Determine all functions satsisfying the functional identity
\[x^{2}f(x)+f(1-x)=x^{4}-2x.\]

Hmm, the way I solved it works, but is there a better one? [hide="My Method"]Note that we must have $f(x)=x^{2}+ax+b$ for some $a$ and $b.$ Hence,
\begin{eqnarray*}x^{4}-2x &=& x^{2}f(x)-f(1-x) \\ &=& x^{2}(x^{2}+ax+b)-[(1-x)^{2}+a(1-x)+b] \\ &=& x^{4}+ax^{3}+(b+1)x^{2}-(a+2)x+(a+b+1). \end{eqnarray*}
For this to be satisfied, $(a,b)=(0,-1),$ so $f(x)=x^{2}-1$ is the only solution.
[/hide]

or you can let $x=1-x$ and just solve the system of equations on $f(x)$ and $f(1-x)$

[quote="4everwise"]Oh, does that work out? I tried it at first, but the arithmetic got a little ugly so I stopped.  :D[/quote]

it does work out though i havent done the arithmetic, so i wouldnt know

[quote="me@home"]It is all functions, not all polynomials, so how can you immediately assume that $f(x)$ is monic second degree?[/quote]

his method is wrong obviously...

as above, just do  1-x-> x, to get

(1-x)^2f(1-x)+f(x)=(1-x)^4-2(1-x)

then, along with the initial statement, you have a linear system for f(x), f(1-x), which is easy to solve


trigonometry
algebra
function
domain
Pythagorean Theorem
geometry
puzzles

For how many values of $\text {n}$ will the equation below be always true? \[\frac {\sin x}{\text {n}} = 6\]

[hide="Answer"]Infinite, assuming $n$ does not have to be an integer. Basically, $\frac{\sin x}{n}=6 \Rightarrow n=\frac{\sin x}{6}$.[/hide]

The answer is indeed infinite, but the working is wrong... you must achieve the answer the "brainteaser" way.   :D

$\frac{\sin x}{n}=six$ ($n$'s cancel out!)

So $six=6$, which is always true ;)

(sin x)/n = 6 ==> n = (sin x)/6 ==> n =  {-1, ..., -0.5, ..., 0, ..., 0.5, .... 1}/6

There are infinite values of n in the closed domain [-1/6, ...., 0, ..... +1/6] that would satisfy sin x = 6*n.

If you need values of n which would make   (sin x)/n = 6, then n will have all the other values except 0 as sinx/0 can never be equal to 6.

n = {-1/6, ......, 0), (0, ......, +1/6}
In any case there are infinite values of n.

"A cockroach cannot be killed immediately by chopping off its head. It takes days before dying of starvation as it cannot make itself feed anymore." :blush:

Well .... it is a brainteaster anyway .... you dont need to work out those step but just cancel off the $n$ (think algebrically :P )  and you will have $six=6$ like what $frt$ said  :)

:first:

But you already knew the question shyong.  :D

#2, because $\sin ^2x+\cos ^2x=1$. :lol:

It depends if you're a scientific calculator or a human, because the scientific calculator will choose #1 any day (because it has less operations)

[quote="mathnerd314"]Here's a well-known one:

Which is easier:
1. Evaluate sin 39 + cos 39
2. Evaluate (sin 39)^2 + (cos 39)^2

You have five seconds to decide.

[hide="hint"]
it's a well-known mathematical joke
[/hide][/quote]I give up.

Here's one. I searched on google and found this, which I copied from a post on another forum:

[quote]The iden***y "sin(2r) = 2sin(r)" applied for a loan
and was turned down.

He needed a cos(r).[/quote] :rotfl:

Why the three stars in "identity"?

There's a well-known trigonometric identity based on the Pythagorean Theorem that states that $\sin^2 x + \cos^2 x = 1$ for any x. I'd choose #2 any day. (What on earth is sin 39?! Degrees OR radians!

[quote="mathnerd314"]Why the three stars in "identity"?[/quote][hide="adults only"]The three stars censor the word "tit."[/hide] :D


number theory
greatest common divisor
relatively prime
number theory proposed

Let $a,b$ be possitive integers such that : the set of prime divisors of $a^n +b^n$ is finite for all n . Prove that : $a=b$

We may assume WLOG that $a,b$ are coprime. Assume Also the contrary to what we want to prove, namely that ther are only finitely many odd prime divisors $p_1,\ldots,p_k$ of the numbers of the form $a^n+b^n$ and $a\ne b$.

If one of $a,b$ is even, then $2$ doesn't divide $a^n+b^n$ for $n\ge 1$, so if we choose $n=\prod_{i=1}^k(p_i-1)$, the number $a^n+b^n$ will be divisible by none of $2,p_i,\ i=\overline{1,k}$, so we have a contradiction.

If $a,b$ are odd, again, for all multiples $n$ of $\prod_{i=1}^k(p_i-1)$ the number $a^n+b^n$ will be divisible by none of $p_i,\ i=\overline{1,k}$, and the only possibility is that $a^n+b^n=2^t$ for some $t$. Since $a,b$ are odd and different, one of them is $\ge 3$, so for large enough $n$ we have $2^n<a^n+b^n$, meaning that $2^n|a^n+b^n$. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=order&t=35430]This problem[/url] says that this is false, i.e. if $a,b$ are odd positive integers, then $2^n|a^n+b^n$ only for finitely many $n$ (there's probably a much simpler way of showing that $a^n+b^n$ is a power of $2$ only for finitely many $n$, but I like the problem in the link :)).

We can assume that $gcd(a,b) = 1$. Suppose that we have that $x,y$ are relatively prime integers and $k,l$ are positive
integers. Then we have that $gcd( x^{k} - y^{k}, x^{l} - y^{l} ) = x^{ gcd(k,l)} - y^{ gcd(k,l)}$. Indeed, if $m = gcd( x^{k} - y^{k}, x^{l} - y^{l} )$ , then $x,y$ are invertible $( mod m )$ and we have that $x^{k} \equiv y^{k} , x^{l} \equiv y^{l} ( mod m )$. Therefore for every choice of integers $t,s$ we have that
$x^{tk+sl} \equiv y^{tk+sl} ( mod m )$, so taking $t,s$ such
that $tk + sl = gcd( k,l )$, we have that $x^{ gcd(k,l)} - y^{ gcd(k,l)} \equiv 0 ( mod m )$. Now, if $k, l$ are odd integers,
taking $x = a$, $y = (-b)$ we get that $gcd( a^{k} + b^{k}, a^{l} + b^{l} ) = a^{ gcd(k,l)} + b^{ gcd(k,l)}$. Now if we take
a sequence of odd primes $p_{1},p_{2}, \ldots$, then we obtain
that $gcd( a^{p_{i}} + b^{p_{i}}, a^{p_{j}} + b^{p_{j}} ) = a + b$. Therefore if we take $x_{i} : = ( a^{p_{i}} + b^{p_{i}} ) / (a+b)$, then $gcd(x_{i},x_{j}) = 1$. Therefore if we assume
that $max(a,b) > 1$, then $x_{i} > 1$ for every $i$, and
they are relatively co-prime, so there are infinite number of
prime divisors for $x_{j}$ for all $j$, and hence for $a^{n} + b^{n}$.

I hope it is correct.

How would you generalize it for $a_1, a_2,...,a_u$ numbers?


i am sure this came up a lot of times... but i could not find using search, so here i ask:

I was on the classroom today, to test some LaTeX and hopefully to find someone i could discuss math with(no one, unfortunately :( ), and somehow i couldnt type in LaTeX. What happened?

[quote="xxreddevilzxx"][quote="towersfreak2006"]Did you remember to use the semicolon?[/quote]

do i use it at the begiining and at the end?[/quote]

beginning

How do you type variables with different bases?

just do [code]a_{base}[/code]

it will show like this $a_{base}$

example: $123_{12}$

[quote="Using TeX in the Classroom handout"]In order to use TeX in a class post, begin the post with a semicolon (; ). This tells the classroom that you are using TeX. If you forget to use the semicolon, your post will simply come up as regular text. ... It may take a little getting used to using the semicolon, but the TeX will make all of your mathematical expressions look much nicer. If you have little experience using TeX it is a good idea to practice using TeX in the forums. You do not need to use the semicolon there. Just be sure to write your math expressions between the dollar signs.[/quote]
So basically, insert your ;, and do what you would normally do here on the forums (dollar signs and all). I assume you know how to get into math mode from your previous post. :)

Hope that helped.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

A (not necessarily commutative) ring R with 1 is local if the sum of an two non-units is a non-unit. (A non-unit is an element which does not have a two-sided multiplicative inverse.)

Here are some problems about on local rings:

1. Let R be a commutative integral domain with P a prime ideal. Let S be the complement of P in R. Let R_P be the subring of the field of fractions of R consisting of fractions of the form r/s where r is in R, s in S.
Show that R_P is a local ring.

2. Let R be a commutative ring with M a maximal ideal. Showt aht R/(M^k) is a local ring for k>0 an integer.

3.  Let R be a ring in which every element is either a unit or nilpotent. Show taht R is a local ring.

4. Let F be a field (or division ring), n>1 an integer and let R be the set of upper triangular nxn matrices over F for which all of the entries on the diagonal are equal. Show that R is a local ring.

5. Let G be a finite p-group where p is a prime. Let F be a field of characteristic p. Show that the group ring F[G] is a local ring.

you should have posted each problem separetely...

for 1) The only maximal ideal is $M=\left\{\displaystyle\frac ps : p\in P, s\in S\right\}$
It is clear that this one is an ideal (because $\underline p$ is prime). And because each element which is not in $M$ is of the form $\displaystyle\frac{s}{s'}$ with $s,s'\in S$ it is thus invertible, so $M$ is maximal and it is the only maximal ideal of $R_{\underline p}$. Hence $R_{\underline p}$ is a local ring.

3) (for $R$ commutative) The maximal ideal will be the Jacobson radical, $J=\{x\in R: x$ nilpotent $\}$. Now it is clear that any other ideal is contained in this ideal, and this one is maximal (because the other elements are invertible, thus they cannot be a part of an ideal).

There is also a result (easy to prove) which says that if $R$ is a ring and $M$ a maximal ideal so that every $x\in R-M$ is invertible then $R$ is local. (it can pe applied to both of this problems).


geometry
circumcircle
trigonometry
function
geometry proposed

Let $ ABC$ be an acute triangle. Take points $ P$ and $ Q$ inside $ AB$ and $ AC$, respectively, such that $ BPQC$ is cyclic. The circumcircle of $ ABQ$ intersects $ BC$ again in $ S$ and the circumcircle of $ APC$ intersects $ BC$ again in $ R$, $ PR$ and $ QS$ intersect again in $ L$. Prove that the intersection of $ AL$ and $ BC$ does not depend on the selection of $ P$ and $ Q$.

[color=darkblue]We have:[/color]

$ \left\| \begin{array}{cc} BPQC \rightarrow \angle{PBQ} \equal{} \angle{PCQ} \\
AQSB \rightarrow \angle{ABQ} \equal{} \angle{ASQ} \\
APRC \rightarrow \angle{PCA} \equal{} \angle{PRA} \end{array} \right\| \Rightarrow \angle{ABQ} \equal{} \angle{ASQ} \equal{} \angle{PCA} \equal{} \angle{PRA}$ [color=brown][i](*1)[/i][/color]

[color=darkblue]Also [/color]

$ \left\| \begin{array}{cc} AQSB \rightarrow \angle{A} \equal{} 180^\circ \minus{} \angle{BSQ} \equal{} \angle{QSC} \\
APRC \rightarrow \angle{A} \equal{} 180^\circ \minus{} \angle{PRC} \equal{} \angle{PRB} \end{array} \right\| \Rightarrow \angle{QSC} \equal{} \angle{PRB}$ [color=brown][i](*2)[/i][/color]

[color=darkblue]From[/color] [color=brown][i](*1)[/i][/color] [color=darkblue]and[/color] [color=brown][i](*2)[/i][/color] [color=darkblue]obtain that triangles $ ARS$ and $ RLS$ are isosceles, so quadrilateral $ ARLS$ is kite-shaped and $ AL\perp BC$. From here their intersection is always constant (foot of the altitude from $ A$).
[/color]
[hide="Image"][img]http://ipicture.ru/uploads/080608/ouKg5V1i2T.png[/img][/hide]

I´ve found another solution:
Let D, E and F the foot of the altitude from A, B and C respectively.
Now, it´s easy to see that DE//QS; because both are antiparalells to BC. Analogly, we have DF//PR and EF//PQ.
Then DEF and LPQ are homothetic from A, then, A, D and L are collinears.
Now we can conclude like the prove above.

here is the sketch for another proof

Let $ P', Q'$ be another pair of points like $ P, Q$. In the same way define $ L'$.
    Then: the sides of the triangle $ PQL$ are parallel to the sides of the triangle $ P'Q'L'$

Let $ T \equal{} AL\cap BC$, $ U\in AC$ such that $ UT\parallel{}LQ$ and $ V\in AB$ such that $ VT\parallel{}AC$.
    Then: the sides of triangle $ UVT$ are parallel to the sides of triangle $ PQL$

Prove that there exist only one triangle with its vertex on the sides of $ ABC$ with sides parallel to a given reference triangle.

This proves that triangle $ UVT$ is "constant", in particular $ T$ is the always the same point.

:)

let $ \angle CAL \equal{} \alpha$. note that $ \angle AQL \equal{} \pi \minus{} B$ and $ \angle APL \equal{} \pi \minus{} C$, therefore $ \dfrac{\sin\alpha}{\sin (A \minus{} \alpha)} \equal{} \dfrac{QL\sin B}{PL\sin C}$

note that $ \angle PQL \equal{} \pi \minus{} 2B$ and $ \angle QPL \equal{} \pi \minus{} 2C$, so, $ \dfrac{\sin\alpha}{\sin (A \minus{} \alpha)} \equal{} \dfrac{\sin 2C\sin B}{\sin 2B\sin C} \equal{} \dfrac{\cos C}{\cos B}$... given that $ \dfrac{\sin x}{\sin (A \minus{} x)}$ is an increasing function, it follows that $ \alpha$ is constant, which is equivalent to the problem statement...

in fact, it follows from $ \dfrac{\sin\alpha}{\sin (A \minus{} \alpha)} \equal{} \dfrac{\cos C}{\cos B}$ that $ \alpha \equal{} \frac {\pi}{2} \minus{} C$, which implies that $ AL\perp BC$.

This problem is mine, my original proof is the following

step1: Triangle LSR is isosceles
step2: A is the L-excenter of triangle PQL, which implies AL bisects angle SLR
step3: conclude that AL is perpendicular to BC.


Euler search quadratics MATHCOUNTS algebra

What are the last 2 digits of the 200th number in the sequence 1, 1, 2, 3, 5, etc. where each number is the sum of the two preceding numbers.

[quote="i_like_pie"][quote="bpms"][quote="jhredsox"]i don't know exactly how to do it, but does it involve dividing the nth term by 6 because of a pattern?
-jorian

edit:sry, ilikepie, we were typing at the same time :wink:[/quote]
No, that's an old state. Just because a problem is posted doesn't mean it has anything to do with an old problem. My solution,
[hide]Look at it mod 100
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 44, 33, 77, 10, 87, 97, 84, 81, 65, 46, 11, 57, 68, 25, 93, 18, 11, 29, 40, 69, 09, 78, 87, 65, 52, 17, 69, 86, 55, 41, 89, 30, 19, 49, 68, 17, 85, 02, 87, 89, 76, 65, 41, 06, 47, 53, 00, 53, 53, 08, 61, 69, 30, 99, 29...[/hide] Someone else can continue. Also, I like pie either used the formula (It's ridiculously ugly, impossible to memorize, and just plain retarded for a problem like this unless you have an 89) or he looked it up. A list can easily be found online. The formula is
$\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}}{\sqrt{5}}$[/quote]
I never knew Fibonacci had a formula... :what?:[/quote]
So basically you looked it up? Or did you actually add all those terms up?

[quote="ragnarok23"]So basically you looked it up? Or did you actually add all those terms up?[/quote]
I didn't know there was a formula and couldn't think of a shortcut, so yes, I looked it up. :ninja:

wait, so someone please explain an easy way to do this
-jorian

I believe the best way is to use the fibonacci formula (Binet's formula) and take each of the individual terms mod 100, either using Fermat-Euler or the Binomial Theorem. I'll post the solution when I have time, maybe in like 20 minutes.

a couple people have tried to explain "mod" to me over aim, but i just don't get it

anyone wanna teach me?
-jorian

Didn't you already post a topic on that, where alot of people gave you answers?

[quote="jhredsox"]ummmmmmmmm........maybe, but posts go so quick that you never know

p.s. do u have the link?
-jorian[/quote]
See, this is where you scroll to the top of the page, and find the button that says search. Then you click on it. Then you type in what you are looking for. Oh so convenient. Or you could use the AoPS wiki. It's a great source ;).

ok bpms, u don't have to wink at me because i don't know this!

when u search, does it find titles, or posts?
-jorian

AFter working through the solution with Binet's, I think this is a little too hard for mathcounts, unless there is a simpler solution that I am overloooking.

[quote="t0rajir0u"][quote="bpms"]It's ridiculously ugly, impossible to memorize[/quote]

Now, that's a bit unfair.  The concepts behind its derivations are very instructive.  And it's not so hard to memorize this way:

Let $\phi, \varphi$ be the (larger and smaller) roots of the quadratic equation $x^{2}= x+1$.  Then

$F_{n}= \frac{\phi^{n}-\varphi^{n}}{\phi-\varphi}$

Doesn't that look nice and neat now?  :)

(This is with the convention $F_{0}= 0, F_{1}= 1$.  It also affords an immediate proof that $F_{n}| F_{mn}$.)[/quote]
Shhh. I needed to back up my point :D.


trigonometry
function
calculus
integration
trig identities
real analysis
real analysis unsolved

a) For each $s,t\in\mathbb{R}$, we have \[\sup_{\mathbb{R}}|\cos(tx)-\cos(sx)|\geq 1\] hint: First consider the case where $s$ and $t$ are linearly dependent over $\mathbb{Q}$.  Otherwise, $s$ and $t$ are not linearly independent over $\mathbb{Q}$, and something like Kronecker's theorem ($\{\{n\alpha\}: n\in\mathbb{N}\}$ is dense in $[0,1]$ for every $\alpha\in\mathbb{R}\setminus\mathbb{Q}$) would come in handy here.

b) The family $\mathcal{F}=\{\cos ax: a\in\mathbb{R}\}$ is an uncountable family of functions satisfying $\|u-v\|_{\infty}\geq 1$ if $u,v\in\mathcal{F}$ with $u\not=v$.

c) $C(\mathbb{R})\cap L^\infty(\mathbb{R})$ is not separable.

[quote="blahblahblah"]
c) $C(\mathbb{R})\cap L^\infty(\mathbb{R})$ is not separable.[/quote]

c) is simpler than a). Let $A=\{ a_{n}\}$ is an arbitrary sequence of integer nubbers. Let us define a piece linear function $F_{A}(x)$:
$F_{A}(x) = 1$ for $x \in A$ and $F_{A}(x) = 0$ for half-integer $x$.
If $A \neq B$ then $\| F_{A}-F_{B}\| = 1$ and hence
$C(\mathbb{R})\cap L^\infty(\mathbb{R})$ is not separable.

A less standard approach to (a): expanding $(\cos tx-\cos sx)^{2}$ and using trig identities, obtain 1 + a linear combination of four trig functions. The integral of this over $[0,A]$ is $A+O(1)$. Let $A\to\infty$.

Of course, we must assume $s\ne t$.

That's a very nice idea.  I don't know if it's less standard; it manages to handle all cases at once though.  In any event, the sup is $2$ for almost all $s$ and $t$ (if they are linearly independent over $\mathbb{Q}$ you can get as close to $2$ as you want, in the linearly dependent case you can assume that they are integers, and even then the sup is usually $2$ with relatively few cases where the sup is $1$).

Of course the usual example to show that $L^{\infty}$ is not separable is characteristic functions, and Alexander Khrabrov's set of functions is the right generalization, but this is what I thought of first and decided to force it through, much to my chagrin.

How can $\sup (\cos tx-\cos sx)^{2}$ be equal to $1$ if the average value is also $1$?  :o

I have no idea why I wrote that.  What I meant to say is that if, say, $s=5$ and $t=1$, then the sup is not $2$  :blush:

Let's assume, as we may, that $t=1$. Two natural questions:

1) for which $s\in\mathbb R$ $\sup|\dots|<2$?

2) what is the smallest possible value of the $\sup$?

1.  The sup is not $2$ for $s\in 2\mathbb{Z}-1$.  It's trivial if $s\in\mathbb{R}\setminus\mathbb{Q}$ by the earlier argument, and if $s\in 2\mathbb{Z}$ is similarly easy.

2.  For $s=2k+1$, the sup is $1-\cos\left(\frac{2k\pi}{2k+1}\right)$, which is minimal for $k=1$, yielding a minimum of $3/2$.

With normalization $t=1$ we cannot assume that $s$ is an integer. It seems that the supremum is less than $2$ if and only if $\{s,1/s\}\subset \mathbb Q\setminus 2\mathbb Z$.  :maybe:


AMC
AMC 12
AIME
calculus
Putnam
USA(J)MO
USAMO

Hi everyone,

I'm a senior in high school, and I'm totally aware that I'm starting this way too late, but I really love math. I was wondering what resources or methods you guys would recommend to get started with math contests and problem solving. Last year I decided on impulse to take the AMC12, and I qualified (albeit with a low score) and scored a 4 on the AIME. I know this isn't exactly a good score, but I was wondering what advice you guys could offer me. I'm a pretty quick learner, so don't hesitate to give me some tough work :T. It just seems like theres so much to learn, I don't even know where to get started!

Thanks!

[quote="JRav"]Read AoPS Volume 2 and practice.[/quote]

Definately. I found volume 1 quite managable but Vol 2 is going to require a lot of hard work and dedication. Also, many problems issued in volume 2 are from AIME (I found even some from USAMO/IMO).

If getting a "bad" score on the AIME is due to lack of knowledge then practicing and learning new concepts is key. However, if you are making mistakes although you understand the concept, then you should go back, and find out why you got the ones that were wrong (careless errors/simple mistakes).

The latter option is much more preferable (less work to be done) and all that remains is simply practicing and getting used to the level of difficulty.

Since you only have a month left to go (for AMC 12) I'd suggest you brush up on AMC 12 for a week or so and hopefully, when you make pass, you can spend the extra 1 or 2 months preparing and reviewing for the AIME.

If you are really into math but don't have the time to get into contest math, just go ahead and start learning math that one could take their first year in college (like calculus, analysis, group theory, linear algebra, etc).

thanks for the help, everyone

when i cant solve problems, its typically because i havent encountered / understood the concept they use to solve it... so i guess aops 2 is the way to go

By the way, it's never too late to start math.  If you love math, then you will get satisfaction from excelling at it your whole life.  Just because you won't have much time to prepare for high-school contests doesn't mean that that's it.  (But if you really love math competitions that much, there's always the Putnam!)

^^ Very correct.  I didn't get into hard-core math until my junior year of high school but now, I am majoring in it in college.  However, I don't think Putnam is a competition that one can really enjoy (6 hours?? too much!!) unless one is knows he/she is going to score well.

But getting to USAMO merits self-esteem.  Anyone can participate in Putnam.

the reason im getting into math right now is because i love it so much! i didn't know about math contests and such until last year... so i'm diving in. i'll be studying applied mathematics in college next year... but math comps touch on a lot of pure math so i wanted to study some :)

The books found in the AoPS Olympiad section of the bookstore can help too. Plus the AoPS Volume 2 and the books in the Intermediate Series would be useful (when the books get released)

[quote="alanchou"]Um, consider the USAMO.  Lots of people enjoy it, even though some of them don't score very high[/quote]
Plus, the Putnam is six hours without break, but you get a break in the middle of the 9-hour USAMO exam.


Find the best $k,l,m,n$ such that for $a,b,c\geq 0$ holds
  $k(a^4+b^4+c^4)+labc(a+b+c)\geq m(a^2b^2+b^2c^2+c^2a^2)+n(a^3b+b^3c+c^3a+ab^3+bc^3+ca^3)$

[quote="silouan"]Prove that  $2(a^4+b^4+c^4)+6abc(a+b+c)\geq 6(a^2b^2+b^2c^2+c^2a^2)+a^3b+b^3c+c^3a+ab^3+bc^3+ca^3$[/quote]

Not true, check a = b = 2, c = 1.

  darij

Ok . I edited the problem now ..... :blush:

What do you mean by "best"? :?
$1,1,1,1$ works

the tightest bound on the inequality... I suspect this is quite dificult to prove...


My first and only thougt was that 7x+2x=180 (since the tetrahedron is inside a circle, and the sum of two opposite angels is 180). Which makes angle A 20 degrees. But that was wrong.
How should I think and what should I do to get the correct answer??

http://yfrog.com/6257684263j

No, what you did was find that $ x\equal{}20^{\circ}$.  What does $ \angle A$ equal, in terms of $ x$?


[quote="myc"]How do you do this problem?  there are three unknowns...

[url]http://www.nspsmo.org/documents/2001trig-02.pdf[/url]

thanks![/quote]
[hide]Since $ ABCD$ is cyclic quadrilaterl, $ \angle C \equal{} 180^{\circ} \minus{} \angle A$. Now you can find $ BD$ using cosine rule.

Let $ \alpha \equal{} \angle BDC$. Using sine rule, you can find $ \alpha$ as  $ \frac{BD}{\sin \angle C} \equal{} \frac{BC}{\sin \alpha}$. So you have $ \angle ADB \equal{} 90^{\circ} \minus{} \alpha$.

Now using sine rule you can find $ AB$ as $ \frac{AB}{\sin \angle ADB} \equal{} \frac{BD}{\sin \angle A}$. Since $ \angle ABD \equal{} 180^{\circ} \minus{} (\angle A \plus{} \angle ADB)$, you can find $ AD$ using sine rule. Now you know, $ CD, AD$, so find $ AC$ using Pythagoras.[/hide]

thanks for answering my question, but for some reason the latex is not showing up on my computer all I see is a bunch of $ and numbers and letters.  What do I need to do to fix this problem?

1. Go to your profile
2. Go to preferences
3. Tick "Always allow BBCode"
4. ???
5. Profit

but how do you know what angle alpha is?  thanks for the latex reply btw  I don't see dollar signs all over the place anymore

Technically you can find closed formulas for each expression... Not that I'd want to do that, because it usually ends in a huge mess :)

[quote="myc"]nevermind my previous post i wasn't reading, new question: how do you calculate all the lengths with such precision without having to store alpha and length BD and all the other necessary lenths into the calculator?

thanks[/quote]

Can you use Excel on the computer. You could have all the calculations plus explanations stored in one spreadsheet. No need to write down and re-enter results of intermediate calculations. Spreadsheets are sp convenient in general.
Should you get cloned problems (same problem with some numbers changed), you may save some time.

[quote="myc"]nevermind my previous post i wasn't reading, new question: how do you calculate all the lengths with such precision without having to store alpha and length BD and all the other necessary lenths into the calculator?  [/quote]

Why would you want to bypass using a calculator?  The numbers involved clearly call for it.

using the calculator without storing all the numbers, it sucks up time.  I'm trying to find the quickest way to solve this problem while getting the answer as precise as possible.  the contest is timed.

It seems like typing in the numbers repeatedly would take more time than pressing "store."


Using digits $ 1, 2, 3, 4, 5, 6$, without repetition, $ 3$ two-digit numbers are formed. The numbers are then added together. Through this procedure, how many different sums may be obtained?

[hide]Choosing the 3 digits to be in the units digits of the 3 numbers will determine the sum. The sums for these range from $ 1 \plus{} 2 \plus{} 3 \equal{} 6$ to $ 4 \plus{} 5 \plus{} 6 \equal{} 15$, giving 10 unique values total.[/hide]
Edit: What's PRMO?

[url=http://ifem.math.uprm.edu/olimpiadas/index.php]Puerto Rican Mathematical Olympiad[/url] (if you're not proficient in Spanish, I discuss it briefly in [url=http://ifem.math.uprm.edu/olimpiadas/index.php]this[/url] blog entry).


[quote="boggart0803"]
.
.
.
16 BRA 152
[b]17 ROU 141
18 PER 141[/b]
19 AUS 140
20 SRB 139
[/quote]

Sorry, but i think it should be:
[b]17 PER 141
18 ROU 141
[/b]

Peru has a gold ... ;)

PS: Good job China.. again.!!

Carlosbr

[quote="BlackMax"]What is ROU? Romania?[/quote]

YES ...

I'm a Taiwanese

Last year, due to some political problem
Our code became TPE

but TWN means Taiwan
      TPE  means Taipei,  the capital of Taiwan

So I think TWN would be more appropriate than TPE

[quote="BlackMax"]Why have Romania, Vietnam and Taiwan changed their codes? Last year they were ROM, VIE and TPE respectively.[/quote]
AFAIK, there is an international standard for country codes.
http://www.worldatlas.com/aatlas/ctycodes.htm
I supposed IMO website follows it starting this year. That's why the changes.


LaTeX
calculus
articles
function
analytic geometry
parameterization
ratio

When I try to print the LaTeX document (by either PDF or just printing it) I get an error message that says "printer resolution 300 does not match selected metafont selection of 600. either select a 300 metafont mode or setup your printer for 300 dpi."  I don't know what this means, but I went to those menus and did what it asked and it still does not work.

I don't know what's wrong with it, but I think the error message is saying that the resolution of your pdf file does not match the resolution of your printer. (In case you don't know, resolution is how many dpi, or dots per square inch, there are).

yeah it's saying that your resolution for pdf is too high, and your printer can't print with such detail.  So maybe (if this is possible) shrink the document and then recover it, decreasing its resolution.  But I suppose pdfs don't shrink, in which case, I don't know.

Okay, so what do I do?  I am not great w/computers past the basics.  Can you tell me what buttons to press and files to go into, etc.?  This is really frusterating since I just spend a few hours making typing up my calculus notes.

When you click print, you should get a print window.  Right next to your printer name, there should be (hopefully) a Properties button.  Click that.

For me, a new window pops up, part of that window has a Print Quality section.  If I change what's in that drop down box, different resolutions occur.  I don't know if this is your problem, but perhaps you can use that to set your print quality to 600 dpi.

Please let us know if this (or anything else) works, so we can guide people in the future.

This is not working.  I can't get to the resolution like rrusczyk said, but I went under "file, print setup" and set the resolution to 600, and under "view, options" I set the printer and the display to be "HP LaserJet4 (600 dpi)" - this is not the printer that I have connected to my computer but that one is not on the list.

What list of printers is this?

Also, can you print other types documents off your printer?

It's a list of printers under view/options/printer in YAP.  Yes, I can make word documents into PDFs easily.  This is REALLY frusterating.  Can someone please help?

LaTeX seemed cool but I think I will just go back to using MathType in word.  *sigh*

In YAP?  Why are you in YAP if you're making PDF's with LaTeX?

Can you provide your LaTeX code?  (Just the preamble will be fine)

I don't understand this.  Okay, here is my document (What is a preamble?  I will just post the whole thing):

\documentclass[11pt]{article}
\begin{document}
\noindent Julie S\\
MTH 141 URI\\
June 8, 2004\\
Professor Kulenovic\\

\noindent \textbf{Parametric Equations}

\noindent Parametric equations represent an implicit function by representing it as two seperate equations.  Each seperate equation represents a coordinate - you have one equation for the x-coordinate, $x=f(t)$, and one for the y-coordinate, $y=f(t)$.  The letter t is called the parameter and usually represents time.\\ \\

\noindent For an example, think of the unit circle.  The implicit equation for the unit circle is $x^2+y^2=1$.  However, if you travel up counter clockwise from the positive x-axis, the point on the unit circle you are at is not only $(x,y)$ but also $(cos(t), sin(t))$.  Therefore, each seperate coordinate can be represented by $x=cos(t)$ and $y=sin(t)$. \\ \\

\noindent For one implicit function there are many different parametric equations that can represent it.  There is an optimal parametrization, but finding it is beyond the scope of MTH 141. \\ \\

\noindent Parametric Equation of a straight line: \\

\noindent $y-y_0=m(x-x_0)$ \\
let $m=\frac{a}{b}$ \\
$\frac{y-y_0}{x-x_0}=\frac{a}{b}=m$ \\
Use the means-extremes property to find that:
$\frac{y-y_0}{b}=\frac{x-x_0}{a}=t$ \\
Since both of the ratios are equal, we allow them to both be equal to t.
$\frac{x-x_0}{a}=t$ thus $x=x_0+at$ \\
$\frac{y-y_0}{b}=t$ thus $y=y_0+bt$ \\ \\

Suppose that an object is moving along a line.  Here, t refers to time.
$\frac{dx}{dt}$ and $\frac{dy}{dt}$ represent their x and y components of its velocity (remember, the first derivative represents instaneous rate of change, i.e. instaneous velocity). \\ \\

\noindent $\frac{dx}{dt}=a$, because: \\
$x=x_0+at$ \\
$\frac{d}{dt}(x)=\frac{d}{dt}(x_0)+\frac{d}{dt}(at)$ \\
${\frac(dx}{dt}=0+a\cdot\frac{d}{dt}(t)$ \\
${\frac(dx}{dt}=0+a\cdot1=a$ \\
Using the same methods one can show that $\frac{dy}{dt}=b$. \\ \\

\noindent The vector sum of their x and y component velocities, i.e. $v=\sqrt{{v_x}^2+{v_y}^2}$, is the overall velocity.  Substituting you get: \\
$v=\displaystyle\sqrt{{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2}$ \\
Though it's hard to show, this formula works for not only a straight line, but any line. \\ \\

\noindent EXAMPLE: Find the velocity of any point ("particle") moving along the curve.  The curve is modeled by the parametric equations $x=t+\sin{t}-2$ and $y=t^2+2t-4$.  Find where, if any, v=0. \\ \\

\noindent $\frac{dx}{dt}=(t+\sin{t}-2)'=1+\cos{t}$ \\
$\frac{dy}{dt}=(t^2+2t-4)'=2t+2$ \\
$v=\sqrt{{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2}$ \\
$v=\sqrt{(1+\cos{t})^2+(2t+2)^2}$ \\ \\

\noindent v=0 iff $\frac{dx}{dt}=0$ \\
$1+\cos{t}=0$ \\
$\cos{t}=-1$ \\
$t=\pi+2k\pi=\pi(1+2k)$ for integral k \\
$2t+2=0$ \\
$t=-1$ \\
Since $\pi(1+2k)$ is never equal to -1, the velocity will never equal 0. \\ \\

\noindent EXAMPLE: Given that $x=t^2+2t+4$ and $y=2t^3-6t+2$: \\
(a) Find an equation for the tangent line at t=2. \\
(b) Find all points where the tangent line is horizontal. \\
(c) Find all points where the tangent line is vertical. \\ \\

\noindent (a) At t=2, $x=2^2+2\cdot2+4=12$ and \\ $y=2\cdot{t^3}-6\cdot2+2=6$.  \\
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ \\
${\frac(dy}{dx}=\frac{6t^2-6}{2t+2}=3\cdot\frac{t^2-1}{t+1}=3(t-1)$ \\
$\frac{dy}{dx}|_{t=2} = 3(2-1)=3$ \\
$y=3x+b$\\
$6=3(12)+b$ substitute in (12,6)\\
$b=-30$\\
$y=3x-30$\\ \\

\noindent(b) $m=0$ when $\frac{dy}{dx}=0$. \\
$3(t-1)=0$ \\
$t=1$ \\
$x(1)=1^2+2(1)+4=7$ \\
$y(1)=2(1)^3-g(1)+2=-2$ \\
The slope of tangent line equals zero when t=1 at (7,-2). \\ \\

\noindent(c)Here, there are no points where slope of tangent line equals infinity.  To find the these points, one lets denominator equal zero and solves.
\end{document}

Also, after I do Ctrl F7 I go to "view output" and then I try making a PDF w/the output by going to file/print and then changing it to PDF.  Am I just making a stupid mistake?  Sorry about my ignorance on the whole thing.

The preamble is the stuff before \begin{document}.  In this case, \documentclass[11pt]{article}.

Ah-ha!

Read [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicFirst.php]this page[/url], the section under Making PDF Documents.  Your problem should go away.  Basically, you make LaTeX build the PDF, not your virtual printer (using the printer to make PDFs that way apparently doesn't work from DVIs - we include it in our instructions to make PDFs that you can then include in LaTeX documents).

FYI, if I do exactly what you describe, I get exactly the error you describe.


Let a,b,c > 0 and m be a positive integer.
Prove that:

    $ \frac{a^m}{b+c} + \frac{b^m}{c+a} +\frac{c^m}{a+b} \geq \frac{2}{3}(\frac{a+b+c}{3})^{m-1}. $

Nesbit+Cebasev+Jensen=solution of the problem.

Thanks YOU!!

when can i find info about this inequalities?

may any web site where a could download *.sp or *.pdf about these "elementary inequalities"? :blush:

Thanks for all!!!

In french at :
http://www.animath.fr

Pierre.


inequalities
algebra
polynomial
calculus
derivative
complex analysis
complex analysis theorems

Assume that $f(z)$ is a real polynomial and thus that its non-real zeros occur in conjugate imaginary pairs. Let us construct the circles  whose diameters are the line-segments joining the pairs of conjugate imaginary zero of $f(z)$. These circles we shall call Jensen circles of $f(z)$.
  We shall state a theorem which was announced without proof by Jensen in 1913. It was proved by Wlsh in 1920 and later by Echols and Nagy.

  Jensen Theorem. Every non-real zero of the derivative of a real polynomial $f(z)$ lies in or on at least one of the Jensen circles of $f(z)$.

Saman Gharib

I'd really like to see that proof =]
does anyone know?
at least a place where I can find it?


algebra polynomial inequalities function calculus derivative

[quote="math wizard"]Solve this inequality:
(2x^2+7x+8)/(x^2+1)[/quote]

maybe there is something else like a > or < sign

[quote="math wizard"]:blush:  oops
well the text of the problem should read
(2x^2+7x+8)/(x^2+1)>0[/quote]

[hide]

well we know that $x^2\ge0$ so $x^2+1\ge1>0$ so multiplying by $x^2+1$ will not need to change the direction of the inequality sign, that gives us

$2x^2+7x+8>0$

and by the discriminant, $49-4(2)(8)=-15<0$ so the solutions are imaginary, so the graph never crosses the x-axis (otherwise there would be real roots), testing $x=0$ we get $8$, so the graph is over the x-axis at x=0, and since it never crosses the x-axis, the whole function is greater than 0, so the answer is

all real x

[/hide]

[quote="math wizard"]Nice Work!
Try this one,
x^22+x+14>10-x
(Note: Using a graphing calculator may be helpful)[/quote]

well, to answer this rigirously, you would need to use calculus, but looking at the graph, we see that this is true for all real x

[hide="with calculus"]

consider
$f(x)=x^{22}+2x+4$
we want to find the minimum, and show that it is above the x-axis
so finding inflection points, and using the derivative, we have
$f'(x)=22x^{21}+2=0$
$x=-\sqrt[21]{\frac{1}{11}}$

and $f''(x)=22*21x^{20}$ is positive for $x=-\sqrt[21]{\frac{1}{11}}$ because 20 is an even power so it makes it positive, making that inflection point a relative minimum, and since it is the only minimum, it is the absolute minimum

so the minimum value occurs at $x=-\sqrt[21]{\frac{1}{11}}$, and $f(-\sqrt[21]{\frac{1}{11}})>0$ (use a calc, or see that 4 is greater than the other 2 terms, so it is positive)
that means that
$f(x)=x^{22}+2x+4 > 0$ for all $x$ so rearranging, we have that

$x^{22}+x+14>10-x$ for all real $x$

[/hide]

I noticed this:
its supposed to be
x^2+x+14>10-x
Sorry  :oops:
However it doesn't matter in the end.
SO since we all love polynomial inequalities, knock yourself out with these:
(once again, a graphing calc is helpful, try not to use calculus)
Number 1!
2-x(not sure how to make less than or equal to sign) sqrt{x+1}
Number 2!
sqrt{x^2-9}(less than or equal to)sqrt{25-x^2}

I know that you are a new member, but in the future do not post more than one problem in one topic


In http://www.tjhsst.edu/~tmildorf/math/Adv.htm under Lectures$\Rightarrow$Esoteric Geometry, pg. 3, it says

[quote]Consider putting weights $w_{1}$, $w_{2}$, and $w_{3}$ on $A$, $B$, and $C$ respectively.[/quote]

And then it goes on to list some facts.

What does it mean to put weights on points?  :ninja:

I didn't know where to put "Mass Points" under, so I just stuck it here...

it is like you have 3 concurrent lines, and you are balancing the verticies on the lines...it is much easier to just assume that the process works...putting weights on a point is just saying like w_A=1, then you find the weights of the other points such that the triangle is balanced about the center...


[quote="perfect_radio"]Under the action of a force, a spring suffers a certain elongation (hopefully this is the right word for "alungire"). Under the action of another force, it suffers the elongation doubles. As a consequence, the potential energy grew  $n$ times. Find $n$.

By looking at other problems in this topic, I think this will be too easy for you. But I don't know to find the answer to this problem. (only the answer is given)[/quote]perfect_radio,

According to Hooke's law, the force applied to a spring, $\vec{F} = -k\vec{x}$

If under $F_1 = -k x_1$ and under $F_2 = -k(2 x_1)$, then since $V(x) = \int kx\: \mathrm{d}x =\frac12 kx^2 + \text{ const.}$,

( One is free to choose $\text{ const. } = 0$ )

$\frac{V(2x_1)}{V(x_1)} = \frac{(2x_1)^2}{(x_1)^2} = 4$

Your $\textstyle n$ then $\textstyle =4 \: \: \blacklozenge$

[quote="perfect_radio"][quote="Dr. No"]If under $F_1 = -k x_1$ and under $F_2 = -k(2 x_1)$, [b]then since $\mathbf{V(x) = \int kx\: \mathrm{d}x =\frac12 kx^2 + \text{ const.}}$,[/b]

( One is free to choose $\text{ const. } = 0$ )[/quote]

how did you find the formula in bold?[/quote]

In classical mechanics the potential energy function is typically called $V$.

For a conservative force field such as is given by Hooke's law:

$\vec{F} = -\nabla V$

Therefore, $V(x) = - \int \vec{F} \cdot \mathrm{d}\vec{x}$

Anyway, I'd consider

$E_p=\frac{1}{2}kx^2$

a perefectly well known formula, we learned it in high school - so if you know that the spring is double it's original length, the energy must have increased $2^2=4$ times.

i have found on the internet the following:

draw the graph $x \mapsto F=-kx$. then the area between graph and $Ox$ is our energy (you don't really need integrals for this)

my question is: why does that area represents the energy (one should note the fact that $\left( \frac12 x \right) \cdot (kx)$ is in $N \cdot m$; but why is it numericaly exactly this)

[quote="perfect_radio"]i have found on the internet the following:

draw the graph $x \mapsto F=-kx$. then the area between graph and $Ox$ is our energy (you don't really need integrals for this)

my question is: why does that area represents the energy (one should the fact that $\left( \frac12 x \right) \cdot (kx)$ is in $N \cdot m$; but why is it numericaly exactly this)[/quote]Okay, let me try a simple minded approach.

First, the 1st Law of Thermodynamics says that energy is conserved.

$P.E. + K.E. = \text{ const. }$

Therefore, $\frac{\partial}{\partial t}(P.E. + K.E.) = 0$

$\frac{\partial K.E.}{\partial t} = mv \frac {\partial v}{\partial t}$

$ \frac {\partial v}{\partial t} = a \Rightarrow \frac{\partial K.E.}{\partial t} = mav$

$ v = \frac{\partial x}{\partial t} \; \&\; ma = F$

$\frac{\partial K.E.}{\partial t} = F \frac {\partial x}{\partial t}$

$\Rightarrow \mathrm{d}K.E. = F\mathrm{d}x$

$\Rightarrow -\mathrm{d}P.E. = F\mathrm{d}x$

$\Rightarrow P.E. = -\int F\mathrm{d}x$

That's why!


function
real analysis
real analysis unsolved

Let $ f$ be a lower semicontinuous function defined on $ [a,b]$. Show that there is a monotone increasing sequence of step functions on $ [a,b]$ that converge pointwise to $ f$.

$ f$ is integrable (or can be approximated by integrable functions at least), and can be approximated by simple integrable functions having constant values on disjoint open sets. Such functions can be approximated by continuous functions, and continuous functions can be approximated by step functions in the usual way  :wink:

Is it possible not to involve the idea of integrability in this problem? Perhaps by constructing suitable step functions based on the definition of semicontinuity?

Define a step function $ h_n(x)$ on each interval $ I_{k}\equal{}[(k\minus{}1)/2^n,k/2^n)$ by setting it equal to $ \inf_{x\in I_k}f(x)$ within $ I_k$. Should work.

So that would mean I have to know that $ f$ is bounded below?

Actually this is only part of a problem in a book and showing that a lower semicontinuous function on $ [a,b]$ is bounded below comes in a later part so I am wondering whether we can do without that fact for the time being.

Could the property that $ \{x: f(x)>\lambda\}$ is open for any real $ \lambda$ be useful here?

Are step functions allowed to assume the value $ \minus{} \infty$? If yes, then my definition of $ h_n$ is OK. If not, then the statement you are asked to prove implies that $ f$ is bounded from below.

[quote] If not, then the statement you are asked to prove implies that $ f$ is bounded from below. [/quote]

Yes I agree with that, but then I wouldn't want to use the latter to prove the statement itself right?

Whichever way you proceed, you [b]will[/b] prove the boundedness from below at some point, since it is (implicitly) a part of the statement that you are to prove.

Why not begin with it? It's a one-liner: open sets $ U_n\equal{}\{x\in [a,b]\colon f(x)>\minus{}n\}$ cover the compact set $ [a,b]$.

Following mlok's idea, I defined each step function $ \varphi_{n}$ on each interval $ I_{k}^{(n)}\equal{}(a\plus{}(k\minus{}1)(b\minus{}a)/2^{n},a\plus{}k(b\minus{}a)/2^{n})$ by setting it to be $ \inf_{x \in I_{k}^{(n)}} f(x)$. I further defined the value at each partition point to be the minimum of the value assumed on the left interval, the value assumed on the right interval and the value of $ f$ at that point, so that the step function is lower semicontinuous. In this way I get a monotone increasing sequence of lower semicontinuous step functions pointwise convergent to $ f$. How can I modify these step functions to get a monotone increasing sequence of continuous functions pointwise convergent to $ f$?

I am thinking of linearising in a small neighbourhood of each partition point but I am not sure whether it works because the way I defined the step functions means that it may take the same value as $ f$ at a partition point while its values in a neighbourhood of the point could be much larger.


What is the smallest three-digit number in Pascal's triangle?

Trick question: $ \binom{100}1\equal{}\boxed{100}$.

[quote="howie2000"]The first number of the $100$th row is $100$[/quote]
The 2nd number
[quote="seeworld75"]??????????[/quote]
The 2nd number in each row of Pascal's triangle except the 0th, is equal to its row number.

The second diagonal is 1,2,3,4,5,6,7,8,9...

So the question is just what is the smallest 3 digit number, which is 100.

[hide=Obviamente!]
Every number appears in the Pascal Triangle. So we're just looking for the smallest 3-digit [I]number[/I]. And that is $\boxed{100}$.
[/hide]


I think he want to calculate:
S(n)=$\sum_{i=0}^n(-1)^{2n-i}*C^i_{2n}$


We see that the 1st pentagon has perimeter 5.

Each following pentagon adds 4 to perimeter.

Therefore the perimeter is 9(4)+5=41

[quote="GameBot"]A pentagonal train is created by adjoining congruent
pentagons with sides of 1 unit each. A pentagonal train of
10 pentagons is constructed. What is the number of units in
the perimeter of the resulting polygon?

[asy]size(200);
real theta = -15, gamma = -50,a = 1;
real length = a*(dir(theta)+dir(-theta)).x- a*(dir(gamma)- (dir(theta)+dir(-theta)+dir(180-gamma))).x ;
path pentagon()
{
return xscale(a)*((0,0)--dir(theta)--dir(theta)+dir(-theta) --dir(theta)+dir(-theta)+dir(180-gamma)--dir(gamma)--cycle);
}
void makepentagon(int n)
{

if(n%2 == 0)
draw(shift((n/2)*length*right)*pentagon());
if(n%2 == 1)
draw(shift(dir(gamma))*  shift(((n+1)/2)*length*right)*rotate(180)*pentagon());
}
makepentagon(0);
makepentagon(2);
makepentagon(5);
makepentagon(1);
makepentagon(3);
label("$\dots$",(2.38*length,0),1.6*S,fontsize(12pt));[/asy][/quote]

Look at the diagram.  The first and last add 4, while the rest add 3.  So we have 8*3+8=32.


Let $f(x)= x^2-C$ where $C$ is a rational constant.
Show that exists only finitely many rationals $x$ such that $\{x,f(x),f(f(x)),\ldots\}$ is finite

My problem at the 2010 Romanian Master in Mathematics (see Contests tab) stated

[quote]Given a polynomial $f(x)$ with rational coefficients, with degree $d \ge 2,$ we define a sequence of sets $f^0(\mathbb{Q}), f^1(\mathbb{Q}), \cdots$ as $f^0(\mathbb{Q})=0, f^{n+1}(\mathbb{Q})=f(f^{n-1}(\mathbb{Q}))$ for $n\ge 0$. (Given a set $S,$ we write $f(S)$ for the set $\{f(x), x\in S\})$

Let $f^{\omega}(\mathbb{Q})=\bigcap_{n=0}^{\infty} f^n(\mathbb{Q})$ be the set of numbers that are in all of the sets $f^n(\mathbb{Q}).$ Prove that $f^{\omega}(\mathbb{Q})$ is a finite set.[/quote]

Clearly now, the rationals $x$ which have finite orbit belong to $f^{\omega}(\mathbb{Q})$, thus there are only finitely many, and this is true for any polynomial with rational coefficients and degree at least $2$.

We write $c=a/b$ in reduced notation and assume $x=k/l\in Q$, also in reduced notation. Also suppose that the forward orbit is a finite set.
Let $p$ be a prime and write $\nu_p(t)$ for the amount of factors $p$ in $t\in Q$, possibly negative.
If $\nu_p(l)>\nu_p(b)$ then $\nu_p(f(x))<\nu_p(x)$. Intuitively, if we apply $f$ then the amount of factors $p$ in the denominator increases strictly.
Hence in that case the forward orbit $x,f(x),ff(x),\cdots$ is infinite, contradiction.
It follows that $l$ can only contain prime factors that are in $b$, and with at most that multiplicity. Hence $l|b$ and $x\in {1\over b}\cdot Z$.
If $|x|$ is large then the forward orbit tends to $\infty$. Hence $x$ lies in a finite set.


geometry
3D geometry
prism
rectangle
number theory
prime factorization

Can you guys post your solutions?

10. Three faces of a right rectangular prism have areas of 48,49, and 50 square units. What is the volume of the prism, in cubic units? Express your answer to the nearest whole number.

13. A group of boys and girls reserved a single row of 120 seats at a theater. The girls were assigned seat numbers in such a way that every boy had to sit next to at least one gril. What is the fewest number of girls necessary to make this happen?

[hide="Explanations"]On #1: Let the sides of the rectangular prism be $ x,y,z$.  Then we can see that the volume is $ xyz$.  We are given that either $ xy,xz,$ or $ yz\equal{}48$, and so on.  Notice, however, that it does not matter which one is which as when we multiply all three together, we come up with a final result of $ x^2y^2z^2\equal{}48*49*50$. Simply take the square root to find your answer.
#2: To minimize the number of girls, we need for each girl to touch two boys.  The answer is then simply $ 40$.
To prove that this is possible, one possible sequence is: $ BGBBGBBGBBGBBG\hdots$[/hide]

Well, 40 is the right answer for #12, and I got it...

But I didn't quite get #10...

I know that you have to divide those areas and multiply somehow...

Help.

[quote="perfect628"][hide="hint"]Well, consider how you'd write the areas of the faces, and the volume. Can you do anything to the area to get the volume?[/hide][/quote]

The areas can be expressed as a prime factorization or factors. But I don't quite get the idea..

It would be helpful if you post a good solution..

I gave a hint because I'm in high school, but here.

[hide="solution"]Let the dimensions of prism be x, y, and z. Then, we can express the areas of the faces as $ xy$, $ yz$, and $ zx$, as they are simply rectangles with these dimensions. Now, we know that the volume can be written as $ xyz$. Note that $ xyz\equal{}\sqrt{(xy)(yz)(zx)}$, or the square root of the area of the faces. Substitute our areas in and we get the answer.[/hide]

Hope that helps!

[quote="frost13"][quote="ZhangPeijin"][hide="My Solutions"]
10) Well, you would just do...
$ (\sqrt48)(\sqrt49)(\sqrt50)$
Which gives you...
$ (4\sqrt3)(7)(5\sqrt2)$
Which gives you...
$ (4)(7)(5)(\sqrt3)(\sqrt2)$
Which simplifies to...
$ 140\sqrt6$
Which you can round by yourself. I don't have a calculator.
13) The order would be...
BBGBBGBBGBBGBBGBBG...
So you do $ 120\div3$
Which gets you 40.
Which is your answer.
I think.  :P
[/hide][/quote]

Peijin, good thinking, but the way you showed your example the very first boy is not sitting next to a girl [hide] You have BBGBBGBBGBBGBBG.....  So that can't be right.  Not saying your answer is wrong, can you change that list so your answe is the same? [/hide][/quote]

Oh...
So it would be
BGBBGBBGBBGB...
 :P
And I'm pretty sure my #10 is correct... Just need to umm, round it to the nearest whole number.


What is the sum of the eight terms in the arithmetic sequence $ \minus{}2, 3, ..., 33$?

the first term is -2+5*0
second term is -2+5*1
third term is -2+5*2
...
eighth term is -2+5*7

so theres total 8 -2's and 28 5's.

so the answer is 140-16=124.

answer : 124

An easier way is to use the formula for an arithmetic sequence

$ (a_1\plus{}a_n)\cdot\frac{n}{2}$

in this case:

$ a_1\equal{}\minus{}2$
$ n\equal{}8$
and $ a_8\equal{}33$

so its $ (33\minus{}2)*4\equal{}124$


calculus
integration
Diophantine Equations

For all $n \in \mathbb{N}$, show that the number of integral solutions $(x, y)$ of \[x^{2}+xy+y^{2}=n\] is finite and a multiple of $6$.

$n=x^{2}+xy+y^{2}\leq |x^{2}|+|xy|+|y^{2}| \leq (|x|+|y|)^{2}$, thus $|x|,|y| \leq \sqrt n$, giving finiteness (and the bound $4n+4\sqrt n$ on the number of solutions).

To each pair $(x,y)$ we assign the number $x+y\zeta_{6}\in \mathbb C$, where $\zeta_{6}=\frac{1+i\sqrt{3}}2$ is a primitive $6$th root of unity. This assignment is unique (bijective in fact if $x,y \in \mathbb R$) and $|x+y\zeta_{6}|^{2}=x^{2}+xy+y^{2}$.

Thus if $|x+y\zeta_{6}y|^{2}= x^{2}+xy+y^{2}=n$ for some integers, we consider the six numbers $a_{k}+b_{k}\zeta_{k}= \zeta_{6}^{k}(x+\zeta_{6}y)$ for $k=0,1,...,5$. By $|\zeta_{6}|=1$ and $\zeta^{k}=1 \iff 6|k$, we get that all of them are different and have the same square of modulus $n$. Simple calculation shows that all $a_{i}, b_{i}$ are integers (we have just to prove $x,y \in \mathbb{Z}\implies a_{1}, b_{1}\in \mathbb{Z}$, but $a_{1}=-y, b_{1}=x+y$).
So we can group the solutions together in couples of six, giving the result.

But there is a much stronger result based on factorisation in Eisensteinian integers:
The number integer $n>0$ can be written as $x^{2}+xy+y^{2}$ ($x,y \in \mathbb Z$) in exactly
\[6\cdot\left(\sum_{\substack{d|n \\ d \equiv 1 \mod 3}}1-\sum_{\substack{d|n \\ d \equiv 2 \mod 3}}1 \right)\]
ways. If there is one or more prime divisor $p \equiv 2 \mod 3$ dividing $n$ an odd number of times, then this is $0$, otherwise it equals
\[6 \prod_{\substack{p \in \mathbb{P}\\ p \equiv 1 \mod 3}}\left(v_{p}(n)+1\right).\]

The discriminant is $4n-3x^2$ which shows there is a finite number of solutions. if (x, y) is a solution then so is (y-x, -x) and (-y, x-y). (-x, -y) is also a solution and this leads to 6 distinct solutions that can be found for every pair (x, y).


[b]first day 27/01/07[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130810]Problem1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130811]Problem2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130812]Problem3[/url]

[b]second day 28/01/07[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130813]Problem4[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130814]Problem5[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130815]Problem6[/url]

Thank you!  :D

Our team leader still hasn't published it on his site  :roll:

No problem. But my LaTex is not very nice, maybe you looked it  :oops:

I was just saying that the pdf (in italian) with all the exercises assigned at the Winter Camp, with official solutions, hasn't been published in internet yet and it's kind from you to write pdf's with all the problems of contests of the world!

But of course, I've been writing all this useless stuff just to practise a bit my English; it can be useful in the future  :D

[quote="edriv"]
Our team leader still hasn't published it on his site  :roll:[/quote]

The reason is trivial .... :P

By the Way

is official that Italy will partecipate at 2007 Balkan MO

so this test is the 2007 BMO Italian TST

Can you post all problems from Italian WinterCamp test ?

There were 6 problems in the final test, and mattigale posted all of them.

During that week we worked on many problems (about six per day), but they all came from old contests, nothing new.


[b]i.)[/b] The polynomial $x^{2 \cdot k} + 1 + (x+1)^{2 \cdot k}$ is not divisible by $x^2 + x + 1.$ Find the value of $k.$

[b]ii.)[/b] If $p,q$ and $r$ are distinct roots of $x^3 - x^2 + x - 2 = 0$ the find the value of $p^3 + q^3 + r^3.$

[b]iii.)[/b] If $r$ is the remainder when each of the numbers 1059, 1417 and 2312 is divided by $d,$ where $d$ is an integer greater than one, then find the value of $d-r.$

[b]iv.)[/b] What is the smallest positive odd integer $n$ such that the product of

\[ 2^{\frac{1}{7}}, 2^{\frac{3}{7}}, \ldots, 2^{\frac{2 \cdot n + 1}{7}} \]

is greater than 1000?

[quote="orl"][b]iv.)[/b] What is the smallest positive odd integer $ n$ such that the product of
\[ 2^{\frac {1}{7}}, 2^{\frac {3}{7}}, \ldots, 2^{\frac {2 \cdot n \plus{} 1}{7}}
\]
is greater than 1000?[/quote]
$ 2^{\frac {1}{7}} \cdot 2^{\frac {3}{7}} \cdot  \ldots \cdot  2^{\frac {2 \cdot n \plus{} 1}{7}} > 1000$
$ \iff$
$ 2^{1\plus{}3\plus{}5\plus{}...\plus{}(2n\plus{}1)} \equal{} 2^{(n\plus{}1)^2} > 10^{21}$
The smallest $ m$ such that $ 2^m > 10^{21}$ is $ m \equal{} 70$, which can be easily calculated with some paper.....
So $ (n\plus{}1)^2 \ge 70$ but the $ n \ge 8$ and thus $ n \equal{} 9$.

[quote="orl"][b]i.)[/b] The polynomial $ x^{2 \cdot k} \plus{} 1 \plus{} (x \plus{} 1)^{2 \cdot k}$ is not divisible by $ x^2 \plus{} x \plus{} 1.$ Find the value of $ k.$

[b]ii.)[/b] If $ p,q$ and $ r$ are distinct roots of $ x^3 \minus{} x^2 \plus{} x \minus{} 2 \equal{} 0$ the find the value of $ p^3 \plus{} q^3 \plus{} r^3.$

[b]iii.)[/b] If $ r$ is the remainder when each of the numbers 1059, 1417 and 2312 is divided by $ d,$ where $ d$ is an integer greater than one, then find the value of $ d \minus{} r.$

[b]iv.)[/b] What is the smallest positive odd integer $ n$ such that the product of
\[ 2^{\frac {1}{7}}, 2^{\frac {3}{7}}, \ldots, 2^{\frac {2 \cdot n \plus{} 1}{7}}
\]
is greater than 1000?[/quote]
i) when $ 3 \mid k$
ii) the answer is $ 4$
iii) $ d\minus{}r \equal{} 179\minus{}164\equal{}15$.

I will post solutions later.. (maybe)

[quote="Mathias_DK"][quote="orl"][b]i.)[/b] The polynomial $ x^{2 \cdot k} \plus{} 1 \plus{} (x \plus{} 1)^{2 \cdot k}$ is not divisible by $ x^2 \plus{} x \plus{} 1.$ Find the value of $ k.$

[b]ii.)[/b] If $ p,q$ and $ r$ are distinct roots of $ x^3 \minus{} x^2 \plus{} x \minus{} 2 \equal{} 0$ the find the value of $ p^3 \plus{} q^3 \plus{} r^3.$

[b]iii.)[/b] If $ r$ is the remainder when each of the numbers 1059, 1417 and 2312 is divided by $ d,$ where $ d$ is an integer greater than one, then find the value of $ d \minus{} r.$

[b]iv.)[/b] What is the smallest positive odd integer $ n$ such that the product of
\[ 2^{\frac {1}{7}}, 2^{\frac {3}{7}}, \ldots, 2^{\frac {2 \cdot n \plus{} 1}{7}}
\]
is greater than 1000?[/quote]
i) when $ 3 \mid k$
ii) the answer is $ 4$
iii) $ d \minus{} r \equal{} 179 \minus{} 164 \equal{} 15$.

I will post solutions later.. (maybe)[/quote]

i) See that the roots of $ x^2 \plus{} x \plus{} 1$ is $ w$ and $ w^2$, where $ w \equal{} e^{i\frac{2\pi}{3}}$. Thus $ w^3 \equal{} 1$. Let $ f(x) \equal{} x^{2 \cdot k} \plus{} 1 \plus{} (x \plus{} 1)^{2 \cdot k}$.
If $ 3 \mid k$ then $ f(w) \equal{} f(w^2) \equal{} 3$. So $ x^2 \plus{} x \plus{} 1 \nmid f(x)$
On the other hand $ x^2 \plus{} x \plus{} 1 \mid x^{2 \cdot k} \plus{} 1 \plus{} (x \plus{} 1)^{2 \cdot k}$ $ \iff$ $ x^2 \plus{} x \plus{} 1 \mid x^{2 \cdot k} \plus{} 1 \plus{} (\minus{}x^2)^{2 \cdot k} \equal{} x^{4k} \plus{} x^{2k} \plus{} 1$ And its easy to see that $ w^{4k} \plus{} w^{2k} \plus{} 1 \equal{} (w^2)^{4k} \plus{} (w^2)^{2k} \plus{} 1 \equal{} 0$. QED

ii) Just use Newton sums or expand $ (p\plus{}q\plus{}r)^3$

iii) We know that $ 1059 \equiv 1417 \equiv 2312 \bmod d$ $ \iff$ $ d \mid 1417\minus{}1059$ and $ d \mid 2312\minus{}1417$ and $ d \mid 2312\minus{}1059$.
But then $ d \mid (358, 895, 1253) \equal{} 179$ which is a prime. So $ d \equal{} 179$. But $ 1059 \equiv 1417 \equiv 2312 \equiv 164 \bmod d$
So the number is $ 179 \minus{} 164 \equal{} 15$


1/ Given triangle $ ABC$ with $ BC \equal{} a, CA \equal{} b, AB \equal{} c$. Prove that:
$ (\frac {4}{x} \minus{} 1)a^2 \plus{} (x \minus{} 1)b^2 \plus{} c^2\geq 4\sqrt {3}\cdot S(ABC)$

Has anyone got a solution to this problem? :huh:

Maybe he means Area of $ \triangle ABC$.

[quote="TFPNH"][quote="Kyoshiro"]$ S(ABC)$ means the area of triangle ABC.
And the inequality is true for all $ x \neq 0$ :roll:[/quote]
Do you mean, $ x > 0$ ?

For negative value of $ x$,
set $ a \equal{} 5$
$ b \equal{} 4$
$ c \equal{} 3$
$ x \equal{} \minus{} 16$
and the inequality is false.[/quote]
Oh yes, I forgot it :blush:


trigonometry
trig identities
Law of Cosines

Round Two December 1991

7. For how many integers $n$, with $3 \le n \le 1991$, can the following $n$ statements be consistent?
1) Statement $2$ and $3$ are false
2) Statement $3$ and $4$ are false
$\vdots$
$n-1$) Statements $n$ and $1$ are false
$n$) Statements $1$ and $2$ are false

8. Quadrilateral $ABCD$ has side lengths $AB = 20$, $BC = 15$, $CD = 7$, and $AD = 24$, and a diagonal with length $AC = 25$. If we write $\angle ACB = \alpha$ and $\angle ABD = \beta$ then calculate $\tan{\alpha + \beta}$

[hide="8"](15,20,25) and (7,24,25) are both Pythagorean triples, so $\angle ABC$ and $\angle ACD$ are both right angles. Then the opposite angles are supplementary and quadrilateral $ABCD$ can be inscribed in a circle.

$\tan \alpha=\frac{20}{15}=\frac{4}{3}$. $\tan \beta=\tan \angle ACD$ (since both intercept arc $AD$) which is $\frac{24}{7}$.

Then $\tan \alpha+\beta=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{4*7+24*3}{3*7-4*24}=\frac{100}{-75}=\boxed{-\frac{4}{3}}$.[/hide]

8. Quadrilateral $ABCD$ has side lengths $AB = 20$, $BC = 15$, $CD = 7$, and $AD = 24$, and a diagonal with length $AC = 25$. If we write $\angle ACB = \alpha$ and $\angle ABD = \beta$ then calculate $\tan{\alpha + \beta}$[/quote]

For #8, the diagnol must be AC, by Triangle Inequality. $\tan{\alpha}=4/3$. So now you have two right angles opposite, meaning cyclic. Use law of cosines to find lenth of DB, find sin&cosABD again by law of cosines, thusfind $\tan{\beta}$
Then solve.

Edit: Wow, scorpious produced a much nicer solution....


Posts regarding the Amadori Rearrangement can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181128]here[/url].

The thread on Stability and pinacol Pinacolone Rearrangemnet can be found [url=http://www.mathlinks.ro/viewtopic.php?p=996885#996885]here[/url]

thread related to [url=http://www.mathlinks.ro/viewtopic.php?t=181814]riemer - Tiemann[/url]
thread related to [url=http://www.mathlinks.ro/viewtopic.php?t=181815]acidity of salicylic acid[/url]


in 2) there's a sign error:

try with
$a=1$
$b=1$
$c=1$
$d=-3$

it results

$-8=\sqrt{64}$

that isn't true...
in my opinion the text would be

$(abc+bcd+cda+dab)^2=|(bc-ad)(ca-bd)(ab-cd)|$

1)

[hide]$(x^4-2x^2yz+y^2z^2)+(y^4-2y^2xz+x^2z^2)+(z^4-2^2xy+x^2y^2)=0$
$(x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2=0$

so
$x^2=yz,y^2=xz,z^2=xy$

$\frac{x^2}{y^2}=\frac{yz}{xz}$

$x^3=y^3$

$x=y$

and similar simmetrically.[/hide]

ehm...you are right(about hiding solutions)...scuse me.
bye bye

frengo,please hide your solutions whenever you post them :)

[quote="DanK"]Why can't we say $-8=\sqrt{64}$?  That's not untrue, just not the more common value for $\sqrt{64}$...  And if we are going to do it your way, why should it be absolute value on the right side if that's already positive 64?[/quote]

i disagree, the square root function only outputs nonegative numbers, so the positive root is not the "common" choice, but it is a convention to have that range thus when you say $\sqrt{16}$ it means $4$ since that is the way the square root function is defined

[hide="3"]
The right side expands to $7x^4+28x^3+42x^2+28x+7$. So subtracting off $1+x^4$ and dividing by 2, $3x^4+14x^3+21x^2+14x+3=0$.

Now divide by $x^2$ and use the substitution $u=x+x^{-1}$. Then $x^2+x^{-2}=u^2-2\Rightarrow 0=3(u^2-2)+14u+21=3u^2+14u+15=(3u+5)(u+3)$.

Solving for $x$ in terms of $u$, $x^2-ux+1=0\Rightarrow x=\frac{u\pm \sqrt{u^2-4}}{2}$. Substituting in the solutions -3 and -5/3 for $u$ gives the solutions $\boxed{\frac{-3\pm \sqrt{5}}{2},\frac{-5\pm i\sqrt{11}}{6}}$.
[/hide]

Did I miss something or is that modulus sign not necessary in 2nd equation?

[hide]\begin{eqnarray*} (abc+bcd+cda+dab)^{2}&=&(cd(a+b)+ab(c+d))^{2}\\ &=&(cd(a+b)-ab(a+b))^{2}\\ &=&(ab-cd)^{2}(a+b)^{2}\\ &=&(ab-cd)(a+b)(ab(a+b)-cd(a+b))\\ &=&-(ab-cd)(a+b)(ab(c+d)+cd(a+b))\\ &=&-(ab-cd)(a+b)(abc+bcd+cda+dab)\\ &=&-(ab-cd)(a+b)(ac(b+d)-bd(b+d))\\ &=&-(ab-cd)(ac-bd)(a+b)(b+d)\\ &=&-(ab-cd)(ac-bd)(ad+b(a+d)+b^{2})\\ &=&-(ab-cd)(ac-bd)(ad-b^{2}-bc+b^{2})\\ &=&(ab-cd)(ac-bd)(bc-ad) \end{eqnarray*}[/hide]

Arctic Monkey, I suggest learning how to use the eqnarray environment for LaTeX :lol: (See the AoPS LaTeX guide.)

[hide="hint for number 3"]expand the right side[/hide]


[hide]He left 1/2 in the house. His brother and sister ate a total of 1/4+1/3=7/12, with 5/12 remaining. the father ate 2/5 fifth of that, or 1/6. So that;s a total of 7/12+1/6=3/4 of the half eaten, was eaten, and 1/4 was left of the half. .

1/4*1/2=1/8 left when he got home [/hide]

[quote="hwenterprise"]I don't know how many times I tried, but I can't get the answer.

Number Four: Micheal bakes a chocolate cake. He left half of the cake on the table for his family and took the rest to a friend's house. Of the amount he left for the family, his brother ate a fourth of it, and his sister ate one third of it. Later that night, his dad ate two-fifths of what was then remaining. When Micheal got home the next morning, how much of the original cake was left?

Express your answer as a common fraction.[/quote]

[quote="anirudh"] There was 1/2 for michaels family. in that, 1/4 + 1/3=7/12. 5/12 remains. 3/5*5/12(because 3/5 is what remains after dad eatsit)=15/60=1/4. the answer is 1/4, i think.

is it 1/4 :?: [/quote]

anirudh, you forgot that it was out of half a cake. Sol it will be 1/4*1/2=1/8.

for some reason im getting

[hide]

$\frac{3}{24}$ because Michael leaves $\frac{1}{2}$ of the cake, and his brother ate $\frac{1}{4}$of the $\frac{1}{2}$which is $\frac{1}{8}$ and his sister ate $\frac{1}{3}$ of the $\frac{1}{2}$ so she ate $\frac{1}{6}$. Subtracting $\frac{1}{6}$and$\frac{1}{8}$ from $\frac{1}{2}$ I am left with $\frac{5}{24}$. The father ate $\frac{2}{5}$ of the remaining portion$(\frac{5}{24})$. Then I get $\frac{1}{12}$ or $\frac{2}{24}$.subtracting that from $\frac{5}{24}$ I am left with $\frac{3}{24}$
[/hide]

Please tell me If I am doing something wrong.....

[quote="nat mc"]anirudh, why are you quoting yourself?[/quote]

That's just what he does  :roll:

but i don't actually make a different post.


A man drives 150 miles to the seashore in 3 hours and 20 minutes.  He returns from the shore to the starting point in 4 hours and 10 minutes.  Let $r$ be the average rate for the entire trip.  Then the average rate for the trip going exceeds $r$, in miles per hour, by:
(A) 5
(B) 4.5
(C) 4
(D) 2
(E) 1

[hide] let's say rate of trip going is s

r=150*2/(3 1/3+4 1/6)=300/7 1/2=40 (miles/hr)

s=150/3 1/3=45

s-r=45-40=5[/hide]

seems a bit too simple. maybe i misunderstood the question.


algebra
polynomial
vector
linear algebra
linear algebra unsolved

Suppose T : V -->V Where V is a finite dimensional vector over R has a
Charateristic Polynomial:
(x - lamda_1)^a1 * (x - lamda_2)^a2 *......(x - lamda_n)^an

Prove
a) det (T) =  lamda_1)^a1 * lamda_2)^a2 *......(lamda_n)^an

b) tr (T) =  a1 * lamda_1 + a2 * lamda_2 +............+ an * lamda_n

Thank u in Advance

By most definitions, the characteristic polynomial is
\[ \det(A-\lambda I)=(-1)^n\prod_i(\lambda-\lambda_i) \ \ \ (*)  \]
Take $\lambda=0$ in $(*)$ to get a)

For b), compute the coefficient of $\lambda^{n-1}$ in $\det(A-\lambda I)$, since you only get terms of degree $n-1$ or higher from the main diagonal, you get that the coefficient is
\[ -\text{Tr}(A)  \]
On the other hand, if you compute the coefficient of $\lambda^{n-1}$ in $(-1)^n\prod_i(\lambda-\lambda_i)$, you get
\[ -\sum_i \lambda_i  \]


Who thinks left-handed people are cool?

This is in response to
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55998]What do u think about left-handed?[/url]

Please justify your feelings below.  Don't be ashamed if you don't like lefties (I'll only hunt you down and kill you...Just kidding  :lol: ).

yes me...i like left handed....because I'm left-handed....but i'm also right handed how can i say it in english in one word :? ?

You are ambidextrous. :)

Southpaws are cool.

Although the proportion of left-handed to right-handed people is about 1:8, I think the proportion of left-handed people to right-handed people on MathLinks will be smaller.

Maybe that's because on the whole people on MathLinks are quite smart and left-handed people have a tendancy to turn out to be geniuses (no offence to right-handed people, I'm simply stating the facts). For example Einstein, Mozart, Bobby Fisher, Faraday, Michealangelo and Leonardo da Vinci were all left-handed. Also, many savants are lefties.

One good thing about being left-handed is that you generally have an advantage in sports. That is because left-handed people are used to playing against right-handed people, but right-handed people are not used to playing against lefties.

Too bad for me that I play most sports with my right hand, I only write with my left.

If your a lefty and can throw over 88 mph, you'll be well set in life.

theres nothing wrong with left-handed people...

I heard left handed people are better at some areas of acedemics because somehow using your left hand stimulates some part of your brain and causes that part to mature, but, of course, how valid is gossip these days?

[quote="Melissa"]
One good thing about being left-handed is that you generally have an advantage in sports. That is because left-handed people are used to playing against right-handed people, but right-handed people are not used to playing against lefties.

Too bad for me that I play most sports with my right hand, I only write with my left.[/quote]
Chyeah, no kidding! Fencing against lefties is hard! (I am right-handed)

[quote="Melissa"]Too bad for me that I play most sports with my right hand, I only write with my left.[/quote]

Same here! I play all sports right-handed except tennis, which I play with both hands.

I'm lefthanded

Are we cool?  I don't know!

But I am extremely proud of being lefthanded!

It has come to my attention that the concentration of lefties in mathematics is a bit higher

I actually thought your poll was gonna ask if I was leftie or not?  Why not do such a poll?

I just made a poll asking this question. Go check it out, it's called Left or right?

i read an article in a magazine once that said that in areas where there was more violence lefties are more common because by natural selectionchooses the left handed trait because lefties are better fighters

I think all people are cool, even if they are left handed.

[quote="Melissa"]...Too bad for me that I play most sports with my right hand, I only write with my left.[/quote]

Whoa! I thought I was the only person that did this. It is complicating my efforts to play Squash - my right is more powerful but does not have the same control as my left. How do you throw a frisbee? My backhand is lefty but everything else is righty, so all of my throws come from my right and spin counterclockwise (from above.)


find an equivalent of:
u_0=a>1,  u_(n+1)=(u_n^2 +1)/2
(is "equivalent" the right word to say a function f so that u_n/f(n)->1 ??)

O, boy, o, boy, this is a hell of a problem! I found a kind of equivalent, but I'm sure it's not what you want. Anyway, I will write:
   Let x_n=(ln u_n)/2^n. It's trivial to see that x_n is decreasing, so it's convergent to L. From u_(n+1)>1/2*(u_n)^2 we find that 0<x_n-x_(n+1)<ln2/2^(n+1). Sum this from k=n to infinity and you will find that 0<x_n-L< ln2/2^n for any n. I made this so that you can aproximate L to see how ugly it is.
    We prove now that u_n/e^(L*2^n) tends to  2 and this will be the equivalent. We have u_n/e^(L*2^n)= e^[ 2^n*(x_n-L)]. Now we compute the limit of 2^n*(x_n-L) using Cesaro-Stolz in case 0/0. That is 2^n(x_n-L) is [x_n-L]/[1/2^n] and has the same limit as (x_(n+1)-x_n)/(1/2^(n+1)-1/2^n), which is 2^(n+1)(x_n-x_(n+1))=ln[ (u_n)^2/u_(n+1)] and this tends to ln2 and consequently u_n/e^(L*2^n) tends to 2. Or at least I hope so.


What is the greatest number of bags that can be used to hold 190 marbles if each bag must contain at least one marble, but no two bags may contain the same number of marbles and if you cannot put the bags inside each other?

good job. now generalize (i.e. try it w/ a number that is not triangular

um...I don't think you have to put the bags inside each other...

[quote="MathFiend"]actually...there was a big debate about a kid who did put that answer down in a competition for putting the bags inside of each other and he plead his case before the MathCounts judges and that answer was made a possibility.[/quote]
That could happen, but with this problem it won't happen.

it DID happen. In one of my MathCount's teams' practices, we did that problem and I came up with the one bag inside of another answer, and our coach told us that story!

er...i think our wonderful moderator modified the problem to make sure that wouldn't happen in this case...

Oh, I see...
*sigh* I feel like I'm not a comoderator :)

if there are k marbles, you just need to solve for least n such that

n(n+1) >= 2k

here that n is 19


algebra polynomial Support vector rotation function geometry

So i am starting a thread strictly for discussion of FIIT-JEE solutions...Here are certain rules
1)Please Control ur emotions :D
2)While posting ur doubt be clear
3)Nothing except discussion of solutions and solutions given by u all would be entertained here
4)while posting the question have it done as in the attachment.
5)Please see before posting if someone lese has done that or not
6)u can use (bull)sh**,hell (only these two) and at my discretion wtf :D

Hindustan Times presents - FIITJEE Open Test, the Pathetic fallacy Before IIT-JEE.

As usual FIITJEE have proved their brilliance in writing solutions.. What the hell, already I'm only a mediocre performer and if the solns are this bad, how can I end up not being bad. :(((

Anyways, sorry for not attaching but its too time consuming. Please tell me if the answer is right in the end. Please check the solns as I'm posting from Set A (Solns)

1) Paper I
Physics
Question No 4
Can anyone explain the answer??

2) Paper I
Chemistry
Question No 9
How the hell does H2S2O8 give one mole of H2So3 and H2SO4

Question 10
Law of equivalents itself contradicted.

4) Paper II  Chemistry match
He puts an isobaric process as isochoric and vroom... 6 marks down the drain.

Too lazy to post more. Will do so later.

i didn't mean u had  tio attach a photocopy but that was the format to answer,,, :D
but which set...posyt  the question :)

PAPER I:-
chemistry question 16.
He has given a superb explanation with a wrong option.
doubt:-Is 1 also a factor of any polynomial function(Rohit don't answer!!!) :?:

can anyone explain the 4th question in physics
 abt that parabolic thingy ?????

shouldnt the answer bew molar conductance???wth :stretcher:

No C(lambda)=1000K
so it should be specific conductance.

assertiom-reasoning...paper1

the projectile sum
FIITJEE answr: D
my answer:B

Ya mine also B but can anyone who supports D pls explain to me :help:  :help:
Ya one more pethals comes in the maths paper2 match 1. on vectors:
(alpha=p,beta=q,gamma=r)
For D axb=i(r-q)+j(p-r)+k(q-p)=j-k
so q-p=-1 or p-q=1 and the option is q

chem paper1-

THat optical activity question...

FIitjee answer:A
My answer:C

That ans should be A so don't worry
[hide="my answer"]C[/hide]

microscope sum...........................
Fiitjee answer:C
My answer:D

And mind explaining that bobbin(yoyo) sum eh?? :huh:  :huh:

same question here

FIITJEE answer:B
My answer:D

[quote="hash_include"]my woeful physics skills are killin me again :(  [/quote]

Dont worry shrids, im there to compete for the worst spot  :D   :(   :o

I am atleast better than u in that aspect of solving some of them rather than doing gumsa all the time :D  :D  :D

amam correct shreyas(Nan onnum srini ella).
It was sappa that's why I think even babies solved it da :rotfl:  :rotfl:  :rotfl:

how would u solve thatfluid mech sum....i didnt do it???

[quote="Cyber-Shreyas"][quote="madness"]The open test was a milestone day, i solved the COM sum fully w/o a single gumsa  :D   :rotfl:[/quote]

Wasn't that really straight forward???[/quote]
Yes it was but have u seen me do phy properly ever?  :D

i think the main problem for me in physics is that i am NOT able to do gumsa, however contradictory that seems.  :oops:  :rotfl:
The sums i'm able to at least get an idea.. but assertion reasoning jus murders me. ive got atleast twice, all 4 wrong :(

Yes and once for me in physics and this time almost close (3 wrong 1 right)  :blush:  and it has happened once in chem as well  :o . But we cannot do anything if he gives stupid answers like that inversion ques  :mad:

[quote="hash_include"]i think the main problem for me in physics is that i am NOT able to do gumsa, however contradictory that seems.
The sums i'm able to at least get an idea.. but assertion reasoning jus murders me. ive got atleast twice, all 4 wrong :([/quote]
Don't worry I am also competing for that.
All square 2-2 da. :D  :D
In open test paper-1: -4 in assertion :(
paper-2: 12 got all right by gumsa. :rotfl:

Can anyone help me solve thaqt flu mech sum in the open test???

Athu nanum srinathum gumsalaye adichitom da :rotfl:  :rotfl:  :rotfl:

nan poda vila.....
atlest method sollu da

[quote="pramod.best"]@Shreyas: Loop I'm talking about is one passing through the pt......... of radius r...[/quote]

If it just below the paraboloid and of radius r (Passing through the pt alone), then how did you get the current cutting the loop??


So how many languages do you people know? I know 4 American,Chinese,French and Spanish fluently.

[quote="Valentin Vornicu"]You mean English, right?[/quote]

That's quite funny actually because we Australians sometimes call our language "Australian" (as opposed to English) because there are words/slangs with cultural meanings that people from other regions might not understand.

In Romania if you got to a different city than your own there will be slangs and expressions that you will not understand. Just because you have a different dialect (and that's a strech even) doesn't mean it's a different language ;)

I know Valentin  :D  I just thought that it's interesting that some people call their language American or Australian when they are just English (even the New Zealanders sometimes call it "Kiwi").

English (my native language, and, yes, my dialect of that is General American), Modern Standard Chinese (I have worked many years as a Chinese-English interpreter and translator), German (for reading), Taiwanese (for family conversations), Biblical Greek (for reading), Biblical Hebrew (ditto), Literary Chinese (ditto), and rather worse proficiency in Japanese, Russian, Cantonese, and Hakka, with moderate reading knowledge of Latin, Aramaic, and most western European Indo-European languages.

I know American fluently (oops I meant english  :D ).

I can speak Taiwanese well enough with my family. (@tokenadult: fellow Taiwanese unite!)

My chinese skills have deteriorated to the point of mindless blabbing.

Spanish is quite fluent (Based on school standards).

[quote="7h3.D3m0n.117"]\
My chinese skills have deteriorated to the point of mindless blabbing.

[/quote]

Same  :oops: . I start talking to my parents in Chinese and I might end up speaking English.

Unlike you language buffs I only know swedish and english. I can read german, but I am afraid of speaking it, and I definitely can not make jokes in it (proficiency in making jokes is the best indicator of how fluent you are ).

I know english since my mother language is spanish. I'd like to learn japanese when having time.

Mother language is English, can only speak Chinese, but not read. Learning Spanish. I notice that sometimes when I speak Chinese, I have forgotten the Chinese expression and instead want to use the Spanish form. It's a little embarassing when that happens in front of my parents.

Portuguese (mother language) and English. Of course, I can understand a lot of Spanish, and I might know one thing or two in French.

I know English, French, and Latin if you count that. I'm learning Hebrew, but I'm just beginning.

Which two?

It depends. I think it's good to learn a new language, as it increases the amount of things you can understand. I like learning Spanish, because I can finally understand the Spanish Channel on my TV.  :D . For those who know Latin, does it really help in vocab comprehension?

I'm Romanian and French  and besides I understand English, a little German and Russian.
 Every new language is for me another window to the world, even if I don't agree often with... "Windows" :D  :D
It's very instructive to make jokes in two languages !

[quote="hm29168"]For those who know Latin, does it really help in vocab comprehension?[/quote]

Not for me. But then, that is not my goal at all in learning Latin.

The only languages I know really well are English and Latin, but I know some of a bunch more. It's very easy for me to learn grammar, but I have a lot of trouble being motivated enough to learn enough vocabulary to really say I *know* a language. I guess fluency has never been my goal anyway in anything but Latin. A consequence perhaps of not trying to learn languages that are actually spoken?

Do all americans know a second language? Isn't there anyone who just knows english?

[quote="Kalle"]Do all americans know a second language?[/quote]

I'm sure there are plenty Americans that know only English, but many are learning a second one. Now, kids all over the nation are given a choice of a foreign language early in middle school to broaden their horizons.

i kno 4--
u mite have heard about only english
others r hindi,sanskrit &  esperanto.............. :)

Meh. Well. I know Chinese. And English. Spanish--half fluently. French--also half fluently. I know bits and pieces of other languages. And I know Latin and Greek fairly well.

...I don't know if I should learn German. Lots of people do. Is this some sort of indicator?

[quote="Kalle"]Do all americans know a second language?

[quote="7h3.D3m0n.117"]I'm sure there are plenty Americans that know only English, but many are learning a second one. Now, kids all over the nation are given a choice of a foreign language early in middle school to broaden their horizons.[/quote][/quote]

All U.S. high school students are required to take 2 years of a foreign language in order to graduate.  Many selective colleges require 3 years of a foreign language for admission.

[quote="cf249"]All U.S. high school students are required to take 2 years of a foreign language in order to graduate.[/quote]

This is not true of every state. Many high school graduates in the United States have had no study of any foreign language.

[quote="cf249"]Many selective colleges require 3 years of a foreign language for admission.[/quote]

Some AoPSers look hard to find the colleges that don't require that.   :|

Anyway, I agree that it's a good idea to study a foreign language while still a teenager or younger. Many countries do better than the United States does in promoting bilingualism.

[quote]Many countries do better than the United States does in promoting bilingualism.[/quote]

Oh, indeed.  My European linguistics colleagues put me to shame... they speak several European languages fluently (I only speak them well enough to get around) plus whatever languages they are actually studying.  Me, I study poorly attested ancient languages, so answering a question about how many languages I know is tricky, because what does it mean to 'know' a language that is only attested by a small PDF file's worth of inscriptions, which we don't even fully understand?

Bilingualism in areas which are home to many different languages is an interesting phenomenon.  The speakers of the language which is the lingua franca tend not to speak other languages, but the people living in remote areas who speak a language with relatively few speakers, and is overlapped by other, more widely spoken languages, tend to be multilingual-- for example, in the Caucusus mountains, even some groups of peasants who live in the mountains who have never left their villiage can speak four or five different languages fluently.

[quote="tokenadult"]This is not true of every state. Many high school graduates in the United States have had no study of any foreign language.[/quote]

I did not know that, and it surprises me to learn that.  I think that studying a foreign language should be a universal high school graduation requirement, especially with the increasing globalization that is occurring at the present time.

English(oops I meant American), and I'm learning Chinese.

"xian zai shi san dian." is true in two parts of the world.

Well... i speak english and chinese fulently... i understand taiwanese and can kinda speak it... and i'm learning french right now..... :icecream:  :jump:  :bye:

[quote="The_Scintillator"]

...I don't know if I should learn German. Lots of people do. Is this some sort of indicator?[/quote]

I don't know any German but I think it is a good language to learn. Many jobs are in Germany and it could come in very useful in the future.

with moderate reading knowledge of Latin, Aramaic, and most western European Indo-European languages.
[url=http://www.math123xyz.com]Interactive Math Resources[/url]

I'm fluent in English and Chinese (for REAL unlike most people who give this claim, after a bunch of years in Chinese school. That does not count).
I'm learning Spanish, and I hate it.

I think the US is pretty bad with the foreign languages thing, in China I think they start English in elementary school or something. I think in Canada you start learning French pretty early on (even though it's not totally foreign)

Why do you hate Spanish? Spanish is really easy for me. It's even easier than my normal math classes.

1.Serbian(my native),English (fluently with minor grammatical mistakes)
2.French (I can read it,know grammar..but my vocabulary is poor because I rarely use it)
3.Ancient Greek,Latin (basic,slowly forgetting)

Well, I know spanish completely, and I also know english... I can read things in portuguese, but I don't know how to pronunciate it.

Why don't you make this a poll?


[quote="mathVNpro"]Consider $ 12$ numbers from $ \{1,2,3,...,12\}$ placed on a circle. Point out a way to arrangle them in the circle such that in any $ (a,b,c)$ wich positions of $ a,b,c$ are right next to each other (Ex: $ a$ next to $ b$, and $ b$ next to $ c$), we have $ 13|bc \minus{} a^2$.

[b]@[/b]Sorry for my mistake, I tend to post this at the [b]Combinatorics section- Unsolved problem[/b]. Can you move it there for me, [b]Mod of Mathlinks[/b]? Thanks a lot.[/quote]
Anybody hlep me with this problem?? :(

Hint: find a primitive root modulo $ 13$. Consider its powers modulo $ 13$.

  darij


do you ever do math jams as previews of certain online classes?

The recent Math Jams were essentially previews of Olympiad courses we will start in a few months.  Our Problem Series courses (AMC/AIME and MATHCOUNTS) are pretty similar to Math Jams of the same/similar names.


[quote]Find all solutions in positive integers x, y, z to the simultaneous
equations
$ x \plus{} y \minus{} z \equal{} 12$
$ x^2 \plus{} y^2 \minus{} z^2 \equal{} 12$.[/quote]

How could I solve it?  :maybe:

I just found the value for xy... But I don't know how to go on...  :)

$ x \plus{} y \minus{} z \equal{} 12$ $ (1)$
$ x^{2} \plus{} y^{2} \minus{} z^{2} \equal{} 12$ $ (2)$

$ (1) \implies z \equal{} x \plus{} y \minus{} 12$ and $ (2) \implies x^{2} \plus{} y^{2} \equal{} z^{2} \plus{} 12$.

Therefore $ x^{2} \plus{} y^{2} \equal{} (x \plus{} y \minus{} 12)^{2} \plus{} 12$

$ \equal{} x^{2} \plus{} y^{2} \plus{} 144 \plus{} 2xy \minus{} 24x \minus{} 24y \plus{} 12$$ 2xy \minus{} 24x \minus{} 24y \plus{} 156 \equal{} 0$

$ xy \minus{} 12x \minus{} 12y \plus{} 78 \equal{} 0$

$ xy \minus{} 12x \minus{} 12y \plus{} 144 \equal{} 66$

$ (x \minus{} 12)(y \minus{} 12) \equal{} 66$

$ 66 \equal{} 2\times3\times11$

Checking the factors, we get eight solutions:$ (x, y, z) \equal{} (13, 78, 79), (14, 45, 47), (15, 34, 37), (18, 23, 29), (23, 18, 29), (34, 15, 37), (45, 14, 47), (78, 13, 79)$

Alternatively, we can use the difference of two squares as follows:

$ x \plus{} y \minus{} z \equal{} 12 \implies x \minus{} z \equal{} 12 \minus{} y$ $ (3)$

$ x^{2} \plus{} y^{2} \minus{} z^{2} \equal{} 12 \implies x^{2} \minus{} z^{2} \equal{} 12 \minus{} y^{2}\implies (x \minus{} z)(x \plus{} z) \equal{} 12 \minus{} y^{2}$ $ (4)$

$ (4)$ divided by $ (3) \implies x \plus{} z \equal{} \frac {12 \minus{} y^{2}}{12 \minus{} y}$ $ (5)$

$ (3) \plus{} (4) \implies 2x \equal{} 12 \minus{} y \plus{} \frac {12 \minus{} y^{2}}{12 \minus{} y} \equal{} \frac {144 \minus{} 24y \plus{} y^{2} \plus{} 12 \minus{} y^{2}}{12 \minus{} y} \equal{} \frac {156 \minus{} 24y}{12 \minus{} y}$

$ \implies x \equal{} \frac {78 \minus{} 12y}{12 \minus{} y}$ which must be a positive integer

$ \equal{} \frac { \minus{} 66 \plus{} 12(12 \minus{} y)}{12 \minus{} y} \equal{} 12 \minus{} \frac {66}{12 \minus{} y}$

Therefore $ (12 \minus{} y)$ must be a factor of $ 66$ and then continue as in first method.


linear algebra
matrix
algebra
polynomial
analytic geometry
geometry
geometric transformation

suppose we have 4 points in the form $ P_{n}=\left\{a_{n},b_{n},c_{n}\right\}$
And we have some plane.

the question is: What is the best plane such as the distance of the 4 points to the plane is minimized?

or in anothers words, a best plane that fits pretty well in the points.

This may involve the pseudoinverse of a matrix. See [url=http://en.wikipedia.org/wiki/Linear_least_squares]Linear least squares[/url].

i dont know how to use a pseudoinverse matrix to solve this problem.

i was mading a project in auto-cad and have 600 points that i made using Mathematica 5.2 and exported to cad.

i need to every 4 points (in some determination) lie in a plane.

so i wanted this interpolation in a usefull formula and the rest i will make with a program.

if you could help me out with the solution i will be glad

more hints

We can write it in a least squares framework; minimize the sum of the squares of the distances to the plane. Unfortunately, it's a lot more complicated than the usual kind; trying to find the minimum leads to a variant of an eigenvalue/eigenvector problem, in which we need the solution to a 3rd degree polynomial equation (in one variable).

Here's the method:
Let the plane be $ Ax+By+Cz+D=0$. If $ A^{2}+B^{2}+C^{2}=1$, the distance from $ (x,y,z)^{T}$ to the plane is $ |Ax+By+Cz+D|$. Let our points be $ (x_{i},y_{i},z_{i})^{T}$ for $ 1\le i\le n$, with $ n\ge 4$.
Applying Lagrange multipliers to find the minimum of the sum of squared distances, we get the system
$ \sum_{i}x_{i}(Ax_{i}+By_{1}+Cz_{i}+D)=\lambda A$
$ \sum_{i}y_{i}(Ax_{i}+By_{1}+Cz_{i}+D)=\lambda B$
$ \sum_{i}z_{i}(Ax_{i}+By_{1}+Cz_{i}+D)=\lambda C$
$ \sum_{i}(Ax_{i}+By_{1}+Cz_{i}+D)=0$
where $ \lambda$ is an additional unknown parameter. This can be put into matrix form as
$ M\cdot\begin{bmatrix}A\\B\\C\\D\end{bmatrix}= \lambda\cdot\begin{bmatrix}A\\B\\C\\0\end{bmatrix}$, where
$ M=\begin{bmatrix}\sum x_{i}^{2}&\sum x_{i}y_{i}&\sum x_{i}z_{i}&\sum x_{i}\\ \sum x_{i}y_{i}&\sum y_{i}^{2}& \sum y_{i}z_{i}&\sum y_{i}\\ \sum x_{i}z_{i}&\sum y_{i}z_{i}&\sum z_{i}^{2}&\sum z_{i}\\ \sum x_{i}&\sum y_{i}&\sum z_{i}&n\end{bmatrix}$

The most useful simplification we can make: the formulas are much nicer if the average of each coordinate is zero, and translation works well. The minimizing plane passes through the mean point; if we translate to there and ignore the last row/column, our matrix $ M$ becomes the covariance matrix of the distribution of points, and the desired vector $ (A,B,C)^{T}$ is one of its eigenvectors. It's specifically the eigenvector corresponding to the smallest eigenvalue, and the minimum value of the sum of squared distances is $ n\cdot \lambda^{2}$.

Oh, and this is definitely linear algebra rather than calculus.

i think this is a calculus problem because is a optimization problem.

we can make another problem just like, we have a table with 4 foots with different lengths.

how can i cut the foots such as the table rest in equilibrium and has the high heigth?

but you solution is not a equation.

i was just looking for a simple solutions.

if the solution is long and hard is useless, because is inapplicable

i think in a think that can simplificate the problem, just a poor approximation of the problem

let we have 4 points, make a plane passing in 2 opposite points.

so we determine the other parametre of the equation of the plane such as the distance from the plane to the other 2 points are minimal.

but still complicate.

Do you want to deal with four points or 600? The least squares approach handles them equally well. I can even go to arbitrarily many dimensions; in $ n$ dimensions, we have to find the smallest eigenvalue of a $ n\times n$ matrix and its associated eigenvalue.
While the statement and solution do invoke a bit of calculus (Lagrange multipliers to find a minimum in my case), there is much more linear algebra.

I would consider reducing the problem to finding the eigenvalues and eigenvectors of a $ 3\times 3$ real symmetric (positive definite) matrix good enough. That is a task that computer algebra systems can handle easily.

no jmerry.

i have 4 points, with those 4 points i make a plane. so i add more 2 points and make on more plane. so i have 4 polygons. etc.

it's the same think as a polyedro. have planar sides.

understand?

but since the figure i was mading is not perfect, is not symetric, so the sides cannot be plannar. but i want to make it plannar. i have a ideia to make a program to fix it.

but first i need to determine the plane.

i can make 4 points lie in a plane in auto-cad. but the result is poor. because is not the planne i explained above.

i can use triangles. but i dont want. because stetical purpose

There are $ \binom{n}{4}$ groups of four you could choose. Which ones would you use, and what would you do with that massive jumble of completely different planes? I don't think you're asking for something that makes sense.

i want 4 points lie in a plane. or in simple words to simple persons understand

the 4 points are already close to a plane.

you that dont make sense. please. dont be silly.

Just run the least squares, and solve that eigenvector problem. It works.


Let $ a,b,c,\geq 0$, so that $ a^3 \plus{} b ^3 \plus{} c ^3 \equal{} 3$. Find maximal value of the following:
$ a^4b^4 \plus{} b^4c ^4 \plus{} c^4a^4$.

Solution by Gabriel Dospinescu:

Maximum is [b]3[/b]

  AM-GM we have $ bc\leq \frac {b^3 \plus{} c^3 \plus{} 1}{3} \equal{} \frac {4 \minus{} a^3}{3}$ then $ b^4c^4\leq \frac {4b^3c^3 \minus{} a^3b^3c^3}{3}$ and similarly.

 Summing up $ \sum {a^4b^4}\leq \frac {4(\sum {a^3b^3})}{3} \minus{} a^3b^3c^3$
Thus is suffices to show that $ 4(\sum {a^3b^3}) \minus{} a^3b^3c^3\leq 9$
Follow Schur's Inequality $ 4(xy \plus{} yz \plus{} zx)(x \plus{} y \plus{} z) \minus{} 9xyz\leq (x \plus{} y \plus{} z)^3$ for $ x \equal{} a^3 ,y \equal{} b^3 ,z \equal{} c^3$ result conclusion.


I found this on a book.
Continue the series:
$ 100 \\
200 \\
300 \\
400 \\
1600 \\
6400$
How is this series is working guys. I think its a mistake  :o