MIT
college
Stanford
Princeton
CMU
Carnegie Mellon

A question (from [b]orl[/b] I believe) came up in the MIT Math Jam earlier tonight.  Since that Math Jam was about undergrad stuff, I thought I'd post orl's question here, and give my 2 cents.  (orl, please feel free to elaborate on your question if you like.)

One question was:
[quote="orl"]I heard it is easier to get accepted as postdoc at MIT because you proved your qualities at several conferences/papers because it seems to get into the Ph.D. programme you must be lucky.[/quote]To some extent this may be true.  I was one of the "lucky" ones -- I got my Ph.D. from MIT.  I went to Carnegie Mellon: a very good school, but (at the time at least) not an elite one in many eyes.  Some of my Ph.D. classmates went to even less "elite" schools: Rice, Montreal, Dalhousie, Brown, William & Mary, etc.  In my opinion the strongest thing you can have on your application is undergraduate research experience.  In the US, you can get this experience by attending a summer REU program.  The next strongest evidence is graduate courses and strong recommendations from "known" professors.

My feeling is that it's the postdoc game where the real "randomness" comes in.  MIT maybe is a special case since they hire a lot of postdocs (relatively speaking), but most places only have a handful of postdocs per year, yet get 100+ applications.  If they're not looking for a postdoc in your field, and if you don't have a faculty member at the place championing your application, you don't have much of a chance.  The moral is: if you want to get a postdoc at MIT, do research with someone at MIT, and/or do research in a field in which MIT definitely wants to hire, and make sure that the relevant professors at MIT know who you are.

[quote="orl"]Why should I choose MIT for Ph.D/Post-Doc and not Berkeley, Caltech, or Stanford?[/quote]For the great Boston weather, of course!   :lol:

Seriously, there's no single good reason to choose MIT over any of those schools.  They're all fantastic.  Berkeley is a bit different in that they have many more grad students and they take more of a "sink-or-swim" approach, in that they expect that not all of their grad students will finish, whereas at MIT (and Caltech and Stanford AFAIK) almost all the students who start end up getting a Ph.D.  Caltech is different in that it's much smaller than the others.

Your best bet is to visit as many as you can, talk to the professors who you think you might like to work with, and decide which one feels best.  Visits can really change your impression -- on paper, I was pretty sure that I wanted to go to Princeton, but when I actually visited Princeton and MIT, I liked MIT much better.  (They have very different characters.)  That's just my personal view -- others prefer Princeton to MIT.

Dave, thanks a lot for your evaluation.  :)  Maybe you should issue a math jam for graduate students?!

I would like that too, but... can it be after 2/10/06?

[quote="tokenadult"]I have read that getting into REUs in math as an undergrad can sometimes be easier for young people who did math summer programs as high school students--does this square with what you have heard?[/quote]
I didn't do any math summer programs in high school, and I did an REU as a freshman that was intended for upperclassmen (or at least, I was the only freshman there). I think my letters of recommendation helped quite a bit. However, it is also possible that the REU to which I applied wasn't as concerned about those kinds of things.


calculus
derivative
function
algebra
domain
real analysis
real analysis unsolved

Let $f(x)$ be differentiable on $[0, 1]$ with $f(0) = 0$ and $f(1) = 1.$ For each positive integer $n,$ show that there exist distinct points $x_1, \dots , x_n$ in $[0, 1]$ such that $\sum_{j=1}^n\frac1{f'(x_j)}=n.$

By the MVT, there exists $x_1$ such that $f'(x_1)=1$.  If there are infinitely many of these, we are done.

Else, consider $\sup f'$.  Derivatives have the intermediate value property, so if $\sup f'>1$ and $\inf f'<1$, we are done.

So WLOG $\sup f'>1$ and $f'(x)\geq 1$ for all $x$.  This means that $f(x)\geq x$ for all $x$ (mean value theorem again?).  Since equality can only occur finitely many times, we can apply the mean value theorem to $(a,f(a))$ and $(1,1)$ where $f(a)>a$.  This implies the existence of some $c$ with $f'(c)<1$.  But this is a contradiction, so we're done.

This post gets the usual 3:24 AM disclaimer - there's probably some mistakes, and I may or may not catch them in the morning.

Yes, MVT works.

Let us prove the generalisation: Let $f$ be differentiable on $[0,1]$  with $f(0)=0$ and $f(1)=1$. $k_1,\dots,k_n$ are $n$ positive reals. Prove that there exists $0\leq x_1<x_2<\cdots<x_n\leq 1$ such that \[ \sum \frac{k_i}{f'(x_i)} = \sum k_i \qquad (\ast)  \].

Let $m = \sum k_i$ and divide both sides of (*) by $m$,
\[ \frac{\lambda_1}{f'(x_1)} + \cdots + \frac{\lambda_n}{f'(x_n)} = 1 \qquad (\ast\ast)  \]
where $\lambda_i = k_i / m$ and $\sum \lambda_i = 1$.

Since $f(0)=0$ and $f(1)=1$ there exists $c_1\in (0,1)$ such that $f(c_1) = \lambda_1$.
Then there exists $c_2\in (c_1,1)$ such that $f(c_2) = \lambda_1+\lambda_2$
there exists $c_3\in (c_2,1)$ such that $f(c_3) = \lambda_1+\lambda_2+\lambda_3$
...

Apply Lagrange's MVT, there exists $x_i\in(c_{i-1},c_i)$ such that
$f'(x_i) = \frac{f(c_i)-f(c_{i-1})}{c_i-c_{i-1}} = \frac{\lambda_i}{c_i - c_{i-1}}$
Thus
$c_i - c_{i-1} = \frac{\lambda_i} {f'(x_i)}$.

So what do you see?

The proof I had was a particular case of liyi's proof. Note that the key is to partition the range of the function (or, at least, the interval from $f(0)$ to $f(1)$) and not to start with a partition of the domain interval.

No, because if $\sup f'>0$ and $\inf f'<0,$ you still may only have a small finite number of points at which $f'(x)=1.$ By "small" we mean "not as many as you need."

OK, maybe you were referring to something more sophisticated there than just finding places where the derivative is 1 - but if so, you need to elaborate more.

If $\sup f'>1$ and $\inf f'<1$, there are $x,y$ such that $1/f'(x)=1-\epsilon_1$ and $1/f'(y)=1+\epsilon_2$ with $\epsilon_1,\epsilon_2>0$.  Since derivatives have the intermediate value property, if we put $\epsilon=\min\{\epsilon_1,\epsilon_2\}$, there exist $a,b$ such that $1/f'(a)=1-\epsilon$, $1/f'(b)=1+\epsilon$.  By the intermediate value property again, we can repeat this for any $\epsilon'$ which satisfies $0<\epsilon'\leq \epsilon$.

For odd $n$, we use the point such that $f'(x)=1$ guaranteed by the MVT and we pair up points using the above process.  The process for even $n$ is obvious.


Could the admins, when they get a chance, consider revising the $\LaTeX$ section of this site?  It seems to be giving everyone who does not know about $\LaTeX$ the impression that they must download a program before they can use $\LaTeX$ on this site.  First someone will tell them to go learn $\LaTeX$.  Then they will click the $\LaTeX$ tab.  On the first page, they see the steps for getting started and they think they must download the program to use $\LaTeX$ at all.  If you could make more clear the difference between using $\LaTeX$ on the forums and downloading software so you can use $\LaTeX$ on your own computer I think it would save a lot of confusion.

Valentin's tutorial for $\LaTeX$ seems to address using $\LaTeX$ on this site well but unfortunately most people go to the other tutorial first.

Thanks

I second the motion.  In the LaTeX section of the AoPS skin, there should be something up front that says you do not have to download anything to use LaTeX on the forum.  Too many newbies are confused on this point.

[quote="joml88"]Valentin's tutorial for $\LaTeX$ seems to address using $\LaTeX$ on this site well but unfortunately most people go to the other tutorial first.[/quote]Yeah I know about that. We will deal with this soon.


[quote="Algadin"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Find maximum of
$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ab}{1 \minus{} ca}$[/quote]
The answer follows from the Vasc's known inequality:
$ \sum_{cyc}\frac{1}{1\minus{}ab}\leq\frac{9}{2}.$

[quote="arqady"][quote="Algadin"]Let $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Find maximum of
$ \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {ab}{1 \minus{} ca}$[/quote]
The answer follows from the Vasc's known inequality:
$ \sum_{cyc}\frac {1}{1 \minus{} ab}\leq\frac {9}{2}.$[/quote]
why?  :(  :(

[quote="Algadin"][quote="arqady"][quote="Algadin"]Let $ a,b,c > 0$ and $ a^2 + b^2 + c^2 = 1$. Find maximum of
$ \frac {ab}{1 - ab} + \frac {bc}{1 - bc} + \frac {ab}{1 - ca}$[/quote]
The answer follows from the Vasc's known inequality:
$ \sum_{cyc}\frac {1}{1 - ab}\leq\frac {9}{2}.$[/quote]
why?  :(  :([/quote]
${ \sum_{cyc}\frac {ab}{1 - ab}\leq\frac{3}{2}\Leftrightarrow\sum_{cyc}(\frac {ab}{1 - ab}+1})\leq\frac{9}{2}\Leftrightarrow\sum_{cyc}\frac {1}{1 - ab}\leq\frac {9}{2}.$ :wink:

[quote="karis"]Can solve it by Chebysep???[/quote]
Now I see only this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230955
which is true without $ a^2\plus{}b^2\plus{}c^2\equal{}1.$ :(

[quote=arqady][quote="Algadin"]Let $ a,b,c > 0$ and $ a^2 + b^2 + c^2 = 1$. Find maximum of
$ \frac {ab}{1 - ab} + \frac {bc}{1 - bc} + \frac {ab}{1 - ca}$[/quote]
The answer follows from [b]the Vasc's known inequality[/b]:
$ \sum_{cyc}\frac {1}{1 - ab}\leq\frac {9}{2}.$
${ \sum_{cyc}\frac {ab}{1 - ab}\leq\frac{3}{2}\Leftrightarrow\sum_{cyc}(\frac {ab}{1 - ab}+1})\leq\frac{9}{2}\Leftrightarrow\sum_{cyc}\frac {1}{1 - ab}\leq\frac {9}{2}.$ :wink:[/quote]
$$ \sum_{cyc}\frac {ab}{1 - ab}\leq 2\sum_{cyc}\frac{ab} {a^2+b^2+2c^2}\leq \frac{1} {2}\sum_{cyc}\left(\frac{a^2} {c^2+a^2}+\frac{b^2} {b^2+c^2}\right)=\frac{3}{2}.$$

[url=http://www.artofproblemsolving.com/community/c6h36873p231824]Crux Mathematicorum,  Apr  2005 , problem 3032 by Vasile Cirtoaje[/url]
https://cms.math.ca/crux/v31/n3/page173-178.pdf
http://www.artofproblemsolving.com/community/c6h230442p1275432
http://www.artofproblemsolving.com/community/c6h225743p1253027

[quote=Algadin]Let $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Find maximum of
$ \frac{ab}{1\minus{}ab}\plus{}\frac{bc}{1\minus{}bc}\plus{}\frac{ab}{1\minus{}ca}$[/quote]

one line down


[b](1) find the value of |Z| if arg(Z^1/3) = 1/2*arg(Z^2+(Z'*Z^1/3))
here 1/2 is multiply by arg. and Z'=congugate of Z and Z'*Z  means Z'is multiply by Z [/b]

Are you sure the answer is unique? Setting $ z=re^{i\theta}$ we get

$ \arctan(\theta/3)=\frac12\arctan(\frac{r^2\sin2\theta-r^{4/3}\sin(2\theta/3)}{r^2\cos2\theta+r^{4/3}\cos(2\theta/3)})$


It might be possible (though I haven't tried it) if you use 8pt as fontsize, assuming that's not too small for you. To do this you need the extsizes package from [url=http://www.ctan.org/]CTAN[/url]. The extsizes classes provide support for 8pt, 9pt, 10pt, 11pt, 12pt, 14pt, 17pt, and 20pt

Can you please post the code on how to use the package?? Thx.

Once you have installed the package (the method depends on your operating and [tex]\LaTeX[/tex] system; see http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy for example) it's very easy to use. Just replace [code]\documentclass{article}[/code] by [code]\documentclass[8pt]{extarticle}[/code]

How can I justify everything to the left?


solve the differential equation for

$\frac{d^2y}{dx^2}=\frac{1}{y^3}$


[hide]We turn these six fractions into two fractions, one for each parenthesis.  We get (a^2b-b^2a+b^2c-c^2b+c^2a-a^2c)/abc.  We factor this into [b^2(c-a)+c^2(a-b)+a^2(b-c)]/abc.  We're going to factor the other's denominator first.  (a-b)(b-c)(c-a) is (ab-ac-b^2+bc)(c-a).  Notice the abc terms cancel so we only have -ac^2-b^2c+bc^2-a^2b+a^2c+b^2a.  Notice this factors into c^2(b-a)+a^2(c-b)+b^2(a-c).  When we compare this to the numerator for the other fraction, we get -1/abc(numerator of second big fraction).  That's c(bc-c^2-ab+ac)+b(ab-b^2-ac+bc)+a(ca-a^2-bc+ba)=bc^2-c^3-abc+ac^2+ab^2-b^3-abc+b^2c+ca^2-a^3-abc+ba^2 but we're taking the negative of that and we're dividing it by abc so we have [a^3+b^3+c^3+3abc-b^2(a+c)-c^2(b+a)-a^2(b+c)]/abc.  Now, we use a+b+c=0 to get (a+b)/c=-1 or c^2(a+b)=-c^3.  Since this equation is symmetric, we can also substitute the other two.  We get 2(a^3+b^3+c^3)/abc+3=x.  Now, let's assume one number is negative.  Without a loss of generatlity, let's say that's c.  So a+b=-c or a^3+b^3+3a^2b+3b^2a=-c^3.  Substituting the c^3 we have -(6a^2b+6b^2a)/abc or -(6a+6b)/c.  We could also shift this to c to get 6(a+b)/-c.  We know (a+b)/c=-1 so (a+b)/-c=1.  So the product equals 6+3=9.  If you look at (a+b)^3=-c^3, the case with two negatives is simply -(a+b)^3=c^3 which is equivalent to (a+b)^3=-c^3.[/hide]

Hmm..it is suprising how much nicer a solution looks in latex.

[quote="maokid7"]i agree. could someone (a mod or the original poster) please put the proof in latex.[/quote]

Sorry, it's not the moderators' job to $\text{\LaTeX}$ solutions for people who don't do that theirself.  Anyone can quote the what the original poster said and $\text{\TeX}$ it up.  I agree, however, that solutions look much, MUCH nice in $\text{\LaTeX}$.

[quote="dengmi"][hide]We turn these six fractions into two fractions, one for each parenthesis.  We get $\frac{a^2b-b^2a+b^2c-c^2b+c^2a-a^2c}{abc}$.  We factor this into $\frac{b^2(c-a)+c^2(a-b)+a^2(b-c)}{abc}$.  We're going to factor the other's denominator first.  $(a-b)(b-c)(c-a)$ is $(ab-ac-b^2+bc)(c-a)$.  Notice the $abc$ terms cancel so we only have $-ac^2-b^2c+bc^2-a^2b+a^2c+b^2a$.  Notice this factors into $c^2(b-a)+a^2(c-b)+b^2(a-c)$.  When we compare this to the numerator for the other fraction, we get $-\frac1{abc}$ (numerator of second big fraction).  That's $c(bc-c^2-ab+ac)+b(ab-b^2-ac+bc)+a(ca-a^2-bc+ba)=bc^2-c^3-abc+ac^2+ab^2-b^3-abc+b^2c+ca^2-a^3-abc+ba^2$ but we're taking the negative of that and we're dividing it by $abc$ so we have $\frac{a^3+b^3+c^3+3abc-b^2(a+c)-c^2(b+a)-a^2(b+c)}{abc}$.  Now, we use $a+b+c=0$ to get $(a+b)/c=-1$ or $c^2(a+b)=-c^3$.  Since this equation is symmetric, we can also substitute the other two.  We get $\frac{2(a^3+b^3+c^3)}{abc}+3=x$.  Now, let's assume one number is negative.  Without loss of generality, let's say that's $c$.  So $a+b=-c$ or $a^3+b^3+3a^2b+3b^2a=-c^3$.  Substituting the $c^3$ we have $-\frac{6a^2b+6b^2a}{abc}$ or $-\frac{6a+6b}c$.  We could also shift this to $c$ to get $\frac{6(a+b)}{-c}$.  We know $\frac{(a+b)}c=-1$ so $\frac{(a+b)}{-c}=1$.  So the product equals $6+3=9$.  If you look at $(a+b)^3=-c^3$, the case with two negatives is simply $-(a+b)^3=c^3$ which is equivalent to $(a+b)^3=-c^3$.[/hide][/quote]

dengmi, you owe me one :D


For all positive real numbers a,b,c prove that:
a2/(a2+2bc) + b2/(b2+2ac) + c2/(c2+2ba) >= 1 >= bc/(a2+2bc) + ac/(b2+2ac) +ba/(c2+2ba) .
I've a solution, and you?

For the first inequality we use the following lemma:

a1,a2,....,an>0    x1,x2...........,xn>0  n>=2,m>=2
m real number, n natural number

x1^m/a1+.......+xn^m/an>=(x1+...+xn)^m/n^(m-2)*(a1+...+an)

So in our case
a^2/(a^2+2bc)+b^2/(b^2+2ca)+c^2/(c^2+2ab)>=(a+b+c)^2/(a^2+b^2+c^2+2(ab+bc+ca))=1

For the second inequality we make the following notations:

a^2/(a^2+2bc)+b^2/(b^2+2ca)+c^2/(c^2+2ab)=E
and
bc/(a^2+2bc)+ca/(b^2+2ca)+ab/(c^2+2ab)=F
We can easily observe that E+2F=3
So F=(3-E)/2 which is <= than 1.
Hope it's the corect solution.

This is my solution:
Applying AM-GM, we have
a2/(a2+2bc) + b2/(b2+2ac) + c2/(c2+2ba) >= (a+b+c)2/ (a+b+c)2 =1., is what we want.
On the other hand, bc/(a2+2bc) + ac/(b2+2ac) +ba/(c2+2ba) <=1 is equivalent: 1/(2+ a2/bc) + 1/(2+ b2/ac) + 1/(2+ c2/ba) <= 1
Put x=a2/bc, y=b2/ca, z=c2/ab, then x,y,z>0 ; xyz=1 and we get:
1/(2+x) + 1/(2+y) + 1/(2+z) <= 1. Multiplying (2+x)(2+y)(2+z), we get (2+x)(2+y) +(2+y)(2+z) + (2+z)(2+x) <= (2+x)(2+y)(2+z).
is equivalent: xy+yz+zx> =3 (1) . Applying Cauchy's inequality, we get xy+yz+zx>= 3.(xyz)2/3 =3. then (1) is true.
Equality holds when a=b=c

I have nice solution for the first one:

\[\sum\frac{a^2}{a^2+2bc}\geq \sum\frac{a^2}{a^2+b^2+c^2}=1\]

My solution:

1) It is an immediate consequence of Cauchy-Schwarz in the Engel form, since $\sum a^2+2\sum bc=(a+b+c)^2$

2) The inequality is equivalent to $\sum \frac {1}{x+2}\leq 1$ with $xyz=1$, which I proved here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23103

EDIT: Now that I look at it again, these solutions are pretty much identical to the previous ones.

it's equivalent to cycsum(bc/(a^2+2bc)) =< 1.
it's homogenous, so WLOG abc=1.
then sub. bc=1/a and you get cycsum(1/(a^3+2)) =< 1.
let x=a^3 etc and you get xyz=1 and RTP: cycsum(1/(x+2)) =< 1.
expand everything and u have to prove 3 =< cycsum(xy), which you can prove using AM-GM on xy,yz,zx.

[quote="hxtung"]For all positive real numbers $a,b,c$ prove that:
$\frac{a^2}{a^2+2bc} + \frac{b^2}{b^2+2ca} +\frac{c^2}{c^2+2ab}\ge 1 \ge  \frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}$[/quote]
I just share my solution with you.
Let $I=\frac{a^2}{a^2+2bc} + \frac{b^2}{b^2+2ca} +\frac{c^2}{c^2+2ab}\/$ and $J=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}$.
We wish to show that $I \ge 1 \ge J\/$.

Since $x^2 +y^2 \ge 2xy\/$. We have $\frac{a^2}{a^2+2bc} \ge \frac{a^2}{a^2+b^2+c^2}\/$.
Similarly, $\frac{b^2}{b^2+2ca} \ge \frac{b^2}{a^2+b^2+c^2}\/$ and $\frac{c^2}{c^2+2ab} \ge \frac{c^2}{a^2+b^2+c^2}\/$.
Then it is clear that $I \ge 1\/$

Next, note that $I+2J=3\/$ or $I=3-2J\/$. By $I \ge 1\/$, it is easy to see that $J \le 1\/$.

Using Cauchy-Schwalz's inequality,

$\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\geq \frac{(a+b+c)^2}{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)}=1$

pipi, $I+2J=3$  Wonderful! ;)


[quote="nutz_for2.718281828"]I usually wake up at 7:15 AM in my time. I overslept an hour or so, and woke up at 7:30 AM. How is that possible?[/quote]
Daylight savings time.

This was especially easy considering DST just happened!

Today was the longest day of the year, 25 hours long :D


probability
videos
Alcumus
LaTeX
counting
distinguishability
Pascal\u0027s Triangle

Oh, combinations.  I'm not sure what those formulas exactly mean, but remember this:  nCr = (nPr)/(rPr).

Example Problems:
Lv. 1: In how many ways can you make a three letter word out of a combination of 4 letters without regard to order?
Lv. 2:  If you toss a coin 7 times, what's the probability you flip tails 3 times or less?
Lv. 3:  What's the value of the 3rd number in the 16th row of Pascal's triangle?

[quote="Cliu0301"]Sorry people, I don't know how to use LaTeX.

Could you give me some example problems for these? And also, some problems to try out.

Here are the formulas.

(n-1) C (k-1)

(n+k-1) C (k-1)

I fail at counting.[/quote]

Those are $ \binom{n-1}{k-1}$ and $ \binom{n+k-1}{k-1}$
[code]$\binom{n-1}{k-1}$ $\binom{n+k-1}{k-1}$[/code]

Take a look at the $ \text{\LaTeX}$ ([code]\LaTeX[/code]) manual online at AoPS.

Also, if you need to improve counting, read [b]Introduction to Counting and Probability[/b] by AoPS, very helpful:)

Generally, nCr=$ \binom{n}{r}=\frac{n!}{r!(n-r)!}$

[quote="Cliu0301"]Sorry people, I don't know how to use LaTeX.

Could you give me some example problems for these? And also, some problems to try out.

Here are the formulas.

$ \binom{n \minus{} 1}{k \minus{} 1}$

$ \binom{n \plus{} k \minus{} 1}{k \minus{} 1}$

I fail at counting.[/quote]

Is it the parenthesis with the numbers and variables on top of each other that you dont get (combinations), or do you not get what the formula is for?

If its the latter, the second formula is a balls and urns formula, and the first one...Im not so sure about, probably something to do with the hockey stick theorem.

If you dont get the parenthesis with the numbers and variables on top of each other (combinations), then I would recommend getting Intro to Counting and probability. I cant really explain its use clearly if we're not in person.

Anyways, you will probably learn it this year at ICAE.

You can view the videos about this here: http://www.artofproblemsolving.com/Alcumus/Videos/VideoList.php . Scroll down to the Pascal Triangle section.

Also click [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.phpurl]here[/url] to learn about $ \text{\LaTeX}$

[quote="Math Champion"]They have to do with the Pascal Triangle, so you can view those videos on Alcumus (And right now you have access to all videos).[/quote]

I knew the answers to all of them.......I was just giving them out as samples.

Also, is there a way I can write combinations using the C thing such as 3C2 using $ \ LaTeX$?

I know what the parentheses are...I just need you to explain what the formulas are used for.

i think he meant like [code] $\binom {3}{2}$[/code]
$ \binom {3}{2}$

[quote="maybach"][quote="Math Champion"]They have to do with the Pascal Triangle, so you can view those videos on Alcumus (And right now you have access to all videos).[/quote]

I knew the answers to all of them.......I was just giving them out as samples.

Also, is there a way I can write [b][size=200]combinations using the C thing such as 3C2[/b][/size] using $ \ LaTeX$?[/quote]

Yes, he DEFINITELY meant that.

HELP ME UNDERSTAND WHAT THESE FORMULAS ARE FOR!!! DON'T POST ALL THIS RANDOM STUFF!!!!!!!!!

Second one is used for balls an urns. Not sure exactly about the first.

yeah, so the first formula is the number of ways to divide "$ n" among "k" people.
It works like this: Say you had$6 and 3 people. Think of inserting 2 "dividers" into 5 "slots":

  \[ |\]$ |$

This arrangement corresponds to giving the first person $ 2, the second person$3, and the third person $ 1. Also note that every arrangement of dividers corresponds to a different allocation of moneys, so there are 5C2 ways to divide$6 among 3 ppls.

yeah, so thats what the formula means  :wink: [/b]

@ged3.14: Wow, maybe you should preview your posts before you actually post them...

[quote="maybach"][quote="Math Champion"]They have to do with the Pascal Triangle, so you can view those videos on Alcumus (And right now you have access to all videos).[/quote]

I knew the answers to all of them.......I was just giving them out as samples.
[/quote]

@maybach: I was actually talking to cliu, not you. Sorry for the misunderstanding :).

@math champion - yeah, sorry about that weird mishap on my computer  :blush:
What i was trying to write was that the formula in question, n-1 C k-1, counts the number ways to distribute n non-distingishable balls into k distinguishable boxes.

The first is balls and urns when each number you use has to be positive.

In other words, the second counts the number of solutions in nonnegative integers to $ a_1\plus{}a_2\plus{}...\plus{}a_k\equal{}n$, while the first counts the number of solutions in positive integers.


Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and (3,-3).


invariant
linear algebra
linear algebra unsolved

Suppose $A,B$ are square complex commuting matrices. Then prove that there are just a finite number of stable invariant subspaces for the family $(A,B)$. Moreover, there is effective bound: there are at most $2^n-1$ stable invariant non-zero subspaces.
 Now, the conventions: for a family $F$ of matrices, an invariant space is a space that is stable by all matrices of the family $F$. Now, an invariant space $V$ is called stable with respect to the family $(A,B)$ if given any $\epsilon>0$ there is $\delta>0$ such that for any two commuting matrices $ (C,D)$ with $ ||C-A||<\delta, ||D-B||<\delta$ we have the existence of an invariant space $V'$ with respect to the family $ C,D$ such that $ max(sup_{x\in V, ||x||=1}{ d(x,V')}, sup_{x\in V', ||x||=1}{d(x,V)})<\epsilon$.

How are these subspaces related to the two matrices, exactly? (i.e. I don't really understand what the problem is asking :)) :?

Upps, sorry, I forgot to add the definitions. Please, a moderator, delete this post. I will edit the problem.


analytic geometry
geometry
graphing lines
slope

Find $ab$ if $(0,0), (a,11), (b,37)$ is a right triangle.

...slopes opposite reciprocals... $\boxed{-407}$

[hide]Assuming that the right angle is at (0,0), the slopes of the sides are $\frac{11}{a}$ and $\frac{37}{b}$.  Because perpendicular lines have slopes that are opposite recipricals, it follows that their product is -1 ($\frac{x}{y}\cdot-\frac{y}{x}=-\frac{xy}{yx}=-1$).  Therefore,
$\frac{11}{a}\cdot\frac{37}{b}=-1$
$11\cdot37=-ab$
$ab=\boxed{-407}$[/hide]


Here's how it works: post a possible "commandment" that relates to math.  Let's do one at a time, folks.

1. What thou doeth to one side of thy equation, thou shalt doeth to thy other.

3. Thou shalt not take thy square root of -1.

5. Thou shalt wisely use $\pi$ and $e$ in thy equations.

7. Thou shall not make useless threads about [i]The Ten Commandments of Mathematics[/i].

I doubt this commandment will have any success.

thou shalt not spend long hours memorizing pi

[url=http://www.math.uiuc.edu/~reznick/]Bruce Reznick[/url] once came up with ten:
[quote]10. Thou shalt not covet thy colleague's office, nor his salary, nor his grants, nor his teaching load.

9. Thou shalt not bear false letters of recommendation nor referee's reports.

8. Thou shalt not plagiarize.

7. Thou shalt not submit the same paper to two journals.

6. Thou shalt not kill a career.

5. Honor thy teachers and thy students.

4. Thou shalt remember the sabbatical and keep it holy.

3. Thou shalt not take the word "proof" in vain.

2. Thou shalt create no unnecessary committees.

1. Thou shalt know the truth and the truth shall maketh thee free. (The truth shall followeth from three lemmas, a routine Mathematica computation left as an exercise for the reader, and a remark of David Hilbert.) [/quote]
BTW, I notice an article "Some thoughts on writing for the Putnam" on his webpage.

[quote="futuremontrealian"]4)  But, if thou really, really, [i]really[/i] wanteth it, thou mayeth expresseth it purely as some unimportant, ambiguously chosen letter of thyne alphabet.  How about, oh, sayeth, $i$?[/quote]
Heh heh... nice.

9. Thou shalt not canceleth addends in thy fractions.

Thou shalt not spam. Or try to raise post counts.

Wait a minute...that has nothing to do with math!

yeah that's no. 1: Thou shalt not spam. Or try to raise post counts.

No. 2?

2. Thou shalt not say anything like this

( don't push yourself too hard just to beat me; i'm supposedly a super genius. not trying to be conceited or anything, but i just don't want your parents to sue me when you drop dead from overexertion. oh, and you most likely won't be able to get it if you didn't get to alg. 1~2 yet. this problem IS solvable and there's no tricks. im a 12 year old and i got it. you can get it too.  )

Courtesy of flamesofdestiny

I suggest Euclid's 5 postulates. And also associative and commutative properties of addition. Also, transitivity. :D

YAY MUSTAFA! I am like a supa geenius so i ii beat u it coz i took calc. in 2nd grade. i'm a 6 yr old and i got it. i'm just sayin thi coz i don't want u to sue me coz i a millionaire.

[quote="ProtestanT"]Thou shalt not use thyne calculator in vain.
Especiallithly, thou shalt not playeth games on thyne calculator during thy exam mathematical.[/quote]

No, that is wrong.

Thou shalt not playeth games on thy calculator whilst thou takes an important mathematical competition.

thou shalt not believe all odd numbers are prime

-jorian

Thou shalt not confuse the numerals 6 and 9 by flipithing themth upsideth down.

Thou shalt not write $-$ and doesth $+$ in thy answerth

if that's even english

Thou shalt respecteth me because thou art reading a commandment by me.


at HCSSIM, i took a mini on Random Walks, which basically covered the Law of Large Numbers, Bernoulli trials, expected value, probability, the Drunkard's Walk, the Ballot problem, etc. Would any of you know a good book with which i could further explore these topics?

You mean all of those in 1 book, or also books that only cover some of those topics?

There are many excellent textbooks on probability theory.  My favorite, and a classic that has survived the test of time, is "An Introduction to Probability Theory and Its Applications", Volume 1, by William Feller.  This book has editions from the 1950's and 1960's.  Despite its age, I think anybody reading it will benefit tremendously.

There are also many free textbooks online.  For example, if you go to http://www.geocities.com/alex_stef/mylist.html you will see many free books on various math topics.  Some of this stuff is pretty advanced, but there are also introductory texts.

Good luck,

Thomas


Let $ y_1,...,y_k\in\mathbb{N}$
Find the general term of $ \sum_{y_{1}\plus{}...\plus{}y_{k}\equal{}n}y_{1}...y_{k}$

what do you mean "the general term"? you want a formula for this sum or something??

But what is variable?

Is $ k$ fixed? Is $ n$ fixed?
On what are you summing? $ n$? the $ k$-uples?

Pierre.

i think he's summing over all $ k$-tuples ($ k$ fixed) with fixed sum $ n$.

If $ k$ is not fixed we are searching for:

$ C_n \equal{} \sum_{(y_{1} \plus{} ... \plus{} y_{k} \equal{} n)\forall k\leq n}y_{1}...y_{k}$
Their recurrence is:
$ C_n \equal{} \sum_{i \equal{} 1}^{n}(n \minus{} i)C_i$
and $ C_1 \equal{} 1$ and $ C_2 \equal{} 3$
As even fibonacci have the same recurrence and the same first to terms:
$ C_n \equal{} F_{2n}$


How come the dvi file I have, instead of going on to a second page, just floods the text into the gray area? Is that a problem? The pdf file works normally, though.

It doesn't matter what happens in the dvi file if you are never going to use it, but if you want to find out why it's happening you need to give a minimal example - a short complete tex file that exhibits the problem, plus details of the system you are using.

ok...i'm using TeXnicCenter

I don't think attatching the .tex file is neccessary.
I just texed a bunch of stuff, built the dvi, then on the dvi, it only has one page of stuff, with the rest of the text coming out of the page. When I print the dvi, it only prints the stuff on the white.

[quote="13375P34K43V312"]ok...i'm using TeXnicCenter[/quote]That's just the editor which runs the command line to produce the dvi/pdf so won't be the source of the problem.[quote="13375P34K43V312"]I don't think attatching the .tex file is neccessary.[/quote]It's absolutely essential since something in that file has caused the problem. Or are you saying that [i]any[/i] tex file has the problem?

It sounds like the text is overspilling the margins which are defined in the tex file and so we would need to see what is there before being able to offer a solution. Of course, you can use a sample tex file which exhibits the problem.


cf: perfect_radio / xy-z^2 / apr 3
How do you write a reference to another post?
[color=green][Moderator edit: he means [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141780]this[/url].][/color]

Let $X,Y,Z\in\mathcal{M}_{2}(\mathbb{R})$ with nonnegative elements. Prove that one of the 3 following equalities aren't satisfied: $XY-Z^{2}=YZ-X^{2}=ZX-Y^{2}=I$.

1) Read in my first post "....one of the 3 following equalities ISn't satisfied:...."
2) Hint : use the AlexFok's relation: $XY^{2}Z^{2}X=...$.

Suppose all the 3 equalities are true.
X,Y,Z invertible
$X^3+X=Y^3+Y=Z^3+Z=XYZ=YZX=ZXY$
Case 1: all of X,Y,Z have only real ev
$X,Y,Z$ share the same spectrum.
$det(X^2+I)=det(YZ)=det(X^2)$
$det(I+X^{-2})=1$
$det(I+X^{-2})>1$(=><=)

Case 2: otherwise X eigenvalues are complex conjugates
X,Y,Z is sim diagonalizable
Case 2.1: all are different
tr(X+Y+Z)=0
tr(X)=tr(Y)=tr(Z)=0
Case 2.2: 2 of X,Y,Z are equal.
$X=Y \implies Z=-2X$
real parts of all ev's are 0
tr(X)=tr(Y)=tr(Z)=0
$0<det(X)\leq 0$ (=><=)

for size n x n
$tr((I+X^2)(I+Y^2)(I+Z^2))=tr(YZ^2X^2Y)=tr(X^2Y^2Z^2)$
$0=tr(I+\cdots)\geq n$ (=><=)
@below: for equal spectrum, i assumed all ev are real.

dominicleejun ,  I don't see why "$X,Y,Z$ share the same spectrum". Indeed, a priori, $X,Y,Z$ don't pairwise commute and $a^3+a=b^3+b$ doesn't imply that $a=b$. Anyway one has the more general result

$\textbf{Proposition.}$ Let $X,Y,Z\in\mathcal{M}_n(\mathbb{R})$ be such that $X^2,Y^2,Z^2$ have non negative entries. Show that one of the $3$ following equalities isn't satisfied: $XY-Z^2=YZ-X^2=ZX-Y^2=I_n$.


Hurt and Heal. Each post you hurt someone and you heal someone (this decreases and increases their health by 1 point, respectively). When someone reaches 0 HP they die.

Also, I will be here as the ever-loving [i]god[/i] that will cause random things to happen at times (don't worry, everything is on notepad -- and I'll try not to favor myself too much)

[b]You cannot heal yourself.
You cannot hurt yourself.[/b]
[b]You cannot double-post in this thread.
No alliances! [/b](God will get angry if the same two people post alternatively for 20 pages! We don't want another TOW!)
[b]Once you hurt and heal, you must recharge (wait) for 30 minutes (at least) before hurting & healing again.[/b]
[u][b]Clarification: You can still hurt & heal even if you're not on the list. Klebian gets in because he's really cool. No one else comes in after this, it will make the game go too long. I can start a second edition of this game later if enough other people join in.[/b][/u]

--------------------------------
randomdragoon: 10
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 10
Treething: 10
ProtestanT: 10
jli: 10
Lena: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 10
-------------------------------
List is G&FF topic starters who have a topic on the first few pages and I have seen post recently. (minus a few) [b]GO![/b]

----===----

[b]Hurt Thales418, heal bubka[/b]

randomdragoon: 10
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 10
ProtestanT: 10
jli: 10
Lena: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 9

[quote="randomdragoon"]
List is G&FF topic starters who have a topic on the first few pages and I have seen post recently. (minus a few) [b]GO![/b]
[/quote]

gee you're mentally obese

hurt thales and heal dragoon

randomdragoon: 11
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 10
ProtestanT: 10
jli: 10
Lena: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 8

I think we should get dragoon out of here asap before he cheats too much. Oh, and I'm in. Heal thales (be nice!), hurt dragoon.

randomdragoon: 10
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 10
ProtestanT: 10
jli: 10
Klebian: 10
Lena: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 9

[quote="jackdillon"]How did you pick the people to play?[/quote]
Hmm. I realize now that you actually did make a topic, but I saw "fabio" in the title and immediately assumed the topic wasn't for real.

Hurt Thales, heal Treething.

randomdragoon: 10
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 11
ProtestanT: 10
jli: 10
Klebian: 10
Lena: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 8

@Klebian: I won't cheat. Every random event that's going to happen is already predetermined and won't trigger for a [i]very[/i] many more posts.

Also, because the average health will always remain the same I don't want too many more people joining in, or else the game will extend way too long. [b]However, you can still hurt & heal even if you're not on the list.[/b]

[quote="randomdragoon"]
Thales418: 10[/quote]

[quote="randomdragoon"]If Thales / Fabiomr annoys you, do not respond to his posts, just pretend that they don't exist and carry on as usual.

Thank you and have a good day.[/quote]

hmmm

This is completely irrelevent to his constant spamming. I just noticed that he was a member here who has a couple topics in G&FF >_>

Hurt Thales418 and heal randomdragoon.

randomdragoon: 11
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 11
ProtestanT: 10
jli: 10
Klebian: 10
Lena: 10
#H34N1: 10
now a ranger: 10
i_like_pie: 10
1=2: 10
Thales418: 7

Okay, this didn't go how I planned.

I'm gonna make a signups for a new improved version of this game.  Hold on.

This game will be suspended for an improved v.2 of this game.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=770589#770589

Someone lock this topic (or keep playing, I don't care)

[quote="randomdragoon"]

@Klebian: I won't cheat. Every random event that's going to happen is already predetermined and won't trigger for a [i]very[/i] many more posts.
[/quote]
I was kidding.

Hurt Treething, Heal i_like_pie
randomdragoon: 11
7h3.D3m0n.117: 10
D3m0n Shad0w: 10
bubka: 11
Treething: 10
ProtestanT: 10
jli: 10
Klebian: 10
Lena: 10
#H34N1: 10
now a ranger: 10
i_like_pie: 11
1=2: 10
Thales418: 7


4 but I can't easily explain the solution.

[hide="Solution"]
Q: An Ausdal bus from Austin to Dallas leaves every hour on the hour. An Ausdal bus from Dallas to Austin leaves every hour on the half-hour. The trip between the two cities takes three hours. How many Ausdal buses going in the opposite direction does an Ausdal bus from Austin to Dallas pass?
A: Draw a diagram that divides the hour before and after the bus starts into 6 parts.
|   |   |   |   |   |   | Train! |   |   |   |   |   |   |
Each gap represents 30 minutes, so the period is 3 hours before and after.
Now shade in each odd one after 1 gap:
|**|   |**|   |**|   | Your Bus |   |**|   |**|   |**|
These are the start times of the trains that will pass by the train, so there are [b]6[/b] buses that it will pass
[/hide]

I think an easier solution is this:

Assume that the bus we are looking at leaves at 4:30 (therefore it will arrive at 7:30):

Therefore it will meet the buses that leave at:

2:00 (because it's arrival time is 5:00, which is between 4:30 and 7:30)
3:00
4:00
5:00
6:00
7:00 (because it's start time is before 7:30)

So the answer is $ \boxed{6}$


find all x;y;z;t are integer if that
$ x^4\plus{}y^4\plus{}z^4\plus{}t^4\equal{}2008xyzt$
(1.2a Secondary contest in Malko)

[quote="foraus"]find all x;y;z;t are integer if that
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4 \equal{} 2008xyzt$
(1.2a Secondary contest in Malko)[/quote]

So too easy !?
This problem has only one root
x=y=z=t=0

This is a basic equation with the way is "infinite backspace "

So that extension with problem

Affirmance  or unaffirmance this problem
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4 \equal{} kxyzt$
has only one root x=y=z=t=0

If this problem is false so find k for equation has only one root x=y=z=t=0

:D

So Who can do this problem with another sollution ?

My english language is too bad :(


geometry
function
radical axis
national olympiad

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112708]Problem 1[/url]
Let $m_{1},m_{2},\ldots,m_{k}$ be integers with $2\leq m_{1}$ and $2m_{1}\leq m_{i+1}$ for all $i$. Show that for any integers $a_{1},a_{2},\ldots,a_{k}$ there are infinitely many integers $x$ which do not satisfy any of the congruences \[x\equiv a_{i}\ (\bmod \ m_{i}),\ i=1,2,\ldots k.\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112709]Problem 2[/url]
Let $ABC$ be an acute triangle and let $k_{1},k_{2},k_{3}$ be the circles with diameters $BC,CA,AB$, respectively. Let $K$ be the radical center of these circles. Segments $AK,CK,BK$ meet $k_{1},k_{2},k_{3}$ again at $D,E,F$, respectively. If the areas of triangles $ABC,DBC,ECA,FAB$ are $u,x,y,z$, respectively, prove that \[u^{2}=x^{2}+y^{2}+z^{2}.\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112710]Problem 3[/url]
Let $A$ be a real number and $(a_{n})$ be a sequence of real numbers such that $a_{1}=1$ and \[1<\frac{a_{n+1}}{a_{n}}\leq A \mbox{ for all }n\in\mathbb{N}.\] $(a)$ Show that there is a unique non-decreasing surjective function $f: \mathbb{N}\rightarrow \mathbb{N}$ such that $1<A^{k(n)}/a_{n}\leq A$ for all $n\in \mathbb{N}$.
$(b)$ If $k$ takes every value at most $m$ times, show that there is a real number $C>1$ such that $Aa_{n}\geq C^{n}$ for all $n\in \mathbb{N}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112712]Problem 4[/url]
In a triangle $ABC$ with $AB\neq AC$, the internal and external bisectors of angle $A$ meet the line $BC$ at $D$ and $E$ respectively. If the feet of the perpendiculars from a point $F$ on the circle with diameter $DE$ to $BC,CA,AB$ are $K,L,M$, respectively, show that $KL=KM$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112714]Problem 5[/url]
Let $t(A)$ denote the sum of elements of a nonempty set $A$ of integers, and define $t(\emptyset)=0$. Find a set $X$ of positive integers such that for every integers $k$ there is a unique ordered pair of disjoint subsets $(A_{k},B_{k})$ of $X$ such that $t(A_{k})-t(B_{k}) = k$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112711]Problem 6[/url]
Find all surjective functions $f: \mathbb{N}\rightarrow \mathbb{N}$ such that for all $m,n\in \mathbb{N}$ \[f(m)\mid f(n) \mbox{ if and only if }m\mid n.\]


If people get the same score on for individuals, how do you decide what place they are in?

See the General Instructions on the mathcounts website.

The tie-breaker system goes something like this:

1. Sprint Round Score.
2. Number correct in sprint 21-30.
3. Look at the following individual problems, in order: 30, 29, ..., 21
4. Look at the following individual problems, in order: Target 8, 7, ..., 1
5. Number correct in sprint 11-20
6. 20, 19, ..., 11
7. 10, 9, ..., 1
8. TIEBREAKER ROUND

The Tiebreaker Round is pretty intense, they give you two questions and when you're done, you raise your hand and they check your answers. If you get them both right, you win, if you don't, the other person has the remainder of the 6 minutes to try to beat you. If not, I believe the first person wins, but I'm not positive. I was hoping I would tie someone with a 46 at chapter last year so I could do a tiebreaker round for once in my life but unfortunately the test was hard and i got a 45 and the next highest was like 41.

There is a similar format for breaking Team Ties. Basically, the computer can break the tie unless you get the exact same problems correct and incorrect; in the case of teams the same number of people on every single problem get each problem correct and incorrect, which is very unlikely (CALCULATE THE PROBABILITY OMG), but it did happen at Nationals this year, and it was pretty awesome because it was a tie for 3rd (NJ/PA).

You really shouldn't worry about these tiebreaker systems, because even if the later problems are SLIGHTLY more important than the earlier problems, the chances that a significant tie (for the last countdown spot, last national spot, or whatever) occurs are pretty low, just try to answer as many as you can correctly.

[quote="gauss1181"]But what about the Countdown Round?[/quote]
The CountDown Rounds are for top 10 people as you might know.

I think from 13375P34K43V312 said, they will probably do it the same way if they have duplicate ranks.

not always top ten. official cds at states must be top ten, but otherwise it can be any format.


function
trigonometry
symmetry
calculus
derivative
algebra unsolved
algebra

I read this problem to day, it is quite ...  :|
Find the maximum and minimum( if exist ) of this function :
 $f(x)=\mid a+bcosx\mid+\mid a+bsinx\mid$
with a and b are given real numbers

[quote="nntrkien"]I read this problem to day, it is quite ...  :|
Find the maximum and minimum( if exist ) of this function :
 $f(x)=\mid a+bcosx\mid+\mid a+bsinx\mid$[/quote]

[hide]

1. Find $\frac{\mathrm{d}f(x)}{\mathrm{d}x}$.

2. Let  $\frac{\mathrm{d}f(x)}{\mathrm{d}x}= 0$.

3. Solve for $x$

4. Substitute $x$ back into $f(x)$ to find your answer(s).

[b] Hint:[/b]

I believe you'll find that the answers are $\pm \frac{\pi}{4}$

[/hide]

[quote="nntrkien"]A pure elementary solutions ? Can you post your complete solution. And i have to say that you missed something.  ;)[/quote]

Sure, here you go...

$f(x)=|a+b\cos x| +|a+b\sin x|$

It follows that

$f'(x) = \frac{(a+b\cos x)(-b\sin x)}{|a+b\cos x|} + \frac{(a+b\sin x)(b\cos x)}{|a+b\sin x|}$

Because of the symmetry of $f'(x)$ my guess is that $x =\pm \frac{\pi}{4}$.

Indeed,

$\frac{(a+\frac{b}{\sqrt{2}})(-\frac{b}{\sqrt{2}})}{|a+\frac{b}{\sqrt{2}}|}+\frac{(a+\frac{b}{\sqrt{2}})(\frac{b}{\sqrt{2}})}{|a+\frac{b}{\sqrt{2}}|}=0$

Ah, yes ... $-\frac{\pi}{4}$ doesn't work, does it?

QED

[quote="nntrkien"]I am not sure. You must discuss $a$ and $b$ and i think f'(x) doesn't always exist such as when $cosx= -\frac ab $  and  $ \mid a \mid  \leq  \mid b \mid$[/quote]

Not to worry...

Starting with $f(x) =|a+b\cos x| + |a +b\sin x|$

The derivative of a function $g(u) = |u|$ is $g'(u) =\frac{u}{|u|} = \pm 1$

Thus, $f'(x) = \pm 1 (-b\sin x) \pm 1 (b\cos x)$

Setting this expression equal to zero, $\pm  b\sin x = \pm b\cos x \Rightarrow \tan x = \pm 1$

$\therefore\, x = \pm \frac{\pi}{4}$

It has already been shown that $ x = -\frac{\pi}{4}$ is not a solution.

Ergo $x = \frac{\pi}{4}$ is the only solution.


[color=darkblue]From the [b][u]Stolz's lemma[/u][/b], is obviously that $p\in N\Longrightarrow\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot\sum_{k=1}^{n}k^{p}=\frac{1}{p+1}\ .$[/color]

Hmmm, first this is not algebra, this is analysis. Secondly, this is anything except olympiad level. Thirdly, this is not even analysis, this is just the definition of Riemann-integrals. So, I'm moving it to its place.

To make what harazi hinted at more explicit, write this as

$\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{4}\cdot\frac1n$

and interpret that as a limit of Riemann sums for an integral.


What's the difference between siempere, algo, and alguien?

[quote="Dave Vather"]You use "algo" for objects, and "alguien" for people. It's a little like "something" and "someone". What did you meant whit "siempere"? "siempre" (always)?[/quote]

Si seguro... al menos siempere nunca lo escuché......


function
calculus
derivative
real analysis
real analysis unsolved

Find all real solutions to the equation $ 2^{x}+5^{x}=3^{x}+4^{x}$.
By inspection, I found that 0 and 1 are solutions but I don't know how to prove that they are the only one of if there are other solutions.

[quote="borislav_mirchev"]1. Write y=2^x+5^x and y=3^x+4^x. Try to sketch their graphs.
2. Or use Newton binoms formulae and or Bernoully's inequality. 5=2.2+1. 3=2+1, 4=2.2.[/quote]
I've made the graph of $ y=2^{x}+5^{x}-3^{x}-4^{x}$. The only solutions are 0 and 1. In fact, for values greater than 1 the function increases. But I want to prove this algebrically.

Put $ F(x)=5^{x}+2^{x}-3^{x}-4^{x}$. Then $ F^{(n)}(x)=(\ln5)^{n}5^{x}+(\ln2)^{n}2^{x}-(\ln4)^{n}4^{x}-(\ln3)^{n}3^{x}$
$ F^{(n)}(1)$ is always non-negative. Then for sufficiently large $ n$ one has $ F^{(n)}(x)$ increasing - the result follows.
(If my method isn't clear try using it to show that for positive $ x$, $ \exp(x)$ is greater than all partial sums of the exponential series - this is obvious but it illustrates the method well.)

Identify the problem as f(x,4) = f(x,2) with f(x,z) = (z+1)^x - z^x for z>=0, x arbitrary. More generally, one can claim that (A) f(x,z) = f(x,w) for z<>w has solutions x=0,1 only.

For the proof, one observes that the derivative of f in the second argument is x * f(x-1,z), ie. (A) implies the existence of y in (z,w) with 0 = x * f(x-1,y)  immediately completing the proof.

We have $ 2^{x}\plus{}5^{x}\equal{} 3^{x}\plus{}4^{x}$ $ \Longleftrightarrow$ $ 3^{x}\minus{}2^{x}\equal{} 5^{x}\minus{}4^{x}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \frac{3^{x}\minus{}2^{x}}{3\minus{}2}\equal{}\frac{5^{x}\minus{}4^{x}}{5\minus{}4}$ $ (1)$

Let $ f(t) \equal{} t^{x}$ .We know that $ f'(t) \equal{} xt^{x\minus{}1}$

Applying the[b] Mean Value Theorem [/b]on $ f$ on$ [2,3]$ and $ [4,5]$ we take that exist's $ \xi_{1}\in(2,3)$ and $ \xi_{2}\in(4,5)$ $ (\xi_{1}\neq\xi_{2})$

such as $ f'(\xi_{1}) \equal{}\frac{3^{x}\minus{}2^{x}}{3\minus{}2}$ and $ f'(\xi_{2}) \equal{}\frac{5^{x}\minus{}4^{x}}{5\minus{}4}$

so $ (1)$  $ \Longleftrightarrow$ $ f'(\xi_{1}) \equal{} f'(\xi_{2})$  $ \Longleftrightarrow$ $ x\xi_{1}^{x\minus{}1}\equal{} x\xi_{2}^{x\minus{}1}$ $ \Longleftrightarrow$  $ x \equal{} 0$ or $ \xi_{1}^{x\minus{}1}\equal{}\xi_{2}^{x\minus{}1}$$ \Longleftrightarrow$ $ \frac{\xi_{1}^{x\minus{}1}}{\xi_{2}^{x\minus{}1}}\equal{} 1$
$ \Longleftrightarrow$ $ x\minus{}1 \equal{} 0$ $ \Longleftrightarrow$ $ x \equal{} 1$ (because $ \frac{\xi_{1}}{\xi_{2}}\neq 1$)

So we have $ x \equal{} 0$ or $ x \equal{} 1$

[color=red]Exercise[/color] Try to solve $ 4^{x}\plus{}25^{x}\equal{} 9^{x}\plus{}16^{x}$

[u]Giannis[/u]

[quote="giannis"]We have $ 2^{x}+5^{x}=3^{x}+4^{x}$ $ \Longleftrightarrow$ $ 3^{x}-2^{x}=5^{x}-4^{x}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \frac{3^{x}-2^{x}}{3-2}=\frac{5^{x}-4^{x}}{5-4}$ $ (1)$

Let $ f(t)=t^{x}$ .We know that $ f'(t)=xt^{x-1}$

Applying the Mean Value Theorem on $ f$ on$ [2,3]$ and $ [4,5]$ we take that exist's $ \xi_{1}\in(2,3)$ and $ \xi_{2}\in(4,5)$ $ (\xi_{1}\neq \xi_{2})$

such as $ f'(\xi_{1})=\frac{3^{x}-2^{x}}{3-2}$ and $ f'(\xi_{2})=\frac{5^{x}-4^{x}}{5-4}$

so $ (1)$  $ \Longleftrightarrow$ $ f'(\xi_{1})=f'(\xi_{2})$  $ \Longleftrightarrow$ $ x \xi_{1}^{x-1}=x \xi_{2}^{x-1}$ $ \Longleftrightarrow$  $ x=0$ or $ \xi_{1}^{x-1}=\xi_{2}^{x-1}$$ \Longleftrightarrow$ $ \frac{\xi_{1}^{x-1}}{\xi_{2}^{x-1}}=1$
$ \Longleftrightarrow$ $ x-1=0$ $ \Longleftrightarrow$ $ x=1$ (because $ \frac{\xi_{1}}{\xi_{2}}\neq 1$)

So we have $ x=0$ or $ x=1$

[color=red]Exercise[/color] Try to solve $ 4^{x}+25^{x}=9^{x}+16^{x}$

[u]Giannis[/u][/quote]

Beautiful solution.
The one you proposed I've tried to fit it in an inequality. $ 25^{x}>9^{x}+16^{x}$ Then $ 25^{x}+4^{x}>9^{x}+16^{x}$. There are no solutions for $ x\geq 1$. I guess the Mean Value Theorem does not apply to this problem. Anyway the solutions are $ x=0$ and $ x=1/2$. I still have no idea how to prove that these are the only solutions.

A generalization of giannis' argument:

Let $ 0<a<b<c<d$. The equation $ a^{x}+d^{x}=b^{x}+c^{x}$ has at most two real solutions.

Proof: Obviously, $ x=0$ is a solution. Suppose that $ x=\lambda$ is also a solution, with $ \lambda\neq 0$.
We can rearrange the equation to $ b^{x}-a^{x}=d^{x}-c^{x}$, and we know that $ b^{\lambda}-a^{\lambda}=d^{\lambda}-c^{\lambda}$. Divide the left hand side by $ b^{\lambda}-a^{\lambda}$ and the right by $ d^{\lambda}-c^{\lambda}$ (which aren't zero):
\[ \frac{b^{x}-a^{x}}{b^{\lambda}-a^{\lambda}}= \frac{d^{x}-c^{x}}{d^{\lambda}-c^{\lambda}}\]
Setting $ x=\lambda y$, this becomes
\[ \frac{(b^{\lambda})^{x}-(a^{\lambda})^{x}}{b^{\lambda}-a^{\lambda}}= \frac{(d^{\lambda})^{x}-(c^{\lambda})^{x}}{d^{\lambda}-c^{\lambda}}\]
We split into two cases. If $ \lambda>0$, $ a^{\lambda}<b^{\lambda}<c^{\lambda}<d^{\lambda}$ and we apply the Mean Value Theorem to $ f(x)=t^{x}$ on the intervals $ [a^{\lambda},b^{\lambda}]$ and $ [c^{\lambda},d^{\lambda}]$, getting $ yt_{1}^{y-1}=yt_{2}^{y-1}$ with $ t_{1}<t_{2}$. As in giannis' post, either $ y=0$ or $ y=1$.
If $ \lambda<0$, we have $ d^{\lambda}<c^{\lambda}<b^{\lambda}<a^{\lambda}$, so we apply the MVT on $ [d^{\lambda},c^{\lambda}]$ and $ [b^{\lambda},a^{\lambda}]$ instead, with the same result $ y=0$ or $ y=1$.

That's the proof that there are at most two solutions. Try this exercise:
If in addition $ ad\neq bc$, show that there are two solutions. If $ ad=bc$, show that $ x=0$ is the only solution.

[quote="jmerry"]A generalization of giannis' argument:

Let $ 0 < a < b < c < d$. The equation $ a^{x}\plus{}d^{x}\equal{} b^{x}\plus{}c^{x}$ has at most two real solutions.

Proof: Obviously, $ x \equal{} 0$ is a solution. Suppose that $ x \equal{}\lambda$ is also a solution, with $ \lambda\neq 0$.
We can rearrange the equation to $ b^{x}\minus{}a^{x}\equal{} d^{x}\minus{}c^{x}$, and we know that $ b^{\lambda}\minus{}a^{\lambda}\equal{} d^{\lambda}\minus{}c^{\lambda}$. Divide the left hand side by $ b^{\lambda}\minus{}a^{\lambda}$ and the right by $ d^{\lambda}\minus{}c^{\lambda}$ (which aren't zero):
\[ \frac{b^{x}\minus{}a^{x}}{b^{\lambda}\minus{}a^{\lambda}}\equal{}\frac{d^{x}\minus{}c^{x}}{d^{\lambda}\minus{}c^{\lambda}}\]
Setting $ x \equal{}\lambda y$, this becomes
\[ \frac{(b^{\lambda})^{x}\minus{}(a^{\lambda})^{x}}{b^{\lambda}\minus{}a^{\lambda}}\equal{}\frac{(d^{\lambda})^{x}\minus{}(c^{\lambda})^{x}}{d^{\lambda}\minus{}c^{\lambda}}\]
We split into two cases. If $ \lambda > 0$, $ a^{\lambda}< b^{\lambda}< c^{\lambda}< d^{\lambda}$ and we apply the Mean Value Theorem to $ f(x) \equal{} t^{x}$ on the intervals $ [a^{\lambda},b^{\lambda}]$ and $ [c^{\lambda},d^{\lambda}]$, getting $ yt_{1}^{y\minus{}1}\equal{} yt_{2}^{y\minus{}1}$ with $ t_{1}< t_{2}$. As in giannis' post, either $ y \equal{} 0$ or $ y \equal{} 1$.
If $ \lambda < 0$, we have $ d^{\lambda}< c^{\lambda}< b^{\lambda}< a^{\lambda}$, so we apply the MVT on $ [d^{\lambda},c^{\lambda}]$ and $ [b^{\lambda},a^{\lambda}]$ instead, with the same result $ y \equal{} 0$ or $ y \equal{} 1$.

That's the proof that there are at most two solutions. Try this exercise:
If in addition $ ad\neq bc$, show that there are two solutions. If $ ad \equal{} bc$, show that $ x \equal{} 0$ is the only solution.[/quote]
How do I use that $ ad\equal{}bc$?

Quoting entire posts is generally not recommended, and quoting the last post in its entirety is always a bad idea- unless you state otherwise, it's assumed you're replying to that post.

As for the content- I'm not giving it away just yet.

you set $ x =\lambda y$ instead of $ y =\lambda x.$

That's what I meant to do, but an error got in anyway- each $ x$ should be $ y$ in the next line.

I pressed the wrong button. My intention was to reply the post. Sorry.


function
real analysis
real analysis unsolved

Find explicitely an injective function $f$ from $R^{2}$ to $R$ for which if $x<y$ then $x<f(x,y)<y$.

You have two numbers $a$ and $b$ whose (proper) decimal representations are $a_{1}\dots a_{n}.a'_{1}\dots a'_{n}\dots$ and  $b_{1}\dots b_{n}.b'_{1}\dots b'_{n}\dots$. If $a<b$, define $f(a,b)$ as follows.

First, pick $N \in \mathbb{N}$ odd such that $a+10^{-N}<b$. Now all the first digits in the proper decimal expansion of $f(a,b)$ are those of $a+10^{-N}$, up to the $N$-th decimal of its fractional part. After this, the digits of the fractional part will alternatively describe those of $a+10^{-N}$ (for the odd ranks) and those of a certain sequence $(x_{n})_{n \geq N+1}$ defined below (for the even ranks).

Let $x_{N+1}=0$, then $X_{N+3}=7$ if $a<0$, $8$ if $a=0$, and $9$ if $a>0$. Starting at $N+5$, we enter the digits of [(the number of digits in the integer part of $a$)+1]. Right after that, $x_{N+2k+1}=0$, and the following digits will be the string of the digits constituting the binary representation of $a$. If this string happens to be finite, you make it infinite by adding infinitely many $6$ to it. That's it.

There may be a few details to work out with the pathologic cases, but I believe it works. (as for the cases where $a>b$, since there is no restriction other than injectivity, it is easy to deal with them by injecting them into those numbers whose decimal representations have no $0$).


linear algebra
matrix
geometry
geometric transformation
rotation
analytic geometry
graphing lines

How do you interpret an equation in parametric form:
\[ \begin{pmatrix}
                x \\ y \end{pmatrix}
      \equal{} \begin{pmatrix}
                t^2\minus{}1\\
                t^2\minus{}2t\plus{}3
          \end{pmatrix}?
\]

This is supposed to represent a curve on the plane. I wonder in general how should one deal with such representations.

Well, for each $ t$, plot the point $ \binom{x}{y}$. It may or may not be practical to solve for $ t$ in terms of $ x$ and then substitute that into the expression for $ y$ to get an equation in $ y$ in terms of $ x$.

In this case, there are a number of things you can see without doing any math:
- There is a lower bound to both $ x$ and $ y$
- For very positive or very negative $ t$, we have, roughly, $ x/y \approx 1$
- With some thought, you realize there are two branches

In this case, the most insightful approach is actually not to to solve for $ t$ and substitute, but to apply a [[Rotation matrix]] to rotate the graph 45 degrees ccw (I picked 45 because it corresponds to a slope of 1, which is the limit of x/y)

$ \textbf{rot}_{\left[ \frac \pi 4 \right]}\binom{x}{y} = \left( \begin{array}{cc} 1 / \sqrt {2} & - 1/ \sqrt {2} & 1/\sqrt {2} & 1 / \sqrt {2} \end{array} \right) \binom{t^2 - 1}{t^2 - 2 t + 3} = \binom{\sqrt {2} (t^2 - t + 1) }{\sqrt {2} (t-2)}$

[b]Now[/b] solve for y in terms of x. You find that you get a parabola that opens upwards! So, your original graph will be this parabla, rotated 45 degrees clockwise about the origin.


geometry
ratio
analytic geometry
inequalities

Let $ ABC$ be a triangle, $M \in (AC), \, N \in (AB)$ s.t. $ MA = m \cdot MC, \, NA = n \cdot NB$ (we're using unsigned distances, of course), where $ m+n = 2$. Prove that if $ P$ is the point of intersection of $ BM$ with $ CN$, then $ [APMN] \geq \frac{mn}3 \cdot [ABC]$.

[i]Note.[/i] $\left[ A_{1}A_{2}\ldots A_{n}\right] =$ the area of the polygon $ A_{1}A_{2}\ldots A_{n}$.

[i]Virgil Nicula[/i]

hmm i think u can do this using coordinates and just making triangle $ABC$ a right triangle on the coordinate plane

but... is there any other way to do it...?

getting the area ratio is pretty trivial with mass points...i get the inequality: \[\frac{mn}{m+n+1}*\left(\frac{1}{m+1}+\frac{1}{n+1}\right)\ge \frac{mn}{3}\] then it reduces to: \[\frac{m+n+2}{(m+1)(n+1)}\ge 1\] Then expand, and cancel to get: \[1\ge mn\] which is true by the am-gm \[\frac{m+n}{2}\ge\sqrt{mn}\Longleftrightarrow 1\ge\sqrt{mn}\Longleftrightarrow 1\ge mn\]

hmmm can u explain how to get area ratios of say... AMPN based off of just mass points?

ok, what's mass point?? (I hear it all the time but I have no idea what it is... I just use area ratios of common height/base and similar polygons...)


Can anyone show  me a nice and elegant proof of this please?

$ \left ( \prod_{i\equal{}1}^{n}(x_{i}\plus{}y_{i}) \right )^{1/n}\geq \left ( \prod_{i\equal{}1}^{n}x_{i} \right )^{1/n}\plus{}\left ( \prod_{i\equal{}1}^{n}y_{i} \right )^{1/n}$

where $ n$ is any positive integer and $ x_{i},y_{i}\geq0$ with $ i\equal{}1,2,...,n$.

[quote="james digol"]Can anyone show  me a nice and elegant proof of this please?

$ \left ( \prod_{i \equal{} 1}^{n}(x_{i} \plus{} y_{i}) \right )^{1/n}\geq \left ( \prod_{i \equal{} 1}^{n}x_{i} \right )^{1/n} \plus{} \left ( \prod_{i \equal{} 1}^{n}y_{i} \right )^{1/n}$

where $ n$ is any positive integer and $ x_{i},y_{i}\geq0$ with $ i \equal{} 1,2,...,n$.[/quote][hide="hint"]Use the following: By AM-GM, we have for $ a,b,c,d\ge0$ \[ (a\plus{}b)(c\plus{}d)\ge (\sqrt{ac}\plus{}\sqrt{bd})^2\][/hide]

[hide]divide the left hand side to the right, and amgm.[/hide]

[quote="james digol"]Can anyone show  me a nice and elegant proof of this please?

$ \left ( \prod_{i \equal{} 1}^{n}(x_{i} \plus{} y_{i}) \right )^{1/n}\geq \left ( \prod_{i \equal{} 1}^{n}x_{i} \right )^{1/n} \plus{} \left ( \prod_{i \equal{} 1}^{n}y_{i} \right )^{1/n}$

where $ n$ is any positive integer and $ x_{i},y_{i}\geq0$ with $ i \equal{} 1,2,...,n$.[/quote]

by AM-GM, we have

\[ \frac {{\sum\limits_{i \equal{} 1}^n {\frac {{a_i }}{{a_i \plus{} b_i }}} }}{n} \ge \sqrt [n]{{\frac {{\prod\limits_{i \equal{} 1}^n {a_i } }}{{\prod\limits_{i \equal{} 1}^n {\left( {a_i \plus{} b_i } \right)} }}}}\]
and

\[ \frac {{\sum\limits_{i \equal{} 1}^n {\frac {{b_i }}{{a_i \plus{} b_i }}} }}{n} \ge \sqrt [n]{{\frac {{\prod\limits_{i \equal{} 1}^n {b_i } }}{{\prod\limits_{i \equal{} 1}^n {\left( {a_i \plus{} b_i } \right)} }}}}\]
add them, get

\[ 1 \ge \sqrt [n]{{\frac {{\prod\limits_{i \equal{} 1}^n {a_i } }}{{\prod\limits_{i \equal{} 1}^n {\left( {a_i \plus{} b_i } \right)} }}}} \plus{} \sqrt [n]{{\frac {{\prod\limits_{i \equal{} 1}^n {b_i } }}{{\prod\limits_{i \equal{} 1}^n {\left( {a_i \plus{} b_i } \right)} }}}}\]
\[ \iff \sqrt [n]{{\prod\limits_{i \equal{} 1}^n {\left( {a_i \plus{} b_i } \right)} }} \ge \sqrt [n]{{\prod\limits_{i \equal{} 1}^n {a_i } }} \plus{} \sqrt [n]{{\prod\limits_{i \equal{} 1}^n {b_i } }}.\]

Thank you Kuing, Soarer and Sparfenkel for your help...c",)


algebra
system of equations
algebra unsolved

hello, do you really mean
$ 2004^{x\minus{}y}\minus{}2.2005^{1\minus{}x}\plus{}2006^{\frac{1\minus{}y}{2}}\equal{}0$ in your second equation?
Sonnhard.

[quote="Hesman"]Solve the following system of equations
\[ \left\{\begin{array}{l} \sqrt {2x} \minus{} \sqrt {y \plus{} 1} \minus{} \left( {1 \minus{} 2x \plus{} y} \right)\left( {1 \plus{} 12x \plus{} 2005{y^2}} \right) \equal{} 0 \\
{2004^{x \minus{} y}} \minus{} {2.2005^{1 \minus{} x}} \plus{} {2006^{\frac {{1 \minus{} y}}{2}}} \equal{} 0 \\
\end{array} \right.\]
[/quote]

First eqiuation may be written $ (\sqrt {y \plus{} 1}\minus{}\sqrt{2x} )(1\plus{} (\sqrt {y \plus{} 1}\plus{}\sqrt{2x} )(1 \plus{} 12x \plus{} 2005{y^2}))\equal{}0$

and so$ \sqrt {2x} \equal{} \sqrt {y \plus{} 1}$ (the other parenthesis is always $ >0$
And so $ y\equal{}2x\minus{}1$ and the second equation becomes then $ \frac{2004^{1\minus{}x} \plus{} 2006^{1\minus{}x}}2 \equal{} \left(\frac{2004\plus{}2006}2\right)^{1 \minus{} x}$ whose solutions are $ x\in\{0,1\}$ (just look at convexity of $ a^x$).

Hence the two solutions $ (0,\minus{}1)$ and $ (1,1)$


$ ABC$ is acute triangle and $ H$ his  orthocenter $ H$.
let $ AH \equal{} R$ with $ R$ is the radius of circonsible circle of $ ABC$.
find $ \widehat{A}$

[color=red]EDIT by moderator: Moved from Geometry / Olympiad section.[/color]

Since ABC is acute, so there is $ AH\equal{}2R\cos{A}$, which gives $ \cos{A}\equal{}\frac12 \Longrightarrow \angle{A}\equal{}60^{o}$.


quadratics
number theory unsolved
number theory

prove taht: such integers $ a,b,c$ do not exist

$ \frac{b\plus{}c}{a}\plus{}\frac{c\plus{}a}{b}\plus{}\frac{a\plus{}b}{c}\equal{}0$

Solve the system wrt $ a$. Then you'll get
$ \delta \equal{} \sqrt{(b^2\plus{}c^2)^2\minus{}4(b^2c\plus{}bc^2)(b\plus{}c)}\equal{} \sqrt{b^4\minus{}4b^3c\minus{}6b^2c^2\minus{}4bc^3\plus{}c^4}\equal{}\sqrt{(b\minus{}c)^4\minus{}12b^2c^2}.$
Hence $ (b\minus{}c)^4\minus{}12b^2c^2\equal{}x^2 \Leftrightarrow 12b^2c^2\equal{}((b\minus{}c)^2\minus{}x)((b\minus{}c)^2\plus{}x).$  This can be done modulo 4.

$ (b\minus{}c)^4 \minus{} 12b^2c^2 \equal{} x^2$

This has solutions:
$ (b,c,x) \equal{} (0,c,c^2)$, $ (b,0,b^2)$ or $ (b,\minus{}b,2b^2)$
These do not lead to valid solutions of the original problem but they provide solutions mod 4 to the discriminant.
Does your argument involve something else using the factorisation?

[quote="aktyw19"]prove taht: such integers $ a,b,c$ do not exist

$ \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c} \equal{} 0$[/quote]

let $ x \equal{} \frac {b \plus{} c}{a}, y \equal{} \frac {c \plus{} a}{b}, z \equal{} \frac {a \plus{} b}{c}$

$ x \plus{} y \plus{} z \equal{} 0 \implies bc(b \plus{} c) \plus{} ca(c \plus{} a) \plus{} ab(a \plus{} b) \equal{} 0$
$ \implies (b \plus{} c)(c \plus{} a)(a \plus{} b) \equal{} 2abc \implies xyz \equal{} 2$
$ \implies xy(x \plus{} y) \equal{} \minus{} 2$
We want rational solutions.

As a quadratic in $ y$ this will have solutions if the discriminant is a square rational i.e.

$ x^4 \minus{} 8x \equal{} z^2$

This is a quartic elliptic curve and quartic elliptic curves can be reduced to cubic elliptic curves.

In this case it is done as follows

$ (z \minus{} x^2)(z \plus{} x^2) \plus{} 8x \equal{} 0$

take $ t \equal{} z \plus{} x^2$

$ t \plus{} \frac {8x}{t} \equal{} (z \plus{} x^2) \minus{} (z \minus{} x^2) \equal{} 2x^2$
$ t^3 \plus{} 8xt \equal{} 2x^2t^2$

let $ u \equal{} xt \minus{} 2$

$ 2u^2 \equal{} t^3 \plus{} 8$

writing $ u \equal{} 2v, t \equal{} 2s$

$ v^2 \equal{} s^3 \plus{} 1$

This is an elliptic curve and we want solutions in rationals. Standard methods can be used to find the rank and torsion group.
It is known that the only rational solutuons are $ (v,s) \equal{} (1,0), ( \minus{} 1,0), (3,2), ( \minus{} 3,2), (0, \minus{} 1)$
I am not sure if there is an easy proof.
Working back through the steps we find these give $ (x,y,z) \equal{} ( \minus{} 1, \minus{} 1,2)$ and permutations
but solving for a,b,c leaves one of the numbers equal to zero which is not allowed because of the divisions.


Can someone list all the famous inequalities I should know?

For example, AM-GM, Jensen's, Cauchy's,...

Refer to the following useful topic, as well as others in the Inequalities forum:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7942

Edit:  The statement of Cauchy in Engel form is incorrect.  It should be

$\sum \frac{a_{i}^{2}}{b_{i}}\ge \frac{ \left( \sum a_{i}\right)^{2}}{\sum b_{i}}$


I think absolutely integrable means: $|f|$ is Lebesgue integrable, right? Then it means that $(f_{+}+f_{-})$ are Lebesgue integrable. But why does it imply that $f_{+}$ and $f_{-}$ are Lebesgue integrable?

Thanks a lot!

After reading around in the library, now I get it:

1. if $f$ is measurable, then $|f|$ is integrable <=> $f$ is integrable.

2. Generally, $f$ is integrable => $|f|$ is integrable, but the reverse is not true.

Both are not hard to prove.


Find all functions such that $ f: \mathbb{R}^\plus{} \rightarrow \mathbb{R}^\plus{}$ and $ f(x\plus{}f(y))\equal{}yf(xy\plus{}1)$ for every $ x,y\in \mathbb{R}^\plus{}$.

If we assume that $ \frac{f(y)\minus{}1}{y\minus{}1}>0$ so we can substitute $ x\equal{}\frac{f(y)\minus{}1}{y\minus{}1}$ and in this case $ y\equal{}\frac{f(x\plus{}f(y))}{f(xy\plus{}1)}\equal{}\frac{f(\frac{yf(y)\minus{}1}{y\minus{}1})}{f(\frac{yf(y)\minus{}1}{y\minus{}1})}\equal{}1$.
So $ \frac{f(y)\minus{}1}{y\minus{}1}<0$ for every $ y\not\equal{}1$. (*)

Now let $ x\equal{}1\minus{}\frac{1}{y}$ for $ y>1$. $ \Rightarrow f(1\minus{}\frac{1}{y}\plus{}f(y))\equal{}yf(y)$.
If $ f(y)>\frac{1}{y}$, then $ 1\plus{}f(y)\minus{}\frac{1}{y}>1\Rightarrow f(1\plus{}f(y)\minus{}\frac{1}{y})<1$ from (*) - contradiction.
If $ f(y)<\frac{1}{y}$, then $ 1\plus{}f(y)\minus{}\frac{1}{y}<1\Rightarrow f(1\plus{}f(y)\minus{}\frac{1}{y})>1$ from (*) - contradiction.
$ \Rightarrow f(y)\equal{}\frac{1}{y}$ for every $ y>1$.

Now, if we take some $ y>1$ and $ x>0$, we will have $ f(x\plus{}f(y))\equal{}f(x\plus{}\frac{1}{y})\equal{}yf(xy\plus{}1)\equal{}\frac{1}{x\plus{}\frac{1}{y}}$.
So, if for some $ 0<a<b<1$ we substitute $ y\equal{}\frac{1}{b\minus{}a}>1$ and $ x\equal{}a$, then we will have $ f(b)\equal{}\frac{1}{b}$.

$ \Rightarrow f(x)\equal{}\frac{1}{x}$ for every $ x>0$.

[quote="far-reaching"]Can someone else post another solution?
BG Yoda just adjusts the right answer. I want some more elegant method, if it's possible? :blush:[/quote]

This is basically the same solution, written in different way:

Take any $x>1$, we have
\[f\left( \frac{x-1}{x}+f\left( x\right) \right) =xf\left( x\right) \] and
\[f\left( x+f\left( \frac{x-1}{x}+f\left( x\right) \right) \right) =\left[ \frac{x-1}{x}+f\left( x\right) \right] f\left( x+xf\left( x\right) \right). \]
Thus $\frac{x-1}{x}+f\left( x\right) =1$ or equivalently $f\left(x\right) =\frac{1}{x}$.

Now for any $x>0$, we have
\[\frac{1}{1+f\left( x\right) }=f\left( 1+f\left( x\right) \right) =xf\left(x+1\right) =\frac{x}{x+1}=\frac{1}{1+\frac{1}{x}}\] or $f\left( x\right) =\frac{1}{x}$.


inequalities
function
LaTeX
inequalities unsolved

[quote="Mary Barton"]If $ xyz \equal{} 8$ then
\[ \frac {1}{\sqrt {x^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {y^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {z^2 \plus{} 1}}\ \geq\ \frac {3}{\sqrt {5}}
\]
I tried doing it using Jensen’s inequality but did not succeed. Is there another way?

Also I think we can assume WLOG that $ x,y,z$ are all positive. Yeah?[/quote]
Try $ x\equal{}y\equal{}6$ and $ z\equal{}\frac{2}{9}.$ :wink:

But it’s true that $ \frac {1}{\sqrt {x^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {y^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {z^2 \plus{} 1}}>1$, though?

[quote="vg23nov"]Let f(x) = 1/sqrt(x)
f(x^2+1)+f(y^2+1)+f(z^2+1) >= 3f((x^2+y^2+z^2+3)/3)
using power mean: x^2+y^2+z^2>=12
since f(x) is always decreasing 3f((x^2+y^2+z^2+3)/3) >= 3f(5)=3/sqrt(5)

Why is this wrong?[/quote]
Because $ 3f((x^2\plus{}y^2\plus{}z^2\plus{}3)/3) >\equal{} 3f(5)$ is wrong. :wink:

In order for that logic to work, you need $ f(x)$ to be increasing, not decreasing.

Let $ x^2\equal{}e^a y^2\equal{}e^b z^2\equal{}e^c$

The function $ f(a)\equal{}\frac{1}{\sqrt{e^a\plus{}1}}$ is the desired function, but is concave when $ a<ln 2$ with $ x\plus{}y\plus{}z\equal{}ln 64$ so a solution may exist when one of the variables is less than $ \sqrt{2}$

As i was looking at my old and unserious post, i couldn't find my mistake. I used the fact that:
$ \frac {a_{1}^{k}}{b_{1}^{l}} + \frac {a_{2}^{k}}{b_{2}^{l}} + ... + \frac {a_{n}^{k}}{b_{n}^{l}}\geq\frac {(a_{1} + a_{2} + ... + a_{n})^{k}}{(b_{1} + b_{2} + ... + b_{n})^{l}}$
when $ k=l+1$
which is equivalent to:
$ ((\frac {a_{1}}{b_{1}^{l/k}})^{k} + (\frac {a_{2}}{b_{2}^{l/k}})^{k} + ... + (\frac {a_{n}}{b_{n}^{l/k}})^{k})^{\frac {1}{k}}*((b_{1}^{l/k})^{k/l} + (b_{2}^{l/k})^{k/l} + ... + (b_{n}^{l/k})^{k/l})^{\frac {1}{k/l}}\geq\(a_{1} + a_{2} + ... + a_{n}$
which should be true by Hoelder's inequality, as:
$ \frac {1}{k} + \frac {1}{k/l} = 1$

This fact would be really useful,but it appears to be mistake somewere. As i saw a lot of really clever people here, i hope those harsh looking fractions won't frighten you and someone will point out my mistake easily :)


Given n points in the plane such that the distance between every pair is at least 1,prove that there are at most 3n pairs of points such that the distance between two points in each pair is 1.


For his birthday, Bert gets a box that holds 125 jellybeans
when filled to capacity.  A few weeks later, Carrie gets a larger box
full of jellybeans.  Her box is twice as high, twice as wide and twice as
long as Bert's.  Approximately, how many jellybeans did Carrie get?

1000. mem.

Volume of first box = $ bwh$
Volume of second box = $ 2b*2w*2h\equal{}8bwh$
So the second box is 8 times as large as the first.
$ 125*8\equal{}1000$

Is it just a coincidence that since the problem doesn't specify that the box is a cube or not, $ \sqrt[3]{125} \cdot 2$ cubed is also $ 1000$?  :maybe:

I said $ \sqrt[3]{125} \cdot 2$ cubed.  :wink:


They are all easy!  :lol:  :blush:
1)
 Let $x, y, z, t\ge 0$. Prove \[(x^{4}+3)(y^{4}+3)(z^{4}+3)(t^{4}+3) \ge 16(x+y+z+t)(xyz+yzt+ztx+txy)\] 2)
Let $a,b,c,d$ be the positive numbers such that $a+b+c+d=1$. Prove \[a^{3}+b^{3}+c^{3}+d^{3}+8abcd \geq 2\sum_{cyc}bcd(b+c+d)\]

[quote="The soul of rock"] \[(x^{4}+3)(y^{4}+3)(z^{4}+3)(t^{4}+3) \ge 16(x+y+z+t)(xyz+yzt+ztx+txy)\] [/quote]
But by holder's inequality we can easily get: \[(x^{4}+3)(y^{4}+3)(z^{4}+3)(t^{4}+3) \ge (x+y+z+t)^{4}\] Which implies your inequality obviously. :)

[quote="The soul of rock"]
2)
Let $a,b,c,d$ be the positive numbers such that $a+b+c+d=1$. Prove \[a^{3}+b^{3}+c^{3}+d^{3}+8abcd \geq 2\sum_{cyc}a^{3}(b+c+d)\] [/quote]
If I'm not wrong,this is equal to: \[2(a^{4}+b^{4}+c^{4}+d^{4})+8abcd \ge (a+b+c+d)(a^{3}+b^{3}+c^{3}+d^{3})\] Which is wrong:Try:$a=b=c=1,d=0$
If you don't believe,you can try $a=b=c=\frac{1}{3},d=0$ in the original one.

Sorry, there is some mistake in my expanding!
The second must be
$a^{3}+b^{3}+c^{3}+d^{3}+8abcd \geq 2\sum_{cyc}bcd(b+c+d)$
Please try this inequality
$(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\ge (a^{3}+b^{3}+c^{3}+24abc)$ where $a,b,c>0$.
I just have an ugly solution for this inequality.

the first one is wrong again d=0 , a=b=c>0
in your last inequality if we take b,c near to 0 we have $a^{2}+a+1\geq a^{3}$ which ...hmm... can't hold for every a :roll

[quote="Albanian Eagle"]the first one is wrong again d=0 , a=b=c>0
[/quote]
Oppps, you forgot that $a+b+c+d=1$.

Nobody like this?

[i]$125 =5^{3}$, what a beautiful number. I'll no longer post any topics.[/i]

[quote="The soul of rock"]
2)
Let $a,b,c,d$ be the positive numbers such that $a+b+c+d=1$. Prove
\[a^{3}+b^{3}+c^{3}+d^{3}+8abcd \geq 2\sum_{cyc}bcd(b+c+d) \]
[/quote]
The following inequality is stronger:
\[a^{3}+b^{3}+c^{3}+d^{3}+20abcd \geq 3\sum_{cyc}bcd(b+c+d) \]
or $a^{3}+b^{3}+c^{3}+d^{3}+32abcd\geq3(abc+abd+acd+bcd).$

Nice ineq, arqady. I have one solution for yours ineq, but It too long. Can you post your solution?


trigonometry
function
algebra
polynomial
induction
limit
algebra proposed

Let $ n\geq 2$ be integers.

(1) Show that there exists polynomials $ P_n(x)$ with degree $ n \minus{} 1,\ Q_n(x)$ with degree $ n$ such that $ \sin (2n\theta) \equal{} n\sin (2\theta) P_n(\sin ^ 2 \theta),\ \cos (2n\theta) \equal{} Q_n(\sin ^ 2 \theta)$ for any real numbers $ \theta$ by induction.

(2) Let $ \alpha _{k} \equal{} \left(\sin \frac {k\pi}{2n}\right)^{ \minus{} 2}$ for $ k \equal{} 1,\ 2,\ \cdots ,\ n \minus{} 1.$

Show that we have $ P_n(x) \equal{} (1 \minus{} \alpha _1x)(1 \minus{} \alpha _2x)\cdots (1 \minus{} \alpha _{n \minus{} 1}x).$

(3) Show that $ \sum_{k \equal{} 1}^{n \minus{} 1} \alpha _k \equal{} \frac {2n^2 \minus{} 2}{3}.$

a) We will prove by induction on $ n$ that the assertion holds and that the leading coefficient of $ P_ n$ (resp. $ Q_n$) is given by $ a_n\equal{}\frac{(\minus{}4)^{n\minus{}1}}{n}$  (resp. $ b_n \equal{} \minus{}2na_n$)

For $ n\equal{}2$ we have $ \sin (4\theta )\equal{}2\sin(2\theta )\cos(2\theta)\equal{}2\sin(2\theta)(1\minus{}2\sin^2(\theta))$ and $ \cos (4\theta )\equal{}2\cos^2(2\theta )\minus{}1\equal{}2(1\minus{}2\sin^2(\theta))^2\minus{}1$, i.e. $ P_2(x)\equal{}1\minus{}2x$ of degree 1 and $ a_2\equal{}\minus{}2$ and $ Q_2(x)\equal{}2(1\minus{}2x)^2\minus{}1$ of degree 2 and $ b_2\equal{}8$.

Let $ n\geq 3$ then we have

$ \sin(2n\theta)\equal{}\sin(2(n\minus{}1)\theta)(1\minus{}2\sin^2(\theta))\plus{}\sin(2\theta)\cos(2(n\minus{}1)\theta)\equal{}\sin(2\theta)\left[(n\minus{}1)P_{n\minus{}1}(\sin^2(\theta))(1\minus{}2\sin^2(\theta))\plus{}Q_{n\minus{}1}(\sin^2(\theta))\right]$, i.e. $ P_n(x)\equal{}\frac{1}{n}\left[(n\minus{}1)P_{n\minus{}1}(x)(1\minus{}2x)\plus{}Q_{n\minus{}1}(x)\right]$ of degree $ \leq n\minus{}1$

$ \cos(2n\theta)\equal{}(1\minus{}2\sin^2(\theta))Q_{n\minus{}1}(\sin^2(\theta))\minus{}\sin^2(2\theta)(n\minus{}1)P_{n\minus{}1}(\sin^2(\theta))\equal{}(1\minus{}2\sin^2(\theta))Q_{n\minus{}1}(\sin^2(\theta))\minus{}4\sin^2(\theta)(1\minus{}\sin^2(\theta))(n\minus{}1)P_{n\minus{}1}(\sin^2(\theta))$

i.e. $ Q_n(x)\equal{}(1\minus{}2x)Q_{n\minus{}1}(x)\minus{}4x(1\minus{}x)(n\minus{}1)P_{n\minus{}1}(x)$ of degree $ \leq n$

the updating of the leading coefficients is given by $ a_n\equal{}\frac{\minus{}2(n\minus{}1)a_{n\minus{}1}\plus{}b_{n\minus{}1}}{n}\equal{}\frac{(\minus{}4)^{n\minus{}1}}{n}\not\equal{}0$ and $ b_n\equal{}\minus{}2b_{n\minus{}1}\plus{}4(n\minus{}1)a_{n\minus{}1}\equal{}\minus{}2na_n\not\equal{}0$, in particular, $ P_n,Q_n$ have the degrees and leading coefficients as claimed.

b) Set $ \theta_k\equal{}\frac{k\pi}{2n}$ for $ k\equal{}1,\ldots,n$, then direct computation yields $ 0\equal{}\sin(2n\theta_k)\equal{}n\underbrace{\sin(2\theta_k)}_{\not\equal{}0}P_n(\sin^2(\theta_k))\Longrightarrow 0\equal{}P_n(\sin^2(\theta_k))\equal{}P_n(\alpha_k^{\minus{}1})$. Since the $ \alpha_k^{\minus{}1}$'s assume $ n\minus{}1$ different values, we must have $ P_n(x)\equal{}C\cdot\prod_{k\equal{}1}^{n\minus{}1}(\alpha_k^{\minus{}1}\minus{}x)\equal{}C\cdot\prod_{k\equal{}1}^{n\minus{}1}\alpha_k^{\minus{}1}\cdot\prod_{k\equal{}1}^{n\minus{}1}(1\minus{}x\alpha_k)$ for some suitable constant $ C$. Applying L'Hospital we observe $ P_n(0)\equal{}\lim_{\theta\rightarrow 0}P_n(\sin^2(\theta))\equal{}\lim_{\theta\rightarrow 0}\frac{\sin(2n\theta)}{n\sin(2\theta)}\equal{}\lim_{\theta\rightarrow 0}\frac{\cos(2n\theta)}{\cos(2\theta)}\equal{}1$ and hence $ P_n(x)\equal{}\prod_{k\equal{}1}^{n\minus{}1}(1\minus{}x\alpha_k)$ as claimed.
More formally, note that we used the continuous extension of the function $ g(\theta)\equal{}\frac{\sin(2n\theta)}{n\sin(2\theta)}$ at zero.

c) Let $ S_n\equal{}\sum_{k\equal{}1}^{n\minus{}1}\alpha_k$. From b) we have $ P_n(x)\equal{} \prod_{k\equal{}1}^{n\minus{}1}(1\minus{}x\alpha_k)\equal{}1\minus{}xS_n \plus{}\underbrace{\ldots}_{higher\;orders}$, i.e. $ P_n'(x)\equal{}\minus{}S_n$ is valid. Away from the (potential) singularities we have $ P_n(\sin^2(\theta))\equal{}g(\theta)$, this yields

$ P_n'(\sin^2(\theta))\equal{}\frac{2n\cos(2n\theta)\sin(2\theta)\minus{}2\cos(2\theta)\sin(2n\theta)}{n\sin^3(2\theta)}$ and applying L'Hospital we compute

$ P_n'(0) \equal{}\lim_{\theta\rightarrow 0}\frac{\minus{}4n^2\sin(2n\theta)\sin(2\theta)\plus{}4\sin(2\theta)\sin(2n\theta)}{6n\sin^2(2\theta)\cos(2\theta)} \equal{}\lim_{\theta\rightarrow 0}\frac{(\minus{}4n^2\plus{}2)\sin(2n\theta)}{3n\sin(4\theta)}$

i.e. $ S_n\equal{}\minus{}P_n'(0)\equal{}\frac{2n^2\minus{}2}{3}$ and the proof is complete.

Similar problem[url=http://reflections.awesomemath.org/2007_6/MR_6_2007_problems.pdf]O71[/url]
Is the problem famous?


geometry
ratio
analytic geometry
calculus

Find the maximum area of a square placed within a regular hexagon with side length $1$. (Placed, i.e. all four vertices may or may not be touching the hexagon, but the whole of the square is within the hexagon.)

[quote="7h3.D3m0n.117"]umm then the area of the square varies...unless I read your question wrong?[/quote]

Yes, the area will vary. Find the [i]maximum[/i] area possible. :wink:

[quote="7h3.D3m0n.117"]umm then the area of the square varies...unless I read your question wrong?

edit: sorry didn't see the maximum...

maximum area is 1[/quote]

The maximum area is not 1. Think about it.... that would mean that  one of the square's sides would be a side of the hexagon; yet you could easily move the square up a bit and expand it, keeping it within the hexagon.

(If you couldnt understand my wording... the hexagon has side length 1, not the square.)

[quote="ajai"]How about $\sqrt{6}$?[/quote]

There's no point in just giving answers like this, especially if you are unsure about them. Give us your reasoning, and show that it is the maximum.

http://s126.photobucket.com/albums/p108/ckck07/?action=view&current=hexagonwithsquare.jpg

[hide]OK so connect a vertice with the midpoint of the opposite side as shown in  the picture. Then do the same for the other side. If you see it you can see that connecting the top and bottom forms a square. Then seeing that the diagnol is the longest diagnol of the hexagon, then it is $2$.(because the longest diagnol is twice the side length).So the side length is going to be $sqrt*2$. The area will then be $2$. [/hide]
(Figure not drawn to size)
This is also the maximum because the square can't be any bigger,because the longest diagnol of the hexagon is also the  diagnol of the square.
Not really sure of my answer.

[quote="ckck"]http://s126.photobucket.com/albums/p108/ckck07/?action=view&current=hexagonwithsquare.jpg

[hide]OK so connect a vertice with the midpoint of the opposite side as shown in  the picture. Then do the same for the other side. If you see it you can see that connecting the top and bottom forms a square. Then seeing that the diagnol is the longest diagnol of the hexagon, then it is $2$.(because the longest diagnol is twice the side length).So the side length is going to be $sqrt*2$. The area will then be $2$. [/hide]
(Figure not drawn to size)
This is also the maximum because the square can't be any bigger,because the longest diagnol of the hexagon is also the longest diagnol of the square.
Not really sure of my answer.[/quote]
HOw does a square have a longest diagonal? Btw, that's not a square in your picture.

http://s126.photobucket.com/albums/p108/ckck07/?action=view&current=Hexagonwithsquare2.jpg
Ughh.. I got a messy answer... But im more positive with this one. [hide]So connect the vertices to the opposite side so it will cut the side into a ratio of 3:1. Then angle BXY will equal 90. So will angle CZY. Then angle CYZ and angle ZCY will equal 45. Cy equals 1/4. So ZY will equal sqrt.*2/8. Since CZ=CY use Pythagoroem to find ZD. ZD will equal $sqrt.*62/8$. So the side length will equal $sqrt.*62/8+sqrt.*2/8$. Sqauring it results in $1+4sqrt*31/64$.[/hide]

I'm pretty sure the maximum area is $12-6\sqrt3$.

I have two methods - the brutal coordinate geometry method and the calculus method.


Suppose that $ \log_xa\plus{}\log_ya\equal{}4\log_{xy}a$. Prove that $ x\equal{}y$

Let $ m = \log_ax\implies\frac1m{\log_xa}$ and $ n = \log_ay=\frac1${\log_ya} so that

$ m + n = \log_ax + \log_ay = \log_a{xy}\implies\frac1{m+n}=\log_{xy}a$ and

$ \frac1m + \frac1n = \frac {m + n}{mn} = \frac4{m + n}$

$ \implies(m + n)^2 = m^2 + 2mn + n^2 = 4mn$

$ \implies m^2 - 2mn + n^2 = 0\implies(m - n)^2 = 0$

$ \implies m = n\implies\log_a x = \log_ ay\implies x = y$.


geometry
ratio
LaTeX
Pythagorean Theorem

[hide]
Call the radius of circle N x. Thus, in triangle MON, MN=6+x, NO=12-x, and OM =6. NOM is a right angle, so using the pythagorean theorem we get $(6+x)^{2}=(12-x)^{2}+36$. Solving for x yields x=4, so the area is $16\pi$[/hide]

[quote="mathgeniuse^ln(x)"]Okay, that is what I got too.

Extension

[hide]The radius of that circle will always be $\frac{1}{3}$, of the original circle.

Prove this.[/hide][/quote]
I'll use the picture from the problem for the proof.
[hide]Call the radius of circle O $a$ (the radius of circle M is then $\frac{a}{2}$. We can name the radius of circle N $x$, then name the sides of triangle MON. MN=$\frac{a}{2}+x$, NO=$a-x$, and OM=$\frac{a}{2}$. Applying the Pythagorean Theorem yields $(\frac{a}{2}+x)^{2}=(a-x)^{2}+(\frac{a}{2})^{2}$ Solving for x in terms of a gets $3ax=a^{2}$, or $x=\frac{a}{3}$[/hide]

[quote="mathgeniuse^ln(x)"]Okay, that is what I got too.

Extension

[hide]The radius of that circle will always be $\frac{1}{3}$, of the original circle.

Prove this.[/hide][/quote]
Another way you could "show" this (but it might not be a rigorous proof) is to notice that as the radius of the large circle changes by some ratio, the radius of the smallest circle will change by the same ratio. In other words, the ratio between the two radii is constant.

[quote="bpms"][quote="mathgeniuse^ln(x)"]Yes, you are right, that is true.

But easyaspi did it correctly, so yeah.[/quote]
You could also notice that circle n and circle o are similar circles.[/quote]
Two things:
1. Circles are always similar.  :roll:  I think you meant to say that their radii have a constant ratio.
2. I just posted that two posts above yours.  :wink:

[quote="anirudh"][quote="nebula42"]Consider a circle of radius $12$ cm with center $O$. Let $AOB$ and $COD$ be two perpendicular diameters of this circle. Consider the two circles of diameters $AO$ and $OB$, respectively (which are tangent to the circle with diameter $AB$). Now construct two circles whose centers are on the segment $CD$ and are tangent to the other three circles. How many square centimeters are in the area of the smallest circle (the one with center $N$)? Express your answer in terms of $\pi$.

[img]http://mathcounts.saab.org/mcgif/p74.gif[/img][/quote]

[hide]Let x be the radius of one of the smallest circle and N be its ceneter. Let M be the center of the circle of diameter AO. Notice the:
ON=19-x, OM=19/2 and MN= x + R/2. In the right triangle MON, one can write:
(19/2 + x)^2 = 19^2/4 + (19 - x)^2.
Solving for x gives x=19/3. Hence the area is:
3.14 * (19/3)^2[/hide]

[size=75][color=darkred]Please hide your answers and (if possible) try to use LaTeX to make your solutions easier to read.
-nebula42[/color][/size][/quote]
Sorry, that's not correct, but it was almost correct. Note that the radius of the large circle is 12, not 19.  :wink:
Also, express your answer in terms of $\pi$ instead of rounding.


A plane leaves an airport from the runway 4 RIGHT, whose bearing is N40degreesE. After flying for 1/4 of a mile, the pilot requests permission to turn 90 degrees and it is given to him. After the plane goes 2 miles in this direction, what bearing should the control tower use to locate this plane?

Might make a diagram later but...

$ b.[40$°$ +tan^{-1}(\frac{2}{0.25})]$
$ \approx b.122$°$ 52'$


algebra
function
domain
logarithms
quadratics
ARML
quadratic formula

1. If $ x$ and $ y$ satisfies $ x^8 \plus{} x^6 y^2 \equal{} (3200000)^2$ dan $ xy \equal{} 1200$, find the value of $ 2x \minus{} 3y$.

2. If $ \frac {(0.12)^4 \text{ }(0.243)^6}{(1.8)^{10}} \equal{} \frac {2^a3^b}{5^c}$, find the value of $ 2a \minus{} b \plus{} 2c$.

3. Calculate the value of $ \left[ a^{\frac {5}{6}}\left( \frac {1}{ab} \right) ^{\frac {1}{3}} \times \left( \frac {ab^{\frac {1}{2}}}{a^{\frac {1}{2}}b^{ \minus{} \frac {2}{3}}} \right)^{ \minus{} \frac {1}{2}} \right]^{\frac {3}{4}}$ if $ a \equal{} 64$ and $ b \equal{} 81$.

4. Find the domain of the function $ f(x) \equal{} \frac {\log (2x^2 \minus{} 9x \plus{} 9)}{\log(x^2 \minus{} x \plus{} 1)}$.

5. The functions $ f(x) \equal{} x^3 \plus{} px^2 \plus{} 2qx \minus{} 24$ and $ g(x) \equal{} x^3 \minus{} 8x^2 \plus{} 19x \minus{} 3r$ have a same linear factor. If $ (x \minus{} 2)$ and $ (x \minus{} 4)$ divides $ f(x)$ but not $ g(x)$, then find all the solutions to $ g(x) \equal{} 0$.

6. It is known that $ f(x \plus{} 2) \equal{} 2x \plus{} 5$. If $ (f\text{ o }g)(x) \equal{} 2x^2 \minus{} 1$ and $ (h\text{ o }f)(x) \equal{} 3 \minus{} 2x$, find $ (g\text{ o }h)(x)$.

7. If $ a \equal{} \log_{245} 175$ and $ b \equal{} \log_{1715} 875$, calculate $ \frac {1 \minus{} ab}{a \minus{} b}$.

8. The product of two real numbers is $ 3$. One of them is $ \left( \frac {1}{2}\sqrt {5} \minus{} \sqrt {\frac {\sqrt {5} \minus{} 1}{\sqrt {5} \plus{} 1}} \right) ^2$. Find the other number.

hello, to 1),setting $ y\equal{}\frac{1200}{x}$ in the equation $ x^8\plus{}x^6y^2\equal{}3200000^2$ we get $ x^8\plus{}x^41200^2\equal{}3200000^2$, let $ x^4\equal{}t$ and solving the quadratic equation $ t^2\plus{}1200^2t\equal{}3200000^2$ then we get $ t_1\equal{}\minus{}4000000$ or $ t_2\equal{}2560000$ from $ x^4\equal{}2560000$ we get the following solutions
$ x_1\equal{}\minus{}40,y_1\equal{}\minus{}30$, $ 2x\minus{}3y\equal{}10$
$ x_2\equal{}\minus{}40i,y_2\equal{}\minus{}30i$,$ 2x\minus{}3y\equal{}10i$
$ x_3\equal{}40i,y_3\equal{}30i$,$ 2x\minus{}3y\equal{}\minus{}10i$
$ x_4\equal{}40,y_4\equal{}30$, $ 2x\minus{}3y\equal{}\minus{}10$.
You can consider the case $ x^4\equal{}\minus{}4000000$ analogously.
Sonnhard.

hello to your second problem , calculating $ \frac{\left(\frac{12}{100}\right)^4\cdot\left(\frac{243}{1000}\right)^6}{\left(\frac{18}{10}\right)^{16}}\equal{}\frac{4782969}{40000000000000000}\equal{}\frac{2^{\minus{}18}3^{14}}{5^{16}}$ so we find $ a\equal{}\minus{}18,b\equal{}14,c\equal{}16$ and $ 2a\minus{}b\plus{}2c\equal{}\minus{}36\minus{}14\plus{}32\equal{}\minus{}18$.
Hm, what a problem!
Sonnhard.

oh I guess I screwed up on #1... the quadratic formula gives $ x^4$ not $ x$.  But i think Sonhard has the signs backwards - $ x^4 \equal{} 4000000, \minus{} 2560000$ and not the other way around.  The problem can easily be solved from there.

#8 note that $ \frac {\sqrt {5} \minus{} 1}{\sqrt {5} \plus{} 1} \equal{} \frac {(\sqrt {5} \minus{} 1)^2}{4} \equal{} \bigg(\frac {\sqrt {5} \minus{} 1}{2}\bigg)^2$ so we have $ \bigg(\frac {1}{2}\sqrt {5} \minus{} \big(\frac {\sqrt {5} \minus{} 1}{2}\big)\bigg)^2 \equal{} \frac {1}{4}$ so the other number is $ \boxed{12}$.

hello, then put $ x\equal{}40,y\equal{}30$ in the equation system $ x^8\plus{}x^6y^2\equal{}3200000^2,xy\equal{}1200$.
Sonnhard.

[b]#6[/b] letting $ x\equal{}y\minus{}2$, we have $ f(y)\equal{}2(y\minus{}2)\plus{}5 \equal{} 2y\plus{}1$.  Now $ f(g(x)) \equal{} 2g(x)\plus{}1 \equal{} 2x^2\minus{}1$ so $ g(x) \equal{} x^2\minus{}1$.  Now $ h(f(x)) \equal{} h(2x\plus{}1) \equal{} 3\minus{}2x$.  Letting $ x\equal{}1/2y\minus{}1/2$ we have $ h(y) \equal{} 3\minus{}2(1/2y\minus{}1/2) \equal{} 3\minus{}y\plus{}1 \equal{} 4\minus{}y$.  Hence $ g(h(x)) \equal{} (4\minus{}x)^2\minus{}1 \equal{} x^2\minus{}8x\plus{}15$.

[b]#5[/b] $ x\minus{}2|f(x)$ and $ x\minus{}4|f(x)$ implies $ f(2)\equal{}f(4)\equal{}0$ which yields the equations $ 4p\plus{}4q\minus{}16\equal{}0$ and $ 16p\plus{}8q\plus{}40\equal{}0$ so $ p\plus{}q\equal{}4$ and $ 2p\plus{}q\equal{}\minus{}5$ implies $ p\plus{}4\equal{}\minus{}5$ and $ p\equal{}\minus{}9$ so $ q\equal{}13$.  Hence $ f(x)\equal{}x^3\minus{}9x^2\plus{}26x\minus{}24$.  Synthetic division yields that $ f(x)\equal{}(x\minus{}2)(x\minus{}3)(x\minus{}4)$.

Therefore, $ g(3)\equal{}0$ so $ 27\minus{}8*9\plus{}19*3\minus{}3r\equal{}0$ so $ r\equal{}4$.  Thus $ g(x)\equal{}x^3\minus{}8x^2\plus{}19x\minus{}12$ so by synthetic division by $ x\minus{}3$ we have $ g(x)\equal{}(x\minus{}3)(x^2\minus{}5x\plus{}4) \equal{} (x\minus{}3)(x\minus{}4)(x\minus{}1)$.  But this is a contradiction since $ x\minus{}4$ does not divide $ g(x)$.

[quote="Dr Sonnhard Graubner"]...we get $ x^8 \plus{} x^41200^2 \equal{} 3200000^2$,... [/quote]
The middle term** is incorrect.  The constant is to be placed
immediately to the left of the variable term (inside of
parentheses or not), or it is to be put inside of grouping
symbols (where you already have it to the right of the
variable term).
There is no multiplication between the variable term** and the
constant after it, as you have typed it.

Okay as:

$ x^8 \plus{} 1200^2x^4 \equal{} 3200000^2$ or

$ x^8 \plus{} (1200^2)x^4 \equal{} 3200000^2$ or

$ x^8 \plus{} x^4(1200^2) \equal{} 3200000^2$

#4 $ f(x) \equal{} \frac {\log (2x^2 \minus{} 9x \plus{} 9)}{\log(x^2 \minus{} x \plus{} 1)}$

$ 2x^2\minus{}9x\plus{}9\equal{}2(x\minus{}3)(x\minus{}3/2)>0\implies x\in(\minus{}\infty,\minus{}3/2)\cup (3,\infty)$
$ x^2\minus{}x\plus{}1>0\forall x\in\mathbb{R}$

[b]#4[/b]

Not quite mathwizard.  The second root is a positive $ 3/2$, so your answer should be $ (\minus{}\infty,3/2) \cup (3, \infty)$

But there is another issue we need to avoid: division by zero.  So we have $ \log(x^2\minus{}x\plus{}1) \neq 0$ which means $ x^2\minus{}x \neq 0$ so $ x \neq 0,1$.

Thus the complete answer is $ (\minus{}\infty, 0) \cup (0,1) \cup (1, 3/2) \cup (3, \infty)$

[b]7[/b] first note that $ 175 \equal{} 5^2 7$, $ 245 \equal{} 5 \cdot 7^2$, $ 1715 \equal{} 5\cdot 7^3$ and $ 875 \equal{} 5^3 7$.

Now let's convert all the logarthms to a common base.  We'll use base $ 5$ for convienence.  Now we have $ a\equal{}\frac{\log_5 175}{\log_5 245} \equal{} \frac{2\plus{}\log_5 7}{1\plus{}2\log_5 7}$ and $ b \equal{} \frac{\log_5 875}{\log_5 1715} \equal{} \frac{3\plus{}\log_5 7}{1\plus{}3\log_5 7}$.

For simplicity sake, let $ x\equal{}\log_5 7$.  Thus we have:
$ \frac{1\minus{}ab}{a\minus{}b} \equal{} \frac{1\minus{}\frac{(3\plus{}x)(2\plus{}x)}{(1\plus{}2x)(1\plus{}3x)}}{\frac{2\plus{}x}{1\plus{}2x} \minus{} \frac{3\plus{}x}{1\plus{}3x}}$
$ \equal{}\frac{(1\plus{}2x)(1\plus{}3x) \minus{} (3\plus{}x)(2\plus{}x)}{(2\plus{}x)(1\plus{}3x)\minus{}(3\plus{}x)(1\plus{}2x)}$
$ \equal{}\frac{6x^2 \plus{} 5x \plus{} 1 \minus{}(x^2\plus{}5x\plus{}6)}{(3x^2\plus{}7x\plus{}2)\minus{}(2x^2\plus{}7x\plus{}3)}$
$ \equal{} \frac{5x^2\minus{}5}{x^2\minus{}1}$
$ \equal{}\boxed{5}$

since $ \log_5 7 \neq \pm 1$

Yeah right forgot about denominator being zero... nice catch...

thanks.  It reminded me of a very similar ARML problem from last year.

btw, in your solution, when you messed up the sign, the interval you accidentally cut out included the division by 0 cases.  I wasn't sure if you ignored division by zero or just thought that it was already not an issue.

Oh, it was a stupid typo (yes i often make those...weren't paying attention when typing)

btw, how come i couldn't find arml problems in the resource section???  :huh:

idk but this page has last year's contest:
[url]http://arml.com/2008contest.php[/url]

The problem I was referring to is #7 on the individual round.  It's a bit more complicated than our problem here, but along the same lines.

Q7: Let $ log_{10} 5 \equal{} x$, $ log_{10} 7 \equal{} y$, then
$ a \equal{} log_{(7^2 \times 5)} (5^2 \times 7) \equal{} \frac {2x \plus{} y}{2y \plus{} x}$
$ b \equal{} log_{(7^3 \times 5)} (5^3 \times 7) \equal{} \frac {3x \plus{} y}{3y \plus{} x}$

So $ \frac {1 \minus{} ab}{a \minus{} b} \equal{} \frac {(2y \plus{} x)(3y \plus{} x) \minus{} (2x \plus{} y)(3x \plus{} y)}{(2x \plus{} y)(3y \plus{} x) \minus{} (3x \plus{} y)(2y \plus{} x)} \equal{} \frac {5y^2 \minus{} 5x^2}{y^2 \minus{} x^2} \equal{} 5$

hello, Q8, we have $ \frac{1}{2}\sqrt{5}\minus{}\sqrt{\frac{(\sqrt{5}\minus{}1)^2}{5\minus{}1}}\equal{}\frac{1}{2}\sqrt{5}\minus{}\frac{1}{2}(\sqrt{5}\minus{}1)\equal{}\frac{1}{2}$, so we get
$ a\cdot\frac{1}{4}\equal{}3 \Leftrightarrow a\equal{}12$.
Sonnhard.


algebra
polynomial
trigonometry
group theory
algebra unsolved

Solve the system of equation:

$ \begin{cases} xy^2 \minus{} 2y \plus{} 3x^2 \equal{} 0 \\
x^2 \plus{} 8y^2 \equal{} 12 \end{cases}$

P/S: Counting down to Friday the $ 13^{\text{th}}$

hello, plugging $ y^2\equal{}\frac{3}{2}\minus{}\frac{x^2}{8}$ in the first equation we have $ y\equal{}\frac{3}{4}x\plus{}\frac{3}{2}x^2\minus{}\frac{x^3}{16}$, inserting this in equation two we can compute $ x$ solving the polynomial
$ x^6\minus{}48x^5\plus{}552x^4\plus{}576x^3\plus{}176x^2\minus{}384\equal{}0$, we get two real solutions and four complex solutions.
Sonnhard.

But the two real roots of the equation you mentioned are all irrational. We have to express them in the closed form. But, may be, by the Group Theory of Galois, the $ 6^{\text{th}}$ grade equation can not be solved generally. :(

Yes, this equation only have Complex root.

I have a hint for this problem, you can try it.

System $ \Longleftrightarrow \begin{cases} xy^2 \minus{} 2y \plus{} 3x^2 \equal{} 0 \\
\frac{x^2}{12} \plus{} \frac{2y^2}{3} \equal{} 1 \end{cases}$

Let $ x\equal{}2\sqrt{3}\sin{\Phi}$, $ y\equal{}\frac{\sqrt{3}}{\sqrt{2}}\cos{\Phi}$.

Hence we get equation: $ 3\sqrt{3}\sin{\Phi}\cos^2{\Phi}\minus{}\sqrt{6}\cos{\Phi}\plus{}36\sin^2{\Phi}\equal{}0$


I've been working with a group of smart middle school and younger high school students, and I'm looking for sources of rougly AMC-10 level math problems to use with them.  So far I've been using the easy parts of old AMC12s, but I'd like to diversify.  Does anyone have suggestions for other problem sources I could use?

im going to have to vouch for my state's competition, the Michigan Mathematics Prize Competition.

The problems from the part I of the contest are about AMC - 10 level, sometimes easier, sometimes harder.

http://www.delta.edu/math/mmpc

If you go to previous exams, there are over 30 years of past exams, and at 40 problems a test, there is a lot of selection.

North Carolina math contest: [url]http://192.154.43.167/goebel/statecon/tests/tests.htm[/url]

South Carolina math contest: [url]http://www.math.sc.edu/contest/problems.html[/url]

Lehigh University contest: [url]http://www.lehigh.edu/~dmd1/hs.html[/url]

The questions towards the end of some of those exams might be too challenging for the AMC-10 level, but there are good problems in those tests.

Canadian Open Mathematics Challenge

http://www.cms.math.ca/Competitions/COMC/

In my experience, a lot of these are easy AMC-level questions.  They get a little harder towards the end.

Thanks a lot all!  I'll have to sort through and see what looks good.