geometry
perpendicular bisector
geometry proposed

[color=darkblue]Let $ABCD$ be a concave quadrilateral where $90^\circ<\measuredangle\ D<180^\circ$. Assume that $\overline{AD}=\overline{CD}=\overline{BC}$ and furthermore $\measuredangle\ C=2\cdot\measuredangle\ A$. If $\measuredangle\ A=\alpha$, then prove that $\measuredangle\ B=120^\circ-\alpha$[/color]

Let bisector of angle $\angle BCD$ meet $AB$ at $E$
$CE$ meets $BD$ at $F$.
The perpendicular of $AB$ through $D$ meets $AB$ at $M$
we have $BC=CD$ so $CF$ is the perpendicular bisector of $BD$. Then $\angle CEB=\angle CED$ $(1)$
$\angle DCF=1/2 \angle BCD=\angle DAM$ and $CD=DA$ and $\angle CFD=\angle AMD=90$ so triangle $CDF$ and $ADM$ are congruent
so $DF=DM$
So $D$ is on the bisector of angle $\angle CEA$.
Thus $\angle CED=\angle DEA$ $(2)$
From (1) and (2) we have $\angle CEB=60$
$\angle CBE=180-\angle CEB-\angle BCE=120-\alpha$ $q.e.d$


Duke college Harvard Stanford Princeton MIT blogs

I know I'm only a 7th grader, but today I took the ACT and I was just thinking about what college I should go to. I am a bit advanced in math and science, so what college would anyone reccomend me to go to? I thought about Duke, because I heard that the math courses over there were good, but I have also heard about other ones too. Help?  :o

(I know I'm young. Everyone was staring at me today as I walked into the room. I'm also only 4 ft 10.... Imagine what kind of sight that is. Everyone was like: "Holy... Azn little kid prodigy... " And I was like... "No.. I'm not!! My mom pushed me into this!!!")

Don't think about it.  You have at [b]least[/b] three more years to enjoy life before you need to worry about making this decision.

Hehe.. You don't know how my mom is... She's always telling me to look up to other people when I'm actually better than them... SIGH... My mom's telling me to study act for 25/7 per week and blech. She's always worried about our financial stuff... Ugh. It's tiring when you have an epileptic sister eh? All those medical bills. A burden enough to  put a ton on my parents back.  :(

Based on College Board's stats, Duke's mid-50% range for SAT/ACT scores don't appear to be very high, at least not high enough that you'd have to stress over them. Personally my SAT scores were in the bottom 50% of most of the schools I applied to and my results weren't bad. Really, there's more productive ways to use your time than for the few hours on a Saturday when you take the SATs/ACTs.

Don't worry, you have 5 years to go before you graduate, things can change a lot within these 5 years. 5 years is a long time. ...Thinking back, I wasn't even interested in math in 7th grade, it wasn't until 8th grade did I really start becoming interested. 5 years ago, I could not have pictured myself where I am now.

You're parents shouldn't be that worried about financial stuff that much, first, because you're in 7th grade; second, because schools will make sure you can afford it (financial aid officers are people too)

But if you really want to think of colleges, please don't get the attitude of "I have to get into this school." Be flexible. I would say good math schools would be: Harvard, Stanford, Princeton, Duke, and MIT, but that's just my opinion. Just a note though, Duke's math department seems to be more leaning towards mathematical physics than pure math.

Personally I think it is a waste of time to spend the whole day doing ACT/SAT prep, people with 36/2400s get rejected from top schools all the time. I would say studying at most 1 hour a day is enough, you have 5 years to get a good score. Spend time doing something else that you like.

studying math or something else productive will help you way more!!

Agreed.  Maxing out your ACT/SAT isn't doing much for you (once you get a 'good enough' score just be happy and move on with life).  Getting a perfect ACT score doesn't earn you any money.  But starting your own company, website, or blog?  These are considerably more challenging but they can be very rewarding in terms of money and college admissions.  You [i]probably[/i] wont start the next google, but even so, you'll come away having learned a lot and would have had fun along the way (hopefully).

That said, I don't expect many people to start their own business in high school.  Getting into web design is certainly more doable.  But, still, these are just two examples.  There's plenty of things to do that are more fulfilling and useful than ACT studying.  Find one or two you like and run with them.

Er.. I really stink at the Reading and Science section, but rule at the essay/writing/math sections. I got a 22 in my 6th grade year. Is that good?  :maybe:  :(

What's the matter with you? Why on earth would you take it in the 6th grade?

CTY and some other programs use the SAT and ACT to determine eligibility; that's why I took the SAT in 6th grade, for example.  If you can handle it, why not?  Just don't get stressed out about it.

@lifesapain: What you mean I wanted to take it? My mom made me! She made me sit down for like 5 hours and then do it!  :mad:

We don't get to choose our parents.  That's life.

On the other hand, the decision to ask about college on this forum was your own decision, wasn't it?  (At least I hope so.)  So the general advice still stands:  Chill out.

@1/(ln x): So you're saying that Duke leans toward science/math more than pure math? What is pure math like?

pure math means that the college focuses more on mathematical theories and equations rather than say mathematical applications on things like physics or engineering
and about taking the act early
you really can't tell what to do with the results until you get them back
most people did i know did not do that well in 7th grade on the ACT/SAT
i only got a 27
if you want to get into an ivy league college easily, you'll need at least a 33 or  2000 on the SAT

MIT MIT MIT MIT

Omg.. 27? Hm.. I had higher expectations of myself. I'm scared now. Study study study!!  :(  :noo:  :weightlift:

are you being satirical?
i can't tell
g2g
i have to go to judo

Don't use big words around me!! I'm kind of stupid... :(

i haven't left yet
what do you mean, big words
expectations is a much larger word than
satirical

i feel dumb asking this, but what's act?
all i know is a 2000 sat alone wont cut it. you need to have a great eq, be actively involved in stuff like music and sports, yada yada yada
thats why the ppl with 2400 sats get rejected - they're nerds.
(btw many of my friends scored higher than 2000-2060,2070,2080,2150,2250. i fell at the bottom with a 2040)

[quote="funtwo"]i feel dumb asking this, but what's act?[/quote]
http://www.actstudent.org/

[quote="funtwo"]thats why the ppl with 2400 sats get rejected - they're nerds.[/quote]
I have a 2400 SAT and I was accepted to every school I applied to except Brown :huh:

I think someone explained something incorrectly to you.  The people with 2400 SATs and [b]nothing else[/b] get rejected from [b]top schools.[/b]  Both of those qualifiers are necessary.  And lots of those top schools happen to love nerds.

oops that's what i meant t0rajir0u (about the only having a 2400 sat)
i should've been more clear
simo14, i think t0rajir0u is in mit right now

Huh?????????????????????? t0rajir0u is in College???????? OMGOMGOMG.... No wonder he's so smart! I thought he was an ordinary high schooler... omg. MIT? Good job!  :)  :P

[quote="jonathanchou711"]I thought he was an ordinary high schooler...[/quote]
Well, I was last year.  What does that have to do with anything?  I know high schoolers smarter than college students.

Oh.. So technically, you're in 13th grade!  :P

Just wanted to clear up a misconception about the Duke math department (and math departments in general).  The pure math vs applied math thing is tricky.  It is true that some departments are mostly pure math with a smaller applied math group (e.g. Stanford), while at other schools there is a more even (e.g. MIT, Duke), and occasionally the applied math group is much bigger (e.g. NYU).  It is also possible for a school's pure math group to have a better reputation than the applied math group, or vice verse.  At Harvard, the two groups don't even share the same department.  However, 95% of the time these issues have no real relevance to a high school student deciding where to go for college.

Regarding Duke, the statement about an emphasis on mathematical physics is not quite accurate.  However, it is true that the Duke math department gives more courtesy appointments to physicists (irrelevant) and is one of the few math departments that has actually had bona fide string theorists working *IN* the math department (relevant).


limit
linear algebra
matrix
function
real analysis
real analysis unsolved

Let $ a_{n}= \frac{b_{n+1}}{b_{n}}$. The recursive formula for $ a_{n}$ is given by $ a_{n}= \frac{a_{n-1}-2}{2a_{n-1}-3}$.

To evaluate $ a_{n}$, let $ A = \begin{pmatrix}1 &-2 \\ 2 &-3\end{pmatrix}$. Then $ A^{n}= (-1)^{n-1}\begin{pmatrix}2n-1 &-2n \\ 2n &-2n-1\end{pmatrix}$ and $ a_{n}= \frac{(2n-1)\alpha-2n}{2n\alpha-(2n+1)}$ where $ \alpha = b_{1}\in (0, 1)$.

Now let $ f(x) = \frac{(x-1)\alpha-x}{x\alpha-(x+1)}= \frac{(1-\alpha)x+\alpha}{(1-\alpha)x+1}$. Easy observation shows that function $ f$ increases for $ x>0$. So $ f(k) < f(k+1) < f(k+2)$ holds for every nonnegative integer $ k$.

Also, $ b_{n+1}= b_{0}a_{0}\cdots a_{n}= f(0)f(2)\cdots f(2n)$ and

$ f(0)^{2}f(1) f(2) \cdots f(2n-1)f(2n) \leq b_{n+1}^{2}\leq f(0)f(1)f(2) \cdots f(2n)f(2n+1)$

or simply

$ \frac{\alpha^{2}}{(2n+1)-2n\alpha}\leq b_{n+1}^{2}\leq \frac{\alpha}{(2n+2)-(2n+1)\alpha}$

This implies that $ b_{n+1}^{2}\to 0$ as $ n \to \infty$. So $ \lim_{n\to\infty}b_{n}= 0$.


geometry
geometric transformation
reflection
calculus
projective geometry

Let $ A_{1}A_{2} A_{3} A_{4}$ be a quadrilateral whose diagonals intersects at $ N$ , and let pairs of
 opposite sides meet in $ P$ and $ Q$; let $ B_{1}$ ,$ B_{2}$ ,$ B_{3}$ ,$ B_{4}$ be the points where the sides of the quadrilateral intersect the lines$ NP$and$ NQ$. Let the sides of$ A_{1}A_{2} A_{3} A_{4}$intersect the sides of the inscribed quadrilateral$ B_{1} B_{2} B_{3} B_{4}$in$ M_{1} , . . . , M_{8}$ . Show that
 (a) Lines $ M_{1} M_{5} , M_{2} M_{6} , M_{3} M_{7} , M_{4} M_{8}$ pass through $ N$ .
 (b) Lines $ M_{2} M_{3}$ and $ M_{6} M_{7}$ pass through $ P$ ; lines $ M_{1} M_{8}$ and $ M_{4} M_{5}$ pass through $ Q$

Make a CP such that the quadrilateral becomes a square ...

and then the prob is almost killed...

can anybody give me some good and simple reading material for this?

The way you talk of attacking a problem reflects on your aggresiveness in problem solving.. :D

Anyways @amaron: What do you mean reading material for this? Do you want some general book for pure geometry or are you asking for problems for Olympiads. Be more specific.

No,I meant notes for projective geometry.

What is CP?

(Im so dead for this year's inmo....what the hell is projective geometry??? i know only pure geometry)

CP is central projection...u need not worry pue geometry is enuf until unless u belong to 12th class :D

So,anybody has notes for projective geometry?any good ebooks that just give basics and some applications?

PS:What is the cutoff in INMO for XII  students generally?
     also,is use of calculus allowed in INMO?

the INMO cut-off for XIIthers is max 55...
(i.e. if you score above 55 -- 3 of the higher marked problems absolutely perfect and a good attempt at the fourth you will be through...)

Use of calculus is permitted but you can never be sure whether the person who checks the answer sheet gives you marks or not...
it completely depends on the discretion of the examiner..

So, if you have a solution without calculus, write that down...
(use calculus as a last resort...)

http://www.bprim.org/inmore06.php

2 Years ago: Cutoff was 42. Dunno about last years cutoffs.

last year looking at the mark lists...and few posts here in AoPS my guess is it should be around 44-48

I guess a score of 45-50 should clearly steer you through.


logarithms
limit
floor function
real analysis
real analysis unsolved

Let $ \{a_n\}$ be a sequence of natural numbers such that $ lim_{n \minus{}> \infty} \sqrt[n]{a_1a_2 \cdots a_n} \equal{} 1$.
Prove that $ lim_{n \minus{}> \infty} \frac{a_1\plus{}a_2\plus{}\cdots \plus{}a_n}{n} \equal{} 1$.

It is convenient to apply Stolz theorem here
[url]http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem[/url]

Supppose $ \{a_n\}$ is strictly increasing, (since finite number of repetition does not affect the convergence), then let
$ p \equal{} \sqrt [n]{n!}\implies \ln p \equal{} \frac {1}{n}\ln n!$, by Stolz theorem, we have $ \lim_{n\to\infty}\ln p \equal{} \lim_{n\to\infty} \ln n \equal{} \plus{} \infty$, $ p$ cannot converge to 1, we deduce that $ \lim_{n\to\infty}a_n \equal{} 1$.

Again by Stolz theorem,
$ \lim_{n\to\infty}\frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}{n} \equal{} \lim_{n\to\infty} a_n \equal{} 1.$

The assumption of monotonicity doesn't work; we can have infinitely many entries that aren't 1.
Let $ a_n\equal{}\begin{cases}2^{2k}&n\equal{}2^k\\1&\text{otherwise}\end{cases}$
Then $ \left(a_1a_2\cdots a_n\right)^{1/n}\equal{}2^{\frac{r^2\plus{}r}{n}}$, where $ r\equal{}\lfloor\log_2(n)\rfloor$. The exponent tends to zero, and $ \lim_{n\to\infty}\left(a_1a_2\cdots a_n\right)^{1/n}\equal{}1$.
Now, $ \frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\equal{}\frac1n\left(1\cdot (n\minus{}r)\plus{}\sum_{k\equal{}0}^r 2^{2k}\right)\equal{}\frac1n\left(1\cdot (n\minus{}r)\plus{}\frac{2^{2r\plus{}2}\minus{}1}{3}\right)$. We have $ n^2<2^{2\lfloor \log_2(n)\rfloor\plus{}2}\le 4n^2$, so the average $ \frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}$ is comparable to $ n$, and tends to $ \infty$.

The original statement is false.

Thx jimerry, I think the problem is wrong somewhere..
How about the reverse ? Is it true or is there any counter example ?

In the other direction, use the AM-GM inequality.


limit
integration
calculus
function
calculus computations

Let $ f:[0,+\infty),\quad f \in C^0 \textrm {and} \displaystyle \lim_{x \to +\infty} f(x)=0 $
Prove
\[
\displaystyle \lim_{x \to + \infty}\displaystyle \int_0^x f(x-t)g(t) \,dt=0
\]
with $ g:[0,+\infty) \to \mathbb {R},\quad g \in C^0. $

Is there something missing? :? Can't we take $g$ to be identically $1$, for instance?

Yes Myth, I forgot the limit. Sorry
\[
\displaystyle \lim_{x \to +\infty}\displaystyle \int_0^x f(x-t)g(t) \,dt=0
\]

Anyway, I realized what grobber's remark means. Even if you return "lim", problem is wrong.

Of course it is :). If $g$ is identically $1$, for instance, and $f$ takes only positive values, then that integral increases when $x$ does, so what did you realy mean, Nene?

Having $f$ and $g$ go to zero at infinity isn't enough. If you want a result here you're going to need some kind of integrability condition on at least one of the functions.

Let $f(x)=g(x)=\frac1{\sqrt{x+1}}.$

Then $f*g(x)=\int_0^x\frac{dt}{\sqrt{(t+1)(x-t+1)}}$

We can show that there is a constant $C>0$ such that $f*g(x)>C$ for all sufficiently large $x.$

Here's an example of a theorem that would be true:

Suppose $f\in C_0([0,\infty)).$ That is, $f$ is a continuous function that tends to zero at infinity.

Suppose $g\in L^1([0,\infty)).$

Then $\lim_{x\to\infty}\int_0^xf(x-t)g(t)\,dt=0.$

In my previous example, it's obvious that $g$ is not integrable.

Nené, your signature causes the width of the page to expand, which causes some readability problems for some of the posts on this topic. Is there any way you could edit that signature so that it goes on two lines and doesn't cause this problem?


analytic geometry
combinatorics unsolved
combinatorics

Let $ S$ be a finite subset of $ \mathbb{R}^n$ such that not all the point in $ S$ in the same line,And colored the point in $ S$ red or blue.A line pass though two red points of $ S$ called a red line, ;and a line pass though two blue points of $ S$ called a blue line.(Note that in this definition a line may be both a red and blue line),Prove:
There is a blue point not on any red line,or there is a red point not on any blue line.

I am sorry,can somebody help to delete the another same topic.(I have edit this one,so please don't delete this one)

Well, I have replied to the other one, so here's a copy of my comment.

The way you express it, it's wrong. Take points of coordinates $ (0,0), (0,1), (1,0)$ to be red, and points of coordinates $ (0,2), (0,3), (2,0), (3,0)$ to be blue. The the $ Ox$ and $ Oy$ axis are both red and blue lines, and all points lie on these two lines.

I am sorry,I think I mistake the problem,here is the right expression,can somebody help me?

Let $ S$ be a finite subset of $ \mathbb{R}^n$ such that not all the point in $ S$ in the same line,And colored the point in $ S$ red or blue.A line pass though two red points of $ S$ called a red line, ;and a line pass though two blue points of $ S$ called a blue line.(Note that in this definition a line may be both a red and blue line),Prove:
There is a red line contain no blue point,or there is a blue line contain no red point.


algebra
polynomial
geometry
system of equations

What is "determining the characteristic polynomial" and how do you do it? (I kind of understand what it is, I just need to learn how to do it).  For example, use the sequence 1, 1, 1, 3, 5, 9, 17, 31...

You can determine it only for [i]linear[/i] recursive equations, i.e. equations where only terms of the form $C_{k}a_{n+k}$ appear, where $C_{k}$ is a constant and $n$ is free index. Instead of every such term just write $C_{k}x^{k}$ and you get the characteristic equation.

So, if you have $a_{n+3}=4a_{n+2}+2a_{n+1}-5a_{n}$, then the c. eq. is $x^{3}=4x^{2}+2x-5$. For your example we'd have $a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$, giving $x^{3}=x^{2}+x+1$.

What's it used for? Solutions to this equation provide a quick way of solving the recurrence, but since the area is extensive, I recommend consulting online resources on the subject.

I'll do a direct closed form example.
For your example, we see that, as Farenhajt said, the characteristic polynomial is
$x^{3}-x^{2}-x-1=0$
Find the roots of this:
The general form of the equation will be
$a_{n}=A(R_{1})^{n}+B(R_{2})^{n}+C(R_{3})^{n}$
where $R_{1}$ is the first root, and so on...
Use the initial values of the series ($a_{1}= 1 ; a_{2}= 1 ; a_{3}= 1$) and set up a system of equations, and solve for constants A, B, and C. In this way, you can find the general equation. I'll work on getting the roots of this equation for you soon, and edit with the solution when I'm done.

Ok, I let mathematica grab the roots of this, and the general form equation is quite complicated. I'll give the roots here:
$R_{1}= \frac{1}{3}+\frac{1}{3}\sqrt[3]{19-3\sqrt{33}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$
$R_{2}=\frac{1}{3}-\frac{1}{6}(1+i\sqrt{3})(\sqrt[3]{19-3\sqrt{33}})-\frac{1}{6}(1-i\sqrt{3})(\sqrt[3]{19+3\sqrt{33}})$
$R_{3}=\frac{1}{3}-\frac{1}{6}(1-i\sqrt{3})(\sqrt[3]{19-3\sqrt{33}})-\frac{1}{6}(1+i\sqrt{3})(\sqrt[3]{19+3\sqrt{33}})$

Clearly, this isn't something you want to be doing by hand. Setting up a system of equations (using $a_{1}= 1, a_{2}= 1$...)
$1 = A(R_{1})^{1}+B(R_{2})^{1}+C(R_{3})^{1}$
$1 = A(R_{1})^{2}+B(R_{2})^{2}+C(R_{3})^{2}$
$1 = A(R_{1})^{3}+B(R_{2})^{3}+C(R_{3})^{3}$, solve for A, B, and C any way you want.
Then the final equation is simply
$a_{n}= A(R_{1})^{n}+B(R_{2})^{n}+C(R_{3})^{n}$

The coefficients on here are nasty to type out (four lines of mathematica, not going to bother typing them, but I'll put them up.)
I'll hide the Mathematica code for these:
[hide="A"] A = \!\(\((\(-162\)\ \[ImaginaryI]\ \@3\ \((\(-9\)\ \[ImaginaryI] + \@11)\) + \((
            627\ \[ImaginaryI] - 11\ \@3 + 180\ \@11 + 44\ \[ImaginaryI]\ \
\@33)\)\ \((
            19 + 3\ \@33)\)\^\(1/3\) + \((\(-354\)\ \[ImaginaryI] + 146\ \@3 \
- 153\ \@11 + 77\ \[ImaginaryI]\ \@33)\)\ \((19 + 3\ \@33)\)\^\(2/3\) + \((
            19 - 3\ \@33)\)\^\(2/3\)\ \((354\ \[ImaginaryI] + 146\ \@3 +
        153\ \@11 + 77\ \[ImaginaryI]\ \@33 + \((207\ \[ImaginaryI] +
              69\ \@3 + 99\ \@11 -
              31\ \[ImaginaryI]\ \@33)\)\ \((19 + 3\ \@33)\)\^\(1/3\))\) + \
\((19 - 3\ \@33)\)\^\(1/3\)\ \((\(-627\)\ \[ImaginaryI] - 11\ \@3 - 180\ \@11 \
+ 44\ \[ImaginaryI]\ \@33 + \((\(-207\)\ \[ImaginaryI] + 69\ \@3 - 99\ \@11 - \
31\ \[ImaginaryI]\ \@33)\)\ \((19 + 3\ \@33)\)\^\(2/
              3\))\))\)/\((3\ \@33\ \((1 + \((19 - 3\ \@33)\)\^\(1/3\) + \
\((19 + 3\ \@33)\)\^\(1/3\))\)\ \((\(-36\)\ \@11 + \((3\ \[ImaginaryI] + \@3)\
\)\ \((19 + 3\ \@33)\)\^\(1/3\) - \((\(-3\)\ \[ImaginaryI] + \@3)\)\ \((19 + \
3\ \@33)\)\^\(2/3\) + \((19 - 3\ \@33)\)\^\(2/3\)\ \((3\ \[ImaginaryI] + \@3 \
+ 2\ \((\(-3\)\ \[ImaginaryI] + \@3)\)\ \((
                      19 + 3\ \@33)\)\^\(1/3\))\) - \((19 - 3\ \@33)\)\^\(1/3\
\)\ \((\(-3\)\ \[ImaginaryI] + \@3 + 2\ \((3\ \[ImaginaryI] + \@3)\)\ \((19 +
                3\ \@33)\)\^\(2/3\))\))\))\)\)
[/hide]
[hide="B"]
B = \!\(\((\(-162\)\ \[ImaginaryI]\ \((\(-5\)\ \[ImaginaryI] + \@11)\) + \((
                  19 + 3\ \@33)\)\^\(1/3\)\ \((\(-335\) - 99\ \[ImaginaryI]\ \
\@3 + 96\ \[ImaginaryI]\ \@11 + 24\ \@33)\) + \((19 + 3\ \@33)\)\^\(2/3\)\ \
\((\(-350\) -
              39\ \[ImaginaryI]\ \@11 + 45\ \@33)\) + \((19 - 3\ \
\@33)\)\^\(1/3\)\ \((\(-335\) + 99\ \[ImaginaryI]\ \@3 + 96\ \[ImaginaryI]\ \
\@11 - 24\ \@33 + \((64 - 3\ \[ImaginaryI]\ \@11 - 3\ \@33)\)\ \((
            19 + 3\ \@33)\)\^\(2/
                3\))\) + \((19 - 3\ \@33)\)\^\(2/3\)\ \((\(-350\) - 39\ \
\[ImaginaryI]\ \@11 -
            45\ \@33 + \((19 + 3\ \@33)\)\^\(1/3\)\ \((64 - 3\ \[ImaginaryI]\ \
\@11 + 3\ \@33)\))\))\)/\((
              9\ \@11\ \((36\ \@11 - \((3\ \[ImaginaryI] + \@3)\)\ \((19 + 3\ \
\@33)\)\^\(1/3\) + \((\(-3\)\ \[ImaginaryI] + \@3)\)\ \((
              19 + 3\ \@33)\)\^\(2/3\) - \((19 - 3\ \@33)\)\^\(2/3\)\ \((3\ \
\[ImaginaryI] + \@3 + 2\ \((\(-3\)\ \[ImaginaryI] + \@3)\)\ \((19 + 3\ \
\@33)\)\^\(1/3\))\) + \((19 - 3\ \@33)\)\^\(1/3\)\ \((\(-3\)\ \[ImaginaryI] + \
\@3 + 2\ \((3\ \[ImaginaryI] + \@3)\)\ \((19 + 3\ \@33)\)\^\(2/3\))\))\))\)\)
[/hide]
[hide="C"]
C = \!\(\(\(1\/\(162\ \@11\)\)\((\(-54\)\ \@11 - 7\ \((
    3\ \[ImaginaryI] + \@3)\)\ \((19 +
      3\ \@33)\)\^\(1/3\) + \((\(-39\)\ \[ImaginaryI] +
            13\ \@3 - 9\ \@11 + 9\ \[ImaginaryI]\ \@33)\)\ \((19 + 3\ \@33)\)\
\^\(2/3\) + \((
      19 - 3\ \@33)\)\^\(2/3\)\ \((\(-39\)\ \[ImaginaryI] - 13\ \@3 - 9\ \@11 \
- 9\ \[ImaginaryI]\ \@33 - 2\ \((\(-3\)\ \[ImaginaryI] + \@3)\)\ \((19 +
          3\ \@33)\)\^\(1/3\))\) + \((19 - 3\ \@33)\)\^\(1/3\)\ \((7\ \((\(-3\
\)\ \[ImaginaryI] + \@3)\) + 2\ \((3\ \[ImaginaryI] + \@3)\)\ \((
                19 + 3\ \@33)\)\^\(2/3\))\))\)\)\)
[/hide]


Let $ 0<a<b<2a, b\not\equal{} a\left ( 1\plus{}\frac{\sqrt{2}}{2} \right )$, and $ (x_n)_{n \ge 0}, (y_n)_{n \ge 0}, (z_n)_{n \ge 0}$ three sequences of real numbers such that:

$ i)$ $ x_0\equal{}b$, $ x_{n\plus{}1}\equal{}a\plus{}\sqrt{x_n (2a\minus{}x_n)}, \forall n \ge 0$

$ ii)$ $ (y_n)_{n \ge 0}$ is a convergent sequence

$ iii)$ $ \forall n \ge 1, z_n\equal{}\frac{1}{n}\displaystyle\sum_{k\equal{}0}^{n} x_k y_k$

Prove that the sequence $ (x_n)_{n \ge 0}$ is divergent, and the sequence $ (z_n)_{n \ge 1}$ is convergent

define t_n = x_n - a  we get,
t_n ^2 = a^2 - t_n-1 ^2

Hence, t_n is divergent (except in a case where t_0 ^2 = a^2 /2, which is not possible because of the condition on b and a). In fact, t_n oscillates between two values.

So, x_n is divergent. It oscillates between two values, and is bounded

z_n - z_(n-1) = (y_n - x_n - z_n )/(n+1) It is easily seen that the numerator in the RHS is bounded, hence, the RHS  approaches zero as n approaches infinity. Hence, z_n is convergent. [/hide]


سلام
يه جلسه با حضورمديران و معاونين 5 دبيرستان برگزار ميشه ,به طوري كه از هر مركز مدير و يكي از معاونين حضور دارن.در بدو ورود به جلسه اين افراد با هم دست ميدن.(مدير و معاون يك مركز با هم دست نميدن).يكي از مديران از هر 9 نفر ديگه مي پرسه "شما با چند نفر دست دادين؟"اگه 9 پاسخ متفاوت دريافت كرده باشه,تعيين كنيد معاون اين مدير با چند نفر دست داده؟
 :blush:  :?:

khob bia sabet kon ke adadaye gofte shode $ \{0,1,2,...,8\}$ hastan bad sabet kon oonai ke javabaye $ \{0,8\},\{1,7\},\{2,6\},\{3,5\}$ dadan harkodoom modiro moavene hamdigan,yeni $ 0,8$ yekishoon modire oonyekishoon moaven,...
bad tanha kasi ke mimoone oonieke ba $ 4$ nafar das dade...


[quote="hungkhtn"]My open form is find the best constant $k$ st

\[ \frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ca}}+\frac{1}{\sqrt{c^2+ab}} \ge \frac{k}{\sqrt{ab+bc+ca}}  \]

I had posted it in http://www.diendantoanhoc.net and Mathlinks but for along time It's still unsolved.[/quote]I found $f(k)=C_{max},k>0$such that$\forall a,b,c>0$we have
                $\sum \frac{1}{\sqrt{ka^2+bc}}\ge \frac{C}{\sqrt{ab+bc+ca}}$
My answer is $f(k)=min(2\sqrt{2},1+\frac{2}{\sqrt{k}},\frac{3\sqrt{3}}{\sqrt{1+k}})$

[quote="hungkhtn"]Oh, It's an incredible result. I wait for the solution.  :)[/quote]
I posted my method in an old topic. :)

[quote="Bluesea"]Can you post the link nthd[/quote]
Click[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55584]here[/url] :)

The problem you mention there of VASC can be solved by SOS, but if we have the square root then everything changes. Can you post some details? I can't understand.


calculus
integration
trigonometry
function
geometry
calculus computations

$ \int_{\minus{}1}^1 (1\plus{}x)^{\frac{1}{2}}(1\minus{}x)^{\frac{3}{2}}\ dx\equal{}\int_{\minus{}1}^1 (1\minus{}x^2)^{\frac{1}{2}}(1\minus{}x)\ dx\equal{}\left\{x\equal{}\sin y\right\}\equal{}$
$ \equal{}\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2} y (1\minus{}\sin y) dy\equal{}\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cos^{2} y \minus{}\cos^2 y \cdot \sin y\right) dy\equal{}$
$ \equal{}\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1\plus{}\cos 2y}{2} \minus{}(1\minus{}\sin^2 y) \cdot \sin y\right) dy\equal{}$
$ \equal{}\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1\plus{}\cos 2y}{2} \minus{}\sin y \plus{} \sin^3 y\right) dy\equal{}$
$ \equal{}\int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1\plus{}\cos 2y}{2} \minus{}\sin y \plus{} \frac{3\sin y \minus{} \sin 3y}{4}\right) dy$

Last is easy :)

Little bit faster (using the same technique)

$ \int_{ \minus{} 1}^1 (1 \plus{} x)^{\frac {1}{2}}(1 \minus{} x)^{\frac {3}{2}}\ dx \equal{} \int_{ \minus{} 1}^1 (1 \minus{} x^2)^{\frac {1}{2}}(1 \minus{} x)\ dx \equal{} \left\{x \equal{} \sin y\right\} \equal{}$
$ \equal{} \int_{ \minus{} \frac {\pi}{2}}^{\frac {\pi}{2}}\cos^{2} y (1 \minus{} \sin y) dy \equal{} \int_{ \minus{} \frac {\pi}{2}}^{\frac {\pi}{2}}\left(\cos^{2} y \minus{} \cos^2 y \cdot \sin y\right) dy \equal{}$
$ \equal{} \int_{ \minus{} \frac {\pi}{2}}^{\frac {\pi}{2}}\frac{1 \plus{} \cos 2y}{2}  dy \equal{} \frac{\pi}{2}$

Using the fact that the other subintegral function is odd.

Here s a slightly simpler method.
We have $ I\equal{} \int_{\minus{}1}^{1} (1\plus{}x)^{\frac{1}{2}} \cdot (1\minus{}x)^{\frac{3}{2}} dx$  ...(1)
Replace $ x \rightarrow \minus{}x$ so that we get, with care taken about limits,

$ I \equal{}  \int_{\minus{}1}^{1} (1\minus{}x)^{\frac{1}{2}} \cdot (1\plus{}x)^{\frac{3}{2}} dx$ ...(2)

Adding (1) and (2) we get
$ 2I \equal{} \int_{\minus{}1}^{1} \sqrt{(1\minus{}x^{2})} (2) dx$
One can cancel two on both sides and the last step is ofc trivial.

Notice also that the integrand of the last integral is even.  :D

The solution gets easier, one more step!

Yes, your answer is right! :)

I made the problem just in my head.

$ x\sqrt{1\minus{}x^2}$ is an odd function, so the evaluation is the area of the semi-circle with centered at the origin and radius 1.

Thus the answer is $ \frac{\pi}{2}$.

The game is over.


analytic geometry
geometry
rectangle
circumcircle

In Cartesian coordinate $Oxy$, given two lines $(d): x=a$ and $(D): y=b$ where $a,b$ are nonzero constant. A right angle $\angle uOv$ around point $O$, cut the lines $(d),\ (D)$ at $A,B$, respectively. Construct point $I \in AB$ such that $OI \perp AB$.
Find equation of locus of point $I$.

this is a very nice problem...first geometrically,

[hide]notice the rectangle formed by $x=a$ and $y=b$, call this

$OMNP$, since $\angle AOB+\angle MNB=180$, 1) is true

so
1)$O$ lies on the circumcircle of $\triangle ANB$, and
2)$OP$, $OM$ and $OI$ are perpendicular to a side of the triangle,
therefore $M,I,P$ are collinear since that is the simpson line of $\triangle ANB$

now to prove that every point on $MP$ is in the locus

start with the rectangle $MNPO$, let $I'$ be on $MP$
let $l$ be perpendicular to $I'O$ at $I$, and interesect $NP$ at $B$ and $MN$ at $A$ respectively (this must occur since lines in a plane intersect unless they are parallel, and when $l//NP$, they are the same line)

then O lies on the circumcircle of $\triangle ANB$, and $\angle AOB=180-\angle BNA=90$

thus every point on $MP$ is in the locus
[/hide]

i have also looked at the problem analytically, but i cannot finish it off...

[hide]

$A: =(a,m)$, $B: =(n,b)$, $I: =(x,y)$, note that $m$, $n$ vary

then $AO\cdot BO=0$

$(a,m)\cdot (n,b)=0$

also, $(A-B)\cdot IO=0$

$(a-n,m-b)\cdot (x,y)=0$

$(x,y)\cdot (a,m)=(x,y)\cdot (n,b)$

we want:

$(x-a,y)\cdot (b,a)=0$, but i don't see how to do this
[/hide]

Yes, I agree that the problem is interesting, and Altheman didn’t finish it with analytic way. Actually, I came to following fact, but there’s nothing finally.

Let $A(\alpha,b)$ and $B(a, \beta)$.
[b]1:[/b] $OA \perp OB \implies \alpha\beta+ab=0$.
[b]2:[/b] $OI \perp AB$, with $I(x,y)$ but there’s too many variables here.

Give me something, everybody!
PS: I’ve thought about theorem $\frac{1}{OI^2}=\frac1{OA^2}+\frac1{OB^2}$, but I don’t know if it would help.


find all real functions such that
1.$f(x)+f^{-1}(x)=2x$
2.$f$ is strictly incereasing.
it is so old and it looks easy but can some one help me?


inequalities
integration
function
absolute value
real analysis
real analysis unsolved

$b>0$, $f\in C^{1}([0,b])$ and $f(0)=f(b)=0$.
I need to show the inequality:
$\int_{0}^{b}|f(x)|^{2}dx\leq (\frac{b}{\pi})^{2}\int_{0}^{b}|f'(x)|^{2}dx$

is equivalent to:
$\int_{0}^{1/2}|f(x)|^{2}dx\leq (\frac{1}{2\pi})^{2}\int_{0}^{1/2}|f'(x)|^{2}dx$

I started with change of variable: $y=2bx$, but I can't absorb the $b^{2}$ term...

May I ask if there is any significance to having the absolute value sign on the squared term?

Please clarify if a function f with the given properties is in addition assumed to satisfy the first inequality or if the given properties imply that it satisfies the first inequality. The latter is impossible and you may have a typo otherwise. I will look at it once more after a siesta

try to use the Wirtinger's inequality: http://en.wikipedia.org/wiki/Wirtinger's_inequality


probability
integration
calculus
function
expected value
binomial coefficients
probability and stats

If you form 1004 intervals by pairing the points 1, 2, ...., 2008 randomly on the number line, what is the probability that among those intervals there is at least one which intersects all the others?

Source: Peter Winkler's puzzle book

By brute force counting, if I made no mistakes:

Pairing the integers $ 1$ through $ 4$, the probability is $ \frac23.$

Pairing the integers $ 1$ through $ 6$, the probability is $ \frac23.$

Pairing the integers $ 1$ through $ 8$, the probability is $ \frac23.$

There's an obvious conjecture there, but without a proof, I'm not ready to claim it.

Let's name the intervals $ [a_i,b_i]$, where $ i \equal{} 1,\dots,n$. Let $ A \equal{} \max a_i$ and $ B \equal{} \min b_i$. We are asked to find the probability of the event "$ \exists i$ such that $ a_i < B$ and $ b_i > A$". Let's count the pairings when this does not happen (call them "bad").

It is clear that $ n\le A\le 2n \minus{} 1$ and $ 2\le B\le n \plus{} 1$. If $ A \equal{} n$, then $ B \equal{} n \plus{} 1$ and there are no bad pairings. It remains to consider the case $ A\ge n \plus{} 1$ and $ B\le n$. Any bad pairing can be constructed as follows:

(1) $ A$ is paired with one of points $ A \plus{} 1,\dots,2n$;
(2) $ B$ is paired with one of points $ 1,\dots,B \minus{} 1$;
(3) Each of the yet-unpaired points $ 1,\dots,B \minus{} 1$ is paired with some point between $ B$ and $ A$;
(4) Each of the yet-unpaired points $ A \plus{} 1,\dots,2n$ is paired with some point between $ B$ and $ A$;
(5) The remaining points are paired in an arbitrary way.

Steps (3)-(4) can be carried out only if $ A \minus{} B \minus{} 1\ge (B \minus{} 2) \plus{} (2n \minus{} A \minus{} 1)$, that is, $ A \minus{} B\ge n \minus{} 1$. Assuming this is the case, how many choices do we have?
(1) $ 2n \minus{} A$
(2) $ B \minus{} 1$
(3) $ \frac {(A \minus{} B \minus{} 1)!}{(A \minus{} 2B \plus{} 1)!}$
(4) $ \frac {(A \minus{} 2B \plus{} 1)!}{(2A \minus{} 2B \plus{} 2 \minus{} 2n)!}$
(5) $ \frac {(2k)!}{2^kk!}$ where $ k \equal{} A \minus{} B \plus{} 1 \minus{} n$.

The product is $ (2n \minus{} A)(B \minus{} 1)\frac {(A \minus{} B \minus{} 1)!}{2^{A \minus{} B \minus{} n \plus{} 1}(A \minus{} B \minus{} n \plus{} 1)!}$. To simplify this, let $ j \equal{} A \minus{} B \minus{} n \plus{} 1$. We know that $ j$ runs over $ 0,1,\dots,n \minus{} 2$. Thus, the number of bad pairings is
\[ \sum_{j \equal{} 0}^{n \minus{} 2}\frac {(n \plus{} j \minus{} 2)!}{2^j j!}\sum_{B \equal{} 2}^{n \minus{} j}(n \minus{} j \minus{} B \plus{} 1)(B \minus{} 1)
\]
Maple somehow simplifies this sum to $ \frac13\frac {(2n)!}{2^nn!}$, and I'm inclined to trust it this time (it's almost 2am here).  :blush:

Of course, there are $ \frac {(2n)!}{2^nn!}$ pairings in total.

Here is a slightly different approach that does not require advanced knowledge of combinatorics by Maple :P. Note that there is nothing sacred about the numbers $ 1,\dots,2n$: the probability will stay the same if we replace them by any other pairwise distinct numbers. From this observation, there is just one (easy) step to the following reformulation of the problem.

Denote by $ I(x,y)$ the interval with the endpoints $ x$ and $ y$ (we do not care which one is the left end and which one is the right one here). Choose $ 2n$ points $ x_1,y_1,x_2,y_2,\dots x_n,y_n$ independently on $ [0,1]$. Find the probability that at least one of the intervals $ I(x_j,y_j)$ intersects every other interval of this kind.

Now we are going to use the good old [[PIE]]. Since everything is completely symmetric, we can say that the expected number of the $ k$-tuples of intervals with this property is just $ n\choose k$ times the probability that first $ k$ intervals have this property. Now let us figure out some formula for the probability that each of the intervals $ I(x_j,y_j)$ ($ 1\le j\le k$) intersects every interval in our collection. Note that this is equivalent to saying that the intersection of these first $ k$ intervals is a non-empty interval and this intersection intersects every of the remaining $ n\minus{}k$ intervals. Conditioning on this intersection being $ [a,1\minus{}b]$ ($ a,b\ge 0$, $ a\plus{}b\le 1$) and using independence, we get (after some routine computations) that the probability in question equals $ k^2\int_{a,b>0, a\plus{}b<1} (1\minus{}a^2\minus{}b^2)^{n\minus{}k}2^k(ab)^{k\minus{}1}\,da\,db$. Thus, using PIE, we obtain that the probability we were asked about in the original problem equals
\[ 2\sum_{k\equal{}1}^n (\minus{}1)^{k\minus{}1}{n\choose k}k^2\int_{a,b>0, a\plus{}b<1} (1\minus{}a^2\minus{}b^2)^{n\minus{}k}2^k(ab)^{k\minus{}1}\,da\,db\]
Swapping the sum and the integral, writing
\[ k^2{n\choose k}\equal{}n{n\minus{}1\choose k\minus{}1}\plus{}n(n\minus{}1){n\minus{}2\choose k\minus{}2}\]
and using Newton's binomial formula (Ok, I admit, there is [i]some[/i] combinatorics here too :) ), we bring the formula for the probability to the form
\[ 2\left[n\int_{a,b>0, a\plus{}b<1} (1\minus{}(a\plus{}b)^2)^{n\minus{}1}\,da\,db\minus{}n(n\minus{}1)\int_{a,b>0, a\plus{}b<1} 2ab(1\minus{}(a\plus{}b)^2)^{n\minus{}2}\,da\,db\right]\]
Now make the change of variable $ a\equal{}a, a\plus{}b\equal{}u$ and integrate with respect to $ a$ first. The expression in brackets will become
\[ n\int_0^1 u(1\minus{}u^2)^{n\minus{}1}\,du \minus{}n(n\minus{}1)\int_0^1 \frac{u^3}3(1\minus{}u^2)^{n\minus{}2}\,du\]
and now only the change of variable $ u^2\equal{}v$ and one integration by parts separate us from the coveted $ 2/3$.

Now I should say that I'm deeply dissatisfied with this solution: it yields the result but doesn't give the slightest clue as to why the probability doesn't really depend on $ n$. Everything is obtained as a result of some miraculous cancellations in sums and integrals, which is not how it should be. There should be some simple underlying reason for such a nice answer but I just fail to see it :(.

This appeared earlier [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161169]here[/url]; I made about as much progress as Kent did :).  A solution to satisfy fedja seems very elusive (or perhaps one does not exist).

I spent some time trying to figure out what Maple did, but :( . The inner sum is easy: $ \sum_{B \equal{} 2}^{n \minus{} j}(n \minus{} j \minus{} B \plus{} 1)(B \minus{} 1) \equal{} (n \minus{} j)\sum_{k \equal{} 1}^{n \minus{} j \minus{} 1}k \minus{} \sum_{k \equal{} 1}^{n \minus{} j \minus{} 1}k^2$, which is equal to $ \frac {(n \minus{} j \minus{} 1)(n \minus{} j)(n \minus{} j \plus{} 1)}{6}$ according to the animations on the left margin of this page. The problem boils down to proving
\[ \sum_{j \equal{} 0}^{n \minus{} 2}\frac {(n \plus{} j \minus{} 2)!}{2^j j!}(n \minus{} j \minus{} 1)(n \minus{} j)(n \minus{} j \plus{} 1) \equal{} 2\frac {(2n)!}{2^n n!}
\]
The most relevant identity I could find (not prove) was $ \sum_{j \equal{} 0}^{n \minus{} 2} 2^{ \minus{} j}\binom{n \minus{} 2 \plus{} j}{j} \equal{} 2^{n \minus{} 2}$.  Incidentally, Maple has a different idea about this sum:
[img]http://www.math.tamu.edu/~lkovalev/files/binom.jpg[/img]
I don't trust Maple anymore.  :|

[quote="mlok"]
The most relevant identity I could find (not prove) was $ \sum_{j \equal{} 0}^{n \minus{} 2} 2^{ \minus{} j}\binom{n \minus{} 2 \plus{} j}{j} \equal{} 2^{n \minus{} 2}$.  Incidentally, Maple has a different idea about this sum:
[img]http://www.math.tamu.edu/~lkovalev/files/binom.jpg[/img]
I don't trust Maple anymore.  :|[/quote]

ZOMG!?  Seriously?  What version are you using?  If true, that is a very obvious bug.  Mathematica 5.0 obtains the correct result (though to be fair, it has its own share of screw-ups).  Goes to show you can't always trust these programs.  *shrug*

This is my attempt to combine [b]fedja[/b]'s approach with mine. Let $ I(x_j,y_j)$ be as in [b]fedja[/b]'s post. I want to find the probability that [u]none[/u] of the intervals intersects all others. Modulo events of probability zero, this happens iff there exist numbers $ 0\le B < A\le 1$ (same meaning as in my post) such that
(i) $ \exists j'$ such that $ \max (x_{j'},y_{j'}) \equal{} B$
(ii) $ \exists j''$ such that $ \min (x_{j''},y_{j''}) \equal{} A$
(iii) for all $ j\notin \{j',j''\}$ we have: $ \max(x_j,y_j)\ge B$, $ \min (x_{j},y_{j})\le A$, and (either $ \min(x_j,y_j) > B$ or $ \max(x_j,y_j) < A$).

Conditions (i)-(iii) mean that $ j'$th interval is the leftmost, $ j''$th is the rightmost, and every other interval fails to intersect one of them. Following [b]fedja[/b], I'll assume $ j' \equal{} 1$ and $ j'' \equal{} 2$, which can be accounted for by multiplying the result by $ n(n \minus{} 1)$. The probability of (i) is $ 2B\,dB$, and the probability of (ii) is $ 2(1 \minus{} A)\,dA$. The probability of (iii) for each $ j$ is $ 1 \minus{} (1 \minus{} A \plus{} B)^2$, which can be found by drawing the unit square and marking its forbidden parts. Thus, the answer is $ n(n \minus{} 1)$ times
\[ \int_0^1\int_0^A (1 \minus{} (1 \minus{} A \plus{} B)^2)^{n \minus{} 2}4(1 \minus{} A)B\,dB\,dA
\]
Letting $ u \equal{} 1 \minus{} A \plus{} B$, we get $ \int_0^1\int_0^{u} (1 \minus{} u^2)^{n \minus{} 2}4(u \minus{} B)B\,dB\,du$, which after integration w.r.t. $ B$ becomes $ \frac23\int_0^1(1 \minus{} u^2)^{n \minus{} 2}u^3\,du \equal{} \frac13\int_0^1 t^{n \minus{} 2}(1 \minus{} t)\,dt \equal{} \frac {1}{3n(n \minus{} 1)}$.
No binomial coefficients were harmed in the writing of this solution. :cool:

[b]atomicwedgie[/b]: it's Maple 11, the latest one. It actually gives correct answers if you put $ n$ or $ n \minus{} 1$ instead of $ n \minus{} 2$. Go figure.

[quote="mlok"][i][Long irrelevant quote removed by a moderator][/i][/quote]

how to simplify your answer with maple, mlok?

Use simplify( ... ) possibly preceded by assume(n, integer) and assume(n>1). It actually gives the answer in terms of $ \Gamma$-function, and you have to use one of basic identities for $ \Gamma$ to get the answer in the desired form.


a,b,c are from C/{0}

a,b,c are distinct

(a+b-c)^3=(b+c-a)^3=(c+a-b)^3

Prove that AB=AC=BC    where  A(a),  B(b) , C(c)

by $a,b,c$ are distinct also $a+b-c,b+c-a,c+a-b$. Therefore $a+b-c,b+c-a,c+a-b$ are all complex roots of a equation $z^{3}=C$, where $C$ is complex.

Suppose for WLOG $a+b-c=1,b+c-a=\omega, c+a-b=\omega^{2}$, where $\omega=\frac{1+i\sqrt{3}}{2}$. Now, this problem is easy :)


function
complex analysis
complex analysis unsolved

let $ f$ be analytic in the unit disk. If $ f(0) \equal{} 0$, show that the series \[ f(z) \plus{} f\left(z^2\right) \plus{} f\left(z^3\right) \plus{} \cdots\] converges locally uniformly to an analytic function in the unit disk.

Since $ f(0) \equal{} 0$, $ f(z) \equal{} z g(z)$ for some analytic function $ g$ on the unit disk (centered at $ 0$).  Then for any $ |z| \leq R < 1$, $ g(z)$ is bounded on $ |z| \leq R$ and we have $ zg(z) \plus{} z^2 g(z^2) \plus{} \ldots$ converging uniformly on $ |z| \leq R$ since $ |g(z^n)| \leq \text{max}_{|z| \leq R} |g(z)|$ for $ |z| \leq R$ (i.e. compare the series to $ z \plus{} z^2 \plus{} z^3 \plus{} \ldots$ since $ R < 1$.


inequalities geometry inradius trigonometry circumcircle

Who can prove the following inequality for triangle:

\[\sum\frac{s(s-a)}{(s-a)^2+P}\geq3\]

(with standard notation)?

Sorry, I'm idiot, but who is s and P. s seems to be the semiperimeter, but P?

Oh no, I'm an idiot, these are our local symbols, I was so stupid.

$s$ is semiperimeter,
$P$ is area,
$a,b,c$ are sides,
and the full inequality goes like this:

\[\frac{s(s-a)}{(s-a)^2+P}+\frac{s(s-b)}{(s-b)^2+P}+\frac{s(s-c)}{(s-c)^2+P}\geq3.\]

I can guess how you came up with this inequality: It is equivalent to the inequality by F. Zeinoulahee which you posted in

  http://mathlinks.ro/viewtopic.php?t=14586&sid=e1e255964d0bb66d054840711bcdefc6

([i]Advanced Section -> Inequalities -> Unsolved Problems -> extraordinary beauty (and complexity)[/i]), and which says that for any three nonnegative reals u, v, w satisfying vw + wu + uv = 1, the inequality

  $\displaystyle \frac{1}{vw+u} + \frac{1}{wu+v} + \frac{1}{uv+w} \geq 3$

holds.

In fact, denoting by r the inradius of triangle ABC, we have P = sr. Now, denoting $u=\tan \frac{A}{2}$, $v=\tan \frac{B}{2}$, $w=\tan \frac{C}{2}$, we can easily find by calculation that

  $\displaystyle vw = \tan \frac{B}{2} \cdot \tan \frac{C}{2} = \frac{s-a}{s}$,

and similarly $\displaystyle wu = \frac{s-b}{s}$ and $\displaystyle uv = \frac{s-c}{s}$, so that vw + wu + uv = 1. Hence, we can apply F. Zeinoulahee's inequality to the three nonnegative reals u, v, w and get

  $\displaystyle \frac{1}{vw+u} + \frac{1}{wu+v} + \frac{1}{uv+w} \geq 3$.

Now, since $\displaystyle vw = \frac{s-a}{s}$ and $\displaystyle u = \tan \frac{A}{2} = \frac{r}{s-a}$, we have

  $\displaystyle \frac{1}{vw+u} = \frac{1}{\frac{s-a}{s} + \frac{r}{s-a}} = \frac{s \left( s-a \right)}{\left( s-a \right)^2 + sr} = \frac{s \left( s-a \right)}{\left( s-a \right)^2 + P}$,

and similarly,

  $\displaystyle \frac{1}{wu+v} = \frac{s \left( s-b \right)}{\left( s-b \right)^2 + P}$ and
  $\displaystyle \frac{1}{uv+w} = \frac{s \left( s-c \right)}{\left( s-c \right)^2 + P}$,

so that

  $\displaystyle \frac{s \left( s-a \right)}{\left( s-a \right)^2 + P} + \frac{s \left( s-b \right)}{\left( s-b \right)^2 + P} + \frac{s \left( s-c \right)}{\left( s-c \right)^2 + P} = \frac{1}{vw+u} + \frac{1}{wu+v} + \frac{1}{uv+w} \geq 3$.

Since Harazi proved Zeinoulahee's inequality, your one is now proved, too.

  Darij

[quote="sbos"]If $ \Delta$ is the area of a triangle $ ABC$, $ a,b,c$ are the sides and $ s$ is the semi-perimeter, then

$ \frac {s(s - a)}{(s - a)^2 + \Delta} + \frac {s(s - b)}{(s - b)^2 + \Delta} + \frac {s(s - c)}{(s - c)^2 + \Delta} > 3$.[/quote]It's equivalent to

$ - 2s^3 + (4R + r)s^2 + 4r(7R + r)s - 3r(4R + r)^2 > 0$,

where $ R$ is the radius of circumcircle and $ r$ is the radius of incircle.

[hide="Proof."]$ - 2s^3 + (4R + r)s^2 + 4(7R + r)sr - 3(4R + r)^2r$

$ = 2\left(s^2 - 16Rr + 5r^2\right)\left[2R + \left(3\sqrt {3} - 4\right)r - s\right]$

$ + r\left[3\left(3 - 2\sqrt {3}\right)s^2 - 2(2R - 7r)s + 16R^2 - 4\left(43 - 24\sqrt {3}\right)Rr + \left(37 - 30\sqrt {3}\right)r^2\right]$

$ = 2\left(s^2 - 16Rr + 5r^2\right)\left[2R + \left(3\sqrt {3} - 4\right)r - s\right]$

$ + \left(\frac {2}{\sqrt {3}} - 1\right)r\left[2R + \left(3\sqrt {3} - 4\right)r - s\right]\left[9s + 2\left(15 + 4\sqrt {3}\right)R - \left(78 + \sqrt {3}\right)r\right]$

$ + \frac {2}{13}\left(11 - 6\sqrt {3}\right)r\left[26R^2 + 4\left(30\sqrt {3} - 23\right)Rr + \left(242 - 141\sqrt {3}\right)r^2\right] > 0$,

which is obviously true for $ s^2\geq 16Rr - 5r^2$, $ s\leq 2R + \left(3\sqrt {3} - 4\right)r$ and $ R\geq 2r$.[/hide]

[quote=Ji Chen][quote="sbos"]If $ \vartriangle$ is the area of a triangle $ ABC$, $ a,b,c$ are the sides and $ s$ is the semi-perimeter, then

$ \frac {s(s - a)}{(s - a)^2 + \vartriangle} + \frac {s(s - b)}{(s - b)^2 + \vartriangle} + \frac {s(s - c)}{(s - c)^2 + \vartriangle} > 3$.[/quote]It's equivalent to

$ - 2s^3 + (4R + r)s^2 + 4r(7R + r)s - 3r(4R + r)^2 > 0$,

where $ R$ is the radius of circumcircle and $ r$ is the radius of incircle.

[hide="Proof."]Left as an [url=http://mp.weixin.qq.com/s/4repHDvv7KN87XHtSdGf4w]exercise[/url].[/hide][/quote]

\[\frac {s(s - a)}{(s - a)^2 + \triangle} + \frac {s(s - b)}{(s - b)^2 + \triangle} + \frac {s(s - c)}{(s - c)^2 + \triangle} \geqslant 3{\left(1+2\prod\frac{{\left(s-b\right)}{\left(s-c\right)}}{{\left(s-a\right)}^2+\triangle}\right)} > 3\textnormal.\] [url=https://artofproblemsolving.com/community/c6h2825241_a_triangle_inequality]a triangle inequality[/url]


How are you gentlemen !!
All your base are belong to us.
You are on the way to destruction.
You have no chance to survive make your time.
Ha ha ha ha

Oh for those who don't understand Kevin's cultural reference: http://en.wikipedia.org/wiki/All_your_base_are_belong_to_us

I play The Game and you lose over 9000 health points.

Yay for combining multiple internet memes!!!!

[img]http://z.hubpages.com/u/785979_f260.jpg[/img]

Dang I'm totally printing that card out and pwning all my friends.  :ninja:

(Jk)

Cat is from lolcats on failblog, laser is from The Impossible Quiz.

Did he die?
Sup dawg!
Yo momma!
etc.

Actually, in the Lazr, the Impossible Quiz merely ripped off another meme, described [url=http://ph.answers.yahoo.com/question/index?qid=20070927063130AARYM9C]here.[/url]


The rest of the integers from one through eight are placed in the remaining seven regions so that no two consecutive numbers are in the regions that border each other, even at the corners. What number must be in region $ b$?
[asy]draw((0,1)--(0,2)--(4,2)--(4,1)--cycle);
draw((1,0)--(1,3)--(3,3)--(3,0)--cycle);
draw((2,0)--(2,3));
label("2",(3.5,1.5));
label("$b$",(.5,1.5));[/asy]

By symmetry we can place 9-a instead of a and still have a valid diagram; similarly we can do the same with 180 degree rotation.

Thus, b=9-2=7.

Cheap, but it works.

Any logical solutions? Didn't understand @james4l's solution.

Guess and check works, but it takes too long.


[hide=" A little messy.."]
After some simplification, from

$\tan\alpha+\tan\alpha\cos2\alpha+\tan\beta\sin2\beta=\tan\beta+\tan\alpha\sin2\alpha+\tan\beta\cos2\beta$
we get:
$\cos \alpha \sin \alpha+\sin^{2}\beta = \cos \beta \sin \beta+\sin^{2}\alpha$
Since $\alpha+\beta = \frac{\pi}{4}$, we have that $\sin \alpha = \sin (\frac{\pi}{4}-\beta)= \frac{\cos \beta-\sin \beta}{\sqrt{2}}$ and $\cos \alpha = \cos(\frac{\pi}{4}-\beta) = \frac{\cos \beta+\sin \beta}{\sqrt{2}}$, so we can substitute to get:

$\frac{ (\cos \beta+\sin \beta)(\cos \beta-\sin \beta)}{2}+\sin^{2}\beta = \cos \beta \sin \beta+\frac{ (\cos \beta-\sin \beta)^{2}}{2}$,
which is true (multiply it out, you get $0=0$).

All steps are reversible, so the proof is done.[/hide]

You don't know $\ \alpha+\beta = 45$ that's what you're trying to prove.

Actually, the way the problem is written, it's utterly unclear what's the direction of the implication. From left to right it's incorrect: The given identity yields $\alpha-\beta=n\pi$ or $\alpha+\beta={\pi\over 4}+n\pi$ where $n\in\mathbb{Z}$. From right to left is already solved.

\[\{\begin{array}{c}\boxed{\ \tan x+\tan x\cos 2x+\tan y\sin 2y=\tan y+\tan x\sin 2x+\tan y\cos 2y\ }\\ \\ \tan x(1+\cos 2x-\sin 2x)=\tan y(1+\cos 2y-\sin 2y)\\ \\ \tan x(2\cos^{2}x-2\sin x\cos x)=\tan y(2\cos^{2}y-2\sin y\cos y)\\ \\ \sin x(\cos x-\sin x)=\sin y(\cos y-\sin y)\ \ |\bullet\ 2\\ \\ \sin 2x-1+\cos 2x=\sin 2y-1+\cos 2y\\ \\ \sin 2x-\sin 2y=\cos 2y-\cos 2x\\ \\ 2\sin (x-y)\cos (x+y)=2\sin (x+y)\sin (x-y)\\ \\ \sin (x-y)=0\mathrm{\ or\ }\tan (x+y)=1\\ \\ \boxed{\ (x-y)\in \pi Z\mathrm{\ or\ }(x+y)\in \pi Z+\frac{\pi}{4}\ }\\\\ (\ x\not\in \pi Z+\frac{\pi}{2}\ ,\ y\not\in \pi Z+\frac{\pi}{2}\ )\end{array}\] [b]Notation :[/b] $aZ+b\equiv\{ka+b\ |\ k\in Z\}$ for $\{a,b\}\subset R$ , $a\ne 0\ .$


Hello guys!
-f(x), g(x), h(x) are quadratic polynomials.  Can f(g(h(x))) = 0 have roots 1, 2, 3, 4, 5, 6, 7, 8?

-Solve cos(cos(cos(cos x))) = sin(sin(sin(sin x)))

Bye! :D

Well for the second question, we can convert the whole left side to sines, and then solve it from there...

answer to 2 is
[hide="answer"]
$45^{\circ}$
[/hide]

but this comes from kalva I think and even kalva agrees.

Yes. Indeed 45 doesn't work.  It makes the innermost sine and cosine equal, but then you still have different functions on the 45 degrees.

[quote="Arne"]Check your solution - it doesn't work.[/quote]
Yes I understand it doesnt work. :blush:  :blush:  :blush:  :blush:

for the second one
you can replace the trig functions by an infinite series then show that there is no common solution for that.
it is a very ugly solution though! ;)

for the first one, i think so because there is nothing about the quadratics having real or imaginary or irrational coefficients, and the highest degree in that composition would be 8, so there is no problem, but if there were real, or rational coefficients, that would might be different


[hide]
For each subset $S$ of $\{ 1,2,3,\cdots ,n-1\}$ with at most $k-1$ elements, we can pair $S$ with $S\cup \{ n\}$. Here exactly one set in the pair has an even number of elements and the other has an odd number of elements.

The binomial sum on the left hand side is the number of subsets of $\{ 1,2,3,\cdots ,n\}$ with at most $k$ elements and of even parity minus those of odd parity. Of the subsets of $\{ 1,2,3,\cdots ,n\}$ with at most $k$ elements, the only ones that do not fall into one of the pairs are those with exactly $k$ elements, all coming from $\{ 1,2,3,\cdots ,n-1\}$. There are $\binom{n-1}{k}$ of them, and they contribute to the even number count (so added) if $k$ even, or to the odd number count (so subtracted) if $k$ odd. Therefore, the binomial sum is equal to $(-1)^{k}\binom{n-1}{k}$, as desired.
[/hide]


algorithm
induction
combinatorics proposed
combinatorics

Can anyone find a neat solution for the following problem?. Mines is defenitely everything but neat :D
A word of lenght $n$ is an ordered sequence $x_1x_2...x_n$ where ${x_i\in \{a,b,c}\}$. Let $A_n$  be the set of words of lenght $n$ such that it doesnt contain any block $x_ix_{i+1}$ of the form $aa$ or $bb$. Also, let $B_n$ be the set of words of lenght n sucht that it doesnt contain any block $x_ix_{i+1}x_{i+2}$ that is a permutation of $\{a,b,c\}$. Show that $|B_{n+1}|=3|A_n|$

Rename the letters and make them ${2, 1, 0}$. Now for any word of length $n+1$, apply the following: if it has ith, i+1th letters $x_i, x_{i+1}$, produce $y_i$ as the residue of $2x_i+x_{i+1}$ mod 3. Now it is easy to check that for any string $x_i, x_{i+1}, x_{i+2}$ which is a permutation of ${2, 1, 0}$, it produces a 2-letter string that is either $1, 1$ or $2, 2$. Also, any string $x_i, x_{i+1}, x_{i+2}$ which is not a permutation of ${2, 1, 0}$ has a digit that appears twice, which generates one $y_i=0$ so the corresponding string of length 2 can't be $1, 1$ or $2, 2$. Thus the application of this algorithm to type B words yields type A words and the application to any non-B word produces a non-A word.

Now the boring parts. Note that if you take a B-word and add 1 to every digit (taking 2+1=0), the resulting A word is unchanged, so there are 3 B-words that produce the same A-word.

It's easy to check that two B-words that can't be obtained from the other by adding 1 or 2 to every digit, won't produce the same A-word, as well: first change the 2nd word by adding 1 or 2 so that the first digits are equal, then it is easy to check the 2nd, ..., last digit must all be equal.

Also, the same algorithm can be easily constructed backwards to get the 3 B-words from an A-word: set an arbitrary first digit and then choose the 2nd, ..., n+1th so that the equation $2x_i+x_{i+1} \equiv y_i$ holds (the choice is unique). So we have established a bijection from A-words of length $n$ to sets of 3 B-words of length $n+1$.

Wow, thats very nice and elegant. My solution is the following.
One defines $A_a(n)$ as the number of A-words of lenght n wich begin in a. Similarly one defines $A_b(n)$ and $A_c(n)$. By simmetry we have that $A_a(n)=A_b(n)$. Now its easy to check that the following recursions hold

\[|A_{n+1}|=2A_a(n)+2A_b(n)+3A_c(n)=4A_a(n)+3A_b(n)\]
\[A_a(n+1)=A_b(n)+A_c(n)=A_a(n)+A_c(n)\]
\[A_c(n+1)=A_a(n)+A_b(n)+A_c(n)=2A_a(n)+A_c(n)=|A_n|\]

After that, one uses induction to prove that both of this recursions hold:

\[A_c(n+1)=2A_c(n)+A_c(n-1)\]
\[2A_a(n)=A_c(n)+A_c(n-1)\]

(its a induction in wich one asumes both recursions to hold for any k<m, and using both one shows that both are truth for m). Moreover, one has that A_c(1)=1, and A_c(2)=3, so by applying the recursion formula one finds that

\[A_c(n)=\frac{1}{2}[(1+\sqrt{2})^n+(1-\sqrt{2})^n]\]
And hence we have that

\[|A_n|=A_c(n+1)=\frac{1}{2}[(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}]\]

After that, one defines $B_{ab}(n)$ as the B-words of lenght n wich begin in ab. Similarly one defines $B_{bc}(n), B_{ca}(n), B_{ac}(n), B_{cb}(n), B_{ba}(n), B_{aa}(n), B_{bb}(n), B_{cc}(n)$ .
And the procedure is almost identical. After some algebra one gets what needed. Not very elegant :blush:


geometry
geometric transformation
reflection
AMC
USA(J)MO
USAMO
Putnam

USAMO 2007 has finally come and past.

After taking this USAMO, I've begun to realize something.  A friend said to me a night ago that: in the beginning, we all saw olympiad as a way to hone our mathematical abilities.  But after a while, we find that it is really a competition, and it really does hurt to lose.

To me, preparation for an olympiad has just become learning tricks and seeing similar problems.  There seems to be very little creative element left (except in several combinatorics problems).  My tutor (41 on IMO04 and 42 on IMO 05), Myth (on this forum), and even several Chinese Fields Medalists I know have felt the same way.

For me, this USAMO has hurt.  Solving 1,5,2(partial),4(1 point) will probably put me right at the MOP selection boundary.  I'm inclined to blame this on the fact that [b]one night[/b] before the USAMO, I was notified that I would be taking it at 12:30 AM-5:00 AM to accomodate USA time.  No offense, but it pains me greatly to see that I may be as qualified (and perhaps moreso) as many MOPpers and quite a few red MOPpers.

I am told there will be another day...but I am already a junior.

I feel that the olympiad has greatly piqued my mathematical interest. Before the one month that I spent preparing for the USAMO, I had not put in much time in preparing for olympiad type problems simply because I was lazy and was preoccupied with different forms of mathematical events and school events. While I was preparing for USAMO in that one month despite it being grueling I found that I really enjoy olympiad style problems. Upon my invitation Red MOP, I have told myself I will not come to the Olympiad next year unprepared starting with Red MOP. I do not see the Olympiad as a competition because I have nothing to lose from not making it to MOP or IMO or even USAMO because I still gain mathematical skills that I have honed.

Although you did not qualify for MOP, in the long run even if you did make it to MOP it would not be the experience you benefited from but the knowledge.

Also, if you live in China(or at least from what I have concluded) then how could you qualify for MOP

You're right that as much as the USAMO is intended to spur interest in math, bottom line is that it's a competition. There are winners and there are losers. There is an element of luck as to whether or not you have solved a similar problem, or whether it's a good subject. As was discussed after the AIME results, there is simply no practical way to measure who is really "better." We can't have 365 USAMOs a year; there will always be someone good who just has a bad day. We just have to prepare as best as we can and hope.

I appreciate the unfairness of your testing situations, but remember that not everybody has such a qualified personal tutor, or other such benefits...

I actually burned out of math and consequently hate math cause of the usamo so..

it didn't relaly work for me? i guess it wasn't my cup of tea

Indeed, the first day of the USAMO was perhaps the most soul-crushing experience of my life. My score fell a good 14 points below what I had predicted as the lower end of my score range.

I'm hoping that if I say "its just a competition" enough times it will make me less depressed, but that seems improbable. I had a great deal of fun preparing for it (I have by now taken for kicks every USAMO, ToT, and IMO as well as a large percentage of shortlisted problems and other national olympiads) but this seems to just make it harder for me to convince myself that I don't care that much.

Congratulations to all who made it, and to everyone else who had a bad day my deepest sympathy.

(sorry if I'm a whiner, but I did I mention the soul crushing?)

For the last two years, I missed the USAMO cutoff by a single AIME point, due to careless mistakes.  Well, yes it was frustrating, but the most frustrating part was knowing that I was capable of making it, and then, after the USAMO problems came out and I took the USAMO subjecting myself to the same conditions, finding that I was actually capable of doing many of the problems.  When I took last year's USAMO in June, I felt I could have done quite well, and now knowing how they grade, I think I would have gotten somewhere around 20-23.  Nevertheless, I grit my teeth and told myself it was just a competition.  Even though I had failed, I kept up my interest in contest math and hoped for better the next year.

I continued to generally do problems that interested me, and learned much more.  My interest for olympiad math grew considerably, in fact.  Qualifying for USAMO was that much more of a goal to attain.  And this year, I went in with a completely different attitude than what I had last year.  Competition math no longer meant the world to me: it was, really, just a competition.  If I made it, great, otherwise, it wasn't the end of the world.  I was happy just to be taking the exam, and hadn't set myself any goals of any sort.

And, well, it's as they say: you always have next year.  I went from sad having missed USAMO last year to a USAMO winner this year.  Even though I might have made honorable mention last year, I don't mind.  I still learned a lot from the experience and had a lot of fun with it.

The point of this story: As painful as it may be to just mess it up, and lose your shot, you do have to remember it's no reason to stop doing math.  These competitions are mainly for fun: how you do is, as Phelpedo said, largely based on luck.  Just pull yourself up, do more olympiad math, and walk into next year's USAMO without expectations.  Not only will it make things much less painful, but you'll enjoy the problems a lot more.

Having had some similar setbacks a couple years ago, the best advice I can offer is that if you view it as a competition, then losing should only encourage you to try harder so that you can perform well the next year. I can appreciate the dissapointment one might have in either possibly losing confidence in their skills, or not making MOP. For the former, it was just one bad day. Look at what happened to Zeb and me on the US Open (CS equivalent of USAMO), or to some past USAMO winners on the USAMO itself. As for not making MOP, remember that there is plenty of other great mathematical opportunities out there. It is probably not too late for late admissions to many of the math programs out there, and if you are a borderling MOPper then I am sure that you can get into any of the programs you apply to if you put some effort into the qualifying quizzes. If this doesn't suit you, you could get a mentorship with a mathematics professor at a local university. There are tons of opportunities, and I'm sure that any of a number of the AoPS staff or alumni of these camps would be more than willing to direct you towards a program that suits your personal interests.

Well I think the key to having a good competition math experience is to only prepare as much as feels comfortable and enjoyable.  If you are enjoying and learning from the problems you are solving, then that is itself the reward regardless of how you actually perform on competition day.  Last year, I studied quite a bit for usamo 2006, and for the most part had fun doing the practice problems.  On the actual test, I didn't do particularly well (I think I got about 13), but I still felt satisfied with the overall experience, since I had enjoyed the preparation and felt good about the problems I had solved.  This year, I didn't study very much because I didn't ever really feel like it (and I was pretty busy with school), so I was very surprised to learn today that I qualified for blue mop (which I will very likely not attend).  But I actually wouldn't say that this year was really any more rewarding from a personal perspective.   I got lucky on this year's test days, and had mediocre luck last year, but I still enjoyed and felt satisfied with my first year.  You shouldn't try to convince yourself that "you don't care" if you don't perform as well as you had hoped.  If you studied hard out of passion and interest, then you should feel satisfied with yourself, independent of your final score.

Some of you guys talk as if just anybody has the ability to qualify for the USAMO at all. People, you are some of the 500 people chosen out of the ENTIRE UNITED STATES plus some places from Canada and other countries. Five hundred people out of I-don't-know-how-many high schoolers who take the AMC. Just think about that a little. Seriously.

Obviously, there's more to math than olympiads, and contests will always have a huge element of luck. There's nothing wrong with taking them seriously, either. But they still don't mean all that much. Contests aren't even great indicators of future math ability. They're fun and they can be educational. That's it. If you had fun and learned, there is nothing to be disappointed about.

i never made usamo in high school.  granted, i've been involved with the physics olympiad a lot, and i didn't really invest the kind of time in math that most imo celebrities do, but i know this is a poor excuse because there are many people who do many different olympiads well (sherry, arnav, steinhardt...)

regardless, i am certain it was worth doing all these academic competitions in high school.  even though it is pretty bitter to suck, the friends you make in the activity more than make up for that.  it is quite corny, but i'm pretty sure most of my closest friends are from math/physics/debate than from school.  saying you don't care or that you've lost your interest in the activity is lying to yourself.  everyone who's done this before, even if you got burned out, or in my case, had to contend with the harsh lesson that you're just not that smart, knows that it is, on the whole, a very good time that you probably couldn't get anywhere else.

[quote="bookaholic"]Some of you guys talk as if just anybody has the ability to qualify for the USAMO at all. People, you are some of the 500 people chosen out of the ENTIRE UNITED STATES plus some places from Canada and other countries. Five hundred people out of I-don't-know-how-many high schoolers who take the AMC. Just think about that a little. Seriously.
[/quote]

Well, this is true on some level, but I don't think its usually very good consolation.  If you think you are in the top [i]30[/i] at olympiad problems but this result is not reflected by your performance on the usamo, then you are not going to be very mollified if somebody tells you that at least you are in the top [i]500[/i].

Thats a good attitude and all...but I would have liked to go to mop [based on feedback, i am pretty sure that i would have gotten 21 to 23 ish...anybody know what partial credit was given for #2,5?]...Obviously it serves no purpose to dwell on a failed goal. The only thing you can do is make the best of the situation and go from there. I am pretty sure that I will either be doing a mathcamp or a research thing. I suppose the only difference is that I will have to pay for the opportunity [well my parents have to pay :P].

Bah, it's just a test. And in fact it's just a [i]high school[/i] test. I can't wait to stop getting owned by the USAMO and get owned by the Putnam instead.

[quote="fourierseries"]To me, preparation for an olympiad has just become learning tricks and seeing similar problems.  There seems to be very little creative element left (except in several combinatorics problems).  My tutor (41 on IMO04 and 42 on IMO 05), Myth (on this forum), and even several Chinese Fields Medalists I know have felt the same way.[/quote]
Honorable mention here with very little systematic preparation, practice or exposure to previous problems. I felt like most of my score came from the creative element; granted, I had to sit on problems for hours to bring it out. I think your comment that preparation for an olympiad is learning tricks and such is spot-on, because it's [b]preparation[/b], and there's not much more preparation can do for you. If you meant performance is based on tricks and experience, I disagree with that comment. If mathematical ingenuity is missing from your repertoire, all the preparation in the world wouldn't get you perfect scores on olympiads.

That said, I don't put that much weight on contests. I had a lot of discussions with my parents about whether I would go to MOP or not if I qualified, as I wasn't sure I wanted to. I don't feel like learning all the olympiad problem solving tricks is going to be very useful for me later in life, and I'm also worried plenty of it won't even interest me (inequalities...). My parents eventually convinced me to go if I made it mostly because it's free, and because it would help fill the many, many holes in my mathematical knowledge.

Go out and make a difference in an area of mathematics, and you'll be better than all of us wasting time in MOP classes learning how to solve problems below calculus under time pressure.

Hey, well, ther's no point in regretting because it's the past, and you really can't change it. The only thing you can change is your future  :wink:

[quote="MellowMelon"]Honorable mention here with very little systematic preparation, practice or exposure to previous problems. I felt like most of my score came from the creative element; granted, I had to sit on problems for hours to bring it out. I think your comment that preparation for an olympiad is learning tricks and such is spot-on, because it's [b]preparation[/b], and there's not much more preparation can do for you. If you meant performance is based on tricks and experience, I disagree with that comment. If mathematical ingenuity is missing from your repertoire, all the preparation in the world wouldn't get you perfect scores on olympiads.

That said, I don't put that much weight on contests. I had a lot of discussions with my parents about whether I would go to MOP or not if I qualified, as I wasn't sure I wanted to. I don't feel like learning all the olympiad problem solving tricks is going to be very useful for me later in life, and I'm also worried plenty of it won't even interest me (inequalities...). My parents eventually convinced me to go if I made it mostly because it's free, and because it would help fill the many, many holes in my mathematical knowledge.

Go out and make a difference in an area of mathematics, and you'll be better than all of us wasting time in MOP classes learning how to solve problems below calculus under time pressure.[/quote]

Hopefully if you want to attend MOP its because you expect the experience to be fun. I know that I really enjoyed it, and for me having fun is pretty much the primary objective during the summer. Also, at least for me some random conversations helped get me interested in toplogy, measure theory, and analysis, all of which have proved to be really interesting topics.

I think the argument that being in the top 500 is already pretty impressive is fairly weak. If there were only 500 highschoolers in America and they happened to be the 500 qualifiers, would that make their accomplishment any less impressive? If there were a billion high schoolers without an ounce of mathematical aptitude, would that make AIME qualification more impressive? The relevant point is your performance relative to your percieved ability. Doing better than expected hopefully makes you feel good about being better than you thought; doing worse than expected makes you depressed for the same reason.

I will be the first to admit there is something unhealthy about caring so much though. While the difference between making the TST and not making blue might have a significant effect on my future participation in olympiad math, it presumably will have a minimal effect on my life.

I don't know about the rest of you, but I do this because I really really like it. I like the math, and I like the feeling of wanting to get up and dance when I come up with a nifty solution to number three. (Unfortunately, in this day of digital video and Youtube, giving in to that impulse would be a bad idea. I've been told I give a bad name to white men who can't dance.) If I didn't get so pleasure from solving the problems, I wouldn't be able to spend time practicing them.
That said, I have found that what I enjoy about math contests has increasingly moved away from the math itself. I enjoy getting results. I enjoy holding contests to design the school math team shirt, and I enjoy convincing people to buy them (155 in a school of 300....) I just dropped off a couple of letters to get a company to sponsor our ARML team, and I enjoy that. I enjoy talking to the other kids on my ARML team, and acting somewhat as a coach. Indeed, I have found that the most rewarding experience I have had with math recently has been acting as somewhat of a coach and organizer for our ARML team. It used to be that I got a rush when I finally understood how to apply a new method, or when I finally solved a problem that had been stumping me for hours. That's still true- that dancing reference was on day one of this year's USAMO- but at the same time, I've found that I now get a rush from teaching my friends things that I know, and watching them have the same moment of illumination that I love myself.
When it comes down to it, problem solving should be something that you do for fun. There are a lot of different ways to approach it. Some of you do it specifically to make it to some goal- whether it be a certain score on the AMC 10 or an invitation to Black MOP. If that's what will make you enjoy it the most, then that is how you should do it. Learn the tricks you need, and best of luck to you. Some of you do it more for enjoyment of the math itself- for the aesthetic value of an elegant solution. That is also a valid way to look at it. My approach is somewhere in between the two of these, but also contains the social value of going to Las Vegas for ARML and spending a month at MOP. At the same time, don't try to take an approach that won't make you happy. If you like results, don't try to convince yourself that what you want is simply the beauty of the math, and vice versa.
I feel like the people in this thread have been trying to come to a right answer about how problem solving should be, when in reality there are a lot of valid ways to approach. Contests are important, but remember that they aren't necessary. If you choose to make the contests an important part of problem solving, then realize that you are also accepting the fact that a lot of it comes down to preparation and past experience, that sometimes you'll have a bad day, and that some people are probably just better than you.
That's about all I have to say, except to point out that the best thing that ever happened for my contest results was spending last year not taking math and instead loving my AP Euro history class.

Again, I'm not saying I'm the only one who feels this way.  I understand that many had high expectations going into the exam and many have come out without results.

I feel no less joy when solving a problem and I feel no less thrill when faced with a challenge.  I just feel that after a year of extensive preparation, I need a way to prove to myself that I've improved.  Receiving less than or equal to the score I received last year makes me doubt myself even more.  I also begin to question my mathematical skill.

I've trained this past year with passion and purpose; USAMO 2006 really opened new doors.  I went through all of MOP notes 2006, all previous USAMO, TST, and Russian problems.  When I won a silver medal in CMO 2007, I felt more ready than ever...

I congratulate all winners, honorable mentions, and MOPpers.  It has been my sincere wish to have joined you!

[quote="fourierseries"]... Receiving less than or equal to the score I received last year makes me doubt myself even more.  I also begin to question my mathematical skill....[/quote]
Hey, as a skilled problem solver, you know those are just false emotions.

While I don't want to minimize your (and many others') disappointment, in your case you must believe that you have earned success because your preparation has brought you challenges and thrills.

Mr. Rusczyk had often spoken of how his rigorous preparation for math competitions had made things easier for him later in college and in life. Your hard work will be similarly rewarding.

By the way, if this could/should make you feel any better: the top score for the 2007 USAMO is 32 (and likely the lone 3x) -- a far cry from the 42 in 2006 USAMO! And yes, as a junior you do have one many year to try to avoid bad days :)!

I feel like your 27 on CMO is a pretty good validation of your skill.  :)

Aagh...I will probably make a 15 and miss blue MOP this year by the same amount that i missed red MOP by last year, unfortunately...but w/e.  i know i have improved.

so...who thinks next year's usamo will be just as hard as this one?

I personally think next year's will be harder as better people take it  :( .

That seems a bit false, since Alex Zhai is the only extremely good person that is not graduating.

Don't sell yourself short, Jacob--I think 11th grade winner counts as extremely good.  Eh, maybe only very good.

Now for the main topic: I scored a 3 on the USAMO twice in a row.  I know what's it like to fail on an olympiad and the pain missing MOP brings.  The cliche "there's always next year" is unlikely to provide much consolation, but, unfortunately, there scarcely seems to be a better answer.  Time, it is said, will heal all wounds, and, in the meanwhile, approach your mathematical studies with renewed focus.  Learn the value of taking a break from olympiads--you might find, as I did, that it broadens your horizons to try content-rich, non-olympiad math.

Always remember: nearly everyone who trains for the AMCs will never do as well as you have.  You'll have your day in the sun.

dude everyone knows class of 07 rocks out

[quote="nealth"][quote="JSteinhardt"]Alex Zhai is the only extremely good person that is not graduating.[/quote]
What about Lars, Sergei, and the eighth grader who made honorable mention, his status is so similar to lars in 2005!
Making both the IMO and the IOI is the real extremely good, i think there is only extremely good person (from TJ).[/quote]

Both Haitao and Jacob made both black mop and USAICO (camp). Of course, Jacob has a better chance at IOI, given his USACO record, but they're both extremely good.

I think by extremely good I meant someone who would get a gold medal on the IMO with reasonable certainty. I.e. Zhai, Blawrence, Sherry Gong, Arnav.

I appreciate the flattery from some people but through personal experience I know that I cannot do that. I'm guessing most of the other people listed can't either, but I'll try not to make too many assumptions.

[quote="MysticTerminator"]yo I would only include maybe two of the people on your list above as "being able to get gold at IMO with reasonable certainty" (no I am not one of them)[/quote]

Dude, I predict a perfect score at IMO for you this year.

Anyways...

Like iostream, I, too, have never made USAMO, but been close in the past 2 years, but have made physics team twice, as well.

I mean, no one from my state "officially" has made it in a while, but the top scorers have been very close to the cutoff.

So... it's very painful to be 2nd in the state last year (when the 1st place person was a senior last year, so I thought I had a good chance of making USAMO by default this year) but then have an even lower AIME score than last year, and tie for 3rd with one other person.

Actually, I know it probably sounds corny, but I was almost MORE disappointed that we didn't tie for first, because then our math team coach would have been very pleased to have TWO qualifiers this year, instead of the usual one (all the DE qualifiers come from my school, btw, as a general rule).

So, I guess the irony here is 2 students from my school scored TOO HIGH for us to have 2 USAMO qualifiers... heh.

(Btw, the person I tied with was another AoPSer, durt to be more specific)

What is the IOI? Is it a math camp or something?

[url=http://olympiads.win.tue.nl/ioi/]IOI[/url]. It's the computer science olympiad.


Two friends get home at 2:30 P.M. They both go to bed 10 P.M. If both friends will go to a park twice for 15 minutes, what is the probability that they will be able to see each other at any random time between the two time intervals? Assume that the travel to and from the park takes no time.

$ \textbf{(A)} \dfrac{25}{225}\qquad\qquad\qquad\textbf{(B)}
\dfrac{8}{15}\qquad\qquad\qquad\textbf{(C)}
\dfrac{27}{225}\qquad\qquad\textbf{(D)}
\dfrac{29}{225}\qquad\qquad\textbf{(E)}
\dfrac{3}{17}$

[quote="#H34N1"]Two friends get home at 2:30 P.M. They both go to bed 10 P.M. If both friends will go to a park twice for 15 minutes, what is the probability that they will be able to see each other [b]at any random time between the two time intervals[/b]
[/quote]
Could you plz explain the line in bold, which two time intervals ?

[hide="solution"]ts a very simple problem if we try to find the area
we divide the total time interval in units of $ 15$

$ 2: 30 : \minus{} 10 \equal{} 7*60 \plus{} 30 \equal{} 15[28 \plus{} 2] \equal{} 30.15$

so we make a square on$ x \minus{} y$ plane with $ x$ axis $ y$ axis $ x \equal{} 30$and $ y \equal{} 30$ lines
now we need the area under the curve
$ |y \minus{} x|\leq 1$

$ \frac {30^2 \minus{} 29^2}{30^2} \equal{} \frac {59}{900}$

this is for one meeting as they come twice we will need some cases
$ |y \minus{} x \plus{} 1| \leq 1$ and $ |y \minus{} x \minus{} 1| \leq 1$
this will take care of all the cases !!

so finally it becomes $ \frac {30^2 \minus{} 28^2}{30^2} \equal{} \frac {58.2}{900} \equal{} \frac {29}{225}$[/hide]


function
complex numbers
complex analysis
complex analysis unsolved

Define [size=150][i]f[/i]: $\mathbb{C}$\{0}$\to$ $\mathbb{C}$[/size] by

[size=150]$f (z)= Ln z = ln|z|+i Arg(z)$[/size]

where -$\pi$ $<$Arg(z) $\lneq$ $\pi$ . Prove that f is not continous along the negative real axis

Thanks

Why not ? $Arg(z)$ will be fixed to $\pi$, and $x \mapsto ln(|x|)$ seems continuous in $]-\infty,0[$ ...

I suppose "not continous along the negative real axis" means that the limits of $f$ on both sides of the axis are different.

I'll give a sketch of the proof a shot (I don't really understand LaTeX/symbols as of yet.)

Suppose a continuous function, without the appropriate branch cut exists. Approach some point z along a unit circle upwards--the imaginary part of the limit is pi.  Now approach the point downwards--the imaginary part of the limit is negative pi.  This contradicts the assumption of continuity, so such a function does not exist.


search
number theory unsolved
number theory

it has sended before:
let make all power of 2 together :124816...
show that evrery integer can be seen in this representation some where.
To reply to t0rajir0u :what you have said isn't true because odd number can't be a begening for any power of two , so we need to search between two consecutif power of 2 !


Help please!!!

Find a 3 by 3 matrix that acts as the identity element for addition of 3 by 3 matrices, and explain your reasoning. I said the matrix with all zeroes in it. Is it right?

Find a 2 by 2 matrix that acts as the identity for multiplication of 2 by 2 matrices, and explain your reasoning.

Define an operation * on people in the world like this.
IF A and B are any two people, then A * B is the shorter of the two people. If A and B are identical in height, then A * B = A. Describe a person in the world who would be an identity element for this operation. Does every person have an inverse under this operation? and explain your reasoning for both.

[hide]EDIT :blush: : This matrix is the 3x3 Identity matrix:

A=
[0 0 0]
[0 0 0]
[0 0 0]

This is because for any matrix 3x3 B, A+B=B+A=B

This matrix is the 2x2 Identity matrix:

A=
[1 0]
[0 1]

This is because for any 2x2 matrix B, AB=BA=B

I'm working on the third problem...

EDIT:

B is the identity if for all A,

A*B = B*A = A

Thus, the tallest person in the world is the identity. (A will always be shorter than B)

The inverse of some person A, is the person C such that

A*C = C*A = B,

where B is the identity.

But this is impossible, since B will never be the shorter of two people.[/hide]


Prove that $ f(x) \equal{} \frac{x \minus{} 1}{x(x\plus{}1)}$is an injective function.

[quote="guile"]Prove that $ f(x) \equal{} \frac {x \minus{} 1}{x(x \plus{} 1)}$is an injective function.[/quote]Are you sure? Take $ f(2) \equal{} \frac{1}{2(3)}\equal{}\frac{1}{6}$ and $ f(3)\equal{}\frac{2}{3(4)}\equal{}\frac{1}{6}$.
In fact, any two reals $ x$ and $ y$ ($ x\ne y$) such that $ (x\minus{}1)(y\minus{}1)\equal{}2$ satisfies $ f(x)\equal{}f(y)$:
Assume $ x\ne y$ and $ \frac{x\minus{}1}{x(x\plus{}1)}\equal{}\frac{y\minus{}1}{y(y\plus{}1)}$. (We need to show that this leads to a contradiction but it instead produces $ (x\minus{}1)(y\minus{}1)\equal{}2$.)
$ y(x\minus{}1)(y\plus{}1)\equal{}x(y\minus{}1)(x\plus{}1)$
$ \implies xy^2\minus{}y^2\plus{}xy\minus{}y\equal{}x^2y\minus{}x^2\plus{}xy\minus{}x$
$ \implies (y\minus{}x)(xy\minus{}x\minus{}y\minus{}1)\equal{}0$
$ \implies xy\minus{}x\minus{}y\minus{}1\equal{}0 \implies (x\minus{}1)(y\minus{}1)\equal{}2$

f’(x)= (-x^2+2x+1)/(x(x+1))^2
f’(x) = 0 has two solutions ,then  the sign of f’(x) changes in two points so f is decreasing or increasing before and after that points .Since f is continuous (except in 0 and -1)  f can’t be injective .

Suppose that the domain is (0,1) can we show that its is an injective function?

f’(x)  is  positive on  (0,1) and the function f  has a “well behavior “ on it  then it is strictly increasing  so it is injective on (0,1).


a) Find all integers $ (m,n)$ such that $ 3^{m}\equal{}2n^3\plus{}1$.

b) Find all integers $ (m,n)$ such that $ 3^{m}\equal{}2n^{q}\plus{}1$, where $ q$ is odd.

b) If q is even, then it had not solution (consider by mod 3).
These problem is case $ p\equal{}3,m\equal{}k\plus{}1$ problem:
$ 1\plus{}p\plus{}p^2\plus{}...\plus{}p^k\equal{}n^q$.
Last problem posted before.


The product of two positive integers is 18. The positive difference of these two integers is 3. What is the sum of the two integers?

Logic + guess check gives $ 3,6$   and sum of $ 3,6$ is $ 3\plus{}6\equal{}9$

So, the question says that the product of two positive integers is 18. One way to solve this problem is to list out the factors of 18 which are 1, 2, 3, 6, 9, and 18, and know that the pairs are 1*18, 2*9, and 3*6. Now all you have to do is trial and error by subtracting the factors in each pair, or 18-1, 9-2, and 6-3. Now that you now the differences of each pair, which are 17, 7, and 3, you can answer the question which asks you to find the sum of the two integers with a positive difference of 3. Since 6-3=3, all you have to do is add 6 and 3 together which gives you your final answer of 9.

Another way to solve this problem is to use guess check and revise. When I think of two numbers that when you multiply together to get 18 and has a positive difference which is not 1 less than the product, I usually think of 9*2 and 6*3 first. You then know by subtraction that 6 and 3 are the two integers and therefore 6+3=9.


Hi. My name is Rafael. I´m brazilian too. Nice to meet you.  :D

Hi! I am Lucas. I know you. You is this forum::D[url]http://www.teorema.mat.br[/url] :D  :theta:  :beta:  :lambda  :cap:  :gamma:  :varepsilon:  :sqrt:  2  :cup:  :downto:  :?  :(  :)  ;)  :lol:  :P  :blush:  :idea:  :|  :roll:  :downto:  :D

Tudo bem Rafael, Lucas,

Desculpen meu portuguese, mas eu nao falo moito em ista bela lingua. O site voce conversa esta moito bom, talvez sea possivel trocar ligacaos com MathLinks.

Any way, welcome to the Mathlinks sites. There is no portuguese site on the forum, but there are spanish sites if you wanted to give it a try. On the other hand, usually there are topics to present yourself on some of the forums, like in this specific forum, you can introduce yourself in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15747]this post[/url].

Seja voce bemvindos.


Hello!

Let $ x_1,x_2,\cdots,x_n$ be positive real numbers. Prove that :
\[ \left(\sum_{k\equal{}1}^nx_k^{{{n\plus{}1}\over{2}}}\right)^n\leq \prod_{k\equal{}1}^n\left(\sum_{k\equal{}1}^n x_k^k\right).\]

[quote="arqady"][quote="yuyang"]Hello!

Let $ x_1,x_2,\cdots,x_n$ be positive real numbers. Prove that :
\[ \left(\sum_{k \equal{} 1}^nx_k^{{{n \plus{} 1}\over{2}}}\right)^n\leq \prod_{k \equal{} 1}^n\left(\sum_{k \equal{} 1}^n x_k^k\right).
\]
[/quote]
It's Holder. :wink:[/quote]
Holder's inequality says:
\[ \sum_{k\equal{}1}^n a_kb_k\leq \left(\sum_{k\equal{}1}^n a_k^p\right)^{1/p}\times \left(\sum_{k\equal{}1}^n b_k^q\right)^{1/q}\]
where $ a_k,b_k>0$ and $ 1/p\plus{}1/q\equal{}1$. How can we apply it to the inequality of this problem?. Thanks

[quote="yuyang"]
Holder's inequality says:
\[ \sum_{k = 1}^n a_kb_k\leq \left(\sum_{k = 1}^n a_k^p\right)^{1/p}\times \left(\sum_{k = 1}^n b_k^q\right)^{1/q}
\]
where $ a_k,b_k > 0$ and $ 1/p + 1/q = 1$. How can we apply it to the inequality of this problem?. Thanks[/quote]
The Holder's inequality says also:
$ \sum_{k = 1}^n \prod_{l = 1}^m a_{kl}\leq \prod_{l = 1}^m\left(\sum_{k = 1}^n a_{kl}^{p_l}\right)^{\frac {1}{p_{l}}}$ $ ,$ where $ a_{kl}\geq0,$ ${ p_l} > 0$ and $ \sum_{l = 1}^m\frac {1}{p_l} = 1.$ :wink:


$\theta = \frac{\pi}{5}$
$n = 4$
$s = \frac{1}{2}$

I did not put theses values in the figure to it become more clear

The Ratio is
$\frac{GH}{IH}= ?$

$\angle ACB=\frac{1}{2}\cdot\frac{180}{5}=18$
$\angle ABC=4\cdot\frac{180}{5}=144$
$\angle BAC=180-(18+144)=18$
$\therefore BA=BC$
$\angle DBC=180-(18+90)=72$
if, $\angle DEC=\frac{180}{5}=36$ then, there is not point E on side BC.

Kouichi Nakagawa, I know that E is not on the side BC, but it doesnt matter, rigth?

because E is on Line that passes through BC.

And we have a little trick on it.

with that angles I set, we have $I = D$  :) can you see that?

I put the image like that on purpose, because is not too ease see that $I=D$

I made another image

$\frac{GH}{DH}=?$

sorry for the last image and for didn't explain  :(


Let a,b,c>0.Prove that
$ \frac{{2\sqrt {ab^3 }  \minus{} ab}}{{\sqrt {a^4  \plus{} b^4 } }} \plus{} \frac{{2\sqrt {bc^3 }  \minus{} bc}}{{\sqrt {c^4  \plus{} b^4 } }} \plus{} \frac{{2\sqrt {ca^3 }  \minus{} ac}}{{\sqrt {a^4  \plus{} c^4 } }} \le \frac{3}{{\sqrt 2 }}$

[quote="TRAN THAI HUNG"]Let a,b,c>0.Prove that
$ \frac {{a^3 \plus{} b^3 \minus{} a^2 b}}{{\sqrt {a^2 b^4 \plus{} a^6 } }} \plus{} \frac {{b^3 \plus{} c^3 \minus{} b^2 c}}{{\sqrt {b^2 c^4 \plus{} b^6 } }} \plus{} \frac {{c^3 \plus{} a^3 \minus{} c^2 b}}{{\sqrt {c^2 a^4 \plus{} c^6 } }} \le \frac {3}{{\sqrt 2 }}$ :weightlift:[/quote]
It's wrong. Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}.$ :wink:

[quote="arqady"][quote="TRAN THAI HUNG"]Let a,b,c>0.Prove that
$ \frac {{a^3 \plus{} b^3 \minus{} a^2 b}}{{\sqrt {a^2 b^4 \plus{} a^6 } }} \plus{} \frac {{b^3 \plus{} c^3 \minus{} b^2 c}}{{\sqrt {b^2 c^4 \plus{} b^6 } }} \plus{} \frac {{c^3 \plus{} a^3 \minus{} c^2 b}}{{\sqrt {c^2 a^4 \plus{} c^6 } }} \le \frac {3}{{\sqrt 2 }}$ :weightlift:[/quote]
It's wrong. Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$ :wink:[/quote]
Yeah,thanks.That is my mistake :blush:.I've edited it. :)

[quote="ocha"]I think this is, Vasc's Inequality

By AM-GM;
$ \sum \frac {2\sqrt {ab^3 } \minus{} ab}{\sqrt {a^4 \plus{} b^4}} \le \frac {ab \plus{} b^2 \minus{} ab}{\sqrt {a^4 \plus{} b^4}} \equal{} \sum \sqrt {\frac {b^4}{a^4 \plus{} b^4}}\le \frac {3}{\sqrt {2}}$[/quote]
Yes, I have tried to make it stronger but it's wrong as Arqady has shown. :wink:


linear algebra
matrix
linear algebra unsolved

Let $A = $$\begin{bmatrix}2&0&x&-1\\1&3&y&0\\0&-1&z&-2\\z&z&xy/11&x\end{bmatrix}$
If $det(A) = x(2z+y+1)-z(y+1-z)$ when $x,y,z$ are distinct reals Find $\frac{3}{2}(x+z)-3y$

The determinant of $A$ is $2zx+3xy-3zy-x^2+2z^2$

Is the answer 3xy - 3zy - (3/2)xx + (3/2)zz + 3z - 3y ??

I think the answer should be in number...

Using moubi's result and factorization, we get (x+z)=2y-1. Put this in the original question, we get the ans $-\frac{3}{2}$

[quote="siuhochung"]Using moubi's result and factorization, we get (x+z)=2y-1.$[/quote]

could you post the details please ?

Expand the determinant of $A$
$det(A)=2xz+3xy-3yz-x^2+2z^2$ (1)

If $det(A)=x(2z+y+1)-z(y+1-z)$ (2)

(1) & (2) give after simplication

$2xy-2yz-x^2+z^2-x+z=0$ or

$(x-z)(2y-x-z-1)=0$ but x,z are distincts so $x+z=2y-1$

And $3/2(x+z)-3y=3/2(2y-1)-3y=-3/2$


How does the method of finite differences work? I searched various places but I couldn't find an explanation I understood... :|

[quote="t0rajir0u"]In simplest terms, we take a polynomial $P_{0}(x)$.  The polynomial

$P_{1}(x) = P_{0}(x+1)-P_{0}(x)$

Is a polynomial with degree one less than that of $P_{0}$ (can you see why?)  Similarly, we take the polynomials

$P_{k+1}(x) = P_{k}(x+1)-P_{k}(x)$

Until we have a polynomial of degree $0$.

In what context exactly do you want to understand how these polynomials are used?[/quote]

Thanks
Some lecture I read showed using this method to find ..for example $P(5)$ knowing $P(1), P(2), P(3),$ and $P(4)$ and knowing what the power of the polynomial was...

Anyways, P(1)=2, P(2)=10, P(3)=30, and P(4)=68...

[quote="ch1n353ch3s54a1l"]and knowing what the power of the polynomial was...[/quote]

This is very important.  The power of the polynomial determines the power of the finite differences.  The idea is this:

Once you have calculated the finite difference down to the linear finite difference, you can easily calculate the next term of the linear polynomial.  Then, by adding, you can calculate the next term of the quadratic, and so forth until you've calculated the next term of the original polynomial.

This is much easier to understand in table format, but I don't possess the patience to $\LaTeX$ an example; my apologies  :(


Okay, I do not actually know the answer, but I would like to. I made it up myself, but I am sure it has already been asked before;

If x and y are positive integers greater than 1, and x is greater than y, then prove that y^x is smaller than or equal to x^y , or give a counter example.

The only reason I say 'smaller than or equal to' is because 2^4 = 4^2

$ \text{\LaTeX}$ help: $ x,y\in\mathbb{N}$ (or $ \mathbb{Z}$)

[quote="ernie"]That just proves it right.

The real proof is:

$ x > 1$, $ y > 1$, $ x$ and $ y$ are integers (idk how to make that in $ \text{\LaTeX}$), $ y^x \geq x^y$

But then a counter example is $ 3^2 > 2^3$[/quote]

No, because 3, is larger than 2, and 3^2 is larger than 2^3. Why do people keep on saying that?

Wickedestjr, what about my post?
It's surely correct.

[quote="Bugi"]Counter example: 5^4<4^5[/quote]

Okay that is a counter example, but can somebody prove that there must be a counter example with out actually looking?

Consider the graphs of $ a^x$ and $ x^a$ for $ a$ slightly over $ 1$.

On one hand, you know that for $ x$ around $ 2$ or whatever, $ x^a$ is gonna be bigger than $ a^x$.

However, you know that since $ a^x$ is an exponential function and $ x^a$ is a power function, $ a^x$ must eventually beat out $ x^a$.

So, there are times for $ x>a$ for which $ x^a>a^x$, and times where $ a^x>x^a$.

Alright, thank you for that proof. Can somebody please tell me what the irrational number e is. What is it?

You've heard this sentence a million times:

Google is your friend.

Really, it's easier to google something, look it up on Wikipedia, AoPS Wiki or Wolfram Mathworld...

And the answer to your question: $ e$ is the base of natural logarithm, known as the Euler/Napier number.

$ e \equal{} \sum_{n \equal{} 0}^{\infty}\frac {1}{n!} \equal{} \lim_{n\to \infty}\left(1 \plus{} \frac {1}{n}\right)^n$.

[quote="ernie"]That just proves it right.

The real proof is:

$ x > 1$, $ y > 1$, $ x$ and $ y$ are integers (idk how to make that in $ \text{\LaTeX}$), $ y^x \geq x^y$

But then a counter example is $ 3^2 > 2^3$[/quote]

  $ y^x \geq x^y$

It is true with $ x;y \ge\ e$ and $ x \ge\ y$

We have $ y^x \geq x^y$

$ \leftrightarrow x.lny \ge\ y.lnx \leftrightarrow \frac{x}{lnx} \ge\ \frac{y}{lny}$

Let $ f(t)=\frac{t}{lnt} \rightarrow f '(t)=\frac{lnt-1}{(lnt)^2} \ge\ 0$

So with $ x \ge\ y$ we have $ f(x) \ge\ f(y)$

We have done!

Alright thanks, but what is the 'ln' mean?

ln is a logarithm.  Once again, [b]Google is your friend[/b].


quadratics
ratio
algebra
number theory unsolved
number theory

Show that the equation

m/n + (n+1)/m = 4 doesnt have any solutions for positive integers m,n.

I think this was in a Romanian TST a long time ago, and that I have posted it on the forum (also a long time ago :)): show that if $\frac mn+\frac{n+1}m$ is an integer, then it's $3$.

Oh, my God! How many similar problem are solved here!
And again a descent method!

...I cant find it!!!  :(  :(  :(

Can sb please post the link or a solution!!! it's very important...

Is it your homework?

Anyway, $\frac mn+\frac{n+1}m=q$ implies $m^2+n^2+n-qmn=0$. If $m=n$ then we immediately obtain $m=n=1$. Otherwise we have two cases:
1. $m>n$. Consider quadratic equation $m^2+n^2+n-qmn=0$ wrt $m$. It has second integer root $m'=\frac{n^2+n}{m}<m$. So $\frac m'n+\frac{n+1}m'=q$ also and $m'+n<m+n$.
2. $m<n$. Consider quadratic equation $m^2+n^2+n-qmn=0$ wrt $n$. It has second integer root $n'=\frac{m^2}{n}<n$. So $\frac m{n'}+\frac{n'+1}m=q$ also and $m+n'<m+n$.
Therefore we obtain a descent method and finally we will come to case $m=n$ and then $q=3$.

Here's a different proof (it reduces this problem to a well-known one, which can also be solved the way Myth solved this one):

Put $n=p_1^{a_1}\cdot\ldots\cdot p_k^{a_k}$. We have $n|m^2,m|n(n+1)$. We then have $m=p_1^{b_1}\cdot\ldots\cdot p_k^{b_k}\cdot q_1^{u_1}\cdot\ldots\cdot q_t^{u_t}$, where $n+1=q_1^{u_1}\cdot\ldots\cdot q_t^{u_t}\cdot N$ for some natural $N$.

After writing $mn|m^2+n(n+1)$ in terms of the $p$'s and $q$'s defined above, we immediately find $a_i=2b_i,\ \forall i$, and then we find that the ratio we want to show is $3$ is $\frac{q_1^{u_1}\cdot\ldots\cdot q_t^{u_t}+N}{p_1^{b_1}\cdot\ldots\cdot p_k^{b_k}}$. If we put $t=p_1^{b_1}\cdot\ldots\cdot p_k^{b_k}$, and $a=q_1^{u_1}\cdot\ldots\cdot q_t^{u_t},b=N$, we have the following problem:

If $ab=t^2+1$ and $t|a+b$, show that $a+b=3t$. In other words, $a|t^2+1,t|a^2+1$, and we have to show that $\frac{a^2+t^2+1}{at}=3$. This one has been solved at least three times, if I remember correctly.

Let $x=2p+r,y=2q+s\ (p,r\in \mathbb{Z},\{r,s\}=\{0,1\})$,then $x^2+y^2=4(p^2+q^2+pr+qs)+r^2+s^2$.
Therefore $4|x^2+y^2\Longleftrightarrow r^2+s^2=0\Longleftrightarrow r=s=0\Longleftrightarrow\  x$ and $y$ are even.

In using this, $\frac{m}{n}+\frac{n+1}{m}=4\Longleftrightarrow m^2+n(n+1)=4mn$ Since $n(n+1)$ is odd,

There aren't existed $m,n\in \mathbb{N}$ satisfying this equation.


Starting at L and proceeding downward diagonally either left or right to each subsequent
letter, how many different paths spell the word LAPS?

                             L
                           A A
                          P P P
                        S S S S

[hide="Solution"]
In each of the first three rows you have two paths you can take, giving $ 2^{3}\equal{}\boxed{8}$ total paths.[/hide]

This is the basic idea of Pacal's triangle. Every entry is equal to the number of paths to that point (how you described it), and the sum of the entries of a row beginning with $ n$ is $ 2^n$.


why has ralph nadar kept on running for every presidential election?

First, it's Ralph Nader. After you correct it, I will answer.

why has ralph nader run for president so consecutively?

If he's so beast, then he keeps winning, right!

No, he gives the voters a choice. :mad:

He is responsible for the War in Iraq, frankly put. :wink:

But Gore easily could have been much worse...