floor function
limit
inequalities
combinatorics unsolved
combinatorics

A is a non-empty set of positive integers. There are positive integers $ b_1, b_2, ... , b_n, c_1, c_2, ... , c_n$ such that each set $ b_iA \plus{} c_i \equal{}$ {$ b_ia \plus{} c_i$ : $ a$ belongs to $ A$} is a subset of $ A$, and the $ n$ sets $ b_iA \plus{} c_i$ are pairwise disjoint. Prove that $ 1/b_1 \plus{} 1/b_2 \plus{} ... \plus{} 1/b_n <\equal{} 1$.

clearly $ A$ is an infinite set. denote $ S_i \equal{} b_iA \plus{} c_i$ and let $ N(i,k) \equal{} |S_i\cap \{1,2,...,k\}|$, $ M(k) \equal{} |A\cap\{1,2,...,k\}|$. note that the $ x$'s in $ A$ that give rise to elements $ b_ix \plus{} c_i$ in $ S_i\cap\{1,2,...,k\}$ are those in $ A\cap \{1,...,\lfloor \frac {1}{b_i}(k \minus{} c_i)\rfloor\}$. so we have

$ N(i,k) \equal{} M(\lfloor \frac {1}{b_i}(k \minus{} c_i) \rfloor )$

and therefore

$ M(\lfloor \frac {1}{b_1}(k \minus{} c_1) \rfloor) \plus{} M(\lfloor \frac {1}{b_2}(k \minus{} c_2)\rfloor) \plus{} \cdots \plus{} M(\lfloor \frac {1}{b_n}(k \minus{} c_n)\rfloor) \leq M(k)$

because of the fact that the $ S_i$ are disjoint.

for $ n \equal{} 1$, this becomes $ M(\lfloor \frac {1}{b_1}(k \minus{} c_1) \rfloor) \leq M(k)$. but if $ \frac {1}{b_1} > 1$ then $ \lim_{k\to\infty}(\lfloor \frac {1}{b_1}(k \minus{} c_1) \rfloor \minus{} k) \equal{} \infty$, which would imply (using the evident fact that $ M(k)$ is monotonic) that $ M$ is eventually constant, contradicting the infinitude of $ A$. so we've proven the statement for $ n \equal{} 1$.

anyone have any ideas??

edit: obviously $ b_1\geq 1$ anyway when $ n \equal{} 1$ because we're given that the $ b_i$ are positive integers, so what i proved was trivial :) , but still... any ideas along these lines? i wanna see this problem solved!

i must admit... $ A$ is necessarily infinite only if $ n > 1$, sorry. anyway, that's the only case that matters! the cardinality of each $ a_iA \plus{} b_i$ is of course the same as the cardinality of $ A$, so if $ n > 1$, $ A$ has at least two disjoint subsets of the same cardinality as its own cardinality, which means that it can't be finite.

First let us define [i]the density[/i] of a subset $ A$ of $ Z_\plus{}$ as $ D(A)\equal{}\lim_{x\to\infty}M(x)$ where $ M(x)\equal{}|A|\cap \{1,2,...,x\},x\in Z_\plus{}$, if the limit exists. The density is additive for 2 disjoint subsets, that is if we have 2 subsets $ A$ and $ B$ with proper densities and $ |A \cap B|\equal{}0$ then $ D(A \cup B)\equal{}D(A)\plus{}D(B)$. Proof: $ D(A \cup B)\equal{}\lim_{x\to\infty}M(A \cup B)\equal{}\lim_{x\to\infty}{M(A) \plus{} M(B)}\equal{}\lim_{x\to\infty}M(A)\plus{}\lim_{x\to\infty}M(B)\equal{}$$ D(A)\plus{}D(B)$.
   We compute that the density of $ b_iA\plus{}c_i$ is $ \lim_{x\to\infty}{\frac{\lfloor \frac {1}{b_i}(k \minus{} c_i)\rfloor}{k}}\equal{}\frac{1}{b_i}$. Making use of the additivity of density, we compute the density of $ B\equal{}B_1 \cup B_2 \cup \dots B_n$ where $ B_i\equal{}b_iA\plus{}c_i$ and get $ D(B)\equal{}\displaystyle\sum_{i\equal{}1}^{n}\frac{1}{b_i}$. All what's left is to say that the density belongs to $ [0,1]$, and it's obvious because $ 0 \leq M(k) \leq k$ for all $ k$. So $ D(B) \leq 1$ and $ \displaystyle\sum_{i\equal{}1}^{n}\frac{1}{b_i} \leq 1$.

Can you say is it true, please?

[quote="pleurestique"]the only trouble is that the density of $ b_iA \plus{} c_i$ should be $ \lim_{k\to \infty}\frac {M(\lfloor \frac {1}{b_i}(k \minus{} c_i)\rfloor)}{k}$, and estimating this is what makes the problem hard.[/quote]
No, I count the density of {b_iA+c_i} in $ Z_\plus{}$ not in $ A$ and use the fact that $ M(x) \leq x$

I'm confused by your notation here.  Suppose $ A\equal{}\{1, 3, 7, 15, ...\}$.  What do you mean by the density of $ 2A\plus{}1$ being $ 1/2$?

[quote="polarb1"][quote="pleurestique"]the only trouble is that the density of $ b_iA \plus{} c_i$ should be $ \lim_{k\to \infty}\frac {M(\lfloor \frac {1}{b_i}(k \minus{} c_i)\rfloor)}{k}$, and estimating this is what makes the problem hard.[/quote]
No, I count the density of {b_iA+c_i} in $ Z_ \plus{}$ not in $ A$ and use the fact that $ M(x) \leq x$[/quote]

i'm aware that you're computing the density of $ B_i \equal{} b_iA \plus{} c_i$ in $ \mathbf{Z}^{ \plus{} }$, but is it not true that $ |B_i\cap\{1,2,...,k\}| \equal{} M(\lfloor\frac {1}{b_i}(k \minus{} c_i)\rfloor)$?

I'm afraid you're right - my solution is not correct, because in one moment I thought, I don't know why, that $ A$ is $ Z_\plus{}$. What a shame...  :(

Here's something that might work, though it all might blow up in my face if I switched an inequality sign somewhere (hidden because it's fairly long and you may want to read the key points first).

[hide]We proceed by contradiction.  Let us assume we can find such a set with $ \sum_{i \equal{} 1}^n \frac {1}{b_i} > 1$.

Let $ f_i(x) \equal{} b_ix \plus{} c_i$.  We are assuming that if $ x \in A$, then $ f_i(x) \in A$ for all $ x$.  Furthermore, we are assuming that the $ f_i(A)$ are all disjoint.

Let $ g(x)$ denote the number of elements of $ A$ which are less than $ x$.  We have the inequality
\[ g(x) \geq \sum_{i \equal{} 1}^n |f_i(A) \cap \{1, 2, \dots n\}| \equal{} \sum_{i \equal{} 1}^n g(\frac {x \minus{} c_i}{b_i}).
\]
Our next goal is to get rid of the pesky $ c_i$.  To compensate, we will slightly increase the $ b_i$.  Let $ b_i' \equal{} b_i \sqrt {\sum_{j \equal{} 1}^n \frac{1}{b_j}}$.  Note that $ b_i' > b_i$ for each $ i$, and furthermore $ \sum_{i \equal{} 1}^n \frac {1}{b_i'} \equal{} \sqrt {\sum_{i \equal{} 1}^n \frac {1}{b_i}} > 1$.

There is a finite $ x_0$ such that for $ x > x_0$, we have $ \frac {x \minus{} c_i}{b_i} > \frac {x}{b_i'}$ for all $ i$.  For any $ x > x_0$, we therefore have
\[ g(x) \geq \sum_{i \equal{} 1}^n g(\frac {x \minus{} c_i}{b_i}) \geq \sum_{i \equal{} 1}^n g(\frac {x}{b_i'}),
\]
where for the last inequality we use the fact that $ g$ is non-decreasing.  Now let $ h(x) \equal{} \frac {g(x)}{x}$.  Substituting in the above gives us
\[ h(x) \geq \sum_{i \equal{} 1}^n \frac {1}{b_i'} h(\frac {x}{b_i'}).
\]
If we knew that $ h(x)$ had a positive limit as $ x$ increased (or even that $ h(x)$ was non-decreasing), we'd be done.  I don't think we necessarily have this though, so our next step will be to "smooth out" our $ h$.  Let $ b$ be the largest of our $ b_i'$, and let
\[ q_n \equal{} \min_{b^{n \minus{} 1} \leq y \leq b^n} h(y).
\]
The point here is that if $ y$ is between $ b^{n \minus{} 1}$ and $ b^n$, then $ \frac {y}{b_i'}$ is between $ b^{n \minus{} 2}$ and $ b^n$.  We therefore have
(for $ n$ chosen so that $ b^{n \minus{} 2} > x_0$)
\[ q_n \geq \min_{b^{n \minus{} 1} \leq y \leq b^n} (\sum_{i \equal{} 1}^n \frac {1}{b_i'} h(\frac {y}{b_i'})) \geq \sum_{i \equal{} 1}^n \min_{b^{n \minus{} 1} \leq y \leq b^n} \frac {1}{b_i'} h(\frac {y}{b_i'}) \geq \sum_{i \equal{} 1}^n \frac {1}{b_i'} \min(q_n, q_{n \minus{} 1}) \\
\equal{} \min(q_n, q_{n \minus{} 1}) \sum_{i \equal{} 1}^n \frac {1}{b_i'}
\]
Now we're just about done.  Comparing the first and last terms of the above equation, we have for sufficiently large $ n$ that $ q_n \geq q_{n \minus{} 1} \sum_{i \equal{} 1}^n \frac {1}{b_i'}$.  Furthermore, $ q_n$ is positive for all sufficiently large $ n$ (as $ A$ is non-empty).  This means that $ q_n$ grows exponentially with $ n$.  On the other hand, $ q_n \leq 1$, so we have a contradiction.  [/hide]

Really the hard part here was not combinatorics so much as analysis.  There are two main technical details:

1. The trick of increasing the $ b_i$ slightly so we no longer have to worry about the $ c_i$.

2.  Using the idea of density without actually using limits.  The intuition, as suggested above, is that the limiting density $ d(A)$ of $ A$ should satisfy $ d(A) \geq \sum_{i \equal{} 1}^k \frac {d(A)}{b_i}$, which seems impossible.  However, the problem is that $ d(A)$ might tend to 0, or might even not have a limiting value in the first place.  So what I did was to define a sort of "smoothed" density by taking the minimum density of $ A$ over some large interval.

The point of this was that we don't need to worry about whether limits exist, since the inequality $ \min(f(x) \plus{} g(x)) \geq \min(f(x)) \plus{} \min(g(x))$ goes the right way for our purposes.

(edited to fix the definition of $ b_i'$, thanks for catching that!)

i think you meant to define $ b_i' \equal{} b_i\sqrt {\sum_{j \equal{} 1}^n\frac {1}{b_j}}$.

[quote="kevinatcausa"]

\[ q_n \geq \min_{b^{n \minus{} 1} \leq y \leq b^n} (\sum_{i \equal{} 1}^n \frac {1}{b_i'} h(\frac {y}{b_i'})) \geq \sum_{i \equal{} 1}^n \min_{b^{n \minus{} 1} \leq y \leq b^n} \frac {1}{b_i'} h(\frac {y}{b_i'}) \geq \sum_{i \equal{} 1}^n \frac {1}{b_i'} \min(q_n, q_{n \minus{} 1}) \\
\equal{} \min(q_n, q_{n \minus{} 1}) \sum_{i \equal{} 1}^n \frac {1}{b_i'}
\]
Now we're just about done.  Comparing the first and last terms of the above equation, we have for sufficiently large $ n$ that $ q_n \geq q_{n \minus{} 1} \sum_{i \equal{} 1}^n \frac {1}{b_i'}$.[/quote]

there seems to be a problem here. it's not true that $ \min(q_n,q_{n\minus{}1})\geq q_{n\minus{}1}$. what if $ q_{n\minus{}1}$ is the larger of $ q_{n\minus{}1},q_{n}$?

If that's the case, then the inequality simplifies down to
\[ q_n \geq q_n \sum_{i\equal{}1}^n \frac{1}{b_i'},\]

which is false since the sum of the $ \frac{1}{b_i'}$ is greater than 1.

right right, i hadn't seen that. it seems then that your proof is correct now that i read all of it. though since i'm not the one who came up with it, it seems to me as the reader like it just came out of thin air, it's one of those proofs.


integration
complex analysis
real analysis
real analysis unsolved

from Rudin's "Real and Complex analysis"

for $ f: R\to R$ let
\[ \varphi(t)\equal{}\frac{1}{\sqrt{2\pi}}\widehat{f}(t)\]
If $ f$ is "good" (satisfies some conditions) then
\[ \sum_{k\equal{}\minus{}\infty}^{\plus{}\infty}f(2k\pi)\equal{}\sum_{n\equal{}\minus{}\infty}^{\plus{}\infty}\varphi(n)\]
(Poisson's summation formula).

What should I take as $ f$ to obtain $ \sum\frac{1}{n^2}\equal{}\frac{\pi^2}{6}$ ?

for $ a > 0$,

$ \sum_{n \equal{} \minus{} \infty}^\infty\dfrac{1}{n^2 \plus{} a^2}\; \equal{} \;\sum_{k \equal{} \minus{} \infty}^\infty\left\{\int_{ \minus{} \infty}^\infty \;e^{2\pi\imath\,k\,x}\cdot \dfrac{1}{x^2 \plus{} a^2}\;\textbf dx\right\}$

LHS is
 $ \dfrac{1}{a^2} \plus{} 2\sum_{n \equal{} 1}^\infty\dfrac{1}{n^2 \plus{} a^2}$

 so you will need to  let $ a\to0$ to get $ \zeta(2)$ at the end.

can you take it from here ?


trigonometry
AMC
AIME
number theory
\/closed

[quote="daermon"]hi
i was wondering when the next time the interm trig/complex numbers class would be offered
thanks[/quote]
Next Fall

oh are classes offered at the same time each year?  that would be useful to know  :D

Hmmm, I think the intermediate number theory class was offered at 2006 summer and 2005-2006 winter. Maybe that was because it's a seminar, not a full class?  :maybe:

[quote="bpms"]Is the entire spring schedule going to be posted? Or is it only intermediate alg., intro to counting, and aime seminar?[/quote]

We don't expect to offer any classes besides those now shown until June.


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Could you give me an example of a ring which is a  unique factorization domain but is not  principal ideal domain

Generally PIDs are precisely the UFDs of dimension one or less (i.e. every nonzero prime ideal is maximal).
For a proof see my AAA post http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2008;task=show_msg;msg=2902.0002


I am hereby announcing that the official Mathlinks chat has been created. This thread was posted in the Questions, Suggestions, & Announcements Forum, but nobody goes there so I will also start the topic here.
MODS DO NOT MOVE THIS PLEASE

So, here are the instructions:
[quote="ZetaX"]First to all who want to chat:

So I proclamate now (later on we can still choose another one) EuIRC:
- irc.euirc.net::6665-6669
- channel: #mathlinks

For those new to that there is some html-client:
- go to http://html.chat.euirc.net/
- enter some nickname, leave the password and Username unchanged
- choose some random channel at the bottom (lets not bother with them ;) ), for example "#euirc - General information channel"
- click on "go"
- in the empty line you see type: "/join #mathlinks" and press enter.
- voila, there you are

Also:
One who doesn't want to chat is not forced to come there.
But everyone is invited to try chatting.[/quote]

Please do not spam too much, but this forum is for chatting on Math-related subjects.
Please join and enjoy!

btw the room does not hold more than 23 ppl, so maybe you could start a "sub-room" if there are too many people.

Have fun!

The room holds as many people an admin sets ;)
And at the moment this is 481, but it could also be 9801...

Edit: ok the last number was I bit to high  :lol:

PS: $\infty$ is also possible...

me too, even though i do not know what it is

:rotfl:  :lol:  :D
But what do you not know¿  :huh:  :)

[quote="rem"]I am posting again so that this topic goes at the top of the page so more people will read it. :)[/quote]

That is called a [url=http://en.wikipedia.org/wiki/Bump_%28internet%29]bump[/url].

I am for the first time since IMO 2002 participating in a mafia game( game 20) , and I immediately notice the difference : you don't know at all who to accuse.  You can simply not post or whatever.  In real life you look at people's faces, how loud they speak, whether or not they look at you, whether or not they look at each other....

Maybe a chat form of mafia would be  :maybe:  possible?
I realize it won't be easy though  :huh:

When these game (I have heard of chat-mafia to exist) can be done in some hours, I would also play from time to time.
It's just totally different due to PM's etc. from real one and cause of small amounts of time to standard online one, too.

Well, to even plan this we would have to take a chat ;)


geometry
geometric transformation
reflection

Lines are drawn through the vertices of triangle ABC, outside the triangle and making  an angle T with AB, BC, CA respectively.
Show that the area of the triangle so formed is largest when

                      TanT = {a^2 + b^2 + c^2 ]/4S , where S is the area of triangle ABC.

[hide="Observation"]
If we take the reflections of the three lines in each respective side of the triangle, they meet at a point iff $T = \omega$ the Brocard angle of $\triangle ABC$. And $\cot \omega = \frac{a^2 + b^2 + c^2}{4\triangle}$.
[/hide]


function
search
algebra unsolved
algebra

Find all functions on reals to reals, that satisfy, for all reals $ x, y$, we have satisfied:

$ g(x\plus{}y)\plus{}g(x)g(y)\equal{} g(xy)\plus{}g(x)\plus{}g(y)$

Searching Mathlinks/AOPS usually helps.  :wink: Take a look at http://www.mathlinks.ro//viewtopic.php?search_id=1167535464&t=178957.


calculus
analytic geometry
function
Princeton
college
derivative
integration

I kind of did it...... I did two chapters of calculus in the beginning of the school year, but stopped in October and didn't start again unil April 20th due to extreme laziness. So in two weeks, I tried to teach myself the remaining 11 chapters.  Needless to say, I didn't learn very well, especially since tennis took up most of my time. I decided that I should skip polar coordiantes and a few other topics, too. But with two to three hours of preparation every day for two weeks, I managed to get a 4 on Calc BC. So this year, I need to retake it. But really, do some Calculus for an hour a day from now on, and you'll easily be prepared to get a 5 on the AP exam. Good luck! :)

Yes, absolutely.

I taught myself material according to the AP Calculus AB and BC syllabus 2 months before the test last year.  Needless to say, the self-motivation proved to be a lot more effective than the false sense of security (not to mention...complacency) I would have had had I taken a Calculus BC class.

Also at the time, I was in Grade 11 Functions and Relations (Canadian equivalent to Precalculus) and I found that a lot of the Mathematics associated with Precalculus could easily be implied while learning Calculus; if you're good at Math, then don't worry about "grasping the concepts."

If you really want a 5 on the BC test, then do some questions from each chapter of any respectable Calculus text in accordance to the syllabus materials.  Then about 1-2 weeks before the test, review all of your knowledge with the Princeton Review book.

The test really isn't that hard.  It makes sure you know your basic differentiation, integration and series techniques, and requires almost no ingenuity to produce your answers.  The full-solutions part is also very textbook; so a review of past Calculus BC tests would be helpful.

The curve is pretty nice, but don't let that give you a false sense of security.  I got a 5 on both the BC test and the AB subscore.

Hope this helps.

yeah, it's definitely possible, this year i wanted to skip Calc AB and go to BC. So the last 2 weeks of summer before school started i "taught" myself Calc AB. Needless to say, i didn't learn everything i was supposed to but it was enough to get an "A" on the qualifying exam. And i spent like 12-13 hours doin calc for those 2 weeks... heh

[quote="nigma"]Is the material covered in an AP Calculus course something you could teach yourself in four months? Assuming, of course, that you really grasped the concepts covered in preceeding courses.[/quote]

The short answer is yes.
The picky but relevant answer is that the "real question" is how many hours of study are needed to learn the material (and to what level of understanding).  I think this would be an interesting question to answer in some detail.   Whether that many hours are available over four months would depend on the rest of your schedule.
The most useful answer may be that (in my opinion, which is hardly an unusual one in this case) you will learn the material much better and more efficiently in a state of relaxation rather than working to a deadline.  To really absorb calculus properly may take much longer than 4 months, even if the particular manipulations needed to pass an AP exam can be learned quickly.

fleeting_guest, that's basically the same answer my physics teacher gave (who is quite qualified to teach calculus). I think because of it I will spend the typical nine months on the material rather than condensing it by five months.

Thanks for the advice.

Well, personally I think that there's nothing wrong with learning it on your own in whatever time it takes you. My guess is that if you have a good understanding of precalculus, at least half the time spent in a classroom learning Calc BC is redundant (most likely applies to AB too).

At the same time, if you want stuff to sink in, go back and rederive stuff you've seen already every once in a while.

[quote="nigma"]Thanks for the advice. I was actually planning on studying the material this summer and challenge testing the course @ my high school so that I could take classes through a university online (epgy, jhcty, etc) next year. But, that would leave a long period of time between the test and when I went over the material, so I don't know.

Once again, thanks for the help, greatly apreciated.[/quote]

hehe, we think alike..well im taking (self study) ap calc right now and its not hard

I am about to do the same thing.  I've just finished the precalculus course at my highschool and I'm going to take the BC test.  I hope I can prepare in time.

i woud have to say most definitely.
i learned most of the calculus i used in the first semester over the summer (less than 4 months). and during this time i was not very focused on studying or even learning the material.
really i think it is possibel, i have a friend who is in AB only because they think he is too young for BC (he is a sophmore) and he know more BC than i do (not including sequences and series or parametric equations) and he had only been in AB for about a month b4 he learned all of this.

AP Calculus AB covers alot of material, but there is a very obvious progression in the course. AP Calculus BC covers a little less material (although 40% of the exam tests you on AB-like material) but it seems to be more divided. The connections are there, just perhaps not so apparent as AB's.

I was in AP Calc AB junior year, then my teacher suggested after midterms that I study for the BC exam and take that instead in May. So i did BC in about 2, maybe 2.5 months, and got a 5 on the exam. The exam also gives you an AB subscore, which sort of tells you how you would have done on the AB test (which of course was a 5 as well).

I STRONGLY suggest doing this if you aren't already taking BC Calc junior year, because I fell in love with calculus as a result. That summer i studied Calculus III (AB is equivalent of I, BC to II) at Columbia, which the Director of Undergraduate Studies offered me for free (but w/o credits, which I don't really mind...I got to learn math and got my first college experience). Then senior year, which I'm in right now, I took Calculus IV (last of the sequence), following Columbia's course outline.

Now I'm taking Ordinary Differential Equations using MIT's OpenCourseWare and I'm falling in love with it, too.

It's pretty cool when you can tell people you finished 4 years of calculus while still in highschool...and then some.

Sort of an unrelated question, but can you take the AP Calculus AB & BC in the same year?

:huuh: i dont see any reason to why they wouldnt let you. but it would really be pointless since in BC you cover all of the material that is covered in AB...and then some.

You can take both the AB and BC test in the same year, but typically people don't.  The exams are given during the same slot so prior arrangements need to be made.

in the bulletin i got for AP's it actually said specifically that you can't take both AB and BC in the year

In any case, because the AP calculus BC test yields an AB subscore, there isn't any point to taking the AB test if you are ready for the BC test.


Determine the sum of the y-coordinates of the four points of intersection of $y=x^{4}-5x^{2}-x+4$ and $y=x^{2}-3x$.

[hide="hint"]
Use Vieta's Formulas. Which equation looks simpler to plug back into/manipulate?[/hide]


Το άλλο σταθερό σημείο είναι το περίκεντρο του τριγώνου $ ABC$

[quote="bilot4"]Η άσκηση είναι λίγο διασκευασμένη (από μένα) αλλά παραμένει στην ουσία της ίδια :)

Εστω δεδομένο τρίγωνο $ ABC$ με $ \angle{A} \equal{} 60$ και $ AB > AC$. Εστω επίσης τα ύψη $ BE,CF$ και το ορθόκεντρο $ H$. Αν $ M,N$ σημεία των ευθυγράμμων τμημάτων $ BE,CF$ αντιστοίχως, έτσι ώστε $ BH > BM \equal{} CN > CH$, να αποδειχθεί ότι καθώς τα $ M,N$ μεταβάλλονται οι περιγεγραμμένοι των τριγώνων $ HMN$ διέρχονται διά σταθερού σημείου (εκτός του $ H$ φυσικά...) και να προσδιορισθεί το σημείο αυτό (συγχωρέστε με αν αυτό το 2ο ερώτημα, που είναι και το βασικό, δεν το εξηγώ καλά, γαιτί ομολογουμένως το "προσδιοριστεί" δεν είναι και ότι πιο σαφές :oops: ).[/quote]

nomizo einai apli efarmogi tou theorimatos mc laurin.

vsk tora pou to ksanavlepo to theorima mc laurin tin kseftilizei entelos tin askisi. kai to a kai to b.

re sy niko se poio 8eorima anaferesai??? mporeis plz na e3igiseis ligo?

Λοιπόν θα κάνω βασικά αυτό που λέει ο  Νίκος αλλά αποδεικνύοντας το ταυτόχρονα (2 σε ένα  :P )
Η άσκηση όντως είναι πολύ απλή .

Λοιπόν έστω ο περιγεγραμμένος κύκλος του HNM . Έστω ότι η διχοτόμος της FHB τέμνει τον κύκλο στο $ O'$ .
Είναι $ MH \plus{} HN \equal{} BH \minus{} BM \plus{} HN \equal{} BH \minus{} CN \plus{} HN \equal{} BH \minus{} HC \equal{} a$ σταθερό .

Από το θεώρημα του Πτολεμαίου έχουμε $ O'N(MH \plus{} HN) \equal{} HO'\cdot NM$  . Τώρα όπως θα έλεγε ο κ Λουρίδας το Ο'NM καθόλη την κίνηση παραμένει όμοιο προς τον εαυτό του  :)  άρα $ \frac {MN}{O'N} \equal{} c$ άρα $ O'N \equal{} \frac {a}{c}$ δηλαδή σταθερό άρα οι κύκλοι αυτοί περνούν από σταθερό σημείο .

Θα δείξουμε ότι αυτό είναι το περίκεντρο $ O$ δηλαδή ότι $ O\equiv O'$ . Πράγματι έστω $ K,L$ οι προβολές του $ O$ στις
$ BE,\ CF$ . Το OHCB είναι εγγράψιμο άρα $ \angle{OBK} \equal{} \angle{OCH}$ δηλαδή $ BK \equal{} CL$  . Άρα συμύφωνα με τα παραπάνω ο περιγεγραμμένος κύκλος του ΗKL περνά από το $ O'$ περνά και από το $ O$ όμως και το $ O$ είναι το μέσο του τόξου άρα $ O\equiv O'$ and so QED

Παιδιά sry :blush: εγώ φταίω που διασκεύασα την άσκηση γιατί δεν είχα καταλάβει ότι το κοινό αυτό σημείο ήταν το $ O$ (δεν είχα κάνει και καλό σχήμα...).
Αν κατάλαβα καλά το θεώρημα mc laurin στο οποίο αναφερθήκατε είναι το ότι αν έχεις σταθερή γωνία και σημεία που κινούνται με σταθερό άθροισμα τότε οι περιγεγραμμένοι τους διέρχονται διά σταθερού σημείου (αρκετά απλό και ήταν και άσκηση της ΕΜΕ).

Παρ' όλα αυτά γαι να καταλάβετε πώς την πάτησα θα σας πω την original άσκηση η οποία ήταν να υπολογίσετε  το $ \frac {MH \plus{} HN}{OH}$ (όχι ότι αυτό είναι τόσο δύσκολο...). Εγώ λοιπόν είχα στο μυαλό μου ότι από αυτό θα προκύψει πόσο κάνει το άθροισμα $ MH \plus{} HN$ (συναρτήσει του $ OH$ που θεωρητικά είναι γνωστό) και μετά θα προσδιοριστεί το σημείο που ζητούσα ως τομή της διχοτόμου με τη μεσοκάθετο του τμήματος $ OP \equal{} MH \plus{} HN \equal{} a$ σταθερό (αλλά και γνωστό). Ομως επειδή δεν κατάλαβα ότι το σταθερό σημείο αυτό ήταν το περίκεντρο κατέστρεψα τελείως την άσκηση.

Και πάλι συγγνώμη :oops:

Where can we discuss this problem in English?


I have a question on the Rabin-Miller primality test:

Given any fixed bound N, does there exist a base "a" such that every composite integer n with gcd(a,n)=1, and n < N, fail the Rabin-Miller test with this base a?

Apparently not - see the last lines of http://primes.utm.edu/prove/prove2_3.html .

  darij


[quote="dvcalculus"]The problem is
8x=-11y-16
2x+3y=-4

How do I do this problem?[/quote]
[hide="Answer"]For the second equation, move the y to the left side. Then, you have 2x = -3y - 4. Multiply by 4 and subtract from the first: y = [b]0[/b]. If y is 0, x is [b]-2[/b].[/hide]

more detailed

[hide="answer"]
8x=-11y-16
2x+3y=-4  => 8x = -12y - 16

8x = -11y - 16 subtract 8x = -12y - 16, 8x's cancel out as well as the -16s so we're left with

0 = y

2x + 3(0) = -4, thus 2x = -4, x = -2[/hide]


triangle ABC is an acute one. M is the mid-point of BC, P is a point on AM such that PM=BM, H is a point on BC such that PH is perpendicular to BC. N on PB such that HN is perpendicular to PB and HN extension intersects AB at Q, S on PC such that HS is perpendicular to PC and HS extension intersects AC at R. Prove that angle RHC=angle HQR.

i see right angled triangle PQR and PCB has some specials.QR's midpoint is on PH while its altitude is on PM...but i dun know how to prove..is there any name for such property??

Let $ d$ the line passes through $ A$ and parallel to $ BC$. Denote by $ E\equal{}BP\cap AC$, $ \equal{}CP\cap AB$, $ E'\equal{}BP\cap d$, and $ F'\equal{}CP\cap d$. We have $ \frac{AE'}{BM}\equal{}\frac{AP}{PM}\equal{}\frac{AF'}{CM}$, it implies $ AE'\equal{}AF'$. Hence, we have $ \frac{AF}{BF}\equal{}\frac{AF'}{BC}\equal{}\frac{AE'}{BC}\equal{}\frac{AE}{CE}$. So we conclude that $ EF\parallel BC$, which follows $ \frac{PE}{PB}\equal{}\frac{PF}{PC}$, i.e., $ \frac{SR}{SH}\equal{}\frac{NQ}{NH}$, i.e., $ NS$ is parallel to $ QR$. The conclusion follows.

[b]ANOTHER RESULT.[/b] $ P$ is the orthocentre of triangle $ AQR$


integration
function
calculus
complex analysis
complex analysis unsolved

Let series $ f(z)\equal{}\sum_{k\equal{}0}^{\plus{}\infty}a_k z^k$ converges on $ K(0,R)$. Prove that for $ 0\leq r<R$;
$ \frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}|f(re^{i\phi})|^2 d\phi \equal{}\sum_{k\equal{}0}^{\plus{}\infty}|a_k|^2 r^{2k}$

It [b]is[/b] Parseval's identity. The functions $ z^k \equal{} r^ke^{ik\phi}$ are orthogonal on each circle, and their coefficients are Fourier coefficients of the sum (Cauchy integral formula). The only thing to note is that for a holomorphic function on the disk, all of the Fourier coefficients with negative index are zero.


[hide]How is the answer supposed to be presented? My calculator tells me the answer is approx. 519.72, but I don't think that's what you're looking for.[/hide]

There's no way to simplify $e^\pi \pi^{e}$ any further...

[quote="Interval"]What is e^pi times pi^e?[/quote]

I don't know where you will ever need this... but
e^pi = 23.1406926
pi^e = 22.4591577
on the other hand,
e^(pi * i) = -1

[quote="Interval"]What is e^pi times pi^e?[/quote]

here is the question in LaTeX

$e^\pi \cdot \pi^{e}$

editted: oops i didn't see the other posts  :oops:  sorry! and cincodemayo5590, why can't you simplify? punch it into the calc

[quote="7h3.D3m0n.117"]why can't you simplify? punch it into the calc[/quote]

That would be evaluation, not simplification.

This topic really should be locked, since the same question was asked in another forum, and I gave a better substitution question there anyway.


well, it's obvious that $x=y=z=1$ is a solution (and probably the only one too :) )
but i can't seem to find a decent proof  :blush:  it just looks nasty with that $ln$  :?
i'm looking forward to see someone solve this ;)

[quote="JBL"]There are two things to do:

First, you must show that any solution must satisfy x = y = z.
Then, you must show that this only happens when x = y = z = 1.

Both parts require some sort of argument based on whether the functions involved are increasing or decreasing.[/quote]
yeah, it's quite "obvious" to show that for every $x,y$ or $z\geq 1$ that $x=y=z=1$
but not for $x,y$ or $z\leq 1$  :|

An old problem.......
But still unsolved....

Can anyone give a solution? warut_suk?? :)

Thanks!!

it is really interesting , and famous solution and method must be used

we consider that:

f(y)=z
f(x)=y
f(z)=x

i solved this kind of equaltion two years ago

we just solve this e:
$x^3+3x-3+ln(x^2-x+1)=x$

so x=y=z=1 is the root
this have a nice method to do this kind of ineq :rotfl:  :rotfl:  :rotfl:  :huh:  :ninja:  :|


John's bike lock is a 4-digit code, where the digits range from 1 to 6. He wrote all possible codes in numerical order. If his code is the 1186th number on the list, what is sum of the ten's digit and the unit's of the square of his code?

Huh? The way I understand the problem is that there are $ 6\cdot5\cdot4\cdot3\equal{}360$ total ways. But $ 360<1186$. Does the list repeat?

no there are 6 x 6 x 6 x 6 = 1296 > 1186 ways. 1296 - 1186 = 110
The last one is 6666 so we need to count backwards 110 only using the numbers 1-6. This is equivilant to counting backwards 110 base 7.

x + 49 = 2058 + 294 + 42 + 6 = 2400 x = 2351 = 6568 which is the 1186th number.
so the sum of the square of the tens and unit digit is [b]15[/b].

@hunter34: I assume the digits are allowed to repeat.

[hide]For simplicity (to be able to work in base 6), I will temporarily change the 6 digits to the digits 0-5.

There are $ 6^4$ or 1296 possible codes for the lock.  The last of these is 5555.  We are looking for the 1186th, or (1296-110)th code.  $ 110_{10}$ is $ 302_6$.  To find the desired code, we can simply subtract $ 302_6$ from $ 5555_6$ and get $ 5253_6$.

Now, since the available digits are actually 1-6, we must add one to each digit in the number, giving a final answer of 6364.[/hide]

i dont know if your allowed to just add one digit. and you didnt give the answer cfc49

I did give the answer.  It is
[hide="Answer"]6364[/hide]
as shown at the end of my first post.

Perhaps you didn't understand my explanation.  To use the base-6 solution to this problem, I had to change the given range of digits, which was 1-6, to 0-5; effectively subtracting 1 from each digit.  When I found my base-6 answer, I had to add that 1 back to the digits in the answer to match the conditions of the original problem.

you didnt give the answer. He wants the sum of the tens and units digit of the square of the code.

OOPS!!!!!!!!  Next time, I will read the whole problem.

[hide="Part 2"]$ 6364^2$ is $ 40500496$.  The sum of the tens and ones digits of this number is 15.  Therefore, the final answer to this problem is $ \boxed{15}$.[/hide]


An inclined plane has an angle of $ 17.458$ degrees. An object of mass M starts at the bottom of the plane with $ v_{i}\equal{} 1.6$ m/s. The coefficient of kinetic friction is $ .1x$ where $ x$ is the distance traveled (ie. the coefficient increases linearly with distance). How far up the ramp does the mass get?

[hide="solution"]
by work-energy theorem finally when the block comes to rest let it have travelled a  distance $ x$ up the incline(assuming the inclined is fixed to it's surfcae of contact)
$ \frac{mv_{i}^{2}}{2}= mg x\sin\alpha+W_{f}$ where $ W_{f}$ refers to the work done by friction
now when the block is ata distance $ x$ from bottom  friction $ f = mgx\cos\alpha\mu_{k}= mgx\cos\alpha (\beta x)$
hence $ W_{f}=\int_{0}^{x}mg\beta\cos\alpha x dx =\frac{mg x^{2}\beta\cos\alpha}{2}$
thus we have $ \frac{g\beta\cos\alpha x^{2}}{2}+gx\sin\alpha-\frac{v_{i}^{2}}{2}=0$
substituting $ \beta = .1$ $ \alpha = 17.458^{\circ}$ $ v_{i}= 1.6 ~m/s$ and solving the quadratic( :rotfl: ) we get the answer

[/hide]


Let a,b,c are three sides of an triangle satisfied: a+b+c=3. Prove that:
\[ \frac{1}{\sqrt[3]{a\plus{}b\minus{}c}} \plus{}\frac{1}{\sqrt[3]{a\plus{}c\minus{}b}} \plus{}\frac{1}{\sqrt[3]{b\plus{}c\minus{}a}}\geq \frac{9}{\sqrt{ab\plus{}bc\plus{}ca}}\]

[quote="smiley4t"]Let a,b,c are three sides of an triangle satisfied: a+b+c=3. Prove that:
\[ \frac {1}{\sqrt [3]{a \plus{} b \minus{} c}} \plus{} \frac {1}{\sqrt [3]{a \plus{} c \minus{} b}} \plus{} \frac {1}{\sqrt [3]{b \plus{} c \minus{} a}}\geq \frac {9}{\sqrt {ab \plus{} bc \plus{} ca}}\]
[/quote]
If $ a\equal{}b\equal{}c\equal{}1$ then $ LHS \equal{} 3, RHS \equal{} 3\sqrt{3}$?

[quote="smiley4t"]Let a,b,c are three sides of an triangle satisfied: a+b+c=3. Prove that:
\[ \frac {1}{\sqrt [3]{a \plus{} b \minus{} c}} \plus{} \frac {1}{\sqrt [3]{a \plus{} c \minus{} b}} \plus{} \frac {1}{\sqrt [3]{b \plus{} c \minus{} a}}\geq \frac {9}{\sqrt {ab \plus{} bc \plus{} ca}}\]
[/quote]

We have:
$ \frac {1}{\sqrt [3]{a \plus{} b \minus{} c}} \plus{} \frac {1}{\sqrt [3]{a \plus{} c \minus{} b}} \plus{} \frac {1}{\sqrt [3]{b \plus{} c \minus{} a}}\geq \frac {9}{\sqrt {3(ab \plus{} bc \plus{} ca)}}$  :oops:

Oh, sorry. Correct the problem:
$ \frac{1}{\sqrt [3]{a\plus{}b\minus{}c}}\plus{}\frac{1}{\sqrt [3]{a\plus{}c\minus{}b}}\plus{}\frac{1}{\sqrt [3]{b\plus{}c\minus{}a}}\geq\frac{9}{\sqrt{3(ab\plus{}bc\plus{}ca)}}$

[quote="smiley4t"]Oh, sorry. Correct the problem:
$ \frac {1}{\sqrt [3]{a \plus{} b \minus{} c}} \plus{} \frac {1}{\sqrt [3]{a \plus{} c \minus{} b}} \plus{} \frac {1}{\sqrt [3]{b \plus{} c \minus{} a}}\geq\frac {9}{\sqrt {3(ab \plus{} bc \plus{} ca)}}$[/quote]

See here http://forum.mathscope.org/showthread.php?t=10424 ;#5

I'm not sure where a+b+c=3 is needed to solve this, which makes me think I made a mistake  :maybe:  but anyway: Keep Ravi substitution in mind (a=x+y, b=x+z, c=z+y).  Applying the AM-GM inequality to the denominator yields on the RHS: \[ RHS\le\frac{3}{\sqrt[3]{abc}}\equal{}\frac{3}{\sqrt[3]{(x\plus{}y)(x\plus{}z)(z\plus{}y)}}\]
As for the LHS the AM-GM inequality gives: \[ LHS\equal{}\sum_{x,y,z}\frac{1}{\sqrt[3]{2x}}\ge\frac{3}{2\sqrt[3]{xyz}}\]
Now we need to show: \[ LHS\ge{RHS}\Leftrightarrow\frac{1}{2\sqrt[3]{xyz}}\ge\frac{1}{\sqrt[3]{(x\plus{}y)(x\plus{}z)(y\plus{}z)}}\]
\[ \Leftrightarrow\sqrt[3]{(x\plus{}y)(x\plus{}z)(y\plus{}z)}\ge{2\sqrt[3]{xyz}}\Leftrightarrow{(x\plus{}y)(x\plus{}z)(z\plus{}y)\ge{8xyz}}\]
Which is certainly true, once again due to the AM-GM inequality.

[quote="GregoryScott"]
$ RHS\le\frac {3}{\sqrt [3]{abc}}$[/quote]

I think you need the constraint for this step.

Actually I did make an error. xyz<=1/8 of course so my second step was incorrect. I did find a proper solution (using the constraint  :roll: ) which I'll try and post later.


group theory
abstract algebra
superior algebra
superior algebra unsolved

With respect to subgroups of a group; Is the trival subgroup the only subgroup of order 1?

Yes. Because the neutral element of a group is also the neutral element of its subgroups.

If there are two different trivial subgroups (subgroups of order 1 according to your definition), there would be 2 distinct neutral elements: say $e_{1}$ and $e_{2}$. But then  $e_{2}= e_{1}* e_{2}= e_{1}$, contrary to our assumption.


function
quadratics
number theory unsolved
number theory

Let $f(c)$ be a function where $f(c)$ is the remainder $c^{2}+2$ divided by $4$. For example, $f(1)=3$ and $f(2)=2$.Find all solutions in positive integers to the equation $a^{2}+(-1)^{b}f(c)=10d$.

[color=red] Edit by Megus: normal font size is enough I think[/color]

What is the equation?  I see no equality sign  :huh:

Since $f(c)=2$ or $3$ and $(-1)^b=1$ or $-1$, then $10d=a^2+3,a^2+2,a^2-2$ or $a^2-3$.
But $-3,-2,2,3$ are not quadratic recidues modulo $5$. Contradiction.


who voted latin? I mean I study it at school but the guy who speak latin as second language must be very funny. :D

Well, actually latin is a death language: it is not spoken anymore

[quote="karthik_k"]Sanskrit's my second language.... :)[/quote]

What's that?

[quote]What's that?[/quote]

Check this link......contains some interesting stuff about the language

http://en.wikipedia.org/wiki/Sanskrit

[quote="Gyan"][quote="karthik_k"]Sanskrit's my second language.... :)[/quote]

[size=150]सत्यम्   ऊवाच् ?[/size]

By the way, you could have given the [url=http://sa.wikipedia.org/wiki/%E0%A4%AE%E0%A5%81%E0%A4%96%E0%A4%AA%E0%A5%83%E0%A4%B7%E0%A5%8D%E0%A4%A0%E0%A4%82]this link[/url]

Or    [url=http://hi.wikipedia.org/wiki/%E0%A4%B8%E0%A4%82%E0%A4%B8%E0%A5%8D%E0%A4%95%E0%A5%83%E0%A4%A4]This one[/url]
Also  (आन्तरजालम्  in) google in  sanskrit at http://www.google.com/intl/sa/[/quote]

Actually, your post doesn't help anybody, beacuse no one can understand it.

[quote="José"][quote="Gyan"][quote="karthik_k"]Sanskrit's my second language.... :)[/quote]

[size=150]सत्यम्   ऊवाच् ?[/size]

/[/quote]

Actually, your post doesn't help anybody, beacuse no one can understand it.[/quote]
(It just said (In Sanskrit) "Is that true? (of being second language)"
(IOW, nothing profound, just spoken in a lighter  way .. should have added ":)")
(I took a couple of years of Sanskrit in school)
(Google is available in many languages including Sanskrit)

[quote="Gyan"][quote="karthik_k"]Sanskrit's my second language.... :)[/quote]

[size=150]सत्यम्   ऊवाच् ?[/size]

By the way, you could have given the [url=http://sa.wikipedia.org/wiki/%E0%A4%AE%E0%A5%81%E0%A4%96%E0%A4%AA%E0%A5%83%E0%A4%B7%E0%A5%8D%E0%A4%A0%E0%A4%82]this link[/url]

Or    [url=http://hi.wikipedia.org/wiki/%E0%A4%B8%E0%A4%82%E0%A4%B8%E0%A5%8D%E0%A4%95%E0%A5%83%E0%A4%A4]This one[/url]
Also  (आन्तरजालम्  in) google in  sanskrit at http://www.google.com/intl/sa/[/quote]

Aam. Aham Satyam eva avadham.  :D (To the others who're unfamiliar with Sanskrit , it just means "Yes, I did tell the truth only." ) :)

I would like to participate in this poll, but please explain the question.

What is second language?  You mean the language I feel most comfortable with, next to my own language?
Or the first other language I was taught at school?

Because living in bilingual Belgium "officially" my second language would have to be French and then English, but my English has become much better in the mean time due to it being easier, more common and amazingly more useful.

My vote was for the second language that i took in school.......  :|

My second language [i]technically[/i] (whatever that means to you) is english, because I was in a second-language class for english. However, it now looks like my second language is chinese or spanish from how well I can communicate in english.

I take Spanish and Latin for foreign languages, though I started Spanish in 5th grade and Latin just this year.

[quote="fredbel6"]So uhm, I repeat my questin, what is the point of the poll, official (like in school) or for yourself?[/quote]

I'll consider the school one

English and this section is here because "The message is too small. Please make the message longer before submitting."

My second language taught in school is English, although I started taking German since I was six, so really it is German. :lol:

I said "other" cause I'm Chinese, but I'm also learning French and Spanish.

I'm much, much better at reading French. This is probably because I didn't have a real French teacher this year and only a bunch of books.

If you are chinese, is Chinese not your first language then?

[quote]My second language taught in school is English, although I started taking German since I was six, so really it is German.
And your original language is?[/quote]
Six is quite young.  My niece in Atlanta is trying to raise her kid bilingual, but he seems to be very quiet (he is about five).  He doesn't say much in any language now actually?!

My original language is Bulgarian. At six I went to a  kindergarden specialized in German and now I have developed a very solid knowledge in the language. Both of my two siblings also speak German fluently.

Das ist sehr gut! :D  (I've only had two years, it is my fourth both officially and for real).

But you still live in Bulgaria (I noticed you don't have a flag)

Danke schon! I am not living in Bulgaria at the moment, which broke my brave dreams to participate in our second
IMO 2003-like performance (check   :D[url=http://www.imo.org.yu/index.php?options=gl|imotres&godzarez=03&p=009311]this[/url]) :D

Aw, too bad. They didn't include C++ as a second language.

[quote="fredbel6"]If you are chinese, is Chinese not your first language then?[/quote]

Perhaps I should have clarified. I'm actually second-generation American, my parents being Chinese, and I don't actually remember whether I learned Chinese or English first.

[color=#000]My second language is German.[/color]

I'm learning Mandarin chinese at school right now.
zai jian!
:)

[quote=alifenix-]Doesn't this violate #1 of the Round Table Posting Guidelines? (not trying to be rude, and this thread SHOULD be kept, but wondering...)[/quote]

[quote=daniellionyang]This thread is so bad it should go in the debate forum...[/quote]

It definitely needs to be moved. But not to the debate forum. There is nothing being argued.  It is odd how it was revived though. The last post before this one [quote=youdonotknowme]मपगलाेा गे सब ेामदल् तोलुहोुा.ेजोलगेप गे सब ूपगी्[/quote]

 was in 2006. And there are no tags to make it easy to find.

Woah this is a REALLY old forum! Let's keep it for nostalgia's sake...

History lesson: long ago there used to be a "Polls" forum on AoPS, that was eventually merged into the Round Table. So some very old (pre... summer 2008?) posts may be poll-only topics that do not involve discussion as they were not originally part of this forum. In fact, this is probably why rule #1 of the Round Table guidelines is disallowing "poll only" topics: they had a specific forum to go to.

In any case there is basically never a good reason to revive a 10+ year old thread.


More simply: "$n$ points lie on a circle.  In how many ways can I choose $k$ of them without choosing two adjacent points?"

Questions of this difficulty probably fit better in the Pre-Olympiad forum.


:( They seem to be gone.... all I see are the restricted (sysop-only) specials, not the public specials. The public specials seem to be deleted: Special:Popular, which appears valid in my browser history, tells me there isn't such a page... :(

Wow, I have NO CLUE of what is up with the [b]Special pages[/b] page.  I sat here reloading it multiple times and sometimes it would show the pages for the sysops only and other it would show the ones for all the users as well. HM.

Well, if they ever don't come up again...here they are as a backup:

[url=http://www.artofproblemsolving.com/Wiki/index.php/Special:Allpages]All pages[/url]

[url=http://www.artofproblemsolving.com/Wiki/index.php/Special:Mostcategories]Articles with the most categories[/url]

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We don't yet know why this is happening, but the list isn't a bad idea.


number theory
prime numbers
arithmetic sequence
number theory proposed

[quote="Sunkern_sunflora"]I don't know whether this problem has a solution or not:

Show that there exist [b]ten [/b]prime numbers in between 1 and 1000 which lie in A.P. Also determine the nos.[/quote]
May I know what is A.P and nos.?

Arithmatic progression and numbers of course!!!

I do not think that is possible. To have 10 primes in arithmetic progression, they must have a common difference divisible by 2, 3, 5, and 7. However, it is not possible, since the numbers will reach over 1000. I believe the largest arithmetic progression possible between 1 and 1000 is 7, 37, 67, 97, 127, 157, for 6 primes total.

If $ p_i\equal{}a\plus{}bi,i\equal{}0,..,n\minus{}1$ are primes, then $ n\#|b$, were $ n\#$ product of all primes less or equal n.
In these case $ 10\#|b$ or $ 7*5*3*2\equal{}210|b$. It give $ p_9\equal{}a\plus{}9b>210*9>1000.$


prove that if $ g \circ f$ is surjective and $ f$ is injective  we found $ f$ surjective .
 :lol:

[quote="greatestmaths"]$ gcerc f$ [/quote]
What is it?


In triangle $ABC$, $\angle A$ equals 120 degrees. A point $D$ is inside the triangle such that $\angle DBC = 2 \cdot \angle ABD$ and $\angle DCB = 2 \cdot \angle ACD$. Determine the measure, in degrees, of $\angle BDC$.

[hide="Answer"]If $\angle ABD=x$ and $\angle ACD=y$, then $3x+3y=60$, so $2x+2y=40$, and $\angle BDC=\boxed{140^\circ}$.[/hide]

[quote="Hokkage"][hide="Fun easy geo problem"]Let $\angle CBA = x$ and $\angle BCA = y$. So then $\angle ABD = 2x$ and $\angle ACD = 2y$. Summing the angles we know $120+3x+3y=180 \Rightarrow 3x+3y=60 \Rightarrow x+y = 20$. Therefore in $\triangle CDB$ we have $m \angle CDB + x + y = 180 \Rightarrow m \angle CDB + 20 = 180$ so therefore $m \angle CDB = 160^{\circ}$.[/hide][/quote]

Hokkage: BDC=140º.

Your mistake (with all my respect, don´t think I´m arrogant):

You wrote 120+3X+3Y=180, but, actually, it was 120+3/2X+3/2Y=180.

You got X+Y=20, but it was X+Y=40.


MIT college email ratio Princeton Harvard

has anyone got any idea regarding which iit has got  the best mathematics department?

IIT Kanpur and Kharagpur.
The Maths and computing branch in IIT Kharagpur is very prestigious and has very good placements.

Delhi offers a M.Tech Course in math and computing. Guwahati a B.Tech and KGP a MsC Course (I think??).

My sincere advice, if you are passionate about mathematics, then take up a course in CMI or ISI, not in an IIT.

I had opted for the maths and computing course at IITK (with second choice the one at KGP) before coming away to ISI Kolkata. I later heard that these courses are horrible. Maths at IITs sucks - be absolutely certain about that. IITs are for Computer Science, not maths. Pay heed to chronoz's wise warning and go for an actual maths course, if you really want to study maths that is.

Chronoz is right. My professor too feels if one want to study mathematics, better go for the [url=http://www.cmi.ac.in/]CMI[/url] rather than an IIT. Also, there are many people who opt for Aerospace in an IIT when there's a separate institue (and better too) for that, the [url=http://www.iist.ac.in/IIST/]IIST[/url].

its not just about the quality of teaching :D
the standard of insti and other facilities too matter :P

atleast for me :rotfl:

Can anybody also tell if I want to study Physics, where should I go?

I have a lot of confusion in it. A B.Sc. course at IIT is not prestigious at all, that's what all say. Moreover, IIT-JEE 2008 topper wanted to study physics but enrolled in B.Tech. at IITK. Did he do so because of the same reason? Also, I wanted to ask is the syllabus of B.Sc. and B.Tech. same for Physics.

for physics too, they have a course at CMI. though I'm not sure if that's the best option.
integrated m.sc. at iitk can't be bad for physics - IPhO gold medallist Ravishankar opted for that course, and was quite satisfied.

I don't want to get stuck at IITs till graduate degree. I want to go abroad either after XII or bachelor degree, so please guide me.

Moreover, which one is the best university or institute for research in Physics? Cambridge or MIT or any other?

Lakhsya, you are in the same position as me ;-)
You'd better go here- http://talk.collegeconfidential.com/
Thats the best site...
If you can tell me exactly what you want to major in, your academic, and extacurricular achievements, plus ability to play big college fees I'll 'try' to help you a bit more. I myself am very lost, but I have done a fair bit of research in this area. You might want to ask rt_08 too. For instance, these are some good universities for nuclear physics- MIT, UC Berkeley, Univ of Michigan, Univ of illinois, Penn State, and Georgia Tech[/list]

I want to do my major in theoretical physics especially related to String Theory. Is MIT good for it? Don't worry regarding fees, there are always scholarship exams. Now, can you guide me? I am moving to XI this year, so till now I have no major achievements despite those NSO and NSTSE ranks. And they are also not under AIR 30 even, so no major academic achievement.

Please guide me on which one to choose for research in theoretical physics.

And is it true that if one turns out on a great rank at IPhO, then MIT provides 100% fee waiver?

Awesome! You are pretty much in my area. You might want to major in nuclear physics too as there is no distinct course for 'string theory'. I have applied to a Canadian program which deals with stuff like that in a 2 week camp. Make sure you apply to these next year-
International Summer School for Young Physicists(Premier Institute Canada)
Summer Science Program(USA)
Reaserch Science Institute(USA-IF they have funding for Indian students in your year)
As far as aid is concerned, IPhO neither grants you admission or financial assistance in MIT. MIT, like the other TOPMOST universities is need-blind. Meaning, they don't give admission on the basis of achievements. Its only given if you need it. If you get selected in MIT, no matter how bad your financial condition is, they will give whatever aid you need upto 100%
Make sure you register at http://talk.collegeconfidential.com/ You'll find everything there. If you have specific questions, you can always ask :) And remember, MIT is VERY VERY VERY tough. EVEN IF you reach IPhO or something. They look at your complete personality. You must be UNIQUE. Coz there will be thousands of super smart students applying to MIT. They want people with a life. Not bookworms ;-)

Can I apply for those programs at such an early stage, I am currently in X and moving to XI this year? Anyways, I heard Sh!tikanth got 100% fee waiver from MIT, is it true?

Apply in 11th. And about Shitikanth, I dunno particular cases. Just the general rules. Go through MIT Admissions site once at least.

hello,
Lakshya i totally agree with nuclear_alchemist.MIT doesn't care about finances as it has huge endowments,the only thing which makes admissions tougher each year is competition and especially for internationals.In a country like India competition is fierce academically as well as liberally.Indians have earned a reputation of scoring highly in SAT'S (around 2200) which is why an asian is a bit of disadvantage.
As per your subject,US universities provide excellent education programs as well as career internships.Also i would also like you to minor in a subject(if you have any queries you may email me).Some universities which excel at this subject are MIT, U Chicago,Arizona State University,UCB(university of california,berkeley)
However the most important factor for most of us is cost.I was going through your previous post and i read that you were talking of scholarships.Well to be honest i couldnt find many scholarships.The research which i have done on US universities basically considers two types of aids:
1:Merit-Based Aid
2:Need-Based Aid
Here Merit-Based Aid is subsided in the form of scholarships,which is given on academic excellence and a merit record.
But here too admissions are not guranteed on academics which is the basic difference between US and UK universites. you need to be very good as well as a hooker(?)to get in the colleges which you are considering.
Also go through the websites of the universities of your choice and make your vitae at mywebstart.
ALL THE BEST(do contact me or @nuclear in case of any problems)
*Register at http://www.collegeconfidential.com and keep in touch with College forum of AoPS.

regards,
romil

I'll still stick to my view about aops college forum. It's pretty much DEAD. Collegeconfidential is waaaaaaaaaay better!
And I assume Romil refer's to the 'best fit for a college' system when he says 'hooker'?  :maybe: OR do you mean THE HOOKS(International Oly medal, patents etc.)?

Thanks to all who have poured in your advices. And yes, Romil, please clear the meaning of word 'Hooker'.
And what do you mean by mywebstart? BTW is Cambridge a good option?

hello,
@nuclear-->college forum is not "dead",it is just inactive due to the lack of user interaction and i would still prefer college forum as your starting point,the reason being college confidential is for those who want to "furnish" their applications and not start from square 1.

@lakshya-->i am really sorry for my lingual misthropy,its "mywebstat" and not mywebstart.Here you get to know the students who have got in top as well as mediocre universities and their relative variations of SAT as well as ACT scores.
The second point is about "HOOKER".Well hook is an uneven advantage you have over other participants.It can be classified on the basis of several points.Some of which are listed below:
1:International Olympiad medalist(even finalist)
2:Legacy
3:EA(applying early)
4:Foundationer's Society
5:Parental administration

Considering your next option about Cambridge(i suppose U.K.)
Well for that you need to fulfill different academic requirements,and focus more on education rather than co-cirricular activities.Also one plus point about studying in U.S. is that you will have a chance to get some good government scholarships.

regards,
Romil

EA is a hook? I didn't know that....
For shich universities? Or do you mean sophomore/junior year apps who apply again in senior year? But that takes place in very few univs like UC as well I guess.

Well if u review US universities carefully then you will come across this fact.It is not publicly distributed but the index of ratio of admissions is quite high for EA students.Especially Princeton,UCSD,UCB,Harvard etc..

[b]NOTE:You need to be careful considering universities coz wrong assumptions may lad to dismal failure[/b]

:omighty: The pretender stands unmasked! I owe all my knowledge to overlord rt!
But really, I'll have to discuss that EA thing with you sometime soon. But my outgoing is deactivated  :blush:

u seem to be funny  :rotfl:  :rotfl:  :rotfl:

And he sent me a message on his forums saying he'll ban me for spam!  :rotfl:

@Lakshya:

You should do a lot more research before calling courses in IIT as junk. IIT's are the best colleges in India and rank amongst the best in the world for engineering and science.
If you really want to pursue physics, then opt for Msc. Physics at IIT-K. It's probably the best course in India for undergraduate physics.
And don't evaluate a college by it's academic courses only, IIT's are the best because, they offer almost unlimited scope to pursue ANY interest. The environment in IIT is the best cuz you're amongst the best and the brightest in the country.

[hide]IIT doesn't offer Bsc. Courses at all.
BTW, I heard that JEE topper of 2008, Shitikanth is fighting to change his dept. to PHysics. :P
[/hide]


Ok, here is the problem.

$2x_1 + \ x_2 - \ 3x_3 = 4$
$4x_1 \ \ \ \ \ \ \ \ +2x_3=10$
$-2x_1+3x_2-13x_3=-8$

If I use the traditional elimination of variables, I get $0=0$ which obviously means that there is no SINGLE point solution.

In other words, if I eliminate $x_1$,
$4x_2-16x_3=-4$
$-2x_2+8x_3=2$

hence, $0=0$

The same logic applies to elimination of other variables.

The problem is I need find the solution set for this system.  I know I could use Gaussian Elimination, making the system in row-echelon form (in which the last line equation of the system would be 0=0, and the first two lines would be the solution set equation.)

But the quesiton is, can I do it without using Gaussian Elimination ?

You don't have to carry it all the way; you've shown that the solution is not unique, and you have a new equation that depends on only two of the variables. Pick one, and express the others in terms of it.

You are absolutely right.  Picking one of the original equations and the new equation will give us the solution set.

Is there a way to prove this rigorously?

Let's consider the matrix A of this system with the following columns (2 4 -2) (1 0 3) (-3 2 -13) .
Det(A)=0 then the solution set for the equation can be describes as follows: S=V+Ker(A) (V is a random solution of the equation)but we can find a simple expression for this set :
dim(ker(A))=1 :
because (2 4 -2) and (1 0 3) are no-linear so dim(im(A))>=2
moreover Det(A)=0 thus dim(im(A))>3
then dim(im(A))=2.
Hence the solution is just a straight line of the space ,you can keep two of the 3 equations you have ,now you have the equation of that straight line(intersection of two plans  ;) )


An equilateral triangle has side length 26 sqrt 3 cm. A point H is placed on the inetrior of the triangle such that its perpendicular distances to two sides of the triangle are 9cm and 14cm respectively. From point H, segments are drawn to the 3 vertices of the triangle to form 3 smaller triangles. What is the area of the largest of the 3 smaller triangles?

[hide="Solution"]

If the triangle is $ ABC$, perpendicular from $ H$ to $ AB$ is $ 10$ and perpendicular from $ H$ to $ AC$ is $ 9$,

 then the area of triangle $ ABH$ is $ 13\sqrt{3} \times 14$ and area of triangle $ AHC$ is $ 13\sqrt{3}\times 9$.

Since the area of triangle $ ABC$ is $ 13\sqrt{3}\times 39$, the largest of the three smaller triangles is $ BHC$ with an area of

$ 39\times 13 \sqrt{3} \minus{} 23\times 13\sqrt{3} \equal{} 16 \times 13\sqrt{3} \equal{} 208\sqrt{3}$.

[/hide]


[b]Problem.[/b] Given a cube of dimensions $2\times 2\times 2$ and spheres of total surface area $97$, prove that there exists a line that intersects at least $7$ of the spheres.

Enjoy!

Before this thread is forgotten, I will post my solution.

[b]Solution.[/b] Notice that if you project a sphere of radius $r$ onto a plane, it will occupy an area of $\pi r^{2}=\frac{S}{4}$, where $S$ is its surface area.  Project all the spheres onto a side of the cube.    We will prove that there exists a region that belongs to at least $7$ of the spheres; the line perpendicular to the side at that (those) point(s) will intersect $7$ spheres.  Since the total surface area of the spheres is $97$, the total area of the projections is $\frac{97}{4}>24=6(2\times 2)$.

And the conclusion follows. $\Box$

Are the spheres inside the cube?

If so, nice problem!

Yes they are; the title says so.  I noticed that the problem did not state that the spheres are inside the cube, but it was too late to edit my post.  Thanks.


geometry
geometric transformation
dilation

Given two circles $C_1$ and $C_2$, and their intersections $A$ and $B$, draw a line from $A$ to a point $P$ on $C_1$, and draw a line from $P$ passing through $B$ and ending after touching $C_2$; draw a line from  $A$ to $B$.  $AB=AQ$, where Q is the intersection of line $PB$ and $C_2$. Prove that $AP=PB$.

[img]http://www.boxedexe.tamranda.com/circles.GIF[/img]

EDIT: Finally, I managed to get up an image.

Maybe I didn't understand something but I don't think it's true:

Basically with any triangle APQ and a point B on PQ, we can construct those circles and work backwards, but the result for an arbitrary triangle (AB = AQ =>  PB = PA) doesn't seem right...

so far, i got nowhere, but the approach i'm going for is that with the lines extended, it forms a quadrics in C2.

Please tell me if you have a solution

You can dilate the length of PB freely without changing the AQ=AB condition. This will clearly change the lenth of AP at a different rate. Are you sure that the question is correct?

the shape u wote is wrong.

and also the problem is wrong...


function
calculus
derivative
integration
parameterization
calculus computations

I know that differentiability requires "smoothness" of a function and that intuitively this means that graph of the function does not have corners. However, what is the formal definition of smoothness? I'd assume that it has some topological interpretation.

The formal definition of smoothness is the existence of sufficiently many derivatives. Not useful here.

What are the conditions for the existence of a derivative then?

That the limit through which the derivative is defined exists.

This can be expressed in many ways. Here's one:

$ f$ is differentiable at $ x_0$ if there exists a number (which we will call $ f'(x_0)$) such that $ f(x_0\plus{}h)\equal{}f(x_0)\plus{}f'(x_0)h\plus{}o(h)$ as $ h\to 0.$ That is, the function is well approximated by a line, which we call the tangent line.

Really, the same thing occurs with anything that is defined by a limit. The derivative is defined by a limit. That limit doesn't always exist; when it does, we call the function differentiable. The integral is defined by a limit (or something like a limit). That limit doesn't always exist; when it does, we call the function integrable. And so on.

It's worth noting that "smoothness" (some geometric or topological consideration) is actually not sufficient: the graphs of $ y \equal{} x^3$ and $ y \equal{} \sqrt[3]{x}$ are graphs of functions and only differ by a rigid transformation, but only one of them is differentiable on all of $ \bf R$.

If you want to formalize that notion of smoothness JBL just mentioned, use parametric curves: there's a parametrization $ r(t)$ with enough derivatives, and $ r'$ is never zero. For $ y\equal{}\sqrt[3]{x}$, one such parametrization is $ r(t)\equal{}(t^3,t)$


[quote="Interval"]We know that any number raised to zero = 1.

We also know that any number raised to 1 = the number itself.

SAMPLES:

100^0 = 1

100^1 = 100

QUESTIONS:

Can you explain THE MATH REASONS why the above is true?

Also, what is zero raised to zero?

What is 0 raised to a power other than zero, say, a negative or positive number?[/quote]

Regarding the thing about $0^{0}$, I say that it is undefined.

The reason: take some logarithms:

If $0^{0}=1$, then $\log_{0}1=0$. No. it can't. Logarithms, I don't think, can have base 0, so therefore it must be undefined?

i'm probably wrong though :(

[quote="anirudh"]Regarding the thing about $0^{0}$, I say that it is undefined.

The reason: take some logarithms:

If $0^{0}=1$, then $\log_{0}1=0$. No. it can't. Logarithms, I don't think, can have base 0, so therefore it must be undefined?

i'm probably wrong though :([/quote]
$1^{1}=1$, obviously.
But $\log_{1}{1}=1$?? Can't have base 1 either.
:maybe:

[quote="The QuattoMaster 6000"][quote="i_like_pie"]1. $x^{0}$ can be expressed as $\frac{x^{1}}{x^{1}}$, which can in turn be expressed as $\boxed{1}$.

2. Using the above method, we obtain $\frac00$, which is undefined. Some argue that $0^{0}=1$, but I maintain that it is $\boxed{\text{undefined}}$.

3. Just like any other number, $0^{x}$ is just $0$ multiplied by itself $x$ times, which is $\boxed{0}$.[/quote]
I agree with one and two, but I would like to add something to 3.  0 Raised to a positive integer is 0 (as i_like_pie said), but i_like_pie forgot about 0 raised to a negative number.  This is undefined because $0^{-n}=\frac{1}{0^{n}}=\frac{1}{0}$, because $0^{n}=0$.  $\frac{1}{0}$ is undefined, and therefore, $0^{-n}$ is undefined assuming n is positive.[/quote]
I stand corrected. In other words, $a^{-n}$ is the same thing as $a$ divided by itself $n$ times. $0^{-n}$ is undefined because you can't divided by $0$.

Before any argument can be made about the validity of such things as $\frac{0}{0}$ one needs a rigorous definition of exponentiation, division, etc. Because most of the time, the real definitions of these concepts are indeterminate in these cases or have exceptions. In fact, the true answer is that the quantity is indeterminate, using limits in fact one can prove using similar logic that if it was determinate, then it could equal any real number...


1.If CosA-CosB = P and sinA-sinB=q express cos(A-B) and sin(A+B) in terms of P and Q
2.Find all the angles between 0degrees and 360degrees which satisfy the equation cosA + sin3A = cos2A
3. Find the general solution of the equation sinA-sin2A=sin4A-sin3A
4. Find all the angles between 0degrees and 360degrees which satisfy the equation cosX+7sinX=5

Please help ive been trying to solve these for 3 hours and couldent get them

[hide="2"]
We always have
\[\cos 4x-\cos 2x=\cos (3x+x)-\cos (3x-x) \]
\[=(\cos 3x\cos x-\sin 3x\sin x)-(\cos 3x\cos x+\sin 3x\sin x) \]
\[=-2\sin 3x\sin x \]
In particular, if $x=\frac{A}{2}$, then $\sin 3A=-2\sin \frac{3A}{2}\sin \frac{A}{2}$. But we always have $\sin 3A=2\sin \frac{3A}{2}\cos \frac{3A}{2}$. We now have two cases.

If $\sin \frac{3A}{2}=0$, then $3A$ is a multiple of $360^\circ$, so we get $0^\circ,\ 120^\circ,\ 240^\circ$. (I assume you want 0 but not 360.)

The other case is $\cos \frac{3A}{2}=-\sin \frac{A}{2}=-\cos \left( 90^\circ-\frac{A}{2}\right)$, then we get
\[\cos \left( 45^\circ+\frac{A}{2}\right) \cos (A-45^\circ)=0 \]
(This can be done by noting that $45^\circ+\frac{A}{2}$ and $A-45^\circ$ have sum $\frac{3A}{2}$ and difference $90^\circ-\frac{A}{2}$ and using trig addition similar to the first few lines.)

Now for this second case, the first factor equalling zero means $45^\circ+\frac{A}{2}$ is equal to an odd multiple of $90^\circ$. The only possible one with $0^\circ \leq A<360^\circ$ is $90^\circ$, in which case $A=90^\circ$. Otherwise, $A-45^\circ$ is an odd multiple of $90^\circ$, and can equal $90^\circ$ or $270^\circ$, giving $A=135^\circ$ or $A=315^\circ$.

All solutions for $A$ are $\boxed{0,90,120,135,240,315}$ degrees.
[/hide]

[hide="1"]
Square both sides of the given equations to get
\begin{eqnarray*}p^{2}&=& (\cos a-\cos b)^{2}= \cos^{2}a-2\cos a \cos b+\cos^{2}b\\ q^{2}&=& (\sin a-\sin b)^{2}= \sin^{2}a-2\sin a \sin b+\sin^{2}b \end{eqnarray*}
Adding both equations,
\[p^{2}+q^{2}= \cos^{2}a+\sin^{2}a+\cos^{2}b+\sin62 b-2(\cos a \cos b+\sin a \sin b), \]
and then using the identity $\sin^{2}\theta+\cos^{2}theta = 1$ to get
\begin{eqnarray*}p^{2}+q^{2}&=& 2-2(\cos a \cos b+\sin a \sin b)\\ \cos a \cos b+\sin a \sin b &=& \frac{2-(p^{2}+q^{2})}{2}\\ \cos(a-b) &=& \frac{2-(p^{2}+q^{2})}{2}\end{eqnarray*}
[/hide]

[hide="3"] We have: \[\sin x-\sin 2x=\sin 4x-\sin 3x\iff\sin 4x-\sin 3x+\sin 2x-\sin x\\ \iff \sin\frac{x}{2}\left(\cos\frac{7x}{2}+\cos\frac{3x}{2}\right)=0\iff\sin\frac{x}{2}\cos x\cos\frac{5x}{2}\] [/hide]

[hide="1.1"] We have: \[\begin{cases}\cos x-\cos y=p\\ \sin x-\sin y=q\end{cases}\implies p^{2}+q^{2}=2-2\cos (x-y)\\ \therefore\cos (x-y)=1-\frac{p^{2}+q^{2}}{2}\][/hide]


using the sum of a geometric series, it simplifies to

$\frac{7^{102}-1}{2400}$, we can do a mod thing, to say find

$7^{102}-1 \mod 2400$, then divide back by 6 to get the remainder mod 400,

we can easily do the above with the euler generalization, and chinese remainder theorem, look at mathworld to learn about those, or search


[hide]Isn't it just the numerator of the other fraction because they are equal?  If that is right, the answer is A[/hide]

No, the sign of the $y$ term in the denominator is negative.

We have $ \frac{x\minus{}1}x\equal{}\frac{y^2\minus{}2y\minus{}2}{y^2\plus{}2y\minus{}1} \equal{} 1\minus{}1/x$
So $ 1/x \equal{} \frac{4y\plus{}1}{y^2\plus{}2y\minus{}1}$
or $ x\equal{}\frac{y^2\plus{}2y\minus{}1}{4y\plus{}1}$, but this doesn't fit into any of the choices...  :huh:

[quote="4everwise"]If $ \frac {x}{x \minus{} 1} \equal{} \frac {y^2 \plus{} 2y \minus{} 1}{y^2 \boxed{\plus{}} 2y \minus{} 2}$, then $ x$ equals

$ \text{(A)}\ y^2 \plus{} 2y \minus{} 1 \qquad \text{(B)}\ y^2 \plus{} 2y \minus{} 2 \qquad \text{(C)}\ y^2 \plus{} 2y \plus{} 2$
$ \text{(D)}\ y^2 \plus{} 2y \plus{} 1 \qquad \text{(E)}\ \minus{} y^2 \minus{} 2y \plus{} 1$[/quote]

Um, this is the original problem, so $ \boxed{\textbf{(A)}\ y^2\plus{}2y\minus{}1}$ is correct.

OK, but how did you know that's the original question... did you make the guess based on the answer choices?  :wink:


Calculate the internal surface area of a bucket 20cm in diameter at the top and 12 cm in diameter at the botton and 16cm in depth                                                                                #2)given that p and q are constants and p^2+q^2=11pq,show that1/2(logp+logq)=#1 log(p-q)/3 #2 log(p+q)/ squareroot of 13.

[quote="21 7 15"]Calculate the internal surface area of a bucket 20cm in diameter at the top and 12 cm in diameter at the botton and 16cm in depth                                                                             [/quote]Your bucket is a truncated cone. Take into account that the 16 cm depth is the height of the truncated cone, not its slant height.

[quote="21 7 15"]#2) given that p and q are constants and p^2+q^2=11pq,show that1/2(logp+logq)=#1 log(p-q)/3 #2 log(p+q)/ squareroot of 13.[/quote]The formulas have log (p+q) and log (p-q). You need to make those sums appear. Start with p^2+q^2=11pq and "complete the square" by adding or subtracting 2pq on both sides. Then write the "completed" square as a square and turn the logs on.


Nhằm thúc đẩy phong trào yêu và làm toán của học sinh, sinh viên tập thể cộng đồng Mathvn cho ra đời tạp chỉ toán với mong muốn là nơi để các bạn yêu và đam mê toán giao lưu và trao đổi. Với mong muốn to lớn nhất là cộng động toán qui tụ lại xây dựng một môi trường làm toán lành mạnh và đầy sáng tạo...

Số đầu tuy đã cố gắng nhưng vẫn chưa thể nào đáp ứng thị yếu bạn đọc rất mong được các bạn đóng góp ý kiến để chúng tôi hoàn thiện hơn.

Để xem số đầu cũng như hiểu rõ hơn về tạp chí này, mời các bạn vào đây: http://mathvn.vn2k.net/viewpage.php?page_id=3

Thay mặt hội đồng biên tập của tạp chí
Vo Quoc Ba Can

Thanks Vo Quoc Ba Can, but your link isn't working

Here is the content of this issue (No01,vol 01)

[quote]Mục lục

+ Lời mở đầu

+ Câu chuyện Toán học

- Cứu sống nhân loại: Toán học trong chụp hình cắt lớp

- Danh ngôn Toán học

+ Bài viết Chuyên đề MathVn

- Định lý Stolz trong các bài toán giới hạn

- Định lý Hall và ứng dụng

- Lớp bài toán sử dụng phương trình Pell

+ Cuộc thi giải toán MathVn

- Đề toán dành cho Học sinh

- Đề toán danh cho Sinh viên

+ Bài viết Chuyên đề dịch thuật

- Đa giác Euler

+ Olympic Học sinh - Sinh viên

- Olympic Sinh viên TashSU 2008

- Olympic Sinh viên NSU 2008

- Olympic Xác suất Kolmogorov 2008

- Olympic Toàn Nga 2008

+ Góc lập trình tính toán

- Lập trình tính toán với Mathematica

+Giải trí Toán học

- Đố vui Toán học.

- Nụ cười Toán học[/quote]

[b]Download:[/b]

[quote][DJVU] (1.35 Mb)

http://www.mediafire.com/file/zxjcgymmmxx/v01n01.djvu

[PDF] (14.8 Mb)

http://www.mediafire.com/file/meitymmiumk/v01n01.pdf[/quote]

And the introduction of editors board:

[quote]Lời nói đầu

Trong những năm gần đây, đất nước chúng ta có sự trở mình đáng kể về mặt giáo dục. Mỗi người dân đều mong muốn nền giáo dục sẽ trở nên tốt đẹp hơn. Đã có nhiều tổ chức, cá nhân nêu ra nhiều ý kiến, nhận xét… Chung quy lại cũng là trăn trở của những người có tâm. Chúng tôi, những học sinh, sinh viên  còn rất trẻ nhưng lòng tràn đầy hăng say, nhiệt huyết, mong muốn đóng góp chúc sức lực nhỏ nhoi của mình cho công cuộc đổi mới giáo dục của đất nước, trong đó cụ thể là giáo dục Toán học. Đây chính là lý do chúng tôi tập hợp nhau lại, cùng nhau trăn trở suy nghĩ và làm việc nghiêm túc để cho ra đời một tạp chí Toán học dành riêng cho học sinh, sinh viên Việt Nam.

Trên thế giới hiện nay, có rất nhiều tạp chí Toán học dành cho học sinh và sinh viên, điển hình như Kvant, Reflections, Crux... Thậm chí, có nhiều khoa Toán của các trường đại học nổi tiếng thế giới cũng có các tập san sinh viên, từng tháng hay từng tuần đều có các chủ đề, các bài toán đưa ra cho những người quan tâm cùng nhau tham gia giải quyết. Cách làm đó đã thu hút được nhiều bạn trẻ trên toàn thế giới. Nhiều bạn học sinh, sinh viên Việt Nam đã tích cực viết bài, giải toán và được xưng tên. Thật đáng tự hào! Những bài viết hay, những lời giải đẹp của các tạp chí Toán học đã thổi vào hồn đam mê làm Toán của bao thế hệ học sinh, sinh viên chúng ta. Điều này thực sự quan trọng cho sự phát triển khoa học nói chung và Toán học nói riêng. Với mục tiêu không quá to lớn, chúng tôi muốn tạo ra một sân chơi cho học sinh, sinh viên nước nhà - một nơi mà các bạn trẻ yêu Toán có thể tham gia giải toán, gửi bài viết là những ghi chép các tìm tòi, ý tưởng sáng tạo trong quá trình học Toán của chính chúng ta. Nói như thế có nghĩa là từ các bạn học sinh phổ thông cho đến sinh viên, học viên cũng như các nhà sư phạm, các bạn yêu Toán nói chung chính là đối tượng mà chúng tôi hướng đến.

Chúng tôi đặt tên chính thức cho tạp chí là: Tạp chí Toán học MathVn. Về ban biên tập, chúng tôi là những sinh viên trẻ còn đang trên ghế giảng đường đại học trong nước và cả ngoài nước, năng lực phần nào còn rất hạn chế. Do đó, chúng tôi rất mong sự tham gia nhiệt tình của các bạn trẻ, sự ủng hộ của các tấm lòng yêu Toán tâm huyết, sự góp sức của các cố vấn khoa học về chuyên môn. Về mặt nội dung tạp chí, bên cạnh những bài viết mang tính chất lý thuyết-bài tập, đề toán hàng tháng, chúng tôi còn đăng những bài viết, bài dịch mang tính chất ứng dụng. Ý tưởng này chúng tôi học hỏi từ tạp chí Kvant của Liên bang Nga. Ở tạp chí Kvant, những bài viết như thế đã trang bị phương pháp tư duy và dẫn đường cho nhiều thế hệ tiếp cận với thế giới khoa học. Nói một cách dễ hiểu hơn, nó đã giúp nhiều bạn trẻ thực sự chọn được con đường đi của riêng mình và từ những bài viết nho nhỏ đó, họ biết rằng: "À! Thì ra lý thuyết, định lý này có ứng dụng thật là tuyệt vời!". Đây chính là cái hay mà chúng tôi cho là cần thiết phải đưa vào. Ngoài ra chúng tôi bổ sung một số chuyên mục như: Danh Ngôn Toán học, Góc lập trình tính toán, Giải trí toán học... Tuy nhiên vạn sự khởi đầu nan, các chuyên mục có lẽ là sẽ chưa được phong phú như mong đợi trong những số đầu tiên. Hy vọng rằng Tạp chí MathVn sẽ ngày càng hoàn thiện hơn, thiết thực hơn, phù hợp với đông đảo bạn yêu Toán.

Chúng tôi phát hành tạp chí điện tử qua website của Cộng đồng Học sinh - Sinh viên yêu Toán Việt Nam tại địa chỉ http://mathvn.org. Website này sẽ là một cổng thông tin, một diễn đàn giao lưu đồng hành song song với Tạp chí Toán học MathVn. Các bạn có thể đề xuất tất cả các ý kiến, góp ý chân tình tại ngôi nhà tinh thần này để chúng tôi có thể điều chỉnh, đổi mới sao cho tạp chí sẽ đa dạng hơn, đầy đủ hơn cả về "chất" và "lượng".

Và cuối cùng, xin trích ngang câu nói nổi tiếng của Giáo sư Hoàng Tụy: "Hãy hết lòng với khoa học, trong khó khăn hãy vươn lên, người tri thức không thể sống hèn. Con người không có cảm xúc, không rung động, vô cảm trước mọi việc thì không thể làm được bất cứ việc gì". Chúng ta hãy tiếp thêm niềm đam mê của mình trên những trang Toán, hãy trả về cho chính bản thân người học Toán những cảm xúc trong sáng và đáng quý thay vì tâm lý bài vở, thi cử hay áp lực thành tích đè nặng lên suy nghĩ của học sinh, sinh viên chúng ta bấy lâu nay để không ngừng vươn lên hăng say sáng tạo, tiếp bước trên con đường làm Toán vẻ vang của các thế hệ cha anh chúng ta.

Thể lệ viết bài cho Tạp chí

A. Thể lệ viết bài

Bài viết được soạn thảo theo một trong các hình thức:
- Sử dụng trình soạn thảo Microsoft Word và Microsoft Equation hoặc MathType, sử dụng font Times New Roman, mã Unicode đựng sẵn (lưu ý: không dùng Unicode tổ hợp).
- Sử dụng chuẩn soạn thảo LaTeX2e, gói UTF-8 Tiếng Việt.
- Bản chụp bài viết tay.

Bài viết được trình bày đơn giản, hạn chế dùng màu, định dạng phức tạp nếu không cần thiết.

Gửi thư kèm file dạng *.doc, *.tex và file hình ảnh hay các file mã nguồn khác nếu có sử dụng trong bài viết.

Ghi rõ đầy đủ thông tin cá nhân bao gồm:
- Họ và tên.
- Số chứng minh thư nhân dân.
- Trường đang theo học hoặc cơ quan công tác.
- Địa chỉ liên lạc.

Bài viết phải chưa được công bố trên bất cứ tạp chí, tập san hay website công cộng nào

B. Các mục dự kiến của tạp chí

-Bài viết chuyên đề: bài viết cho toán học sơ cấp, toán học cao cấp, bài viết chuyên đề dịch từ tạp chí nước ngoài.

-Câu chuyện toán học: các ứng dụng thực tế của toán học vào đời sống, chuyện kể về các danh nhân, giai thoại toán học, danh ngôn toán học.

-Đề ra kì này: bao gồm đề cho học sinh phổ thông và đề cho sinh viên đại học.

-Góc lập trình tính toán: sử dụng các phần mềm toán học và các thuật toán có ứng dụng trong học tập nghiên cứu.

-Giải trí toán học: các câu chuyện vui, đố vui toán học, thơ văn tự sáng tác của các độc giả.

C. Nhuận bút và giải thưởng

Sẽ có nhuận bút dành cho tất cả tác giả các bài viết và các bạn tham gia giải toán (được tính từ số đầu tiên), thông tin chi tiết sẽ được công bố khi chúng tôi đủ điều kiện về nguồn quỹ.

Tất cả các bài viết, bài giải các đề toán và ý kiến đóng góp của độc giả xin gửi về gửi về địa chỉ thư điện tử mathvn2008@gmail.com. Thông tin thêm, xin tham khảo ở website http://mathvn.org[/quote]

More information, visit at the site: http://mathvn.org/

[quote="abhilash"]Anybody translates some problems to english, please!!

I only read some inequalities, these're so nice   :roll:[/quote]
Hello my friend,
At the first issue, there are some inconveniences so we just publish the magazine with only Vietnamese. But from the second issue, we will publish it in both languages: Vietnamese and English. We hope you and the others will enjoy it. :)
Regards

[quote="abhilash"]Anybody translates some problems to english, please!!

I only read some inequalities, these're so nice   :roll:[/quote]
Hello, this is the English version of the problems. We hope you'll like it. :)

[quote=Lengendary]Anybody translates some problems to english, please!!

I only read some inequalities, these're so nice   :roll:[/quote]

http://molympiad.ml/2017/01/mathvn-journal/


[i][b]P[/b][/i] is a polynomial with integer coefficients. For four distinct integers
[i][b]x[/b][/i], we have [b][i]P(x) = 9[/i][/b]. Show that there are no integers [i][b]x[/b][/i] with [b][i]P(x) = 16.[/i][/b]

Suppose otherwise.  We have $ Q(x)\equal{}P(x)\minus{}9\equal{}(x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d)R(x)$.  But this is impossible since $ 16\minus{}9\equal{}7$ cannot be the product of at least 4 distinct integers.

Ok, I solved it the same way. I thought probably some other solution too existed to this.
Anyways, thanks man.  :lol:


Prove that if an arithmetic sequence has 2 integer terms, then it must have infinitely many.

Let d be the difference between each arithmetic term.  Let k be the first integer in the sequence.

Each term after that will be k+d, k+2d, k+3d, ...

Eventually, one of those other terms will be an integer.  We'll say it's x terms after the first, or m=k+xd. (m, being the second integer)

So m-k=xd.  Since m and k are integers, xd must be integral as well.  Adding an integer to an integer will always produce another integer, so just keep adding xd to get the rest of your integers.

That will go on forever and that's why there are infinitely many.

[quote="4everwise"]That's what I was going to type.[/quote]
Then type it! :P


Suppose the positive integers have been expressed as a disjoint union of arithmetic progressions $\{a_i+nd_i\}_{n=0}^{\infty}$,$i=1,2,\ldots,k$.Show that
$\sum_{i=1}^k\frac{a_i}{d_i}=\frac{k+1}{2}$.

[quote="jin"]Anyone can solve it? :?:[/quote]

Patience! Give people more time. This is holiday season.

On first thoughts, generating functions seem useful. For instance we can show that  if $k > 1$ then  $d_i = d_j$ for some $i \neq j$.

Very interesting question btw. Thanks for posting this.

[quote="harazi"]Check the number theory forum. I solved it there.[/quote]

Link please...

sorry i can't write with english.

d'abord on remarque que les progression arithmétiques soient disjointes que lorsque di=d pour tous i appartient à {1..k}.
deuxiément supposons d>k alors pour des arguments de congruences on remarque facilement que les progressions ne couvre pas IN* tous entier.
si d<k alors on remarque qu'il existe i et j tel que ai=aj mod(d) d'ou deux progression sont presque partous égales.d'ou le seul cas qui reste est d=k.et on remarque que les ai ne peut etre que les 1,2,..,k pas necessairement dans le meme ordre.
d'ou sum(ai/di)=1/k+2/k+...+k/k=(k+1)/2

Anass

[quote="RAISSOUNI Anass"]
d'abord on remarque que les progression arithmétiques soient disjointes que lorsque di=d pour tous i appartient à {1..k}.[/quote]

No, this is untrue.
For exemple, use the three arithmetic sequences $(2k+1),(4k),(4k+2)$.

Pierre.


Insert two real numbers (one positive and one negative) between $ 2$ and $ 9$ so that the first three terms form an arithmetic sequence and the last $ 3$ terms form a geometric sequence. Find these two numbers.

Let the terms be $ 2,9r^2,9r,9$ so that the ratio is $ \frac1r$ (this is negative as well!) in the last three terms. Then

$ 9r^2\minus{}2\equal{}9r\minus{}9r^2\implies18r^2\minus{}9r\minus{}2\equal{}(6r\plus{}1)(3r\minus{}2)\equal{}0,r<0$

$ \implies r\equal{}\minus{}\frac16\implies2,\boxed{\frac14,\minus{}\frac32},9$.


inequalities
geometry
trigonometry
area of a triangle
geometry proposed

Prove the inequality

$\frac{1}{h_{a}^{2}}+\frac{1}{h_{b}^{2}}+\frac{1}{h_{c}^{2}}\geq\frac{\sqrt3}{S}$,

where $h_{a}$, $h_{b}$, $h_{c}$ are the altitudes and $S$ is the area of a triangle $ABC$.

[color=blue][b]Problem.[/b] Prove the inequality

$\frac{1}{h_a^2}+\frac{1}{h_b^2}+\frac{1}{h_c^2}\geq\frac{\sqrt3}{S}$,

where $h_a$, $h_b$, $h_c$ are the altitudes and S is the area of a triangle ABC.[/color]

[i]Solution.[/i] Since the area of a triangle equals half the product of one of its sides with the corresponding altitude, we see that the area S of triangle ABC equals $S=\frac12ah_a$. Thus, $h_a=\frac{2S}{a}$, so that $h_a^2=\left(\frac{2S}{a}\right)^2=\frac{4S^2}{a^2}$ and $\frac{1}{h_a^2}=\frac{a^2}{4S^2}$. Similarly, $\frac{1}{h_b^2}=\frac{b^2}{4S^2}$ and $\frac{1}{h_c^2}=\frac{c^2}{4S^2}$. Hence, the inequality $\frac{1}{h_a^2}+\frac{1}{h_b^2}+\frac{1}{h_c^2}\geq\frac{\sqrt3}{S}$ becomes

$\frac{a^2}{4S^2}+\frac{b^2}{4S^2}+\frac{c^2}{4S^2}\geq\frac{\sqrt3}{S}$.

Multiplication by $4S^2$ transforms this into $a^2+b^2+c^2\geq 4\sqrt3S$. But this is the famous Weitzenböck inequality, proven at http://www.mathlinks.ro/Forum/viewtopic.php?t=14388 , http://www.mathlinks.ro/Forum/viewtopic.php?t=45611 and http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 .

  Darij

Well, here is an alternative solution. First let's write the problem again.

 [color=green][b]Problem[/b]

    Prove that in any trinagle $ABC$ the following inequality holds:

                          \[ \frac{1}{h_{a}^{2}}+\frac{1}{h_{b}^{2}}+\frac{1}{h_{c}^{2}}\geq\frac{\sqrt{3}}{S},  \]
    where $h_{a}, h_{b}, h_{c}$ are the altitudes of the triangle $ABC$.
  [/color]
 [i]Solution[/i]

     Let us apply Cauchy-Schwarz inequality. Well, we get

                                \[ \frac{1}{h_{a}^{2}}+\frac{1}{h_{b}^{2}}+\frac{1}{h_{c}^{2}}\geq\frac{1}{3}(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}})^{2}  \]

     But, since $a\cdot h_{a}=2S$ and the analogues, we get \[ \frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{2p}{S}.  \]

     Now, our problem reduces to Mitrinovic inequality \[ p^{2}\geq 3\sqrt{3}S,  \]
     which can be easily solved by applying AM-GM like this:

                   \[ p=(p-a)+(p-b)+(p-c)\geq3\sqrt [3] {(p-a)(p-b)(p-c)}.  \]
     From here you know how to proceed. :)  ;)

[i]Mapedd[/i], I think must "[u]Thanks Darij Grinberg and Cezar Lupu[/u]".

yes of course  thanks for darij and cesar,  i'm very thankful for that prove  :blush:

I am sorry for my words. You are very kind young. It is O.K.

sorry but what is the am gm?

edit:
 oh it's an average irregulary! :blush:

i dont understand this prove, can somebody explain it for me?

[quote="fuzzylogic"]$a^2+b^2+c^2-4\sqrt{3}S = a^2+b^2 + (a^2+b^2-2ab\cos{C})-2\sqrt{3}ab\sin{C}$ $= 2(a-b)^2 + 4ab(1-\cos(C+\pi/3)) \geq 0$[/quote]

cos of the sum? from where?

[quote="mapedd"][quote="fuzzylogic"]$a^2+b^2+c^2-4\sqrt{3}S = a^2+b^2 + (a^2+b^2-2ab\cos{C})-2\sqrt{3}ab\sin{C}$ $= 2(a-b)^2 + 4ab(1-\cos(C+\pi/3)) \geq 0$[/quote]

cos of the sum? from where?[/quote]

It actually should be $C-\frac{\pi}{3}$ instead of $C+\frac{\pi}{3}$.

You have

$a^2+b^2+\left(a^2+b^2-2ab\cos C\right)-2\sqrt3ab\sin C$
$=2\left(a^2+b^2-2ab\right)+4ab\left(1-\left(\frac12\cos C+\frac{\sqrt3}{2}\sin C\right)\right)$
$=2\left(a^2+b^2-2ab\right)+4ab\left(1-\left(\cos\frac{\pi}{3}\cos C+\sin\frac{\pi}{3}\sin C\right)\right)$.

Now, by the formula $\cos x\cos y+\sin x\sin y=\cos\left(x-y\right)$, applied to $x=\frac{\pi}{3}$ and y = C, we get $\cos\frac{\pi}{3}\cos C+\sin\frac{\pi}{3}\sin C=\cos\left(\frac{\pi}{3}-C\right)=\cos\left(C-\frac{\pi}{3}\right)$, so that this becomes

$a^2+b^2+\left(a^2+b^2-2ab\cos C\right)-2\sqrt3ab\sin C=2\left(a^2+b^2-2ab\right)+4ab\left(1-\cos\left(C-\frac{\pi}{3}\right)\right)$.

I guess this is what you wanted to show.

  Darij

and again i have to thank you, darij, you helped me a lot, happy new year!


advanced fields
advanced fields unsolved

The Banach-Mazur distance between c and c_0 is 3,that is d(c,c_0)=3 but d(c',c_0')=1 (in fact c' and c_0' are isometrically isomorphic).
Question:Let n be a natural number,does there exist two Banach spaces X,Y so that X',Y' are isometrically isomorphic but d(X,Y)>=n?
Does there exist two Banach spaces X,Y so that d(X,Y)>=nd(X,Y)?
Patrick

[quote]Question:Let $ n$ be a natural number, does there exist two Banach spaces $ X,Y$ so that $ X',Y'$ are isometrically isomorphic but $ d(X,Y) \ge n$? [/quote]

Take $ X\equal{}C(\omega^n)$, $ Y \equal{} c_0$. The dual spaces to $ X$ and $ Y$ are both isometric to $ \ell_1$. $ X$ and $ Y$ are isomorphic, but $ d(X,Y) \ge n \plus{} 1$.

[b]Remark.[/b] I don't know the exact value of $ d(C(\omega^n), c_0)$, but I can show the estimate $ d(C(\omega^n), c_0) \ge n \plus{} 1$.


For #1, I'm too lazy to write my solution, but I believe that $\frac{1}{6}$ is the only solution.[/hide]

[quote="computer_nuke"][quote="xxxyyyy"]$1.$ consider $1/6 = .16666\cdot\cdot\cdot$
Find out another fraction with a similar property that is to say $a/b = .abbbb\cdot\cdot\cdot$
Please, present support for your answer.

$2$ Have a look at the number $53$ and analyse it. It is a $2$-digit prime but also has a few notable characteristics. Its digits $5$ and $3$ are primes themselves. What about $173$? It's a $3$-digit prime where the digits $1$ (treat it as a prime for this problem), $7$, $3$, $17$, and $73$ are also primes. Find out the largest number, say a $n$-digit number, which is prime, each pair of consecutive digits are prime themselves, each triplets of consecutive digits is prime, and so on.
If the answer is $abcd$, then $a$, $b$, $c$, $d$, $ab$, $bc$, $cd$, $abc$, $bcd$ and $abcd$ are all primes. (Do not forget to treat the single digit $1$ as a prime though it is not!!!!)[/quote]

For the first question, I assumed (from "$a/b = .abbbb\cdot\cdot\cdot$") that:

1) $a$ and $b$ are $1$ digit integers in base $10$

2) $b>a$

And from the above, I got that only $\frac{1}{6}$ works for the condition.[/quote]

Can you show the method by which u have arrived at the conclusion?

Well you just set it up:

$\frac{a}{b}=\frac{a}{10}+\frac{b}{90}=\frac{9a+b}{90}$

Simplify by cross multiplying:

$90a=(9a+b)(b)=9ab+b^{2}$

Move all but $b^{2}$ to the LHS to get: $(9)(a)(10-b)=b^{2}$

From the assumptions I listed above, you know that there are only so many possibilities for $a$ (there should be only 10: 0, 1, 2...9). List them and then next to each list the possible values of $b$. Now cross all the ones where $b<a$ which is against the rules listed above. Then plug each value left (there shouldn't be much more left) to check which work and which don't work. You see that only when $b=6$ and $a=1$ the problem works out.

Thus there is one and only one fraction that has the property $\frac{a}{b}= .abbbb\cdot\cdot\cdot$

[quote="miyomiyo"]As for # 2,
[hide="Step 1"]
If you ever get to a number that is congruent to 2 mod 3, the following number cannot be 1 or 7 (this would be a multiple of 3) or 5 (this would be a multiple of 5).  If it is followed by a 3, we remain congruent to 2 mod 3.  We still cannot tack on another 1, 5, or 7, and now not even a 3 (33 is a multiple of 11).
[/hide]

[hide="Step 2"]
By this logic, the greatest number with a 5 in it is 53.  But even 137 is greater than that, so we rule out 5.
[/hide]

[hide="Step 3"]
The only numbers left are either congruent to 1 mod 3 or 0 mod 3.  The largest possible number then is 5 digits because 3's cannot follow themselves, therefore any 6 digits number has 3 numbers that are congruent to 1 mod 3, therefore this number is a multiple of three.
[/hide]

[hide="Step 4: The final step!"]
There are many 5 digits numbers that work; let's find the largest.  We need to start with a 7.  Next we need a 3.  Then another 7.  Then 3.  Then 7.  If you want to, you can work it out, but it is true that 73737 is prime.  Therefore are answer is...
[/hide]

[hide="Answer"]
$\boxed{73737}$
[/hide]

Nice problem, xxxyyy.[/quote]

$73737$ is not valid as $737$ is not a prime. $737 = 11\times67$.

For first problem, I don't see why one should consider $b>a$.. if you don't hold that condition, then you hae the solution $a=b=9$
Because : $\frac{9}{9}= .9999\dots$ :)

[quote="miyomiyo"]Then it would be 73137[/quote]

$7$ is a prime; $73$ is a prime; $731$ is NOT; $13$ is a prime; $137$ is a prime; and $73137$ is NOT.

$731 = 17 \times 43$
$73137 = 3 \times 24379$

[quote="miyomiyo"]Sorry.  There was an error in my solution.  It needs to go 3_3_3, not _3_3_.  The answer is 37373, then.  Is that prime?[/quote]
Not if it is divisible by 7  (or 19 for that matter) :)

(Not to mention that 3737 (or 7373) are divisible by 11)

731 isn't prime.

Sorry if we're annoying you.

31373=229*137...

If the two blanks are the same, then some of the four digit numbers are divisible by 11....

So that eliminates 3_3_3.

It can't be 5_5_5 or _5_5_.


I think my eyes are bleeding.

Can someone move this to HSB?

My eyes are bleeding, too...but it looks fun. :P
Hmm. I noticed something strange that may help or just be completely wrong. (I just looked at it and figured it out and haven't quite thought it through so, I don't know...)
But here goes. :P
(And this is only working with one part of the system, assuming it's all one system.)

If $ \frac {x}{y} \equal{} \frac {z}{x}$,

and $ \frac {z}{y} \equal{} \frac {x}{z}$,

$ \frac {y}{x} \equal{} \frac {z}{y}$,

Then shouldn't these be true?

$ x^2 \equal{} yz$

$ z^2 \equal{} xy$

$ y^2 \equal{} zx$

(By cross-multiplication)

Now we manipulate some equations.

If $ y^2 \equal{} zx$, then $ (y^2)^2 \equal{} (zx)^2$

$ y^4 \equal{} {z}^2{x}^2$

Now we solve for $ x^2$:

$ x^2 \equal{} \frac {y^4}{z^2}$

Then, since $ x^2 \equal{} yz$, we can set up this equation:

$ yz \equal{} \frac {y^4}{z^2}$

Solve:

$ y z*z^2 \equal{} y^4$

$ z^3 \equal{} y^3$

Now shouldn't $ z \equal{} y$?
Now assuming this is true, (which it might not be, I'm not sure how sound my logic is..),
using a previous formula, we can then sub $ y$ in for $ z$.

$ x^2 \equal{} yz$

$ x^2 \equal{} yy \equal{} y^2$

$ x^2 \equal{} y^2$

Now shouldn't $ x \equal{} y$ by taking the square root of both sides?

So, shouldn't $ x \equal{} y \equal{} z$?

If this is all completely false....I wouldn't be surprised.
But, um, this isn't the only place where I saw an "identity"-like problem. The ones that look evilly-hard, but are actually easy due to some sort of property thing.
Like where you have:

$ x[(z \minus{} y)(z \plus{} y)] \equal{} 5y$

$ y[(z \plus{} x)(z \minus{} x)] \equal{} 10x$

$ z[(x \plus{} y)(x \minus{} y)] \equal{} 10x \minus{} 5y$

I don't know. My gut just screams identity. Perhaps I'm missing something?

My questions:
A) Where did you find this?
B) What do you mean by: X^2+ 2(x) +x^4=x^-4(19)
Do you mean-
Option Alpha) $ x^2 \plus{} 2x \plus{} x^4 \equal{} \frac{19}{x^4}$
---------or----------
Option Beta) $ x^2 \plus{} 2x \plus{} x^4 \equal{} \frac{1}{x^76}$
(It's supposed to say x^-76. :mad:)

YEAH SRY my first line was wrong ,i edited sry hahahaha, its all in one system. I did all the cross multiplication work ,and its so confusing.........its option B

Replying to Question A= Erm.............from some stupid book i guess?


(ABCD)=-1.and M is the mid point of AB.prove that:

1/(CA.CB)+1/(DA.DB)=1/(MA.MB)

for the ones who dont know what harmonic is:{(ABCD)=-1,means AD.BC=AC.BD,and A,B,C,D are for colinear

points}

may be i should clear it:

1:(CA.CB)+1:(DA.DB)=1:(MA.MB)

i think it can be solved only by using similarity,and harmonic just make it looks hard :(


Salutations!

Mods:
I wasn't exactly sure where to put this--so I put ti here. If you need to move it, please help me do so.

I'm thinking about starting a math competition at my school.  The first year, as I've experienced with hosting our first forensics tournament, is the rockiest.  Thus, I am not exactly sure what to expect.  I've been to some high school math contests.  But I'm not exactly sure what to include in our math contest.  So I'm asking you guys for help--I bet you would know a lot more about this then I do, and would appreciate any sort of help possible.

Here are some of the things I think I need to consider:
What time of year? During the week right after school? During school? During the weekend?
Food/concessions? Meal ticket?
Location within my school?
What kind of test? Levels of tests? Difficulty? Multiple choice or written answer? Scoring?
Volunteers? Am I going to run it all?
Number of people to host? Limit the number of kids from each school?
Coaches' lounge?
Quiz bowl math? Layover time?
Entrance fee? Registration Fee? Registration Check-in?
Map? Signs?
Award stuff? Medals, certificates?

My math club's sponsor hasn't really done much...and I don't think he intends to do much for the competition, so the head of the math department (who used to run math club) told me to draw up the prospectus, the proposal for what exactly we're going to do. :ninja:  :o

Help?! Any would be much appreciated! :D

Advice: [url=http://web.mit.edu/hmmt/www/datafiles/coordinators/2009.shtml]Here[/url] are the coordinators for the 2009 HMMT. Contact them and see if they could give you any tips.

Also go to competitions like NYSML, PUMaC, and ARML, and ask "How the hell do you guys manage all this?"

:lol:
Got it. Next competition I go to, find the director, shake his hand and then go, "So...how the heck do you manage all this? :ninja:"

Thanks! I am off to go bug the HMMT coordinators.