A school has 7 periods a day. Six are 48 minutes in length and one
is 1 hour. The school day starts at 7:35 AM and ends at 2:13 PM
with 4 minutes passing time between periods and one-half hour for
lunch. What percent of the school day is spent in class? Round
your answer to the nearest whole percent.

All we have to do is find the number of minutes in their school day and how many minutes they're in classes.  We have $ 6$ classes, each $ 48$ minutes, so we have $ 6 \times 48\equal{}288$.  Then, we add $ 1$ hour, which is $ 60$ minutes, so then we have $ 348$.  We see that from $ 7: 35$ AM to $ 2: 13$ PM, it is $ 6$ hours and $ 38$ minutes, which is a total of $ 398$ minutes.  So we have $ \frac {348}{398}\approx \frac {348}{400}$, which is $ \frac {87}{100}$, or $ \boxed {87}$%


function
linear algebra
matrix
LaTeX
complex analysis
linear algebra unsolved

It's classical but  couldn't find it on this forum
Show that $A\mapsto A\exp(A)$ is surjective from $\mathcal{M}_{n}(\mathbb{C})$ to $\mathcal{M}_{n}(\mathbb{C})$.

is there an entire surjective function $f: \mathbb{C}\to \mathbb{C}$ such that $f: M_{n}(\mathbb{C}) \to M_{n}(\mathbb{C})$, defined in the obvious way, is not surjective ? (I think I can prove the easy case when $f'$ never vanishes)

For example, $f(z)=z^{2}$: there is no square root of $\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

Indeed, if $\begin{pmatrix}a&b\\c&d\end{pmatrix}^{2}= \begin{pmatrix}0&1\\0&0\end{pmatrix}$, then $a^{2}=d^{2}=-bc$ and $(a+d)c=0$. This implies $a+d=0$, contradicting $(a+d)b=1$.

:blush:  I don't know why I've posted such an idiot thing ? Anyway, try the case where $f'$ never vanihes ..

Hi JC_math; your result is false.
i) Here $f(z)=ze^{z}$ and $f'(z)=e^{z}(z+1)=0$ iff $z=-1$.
Moreover $f(z)=-\frac{1}{e}$ iff $z=-1$ (the converse is easy: $z=x+iy$.....)
\[\text{ Let }A=\begin{pmatrix}-\frac{1}{e}&1\\0&{}-\frac{1}{e}\end{pmatrix}\]
If $f(B)=A$ then $spectrum(B)=\{-1,-1\}$ and $B=I+N$ where $N$ is nilpotent. ${f(B)=f(-1)I+f'(-1)N=}-\frac{1}{e}I\neq{A}$.
Contradiction and $f$ isn't surjective over the complex matrices.
ii) Now we assume, as Alekk did, that $f$ is surjective over $\mathbb{C}$ and that $f'(z)$ is never 0. We prove that $f$ is surjective over the matrices.
Let $A\in\mathcal{M}_{n}(\mathbb{C})$; it's sufficient to assume that $A$ is a Jordan matrix ($A=\lambda{I}+N$ where $N$ is a nilpotent matrix s.t. $N^{n-1}\neq0)$. There exists $\mu$ s.t. $f(\mu)=\lambda$ and $f'(\mu)\neq0$. Let $B=\mu{I}+P$ where $P$ is nilpotent.
$f(B)=\lambda{I}+f'(\mu)P+...+f^{(n-1)}(\mu)\frac{1}{(n-1)!}P^{n-1}=A$ iff $N=f'(\mu)P+...+f^{(n-1)}(\mu)\frac{1}{(n-1)!}P^{n-1}$.
We seek $P$ in the form: $a_{1}N+...+a_{n-1}N^{n-1}$.
We are brought back to a triangular system easy to solve:
$a_{1}=\frac{1}{f'(\mu)}, a_{2}=-\frac{f''(\mu)a_{1}^{2}}{2f'(\mu)}$,....

[quote="loup blanc"]
Moreover $f(z)=-\frac{1}{e}$ iff $z=-1$ [/quote]
This is not true. If it were true, $f(z)+\frac{1}{e}=ze^{z}+\frac{1}{e}$ would be representable as $c(z+1)^{m}e^{Cz}$, which is, clearly, not the case. It seems like what we actually need is that for every $\lambda\in \mathbb C$ there exists $z\in\mathbb C$ such that $f(z)=\lambda$ and $f'(z)\ne 0$. The function $ze^{z}$ does satisfy this property, so the proof given in the second part of loup blanc's post works for this case.

Apologies JC_math.
Fedja is right; there exist others solutions than -1 for the equation $ze^{z}=\frac{1}{e}$ for example
$z_{0}=-3.089...+i7.4615...$; $f'(z_{0})\neq0$.
As Fedja said, the existence of this $z_{0}$ is sufficient to prove the *f*'s surjectivity with my second proof.

No worries loup blanc.
I think I have 2 different (correct :?: ) proofs of that result, so I think it is right.

PS: Good to see that you are now using $\LaTeX$, it much easier to read your threads.

Hi JC_math, your problem is solved; do you want another solution?

Sorry loup blanc, I only read the first part of your message "Hi JC_math; your result is false. ", since the result was true I stopped reading and didn't realize that you solved part of the problem.

P.S.: You didn't solve the case $n=1$ which is not the simplest part of the question.

OK JC_math:
Let  $a=-3.089+7.4615i, g(z)=-1-ln(z)+3i\pi$, where $ln(re^{i\theta})=ln(r)+i\theta$ if $r>0$ and $\theta\in]-\pi,\pi[$. $g(z)=z$ implies $ze^{z}+\frac{1}{e}=0$.
 Let  $D$ be the closed convex subset $\mathcal{B}'(a,0.1)$.
 $|a-g(a)|<2.10^{-4}$ and $|g'(z)|=\frac{1}{|z|}<\frac{1}{7}$ over $D$.
Thus  $g: D\rightarrow{D}$ and is a holomorphic function. We conclude, with the help of Banach fixed point theorem, that there exists a complex $z\not={-1}$ s.t. $ze^{z}+\frac{1}{e}=0$ and $(ze^{z}+\frac{1}{e})'\not=0$

It remains to show that $f: z\rightarrow{ze^{z}}$ is a surjective map from $\mathbb{C}$ over $\mathbb{C}$.
According to the little Picard's theorem $f(\mathbb{C})=\mathbb{C}$ or $\mathbb{C}\setminus{u}$ (where $u\notin{R}$ or $u<\frac{-1}{e})$.
Assume that we are in the last case. Let $\epsilon>0$ and $v=f(a)\notin\mathbb{R}$ s.t. $|v-u|<\epsilon$. $\exists{k}\in \mathbb{Z}$ s.t. $ln(f(a))-ln(a)+2ki\pi=a$ (with the same $ln$ than in my last post). Let $g(z)=ln(u)-ln(z)+2ki\pi$. $\{b|f(b)=v\}$ is unbounded (Picard again); thus we may assume $|a|>M+1$,$M$ as you want.
Remark: if $u<\frac{-1}{e}$ then choose another $ln$.
Let $D$ be the closed convex subset $\mathcal{B}'(a,\alpha)$ where $\alpha<1$ is s.t. $D$ doesn't intersect with $\mathbb{R}^{-}$. $g: D\rightarrow{D}$ is an holomorphic function and $|g'(z)|=|\frac{1}{z}|<\frac{1}{M}$ over $D$.
Therefore you can conclude as in my last post.


A game is played with 2006 chips.  Two players take turns pulling 2, 3, 4, 5, or 6 chips out of the bag.  No players are aloud to skip their turn.  To win, you make your opponent take out the last chip.  If there is one chip left, the person who just took out the $2{}^n{}^d$ to last chip loses.  I know I can win against my opponent by taking out [i]n[/i] chips on my first turn.  Find the two correct values of [i]n[/i].  Show all work in hidden text, then read my solution.
Also, open the hidden text if you just want the answer (no work).
Solution:
[hide]
Here's for people who don't want the solution, just the answer.
[hide]
To win, I need to leave my opponent with 2 or three chips; this way, he cannot force me to take the last chip.  So I must make him take out...
a. The $2005{}^t{}^h$ and $2006{}^t{}^h$ chips.
b. The $2004{}^t{}^h$ and $2006{}^t{}^h$ chips.
c. The $2004{}^t{}^h$, $2005{}^t{}^h$, and $2006{}^t{}^h$ chips.
Meaning I must take out the $2004{}^t{}^h$ or $2003{}^r{}^d$ chip.  Every turn, no matter what my opponent plays, I can take out some amount of chips so that the sum of the chips taken out in that round is 8.  So on my first turn, I should take out the 2004(mod 8) chip or the 2003(mod 8) chip.  Which means I should take the[/hide]$4{}^t{}^h$ or $5{}^t{}^h$ chip.
[/hide]

uhh i dont think ur making some of the rules of this game clear

1) how do you win exactly? do i want to take out the last chips or do i want him to take out the last chips?

2) wat if there is 1 chip left? for example, if there are 7 chips left and i take out 6, what happens to the 1 left over? since we're not allowed to only take out 1 chip how does that work?

[quote="gotztahbeazn"]
wat if there is 1 chip left? for example, if there are 7 chips left and i take out 6, what happens to the 1 left over? since we're not allowed to only take out 1 chip how does that work?[/quote]

I think that is probably going to be a tie, so you have to find a way to force it not to be 1 chip left over inorder to win.


search
inequalities proposed
inequalities

I search but I didnot found it. So

Let $n \geq 3$ be an integer. Find the minimal value of the real numbers $k_n$  such that for all positive numbers $x_1,x_2,...,x_n$ with product $1$ we have

$\frac{1}{\sqrt{1+k_nx_1}}+\frac{1}{\sqrt{1+k_nx_2}}+...+\frac{1}{\sqrt{1+k_nx_n}} \leq n-1$


invariant
geometry
geometric transformation
vector
combinatorics unsolved
combinatorics

Let $ M$ denote the set of all points on the plane with have integer coordinates.

a) Prove that it is possible to color each point from $ M$ with one of the two colors in such a way that any infinite set consisting of the points having the same color is not symmetrical with respect to either $ A(0,0)$ or $ B(1,0)$

b) Prove that for any coloring of $ M$ which satisfies condition of the case a) and for any point $ C(x,y)$ with nonzero $ y$, there is infinite set consisting of the points having the same color which is symmetrical with respect to $ C$.

I think I got it...

In the following, each point is supposed to be an integer point.

a) Let's color black each point $ M(x,y)$ with $ y<0$ or with $ y\equal{}0,x\geq 1$.
And color white $ M(x,y)$ for $ y>0$ or $ y\equal{}0,x \geq 1$.
This gives an appropriate colouring because no two points symmetric with respect to $ A$ or $ B$ have the same colour (except $ A$ and $ B$...).

b) Let $ C(x,y)$ with $ y \neq 0$. Wlog, we may assume that $ y > 0$.
Assume that we have a $ 2$-colouring such that no infinite monochome set is symmetric with respect to $ A$ or $ B$ or $ C$.

For any point $ M$, let $ M_A,M_B,M_C$ be the symmetric of $ M$ with respect to $ A,B$ and $ C$, respectively.

It follows that there is a finite number of points $ M$ such that $ M_A$ has the same colour than $ M$ (otherwise, using pigeon-hole principle, there would be an infinite monochrome set symmetric with respect to $ A$). The same for $ B$.
Thus, for all integers $ a,b$ such that $ |a|$ or $ |b|$ is sufficientely large, the point $ M(a,b)$ has opposite colour than $ M_A$ and $ M_B$. It follows that $ M_A$ and $ M_B$ have the same colour. But $ P_B\equal{}M_A$ for $ P(a\plus{}2,b)$ thus $ M$ and $ P$ have the same colour.
This means that the two-colouring is invariant by translation of vector $ (2,0)$, except perhaps for a finite number of points which are in some given rectangle.

The same reasoning but with $ A$ and $ C$ gives that the $ 2$-colouring is invariant by translation of vector $ (2x,2y)$, except for a finite number of points, which are in some other given rectangle.
Thus, we may assume that if we restrict ourselves to the set $ S$ of all points $ M(a,b)$ with $ |b| > k$ for some given constant $ k$, the $ 2$-colouring is invariant by the translation of vector $ (2n\plus{}2xm,2my)$ for any integers $ m,n$ (according that the translation of the considered point stay in $ S$).

Let $ a,b$ be any integers with $ b$ such that $ k < 2by$ and $ 2y(1\minus{}b) < \minus{}k$.
Wlog, we may assume that $ (2ax,2by)$ is black and that $ (2ax\plus{}2t,2by)$ is black for each integer $ t$ (invariance by translation $ (2,0)$).
Now, choose $ m\equal{}1\minus{}2b,n\equal{}2xb\minus{}2t\minus{}2ax$. Thus, the translation with vector $ (2n\plus{}2xm,2my)$ transforms $ M_t(2ax\plus{}2t,2by)$ into $ P_t(2x\minus{}2ax\minus{}2t,2y\minus{}2by)$. Note that $ M_t$ and $ P_t$ belong to $ S$. Thus, using the invariance of the $ 2$-colouring, we deduce that $ M_t$ and $ P_t$ are blacks for all integer $ t$.
But clearly, $ P_t$ and $ M_t$ are symmetric with each other with respect to $ C$, then we have found an infinite monochrome set which is symmetric with respect to $ C$. A contradiction.

Pierre.


Prove that
$\frac{\left(C_{n+1}^{n-1}\right)!}{\prod_{k=1}^{n}k!}$ is natural number, for any $n \in \mathbb{N}$

We use induction on $n$ to prove our result. There are two cases to consider while working on the induction case, only the first of which I have worked out in detail.

In the solution below, we will use the well known fact that the product of $n$ consecutive positive integers is divisible by $n!$

[b][u]Base case[/u][/b]: For $n=1$, we can verify that the expression is equal to $1.$

[b][u]Induction case[/u][/b]: Let the expression be an integer for some arbitrary $n > 1.$

[b]Case 1[/b]: $n =2m,$ for some $m \in \mathbb{N}.$

$\left(C_{n+1}^{n-1}\right)! = \left(\frac{n(n+1)}{2!}\right)! = (2m^{2}+m)!$ So, $\frac{(2m^{2}+m)!}{\prod_{k=1}^{2m}k!}$ is an integer (by [b]induction hypothesis[/b].)

Now, $\left(C_{n+2}^{n}\right)! = \left(\frac{(n+1)(n+2)}{2!}\right)! = (2m^{2}+3m+1)!$

$\therefore \frac{\left(C_{n+2}^{n}\right)}{\prod_{k=1}^{n+1}k!}= \frac{(2m^{2}+3m+1)!}{\prod_{k=1}^{2m+1}k!}= \frac{(2m^{2}+m)!}{\prod_{k=1}^{2m}k!}\cdot\frac{(2m^{2}+m+1)(2m^{2}+m+2)...(2m^{2}+m+2m+1)}{(2m+1)!}$

Note that in the last expression, the first term of the product is an integer by [b]induction hypothesis[/b], and the second term of the product is also an integer [ the product of $(2m+1)$ consecutive integers is divisible by $(2m+1)!$]

$\therefore \frac{\left(C_{n+2}^{n}\right)}{\prod_{k=1}^{n+1}k!}$ is also an integer.

[b]Case 2[/b]: Let $n =2m+1$ for some $m \in \mathbb{N}.$

The method used in [b]case 1[/b] works in this case too. I have verified it for myself though I feel lazy writing down all the necessary steps. My apologies for the same -- I do understand this is not a very good substitute for writing down actual proofs! :)

The above two cases thus complete our proof.

[quote="libra_gold"]$\frac{\binom{n+1}{n-1}!}{\prod_{k=1}^{n}k!}=\frac{(1+2+\cdots+n)!}{1!2!\cdots n!}$ which is a natural number.[/quote]
This is indeed a very good solution! The above expression counts the number of certain objects -- I don't remember what. Could someone point this out?

It's a multinomial coefficient.
I'll give you some links:
[url=http://mathworld.wolfram.com/MultinomialCoefficient.html]Mathworld[/url]
[url=http://en.wikipedia.org/wiki/Multinomial_theorem]Wikipedia[/url]


$ (a\plus{}b\plus{}c)^2\geq 3(ab\plus{}bc\plus{}ca)\Longleftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\geq 0$.

[quote="kunny"]$ (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)\Longleftrightarrow (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2\geq 0$.[/quote]

you' re right, thx. My solution is by this way:

3(ab+bc+ca)<=a^2+b^2+c^2 +2(ab+bc+ca)
ab+bc+ca<=a^2+b^2+c^2, It's solved by amgm inequality.


O, H are the circumcentre and orthocentre of triangle ABC. $OD \perp BC$.
Prove: $AH=2OD$.

What is this point D? It is the midpoint of BC?

If yes:

This is a very interesting lemma to prove the Euler Line.

AH // OD, because both are perpendicular to BC
If E is the midpoint of AB, CH // OE.

Then the riangles ODE and HAC are similar, and thus OD : AH = DE : AC.
The triangles EBD and ABC are also similar (The angle B is common, and 2*BE = BA, 2*BD=BC).
Then ED : BC = 1 : 2.
And now, OD : AH=1 : 2.
This is the end!


inequalities
geometry
inradius
inequalities unsolved

Denote by $r(ABC)$ the inradius of a triangle $\Delta ABC .$ Let $k$ be a  positive number and suppose that in a triangle $\Delta ABC$ we have a cevian $AA^{\prime} \; ,\; A^\prime \in (BC) ,\;$ such  that $r(ABA^{\prime}) = k\cdot r(AA^{\prime}C)$.
 [b] i).   Find all positive $k$ such that   $\; \; \; (*)\; \; \; \; \; \; \; ||AA^{\prime}||\le \sqrt{r_br_c}\; \; . \;$
 ii).  Prove that  for $k=1 \; \;$  inequality  (*) is true. [/b]

[quote="mecrazywong"][quote="siuhochung"][quote="flip2004"]Denote by $ r(ABC) $ the inradius of a triangle $\Delta ABC .$ Let $  k $ be a  positive number and suppose that in a triangle $\Delta ABC $ we have a cevian $ AA^{\prime} \; ,\; A^\prime \in (BC) $ such  that $  r(ABA^{\prime}) = k\cdot r(AA^{\prime}C) $.
 [b] i).   Find all positive $ k $ such that   $ \; \; \;  (*)\; \; \; \; \; \; \;  ||AA^{\prime}||\le \sqrt{r_br_c}\; \;  . \; $
 ii).  Prove that  for $ k=1 \; \; $  inequality  (*) is true. [/b][/quote]
what is $r_b$? is it the exradius?[/quote]
The exradius wrt B.[/quote]
Yes ! /proposer.

for k=1,suppose AB=c,BC=a,CA=b.then AA'=[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a9c2937da08e8b956f0265dcd26fdd5a.gif[/img]
What  I can do is only this. :blush:


Prove that the following inequality holds for a,b,c positive reals such that a+b+c=1
$ \frac{ab}{\sqrt{ab\plus{}bc}} \plus{} \frac{bc}{\sqrt{bc\plus{}ca}}\plus{}\frac{ca}{\sqrt{ca\plus{}ab}} \le \frac{1}{\sqrt{2}}$

[quote="geniusbliss"]Prove that the following inequality holds for a,b,c positive reals such that a+b+c=1
$ \frac {ab}{\sqrt {ab + bc}} + \frac {bc}{\sqrt {bc + ca}} + \frac {ca}{\sqrt {ca + ab}} \le \frac {1}{\sqrt {2}}$[/quote]

${ \sum{\frac {ab}{\sqrt {ab + bc}}} \le \sum{\frac {\sqrt{2}ab}{\sqrt {ab} +\sqrt{bc}}}}$

just need to prove
${ \sum{\frac {ab}{\sqrt {ab} +\sqrt{bc}}}} \le \frac{a+b+c}{2}$

note that
${{ \sum{\frac {ab}{\sqrt {ab} +\sqrt{bc}}}} = \sum{\frac {bc}{\sqrt {ab} +\sqrt{bc}}}}$

so equivalent to
${ \sum{\frac {ab+bc}{\sqrt {ab} +\sqrt{bc}}}} \le {a+b+c}$

S.O.S that, equivalent to
$ \sum{\frac{\sqrt{ab}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}(\sqrt{a}-\sqrt{b})^2}\ge0$

obvious true.

[quote="geniusbliss"][quote="kuing"].........
note that
${{ \sum{\frac {ab}{\sqrt {ab} + \sqrt {bc}}}} = \sum{\frac {bc}{\sqrt {ab} + \sqrt {bc}}}}$

[/quote]
er.. how is that true ?[/quote]
Yes it is true and it's nice!!!

$ \sum \frac{ab-bc}{\sqrt{ab}+\sqrt{bc}}=\sum \frac{(ab-bc)(\sqrt{ab}-\sqrt{bc})}{ab-bc}=\sum(\sqrt{ab}-\sqrt{bc})=0$ :wink:

[quote="geniusbliss"][quote="kuing"].........
note that
${{ \sum{\frac {ab}{\sqrt {ab} + \sqrt {bc}}}} = \sum{\frac {bc}{\sqrt {ab} + \sqrt {bc}}}}$

[/quote]
er.. how is that true ?[/quote]
because $ \sum{\frac {ab-bc}{\sqrt {ab} + \sqrt {bc}}}} =\sum{({\sqrt {ab} - \sqrt {bc}})=0$ ;)

sry for my stupid question and nice solution kuing..


email
geometry
geometric transformation
analytic geometry

I asked radev through email if contestants can talk yet and he sent back "Yes, the problems are available publicly now."

So how'd people do? Most of the people at my site got about 3 or 3.5 problems done.  Personally, I thought 3 and 4 were the most interesting (haha I was relieved to find out my answer to 4 wasnt completely crazy)

I got the first 3, most of 4, and some of 5.

I thought 5 was really hard. The presence of "ma" vs "kimsa" vs "paya" along with "hacha'a" was really confusing. I at least figured out that the "chall" words were the types of fish, and which sentences had more than one type of fish.

Where are answers?

I also got the first 3, most of 4, and some of 5. For 5, I just said the chall words were "big", "middle", "small" fish, singular and plural, said that one of the first two was wrong, and "ma", "kimsa", "paya" were numbers (woah awesome I just looked somewhere and realized I correctly guessed ma=1, paya=2, kimsa=3), and then threw out some guesses for the first two parts. I've heard a lot about Aymara recently... now I just wish I had looked more at it.

For 4, I got Little Red Riding Hood!!! For a long time I was trying to make it translate into Romeo & Juliet... (varfor har du). Knowledge of Old English ended up REALLY helping me on that.
-My grandmother, why do you have such big eyes?
-They are so that I can see you better, my girl.
There were a few on the chart I wasn't sure of. My reasoning was kind of rushed and not as good as it could have been, and I wasn't sure what they meant for D6, but I put a lot of minor stuff.

3 was really cool. I almost forgot to figure out the diacritics... I didn't like 2 at all, since it was easy, tons of people know Hindi, and it was pretty contrived. And what is up with them not making us write in Devanagari, but making us write in the infinitely more complicated whatever-alphabet-that-is.

So for me I predict maybe 70-80ish. Maybe higher seeing my correct guesses on E. What about other people? Any ideas for cutoffs for the next round?

I got the first three pretty easily. On the fourth, I got Little Red Riding Hood by just reading it as if it were English. However, I was very unsure about many of the translations between X and Y.

On the fifth, I got all of it except the suffixes associated with plural nouns (lose some points on the practical at least - used singular for 2 big fish) and what hach'a meant (did it need to be used in the practical? It was actually really easy to figure out everything on E. Chall means fish since Challwataxa is in every sentence. Then ma/ma is the only duplicate numbers, so ma = 1. Then you look at other stuff and notice thal Challwa is big fish. Then you look at the 1/2, 2/3, 3/1. Only one of them had a big fish, so you could then get what little fish was and thus medium fish. Then getting the numbers is simple.

I have no idea how strictly they score, so I have no idea what I'll get.

Ah so those are called diacritics... I had no idea what they were so on the test I called them "peripherals."

EDIT: On further review of E, I have no idea if I got it right. I put down: "Paya challwa kimsa challwallampiwa challwataxa." Anyone know the right answer?

Yay I got top 20. Now I just need to up my game a tiny bit for the next round. So if the first test is the NACLO, what is the second test? The USALO? The ILO TST? They need to have nice succinct names for these things.

About #5. I suppose it makes sense that -mpi is "and", but what is up with -wa? Some objective suffix that only needs to occur on one word, perhaps? When I read through the text slowly left to right, I just don't understand how a native could find the system natural. Bleh. I always get confused on languages that are agglutinative instead of inflectional, even though I know perfectly well that American languages are mostly agglutinative.

I'm not really ranting, if it sounds like it :roll: . I'm quite happy. Congrats to everyone who made the next round, whatever it is called.

I got 12th, yay! Yeah I'd kinda like making the team, but I've got time.

So hach'a means big. Challwa = medium/big, Challwalla = small. Then if it ends with -wa, it's the only type of fish. If it ends in -mpiwa, it's the second of two types of fish. So the answer was supposed to be

Paya hach'a challwa kimsa challwallampia challwataxa.
(I caught two big fish and three little fish)

Oh and I think both tests are called the NAmCLO / NACLO.

To me, Little Red Riding Hood was by far the easiest practical. jb05, I too thought "Romeo and Juliet" for some moments! Then I realized that they didn't go on in regular patterns for so long. But really, how should we respond to a question like "Say as much as you can about languages X and Y?" I possibly wasted my time talking about cases and subtle structural aspects, since those weren't mentioned in the published solutions.

Hindi and adverb/verb ones were smooth enough to be completely enjoyable. Mayan fish and diacritic wonderland were painful, but I'm not well-familiarized with agglutinative languages. I should get on that before 3/11, I guess.

Is there any way to learn to categorize (semantically, etc.) unfamilar word forms, as in #3 (summit/skeleton/appearance)? I ended up categorizing by verb/non-verb and conjecturing wildly from there, drawing from latin and mandarin in the same sentence (short on time).

What did you guys think about the second round? I actually thought it was easier. Since I apparently predicted my score right for the first round, I would say I probably got upper 80s-ish. I kind of had a little issue with thinking an-h, an, and na were three different genders on #5, and I'm not sure if the thing in #4 was meant to contain all the legitimate words from the example (I said t could not precede i and s had to precede i), but other than that... I was very happy.

By the way, for the last one...
[hide]lice/lease, ash, sheep, louse/lass ?[/hide]

I think I put lice, pass, sheep, lass

On thinking back, ash works better for the second one, but I'm pretty sure about the other three.

For the volcano one, is the general consensus that is was rectangular or polar coordinates? I solved it with N/E/S/W, but apparently polar works better?

[quote="Hamster1800"]
On thinking back, ash works better for the second one, but I'm pretty sure about the other three.

For the volcano one, is the general consensus that is was rectangular or polar coordinates? I solved it with N/E/S/W, but apparently polar works better?[/quote]

I think I had pass for the second one, but changed my mind to ash. I hope these linguists don't shoot themselves for my inane-sounding rationale: "the lines weren't dark enough."

And I used polar for the volcano, since rectangular chucked one of the islanders (?) into two positions at once. I made the assumption that any shoreline is better than being caught midway between house and shore-closest-to-mainland. Did anyone else notice the pseudo-(likely not)-cognates in that problem? Awa, ata, ilau, auta ~ away, at it, inland, outer?

And what was the intelligent way to do the Mayan calendar problem? 3 buns = baguette, lion --> eagle --> bulldog --> butterfly..?

Are you sure about awa/ata/ilau/auta being in/out/whatever? The NSEW system works perfectly except for one slightly off person who is not quite north (but almost). I am fairly certain there is no logical way to do the calendar thing except brute force. You did have to guess once for the first part of the glyph... all 20 second parts were provided though.

How are you so sure about lice and lass over lease and louse? I am actually quite sure of ash and sheep (sh and s are pretty easy to differentiate for me).

I, too, treated awa/ata/ilau/auta as the four cardinal directions, despite one dwelling not being directly north of the other.  Did anyone else find the Irish Gaelic to be on the easy side?  I was actually fairly confident I had done well until I got to the calendar, at which point I for some reason had difficulty.  After looking at a web site about the Mayan calendar, it seems that "guessing" the glyph we needed for the first part would have been easier if the glyphs had been depicted more closely to their original forms.

I'm expecting about 70-75% on this round.  Anyone know when we'll hear the results?

[quote="calc rulz"]The guy at Brandeis told me that it would probably take about 10 days, but it has already been around 16 days.[/quote]

I took it offsite, and I seem to recall something being read at the end of the test saying results would come out March 27 or something like that.  I remember it was very close to when colleges send out responses.  But given what happened last round, I'm not holding my breath for them today.

Looks like we're getting results tomorrow (April 2)
[url]http://www.naclo.cs.cmu.edu/[/url]

Results are up! I got 19th =)

The scores were a lot lower than I had thought. I thought that I had got around 90, but the highest score was 87.5, and I only got a 66.

I did much worse than expected.... So be it. I wish I could see feedback on this one - seems like they graded it MUCH harder than last time.

I had 59.5, not enough to get on the named rankings. Barely beat out the other person from my school :)

So it turns out that they weren't north/south/east/west.  I got 12th, though that was the difference between 12th and 8th.  Congratulations to those who will be representing the United States!


let $ x,y,z\in\mathbb{R}^\plus{}$, satisfying that:

$ \frac{x}{\sqrt{y\plus{}z}}\plus{}\frac{y}{\sqrt{z\plus{}x}}\plus{}\frac{z}{\sqrt{x\plus{}y}}\leq k\sqrt{x\plus{}y\plus{}z}$

find the mininum of $ k$

There is no such $ k$,because if we take $ y \plus{} z\rightarrow 0$ then RHS will become very large and LHS will be less than RHS.

there is no $ k$ for which the inequality is true for all $ x,y,z$. For example $ z\equal{}0$ and $ y\equal{}x^2$ shows that it is impossible. However, the reverse inequality holds for $ k\equal{}\sqrt{6}/2$.


In a party with 1982 persons, amongst any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?

If Person A knows Person B, is it necessarily true that Person B knows Person A?

[quote="igiul"][quote="t0rajir0u"]Yes.  "Knowing" is a relation that is symmetric but not transitive.[/quote]

I would like to state that this is creepy. [i]How[/i] do you know that?[/quote]

Ummm, it's trivial?

Let $ a\approx b$ denote $ a$ knowing $ b$.
$ a\approx b$ implies $ b\approx a$, which is the definition of a symmetric relation.
$ a\approx b$ and $ b\approx c$ does not necessarily imply $ a\approx c$, so it is not transitive.

Strangely I got [hide]$ 1979$[/hide]. Is that correct?


logarithms
function
real analysis
real analysis unsolved

Hi

i am trying to implement a reliability mathematical model, however, i need to solve to following equation for b:

n/b - (n*a)/( exp(b*a)-1 ) - t = 0

where n, a and t are all known variables.

I have tried to solve be by estimating the exp, by using exp(x) is approximatly equal to (1+x), however b*a is known to be larger than one so this solution fails.

It is known that for example, n=10, a=222 and t = 870, b would be equal to 0.006.

I have also tried to apply logs on both side but always got stuck.  Any help would be greatly appreciated, as i really am out of ideas!

Thanks alot in advance!

MA

thanks but dont really understand all of that :S will try to make some research ... thanks for the guidance

$\frac{n}b-\frac{na}{e^{ab}-1}-t = 0$

Unfortunately, this equation does not have elementary solutions, i.e. no amount of algebraic manipulation, logarithms, etc. will allow you to express the answer exactly.  It's probably possible to express it in terms of the Lambert W function or something, but that won't do you much good.

If you know that $a\cdot b$ will be near to some other value (say, 1, or 2, or whatever), you could try to use the Taylor series approximation for the exponential centered around that value.  In general, the linear approximation to $e^{x}$ when $x$ is near some constant $c$ is $e^{x}\approx e^{c}+e^{c}(x-c)$.  (Your approximation is this approximation when $c = 0$.)


limit
number theory unsolved
number theory

Write $ [x]$ is the largest integer not exceeding x and {x} = x - [x].
Find $ lim\limits_{n \to \infty} {(7 + 4\sqrt {3})^n}$

[quote="thesunflower"]Write $ [x]$ is the largest integer not exceeding $ x$ and $ \left\{x\right\} \equal{} x \minus{} [x]$.
Find $ \lim_{n \to \infty} {(7 \plus{} 4\sqrt {3})^n}$[/quote]
Maybe $ \left\{\begin{array}{c}
\{a,b,c\}\subset\mathcal N^*\\\\\
\sqrt c\not\in\mathcal Q\ ,\ a\equal{}1\plus{}\left[b\sqrt c\right]\end{array}\right\|$ $ \implies$  $ \lim_{n \to \infty} \left\{(a \plus{} b\sqrt {c})^n\right\}\equal{}1\ !$

It's not clear! [b]Virgil Nicula[/b], can you say that more clearly?

[quote="Virgil Nicula"][quote="thesunflower"]Write $ [x]$ is the largest integer not exceeding $ x$ and $ \left\{x\right\} \equal{} x \minus{} [x]$.
Find $ \lim_{n \to \infty} {(7 \plus{} 4\sqrt {3})^n}$[/quote]
Maybe $ \left\{\begin{array}{c} \{a,b,c\}\subset\mathcal N^* \\
 \\
\ \sqrt c\not\in\mathcal Q\ ,\ a \equal{} 1 \plus{} \left[b\sqrt c\right]\end{array}\right\|$ $ \implies$  $ \lim_{n \to \infty} \left\{(a \plus{} b\sqrt {c})^n\right\} \equal{} 1\ !$[/quote]
$ \lim_{n \to \infty} {(7 \plus{} 4\sqrt {3})^n}\equal{}\infty\ !$ Maybe$ \boxed {\lim_{n \to \infty} \left\{(7 \plus{} 4\sqrt {3})^n\right\}}\ !$

Here $ \{x\}$ is $ x\minus{}[x]$, where $ [x]$ is the largest integer not exceeding $ x$.

Now is more clearly ? I said that this last (boxed) limit is equally with $ 1\ !$

Indeed, $ 7\equal{}1\plus{}[4\sqrt 3]\ .$ I proposed you to solve a general case.

I tried to solve the general problem but i can't... I think it's new and very hard with me and it's a nice conclusion! Please share the solution!

[quote="thesunflower"][color=darkred]I tried to solve the general problem but i can't ... I think it's very hard.
It's a [u]nice conclusion[/u] ! [b]Please share the solution ![/b][/color][/quote]
[color=darkblue][b]O.K.[/b] I like very much this limit. I hope to be and a nice and clear proof ![/color]

[quote="Virgil Nicula"]$ \left\{\begin{array}{c} \{a,b,c\}\subset\mathcal N^* \\
 \\
\ \sqrt c\not\in\mathcal Q\ ,\ a \equal{} 1 \plus{} \left[b\sqrt c\right]\end{array}\right\|$ $ \implies$  $ \lim_{n \to \infty} \left\{(a \plus{} b\sqrt {c})^n\right\} \equal{} 1\ .$[/quote]

[color=darkblue][b]Proof.[/b] $ \left(a \plus{} b\sqrt c\right)^n \equal{} A_n \plus{} B_n\sqrt c$, where $ \left\{A_n,B_n\right\}\subset N^*$ $ \implies$

$ \left(a \minus{} b\sqrt c\right)^n \equal{} A_n \minus{} B_n\sqrt c$. Thus, $ a \minus{} 1 \equal{} [b\sqrt c]\le b\sqrt c < a$ $ \implies$

$ 0 < a \minus{} b\sqrt c\le 1$. But $ \sqrt c\not\in \mathcal Q$ $ \implies$ $ a \minus{} b\sqrt c\ne 1$. Hence $ \boxed {0 < a \minus{} b\sqrt c < 1}$.

Therefore, $ B_n\sqrt c < A_n$ and $ \lim_{n\to\infty}\left(a \minus{} b\sqrt c\right)^n \equal{} 0$,  i.e. $ A_n \minus{} B_n\sqrt c\rightarrow \plus{} 0$.

Exists $ n_1\in N^*$ so that for any $ n\ge n_1$, $ 0 < A_n \minus{} B_n\sqrt c < 1$, i.e.

$ A_n \minus{} 1 < B_n\sqrt c < A_n$. Thus, $ \boxed {\left[B_n\sqrt c\right] \equal{} A_n \minus{} 1}$ for any $ n\ge n_1$.  Therefore,

$ \left\{(a \plus{} b\sqrt c)^n\right\} \equal{}$ $ \left(a \plus{} b\sqrt c\right)^n \minus{} \left[\left(a \plus{} b\sqrt c\right)^n\right] \equal{}$ $ A_n \plus{} B_n\sqrt c \minus{} \left[A_n \plus{} B_n\sqrt c\right] \equal{}$

$ A_n \plus{} B_n\sqrt c \minus{} A_n \minus{} \left[B_n\sqrt c\right] \equal{}$ $ B_n\sqrt c \minus{} A_n \plus{} 1 \equal{}$ $ 1 \minus{} \left(A_n \minus{} B_n\sqrt c\right) \equal{}$

$ 1 \minus{} \left(a \minus{} b\sqrt c\right)^n$. Thus, exists $ n_1$ so that for any $ n\ge n_1$,

$ \left\{(a \plus{} b\sqrt c)^n\right\} \equal{} 1 \minus{} \left(a \minus{} b\sqrt c\right)^n$. In conclusion, $ \lim_{n\to\infty}\left\{\left(a \plus{} b\sqrt c\right)^n\right\} \equal{} 1\ .$[/color]

I thinks it can train as follow
1)$ (a \plus{} b\sqrt {c})^n \plus{} (a \minus{} b\sqrt {c})^n$ is an integer.
2)$ \{)a \plus{} b\sqrt {c})^n\} \equal{} 1 \minus{} \{(a \minus{}b \sqrt {c})^n\} \equal{} 1 \minus{} (a \minus{} b\sqrt {c})^n\to 1$
It is an lovely problem.


Hola amigos... aqui de vuelta... la semana pasada estuve en Buenos Aires en la Olimpiada Rioplatense  como tutor de la delegación peruana y tambien como jurado, fue una nueva y bonita experiencia para mi, a pesar de que nos quedamos algunos días hasta muy tarde en la corrección valió la pena el esfuerzo, la olimpiada se realizó con normalidad, Perú obtuvo cuatro medallas tres de bronce y una de plata, y en términos generales el país con mejores puntajes fue Argentina. Tengo los exámenes que se tomaron a los cuatro niveles, pero no como archivo (osea ni en Word, ni en pdf), creo que no hay problema si pongo los examenes escaneados, es que la verdad no tengo tiempo para pasarlo a LaTeX. En mi siguiente post les envio los exámenes, respondan los que saben si se puede poner los exámenes escaneados.
$Tipe$

Este es el examen del Nivel 1-primer día

$Tipe$

Una pregunta. ¿Cómo son los niveles de la rio? ¿EL nivel 1 es para gente de que edad? ¿Alguien podria poner los otros examenes?

Bueno pues muchas gracias

La olimpiada rioplatense se realiza en cuatro niveles... comenzando desde los menores a los mayores el orden es Nivel A, 1,2, 3 osea el nivel 3 es para los mayores(últimos grados de escolaridad)

aqui les mando el examen del nivel 2, primer día

$Tipe$

[quote="carlosbr"]Ok Type,
gracias por el examen.....

 pero tienes los resultados a detalle de este evento??

o bueno, hay algun miembros aki q tenga esa informacion??
les pediria publicarlos....
gracias..

Carlos Bravo :)
Lima -PERU[/quote]

Por si alguien está interesado: Los resultados: http://www.oma.org.ar/internacional/resultados-omr14.htm


What is the area of the triangle with vertices (1, 4), (3, 1) and (11, 1)?

Either you could do the fast way of noticing that the length of the triangle would be 11-3=8 and the height would be 4-1=3 so the area would be (8x3)/2=12, or you could use the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282647]shoelace theorem[/url].


function
algebra
domain
limit
real analysis
real analysis unsolved

I searched the forums and did not find this (something close but not quite this)...
Let $ 0<a\le 1$.  Show that $ X=Lip_{a}(A)$ is complete where $ Lip_{a}(A)$ is the set of continuous functions on the compact domain $ A$ such that $ \|f\|_{X}=\|f\|_{C}+\sup_{t>0}\frac{\omega(f,t)}{t^{a}}$ is finite ($ C=$ continuous functions and $ \omega(f,t)=\sup_{|h|\le t}|f(x+h)-f(x)|$), could someone please help me prove this.

Here is how I started:
Let $ \{f_{n}\}$ be a Cauchy sequence in $ X$.  As $ \{f_{n}(x)\}$ is Cauchy in $ \mathbb{R}$ for each $ x\in A$, we have that for each $ x$, $ \lim_{n}f_{n}(x) = f(x)$.  I showed that $ \|f_{n}-f\|_{C}\rightarrow 0$ as $ n\rightarrow \infty$.  Also, I can see that $ f \in X$.  Now, I am stuck on showing that $ \sup_{t>0}\frac{\omega(f_{n}-f,t)}{t^{a}}\rightarrow 0$ as $ n\rightarrow \infty$.

Thanks for any help!

Choose a subsequence $ (f_{n_{k}})$ such that $ \|f_{n_{k+1}}-f_{n_{k}}\|_{X}<2^{-k}$. Estimate $ \|f-f_{n}\|_{X}$ using the triangle inequality.
If a Cauchy sequence has a convergent subsequence, then the whole sequence converges.

I think that in the first version of your post you misspelled "Lipschitz" in a way that caused the name to be replaced with "-". (I deleted that copy.)

[quote="baz"]Now, I am stuck on showing that $ \sup_{t>0}\frac{\omega(f_{n}-f,t)}{t^{a}}\rightarrow 0$ as $ n\rightarrow \infty$.[/quote]

mlok,
Thanks for the help, but could you elaborate just a little.  I am not following.  Specifically, if I started the problem correctly, then are you finishing this above idea?

You want to prove that $ \|f_{n}-f\|_{X}\to 0$. Since $ (f_{n})$ is a Cauchy sequence in $ X$, it suffices to prove that $ \|f_{n_{k}}-f\|_{X}\to 0$ for some subsequence $ (f_{n_{k}})$. For any two points $ x',x''$ you have
$ (f(x')-f_{n_{k}}(x'))-(f(x'')-f_{n_{k}}(x''))$ $ =\sum_{j=k}^{\infty}(f_{n_{j+1}}(x')-f_{n_{j}}(x'))-(f_{n_{j+1}}(x'')-f_{n_{j}}(x''))$

Dividing by $ |x'-x''|^{\alpha}$ and taking $ \sup$, you will find that $ \sup_{t>0}t^{-\alpha}\omega(f-f_{n_{k}},t)\le \sum_{j=k}^{\infty}\sup_{t>0}t^{-\alpha}\omega(f_{n_{j+1}}-f_{n_{j}},t)$. The latter sum goes to zero as $ k\to \infty$ provided that $ \sum_{j=1}^{\infty}\|f_{n_{j+1}}-f_{n_{j}}\|_{X}<\infty$.


geometry
geometric transformation
reflection
geometry unsolved

Let $ (O)$ is a circle, $ A, B$ is two points on $ (O)$. Let $ I$ is midpoint of $ AB$, $ (d_1)$ and $ (d_2)$ are two lines pass point $ I$, $ (d_1)$ meet $ (O)$ at $ M$ and $ N$, $ (d_2)$ meet $ (O)$ at $ P$ and $ Q$. Let $ MP$ meet $ AB$ at $ J$, let $ NQ$ meet $ AB$ at $ K$. Prove $ I$ is midpoint of $ JK$.

[quote="plane geometry"]It is butterfly theorem
if we change  P,Q    the conclusion also holds[/quote]
SOLUTION:
Denote Q’ , N’  the reflections of Q , N  w.r.t  OI  respectively
Q’I∩⊙O=P’
It is easy to show  P is the reflection of P w.r.t  OI
Y,N,Q’,I are concyclic
From butterfly thereom we know IX=IY
∠NYB=∠NQ’I =∠NQ’P’=∠N’QP=∠JMN’
But ∠N’XJ=∠NYK   we obtain J,M,X,N’ are concyclic
∠N’JX=∠N’MX   ∠YKN=1/2*(arcQA-arcBN)= 1/2*( arcQA-arcAN’)= 1/2*arcQN’=∠N’MQ
=>  ∠N’JX=∠NKY  =>  XN'=NY  △N’XJ  and  △NYK are congruence  => XJ=KY
=>  IJ=IK

Dear Mathlinkers,
Darij Grinberg wrote a nice article on the Butterfly theorem on his website: http://www.cip.ifi.lmu.de/~grinberg/
Leon Bankoff wrote also an article that can be found in the Monthly.
Sincerely
Jean-Louis


geometry
3D geometry
sphere
parameterization
trigonometry
vector
integration

Hi!

what is the shortest path to go from the northern hemisphere to southern hemisphere? :P  :P

[quote="armpist"]Via Copenhagen.

   Thank you for a wonderful problem!

   M.T.[/quote]

More seriously: do you have a mathematical proof

[quote="grobber"]What do you mean by "path from the northern hemisphere to the southern hemisphere", exactly? :? Did you mean "path from the north pole to the south pole"?[/quote]

Yes, I mean short path from the north pole to the south pole.

Any path that when laid down in 2-dimensions, it appears to be straight. All teh straight paths will be equal in length on a sphere if you have a main primary and terminal point.

Masoud Zargar

In fact, we can use some basic knowledge on differential geometry to prove this. :) (I think this would be a "mathematical" proof. :P )

[u][b]Theorem[/b][/u]
The shortest path between two points on $\mathbb{S}^2$ is an arc on the great circle passing through these two points.

Note: A great circle is a circle obtained by intersecting the sphere with a plane passing through the centre.

[u][b]Proof[/b][/u]
Firstly, a curve on $\mathbb{S}^2$ has the regular parametrization
$\begin{cases}x(t)=\cos\phi(t)\cos\theta(t) \\ y(t)=\cos\phi(t)\sin\theta(t) \\ z(t)=\sin\phi(t) \end{cases}$
where $t\in [a,b]$.

By differentiating  the parametrization with respect to $t$, we get the tangent vector
$\begin{cases}x'(t)=-\theta'(t)\cos\phi(t)\sin\theta(t)-\phi'(t)\sin\phi(t)\cos\theta(t) \\ y'(t)=\theta'(t)\cos\phi(t)\cos\theta(t)-\phi'(t)\sin\phi(t)\sin\theta(t) \\ z'(t)=\phi'(t)\cos\phi(t) \end{cases}$

Therefore, by grouping term, we get
$ds=\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}dt=\sqrt{\phi'(t)^2+\theta'(t)^2\cos^2\phi(t)}dt$

Now, assume that $\theta(a)=\theta(b)=k$, for some real number $k$, by suitably rotating the curve. That is, the end points lie on a plane perpendicular to the $xy$ plane.

If $L$ is the arc length of the curve, then
$L=\int_a^b ds$
$=\int_a^b \sqrt{\phi'(t)^2+\theta'(t)^2\cos^2\phi(t)}dt$
$\geq \int_a^b \sqrt{\phi'(t)^2}dt$
$\geq \int_a^b \phi'(t)dt$
$=\phi(b)-\phi(a)$

Therefore, we have shown that the arc length joining two points on the unit sphere is greater than or equal to $\phi(b)-\phi(a)$, which is the arc on the great circle passing through these two points. (Since the radius of the sphere is $1$.)

Since the curve is arbitrary, the arc on the great circle passing through these two points is the shortest curve between these two points.

On the other hand, if $L=\phi(b)-\phi(a)$, then we have
$\begin{cases} \theta'(t)\cos^2\phi(t)=0, &\forall t\in [a,b] \\ \phi'(t)\geq 0, &\forall t\in [a,b] \end{cases}$

The first case happens if $\theta(t)=C, \forall t\in [a,b]$, for some constant $C$ or $\phi(t)=\frac{\pi}{2}$ or $-\frac{\pi}{2},\forall t\in [a,b]$.

If $\theta(t)$ is a constant for all $t\in [a,b]$, the intersection of the plane $\theta(t)=C$ and the unit sphere gives the great circle passing through the two points. For $\phi(t)=\frac{\pi}{2}$ or $-\frac{\pi}{2},\forall t\in [a,b]$, the curve is degenerated to a single point, and the shortest distance between the two points is $0$. Obviously, the great circle passing the points is the only curve that gives zero diatance.

The second case happens if $\phi(t)$ is non-decreasing $\forall t\in [a,b]$.

Thereofre, we have proved that the shortest path between two points on $\mathbb{S}^2$ is an arc on the great circle passing through these two points. :)

In view of the latest events concerning Danmark,

   I have to retract my solution to the problem.

   In a show of solidarity with Muslim religion, from now

   until Danmark issues a proper apology, the shortest path

   in my atlas will temporarily be via some  neutral and

   non-religious place,  say  the capital of Noth Korea.

   T.Y.

   M.T.


I'm going to mod a rumble game with 8 players.

Here are the [url=http://kevan.org/rumble.cgi?genre=hero&mode=rules]rules[/url]. This game will be [url=http://kevan.org/rumble.cgi?genre=hero&mode=viewall]hero[/url] themed. That means the superpowers you want to pick must come from that 2nd link.

[u]Main Rules[/u]
Discussion can occur anywhere, publically in thread, privately by PM or metaphysically by ESP or offline.

Bids will be done in a blind auction type of way -- send in how much you are willing to pay; if you bid the most, you get that power.

There will be a 7 day deadline for each round. Players not sending in an action will use 75% of their energy (rounding down) to defend themselves. (and if this seems to be happening too much, I'll repace you)

Signed up:
1. perfect628
2. daine
3. ignite168
4. protestant
5. mustafa
6. ars
7. koda
8. [color=white]mysmartmouth[/color] BoesFX

Replacements:
1. me@home
2.

If you have any questions, ask in the thread!

Currently: [b]The game is completed.[/b]

Updated information. Game over.

[img]http://img397.imageshack.us/img397/672/rumble1cg9.png[/img]

I wanna withdraw.
don't feel like playing anymore

Thanks!  :)

IT IS NOW TIME TO SEND IN THE SUPERPOWERS.

Remember, the powers you choose will not be yours, they will be added to the list of powers. You must choose 2, and they must be from [url=http://kevan.org/rumble.cgi?genre=hero&mode=viewall]this[/url] website.

You have 5 days to do so, and I will send in reminder PMs in 3 or 4 days.

All people have sent in powers.

NOTE: You must select TWO POWERS. If you select 1 or 0, I will randomly choose 1 for you, using a random number generator.

If someone posts that they would like to take my spot, I will relinquish it, because I am VERY busy.  If no one wants to take my spot, I'll keep playing, I guess....

awesome.
I don't have much time, but me@home officially replaces koda.
Please send in your powers asap. Thanks  :)

Edit: oh, wait, koda already sent in his stuff....
er, me@home is the first replacement.
Sorry

So the person you were waiting for already sent in?

Well, he sent it in at 3 am this morning (13 hours before this post)
So, yes, the person I was waiting for sent it in, but it's not like i missed his PM or anything.
btw, I will get powers up in a little bit.

Koda has less than 36 hours to give me his submissions.

I'm sorry, imageshack wasn't working well for me last night...
I will upload the pictures when I get home... until then:
The remaining energies are

Ars: 6
Koda: 7
Ignite168: 68
Mustafa:0

[img]http://img172.imageshack.us/img172/8516/rumble5pz2.png[/img]

Remaining players are Ars, ignite168, koda.

By the way:
-A person who has the skill of sword chucks must use sword chucks every attack. It is not optional.
-Koda will not be able to gain any points this round.

Why can't Koda gain any energy? That's unfair

We won't be reaching 700 energy used in the game this turn (more like 690 or so)

Round 3 is over!
[img]http://img213.imageshack.us/img213/183/rumble6qd8.png[/img]

Down to our final 2.
Koda, ignite, I need your submission in quickly.

By the way, my last post was a bit mxied up. I meant we won't be reaching 800, more like 790 (791 to be exact)

see ignite you cannot win you have to attack Koda otherwise he wins but if you do you die which I knew would happen which is why I attacked last time

[img]http://img428.imageshack.us/img428/663/rumble7bz9.png[/img]

And ignite uses only 1 point of energy, to bring the total energy total to 799!!! Which means that the game ends in a tie!

Co-winners: Koda and Ignite!

Good game all!

Is there going to be another game? If so, I sign up!

The only problem with this game is that it takes 2+ weeks before we can even start. After that, we finish fairly quickly...

It seems pretty popular, so it might go faster than 2 weeks.

That was a rummy game.

Now could somebody teach me the rules?


Jeni invested $ \$180.00$ in jeweled pins at the antique sale. She sold them to friends at a $ \$3.00$ discount each. In doing so, she lost $ 25\%$ of her investment. How many pins did she buy?

Since she lost $ 25$% of her $ \$180$ investment when she sold them for $ \$3$ less, she only earned $ \$135$ from selling the pins. Now we know she lost $ \$45$ from selling the pins, and she lost $ \$3$ from each, we know she bought $ \boxed{15}$ pins.
As a check, we see that she bought each one for $ \$12$ and sold each one for $ \$9$.

Wait, this doesn't make sense.
Or rather, this is ambiguous.
Sorry for bumping up this 11 year old thread but "each" is ambiguous.
Like I thought it was each friend but apparently it is each pin.


geometry
circumcircle
trigonometry
geometry proposed

Let $ M$ be a point in triangle $ ABC$ such that $ \angle AMC\equal{}90^{\circ}$, $ \angle AMB\equal{}150^{\circ}$, $ \angle BMC\equal{}120^{\circ}$. The centers of circumcircles of triangles $ AMC,AMB,BMC$ are $ P,Q,R$, respectively. Prove that the area of $ \triangle PQR$ is greater than the area of $ \triangle ABC$.

[color=darkblue]Denote $ \| \begin{array}{ccc} E = QR\cap BM \\
H = RP \cap MC \\
G = AM \cap QP \end{array}\|$. It is clear that $ P \in AC$.

 Then quadrilaterals   $ \begin{array}{cc} %Error. "SMALL" is a bad command.
{QMPA} \\
RMPC \\
QMRB \end{array}$ are all kite-shaped, so:

$ \| \begin{array}{cc} \triangle{AQP}\equiv \triangle{MQP} \\
\triangle{CPR}\equiv \triangle{MPR} \\
\triangle{BQP}\equiv \triangle{MQR} \end{array} \| \Rightarrow \| \begin{array}{cc} S_{MQP} = \frac {1}{2}S_{QMPA} \\
 \\
S_{MPR} = \frac {1}{2}S_{RMPC} \\
 \\
S_{MQR} = \frac {1}{2}S_{QMRB} \end{array} \| \Rightarrow S_{PQR} = \frac {1}{2}S_{AQBRC}$

Then we should prove:

$ \begin{array}{cc} S_{PQR} > S_{ABC} \\
\frac {1}{2}S_{AQBRC} > S_{ABC} \\
\frac {1}{2}(S_{AQB} + S_{BRC} + S_{ABC}) > S_{ABC} \\
S_{AQB} + S_{BRC} > S_{ABC} \end{array}$

But we have $ \| \begin{array}{cc} \triangle{AQB} \| \begin{array}{cc} \angle{AQB} = 60^\circ \\
 \\
\frac {AB}{\sin{150^\circ}} = 2R \end{array} \| \Rightarrow \| \begin{array}{cc} \angle{AQB} = 60^\circ \\
 \\
QA = QB = AB \end{array} \| \\
 \\
\triangle{BRC} \| \begin{array}{cc} \angle{BRC} = 120^\circ \\
 \\
\frac {BC}{\sin{120^\circ}} = 2R \end{array} \| \Rightarrow \| \begin{array}{cc} \angle{BRC} = 120^\circ \\
 \\
BR = RC = \frac {BC}{\sqrt {3}} \end{array} \| \end{array} \|$

 So $ \| \begin{array}{cc} S_{QAB} = \frac {\sqrt {3}}{4}AB^2 \\
 \\
S_{BRC} = \frac {1}{4\sqrt {3}}BC^2 \end{array}$

Then:
$ \\
S_{AQB} + S_{BRC} = \frac {\sqrt {3}}{4}AB^2 + \frac {1}{4\sqrt {3}}BC^2 \stackrel{AM - GM}{\ge} \frac {1}{2} AB\cdot BC > S_{ABC}$

$ \mathrm{\LARGE{Q.E.D.}}$[/color]

Let $ M\in Int{\triangle {ABC}}$ and P, Q, R centres of circumcircles of triangles BMC, CMA, AMB. Let $ QR\cap MA=\{A_1\}, RP\cap MB=\{B_1\}, PQ\cap MC=\{C_1\}.$
$ A_1$=midpoint to AM, $ B_1$=midpoint to BM and $ C_1$=midpoint to CM. Results $ \triangle{A_1B_1C_1}=$podar triangle of point M
$ \Rightarrow Area{\triangle{A_1B_1C_1}}=\frac{R^2-OM^2}{4.R^2}.Area{\triangle{PQR}}\Rightarrow \\
 \Rightarrow Area{\triangle{A_1B_1C_1}}\le \frac{1}{4}.Area{\triangle{PQR}}$    (1)
And $ Area{\triangle{A_1B_1C_1}}=\frac{1}{4}. Area{\triangle{ABC}}$                (2)
 beaucause $ \triangle {A_1B_1C_1} \sim \triangle{ABC}$, and $ \frac{A_1B_1}{AB}=\frac{A_1C_1}{AC}=\frac{B_1C_1}{BC}=\frac{1}{2}.$
(1) and (2)$ \Rightarrow Area{\triangle{ABC}\le Area{PQR}.}$


calculus
integration
number theory
least common multiple

Prove that there are no rational numbers u, v,w such that u^2+v^2+w^2 = 7.

$ u=\frac{x}{a}, v=\frac{y}{b}, w=\frac{z}{c}$, where $ a,b,c,x,y,z$ integral and the fractions are in lowest terms.

$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}= 7$

If $ a=b=c$, then $ x^{2}+y^{2}+z^{2}=7a^{2}$. Looking at it mod 4, there are no solutions ($ x,y,z,a$ cannot all be even).

However, if $ a=b=c$ is not true, then... blah I can't explain it. <_< I fail.

if the fractions aren't necesssarily in lowest terms, then you can assume that they all have the same denominator

I don't think your solution works though.
I'm still stumped.

Combine your two ideas... but try it modulo a larger number...

[quote="Lazarus"] $ x^{2}+y^{2}+z^{2}=7a^{2}$. [/quote]
If you can prove there are no solution for the  above, you are done.[hide] (as some one said, assume $ a$ is LCM of all the denominators etc)

Remember for an odd number, its square is always 1 modulo 8.
if $ a$ is odd, no solutions
[hide]Look at  mod 8, LHS can be only 1 or 3   while RHS is 7[/hide]
if $ a$ is even , all (x,y,z) have to be even  too

[hide](else, LHS is 2 mod 4 and RHS is 0 mod 4) [/hide]
Infinite decent ...[/hide]

Oh Christ wow the solution was right in front of me :[

Thanks Gyan!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $G$ is a groupe and $H$ is a subgroup of infinite index. Suppose that there are $H_1,...,H_k$ subgroups of $G$ such that $G$ can be covered by the union of $H$ and all $H_i$. Prove that $G$ can be also covered by the union of $H_i$.

that would be becuse  the following is true

if $x \in H$ then also $x \in$ reunion of $H_i$

hah, well that's exactly what the problem is asking us to prove

You can an answer here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=667775#667775


How do you create an arc without drawing an entire circle? Thanks.

Use either arc() or Arc() command (Arc() is a bit more accurate but requires graph package):
[asy]pair center=(0,0);
real radius=3, startangle=30, endangle=120;
draw(Arc(center,radius,startangle,endangle ));[/asy]

Based on your code, the center is the center of the imaginary circle, the radius is the radius of the imaginary circle, and startangle and endangle are the points on the imaginary circle at 30 degrees and 120 angles Am I right?


$ n \ge 2$
Prove

$ \sum_{k\equal{}0}^{n\minus{}2} \frac{1}{k!} \left( \frac{1}{2!} \minus{} \frac{1}{3!} \plus{} \ldots \plus{} (\minus{}1)^{n\minus{}k} \frac{1}{(n\minus{}k)!}\right)\equal{}1\minus{} \frac{1}{n!}$

Simple induction. The difference between LHS for $ n\plus{}1$ and $ n$ is $ (1\minus{}1)^{n\plus{}1} \plus{} \frac {1} {n!} \minus{} \frac {1} {(n\plus{}1)!}$.


If $x$, $y$, and $z$ are positive real numbers and $x=\frac{12-yz}{y+z}$, find the maximum value of $xyz$.

[hide]
Rearrange the equation as
\[xy+xz+yz = 12 \]
From AM-GM,
\[\frac{xy+xz+yz}{3}\geq (xyz)^{\frac{2}{3}}\Longrightarrow xyz \leq 8 \]
So the maximum value is $8$.
[/hide]


Find all strictly increasing functions $f: R\rightarrow R$ satisfying
\[f(x+f(y))=f(x+y)+2005\]

Since $f$ is injective and $f(x+f(y))=f(y+f(x))$ we conclude $x+f(y)=y+f(x)$ for all $x,y$. I.e. $f(x)-x=const$.


integration
real analysis
function
probability
calculus
real analysis unsolved

Let $ \alpha,\beta$ be two strictly positive real numbers such that $ \alpha \plus{} \beta < n$ for some positive integer $ n$. Prove that there exists a constant $ c$ which depends on $ \alpha,\beta,n$ only such that for every $ y\in\mathbb{R}^n$,
\[ |y|^{\alpha \plus{} \beta \minus{} n} \equal{} c\int_{\mathbb{R}^n}dx |x|^{\alpha \minus{} n}|x \minus{} y|^{\beta \minus{} n}.
\]

Since that is a convolution, one possible path is to look at Fourier transforms. Now, the function $ |x|^{\alpha\minus{}n}$ isn't in $ L^1$ (or any $ L^p$ space, for that matter) but its Fourier transform does make sense as a tempered distribution, and that transform is $ c_{\alpha}|\xi|^{\minus{}\alpha}.$

Accept that result, and the identity above is formally true.

We're convolving the functions with Fourier transforms $ c|\xi|^{\minus{}\alpha}$ and $ c|\xi|^{\minus{}\beta}.$ The transform of the convolution is the product of the transforms, namely $ c|\xi|^{\minus{}(\alpha\plus{}\beta)}.$ The inverse transform of that is then $ c|x|^{\alpha\plus{}\beta\minus{}n}.$

But as I said, this is a formal computation (and with care we could supply the exact values of the constants); some work on the actual convergence would probably be in order.

Yes I thought it could work that way (with the obvious fact that it is a convolution, the Fourier transform behaves well).
But, I used an approximation argument to ensure the computations make sense. There must exist a more straightforward way to prove this formula, and I(Well, it is the hint proposed in Analysis by Lieb and Loss) possibly found an "elementary" solution suggested by the following identity :
\[ c_{\alpha}|x|^{\minus{}\alpha}\equal{}\int_{0}^{\infty}d\mu\exp\{\minus{}\mu\pi |x|^2\}\mu^{\alpha/2\minus{}1}.\]
But I'm not quite sure I then apply Fubini in the easiest way. Can you take advantage of this identity to provide a "natural" proof of the statement?

If $ f(x)\equal{}e^{\minus{}a|x|^2}$ and $ g(x)\equal{}e^{\minus{}b|x|^2},$ then $ f*g(x)\equal{}ce^{\minus{}d|x|^2}.$ (In probability, the sum of independent normal distributions is normal.) I'll leave to you the computation of how $ c$ and $ d$ depend on $ a$ and $ b.$

Yes, I got that point,
\[ c \equal{} \left(\frac {\pi}{a \plus{} b}\right)^{n/2}\,\,\textrm{and}\,\,d \equal{} \frac {ab}{a \plus{} b}.
\]
But I don't see how to compute the integral
\[ \int_0^\infty d\mu \int_0^\infty d\lambda \left(\frac {\lambda\mu}{\lambda \plus{} \mu}\right)^{n/2}\lambda^{ \minus{} \alpha/2 \minus{} 1}\mu^{ \minus{} \beta/2 \minus{} 1}\exp\left[\frac { \minus{} \lambda\mu\pi}{\lambda \plus{} \mu}|y|^2\right].
\]
EDIT : Sometimes, I should realise that my problems are trivial... Ok, it is just a matter of rescaling... Anyway, the value of the constant is not obvious.
EDIT2 : Thank you!


function
real analysis
real analysis solved

determine all continuous functions f:R-->R with the property that:
f(f(x))+f(x)=2x+a; for all x real and a is a real constant which is given.

cheers!

A very hard problem! Let me denote f^n be the iterated composition of f with itself if n>0 and of the inverse of f with itself if n<0. First, let us prove that f is bijective. We have f^4(x)=5f^2(x)-4x-2a (1). Of course, f is injective, so f^2 is strictly increasing. From (1) and the monotony it follows that f^2 has infinite limit at infinity and minus infinit limit at minus infinity. THus, f^2 is surjective and f is also surjective. Thus, f is bijective.
   The case a<>0 is not very difficult. Indeed, in this case f doesn't have fixed points. Let g(x)=f(x)-x. Then g has constant sign on R. Define x_n=g(f^n(x)) for a certain x. It is easy to see that x_n=(a/6-x_1/2)(-2)^n+a/3 and since all terms of the sequence have the same sign we must have a/6=x_1/2. Thus, g(f(x))=a/3 for all x and also g(f(x))=a-2g(x). It follows that g(x)=a/3 for all x and thus f(x)=x+a/3 for all x.
    Suppose now that a=0. The first case is when f is increasing. In this case it is easy to prove that f(x)=x for all x( if f(x)>x, then f((f(x))>x, false; the same if f(x)<x). So, assume f is decreasing. Let g denote the inverse of f. Then g^2(x)=(x+g(x))/2 and g is decreasing. From here we find that g^n(x)=(2^n-(-1)^n)/[3*2^(n-1)] g(x)+(2^(n-1)-(-1)^(n-1))/[3*2^(n-1)]x. For fixed x the above sequence (let us call it a_n)tends to (2g(x)+x)/3. But g(a_n)=a_(n+1), so the limit must be a fixed point of g. Since g is decreasing, it has at most one fixed point. Thus, the function h(x)=(2g(x)+x)/3 is constant, let it be t. So, g(x)=(3t-x)/2 and also f(x)=3t-2x. So, in this case the function is of the form b-2x, with real b. These are all the functions required.


The square 0 ≤ x ≤ n, 0 ≤ y ≤ n has (n+1)^2 lattice points. In how many ways can each of these points be colored red or blue, so that each unit square has exactly two red vertices?

Can it be solved by considering for example the strip 0 <= x <= n, 0 <= y <= 1. Now one finds the number of ways to color that strip of unit squares and then one translates that coloring two steps upwards through the rest of the grid. One does that same for the strips
0 <= x <= 1, 0 <= y <= n; 0 <= x <= n, n-1 <= y <= n; n-1 <= x <= n, 0 <= y <= n (by translating two steps in those other directions through the whole grid).

Am I exhibiting all colorings in that way?

When I say coloring, I mean the coloring as in the problem: so that a unit square has exactly two vertices of one colour.


Just expand and amgm. How refined is it?!

generalization:
Let
$x_i > 0(i = 1,2, \cdots ,n),n \in N,n > 1,r \ge \frac{n}{{n - 1}}$,
that$(\sum\limits_{i = 1}^n {x_i } )^r \ge \sum\limits_{i = 1}^n {x_i^r } + (n^r - n)(x_1 x_2 \cdots x_n )^{\frac{r}{n}}$


Please discuss Harry Potter stuff in the Game and Fun factory. There is thread for such topics.

Mod: Please lock it.

[quote="Z = mod 2Z"]If you have finished reading the latest Harry Potter book, do you have any idea who R.A.B. is?It's driving me crazy.[/quote]
i think it is:
[hide]Regulas Black[/hide]

My theory is that RAB is regulus black...sirius's brother. As he was a Death Eater, the note said "to The Dark Lord". Also, in the 5th book, while cleaning out the drawing room in Grimmauld Place, they found a heavy locket that none of them could open, among many other things. This would mean that one of the horcruxes is in Harry's possesion.

Ok. Post this kind of things [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=294927#p294927]here[/url]

[hide]I hope it isn't Remulus, cause then Hermione shoulda got that....[/hide]

it could be a group of people, or that amy b....something that tom tookl into the cave.  there is a whole bunch of stuff on mugglenet

ok i dont think its regulas r however u spell his name because if u think about it the note says he will b dead long before voldmort reads it and he was in prizon for a while so if he was in prizon he couldnt have done it and before he was out of prizon he wouldnt of been dieing so..............

this is just a detailed discussion on one area of HBP

[quote="star99"]looks like you haven't read the book.........an anti-witchcrafter?[/quote]
looks like who hasn't read the book? me?

[quote="star99"]oh.  so who do you think RAB is?  I've also been wondering, what is everybody's favorite harry potter fan site?[/quote]
Black or Bones. Or Rowling will creat a new figure.

in the third post i said [hide]Regulas Black[/hide]

[quote="star99"]she said in an interview that no new characters will be introduced in book 7[/quote]
In [b]which[/b] interview?

Regulus didn't go to Azkaban as far as I remember. He hid in a remote cabin after betraying Voldemort. He was killed by death eaters in that cabin. My guess is that his betrayal was, if only in part, the taking of the locket horcrux.

[quote="star99"]it's an awesome HP fan site.  http://www.mugglenet.com  you can even join forums to chat about the books.  there is also some hogwarts/hogsmeade thing, but I don't quite get that[/quote]

Yeah mugglenet's my favorite HP website too soemtimes i check out [url]http://www.jkrowling.com[/url] as well. She has a place where it lists a bunch of fansites..

mungdungus could have actually stole and sold the horcrux which could possibly be a death eater but i believe it is regulus black we have reached a general assumption of that but the question is how did he find out about horcruxes and why did he betray voldy

i dont think its regulus... What if R.A.B. is 3 people? 3 Death Eaters... Rockwood, Avery, and Narcissa. She seemed pretty angry with Voldemort, Avery was tortured by voldy in book 5, and rockwood was, like snape, pretending to pass info to voldemort and passing info to the order instead? theres also the possibility that it is Regulus, and Crookshanks is Regulus. Sirius did say in book 3 that it took Crookshanks some time to trust him as a dog, and when harry wanted to kill black in the 3rd book crookshanks jumped up on his back. What if, like Pettigrew, he was believed killed by voldemort and turned into an animagus?

Animagus or not evidence points to regulus.

The initials in the books translated into other languages always correspond to Regulus.
Moreover for every other name, there is some translation or another where initials do not correspond to that name.

More than one qualified wizard could not have gone together (Remeber the Boat)

Two were atleast needed(Remeber the Potion)

So probably Kreacher went with Regulus

I think you are in for a major disappointment. Let's get this  straight: Rowling has definitely lost control, what with the death of poor old Dumbledore and all. Let's get this straight: TOOTP was a bad book, HBP was a horrible book, and the last one will be downright stinking. Why at the rate at which she is handing out gimmicks, I would not be surprised if RAB was Ron Weasley, or Dudley Dursley, or Viktor Crum, or Rufus Scrimgeor, or Tom Riddle himself! If you are expecting the 'best book ever written', prepare yourself for a shock.

I read somewhere like couple of weeks after HBP came out that Rowling had debunked the rumor that it was Regulus Black. I am not sure if it was on TV or on mugglenet. I think RAB is most likely one of the people in old Order of Phenix in the pic that Moody had. MOst likely RAB is either dead or has been tortured. OR and even greater suggestion, Voldermort may have finally gotten smart ( :o  :rotfl:  ) and figured out after MR.MAlfoy distroyed the first Horocrux that people will try to find his Horucrux and made that fake one just to send people like Dumbedore on a wild goose chase.

BTW: Did anyone notice that how all the Time Turner that [u]ever[/u] existed in the Magical world were all in the same place and were all distroyed at the same time when everyone was in the Department of Mysteries? I don't know about you, but it seems to me that this one way too convenient plot device, hence it squashes all time turning theories.

BTW2: Has anyone realized that if HArry had more than maybe half a brain he would realize by know he could just go to the Room of REquirment and just wish for a room with equipment to find and Destroy Voldermort, or better yet a room full of time turners so he could just warp back in time and makes sure Voldermort died before he could do harm??[/u]

I think that it is Regulas Black.

You should go to [url]http://www.dumbledoreisnotdead.com/[/url]

[quote="thatin"]Regulus Black or any other character (using a nickname)[/quote]

You basically just said "anyone"... Thanks for that insightful bump!

I saw on Yahoo news that two major characters will get killed in the last book. (Yay! :P ) If they were Ron and Hermione, do you think Harry would kill himself?


[img]http://i23.photobucket.com/albums/b370/survivorxxi/chocolate.png[/img]

Use the numbers 1 through 16 to complete the equations.

Each number is only used once.
Each row is a math equation. Each column is a math equation.
Remember that multiplication and division are performed before addition and subtraction.

Thanks!

It should. This one looks pretty hard with the use of the order of operations.

how did you do that? DId you use guess and check?

I was also in the process to solve the same. First, I found out 8 equations in 16 variables with the additional regulation of that all the 16 variables take diferent integral values for 1 to 16 and none of them should repeat. Second, I wrote a computer program to arrive at the result.

And, last but not the least. How did you do it?  :?

                                                                            - Shamik
********************************************************************
"IT is raining." I know 'IT' is a pronoun. Can someone let me know -- 'WHAT' is raining?
********************************************************************

That was just a question to keep guys confused and interested.

"I need IT."  If someone asks, "What do you need?" then there can be specific replies like 'I need a pen' or "I need a book' et al.

Similarly, if someone says, 'IT is raining'. Can you tell me 'WHAT' is raining?

I am still waiting for the efficeint process to solve the problem!!!!!

                                                                     -- Shamik

[quote="Shamik Banerjee"]Second, I wrote a computer program to arrive at the result.[/quote]
I also used a computer program to find the answer. I don't know any easy method of solving without a computer.

[quote="prototypex24"]Ithink this should be in the Puzzles and Brainteasers Forum[/quote]

it is.  :)

It was moved. Try doing some thinking before posting something completely worthless. I don't think that anyone is stupid enough to say "This should be in the Puzzles & Brainteasers Forum" if it's already there.


hallo,
Let $x$ and $y$ be two coprime integers. solve the system:
$(1)$ $2x+y=3z$ and $(2)$ $x^{3}+2y^{3}=3z^{3}$. :)

From (1) we have y=3z-2x, from (2) $-5x^{3}+24zx^{2}-36z^{2}x+17z^{3}=0$.  Let $t=5x/z$, then $t^{3}-24t^{2}+180t-425=0$. Because t is algebraic integer and rational, t is integer and $t|425$. $t=5$ is root, another roots irrational. It give $x=z=y$.

But because it's said that $x,y$ are coprime, the only solution is $x=y=z=1$.


function
calculus
derivative
calculus computations

What's the sum of $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$?

[quote="Danbert"]\[ A=\frac12+\frac44+\frac98+\frac{16}{16}+\dots+\frac{n^2}{2^n}+\dots  \]
\[ 2A=1+\frac42+\frac94+\frac{16}8+\frac{25}{16}+\dots+\frac{(n+1)^2}{2^n}+\dots  \]
\[ 2A-A=A=1+\frac{4-1}2+\frac{9-4}4+\frac{16-9}8+\dots+\frac{(n+1)^2-n^2}{2^n}+\dots  \]
\[ A=\sum_{n=0}^\infty {\frac{2n+1}{2^n}}=\sum_{n=0}^\infty{\frac n{2^{n-1}}}+\sum_{n=0}^\infty{\frac1{2^n}}=\sum_{n=0}^\infty{\frac n{2^{n-1}}}+2  \]

Repeat this process with the last summation above.

\[ \text{Sum}=2+\left(2+\sum_{n=0}^\infty {\frac1{2^n}}\right)=2+(2+2)=\boxed{6}  \]

Any ideas on a [i]nice[/i] way to generalize this to $\sum_{n=0}^\infty {\frac{n^a}{b^n} }$ ?[/quote]

Thanks. I've got the same answer with you...though the answer in my book is 10. I thought we are right! :)

Another path to the same conclusion - a generating function idea.

In everything that happens below, assume that $|x|<1,$ justifying all of the power series manipulations.

Let $f_0(x)=\sum_{n=0}^{\infty}x^n=\frac1{1-x}.$

Differentiate, then multiply by $x.$

$f_1(x)=xf_0'(x)=\sum_{n=1}^{\infty}nx^n=\frac x{(1-x)^2}.$

$f_2(x)=xf_1'(x)=\sum_{n=1}^{\infty}n^2x^n=x[(1-x)^{-2}+2x(1-x)^{-3}]$

$=(x+x^2)(1-x)^{-3}.$

We can keep going. $f_{k+1}(x)=xf_k'(x)$ and $f_k(x)=\sum_{n=1}^{\infty} n^kx^n.$

The derivatives must be calculated recursively, and it does get messy.

The next step should be $f_3(x)=x(1+4x+x^2)(1-x)^{-4}.$

$\sum_{n=1}^{\infty}\frac n{2^n}=f_1\left(\frac12\right)=2.$

$\sum_{n=1}^{\infty}\frac{n^2}{2^n}=f_2\left(\frac12\right)=6.$

$\sum_{n=1}^{\infty}\frac{n^3}{2^n}=f_3\left(\frac12\right)=26.$


The City Council has nine members, who serve on different committees. Each committee has three members and no two members serve together on more than one committee. Determine the largest possible number of committees.

[hide]For each pair of people, there can be only one committee. There are 9 choose 2 = 36 pairs of people. Since each committee is triple-counted (if there is a committee XYZ, we are counting it once for each of three pairs: XY, XZ, and YZ), there can be at most 12 committees. An example, with people A-I, is the following:
ABC, ADE, AFG, AHI, BDH, BEF, BGI, CDG, CEI, CFH, DFI, EGH[/hide]

EDITED TO ANSWER THE BELOW QUESTION

[quote="jb05"][hide]For each pair of people, there can be only one committee. There are 9 choose 2 = 36 people. Since each committee is triple-counted, there can be at most 12 committees. An example, with people A-I, is the following:
ABC, ADE, AFG, AHI, BDH, BEF, BGI, CDG, CEI, CFH, DFI, EGH[/hide][/quote]

Can you explain why each committee is triple-counted?  I see where you get the logic "For each pair of people, there can be only one committee."  But I don't see why we divide by 3.  36 is obviously too large, but why do we divide by 3, instead of some other number?


floor function
limit
real analysis
real analysis unsolved

find all the limit points of the $n\pi-\lfloor n\pi \rfloor$

Lets do something more general, that is, instead of $\pi$ let $\alpha$ is not $\mathbb{Q}$. We will show that every real number $x \in [0,1]$ is a limit point of the sequence $\{n\alpha - [n\alpha]\}$. I assume that you know that there exist $p_{n} \in \mathbb{Z}$ and $q_{n} \in \mathbb{N}$ such that  $0 < \alpha - \frac{p_{n}}{q_{n}} < \frac{1}{q_{n}^{2}}$. Since $\lim_{n \to  \infty} q_{n} = +\infty , \lim_{n \to  \infty}(\alpha q_{n} - p_{n}) = 0$. Let $x \in (0,1)$ and let $\epsilon > 0$ be so small that $0 < x - \epsilon < x + \epsilon < 1$. Now suppose that $n_{1}$ is so large that
$0 < \alpha q_{n_{1}} - p_{n_{1}} < \frac{1}{q_{n_{1}}} < \epsilon$
There is $n_{0} \in \mathbb{N}$ satisfying
(1) $n_{0}(\alpha q_{n_{1}} - p_{n_{1}}) \in (x - \epsilon, x + \epsilon)$.

It follows from (1) that $[n_{0} \alpha q_{n_{1}} - \n_{0} p_{n_{1}}] = 0$ thats is $n_{0} p_{n_{1}} = [n_{0} \alpha q_{n_{1}}].$ Therefore the term $n_{0} \alpha q_{n_{1}} - [n_{0} \alpha q_{n_{1}}]$ from the range of our sequence belongs to the interval $(x - \epsilon, x + \epsilon)$ wich means that $x$ is a limit point of the sequence under consideration. Similaryly, you can show that 0 and 1 are also limit points.


If the product
$ \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdot\ldots\cdot\frac{a}{b}\equal{}9$,
what is the sum of $ a$ and $ b$?

This cancels to $ \frac{a}{3}\equal{}9 \implies a\equal{}27$, so $ a\plus{}b\equal{}27\plus{}26\equal{}\boxed{53}$.

Should be $ \frac{a}{2}\equal{}9\Rightarrow a\equal{}18$, so $ 18\plus{}17\equal{}\boxed{35}$


algebra
function
domain
number theory
Ring Theory
absolute value
superior algebra

Prove that the following rings are euclidian:

       $\mathbb{Z}[\sqrt{3}]$, $\mathbb{Z}[\sqrt{2}]$, $\mathbb{Z}[i\sqrt{2}]$, $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$.

I can't do the last.  I'll do the two first.  The third one is almost the same but easier.

Suppose you want to do euclidean division with a and b\neq 0
we want$a=b q+r$ with $N(r) < N( b)$

due to multiplicativity this means $\frac{a}{b}=q+z$ with $N(z)<N(b)$

suppose $\frac{a}{b}=e+f\sqrt{m}$ with $e,f\in \mathbb{Q}$

now you can approximate e and f with integers r and s respectively, with distance at most $\frac{1}{2}$

now call $q=r+s\sqrt{m}$

now is $\frac{a}{b}-q=u+v\sqrt{m}=z$ with u and v at most a half

this means $N(z)\geq -\frac{m}{4}$
and $N(z)\leq \frac{1}{4}$

if you take m=2 or 3 you get the two first questions

[quote]we want with  a=bq+r with N(r)<N(b)
due to multiplicativity this means  a/b=q+z with  N(z)<N(b)
[/quote]

Why? I am a little confused with the relation between z and r

I'll stop running away from the fact and try to face it like a man  :( .
My whole understanding of number theory is falling to pieces.

I checked Hardy and Wright but to no avail .  So yeah, please anyone, who can repair the hole in my proof

when i can write $\frac{a}{b}$ as 'integer' q +r with $|N(r)|<1$
that still doesn't mean
$a=b q+b r$ with $b r$ 'integer'

 :|  i had never noticed this problem

emm, why do you need to br to be integer? Isn't it okay as long as N(r) < N(b)? Anyway may I know what is your N?

[quote="fredbel6"]definition of euclidean domain :

a ring without zero divisors (let us assume commutative ) D such that there is map V from the nonzero elements D to the naturals

such that :
for all a and b in D, b nonzero, there is a q and r  such that $a=b q + r$ with V(r)<V(b) of r=0

you really need r, and thus in my previous post b r , to be 'integer', thus in the ring D

the N i mean is , , when working with elements of form $a+b \sqrt{m}$ equal to $a^2-b^2 m$
when taking its absolute value i intend to make it the map V[/quote]

I still don't quite understand.... like for the first ring, isn't it composed of $a+b\sqrt{3}$? If yes, can't the remainder be of the same form? Maybe I misunderstood your meaning of integer.
For N, I came up the same map, but it doesn't seem like N(r) < N(b).... (even for the case the divisor is integer, or maybe I'm missing something trivial?)

[quote="Soarer"]

I still don't quite understand.... like for the first ring, isn't it composed of $a+b\sqrt{3}$? If yes, can't the remainder be of the same form? Maybe I misunderstood your meaning of integer.
For N, I came up the same map, but it doesn't seem like N(r) < N(b).... (even for the case the divisor is integer, or maybe I'm missing something trivial?)[/quote]

yes for the first ring it is composed of $a+b\sqrt{3}$ with a, b in $\mathbb{Z}$ ( so not rational, integer!!)
let me call that ring R

well, if i give you a and b, b nonzero, both in R, can you give me q and r in R (so in R both of them!)

such that $a=b q +r$

now my trick was : calculate :$\frac{a}{b}$ , this will still be of form $u+v\sqrt{m }$, but u  and v will probably not be in $\mathbb{Z}$ anymore
I wanted to approximate it with an element $q$, such that $|N(a/b-q)|< 1$
now realising that $N(a b) = N(a) N(b)$

this implies $a=b q+j$ with $|N(j)|<|N(b)|$ or $j=0$

but now the problem is : j is of form $a+b\sqrt{3}$ alright, but is it in R

i apologise for the confusion, one can either be talking in algebraic number theory about the ring of integers of a field, or just plainly$\mathbb{Z}$

So uhm has anyone managed to see what I did wrong?  I would really like to know  :oops:


is there chicken mcnugget theorem with 3 numbers?

if so, what is it?

The chicken mcnugget theorem means that the greatest number that cannot be produced by x or y. The formula is xy-(x+y).  Example: if x=4 and y=6, 24-10=14 is the largest number that can't be produced by a combination of 4's and 6's.

Fantasy, for 3 numbers, I think that try it with 2 numbers and use the answer and plug it in again with the 3rd number. :maybe:

i know that, jongao...

i tried 4, 6, and 10.

4*6-10=14

14*10-24=116

but 116 can be made with bunch of 4's.

@nick

see the izzy's question...

was it really necessary to start a new topic?
I just asked that in the other one.....
See Frobenius number in Mathworld.
4,6,10
any number greater than 24 -10 = 14 can be made with 4,6, so our number < 14. 14 = 10 + 4 so we test one less which clearly works. the answer is then [b]13[/b]

On a random note when are USAMTS registration and soln's due again??

oh...

thanks!

sorry for creating a new topic  :blush:


induction
Ramsey Theory
combinatorics
pattern
Poland
Permutation patterns
IMO Shortlist

Let $S_n$ be the number of sequences $(a_1, a_2, \ldots, a_n),$ where $a_i \in \{0,1\},$ in which no six consecutive blocks are equal. Prove that $S_n \rightarrow \infty$ when $n \rightarrow \infty.$

We use the following lemma: there's an infinite sequence $a_1a_2a_3\ldots$ with no block of the form $bbb$, where $b$ is a block (this has appeared before in the forum).

If the problem is false, then $S_n<C$ for arbitrarily large $n$ and some constant $C$. Now take a sequence with no $bbb$, and look at its blocks of length $n$ for some large enough $n$ such that $S_n<C$ (in particular $n>4C$, for example). We say that $b_1=a_1 a_2 \ldots a_n, b_2=a_2 a_3 \ldots a_{n+1},$ etc. Obviously, among $b_1, b_2, \ldots, b_{C+1}$ there must be two identical sequences, by pigeonhole. So there are two sequences at most $C$ elements apart. In other words, we have $b_i, b_j$ with $j \leq i+C$, so $a_j=a_i, a_{j+1}=a_{i+1},$ etc. The first part $a_i a_{i+1} \ldots a_j=X$ repeats as long as the two sequences overlap, so the sequence has the form $XX \ldots XR$ where $R$ is of length at most $C$. Since $n$ can be as large as we want, we can get as many $X$'s as we want, in particular at least $3$, which contradicts the fact that our sequence had no block of the form $bbb$.

Note that this is much stronger (we have proved $S_n \rightarrow \infty$ for $S_n$ with no three consecutive blocks, instead of $6$).

Note that you can prove by induction that$S_n\geq\ \frac{3}{2}S_{n-1}$,using the recurence:$S_n=2S_{n-1}-\sum 6^kS_{n+1-6k}$

Denote by $A_n$ the number of sequences with the properties given in the problem. Their number will be $S_n$. It is enough for us to prove that  $S_n \geq (\frac{3}{2})^n$, as $\lim_{j \to \infty}i^j= \infty$ when $i>1$. Or in another way said we have to prove the following recursion $S_n \geq \frac{3}{2}S_{n-1}$.

We will use induction on $n$ to prove that. It is obvious for $n=1$. Let $S_k \geq \frac{3}{2}S_{k-1}$ for $\forall k=2, 3...n$. Then $S_{n-1} \leq \frac{2}{3}S_n \Rightarrow S_k \leq  (\frac{2}{3})^{n-k}$.

Let $B_{n+1}$ be the set of all sequences formed up by adding 0 or 1 to a sequence from the set $A_n$. Their number is $2S_n$. These sequences that are not in $A_{n+1}$ will certainly end on six equal consecutive blocks. That is to say their structure will be the following: They will begin with the sequence $A_{n+1-6k}$ for some $k \in \{ 1,2...n \}$ and continue with k-member sequence from 0's and 1's  repeated 6 times. Thus for $\forall$ fixed
$k \in \{ 1,2...n \}$ the number of sequences with the upper property will be $S_{n+1-6k}.2^k$. Hence the number of sequences in $B_{n+1}$ which won't have the property, given in the problem, won't surpass:
$\sum_{k=1}^{n} S_{n+1-6k}.2^k \leq \sum_{k=1}^{n}  (\frac{2}{3})^{n-(n+1-6k)}.S_n.2^k \leq S_n \sum_{k=1}^{\infty} (\frac{2}{3})^{6k-1}.2^k= 3. \frac{1}{(\frac{3}{2})^6-2}.S_n < \frac{1}{2}S_n$.

Hence $A_{n+1}$ contains at least $2S_n - \frac{1}{2}S_n= \frac{3}{2}S_n$ different sequences that satisfy the problem with which the induction is completed and so the problem is solved.

[quote=xirti][quote="Severius"]We use the following lemma: there's an infinite sequence $a_{1}a_{2}a_{3}\ldots$ with no block of the form $bbb$, where $b$ is a block (this has appeared before in the forum).

[/quote]

I searched for it and I found nothing. Can you post a link for this lemma?[/quote]

Even I did not find it here (though I must say that I never searched for it extensively).
But there is something where you can read about it, and this was told to me by my teacher.

Such words where a block $b$ doesn't occur more than twice consecutively are called [i]cube free words.[/i]

Axel Thue's document on cube free words and further extensions of it describe a lot of these things.

So on internet you can search for : [u]"[i]Axel Thue cube free words[/i]"[/u] and hopefully it will show you many results from which you can read more about the topic.

Here is an attached document on [u]cube free words on 2 alphabets[/u], and the theorem which is apparently of most interest in the view of this problem is [b]Theorem 3[/b] from the paper


Αν $ a,b,c,d,e,f>0$ αποδείξτε ότι $ \frac{ab}{a\plus{}b}\plus{}\frac{cd}{c\plus{}d}\plus{}\frac{ef}{e\plus{}f}\leq\frac{(a\plus{}c\plus{}e)(b\plus{}d\plus{}f)}{a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f}$

Nice and easy one  :)  Just a hint

[hide]$ \frac{xy}{x\plus{}y}\plus{}\frac{zw}{z\plus{}w}\leq\frac{(x\plus{}z)(y\plus{}w)}{x\plus{}y\plus{}z\plus{}w}$ [/hide]


Find a  polynomial $P(x,y)$ with real coefficients which satisfies the following properties:
(1) $P(x,y)\geq 0$ for all $x,y \in \mathbb{R}$, and
(2) $P(x,y)$ cannot be expressed as a sum of squares of real polynomials.

$x^2(x^2 - y^2)(x^2 - 1) + y^2(y^2 - 1)(y^2 - x^2) + (1 - x^2)(1 - y^2)$ - that's quite a well known example.

I'm wondering how to come up with it though :D

Need to clarify why (1) and (2) are satisfied, especially (2).  :)


Find the units digit of the decimal expansion of \[(15 \plus{} \sqrt{220})^{19} \plus{} (15 \plus{} \sqrt{220})^{82}.\]
$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 9\qquad
\textbf{(E)}\ \text{none of these}$

[hide="Solution"]
Since $ (15 \minus{} \sqrt {220})^{19} \plus{} (15 \minus{} \sqrt {220})^{82}$, we add this to the given expression to cancel out the irrational terms and get

$ \left(2\cdot15^{19} \plus{} 2\dbinom{19}215^{17}\cdot220^2 \plus{} \cdot\right) \plus{} \left(2\cdot15^{82} \plus{} 2\dbinom{82}215^{80}\cdot220^2 \plus{} \cdots\right)$

$ \equiv0\pmod{10}$ (since the first terms of both parentheses have a factor of $ 2$ and many factors of $ 5$, and the rest have many powers of $ 10$ from the $ 220$'s).

This means that $ (15 \plus{} \sqrt {220})^{19} \plus{} (15 \plus{} \sqrt {220})^{82} \equal{} 10k \minus{} ((15 \minus{} \sqrt {220})^{19} \plus{} (15 \minus{} \sqrt {220})^{82})$, and since

$ 0 < (15 \minus{} \sqrt {220})^{19} \plus{} (15 \minus{} \sqrt {220})^{82} < 1$, the units digit is $ \boxed{\textbf{(B)}\ 9}$.
[/hide]

[hide="Alternately"]$ S_n \equal{} (15 \plus{} \sqrt {220})^n \plus{} (15 \minus{} \sqrt {220})^n$ satisfies the recurrence $ S_{n \plus{} 1} \equal{} 30 S_n \minus{} 5 S_{n \minus{} 1}$ with initial conditions $ S_0 \equal{} 2, \;S_1 \equal{} 30$.

It is easy to see that $ 10|S_n$ for $ n \geq 2$ via the recurrence.[/hide]

[quote=candyofthesun][quote=TZF][hide="Alternately"]$ S_n \equal{} (15 \plus{} \sqrt {220})^n \plus{} (15 \minus{} \sqrt {220})^n$ satisfies the recurrence $ S_{n \plus{} 1} \equal{} 30 S_n \minus{} 5 S_{n \minus{} 1}$ with initial conditions $ S_0 \equal{} 2, \;S_1 \equal{} 30$.[/hide][/quote]
How is this figured out?[/quote]

Since $15\pm\sqrt{220}$ are solutions to the equation $x^2=30x-5$

[quote=Afo]A better way is to see as $a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$

Here $a+b=15+15=30,ab=15^2-220=5$[/quote]

I am a bit confused about why $a=b=15$ and how $a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$ can be used to figure out the recurrence formula.

Another problem involving this technique of binomial expansion cancellation as shown in post #2 is [hide=this problem.]What is the smallest integer that is greater than $(\sqrt{3}+\sqrt{2})^6?$[/hide]

They weren't saying a=b=15, they were saying wlog a = 15 + sqrt(220), b = 15 - sqrt(220)

About the recurrence formula (I'm not that familiar with recurrence relations, so please correct me if I messed up/was wrong), we let $S_n = a^n+b^n$, so $S_{n-1}$ would be $a^{n-1}+b^{n-1}$, same idea for $S_{n+1}$. Then we substitute into $a^n+b^n=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$ to get $ S_{n + 1} = 30 S_n - 5 S_{n - 1}$.


I am trying to find a point on the surface $z=16-4x^{2}-y^{2}$ at which the tangent plane is perpendicular to the line $x=3+4t, y=2t, z=2-t$.

So I take the point where the line and z intersect.  Which is:
$2-t=16-4(3+4t)-(2t)^{2}$ = $-4t^{2}-15t+2=0$ which makes:
$t=\frac{\pm(\sqrt{257}+15)}{8}$

But that seems over complex of a value for $t$. What am I doing wrong?

(I know that once I have the point of intersection $P(x_{0},y_{0},z_{0})$ that the tangent plane is $z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$

You don't need to find the point where the line intersects the surface.  If you look at the equation of a tangent plane, you know that <a,b,c> is perpendicular to the plane:

$a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0$

This tells you that <a,b,c> must be parallel to the line you are given.

Does this help at all?

Okay let me see if I get this:
Generically the tangent plane is:
$z-z_{0}=-8x_{0}(x-x_{0})-2y_{0}(y-y_{0})$
I need to convert that to the form:
$a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0$
Such that
$\left<a,b,c\right>\times\left<3+4t,2t,2-t\right> =0$
So:
$\left<-8x_{0},-2y_{0},1\right>\times\left<3+4t,2t,2-t\right> =0$
$\left< t(2y-2)-4y, t(4-8x)+16x+3, t(8y-16x)+6y \right> =0$

Okay, this is looking hopelessly more complex than before.....
:wallbash_red:

My solution is still incomplete. I am not getting this. Another hint would be nice, please.

The only formulae I find for tangent planes are:
$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
and
$F_{x}(x_{0},y_{0},z_{0})(x-x_{0})+F_{y}(x_{0},y_{0},z_{0})(y-y_{0})+_{z}(x_{0},y_{0},z_{0})(z-z_{0})=0$

Both of which seem to require that I have a point.

the tangent plane at $(x,y,z)$ to the surface $F(x,y,z)=0$ has normal vector $\nabla F(x,y,z)$.
the line $x=a+mt, y=b+nt, z=c+pt$ is parallel to the vector $(m,n,p)$
a plane is perpendicular on a vector if the normal is parallel  to (multiple of) that vector.

question: is it ok if people get help on this forum for their exams? :)

Okay, so applying that I get:
$\nabla F=\left<8x, 2y, 1\right> = k\left<4,2,-1\right>$
$k=-1$ $8x=-4$ $2y=-2$
$x=-\frac{1}{2}$ and $y=-1$
Therefore:
The point $P=\left(-\frac{1}{2},-1,0\right)$
Therefore:
The plane: $4(x+\frac{1}{2})-2(y+1)+z=0\Rightarrow4x+2-2y-2+z=0\Rightarrow4x-2y+z=0$


Hello,

I have a theory that you could not go back in time, but see back into time, if you could:
a) travel backwards faster than the speed of light, and
b) see something and make sense out of it while you are travelling faster than the speed of light

What is your opinion on this?

Where did you get the idea of this?  :huh:

What were your observations?

That is very interesting. I'm also confused as to why that happens. There isn't any way to experiment this as no rocket on this Earth is faster than the speed of light, and we can't see babies on other planets.

I'm guessing that it indeed happens, but we can't notice because our normal speeds are only millionths of the speed of light. But the current theory states that if you go faster than the speed of light, you're supposed to go back in time. Your theory states that light only goes one way, which is from Earth to Planet X, since when you went from Planet X to Earth, the baby got older faster to the vision, and when you went from Earth back to Planet X, it got younger.

I can't say anything further.

No I don't say that light goes only one way. If you are on Earth it becomes younger only because you are travelling backwards. If you were to turn around and watch something on planet X while travelling faster than the speed of light towards it, you would see something in fast forward, just like you did when travelling towards earth.

Let me just note something that when we speak of time as going backwards, it does not mean that it is travelling towards the negative, time just slows down or you enter another dimension back in time (you will stay the same age as you left). Another thing is that all of these twin paradox and going "back in time" ideas are [b]hypothetical[/b], meaning that they can never happen in any universe such as ours.  All of these discoveries/theories are given for the excitement of the human mind, so we are not meant to question hypothetical things that do not happen, or even involve "delta-age."

Masoud Zargar

Practical observation: If you travel faster than light your mass will become infinity and your length will contract to zero.

[quote="boxedexe"]Let me just note something that when we speak of time as going backwards, [/quote]

I did not mean that you could go back into time, I said that theoretically you could see back into time when travelling faster than the speed of light. But mathnerd314 has a point, it would be technically impossible to travel faster than the speed of light.

We cannot go faster than teh speed of light because if you travel at the speed of light (or faster), your mass will become infinite and therefore, you'll need infinite energy to travel that fast; infinite energy is impossible, so going faster than the speed of light is also impossible.

Correct.  Our discussion is about what happens if someone goes back to a point that he or she has not been born yet, not about whether a velocity of c is possible.  Well, said that, I will repeat my response: we are not meant to question hypothetical factors that are impossible.  Even if we are able to travel faster than teh speed of light, you are basically taking your current body and shifting it to a different time period.  It doesn't matter whether you were born or not.  Keep in mind that time does not move towards the negatives.

Masoud Zargar

[quote="mathnerd314"]Going off on tangents (sorry):

Do photons have mass? Because if they did, they would have infinite mass. But if they didn't, there would be no hope for things like solar sails.[/quote]

There is something called particle-wave duality (even though everythign are waves....for brevity's sake, I would not explain that in here).  Particle-wave duality means that everything are both waves and particles.  Photons rely on there momentum.  It's momentum is calculated using $E^2=p^2c^2+m_0^2c^4$...you should understand the rest...Hint: use mathematics.  Read about Louis de Broglie's ideas and theories.

Masoud Zargar

If all you want to do is see back in time, look no further than around you. You see everything as it was fractions of a second earlier.

What about the fact that clocks go slower if they are moving at the speed of light? Can anyone explain why that happens?

[quote="MysticTerminator"][quote="Melissa"]Well, let's take a little long, but explanitory example (this is the way that I discovered my theory):

Suppose you were on a planet X exactely 3 light years away from earth, and suppose you could see what is happening on earth. Then if you see a baby being born, how old would it be? It is actually 3 years old, because the light took 3 years to reach you. Now this is nothing new, as we all know that some of the stars we see doesn't even exist anymore, we just still see their light.

Now, suppose you were to travel from this planet X at the speed of light, towards the earth, starting the moment that you see the baby being born. How old will it be when you reach the earth? It will be 6 years old, because it had been 3 when you departed and it took you 3 years to get there. But, assuming that you can actually see what is happening on earth, what would you have seen during your journey? You would have seen the child grow 6 years older in 3 years' time. Thus you see everything happening twice as fast as it actually is happening.

Let's start again at planet X, but this time you travel towards the earth at double the speed of light, also starting the moment that you see the baby being born. How old will it be when you reach the earth? It will be 4.5 years old. But it took you 1.5 years to get to the earth. Thus this time you saw everything happening three times as fast as it actually happened.

Generalizing this, we can see that if you go n times the speed of light, you will see everything happening n+1 times as fast.

OK, so suppose we start at the earth when the child is 3 years old, and we travel backwards towards planet X at the speed of light. It will seem as though the child stays the same age until we stop travelling and reach planet X.

AND NOW, THE REAL SHOCK: what would happen if you start at the earth when the child is 3 years old, and you travel backwards towards planet X at TWICE the speed of light? Well of course you will see it becoming younger, until you reach planet X and it seems as though it is 1.5 years old.

Well I hope this makes sense, I tried to explain it the loooong way round...[/quote]

er...doesn't this just completely ignore special relativity? particles traveling at the speed of light travel instantaneously[/quote]

Well, no particle travels at the speed of light, thus, we should call them hypothetical or virtual particles.

Masoud Zargar

I don't know if anyone has said this, but how could you see something and make sence out of it if you are traveling faster than light; we see things by light reflecting off the object and traveling to us.

Yes, exactly. Therefore both assumptions are faulty. Plus, what is desired (seeing back in time) can be achieved by much simpler means. See previous posts.

ya this should be moved to the physics forum where they can more efficiently poke holes in it :)

Ok, I my idea was very very stupid. But I didn't say that you could actually travel faster than the speed of light or observe anything while doing so. I just said that if you could have done that, you would have been able to see back into time.

Melissa,

Have you seen the movie "Paycheck"? There is another interesting idea for a way to view different times in it.

[quote="mathnerd314"]I'm confused. Photons go at the speed of light, don't they? And if something travels at the speed of light, it doesn't move instantaneously. It just seems instanteous to the particle.

The clock goes more slowly because time slows down.[/quote]

Photons travel at the speed of light. They don't have mass.

:stretcher:  :stretcher:  :stretcher: oww..this hurts my head..

what happens, theoretically, in a black hole?

That's a little off topic. Go to the physics forum.

Yes, photons have mass. "My bad"... I asked in the physics forum and I asked my science teacher and I finally got an answer. Photons have momentum because of their energy...

THEREFORE, anything with mass can't travel at the speed of light. If something can't travel at the speed of light, it can't travel faster than the speed of light. Problem solved.

but perhaps, when nearing the speed of light, you begin to lose mass, and when you are traveling at the speed of light, you have no mass, and when you are traveling faster than the speed of light, you being to have negative mass because you have ripped a hole in time-space continum, therefore violating all known rules of physics, and thus nothing applies to you any longer. then, you could travel back and forth in time.
 :ninja:

[quote="elston"]but perphaps, when nearing the speed of light, you begin to lose mass, and when you are traveling at the speed of light, you have no mass, and when you are traveling faster than the speed of light, you being to have negative mass because you have ripped a hole in time-space continum, therefore violating all known rules of physics, and thus nothing applies to you any longer. then, you could travel back and forth in time.
 :ninja:[/quote]
sounds made up lol :P

Nice 100 posts, btw.