We also have the following result
\[ \sum\frac{a\plus{}bc}{1\plus{}ab} \ge \frac{3(\sqrt{3}\plus{}1)}{4}
\]
for any $ a,b,c \ge 0,ab\plus{}bc\plus{}ca\equal{}1$ :)

[quote="vodanh_tieutot"][quote="NguyenDungTN"]Let $ a,b,c \ge 0, ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \frac {ab \plus{} 1}{a \plus{} b} \plus{} \frac {bc \plus{} 1}{b \plus{} c} \plus{} \frac {ca \plus{} 1}{c \plus{} a} \ge 3$

[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php
[b]Welcome to VI![/b][/quote]
ineq <=> $ p^2 \plus{} (r \minus{} 3)p \plus{} 2r \plus{} 3\ge 0$
$ \delta \equal{} r^2 \minus{} 18r \plus{} 1\le 0$ (with $ r\le \frac {1}{3.\sqrt {3}}$)
so ineq was proved  :D[/quote]
When equality holds?
[quote="can_hang2007"]We also have the following result
\[ \sum\frac {a \plus{} bc}{1 \plus{} ab} \ge \frac {3(\sqrt {3} \plus{} 1)}{4}
\]
for any $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 1$ :)[/quote]
Nice problem, Vo Quoc Ba Can!

[quote="vodanh_tieutot"][quote="NguyenDungTN"]Let $ a,b,c \ge 0, ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \frac {ab \plus{} 1}{a \plus{} b} \plus{} \frac {bc \plus{} 1}{b \plus{} c} \plus{} \frac {ca \plus{} 1}{c \plus{} a} \ge 3$

[b]Vietnam Inequality Forum:[/b] http://www.batdangthuc.net/forum/index.php
[b]Welcome to VI![/b][/quote]
ineq <=> $ p^2 \plus{} (r \minus{} 3)p \plus{} 2r \plus{} 3\ge 0$
$ \delta \equal{} r^2 \minus{} 18r \plus{} 1\le 0$ (with $ r\le \frac {1}{3.\sqrt {3}}$)
so ineq was proved  :D[/quote]

Why you can say that $ p^2 \plus{} (r \minus{} 3)p \plus{} 2r \plus{} 3\ge 0$

Can you please explain it to me?

Thank you very much.

[quote="vodanh_tieutot"]Let $ a,b,c \ge 0, ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \frac {ab \plus{} 1}{a \plus{} b} \plus{} \frac {bc \plus{} 1}{b \plus{} c} \plus{} \frac {ca \plus{} 1}{c \plus{} a} \ge 3$

ineq <=> $ p^2 \plus{} (r \minus{} 3)p \plus{} 2r \plus{} 3\ge 0$
$ \delta \equal{} r^2 \minus{} 18r \plus{} 1\le 0$ (with $ r\le \frac {1}{3.\sqrt {3}}$)
so ineq was proved  :D[/quote]

Why $ \delta \equal{} r^2 \minus{} 18r \plus{} 1\le 0$  with $ r\le \frac {1}{3.\sqrt {3}}$????.

Sorry, but I don't see it.  :maybe:

Can anyone,please,  explain it to me.  :wink:

Thank you very much.  :)

[quote="manlio"]NguyenDungTN, can you please post your solution?

I cannot solve it :(

Thank you very much.[/quote]
[hide="Hint"]We have:
$ \frac {ab \plus{} 1}{a \plus{} b} \plus{} \frac {bc \plus{} 1}{b \plus{} c} \plus{} \frac {ca \plus{} 1}{c \plus{} a} \ge 3$
$ \Leftrightarrow \sum_{cyc} (ab\plus{}1)(c\plus{}a)(c\plus{}b) \ge 3(a\plus{}b)(b\plus{}c)(c\plus{}a)$
$ \Leftrightarrow \sum_{cyc} (ab\plus{}1)(c^2\plus{}1) \ge 3[(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc]$
$ \Leftrightarrow (a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca\plus{}abc(a\plus{}b\plus{}c)\plus{}3\plus{}3abc \ge 3(a\plus{}b\plus{}c)$
$ \Leftrightarrow (a\plus{}b\plus{}c)^2\plus{}abc(a\plus{}b\plus{}c\plus{}3)\plus{}2 \ge 3(a\plus{}b\plus{}c)$
Now let $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca\equal{}1,r\equal{}abc$
The inequality becomes:
$ p^2\plus{}r(p\plus{}3)\minus{}3p\plus{}2 \ge 0$
$ \star p>2\Rightarrow p^2\minus{}3p\plus{}2\plus{}r(p\plus{}3)\equal{}(p\minus{}1)(p\minus{}2)\plus{}r(p\plus{}3)>0$
$ \star p \le 2$
By Schur inequality, we have:
$ p^3\plus{}9r \ge 4pq$ or $ r \ge \frac{4p\minus{}p^3}{9}$
So we need to prove that:
$ p^2\minus{}3p\plus{}2\plus{}(p\plus{}3)\frac{4p\minus{}p^3}{9} \ge 0$
$ \Leftrightarrow p^4\plus{}3p^3\minus{}13p^2\plus{}15p\minus{}18 \le 0$
$ \Leftrightarrow (p\minus{}2)(p^3\plus{}5p^2\minus{}3p\plus{}9)$
Because $ p \le 2$ and $ p^3\plus{}5p^2\minus{}3p\plus{}9\equal{}p^3\plus{}4p^2\plus{}\left(p\minus{}\frac{3}{2}\right)^2\plus{}\frac{27}{4}>0$
so we are done.
Equality hold if and only if $ a\equal{}b\equal{}1,c\equal{}0$ and permutations.[/hide]

Dear NguyenDungTN,

thank you very much for your nice solution and inequality.  :)

[quote="can_hang2007"]We also have the following result
\[ \sum\frac {a \plus{} bc}{1 \plus{} ab} \ge \frac {3(\sqrt {3} \plus{} 1)}{4}
\]
for any $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 1$ :)[/quote]

Dear can_hang2007 have you a solution for this very hard inequality? I tried with full expansion by applying p,q,r method but it is almost impossible for me. :oops:

Thank you very much.

Can't we see the solution of the first problem without using $ pqr$ method?

[quote="can_hang2007"]We also have the following result
\[ \sum\frac {a \plus{} bc}{1 \plus{} ab} \ge \frac {3(\sqrt {3} \plus{} 1)}{4}
\]
for any $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \equal{} 1$ :)[/quote]
My urgly solution.

Dear can_hang2007,

thank you very much for your solution at this very hard inequality. :)


Hi,
I looked on the latex guide and couldn't find much.
Is there any way to page break?
Fore xample, say im doing texniccenter latex, and have 1.5 pages for problem 1. And i want problem 2 to be on the third page. How do i do that without doing "\linbreak" the needed number of times?

Is there a \pagebreakuntilendofpage type function?

Thanks

If you use the USAMTS template (which can be found on the USAMTS website), page breaks will be added automatically between solutions.

If you don't want to do that, you can use \newpage

Also, it says "You only need to submit this file and the PDF file with your solutions (and please attach any images you include)."

Why can't I just submit the pdf file?

That's outdated...you can just upload the PDF file.

Hmmm I've always used \begin{newpage} to make a new page.  It typically produces a few bad box errors and stuff though.  Is there another \lpagebreakish command?


Given $ 2x^{2} \plus{} y^{2} \plus{}2xy \minus{}12x \minus{}8y \minus{}5\equal{}0$, find the maximum and minimum values of $ 7x\plus{}4y$.

hello, the maximum ist $ 47$ and will be attained for $ x\equal{}5,y\equal{}3$ , the minimum will reached for $ x\equal{}\minus{}1,y\equal{}1$ and is $ \minus{}3$.
Sonnhard.

This problem is readily solved by geometric means after [hide="a bit of algebra"]
Write as:
[color=darkblue][i]Given $ x^2 \plus{} (x \plus{} y)^2 \minus{} 4x \minus{} 8(x \plus{} y) \plus{} 5 \equal{} 0$, find the bounds on $ 3x \plus{} 4(x \plus{} y)$[/i][/color]

Perform the change of variables $ z \equal{} x \plus{} y$ to rewrite as:
[color=darkblue][i]Given $ x^2 \plus{} z^2 \minus{} 4x \minus{} 8z \plus{} 5 \equal{} 0$, find the bounds on $ 3x \plus{} 4z$[/i][/color]

Complete squares and define an auxiliary variable $ M$ to rewrite as:
[color=darkblue][i]Given $ (x \minus{} 2)^2 \plus{} (z \minus{} 4)^2 \equal{} 15$ and $ 3x \plus{} 4y \equal{} M$, find the bounds on $ M$.[/i][/color][/hide]


superior algebra
superior algebra unsolved

Let $ (G,.)$ is a group .

$ \exists n \in N \ ; \ n >1 : (a.b)^n\equal{}a^n.b^n \ , \forall a,b \in G$ .

Prove that :

i ) $ (a.b)^{n\minus{}1}\equal{}b^{n\minus{}1}.a^{n\minus{}1}$

ii) $ a^{n}.b^{n\minus{}1}\equal{}b^{n\minus{}1}.a^{n}$

$ \forall a,b \in G$ .


algebra
polynomial
limit
Functional Analysis
advanced fields
advanced fields unsolved

Let $A$ be a contraction ($\|A\|\le 1$) on a complex Hilbert space, and let $D$ be the unit disc in the plane. Then, for every complex polynomial $p$, the following equality holds: $\|p(A)\|\le\sup_{z\in D}|p(z)|$.

I think this was used in my analysis class as a step in the proof that every linear constant coefficient PDE has a weak solution. However my professor went over the proof quite quickly and I didn't understand very well... I will think on it though :)

Spectral radius $r(A)=\lim\|A^{n}\|^{1/n}\le1$ so $\mathrm{Sp}\,A\subset D$ and thus $\|p(A)\|=\max|p\,(\mathrm{Sp}\,A)|\le\max|p\,(D)|$.

[quote="Vladimir Lubyshev"]$\|p(A)\|=\max|p\,(\mathrm{Sp}\,A)|$[/quote]

In other words, the norm of $p(A)$ is equal to its spectral radius? This certainly isn't true for all operators.

Oh... yes it is true for $A$ self-adjoint and $p$ having no poles inside $\mathrm{Sp}\,A$. I forgot that.

I'll think on the problem when I have time. But do you know for sure the statement is true?

I even have a proof. It's only here so that you guys can enjoy it :).


pigeonhole principle
number theory solved
number theory

The real number x between 0 and 1 has decimal representation .a1a2a3a4... with the following property: the number of distinct blocks of the form akak + 1ak + 2...ak + 2003 as k ranges through all positive integers, is less than or equal to 2004. Prove that x is rational.

Did anyone think of this yet? Quite a nice question, but not so easy I think... I got somewhere using Pigeonhole Principle but I didn't manage to finish it off so far.


Yup, I do suscribe to it.  How the hell do you find tension in a magazine?  And what illegal stuff are you referring to?

[quote="nutz_for2.718281828"]Using preformance enhancing drugs to improve your preformance. One guy hit farther than Babe Ruth, all thanks to drugs.[/quote]
That was an article a couple of issues back, wasn't it?  A magazine reporting about illegal stuff, I don't see how that's particularly unusual...

[quote="nutz_for2.718281828"]I do. I've found a lot of tension between a few countries. I've also found a few illegal things in there.[/quote]
What sort of tensions,and what sort of illegal things are we speaking about?

[quote="star99"]its ok, but it has too much stuff.  i like The Week and Time better[/quote]

I think it is a great magazine and extremely insightful when compared to those two magazines (the quality of Time has decreased dramatically, IMO, it is now nothing more than a pile of cultural biases plopped on top of several events and alleged "trends").  I do not usually read as much of it as I'd like though, just cause it's kinda dense.


The baseball catcher needs to stop the player on first base from stealing second base. Approximately how far must the catcher throw the ball to get to second base? A regular baseball diamond is a square 90 feet on each side.

A) 90 ft B) 127 ft C) 180 ft D) 64 ft E)100 ft

Off the top of my head, I know the answer is 127.  The distance the catcher has to throw is the hypotenuse of the right triangle formed by the first and second base paths.  So it's a triangle, 90 feet on two sides.  The third side is the desired distance.  The Pythagorean Theorem works.


group theory
superior algebra
superior algebra unsolved

I just started my group theory class:

$ \mathbb{Q}'$ of $ \mathbb{Q}$ is the collection of cauchy sequences in $ \mathbb{Q}$ mod the equivalence relation of being co-Cauchy. The arithmetic is defined by:

$ A\plus{}B \equal{} (a_{n} \plus{}b_{n})$
$ A\minus{}B \equal{} (a_{n} \minus{}b_{n})$
$ AB \equal{} (a_{n} *b_{n})$
$ A/B \equal{} (a_{n} /b_{n})$

How do you check if this is a field? Can someone show an example? Thanks!

Presumbably you have the definition of "field" ... check it.
Part of this will be to check that that your operations are well-defined.  For example, to check that + is well defined, you need:  if $ (a_n)$ and $ (u_n)$ are co-Cauchy and $ (b_n)$ and $ (v_n)$ are co-Cauchy, then show $ (a_n\plus{}b_n)$ and $ (u_n\plus{}v_n)$ are co-Cauchy.

How come "well definedness" is not included amongst the other definitions of the field that I see such as closure, associativity, distributivity, and commutativity?

How do you show co-Cauchy with the above? Do you just add each element together and just show that each resulting sum still forms a Cauchy sequence?

Well-definedness is more likely a part of "the addition/etc. can be defined that way". That is, you need to show that the result is again a Cauchy sequence and that it doesn't depend on the equivalence class.


ratio
geometry
circumcircle
conics
geometric transformation
reflection
homothety

Hi, everyone  :roll:

Prove the following:
if R be any point on a circle, A and B fixed points on a diameter and equidistant from the center,
the envelope of a line which cuts harmonically the two circles with A, B as centers and AR, BR as radii
is independent of the position of R on the circle.

Thanks  :)
Please your answers. I am studying materials related to this post nowadays.

Let $ (O)$ be the circle with $ R$ on it. Let a line $ l$ cut the circle $ (A)$ at $ X, Y$ and the circle $ (B)$ at $ Z, T.$ Assume that the cross ratio $ \frac {\overline{XZ}}{\overline{XT}} \cdot \frac {\overline{YZ}}{\overline{YT}} \equal{} \minus{} 1$ is harmonic. Let $ M, N$ be midpoints of $ XY, ZT$ and $ (M), (N)$ circles with diameters $ XY, ZT,$ intersecting at $ P, P'.$ Since $ PX \perp PY,$ these lines bisect $ \angle ZPT$ internally and externally, $ (M)$ is P-Apollonius circle of $ \triangle PZT$ with circumcircle $ (N)$ $ \Longrightarrow$ $ (M) \perp (N).$ Inversion in $ (M)$ takes $ (N)$ to itself and $ Z$ to $ T$ $ \Longrightarrow$ $ \overline{MZ} \cdot \overline{MT} \equal{} \minus{} \overline{MX} \cdot \overline{MY}$ $ \Longrightarrow$ $ \frac {p(M, (B))}{p(M, (A))} \equal{} \minus{} 1.$ In exactly the same way, $ \frac {p(N, (B))}{p(N, (A))} \equal{} \minus{} 1.$ Since powers of $ M, N$ to $ (A), (B)$ are in the same ratio $ r \equal{} \minus{} 1,$ $ M, N$ are on a circle coaxal with $ (A), (B),$ centered on $ AB$ and passing through $ R.$ As $ \frac {\overline{AO}}{\overline{BO}} \equal{} r \equal{} \minus{} 1,$ $ \odot(MNR) \equiv (O).$ The feet $ M, N$ of perpendiculars $ AM \perp l, BN \perp l$ from the fixed points $ A, B$ to the arbitrary line $ l$ are therefore on the fixed circle $ (O)$ $ \Longrightarrow$ $ l$ is tangent to a conic with foci $ A, B$ and pedal circle $ (O).$

Suppose that $ A,B$ lie inside the circle $ (O)$ centered at the midpoint of $ AB$ with radius $ OR \equal{} \varrho.$ Let the variable line $ \ell$ cut $ (A)$ at $ P,Q$ and $ (B)$ at $ X,Y.$ Let $ M$ denote the orthogonal projection of $ A$ on $ \ell,$ which is the midpoint of the chord $ PQ.$ Since $ (P,Q,X,Y) \equal{} \minus{} 1,$ by Newton's theorem we have $ MP^2 \equal{} MX \cdot MY$ $ \Longrightarrow$ $ p(M,(A)) \equal{} p(M,(B)).$ Hence $ M$ lies on the circle centered at the midpoint of $ AB$ and coaxal with $ (A)$ and $ (B),$ i.e. $ M \in (O).$ Let $ A'$ be the reflection of $ A$ across $ \ell$ and let $ N \equiv \ell \cap BA'.$ $ A'$ moves on the homothetic circle $ (B, 2\varrho)$ of $ (O)$ through the homothety centered at $ A$ with coefficient $ 2.$ Therefore $ NA \plus{} NB \equal{} NA' \plus{} NB \equal{} 2\varrho \equal{} \text{const}$ $ \Longrightarrow$ Locus of $ N$ is the ellipse $ \mathcal{E}$ with foci $ A,B$ and major axis $ 2\varrho.$  Since $ \ell$ is the external bisector of $ \angle ANB,$ it follows that $ \ell$ is tangent to $ \mathcal{E}.$ When $ A,B$ lie outside $ (O),$ analogous reasoning yields that $ \ell$ is tangent to the hyperbola with foci $ A,B$ and major axis $ 2\varrho.$


$ a,b,c,p,q\in R^\plus{}$
$ abc\equal{}1$ and $ p\ge q$
find minimum value of $ \dfrac{a^p}{b^q\plus{}c^q}\plus{}\dfrac{b^p}{c^q\plus{}a^q}\plus{}\dfrac{c^p}{a^q\plus{}b^q}$

I think that p and q are in IN and $ p > q$
Chebychev:
$ S \equal{} \dfrac{a^p}{b^q \plus{} c^q} \plus{} \dfrac{b^p}{c^q \plus{} a^q} \plus{} \dfrac{c^p}{a^q \plus{} b^q}\geq \frac {1}{3}(a^p \plus{} b^p \plus{} c^p)\left(\sum \frac {1}{a^q \plus{} b^q}\right)$
caushy:
$ \left(\sum \frac {1}{a^q \plus{} b^q}\right)(2(a^q \plus{} b^q \plus{} c^q))\geq9$
$ \Leftrightarrow \left(\sum \frac {1}{a^q \plus{} b^q}\right)\geq\frac {9}{2(a^q \plus{} b^q \plus{} c^q)}$
therefor:
$ S \geq\frac {3}{2}\times\frac {a^p \plus{} b^p \plus{} c^p}{a^q \plus{} b^q \plus{} c^q}$
with Chebychev:
$ S \geq \frac {3}{2}\times\frac {a^p \plus{} b^p \plus{} c^p}{a^q \plus{} b^q \plus{} c^q} \geq \frac {3}{2}\times\frac {1}{3}\times\frac {(a \plus{} b \plus{} c)(a^{p \minus{} 1} \plus{} b^{p \minus{} 1} \plus{} c^{p \minus{} 1})}{a^q \plus{} b^q \plus{} c^q}\geq \frac {3}{2}$
because with AM-GM:
$ a \plus{} b \plus{} c \geq3$
then:
$ S\geq \frac {3}{2}$
with equality only if $ a \equal{} b \equal{} c$.

[quote="pqrs"]$ a,b,c,p,q\in R^ \plus{}$
$ abc \equal{} 1$ and $ p\ge q$
find minimum value of $ \dfrac{a^p}{b^q \plus{} c^q} \plus{} \dfrac{b^p}{c^q \plus{} a^q} \plus{} \dfrac{c^p}{a^q \plus{} b^q}$[/quote]
Let $ r \equal{} \frac{p}{q}$. And $ x \equal{} a^q$, $ y \equal{} b^q$, $ z \equal{} c^q$.
Then we have to find the minimum of $ \frac{x^r}{y\plus{}z} \plus{} \frac{y^r}{z\plus{}x} \plus{} \frac{z^r}{y\plus{}x}$ when $ xyz\equal{}1$.
We prove that $ \frac{x^r}{y\plus{}z} \plus{} \frac{y^r}{z\plus{}x} \plus{} \frac{z^r}{y\plus{}x} \ge \frac{3}{2}$ $ \iff$
$ \sum_{cyc} 2x^r(x\plus{}y)(x\plus{}z) \equal{} \sum_{cyc} 2x^{r\plus{}2} \plus{} 2x^{r\plus{}1}y \plus{} 2x^{r\plus{}1}z \plus{} 2x^ryz \ge 3(x\plus{}y)(y\plus{}z)(z\plus{}x)$.
Written by muirhead means:
$ 6[r\plus{}2,0,0] \plus{} 12[r\plus{}1,1,0] \plus{} 6[r,1,1] \ge 12[1,1,1] \plus{} 18[2,1,0]$. Which is obviously true when homogenizing since $ r \ge 1$.

More directly, by Chebyschev

$ \sum \frac {a^{p \minus{} q \plus{} q}}{b^q \plus{} c^q} \ge (\sum \frac {a^{p \minus{} q}}{3})(\sum \frac {a^q}{b^q \plus{} c^q})$

By AM - GM,

$ \sum \frac {a^{p \minus{} q}}{3} \ge 1$.

But $ \sum \frac {a^q}{b^q \plus{} c^q} \ge \frac {3}{2}$ is Nesbitt.


$a,b,c \in \mathbb{R}$
For all $n \in \mathbb{N}$ , we have:
$[na] + [nb] = [nc]$
Prove that at least one of those numbers (a,b,c) is an integer.

Hmm... I have solution but it is messy, I have another idea, so I'll try to solve it in easier way and then post the solution.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=density&t=4568

Pierre.


complex analysis
function
integration
calculus
derivative
complex analysis unsolved

Hello to all!

I'm reading an article about series expansions and I found, what I think is a really strange result:

We have

$ \Theta(x)$ an holomorphic function inside a contour S, having only a root $ a$ inside S;

$ x$ the affixe of a point inside S;

For every point $ z$ of the contour S:

\[ |\Theta(x)|<|\Theta(z)|\]

and the author states that the following:

[i]The equation [/i]

\[ \Theta(z)\minus{}\Theta(x)\equal{}0\]

[i]has only one root $ z\equal{}x$ inside $ S$, as we can see in this equality[/i]
\[ \frac{1}{2i\pi}\int_{S}\frac{\Theta'(z)dz}{\Theta (z)\minus{}\Theta(x)}\equal{}\frac{1}{2i\pi}\left[\int_{S}\frac{\Theta'(z)dz}{\Theta (z)}\plus{}\Theta(x)\int_{S}\frac{\Theta'(z)dz}{\Theta^{2}(z)}\plus{}\ldots\right]\equal{}\frac{1}{2i\pi}\int_{S}\frac{\Theta'(z)dz}{\Theta(z)}\]

[i]where the first and last members represent respectively, the number of roots of the considered equation and the number of roots of $ \Theta(z)\equal{}0$.[/i]

I understand the first step of the equality (he used the same expansion as $ \frac{1}{1\minus{}x}$), but not the second... it looks as if he just made $ \Theta(x)\equal{}0$

Does anyone know what is happening there?

Thank you

Key words to look for: [url=http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem]Rouche's theorem[/url], [url=http://en.wikipedia.org/wiki/Argument_principle]argument principle[/url].

In the second term, $ \frac{\Theta'}{\Theta^{2}}$ is the derivative of $ \minus{}\frac{1}{\Theta}$, and we're integrating around a closed curve. It evaluates to zero by the Fundamental Theorem.

Thanks a lot!!!   :)

I thought I was never going to figure it out!! :oops:


The sequence of the Human Genome was published 2003. Some people are using this code to generate music or to cipher messages or use DNA to do computing.

I was wondering if there is any correlation between the DNA code and the sequence of prime numbers. I am not a mathematician but generally interested in Science, so forgive me if this  a "weired" idea.
But would it be not funny to be able to predict a DNA motif with number, especially prime number, theory?

Thanks!

Why primes and not composite, or Fermat numbers ?
Do you have any reasonable reason to believe there is any link with [b]any[/b] known number ?
To give my sceptical answer:  You can "find" prime numbers anywhere in nature, but that's mostly (and most probably) numerology not mathematics.

P.S: This post does not belong to the number theory section (tough it's "related" to prime) it should go in the Round Table section.
P.S.S: Why do you need the human genome to generate music ? You could aswell (and with the same sucess) use the inscription on the butter you bought (let's suppose) today :).


I was thinking to put this in SC competition but I decided not to because it might be hard for some beginners. Anyway, this is quite good thinking question so here it goes:

d, a positive integer, is a nice number that can be resulted from either of expression a+b+c or a*b*c where a,b, and c are each distinct positive number.  Prove that (a+b+c)*d equals to a*b*c*d by defining each value for a,b,c, and d (a,b,c, and d are each different numbers) if a,b,c are consecutive numbers while a,b,c,d aren't.

Wording might be pretty confusing but you will find this easy.

But I thought:

[hide]a+b+c = d

d = a*b*c

By transitivity,

a+b+c = a*b*c

Multiplying both sides by d,

(a+b+c)*d = a*b*c*d[/hide]

[quote]Wording might be pretty confusing but you will find this easy.[/quote]

No offense, but I'll say.

Remember, you have to define the value for a,b,c, and d.

Excellent. Now, prove that it is only positive pair that's possible. Also, state the another pair that isn't all positive but satisfies the requirements I asked.


Probably there is a more astute solution but log (1/x-1) is convex on ]0..1/2] + Jensen.


Let $O =(0,0), A = (1,3), B = (4,1)$ $D$ is on $OB$ such that $AD$ is perpendicular to $OB$. Find the area of triangle $OAD$.

The line through $O$ and $B$ is $y=\frac{1}{4}x$

So let $D=(t, \frac{1}{4})t$ $\vec{AD}=[t-1 ; \frac{1}{4}t-3]$ $\vec{OB}=[4;1]$

$AD$ is perpendicular to $OB$ so : $4t-4+\frac{1}{4}t-3=0$ $t=\frac{28}{17}$

Now we know coordinations of $A,O,D$ so it's easy to calculate $AD$ and $DO$,
Area of $\triangle{ADO}= \frac{1}{2}AD \cdot DO$


probability
parameterization
limit
integration
calculus
probability and stats

$ X_1, ..., X_5$ are iid random variables having exponential distribution with parameter $ \lambda$. Find $ P(X_1 \geq 3|X_1 \plus{} ... \plus{} X_5\equal{}5)$

[quote="roah"]You may think about $ X_{1}$ as about minimum of five uniformly selected on $ [0;5]$ points.[/quote]
Correct answer is $ 0,0256$, but I don't know how to find it  :wink: If we think like You said we get $ 0,01024$ which is wrong. When I think about $ X_{1}$ as about minimum of FOUR uniformly selected on $ [0;5]$ points we get $ 0,0256$. But, could You explain this thinking? Why should I think about $ X_1$ like You wrote? I would be appreciate :)

it should be four, since the FIFTH point is AT 5 really, therefore we dont need to think about it.

But I still don't understand why these point are uniformly selected from $ [0,5]$, I wrote in my first post that $ X_1,...,X_5$ are exponential distributed random variables  :wink:. Any ideas to explain this fact?  :blush:

thinking about it today i came across the same issue you just asked...

the exponential thing doesnt seem to play a role if we can generalize like this.. as long as its a continuous distribution and $ X_n \ge 0$ i think we can use the same argument..?

why don't we try to compute for example the exact conditional law of $ X_1$, give that $ X_1 + \ldots, X_n = A$:
$ X = X_1$ is independent of $ Y = X_2 + \ldots + X_n$, and $ Y$ has a Gamma probability distribution with parameter $ (\lambda, n - 1)$:
\[ P(Y \in [y; y + dy]) = G(y) dy = \frac {\lambda^{n - 1}}{\Gamma(n - 1)} y^{n - 2} \exp( - \lambda y) dy
\]
Now, what we are trying to compute is:
\[ F(b,a) = \lim_{\epsilon \to 0} \frac {P[ (X \leq b) \cap (X + Y \in [a; a + \epsilon])] }{ P[ (X + Y \in [a; a + \epsilon])] } = P(X \leq b | X_1 + \ldots + X_n = a)
\]
where
\[ P[ (X \leq b) \cap (X + Y \in [a; a + \epsilon]) = A(b,a + \epsilon) - A(b,A)
\]

\[ P[ (X + Y \in [a; a + \epsilon]) = A(\infty,a + \epsilon) - A(\infty,A)
\]

\[ A(b,a) = P[(X \leq b) \cap (X + Y \leq a)] = \int_{x = 0}^b \lambda \exp( - \lambda x) [\int_{y = 0}^{a - x} G(y)dy] dx
\]
hence
\[ F(b,a) = \frac {\partial_{a} A(b,a) }{\partial_a A(\infty, a)} = \frac {\int_{x = 0}^b \lambda \exp( - \lambda x) G(a - x) dx }{\int_{x = 0}^{a} \lambda \exp( - \lambda x) G(a - x) dx } = \frac {\int_{x = 0}^b x(a - x)^{n - 2} dx }{\int_{x = 0}^{a} x(a - x)^{n - 2} dx }
\]
whence $ P(X \leq b | X_1 + \ldots + X_n = a) = (\frac {b}{a})^n \frac {\int_{u = 0}^1 u(\frac {a}{b} - u)^{n - 2} du }{\int_{u = 0}^{1} u(1 - u)^{n - 2} du } = \phi(\frac {b}{a})$.
This shows that the conditional density of $ X_1$ given that $ X_1 + \ldots + X_n = a$ is something like $ Cx(a-x)^{n-2} dx$
(which is a little weird for $ n=2$, so I may have made a computation mistake somewhere  :blush: )

Let me try to pick up some loose ends and give some more details to the above posts.

Look at the more general problem $ P(X_1 \le b| X_1 \plus{} \ldots \plus{} X_n \equal{} a) \equal{} ?$, where $ n \ge 2$, the $ X_i$ are iid exponential random variables (parameter $ \lambda >0$) and $ 0 \le b \le a$ (as considered by alekk).

The connection to the uniform distribution is the following: Because of the exponential distribution (which is crucial for this) the points $ Y_n : \equal{} X_1 \plus{} \ldots \plus{} X_n$ form a Poisson point process on the positive part of the real line. The random set of points $ Y \equal{} \{ Y_n: n \ge 1\}$ has the following properties:
For every interval $ I$ of length $ L$ the number of points of $ Y$ in $ I$ is Poisson-distributed (with parameter $ \lambda L$) and for a fixed number $ n$ of points the conditional distribution of the positions of these $ n$ points in $ I$ is: The positions are independent and uniformly distributed.  The condition $ X_1\plus{} \ldots \plus{} X_n \equal{} a$ is the same as "in the interval $ [0,a)$ there are exactly $ n\minus{}1$ points of $ Y$". The condition $ X_1\plus{} \ldots \plus{} X_n \equal{} a$ and $ X_1 \le b$ is the same as "in the interval $ [0,a)$ there are exactly $ n\minus{}1$ points of $ Y$ and at least one of them is in $ [0,b]$". (This elaborates the idea of roah, and the comment of kenn4000 that really only $ n\minus{}1$ points are uniformly chosen, and $ X_1$ is indeed the minimum of $ n\minus{}1$ independent uniformly distributed random variables.)

To apply this to the above problem: $ P(X_1 \le b| X_2 \plus{} \ldots \plus{} X_n \equal{} a) \equal{}$
$ 1 \minus{}P(Y$ has no point in $ [0,b]$ | $ Y$ has exactly $ n\minus{}1$ points in $ [0,a)$)
$ \equal{} 1 \minus{} P($of $ n\minus{}1$ uniformly chosen independent points in $ [0,a)$ no point is in $ [0,b]) \equal{} 1 \minus{} (1 \minus{} \frac{b}{a})^{n\minus{}1}$.

The calculation of alekk gives the same formula (small mistake in the last step of the calculation of $ F$):
$ F(b,a) \equal{} \frac {\partial_{a} A(b,a) }{\partial_a A(\infty, a)} \equal{} \frac {\int_{x \equal{} 0}^b (a \minus{} x)^{n \minus{} 2} dx }{\int_{x \equal{} 0}^{a} (a \minus{} x)^{n \minus{} 2} dx } \equal{} 1 \minus{} (1 \minus{} \frac{b}{a})^{n\minus{}1}$,

Applied to the original problem this gives $ P(X_1 \geq 3|X_1 \plus{} ... \plus{} X_5\equal{}5) \equal{} (1 \minus{} \frac{3}{5})^4 \equal{} 0.0256$.

Thank You very much. I have only one question. [b]alekk[/b] wrote
$ A(b,a) \equal{} P[(X \leq b) \cap (X \plus{} Y \leq a)] \equal{} \int_{x \equal{} 0}^b \lambda \exp( \minus{} \lambda x) [\int_{y \equal{} 0}^{a \minus{} x} G(y)dy] dx$
so, how should I understand $ A(\infty ,a)$ ?
I think it is $ P[(X \leq \infty) \cap (X \plus{} Y \leq a)]\equal{}P(X \plus{} Y \leq a)$?  But in integral form? I guess it isn't
$ \int_{x \equal{} 0}^\infty \lambda \exp( \minus{} \lambda x) [\int_{y \equal{} 0}^{a \minus{} x} G(y)dy] dx$ because this integral doesn't coverage on $ (0,\infty)$. So my question is, how should I understand $ A(\infty ,a)$? How to write this expresion using integrals?

[quote="zuraw"]
$ \int_{x \equal{} 0}^\infty \lambda \exp( \minus{} \lambda x) [\int_{y \equal{} 0}^{a \minus{} x} G(y)dy] dx$ because this integral doesn't coverage on $ (0,\infty)$ So my question is, how should I understand $ A(\infty ,a)$? How to write this expresion using integrals?[/quote]

yes, as you said:
$ A(\infty,a)\equal{}P[X\plus{}Y \leq a]\equal{}\int_{x \equal{} 0}^\infty \lambda \exp( \minus{} \lambda x) [\int_{y \equal{} 0}^{a \minus{} x} G(y)dy] dx$
which is equal to:
$ \int_{x \equal{} 0}^a \lambda \exp( \minus{} \lambda x) [\int_{y \equal{} 0}^{a \minus{} x} G(y)dy] dx$


MIT college Harvard Princeton parameterization topology Yale

Hi,

What are the differences between top US Universities such as Harvard, Princeton, MIT with top UK Universities such as Cambridge and Oxford in terms of quality/character of students, environment, community, overall college experience (academically and non-academically), quality of teaching and X,

where X is any parameter you may choose and any operation may be enclosed (If you know what I mean. I don't even sure whether I'm using the correct words! Haha! Never mind. You get the point. Just having a little fun here. Btw, it is not profane!)?

Please give your opinions. Thanks!

Honestly, you can't go wrong choosing any of those.  Do you want to have a chance to be around Stephen Hawking?  Go to Cambridge.  Do you want to be taught by the great Andrew Wiles?  Go to Princeton.  Do you want to say that you graduated from Harvard?  Go to Harvard.  You can't go wrong.

Oxford isn't really advisable - it's a reasonable amount worse than Cambridge for maths and sciences in general. (This is pretty widely accepted; I'm not just being all anti-Oxford because I'm at Cambridge.)
There are loads of great mathematicians here - Tim Gowers, Alan Baker, Ben Green, Peter Swinnington Dyer etc. - I imagine this is a trait shared with all the major U.S. ones mentioned.
(A nice thing about Oxbridge - supervisions/tutorials - you could reasonably end up having a Field's medallist go over your solutions to example sheets once a week.)
In truth I'm just saying all that famous people stuff just to wow you - it really matters what they're like as teachers and I can't really comment on that.
One more advantage about the U.K; you only have to do maths, so none of all those other classes you might end up going to at a U.S. school.
(BTW - Cambridge and Oxford are stunning towns - I don't want to offend anyone but I'm willing to go ahead and say they are nicer places than the locations of the U.S. ones mentioned.)

Don't make your decisions about what famous people are at the university. Take note of these key factors:

1) Famous mathematicians and physicists might not actually have to teach undergrads anyway (Wiles rarely teaches undergrad classes here). If you're thinking grad school, it may be really competitive to get one of them for an advisor. Also some of the really famous people are also ancient and aren't doing much (like someone mentioned Swinnerton-Dyer before -- isn't he like 80 yrs old? and Stein at Princeton is almost that old, and he might stop teaching after this year.)

2) Famous mathematicians and physicists aren't necessarily good teachers. For example, earlier this year I went to a lecture by Endre Szemeredi, who is one of the most brilliant combinatorialists alive. But his lecture was not good at all. And Chris Skinner (a great algebraic NT guy at princeton) is known for giving really tedious and painful and annoying assignments. And Stein is kinda boring and basically says exactly what's in the book (not surprising because he wrote it, but still).

So basically, my point is that you shouldn't base your decision on celebrities. That's not to say you won't benefit from them at all -- my friend at MIT has Munkres for topology and Artin for algebra and he says they're really good. And last year I did a reading course with Fefferman (fields medalist) and I got to have one-on-one discussions with him every week about geometric measure theory.

Now I haven't been to Oxford or Cambridge and I don't know many people there, but I can say that the top USA universities are all pretty similar in overall quality, and you should choose between them based on your personal preferences. Do you want to live in a city? Then MIT or Harvard might be preferable over Yale and especially Princeton. If you prefer a more quiet setting with actual trees and green stuff and old buildings, reverse that order (and probably Oxbridge go ahead of all of the USA places).

You might also want to wait and find out what kind of financial aid you get (the USA places are all pretty generous but Harvard and Princeton are the best).

Other factors might be the specific topics you want to study (for example, Princeton has a lot of people in analysis, number theory, and low dimensional topology, and I don't know the reputations of other schools' departments that well but you can find out easily enough), what options the schools offer (e.g. Princeton has a lot of interesting certificates you can get, but they won't let you double major), and what kind of classroom environment you want. Or extracurriculars -- do you want to do sports, join a performance group, join a math or physics club? Are you interested in frats and partying? You can ask a student at whichever university you're interested in to find out what the extracurriculars are like there.

There are tons of other factors too. But don't make 'what big name professor does the school have' one of them.

Hello everyone!

  Im a student about 16 years old and im very interested about maths. What i mean is that i participate in mathematical competitions study hard maths especially number theory< i love Andreescu> but i think i dont have the <big> brain to go to I.M.O. My dream is to join a high college like Princeton, M.I.T and others. Can i do that withoiut taking place in an I.M.O??

You don't need to go to IMO in order to get into that kind of university. You just need to be a good student. In your application, you should stress the following points:

(1) You have the necessary English skills and a good enough secondary school education to succeed in the university you're applying to.

(2) You fall into one of the three major categories of students that are generally accepted to this type of university. They are: (i) Students who are amazingly good at one particular thing and not much else (you would probably need to go to IMO at least in order to qualify as this for math). (ii) Students who are well-rounded, i.e. good at a variety of things (and I don't mean maths, physics, and chemistry). If you do stuff like organizing a group, writing a lot, playing music, etc -- that will help in this category. (iii) Students who contribute significantly to the diversity of the community (you are Greek? Maybe you should emphasize Greek culture in your application. Do you play an instrument -- can you play Greek folk music? Would you want to get involved with (or create) a Greek culture student group at your university? etc...)

(3) You have reasons for going to the university besides the name recognition. Learn some specific things about the university you're applying to and explain why those things make you want to go there. They don't want to be filled by a bunch of students who just want the name MIT or Princeton on their resumes (although it's inevitable that most people really are there for that reason). If you like number theory, in your Princeton application you could talk about the great number theorists in the Math Dept, like Sarnak and Skinner and Wiles and Katz.

[quote="BanishedTraitor"][quote="Xevarion"] (i) Students who are amazingly good at one particular thing and not much else (you would probably need to go to IMO at least in order to qualify as this for math).  [/quote]

Aren't there other ways of showing this?  :([/quote]
Publish something? My impression about math is that there are so many international students who are extremely good at math (but not particularly special in any other way) that you have to basically be one of the best in order to get in to a top university for that reason alone. Maybe if you don't go to IMO but show knowledge of advanced math and very high potential, that would be enough. Who knows, I'm not an admissions officer.

[quote="BanishedTraitor"]Well, I don't really mean Mathematics (too many good people  :D ) but what about Sciences?[/quote]
I don't know enough about it. Sorry. I imagine anything beyond what a typical "smart student" would do in school in your country is at least a useful step in that direction.


Anyone interested in a chess tournament between mathlinks.ro members?

I would be if I had a clue besides the basic rules of how to play that game :)

the basic rules are everything you need  :D

I'm up for it, altough it has been a long time since I played "seriously" -- playing Go now instead.

Maybe someone could put up a poll regarding this.

Done.
Also,post which chess network would you like to play on.

im up for it, but i dont have a mathlinks account, so if you go to funorb.com, my username is hyperhavoc5. funorb has many different games


Let $a,b,c,d > 1$ natural numbers which satisfy $a^{b^{c^d}}=d^{c^{b^a}}$ . Prove that $a = d$ and $b=c$.

Who can post solution of this problem? Please post it!

I'm not sure if I'm right anway I'll post my solution since nobody else does...

Since LHS=RHS they must contain the same prime factors, especially being $ a,b,c,d \in \mathbb N$ then $ a | d , d| a$

which means $ a=d$. With substitutions we get $ {a}^{b^{{c}^{a}}}=a^{c^{b^{a}}}$$ \Rightarrow  b^{c}=c^{b}$ $ \Rightarrow b|c,c|b\Rightarrow b=c$.

Please tell me if I'm wrong  :)

Your solution is based on the argument that, in general $ x^{y}=y^{x}\Longleftrightarrow x|y\wedge y|x\Longleftrightarrow x=y$. The first equivalence is wrong. We have indeed $ 2^{4}=4^{2}$ and $ 4\not=2$.

however, i've been trying to find a second counter-example, but i can't find one. we can indeed use the following argument:

It is evident that (or quickly established), supposing $ x\geq y$ by symetry, $ x^{y}=y^{x}\Longleftrightarrow x=y^{k}$, $ k$ being a natural number. Now, we may say that the relation is equivalent to $ \left(y^{k}\right)^{y}=y^{y^{k}}\Longleftrightarrow y^{k\cdot y}=y^{y^{k}}\Longleftrightarrow k\cdot y=y^{k}$ which implies either $ k=1\Longleftrightarrow x=y$ or $ y=2\wedge x=4$. This lemma shows that, supposing we have established the relation $ a=d$, we also have the relation $ b=c$ (or the special case stated above).


If $ 60$ miles per hour is $ 88$ feet per second, how many feet per second is $ 66$ miles per hour? Express your answer as a decimal to the nearest tenth.


1)Given sequence $ x_n$ such that:$ x_n  \in R;x_{n + 1}  = x_n  + \frac{1}{2}(c{\rm{os}}\,x_n  + \sin \,x_n )(\forall n \in N^* )$.Find limited of sequence depending on $ x_1$.
2)Given p be a prime and $ p \ge 5$.Prove that exist $ q_1$,$ q_2$ such that $ 1 < q_1  < q_2  < p$ simultaneously $ q_1 ^{p - 1}  - 1$ and $ \q_2 ^{p - 1}  - 1$ non divisible $ p^2$


If $a$, $b$, $c$ are positive real numbers, prove that
$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}> 2$.

Using Cauchy:
$a+(b+c) \geq 2\sqrt{a(b+c))}$
=> $\sqrt{\frac{a}{b+c}} \geq \frac{2a}{a+b+c}.$
Similar:
     $ \sqrt{\frac{b}{a+c}} \geq \frac{2b}{a+b+c}.$
     $ \sqrt{\frac{c}{a+b}} \geq \frac{2c}{a+b+c}.$
=> $L.H.S > \frac{2a+2b+2c}{a+b+c} = 2.$
(Because a,b,c>0).

The inequality very easy.

It was posted but i will give a solution
$(a+b+c)^2\geq4a(b+c)$ and equivalently $\sqrt\frac{a}{b+c}\geq\frac{2a}{a+b+c}$
Same $\sqrt\frac{b}{a+c}\geq\frac{2b}{a+b+c}$ and
$\sqrt\frac{c}{b+a}\geq\frac{2c}{a+b+c}$
we add the three ineqs and we have
$\sqrt\frac{a}{b+c}+\sqrt\frac{b}{a+c}+\sqrt\frac{c}{b+a}\geq2$
the equality holds when a=b+c ,b=c+a, c=a+b
If we add these we have that a+b+c=0 contradiction.
So $\sqrt\frac{a}{b+c}+\sqrt\frac{b}{a+c}+\sqrt\frac{c}{b+a}>2$

Metru, excuse me for saying this, but this inequality has been posted for many times and solved as well. ;)

Disprove or Prove that:
If $a,b,c>0$ and $k \in N$ then:

$ \large \sqrt[2k]{\frac{a}{b+c}} + \sqrt[2k]{\frac{b}{a+c}} + \sqrt[2k]{\frac{c}{a+b}} > 2.$

[quote="Gohan"][quote="tranvinhphuctk14"]Using Cauchy:
$a+(b+c) \geq 2\sqrt{a(b+c))}$
=> $\sqrt{\frac{a}{b+c}} \geq \frac{2a}{a+b+c}.$
[/quote]
can you explain this step in more details ... i dont know this cauchy ineq ;<[/quote]

you can treat it as AM-GM by noticing that $x+y\geq 2\sqrt{xy}$

cezar lupu
[url=http://kvant.mirror0.mccme.ru/pdf/2002/06/18.pdf]in another way:
[/url]

[quote="siuhochung"]in vietnam, amgm is called cauchy, cauchy is called bun???ski[/quote]
Yes, i'am sorry for mistake.


function
inequalities
calculus
derivative
algebra
polynomial
inequalities unsolved

Hi :D ! $ x,y,z$ are positive real numbers and $ 3x\plus{}2y\plus{}z\equal{}6$ . prove that :
$ x\plus{}xy\plus{}xyz\leq 3$

i have an ugly soloution for this one but it seems too easy !! plz help me ?!?!

Let's even allow zero values, and try Lagrange multipliers. Build function $ L(x,y,z,\lambda) \equal{} x \plus{} xy \plus{} xyz \minus{} \lambda(3x\plus{}2y\plus{}z\minus{}6)$. Then we get the system $ \frac {\textrm{d}} {\textrm{d}x} L(x,y,z,\lambda) \equal{} 1\plus{}y\plus{}yz\minus{}3\lambda \equal{} 0$, $ \frac {\textrm{d}} {\textrm{d}y} L(x,y,z,\lambda) \equal{} x\plus{}xz\minus{}2\lambda \equal{} 0$, $ \frac {\textrm{d}} {\textrm{d}z} L(x,y,z,\lambda) \equal{} xy\minus{}\lambda \equal{} 0$. Plug $ \lambda \equal{} xy$ in the second equation to get $ x(1\plus{}z\minus{}2y) \equal{} 0$, so $ z\equal{}2y\minus{}1$ ($ x\equal{}0$ is on the contour, and provides a minimum, not a maximum). Plug again in the first equation to get $ 1\plus{}y\plus{}y(2y\minus{}1)\minus{}3xy \equal{} 0$, so $ x \equal{} \frac {2y^2\plus{}1} {3y}$. Back into the restriction, all these yield $ \frac {2y^2\plus{}1} {y} \plus{} 2y \plus{} (2y\minus{}1) \equal{} 6$, hence $ 6y^2 \minus{}7y \plus{} 1 \equal{} 0$, so $ y\equal{}1$, when $ x\equal{}y\equal{}1$, and $ y\equal{} \frac {1} {6}$, when $ z\equal{}\minus{}\frac {2} {3}$ is not convenient. One can check that for $ x\equal{}y\equal{}z\equal{}1$ the critical value of $ L \equal{} 3$ is a maximum (on the contour, we have $ x\equal{}0$, with $ L\equal{}0$, or $ y\equal{}0$, with $ L \equal{} x \leq 2 < 3$, or $ z\equal{}0$, with $ L \equal{} x\plus{}xy \leq \frac {8} {3} < 3$).

Thanks a lot mavropnevma !! but is there any easier soloution (by AM-GM , Couchy ,...) ?

Ok, here is an elementary solution. Let's denote $ x \equal{} a \plus{} 1, y \equal{} b \plus{} 1, z \equal{} c \plus{} 1$. Then you can rewrite the condition of the problem as follows:
$ 3a \plus{} 2b \plus{} c \equal{} 0$ and you want to prove $ (a \plus{} 1) \plus{} (a \plus{} 1)(b \plus{} 1) \plus{} (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)\le 3$ or (after simplification) it's enough to verify
$ 2ab \plus{} bc \plus{} ac\plus{}abc\le 0$. But that's almost obvious. Just write $ c \equal{} \minus{} (3a \plus{} 2b)$ and you will get the inequality $ 3a^2 \plus{} 2b^2\plus{}ab(3a\plus{}2b)\plus{}3ab\equal{}3a^2(b\plus{}1)\plus{}2b^2 (a\plus{}1)\plus{}3ab\ge0$ for $ a,b,c\ge \minus{}1$ and $ 3a\plus{}2b\le 1$. Indeed, if $ ab\ge 0$ we are done, and if not, then $ ab(3a\plus{}2b)\ge ab$ and we get that $ RHS\ge 3a^2\plus{}2b^2\plus{}4ab\ge 0$.

More elementary solution (but, yes, ugly).

Keep $ x$ constant, then $ L \equal{} x(1\plus{}y\plus{}yz)$, with $ z \equal{} 6\minus{}3x\minus{}2y$, so $ L\equal{} x(1\plus{}y\plus{}6y\minus{}3xy\minus{}2y^2) \equal{} \minus{}x(2y^2 \plus{}y(3x\minus{}7)\minus{}1)$. The maximum for this occurs for $ y \equal{} \frac {7\minus{}3x} {4}$, and amounts to $ \frac {1} {8} x(9x^2 \minus{}42x \plus{} 57)$. Now revert to considering $ x$ variable in the interval $ [0,2]$. The derivative of the above polynomial annuls in $ x\equal{}1$ and $ x\equal{}\frac {57} {27} > 2$, so its maximum is obtained for $ x\equal{}1$, when we get value $ 3$. Then $ y$ is also equal to $ 1$, and finally $ z\equal{}1$.

leshik, you wrote "enough to verify $ 2ab \plus{} bc \plus{} ac\le 0$", but you forgot the term $ abc$, so one needs $ abc \plus{} 2ab \plus{} bc \plus{} ac\le 0$, with more work...

Thanks for the corrections.I've edited :blush:


I was wondering where everybody in AoPS lived.  Because there are 195 countries in the world, I'm just going to do continents on the poll(but include a few big countries).  Post where you live here and answer the poll.

Note: I have to delete a few poll options because the settings won't allow it, so if your country doesn't get separately represented and you think it should, that's probably why.

Similarly, I live under a rock under the sea, and I am quite stupid when compared to mewto's spongy intellect.

hee hee.  THis isn't what I was thinking what your posts are like, but this is more fun than what I had in mind.  of course, no being silly in the polls!

If you would like me to be more formal...
my female clone lives in Washington... one of the states with the most rain. Sad part is we haven't had rain in... I don't know a month? And our temperatures are some what close to Nevada, :( I am NOT used to this type of weather.

I live in an igloo, and I hunt caribou for dinner. It should be obvious what country that is.

I live in an anchor under the sea, and I'm very cheap. I'm a big miser. I would even sell my soul for money :P . I make mewto work for me :P

So far, 75% of the people here(6 out of 8) live in the U.S.A.  I'm surprised there aren't any more from say, Eurpope and Asia.

[quote="meewhee009"]And I live in the middle of mewto and battered in a house that resembles my face. I am very grouchy compared to mewto and battered's two-legged kind....[/quote]

I live in a dome and i am a scientist with a white suit.

I live in your wildest dreams.
I am stuck in a temporal tunnel with my computer.

What the hell, am I the only Canadian here? (Besides dnkywin, of course. But then again, he's also American.)

[quote="PowerOfPi"]I am Canadian, but I don't live in Canada.[/quote]

Same deal with dnkywin. Why do people claim nationalities where they aren't a resident...? I mean, I'm Chinese both by origin and ethnicity, but I consider myself Canadian, because I've lived here for 12 years...

Looking at the polls, the AoPS population is no where near as diverse as I thought(at least in the FF).  I'm incredibly surprised I don't see any "China's" or "India's" considering the huge populations of the countries.

Maybe it is because people in those countries study harder and don't have the time for this stuff.

@sudoku dragon: I consider myself Canadian because I was born in Canada. I moved a year ago, and not very far either. I still visit every few months.

[quote="meewhee009"]And[b] I live in the middle of mewto[/b] and battered in a house that resembles my face. I am very grouchy compared to mewto and battered's two-legged kind....[/quote]

*loud coughing*

Well, maybe we have so many people living in the US because, hmm, well, this is a forum made for Americans and written in English.

And those asians do study all day.
They don't have time for all this stuff.

No wonder the Chinese come first every year. :huh:
Don't know about Indians, they rarely post outside their national forum on Aops :rotfl:

[b][color=red]Edited to avoid any misunderstanding[/color][/b]
 :wink:  :)

[quote="Xaenir"]And those asian nerds do study all day.[/quote]
Considering AoPS is a worldwide forum, and this forum is frequented by people internationally, that's not a very smart thing to say.  :wink:

Xaenir, that's the second time you insult somebody. Do it once more, this goes to mods.

I am the only one voted to live in Europe...

as you may have seen , I have long edited my post, so your post qualifies as spam :D
Besides i didn't mean to insult anybody, just implied that asians work harder and are intellectually superior, you got it the wrong way. :wink:

No, wasn't it ''asian nerds''? Maybach's post tells it all. And that ''girls are lazy, girls are dumb''?

I say again, you can say anything without insults. If you want to continue the discussion, PM me. I'm not against you, just giving you a little piece of advice :)

[b]nerd[/b] refers to a person who passionately pursues intellectual activities, esoteric knowledge, or other obscure interests rather than engaging in more social or popular activities.

That isn't an insult :!:

um.........look at this(Source:  Merriam Webster Dictionary&Thesaurus):
nerd \'nərd\  n [perh. fr. nerd, a creature in the children's book If I Ran the Zoo (1950) by Dr. Seuss (Theodor Geisel)] (1951) : an unstylish, unattractive, or socially inept person ; esp: one slavishly devoted to intellectual or academic pursuits ‹computer ~s›
— nerd·i·ness \'nər-dē-nəs\ n
— nerd·ish \'nər-dish\ adj
— nerdy \-dē\ adj

Your definition doesn't tell it all.  I'm not saying "nerd" is a terrible word to say, but it usually has a very negative connotation, which is why you certainly shouldn't call someone that, let alone steryotype billions of people considred "Asians".  Bottom line:  This is your second strike.  Don't do it again.

Well I didn't mean it as an insult.I apologize if I offended someone, is it alright now?
And I think I would be more careful in the future

@Isabella,
Ha. Grammar gone wrong. I didn't think anyone would find that so I left it be. You proved me wrong  :lol:

to me, nerds are a type of candy
i like nerds
they do taste quite sweet
i live in a Gullible


I want a recent national competition.
I don't have anything to trade.
Well, my teacher has 2005 (maybe?) state competition.
Also A few of the recent work books.
Please(If there is any chance of avoiding the actual trade, and I could just get the competition, It would be highly appreciated. I don't know if just giving away a nat competition is legal, though.) :help:

I have 2008 or 2006 nats, if you could trade anything it would be most helpful.


algebra
polynomial
factorial
function
domain
Ring Theory
superior algebra

Let $ f_1 ,f_2 \in k[x_1,...,x_n]$ ($ k$ is a field) be polynomials with no common irreducible factors. Show That
$ Syz(f_1 ,f_2)\equal{}(f_2 ,\minus{}f_1)k[x_1,...,x_n] \subset k[x_1,...,x_n]^2$.
What happens if $ f_1$ and $ f_2$ have common irreducible factors?

This sounds very much like homework, so I'd just give a few hints, but there is no much to hint here - if you can't solve the problem, you have either not understood the meaning of "syzygy", or you don't know some basic facts about factorial domains (also called unique factorization domains, UFDs, ...). If any of these is not familiar to you, read them up first!

Do you agree that $ \left(g_1,g_2,...,g_n\right)\in\text{Syz}\left(f_1,f_2,...,f_n\right)$ if and only if $ g_1f_1 \plus{} g_2f_2 \plus{} ... \plus{} g_nf_n \equal{} 0$ ? Now think about what this means for $ n \equal{} 2$. That is, $ g_1f_1 \plus{} g_2f_2 \equal{} 0$. That is, $ g_1f_1 \equal{} \minus{} g_2f_2$. If $ f_1$, $ f_2$, $ g_1$, $ g_2$ were integers, this would yield some divisibility... now, $ f_1$, $ f_2$, $ g_1$, $ g_2$ are not integers, but polynomials. What can you say about divisibility among polynomials?

If $ f_1$ and $ f_2$ have common irreducible factors - worst case: $ f_1 \equal{} f_2$, there are many more syzygies. Do you see a very obvious one?

  darij


inequalities
induction
number theory unsolved
number theory

Show that:
$F_{2n}\leq \frac{(2n+2)^n}{n!}$
Denote :$F_{2n}$ is $2n$-th Fibolnaxi number
It's itshelf Mathlinks Contest Round 4 ,after some conclusions ,of course !
But I have no final solution.Anybody has one??

[quote="Rust"]It is trivial. Let Z(n) is right expression. We have Z(1)=4, Z(n)/Z(n-1)=2*(1+1/n)^n>4 if n>=2. Therefore Z(n)>=4^n.
If F(0)=a>=0,F(1)=b>=0,F(n+1)=F(n)+F(n-1) we have F(2n+2)=3F(2n)-F(2n-2)<=3F(2n+2). Therefore F(2n)<=F(2)*3^(n-1)<4^n if F(2)=a+b<4.[/quote]
Please, use LaTeX...

It's easy to know $F_m\geq F_{m-1}(m=1,2,\ldots)$.So $F_{m+1}=F_m+F_{m-1}\leq2F_m$.
So $F_{2n}=F_{2n-1}+F_{2n-2}\leq2F_{2n-2}$.
First,$F_2<4$,so the inequality is correct for $n=1$.
By induction,suppose $F_{2n}\leq\frac{(2n+2)^n}{n!}$
So $F_{2n+2}\leq3F_{2n}\leq3\frac{(2n+2)^n}{n!}$.
We only need to show $3\frac{(2n+2)^n}{n!}\leq\frac{(2n+4)^{n+1}}{(n+1)!}$.
It can be writed like this:$3(n+1)\leq(2n+2)(1+\frac{1}{n+1})^{n+1}$.
$\frac{3}{2}\leq(1+\frac{1}{n+1})^{n+1}$.
And we can easily know $(1+\frac{1}{n+1})^{n+1}\geq1+(n+1)\frac{1}{n+1}=2>\frac{3}{2}$.
So the original inequality is correct. ;)


There are 20 people in my club.  8 of them are left-handed.  15 of them like jazz music.  2 of them are right-handed and dislike jazz music.  How many club members are left-handed and like jazz music?

Draw a Venn diagram: one category is for left-handed, the other for likes jazz.  We are told 2 are right-handed and don't like jazz, so they are outside, meaning there are 18 inside.  Let $ x\equal{}\text{number of left\minus{}handed people}$, $ y\equal{}\text{left\minus{}handed and like jazz}$ and $ z\equal{}\text{like jazz}$.  Then $ x\plus{}y\plus{}z\equal{}18$, $ x\plus{}y\equal{}8$, and $ y\plus{}z\equal{}15$.  Then $ y\equal{}15\minus{}z\Rightarrow x\plus{}15\minus{}z\equal{}8\Rightarrow x\equal{}z\minus{}7$.  This means that $ z\minus{}7\plus{}15\minus{}z\plus{}z\equal{}18\Rightarrow z\equal{}10\Rightarrow x\equal{}3,y\equal{}5$.  So there are 5 people who like jazz and are left-handed.

Or for a faster way:

We want the lefties and those who like jazz, so we can disregard those 2 righties who don't like jazz. That leaves 18. Since 15 like jazz and 8 are leftie, our answer is simply $ (15 \plus{} 8) \minus{} 18 \equal{} \boxed{5}$.

EDIT: Err...I knew that... :P Thanks, Coptie.


The fraction $ \frac{x}{5}$ does not change its value when 3 is added to both its numerator and its denominator. What is the value of $ x$?

hello
u can simply say it should be $ 5$
$ \frac{x\plus{}3}{8 }\equal{} \frac{x}{5}$
$ x \equal{}5$
thank u

[hide="Clearer solution"]
We have the equation \[ \frac{x}{5}\equal{}\frac{x\plus{}3}{5\plus{}3}.\] Cross multiplying gives \[ 8x\equal{}5x\plus{}15\Rightarrow3x\equal{}15\Rightarrow x\equal{}\boxed{5}.\][/hide]


The circumference of a circle is $ 24\pi$ cm. What is the number of square centimeters in the area of the largest square that can fit inside the circle?


I need help on this problem.

Let $a_1 = 1$ and for $n>1$, $a_n = 2a_{n-1}$ if $a_{n-1}$ is odd and equals $a_{n-1}+1$ if it's otherwise. Find $a_{2004}$.

The problem I'm facing is writing $a_n$ without using $a_{n-1}$. Like I see that:

$a_4 = 2(2a_1+1)$
$a_6 = 2[2(2a_1+1)+1]$

and so forth but I'm having trouble writing $a_n$. Can anyone help me on how to do this?

I posted in AMC forum because this type of problem appears quite often in AMC.


Inscribed in a circle is a quadrilateral having side lengths 25,39,52 and 60 taken consecutively. What is the diameter of this circle?

[hide]just by luck, if you draw the diagonal, they're right triangles (and Pythagorean tripleets at that!) so the hypotenuse (and the diameter) is [b]65.[/b][/hide]

Or...
The magic of....
[hide="Super Awesome Theorem of pwnage"]
Ptolemy's Theorem!!!
[/hide]

Of course, it takes some more manipulation, because the diagonals of the quad aren't actually the diameters...

How would you use Ptolemy's Theorem on this problem?

[hide]Bhramagupta's formula says that the answer is $\sqrt{63 \cdot 49 \cdot 36 \cdot 28}$
$84 \cdot 21$
$1764$[/hide]

Random trivia:
We can generalize Brhamagupta's Formula (or however you spell his name) to any quadrilateral, cyclic or non-cyclic. This formula is sometimes called Bretschneider's formula:
$\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}{\frac{A+C}{2}}}$, where $s$ is the semiperimeter, $a, b, c, d$ are the sides of the quadrilateral, and $A, C$ are opposite angles. In fact, any set of opposite angles will do.


algebra
polynomial
search
algebra unsolved

Hi all!!!!!!!!! :wink:
Here is a problem from VIETNAM 2003 olimpiad.
$ B2.$  Define $ p(x) = 4x^{3}-2x^{2}-15x+9, q(x) = 12x^{3}+6x^{2}-7x+1$. Show that each polynomial has just three distinct real roots. Let $ A$ be the largest root of $ p(x)$ and $ B$ the largest root of $ q(x)$. Show that $ A^{2}+3B^{2}= 4$.

Please search it on this forum. It is posted before!


function
inequalities
integration
triangle inequality
real analysis
real analysis unsolved

let $ f_n$ be a sequence of non-negative measurable functions in $ L^p(\Bbb R)$ for some $ 1 <p<\infty$. show that $ f_n \to f$ in $ L^p$ iff $ f_n^p \to f^p$ in $ L^1(\Bbb R)$.

Here's one direction:

We can show that for $ p\ge 1$ and $ x\ge 0,$ $ (1 \plus{} x)^p\ge 1 \plus{} x^p.$

For $ a\ge b > 0,$ we can let $ x \equal{} \frac ab \minus{} 1$ and rearrange this to

$ \left(\frac ab\right)^p \minus{} 1\ge\left(\frac1b \minus{} 1\right)^p$ or $ a^p \minus{} b^p\ge(a \minus{} b)^p$

So we have that $ |f_n \minus{} f|^p\le |f_n^p \minus{} f^p|$ and hence $ \|f_n \minus{} f\|_p^p\le \|f_n^p \minus{} f^p\|_1.$

So this gives us $ f_n^p\to f^p$ in $ L^1$ implies $ f_n\to f$ in $ L^p.$

Here is a somewhat weaker (I think, I very well could be wrong) version of the converse:

Since $ f_{n}\rightarrow f$ in $ L^{p}$, then $ \left\|f_{n}\minus{}f\right\|_{p}\rightarrow 0$ as $ n\rightarrow\infty$.  By the triangle inequality, $ \left\|f_{n}\right\|_{p}\leq\left\|f_{n}\minus{}f\right\|_{p}\plus{}\left\|f\right\|_{p}\Rightarrow\left\|f_{n}\right\|_{p}\minus{}\left\|f\right\|_{p}\leq\left\|f_{n}\minus{}f\right\|_{p}$ and $ \left\|f\right\|_{p}\leq\left\|f_{n}\minus{}f\right\|_{p}\plus{}\left\|f_{n}\right\|_{p}\Rightarrow\left\|f\right\|_{p}\minus{}\left\|f_{n}\right\|_{p}\leq\left\|f_{n}\minus{}f\right\|_{p}$.  Hence $ (\left|f_{n}\right|^{p})$ converges to $ \left|f\right|^{p}$ in $ L^{1}$.

[quote="JRav"]Hence $ (\left|f_{n}\right|^{p})$ converges to $ \left|f\right|^{p}$ in $ L^{1}$.[/quote]

are you sure that this is what you've shown?

By Minkowski's we have:
$ \left\parallel{}f^{p}_{n}\right\parallel{}_{1} \equal{} \parallel{}f^{p}_{n} \minus{} f^{p} \plus{} f^{p}\parallel{}_{1} \leq \parallel{}f^{p}_{n} \minus{} f^{p}\parallel{}_{1} \plus{} \parallel{}f^{p}\parallel{}_{1}$

Then we get:
$ \left| \parallel{}f^{p}_{n}\parallel{}_{1} \minus{} \parallel{}f^{p}\parallel{}_{1} \right| \leq \parallel{}f^{p}_{n} \minus{} f^{p}\parallel{}_{1} \equal{} \int \left|\left(f^{p}_{n} \minus{} f^{p}\right)^{\frac{1}{p}}\right|^{p} \leq \left\parallel{}f_{n} \minus{} f \right\parallel{}^{p}_{p} \plus{} \parallel{}f_{n}\parallel{}^{p}_{p}$

for other direction : let $ M \equal{} max\{\parallel{}f\parallel{}_{p}, \parallel{}f_{n}\parallel{}_{p}: n\in\mathbb{N}\}$, then $ \int\parallel{}f|^{p} \minus{} |f_{n}|^{p}|d\nu\leq 2pM^{p \minus{} 1}\parallel{}f \minus{} f_{n}\parallel{}_{p}$

Prof. Kent proposed very nice solution but I think we can apply Fatou's lemma :

we have $ |f \minus{} f_{n}|^{p}\leq 2^{p}(|f|^{p} \plus{} |f|_{n}^{p})$, so consider : $ 2^{p}(|f|^{p} \plus{} |f|_{n}^{p}) \minus{} |f \minus{} f_{n}|^{p}\to 2^{p \plus{} 1}|f|^{p}$

and apply the Fatou's lemma we have $ \overline{lim}\int|f \minus{} f_{n}|^{p}\leq 0$ .

1234567a: You have to pass to a subsequence first, though. So you get that every subsequence of $ f_n$ has a further subsequence that converges to $ f$ in $ L^p$, which is good enough.

[quote="blahblahblah"]are you sure that this is what you've shown?[/quote]
I showed that $ \left|\left\|f\right\|_{p}\minus{}\left\|f_{n}\right\|_{p}\right|\leq\left\|f_{n}\minus{}f\right\|_{p}$, which means that $ \left|\int\left|f_{n}\right|^{p}\minus{}\int\left|f\right|^{p}\right|\rightarrow 0$, which I thought meant that $ \left|f_{n}\right|^{p}$ converged to $ \left|f\right|^{p}$ in the $ L^{1}$ norm but apparently I am mistaken.  Could you elaborate?

[quote="ficsur"]let $ f_n$ be a sequence of non-negative measurable functions in $ L^p(\Bbb R)$ for some $ 1 < p < \infty$. show that $ f_n \to f$ in $ L^p$ iff $ f_n^p \to f^p$ in $ L^1(\Bbb R)$.[/quote]

I will prove that " if $ f_n\to f$ in $ L^p$, then $ f_n^p\to f^p$ in $ L^1$".
Indeed, bacause the function $ \varphi(x)\equal{}x^p$ is convex with $ p>1$. By the convexity we have
$ |x^p\minus{}y^p|\leq p(\max\{x,y\})^{p\minus{}1}|x\minus{}y|$ with $ x,y\geq 0$.
From inequality $ \max\{x,y\}\leq x\plus{}y$, we have $ |x^p\minus{}y^p|\leq p(x\plus{}y)^{p\minus{}1}|x\minus{}y|$ (by $ p\minus{}1>0$). Apply this inequality for $ f_n, f$, we obtain
$ |f_n^p\minus{}f^p|\leq p(f_n\plus{}f)^{p\minus{}1}|f_n\minus{}f|$
Hence, by using the Holder's inequality, we have
$ \parallel{}f_n^p\minus{}f^p\parallel{}_1\leq p\parallel{}(f_n\plus{}f)^{p\minus{}1}\parallel{}_q\parallel{}f_n\minus{}f\parallel{}_p(*)$
here, $ 1/p \plus{} 1/q\equal{}1$ therefor $ q(p\minus{}1)\equal{}p$.
 Thus, $ \parallel{}(f_n\plus{}f)^{p\minus{}1}\parallel{}_q\equal{}\parallel{}f_n\plus{}f\parallel{}_p^{p/q}\leq 2^{p/q}(\parallel{}f_n\parallel{}_p\plus{}\parallel{}f\parallel{}_p)^{p/q}$, this term is bounded, by the hypothesis $ f_n\to f$ in $ L_p$
The proof is completed by the inequality (*).

[quote="JRav"][quote="blahblahblah"]are you sure that this is what you've shown?[/quote]
I showed that $ \left|\left\|f\right\|_{p} \minus{} \left\|f_{n}\right\|_{p}\right|\leq\left\|f_{n} \minus{} f\right\|_{p}$, which means that $ \left|\int\left|f_{n}\right|^{p} \minus{} \int\left|f\right|^{p}\right|\rightarrow 0$, which I thought meant that $ \left|f_{n}\right|^{p}$ converged to $ \left|f\right|^{p}$ in the $ L^{1}$ norm but apparently I am mistaken.  Could you elaborate?[/quote]

well, first of all, you've left out the $ 1/p \minus{}$th roots.  Second of all, showing that $ |f_n|^p\to |f|^p$ in $ L^1$ is the same as showing
\[ \int \parallel{}f_n|^p \minus{} |f|^p|dx\to 0,\]
which is harder, since $ \int |u_n|dx \to 0$ is stronger than $ |\int u_ndx|\to 0$.


trigonometry
trig identities
Law of Cosines
geometry proposed
geometry

Let α+β= Fixed Value
a , b is Fixed Value respectively

Find  R , α , β

Let $O$ be the center of the circle, and the segments $b=BC,a=CA$.
Also let the given angle $x=\angle AOB$

At first, we can notice that we can easy construct this figure geometricaly.

The angle $\phi=\angle ACB$ is such that $\phi+\frac{x}{2}=\pi\Rightarrow \phi = \pi -\frac{x}{2}$

from the triangle $ABC$ we can get that

$AB = 2Rsin \phi\Rightarrow AB^2 = 4R^2 sin^2 \phi$

Law of cosines in $\triangle ABC: AB^2 = a^2+b^2-2ab\cdot cos \phi$

So we have

$4R^2 sin^2 \phi = a^2+b^2-2ab\cdot cos \phi \Rightarrow$

$R^2 = \frac{a^2+b^2-2ab\cdot cos \phi}{4sin^2 \phi}$

$R = \frac{\sqrt{a^2+b^2-2ab\cdot cos \phi}}{2sin \phi}$

Once we have found $R$, we can find the angles on the center using the formulas

$\boxed{ sin \frac{\widehat{\alpha}}{2} = \frac{a}{2R} } \boxed{sin \frac{\widehat{\beta}}{2} = \frac{b}{2R}}$

We can use sine and cosine of $\ \frac{x}{2}$ instead of $\phi$:

$\phi = \pi -\frac{x}{2} \Rightarrow$

$sin \phi = sin \frac{x}{2}$

$cos \phi = -cos \frac{x}{2}$


linear algebra
matrix
linear algebra unsolved

Let an invertible square complex matrix A have Jordan form X. Find the Jordan form of the inverse of A.

This comes down to this question: can you find the inverse of one Jordan block?

Let's do this in a slightly more general way. Suppose $ A\equal{}\lambda I\plus{}N$ where $ N$ is nilpotent and $ \lambda\ne 0.$ What is $ A^{\minus{}1}?$

Write $ A\equal{}\lambda(I\plus{}\lambda^{\minus{}1}N).$ I claim that
\[ A^{\minus{}1}\equal{}\lambda^{\minus{}1}\sum_{k\equal{}0}^{\infty}(\minus{}1)^k\lambda^{\minus{}k}N^k\equal{}\sum_{k\equal{}0}^{\infty}(\minus{}1)^k\lambda^{\minus{}k\minus{}1}N^k.\]
Note that this apparent infinite series is in fact a finite sum, since $ N$ is nilpotent.

An example of this theorem:
\[ \begin{bmatrix}\lambda&1&0&0\\0&\lambda&1&0\\0&0&\lambda&1\\0&0&0&\lambda\end{bmatrix}^{\minus{}1}\equal{}\begin{bmatrix}\lambda^{\minus{}1}&\minus{}\lambda^{\minus{}2}&\lambda^{\minus{}3}&\minus{}\lambda^{\minus{}4}\\0&\lambda^{\minus{}1}&\minus{}\lambda^{\minus{}2}&\lambda^{\minus{}3}\\0&0&\lambda^{\minus{}1}&\minus{}\lambda^{\minus{}2}\\0&0&0&\lambda^{\minus{}1}\end{bmatrix}\]
One you figure out how to do that to any Jordan block, you should be able to figure out how to handle several blocks put together.

Hi Kent. Your answer is not complete.
1) $ \lambda,\mu\not\equal{}0,\lambda\not\equal{}\mu$ imply $ 1/\lambda\not\equal{}1/\mu$.
 2) Let $ A\equal{}\lambda{I}\plus{}J$ where $ J$ is a nilpotent Jordan block ($ J^n\equal{}0,J^{n\minus{}1}\not\equal{}0$). $ A^{\minus{}1}\equal{}(1/\lambda)I\plus{}N$ where $ N^n\equal{}0,N^{n\minus{}1}\not\equal{}0$. Then $ A^{\minus{}1}$ is similar to $ (1/\lambda)I\plus{}J$, its Jordan form.
Conclusion: Let $ A\in\mathcal{M}_n(\mathbb{C})$. If in the Jordan form of $ A$ , we change, for all $ i$, $ \lambda_i$ with $ 1/\lambda_i$ then we obtain the Jordan form of $ A^{\minus{}1}$.


Sorry this is a pretty standard question but when trying to come up with the formula for the area ratio asked in the following:

Suppose $ ABC$ is a triangle and let $ D$, $ E$, $ F$ are on $ BC$, $ CA$, and $ AB$ respectively s.t.
$ \frac {BD}{DC} \equal{} \frac {1}{x}$

$ \frac {CE}{EA} \equal{} \frac {y}{1}$

$ \frac {AF}{FB} \equal{} \frac {1}{z}$

If $ AD$, $ BE$, $ CF$ are drawn then you get a a smaller triangle $ PQR$ where $ P$ is intersection of $ AD$ and $ BE$, $ Q$ is intersection of $ CF$ and $ BE$, and $ R$ is intersection of $ AD$ and $ CF$.

Compute $ \frac {[PQR]}{[ABC]}$

I know this uses mass-point and is not pretty,as it looks at only two cevians at a time, etc (i've done it before, but can't seem to remember the sequence of steps i used) could somebody help?

Check out [url=http://www.cut-the-knot.org/triangle/RouthTheorem.shtml]Routh's Theorem[/url].


Given $\sum^{n}_{i=1} x_i=n$ for $x_i\in \mathbb{R}$ and

$\sum^{n}_{i=1}x_i^4=\sum^{n}_{i=1}x_i^3$

Solve the system of equation for $x_i$ .

 :)

Two of the power means turn out to be equal, so we must have equality everywhere, and hence $x_1 = x_2 = \dots = x_n = 1$.

You could also notice that $\sum_{i = 1}^n\left(x_i - 1\right) = 0$ and $\sum_{i = 1}^nx_i^3\left(x_i - 1\right) = 0$.

Subtract the first equation from the second to get $\sum_{i = 1}^n\left(x_i - 1\right)^2\left(x_i^2 + x_i + 1\right) = 0$. Hence $x_i = 1$ for all $i$.

Arne , I dont undestand what you mean by

[quote]
Two of the power means turn out to be equal, so we must have equality everywhere,
[/quote]

Can you explain more detail ?
 :)

Let's see... We have \[\sqrt[4]{\frac{\sum_{i = 1}^nx_i^4}{n}} \geq \sqrt[3]{\frac{\sum_{i = 1}^nx_i^3}{n}}\] by Power means, which gives \[{{n \cdot (\sum_{i = 1}^nx_i^4})^3 \geq (\sum_{i = 1}^nx_i^3})^4\] or \[n \geq \sum_{i = 1}^nx_i^3.\] So, \[1 = \frac{\sum_{i = 1}^nx_i}{n} \geq \sqrt[3]{\frac{\sum_{i = 1}^nx_i^3}{n}}\] which is the power mean inequality "the wrong way around".

So, equality must hold, and all variables must be equal.


[quote="chien than"]Prove that:$\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\leq 1$
where $a;b;c$ are non-negative and satisfying $abc=1$[/quote]

This is a particular case (n = 3) of the following fact:

If $x_{1}$, $x_{2}$, ..., $x_{n}$ are n nonnegative reals that satisfy $x_{1}x_{2}...x_{n}=1$, then $\sum_{i=1}^{n}\frac{1}{n-1+x_{i}}\leq 1$.

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=4951
http://www.mathlinks.ro/Forum/viewtopic.php?t=27900
http://www.mathlinks.ro/Forum/viewtopic.php?t=164
http://www.mathlinks.ro/Forum/viewtopic.php?t=64349
http://www.mathlinks.ro/Forum/viewtopic.php?t=116016

(Someone should really moderate this forum...)

  darij

It was also discussed here http://www.mathlinks.ro/Forum/viewtopic.php?t=131882


[hide="Took me around 60 - 80 seconds."]Break up the original $\log_8 \frac{216}{81} = 3 \log_8 6 - 4 \log_8 3$. Now change the a and b expressions to base 8, so $a= \log_4 36 = \frac{\log_8 36}{\log_8 4} = \frac{\log_8 36}{\frac{2}{3}} = \frac{3}{2} 2 \log_8 6 = 3 \log_8 6$ then doing the same for b yields $b=3 \log_8 3$. Plugging back in appropriate a and b values into the simplified original, $3 \log_8 6 - 4 \log_8 3 = a- \frac43 b$.[/hide]

[quote="weatherbottle"]Source: MAML 2006 States!!

If $\log_4{36}=a$ and $\log_4{9}=b$, find $\log_8{\frac{216}{81}}$ in terms of $a$ and $b$.

Oh yea, you should have 2.5 minutes to do this question.[/quote]
[hide="logs are fun"]
Lets start with $\log_4{36}$
$\log_4 {36}=\log_4 {4\cdot9}=\log_4 {4}+\log_4 {9}$
$a=1+\log_4{9}$
$\log_4 {9}=a-1$
$a-1=b$
So now we have to break down $\log_8{\frac{216}{81}}$.
Immediately we see that that equals $\log_8{216}-\log_8 {81}$
Using the exponent rules, we get $3\log_8 {2}-4\log_8 {2}$
After converting to base 4, we get $a-4b/3$
[/hide]


[b](1) Is there is any Geometrical method to find the value of cos(18 degree), cos(36 degree) ,cos(54 degree).[/b]

hello, maybe this will help you
http://2000clicks.com/mathhelp/GeometryTrigSpecialAngles.htm
Sonnhard.

Sir have found no. idea how to find the value of sin(18 degree) or tan(18degree).
so sir plz explain me.

we have $ \sin 36^o \equal{} \cos 54^o$
use Triple Angle Formula  on $ RHS$ and Double Angle Formula on $ LHS$
and it will solve the problem


geometry
circumcircle
incenter
angle bisector

Right triangle $ ABC,$ with $ \angle A\equal{}90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC \equal{} \angle CAF.$ Segment $ BF$  meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.

[hide="click text"]Let I be the incenter of AEF. you should prove <EIF=<EDF. or in other words <EDF=90°+(1/2)(<EAF).
Just see that <EDB=<EAB=90°-(1/2)(<EAF) => <EDF=90°+(1/2)(<EAF).DONE[/hide]

let angle CAE=y, ACB=x
then ACF=90+y-x, EDF=90+y, AFE=90+x-2y, AEF=90-x
hence OEC=y
then EC/ CF=AE/AF=sin(90+x-2y)/cos x
we get EM/MC=(cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx) = tan ECO / tan y
so tan ECO = tany( (cos x +sin(90+x-2y))/(sin(90+x-2y)-cosx)) = tan(90+y-x)
so ECO = 90+y-x hence ACF,ECO are vertical angles which means A,C,O collinear.

[hide]
We can deduce that \[ \angle EOF \equal{} 2(180 \minus{} \angle EDF) \equal{} 2 \angle EDB \equal{} 2 \angle EAB \equal{} 2(90 \minus{} \angle EAD) \equal{} 180 \minus{} \angle EAF\], hence $ AFDE$ is cyclic. Since $ \angle OEF \equal{} \angle OFE$, we can deduce that $ AO$ is the angle bisector of $ \angle EAF$, hence $ A, C, O$ are collinear.
[/hide]

Let $ AF$ meet the circuncircle of $ EDF$ again at $ E'$. Angle chasing shows that $ AEE'$ is isosceles, that is, $ E$ and $ E'$ are symmetrical wrt to $ AC$ and the result follows.

We solve it just by angles chasing : denote $ {\angle ABC \equal{} N}$；$ {\angle BCA\equal{}90\minus{}N}$ ;$ {\angle ACE\equal{} M}$; then we get :$ {\angle EAC\equal{}180\minus{}M\minus{} N}$ $ {\longrightarrow}$ $ {\angle EAF \equal{}360\minus{}2M\minus{}2N}$;and $ {\angle BCE\equal{}\angle BCE\equal{}M\minus{}(90\minus{}N)\equal{}M\plus{}N\minus{}90}$; $ {\longrightarrow}$  $ {\angle EDF \equal{}270\minus{}M\minus{}N}$ ;$ {\angle EOF\equal{}[180\minus{}(270\minus{}M\minus{}N)]*2\equal{}2M\plus{}2N\minus{}180}$;
 thus $ {\angle EAF\plus{}\angle EOF\equal{}180}$; $ {\longrightarrow}$  $ {A,E,F,O}$ are con-cyclic; since $ {EO\equal{}FO}$ ;$ {O}$ is on the angle bisector of $ {\angle EAF}$; that is $ {A,C,O}$ are collinear.....$ {Q.E.D.}$[ :) ]

[quote]Right triangle $ ABC,$ with $ \angle A \equal{} 90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA > EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC \equal{} \angle CAF.$ Segment $ BF$  meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.[/quote]
Here is my solution

Since $ \angle OED$ $ \equal{}90^{\circ}\minus{}$ $ \frac {1}{2}$ $ \angle EOD$ $ \equal{}90^{\circ}\minus{}$ $ \angle DFE$ $ \equal{}\angle DCF$ $ \equal{}\angle DBE$, implying $ OE$ is a tangent from $ O$ to $ (\Gamma)$. The same argument holds for $ OD$, which yields that $ (O)$ and $ (\Gamma)$ are orthogonal. But in the other hand, due to the fact that $ (AB,AC,AE,AS)\equal{}\minus{}1$, then if $ W$ is the intersection of $ AB$ with $ EF$ then $ (EFCW)\equal{}\minus{}1$, which implies $ (BE,BF,BC,BW)\equal{}\minus{}1$ or we can say $ (BE,BD,BC,BA)\equal{}\minus{}1$ $ \Longrightarrow$ $ ADCE$ is a [i]harmonic quadrilateral[/i]. As the result, $ AC$ with then tangents at $ D$ and $ E$ will concur, hence, $ AC,$ $ OD,$ $ OE$ will concur at $ O$, which implies $ A,$ $ C,$ $ O$ are collinear. $ \square$

Notice $$-1=(E,F;C,\overline{AB}\cap\overline{EF})\stackrel{B}=(E,D;C,A)$$ by the Right Angles and Bisectors Lemma. Since $$\angle CEO=90-\tfrac{1}{2}\angle FOE=90-\angle EDB=90-\angle EAB=\angle CAE,$$ we know $\overline{OE}$ is tangent to $\Gamma$ at $E.$ Similarly, $O=\overline{DD}\cap\overline{EE}$ so if $A'=\overline{OC}\cap\Gamma,$ we see $(A',C;DE)=-1.$ Hence, $A=A'.$ $\square$


trigonometry
real analysis
real analysis unsolved


inequalities
logarithms
calculus
calculus computations

Let $a,\ b$ be real numbers such that $\log_{a}8=\log_{\sqrt{b}}2.$

(1) Find the values of $\log_{ab}a+\log_{b}{\frac{1}{\sqrt{a}}}+(\log_{a}b)^{2}.$

(2) Solve the following inequality with respect to $x.$ \[a^{3x}+a^{x+2}+a^{x}+a^{x-2}<b^{3x+3}+b^{3x}+b^{3x-3}+1\]

$\log_{a}8 = \log_{\sqrt{b}}2 \Leftrightarrow \frac{\ln 8}{\ln a}= \frac{\ln2}{\ln \sqrt{b}}\Leftrightarrow 3 = \frac{\ln 8}{\ln 2}= \frac{\ln a}{\ln \sqrt{b}}\Leftrightarrow a^{2}= b^{3}$

[b](1)[/b]  Hence the term is equal to $\frac{3}{5}-\frac{3}{4}+\frac{4}{9}= \frac{53}{180}$


Find all positive integers n such that 2^n-1 is a multiple of 3 and 1/3*(2^n-1) is a divisor of 4m^2+1 for some integer m.

I think it is not hard !
n=2^k.
First, We prove that n is power of 2 :
  from 2^n -1 is a mutilple of 3 hence n is even
  assume n has an odd prime divisor d
  2^d -4 is a mutilple of 4 that is 2^d-1 has prime divisor p is of the from  4k+3 .
  then p divides 2^d-1 , which divides 2^n-1,which divides 4m^2+1
  hence impossible
  so n=2^k
  consider f_k = 2^2^k +1
  by the chinese remainder theorem we will see n=2^k satisfies .

nice problem
i reached the same solution but once agen sum1 beat me to it
 :)

i can't understand that why p=3(mod4) can't divide 4m^2+1

sorry but could anyone explain why  numbers of the form $ n\equal{}2^k$ satisfy conditions of the problem? J understand why other cannot be solutions but I can't understand why result follows from chinese remainder theorem.

Thanks for Your explanation.

I also think it is not very clear (actually, I think it is wrong)...

Let $ n \equal{} 2^k$, with $ k \ge 1$. To prove that $ 4m^2 \equiv \minus{} 1 mod \frac {2^n \minus{} 1}{3}$ has a solution, we need to show that, for prime $ p$,

$ (\frac { \minus{} 1}{p}) \equal{} 1 \forall p: p|\frac {2^n \minus{} 1}{3}$,

i.e., $ p \equiv 1 mod 4$ if $ p|\frac {2^n \minus{} 1}{3}$. It is easy to see with congruences that $ 2^n \not\equiv 1 mod 9$ (remembering that $ n \equal{} 2^k$), so $ p \not \equal{} 3$. Now, if $ p|2^n \minus{} 1$, we have $ ord_p2|n$. Thus, $ ord_p2$ is a power of $ 2$, say $ 2^t$. Since $ 2^{2^t} \equiv 1 mod p$ and $ p > 3$, we have $ t > 1$. Then, as $ 2^{p \minus{} 1} \equiv 1 mod p$, $ 2^t|p \minus{} 1 \Rightarrow 4|p \minus{} 1$, q.e.d.


greatest common divisor
number theory
least common multiple

[hide] The product of the GCF and LCM is the product of the two numbers.  When you multiply the GCF and LCM, you get $2^{3}*3^{5}*5^{2}*7$.  Call the other number n.  This means that 210n=8*243*25*7, so n=1620.  [/hide]

[quote="espark52"]the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]

[size=50][hide="answer"]3240[/hide][/size][/color][/quote]
[hide]
LCM*GCF=Product
LCM*GCF/1number=theother number
$210=2*3*5*7$
LCM*GCF$=2^{3}*3^{5}*5^{2}*7$
LCM*GCF/210$=2^{2}*3^{4}*5$
$=1620$
[/hide]

[quote="davidlizeng"][quote="espark52"]the least common multiple of two nmbers is $2^{2}*3^{4}*5*7$. The greatest common divisor of the same two numbers is $2*3*5$. One of the numbers is 210. What's the other?[color=cyan]

[size=50][hide="answer"]3240[/hide][/size][/color][/quote]

uhh... espark, i think the answer is 1620, not 3240 as you have posted.[/quote]

oh woops, ur right :blush: the reason i put 3240 is cause i mis-typed the problem it's supposed to be $2^{3}*3^{4}*5*7$


geometry
circumcircle
inradius
incenter
geometric transformation
reflection
homothety

[color=darkblue]Given are two parallel lines $d_{1}$ , $d_{2}$ and the circle $w$ which touches the line $d_{1}$ in the point $A$ and $d_{2}\cap w=\emptyset\ .$
For a mobile point $M\in d_{2}$ denote the intersections $X$ , $Y$ between the line $d_{1}$ and the tangents from the point $M$ to the circle $w\ .$
Prove that exist two constants $k$ , $l$ so that the product $AX\cdot AY\in \{k,l\}\ .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=115606 (seek and will find ... the hint !).[/color]

The problem was also posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=85336[/url].

Assume the circle center [color=blue][b]is[/b][/color] between the lines $d_{1},\ d_{2}.$ Then $w$ is the incircle of $\triangle MXY,$ so we denote its center I.  Let (O) be the circumcircle of this triangle, $(I_{m})$ its M-excircle, T the tangency point of $(I_{m})$ with $d_{1}\equiv XY,$ and let $r,\ r_{m},\ h_{m}$ be its inradius, M-exradius, and M-altitude. Since $\frac{r_{m}}{r}= \frac{h_{m}}{h_{m}-2r}$ and since $r,\ h_{m}$ are constant, $r_{m}$ is also constant. The bisector $MI \equiv MI_{m}$ of $\angle XMY$ meets (O) again at the midpoint P of the arc XY of (O) opposite to M. The incenter I and the M-excenter $I_{m}$ are both on a circle (P) centered at P and passing through the triangle vertices X, Y. Since this circle is centered on the perpendicular bisector p of XY and $I_{m}\in (P),$ the reflection $I_{m}'$ of $I_{m}$ in p is also on (P). In addition, the reflection of T in p is A, the incircle tangency point with XY. Thus $I_{m}'A \perp XY,\ I_{m}'A = r_{m},$ and the points $I,\ I_{m}' \in P$ are both fixed. Hence the power of A to (P) is $AX \cdot AY = AI \cdot AI_{m}' = rr_{m}= \frac{h_{m}r^{2}}{h_{m}-2r}$ is constant. If we substitute the distance  $d = h_{m}-r$ of the line $d_{2}$ from the center I of the circle $w \equiv (I),$ we get

$\frac{AX \cdot AY}{r^{2}}= \frac{d+r}{d-r}$

Similarly, assuming the circle center [color=blue][b]is not[/b][/color] between the lines $d_{1},\ d_{2},$ $w$ is the M-excircle of $\triangle MXY$ and can get

$\frac{AX \cdot AY}{r^{2}}= \frac{d-r}{d+r}$

This proof is easily modified for the case $w \cap d_{2}\neq \emptyset$ and M is still outside of $w.$ WLOG assuming AX > AY, $w$ is the X-excircle of $\triangle MXY.$ For the proof, we use the Y-excircle, the radius of which is fixed. For example, if $r_{x}> h_{m}$ are fixed, since M is the internal homothety center of $(I_{x}),\ (I_{y})$, $\frac{r_{x}-h_{m}}{r_{x}}= \frac{h_{m}-r_{y}}{r_{y}}$ and $r_{y}= \frac{r_{x}h_{m}}{2r_{x}-h_{m}}$ is also fixed. The X, Y-excenters are both on a circle (P) centered at the midpoint P of the arc XY of (O) containing M. Similarly as before, (P) passes through the reflection of the Y-excenter in the perpendicular bisector of XY, which is fixed, etc.

[quote="Virgil Nicula"][color=darkred]Given are two parallel lines $d_{1}$ , $d_{2}$ and the circle $w$ which touches the line $d_{1}$ in the point $T$ and $d_{2}\cap w=\emptyset\ .$
For a mobile point $A\in d_{2}$ denote the intersections $B$ , $C$ between the line $d_{1}$ and the tangents from the point $A$ to the circle $w\ .$
Prove that exist two constants $k$ , $l$ so that the product $TB\cdot TC\in \{k,l\}\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] I"ll use the standard notations for the triangle $ABC\ .$ Prove easily that the relations $(p-b)(p-c)=\frac{r^{2}h_{a}}{h_{a}-2r}=\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}\ .$
[hide="Here is prove."]
$\{\begin{array}{cc}|\begin{array}{c}h_{a}=\frac{2pr}{a}\\\\ (p-a)(p-b)(p-c)=pr^{2}\end{array}& (p-b)(p-c)(h_{a}-2r)=(p-b)(p-c)(\frac{2pr}{a}-2r)=(p-a)(p-b)(p-c)\cdot \frac{2r}{a}=r^{2}h_{a}\\\\ |\begin{array}{c}h_{a}=\frac{2(p-a)r_{a}}{a}\\\\ p(p-b)(p-c)=(p-a)r^{2}_{a}\end{array}& (p-b)(p-c)(h_{a}+2r_{a})=(p-b)(p-c)[\frac{2(p-a)r_{a}}{a}+2r_{a}]=p(p-b)(p-c)\cdot\frac{2r_{a}}{a}=r_{a}^{2}h_{a}\end{array}$[/hide]
 $TB\cdot TC=(p-b)(p-c)\Longrightarrow$ $\{\begin{array}{cccccc}\mathrm{Case\ 1.\ : \ }& w=C(I,r) & \Longrightarrow & (p-b)(p-c)=\frac{r^{2}h_{a}}{h_{a}-2r}& \Longrightarrow &{ k=\frac{r^{2}h_{a}}{h_{a}-2r}\ }\\\\ \mathrm{Case\ 2.\ : \ }& w=C(I_{a},r_{a}) & \Longrightarrow & (p-b)(p-c)=\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}& \Longrightarrow & l=\frac{r_{a}^{2}h_{a}}{h_{a}+2r_{a}}\ . \end{array}$[/color]

that's a nice problem I see follow other way
Given $(I)$ touch line $d$ at $T$, find locus $A$ such that $(I)$ is incenter of $\triangle ABC$ with $B,C$ move on $d$ and $TB.TC=const$.
[hide="Solution"]
a line $// BC$:D :D
[/hide]


I'll post COPSD 3rd B soon to encourage more middle schoolers to attend my competition.

Anyone interested? (Even you qualified on COPSD 3rd A, you're welcome to just take it again for fun --> no prize for double qualifying though  :P )

When do you get when you qualify for the A one? (I'll do this one too  :D )

I'm interested, but could you please explain what COPSD 3rd A is? :? And what COSPD 3rd B is?

COPSD 3rd A was this thread 236factorial and Drunken_Math did:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50030

COPSD 3rd B is just like COPSD 3rd A but with different set of problems that are in same level. The reason that I have A and B is because on the last time, some of people couldn't compete because of AP tests.

Anyway, if you guys know AMC, this is same. Same level problems, just in different date.

(Although you can do both  :P )

To qualify, you must send at least 7 SOLUTIONS (answers + explanation) to the problems that are correct. In another word, if you get 7 right, you're qualified. If you get more, good to you. If you get less,  :( .

Anyone who qualifies from COPSD will go to COP Round 2, along with people from COP Round 1 (who are mostly high schoolers ---> This means you can go against the high school students!!).

If you guys have any more questions, please post!!

I guess I will do it, it sounds fun.  Are the problems like AMC 10 level?  When are you going to post the questions?

[quote="236factorial"]Oh, so now I get to participate in COP 5th now?  :D And can I do the COPSD B round at the same time?[/quote]

Yes. You're ALREADY qualified to COP Round 2 which is eh.. about level of later AMC-12 and possibly easy AIME.

[quote="b-flat"]Are the problems like AMC 10 level?[/quote]

No. They're level of mathcounts varying from chapter to national level.

[quote="b-flat]When are you going to post the questions?[/quote]

Soon. But I can't give you an exact date since I'm very busy this year (about 3 times more than last year). I'll announce it when it's ready.

We need more people to sign up!! Come on, guys! This is a VERY good practice for mathcounts!

Alright I think I'll have time to participate in this one. I'm sorry i didn't do 3rd A... I didn't have the time.

Here are the problems, folks! Enjoy!

Note: These are VERY good practices for local, state, national mathcounts. Everybody who visit MC forum should give a shot. Also, some problems may require brute forcing and use of calculator so don't be afraid on this fact.

[b]Problems[/b]

1. Find the maximum or minimum height defined in equation $y = x^2-12x+2$.

2. Find $a+b+c$.

\[2a+b+c = 10\\
a+2b+c = 12\\
a+b+2c = 14\]

3. In a 10-questioned test, the student can answer either true or false but must answer all of the 10 questions (no omitting!). How many ways can the student do this?

4. Find $a$ and $b$ in:

$\sqrt{4 + \sqrt{4 + \sqrt{4 + \sqrt{4 + \cdots}}}} = \frac{a+ \sqrt b}{2}$

5. The equilateral triangle $ABC$ has side length of 10. Connecting triangle $ABC$'s midpoints, we form triangle $A'B'C'$ with side length 5. In same manner, we form $A''B''C'', A'''B'''C'''$, and so on. Find the area of all equilateral triangles formed this way (including triangle $ABC$). In another word, what is $[ABC]+[A'B'C']+[A''B''C'']+[A'''B'''C''']+[A''''B''''+C''''] + \cdots $? ($[z]$ denotes the area)

6. Find the EXACT value of

\[\sqrt{4 + 2 \sqrt 3} + \sqrt{4 - 2 \sqrt 3}\]

7. The polynomial below has four roots. One of them is $4i$. Find the three other roots.

\[x^4-5x^3+22x^2-80x+96\]

Note on #7. This type of problem won't appear on Mathcounts history because $i$ indicates $\sqrt{-1}$ which is a complex number. But here is a big hint on this problem. If a polynomial contains the $a+bi$ as one of root, it also contains $a-bi$ as another root.

8. For non-real number $r,s$, we have:

\[r+s = 5\\
rs = 10\]

Find $\frac{1}{r^2} + \frac{1}{s^2}$.

9. How many number less than 500 satisfy:

(i) has perfect square as one of its factors
(ii) divisible by 7
(iii) sum of its digit is 18

10. In how many ways can you pick two numbers in a set $\{1,2,3,4,5 \}$ such that their sum is even?

All SOLUTIONS must be turned to ME by PM today (11:59:59).

Thank you and good luck!

[quote="nonie"]I'm too late :([/quote]

Here's a very good news for you.

I decided to extend the deadline since I only received solution from only one contestant (can't believe only one in COP and only one in COPSD  :( ).

So, the new deadline is 10/3/05. Enjoy problem solving till then!

Remember: I need the solutions to at least 7 of the problems.


li m x----8           sin{ x-10}/{10-x}

{} denotes fractional part function
pl also explain little bit about fractional part function.i dont  know any thing about it

Q2 lim X---0   Sec-(x/sin x)=l and limx---0 sec-(x/tan x)=m then which limit exist and which will not pl explain

The [url=http://mathworld.wolfram.com/FractionalPart.html]fractional part of a function[/url].


This Problem is marvelous ,i shall not say the source ,cause then you will not think on it ,but here it is [hide="Dont look"]IMO 1991,Problem 6[/hide]

[color=blue]Problem[/color]
The real number $a>1$ is given .
A sequence like  \[ \{x_i \}_{i=0}^{ \infty}  \]
is called $LHTB(\ little \ hard \ to \ build )$ if for every two not-negative  integers like $i,j,that, i \neq j$ \[ |x_i-x_j| \geq \frac{1}{|i-j|^a}  \]

1.Prove that there exists a[b] bounded[/b] $LHTB$ sequence.

2.[hide="The Original problem,"]IMO1991,6[/hide] Built a[b] bounded[/b] $LHTB$ sequence.

@lomos_lupin: Isn't this sequence supposed to be bounded in the original statement ?

If you want bounded sequence, then it can be build:
Let $n=a_0+a_1*2+\dots+a_k*2^k, a_i\in \{0,1\}$.
$x_n=2(a_0+a_1*2^{-1}+\dots+a_k*2^{-k}).$

:blush:  Yes its bounded ,All of its fun was for being bounded.My bad

A prove for existance of such a sequence:

[color=blue]Solution[/color]Assume that we have built $x_0,...,x_{n-1}$ and we want to built the $x_n$ . Let $I$ be a bound for the sequence .for satisfying the condition of the problem ,we should have :

 \[ |x_n-x_i| \geq \frac{1}{|n-i|^a} ,0 \leq i \leq n-1  \]

So for existance of $x_n$ we should have :

\[ \sum_{i=0}^{n-1} \frac{1}{|n-i|^a} < I  \]

but we know that  :
\[ \sum_{i=1}^{n} \frac{1}{i^a} < M= \sum_{i=1}^{\infty} \frac{1}{i^a} < \infty  \]
(Actually $M$ is a rational product of $\pi^{a})$
By choosing $I=2M$ ,such a $x_n$ satisfying the condition of the problem exists.

To Rust : NO that sequance doesnt work ,Your sequance is not related to $a$,but it should be related.
A book that i have present this sequence :

Let the $x_n$ in the Rust sequence be $y_n$ then define your sequence as following
\[ x_n=\frac{2^a-2}{2^a-1}y_n  \]


calculus
integration
trigonometry
complex analysis

For the first, how about first just expanding $ e^{1/z}$ in a Laurent expansion:

$ e^{1/z}\equal{}1\plus{}1/z\plus{}1/2(1/z)^2\plus{}1/6(1/z)^3\plus{}...$

Alright, now expand $ Exp[1\plus{}1/z\plus{}1/2(1/z)^2\plus{}1/6(1/z)^3\plus{}...]$

Can you show that I get for the residue:

$ \mathop\text{Res}\limits_{z\equal{}0}\left\{e^{e^{1/z}}\right\}\equal{}
\sum_{n\equal{}1}^{\infty}\frac{n}{n!}$

[quote="sailorsenshi"]How would I use Residue Theorem and Laurent expansions to do the following problem?

[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob1-1.jpg[/img]

Thanks![/quote]
I don't known.
Is it right?
${ e^{ix}=\cos{x}+i\sin{x}}$
${ e^{x}=\cos{\frac{x}{i}}+i\sin{\frac{x}{i}}}$
Use it , we  will ...


[quote="Trisha"]Hi,
I need help in solving the equation.

5x-7y=31
-4x+2y=-14

Please solve by substitution and show steps.  Thanks[/quote]

[hide="hint"]

solve for either variable in either equation.

then substitute that equation into the corresponding variable in the other equation

solve for the other variable

plug the integer into an equation and solve for the original variable.
[/hide]

[hide]
1.$5x-7y=31$
2.$-4x+2y=-14\implies-2x+y=-7\implies y=2x-7$

Substitute into 1

$\Rightarrow 5x-7(2x-7)=31\implies x=2$

substitute $x=2$ into 2

$\Rightarrow y=2(2)-7=-3$

$\boxed{x=2,y=-3}$
[/hide]

Thanks SM4RT this is exactly what I wanted to see. :lol:

Another option is to convert into the coefficient matrix into reduced row echelon form...

Or use cramer's rule, but both of these solutions are overkill on a two equation system.

The substitution method in general is the fastest way of solving small (2x2 or 3x3) systems.

Anything larger, and matrix transforms are your friend.


real analysis
topology
real analysis unsolved

This question is from royden's real analysis from the chapter on lebesgue measure. Should be quite easy to solve. But i cant  :(

Let $E$ be a set. Denote $m(E)$ to be the outer measure of $E$. Assume further that $m(E)$ is finite. Prove that the following are equivalent.

(i) For all $\epsilon>0$, there exists an open set $O\supset E$ s.t. $m(O\backslash E)<\epsilon$.
(ii) For all $\epsilon>0$, there exists a finite union of open intervals $U$ s.t. $m(U\Delta E)<\epsilon$.

NB: $U\Delta E= (U\backslash E) \cup (E\backslash U)$.

You missed a condition in (i). In royden's book, we also have $O\supseteq E$.

(i) => (ii)

Let $\epsilon>0$, by (i) there exists $O=\bigcup I_{n}$ such that $m^\ast(O\setminus E)<\epsilon/2$, where $I_{n}$ are open intervals.
Then
$\sum|I_{n}| = m(O) \leq m^\ast(O\setminus E)+m^\ast(E) < \frac \epsilon2+m^\ast E$ is finite.

Hence $\sum |I_{n}|$ converges, so there exist $N$ such that $\sum_{n=N}^\infty |I_{n}| < \frac\epsilon 2$. Let $U = \sum_{i=1}^{N}I_{n}$ then $m^\ast(U\Delta E) \leq m^\ast(U\setminus E)+m^\ast(E\setminus U)<\frac \epsilon 2+\frac\epsilon 2 = \epsilon$. (Note that $E\setminus U$ is a subset of $\sum_{i=N}^\infty I_{n}$)

(ii) => (i)

Let $\epsilon>0$, by (ii) we have a finite union of open intervals $U$ such that $m^\ast(U\Delta E)<\frac\epsilon 3$. By definite of outer measure, we have an open covering of $U\Delta E$ thus an open set $G\supseteq G\Delta E$ such that $m(G)<m^\ast(U\Delta E)+\frac\epsilon 3 < \frac{2}{3}\epsilon$.

Let $O=G\cup U$, it is clear that $E\subseteq O$, and

$m^\ast(O\setminus E)\leq m^\ast(U\setminus E)+m^\ast(G\setminus E)<\frac\epsilon3+\frac23\epsilon = \epsilon$.

[quote="liyi"]You missed a condition in (i). In royden's book, we also have $O\supseteq E$.
[/quote]

ooops.  :blush: i've just edited the post.

[quote="liyi"]
Let $\epsilon>0$, by (i) there exists $O=\bigcup I_{n}$ such that $m^\ast(O\setminus E)<\epsilon/2$, where $I_{n}$ are open intervals.
[/quote]

think i'm forgetting some facts. But is it always true that any open set $O$ can be written as a countable disjoint union of open intervals? (i think you made use of the fact that it's disjoint when you assumed $\sum|I_{n}|=m^{*}(O)$.)

Consider an open subset of the reals. Its connected components must be open, as given any element of a particular connected component, there exists a neighborhood of that element contained in the set, and the union of this neighborhood with the connected component is readily seen to still be connected. Thus, as open, connected subsets of the reals are simply open intervals, any open subset is a union of open intervals; they are disjoint as connected components are disjoint and there are a countable number as we can choose a rational in each of the open intervals.

[quote]think i'm forgetting some facts. But is it always true that any open set $O$ can be written as a countable disjoint union of open intervals? (i think you made use of the fact that it's disjoint when you assumed $\sum|I_{n}|=m^{*}(O)$.)[/quote]
Yes, i think you can find the standard proof in Royden's book. But it is for sets on the real line only.


a,b,c>0
 ab+bc+ac = k

 ????
prove a+b+c>=k

a=1, b=1, c=3
And I have proved it wrong.

For your information,
$(a+b+c)^2\ge 3(ab+bc+ca)$
This can be proved using AM-GM

See my above post, I've solved it.

[hide="Solution"]$(a+b+c)^2\ge 3(ab+bc+ca)$
$\Rightarrow (a+b+c)^2 \ge 9$
Thus proved.[/hide]

You may like to prove the inequality yourself.
[hide="Big Hint"]
Use AM-GM[/hide]

[hide="Is this it?"]Expaninding gives us \[ a^2+b^2+c^2+2ab+2ac+2bc \geq 3ab+3ac+3bc \] So we need to proof that \[ a^2+b^2+c^2 \geq ab+ac+bc \] Apply AM-GM on $\{a^2,b^2\}$, $\{a^2,c^2\}$ and $\{b^2,c^2\}$, getting $\frac{a^2+b^2}{2} \geq ab$, $\frac{a^2+c^2}{2} \geq ac$ and $\frac{b^2+c^2}{2} \geq bc$. Adding those three: \[ \frac{a^2+b^2+a^2+c^2+b^2+c^2}{2} = a^2+b^2+c^2 \geq ab+ac+bc\ \ \square \] [/hide]

A simpler way to prove $a^2 + b^2 + c^2 \geq ab + bc + ac$ is to re-arrange and factor it as

$1/2 (a - b)^2 + 1/2 (b-c)^2 + 1/2(c -a)^2 \geq 0$, which is always true.