Hi,

I considered many of the olympiads problems that appeared in the last years. And I wondered how many problems are already known ? I mean I noticed many problems which go around the world. Others are slightly modified, e.g. the (year) number. Particularly in the first rounds of the olympiads there are lots of similar problems. I have to admit is difficult to define what similar means, but I think you know what I mean. So my question refers to the NMOs and TSTs.  What do you think how many problems that appeared on olympiads in 2003, (almost) identical were already published in different olympiads after 1990 ? Me and Yue Yang guess about 90-95 % ? What do you think ? Is it possible to invent some quite different new problems even today ?

I've never really thought about this problem, but it seems impossible to reach a point from which we can't invent any new problems. There are always new things to be discovered (not necessarily nasty, non-elementary things, but elementary as well).

Of course, the whole question depends on the definition you give to "similar". I think that every teacher which composes a problem is inspired, more or less, by something he's seen before, and I think this is a natural process. Problems shouldn't be copied, but some similarities will always be found between new problems and old ones.

Something that is also important to note is that the creation of original (really original, not variation of others) olympiad problems in not an easy task, and the number of ideas in some levels are somehow limitated.

During sometime in 1999 and 2000 I decided I was going to create a lot of problems, and the truth is that after a lot of time invested (and I mean, A LOT), I had created something as seven problems of IMO or a little lower level, and just one of them I really considered it TOTALLY ORIGINAL (unfortunately the statments are handwritten in a paper, in a box, three thousand miles away!). It is not something most of the people (I include myself) can do very often, and many people don't even try (Here I don't include myself).

On the other hand, I started in Olympiads in 1997, and by now, I believe that easely I have solved at least 2000 or more Olympiad kind problems, then, of course I haven't seen it all, but if you give me a problem, chances are that I have seen some time something that at least look very similar (Even when this doesn't mean that I will be able to solve it  :oops: ) Now, it is a shame, because at least the four years that I went to the national olympiad in Costa Rica as a tutor, I new most of the problems, because they were either variations or the same problem that I have seen before, and I have listen this from many people of many other countries (eg: Turkey, Hugary, Venezuela, Argentina)

Best,

[quote="djimenez"]will be able to solve it  :oops: ) Now, it is a shame, because at least the four years that I went to the national olympiad in Costa Rica as a tutor, I new most of the problems, because they were either variations or the same problem that I have seen before, and I have listen this from many people of many other countries (eg: Turkey, Hugary, Venezuela, Argentina)
[/quote]

Maybe you know that during IMO there is a tradition of distributing its olympiads problems (or a subset of them), that led to the IMO qualification that year, to the delagation leaders of other countries. But today several less strong IMO countries don't do that anymore because they do not invent new problems but rather take them from other olympiads. And giving such booklets to other nations would be rather a shame. Just look at ML/AoPS and observe how many problems get reposted.

Somewhere on the forum I also commented on the cycle of problems from olympiads to journals via math circles and vice versa.

With the acceleration of the distribution of the internet creating really new problems gets harder and harder. But on the other hand it is impossible for a young student to know all these problems at his/her age. And if he does he will deserve to win any contest.  ;)


LaTeX MIT college Harvard Yale Princeton articles

I used calc2latex to turn a table into latex. That worked fine.
But when I went to insert it into the latex, I got the same problem as when I tried to manually make the (ugly) table. It's way too large. It doesn't fit on the width of a page. It's not that way openoffice. There it fits fine. Is there a way to just resize it so it fits?
Thanks.

Try using \small before \begin{tabular} which makes the text smaller or \begin{tabular*}{\textwidth} to force the widht of the table to be no larger than the width between the margins. Or you can use \scalebox or \resizebox and other ways.
It's impossible to tell what's required unless you give a complete [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url].

Hhere's my table (as a minimal example):

\begin{table}[htbp]
\caption{}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
\textbf{School} & \textbf{Undergraduates} & \textbf{Meal Plan} & \textbf{Types of plans} & \textbf{Mandatory?} & \textbf{Costs} & \textbf{Dining Halls} & \textbf{BRDC Estimates (14 meals/wk)} & \textbf{} \\ \hline
\textbf{MIT} & 4172 & No & Preferred Dining & For students in dining hall dorms & \$600/year with a 50\% discount on all food puchased in dining halls (BRDC estimates students in dining halls spend \$2250/year) & 4 residential & \$4400/year & freshmen required to live on campus \\ \hline
 &  &  &  &  &  &  &  &  \\ \hline
\textbf{Harvard} & 6650 & Yes & 21 meals/wk & For all residents & \$4892/year & 13 residential & \$4982/year & freshmen required to live on campus \\ \hline
 &  &  &  &  &  &  &  &  \\ \hline
\textbf{Yale} & 5332 & Yes & Anytime & Some meal plan required for all residents & \$5122/year & 13 (12 residential, 1 common) & \$4860/year & freshmen required to live on campus \\ \hline
 &  &  & 21 meals/wk & For freshmen & \$4860/year &  &  &  \\ \hline
 &  &  & Kosher &  & \$4860/year &  &  &  \\ \hline
 &  &  & Any 14 &  & \$4860/year &  &  &  \\ \hline
 &  &  & Any 10 &  & \$3780/year &  &  &  \\ \hline
 &  &  & 5 lunches & non-residents only & \$1430/year & Only used in Commons &  &  \\ \hline
 &  &  &  &  &  &  &  &  \\ \hline
\textbf{Princeton} & 4635 & Yes & Unlimited & Freshmen/Soph & \$5340/year & 6 dining halls &  &  \\ \hline
 &  &  & Block 235 & Freshmen/Soph & \$5127/year &  &  &  \\ \hline
 &  &  & Block 190 & Freshmen/Soph & \$4913/year &  &  &  \\ \hline
 &  &  & Block 95 & upperclassmen & \$2937/year &  &  &  \\ \hline
 &  &  & Eating Club shared meal plan & upperclassmen &  & 10 Eating clubs &  & also includes use of facilities, parties, etc. \\ \hline
 &  &  & Eating Club membership & upperclassmen & \$7200/year & 10 Eating clubs &  & also includes use of facilities, parties, etc. \\ \hline
 &  &  & Independents & upperclassmen &  & Dorms with kitchens &  &  \\ \hline
 &  &  & Coops & upperclassmen & \$1000-1200/year & 3 co-ops &  &  \\ \hline
\end{tabular}
\label{Benchmarking Table}
\end{table}

I asked for a COMPLETE minimal example. You haven't put in your preamble so it's impossible to tell what paper, fonts margin etc you are using.
Having said that, you are very unlikely to fit the table on a sheet of normal paper, even landscape and tiny fonts, unless you put some of the text on more than one line. After all even outside the table
[code]\$600/year with a 50\% discount on all food puchased in dining halls (BRDC estimates students in dining halls spend \$2250/year  \$4400/year freshmen required to live on campus[/code]wouldn't fit on one line anyway.
How did you get it to fit onto a page in OpenOffice?

I think with the following, that includes everything. Please tell me if I'm wrong.
As for open office, it was made in openoffice and then I used calc2latex to generate the latex. I had tried making it originally, but was running into the same problem. Someone suggested I use a macro and it seemed to work well.
Thanks for your help.

\documentclass[12pt]{article}
\usepackage[pdftex]{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{enumerate}
\usepackage[pdftex]{graphicx}

\setlength{\oddsidemargin}{.25in}
\setlength{\evensidemargin}{.25in}
\setlength{\textwidth}{6.25in}
\setlength{\topmargin}{-.4in}
\setlength{\textheight}{8.5in}
\setlength{\parindent}{0in}
\setlength{\parskip}{0.1in}

\usepackage{hyperref}
\urlstyle{same}
\begin{document}

I am still guessing as you haven't answered how you managed to get it on one page in OpenOffice (you said it fitted there). I guess that text is NOT on one line but on several lines.
You will also have to use landscape of course. Did you use 12pt in OpenOffice? Did you use the same margins there as in your LaTeX document?
You can get text to use more than one line by replacing l in \begin{tabular}{|l|l|l|l|l|l|l|l|l|} by p{[i]width[/i]} where [i]width[/i] is the width of the column and can be different for each column..
So print out your table in OpenOffice, then measure the width of each column and use that for p{[i]width[/i]} making sure that the margins and font size is the same.

In openoffice, it was just in a table. It was made in a spreadsheet, then exported to a document. In texing up the document, I need to also tex the table. So I put the table back into the spreadsheet and used the macro to output the result. I don't know what else I could say about how it fit. It just fits.
I'll try your suggestion.
Thanks.

Exporting to a document would have shrunk the table and you need to copy the way it looks there. It's that document you should print out and use as a guideline for column widths, font sizes and margins.


geometry parallelogram rectangle rhombus trapezoid

A figure is an equiangular parallelogram if and only if it is a:

$ \textbf{(A)} \text{ rectangle} \qquad \textbf{(B)} \text{ regular polygon} \qquad \textbf{(C)} \text{ rhombus} \qquad \textbf{(D)} \text{ square} \qquad \textbf{(E)} \text{ trapezoid}$

[hide="My Solution"]
It's $ \boxed{(A)}$ because a rectangle has parallel sides and equal angles. A square wouldn't work in all cases.
[/hide]


[quote="ashrafmod"]The equation
 $2000x^{6}+100x^{5}+10x^{3}+x-2 = 0$
has just two real roots. Find them.[/quote]
Okay, you don't need to post AIME problems. When you do the problem, instead just click on the problem number and it'll take you to the link.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=385899#p385899


geometry
inequalities
vector
analytic geometry
trigonometry
geometry solved

i have a new solution for fisaghoures problem(but I think it is new and i dont know if it is or not). and i want to know if any body prove it like me or not.
to prove it i used power of the point. and also it:the locus of a point that :the power of it to a circle + the power of it to another circle=k(stable) is a circle.
i think if u think about it u can prove it by what i said.
and if u want the complete  soloution i can tell it.

yes i want your solution because i think it is very nice and cool,please post your solution as soon as you can.
and i am glad to see you in this site,because i know you are so cool for all geometry problems and your solution will be the best. :D

What is the problem you are talking about?

im talking about it arne : AB^2=AC^2+BC^2

dear jensen as u know i cant speak english so good .i have to translate my solution to english and it may take time.im trying to tell it to u as soon as i can.

[quote="ashegh"]AB^2=AC^2+BC^2[/quote]

Fisaghoures = Pythagoras or what? :huh:

exactly is what u said.i couldnt write it  right.

But there exists thousands of proofs already! :?

ofcourse i know it arne. but if i write my solution ,i want to know if its new or not :roll:

ok .finaly icould do it.
as i said we know that :the locus of a point like p that:the power of p to circle w+the power of p to circle q=k(stable)
is a circle that the center of it is in the line betwin the 2 centers of the circlesw and q.
and also the location of this center is the midpoint of the line that contains centers of the circles w and q.
any way ,i think it doesnt need proof .because u can prove it so easy by using (power of the point).
and also because our discussion isnt to prove it.
now we use the problem that i said in the top.
we want to change this problem in to an special form.we will look at the circles w and q from extreme.i mean that it doesnt injure or spoil the problemif we change the circles w and q to 2 pointsWandQ.does it?
and the problem will change to this one:the locus of apoint like p that:pQ^2+pW^2=k(stable)
and again we can find that pythagoras is an speciall form of this problem.
i think the solution is complete .because if we choosek some thing that we like,for example PQ^2 ,easily we can find that it needs :pn(n is the midpoint of PQ)should be equal to np and nq .
and its exactly what we want. :D

[quote="ashegh"]to prove it i used power of the point. and also it:the locus of a point that :the power of it to a circle + the power of it to another circle=k(stable) is a circle.[/quote]

Ha-ha-ha...  :rotfl:

A better problem would be: How many times is the Pythagoras theorem used to establish the above result about powers of points?

  Darij

do u mean i use pythagoras theorem in my proof?
if u think so ,plz tell me why?

:D  :D  :D  :rotfl:  :rotfl:  :rotfl:  i find that why.

[quote="ashegh"]do u mean i use pythagoras theorem in my proof?[/quote]

Well, try to prove the result you use (the locus of all points with fixed sum of powers with respect to two circles is a circle) without Pythagoras. Of course, this also means without vectors, without coordinates and without trigonometry, since any of these would make Pythagoras trivial.

  Darij

thanks darij. i found that where was my mistake. :D

after some thinking i again have question.

i think power of points as usaid can also be proved by pytagoras theorm.but if i prove them without pytagoras,i think

my solution will be true.power of points can be proved by similarity.

then i want to know will my solution be true ,if i prove power of points without using pytagoras or not.


There seems to be some debate as to which of the two AIME's this year was easier.  Also, many people have said that people are more likely to say that the test that they took was harder.

Please only vote if you took the AIME I or the AIME II this year.

I did the AIMEII and thought the AIMEII was easier, well at least in terms of the easy problems were easier, but the hard problems were harder. But I always get lower scores in practice.

Funnily enough, I think the easier problems were harder. Compare #1 against #1, #2 against #2, etc. #1 on the AIME I was a complete joke. #1 on the AIME II was pretty easy, but not as blatantly simple as on the AIME I. #2 on the AIME I was also really simple, while #2 on the AIME II was a similar problem but involved logs, which cuts out a good portion of the early high school test-takers.

Yeah the easy problems on II were harder than the easy problems on I.

Even if the easier problems are harder on the II, they are still easily solved by the top scorers, so that can't factor into the difficulty of the aime.  But to the to-be-qualifiers, there are more "less solvable" problems on the I than the II at the end.
Resolved: 2006 AIME II was easier than 2006 AIME I  :o
Q.E.D. whatever  :lol:

[quote="matt276eagles"][quote="pieterminate"]Even if the easier problems are harder on the II, they are still easily solved by the top scorers, so that can't factor into the difficulty of the aime.[/quote]

Who says difficulty only applies to the top scorers?[/quote]

When you are determining the floor/index and those who qualify probably wont miss the first few then obviously......... :?  :?

[quote="Phelpedo"]Well the questions is "Which AIME was easier?" not "Which AIME was easier to top-scoring people who are likely to qualify for the USAMO." I am also not very convinced by your attempt at a proof of why the problems on the AIME I were "less solvable." It's been pointed out for this sake many times already, but I'll mention #13, which is extremely easy for a #13 problem, and much easier than #13 on the AIME II.[/quote]

I agree. I'm a first time AIME taker and I got #'s 1,2,3,4,5,7,13 right.
Looking at that, It seems that #13 was too easy to be #13.
Then again 13 is a lucky number for penguins like me.  :D

[quote="Phelpedo"]. #2 on the AIME I was also really simple, while #2 on the AIME II was a similar problem but involved logs, which cuts out a good portion of the early high school test-takers.[/quote]

believe it or not, the best person in my state, missed #2 on AIMEI ;)

[quote="Phelpedo"]Which contradicts pieterminate's earlier statement that top scorers don't miss easy problems.[/quote]
yeah, top scorers don't miss hard problems and hate the easy ones cause they mess them up a lot

[quote="Phelpedo"]Which contradicts pieterminate's earlier statement that top scorers don't miss easy problems.[/quote]

Ok, I meant [i]usually[/i].  Tops scorers don't [i]usually[/i] miss easy problems.
 But u missed my point anyways....if a top scorer would miss an easy problem on the II there's about the same chance that they'd miss it on the I.

[quote="Magixmaniak"][quote="amitavadhar"]I thought both tests were really hard. I took both and I really sucked on both of them. :([/quote]

I thought you got a 9 on the AIME II!! How does that qualify as "really sucked"??[/quote]

i thought you said the floor was going to be 10 with your secret amc contacts.

AIME II is really easier than AIME I. see the Stats for both AIMEs.
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/06aimestats.html
AIME I taker: 18280,  to score above 10: 158 people, above 9: 303 people
AIME II taker: 4484, to score above 10: 150 people, above 9: 250 people
See, there are 4 times people in AIME I than II, but the above 10 scorers are almost the same numbers, and above 9 very close, so AIME II is easier accually.
Those who take II are lucky, but squeeze AIME I takers down.  ;)
I took I, but luckily made it. :P

So in this case, they should have made 2 indexes for I and II.
Then those people who took I and made 216.5 index, may qualify, and those who took II and made 217.5 index, may not.

[quote="pieterminate"][quote="mkkool64"]This ignores the fact that many of the large magnet and private schools took the AIME2 this year.  The stats do not necessarily tell the whole story.  I could claim the AIME2 was harder because 8.7% of takers got a zero, compared to only 3.2% of the AIME1 takers getting a zero, and this stat would be almost meaningless.  Note how close the average scores on the AIME's are, and the fact that the AMC average of the AIMEII takers is slightly higher, which could help account for this.   Last year, the differences among all of these stats were much more dramatic and the difference in index was easily justified.[/quote]

Hmm, but the magnet schools do not contain all the qualifiers and there are magnet schools taking the I as well, and 2 point different is fairly large.[/quote]

The whole issue of which AIME was easier is irrelevant now anyway.  The point I was trying to make is that there are many reasons that the AIMEII could have appeared easier and many ways to use the statistics without having all of the detailed information on which schools took each.  I am sure the people at the AMC did not have 2 floor values because they felt based on the results that the tests were similar enough to have 1 index value.  Had they thought the tests were really different by 2 full questions, they would have had 2 different indexes.  The difference in indexes last year was only 12.5, far less than 2 AIME questions.  This years tests were unquestionably closer in difficulty than last years tests.  The AMC has made there decision and arguing about it further only means more criticism towards the AMC.


Let $ x_n$ denote the number obtained by writing natural numbers from $ 1$ to $ n$ one after another. For example, $ x_1\equal{}1,x_2\equal{}12,...,x_{13}\equal{}12345678910111213$. Prove that the sequence $ x_1,x_2,...$ contains infinitely many terms divisible by $ 2005$.

Consider $ x_M$, were $ M\equal{}10^{400k\minus{}1}$, let $ x_M\equal{}a\mod 401$, then for $ M<n<10M\minus{}1$ we get $ x_{n\plus{}1}\equal{}10^{400k}x_n\plus{}n\plus{}1\mod 401\equal{}x_n\plus{}n\plus{}1\mod 401$.
Therefore for $ n\equal{}M\plus{}m, 0\le m<9*M$ we get $ x_n\equal{}a(m\plus{}1)\plus{}\frac{m(m\plus{}1)}{2}\mod 401\equal{}(m\plus{}1)(a\plus{}m/2)\mod 401.$
It give $ 2005|x_{M\plus{}2005l\plus{}400}, 0\le l<(9M\minus{}401)/2005$.


[quote="Derek"]Hi,

Just today I was playing around with sine of 3x out of boredom, here's what I get:

$\sin 3 \theta = \sin (2\theta + \theta)$
$= \sin \theta \cos (2\theta) + \cos \theta \sin (2\theta)$
$= \sin \theta (1-2\sin^2 \theta) + \cos \theta 2 \sin \theta \cos \theta$
$= \sin \theta (1-2\sin^2\theta+2\cos^2 \theta)$
$= \sin \theta (1-2)$
$= -\sin\theta$

This looks a little bit wrong to me... Maybe I did something wrong? Because I am pretty sure sine of 180 doesn't equal negative sine of 60... I thought I might've made a stupid mistake somewhere but I couldn't find it... maybe I am on crack[/quote]

$-2\sin^2\theta+2\cos^2 \theta \not= -2$

Notice the subtraction sign before the $2\sin^2\theta$.

If you were using hyperbolic trig functions, this would simplify to $2\cosh^2\theta - 2\sinh^2\theta=2$

look at the sticky: Trig formulas. There is a post about $\sin{3x}$.

another way to approach this is by using DeMoivre's Theorem which states $(\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$

By equating real and imaginary parts on both sides and saying $n=3$, you can get both $\sin{3\theta}$ and $\cos{3\theta}$.


inequalities
function
inequalities theorems

[quote="dima ukraine"]Can someone write Schur for n var in form of Murhed functions please?[/quote]
You mean $ \sum_{cyc} a^t(a\minus{}b)(a\minus{}c) \ge 0$ $ \iff$ $ 3[t\plus{}2,0,0] \plus{} 3[t,1,1] \ge 6[t\plus{}1,1,0]$ ??

[quote="Mathias_DK"][quote="dima ukraine"]Can someone write Schur for n var in form of Murhed functions please?[/quote]
You mean $ \sum_{cyc} a^t(a \minus{} b)(a \minus{} c) \ge 0$ $ \iff$ $ 3[t \plus{} 2,0,0] \plus{} 3[t,1,1] \ge 6[t \plus{} 1,1,0]$ ??[/quote]
Yes but it for 3 var, i want to see it for n var

Schur inequality doesn't hold for 4 variables..


For a cubic, choose a line which crosses it in 3 points. Add the x-coordinates of these points. What do you notice?

For a line with equation y = mx + c, I get -3c/m.

If the cubic is of the form $ ax^3\plus{}bx^2\plus{}cx\plus{}d$ then the sum of the x-coordinates is just -b/a from Vieta's formula... is this a valid answer???

(btw, how did you get -3c/m?)

To elaborate on the above --

Let the cubic be $ y\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ and the line be $ y\equal{}mx\plus{}k$

The x-coordinates of the points of intersection are the real roots of $ (ax^3\plus{}bx^2\plus{}cx\plus{}d)\minus{}(mx\plus{}b) \equal{} ax^3\plus{}bx^2\plus{}(c\minus{}m)x\plus{}(d\minus{}k)$

We are given that there are three points of intersection, so there are three real roots. So, we just want the sum of the roots of $ ax^3\plus{}bx^2\plus{}(c\minus{}m)x\plus{}(d\minus{}k)$. This is known to be $ \boxed{\minus{}b/a}$. Notice that it does not depend on the particular line we choose, if we are told that it intersects the cubic in three places.

Yeah thanks TZF for elaborating... that's exactly how I got my answer, but the thing that puzzles me is that AndrewTom's answer (-3c/m) seems to imply that the sum is dependent on the choice of lines...
So AndrewTom, did you make these problems up (w/ possible solutions)?

I made a silly mistake: I substituted mx+c for x!


geometry
parallelogram
limit
inequalities
Functional Analysis
real analysis
real analysis unsolved

Prove that a nonempty closed convex subset of a (real or complex) Hilbert space must have a unique element of least norm.

(Possible motivation for and application of this: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227031]this topic.[/url])

I barely know what a Hilbert space is, but let's see if I can do half: I will try to show that the least element, if it exists, is unique.  Suppose there were two different elements $ x$ and $ y$ of least norm.  Let's use the parallelogram law:
\[ \left\vert \frac{x\plus{}y}{2} \right\vert^2 \plus{} \left\vert \frac{x \minus{} y}{2} \right\vert^2
  \equal{} \frac{\left\vert x \right\vert^2 \plus{} \left\vert y \right\vert^2}{2} \, .
\]
That identity shows that $ \frac{x \plus{} y}{2}$ has even smaller norm, which is a contradiction.

I made sure I could do this a few weeks ago before prelims.

Let $ E\subset H$ be a nonempty closed convex subset of a Hilbert space.

Let $ \delta\equal{}\inf \{\|x\| : x\in E\}$. Now apply the Parallelogram Law to $ \frac{1}{2}x$ and $ \frac{1}{2}y$ where $ x,y\in E$. Thus $ \frac{1}{4}\|x\minus{}y\|^2\equal{}\frac{1}{2}\|x\|^2\plus{}\frac{1}{2}\|y\|^2\minus{}\|\frac{x\plus{}y}{2}\|^2$.

Now by convexity $ \frac{x\plus{}y}{2}\in E$, so $ \|x\minus{}y\|^2\leq 2\|x\|^2\plus{}2\|y\|^2\minus{}4\delta^2$ (*). So if $ \|x\|\equal{}\|y\|\equal{}\delta$ then $ x\equal{}y$ which is uniqueness.

By definition there is a sequence $ \{y_n\}\subset E$ such that $ \|y_n\|\to \delta$ as $ n\to\infty$. Plug in $ y_n$ and $ y_m$ to (*). Let $ n,m$ tend to infinity and then the RHS will go to 0. Thus $ \{y_n\}$ is Cauchy.

By completeness of $ H$ there is $ x_0\in H$ such that $ y_n\to x$. Since $ y_n\in E$ and $ E$ is closed $ x_0\in E$. Now by continuity of norm $ \|x_0\|\equal{}\lim_{n\to\infty}\|y_n\|\equal{}\delta$.

The assertion generalizes to uniformly convex spaces where only the completeness of the convex set is needed.
Hilbert spaces are uniformly convex.

That was quick. Glad to see you've all got it under control. Ravi - do you see now how the same parallelogram law inequality gives you both halves of the argument, not just the uniqueness?

Combining this argument with the link in my first post gives you that a Hilbert space is self-dual. That is, assume $ L$ is a nonzero bounded linear functional on a Hilbert space. Let $ A\equal{}L^{\minus{}1}(\{1\}).$ A is closed (since I did specify a bounded functional) and convex, so it has a unique element of least norm, $ z,$ from which (by the proposition on the link) $ L(x)\equal{}\frac1{\|z\|^2}\langle x,z\rangle.$ Or, for notation's sake, we write $ y\equal{}\frac{z}{\|z\|^2}$ and $ L(x)\equal{}\langle x,y\rangle.$ And we can show that $ \|L\|\equal{}\|y\|.$

Yes, HilbertThm (apt name) gave a clear demonstration of the harder half.


Solve for $ x$:

\[ \log \left (2^{2x\plus{}3} \right )\equal{}x
\]

If $ log x \equal{} \ln x$ then

$ (2x\plus{}3)\ln 2 \equal{} x$

$ (2\ln 2 \minus{} 1)x\equal{} \minus{}3\ln 2$

$ x\equal{}\frac{\minus{}3\ln 2}{\ln 4 \minus{}1}$

I think (s)he means [b]log[/b], not [b]ln[/b], although the solution is essentially the same.


Find all functions $f: R \rightarrow R$ such that

$xf(y)+yf(x)=(x^{2}+y^{2})f(xy)$

[hide="Solution"]Put $y=1$ to obtain

$xf(1)+f(x)=(x^{2}+1)f(x)\iff xf(1)=x^{2}f(x)\iff f(x)=\frac{C}{x}$

where $C=f(1)$.

Plugging $f(x)={C\over x}$ back in the initial equation, we get the valid identity for any $C$. Hence all the solutions are given by $f(x)={C\over x},C\in\mathbb{R}$[/hide]

One thing that I never know about problems like this... When are you sure that you have found _all_ the functions?


1)If in the expansion of $ (1\plus{}x)^{m}(1\minus{}x)^{n}$,the coefficients of $ x$ and $ x^{2}$ are $ 3$ and $ \minus{}6$ respectively then find out the values of $ m$ and $ n$.

2)For any positive integers $ n\ge m$.Prove that $ {n \choose m}\plus{}2{n\minus{}1 \choose m}\plus{}3{n\minus{}2 \choose m}\plus{}...\plus{}(n\minus{}m\plus{}1){m \choose m}\equal{}{n\plus{}2 \choose m\plus{}2}$

3)Find the coefficient of $ x^{24}$ in $ (1\plus{}x^{2})^{12}(1\plus{}x^{12})(1\plus{}x^{24})$.

[hide]Please try to give detailed solutions [/hide]

[hide="1"]
$ (1\plus{}x)^m$ can be expanded to $ 1\plus{}\dbinom{m}{1}x\plus{}\dbinom{m}{2}x^2 ...$
$ (1\minus{}x)^n$ can be expanded to $ 1 \plus{} \dbinom{n}{1} (\minus{}x) \plus{} \dbinom {n}{2} (\minus{}x)^2 ...$
Note that terms with degree higher than 3 can be ignored for this problem, so we can write this as
$ (1\plus{}mx\plus{}\dbinom{m}{2}x^2...)(1\minus{}nx\plus{}\dbinom{n}{2}(\minus{}x)^2...)$=
$ 1\plus{}(m\minus{}n)x\plus{}(\dbinom{m}{2}\plus{}\dbinom{n}{2}\minus{}mn)x^2...$

And setting the coefficients equal to those described in the problem,
$ m\minus{}n\equal{}3$
and
$ \dbinom{m}{2}\plus{}\dbinom{n}{2}\minus{}mn\equal{}\dfrac{m^2\minus{}m}{2} \plus{} \dfrac{n^2\minus{}n}{2}\minus{}mn\equal{}\minus{}6$

From the second equation,
$ m^2\minus{}m\plus{}n^2\minus{}n\minus{}2mn\equal{}\minus{}12$
But note that this is
$ (m\minus{}n\plus{}1)^2\minus{}3m\plus{}n\minus{}1\equal{}\minus{}12$
$ 4^2\minus{}3m\plus{}n\minus{}1\equal{}\minus{}12$
$ \minus{}3m\plus{}n\equal{}\minus{}\minus{}27$
And adding to this m-n=3,
$ \minus{}2m\equal{}\minus{}24$
$ m\equal{}12$
And
$ n\equal{}9$
[/hide]


geometry
geometric transformation
reflection
percent

This is a thread dedicated to the discussion of the mind-boggling question:  Iran: Axis-of-Evil or Mathematical Superpower?

Well, can you tell me the connection between Axis-of-Evil and Mathematical Superpower?

Also, I think this title is not very good. If you are a Iranian, then what will you feel if people call your country the Axis-of-Evil?

I think Bush categorized Iran as being in the Axis of Evil in a State of the Union address some years ago. G-Unit is apparently trying to reconcile this image with that of Iran regularly being a top 10 country at IMO.

Coincidentally, I was pondering the same question.
I heard on the news that Iran might be supporting the Hazbollah(sp?) (although that might be a rumor, I didn't hear everything sure).
And that's also knowing the fact that Iran has been doing so well at IMO, could it just be that the government doesn't fully reflect the best interests of its people?

Iran is currently a vastly developing country with a reasonable education system. However, there is a huge proportion of the Iranian public that are not educated and very religious. The religious proportion of the population elected Ahmadinejad, who is strengthening the "Axis of evil" view of Iran.

At the same time, the higher societal classes in Iran are usually very educated and don't care much for the silly religious side of the government, but these classes don't represent the majority of the population.

The Iranian revolution happened 25 years ago. Iran has the youngest population in the world. Over 40 percent of the population is below the age of 15 while some 4.37 percent are above 65. The remaining portion, that is 55.62 percent of the population, is aged 15 to 65. A further breakdown reveals that some 19.44 percent of the population is aged 15 to 24 and some 32.79 percent is aged 25 to 46.

The problem G-UNIT stated is irrelevant, Iran is a divided country whose future will be determined fairly well in the upcoming decade after the older, religious generation retire and the younger population take over. The activities surrounding the Israel-Lebanon war will have a big impact.

--an Iranian.

Well, please forgive me, but I think this picture illustrates the painful division in the world :

[img]http://imo2006.dmfa.si/images/20060717/IMG_1306.JPG[/img]

It's like the Israeli girl is feeling a bit uneasy and rather refrains from folding out her own flag standing next to that Iranian participant.
Which is a shame as I noticed when I went to IMO that whatever you think of countries like China or Belarus, I talked to their participants and they are just kids like you, who like math and fun.
Also interesting is that I think there have been more female contestants from Iran than from my own country (I can only think of one)

[u]But more on topic :[/u]

I don't know what to think of Iran (I highly appreciate the reactions from Iranian people).
How democratic is it?  As I understand it, presidential candidates can still be refused by the religious leaders?  And I also heard Khamenei holds a lot of power as well?
It is interesting that Iran is such a young country.  You think Iran will be really different concerning wealth,technology,religion and democracy in a few decades?

About Hezbollah, there are so many theories, one of them is that Iran orchestrated this conflict in Lebanon to divert attention from its nuclear programme at the G8 summit.

About the nuclear programme, well I can only think of one country that actually used nuclear weapons against another nation, it's despicable when that nation starts checking up on everyone else.
I personally think USA is trying to make a joke of their policies towards Iran  :huh:  : first we had the Axis of evil : Iran-Iraq-North Korea(apparently the Ministry of Truth is erasing the Iran Iraq war) and now there are rumors that 'at least one Iranian watched the North Korean missile tests'.

what I think she's fine she just doesn't like pictures much I forget why though but she's pretty good at charades? yeah that was sort of random it's just like what

you know that game where you have to guess the word by gestures only; no talking allowed

Huh, aah that game, I didn't know there was such a nice English word for it.  So...you mean you know her or something? :|


floor function
number theory unsolved
number theory

A rational number $ x$  is called [i]good[/i] if it satisfies: $ x\equal{}\frac{p}{q}>1$ with $ p$, $ q$ being positive integers, $ \gcd (p,q)\equal{}1$ and there exists constant numbers $ \alpha$, $ N$ such that for any integer $ n\geq N$, \[ |\{x^n\}\minus{}\alpha|\leq\dfrac{1}{2(p\plus{}q)}\] Find all the good numbers.

[quote="tringo"][quote="April"]A national number $ x$  is called [i]good[/i] if it satisfies: $ x \equal{} \frac {p}{q} > 1$ with $ p$, $ q$ being positive integers, $ \gcd (p,q) \equal{} 1$ and there exists constant numbers $ \alpha$, $ N$ such that for any integer $ n\geq N$,
\[ |\{x^n\} \minus{} \alpha|\leq\dfrac{1}{2(p \plus{} q)}
\]
Find all the good numbers.[/quote]

Could you tell me what \{x^n\}  means?[/quote]

{x} means the fraction part of x. i.e. {x}=x-[x]. For example, {3.6}=0.6, {e}=e-2.. is this what you want to know?

Very nice problem!

Here is my solution. Thesis:q=1

For $ n\ge N$ we can write $ x^n \equal{} a_n \plus{} \alpha \plus{} \beta_n$ where $ a_n\in\mathbb{Z}$ and $ |\beta_n|\le \frac {1}{2(p \plus{} q)}$

It's easy to see that the equality case (i.e. $ |\beta_n| \equal{} \frac {1}{2(p \plus{} q)}$) is possible only a finite number of times if q>1, so for n large enough we have $ |\beta_n| < \frac {1}{2(p \plus{} q)}$

Now $ qa_n \minus{} pa_{n \minus{} 1} \equal{} \alpha (p \minus{} q) \plus{} \beta_{n \minus{} 1}p \minus{} \beta_n q$ so $ \alpha (p \minus{} q) \plus{} \beta_{n \minus{} 1}p \minus{} \beta_n q$ is an integer.

For n large enough $ |\beta_{n \minus{} 1}p \minus{} \beta_n q| < \frac {1}{2}$ so, since $ \alpha (p \minus{} q) \plus{} \beta_{n \minus{} 1}p \minus{} \beta_n q$ is an integer, it must be the nearest integer to $ \alpha (p \minus{} q)$ (which is unique since it differs from $ \alpha (p \minus{} q)$ of less than $ \frac {1}{2}$), so it doesn't depend on n. Therefore, if we call that integer c we have that, for n large enough, $ qa_n \equal{} pa_{n \minus{} 1} \plus{} c$.

If we write $ b_n \equal{} (p \minus{} q)a_n \plus{} c$ we have that $ \{ b_i\}$ (for $ i\ge n_0$) is a sequence of positive integers such that $ qb_n \equal{} pb_{n \minus{} 1}$

So, for a given n, $ q^k$ divides $ b_n$ for all $ k$ (because $ q^kb_{n \plus{} k} \equal{} p^kb_n$ and $ (p,q) \equal{} 1$) so $ q \equal{} 1$.

OTOH, if x is integer (and greater than 1), then x is good, so good numbers are integers >1.

[quote="feliz"]How did you create the sequence $ b_n$?[/quote]

I will try to explain my solution better:

I define $ a_n\equal{}\lfloor x^n \rfloor$ and then I define $ \beta_n\equal{} \{ x^n\} \minus{}\alpha$ (in my previous post I had only claimed the existance of $ a_n$ and $ \beta_n$ such that $ x^n \equal{} a_n \plus{} \alpha \plus{} \beta_n$, $ a_n\in\mathbb{Z}$ and $ |\beta_n|\le \frac {1}{2(p \plus{} q)}$ but it's actually better to simply define them).

I proved (in the above post) that it exists an integer $ c$ such that, for n large enough, $ qa_{n}\equal{}pa_{n\minus{}1}\plus{}c$

I wanted to prove that this is impossible if q>1, so I defined $ b_n\equal{}(p\minus{}q)a_n\plus{}c$ and now, since $ qa_{n}\equal{}pa_{n\minus{}1}\plus{}c$, we have that $ qb_n\equal{}pb_{n\minus{}1}$ which implies q=1

Let me know if there is still something that is not clear.

[quote="feliz"]Actually I understood your solution... I am just asking what motivated you to create such sequence.[/quote]

I have not learnt how to write a solution properly, so it is quite common that nobody understand my solutions  :lol:  (even for very easy problems)

Anyway I think that the main idea of the solution is to consider the integer $ q\lfloor x^n \rfloor\minus{}p\lfloor x^{n\minus{}1} \rfloor$ (this is probably the only way to use the hypothesis that x is a rational number)

It is quite logical to define the sequence $ a_n\equal{}\lfloor x^n \rfloor$

Once you have proven $ qa_n\equal{}pa_{n\minus{}1}\plus{}c$ you migh notice that "c" is extremely annoying (without "c" it would be easy to write a_n explicitly). There is a standard way of solving this problem: you create another sequence $ b_n\equal{}a_n\plus{}\lambda$ choosing $ \lambda$ so that in the formula for $ b_n$ you don't have that annoyng "c" anymore.

I hope that this answers your question.

[quote="April"]A national number $ x$  is called [i]good[/i] if it satisfies: $ x \equal{} \frac {p}{q} > 1$ with $ p$, $ q$ being positive integers, $ \gcd (p,q) \equal{} 1$ and there exists constant numbers $ \alpha$, $ N$ such that for any integer $ n\geq N$,
\[ |\{x^n\} \minus{} \alpha|\leq\dfrac{1}{2(p \plus{} q)}
\]
Find all the good numbers.[/quote]

From this problem, could we generalize it as follow "For any positive non-integer number x and a given real number c, 0<c<1, there always exists a positive integer n such that {x^n} can get as close to c as desired"?

In other words,for rational,but not integer $x$ the set $A=\{\{x^n\}:n=1,2,3,...\}$ is dense.But this is not trivial.Can someone prove it?


inequalities
vector
abstract algebra
linear algebra
linear algebra unsolved

Let $ V$ be a finite-dimensional vector space, and let $ f: V\rightarrow V$ be an endomorphism with $ f^k\equal{}0$.  Prove that $ \dim\ker f\geq \frac{\dim V}{k}$.

This will follow from $ \dim \ker gh\leq \dim \ker g\plus{}\dim \ker h$ (for all endomorphisms $ g,h$) because $ V\equal{}\ker f^k$.

[hide]
Restrict the endomorphism $ h$ to the subspace $ \ker gh$ of $ V$. Under this restriction, the kernel is still $ \ker h$, because every element in $ \ker h$ is an element of $ \ker gh$. The image under the restriction is a subspace of $ \ker g$. So we get
\[ \dim \ker gh \equal{}\dim h(\ker gh)\plus{}\dim \ker h\leq \dim \ker g\plus{}\dim \ker h\]
as desired.
[/hide]


Describe suitable procedures for the synthesis of each of the following two compounds, starting with Xenon and other reagents of your choice:

- xenon difluoride

- potassium perxenate

I think $ {Xe}$ + $ {F_2}$----->$ {XeF_2}$
   with $ {Xe}$ in excess....

dont know how to pruduce the second one...

Yes, that's right, although you could hide your answers.


Hi,

I need to find a solution for all intergers that satifies the following 2 eqns. going to use {} as absoulte value symbols.

{x} + {y} = 3
{x-y} =3

I thought maybe if you squared the two eqns you could work from there, but i wasnt sure how to remove one of the terms to solve for the other.

Cheers

Hello,Elrojo!

Let's draw the graphs of $|x|+|y|=3,\ |x-y|=3$, then you will find eight pairs of integers $(x,y)$. :)

kunny

ok so drawing the eqn {x} + {y} = 3 i get a square that has intercepts at (-3,0), (3,0), (0, -3) & (0, 3).  from the first eqn x and y lie between -3 and 3 and cannot be greater or less.  Now these numbers satisfy the second eqn also {x-y}=3 from then on i have trouble drawing how the graph would look.  from my drawings there are only 4 points where they meet and those are mentioned above.

I need a little more guidance if i am to find all 8 pairs of integers.

Cheers

Therefore, |x|+|y|=|x-y|=3

Squaring, $x^2+y^2+2|x||y|=x^2+y^2-2xy$ which gives:
2|xy|+2xy=0 and |xy|+xy=0. Therefore, one of x and y is nonpositive. The other is nonnegative.

We will assume that x is nonnegative and then reverse. Therefore, x-y=3, x is nonnegative, and y is nonpositive. So (3,0), (2,-1), (1,-2), and (0,-3) all work. Now we can negate each pair to get (-3,0), (-2,1), (-1,2), and (0,3).

(-1, 2) works, just from plugging in numbers for a bit.

The graph for |x-y|=3 is two lines: x-y=3, and x-y=-3. The graphs have two line segments in common. There are 4 points in each segment, so that's eight ordered pairs.

Oops; I fixed my earlier post. I was doing my final calculations wrong. I am proud to announce that I have the only algebraic proof of the only 4 responses here :) . Elrojo:

Keep in mind that $|x|^2=x^2$. This will be very helpful in solving problems like this. Also, |x||y|=|xy|. Finally, the triangle inequality states that |x|+|y| >= |x+y|. All of these are trivial to prove, but it is important that you think to use them (especially the first) when dealing with equations like these.

thanks for that, i dont know what i was drawing when i was doing {x-y} = 3.  it looked funny.  but all is good now


trigonometry
function
calculus
calculus computations

My friend Akym Rinkovsky and I came up with this diagram:

[img]http://i291.photobucket.com/albums/ll282/benorin/AkymsTrigDiagramB.jpg[/img]

What may not be new to you hsiljak may be new to someone else.

It is true though that this relationship between the trig functions and the unit circle has been found before and can be found in some (usually older) textbooks.


MATHCOUNTS
AMC
AIME
Pythagorean Theorem
geometry

solve the following problem using pythagorean triples but not the pythagorean theorem
i know its weird but my teacher wants our mathcounts team to find a pythag triple way to do it?

The medians of a right triangle are drawn from the vertices of the acute angles are 1.2 and .5. What is the length of the hypotenuse?

thats the question word for word.

What is that supposed to mean?  You can't define Pythagorean triples without the Pythagorean theorem.

It's likely that what your teacher wants you to notice is that $ (5, 12, 13)$ is a Pythagorean triple.  Looking for Pythagorean triples in a contest problem can be a good way to guess the form of the solution (this is especially true of certain AMC and AIME problems).

I don't really think this should go here...


ABC  is an  acute  triangle. Prove  that :
min(A,B,C)>30 <==>for  any point  M  in  the  interior  of  the  triangle  ABC  we  have  min(MA,MB,MC)<min(a,b,c).

Thanks!!

[b]Lemma[/b]

min(MA, MB, MC)<=R. Let D, E, F be the midpts of BC, CA, AB respectively and let O be the circumcenter. The point M is in the interior or on the sides of at least one of the quadrilaterals OEAF, OFBD, ODCE. Let's assume M is in ODCE. From the fact that ODC and OEC are right triangles we can deduce that max(CX) with X in ODCE is R, which is the common hypothenuse of the 2 right triangles mentioned above, so MC<=OC=R, so the lemma has been proved.

[b]"=>"[/b] We have min(A, B, C)>30 Let's assume A<=B<=C, so BC<=CA<=AB. We have min(MA, MB, MC)<=R (from the lemma) (*). Let's look at the right triangle ODB. Angle DOB=A>30, so sin(DOB)>1/2, so BD/OB>1/2, so BC/2R>1/2, so BC>R, and by combining this with (*) we get min(MA, MB, MC)<BC=min(a, b, c) for any M in the interior of the triangle.

[b]"<="[/b] We have min(A, B, C)<=30. We take M=O and, for the same reasons as in the previous proof, min(MA, MB, MC)=R>=min(a, b, c).

Nice problem :D


Solve the following equations given that 0 < x <2x.  (0 is less than or equal to).  Give exact answers whenever possible.
a. 1+sin^2(x)=3sinxcosx
b. 4sin^2(x)x+csc^2(x)=5

[hide="Hints"]a) $ 1\plus{}\sin^2x\equal{}3\sin x\cos x$Recall half-angles
b) $ 4x\sin^2x\plus{}\csc^2x\equal{}5$ has no "nice form" solution. Use a calculator  :read:
However, $ 4\sin^2x\plus{}\csc^2x\equal{}5$ is solvable. Convert it into a polynomial.[/hide]

[hide]$ 1\plus{}sin^2x\equal{}3sinxcosx.\\
     \Rightarrow  1\plus{}\frac{1}{2} \minus{}\frac{cos2x}{2}\equal{}\frac{3sin2x}{2}\\
\Rightarrow(1\minus{}sin2x)\equal{}\frac{cos2x}{3}. \square both sides
\Rightarrow(1\minus{}sin2x)(1\minus{}sin2x\minus{}1/9\minus{}sin2x/9)\equal{}0
\Rightarrow(1\minus{}sin2x)(8\minus{}10sin2x)\equal{}0
\Rightarrow sin2x\equal{}1 \Rightarrow x\equal{}\frac{\pi}{4}\ or \ \frac{5\pi}{4}\\
or sin2x\equal{}\frac{8}{10}
\Rightarrow x\equal{}\frac{sin^{\minus{}1}\frac{8}{10}}{2} \ or\  \pi \plus{}\frac{sin^{\minus{}1}\frac{8}{10}}{2}$[/hide]

Another Hint for a) Divide the both sides by $ \cos ^ 2 x$.

[quote="123456789"][hide]Rewriting the equation, we have:
\[ 4\sin^2 x \plus{} \frac {1}{\sin^2 x} \equal{} 5
\]
Making the substitution $ y \equal{} \sin^2 x$, we have:
\[ 4y \plus{} \frac {1}{y} \equal{} 5 \implies 4y^2 \minus{} 5y \plus{} 1 \equal{} 0 \implies \left(4y \minus{} 1\right)\left(y \plus{} 1\right) \equal{} 0 \implies \sin^2 x \equal{} \frac {1}{4},\ \minus{} 1
\]
We discard $ \minus{} 1$, and solving $ \sin^2 x \equal{} \frac {1}{4}$, we have $ \sin x \equal{} \pm \frac {1}{2}$.[/hide][/quote]

why do we discard -1?

Because it's $ \sin^2 x \equal{}(\sin x)^2\equal{}\minus{}1$, and a square of a real number can never be negative.


find the number of $(x,y)$ solutions such that $\frac{2^x+3^x+4^x}{5^y}$ is an integer.

this isnt too difficult but i just want to see if some people will do it differently.

[hide]Infinite solutions for $y=0$. Done[/hide]

We know that if x is odd, then the top comes out to $1\mod{5}$.  Hence, $x$ is even.  So we then have on the top something like $(2^x+(-2)^x+(-1)^x)\mod{5}\equiv(2^x+2^x+1)\mod{5}\equiv(2^{x+1}+1)\mod{5}$.  Hence, $2^{x+1}\mod{5} \equiv 4\mod{5} \implies 2^x\mod{5} \equiv 2\mod{5}$.  So, wecan have $x=1, 5, 9, 13, ...$.  I guess we could then find corresponding values for y...

[hide]
I am tired and should sleep, so if this looks stupid, it probably is.

The last digit in $2^x$ follows the pattern $2, 4, 8, 6, 2, 4, 8, 6...$.
The last digit in $4^x$ follows the pattern $4, 6, 4, 6...$
The last digit in $3^x$ follows the pattern $3, 9, 7, 1, 3, 9, 7, 1...$

So we have four possible "last digits" in one period.
$2+4+3=9$
$4+6+9=19$
$8+4+7=19$
$6+6+1=13$

The last digit in $2^x+3^x+4^x$ will always either be a 9 or a 3. We know that the last digit has to be eiher a 0 or a 5 to make a number divisible with 5. Therefore no positive integer solutions exists.

I apologize for my wrong use of the word number. Should I replace it with digit? Apparently I should, now it is changed.
[/hide]

yes i think last digit would be more correct but we should still be able to know what u mean.
well u got the right answer. I'm some what dissapointed, only because you did it the same way i did.  :blush:  o well.

[quote="dakyru"][quote="K81o7"]We know that if x is odd, then the top comes out to $1\mod{5}$.  Hence, $x$ is even.  So we then have on the top something like $(2^x+(-2)^x+(-1)^x)\mod{5}\equiv(2^x+2^x+1)\mod{5}\equiv(2^{x+1}+1)\mod{5}$.  Hence, $2^{x-1}\mod{5} \equiv 4\mod{5} \implies 2^x\mod{5} \equiv 2\mod{5}$.  So, wecan have $x=1, 5, 9, 13, ...$.  I guess we could then find corresponding values for y...[/quote]

1, 5, 9, 13... are not even.   ;)[/quote]

Oh yeah.  So then there are no solutions....


Is it true that mods have no smiley limit in each post?

And why IS there a smiley limit in blog entries but not shoutboxes?

so then does that mean they get to post a million smilies in one post? :o
isn't that unfair? cuz if they have the right to do that and they do that then things get out of control :o

[quote="gauss1181"][quote="program4"]mods keep things in control, its [i]regular[/i] users that are out of control.
[size=200] :o [/size][/quote]
but then why would that entitle mods the right to post as many smilies as they want?[/quote]Because I decided so.

Is it really such a big deal?  Although it does get REALLY annoying when people post a bunch of smilies in a shoutbox.

[quote="isabella2296"]I've never seen a mod go overboard on smileys, with the exception of Valentin Vornicu once, when he was demonstrating why there was a smiley limit.[/quote]I'm not a mod  :ninja:


Let $q$ be a prime factor of $a^2 + b^2$. Prove that if $q=3$(mod 4) then $q^2|a^2 + b^2$.

A similar question has already been posted!!

Was the question exactly the same or similar?  :huh:

Please keep posts at the level of the forum.  I am moving this from Getting Started to Intermediate.

Isn't it just because a square number is either 1 or 0 (mod 4), so when you sum two squares, it is 0,1, or 2 (mod 4). Since q is 3 mod 4, we must take it to some power to make it 0,1, or 2 (mod 4). we see that we can only take it to an even power because odd powers will give 3 (mod 4)

Yes but $q^4$ doesn't necessarily divide $a^2 + b^2$.


superior algebra
superior algebra solved

I didn't really know where to post this, so I apologize in advance if it's in the wrong place.

I just have a few questions regarding set products and equivalence relations.  For the set $A=\left\{1,2\right\}$, would the product $A\times A=\left\{\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\}$?

Also, I know that $\neq$ is not an equivalence relation, but am unsure of the properties it satisfies.  Well, I know that it doesn't satisfy the reflexive property and it DOES satisfy the symmetric property, but I am having trouble on the transitive property.  I am saying that it DOESN'T satisfy the transitive property for the following reason:

$a,b,c\in\mathbb{R}$

Consider the case where $a=c$.  Then $a\neq b$ and $b\neq c$ does not necessarily mean that $a\neq c$.  Therefore it doesn't satisfy the transitive property.  Is my reasoning correct?  Am I allowed to assume the case where $a=c$?

Thank you very much :)

The Cartesian product is the set of all ordered pairs- in your example, that's $\{(1,1),(1,2),(2,1),(2,2)\}$

On the equivalences: You've just constructed a counterexample to the transitive property, so it doesn't hold (as long as the set has at least two elements).

Alright, so know if I wanted to know $A\times A\times C$, where $C=\left\{3,4\right\}$, the answer would be:

$A\times A\times C=\left\{\left(1,1,3\right),\left(1,1,4\right),\left(1,2,3\right),\left(1,2,4\right),\left(2,1,3\right),\left(2,1,4\right),\left(2,2,3\right),\left(2,2,4\right)\right\}$

Is this correct?

Thank you.

Yes, assuming the identification of the ordered pair of ordered pairs $((a,b),c)$ (or $(a,(b,c))$) with the ordered triple $(a,b,c)$; we might as well take that as a definition.

OK, so assuming we take that as a definition, I can say that $A\times A\times C=A\times\left(A\times C\right)$, right?

Thanks again :)


Today I overheard a yoga instructor asking, "what's a mind look like?".  I'm pretty sure he hadn't a clue.  Someone once told me "all the secrets of the universe can be found in Complex Analysis".  I disagree, I think they're all in differential equations.  Once I saw a sihouette of a man looking at something in his hand. The sihouette was made up of hundreds of small ball-and-stick molecules.  In his hand . . . was another molecule.

A mind is whatever you want to believe it is.  Similar question is "What does god look like"?  In my case, I don't believe in god so there is no answer.  Some people might say the typical old guy with a white beard.  A Hindu might say there are thousands.  In other words, it is a completely subjective question.

And anyway, we all strings are the key to the universe.  :lol:

I dug up a pretty decent quote from one of the Halo novels.

[quote="Dr. Halsey"]What is the mind, really? Intuition, reason, emotion-we acknowledge they exist, but we still don't know what makes the human mind [i]work[/i]...[/quote]

I believe the human mind is a fabrication as a result of the evolution of self-consciousness. Humans are of one of the most evolved species to have ever existed on Earth, and the result of that is a superior central nervous system, including the brain. With our brain, we are able to store memories, emotions, and automated intuition, something that was not possible with the Platyminthes that evolved on Earth millions of years ago.

Hey JRav.  I'd have to disagree with you. God and the mind are not in the same category:  God is make-believe, the mind however is a real phenomenon accessible to scientific inquiry and more and more I see it becoming accepted that the mind is an emergent property of neural assemblies.

And 7h, I think in order to know what makes the mind work, we first have to agree on what the mind is and there's a lot of disagreement on that although personally I'm quite convinced it is a dynamic phenomenon emerging from the global firing patterns of neurons and those dynamics are potentially describable by . . .

I don't think the yoga class would have been into differential equations though, not hardly anyway.

[quote]God is make-believe, the mind however is a real phenomenon accessible to scientific inquiry and more and more I see it becoming accepted that the mind is an emergent property of neural assemblies.[/quote]

Thank god we have atheists among us (pun intended!).

i hate how everyone (except some of you guys) around me arent atheists, and then im so lonley bkuz they always keep talking about god!!
yeah...

Actually, part of being a good atheist is knowing religion, the Bible,...  Because if you are well-versed in religion, then you can be well-versed in its flaws and hence when others discuss God, you can join in and say "No, no, no.  Here's the problem with that..."

I knew I should have never left this topic open in G&FF.  :noo:

(Extraneous posts deleted by the way)


find all functions f defined from non zero real numbers to real numbers

such as for all x ,y  we have:
     f(f(y)/x)=3xy+f(x)

Manifestly the function is surjective, and there is $y\neq 0$ such that $f(y)\neq 0$
$x= 1 \Longrightarrow f(f(y))=3y + f(1)$
$x= f(y) \Longrightarrow f(1)= 3yf(y)+ f(f(y))$
Thus $f(y)\neq 0 \Longrightarrow f(y)=-1$, and that's non surjective , so there is contradiction !
:cool:  A such function does not exist ! :cool:


vector
linear algebra
linear algebra unsolved

Let $V$ be a finite-dimensional vector space over $\mathbb R$. To any nondegenerate* positive symmetric bilinear form $B\colon V\times V\to\mathbb R$ we can associate a norm on $V$, namely $\|u\|=\sqrt{B(u,u)}$. The map $v\mapsto B(\cdot,v)$ is an isometry between $(V,\|\cdot\|)$ and the dual space $V^{*}$.

My question is: If $B$ is a nondegenerate [b]skew-symmetric[/b] bilinear form, can we find a norm on $V$ for which $v\mapsto B(\cdot,v)$ is an isometry between $V$ and $V^{*}$?

(*) "nondegenerate" means "$\forall u\,B(u,v)=0\implies v=0$".

So, we may just think that $V=\mathbb R^{n}$ and $B$ is given by an antisymmetric matrix. Then the eigenvalues of $B$ are purely imaginary and there is an orthonormal basis in $\mathbb R^{n}$ consisting of pairs of vectors $u_{k}, v_{k}$ such that $Bu_{k}=\lambda_{k}v_{k}$, $Bv_{k}=-\lambda_{k}u_{k}$ with some $\lambda_{k}>0$. The norm $\|\sum \alpha_{k}u_{k}+\sum\beta_{k}v_{k}\|=\sqrt{\sum \lambda_{k}(\alpha_{k}^{2}+\beta_{k}^{2})}$ seems to satisfy the requirements.


[quote="GameBot"]Abby, Bart, Cindy and Damon weigh themselves in pairs. Together Abby and Bart weigh 160 pounds, Bart and Cindy weigh 180 pounds, and Cindy and Damon weigh 200 pounds. How many pounds do Abby and Damon weigh together?[/quote]

Let $ a$ = Abby's weight, $ b$ = Bart's weight, $ c$ = Cindy's weight, and $ d$ = Damon's weight. We have:

\[ a \plus{} b \equal{} 160 \]

\[ b \plus{} c \equal{} 180 \]

\[ c \plus{} d \equal{} 200 \]

Add the equations together to get:

\[ a \plus{} 2b \plus{} 2c \plus{} d \equal{} 540 \]

We want $ a \plus{} d$, so let's try to isolate that. $ 2b \plus{} 2c$ is in the way, however. Remember that $ 2b \plus{} 2c \equal{} 2(b \plus{} c)$...but we know what b + c is, don't we? According to our second equation, it's 180.

\[ a \plus{} 2(180) \plus{} d \equal{} 540 \]

\[ a \plus{} d \equal{} 540 \minus{} 2(180) \equal{} \boxed{180} \]


$(a,b,c,d)=(a,a,a,{2\over a^{3}})\Rightarrow e=3a+{2\over a^{3}}\mathop{\longrightarrow}_{a\to\infty}\infty$


Given triangle ABC, and any segment DE, D on AC, E on AB, with circumcenters O1 O2 of ABC and ADE respectively, angle BO1C = angle DO2E.

not really hard, but it looks kinda cool

[hide="Answer"]
By inscribed angles, $ \angle BO_1C \equal{} 2\angle BAC \equal{} \angle DO_2E$.
[/hide]


function
logarithms
calculus
derivative
parameterization
limit
absolute value

If the functional relationship f(xy) = f(x)/y + f(y)/x holds for any real x , y > 0 and f(x) is a differentiable function for all real x > 0 such that f(e) = 1/e , then  find f(x).

[hide="Hint"]
Look at $ f(1), f(2)$, what does that tell you about $ f(1)$
[/hide]

[hide="Solution"]
Multiply through by $ xy$ to get $ xy \cdot f(xy) \equal{} x \cdot f(x) \plus{} y \cdot f(y)$

Define $ g(x) \equal{} x \cdot f(x)$. Then, we have that $ g(xy) \equal{} g(x) \plus{} g(y)$

By properties of logarithms, we know that any function of the form $ g(x) \equal{} a \ln |x|$ will satisfy the above.
[hide="Demonstration"]
f is differentiable, thus g is differentiable.

Taking the derivative wrt. y, we have:
$ x g'(xy) \equal{} g'(y) \implies \frac {g'(xy)}{g'(y)} \equal{} \frac {1}{x}$

Setting the parameter $ y$ equal to $ 1$, we have that $ g'(x) \equal{} \frac {g'(1)}{x} \implies g(x) \equal{} g'(1) \ln |x| \plus{} C$

By inspection, $ C \equal{} 0$, but $ g'(1)$ is free, so we will define $ a : \equal{} g'(1)$. Hence, $ g(x) \equal{} a \ln |x|$
[/hide]
Hence, $ f(x) \equal{} \frac {g(x)}{x} \implies$ $ \boxed{ f(x) \equal{} \frac {a \ln |x|}{x} \; : \; a \in \mathbb{R}}$[/hide]

[quote="The Zuton Force"]By properties of logarithms, we know that any function of the form $ g(x) \equal{} a \ln |x|$ will satisfy the above.[/quote]

In fact, we can say more strongly that the only functions differentiable on $ {\bf R}_{ > 0}$ that satisfy the equation for $ g$ are the logarithms (no absolute value needed, since we're only living over the positive reals).  Proofs of this can be found in the Calculus fora, which is where this problem should have been posted.

$ h(x) \equal{} e^x f(e^x)$ satisfies the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url], which as JBL said has been discussed in some detail repeatedly.

[quote="scorpius119"]This can be done directly, although it's sort of ugly. Also, we only need differentiability at $ x \equal{} 1$.
[hide]
First plug in $ x \equal{} y \equal{} 1$ into the functional equation to get $ f(1) \equal{} f(1) \plus{} f(1)\Rightarrow f(1) \equal{} 1$. The definition of the derivative then says $ \lim_{h\rightarrow 0} \frac {f(1 \plus{} h)}{h}$ exists; call this limit $ A$.

We now determine the derivative for positive $ x$: since $ x \plus{} h \equal{} x\left( 1 \plus{} \frac {h}{x}\right)$, we get
\[ f(x \plus{} h) \equal{} \frac {f(x)}{1 \plus{} \frac {h}{x}} \plus{} \frac {f\left( 1 \plus{} \frac {h}{x}\right)}{x}
\]
Subtracting off $ f(x)$ from the first term in the right hand side gives $ f(x)\cdot \frac { \minus{} \frac {h}{x}}{1 \plus{} \frac {h}{x}}$. So we get
\[ \frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{} \minus{} \frac {f(x)}{x}\cdot \frac {1}{1 \plus{} \frac {h}{x}} \plus{} \frac {1}{x^2}\cdot \frac {f\left( 1 \plus{} \frac {h}{x}\right)}{\frac {h}{x}}
\]
As $ h\rightarrow 0$, $ \frac {h}{x}\rightarrow 0$ for positive $ x$, so that $ f$ is differentiable for positive $ x$ and satisfies
\[ f^\prime (x) \equal{} \minus{} \frac {f(x)}{x} \plus{} \frac {A}{x^2}
\]
This leads to the usual integrating factor trick: substituting $ g(x) \equal{} xf(x)$ gives
\[ g^\prime (x) \equal{} f(x) \plus{} xf^\prime (x) \equal{} \frac {A}{x}
\]
so $ g(x) \equal{} A\ln x \plus{} B$ for some constant $ B$.

This leads to $ f(x) \equal{} \frac { A\ln x}{x} \plus{} \frac {B}{x}$. Plug in $ x \equal{} 1$ to get $ B \equal{} 0$, then plug in $ x \equal{} e$ (we're given $ f(e) \equal{} \frac {1}{e}$) to get $ A \equal{} 1$, so that $ f(x) \equal{} \boxed{\frac {\ln x}{x}}$.
[/hide][/quote]

Errm

You have made a mistake
[hide]
$ f(1) \equal{} f(1) \plus{} f(1)\Rightarrow f(1) \equal{} 1$
If this was true
$ f(1)\equal{}f(1)\plus{}f(1)\equal{}2 \Rightarrow f(1)\equal{}2$
and if that was true
$ f(1)\equal{}f(1)\plus{}f(1)\equal{}4$
...

Using $ f(1)\equal{}f(1)\plus{}f(1)\equal{}2f(1)$ and $ f(2)\equal{}f(2)\plus{}\frac{f(1)}{2}\equal{}f(2)\plus{}f(1)$ Therefore $ f(1)\equal{}0$
[/hide]

[quote="SimonM"][quote="scorpius119"]This can be done directly, although it's sort of ugly. Also, we only need differentiability at $ x \equal{} 1$.
[hide]
First plug in $ x \equal{} y \equal{} 1$ into the functional equation to get $ f(1) \equal{} f(1) \plus{} f(1)\Rightarrow f(1) \equal{} 1$. The definition of the derivative then says $ \lim_{h\rightarrow 0} \frac {f(1 \plus{} h)}{h}$ exists; call this limit $ A$.

We now determine the derivative for positive $ x$: since $ x \plus{} h \equal{} x\left( 1 \plus{} \frac {h}{x}\right)$, we get
\[ f(x \plus{} h) \equal{} \frac {f(x)}{1 \plus{} \frac {h}{x}} \plus{} \frac {f\left( 1 \plus{} \frac {h}{x}\right)}{x}
\]
Subtracting off $ f(x)$ from the first term in the right hand side gives $ f(x)\cdot \frac { \minus{} \frac {h}{x}}{1 \plus{} \frac {h}{x}}$. So we get
\[ \frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{} \minus{} \frac {f(x)}{x}\cdot \frac {1}{1 \plus{} \frac {h}{x}} \plus{} \frac {1}{x^2}\cdot \frac {f\left( 1 \plus{} \frac {h}{x}\right)}{\frac {h}{x}}
\]
As $ h\rightarrow 0$, $ \frac {h}{x}\rightarrow 0$ for positive $ x$, so that $ f$ is differentiable for positive $ x$ and satisfies
\[ f^\prime (x) \equal{} \minus{} \frac {f(x)}{x} \plus{} \frac {A}{x^2}
\]
This leads to the usual integrating factor trick: substituting $ g(x) \equal{} xf(x)$ gives
\[ g^\prime (x) \equal{} f(x) \plus{} xf^\prime (x) \equal{} \frac {A}{x}
\]
so $ g(x) \equal{} A\ln x \plus{} B$ for some constant $ B$.

This leads to $ f(x) \equal{} \frac { A\ln x}{x} \plus{} \frac {B}{x}$. Plug in $ x \equal{} 1$ to get $ B \equal{} 0$, then plug in $ x \equal{} e$ (we're given $ f(e) \equal{} \frac {1}{e}$) to get $ A \equal{} 1$, so that $ f(x) \equal{} \boxed{\frac {\ln x}{x}}$.
[/hide][/quote]

Errm

You have made a mistake
[hide]
$ f(1) \equal{} f(1) \plus{} f(1)\Rightarrow f(1) \equal{} 1$
If this was true
$ f(1) \equal{} f(1) \plus{} f(1) \equal{} 2 \Rightarrow f(1) \equal{} 2$
and if that was true
$ f(1) \equal{} f(1) \plus{} f(1) \equal{} 4$
...

Using $ f(1) \equal{} f(1) \plus{} f(1) \equal{} 2f(1)$ and $ f(2) \equal{} f(2) \plus{} \frac {f(1)}{2} \equal{} f(2) \plus{} f(1)$ Therefore $ f(1) \equal{} 0$
[/hide][/quote]

I see, I put $ f(1)\equal{}1$ instead of $ f(1)\equal{}0$ by mistake.  :blush:

Oh well, I use $ f(1)\equal{}0$ in the rest of my solution anyway.


Two circunmferences $ \Gamma_1$ $ \Gamma_2$ intersect at $ A$ and $ B$
 $ r_1$ is the tangent  from $ A$ to $ \Gamma_1$  and $ r_2$ is the tangent from $ B$ to $ \Gamma_2$
  $ r_1 \cap r_2\equal{}C$
 $ T\equal{} r_1 \cap \Gamma_2$  ($ T \neq A$)
 We consider a point $ X$ in $ \Gamma_1$ which is distinct from $ A$ and $ B$.
 $ XA \cap \Gamma_2 \equal{}Y$  ($ Y \neq A$)
    $ YB \cap XC\equal{}Z$
Prove that $ TZ \parallel XY$

this exercise is very easy
$ CB.CD\equal{}CA^{2}$ and $ CB^{2}\equal{}CT.CA$ so $ CT/CA\equal{}CB/CD$
$ \Rightarrow (TA)//(BD)$
now we need to prove that $ (YB)//(XD)$
it's clair because $ \angle AYB\equal{}\angle ABD\equal{}\angle AXD$
now it's done with THALES...


calculus
integration
logarithms
trigonometry
calculus computations

good day nutties.  :D

show

$ \int_0^{\dfrac{\pi}{2}}\;\ln\,\sin\,\theta\;\cdot\;\ln\,\cos\,\theta\;\;\textbf d\theta\;\equal{}\;\boxed{\dfrac{\pi}{8}\left(4\,\ln^2\,2\;\minus{}\;\zeta(2)\right)}$

[quote="Black"][url=http://www.mathlinks.ro/viewtopic.php?t=193492&search_id=1248903]Here[/url].[/quote]

oh !  :blush: [size=59] ummm.....  sorry...[/size]  :(

let me instead ask for the solution to this integral:

$ \int_0^{\dfrac{\pi}{4}}\;\ln\,\sin\,\theta\;\cdot\;\ln\,\cos\,\theta\;\;\textbf d\theta$


geometry
conics
ellipse
function
linear algebra
matrix
calculus

Hello,
 I need to find the volume inside the $ egg$ which has the shape $ 2x^2\plus{}3y^2\plus{}z^2\equal{}1$, using cavalieri's principle.
I have found that, by keeping z constant, the area that lies inside the ellipse in the $ (x,y)$ plane that is given by the section $ 2x^2\plus{}3y^2\equal{}1\minus{}z^2$ is equal to $ \frac{\pi (1\minus{}z^2)}{\sqrt{6}}$. I don't know how to continue to find the whole volume. Are there any hints?Thanks.

Just integrate that area (as a function on $ z$) between $ \minus{} 1$ and $ 1$; it's as simple as that.

A slightly more multivariable approach to this: the volume is
\[ V\equal{}\iiint\limits_{2x^2\plus{}3y^2\plus{}z^2\le 1}1\,dx\,dy\,dz\]
Now make the changes of variable $ u\equal{}\sqrt{2}\,x,v\equal{}\sqrt{3}\,y,w\equal{}z.$ As part of that change of variables, we get that $ dx\,dy\,dz\equal{}\frac1{\sqrt{2}}\cdot\frac1{\sqrt{3}}\,du\,dv\,dw.$ (While I could call that a Jacobian, the matrix language seems like overkill for a diagonal matrix.) Then
\[ V\equal{}\frac1{\sqrt{6}}\iiint\limits_{u^2\plus{}v^2\plus{}w^2\le 1}1\,du\,dv\,dw.\]
But that integral is the volume of the unit sphere, and we know that number.


Yes, the answers to the second one are always double the answers to the corresponding question on the first one. Everybody should get a perfect by staring really hard at the answers to the first AIME right before the test. There is also the strange coincidence that each test has the same number of questions each year.

But seriously, if there were any correlation, there would be too big of an advantage to the AIME II people. It sorta defeats the purpose of taking a challenging math exam.

[quote="Bob"]Is there any relation between the two AIMEs? In other words, can you tell anything about the alternate one from looking at the first one. Thanks.[/quote]

The best you can say is like they are at similar difficulty level, and that's not even true sometimes.

Some say that the AMC people that make the AIME II purposely design it to be more difficult becuase the people that take it have more time (2 weeks) to prepare for it.  Then again, difficulty is opinionated.

I've noticed though... the AIME II always seems to have more geometry than the AIME I, and seems less "straightforward". It seems like they try to balance out the questions in each subject throughout the two AIMEs. So if the first AIME had very few geometry questions, then the second with have more to make up? Did anyone else notice this?

[quote="h_s_potter2002"]Two weeks isn't that much of a big difference, considering that most takers are just happy to be taking the AIME and don't try hard to prepare, or have been trying to get a high score for a couple of months, or maybe even a year or two, so two weeks won't help them that much.[/quote]

Hey there are people like me that don't prepare at all for the amc and just the aime and end up having 2 weeks total of preparation  :(  Whereas having taken the aime II would have given me twice as much preparation and doubled my score :lol: .


Dear everyone.

please think of the following.

for a square ABCD, let a point P in it such that $ \angle{PAB}\equal{}\angle{PBA}\equal{}15^{\circ}$.
prove that $ \triangle{PCD}$ is equilateral.

thanks.

Just draw a square DCEF with M in it,such that <MDC=<MCD=60.
=> <MDF=<MCE=30  => <MFD=<MEC=75 => <MFE=<MEF=15.
=> tr.MFE=tr.APB => AP=PB=ME=MF => tr.APD=tr.DMF => tr.PDC is equilateral. :)

Alternatively, let $ AB \equal{} 2x$. If $ M$ is the mid-point of $ AB$ and $ PM \equal{} d$, then $ \tan 15 \equal{} \frac{d}{x}$, so $ d\equal{}x\tan 15$.

Draw $ EF$ parallel to $ AB$ where $ E$ is on $ DA$ and $ F$ is on $ CB$. Then $ CF \equal{} x(2\minus{}\tan 15) \equal{} 2 \minus{} \sqrt{3}$ and $ \tan 45 \equal{} \tan (60 \minus{} 15)$. Therefore $ 1\equal{} \frac{\sqrt{3} \minus{} t}{1\plus{}\sqrt{3}t}$ where $ t \equal{} \tan 15$.

$ t \equal{} \frac{\sqrt{3}\minus{}1}{\sqrt{3}\plus{}1} \equal{} 2\minus{}\sqrt{3}$

Therefore angle $ CPF \equal{} 60$ degrees nand so angle $ PCF \equal{} 30$ degrees and so triangle $ PDC$ is equilateral.


There's a typo in the book ([color=darkred][size=75]AoPS vol. 1, page 161[/size][/color]).

The area of a parallelogram always equals $\frac{1}{2}b_1b_2sin\theta$.

The book says "A parallelogram has two sides of length 3 and 6.  The angle opposite

the angle included between these sides is 30 degrees.  Find the area of the

parallelogram."

$(frac{1}{2})(\frac{1}{2})(3)(6)=\frac{9}{2}$

The book says 9.

 :coolspeak:
 :coolspeak:
 :coolspeak:
 :coolspeak:
 :coolspeak:
 :coolspeak:

[quote="mgao"]There's a typo in the book.

The area of a parallelogram always equals $\frac{1}{2}b_1b_2sin\theta$.

[/quote]

Where in the book did it say that? They said that for the diagonals, but for the sides, the 1/2 isn't there. So there wasn't a typo.

[quote="A+MATH"]It might help to be more descriptive than just "The Book."  ;)[/quote]

I wanted to address two concerns in two different posts, so excuse this double post. The "book" was from AoPS book 1, page 161. I have added it in his original post.


geometry
geometric transformation
homothety
geometry proposed

This is a nice problem.. Enjoy!

A circle C_0 of radius 1 km is tangent to a line L at Z. A circle C_1 of radius 1mm is drawn tangent to both C_0 and L, on the right-hand side of C_0. A family of circles C_i is constructed outwardly to the right side so that each C_i is tangent to C_0, L, and to the previous circle C_i-1. Eventually the members become so large that it is imposible to enlarge the family any further. How many circles can be drawn before this happens?

The radii of the circles $(C_0), (C_1)$ are $r_0 = 10^3\ \text m, r_1 = 10^{-3}\ \text m$. The segment length $t_1$ of the common external tangent of the touching circles $(C_0, r_0), (C_1, r_1)$ is

$t_1 = 2 \sqrt {r_0 r_1} = 2\ \text m$

Invert the figure in a circle $(Z, 2r_0)$ tangent to the circle $(C_0)$ at a point $A$. The line $l$ passing through the inversion center $Z$ is carried into itself. The tangency point $A$ of the circles $(Z), (C_0)$ is on the inversion circle and as such, it stays in place. The angle between the circle $(C_0)$ and the line $l$ is zero. The circle $(C_0)$ is carried into a line forming zero angle with the inverted line $l$, i.e., with the line $l$ itself, and passing through the point $A$, i.e., into a parallel line $l_0 \parallel l$. The chain of inverted circles $(C_1'), (C_2'), ...$ must touch the inverted circle $(C_0)$ and the inverted line $l$, i.e., 2 parallel lines $l_0 \parallel l$. Hence, the circles of the inverted chain $(C_1'), (C_2'), ...$ have all the same radii $r_1' = r_2' = ... = r_0$. The circle pairs $(C_n', C_n),\ n = 1, 2, ...$ are centrally similar with common homothety center identical with the inversion center $Z$ and homothety coefficients equal to the power of the inversion center to the inverted circle divided by the power of inversion, i.e., to square of the tangent length $t_n'$ from the inversion center $Z$ to the inverted circle $(C_n')$ divided by square of the radius of the inversion circle:

$h_n = \frac{t_n'^2}{4r_0^2},\ n = 1, 2, ...$

Consequently,

$h_n = \frac{r_n'}{r_n} = \frac{t_n'}{t_n} = \frac{t_n'^2}{4r_0^2}$

For $n = 1$, we get $t_1' = \frac {4r_0^2}{t_1} = 2 \cdot 10^6\ \text m$. Since the inverted circles $(C_n'),\ n = 1, 2, ...$ are tangent to the 2 parallel lines $l_1 \parallel l$ and to each other, the tangent lengths $t_n',\ n = 2, 3, ...$ are

$t_n' = t_1' - 2(n - 1)r_n' = \frac {4r_0^2}{t_1} - 2(n - 1) r_0 = \frac {2r_0^2}{\sqrt{r_0 r_1}} - 2(n - 1) r_0$

The problem condition, as I understand it, requires $t_n' > 0$. Hence

$t_n' = \frac {4r_0^2}{2 \sqrt{r_0 r_1}} - 2(n - 1) r_0 > 0$

$n < 1 + \sqrt{\frac{r_0}{r_1}} = 1 + \sqrt{10^6} = 1001$

$n \le 1000$

Indeed, the tangent length of the 1001-st inverted circle $(C_{1001}')$ is $t_{1001}' = \frac{4 \cdot 10^6\ \text m^2}{2\ \text m} - 2 \cdot 1000 \cdot 10^3\ \text m = 0$, which means that it passes through the inversion center $Z$. Hence, this circle is an image of a line through the point $A$ parallel to the line $l$, i.e., of the line $l_0$. If we require that all members of the defined chain are proper circles, i.e., if we exclude lines, the final member of the chain is the circle $(C_{1000})$.


algebra
polynomial
Vieta
arithmetic sequence

Suppose $ P(x)\equal{} 11x^3\minus{}4x^2\plus{}D$. If the three real roots of $ p
(x)$ form an arithmetic sequence then what is the value of $ D$.

Let $ \alpha ,\ \beta ,\ \gamma\ (\alpha <\beta <\gamma )$ be the roots the equation, Vieta's formula gives

$ \alpha \plus{}\beta \plus{}\gamma \equal{}\frac{4}{11},\ \alpha \beta \plus{}\beta \gamma \plus{}\gamma \alpha \equal{}0,\ \alpha \beta \gamma \equal{}\minus{}\frac{D}{11}$.

$ \alpha ,\ \beta ,\ \gamma$ form arithmetic sequence $ \Longleftrightarrow 2\beta \equal{}\gamma \plus{}\alpha$. Use this, we have $ D\equal{}22\beta ^3$,
since $ \beta \equal{}?$ the answer is ....

[quote="kunny"]Let $ \alpha ,\ \beta ,\ \gamma\ (\alpha < \beta < \gamma )$ be the roots the equation, Vieta's formula gives

$ \alpha \plus{} \beta \plus{} \gamma \equal{} \frac {4}{11},\ \alpha \beta \plus{} \beta \gamma \plus{} \gamma \alpha \equal{} 0,\ \alpha \beta \gamma \equal{} \minus{} \frac {D}{11}$.

$ \alpha ,\ \beta ,\ \gamma$ form arithmetic sequence $ \Longleftrightarrow 2\beta \equal{} \gamma \plus{} \alpha$. Use this, we have $ D \equal{} 22\beta ^3$,
since $ \beta \equal{} ?$ the answer is ....[/quote]

uh........

$ \alpha\plus{}\gamma\equal{}2\beta$ and $ \alpha\plus{}\gamma\plus{}\beta\equal{}\frac{4}{11}$

$ 3\beta\equal{}\frac{4}{11}$


calculus
integration
combinatorics unsolved
combinatorics

Find out the number of permutations of $ n$ first positive integer such that they contain no block $ (i,i \plus{} 1)$ or $ (i \plus{} 1,i)$ for all $ i$.

By inclusion-exclusion it is
\[ n! + \sum_{k=1}^n (-1)^k \sum_{t=1}^k {k-1\choose t-1}{n-k\choose t} 2^t (n-k)!  \]
See also sequence [url=http://oeis.org/A002464]A002464[/url] for recurrent formula and other references.

[quote="maxal"]By inclusion-exclusion it is
\[ n! + \sum_{k=1}^n (-1)^k \sum_{t=1}^k {k-1\choose t-1}{n-k\choose t} 2^t (n-k)!  \]
.[/quote]
Could you explain me how you got it???

k stands for the number of pairs (i,i+1) that are adjacent in the permutation, such pairs form chains (e.g., if i,i+1 are adjacent and i+1,i+2 are adjacent then i,i+1,i+2 is a chain in the permutation), and t stands for the number of such chains. Summands are the number of permutations with k adjacent pairs forming t chains. In particular,
${k-1\choose t-1}$ is the number of distributions of k adacent pairs among t chains such that each chain contains at least one pair;
${n-k\choose t}$ is the number of ways to select t chains of total length k+t out of elements $1,2,\dots,n$;
$2^t$ is the number of ways to orient the chains in the permutation (e.g., a chain i,i+1,i+2 can appear in the permutation as i,i+1,i+2 or as i+2,i+1,i);
$(n-k)!$ is the number of permutations of n-k-t elements not included into the chains and t chains (chains here can be viewed as single elements, hence, it's simply the number of permutations of n-k-t+t=n-k elements).

it's the way to divide k pairs into t chain.I use alsi this way to solve this pro:
http://www.mathlinks.ro/Forum/viewtopic.php?t=4110

[quote="keira_khtn"]As I understand now:$(^{k-1}_{t-1})$ is number of oisitive intergral roots of the Diophant equation:$x_1+x_2+...+x_t=k$.
  Is it right??[/quote]
That's correct.


[quote="tang zy"]$ 1\leq n\leq 2008$
Prove:$ \sqrt [n]{2008\cdot 2007\cdot ...\cdot (2009 \minus{} n)}\leq \sqrt [n]{2009\cdot 2008\cdot ...\cdot (2010 \minus{} n)} \minus{} 1$[/quote]
We can also prove that:
If $ 1\leq n\leq m\minus{}1$
$ \sqrt [n]{(m\minus{}1)\cdot (m\minus{}2)\cdot ...\cdot (m \minus{} n)}\leq \sqrt [n]{m\cdot (m\minus{}1)\cdot ...\cdot (m\plus{}1 \minus{} n)} \minus{} 1$
[b]Can any one prove it?[/b] :?:

[quote="tang zy"][quote="tang zy"]$ 1\leq n\leq 2008$
Prove:$ \sqrt [n]{2008\cdot 2007\cdot ...\cdot (2009 \minus{} n)}\leq \sqrt [n]{2009\cdot 2008\cdot ...\cdot (2010 \minus{} n)} \minus{} 1$[/quote]
We can also prove that:
If $ 1\leq n\leq m \minus{} 1$
$ \sqrt [n]{(m \minus{} 1)\cdot (m \minus{} 2)\cdot ...\cdot (m \minus{} n)}\leq \sqrt [n]{m\cdot (m \minus{} 1)\cdot ...\cdot (m \plus{} 1 \minus{} n)} \minus{} 1$
[b]Can any one prove it?[/b] :?:[/quote]
It's a direct consequence of Holder's Inequality:
$ (a_1a_2\cdots a_n\plus{}b_1b_2\cdots b_n)^n \le (a^n_1\plus{}b^n_1)(a^n_2\plus{}b^n_2)\cdots (a^n_n\plus{}b^n_n),$
for $ a_i,b_i>0.$

[quote="can_hang2007"]
It's a direct consequence of Holder's Inequality:
$ (a_1a_2\cdots a_n \plus{} b_1b_2\cdots b_n)^n \le (a^n_1 \plus{} b^n_1)(a^n_2 \plus{} b^n_2)\cdots (a^n_n \plus{} b^n_n),$
for $ a_i,b_i > 0.$[/quote]
Very nice! Sorry,but I have not thought it quiet easy,but actually it was :blush:


geometry
3D geometry
frustum
rectangle
pyramid
circumcircle
ratio

How do you find things like the circumsphere of a frustum?

$ ABCD$, and rectangle with $ AB \equal{} 12$ and $ BC \equal{} 16$, is the base of pyramid $ P$, which has a height of $ 24$. A plane parallel to $ ABCD$ is passed through $ P$, dividing $ P$ into a frustum $ F$ and a smaller pyramid $ P'$.  Let $ X$ denote the center of the circumsphere of $ F$, and let $ T$ denote the apex of $ P$.  If the volume of $ P$ is eight times that of $ P'$, then the value of $ XT$ can be expressed as $ \frac {m}{n}$, where $ m$ and $ n$ are relatively prime positve integers.  Compute the value of $ m \plus{} n$.

I usually find two ways to represent the circumradius, and use that to find where the circumcenter is.  Let's call the other 4 verticies of the frustum A', B', C', and D' (A' being situated above A, B' above B, etc.)  We can draw a segment from A to a point raised above the center of ABCD, and another segment from A' to the same point.  Set the lengths of these two segments equal, and the point will be the circumcenter.  Then, it's a simple matter of finding the distance from the circumcenter to ABCD.  This is simple:  we have two right triangles, both with the same hypotneuse, and we can find the length of one leg for each triangle (cut ABCD and A'B'C'D' each into 4 congruent rectangles, then use the Pythagorean Theorem).  I think you can get it from there.   :lol:

[hide="Solution"]
As we are dealing with volumes, the ratio of the volume of $ P'$ to $ P$ is the cube of the ratio of the height of $ P'$ to $ P$.

Thus, the height of $ P$ is $ \sqrt[3]{8} \equal{} 2$ times the height of $ P'$, and thus the height of each is $ 12$.

Thus, the top of the frustum is a rectangle $ A'B'C'D'$ with $ A'B'\equal{}6$ and $ B'C' \equal{} 8$.

Now, consider the plane that contains diagonal $ AC$ as well as the altitude of $ P$. Taking the cross section of the frustum along this plane gives the trapezoid $ ACC'A'$, inscribed in an equatorial circular section of the sphere. It suffices to consider this sphere.

First, we want the length of $ AC$. This is given by the Pythagorean Theorem over triangle $ ABC$ to be $ 20$. Thus, $ A'C' \equal{} 10$. Since the height of this trapezoid is $ 12$, and $ AC$ extends a distance of $ 5$ on either direction of $ A'C'$, we can use a 5-12-13 triangle to determine that $ AA'\equal{}CC'\equal{}13$.

Now, we wish to find a point equidistant from $ A$, $ A'$, and $ C$. By symmetry, this point, namely $ X$, must lie on the perpendicular bisector of $ AC$. Also, $ AXA'$ must be an equilateral triangle, so $ X$ must also lie on the perpendicular bisector of $ AA'$. If we set $ A$ as $ (10,0)$, $ C$ as $ (\minus{}10,0)$ and $ A'$ as $ (5,12)$, we have by the first requirement that $ X$ is of the form $ (0,x)$, and by the second requirement that $ X$ is of the form $ \left( \frac{15}{2}, 6 \right) \plus{} (12,5)t$

Thus, $ (0,x) \equal{} (7.5\plus{}12t, \; 6\plus{}5t)$. By equating first coordinates, $ t\equal{} \minus{} \frac{5}{8}$, and by equating second coordinates, $ x\equal{}\minus{}2 \implies X \equal{} (0,\minus{}2)$

The apex of the pyramid is $ (0,24)$, so $ XT \equal{} 26$.[/hide]

I must have messed up somewhere... listen to Zuton.

[quote="The Zuton Force"]Also, $ AXA'$ must be an equilateral triangle[/quote]

I don't think so; otherwise the radius is $ 13$. :wink:

[hide="Minor correction"]
Let $ X$ be $ h$ units from $ A'C'$ in $ ACC'A'$. By the Pythagorean Theorem twice,
\begin{align*}
5^2 + h^2 &= r^2\\
10^2 + (12-h)^2 &= r^2
\end{align*}
Subtracting gives $ 75 + 144 - 24h = 0 \Longrightarrow h = \frac{73}{8}$. Thus $ XT = h + 12 = \frac{169}{8}$ and $ m+n = 177$.
[/hide]


vector
geometry
rectangle
linear algebra
combinatorics proposed
combinatorics

Let $X$ be a set with $n$ elements, and let $(A_k)_{1\le k\le m}$ be a family of subsets of $X$ s.t. $|A_i\cap A_j|=1,\ \forall i\ne j\in \overline {1,m}$. Prove that $m\le n$.

Sorry if you guys find this too easy ot anything (I somehow seem to misjudge problems lately :)).

Kvant magazine, problem [b]5[/b]:
Let $X$ be set of $n$ elements, and let $\{A_k\}_{1\leq k\leq m}$ be a family of proper subsets of $X$ s.t. $\forall\,a,b\in X\ \exists!\,k$ s.t. $a,b\in A_k$. Prove that $m\geq n$.

assume that $v_1,v_2,\dots,v_m\in \{\ 0,1\}\ ^n$ be incidence vectors of $A_1,A_2,\dots,A_m$ now u should prove that they are linearly independnt.assume the contrary so $\sum \alpha_i.v_i=0$ now we have $0=(\sum\alpha_i.v_i)(\sum\alpha_j.v_j)=\sum\alpha_i^2(|A_i|-1)+(\sum\alpha_i)^2$ contradiction. ;)

Excellent!
Kvant's solution (1970), which was proposed by N. Burbaki :), takes 4 pages!
However, your approach was mentioned, but linear algebra isn't included in school educational program, so editor rejected it.

This problem and some similar ones (including myth's above) are solved in the paper I attached here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=13941&highlight=linear+algebra

Pierre.

Thanks for the solution. That's what I did.

[quote="grobber"]Wow! Is this stupid or what: only a few minutes ago did I realize that the problems were, in fact, one and the same (or dual, if you prefer) :).[/quote]
It was the reason why I put it here. :D

[quote="pbornsztein"]This problem and some similar ones (including myth's above) are solved in the paper I attached here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=13941&highlight=linear+algebra
[/quote]
Is attached paper in french?

Yes.
The one at the end of the thread, because the first link does not work anymore.

Pierre.

[quote="pbornsztein"]Yes.[/quote]
:(  :(  :(

Well, you know, it is math too...you will understand it easily.

Pierre.

Actually, Pierre, there's quite a lot of explanation in French in the document (of course, it helps that every alternate keyword looks similar to that in English ;) )...

Along the way I realised that I have not only forgotten what "rectangle" is in Chinese, I've also forgotten what "square" is in French :blush: .

Yes, as I found out when I consulted the dictionary... :(

If you want, you can find here a
simple proof of this problem and the dual one (exchanging sets with elements)
http://www.mathlinks.ro/Forum/viewtopic.php?t=15053&highlight=sets


How many isosceles triangles with whole-number length sides have
a perimeter of 20 units?

[quote="arb95"]I got 5 pairs of:
(5,5,5)
(7,7, 6)
(8,8, 4)
(9, 9, 2)
(6, 6, 8)

The answer though was given as 4. Is one of these wrong?[/quote]

hmm... $ 5\plus{}5\plus{}5 \neq 20$.

If you meant 5,5,10, this doesn't satisfy the triangle inequality.


Suppose(as is roughly true) that 88% of college men and 82% of college women were employed last summer. A sample survey interviews independent SRSs of 500 college men and 500 college women.

(a) Find the probability that fewer than 85% of the men in the sample were employed last summer.
Which setting is this? Would this be a binomial setting where I use np for the mean and $ \sqrt{npq}$ for the standard deviation? Or would I use the normal setting for sample distributions where p is the probability and $ \sqrt{\frac{pq}{n}}$ is the standard deviation?

The binomial method is more exact, but there's no real reason why you can't use the normal approximation: sample size is 500 and event probability is not that rare (88% & 12%).


Prove that for all matrices A, B, C from M_2(R) we have

       (AB-BA)^2004.C = C(AB-BA)^2004.

This seems extremely easy: from the Hamilton-Cayley theorem and the fact that $Tr(AB-BA)=0$ we get that $A^2=tI_2$ for some real $t$. In this case $(AB-BA)^{2004}=\alpha I_2$ for some real $\alpha$, so it commutes with any matrix $C\in M_2(R)$.


Let $\displaystyle a+b+c = 1$

What is the maximum value of $\displaystyle (ab)^2+(bc)^2+(ac)^2$ ?

[quote="arqady"]I think you meant that $a,b,c$ are non-negative real numbers.
(if no, then $\max_{a+b+c=1}(a^2b^2+a^2c^2+b^2c^2)=+\infty.$)
$(ab)^2+(bc)^2+(ac)^2\leq\frac{1}{16}\Leftrightarrow(a+b+c)^4\geq16(a^2b^2+a^2c^2+b^2c^2)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^4+4a^3b+4a^3c-10a^2b^2+12a^2bc)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+5\sum_{cyc}(a^3b+a^3c-2a^2b^2)+11\sum_{cyc}a^2bc\geq0.$
The last inequality is true by Schur and Muirhead.
Hence, $\max_{a+b+c=1,a\geq0,b\geq0,c\geq0}(a^2b^2+a^2c^2+b^2c^2)=\frac{1}{16}.$ :)[/quote]
Ahh i see.  I don't see how you prove the last statement with Schur and Muirhead though.

[quote="JellyWorld"]
 I don't see how you prove the last statement with Schur and Muirhead [/quote]
$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq0.$ This is Schur.
$\sum_{cyc}(a^3b+a^3c-2a^2b^2)\geq0.$ This is Muirhead. ;)


x,y,and z are positive numbers such as xyz>=1
  prove that:
        x^5+y^5+z^5>=x^2+y^2+z^2

Follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=32719 .

  darij

i don't understand
can you solve it please

Do you mean "such [b]that[/b] $xyz\geq 1$"? :huh: If so:
By Muirhead Inequality, we have $x^5+y^5+z^5\geq x^3yz+y^3zx+z^3xy$.
Therefore, $x^5+y^5+z^5\geq xyz(x^2+y^2+z^2)\geq x^2+y^2+z^2$

This problem likes on problem from the IMO 2005.

[color=blue][b] Problem.[/b]

    Let $x, y, z$ three nonnegative real numbers such that $xyz\geq 1.$ Prove that the following inequality is true:

                                        \[ x^{5}+y^{5}+z^{5}\geq x^{2}+y^{2}+z^{2}.  \]

  [/color]

  [i] Solution. [/i]

    Let us assume, without loss of generality, that $x\leq y\leq y\leq z$. So we have $x^{2}\leq y^{2}\leq z^{2}$ and $x^{3}\leq y^{3}\leq z^{3}$. By Chebyshef's inequality, we have

                                         \[ x^{5}+y^{5}+z^{5}\geq\frac{1}{3}(x^{2}+y^{2}+z^{2})(x^{3}+y^{3}+z^{3}).  \]

    But, from AM-GM, we get \[ x^{3}+y^{3}+z^{3}\geq 3xyz\geq 3.  \]
    The conclusion follows immeaditely. :)  ;)

[quote="cezar lupu"]from AM-GM, we get \[ x^{3}+y^{3}+z^{3}\geq 3xyz\geq 1.  \][/quote]

It should be $x^{3}+y^{3}+z^{3}\geq 3xyz\geq 3$

so IMO 2005 is solved easly like this
thank you

[quote="Ezbakhe Yassin"][quote="nawram88"]can somebody explain to me when we have abc superior to ab+bc+ac
a b c positive reals[/quote]

I do not understand what you are asking for :([/quote]

He maybe(?) wants to know  circumstances when
$abc>ab+bc+ac$ "holds the water".:)
 :trampoline:

Hmm, no responses yet as regards $abc>ab+bc+ac$ ..  :? That isn't so stupid question.Sharp minded answer should be ,pressumably,to state simple that the inequality is [b]always[/b]  true if $a>3,b>3,c>3$. ;)
Otherwise,the terms of "superiority" can hardly be configured in more simple form than the inequality form itself.
Also,it's easy to see,the inequlity is  [b]never[/b] true if any of three factors is smaller than $1$.


calculus
derivative
analytic geometry
graphing lines
slope
algebra proposed
algebra

The equation $x^3+px+q=0$ has three distinct real roots. Show that $p<0$

typical method... first derivative, $3x^2+p =0$, must have two distinct real roots, which obviously implies $p < 0$.

Here is another method. Let $a$, $b$ and $c$ be the roots of the equation, For $a=b=c=0$ we obviously have $p=0$, so let's suppose that the roots are not all zero. Then we have, from Cardano-Viète relations: $p=ab+bc+ca$ and $a+b+c=0$. From the last equation we get $(a+b+c)^2=0$, or $a^2+b^2+c^2+2(ab+bc+ca)=0$, or $a^2+b^2+c^2=-2(ab+bc+ca)$. The left hand side of this equation is strictly positive (it cannot be zero because $a$, $b$, $c$ are not all zero); hence we have $p=ab+bc+ca<0$ as desired.

Another method(a bit more graphical).
$f(x)=x^3$ and $g(x)=-px-q$
can intersect at three distinct points iff $g(x)$ has positive slope ,or $-p>0$, or $p<0$

Let the roots of $x^3+px+q=0$ be $a,b,c.$  Then, we have \begin{align*} a^3+pa+q=0\\b^3+pb+q=0\\c^3+pc+q=0.\end{align*}  Subtracting pairwise equations gives \begin{align*} (a-b)(a^2+ab+b^2+p)=0\\(b-c)(b^2+bc+c^2+p)=0\\(c-a)(c^2+ca+a^2+p)=0.\end{align*}  Since $a\neq b\neq c$, we have \begin{align*}a^2+ab+b^2+p=0\\b^2+bc+c^2+p=0\\c^2+ca+a^2+p=0.\end{align*}.  Adding them up gives $$(a^2+b^2+c^2)+(ab+bc+ca)+3p=0.$$  By Vieta's formulas, $p=ab+bc+ca$ so we have $$a^2+b^2+c^2=-2p.$$  Since the sum of distinct squares is positive, then $q<0.$ $\blacksquare$


Is there a reccurence to solutions of Pells equation?
 At exemple $ 5y_{n\plus{}1}^{2}\minus{}x_{n\plus{}1}^{2}\equal{}4$ reccurence is $ x_{n\plus{}1}\equal{}\frac{5y_{n}\plus{}3x_{n}}{2}$ and $ y_{n\plus{}1}\equal{}\frac{3y_{n}\plus{}x_{n}}{2}$
How to get it in general case?

[quote="uNc"]Is there a reccurence to solutions of Pells equation?
 At exemple $ 5y_{n \plus{} 1}^{2} \minus{} x_{n \plus{} 1}^{2} \equal{} 4$ reccurence is $ x_{n \plus{} 1} \equal{} \frac {5y_{n} \plus{} 3x_{n}}{2}$ and $ y_{n \plus{} 1} \equal{} \frac {3y_{n} \plus{} x_{n}}{2}$
How to get it in general case?[/quote]

$ a(ux \plus{} vy)^2 \minus{} (avx \plus{} uy)^2 \equal{} (ax^2 \minus{} y^2)(u^2 \minus{} av^2)$

So, if you got a solution $ (u,v)$ such that $ u^2 \minus{} av^2 \equal{} 1$, then :

$ ax^2 \minus{} y^2 \equal{} b$ $ \implies$ $ a(ux \plus{} vy)^2 \minus{} (avx \plus{} uy)^2 \equal{} b$

(in your example : $ a\equal{}5$ and $ u\equal{}\frac 32$ and $ v\equal{}\frac 12$)


I don't get it. I'm in 8th grade, and this was my first time with the MAML. Do you qualify for level 2 through level 1? And would a 109 be a good enough qualifying score?

Yeah. Anyone who answers at least 15 questions is definitely in top 100.

Do you mean get them right?

Anyway you still haven't answered my question about qualifying for Level 2.

I mean get them right.

'Yeah' was my answer to your question about qualifying.

You never know what the cutoff is, but I'll be surprised if the cutoff is >105


geometry
incenter
angle bisector
geometry proposed

Let D,E,F,D',E',F' be the midpoints and feet of perp. of BC,CA,AB of $\triangle ABC$. Prove that there exists a point P s.t. $\frac{PD}{PD'}=\frac{a}{b+c-a},\frac{PE}{PE'}=\frac{b}{a+c-b},\frac{PF}{PF'}=\frac{c}{a+b-c}$.

[color=red]  [edit]: What I meant by feet of perp. of BC,CA,AB is the feet of altitudes.[/color]

Let Ca be the circle locus of M such as MD/MD' = a/(b+c+a); similarly Cb and Cc.
As the NP-circle goes through D and D', Ca is orthogonal to the NP-circle.
If U is the projection of the incenter upon BC, we have 2.UD = |b-c| and 2.UD' = |b-c| (b+c-a)/a. Hence Ca goes through U and, as Ca is centered on the line BC,  Ca is orthogonal to the incircle.
Thus, Ca, Cb, Cc are members of the pencil orthogonal to the pencil generated by the NP-circle and the incircle; hence they touch eachother at the Feuerbach point.

Why is $C_a$ orthogonal to the nine-point circle?

Any circle on which MD/MD' is constant is orthogonal to any circle going through D and D' .

I see :)
Here is my proof:
Let F be the Feuerbach point. Since it is the tangency point of the nine-point circle and incircle, FD" is the bisector of $\angle DFD'$, where D" is the tangency point of the incircle and BC. By angle bisector theorem, $\frac{DF}{DF'}=\frac{DD''}{D''D'}$. Now the conclusion follows since $\frac{DD''}{D''D'}=\frac{a}{b+c-a}$.

jpe, your idea of orthogonality is really great ;)

Your proof is great too.
Anyway, the big problem is to guess that the common point is the Feuerbach point.


probability
conics
parabola
geometry
analytic geometry
combinatorics unsolved
combinatorics

:arrow: Compute the probability of the equation $ ax^2\plus{}bx\plus{}c\equal{}0$ present a solution $ x_n$ such as $ x_n\in\mathbb{R}$.

What is random and what randomness is that?

Well, the original form of this problem proposed the computation of the probability of the equation $x^2+2bx+c=0$.

Considering the two-dimensional, suppose that he point $(b,c)$ is randomly chosen fron a uniform distribution over a large enough square centered at the origin, with side $2k$. For real roots there has to be $b^2-c\geq0$.

Supposing $k=4$, wich is big enough to form a square that includes the region where the original equation has real roots and plotting the parabola $b^2=c$ we wil have an area at the internal part of the parabola limited by the arbitrary square wich is given by $\frac{4}{3}k^{\frac{2}{3}}$ for $k\geq1$. Consequently the probability of getting complex roots is $\frac{1}{3\sqrt{b}}$.

Then it is easy to give proof that when we take the whole coordinate system, $k\longrightarrow\infty$, the probability of the equation having real roots tends to $1$.

Is this approach correct?


Once was enough:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=870814854&t=306676

hello, you've got several nice answers, why do you post your problem twice?
Sonnhard.

Hi Sonnhard. It was an accident. As I'd posted the question for someone else, I didn't get back to it and had forgotten about it.


Let $ a,b,c,d\in \mathbb{R}$ such that $ ac\equal{}bd$ and $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{d}{a}\equal{}4$.

Find the maximum of $ \frac{a}{c}\plus{}\frac{b}{d}\plus{}\frac{c}{a}\plus{}\frac{d}{b}\minus{}\frac{abcd}{(ab\plus{}cd)^2}$

[quote="outback"]Let $ a,b,c,d\in \mathbb{R}$ such that $ ac \equal{} bd$ and $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a} \equal{} 4$.

Find the maximum of $ \frac {a}{c} \plus{} \frac {b}{d} \plus{} \frac {c}{a} \plus{} \frac {d}{b} \minus{} \frac {abcd}{(ab \plus{} cd)^2}$[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=139208