so we can still use Cauchy Goursat right? My book asked me to use Cauchy Goursat to show that integral is $ 0$.

And is this an example in which the Cauchy Goursat does not apply to the real and imaginary parts of $ f(z)$?


group theory
abstract algebra
superior algebra
superior algebra theorems

If A and B are subgroups of C, and |A| divides |B|, then A is a subgroup of B? Is this true for all groups, or only cyclic groups?

Certainly not. In fact, every non-cyclic finite group contains at least one counterexample (two distinct subgroups with the same order).


Fie A o matrice de ordin 4 cu elemente numere[size=150] intregi[/size] astfel incat $ A^3\equal{}A\plus{}I_4$. Demonstrati ca $ det(A^2\minus{}I_4)\equal{}1$.
Oare exista o astfel de matrice? Ma gandesc la valorile ei proprii.

[b]INDICATIE:[/b]
[b]1).[/b] $ A^3\minus{}A\equal{}I_4\Leftrightarrow A.(A^2\minus{}I_4)\equal{}I_4\Rightarrow det[A.(A^2 \minus{}I_4)]\equal{}1\Leftrightarrow det A.det(A^2\minus{} I_4)\equal{}1.$
[b]2).[/b]$ \mbox{Daca } A\in M_4(\mathbb{Z}), \mbox{ atunci: } det A, \ det(A^2\minus{} I_4) \in \mathbb{Z}\Rightarrow det A\equal{}det(A^2\minus{} I_4)\equal{}\pm 1.$

Daca calculez valorile proprii ajung la contradictie cu faptul ca are elemente nr intregi. Nu cred ca exista o astfel de matrice.

$ A^3 \minus{} A \minus{} I_4 \equal{} O_4$$ \Rightarrow$ polinomul minimal $ m_A (X)|f(X),f(X) \equal{} X^3 \minus{} X \minus{} 1;$;f e continua si f(1)f(2)<0$ \exists !c \in (1,2)$a.i.f(c)=0 si cum f e crescatoare implica f(X)=0 are o singura radacina reala$ f(X) \equal{} (X \minus{} \alpha )(X^2 \plus{} pX \plus{} q), \\
p^2 \minus{} 4q < 0; \\
cu \alpha \notin Q$
De asemenea polinomul caracteristic $ p_A (X) \equal{} a_0 \plus{} a_1 X \plus{} a_2 X^2 \plus{} a_3 X^3 \plus{} a_4 X^4$ are aceeasi divizori ireductibili cu polinomul minimal.
Daca $ m_A (X) \equal{} X \minus{} \alpha \Rightarrow A \equal{} \alpha I_4 \Rightarrow \alpha \in Z$,dar radacina ecuatiei e intre 1 si 2,fals.Daca $ m_A (X) \equal{} X^2 \plus{} pX \plus{} q \Rightarrow \det (A \minus{} \alpha I)(A^2 \plus{} pA \plus{} qI) \equal{} 0$ si cum $ {\rm det}(A^2 \plus{} pA \plus{} qI) > 0 \Rightarrow \det (A \minus{} \alpha I) \equal{} 0 \Rightarrow p_A (\alpha ) \equal{} 0;$.Dar $ p_A(\alpha) \notin Z$ deoarece $ \alpha \in (1,2)$ si coeficientii polinomului caracteristic sunt intregi.
Deci $ m_A(X) \equal{} X^3 \minus{} X \minus{} 1 \Rightarrow p_A(X) \equal{} (X \minus{} \alpha)^2*(X^2 \plus{} pX \plus{} q).$
Din identificarea coeficentilor celor doua polinoame cu $ a_0,a_1,...,a_4 \in Z$ vom obtine contradictie deci nu exista astfel de matrici.


Show that no integer of the form $xyxy$ in base $10$ (where $x$ and $y$ are digits) can be the cube of an integer.
Find the smallest base $b > 1$ for which there is a perfect cube of the form $xyxy$ in base $b.$

[hide]
$xyxy=101(10x+y)=k^{3}$

$\implies 10x+y=101^{2}\cdot m^{3}$ as 101 is prime

minimum is when $m=1 \implies 10x+y=10201$ but $0 \leq x, y \leq9$

now let it be in base b

$\implies x \cdot b^{3}+y \cdot b^{2}+x \cdot b+y = xb(b^{2}+1)+y(b^{2}+1)=(xb+y)(b^{2}+1)$

from here, i just substituted in $b=2,3,4,5,6$, and arrived at no solutions for $x, y$, where $0 \leq x, y \leq (b-1)$

$b=7$

$\implies (7x+y)50=k^{3}$ and a solution can be found $x=2, y=6$

hence base 7 is smallest.[/hide]


Bow to solition the number of equation on $ [\minus{}2\pi;2\pi]$
$ \sin(1\cdot x)\cdot \sin(1\cdot 3x) \cdot \sin(1\cdot 3\cdot 5x)...\sin(1\cdot3\cdot ...\cdot 2009x)\equal{}1$

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :ninja:

[quote="Pirkuliyev Rovsen"]Bow to solution the number of equation ... [/quote]

What does this mean ?

[quote="Pirkuliyev Rovsen"]Bow to solition the number of equation on $ [ - 2\pi;2\pi]$
$ \sin(1\cdot x)\cdot \sin(1\cdot 3x) \cdot \sin(1\cdot 3\cdot 5x)...\sin(1\cdot3\cdot ...\cdot 2009x) = 1$
[/quote]

We need all these factors be $ 1$ or $ -1$ and so $ x\in\{-\frac{3\pi}2,{-\frac{\pi}2,{\frac{\pi}2,{\frac{3\pi}2\}}}}$

For $ x\in\{-\frac{3\pi}2,{\frac{\pi}2\}}$, we have $ 502$ factors equal to $ -1$ and $ 503$ equal to $ 1$, and so these values are solution.

For $ x\in\{-\frac{\pi}2,{\frac{3\pi}2\}}$, we have $ 503$ factors equal to $ -1$ and $ 502$ equal to $ 1$, and so these values are not solution.

And so only two solutions in $ [-2\pi,2\pi]$ : $ -\frac{3\pi}2$ and $ \frac{\pi}2$


See http://www.mathlinks.ro/Forum/viewtopic.php?t=44547 or http://www.mathlinks.ro/Forum/viewtopic.php?t=19870 .

  darij


I am making a pdf from TeXnicCenter and everytime I build, it shows 0 errors, 0 warnings, and 0 bad boxes.  But when I press F5 to view the output it brings up a box that says [DocOpen("%bm.pdf")][FileOpen("%bm.pdf")] Cannot execute the command.
Why can't I view the output?

It may be that the path to Adobe Reader is not correct. Go to Build, Define Ouput Profiles (or just press Alt-F7), click on LaTeX=>PDF, then on Viewer tab and check path of executable is correct and if necessary change it*. Then do the same for LaTeX=>PS=>PDF.

*It should be something like C:\Program Files\Adobe\Acrobat 7.0\Reader\AcroRd32.exe

Actually, the paths should be fine, because this error just happens like half the time.  Sometimes it works, sometimes it doesn't, but I'll check though.

Sometimes, I gett this error when I have stuff open.  I'm not sure why it happens.  Usually, a quick fix is to save my work, close out of EVERYTHING, then open up TeXnicCenter again and give it another shot.  That usually works for me.

Well, I just rebooted my laptop and it started working.  I don't know what was wrong.


Given f(x) = (1/x) and (f/g)(x) = (x + 1)/(x^2 - x), find the function g.

could someone please verify my answer..i'll post a solution if its correct..

[hide]
$g(x)=\boxed{\frac{x-1}{x+1}}$[/hide]

i'll hide my solutions regularly now..sorry for the inconvenience

[quote="SM4RT"]could someone please verify my answer..i'll post a solution if its correct..

[hide]
$g(x)=\boxed{\frac{x-1}{x+1}}$[/hide][/quote]
That's correct.

[b]@Interval[/b]: Please try to use $\LaTeX$.

Look how nice your question would have looked:

Given $f(x) = \frac{1}{x}$ and $\left(\frac{f}{g}\right)(x) = \frac{x+1}{x^{2}-x}$, find the function $g$.

Hover over the text to see what was used to produce it. In general, put dollar signs around all your math.

so my solution

[hide]
$\Rightarrow\left(\frac{f}{g}\right)x=\frac{f(x)}{g(x)}=\frac{1}{x\cdot g(x)}\implies g(x)=\frac{x(x-1)}{x(x+1)}=\boxed{\frac{x-1}{x+1}}$[/hide]

I thank all of you for your help.

In terms of using LATEX, I have to find time to review the material.  When I learn how to use it, I will.  For now, I will continue typing the old fashioned way.

This is what you should type:

[code]Given $f(x)=\frac{1}{x}$ and $\left(\frac{f}{g}\right)(x)=\frac{x+1}{x^{2}-x}$, find the function $g$.
[/code]


Do any of you guys know a website where I can find old mathcounts tests in their entirety?

I am specifically looking for Chapter and State round tests from 2001-2003.

Thank You! :)

http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm
Goodly Maths - MATHCOUNTS Archive

http://www.yashar.biz/mathcounts/

Similar, but who knows?


algebra
polynomial
complex numbers
algebra proposed

Let $n\geq 2$ and $u_{1}=1 , u_{2},...., u_{n}$ be complex numbers with absolute values of at most $1$. and let
$f(x)=(x-u_{1})(x-u_{2})...(x-u_{n})$
Prove that the $\frac{d }{dx}f(x)$ polynomial has a complex root whose real part is non-negative.

Because roots of $f'(x)$ will lies convex-hull of points $u_{1},u_{2},...,u_{n}$ therefore we are done.


calculus
integration
calculus computations

Integral from 0 to y of  root[1+(coshx)^2] dx

I was wondering if it was meant to be $ \sinh^2 x$ instead of $ \cosh^2 x$. If that is the case, then it will be a matter of using: $ \boxed{\cosh^2 x \equal{} 1 \plus{} \sinh^2 x}$.


search
number theory proposed
number theory

Find all finite sets $S$ of positive integers such that for any $i,j\in S$ we have $\frac{i+j}{(i,j)}\in S$.

i have the following solution but before reading it come up with your own one. :D and tell me if my soluiton is right or wrong.

[hide]first we prove that if $i,j\in S, (i,j)>1$. if its not true then assume that $(a_{1},a_{2})=1$for some $a_{i},a_{j}\in S$. now we set $a_{k}=a_{k-1}+a_{1}=(k-2)a_{1}+a_{2}$. since $(a_{k},a_{1})=1$, we have$a_{k}\in S \forall k$. but this contradicts that$S$is finite. so$i,j\in S\Rightarrow (i,j)>1$

now assume that$|S|\ge 3$. let$a,b,c\in S$such that$a<b<c$and for any other$k\in S, k>c$. we have$\frac{a+b}{(a,b)}\in S$. but$\frac{a+b}{(a,b)}\le \frac{a+b}{2}<b$.so$\frac{a+b}{(a,b)}=a\Rightarrow b=a((a,b)-1)\Rightarrow (a,b)=a\Rightarrow b=a(a-1)$. now$\frac{a+c}{(a,c)}\in S\Rightarrow\frac{a+c}{(a,c)}=a$or$b$. but$\frac{a+c}{(a,c)}=a\Rightarrow b=c$so$\frac{a+c}{(a,c)}=a\Rightarrow c=a(a(a-1)-1).$

but note that $\frac{b+c}{(b,c)}\in S\Rightarrow\frac{a^{2}-a+a^{3}-a^{2}-a}{a}=a^{2}-2\in S$.so $a^{2}-2=a,b$or$c$. but from these we don't get any solutions.(show this yourself :wink: )

so all solutions are $S=\{n\}$or$\{n,n(n-1)\}$ with $n\in \mathbb N$. :P
[/hide]

okey, even if its posted before then you can at least tell me if my soluitons right or wrong.

please someone tell me if my solution is correct or not.

EDIT: okey, i thought that i and j had to be distinct. :blush: now i find my ans


[quote="Namdung"]Mấy hôm nay mình không vào được. Có ai biết nguyên nhân không?[/quote]
Dien dan toan hoc tam thoi dung dia chi: http://diendantoanhoc.net/ :)

Hom nay ten mien nay cung khong vao duoc,chang biet tai sao nua?

[quote="keira_khtn"]Co le la sua chua gi do! :lol: !Bay gio thi on roi!! :D[/quote]

Van vay ,hom nay khong vao duoc :D

Dung that,hom nay lai ko vao duoc.Co ve co nhieu loi phai sua qua nhi?? :( .Chac dien dan cua minh phai sua nhieu nhieu nua moi bang duoc mathlinks :D

[quote="keira_khtn"][quote="nttu"]Cac ban cho minh hoi tai sao khi dung Internet duong truyen binh thuong thi vao duoc trang web diendantoanhoc.net , con khi dung dich vu ADSL thi chang co noi nao vao duoc .
  Minh khong doc duoc Latex o trang  diendantoanhoc.net .
 Neu co phai chinh sua hay cai dat them phan mem nao khac , cac ban mach cho minh voi . Thank you ;)[/quote]
Minh ra quan co ADSL suot ma van vao duoc ma ..Con su co ko doc duoc latex thi co le la do dd minh bi hong thoi....[/quote]

Bay gio chac lai khong vao duoc roi dung khong? :D

Dien dan minh qua thieu tinh on dinh.

Hom nay lai ko doc duoc cong thuc... :( (10/7)

:lol: Minh cung y chang nhu cac ban vao mai ma cung ko duoc. Hy vong trang web toan hoc viet nam minh se ngay cang tot va hoan thien hon

[quote="nttu"]Cac ban cho minh hoi bay gio dien dan toan hoc cua VN dung dia chi nao vay ? Co vao duoc khong ?  :roll:[/quote]

[url]diendantoanhoc.net[/url]

Em cung khong the vao duoc diendan,ngay ca khi da click vao link cua google.Co ai biet phai lam the nao khong a?

3 hom nay cung khong vao duoc,chan that,chac tai mang o VietNam

Tôi thấy có sao đâu. Ngày hôm qua (thứ năm) vẫn vào được bình thường mà. :)

[quote="vietnamesegauss89"]Minh van vao duoc ma:[url=http://http://diendantoanhoc.net/]http://diendantoanhoc.net/[/url][/quote]

Chang thay gi ca? :D

trời ạ!topic này chỉ để thông báo là "hôm nay mình vào được,ngày mai mình vào không được" thôi à.Đơn giản là vì hệ thống server của mathlink mạnh hơn diendantoanhoc nhieu, :roll:

Em vào mãi mà cũng ko được, chắc là có sửa chữa phải ko? Thế bao giờ thì lên được ?

Dung la the that ?tai thoi diem nay em cung khong vao duoc diendantoanhoc,cu thay thong bao 'this account has been suspended"

Hì, chắc Admin chưa trả tiền cho mấy thằng bán Host!

Ghé thăm nhà mới đi các bạn http://diendantoanhoc.net/forum   :wink:

[quote="N.T.TUAN"][quote="tunganh"]Diễn đàn đổi tên miền:
http://diendantoanhoc.org/forum[/quote]
Không phải thay đổi, mà theo Admin Thuận thì chỉ là tạm thời thôi :)[/quote]
Vâng, nhưng thực ra thì tên miền này cũng song hành với tên miền cũ từ lâu rồi nhưng nhiều người không để ý :)

[quote="N.T.TUAN"]Điều này anh biết, từ khi còn làm mod bên ddth :D[/quote]
Nick [b]anh[/b] bên ddth là gì ạ? :P

Trước kia anh dùng nick QUANVU, bây giờ dùng Vưu Hồi :D . Mà thôi, không nhắc lại chuyện cũ nữa, không các bác ấy lại bảo là spam, mặc dù là mình không cố ý :P

Bên http://diendantoanhoc.net/forum vẫn vô được bình thường thui mà, có sao đâu. :D

Thằng cu admin Thuận mới sửa đấy!  :lol:

Em cũng vào được bình thường  :( . Ở chỗ khác chắc là mạng đắt tiền quá nên nó kiêu  :rotfl:

Đọc kĩ rồi post, anh nói nó bị kẻ phá hoại nghịch tý, bây giờ thì ok rồi.


What is this question referring to when it says "upper bound"?

What is the smallest positive integer, that is an upper bound of zeros at $ P(x)\equal{} 3x^{3}\plus{}2x^{2}\plus{}4x\minus{}6$

I'm guessing it want syou to use th rational root theorem. The upper bound means that there is no root higher than that integer. So if 3 was an upper bound, this is not necessarily saying that 3 is a root, but just that there is no root bigger than 3.

The answer is $ \boxed{1}$. This is because $ 3x^3 \plus{} 2x^2 \plus{} 4x \minus{} 6 > 0$ for all $ x > 1$, and hence it can never achieve zero.

Thank you Tycoon. The SAT IIC book I got this out of didn't provide an accurate explanation as to what the term upper bound meant. Thanks again guys.


Is there a way to quickly find out how many complex zeroes there are in a cubic equation?  For a quadratic, we can just use the discriminant; can that be adapted for a cubic?

[quote="Aznwithabrain"]I think there is usually $3$... :ninja:[/quote]
Well, if the cubic has real coefficients, then there are either 0 or 2.

Yeah, but I think there's some inequality that gives you a definite answer.  It was something like

If $f(x) = ax^{3}+bx^{2}+cx+d$, and if $9d > ab$ then the cubic has 2 complex zeroes....or something like this...does anyone know it?

see http://en.wikipedia.org/wiki/Cubic_equation#The_nature_of_the_roots

btw, there are always 3 complex roots (counting multiple roots 2 or 3 times), but there might be 0 or 2 nonreal ones.


Find the value os $a$ so that two following equations are equavalent:
$x^{2}+y^{2}=a$ and $x^{4}-y^{4}=a^{2}$

What do you mean by the two equations being equivalent?

All I can think of is

[hide="Is this right?"]

If $a=0$ then $x=y=0$. Given that $|a|>0$ then $x^{4}-y^{4}=(x^{2}+y^{2})(x^{2}-y^{2})=a(x^{2}-y^{2})=a^{2}$

Therefore we have $x^{2}-y^{2}=x^{2}+y^{2}=a$

Which implies that $y=0$. Since $x^{2}=a$ which shows that $a$ is non-negative. So we have $a>0$ then $a$ can be any positive real number.[/hide]

Two equations are called equivalent iff the sets of solutions are the same.
[hide="No such $a$ exist"]The 1st equation has
no solution if $a<0$,
one solution  if $a=0$,
has bounded set of solutions (circle)
The 2nd equation has unbounded set of solutions always.[/hide]
Any typo?

I don't think so. If the first equation has bounded set of solutions (circle)
The 2nd equation has unbounded set of solutions always, then we have to find valus of $a$ so that two sets of solutions have the same one

[quote="hien"]I don't think so. If the first equation has bounded set of solutions (circle)
[b]The 2nd equation has unbounded set of solutions always[/b], then we have to find valus of $a$ so that two sets of solutions have the same one[/quote]
Fix any $a$. Then the set of solutions of the 1st equality is bounded. For this $a$ those of the 2nd one is unbounded. There is no set which is bounded and unbounded simultaneously.


geometry
geometric transformation
rotation
puzzles

Count the different arrangements, topologically speaking, to ignore handedness, flipping over, or rotations of the following later description.

One circle --> You have one circle.  {There's nothing to attach it to.}

Two or more circles   .......... Note:  "They" refers to "The circles."
-------------------------
Rules:

1) They can differ in size.
2) They can be apart on the outside.
3) They can be apart on the inside.
4) They can be internally tangent.
5) They can be externally tangent.
6) They can intersect each other.
7) They cannot coincide with another circle.
8) If enough circles are present, two or more
    of the above numbered conditions may apply.

Question 1:  How many different ways are there of putting two circles together
with these rules?  Just a number is sought. -- No diagrams/work is OK.

Question 2: (Unknown answer) -- low priority then // As the number of circles
increases, do you know roughly how fast (formula-wise) the number of ways of the circles can be displayed together increases?

Question 1
[hide]5 ways --> i. a circle inside of a circle (or look at it as a circle on the outside of a circle), ii.  circles internally tangent to each other, iii. circles externally tangent to each other, iv. circles not touching each other and one of them is not inside the other, v. intersecting circles (which happen to cross at two places for each pair)

[/hide]
Question 2

No, I don't have an idea at this time for how fast it's growing.
(I tried to diagram cases for 3 circles for a count to get a sense of a pattern, but more cases kept "cropping up" that I didn't expect, so I stopped trying to list them.)


I heard  "The Karush-Kuhn-Tucker Theorem "is

similar formula with Lagrange multiplier(It can be perfectly different, but just I 'heard')

Do you know about this theorem??;

I have heard this theorem but I haven't needed it anywhere because I'm not doing that kind of maths. For example [url=http://webrum.uni-mannheim.de/vwl/mokuhn/public/KarushKuhnTucker.pdf]this document[/url] explains what is the Karush-Kuhn-Tucker Theorem.


John was three times as old as his sister 2 years ago and five times as old 2 years before that. In how many years will the ratio be 2 to 1 ?

[hide="Solution"]Let $ j$ be John's current age and $ s$ be his sister's current age. We have:

$ \frac {j \minus{} 2}{s \minus{} 2} \equal{} 3 \\ \\
\frac {j \minus{} 4}{s \minus{} 4} \equal{} 5$

Solving the first, we have:

$ j \minus{} 2 \equal{} 3s \minus{} 6 \\ \\
j \equal{} 3s \minus{} 4$

And the second:

$ j \minus{} 4 \equal{} 5s \minus{} 20 \\ \\
j \equal{} 5s \minus{} 16$

Substituting:

$ 3s \minus{} 4 \equal{} 5s \minus{} 16 \\ \\
\minus{} 2s \equal{} \minus{} 12 \\ \\
s \equal{} 6$

Plugging back in:

$ j \equal{} 3s \minus{} 4 \\ \\
j \equal{} 3 \cdot 6 \minus{} 4 \\ \\
j \equal{} 14$

So John is 14 years old and his sister is 6 right now. To see when the ratio becomes double, we will let $ x$ be the number of years from now:

$ \frac {14 \plus{} x}{6 \plus{} x} \equal{} 2 \\ \\
14 \plus{} x \equal{} 12 \plus{} 2x \\ \\
x \equal{} 2$

So the ratio is 2 to 1 in $ \boxed{2}$ years.[/hide]


This series is simply $\sum_{k=1}^n\frac1{k(k+1)(k+2)(k+3)(k+4)}$
Someone can do the algebraic part :P

Partial fraction decomposition, then the series telescopes, right?

basically but its still alot of work, but doable   :P
you just have to solve a system w/ 5 variables...and then telescope.

[quote]basically but its still alot of work, but doable Razz
you just have to solve a system w/ 5 variables...and then telescope.[/quote]
Perhaps less work  than one may  think.

[hide]One   wrt to middle term k+3=n it may  look like $(n^2-4)(n^2-1) n$

In any case, it is easy to figure out the coeffcients,

Put k=0 (excluding the factor k) you get $\frac 1 {((k)(1.2.3.4)}$
put k=-1 (excluding the factor (k+1) you get  $\frac 1 {(k+1)(-1.1.2.3)}$ etc

You get something like:
$\frac 1 {24} (\frac 1 k + \frac 1 {k+4} - \frac 4 {k+1} - \frac 4 {k+3} + \frac 6 {k+2})$[/hide]


algebra
polynomial
modular arithmetic
function
Congruences

Let $p$ be a prime number. Determine the maximal degree of a polynomial $T(x)$ whose coefficients belong to $\{ 0,1,\cdots,p-1 \}$, whose degree is less than $p$, and which satisfies \[T(n)=T(m) \; \pmod{p}\Longrightarrow n=m \; \pmod{p}\] for all integers $n, m$.

As $ T(x) \equal{} x^{p \minus{} 2}$ shows, there is a polynomial of degree $ p \minus{} 2$ with the required property.
Thus we only have to exclude ones with degree $ p \minus{} 1$:
Let $ T(x) \equal{} \sum_{k \equal{} 0}^{p \minus{} 1} a_k x^k$.
Define $ s_k \equal{} \sum_{i \equal{} 1}^p i^k$, then it's known that $ p|s_k$ if $ 0\leq k \leq p \minus{} 2$ (posted by me on PEN somewhere) and $ s_{p \minus{} 1} \equiv \minus{} 1 \mod p$ (simply little Fermat).
As $ T$ is supposed to be injective as a function $ \mathbb Z / p \mathbb Z \to \mathbb Z / p \mathbb Z$, it is bijective. Especially $ 0 \equiv \sum_{i \equal{} 1}^p i \equiv \sum_{i \equal{} 1}^p T(i) \equal{} \sum_{i \equal{} 1}^p \sum_{k \equal{} 0}^{p \minus{} 1} a_k i^k \mod p$.
Using what we know, we get $ 0 \equiv \sum_{k \equal{} 0}^{p \minus{} 1} a_k \sum_{i \equal{} 1}^p i^k \equal{} \sum_{k \equal{} 0}^{p \minus{} 1} a_k s_k \equiv \minus{} a_{p \minus{} 1} \mod p$. This gives $ p|a_{p \minus{} 1}$, thus $ a_{p \minus{} 1} \equal{} 0$, in other words: $ T$ has degree strictly less than $ p \minus{} 1$.

[color=red][thanks ZeTaX for rewriting nicely][/color]

You have probably finished learning the English meanwhile, can you please write your solution in good english now? I get what you mean, but as it stands there it's quite confusing for people unfamiliar with the matter.


In professional baseball, the distance from the pitching mound to home plate is 60 feet, 6 inches. How many seconds does it take a Roger Clemens’ 98 mile per hour fast ball to arrive at home plate?  Round your answer to the nearest hundredth of a second.   (There are 5280 feet in one mile.)

A.0.04 	B.0.42 	C.0.52 	D.0.64	E. 0.91

[hide=" click to reveal hidden content"]

$ 98 \times 5280$ feet in $ 60 \times 60$ seconds $ \equal{} 143.7333333...$ feet per second. Therefore $ 60.5$ feet in $ \frac{60.5}{143.733333....}$ seconds, that is approximately $ 0.4209$ seconds. Therefore $ B$.

[/hide]


The list 1,2,4,5,8,10,11,13 $ \ldots$ contains every positive integer less than 1000 that is not a multiple of 3 or a multiple of 7. How many integers are in the list?

[quote="AIME15"]NOTE: For a real number $ x$, $ \lfloor x \rfloor$ denotes the greatest integer less than or equal to $ x$.  For example: $ \lfloor \pi \rfloor=3$, $  \lfloor \frac{9}{2}  \rfloor=4$, and $  \lfloor 2  \rfloor=2$.

We can use the [url=http://en.wikipedia.org/wiki/Inclusion-exclusion_principle]Principle of Inclusion-Exclusion[/url]:

${ 1000-  ( \lfloor \frac{1000}{3}  \rfloor +  \lfloor \frac{1000}{7}  \rfloor } -  \lfloor \frac{1000}{3 \times 7}  \rfloor  )=1000-(333+142-47)=\boxed{572}$

The $ 1000$ comes from the original number.  The $  \lfloor \frac{1000}{3}  \rfloor$ is the number of integers less than $ 1000$ that are divisible by $ 3$.  The $  \lfloor \frac{1000}{7}  \rfloor$ is the number of integers less than $ 1000$ that are divisible by $ 7$.  The $  \lfloor \frac{1000}{3 \times 7} \rfloor$ is the number of integers less than $ 1000$ that are divisible by both $ 3$ and $ 7$.  We add these back in because we have double-counted them in the original count.[/quote]

572 is not the right answer, because you don't include 1000. Therefore it should be $\boxed{571}$

The problem specifically asked for integers [i]less[/i] than 1000.  Even if you were right, you would've added 1 to get 573. ;)


induction
number theory proposed
number theory

For each subset $C$ of $\mathbb N$, Suppose $C\oplus C=\{x+y|x,y\in C, x\neq y\}$. Prove that there exist a unique partition of $\mathbb N$ to sets $A$, $B$ that $A\oplus A$ and $B\oplus B$ do not have any prime numbers.

Iranian Mathematical Society competition (for University Students)

very nice problem!

we can show that the only possible partition is $A=\{1,3,5...\}$and $B=\{2,4,6...\}$
this can be easily done by induction and Bertrand's postulate :wink:

Hm..
What's Bertrand's Postulate ??
I've seen one's using that in SMO solution...
But I think it's weird and Postulate isn't Theorem, is it ???  :|

Does anybody know solution without this postulate? What about authors solution?

[quote=Albanian Eagle]The solution with Bertrand is very short and plus remember that this is a university level competition so you actually can use such theorems...[/quote]
I can't underestand how u prove the uniquenes set..please explain
but u can use Dirichlet theorem and solve easily i think
mercy


hello
[hide="solution"]
well the characteristic equation is $ r^2 \plus{} 1\equal{}0$
thus the general form of the equation is:
$ Acosx \plus{} Bsinx$
now putting the condition::
$ g(0) \equal{} A \equal{} a$
$ g'(0) \equal{} B \equal{} b$
thus ur answer is $ g(x)\equal{}(acosx \plus{} bsinx)$
[/hide]
Thank u

[quote="kabi"]
well the characteristic equation is $ r^2 \plus{} 1 \equal{} 0$
thus the general form of the equation is:
$ Acosx \plus{} Bsinx$[/quote]
Could you please explain what you did there in detail?? :|

[quote="Rajiv"]
Could you please explain what you did there in detail?? :[/quote]
hello rajiv
I know its not in our course.I learnt these things with some of the questions of Kunny's Digest of differential equation.So guys if i m writing any thing wrong kindly correct me.
    Consider the homogeneous linear second-order differential equation with constant coefficients
$ x''(t) \plus{} ax'(t) \plus{} bx(t) \equal{} 0$
The general solution of this equation depends on the character of the roots of the characteristic equation$ r^2 \plus{} ar \plus{} b \equal{} 0$
now for complex roots($ a^2 < 4b$):
the general solution of the equation is

   $ (Acos(\beta t) \plus{} Bsin(\beta t))e^{\alpha t}$,
where $ \alpha \equal{} \minus{} \frac {a}{2},and \beta \equal{} \sqrt {b \minus{} \frac {a^2}{4}}$
go for Kunny's digest.U will find some good questions there.
Thank u


group theory
superior algebra
superior algebra unsolved

Doesn't there exist some database with all the numbers 1..1000 worked out that if there's no simple group of that order, why that is? Or any alike resource?

Would be very handy, as we get one of those every year on our exam. :P

you mean aside from the ATLAS or the GAP libraries?

up to 1000 it is easy

you only have
$A_5, L_3(2)=L_2(\&), A_6=L_2(9), L_2(8), L_2(11)$ and so the sizes are 60,168,360,504,660

Yes but I think Peter meant not only a list , but also proof that for the other ordes there are no nonabelian simple groups, am I right Peter?

Wouldn't that be nice?  In the long run I think we wil go over thousand?


[quote="manlio"]For $ a,b,c$ positive reals

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \plus{} 4(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq 6(a \plus{} b \plus{} c)^2$[/quote]
I think you should change another name for this topic since I dont use ABC, it is just come from my pqr techniques And I dont like the other always say these words in front of me!

Dear can_hang,

unfortunately I don't know vietnamese language, so when I read in your solution of problem 14 " theo abc" I supposed you referred to ABC theorem, and not to your very nice pqr tecnique. Sorry for my misunderstanding of your solution. I have corrected the title of the topic :blush:  :blush:

P.S. Can you please explain why it is sufficient, using your pqr tecnique, to prove only the two cases b=c=1 and a=0, b=1.  :blush:

Thank you very much.


geometry
geometric transformation
reflection
rotation
Asymptote

What is the fewest number of L-shaped tiles taken from a supply of tiles in the given shape which can be placed entirely on the 5 by 5 grid so that no more tiles can be placed on the board without overlapping? Assume each square of a tile completely covers a square on the grid and reflections and rotations of tiles are allowed.

[asy]size(0.9cm);
draw((0,0)--(2,0)--(2,1)--(1,1)--(1,3)--(0,3)--(0,0)--cycle);
draw((1,0)--(1,1));
draw((0,1)--(1,1));
draw((0,2)--(1,2));[/asy]
[asy]size(1.5cm);
draw((0,0)--(5,0)--(5,5)--(0,5)--cycle);
draw((1,0)--(1,5));
draw((2,0)--(2,5));
draw((3,0)--(3,5));
draw((4,0)--(4,5));
draw((0,1)--(5,1));
draw((0,2)--(5,2));
draw((0,3)--(5,3));
draw((0,4)--(5,4));[/asy]

I believe you can get 6 if you only leave the middle square blank.

Whoops. :blush:

Sorry, that is a perfect example of how I make stupid mistakes all too often in MC.

answer is 3.
(see attachment)
the way i constructed it was by finding the squares shared by most of the L's that shared a side with the outer ring of 16 squares. and then i covered them up with my L's, leaving the bottom line uncovered but blocked off.

It's impossible to do with 2, can't think of any clever proof atm.

edit: sry for bad diagram, can someone asymptote it? (too lazy to make good one)

The answer is three. I just tried to make as many 2x2s as possible because that optimizes the blank space. I also tried leaving most of the corners empty because I thought that helped.  :ddr:


AMC
AIME
number theory
relatively prime
geometry
angle bisector

Don't answer if you have seen this before.

In quadrilateral ABCD, :ang: BAD :cong: :ang: ADC and :ang: ABD :cong: :ang: BCD, AB = 8, BD = 10, and BC = 6.

The length CD may be written in the form m/n, where m and n are relatively prime positive integers. Find m+n.

[hide]Call measure of angle DAB x, angleABD y, and ADB k. y=180-x-k. Also looking at triangle DBC, we have x-k+y+m(DBC)=180, so sub y in, we have m(DBC)=2k. Draw the angle bisector of DBC to DC, call the point they intersect E.  so triangleBCE :sim: triangleDBA. Call EC m and ED n. 6/10=m/8, so m=24/5. Using triangle angle bisector theorem, m/n=6/10, so n=8. 8+(24/5)=64/5. 64+5=69[/hide]