I have this fear that if I study what I enjoy (i.e. Mathematics or Science) that I will end up with no job. My parents (not so much my dad) and their friends (all Asian so certain cultural contexts are introduced) are always on about how if you become a Doctor you are almost certainly guaranteed a job etc. and if you study Science, Computing, Mathematics etc. than you have to be the top of your field or else you end up making little/no money.

Well money is not something I am concerned with too much. For me, it is enough that I earn a living. I don't need to earn a lot so I can pursue an elaborate life style or anything. My passions are their own reward because I just love Mathematics and Science. I mean I would rather spend an hour working on a Math problem. At the same time, a large percentage of the graduates end up not doing Mathematics or Science (in Australia anyway) and I don't want to fall into that category.

How do I convince myself that my fears may just be theresult of someone who is stressed and indecisive about what he wants to do after University.

I read in another post that you got accepted into UNSW at 16! That's gotta be indication of significant genius! The smartest people at my school (Scotch College, Melbourne) would likely be less suited than you to a career in mathematics! You've got way too much doubt.

Glad to see a Scotch boy browsing AoPS  :D

VCE is almost over! :)

Oh, and I too find it difficult to choose a career path. Deep down I'd love to do mathematics, but will I really be able to get a job, raise a family? It's something you must consider. Right now I'm sort of settling for actuarial while I learn maths as a hobby. I hope it works out!

[quote="Identity"]I read in another post that you got accepted into UNSW at 16! [/quote]

Well I am not technically in. I need to be interviewed for my 'maturity'. But I reckon that won't be too hard.

Go look at the "Jobs in Mathematics" forum (or whatever it's called).  A mathematics education is extremely broadly applicable, not because of the higher mathematics involved but rather because of the analytic and problem-solving skills taught.  (It's also true that many people studying mathematics learn in addition some economics, statistics and other more "applicable" knowledge.)  Also, there have been lots of posts like this before -- I recall one recently in the College forum.

I forgot to mention I'm going to get a diploma in mathematical sciences while studying actuarial!  :lol:

if money's not the problem, then u should definitely go for it

The feelings of stress/indecisiveness/depression- that's very common around here, whether justified or not. It's linked to the same personality types that enjoying math is. You may not be able to get rid of those fears- but know that there's a very good chance you'll be able to spend your life getting paid for doing something you enjoy. That's a good place to start on happiness.
I'm following the academic path myself.

You should pursue career in mathematics + science. You are good at it and also enjoy it. Who said you can't be at the top of your field anyways? I don't know many people who do 4unit at an early age.......
In fact i dont know anyone ....................

Do what you like

Where I live, most academic degrees tend to be mostly an indication of "none-dumbness".  Meaning that if you were able to become an engineer or a mathematician or a physicist, you are probably disciplined and clever enough to do a job involving science, calculations,......

I'd definitely go for what you like, take it from someone who thought just like you and decided to do mathematics anyway after less than one year. :|

Math is a perfectly good major to pursue, even if you are interested in making money.  Quants make perfectly good money (this is an understatement).  For example, show your parents this (somewhat silly) article:

http://news.bbc.co.uk/2/hi/business/7109805.stm

But really, a math major leaves the door open for a variety of careers, *including* medicine.  One thing you could do is tell your parents that you are majoring in math (or whatever) but still planning to go to medical school.  You might even claim that it will look good on your medical school application.

Just thought I'd mention that I've changed my preferences and will be majoring in maths next year, despite all the "won't find job yada yada, no money" ranting. Go for it!  :D

Thanks for all the feedback. I feel so much better now.  :lol:


function logarithms trigonometry vector inequalities probability conics

Hey guys, what's a good pre-calc textbook? I'm going to get one just for reference and whatnot.

For my summer self study I chose "Precalculus - Fuctions and graphs - A graphing approach" third edition. I think it was published by the Houghton Mifflin Company. The authors are Ron Larson, Robert P. Hostetler, and Bruce H. Edwards.

It pretty much covers functions, polynomial/rational functions, exponents and logarithms, trig functions, analytic trig, vectors, system of equations and inequalities, matrices and determinants, proofs, series and sequences and probability, and conic sections. They also have important theorems and their proofs.

IMO it's pretty good.

OK, I remember I used around eight years ago the book

Algebra and Trigonometry with Analytic Geometry, Earl W. Swokowski.

Now, it is not the new editions. I don't remember exactly what edition I used, it was in Spanish, with a Rollercoster in the front. Unfortunatelly, latest editions are not that good. The problems have been simplified and the theory discussion was more vague (believe me, I had to tought from the book a few years after that, and I prefered to use the old version).

Therefore, if you have a library available, well, take a look on it. The edition has to be prior to 1997.

Best,

Hey, that's the book I'm studying from right now! Must skip precalc... concentrate efforts... no life studying... aghh. The person that writes the tests hates ppl skipping. She's going to put problems like: Define dot product word for word from your book.

[quote="ankur87"][quote="Derek"]For my summer self study I chose "Precalculus - Fuctions and graphs - A graphing approach" third edition. I think it was published by the Houghton Mifflin Company. The authors are Ron Larson, Robert P. Hostetler, and Bruce H. Edwards. [/quote]

Is that the orange one?[/quote]

Well it's kind of goldenish, orangish, and it has some kind of rectangular grids on the cover.

David A. Cohen's Precalculus with Unit-Circle Trigonometry third edition. It has good problems, good coverage of all the important topics, and excellent historical information about the development of mathematics. We have about six other precalculus textbooks at home, but that's the one I bought with the most gladness, and it's the one used right now in my son's accelerated precalculus course.

This year, I used Advanced Mathematics: Precalculus with Discrete Mathematics and Data Analysis by Richard G Brown, published by McDougal Littel Houghton Mifflin

I liked it, but I'm not sure how good it would be for self-studying. I think it's pretty useful, and I'm keeping it in the future in case I want to review Precalc or Trig for any reason.

I used the same book as Yifman, for self study to skip Precalc. Great book. Real good book to use for self studying.

[quote="yif man12"]I liked it, but I'm not sure how good it would be for self-studying.[/quote]

Oh sorry. Um...I took pre-calc years ago. I just need a textbook for reference purposes now. So I'm not actually looking for something with good practice problems (since I won't be wasting time doing them) or good proofs or whatnot but rather a wide range of topics and formulas, etc.

Oops, sorry. I looked at Derek's post and sort of just got confused.  :blush:  :D

I like it for reference though, since that answers the question.  :D

I like the UCSMP series (except for their geometry section)

[quote="MithsApprentice"] So I'm not actually looking for something with good practice problems (since I won't be wasting time doing them) or good proofs or whatnot but rather a wide range of topics and formulas, etc.[/quote]

Well, if you just want lots of topics and formulas you can't go wrong with Larson, Hostetler, and Edwards. I didn't particularly like it when I studied from it last year, but if you're not going to be trying to learn from it there's no harm done.


algebra
polynomial
linear algebra
matrix
number theory
least common multiple
superior algebra

Let  $ r,s$,and $ t$ be positive integers greater than $ 1$.Show that there is a finite group $ G$ haveing elements $ x$ and $ y$ such that $ x$ has order $ r$,$ y$ has order $ s$,and $ xy$ has order $ t$.

Ok, this took me forever to find, but I found them in the finite group $ PSL_2(q)$ for some odd prime power $ q$.

Take some odd prime $ p\nmid r,s,t$ and some extension field $ \mathbb{F}_{p^n}$ of $ \mathbb{F}_{p}$,
in which $ X^{2r} \minus{} 1,X^{2s} \minus{} 1,X^{2t} \minus{} 1\in \mathbb{F}_p[X]$ split into linear factors.
But $ q: \equal{} p^n$ is coprime to $ 2r,2s,2t$, so these polynomials have no multiple roots
and we find primitive $ 2r,2s,2t$-th roots of unity $ \zeta_{2r},\zeta_{2s},\zeta_{2t}\in\mathbb{F}_q$.

Take $ A \equal{} \begin{pmatrix}\zeta_{2r} & \\
& \zeta_{2r}^{ \minus{} 1}\end{pmatrix},B \equal{} \begin{pmatrix}\sigma & (\zeta_{2s} \plus{} \zeta_{2s}^{ \minus{} 1} \minus{} \sigma)\sigma \minus{} 1 \\
1 & \zeta_{2s} \plus{} \zeta_{2s}^{ \minus{} 1} \minus{} \sigma\end{pmatrix}\in SL_2(q)$ for some nice $ \sigma\in\mathbb{F}_q$
and let $ x,y$ be the corresponding elements of $ PSL_2(q) \equal{} SL_2(q)/\{\pm I_2\}$.

Then $ A\in SL_2(q)$ has obviously order $ 2r$, which gives $ x\in PSL_2(q)$ of order $ r$,
as $ A^r\equal{}\minus{} I_2$ is the only involution in $ SL_2(q)$.

$ B$ has characteristic polynomial $ X^2 \minus{} {\rm tr} B\cdotp X \plus{} \det B \equal{} X^2 \minus{} (\zeta_{2s} \plus{} \zeta_{2s}^{ \minus{} 1})X \plus{} 1$ and eigenvalues $ \zeta_{2s}\not \equal{} \zeta_{2s}^{ \minus{} 1}$.
So $ B$ must be diagonalizable, which analog to $ A$ gives $ B\in SL_2(q)$ of order $ 2s$ and $ y\in PSL_2(q)$ of order $ s$.

And now $ AB$ has characteristic polynomial $ X^2 \minus{} {\rm tr} AB\cdotp X \plus{} \det AB \equal{} X^2 \minus{} {\rm tr} AB\cdotp X \plus{} 1$,
and if $ {\rm tr} AB \equal{} \zeta_{2t} \plus{} \zeta_{2t}^{ \minus{} 1}$, then analog to $ B$ we get $ AB\in SL_2(q)$ of order $ 2t$ and $ xy\in PSL_2(q)$ of order $ t$.

But $ \zeta_{2t} \plus{} \zeta_{2t}^{ \minus{} 1} \equal{} {\rm tr} AB \equal{} \zeta_{2r}\sigma \plus{} \zeta_{2r}^{ \minus{} 1}(\zeta_{2s} \plus{} \zeta_{2s}^{ \minus{} 1} \minus{} \sigma)$ holds in fact for $ \sigma \equal{} {\zeta_{2t} \plus{} \zeta_{2t}^{ \minus{} 1} \minus{} \zeta_{2r}^{ \minus{} 1}(\zeta_{2s} \plus{} \zeta_{2s}^{ \minus{} 1})\over \zeta_{2r} \minus{} \zeta_{2r}^{ \minus{} 1}}\in \mathbb{F}_q$.
And we are done. :roll:

EDIT in fact we can take $ n \equal{} \varphi ({\rm lcm}(2r,2s,2t))$.
Then $ |\mathbb{F}_{p^n}^\times| \equal{} p^n \minus{} 1$ is divisible by $ 2r,2s,2t$ and we find $ \zeta_{2r},\zeta_{2s},\zeta_{2t}\in\mathbb{F}_{p^n}^\times$ of orders $ 2r,2s,2t$.


Please tell me :) I wana participate in da contest

I am also interested in this question :!:


inequalities
function
integration
logarithms
real analysis
real analysis unsolved

[quote="hpe"]Let $f: [0,1] \to \mathbb{R}$ be a $C^1$ function for which $\int_0^1 x |f'(x)|^2 dx = 1$ and $f(1) = 0$. Prove that there exists a constant $C$, independent of $f$, such that
\[ |f(0)| \le C\sqrt{\ln \left(1 + \int_0^1 x |f'(x)|^3 dx \right)} .  \]
:)[/quote]Since nobody took it, here's a solution.

Let $K = \left(\int_0^1 x |f'(x)|^3 dx\right)^{1/3}$. We first show that
\[ |f(x) - f(0)| \le 2K x^{1/3} \]
for all $x>0$. Indeed,
$|f(x)-f(0)| = |\int_0^x f'(t)dt| = \int_0^x t^{1/3}f'(t)t^{-1/3} dt| \le \left(\int_0^xt|f'(t)|^3 dt\right)^{1/3} \left(\int_0^x t^{-1/2} dt \right)^{2/3}$
by Hoelder's inequality. This turns out to be bounded by $2^{2/3}x^{1/3}\left(\int_0^xt|f'(t)|^3 dt\right)^{1/3} \le 2Kx^{1/3}$.

Next, for any $\epsilon > 0$,
$|f(0)| \le |f(0)-f(\epsilon)| + |f(\epsilon)| \le 2K \epsilon^{1/3} + \int_\epsilon^1 |f'(t)|dt \le 2K + \left(\int_\epsilon^1 t |f'(t)|^2 dt\right)^{1/2} \left(\int_\epsilon^1 \frac{dt}{t} \right)^{1/2}$
this time by the Cauchy Schwarz inequality. It follows that
\[ |f(0)| \le 2K\epsilon^{1/3} + \sqrt{|\ln \epsilon|} \]
for all $0 < \epsilon < 1$.

Now set $\epsilon = \min \{ 1, \frac{\ln (1 + K)^{3/2}}{(2K)^3} \}$.
If $2K \le \sqrt{\ln(1+K)}$, i.e. $\epsilon = 1$, then $K \le K_0$ for some $K_0$ and therefore $|f(0)| \le 2K \le C (\ln(1+K))^{1/2}$ for a suitable $C$.
If on the other hand $2K > \sqrt{\ln(1+K)}$, then $\epsilon <1$ and therefore $|f(0)| \le \sqrt{\ln(1+K)} + \sqrt{3|\ln(2K - \frac{1}{2} \ln(1+K))|}\le \sqrt{\ln(1+K)} + \sqrt{3 \ln (2K)}$.

In both cases one reads off $|f(0)| \le C \ln(1+ K)^{1/2}$, which is the desired estimate.

[i]This works in $\mathbb{R}^n$ and is known as [b]Brezis-Wainger-Gallouet[/b] inequality.[/i]


I was just wondering if anyone had any copies of any Mathcounts Sprint or Target Rounds that are not 2001-2008 Chapters or States and are willing to share. Pretty please? Anything is appreciated. =)

I'm also willing to trade...I have 2001-2008 Chapters and States. (My collection is poor) =(

If you want to be a good person, (sharing is caring), email me at ecstaticpotter@gmail.com =) Thanks

love too dude. I have a ton of nationals. Btw his stuff belongs on either this thread, [url]http://www.mathlinks.ro/viewtopic.php?t=133189[/url], or on AIME's Blog. Ill email u.

Er, I deleted my blog, citing possible copyright issues.


let's get this forum active guys. let's start with me:

hi I'm Yuqing. I'm moving here on the first of september and i will be oging to Longfellow MS. i am in 8th grade.

anybody here in my grade at Longfellow?

you can see my "achievements" in the introductions topic.

if you want an active forum, you should either spam or do problems. This is neither. Therefore, I suspect that any activeness in your forum will be short-lived. And I'm not joking... for example, the NC forum has lost its spam and problems, so nothing's really happening... except for the post I just posted like 1 minute ago, it has been inactive for like a week :wink:

Yeah, he's right. It's like a ghost town in here most of the time. I think that that's fine since there are so many other active forums and there's no point in sheltering ourselves, although it's nice once in a while to know some aops users in your state.

This is no longer the oldest post. Get trolled.
It is a ghost town.

lol, Virginia forums been dead 15 years ago, and its still dead today lmao

that's only 2 years after the creation of 109...

Dayum that's so weird to think about. Didn't they only lhave like 4 students or something XD


If there are six balls and they must all go into 5 buckets, how many possible combinations is there?

eg, one combination is all six balls go into bucket one. Another combination is ball 1 and 2 go into bucket 1, ball 3 into bucket 2, ball 4 into bucket 3, ball 5 into bucket 4 and ball 6 into bucket 5.

All six balls must be used and not all buckets have to be used in any combination set.

Please include the mathematical formula involved and list of combinations if possible

ok, when I work on problems like this I generally use the formula $x_1+x_2+....+x_n=r$ where $x_i$ denotes the basket a ball goes into and $r$ denotes the number of balls.  So we can say, in this case that $x_1+x_2+x_3+x_4+x_5=6 ; 0\leq x_i \leq 6$

Generally we use the formula :

$\frac{(n+r-1)!}{r!(n-1)!}$  Where each container is distinct

The answer to your problem is

$\frac{10!}{6!(4!)}$

There are $210$ ways to arrange 6 objects into 5 distinct containers.

Are the balls distinct? That is, if there are five balls in the first bucket and one in the third, does it matter which ball went into the third bucket or not?

If it doesn't matter, then Petoria got it (at least I think - I didn't follow it, but looks like they know what they're talking about)

If it DOES matter, ball 1 can go 5 different places, then ball 2 can go 5 different places, etc. and the answer is $5^6$


Ανακάλυψα και αυτή την άσκηση σε ένα βιβλίο ανισοτήτων τελευταία. Αν και φαίνεται δύσκολη είναι πολύ απλή.

Αν $ n >\equal{} 2$ και $ a1, a2, a3, ...a(n\minus{}1) > 0$ με άθροισμα ένα και $ b1, b2, b3, ...bn$ πραγματικοί,
να δειχτεί ότι $ b1^2 \plus{} \frac{b2^2}{a1} \plus{} \frac{b3^2}{a2} \plus{} ...\frac{bn^2}{a(n\minus{}1)} >\equal{} 2b1(b2 \plus{} b3 \plus{} ... bn)$

Θα παραθέσω και την απόδειξη:
[hide]
Από την ανισότητα Αντρέσκου βγαίνει ότι:
$ b1^2 \plus{} \frac{b2^2}{a1} \plus{} \frac{b3^2}{a2} \plus{} ...\frac{bn^2}{a(n\minus{}1)} >\equal{} \frac{( b1 \plus{} b2 \plus{} b3 \plus{} ...bn)^2}{1 \plus{} a1 \plus{} a2 \plus{} a3 \plus{} ...a(n\minus{}1)} >\equal{}$
(από την ανισότητα αριθμητικού - γεωμετρικού μέσου)
$ \frac{(2\sqrt{b1(b2 \plus{} b3 \plus{} b)})^2}{2} \equal{} 2b1(b2 \plus{} b3 \plus{} ... bn)$
[/hide]

Όμορφη ασκησούλα....

Πολλές εφαρμογές της Andreescu μπορείς να βρείς στο βιβλίο για μαθηματικές ολυμπιάδες Α Λυκείου  του κυρίου Στεργίου. Αν πάλι σου αρέσει αυτό το πεδίο μπορείς να πάρεις το τελευταίο βιβλίο των κ. Στεργίου -  Σκομπρή .  Πραγματικά είναι θησαυρός για την Ελληνική βιβλιογραφία.

Ναι, αυτό έχω. Είναι πολύ δυνατό βιβλίο.


Five people are sitting around a table. A name tag is made for each person. In how many ways can these name tags be passed out so that every person gets the wrong name tag?

Ugly casework... I'll post the solution if it's correct, is the answer

[hide]30[/hide]

[hide]
Overall, we know that there are $5!$ ways to hand out nametags. Of these, we can look at the cases where a certain number of people gets their own nametag.

Everybody: 1 way

4 people: 0 ways (if 4 people have their own nametag, then the 5th person must as well)

3 people: $_{5}C_{3}$ ways. If 3 people have their own, then the other 2 must be switched (there's only one way to do this), and there are $_{5}C_{3}$ ways to pick any 3 people.

2 people: $_{5}C_{2}\cdot{2}$ ways. Obviously, there are $_{5}C_{2}$ ways to pick the 2 people, and from the remaining three, the first to get the nametag has 2 options. The other two have only one choice.

1 person: $5\cdot{3}\cdot{3}$ ways. There are five ways to chose which one you want. Then, the first of the remaining 4 people has 3 ways to pick the wrong nametag. For each of these 3 ways, there are 3 arrangements that result in at least one person getting their own nametag.

Finally, we get:

$5!-1-0-_{5}C_{3}-_{5}C_{2}\cdot{2}-5\cdot{3}\cdot{3}= \fbox{44}$

[/hide]

Oh, I messed up somewhere apparently. I'll go back and check my work.

it sounds easier than it is :P

its a little confusing!

-jorian

Most good problems are!!  Nice solution Humble!


geometry
3D geometry
sphere
function
trigonometry

Find the potential in the interior of a sphere of unit radius when the potential on the surface f(θ)=cos^2(θ).

Should it be Like this:f(r,θ)=(r/R)cos^2(θ)?
Likely.
But I am writing this as a trial and error...
How to show this is the answer...i.e. what should be the correct procedure of approach?

Is $\theta$ the vertical angle, or the angle in the $x-y$ plane?

ie. what is $\theta$ in terms of $x, y, z$?

err if it's a sphere, your function can't be just in $r$ and $\theta$.  you need three coordinates.

maybe he means a 2D ring?

θ is the angle between r direction and z axis.

A 2nd possiblility is to apply uniqueness theorem.We know when the potential at every point of the surface is given,and the potential in that region obeys Laolace's (Here, Poisson'sequation),the potential function is inique.

So,I think it would be the same inside the sphere.

I was a bit wrong.

We are to employ Laplace's equation's solution in spherical polar co-ordinates.Then we are to feed the appropriate boundary conditions.

Ok, so $\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\frac{\partial f}{\partial r})+\frac{1}{r^{2}\sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial f}{\partial \theta}) =-\frac{\rho}{\epsilon_{0}}$.

But, to simplify this, realize that $z = \cos^{2}\theta$ at the surface.

So, I think the potential is just given by $f(x, y, z) = z^{2}$.

I was thinking on it.Should we consider Poisson's equation,or, Laplace's?After all we are given no "rho".
If it be Laplace's equation there are standard expression of potential in terms of (r,θ)...There we are to feed at V=cos^2(θ)
at r=R.
But what would be if it is Poisson's eqation?The separation of variables will work,but the form of potential function would change.


geometry
analytic geometry
graphing lines
slope
trigonometry
Pythagorean Theorem
area of a triangle

Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C$. Given that the length of the hypotenuse $AB$ is 60, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$.

Hmm, does somebody know how to solve this one?  :?

I'm stuck.  :(

[quote="Hokkage"][hide="Wow, did I misread?"]Quite an easy #15, or I might have done something wrong...
Since the median through $A$ is on $y=x+3$, the slope is 1, and therefore the length of side $AC$ is half of the length of $BC$. Let the length of side $AC = x$, and $BC = 2x$. By the Pythagorean theorem, $x^2 + (2x)^2 = 60^2 \Rightarrow 5x^2 = 3600 \Rightarrow x^2 = 720$ At this point we can stop, because the area is going to be $\frac 12 \cdot AC \cdot BC = \frac 12 \cdot x \cdot 2x = x^2 = 720$. Therefore, Area $= \boxed{720}$.[/hide][/quote]

this is wrong because you assumed that the sides were parallel to the cooridnate axies

[quote="Altheman"][hide="solution"]this is kind of messy, but here it goes, translate so the medians are $y=x$, and $y=2x$, then model the points $A: (a,a)$ and $B: (b,2b)$, $(0,0)$ is the centroid, and is the average of the verticies, so $C: (-a-b,-a-2b)$
$AB=60$ so
$3600=(a-b)^2+(2b-a)^2$
$3600=2a^2+5b^2-6ab$  (1)

AC and BC are perpendicular, so the product of their slopes is -1, giving
$\left(\frac{2a+2b}{2a+b}\right)\left(\frac{a+4b}{a+2b}\right)=-1$
$2a^2+5b^2=-\frac{15}{2}ab$  (2)

combining (1) and (2), we get $ab=-\frac{800}{3}$

use the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is $\left|\frac{3}{2}ab\right|$, so we get the answer to be $\boxed{400}$[/hide][/quote]Thanks. :)

I have another question. If the angle between the medians was $\theta$, why would $\tan(\theta + 45^\circ)=2$?

the tangent of the angle that a line makes with the x-axis is the slope, we know the angle for y=x is 45, and the angle associated with y=2x is greater than 45 since it is a larger slope, and that angle is \theta+(the other angle), so it is tan( \theta+45 )=slope of the other line=2

Call the midpoint of AC E, and that of BC F. Let AC=2m, BC=2n. We know $ m^2 \plus{} n^2 \equal{} 900$.
The angle between the two medians is $ tan^{ \minus{} 1}2 \minus{} 45$. So $ \angle FAC \equal{} 90 \minus{} \angle AFC \equal{} 90 \minus{} (\angle EBC \plus{} tan^{ \minus{} 1} 2 \minus{} 45)$. In other words,
$ tan^{ \minus{} 1}(n/2m) \equal{} 90 \minus{} (tan^{ \minus{} 1}(m/2n) \plus{} tan^{ \minus{} 1} 2 \minus{} 45)$, so
$ tan^{ \minus{} 1}(m/2n) \plus{} tan^{ \minus{} 1} 2 \minus{} 45 \equal{} 90 \minus{} tan^{ \minus{} 1}(n/2m)$.
Taking the tan of both sides, and doing a buch of algebra, gives $ 9mn \equal{} 2n^2 \plus{} 2m^2 \equal{} 1800$, so mn=200 and the answer is 400.

Nice problem.

[quote="timwu"][hide]
Call the midpoints on $ BC, AC$ as $ M, N$. Let $ m \equal{} \tan \angle ABC$, then $ \tan \angle AMC \equal{} 2m$, $ \tan \angle NBC \equal{} \frac {m}{2}$. Now, by tangent addition formula
\[ \dfrac{2m \minus{} \dfrac {m}{2}}{1 \plus{} 2m\cdot \dfrac {m}{2}} \equal{} \tan \theta_{dif} \equal{} \frac {2 \minus{} 1}{1 \plus{} 1\cdot 2}
\]
which simplifies to
\[ 2m^2 \minus{} 9m \plus{} 2 \equal{} 0 \\
 \\
m \plus{} \frac {1}{m} \equal{} \frac {9}{2}
\]
The area of $ \triangle ABC$ is then
\[ S \equal{} \frac {60^2\sin B\cos B}{2} \equal{} 900\cdot \sin 2B \equal{} 900\cdot \dfrac {2m}{1 \plus{} m^2} \equal{} 900\cdot \dfrac {2}{m \plus{} \dfrac {1}{m}} \equal{} 400
\]
[/hide][/quote]
In calculating $ \sin 2B$ I did $ \sin B$ and $ \cos B$ separately, but is there a way to calculate $ \sin 2B$ directly without using the double angle identity (it appears that you calculated $ 2m/(1\plus{}m^2)$ directly from $ \sin 2B$ in your solution)?

[hide]
Let the sides be $x$ and $y$. Let the acute angles between the medians and the legs of the triangle be $\alpha, \beta$. We have \[\tan\alpha = \frac{x}{\frac{y}{2}} = \frac{2x}{y}\]
Similarly, \[\tan\beta = \frac{2y}{x}\]

Let the acute angle between the two medians be $\theta$.
We have (using simple angle chasing) $\theta + 90^{\circ} = \alpha + \beta$.

Using the given equations in the coordinate plane, we can use the tangent sum formula \[\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\]
The tangents angles formed by the lines and the x-axis are just their slopes.
We get $\tan\theta = \frac{2 + (-1)}{1 - (2)(-1)} = \frac13$.

Now we plug this in to our previous equation relating $\theta$ to $\alpha,\beta$. \[\tan(\theta+90^\circ) = \tan(\alpha+\beta)\\
-3 = \displaystyle{\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}}\\
-3 = \displaystyle{\frac{\frac{2y}{x} + \frac{2x}{y}}{1 - \frac{2y}{x}\cdot\frac{2x}{y}}}\\
\frac{2y}{x} + \frac{2x}{y} = (-3)(1-4) = 9\\
2x^2 + 2y^2 = 9xy\]

The area of the triangle is then \[\frac12xy = \frac{2x^2+2y^2}{18} = \frac{60^2}{9} = 20^2 = \boxed{400}\][/hide]

We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$. [asy] import graph;  size(150);  Label f;  f.p=fontsize(6);  xaxis(-8,8,Ticks(f, 2.0));  yaxis(-8,8,Ticks(f, 2.0));  real f(real x)  {  return (x + 3); }  real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g,-6,2),blue+linewidth(1)); [/asy] Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$. Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$. Note that the acute angle between the red and blue lines is clearly $\beta - \alpha$. Therefore, we have that: \[\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}\]It follows that $\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}$ and $\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}$. From now on, refer to $\theta = \beta - \alpha$ [asy] import graph; size(4cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -5.96, xmax = 24.22, ymin = -10.06, ymax = 14.62;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);   draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1));   draw((0,8)--(0,0), linewidth(1) + wrwrwr);  draw((0,0)--(6,0), linewidth(1) + wrwrwr);  draw((6,0)--(0,8), linewidth(1) + wrwrwr);  draw((0,8)--(3,0), linewidth(1) + wrwrwr);  draw((0,4)--(6,0), linewidth(1) + wrwrwr);   dot((0,0),dotstyle);  label("$A$", (0.08,0.2), NE * labelscalefactor);  dot((6,0),dotstyle);  label("$B$", (6.08,0.2), NE * labelscalefactor);  dot((0,8),dotstyle);  label("$C$", (0.08,8.2), NE * labelscalefactor);  dot((3,0),linewidth(4pt) + dotstyle);  label("$D$", (3.08,0.16), NE * labelscalefactor);  dot((0,4),linewidth(4pt) + dotstyle);  label("$E$", (0.08,4.16), NE * labelscalefactor);  label("theta ", (2.14,2.3), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   [/asy] Suppose $ABC$ is our desired triangle. Let $E$ be the midpoint of $CA$ and $D$ be the midpoint of $AB$ such that $CE = EA = m$ and $AD = DB = n$. Let $G$ be the centroid of the triangle (in other words, the intersection of $EB$ and $CD$). It follows that if $CG = 2x$, $GD = x$ and if $GB = 2y$, $GE = y$. By the Law of Cosines on $\triangle GEB$, we get that: \[GE^2 + GC^2 - 2GE \cdot GC \cdot \cos \theta = CE^2 \Longleftrightarrow\]\[(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2\]By the Law of Cosines on $\triangle GDB$, we get that: \[DG^2 + GB^2 -2 DG \cdot GB \cdot \cos \theta = DB^2 \Longleftrightarrow\]\[(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2\]Adding yields that: \[5x^2+5y^2 - 8xy\cos \theta = m^2 + n^2\]However, note that $(2m)^2 + (2n)^2 = 60^2 \Longleftrightarrow m^2+n^2 = 30^2$. Therefore, \[5x^2+5y^2 - 8xy\cos \theta = 30^2\]We also know that by the Law of Cosines on $\triangle CGB$, \[CG^2 + GB^2 - 2CG \cdot GB \cdot \cos (180 - \theta) = CB^2 \Longleftrightarrow\]\[(2x)^2 + (2y)^2 + 2(2x)(2y) \cos \theta = 60^2 \Longleftrightarrow\]\[x^2+y^2+2xy \cos \theta = 30^2 \Longleftrightarrow\]\[5x^2+5y^2 + 10xy \cos \theta = 30^2 \cdot 5\]Subtracting this from the $5x^2+5y^2 - 8xy\cos \theta = 30^2$ we got earlier yields that: \[xy \cos \theta = 200\]but recall that $\cos \theta = \frac{3}{\sqrt{10}}$ to get that: \[xy = \frac{200 \sqrt{10}}{3}\]Plugging this into the $x^2 + y^2 + 2xy \cos \theta = 30^2$, we get that: \[x^2 + y^2 = 500\]Aha! How convenient. Recall that $(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2$ and $(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2$. Then, we clearly have that: \[m^2n^2 = ((2x)^2 + y^2 - (2x)(y)(2)\cos \theta)((2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2) \Longleftrightarrow\]\[m^2n^2 = 17x^2y^2 - 4xy \cos \theta (5x^2 + 5y^2) + 16x^2y^2 \cos ^2 \theta + 4x^4 + 4y^4 \Longleftrightarrow\]\[m^2n^2 = 17 \cdot \frac{200^2 \cdot 10}{9} - 800(5 \cdot 500) + 16 \cdot 200^2 + 4 \left( (x^2+y^2)^2-2x^2y^2\right) \Longleftrightarrow\]\[\frac{m^2n^2}{100^2} = \frac{680}{9} - 8 \cdot 5 \cdot 5 + 16 \cdot 2^2 + 4 \left( 5^2 - 2 \cdot \frac{2^2 \cdot 10}{9}\right) \Longleftrightarrow\]\[\frac{m^2n^2}{100^2} = 4\]But note that the area of our triangle is $2m \cdot 2n \cdot \frac {1}{2} = 2mn$. As $mn = 200$, we get a final answer of $\boxed{400}$.

~AopsUser101

P.S. I over-explained.


It could probably mean whatever you want, as long as you define it and use in a reasonable context ;)
Ie, in what context?

i think he is refering to data structure.

you can find it at any good data structure book! it is something like binary search. but u know we make binary search on 1 dimension. but we make it here in 2 dimension!


group theory
abstract algebra
superior algebra
superior algebra solved

Suppose that $H$ is a subgroup of $G$ such that whenever $Ha \not= Hb$ then $aH \not= bH$. Prove that $_gH_{g^{-1}} \subset H$ for all $g \in G$.

$aH$ is left coset and $Hb$ is right coset.
 :lol:

$gH = ghH$, so $Hg = Hgh$, then $H =Hghg^{-1}$, so $ghg^{-1} \in H$ for all $g \in G$, $h \in H$

You're incredible, Fast Reply. My solution is the same.


I didn't post this in the Computer Informatics forum because I thought that in there they talk about other stuff, not these simple things.

Main thing is, I have this adware on my system displaying false alerts for other malware, and I don't know its exact name.

But it does "advise" me to download its software, which I will not, and what do I do?

Some examples of viruses that it "found":

Trojan_SPM or something

SpyBot

I had sometime like this recently. Google for "virusbusters" to see if you are infected by it, it is most likely this. That requires a special fix, but in the meantime run malware cleaners to get any other nasties out, to isolate the problem.

[quote="mgao"]don't know how to get rid of VirusBursters.

I thought I had uninstalled it.[/quote]
Google for a webpage that has smitfraudfix.exe and follow the instructions.

BY THE WAY:

Dad has decided to install (upgrade) Vista on our computer in January, do you think MiKTeX 2.5 will be able to run?

Yes. It does. THe Beta version does. Not sure about the stable one. I have Vista. :)

you mean Beta 2 or RC 1?

(whichever one)

I don't know too much about vista, but one way to kill the virus would be system restore to a day or few befoer you got the virus

[quote="Ignite168"]Only people who WANT to keep Viruses have Windows Restore on.[/quote]
Stop spreading FUD.  I hope you're not advocating turning system restore off, as it's quite useful in system recovery.

[quote="Ignite168"]Only people who WANT to keep Viruses have Windows Restore on.

Anyways, if your problem isn't solved yet, what's the name of the Software it wants you to download? It wouldn't happen to be WinFixer 2006 or something like that would it? As that's a very common outbreak and you can download a simple VirtumundoBeGone.[/quote]

actually system restore is pretty good, as I've used it before. It is solved once you go back in computer time.
Try getting spybot search and destroy, and search the internet for cracks and warez for the full version.

I would strongly recommend taking a look at [url=http://wiki.castlecops.com/Malware_Removal_and_Prevention:_Overview]Malware Removal and Prevention[/url] and following the instructions there. About a year ago, my computer was infected with malware too, and I found the way to fix it on this website.
If you still don't find a way to solve your problem after going through their recommended fixups, you should post your HijackThis log (more info can be found on the website; it's step #10) in one of their forums, where an expert will be able to help you.  :)

Also, once you've solved your problem, be sure to check out their article on [url=http://wiki.castlecops.com/Malware_Prevention:_Prevent_Re-infection]How to Prevent Re-infection[/url] so that something like this won't happen again.


Let ABC a triangle and I a point in its plane. A line trough I intersects the circumcircles of BIC,CIA, AIB in L,M,N. A', B', C' are three collinear points on the sides of ABCand A'', B'' ,C'' are the intersections of LA', MB' and NC' with the circles BIC, CIA, AIB. Prove that the tangents in A",B",C" at the circles BIC, CIA, AIB intersect BC,CA,AB in three collinear points.

Hint: the tangent at A the sides AB and AC and the simedian form an armonic sheet
 Any solution now? :)


function integration Support calculus derivative algebra polynomial

For a function $ f(x) \equal{} e^x$, we write
\[ e^x \equal{} 1 \plus{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \cdots
\]
Thus by applying FT we get
\[ \mathfrak{F}[e^x] \equal{} 2\pi\left[\delta(t) \plus{} i\delta^{(1)}(t) \plus{} i^2\frac {1}{2!}\delta^{(2)}(t) \plus{} i^3 \frac {1}{3!}\delta^{(3)}(t) \plus{} \cdots\right]
\]
By using the direct FT equation we get
\[ \mathfrak{F}[e^x] \equal{} \int_{ \minus{} \infty}^{\infty}e^xe^{ \minus{} ixt}dx \equal{} \int_{ \minus{} \infty}^{\infty}e^{(1 \minus{} i t)x}dx \equal{} 2\pi \delta(1 \minus{} it) \equal{} 2\pi\delta(t \plus{} i)
\]
So how are these two FTs related? I mean can I say that
\[ \delta(t \plus{} i) \equal{} \sum_{k \equal{} 0}^{\infty}\frac {i^k}{k!}\delta^{(k)}(t)??
\]
here $ i \equal{} \sqrt { \minus{} 1}$ and $ \delta(t)$ is Dirac delta and $ \delta^{(k)}(t)$ is its $ k^{th}$ derivative.

Thank you
Regards
bhupala

To attempt to define the Fourier transform of generalized functions (distributions), we need a space of test functions that is preserved by the Fourier transform. We need that because we're basically going to use the idea that $ \int_{\mathbb{R}}f\overline{g}\equal{}
\int_{\mathbb{R}}\hat{f}\overline{\hat{g}}$ for nice enough $ f$ and $ g$ and generalize it to a definition of the Fourier transform of a distribution.

With this, we have a problem with the space $ \mathfrak{D}\equal{}C_0^{\infty}$ as a space of test functions: the Fourier transform of a function of compact support is analytic (in fact, entire) and hence can't have compact support (except for the zero function). This leads us to a different space of test functions: $ \mathfrak{S},$ the Schwartz space, which is the space of all infinitely differentiable functions all of whose derivatives go to zero faster than any reciprocal polynomial.

The dual space $ \mathfrak{S}'$ to $ \mathfrak{S}$ is called the [b]tempered distributions.[/b] We can define the Fourier transform on the space of tempered distributions.

When is an ordinary locally integrable function a tempered distribution? Roughly speaking, when it grows no faster than some polynomial (that's the "tempered" part.)

I need all of that background to discuss what is wrong with your problem.

$ e^x$ is not a tempered distribution. As such, it does not have a Fourier transform (at least, not within the tempered distributions). As for that power series: each polynomial partial sum is a tempered distribution, and hence has a Fourier transform which is a tempered distribution - as you have indicated, a linear combination of derivatives of the Dirac delta. However, the power series does not converge in $ \mathfrak{S}',$ and the series of derivatives of $ \delta$ does not converge in $ \mathfrak{S}'.$


function
algebra
domain
real analysis
real analysis unsolved

Find two sequences functions $(f_n)_{n\ge 0}$ and $(g_n)_{n\ge0}$ so that they uniformly converge to $f$ respectively $g$ but their product does not converge uniformly to $fg$. (it's free to choose the domain and the co-domain of the functions...)

$f_n(x)=g_n(x)=x+\frac 1n ,\ x\in [0,+\infty[$
 :diablo:  :alien:


For a power series $\sum a_{n}x^{n}$ in $\mathbb{R}$, how do you prove that the radius of convergence is $1/\alpha$, where $\alpha=\limsup\left|a_{n}\right|^{\frac{1}{n}}$?

Suppose $|x|<1/\alpha$. Choose a number $r$ such that $|x|<r<1/\alpha$. There is $N$ such that $|a_{n}|^{1/n}<1/r$ for all $n\ge N$. Hence $|a_{n}x^{n}|<(|x|/r)^{n}$, and the latter series converges.

Now suppose $|x|>1/\alpha$. Choose a number $r$ such that $|x|>r>1/\alpha$. There is an infinite subsequence... and stuff.


group theory
abstract algebra
superior algebra
superior algebra theorems

Let $ G$ be a group with subgroups $ H$ and $ K$. Is it possible to give an iff condition on when $ G$ and $ H \times K$ are ismorphic.

I know the following only if theorem:
If $ H$ and $ K$ are normal, have trivial intersection and generate $ G$ then they are isomorphic.
(I guess you could also replace normality with commuting).

The conditions of that theorem seem to be designed so that $ \phi(h,k) \equal{} hk$ will be an isomorphism.

Is there a general iff condition, or are there some other "constructive" (i.e. I can check the conditions and use it) only if conditions?

whether $ G$ and $ H \times K$ are somehow isomorphic is not a natural question. but rather whether $ G$ is the internal direct product of $ H$ and $ K$, and there you have the mentioned criterion which is sufficient and necessary. in general, it cannot be simplified, of course.


Let $ a,b,c,d$ be nonnegative real numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$. Prove that

$ \sum_{sym}{a^3b^3} \ge 6abcd$.

It's equivalent to
\[ 2\sum_{sym}a^3b^3\ge 3abcd(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\]
which follows from adding the following inequalities
\[ a^3b^3\plus{}a^3c^3\plus{}a^3d^3\ge 3a^3bcd\\
a^3b^3\plus{}b^3c^3\plus{}b^3d^3\ge 3ab^3cd\\
a^3c^3\plus{}b^3c^3\plus{}c^3d^3\ge 3abc^3d\\
a^3d^3\plus{}b^3d^3\plus{}c^3d^3\ge 3abcd^3\]

Nice,Nayel.

And now, the general form:

If $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1^{n\minus{}2} \plus{} a_2^{n\minus{}2} \plus{} ... \plus{} a_n^{n\minus{}2} \equal{} n$, then

$ 2\sum_{sym}a_1^{n \minus{} 1}a_2^{n \minus{} 1} \ge n(n \minus{} 1)a_1a_2...a_n$.

[quote="Vasc"]Let $ a,b,c,d$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$. Prove that

$ \sum_{sym}{a^3b^3} \ge 6abcd$.[/quote]
Vasc, maybe $ \sum_{sym}{a^3b^3}\equal{}4(a^3b^3\plus{}a^3c^3\plus{}a^3d^3\plus{}b^3c^3\plus{}b^3d^3\plus{}c^3d^3)$ $ ?$
Here is $ 4\equal{}4!: 6.$ Where I am wrong? Thank you!

[quote="arqady"][quote="Vasc"]Let $ a,b,c,d$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$. Prove that

$ \sum_{sym}{a^3b^3} \ge 6abcd$.[/quote]
Vasc, maybe $ \sum_{sym}{a^3b^3} \equal{} 4(a^3b^3 \plus{} a^3c^3 \plus{} a^3d^3 \plus{} b^3c^3 \plus{} b^3d^3 \plus{} c^3d^3)$ $ ?$
Here is $ 4 \equal{} 4!: 6.$ Where I am wrong? Thank you![/quote]
I assumed here that $ \sum_{sym}{a^3b^3} \equal{} (a^3b^3 \plus{} a^3c^3 \plus{} a^3d^3 \plus{} b^3c^3 \plus{} b^3d^3 \plus{} c^3d^3)$.

[quote="dima ukraine"]For a first ineq $ (3,3,0,0)\succ(3,1,1,1)$[/quote]
my proof is also similar
For the genaral form.I think
$ (n-1;n-1,0,..0)\succ{3,1,1...1)}$ for $ n\ge 4$


superior algebra
superior algebra solved

Prove that if an element of a ring has more than one right inverse, then it has infinitely many.

Assume $ab=ac=1$ for some $b \neq c$. Then $a(k(b-c)+b)=1$ for every integer $k$. Let $I_{k}=k(b-c)+b=(k+1)b-kc$.

Note that $I_{k}=I_{k'}$ yields $(k-k')(b-c)=0$, and if $I_{k}=I_{k'}=I_{k''}$, where $k,k',k''$ are all distinct, then by Bezout's theorem, one must have $k'' \neq k'-k+p$ for every $p$ such that $gcd(p,k'-k)=1$.

I'm pretty sure it's not too hard to conclude from there, but I need to go because it's past 11pm and my girlfriend is going to scream at me (I can already imagine it  :oops: ) ...

I'll try to finish the proof tomorrow if noone does it before !

Thank you. As I thought a little more about this problem yesterday, I realized my approach (although correct) doesn't allow to conclude. I didn't pick the right form for my inverses ...


[quote="GameBot"]Twelve of the 40 students in a class are boys. What percent of the class is girls?[/quote]

If 12 are boys, then 28 are girls. Our answer is then $ \frac{28}{40} \implies \frac{7}{10} \equal{} \boxed{70}$ percent.


algorithm function calculus integration inequalities trigonometry algebra

Well a couple of weeks ago nukular asked for some enumerative combinations questions and I finally got around to finding some, of varying difficulty.

1) How many subsets of {1, 2, .., n} have no two successive numbers?

2) Can you arrange the numbers 1 through 9 along a circle so that the sum of two neighbours are never divisible by 3, 5, or 7?

3) Four noncoplanar points are given. How many boxes (a box is bounded by three pairs of parallel planes) have these points as vertices?

4) In how many ways can you choose 2 disjoint subsets from a set of size n?

5) Find the general term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4,... in closed form (allowed to use floor and ceiling functions).

6) Does the set {1, 2, .., 3000} have a subset of 2000 elements so that no number in the subset is double another in the subset?

7) What is the most number of binary n-words that differ in at least 3 places?

8) Call a permutation of {1,2,..,2n}, {x1,..,x2n}, pleasant if |xi - xi+1| = n for at least one i of the values 1 through 2n-1.
For each n, prove that more than half of all permutations are pleasant.

1. [hide]We can find a recursion. For n=1, the number is clearly 2, either {1} or the null set. For n=2, the number is 3, the null set, {1}, or {2} (but not {1.2}). For n>2, we define f(n) to be this number that we're looking for. Then we have f(n)=(f(n-1)+f(n-2) because any set either contains n (and thus does not contain n-1, so there are f(n-2) ways) or does not (f(n-1) ways). So now we have a recursion that is the fibonacci sequence shifted over some. We can get the closed form explicity by Binet's formula if we really wanted to. Let's now see if we can find fk(n), where k is the number of consecutive integers that we cannot have. In the previous problem, k was simply 2. Now, if k is 1, we can't have any consecutive integers i.e. any integers at all. So the only acceptable subset for any n would be the null set, i.e. f1(n)=1 for any n. We already found f2(n) somewhat. Now, let's see what f3(n) is. We find the base cases => f(1)=2, f(2)=4, f(3)=7. So now we get our recursion: f3(n)=f3(n-1)+f3(n-2)+f3(n-3) by the same reasoning: either the last term is not in the set, the last term is in the set but the penultimate term is not, or the last term and the penultimate term are in the set and thus the antepenultimate term cannot be. So now we attempt to solve the characteristic equation r^3-r^2-r-1=0. Unfortunately, this does not have any rational roots (rational root thm.) and I don't know the algorithm for finding all roots, so it looks like a generalization for the closed form is doomed. However, it is easy to find a recursion for fk(n). The base cases will be fk(n)=2^n for 1:le:n:le:(k-1), fk(k)=2^k-1, and then the recursion which is fk(n)=:Sigma:(i=1 to k) fk(n-i)[/hide]

there...that is my so-so solution and bad generalization.

4. [hide]after doing something stupid that I will outline next, I came up with 3^n. Each element can either be in the first subset, the second subset, or not in either. If order does not matter, i.e. {1,2,4} and {3,5} is the same as {3,5} and {1,2,4}, then it would be (3^n+1)/2, as we take off the thing corresponding to both sets being empty, divide by 2 (for order), and then add back in the double-null. Okay, the stupid way I did it was :Sigma: (i=0 to n) C(n,i)*2^(n-i), because I was choosing i of the elements to put in my first set, and then just using normal 2^a to find the number of subsets of a set with a elements. Of course, that works out to the same thing with the binomial theorem[/hide]

5. [hide]f(n(n+1)/2)=n, g(n(n-1)/2+1)=n. If we could find such functions, and if they were nice enough, we could give the solution: ceiling(f(n)) or floor(g(n)). Nice enough means increasing on the positive reals, in this case. In this case, for any n that is a triangular number, we would get an exact integer output. For the next such n that is a triangular number, we would get the next integer. For all n in between, we would get non-integral output in between. Taking either the floor or the ceiling (depending on the function) will give us the right value. However, it is not hard to find these functions and verify that they are increasing on the reals: f(n)=m gives us n=m(m+1)/2 => 2n+1/4=m^2+m+1/4=(m+1/2)^2 => (sqrt(8n+1)-1)/2=m => f(n)=(sqrt(8n+1)-1)/2. Similarly, we find g(n): g(n)=m => 2n-7/4=m^2-m+1/4=> g(n)=(sqrt(8n-7)+1)/2. So my two functions are ceiling((sqrt(8n+1)-1)/2) and floor((sqrt(8n-7)+1)/2). [/hide]

6. [hide]let's make a table with a lot of columns. at the head of every column, we put an odd number from 1 to 2999. So we will have 1500 columns. Now, below each number, we put the powers of 2 times that odd number. We only put numbers :le: 3000 though, so not every column will have the same length. In particular, the last 750 columns will only have 1 number in them. The beginning of the table will look somwhat like this:

1  3   5   7   9  ...

2  6  10 14 18 ...

4 12 20 28 36 ...

8 24 40 56 72 ...

...

Now, in this table, we can only include every other number in each column. So, we need to find the lengths of each of the columns. The last 750 columns have length 1, as already noted. The first column has length 12, the next two columns have length 10, the next three have length 9, the next 6 have length 8, the next 11 have length 7, the next 24 have length 6, the next 47 length 5, the next 94 length 4, the next 187 length 3, the next 375 length 2, and of course the next 750 length 1. So the maximal magnitude of a subset would be 1(6)+2(5)+3(5)+6(4)+11(4)+24(3)+47(3)+94(2)+187(2)+375(1)+750(1)=1999, which is just shy of our goal, so the answer is no.[/hide]

8.

[hide]

Call a permutation that is not pleasant "unpleasant"

if |x - y| = n and x, y belong to {1, 2.... 2n} call x and y "partners"

call xi+1 the "left neighbor" of xi+2 and the "right neighbor" of xi.

It is clear that each number x between 1 and 2n has a unique partner.

If n = 1, the only permutations are {1, 2} and {2, 1} both of which are pleasant.

If n>1:

We will show that for each unpleasant permutation, there exists a unique pleasant permutation with exactly one pair of neighbors who are partners. Since {1, n+1, 2, n+2, .... , n, 2n} has n such pairs, and since n>1 that means it remained uncounted and there are more pleasant permutations than unpleasant permutations. Since every permutation is either pleasant or unpleasant, more than half of the permutations are pleasant.

Now to define the function (between unpleasant and pleasant with one pair of neighbors who are partners)

unpleasant-> pleasant:

Let's define this algorithmically since it works best:

1. Pick the first term in the permutation.

2. Find it's partner.

3. Insert the term after its partner(making it a right-neighbor).

Note two things about this:

1. Not all the pleasant permutations with one pair of neighbors that are partners are counted, since if the FIRST two terms are partners this process cannot be reversed (nor is this in the range of this function)

2. The function is COMPLETELY reversible for all other pleasant permutations with one pair of neighbors as partners.

So not only are there more than half, there are MUCH MORE than half...

[/hide]

7. Do you mean pairwise differ in three places or globally differ in at most three places ?

(like gcd(a,b,c) as opposed to gcd(a,b) gcd(a,c) and gcd(b,c) )

Be prepared for disgusting algebra:

5) Find the general term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4,... in closed form (allowed to use floor and ceiling functions).

[hide]

We try to count n: the number one less than the value of the xth term. If we consider the terms to be blocks of value n and length n (since there are 2 2's, 3 3's.. .etc) then the value of the block ending BEFORE the block containing x is n'(n'+1)/2, where n' is the largest such integer for which n'(n'+1)/2 is smaller than x.

So now we use a greedy inequality:

n'(n'+1)/2 < x <= (n'+1)(n'+2)/2

n'2 + n' < 2x <= n'2 + 3n' + 2

(n'+1/2)2 < 2x + 1/4 <= ((n'+1)+1/2)2

n'+1/2 <  :sqrt: (2x+1/4) <= n'+3/2

n' < :sqrt: (2x+1/4) - 1/2 <= n'+1

so n'+1 = ceil(:sqrt: (2x+1/4)-1/2)

And n'+1 is the value of the xth term, as before stated.

so the value of the xth term is ceil(:sqrt: (2x+1/4)-1/2)

[/hide]

2)

[hide]

Put these around a circle however you would like (just in this circular order)

1, 7, 4, 9, 2, 6, 5, 8, 3

Not 4, 8, 11, and 13 are not divisible by 3, 5, or 7 (two are prime, the others are powers of two)

1 + 7 = 8

7+ 4 = 11

4 + 9 = 13

9 + 2 =11

2+ 6 = 8

6 + 5 =11

5 + 8 + 13

8 + 3 = 11

3+ 1 = 4

1 + 7 = 8

So this fits the conditions

How I got this: Make a huge table with all the possible neighborings of each number (well its not that huge)... or you can use a graph (which is how i started it) and find a hamilton circuit.

[/hide]

For number 7:

I actually changed this from the original, but shouldn't have. Firstly i'm pretty sure its pairwise. Secondly you have to prove that there are  :le: (2n)/(n+1) words.

2^n/n+1. Hmm that reminds me of that cominatorial identity :Sigma: (i=0 to n) C(n,i)/(i+1). Now if I could get from there to the whole binary word thing, I'd be happy.

grobber wrote:Here's something similar to problem (5):

Find the general term of the sequence 1, 2, 4, 3, 5, 7, 6, 8, 10, 12, etc.

I assume you can see the pattern. This was used in a Romanian IMO selection test some time ago.

w000t i solved it. [hide]first i took the difference between that and n, and got 0, 0, 1, -1, 0, 1, -1, 0, 1, 2, -2, -1, 0, etc. which didn't really help. I tried to get a sine function to nail those values, but it didn't work. So I took the difference between this function and 2n, which got me 1, 2, 2, 5, 5, 5, 8, 8, 8, 8, 13, 13, 13, 13, 13, 18, 18, 18, 18, 18, etc. So, I knew that I could use the original function in the original problem 5 to get me the sequence: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, etc. so if I could get a function that would map 1 to 1, 2 to 2, 3 to 5, 4 to 8, 5 to 13, etc. I would be very happy indeed. However, the differences between the values is just a normal arithmetic sequence, once we fiddle a bit: the differences are 1, 3, 3, 5, 5, 7, 7, etc. where each value occurs twice. However, the original values are the fllors of 1, 2.5, 5, 8.5, 13, etc. and the differences between these are 1.5, 2.5, 3.5, 4.5, etc. which is a straight-out arithmetic sequence. So the function that maps 1 to 1, 2 to 2, 3 to 5, etc. can be found to be floor((n^2+1)/2). We plug this into the original #5 function and subtract from 2n to get the final function as :

2n-floor((floor((sqrt(8n-7)+1)/2)^2+1)/2)[/hide]

3. [hide]C(C(4,2),3)-C(4,1)-16. The first term comes from choosing 3 lines out of the 6 that are generated by connecting 2 of the 4 pts. All of these will determine a unique box, except for the triples of lines that are coplanar, of which there are 4.[/hide]

I noticed that #7 was the only one which remained unanswered, so I decided to work it.

7.[hide] I will construct a set of pairwise disjoint subsets of binary n-words, each with n+1 elements, such that for each such subset, exactly one of these elements is in our 3-differences set.  This clearly is sufficient to prove the claim.

Suppose that we had a set A of binary n-words satisfying the required condition; that is, they pairwise differ in at least three elements.  For each element e of A, consider the set B consisting of: e and all n binary n-words created by flipping one of the n bits in e.  Of the elements B only e can be in A, since the other elements of B differ from e in only one place.  Since this set has size n+1, it remains only to show that for any e and e' in A, their corresponding sets B and B' are disjoint.  To do this consider an element p in B.  It is, recall, different in only one place from e.  If p is also in B', then it is only different in one place from e'.  But this means that e and e' are only different in two places (since we can, via p, move between them with two bit flips), and this contradicts our assumption that e and e' are different in at least three places.

Thus we have identified pairwise disjoint subsets of the set of binary n-words, each with n+1 elements, such that exactly one of these elements is in our set A.  Hence A has maximal size (2^n)/(n+1), which is what we wanted.[/hide]


[hide]The number that occurs twice is 45. The 10 numbers I found are 41,42,43,44,45,45,46,47,48,49. There may be more possible sets. [/hide]

You clearly made a good start.  However, read the question again:  the sum of any [i]9[/i] should be one of the numbers from 86 to 94.

darkquantum wrote:Will is given a list of 10 positive integers, which together have exactly 9 distinct values. Will finds that by adding 9 of the numbers, he can make any integral sum from 86 to 94, inclusive. Find the integer that occurs twice.

[hide]let our repeated element be x, and the sum of all the other elements be S. well, then, there will be eight equations of the form 2x+(7 of the other 8 numbers)=(a number between 86 and 94), and one equation of the form x+S=(a number between 86 and 94 which we will call z). Multiplying this last equation by 7 and subtracting it from the sum of all the others, we get 9x=810-8z. Since z has to be divizible by 9, it is 90, and thus x=10, so yeah, you're right. Um. why are you answering your own question though?[/hide]

It's not my question. All these questions originally came from Joel's post "stuff"
I just broke up his 15 quesitons in one post into 15 different threads


I just noticed that there's no thread about this yet, and it's 9 days until the competition.
Who's going?

I know that Edward was 1st, "Jason Zheng" was 2nd, and I was 3rd. (Abubakar was 7th.) I'm pretty sure that Robert and Oliver were somewhere in between, I think 4th and 5th? Monica ranked somewhere in the top 10, I believe as 6th. Gil won JV. Walton won Varsity and JV.

Also, the problems improved significantly from last year ;), although no one in JV got any relay (besides Walton who got all 3) and only one question in Varsity relay was answered correctly.  Even if the problems were a bit difficult, congrats to Northview for holding a successful tournament.

Thanks Allen!
Congratulations to all of the competitors! The results I can remember that haven't been posted already are as follows:
Team Results:
JV:
1. Walton
2. Chattahoochee
3. Chamblee
4. Norcross
5. Lassiter

Varsity:
1. Walton
2. Chamblee
3. Pace Academy
4. Grayson
5. Norcross

i was there with norcross jv. It was pretty fun, especially the ciphering round in my opinion


I have a graph of a parabola with the graph only showing +x and +y axes.
The parabola starts at (0,0) and has a vertex at (5.0, 2.5), concave down.
Some other points are (1.5, 1.25) and the final point shown is (8.0, 1.6).
In the answer to the problem regarding this graph, the first line states:
"We can easily fit the curve to a concave-downward parabola: $ y \equal{} 1/10x(10 \minus{} x).$"

My question is: how do you get that equation please?

Let the equation be $ y \equal{} ax^2 \plus{} bx \plus{} c$

From the point (0,0) we have $ 0 \equal{} c$.

From the two other points we may then solve for a and b by plugging them in.

Or you may use the definition of the vertex to reduce the number of calculations.
Hint: Vertex is at $ (\minus{}\frac{b}{2a},\minus{}\frac{b^2}{4a})$

Thanks very much Iniquitus!  I've forgotten most of what I learned about
this type of math - do you use simultaneous equations to find a solution?

When you said that the vertex is at $ \left(\minus{}\frac{b}{2a},\minus{}\frac{b^2}{4a}\right)$, I hope you only meant for this instance.

Otherwise it's $ \left(\minus{}\frac{b}{2a},f\left(\minus{}\frac{b}{2a}\right)\right)$

Yes, I meant for this instance...

Plugging in some stuff, we get

$ 2.5 \equal{} 25a \plus{} 5b$, or
$ 1 \equal{} 10a \plus{} 2b$ (1)

and

$ 1.6 \equal{} 64a \plus{} 8b$, or
$ 1 \equal{} 40a \plus{} 5b$ (2)

4x(1) - (2):
$ 3 \equal{} 3b$, $ b \equal{} 1$, $ a \equal{} \minus{} \frac {1}{10}$

So the equation is

$ y \equal{} \minus{} \frac {1}{10}x^2 \minus{} x$

OR

Using the information about the vertex, we have $ \frac{y_{vertex}}{x_{vertex}} \equal{} \frac{2.5}{5} \equal{} \frac{b}{2}$, so $ b \equal{} 1$, $ a \equal{} \minus{} \frac{1}{10}$

Write $ y \equal{} a(x \minus{} h)^2 \plus{} k$ where $ (h, k)$ is the vertex, which is given.  Then $ y \equal{} a(x \minus{} 5)^2 \plus{} 2.5$ and we know that $ 0 \equal{} a( \minus{} 5)^2 \plus{} 2.5 \implies a \equal{} \minus{} \frac {1}{10}$.  This form can be more useful than standard form, especially if the vertex is known.

Yet another solution:  parabolas are symmetric with respect to the vertex, so if $ (0, 0)$ is a point and the $ x$-coordinate of the vertex is $ 5$ then $ (10, 0)$ is also a point.  This tells us $ y \equal{} ax(10 \minus{} x)$ and $ a$ can be calculated using any of the other points.

This problem can be done without knowledge of the vertex using the other three points, but the vertex is extremely important information.  As the vertex form of the parabola I gave above demonstrates, a parabola is uniquely described by the location of its vertex and its value of $ a$.


geometry
analytic geometry
geometry unsolved

Let ABC is a acute triangle.M is in the triangle.Determine A1,B1,C1 satisfy that MA1=BC;MA1 is perpendicular to BC
MB1=CA;MB1 is perpendicular to CA;MC1=AB;MC1 is perpendicular to AB.
   Prove that M is the centre of gravity of triangle ABC

Yes ,center of gravity there is centroid

I think $M$ should be the centroid of triangle $A_1B_1C_1$ . Is this problem from Russia ?

Yes,Iam sorry .Prove that M is the centroid of the triangle A1B1C1.  HELP ME

Let each point $(x,y)$ in the Cartesian plane represent the complex number $x+y\mbox{i}$.And the complex number that $A,B,C,M,A_1,B_1,C_1$ represent is $a,b,c,m,a_1,b_1,c_1$.Then we know
$m-a_1=(b-c)\mbox{i}$,$m-b_1=(c-a)\mbox{i}$,$m-c_1=(a-b)\mbox{i}$,
So $a_1+b_1+c_1=3m$,that means $M$ is the centroid of triangle $A_1B_1C_1$.

another idea would be to prove that the triangles $B_1MC_1$, $C_1MA_1$, $A_1MB_1$ have the same area (which equals to that of triangle ABC). this easily yields that M is the centroid of $\triangle A_1B_1C_1$.

if required, i could detail this better after the seminar is over.

  darij from the pentacracy


In a Bulgarian Olympiad problem, we are asked to prove that there are infinitely many integer solution quadruples $ (x,y,z,t)$ to $ x^{3}+y^{3}+z^{3}+t^{3}=1999$.

I solved this by a transforming it into a Pell Equation, so we obviously have infinite solutions. The Pell Equation to which I transformed the original equation said there were infinitely solution quadruples $ (x,y,z,t)$ such that $ x+y+z+t=19$. Can it be proven (I don't know too much about Pells - just the 1.5-hour lecture at AMSP) that $ x^{3}+y^{3}+z^{3}+t^{3}=1999$ only if $ x+y+z+t=19$?

Yes ${ (x,y,z,t)=(a+b,a-b,\frac{c+d}{2},\frac{c-d}{2}}$
Now try to find out an special $ (a,c)$ and $ (b,d)$ is root of the Pell equation.
If you don't solve, it I will post solution. :wink:


The 2006-07 Worldwide Online Olympiad Training (WOOT) class starts on Monday, September 11.

There is a Math Jam today (September 5) at 7:30 PM Eastern / 4:30 PM Pacific, where you can learn more about WOOT and ask questions about the program.

[url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]Click here for more information about WOOT.[/url]

[quote="Kileel"]If paying for WOOT by credit card may be problematic, is there any way payment by cheque would be accepted?[/quote]

Absolutely.  [url=http://www.artofproblemsolving.com/Classes/AoPSClasses.pdf]Click here for an order form[/url] you can print out and mail with the check.


calculus
trigonometry
ratio
geometry
inradius
geometry proposed

let  a  be the apothem,  r  be the radius,  C  be a central angle, and  d  be the sum of  a  plus  r.
Given  d  and  C, find  r.

In other words, for any regular  n-gon with an ODD number of sides, given the number of sides (n), and the distance (d) from one vertex to the middle of the opposing side, calculate the radius.

I imagine that this is a simple matter, and I apologize if this is the incorrect forum to ask such a question.

Feel free to remove this topic from this forum.
I realized that this belongs in the Unsolved Problems forum, and I've reposted the topic there.

Don't repost questions: the moderators will move your posts to the appropriate forum.

And indeed, this should be Pre-Olympiad or something like that.

Thank you Arne, I'll leave it up to the moderators next time.

I didn't realize my question was so trivial, and after a quick look at some of the other posts in this forum I realize that I've googled into a community who are way beyond my current knowledge.

I am 22 years old and I just started studying calculus last month (on my own).  I learned a lot of trigonometry earlier this year (also on my own), and I've had no formal mathematical education beyond 9th grade algebra (when I left highschool and eventually acquired my GED).  I've been stumped for days by the problem mentioned above, and I can't find anything in my books or online that tackles the ratio between the radius and inradius of a regular Ngon (just triangles) and I really want a solution.

Obviously the apothem approaches the radius as the central angle approaches 0, and the apothem approaches 0 as the central angle approaches pi.

If the question is fairly novice, perhaps you might offer up a solution, or even some hint to push me in the right direction.


arithmetic sequence
number theory proposed
number theory

Prove that $n \geq 3$ be a prime number and $a_{1}<a_{2}<...<a_{n}$ be integers.Prove that $a_{1},...,a_{n}$ is arithmetic progression if and only if there exist a partition of $\{1,2,...,n\}$ into n classes $A_{1},A_{2},..,A_{n}$ such that :
                   $a_{1}+A_{1}=a_{2}+A_{2}=...=a_{n}+A_{n}$,
where $x+A$ denotes the set $\{x+a|a \in A\}$.
Let's go for it .It isn't hard and when you have the answer,you will discover some secret from this problem.Goodluck :)  :)

I believe that ,it's a very very hard and beautiful problem  :?  and I think it should be in this form:  ;)

[quote] if and only if there exist a partition of  ${\{0,1,2,...,n}\}$ into n classes ....[/quote]

Isn't it?  :P


Suppose I need to insert an [u]array of images[/u] in a $\text{\LaTeX} $ document and label them: (a) This is "Plot A" ; (b) This is "Plot B"; etc. How do I actually make it?

I heard [i]Miniplot[/i] package allows one to prepare such things at ease. I downloaded it and followed the instructions on the documentation, but it never really worked.

Does anybody know a way to do this?

Thanks.

[quote="thiago_j"]Suppose I need to insert an [u]array of images[/u] in a $\text{\LaTeX} $ document and label them: (a) This is "Plot A" ; (b) This is "Plot B"; etc. How do I actually make it? […][/quote]
What you are searching for is a figure that is split to sub-figures. A package that is capable to do such a thing is »[url=http://ctan.org/pkg/subfig]subfig[/url]«, not only for external graphics files.

[quote="thiago_j"]I heard [i]Miniplot[/i] package allows one to prepare such things at ease. I downloaded it and followed the instructions on the documentation, but it never really worked. […][/quote]
Then you can for sure show your tries in form of a [i]proper[/i] [url=http://www.dickimaw-books.com/latex/minexample/index.html]minimal example[/url]. If you catch up on that a solution will be in transit. In case you want to produce plots of datasets or mathematical functions, try the »[url=http://ctan.org/pkg/pgfplots]pgfplots[/url]« package in combination with the already mentioned »[url=http://ctan.org/pkg/subfig]subfig[/url]« package.


Find the ordered pair $\ (x,y)$ if $\ \sin x + \sin y = \sin 72$

[hide="my progress"]
$\sin(x) + \sin(y) = 2\sin\displaystyle \left(\frac{x+y}{2}\right)\cos\displaystyle \left(\frac{x-y}{2}\right)$

$\sin(72) = 2\sin(36) \cos(32)$

We have:
$2\sin\displaystyle \left(\frac{x+y}{2}\right)\cos\displaystyle \left(\frac{x-y}{2}\right) = \sin(72) = 2\sin(36) \cos(32)$

From this, we can write several systems of equations:

$\frac{x+y}{2} = 36, \frac{x-y}{2} = 36$

or

$\frac{x+y}{2} = 36, \frac{x-y}{2} = 36+360n$

But all these yield trivial solutions where one of the $\sin$ terms is equal to zero...

[/hide]

EDIT: The $\sin$ and $\cos$ terms don't have to equal each other for the whole expression to be equal, so ignore what I wrote  :blush:


superior algebra
superior algebra unsolved

Given a group $ G$ that acts sharply $ 2$-transitive on a set  $ X$ with $ \left|X\right| \geq 3$.
Prove that $ G_x$ contains at most $ 1$ involution for all $ x \in X$.

Note that $ G_x \equal{} \left\{g \in G\ |\ g(x) \equal{} x\right\}$ denotes the stabilizer of $ x$.

Could you briefly remind me what ``sharply 2-transitive'' means?

Thanks!

That the element $ g \in G$ such that $ gx_i \equal{} y_i$ of the 2-transitivity is unique (I needed to look that one up myself).

Indeed. I should also mention that the problem states some (trivial) observations first, probably to be used in solving it.
They are:
[list]
[*]Show that for every two distinct elements $ y,z$ in $ X$, there exists a unique involution that interchanges these two elements.
[*]Let $ x \in X$. Show that $ G_x \cap (G_x)^h \neq {e}$ if and only if $ h \in G_x$.
[*]Show that every two involutions are conjugates.[/list]

Assume that we have some involution in a stabilizer, then we must have $ |X|\equal{}2n\plus{}1$.  Then every involution is the product of exactly $ n$ 2-cycles. (note that any element with more than one fixed point is the identity)

By the first observation, we can now count the number of involutions: $ {{2n\plus{}1}\choose 2}/n\equal{}2n\plus{}1$.  But there are also $ 2n\plus{}1$ stabilizers, which are conjugate to each other but have trivial intersection.  Hence each contains exactly one involution.

Of course, very elegant solution! Thank you LydianRain.

Dear LydianRain,

we both overlooked the fact that $ G$ can also be infinite.
What now?  :)

Jan


Given a real number $ c > 0$, a sequence $ (x_n)$ of real numbers is defined by $ x_{n \plus{} 1} \equal{} \sqrt {c \minus{} \sqrt {c \plus{} x_n}}$ for $ n \ge 0$. Find all values of $ c$ such that for each initial value $ x_0$ in $ (0, c)$, the sequence $ (x_n)$ is defined for all $ n$ and has a finite limit $ \lim x_n$ when $ n\to \plus{} \infty$.

I'm pretty sure you aren't supposed to revive 11-year-old threads.

Although, why can't you double post a thread like this?

Why is that? I'm pretty sure I wasn't supposed to double post threads, so I kind of had to...

The sequence is defined only when $c-\sqrt{c+x_n}\geq 0 \Rightarrow x_n \leq c^2-c$. Since $x_0\in (0,c)$, then the constraint $c \leq c^2-c \Rightarrow 2 \leq c$ must be satisfied. As for all other $n$, notice that $x_{n+1}=\sqrt{c-\sqrt{c+x_n}}<\sqrt{c}$, which is definitely less than $c^2-c$ for $c\geq 2$; hence, this is the only condition.
Suppose $(x_n)$ converges to a limit $L$. Since $0<L< \sqrt{c}$, the function $f(x)=\sqrt{c-\sqrt{c+x}}$ is analytic in that bound, so we let $L=\sqrt{c-\sqrt{c+L}}$. Expanding yields $L^4-2cL^2-L+c^2-c=0$. We find that $L\notin \{\pm 1, \pm c, \pm (c-1), \pm c(c-1)\}$, so we factor into a pair of quadratics with respect to $L$: $(L^2+L-c+1)(L^2-L-c)=0$. Keeping in mind that $0<L< \sqrt{c}$, $L=\frac{\sqrt{4c-3}-1}{2}$ is the only possible solution (which came from the factor $L^2+L-c+1=0$; we will use this fact a lot). We now need to show that $(x_n)$ converges to that limit.
$|x_{n+1}-L| = \left|\sqrt{c-\sqrt{c+x_n}}-L\right| =\left|\frac{c-\sqrt{c+x_n}-L^2}{\sqrt{c-\sqrt{c+x_n}}+L}\right|=\left|\frac{L+1-\sqrt{c+x_n}}{\sqrt{c-\sqrt{c+x_n}}+L}\right|=\left|\frac{L^2+2L+1-c-x_n}{(\sqrt{c-\sqrt{c+x_n}}+L)(L+1+\sqrt{c+x_n})}\right|=\frac{|L-x_n|}{(\sqrt{c-\sqrt{c+x_n}}+L)(L+1+\sqrt{c+x_n})}\leq \frac{|x_n-L|}{L(L+1+\sqrt{c})}=\frac{|x_n-L|}{L^2+L+L\sqrt{c}}=\frac{|x_n-L|}{c-1+L\sqrt{c}}.$ Since the denominator is greater than $1$, the bound can be arbitrarily small, and hence the sequence approaches the finite limit $L$. So $c\geq 2$ is our solution.


function
integration
calculus
calculus computations

For $x>0$ define $\Gamma(x)=\int_{0}^{\infty} e^{-t} t^{x-1} dt$

1. Show that the improper integral $\Gamma(x)$ is converget

2. The function $\Gamma: (0,\infty) \rightarrow R$ so defined is continuous

3. For $x\in(0,\infty)$, $\Gamma(x+1)=x\Gamma(x)$

4. If $n\in N^{*}$, then $\Gamma(n)= (n-1)!$

1) There are two cases: $x<1$ (in which case you use that $e^{-t}t^{x-1} \leq e^{-t}$ on $[1,+\infty[$, and $e^{-t}t^{x-1} \sim t^{x-1}$ as $t \rightarrow 0$) and $x \geq 1$ (use that $e^{-t}t^{x-1} \leq e^{-t}$ on $]0,1]$ and $e^{-t}t^{x-1} = o(t^{-2})$ as $t \rightarrow +\infty$).

2) For any $\epsilon >0$, you can find $a,b$ such that $\int_{0}^{a}{e^{-t}t^{x-1}dt}+\int_{b}^{+\infty}{e^{-t}t^{x-1}dt} \leq \frac{\epsilon}{2}$. But then for any $x \in ]0,+\infty[$ and for any $x'$ close enough to $x$, $\int_{0}^{+\infty}{e^{-t}|t^{x-1}-t^{x'-1}|dt} \leq \frac{\epsilon}{2}+\int_{a}^{b}{|t^{x-1}-t^{x'-1}|dt} \leq \epsilon$.

3) Integration by parts.

4) Use 3) and $\Gamma{(1)}=1$.

Thx for explaining the first 2 parts, but when you mentioned (4) by using (3) and with $\Gamma(1)=1$, this I dont get it.
And the (3), U suggest to use the integration by parts that's what I thought, too. But I didnt get it right, Could you kindly show me how, Thx alot.

Using 3) repeatedly you have $\Gamma{(n+1)}=n (n-1) \dots \Gamma{(1)}$. Now you need to evaluate $\Gamma{(1)}=1$, whence $\Gamma{(n+1)}=n!$.
Integration by parts: $\int_{a}^{b}{e^{-t}t^{x-1}dt}=[e^{-t}\frac{t^{x}}{x}]_{a}^{b} -\int_{a}^{b}{-\frac{1}{x}e^{t}t^{x}dt}=\frac{1}{x}\int_{0}^ {+\infty}{e^{-t}t^{x}dt}=\frac{1}{x}\Gamma{(x+1)}$ when taking limits.


modular arithmetic
blogs
AMC
USA(J)MO
USAMO
number theory
prime factorization

Prove that there are infinitely many primes of the form $ 4k\minus{}1$; that is, congruent to $ 3\mod 4$.

http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions    with $ a\equal{}3,d\equal{}4$.

[hide="or..."] Suppose there are finitely many.  Let them be $ p_1,p_2,...p_n$.  Consider the number $ x\equal{}2p_1p_2\cdots p_n\plus{}1$.  Since $ 2p_1p_2\cdots p_n$ is even but not a multiple of four, we can conclude that it leaves remainder  $ 2\mod 4$.  Thus, we have $ x\equiv \minus{}1\pmod{4}$.  Consider the prime factorization of $ x$.  It can only contain primes $ 1\mod 4$ since all primes$ \minus{}1,2\mod 4$ leave remainder $ 1$.  But since a product of integers $ 1\mod4$ is equal to $ 1\mod 4$, this contradicts the fact that $ x\equiv\minus{}1\pmod{4}$.[/hide]

Or you could be like Hamster and prove that there are infinitely many primes that are 1 mod n with cyclotomic polynomials. :wink:

$ 4k\minus{}1$ is pretty easy mostly because ... -1 is not 1. Actually, more because (-1)^2=1. But like 4k+1 and 3k+1 are not as easy.

The proof is similarly easy for primes of the form $ 6k \minus{} 1$; after that it gets hard to do (without actually using Dirichlet's theorem).  Consult [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1120170#1120170]my blog post about USAMO 2008 #1[/url] and [url=http://www.math.uconn.edu/~kconrad/blurbs/dirichleteuclid.pdf]this paper[/url] for some interesting elementary generalizations.


For how many pairs of positive integers is the sum less than 14?

let's use a little casework

case 1: the smallest number is one
we can have 1-1, 1-2, 1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9, 1-10, 1-11, 1-12
there are 12 ways

case 2: the smallest number is two
we can have 2-2, 2-3, 2-4, 2-5, 2-6, 2-7, 2-8, 2-9, 2-10, 2-11
there are 10 ways

case 3: the smallest number is three
we can have 3-3, 3-4, 3-5, 3-6, 3-7, 3-8, 3-9, 3-10
there are 8 ways

we can see that for each case, there are 2 ways less than the case above it

therefore, there are $ 12\plus{}10\plus{}8\plus{}6\plus{}4\plus{}2\equal{}42$ ways
our answer is 42

@ebob, in your counting, you have counted 1-12 and 12-1 as different things, but the question implies they are the same

[quote="Ebob"]Take this as selecting 2 numbers. Starting with 12, there is 1 number that works, 1. With 11, there are 2, 1 and 2. With 10 there are 3, etc. making the answer 1+2+3...+12= $ \frac{12 \cdot 11}{2}$ = $ \boxed{66}$[/quote]
It should be 12*13/2=78.  However, it is possible that the answer is 42 because the question does not say ordered pairs.  But the FTW answer (91) is DEFINITELY WRONG.

[hide="Solution"]Hm. I would split into cases:

$a+b=13$: WLOG assume $a \leq b \implies 6$ possibilities.
$a+b=12$: Same as above, 6 possibilities
$a+b=11$: 5 possibilities.
Etc.

$6+6+5+5+4+4+3+3+2+2+1+1=2(6+5+4+3+2+1)=2(21)=42$.[/hide]


real analysis
function
real analysis solved

Let $f:[0,1]\rightarrow[0,1]$, continuously differentiable and satisfy $f(0)=0$ and $f(1)=1$. Show that the Lebesgue Measure of $f\left(\left\{x\in[0,1] : |f'(x)|<\frac1m\right\}\right)$ is less than or equal to $\frac1m$.

[quote="Buffalo"]This can't be true. Try $f\left(x\right)=\{_{4\left(x-\frac{1}{2}\right)^2\  if\ x\ge\frac{1}{2}}^{0\   if\  x<\frac{1}{2}}$[/quote]

I don't see why this example contradicts the statment. Read it carefully, I believe you are missunderstanding something. Just in case, it is asking for the measure of the image of the set, not the measure of the set itself.

In your specifical example, if $|f'(x)|<\frac1m$, then the measure of the image of such $x$'s is $\frac{1}{16m^2}<\frac1m$, then, there is no contradiction to the problem.

Best,

The set $\{x\in[0,1]||f'(x)|<\frac 1n\}$ is open (because of the continuity of $f'$), so it's a union of mutually disjoint open intervals (which must necessarily be countably many). It suffices then to show that if a function $f$ is continuously differentiable on $[a,b]$ and $|f'(x)|<\frac 1n,\ \forall x\in(a,b)$, then the measure of $f([a,b])$ is $\le\frac 1n\cdot(b-a)$.

Let $M,m$ be the maximum and, respectively, minimum value of $f$ on $[a,b]$. There exist two points $x_m,x_M$ s.t. $f(x_m)=m,f(x_M)=M$, and $M-m=(x_M-x_m)\cdot f'(c)$ for some $c$ between $x_m$ and $x_M$. We have thus shown that $|M-m|\le\frac 1n\cdot(b-a)$, and, since $M-m$ is the measure of $f([a,b])$, we're done.


search
symmetry
set theory
combinatorics unsolved
combinatorics
santa claus

there are 9 reindeers for santa claus in that position:

A-B-C
D-E-F
G-H-I

where the reindeers are called A,...,I. now, after they had to get santa claus to all the houses at one year, the neighboring reindeers don't like each other anymore (so, A and B, B and C, D and E, E and F, G and H and H and I don't like each other any more; the others still do). and it lasts long time that two reindeers don't like each other and it's impossible for santa claus that he uses them the next years again as two neighboring reindeers. how many years can santa claus arrange the reindeers s.t. no two reindeers are neighbored which dislike each other?

Peter

I managed (finally :)) to find arrangements for $5$ years, but I can only show that there are at most $6$ years, so I don't know the answer yet.. :(

ABC
DEF
GHI

DBF
EAH
GCI

HBE
ADG
CFI

BID
CEH
FAG

BGF
AIE
CHD

I hope the example is Ok. You can imagine that it's very easy to make mistakes :).

As for the proof that there are at most $6$ years, assuming thare are more than $3$ years, at least one reindeer must appear at least twice in the middle of a row, meaning that this reindeer can appear in at most $6$ years (it's the neighbour of $4$ different reindeers in two years in which it's in the middle, and there are $4$ reindeers left the neighbour of which it can be, so there are at most $4$ years except for those two in which it was in the middle).

Hey, this is much to my relief!

Well knowing that there might be solutions for six years (at most), I couldn´t figure out any of those either.

Nor could any of my students or colleges in school. So I had this one wrong in the end. :(

If someone finds a method to generate a solution systematically (e.g., marking the cells of a 9x9-table in a certain order), I would still be interested in that.

ostfriese -- a simple mathematics teacher from Germany :blush:

[quote="ostfriese"] If someone finds a method to generate a solution systematically (e.g., marking the cells of a 9x9-table in a certain order), I would still be interested in that.[/quote]

Back in a graduate class in extremal set theory, I discovered that working at the extreme often forces one into the solution. The 6-year solution is the extreme, so just start playing and see what ends up forced into place.

In each year, the three lines of reindeer count as a parallel class (do not have an element in common, yet partition the entire set). Let us start with some parallel classes and try to turn them into lines. An easy way to find parallel classes is to arrange the nine reindeer in a 3-by-3 square and take the rows, columns, diagonals, and antidiagonals as the parallel classes:

A B C
D E F
G H I

Rows: ABC DEF GHI
Columns: ADG BEH CFI
Diagonals: AEI BFG CDH
Antidiagonals: AFH BDI CEG

We are short two years and we have not decided which reindeer is the center of each set. When we choose the center, the two reindeer at the sides could be together in another year. These leftover pairings will be available to create the arrangements for years 5 and 6.

grobber said that at least one reindeer has to be in the center twice. Given that there are 18 center locations over 6 years, every reindeer is in the center on average twice -- exactly twice since each reindeer can have only 8 enemies at the end of the 6 years.

Attempt 1
For our first attempt, let us choose the centers as evenly distributed as possible. For the rows, let the centers be the reindeer in the middle column BEH. For the columns let the centers be the reindeer in the middle row DEF. For the diagonals let the centers be the reindeer in the first row ABC. For the antidiagonals, let the centers be the reindeer in the last row GHI. That gives:

Year 1:  A-B-C  D-E-F  G-H-I
Year 2:  A-D-G  B-E-H  C-F-I
Year 3:  E-A-I  F-B-G  D-C-H
Year 4:  A-H-F  B-I-D  C-G-E

Leftover pairings: A-C, D-F, G-I, A-G, B-H, C-I, E-I, F-G, D-H, A-F, B-D, C-E.
B, E, and H have been used as the center twice already, so they cannot be the center again. However, B-H is one of our unused pairings, so one of B and H would have to be the center again. That does not work.

Attempt 2
Let us back up and try again by altering the center of GHI in year 1 to no longer be H. Letting I be the center is just as bad, because E-I is an unused pairing. So the center should be G.

Year 1:  A-B-C  D-E-F  H-G-I
Year 2:  A-D-G  B-E-H  C-F-I
Year 3:  E-A-I  F-B-G  D-C-H
Year 4:  A-H-F  B-I-D  C-G-E

Leftover pairings: A-C, D-F, H-I, A-G, B-H, C-I, E-I, F-G, D-H, A-F, B-D, C-E.
B, E, and G have been used as the center twice already, so they are sides. That means our six lines for years 5 and 6 are B-H-?, B-D-?, E-I-?, E-C-?, G-A-?, G-F-?.
The possible ends of B-H-? are D and I. Let's pick B-H-D.
H has now been a center twice, so the unused H-I pairing has I in the center. It has to go with the E-I-? line, to give E-I-H.
Likewise, we get E-C-I, C-A-G, A-F-G, B-D-F.

Year 5:  B-H-D  E-C-I  A-F-G
Year 6:  E-I-H  C-A-G  B-D-F

Or we could have picked B-H-I to get:
Year 5:  B-H-I  D-F-G  A-C-E
Year 6:  H-D-B  F-A-G  C-I-E

Thus we found two solutions for the six-year case.

Erin Schram

Thank you!!

[quote]let us choose the centers as evenly distributed as possible.[/quote]
That seems to make sense. But...

[quote]That does not work.

Attempt 2
Let us back up and try again by altering the center of GHI in year 1 to no longer be H.[/quote]
The point I didn´t get yet. Why should I try exactly this? How can I expect this little change leading to success?

[quote="ostfriese"][quote]let us choose the centers as evenly distributed as possible.[/quote]That seems to make sense. But...[/quote]
That was merely intuition. I have not proven that it is a good idea, but I decided that I wanted flexibility in choosing centers for years 5 and 6.

[quote="ostfriese"][quote]That does not work.

Attempt 2
Let us back up and try again by altering the center of GHI in year 1 to no longer be H.[/quote]The point I didn´t get yet. Why should I try exactly this? How can I expect this little change leading to success?[/quote]
I did not expect to find a solution so soon. I was preparing for an exhaustive search of all 27 choices for the centers of the three sets in the first year. Luckily, I came across a solution on the third try (trying "H" as the center of GHI was my first try, trying "I" was my second try, and trying "G" was my third try).

By making a small change for my second and third tries, I saved myself some effort in recalculating the unused pairings. Also, small changes lead to a more orderly search of all possibilities.

In retrospect, however, because of the symmetry of the parallel classes I chose, those 27 cases I intended to examine are largely equivalent to each other under some permutation of the nine letters A,B,C,...,I. I haven't calculated the number of equivalence classes, but it is at most 9. So a small change was sufficent because there were few meaningful changes that I could make.

Erin Schram

[quote]By making a small change for my second and third tries, I saved myself some effort in recalculating the unused pairings. Also, small changes lead to a more orderly search of all possibilities.[/quote] Sounds good. The "horrible reindeers" won´t scare me anymore! :)

Thanks again, ostfriese

[quote=Peter Scholze]hey, then you're equally good as me; i didn't manage to find an example for 6 years as well. however, the official solution says that there are 332 of those; however, the one which is given there doesn't like there's any system behind it, i don't know how they constructed it. see https://www.mathekalender.de/aufgaben/AdventGesamtLoesungen.pdf problem 17 for that example.[/quote]

The link is not working.

[quote=adityaguharoy]The link is not working.[/quote]
You can find the file [url=https://www.mathekalender.de/info/Loesungsheft2004.pdf]here [/url](but it will be in German anyway).

it says in the solution that:
[quote]For example, if there is a group (1, 2, 3), it may later be the group (1, 3, 2)
give. This makes 9 reindeer a total of 6 possibilities of permissible
Groupings. One of 332 possible solutions is shown in the table.[/quote]
but 3 and 2 would still dislike each other in the next year so it wouldnt be possible or maybe i understood wrong since im using google translator.anyway, assuming that A will always have at least one neighbor, then there will be, at most, 7 years


floor function
LaTeX
number theory unsolved
number theory

Given positive reals $a$ and $b$ such that $\lfloor a \rfloor, \lfloor 2a \rfloor, \lfloor 3a \rfloor, \ldots$ and $\lfloor b \rfloor, \lfloor 2b \rfloor, \lfloor 3b \rfloor, \ldots$ together contain each positive integer exactly once. Is it necessary that $\frac{1}{a} + \frac{1}{b} = 1$?

A necessary and sufficient condition is that $\frac 1 a + \frac 1 b = 1$ and $a,b$ are positive irrational numbers.

This is known as Beatty's theorem.
See the link here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=beatty&t=22013

In another hand, for the necessary condition you are asking for, this is just the fact that the sum of the respective densities of the two sets is $1$ (since they have to cover the positive integers without overlapping).

Pierre.

$Pierre$,can you detail your idea better?
and here is my solution.first,it is easy to see $a,b>1$
let $f(N),g(N)$ be the number of $\lfloor ax \rfloor  and \lfloor bx \rfloor$ can express.
then we get  $f(N)+g(N)=N,af(N) \leq N+1,bg(N) \leq N+1$
and $a(f(N)+1) \geq N+1,b(g(N)+1) \geq N+1 $
then we have $N=f(N)+g(N) \leq (N+1)(a^{-1}+b^{-1})$
and  $(N+1)(a^{-1}+b^{-1}) \leq N-2 $
then let $N$ be large enough.we can prove $a^{-1}+b^{-1}=1$

[quote="zhaobin"]$Pierre$,can you detail your idea better?/quote]

I'm not a latex man  ;)  I hope being a real human.

Well, for each integer $N$, the number of integers not greater than $N$, which can be expressed in the form $[an]$ (resp $[bn]$) is $[\frac N a]$ (resp. $ [\frac N b]$ ). Since each of the positive integers not greater than $N$ belongs to exactly one of the two sets, it follows that $N = [\frac N a] + [\frac N b].$
Using that $lim_{x  \rightarrow + \infty} \frac {[x]} x = 1$ the desired result follows.

Pierre.


probability
number theory
greatest common divisor
geometry
circumcircle
function
vector

I was wondering if someone could give a solution to Q3 in Mathlinks Contest 2008 Round 1. It seems like a very beautiful problem, but I didn't manage to get anywhere on it: even the case $ n \equal{} 3$ is already non-trivial. I guess it's easy to justify intuitively using a probability argument (once the first number in our set has been chosen, the chance of choosing the second number in a satisfactory way is $ \frac {1}{ | X_{1} |}$, and after that the third may be chosen in a satisfactory way with probability $ \frac {1}{ |X_{2}| }$, etc ; so total number of ways is approx. $ \frac {|X|^n}{ |X_{1}\parallel{}X_{2}| ... |X_{n}| }$.

I can't participate in the contest, but Q1 and Q2 seemed much more approachable:
Q2 seemed standard, first using some gcd arguments prove that the 4 numbers are either squares or small multiples of squares, which reduces the problem to either proving that 4 squares cannot be in arithmetic progression ( a fairly well-known problem, can be solved using standard infinite descent arguments ), or a very similar problem with numbers which are small multiples of squares (e.g., $ 3a^2, 2b^2, c^2, 6d^2$) or arrangements of that, which can be handled in the  very similar way.
Q1 seemed like a very nice problem: my idea was to first use radical axis theorem to prove that the tangent of the point in the circumcircle, a sideline, and the line joining two feets of altitudes were concurrent; from there some simple pole/polar arguments showed that the lines that we wanted to prove are concurrent, pass not only through feet of altitudes but also through intersections of tangents at the respective vertices; it's easy to angle chase that the orthic triangle, and the triangle enclosed by tangents at vertices are similar with parallel sides, so proving the lines intersect on a point, which is collinear with $ H$ and $ O$ was simply a dilation.

epitomy01, could you please post the 2 solutions(especially the 1st) [b]in detail[/b]?
You might have an exact copy of your posted solutions, I think.
Anyone else... the 3rd problem... or new solutions to the 1st or 2nd...
It looks like Valentin is too busy(See [url=http://www.mathlinks.ro/viewtopic.php?t=63224]here[/url])

P.S. I couldn't solve any.

Q3 did seem to fall down to a probability argument as you had guessed.

Basically you proved the n-1 events that had to be true were independent, which was quite fiddly, and then you can just multiply them and use Chebyshev's inequality.

For Q1, radical axis theorem and messing around with poles and polars worked (the lines are concurrent at the pole of the radical axis of the nine-point circle and the circumcircle.

I didn't get Q2.

[quote="I"]You might have an exact copy of your posted solutions, I think.[/quote]
I think I was wrong. Else why not care to post them for ppl like me who don't get everything from hints?

First of all when the results of the first round will be announced and second could Valentin post the solutions ?

I will attach my solutions (to the first two problems ) ( I didn't solve the third )

[quote="Ilthigore"]Here is my solution to Q3. It is an almost purely probabilistic approach, though could probably be rephrased in terms of counting sets.[/quote]

Sorry, Ilthigore, but I don't understand some (apparently important) parts of your probability arguments. Moreover, I don't quite believe in Lemma 1.

Take the example $ n \equal{} 3$, $ X \equal{} \left\{1,2,...,k\right\}$, $ A_1 \equal{} A_2 \equal{} A_3 \equal{} \left\{1,2\right\}$, and let $ f_1 \equal{} f_2$ be the function that maps $ 1$, $ 2$, ..., $ k \minus{} 1$ to $ 1$ and $ k$ to $ 2$. You claim that the events $ f_1\left(x_1\right) \equal{} f_2\left(x_2\right)$ and $ f_2\left(x_2\right) \equal{} f_3\left(x_3\right)$ are independent (for random vectors $ \left(x_1,x_2,x_3\right)\in\left\{1,2,...,k\right\}^3$). This would mean that

$ P\left(f_1\left(x_1\right) \equal{} f_1\left(x_2\right)\wedge f_2\left(x_2\right) \equal{} f_2\left(x_3\right)\right)$
$ \equal{} P\left(f_1\left(x_1\right)\equal{} f_1\left(x_2\right)\right)\cdot P\left(f_2\left(x_2\right) \equal{} f_2\left(x_3\right)\right)$.

But the left hand side of this equation is $ \frac {\left(k \minus{} 1\right)^3 \plus{} 1}{k^3}$, while the right hand side is $ \left(\frac {\left(k \minus{} 1\right)^2 \plus{} 1}{k^2}\right)^2$. Are you sure Lemma 1 holds?

I may have totally misunderstood you as well...

  darij

If such events are independent, then we have equality instead inequality, yes? But of course equality holds not always.

Thank you for posting your solutions here :)

Q3 can be solved by using Cauchy - Schwartz inequality. But when the day I found this solution in my school, school was ended too late.. so I couldn't post the solution... I get a information about this contest only 3 days ago!! :(
I sent a private message to Valentin, but he seems busy..

Round 2 problems seems too hard for me... :blush:

Indeed round 2 has quite difficult problems . The third problem seems beautiful ...(and it has also three tangent circles as round 1  :D )

BTW , could someone check my solution above ? Thank you
I am also interested in a quicker solution to the second problem of the round 1

I only can solve geometry problem in Round 1 :D. My solution is as same as yours. Yes, the geometry problem in second round seems strangely and difficult  :D

Problem Q3 is just Lemma 2.1 from here: http://xxx.lanl.gov/abs/math/9906097.

Enjoy the solution!

Wow, What a clear proof!!
How did you found such a nice article??

While preparing to my course on additive combinatorics.

Ah, yes, my proof was indeed wrong. My definition of independence turns out not to be equivalent to actual independence, despite intuitively looking like it was. However, Fedor, I only predicted equality in the case where the f_i are all k_i-to-one functions for some fixed naturals k_i (everything in the image of a given f_i is mapped to the same number of times), because I invoked the Chebyshev inequality in my proof of the main result.

[quote="Little Gauss"]Q3 can be solved by using Cauchy - Schwartz inequality.[/quote]

Could you please post this solution? I can show $ n\equal{}3$ using Cauchy-Schwarz (but not only).

  darij

[quote="darij grinberg"][quote="Little Gauss"]Q3 can be solved by using Cauchy - Schwartz inequality.[/quote]

Could you please post this solution? I can show $ n \equal{} 3$ using Cauchy-Schwarz (but not only).

  darij[/quote]He has his solution posted, but I was *extremely* busy, and I didn't have time yet to complete the grading tool. You'll see it then ;)

Here are my solutions. I hope they are correct. Remembered about the third too late and then I hadn't enough time to finish my approach with Linear Algebra (assuming it could be finished...) :)

[quote="pohoatza"]@ Problem 1: Silouan, your solution is (as expected) similar to the "official" one.

@ Problem 2: See [url=http://www.math.fau.edu/yiu/NT2007notes.pdf]here[/url] (pp. 21) a solution mentioned by Paul Yiu in some of his NT lecture notes from 2007 (apparently again similar to Silouan's).[/quote]

Happy for this  ... :)

At the number theory it was the first way that came up to my mind ... Paul Yiu's proof is better than mine and he has also better writing ...


Wendy, a recent retiree, requires 6000 dollars per year in extra income. She has 50,000 dollars to invest and can invest in B-rated bonds paying 15 percent per year or in a Certificate of Deposit paying 7 percent per year.

How much money should be invested in each to realize exactly 6000 dollars in interest per year?

The answer for x and y is an approximation?


It seem to become easier if you let $a_5=b_1, a_6=b_2, a_7=b_3, a_8=b_4$.  [hide]Then let the sum of the a's be $a$.  Anyhow, then $(a_1+a_3)(a_2+a_4) \le (\frac{a}{2})^2$, and $(b_1+b_3)(b_2+b_4) \le (\frac{1-a}{2})^2$, and $a_1b_1+a_2b_2+a_3b_3+a_4b_4 \le 2(\frac{a}{2}\frac{1-a}{2})$, ie with  $a_1=a_2=b_1=b_2=\frac{1}{4}$.  Then we have the total sum $\le (\frac{a}{2} + \frac{1-a}{2})^2=\frac{1}{4}$...[/hide]
Anyways, that's all I can say right now...i'm tired...

I don`t understand the reason why $a_1b_1+a_2b_2+a_3b_3+a_4b_4 \le 2(\frac{a}{2}\frac{1-a}{2})$
It `s  not clear

If $a_2+a_4=a_1+a_3=b_1+b_3+b_2+b_4=\frac{1}{4}$, then the largest possible value for $a_1b_1+a_2b_2+a_3b_3+a_4b_4$ occurs when $a_2=a_1=b_2=b_1=\frac{1}{4}$, and the others are zero.  As an example to demonstrate this,

What is the maximum value of $x^2+(1-x)^2$?
It occurs when $x=0, 1$, correct?


number theory
greatest common divisor
algorithm
Euclidean algorithm
prime factorization

For integers $a,b$? That's pretty obvious... if the two numbers don't have a common factor, why would they after squaring?

(Or you could use the Euclidean algorithm and factorize $(b+a)(b-a)$, then logic it out from there.)

squaring does not add new prime factors to a number, this is obvious by looking at the prime factorization of a number, and noting that representation is unique, so if they have no prime factors before being squared, they have no prime factors after squared

Just square them.  Easy like spelling Mississippi.

[quote="1=2"]you spelled mississippi correctly, and, for the record,

a and b won't have new prime factors after they are squared, and you need to have two numbers that DO have a common prime factor to have a GCD of 1.[/quote]So two numbers DO need to have a common prime factor to have a GCD of 1?

i'm sure that was a typo, because they need to have a common factor other than one to have a gcd greater than one

[quote="mna851"]Everyone has gotten the gist.  To "rigorize" this I guess you could just assume that there exists a prime $p \geq 2$ that divides $a^{2}$ and $b^{2}$ but not $a$ and $b$ and it's pretty easy to arrive at a contradiction from here.[/quote]

huh? my proof works...


【Help】How can i show the "cursive" font??

The "cursive" font I want [color=red]is not [/color]the "[color=red]\mathcal[/color]”or“[color=red]\mathfrak[/color]" .

The "cursive" font [color=blue]I want [/color]often is used to denote:

[1] the [color=blue]Topology[/color] "[color=blue]T[/color]" of a  Topological Space (X,T);

and [2] the [color=blue]Power Collection [/color]"[color=blue]P[/color](X)" of all the subsets of a given set X.

The font often used to show "T" and "P" is what i want.

If someone know how to solve it, please reply me.

Thanks a lot !!

Use:
\usepackage{mathrsfs}

$[color=blue]\mathscr[/color]{ABCDLPT...}$

But, it does not show the "Bold" style in "\chapter".
Dose any one know how to show the [color=blue]"Bold" style of "\mathscr"[/color]?

eh, i'm not sure what you want...

you want to:

use the cursive font;

make the name of chapter in bold.

right?

ledurt


[quote="Bakkan"]You ever see a sports team name and it immediately reminds you of something in mathematics that has nothing to do with the situation at hand?  I've felt this one too many times to ignore it any more, so I present to you the newest U.S. football team, the Eulers!

[img]http://i46.photobucket.com/albums/f111/wdtboss/Eulerslogo.jpg[/img][/quote]

YAAAAAAAAAAAAAAAAAAAAAAAAAAAY!


[quote="math92"]If a plane can fly 1200 miles in 2 hours, how long will it take if you traveled 1200 miles at 60 miles per hour?[/quote]

[hide="Solution"]If it flies 1200 miles in 2 hours, then it has a speed of 600 MPH. Of course, this has nothing to do with the quiestion being asked, but that's ok. The question:
$\frac{1200}{60}=20$. 20 hours.[/hide]

[hide] The first part has nothing to do with it.  1200/60=[b]20 hours [/b][/hide]

[quote="math92"]If a plane can fly 1200 miles in 2 hours, how long will it take if you traveled 1200 miles at 60 miles per hour?[/quote]

[hide]$D=R\cdot T$

$1200=60\cdot T$

$T=20$[/hide]