$x,y,z$ are positive reals with product $1$. Prove that: \[(xy+yz+zx)+(x+y+z)-(x^2y+y^2z+z^2x) \leq 3.\] [u]Rem:[/u] I am not sure that the inequality is true! :D

Let $a=x/y$, $b=y/z$, $c=z/x$

Then we get

$x^3+y^3+z^3+3xyz\geq xy(x+y)+yz(y+z)+zx(z+x)$

which is quite a well known inequality (Schur's, in fact)

Now that you've solved this, see if you can solve the following (IMO A2 2000)

Let $a,b,c$ be positive real numbers such that $abc=1$

Prove that

$\left(a-1+\frac 1b\right)\left(b-1+\frac 1c\right)\left(c-1+\frac 1a\right)\leq 1$

(hint: the solutions are almost exactly the same!)

It becomes $(x+y-z)(x+z-y)(y+z-x) \leq xyz$
 Let $x \geq y \geq z$ if $y+z \leq x$ then LHS becomes negative
 If $y+z \geq x$  put $x=a+b, y=a+c, z=b+c $

Or this:
$x^2\geq x^2-(y-z)^2=(x+y-z)(x-y+z)$
$y^2\geq y^2-(z-x)^2=(x+y-z)(-x+y+z)$
$z^2\geq z^2-(x-y)^2=(x-y+z)(-x+y+z)$
Multiply these three to get
$x^2y^2z^2\geq (x+y-z)^2(x-y+z)^2(-x+y+z)^2$
then take the square root of both sides and we're done.

[quote="seungjun"]For the first problem, how did you use a, b, and c? :blush:[/quote]

My mistake, you get $a^3+b^3+c^3+3abc\geq a^2b+b^2c+c^2a+a^2c+c^2b+b^2a$


Let a,b,c be real numbers such that a+b+c = 6. Prove that
$\sqrt{a^{3}+1}$ + $\sqrt{b^{3}+1}$ + $\sqrt{c^{3}+1}$ :gep 9

[quote="Thao"]Let a,b,c be real numbers such that $a+b+c = 6.$ Prove that
$\sqrt{a^{3}+1}$ + $\sqrt{b^{3}+1}$ + $\sqrt{c^{3}+1}\geq9.$  [/quote]
If $a\geq0\geq b\geq c\geq-1$ then $a=6-b-c\geq6$ and $\sum_{cyc}\sqrt{a^{3}+1}\geq\sqrt{a^{3}+1}\geq\sqrt{217}>9.$
If $a\geq b\geq0\geq c\geq-1$ then $a+b=6-c\geq0$ and $\sum_{cyc}\sqrt{a^{3}+1}\geq\sqrt{a^{3}+1}+\sqrt{b^{3}+1}\geq$
$\geq\sqrt{a^{3}}+\sqrt{b^{3}}\geq\frac{\sqrt{(a+b)^{3}}}{\sqrt2}\geq6\sqrt3>9.$
If $a,$ $b$ and $c$ are non-negative nambers. Then
$\sum_{cyc}\sqrt{a^{3}+1}\geq9\Leftrightarrow\sum_{cyc}\left(\sqrt{a^{3}+1}-3-2(a-2)\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{a(a-2)^{2}}{\sqrt{a^{3}+1}-1+2a}\geq0.$ :)


calculus
integration
function
logarithms
calculus computations

What is the integral of function (a^x)/x, where a is constant. Thanks!

Not an elementary function unless $a=0$ or $a=1$. Can be expressed in terms of the [url=http://mathworld.wolfram.com/ExponentialIntegral.html]Exponential integral[/url].

$\int \frac{a^{x}}{x}\cdot dx = Ei (x \ln a)$


[quote="flip2004"]Let $n\in{\mathbb N },\; (n\ge 3),$ and assume $a_{1},a_{2}$ are given in ${\mathbb R}$ such that $\Delta: =(n-1)a_{1}^{2}-2na_{2}>0 .$ Consider two integers $a, b$  with $1\le b<a\le n , \; (a-b \ge 2) , \;$ and let
\[f(x): =x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\sum_{k=3}^{n}b_{k}x^{n-k}=\prod_{j=1}^{n}(x-x_{j}). \]
with $+\infty >x_{1}\ge \ldots\ge{\mathbf x}_{\mathbf b}\ge \ldots \ge{\mathbf x}_{\mathbf a}\ge \ldots \ge x_{n}>-\infty .$ [b][color=green] If ${\mathbf x}_{\mathbf b}-{\mathbf x}_{\mathbf a}\ge \sqrt{\frac{n-a+b+1}{nb(n-a+1)}\Delta}\; ,\;$ solve the equation $f(x)=0\; ,\; \; x\in{\mathbb R}.$ [/color][/b]

[size=75]
[b]see:  [/b] [/size]
http://www.mathlinks.ro/Forum/viewtopic.php?t=151999[/quote]
This one is for pco.so I hide my post.
[hide]Actually, I think this is an inequality problem.Let $f(x_{1},...x_{n})=\sum_{i=1}^{n}{x_{i}^{2}}$
then when $\sum x_{i}=n,x_{b}-x_{a}\geq t$($+\infty >x_{1}\ge \ldots\ge{\mathbf x}_{\mathbf b}\ge \ldots \ge{\mathbf x}_{\mathbf a}\ge \ldots \ge x_{n}>-\infty .,t>0$)
then we can prove that $f(x_{1},...x_{n})$ get the minimum at
$x_{1}=...=x_{b},x_{b+1}=.....x_{a-1},x_{a}=...=x_{n}$or $x_{1}=...=x_{b-1},x_{b}=x_{b+1}=.....x_{a-1},x_{a}=...=x_{n}$
Or
$x_{1}=...=x_{b},x_{b+1}=.....x_{a-1}=x_{a},x_{a+1}=...=x_{n}$by only one property of $f$
$f(x_{1},...x_{n})\geq f(x_{1},...x_{i-1},\frac{x_{i}+x_{j}}{2},x_{i+1}.....,\frac{x_{i}+x_{j}}{2},x_{j+1}...)$[/hide]

[quote="Hawk Tiger"]....This one is for pco.so I hide my post....
[/quote]
No ! it's for every interested collaborator at this site !
But, you write :[b]then we can prove that $f(x_{1},...x_{n})$ get the minimum at
$x_{1}=...=x_{b},x_{b+1}=.....x_{a-1},x_{a}=...=x_{n}$or $x_{1}=...=x_{b-1},x_{b}=x_{b+1}=.....x_{a-1},x_{a}=...=x_{n}$
Or
$x_{1}=...=x_{b},x_{b+1}=.....x_{a-1}=x_{a},x_{a+1}=...=x_{n}$by only one property of $f$
$f(x_{1},...x_{n})\geq f(x_{1},...x_{i-1},\frac{x_{i}+x_{j}}{2},x_{i+1}.....,\frac{x_{i}+x_{j}}{2},x_{j+1}...)$[
[/b]
[b][color=blue]
In order to complete the solution, please prove this assertion ! Thanks,Alex [/color][/b]


[quote] [/quote]
how can I exchange "quote:"  to "$X$ wrote:"?

I'm pretty sure that you can't, but perhaps I'm wrong.

By the way, this has nothing to do with LaTeX, so I've moved this to the general support forum.

[quote="MCrawford"][quote="MCrawford"][quote="MCrawford"]DPatrick hasn't seen this yet, but...[/quote]

Once you slap the quote tags around something, type
="username"
after the beginning quote tag.  It's just like inserting URLs except you need the quotes around the username.[/quote][/quote]Duh.  :fool:

[quote="amir2"][/quote][quote="MCrawford"]][/quote][quote="MCrawford"]][quote="MCrawford"]][quote="MCrawford"]Once you slap the quote tags around something, type
="username"
after the beginning quote tag. It's just like inserting URLs except you need the quotes around the username.[/quote][/quote][/quote][quote="MCrawford"][/quote][quote="amir2"] I could :lol: [/quote] :clap:  :clap2:  :trampoline:  :thumbup:

Now this doesn't mean you should start building ziggurats like the one in your previous post (quote in a quote which is in a quote etc.) :)
Topic closed!


If

$ a,b,c>0$

then

$ \sqrt{\frac{a\plus{}b}{c}}\plus{}\sqrt{\frac{b\plus{}c}{a}}\plus{}\sqrt{\frac{c\plus{}a}{b}}\geq2\sqrt[6]{\frac{9(a\plus{}b\plus{}c)^4}{\sqrt[3]{abc}(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$

If $ a,b,c>0 $   then:
$ \sum_{cyc}\sqrt{\frac{b+c}{a}}\ge\sqrt{6}\sqrt[6]{\frac{(a+b+c)^3}{abc}} $
______
Marin Sandu


if $n$is a postive integer  $\{\sqrt[k]{n}\}>0$ prove that
$\{ \sqrt[k]{n}\}$
$> \frac{1}{k \sqrt[k]{n^{k-1}}}$
where ${x}$is the fractional part of the nmber $x$

Continue in http://www.mathlinks.ro/Forum/viewtopic.php?t=98109 please.


geometry
3D geometry
tetrahedron
geometry unsolved

We have a convex quadrilateral so that the sum of squares of its sides projections on any line is constant. Prove that the quadrilateral is  the plane projection of an equilateral tetrahedron with isosceles faces having the bottom angle $\arctan \sqrt 2$.

[quote="Myth"]Is it true? :?[/quote]
    I believe that it is right but I can't prove it    :lol:

Very interesting problem... but I'm not sure what "equilateral tetrahedron with isosceles faces having the bottom angle $\arctan \sqrt 2$" means.

It's not hard to show that the only triangle with the given property is the equilateral triangle.  The square also has the given property.  Are there any other quadrilaterals?


If $p(x)$ cannot be factorized $\implies p(x+1)$ cannot either.

When you expand $(+-)(x+1)^{1999} (+-)(x+1)^{1998} (+-) ... (+-) (x+1)^1 (+-) 1$ there exists a prime $1999$ such that it does not divide the leading coefficient, it divides all the other coefficients except for the constant and the square of 1999 does not divide the constant. By Eisenstein's criteria, this polynomial cannot be split into $p(x)$ and $q(x)$, polynomials with integer coeffients.

Uhh....I believe I mean that for 1999 distinct numbers $a_1, a_2, a_3, ...$, $P(a_k) = \pm1$, instead of the coefficients being either plus or minus 1.  I think this makes the problem easier, tho.

[quote="pkerichang"]Have fun:

If $f(x)$ is a polynomial with integer coefficient, prove that $f(a + f(a))$ is always a multiple of $f(a)$.

let $p(x)$ be a polynomial with $p(0) = p(1) = 1999$.  Prove that $p(x)$ does not have integer zeros.  Can you generalize?

I think the following is classic:

A polynomial $f(x)$ has degree of 1999 with integer coefficient and has a value of 1 or -1 for 1999 different values of $x$.   Prove that $f(x)$ cannot be factored into two polynomials with integer coefficient that both has degree greater than 0.[/quote]
1) consider the polynomial f(a+f(a)) mod f(a)
Since (x1+x2+x3+...)mod (n) is congruent to x1(mod n)+x2(mod n)+...+xn(mod n)
So f(a+f(a)) mod f(a) is congruent to f(a) mod f(a) which is congruent to 0 mod f(a)
2) Since p(0) is odd, then p(0 mod 2) is odd as well. Since p(1) is odd, then p(1 mod 2) is odd as well. Since 0 is not odd, there can exist no integer roots

3) Assume it could be factored as such. Then, since the polynomial has integer coefficients, it can be factored into two different polynomials, each of which, when you plug in the values that make it 1 or -1, has one of them output +-1 and one of them output +-1. More than half the values of such when you plug them in have output 1. Thus we must have P(x)=f(x)(x-a1)(x-a2)(x-a3)...(x-a(n))+1, where n>=1000 and a1, a2, ... an are some of of the values that make our polynomial output +-1. That clearly cannot be multiplied to get a polynomial that fits our constraints.

[quote="pkerichang"]
let $p(x)$ be a polynomial with $p(0) = p(1) = 1999$.  Prove that $p(x)$ does not have integer zeros.  Can you generalize?[/quote]

[hide="Generalization (Canadian MO, AoPS V2)"]
Generalization:  If $p(0), p(1)$ are both odd, then $p(x)$ has no integer zeroes.

We prove by proving the converse:  If $p(0)$ is odd and $p(x)$ has at least one integer zero, then $p(1)$ is even.

The proof is immediate:

$p(x) = (x-r) q(x)$ where $q(x)$ is also a polynomial with integer zeroes.
$p(0) = -r q(0)$ is odd, implying $r$ odd and $q(0)$ odd (actually unnecessary).
$p(1) = (1-r) q(1)$ must be even.  QED.
[/hide]


Let $ C_n$ be the subspace of $ \mathbb{R}^2$ defined by :

$ C_n \equal{} \{ (x,y) : (x \minus{} 1/n)^2 \plus{} y^2 \equal{} 1/n^2\}$ , let $ Y \equal{} \bigcup_{n > 0} C_n$, and let $ X \equal{} C_{1}\times \mathbb{Z}_{ \plus{} }$  be a subspace of $ \mathbb{R}^2 \times \mathbb{R}$ , define : $ g : X \to Y$ by $ g[(x,y), n] \equal{} (x/n; y/n)$

Could some one please explain for me that why g is not a quotient map ?  Thanks

We need to show that the topology on $ Y$ induced by $ \mathbb R^2$ is not the quotient topology w.r.t. $ g$, i.e. we need to construct a subset $ V$ of $ Y$ which is not open, but $ g^{ \minus{} 1}(V)$ is open in $ X$. The crucial point will be the origin $ (0,0)\in Y$ where all the circles are patched together. Note that $ g$ induces a bijection $ g'$ between $ X\setminus(0,0)\times\mathbb Z_ \plus{}$ and $ Y\setminus(0,0)$.

So take an open ball $ U \equal{} B_0(r)$ in $ \mathbb R^2$ centred at 0 with radius $ r > 0$. Then all but finitely many circle lines $ C_n$ are contained in $ U$, since the circles $ C_n$ accumulate to $ (0,0)$.

Denote by $ H$ the open half circle line in $ C_1$ containing $ (0,0)$. Then $ V' \equal{} H\times\mathbb Z_ \plus{}$ is open in $ X$. For $ V \equal{} g(V')$ we have $ V' \equal{} g^{ \minus{} 1}(V)$. If $ V$ was a neighbourhood of $ (0,0)$, then it would contain all but finitely many of the circle lines. But this is not the case by the reasoning above, since open balls centred at $ (0,0)$ induce a basis of the open sets of $ Y$ containing $ (0,0)$.

BTW: The claim is equivalent to show that $ g'$ is a continuous bijection, but not a homeomorphism. ... and this is what was done above.


Asymptote
calculus
derivative
function
conics
hyperbola
calculus computations

Find the slant asymptote(s) for $ f(x) \equal{}\frac{x^{2}}{\sqrt{x^{2}\plus{}9}}\plus{}x\minus{}2$ and justify your answer by computing the appropriate limit(s).

[hide="I believe the answer is"]

$ 2x\minus{}3$ (used derivatives and guessed the initial value)

[/hide]

No, it's $ y\equal{}2x\minus{}2$ for $ x\to\infty$ and $ y\equal{}\minus{}2$ for $ x\to\minus{}\infty$. $ \sqrt{x^{2}\plus{}9}\equal{}|x|\sqrt{1\plus{}\frac{9}{x^{2}}}\equal{}|x|(1\plus{}O(\frac{1}{x^{2}})$ for large $ |x|$, so $ \frac{x^{2}}{\sqrt{x^{2}\plus{}9}}\equal{}|x|\plus{}O(\frac{1}x)$, and $ f(x)\equal{}x\plus{}|x|\minus{}2\plus{}O(\frac{1}x)$.

Your first answer is wrong, just plug in 0 and you see that the function and the "asymptote" intersect. Also, I can't follow what $ O$ represents.

It's not wrong. The definition of an asymptote does not include "doesn't intersect". The graph actually approaches both of its asymptotes from below.
For the notation, there's a sticky you might want to look at.

Prove the slant asymptotes of a hyperbola $ \frac{x^{2}}{a^{2}}\minus{}\frac{y^{2}}{b^{2}}\equal{} 1$ are $ \pm\frac{b}{a}x$

The solution I got is in the attachment.

Hey prime...I don't get how you went from the whole equation to just b/ax after plugging in +/- inf...Whouldn't you have to plug in the infinity on both x's, not just one?

here's how I'd do it

let $ g(x) \equal{} |x|\sqrt {1 \minus{} \frac {9}{x^2 \plus{} 9}} \equal{} f(x) \minus{} x \plus{} 2$
$ (1 \plus{} a)^n \equal{} 1 \plus{} na \plus{} n\frac {n \minus{} 1}{2}a^2 \plus{} ...$
$ \frac {g(x)}{|x|} \equal{} 1 \minus{} \frac {9}{2(x^2 \plus{} 9)} \plus{} ...$
$ g(x) \equal{} |x| \minus{} \frac {|9x|}{2(x^2 \plus{} 9)} \plus{} ...$
as $ x\to\infty$
$ g(x)\to x$ and
as $ \minus{}x\to\infty$
$ g(x)\to \minus{}x$ and
$ f(x)\to x \plus{} |x| \minus{} 2$


let ABC be a triangle.draw an equaliraterall triangle,inwich its vertecs are on the sides of ABC.and name it EFG.

and also find that:

1)when is the (S) of EFG,min?

2)when is the (P) of EFG,min?

i suppose you denote by (S) and (P) the area and the perimeter, respectively.

in case i have supposed well, we have that (P) is the minimum (in fact, in any triangle) when EFG is the orthic triangle.

i think that about the area, we may have its minimum when two of those points "tend" (if that word can be used in geometry) to one vertex and the other point is the midpoint of the opposite side. am i wrong, ashegh?

about (P),

campos,i said EFG is equaliraterall...

about(S),

campos,i said EFG is equaliraterall

but do u know this fact that(about (P)) : if u find how to draw EFG in ABC.

u also will find the min(S)...

can u understand?

think about what i said.its correct.

then try to solve number (2).

i'm so sorry ashegh,  :blush: , i understood that ABC was equilateral, i'll give it a try.


The only number with this property is $1989$.

Notice that we must have at most 3 primes dividing the number $n$.
By the upper relation we must have that $m > 8$, implying that $m \in \{ 9,10,11,12 \}$.

By case bashing we see that $m = 12$, thus implying that $d_4=13$.
Giving us that $d_{12}=(14+d_2)d_8$.
Thus we must have that $d_5=14+d_2$, since $d_5d_8=d_{12}$.

Since $d_2$ is the first number that divides $n$ it must be prime, thus we must have that $d_2 \in \{ 2,3,5,7 \}$, it can't be $11$, since then we wouldn't have a $d_3$.

By bashing we have that $d_2=3$ and that $n=3^2.13.17=1989$, hence we are done :D


trigonometry
logarithms
geometry
ARML
arithmetic sequence
similar triangles

1. Compute $n$ if $(10^{12} + 25)^2 - (10^{12} - 25)^2 = 10^n$.

2. If $0^\circ < x < 90^\circ$ and $\cos{x} = \frac{3}{\sqrt{10}}$, compute $\log{\sin{x}} + \log{\cos{x}} + \log{\tan{x}}$.

3. The integers $x$, $y$, and $z$ are each perfect squares, and $x > y > z > 0$.  If $x$, $y$, and $z$ form an arithmetic progression, compute the smallest possible value of $x$.

4. In triangle $ABC$, altitude $AH$ and median $BM$ intersect inside the tirangle and are equal.  If $\angle{ACB} = 41^\circ$, find the measure of $\angle{MBC}$.

5. There are exactly four positive integers $n$ such that $\frac{(n+1)^2}{n+23}$ is an integer.  Compute the largest such $n$.

6. In triangle $ABC$, $P$ and $Q$ are midpoints of $AB$ and $AC$ respectively; $R$ and $S$ are the trisection points of $BC$ (with $R$ closer to $B$).  $PS$ and $QR$ intersect in $T$.  If the area of triangle $ABC$ is 60, compute the area of triangle $PQT$.

I've been slacking off lately in math...but I'll try these.

1. [hide]Call 1012 "A" and 25 "B".

So (A+B)2 + (A-B)2 = 10n

Everything cancels out except 4ab=10n

4(1012)(25)=10n

(100)(1012)=10n

1014=10n

N = 14.

[/hide]

2. [hide]

Well all you have to do is to draw a triangle to figure out that sinx= 1/:sqrt:10, and tanx=1/3. Since LogA+LogB=LogAB, logsinx+logcosx+logtanx=log(sinxcosxtanx)=log(1/:sqrt:10)(3/:sqrt:10)(1/3)=log(1/10)= -1?

[/hide]

hmm. I think I'll do some more later.[/hide]

I'll just follow the precedence set by MagnusRex. I'm John Hu from Grissom High School, where I am a sophomore this year. You might have seen me at ARML and MAO this past summer; if we haven't met yet, I hope we'll encouter each other one day.

Spoiler:

[hide]

3. Setting variables:

  Since the three numbers are in an arithmetic sequence:

	The first time when this occurs is when b=5, a=7, and c=1.

	Therefore, x=49.

4. Drawing the perpendicular line from  triangle . If we set  to be of length

, the newly drawn side has a length of  due to similar triangles.

.

Therefore, .

5. Using synthetic division, the fraction is equivalent to n-21+ . Therefore, the largest value of n that would produce an interger would be 461.

6. Letting side  be  and the altitude of  from  be ,  is now equal to 20 since the area of triangle  is equal to 60.  is equal to  from similar triangles  and ; the length of the portion of the altitude from  between  and  is equal to . From similar triangles  and , we can infer that the altitude of triangle  from  if equal to . The area of triangle  is equal to , which has the value of 9.

[/hide]


The x-intercepts of the graph of $f(x)=||x-a|+b|$ are 15 and -3.  What is the value of $b$?

[hide]From the function, we know that when $f(x)=0$, it is implied that $|x-a|=-b$ which means that $x=a\pm b$.

We also know that the x-intercepts are 15 and -3, which leads to two possible systems which might satisfy the conditions.

(i) $a+b=15$ and $a-b=-3$
(ii) $a+b=-3$ and $a-b=15$

Solving (i), we get $a=6$ and $b=9$, which does not satisfy our given conditions.  However, solving (ii), we get $a=6$ and $b=-9$, which works.

Therefore, our answer is $\boxed{-9}$.[/hide]


in a 52-card deck you pick 4 cards
what is the probability of getting one heart
a)with
b)without replacement

$13/52*12/51*11/50*10/49$

but for the other one, is it

$3/4^{3}$????
im not sure

Do you mean 4 hearts? Because in that case both your answers would be wrong. Here are the solutions for picking exactly 1 heart.

[hide="with replacement"]There are 4 places in which we can choose the first heart. WLOG let it be the first. The probability of a heart is $\frac{1}{4}$. Then, we want non hearts. The probability is $\frac{3}{4}$ for each time. Thus, the probability of this situation is $\frac{1}{4}(\frac{3}{4})^{3}=\frac{27}{256}$. However, we WLOGged in the beginning, so we have to multiply by $4$. Thus, the final probability is $\frac{27}{64}$.[/hide]
[hide="without"]It's more complicated here, we have to use casework on when the heart shows up, I don't think we can WLOG this time. I don't feel like doing it now.[/hide]


I have been studying algebra 2 by myself over the summer, and was planning to skip it, but my highly talented friend says that this is a very important math course. If i were to skip it, i should learn it thoroughly.

I was wondering if i should:

1. skip algebra 2 and re-learn it by myself next year while learning pre-cal
2. skip algebra 2 and re-learn it with tutor next year while learning pre-cal
3. Take alegbra 2 and learn pre-cal next year with tutor
4. Only take algebra 2 next year and continue without skipping anything else
5. Forget perfecting algebra 2 and move on
6. I'm open to more suggestions

please note i am not a genious in math, just highly motivated, so I may not be able to perform as well as many of you guys[/list]

It depends on how comfortable you are that you understand the stuff in Algebra II.  I would personally be hesitant to take pre-cal without thoroughly understanding the Alg II stuff, but 95% of the kids at my school don't thoroughly understand the stuff yet gradewise do fine... :?

If you are comfortable with your Algebra II skills, I think you should take it as an online course for credit, which shouldn't take much time if you've understood most of the concepts. Take Pre-Cal once school starts.

Teachers often allot a few weeks for review once school starts anyhow.  ;)

Actually, after looking at an Algebra II book, it looked like it was just a repeat of Algebra I, plus conic sections.  I don't think I saw anything knew besdies that.  But I dunno if I'm right, since I'm taking Algebra 2 this year too.

Don't worry about perfecting Algebra II. Just skip it. :) Anyways, pre-calculus is basically a review of AlgII, plus trig (sometimes) and vectors. That's about it. You can relearn AlgII when you do pre-calc. Good luck! :)

If I was you, I would take Algebra II and make sure that I know everything in Algebra II. It will be okay for you to take pre-calc now but my point is, Algebra II isn't really difficult class. It's easy to get A and keep your GPA in shape. Not saying that you won't get A in pre-calc but it would be definitely easy to get A in Algebra II (because I took it 2 years ago :D ).

My math administrator won't let me skip it. We got into a big argument. She says that colleges want to see it on my transcript.

And Algebra II is easy. It's covered in detail in the AoPS Volume 1. With conic sections in Volume 2 of course.

What I did at my school was take Algebra II but then I skipped pre cal.  The only reason they let me do this was because (a) I had a 100 in Alg II and (b) I sat in Pre cal the final quarter and got a high A for the quarter with a 99 on the final exam.  If you were to skip one or the either, I would make it Algebra II to skip.  The American school system is set up such that you only learn very little new material each year.  So an ambitious, eager math students (like most of those on AoPS) often times need to be in a more challenging environment.

Also, I don't think colleges will care that you don't have an Alg II class if have A's in pre cal, and AP Calc (as you probably will).

It totally depends on you.

Though building the basics now, lets you handle heavier stuffs later.  Even though if AlgebraII is a repeat of AlgebraI with additional stuff (also about Calulus being a repeat of Algebra with additional stuff).

If you are confident and a quick study though, i guess its alright if you do away with Algeb I and go to Algeb II or go directly to Calculus without going through Algebra.

(this site is really nice, i hope to brush up my mathematics skills... its really rusty as of the moment)...

alright, thanks everyone.

since it seems like precal is a repeat of algebra 2, ill try skipping algebra 2 and relearn the algebra not covered in the precal book with a teacher (or take an online course to perfect it)

 :)  :)  :)  :) thanks for the help, i really appreciate it!!!!!

Sincerely,
Armin Samii


The choices are both laptop and whatever you call the ones you put on the desk.

[quote="7h3.D3m0n.117"]hmm don't forget the iPod. but yeah the Macs do suck. my middle school is filled with them (either that or extremely ancient PCs)[/quote]

haha ditto.

My school is filled with PCs that make Macs look good(even if that isn't saying much, lol!) (I have a mac laptop with Windows installed, its faster than my old HP)

I think in generall, both are good. However, Macs can get ultra slow, like the ones in my school. Why doesn't dell offer lower prices and promotions for educational purposes like Apple (all the time)?
The leapord sounds pretty good, excluding that parental thingy mabobber.
The iphone, which is coming out tomorrow :) is pretty good.
The old desktops however are bad for both.

PC!PC!PC!PC!PC!PC!PC!PC!PC!PC!PC!PC!

yeah!! :rotfl:

PCs own Macs. The only reason schools use Macs is because they cost less and they think they are the future of computers. I don't want to be part of that future. :P

@ dingzhou - The iPhone, just like all other Mac products, are extremely advertised to attract attention, but in fact aren't that good. Some other phones with much better features and better looks are generations ahead of the iPhone.

It's funny how that in my school library, the iMacs are installed with Word and PowerPoint  :rotfl:

I think most of the AoPSers need not an iMac.

[quote="dingzhou"]
the only good thing apple made is the iphone[/quote]
eww, no.

My school uses lots of huge Macs, too.  They are very noisy when all together.

BTW, the ones that go on a desk are called desktops...
Of the 4 computers in my house, 3 are windows and my mom's new laptop is *shudder* a mac.

[quote="dingzhou"]why likes mac anyway
the only good thing apple made is the iphone[/quote]I dare contradict you on this one. Iphone is by far the most overpriced phone ever and none of its features are unique (except maybe running a MacOS :) which is bad because there are not that many applications compared to Windows Mobile 5-6 or Symbian OS).

I heard the iphone can do everything but be a phone.

I has a pc. it pwns

[quote="hunter34"]I heard the iphone can do everything but be a phone.[/quote]It cannot navigate you around. Why? Because unlike Nokia N95 and a bunch of other phones it doesn't have a GPS chip inside. It cannot shot videos. Why? God knows, it's got a (pathetic) camera, but it can only shoot (really) crappy pictures. And I can go on with this.

The only thing Iphone is good at is the screen. Even there it has competition, as people are starting to come out with VGA 2.8'' diagonal screens for Windows Mobile 6 (which can be easily programmed as proven before to "act" like the iPhone with the whole finger commands and all).

And up until now there are NO 3rd party applications (so all applications you can possibly have are from Apple). Steve Jobs has announced the opening of some APIs but who knows when and exactly what will be available for 3rd party developers.

Anyway the iPhone is the perfect example of how marketing can sell stuff that is way overpriced for what it can actually do.

PC PWNS MAC!!! BIG TIME, TOO!

GO BILL GATES!! :D

Stop being Windows fanboys!
Be a Linux fanboy  :D
Well, seriously I think Macs pwn, because they run Windows too, and they're overpriced if you select a base system, but they aren't if you select more options.  The additional items are generally discounted .

I run a Mac. I like it. Judging by Apple's share price, so do the investors.

[quote="LevatorNasolabialis"]Go PC  My school is filled with ones from 1995[/quote]

And that's a good thing?

Our elementary school is raising $ \$$30,000 to buy laptops for the staff and the computer lab. What sucks is that by the time they raise all that money I'll be out of the school..(and into 6th grade!)

We're raising the money by making Boo pops, Saint Patty's Day Pops, bake sales, picking carnations and selling them (I think that's for our Philadelphia trip though) and selling cheap Spirit Wear. Honestly, I think we're never going to get lap tops.

You [i]can[/i] right click on a Mac. In fact, that's what I've been doing since I first bought it. I think it was true Once Upon A Time that you didn't have an obvious way to right click, but, thankfully, Apple seem to be phasing that general idea out.

[quote="ProtestanT"]Macs can't right click.
I use a PC :D[/quote]

It's been a while since I used one but you pressed a key on the keyboard while clicking and that opens up a few options.

[quote="Ihatepie"]Every single person I know who has an iPhone says it is the best phone that they have ever had. I know 21 who have one. Btw, Macs can right-click.

A couple things, one, PCs (especially with Microsoft on) crash ALOT compared to the mac. Most of my friends have needed to get their computers fixed at least twice a year. My mac hasn't froze or broke and no one has needed to fix it for 5 years.

@ Ilikepie, usually macs cost more than pcs, so schools don't use them because they are cheap, they use them because they are good.

@archimedes my 5-year-old mac doesnt make a sound[/quote]

PCs usually don't crash or need to be fixed over many years. The only problem is that they get SLOW over years (like after one or three years it gets really slow and takes a while to load things up)
I've experienced that in both the computers in my school and computers at home. Only one computer has been repeatedly crashing, and that's one out of over 30 computers i've looked at.

[quote="ProtestanT"]
PCs usually don't crash or need to be fixed over many years. The only problem is that they get SLOW over years (like after one or three years it gets really slow and takes a while to load things up)
I've experienced that in both the computers in my school and computers at home. Only one computer has been repeatedly crashing, and that's one out of over 30 computers i've looked at.
[/quote]
You can fix this by disabling Virtual Memory. That is what causes it to slow down.

Er- no. The main reasons for a computer to slow down are
(1) Lots of stuff running in the background, consuming system resources. On Windows systems, this is usually viruses/spyware/assorted other names for stuff you would get rid of if possible. For obvious reasons, it isn't an issue on a Mac unless you're deliberately running programs your system can't handle.
If you are running programs your system can't handle, you probably need virtual memory just to run them in the first place.
(2) Serious hardware problems. This can happen to any machine- it's an early symptom on the way to total failure.

As for the mouse stuff- A one-button mouse is a mild inconvenience on a Mac and crippling in Windows. A mouse with at least two buttons works fine for either.
I think Apple is the only manufacturer that still sells one-button mice.

The only real gripe I have with my Mac laptop is its cursed hardware. In a little over two years, I've had to replace both the hard drive and the logic board. There was also a battery recall- that part wasn't even giving me trouble. Fortunately, I'm within walking distance of an Apple store, and the support service there is great.

There's one more possible cause of slowdown: updates of application programs, and the files that get made or sent around in newer versions of application programs. (Yes, I'm looking at you, Microsoft.) The updates almost always have more complicated file structures, have bigger programs with more options, and occupy more memory space than their predecessors. Hardware which was perfectly adequate to run the previous version may struggle with the more bloated update.

The first computer I had was a Mac Plus: an 8MHz CPU with no FPU, 1M of RAM, and no hard drive at all. And you could fit an operating system and an application program (such as Word or Excel) on a single 800K hard-shell "floppy" disk. Needless to say, those programmers had to think more about whether it would fit into a certain amount of storage or whether it the commands could be executed in a reasonable amount of time using the available RAM than do today's programmers.

By the way, could anyone explain or provide a link to what is Virtual Memory?

Virtual memory is swap-file. This means that when you computer runs out of physical memory to put things on, which is faster, it swaps the less used stuff to harddisk. If you have to use a lot of virtual memory your computer will slow down. However, if you tell the OS to use no virtual memory at all - and Windows plasters you with warnings about this, if I recall - then when you run out of physical memory it will tell you that you simply can't run any more applications, and possibly things will start to crash.

You won't really get any speed bonus from disabling virtual memory, as a consequence. Really you just make the OS' job harder, and free up a little space on the harddisk.

PC are better in almost every aspect and can be upgraded as well (not like macs can't though)
G5s are pretty nice though.

But can PCs be downgraded? I think you'll find the answer is no. What happens when you have a 3-5 year old computer but your windows is up to date? Do you really think that your processor is capable of running all these new programs at the same speed it ran the old programs years ago?

Seriously, I think the best choice is linux. Sure, there's fewer games, but it gives you much more control over your system. It's also more secure and everything is free. No more downloading shareware illegally in order to simply burn a CD. Just download and install the open source tools from your distro's package manager. People say it's not compatible, but that is a complete lie. Openoffice has support for every Microsoft Word format up until office 2007. Pretty much every file that you want to open can be opened. (A few online apps won't work because the people who made them didn't consider the linux people, but most of them are fine)

I still personally find that Linux being user-friendly is something of a lie. The UI consistency is laughable (OpenOffice looks like something from the 90s, and behaves like it too) and I have yet to find a distro that does not require [i]some[/i] editing of configuration files with something other than a pretty GUI, or [i]some[/i] use of the terminal at some stage.

Something I personally don't mind (I'm quite capable of using Linux and did for a year, thankyou very much), but somehow I don't think Joe Bloggs is going to agree when his Windows computer can do it for him, and I always felt I was spending more time trying to get my super-customisable computer to do things rather than actually using it.

Sure, Windows will do what an average person will want. The thing is that windows will stop being practical on a computer after 2-3 years. It doesn't matter if you buy the most up to date high tech 64 bit 3GHz processor that you can find. In 2-3 years it will be obsolete and windows will run slowly on it.

Also, OpenOffice has no reason not to look and act like something from the 90s. It's a clone of microsoft word, so how could it be better than word?

Although yes, linux being user friendly is a bit of a stretch.

Sorry, but I think PCs becoming slower, obsolete and less able to keep their pace with modern OS upgrades is not something that's confined to any particular operating system. One can only get so much bang for a buck in terms of hardware: very much the same issue can be said of mainstream Linux distributions (you will not run Beryl/Compiz/whatever very well without a relatively decent graphics card, for example) and of the Mac. With their slightly different upgrade patterns or the fact Linux doesn't as of yet tend to incorporate as much flashy stuff as Windows or Mac, the effect may be a little less pronounced, but it's just as existent. To run the same computer for several years what with the current pace of technology is something that isn't viable yet. Hardware [i]always[/i] becomes outdated, and until Moore's Law is smashed to pieces OS vendors will (and I believe should) take advantage of the gains in speed.

What Windows doesn't have, however, is a very good upgrading system, a problem that will not plague Linux users in the slightest and an issue I think is really giving Microsoft some headaches. While Ubuntu for example will forever be free, and the next OS release is therefore something to look forward to, paying £200 for the next copy of Windows every two years is not so ideal, and stings no matter how many features they ladle into the thing. People are still sour about Windows 2000, Windows ME and those other suspiciously money-spinner releases.

As for OpenOffice being fine looking like something from the 90s, that's mad. If it ever wants to be taken seriously - and frankly I think right now it's more of a joke than a serious alternative to Office - it needs to get up to speed. It's painful to use. [url=http://www.openoffice.org/dev_docs/features/1.1/images/open_xml_file_format.jpg]This[/url] compared to [url=http://officeblogs.net/UI/WordVista.png]this?[/url]. [url=http://www.codinghorror.com/blog/archives/000376.html]Usability is the key to any program[/url], and while it sticks to You Can Never Have Too Many Toolbars metaphor, it's never going to be as mainstream as it could be, unless Microsoft - and this is looking ever more possible - price themselves out of the market.

That's my three cents, anyway. Don't feel that I'm saying "YOU'RE WRONG!" or something, because this is obviously more a matter of opinion than fact.

I didn't say that the hardware wouldn't become outdated. I was saying that it is reasonable to use linux on an old computer because you can find the old software around still. I agree that beryl won't run as fast on an old computer, but less eye-candy rich desktop environments will. My point is that linux supplies old software, whereas most people that sell for Windows only sell the most up-to-date version. Please, show me where I can find Windows XP. It won't be from Microsoft, most likely. Also, if you haven't noticed, people are [b]already[/b], and have been sour about Vista. Really amazing - almost everyone who can either switches to linux or XP. As for Moore's Law, I'm guessing you haven't heard about [url=http://en.wikipedia.org/wiki/Wirth%27s_law]Wirth's law[/url].

Also, as I said before, OpenOffice is a [b]clone[/b]. Comparing it to vista is simply unfair. It can't progress until after the newest version comes out.

I never said anything about Vista being good or having high levels of customer satisfaction. I don't even use Vista myself. But I think to say "almost everyone" is switching is kind of like saying "9 out of 10 cats prefer". It really is very vague, and hardly grounded in any fact that's more than anecdotal. Microsoft hasn't done itself any favours recently, but it is certainly not losing customers in droves no matter how much the activists would like it to be, and the [url=http://finance.google.com/finance?client=ob&q=MSFT]stock isn't dropping like a stone[/url] either. Whatever you're saying about XP, [url=http://www.amazon.com/Microsoft-Windows-Home-FULL-VERSION/dp/B00022PTRU/ref=sr_1_1/103-8596014-8310252?ie=UTF8&s=software&qid=1194192746&sr=1-1]it's not exactly hard to buy[/url], having found this Amazon link off of Microsoft's own site.

Once again: this is more opinion than fact, so do what you want, I've had this discussion before and they never wind up with anyone changing their opinions on anything, although often someone gets upset. I'm done with this.

[quote="1=2"][quote="DonkeyKong"]MAC PWNS PC!!! BIG TIME, TOO!

GO STEVE JOBS!! :D[/quote]

That's the spirit! :)[/quote]
PC's are okay but Microsoft really needs to make Vista better.
I mean, half  the time when people who HAVE a problem with Vista the troubleshooting center suggested that they switch back to XP.
That's just LAME!

Computer.. my brother has a laptop  :lol:


number theory
greatest common divisor
number theory unsolved

show that if $ (a,b)\equal{}1$ and $ p$ is an odd prime then
$ (a\plus{}b,\frac{a^p\plus{}b^p}{a\plus{}b})\equal{} 1 or p$

If $ (a,b) \equal{} 1$ and $ n$ is odd, then
$ (a \plus{} b,\frac {a^n \plus{} b^n}{a \plus{} b})|n$.
Prove. Let $ q^k|a\plus{}b$, q prime, then $ \frac{a^n\plus{}b^n}{a\plus{}b}\equal{}a^{n\minus{}1}\minus{}a^{n\minus{}2}b\plus{}...\plus{}b^{n\minus{}1}\equal{}na^{n\minus{}1}\mod q$. Because $ (a,b)\equal{}1$, we get $ q\not |a$, therefore $ q^k|(a\plus{}b,\frac{a^n\plus{}b^n}{a\plus{}b})\to q^k|n$. It mean $ (a\plus{}b,\frac{a^n\plus{}b^n}{a\plus{}b})|n$ for $ (a,b)\equal{}1,n\minus{}$ odd.


[quote="7h3.D3m0n.117"]There has been a ton of flooding going on recently (due to the AMC season) in this forum.

[size=150][b]Please do not create multiple threads at any given time. [/b] [/size]

It is suggested that you compile all similar grouped problems in one thread (all AMC 2002A questions all in one, for example).

From here on out, if any of the mods see a Flood they are authorized to delete the posts from experienced users. Recent users however, will be warned once [i]and only[/i] once, and their posts merged.[/quote]

This is rather limiting...

I think that a better rule would be to say to try to avoid posting many different threads with similar problems / subjects.  It is fine to create multiple threads, just try to avoid creating an excess amount of threads.

As a related note, you probably don't need to post about every single AMC question from a given test  :roll:


Does there exist continuel real function which has each it's value uncountably many times?

[quote="t0rajir0u"]Any periodic continuous function has this property.[/quote]

Not so. Periods are countable by their nature. (For example, $f(x)=\sin x$ takes zero value in countably many points $x=k\pi,k\in\mathbb{Z}$)

[quote="Farenhajt"]For example, a constant function.[/quote]

Sorry, I have missed trivial case.

Well, let in addition a function to have uncountably many values.

Now that's a bit strong.  Then the domain of the function is the union of uncountably many uncountable disjoint subsets of the reals... which, if I recall correctly, has cardinality greater than $\mathbb{R}$.

Yes, you are right.

Added: ... but it does not mean the affirmation of the existence of such function.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $w$ be a circle and $P$ an exterior point and $A,B$ the points where the tangents from $P$ to $w$ intersect the circle. Denote $M$ the midpoint of $PA$, and $N \in BM \cap w$. Prove that $PN= 2MN$.

[img]8482[/img]

It is indeed interesting.
Let $\omega'$ be the circumcircle of $PNB$. Then $BN$ is the radical axis of $\omega, \omega'$: M is on the radical axis. But we already know that $MA^{2}= MP^{2}$, and MA is a tangent, therefore also MP is. But then $\angle MPN = \angle MBP$, MPN and MBP are similar, and $\frac{PN}{MN}= \frac{PB}{MP}= \frac{PA}{MP}= 2$.

Remark: you can easily show that N is a Brodcard point of the isosceles triangle PAB. We conclude: a cevian of a Brodcard point of an isosceles triangle is a median, isn't it?

indeed, you are right, but, oh my god, what a simple solution, mine is longer with a complicated angle chasing and a construction, congrats.

[quote="Virgil Nicula"][color=darkred][b][u]Generalization I.[/u][/b] Let $w$ be a circumcircle of the triangle $AB_{1}B_{2}$. Denote : the intersection $P\in AA\cap B_{1}B_{2}$ ; the middlepoint $M$ of the segment $[AP]$ ; the second intersection $N_{1}$, $N_{2}$ of the circle $w$ with the lines $MB_{1}$, $MB_{2}$ respectively. Prove that $\boxed{PN_{1}\cdot PN_{2}=4\cdot MN_{1}\cdot MN_{2}}$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] $MP^{2}=MA^{2}=MB_{1}\cdot MN_{1}=MB_{2}\cdot MN_{2}$ $\Longrightarrow$ $\left\{\begin{array}{ccccc}MP^{2}=MB_{1}\cdot MN_{1}& \Longrightarrow & \triangle N_{1}MP\sim\triangle PMB_{1}& \Longrightarrow & \frac{PN_{1}}{MN_{1}}=\frac{B_{1}P}{MP}\\\\ MP^{2}=MB_{2}\cdot MN_{2}& \Longrightarrow & \triangle N_{2}MP\sim\triangle PMB_{2}& \Longrightarrow & \frac{PN_{2}}{MN_{2}}=\frac{B_{2}P}{MP}\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}\frac{PN_{1}\cdot PN_{2}}{MN_{1}\cdot MN_{2}}=\frac{PB_{1}\cdot PB_{2}}{MP^{2}}\\\\ PB_{1}\cdot PB_{2}=PA^{2}=4\cdot MP^{2}\end{array}\right\|$ $\Longrightarrow$ $\boxed{PN_{1}\cdot PN_{2}=4\cdot MN_{1}\cdot MN_{2}}$.

[b]Remark.[/b] If $B_{1}\equiv B_{2}\equiv B$, i.e. $P\in BB$ then $N_{1}\equiv N_{2}\equiv N$ and obtain $PN=2\cdot MN$ - the [u]Pohoatza's proposed problem[/u].[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Generalization II.[/u][/b] Let $w=C(O,R)$ be a circumcircle of the convex quadrilateral $A_{1}A_{2}B_{1}B_{2}$. Denote : the intersection $P\in A_{1}A_{2}\cap B_{1}B_{2}$ ; the point $M\in \overline{A_{1}A_{2}P}$ for which $MO^{2}-MP^{2}=R^{2}$ ; the middlepoint $R$ of the side $[A_{1}A_{2}]$ ; the second intersections $N_{1}$, $N_{2}$ of the circle $w$ with the lines $MB_{1}$, $MB_{2}$ respectively. Prove that $\boxed{\frac{PN_{1}\cdot PN_{2}}{MN_{1}\cdot MN_{2}}=2\cdot\frac{PR}{PM}}$.[/color][/quote]
[color=darkblue][b]Remark.[/b] If $A_{1}\equiv A_{2}\equiv R\equiv A$, then $PR=PA$ and obtain Generalization I.[/color]

Excuse me guys, but we shouldn't discuss the current issue's problems until their solutions are published in the next issue, should we? :maybe:


Prove that:

  p is [b]prime[/b] $\Longrightarrow$ one only of $(p-1)$, $p$ or $(p+1)$  $\in$ $2N+3N$

[quote="ciprian"]What do you mean by 2N+3N? Every number grater than 5 can be written in the form 2a+3b with a,b positive integers...Maybe i don't understand the question... :?[/quote]

$2N+3N$ is the set of numers written $2a+3b$ with a and b non zero positive intergers.


1) Would you trade some of your mathematical skills for other valuable things in life, for example: social skills, more closer friends, being less introvert, being more involved in non-mathematical discussions, doing more sports, finding a significant other, etc...

2) Why not? /  Why don't you?  / Why can't you?

[quote="evilhamster"] My suggestion for rewording the first question would be "How much do you value math? Would you trade it for proficiency in a sport, or in society, or in another field, etc..." or "Do you think your math skills hogged too much time, and that you have neglected other aspects of your life?".
[/quote]

Based on what you suggest to trade mathematical capabilities for: Is it possible that by default you assume skilled math students (researchers etc.) are socially inept, quixotic guys locking themselves in their rooms all day?

I agree with orl. There are many AoPSers here that are skillful in mathematics and still keep up with society.

As for your question, I would not trade my mathematical skills because

1. I lack them.

2. I already have all the valuable things you describe  :) .

orl: No, I didn't assume that stereotype, I know plenty of social, very friendly, mathematicians who live a happy life.

However, I'm convinced to be successful in mathematics, as a researcher for example, you need to spend a lot of time on the subject. So you have merely any time left for other things to take up in your life project. Therefore my actual question was: do you regret that you have so little time for the other things in life, which you give too little attention.

Of course, this questions is not representative for everybody. But I think it's a common fact that mathematicians, or scientists in general, get really caught up in their subject, while ignoring the many other great things one can do in life.

n.b. A fellow PhD student wrote in his foreword that his girlfriend was his "favorite waste of time" while working on his thesis. So yeah, I think it's clear that not all researchers have a very broad vision on life, and how one can (and must) live a 'good' life.

[quote="Jacobson"]orl: No, I didn't assume that stereotype, I know plenty of social, very friendly, mathematicians who live a happy life.[/quote]Yes, you did. Your post is simply begging the question.

[quote="Jacobson"]However, I'm convinced to be successful in mathematics, as a researcher for example, you need to spend a lot of time on the subject. So you have merely any time left for other things to take up in your life project. Therefore my actual question was: do you regret that you have so little time for the other things in life, which you give too little attention.[/quote]You can replace "mathematics" with pretty much anything. To be really successful, you have to input time. I don't think there is a correlation there. And I think especially that the people who post on AoPS tend to be the most social of the bunch, so you're really talking to the wrong crowd here.

[quote="Jacobson"]n.b. A fellow PhD student wrote in his foreword that his girlfriend was his "favorite waste of time" while working on his thesis. So yeah, I think it's clear that not all researchers have a very broad vision on life, and how one can (and must) live a 'good' life.[/quote]Not all researchers have a very broad vision on life, just as not all musicians, politicians, doctors, lawyers, etc. have a very broad vision on life.

:rotfl: 2 assumes the answer to 1 is no

I value mathematical skill, but I am extroverted. Friends are more than half an hour away, and I see some of them once a week and others twice. Sports won't matter a nanotube in a few years probably, I'd like to be an interesting guy who participates in non-mathematical discussions, and I'm planning on [i]not[/i] finding a significant other. All in all, some mathematics for some social skill and some more of being involved in mathematical discussions would be in order.

I'm applying to a boarding school, so yay I might get a chance! :D
Of not losing mathematical skill and gaining all the other things(except a significant other).

I would like to contribute to the discussion, but I don't have personal experience. I myself am an introvert, though that is by nature, not because of personal activities. But back to the discussion:
I think the thread is drifting off. The point was to discuss the question, not the outlook of people on life. My suggestion for rewording the first question would be "How much do you value math? Would you trade it for proficiency in a sport, or in society, or in another field, etc..." or "Do you think your math skills hogged too much time, and that you have neglected other aspects of your life?" This probably gets to the heart of the question and avoids this kind of argument.
But the current discussion also merits attention:
[quote="worthawholebean"]And I think especially that the people who post on AoPS tend to be the most social of the bunch, so you're really talking to the wrong crowd here. [/quote]
Well, on AoPS, it's likely that some people express themselves much more than in real life, for two reasons:
1. You're anonymous, so you don't have to be haunted by what you say
2. You are communicating with people who share similar interests. This tends to form, at the very least, some amiability.

And my answers:
1. No.
2. Because I don't value social skills and sports that much.

Sorry if my post is rather disjointed and hypocritical. I just had to say everything, and I didn't know how to express it smoothly.

[quote="evilhamster"]Well, on AoPS, it's likely that some people express themselves much more than in real life.[/quote]
Very true. I, even being not anonymous find it more comfortable to express more.

Sure, I'd trade my math-[i]specific [/i]skills for many things. It would be fun. If you think about it, math is just as arbitrary an interest as poetry or pottery or psychology.

But much of what I actually use for doing math is not really "math skills" but "reasoning skills". Logic, analytical thinking, problem solving, pattern recognition, drawing analogies, spatial visualization, concise communication in proof-writing, etc. And these aren't things I'd trade. I'd say that these skills have a positive impact on all the alternatives you mentioned -- social skills, friendship, varied interests, sports, love...

For example, just like if someone has a good feel for mathematical concepts, then they can apply logic, analytical thinking, problem solving, and all those other general skills very fluidly, someone who has developed a good understanding of human personalities can apply these reasoning skills to become a great people-person. The ability to recognize and extrapolate patterns and quickly analyze observations can, for example, give you a great read on new people you meet, which is a very helpful life skill.

I don't spend that much time reading books to learn math. I sort of knock around and absorb new material as it comes up. So a lot of the time I spend practicing math goes more towards developing those general skills. I don't think that time is wasted, even outside of the context of mathematics.

[quote="TZF"]Sure, I'd trade my math-[i]specific [/i]skills for many things. It would be fun. If you think about it, math is just as arbitrary an interest as poetry or pottery or psychology.
[/quote]

 :o  :o  :o I could write essays for the rest of my life trying to refute this statement, but that's getting philosophical, besides the point, and maybe biased (but probably not  :lol: ).

I have a hard time imagining someone [i]wanting[/i] to change their inherent desires. "You like apples better than oranges. Would you like to switch that around so that you would like oranges better than apples?" is not so different from "You're an introvert. Would you like to be an extrovert?"

First, the nitpicking part:
Fruits are foodstuffs, and thus essential to our health, and thus more important than introversion and extroversion. In addition, apples and oranges have different nutrients, and thus a preference could entail a loss of some nutrients and a gain of others, which would have very real consequences on your health.

Second, I'm going to try to refute the idea of extroverts being fundamentally happier. Introverts are definitely the minority, but that's because of the social nature of society. It is definitely possible to live as an introvert and still have a happy life. Besides, introverts are less dependent on external factors because they isolate themselves from outward influences. Everybody has their advantages and disadvantages. Finally, introverts are able to work out fundamental problems with themselves, because they think about themselves.
And then there's the loneliness of power. When you're at the top, there's no one with you. Introverts don't have to adjust.
Mind you, I'm comparing pure introversion and extroversion. Everyone needs to be somewhere in the middle. Confucius preached moderation; this is a good example. You may argue that "But I do take time for myself." Well that's introversion.

And from a purely biological point of view, it's interesting to note how many species are loners-much more than the socialites. This means that self-reliance tends to be evolutionarily better because more species survived.

Speaking of fruits, if you want a more relevant and meanignful example of someone feeling uncomfortable with their desires, think about homosexuals. Society makes them feel like their desires are inferior and immoral. Similarly, I'm sure some math-oriented youths are sometimes made to feel like their interests and priorities are wrong.

Having said that, I'm going to throw out some concerns on the various posts above mine; mainly, the above three.

First, there is no fitness advantage in being an introvert rather than an extrovert. In today's society, it's not like all the introverts (say, software engineers, for a concrete image) are unable to form a family and raise kids.

Second, nobody in this thread says that extroverts are happier than introverts. In fact, people who have mentioned happiness have said something along the lines of, "Hey, I'm a happy introvert," and not in a defensive way.

Third, introversion vs extroversion is not really about how much time you leave for yourself. It's about how comfortable and effective you are in a social setting, and how important social dealings are to you and to your life. Personally, I don't think that introversion and extroversion are mutually exclusive, or that the sum of their levels is invariant. If you choose to connect more with yourself, you are becoming more introverted, and if you choose to connect more with others in a social environment, you are becoming more extroverted. You can do both at the same time.

Fourth, there are plenty of people at the top to chill with if you're in power.

Fifth, who says that species of loners tend to do better evolutionarily? I see animals in groups all the time. Most importantly, the species with the highest fitness level, namely humans, is very social. Even introverts interact a lot with society.

Finally, TPAM used the word "socialite" incorrectly.

[quote="jmerry"]Are you unhappy with yourself? Do you want to be a completely different person?
That's how these questions read to me, and I have to say no.

I'm a very strange person for this, but I generally don't want. I embody that introverted, asocial stereotype - and I'm happy. I don't have a significant other. I've never had one, never looked for one, and never wanted one.[/quote]
How do u know u would not be more happy if you were more social/married? It is something to atleast try i think.

1) First of all, I don't even know what "extroverted" means.  :|

2) Secondly, I have no math skills whatsoever. (I got a 10 out of 30 on the MathCounts school sprint.)  :(

3) And finally, I am totally and completely satisfied with my life right now.  :cool:

although if anything, I might trade the little math talent I have for maybe, more technology in my house? I dunno, I'm kinda a tech freak even though my parents don't like technology. (I'm also a Japan Fan even though my parents hate the topic.) :D

Besides, sometimes it's good to be dumb.  :wink:

The personality terms are discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=88374]here[/url]. Introvert/extrovert is the scale for how much you like social interaction.

Now, I would like to bring my insane perspective to this. I have followed this thread from the beginning, with several different reactions- including that I was somewhat offended by the evolutionary tangent. You talk of being comfortable or not with desires- but what about a lack of desires? Can you even imagine that state?

[quote="jmerry"] but what about a lack of desires? Can you even imagine that state?[/quote]
Yes, very easily :P:(. That is a state that one has to fight sometimes. Its great advantage is that you can consciously choose what you would like to want but its great disadvantage is that it is quite hard to make the first step towards what you chose from it. Actually, I wish that all people would come to that state [i]once in a while[/i], with all ambitions, lust, envy, etc. going away, but only once in a while ;).

Now talking about desires, Buddhism teaches the 4 Noble Truths, one being that to end suffering you must end desires for temporary earthly pleasures.

Unfortunately, I don't think I could make up my mind about the trade-off.  Also, these things can't really be quantified either.  I would probably have to experience the other side before making a decision, living the life of some1 less proficient at math but maybe more popular/better at sports/etc.

As a "math geek" I realize that I don't represent the normal segment of the population, but I never feel like what I'm doing is wrong/desiring math is bad.

also in buddhism they teach moderation is key. Having only math in your life is an empty life in my POV.

[quote="7h3.D3m0n.117"]There's one flaw in your argument, leoxnlin. Fruits have no meaning in society other than for consumption and health. Introverts and extroverts are part of society, with the extroverts with the clear biological selective advantage.[/quote]

I think before you make such a bold claim you first should be reading on things like evolutionary psychology a bit more whether your statement is true for any arbitrary mate:

"[i]...If extraverts attract more potential mates, do they also enjoy higher reproductive success than introverts? If so, has there been consistent positive selection for extraversion? Our trade-off reasoning suggests not—if extraversion had mating benefits and no costs, we might all have evolved to be highly extraverted, and extraversion would no longer be variable or heritable across individuals. Indeed, extraversion does have costs: extraverts, through their risk-taking behaviors, suffer more accidents and hospitalizations (Nettle, 2005), and their promiscuity may undermine long-term relationships, reducing the parental investment available to their offspring. For example, extraverted women more often end up having their children raised with stepfathers (Nettle, 2005), who much more often neglect, abuse, and even kill their stepchildren (Daly & Wilson, 1985). Thus, extraverts may live fast and die young; introverts may live longer and stay married longer, to the benefit of their children. The continuing variability and heritability of the extraversion dimension suggests some kind of balance between such costs and benefits...[/i]" ([b]From: Nettle, D. and Clegg, H. (2007). Personality, mating strategies and mating intelligence. In Geher, G. (ed)., Mating Intelligence: Theoretical and Empirical Applications. New Jersey: Erlbaum.[/b])

I'm gonna answer the original questions instead of commenting on some of the other interesting posts.

I do wish I could trade some of my math skills for other things.

I don't do much math anymore, though I'm studying it in university, it's not the same as when I was in high school working on improving my skills.  I actually do wish I could have traded some of what I did for something else, in particular, music.  I discovered music late in my life, and if I had more time for it in high school, I am confident I could have at least competed at Nationals, as opposed to just making it to provincial competitions.

And now, at this point in my life, my mathematical interests have taken a very different turn from what I used to do.  So that extra work I did do in the past...could have gone towards something else since it hasn't benefited me.  However, I don't regret it...though more in a social than academic aspect.  I met some amazing people along the way, though we are still good friends not because of the math that introduced us, but because of other ties we found as a result.  I'm extremely happy with my life and work, especially my boyfriend.  If only I had a little more time for music =)


I believe that it is still pacific time just like last year. Otherwise, some of my submissions would have been late.

[quote="nameless4365"]If the postmark deadline is the 9th, then emailed submissions are due at midnight that night. correct? using what time though, Pacific daylight, or eastern daylight, or something else?[/quote]

Pacific.


algebra
polynomial
Vieta
algebra unsolved

[quote="makar"]Problem 2:
Let $ a,b,c$ and $ d$ be any four real numbers, not all equal to zero. Prove that the roots of the polynomial $ f(x) \equal{} x^{6} \plus{} ax^{3} \plus{} bx^{2} \plus{} cx \plus{} d$ can't all be real.[/quote]

If $ f(x)$ has 6 real roots, $ f'(x)$ has 5, $ f^{(2)}(x)$ has 4, and $ f^{(3)}(x)$ has 3 real roots.

$ f^{(3)}(x)\equal{}120x^3\plus{}6a$ can have 3 real roots only if $ a\equal{}0$
Then $ f^{(2)}(x)\equal{}30x^4\plus{}2b$ can have 4 real roots only if $ b\equal{}0$
Then $ f'(x)\equal{}6x^5\plus{}c$ can have 5 real roots only if $ c\equal{}0$
But then $ f(x)\equal{}x^6\plus{}d$ can have 6 real roots only if $ d\equal{}0$

So $ f(x)$ can have  6 real roots only if $ a\equal{}b\equal{}c\equal{}d\equal{}0$
Hence the result.

What about $ (x \minus{} 1)^6$? Or what if we just multiply $ (x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)(x\minus{}6)$?

[quote="dgreenb801"]What about $ (x \minus{} 1)^6$? Or what if we just multiply $ (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)(x \minus{} 5)(x \minus{} 6)$?[/quote]

In the original problem, the key is that there are no $ x^5$ and $ x^4$ terms.

In your two examples, we have such terms.

[quote="pco"][quote="earldbest"][quote="pco"]
If $ f(x)$ has 6 real roots, $ f'(x)$ has 5, $ f^{(2)}(x)$ has 4, and $ f^{(3)}(x)$ has 3 real roots.

[/quote]

why?  :blush:[/quote]

There is always at least a zero of $ f'(x)$ between two zeroes of $ f(x)$ (Rolle's theorem)[/quote]

oh, thx :D

[b]my solution[/b]

suppose all the roots are real.

let $ r_i$ $ (i \equal{} 1,2,...,6)$ be the roots.

then
\[ \sum_{i \equal{} 1}^{6} r_{i} \equal{} 0\]

\[ \sum_{1 \le i < j \le 6} r_{i}r_{j} \equal{} 0\]
combining these two yields $ \sum_{i \equal{} 1}^{6}r_{i}^{2} \equal{} 0$ :pilot:

[quote="earldbest"][/quote][quote="pco"][quote="earldbest"][quote="pco"]
If $ f(x)$ has 6 real roots, $ f'(x)$ has 5, $ f^{(2)}(x)$ has 4, and $ f^{(3)}(x)$ has 3 real roots.

[/quote]

why?  :blush:[/quote]

There is always at least a zero of $ f'(x)$ between two zeroes of $ f(x)$ (Rolle's theorem)[/quote][quote="earldbest"]

oh, thx :D

[b]my solution[/b]

suppose all the roots are real.

let $ r_i$ $ (i = 1,2,...,6)$ be the roots.

then
\[ \sum_{i = 1}^{6} r_{i} = 0\]

\[ \sum_{1 \le i < j \le 6} r_{i}r_{j} = 0\]
combining these two yields $ \sum_{i = 1}^{6}r_{i}^{2} = 0$ :pilot:[/quote]
why?  :blush:

[quote="zool007"][quote="earldbest"]
[b]my solution[/b]

suppose all the roots are real.

let $ r_i$ $ (i \equal{} 1,2,...,6)$ be the roots.

then
\[ \sum_{i \equal{} 1}^{6} r_{i} \equal{} 0\]

\[ \sum_{1 \le i < j \le 6} r_{i}r_{j} \equal{} 0\]
combining these two yields $ \sum_{i \equal{} 1}^{6}r_{i}^{2} \equal{} 0$ :pilot:[/quote]
why?  :blush:[/quote]

by Vieta's relations, and the fact that the square of a real number is non-negative. :pilot:

Assuming that all the roots of $f(x)$ are real. Call them $\alpha_1,\alpha_2\dots , \alpha_6$.
By Vieta's relations on $f(x)$, we get that \( \sum_{i = 1}^{6} \alpha_{i} = 0\). Also, \(\sum_{1 \le i < j \le 6} \alpha_{i}\alpha_{j} = 0\).

Now, we know that \( {(\sum_{i = 1}^{6} \alpha_{i})}^2\) = \( \sum_{i = 1}^{6} \alpha_{i}^2 + 2(\sum_{1 \le i < j \le 6} \alpha_{i}\alpha_{j})\)
\(\Rightarrow  \sum_{i = 1}^{6} \alpha_{i}^2 = 0\). This is only possible when $\alpha_i=0$ for all $i \in \{1,2,3,4,5,6\}$. But if that is the case, we get $a=b=c=d=0$, again by Vieta's relations. As this contradicts the problem, our assumption was wrong. Thus, all roots of $f(x)$ cannot be real. $\boxed{Q.E.D!}$


On a table there is a pile with $ T$ tokens which incrementally shall be converted into piles with three tokens each. Each step is constituted of selecting one pile removing one of its tokens. And then the remaining pile is separated into two piles. Is there a sequence of steps that can accomplish this process?

a.) $ T \equal{} 1000$ (Cono Sur)
b.) $ T \equal{} 2001$ (BWM)

Let n, m be the number of disks and the number of piles at a moment.
Each movement will rduce n by one, but increase m by one.
So n+m = 1000 + 1 = 1001
Since there are no empty piles, when only 3-disk piles exist:
3m + m = 1001
4m = 1001
Which is contradictory.

Similarly, since 2002 is not a multiple of 4, point b) is solved.

The same solution applies to point B (edited in the above.)

Another solution is possible.We prove that all $n$ for which it is possible to make all piles of size $3$ are of the form $n=4k+3$.The base case is $k=0$.Also,we have for our base that it is not possible for $n=0,1,2,4$.Now,we prove this claim by induction.Suppose that it is possible for some pile.Let its size be $n$.Denote the size of $2$ piles in which it is separated by $p,q$.We have by our base that $p$ and $q$ are of the form $4k+3$,so $s=p+q+1$ is also of the form $4k+3$,so we are done.


pigeonhole principle
induction
number theory proposed
number theory

Show that there is a unique positive integer which consists of the digits $2$ and $5$, having $2005$ digits and divisible by $2^{2005}$.

It is obviosly proved by induction.
1)2|2
Let $2^{n}|A_{n}$, we must find $A_{n+1}=A_{n}(mod \ 2^{n})$. Because it had unique solution $A_{n+1}=A_{n}+c*10^{n}$, were $c=2$ or $c=5$. Obviosly $c=2$ if $2^{n+1}|A_{n}$ else $c=5$.

We use pigeonhole and induction:
Let
\[ S_n \equal{} \{s|s_{10}\text{ consists of only } 2 \text{ and }5\text{ and is }n\text{ digits long}\}\]
Consider $ a,b\in S_{2005}$ that are equal $ \mod{2^{2005}}$. Then consider $ a \minus{} b\mod{10}$; it must be $ 0$ or $ \pm3$. Obviously the latter case fails. So, $ a \minus{} b$ can be written as $ 5$ times a multiple of $ 2^{2005}$, that is $ 10$ times a multiple of $ 2^{2004}$. Now we take $ a \minus{} b$, divide by $ 10$, and get a new number that can be written as the difference of some $ a_2,b_2\in S_{2004}$. By induction we continue in this way until we are left with two numbers in $ S_1$ that are congruent mod $ 2$. Thus the two numbers are equal, either both 2 or both 5, and the two numbers where forced to be equal at every step along the induction. Hence $ S_{n}$ consists of the entire residue class $ \mod{2^n}$. In particular, it contains the class 0, and we are done.

Well, here 2 and 5 are dummy—they can be replaced by an odd number and an even number (except 0).


Hi, I have kind of a silly question.
When I click on other people's name, I see their number of posts and their number of posts each day

Is there a way to see that window of myself, logging out does not do the trick, and creating a second account is a bit overkill for something that silly

If not possible, please be a dear and tell me my statistics :blush:

Doesn't just clicking on your username work?  (It does for me, even when I am logged in.)

[quote="rrusczyk"]It does for me, even when I am logged in.[/quote]It will only work when you are logged in :)

In plus, on all pages you posted you will see the stats for yourself, just as you see other people's stats :D


calculus
integration
calculus computations

$ \ldots \equal{} \int_0^1 \frac{t^4 \plus{} t^2}{(t^2 \plus{} 1)^2} \, dx \minus{} \int_0^1 \frac{t^2}{(t^2 \plus{} 1)^2} \, dx$
The first one is simply
$ \int_0^1 \frac{t^2 \, dt}{t^2 \plus{} 1}$
And now you can easily integrate both interals by parts taking u=t...
Answer is $ \frac{1}{8} (10 \minus{} 3 \pi)$.


the tens digit of a number is twice the units digit.
if the digits are reversed, the new number is 27 less than the original
what is the original number??

[hide] Let our number be $ 10x\plus{}y$. Then,
$ x\equal{}2y$
$ 10x\plus{}y\equal{}10y\plus{}x\plus{}27$

Solving the second for $ y$, (we substituted the first equationinto the second), we find that $ y\equal{}3$. Then $ x\equal{}6$ and our number is $ \boxed{63}$.[/hide]

[quote="yildobekir"]$ (63)$[/quote]
This is the only right one because the tens digit must be twice the ones digit.


sin x + cos x =a
sin^3 x + cos^3 x=b

GIVEN:-

w*a^3 + u*b + v*a=0

Find the value of w^3 + u^3 +v^3.

[hide]Let $ \sin(x)=p$, $ \cos(x)=q$. Then we are given:\begin{align}p+q&=a\\p^2+q^2&=1\\p^3+q^3&=b\end{align}

Now the final given states, expanded, \[ w(p^3+q^3+3pq(p+q))+u(p^3+q^3)+v(p+q)=0\]

Collecting like terms, we have\[ (p^3+q^3)(w+u)+(p+q)(3wpq+v)=0\]

Now $ p^3+q^3$ has the same sign as $ p+q$, so either $ p+q=0$ or $ w=u$ and $ 3wpq+v=0$. In the first case, satisfied by $ x=135^\circ$, $ w,u,v$ may have any values, so $ w^3+u^3+v^3$ may take any value. In the second, we must have $ pq=0$, satisfied by $ x=180^\circ$ for instance, with $ w^3+u^3+v^3=v^3$, or $ w=u=0$, $ v=0$, for $ 0$ as the sum in question.

Here ends this ugliness.[/hide]

[quote="solafidefarms"]Now $ p^3 \plus{} q^3$ has the same sign as $ p \plus{} q$, so either $ p \plus{} q \equal{} 0$ or $ w \equal{} u$ and $ 3wpq \plus{} v \equal{} 0$.[/quote]

Why does $ p^3\plus{}q^3$ and $ p\plus{}q$ having the same sign imply $ p\plus{}q\equal{}0$ or $ w\equal{}u$ and $ 3wpq\plus{}v\equal{}0$? All I see is that either $ p\plus{}q\equal{}0$ or $ (p^2\minus{}pq\plus{}q^2)(w\plus{}u)\plus{}3wpq\plus{}v\equal{}0$. And in the second case why must we have $ pq\equal{}0$? Also, why does $ w^3\plus{}u^3\plus{}v^3\equal{}v^3$ imply $ w\equal{}u\equal{}0$, $ v\equal{}0$?

Bah, I should've known that using 'we must have ...' was an indication I was making a mistake. I assumed w,u,v were positive, which isn't true and fails.


the tangent to a circle at P and Q intersect at A and the diameter line BC passes through A. show that A and Q are harmonically separated by the points at which your line is intersected by PB AND PC

Let the intersection of $ PQ$ and $ CA$ be $ F$.note that $ PQ$ is the polar of $ A$ wrt circle with diamater $ BC$.
then $ P(AFBC)\equal{}\minus{}1$.
since four lines $ PA,PB,PF,PC$ intersects the line $ AQ$ , for these points of intersections (let them be $ A,B',Q,C'$) we'll have:
$ P(AQB'C')\equal{}\minus{}1$.
so we r done!


linear algebra
matrix
geometry
geometric transformation
rotation
rotation matrix

so from my understanding the lambda part could be found using the concept (A-lambda I)  but what i cant figure out is the Q part.... it has something to do with projection matrices but i really don't understand this part.... please can someone explain....

for example A=[12
                        24]     <== 2x2 matrix  how do i get the perpendicular eigenvectors......

First let's diagonalize $ A.$ Write  $ A\equal{}PDP^T$ for orthogonal $ P.$

$ P$ is a matrix whose columns consist of eigenvectors of $ A.$ If you also want $ P$ to be orthogonal, you need to normalize those eigenvectors. For eigenspaces of dimension 1, normalization is all you need to do; for eigenspaces of larger dimension, you may need to execute Gram-Schmidt within that eigenspace.

$ A \equal{} \begin{bmatrix}1 & 2 \\
2 & 4\end{bmatrix}.$ This is a real symmetric matrix, which guarantees that it is orthogonally diagonalizable.

I can see at a glance that it is singular - after all, the second row is twice the first row. So $ 0$ is one eigenvalue. The trace of $ A$ is $ 5,$ so the other eigenvalue is $ 5.$

For each $ \lambda,$ write $ A \minus{} \lambda I,$ then row eliminate to find the row space of that.

$ \lambda \equal{} 0:$ $ A \minus{} 0I \equal{} \begin{bmatrix}1 & 2 \\
2 & 4\end{bmatrix}\sim \begin{bmatrix}1 & 2 \\
0 & 0\end{bmatrix}.$ Eigenvector $ \begin{bmatrix}2 \\
\minus{} 1\end{bmatrix}.$

$ \lambda \equal{} 5:$ $ A \minus{} 5I \equal{} \begin{bmatrix} \minus{} 4 & 2 \\
2 & \minus{} 1\end{bmatrix}\sim \begin{bmatrix} \minus{} 2 & 1 \\
0 & 0\end{bmatrix}.$ Eigenvector $ \begin{bmatrix}1 \\
2\end{bmatrix}.$

We do need to normalize both of those eigenvectors. By coincidence, each has to be divided by the same amount, $ \sqrt {5}$ So we have

$ P \equal{} \frac1{\sqrt {5}}\begin{bmatrix}2 & 1 \\
\minus{} 1 & 2\end{bmatrix}.$

$ P$ is a rotation matrix - rotation by $ \arctan\left( \minus{} \frac12\right).$

$ P$ is not unique; I had choices available about both order and sign.

Now: We want to write $ A \equal{} PDP^T \equal{} QQ^T.$ We can only do that if $ A$ is positive semidefinite - that is, if all of the eigenvalues of $ A$ are nonnnegative. That is the case here. With that being the case, we can readily take the square root of $ D.$

Let $ Q \equal{} P(D^{1/2}).$ Then $ Q^T \equal{} D^{1/2}P^T$ and $ QQ^T \equal{} A.$

In this case,

$ Q \equal{} P(D^{1/2}) \equal{} \frac1{\sqrt {5}}\begin{bmatrix}2 & 1 \\
\minus{} 1 & 2\end{bmatrix}\begin{bmatrix}0 & 0 \\
0 & \sqrt {5}\end{bmatrix} \equal{} \begin{bmatrix}0 & 1 \\
0 & 2\end{bmatrix}.$

Check it out: $ \begin{bmatrix}0 & 1 \\
0 & 2\end{bmatrix}\begin{bmatrix}0 & 0 \\
1 & 2\end{bmatrix} \equal{} \begin{bmatrix}1 & 2 \\
2 & 4\end{bmatrix},$ as advertised.

===========

That was inefficient. You don't really have to diagonalize $ A.$ I got started diagonalizing $ A$ because I started working the wrong problem. Why was I working the wrong problem? Because you used letters against type.

The letter $ Q$ is in fact most commonly used for an orthogonal matrix - that is a matrix such that $ Q^TQ \equal{} I.$ So your use of $ Q$ to mean a matrix that cannot be orthogonal was something unexpected to my eyes.

============

Edit: for efficiency, see the [url=http://en.wikipedia.org/wiki/Cholesky_decomposition]Cholesky Decomposition.[/url]


How many subsets of the set $ {1,2,3,4,\ldots ,30}$ have the property that the sum of the elements of the subset is greater than $ 232$?

(Sorry if this is too easy or hard.  :oops: )

This is from Zeitz.  Problems from books like that tend to get posted a lot.  Here are a few where this has been posted:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1752077506&t=247829
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1752077506&t=196271
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1752077506&t=142450
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1752077506&t=108777

(I found these by searching for the words "subset 232" without quotes.)

Sorry, I wasn't sure what keywords to search.  :oops:


[quote="ch1n353ch3s54a1l"][quote="Silverfalcon"]First $a$ is chosen at random from the set $\{1,2,3,...,99,100\},$ and then $b$ is chosen at random from the same set. The probability that the integer $3^a + 7^b$ has units digit 8 is

$\text{(A)} \ \frac{1}{16} \qquad \text{(B)} \ \frac{1}{8} \qquad \text{(C)} \ \frac{3}{16} \qquad \text{(D)} \ \frac{1}{5} \qquad \text{(E)} \ \frac{1}{4}$[/quote]

[hide]i get (B) $\frac{1}{8}$
since $3^a$ has units digit 3,9,7,1 etc. and $7^b$ has units digit 7,9,3,1 etc. there are $4\times4$ different unit sums and two ways to choose 8: 1+7 or 7+1. $\frac{2}{16}=\frac{1}{8}$[/hide][/quote]

You missed
[hide]9+9.[/hide]

[hide="Answer"]Both cycle every 4 numbers. If $a=4k+1$, then we have no solutions. If $a=4k+2$, then we have $b=4m+2$, so a total $\frac{1}{16}$ chance. If $a=4k+3$, then $b=4m$, so a total $\frac{1}{16}$ chance. Finally, if $a=4k$, then $b=4m+1$, a total $\frac{1}{16}$ chance. Therefore, our answer is $\frac{3}{16}\Rightarrow \boxed{C}$.[/hide]


inequalities
algebra
polynomial
LaTeX
inequalities solved
3-variable inequality

Let $x, y, z$ be real numbers satisfying  $x+y+z = 5$ and $x^2+y^2+z^2=9$. Find the minimum and maximum of $P = x^2y + y^2z + z^2x$.

Well, without the constraint x2 + y2 + z2 = 9, assuming x + y + z = 5,
the maximum would be (5/3)(10/3)2 + (10/3)(0)2 + (0)(5/3)2 = 500/27.
Right now, I'm thinking of what to do with the condition x2 + y2 + z2 = 9.

Splendid and very difficult problem.
Have you a solution, Namdung?

Thank you very much.

Firstly, for an easier computation, we put
     $ x = \frac53 + a, y = \frac53 + b, z = \frac53 + c$
then
    $ a + b + c = 0, a^2 + b^2 + c^2 = \frac23  \ \ \ (1)$
and
    $ P = x^2y + y^2z + z^2x = a^2b + b^2c + c^2a + \frac {125}{9}$

So, we have to find maximum and minimum of  $ S = a^2b + b^2c + c^2a$ with condition (1)

From (1), we have $a+b+c=0, ab+bc+ca=-\frac13$. If we put $abc = t$  then all symmetric polynomial of $(a, b, c)$ can be expressed via $t$. $S$ is not symmetric, but if we take $T = ab^2+bc^2 + ca^2$ then $S+T, ST$ are symmetric. Easily to find that:
\[ S + T = - 3t, \ \ \ \  ST = 9t^2 - \frac { 1}{27}\]
From that, we can compute that
\[S = \frac {3t \pm \sqrt {\frac4{27} - 27t^2}}{ 2}\]
Now, for example, we try to find the maximum
By Cauchy-Schwarz:
\begin{align*}3t + \sqrt {\frac4{27} - 27t^2} & = \frac {1}{\sqrt {3} }(3\sqrt {3}t) + \sqrt {\frac4{27} - 27t^2} \\ & \le \sqrt {\left(\frac13 + 1\right)\left(27t^2 + \frac4{27} - 27t^2\right)} = \frac49\end{align*}
The equality holds when  $ t = \frac1{27}$. So $ \max S = \frac29$ and $ \max P = \frac{127}9$.

Similiarly we can find that $ \min P = \frac{41}3.$

Splendid and very difficult solution.

Thank you, Namdung.

[quote="Namdung"]The main idea here (I really like it) is to make the symmetry! Firstly, for an easier computation, we put
     $ x \equal{} 5/3 \plus{} a, y \equal{} 5/3 \plus{} b, z \equal{} 5/3 \plus{} c$
then
    $ a \plus{} b \plus{} c \equal{} 0, a^2 \plus{} b^2 \plus{} c^2 \equal{} 2/3 (1)$
and
    $ P \equal{} x^2y \plus{} y^2z \plus{} z^2x \equal{} a^2.b \plus{} b^2.c \plus{} c^2.a \plus{} \frac {125}{9}$

So, we have to find maximum and minimum of  $ S \equal{} a^2.b \plus{} b^2.c \plus{} c^2.a$ with condition (1)

From (1), we have a+b+c=0, ab+bc+ca=-1/3. If we put abc = t  then all symmetric polynomial of (a, b, c) can be expressed via t. [b]S is not symmetric, but if we take T = a.b^2+b.c^2 + c.a^2 then S+T, S.T are symmetric.[/b] Easily to find that:

       $ S \plus{} T \equal{} \minus{} 3t, S.T \equal{} 9t^2 \minus{} \frac { 1}{27}$
From that, we can compute that
      $ S \equal{} \frac {3t \pm \sqrt {4/27 \minus{} 27t^2}}{ 2}$
Now, for example, we try to find the maximum

By Cauchy-Schwartz   :

$ 3t \plus{} \sqrt {4/27 \minus{} 27t^2} \equal{} \frac {1}{\sqrt {3} }(3\sqrt {3}.t) \plus{} 1.\sqrt {4/3 \minus{} 27t^2} \le \sqrt {(1/3 \plus{} 1)(27t^2 \plus{} 4/27 \minus{} 27t^2)} \equal{} 4/9$.

 The equality holds when  $ t \equal{} 1/27$. So $ maxS \equal{} 2/9$ and $ maxP \equal{} 127/9$.

Similiarly we can find that $ minP \equal{} 41/3.$

Have joy

Namdung[/quote]

  [i]since it's a rather good problem , I have edited the post of Namdung in order to be clear to understand

  Ps namdung : Using Latex doesn't take many time . I think that you should use it , Your post always look very bad and ugly , very hard to

understand your idea  :mad:

Due to the fact that you have been advised many times but you continue to be so careless , I have to warn you again .[/i]

[quote="Namdung"]Let x, y, z be real numbers satisfying  x+y+z = 5, x^2+y^2+z^2=9. Find the minimum and maximum of P = x^2*y + y^2*z + z^2*x.[/quote]
What do you think about a nice solution by Cauchy-Schwarz for this problem? Such solution exists, let's try!

[quote="Namdung"]Let x, y, z be real numbers satisfying  x+y+z = 5, x^2+y^2+z^2=9. Find the minimum and maximum of P = x^2*y + y^2*z + z^2*x.[/quote]
mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

I also thought that cauchy might work but I could not get that far.
Here's what I did:
(x^2(y)+y^2(z)+z^2(x))(1/y+1/z+1/x)>=(x+y+z)^2
Clearly minimum is attained when LHS equals to 25.
After expanding we get:
x^2+y^2+z^2+(x^2+y^2)/z+(z^2+y^2)/x+(x^2+z^2)/y+xy+xy+yz+xz>=25
we know that x^2+y^2+z^2=9.
[(x+y+z)^2-x^2-y^2-z^2]*1/2=8=xy+yz+xz.
(x^2+y^2)/z+(z^2+y^2)/x+(x^2+z^2)/y>=8
Now, I am not so sure about what to do..


Aratam ca Z[i\sqrt{3}]/(2+i\sqrt{3}) este corp de unde rezulta 2+i\sqrt{3} e prim. Fie z \in Z[i\sqrt{3}], z=(2+i\sqrt{3})*q+r, N(r)<N(2+i\sqrt{3})=7. Trecem la clasa mod (2+i\sqrt{3}) si avem ca \hat{z}=\hat{(2+i\sqrt{3})*q}+\hat{r}, deci \hat{z}=\hat{r}.Fie r=a+b*i\sqrt{3}, N(r)<7 \leftrightarrow a^2+3b^2<7 deci r\in{0,-1,1,i\sqrt{3},-i\sqrt{3},1+i\sqrt{3},1-i\sqrt{3},-1+i\sqrt{3}, -1-i\sqrt{3} } Se observa ca \hat{1}=\hat{-1-i\sqrt{3}} si \hat{-1}=\hat{1+i\sqrt{3}} deci Z[i\sqrt{3}]/(2+i\sqrt{3})={0,1,-1,i\sqrt{3},-i\sqrt{3},1-i\sqrt{3},-1+i\sqrt{3}} \approx Z7 prin f cu legaturile: f(0)=0, f(1)=1, f(2)=-i\sqrt{3}, f(3)=1-i\sqrt{3}, f(4)=-1+i\sqrt{3}, f(5)=i\sqrt{3}, f(6)=-1. Deci Z[i\sqrt{3}]/(2+i\sqrt{3}) corp.

Hey english please. We don't understand a word of what you've posted.


[quote="Maldo"]:|  man!! are you crazy?!   :lol:[/quote]
   [b]Crazy[/b] ...  If you caculate it , It is very interesting  ....  :play_ball:

a) Call $ D_n$ the determinant (regardless of any property that the $ x_n$ may satisfy).  It follows easily that $ D_n \equal{} 2009^{n \minus{} 1}x_n^2 \plus{} 2009D_{n \minus{} 1}$, by considering the determinant as the sum of two, one of which has last row equal to $ (0,0,\dots,0,2009)$ (contributing $ 2009D_n$ after developing by the last row), and one which has last row equal to $ (x_1x_n,x_2x_n,\dots,x_{n \minus{} 1}x_n,x_n^2)$; in this last determinant, take $ x_n$ as a common factor in this last row out of the determinant, and add $ \minus{}x_k$ times the last row thust obtained to the $ k$-th row ($ k \equal{} 1,2,\dots,n \minus{} 1$), thus obtaining a triangular matrix with zeros above a main diagonal formed by $ n \minus{} 1$ times $ 2009$, and $ x_n$.  Since $ D_n \equal{} 2009^{n \minus{} 1}x_n^2 \plus{} 2009D_{n \minus{} 1}$, and clearly $ D_1 \equal{} x_1^2 \plus{} 2009$, it follows easily by induction that $ D_n \equal{} 2009^n \plus{} 2009^{n \minus{} 1}(x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2)$.  In our case, Cardano-Viete relations ensure that $ x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2 \equal{} 2009^2 \minus{} 2\cdot2008 \equal{} 2009^2 \minus{} 2\cdot2009 \plus{} 2$, or the determinant is $ 2009^{n \minus{} 1}(2009^2 \minus{} 2009 \plus{} 2)$.

b) Same deal, except this time you obtain $ D_n \equal{} 2009^{n \minus{} 1}x_n^3 \plus{} 2009D_{n \minus{} 1}$ because the triangular matrix has zeros above a main diagonal formed by $ n \minus{} 1$ times $ 2009$ and $ x_n^2$.  Thus  $ D_n \equal{} 2009^n \plus{} 2009^{n \minus{} 1}(x_1^3 \plus{} x_2^3 \plus{} \dots \plus{} x_n^3)$.  Now,
\[ (x_1 \plus{} x_2 \plus{} \dots \plus{} x_n)(x_1^2 \plus{} x_2^2 \plus{} \dots \plus{} x_n^2) \minus{} (x_1x_2 \plus{} x_1x_3 \plus{} \dots \plus{} x_{n \minus{} 1}x_n)(x_1 \plus{} x_2 \plus{} \dots \plus{} x_n) \plus{} 3(x_1x_2x_3 \plus{} x_1x_2x_4 \plus{} \dots \plus{} x_{n \minus{} 1}x_{n \minus{} 1}x_n) \equal{} x_1^3 \plus{} x_2^3 \plus{} \dots \plus{} x_n^3,
\]
and by Cardano-Viete relations,
\[ x_1^3 \plus{} x_2^3 \plus{} \dots \plus{} x_n^3 \equal{} 2009(2009^2 \minus{} 2\cdot2008) \minus{} 2008\cdot2009 \plus{} 3\cdot0,
\]
substitute and done, result $ 2009^{n\plus{}2}\minus{}3\cdot2009^{n\plus{}1}\plus{}4\cdot2009^n$.

i repeat it again: man are you crazy     :P


calculus
integration
trigonometry
real analysis
real analysis unsolved

$ \int_{0}^{\infty}\frac {e^{x} \plus{} 2e^{\frac { \minus{} x}{2}}\cos{\frac {\sqrt {3}x}{2}} \minus{} 3x \minus{} 3}{x(e^{x} \minus{} 1)}dx$

[color=green][Edited by moderator. The O.P. should check to see if this is what he meant.][/color]

hello, can you please correct your post?
Sonnhard.