linear algebra
matrix
group theory
superior algebra
superior algebra unsolved

How do I even start to prove that something is a presentation for a group? (in my case, it's showing that $ \{x, y|y^{2}, (xy)^{2}\}$ is a presentation for the infinite dihedral group)

What do I need to do?

That depends a great deal on what you already know about the given group, or how it is described to you. For example, if you know some other finite presentation of the infinite dihedral group, then you just need to show that the two presentations describe the same group, which can be done via Tietze transformations. It's not necessarily easy, but it gives you a concrete object to look for.

If the given group is described by some other means - as a permutation group, matrix group, whatever - then the process may be different. A good start is to identify the generators in the presentation ($ x$ and $ y$ in your case) with some particular elements of the group so that the relations are indeed satisfied. You then have to work out some argument that there is no further "collapsing" of the elements due to the given relations.

In general, of course, the problem is not just "hard", it's downright unsolvable (even deciding if a given presentation describes the trivial group $ G \equal{} \{1\}$ is not algorithmically solvable).

The presentation $ \langle x,y | y^2, (xy)^2 \rangle$ is, as far as I remember, one of the standard [i]definitions [/i]of the infinite dihedral group. Presumably, then, you are being asked to use some other definition of that group - which is it?


inequalities
geometry
perimeter
circumcircle
inequalities unsolved

Let $x,y,z$ be length of the sides of triangle satisfy $x^2+y^2+z^2=3$.
Prove that:
$\displaystyle \sum_{cyc} \sqrt{\frac {x^2}{(2-y^2)(2-z^2)}} \geq x+y+z $
I've a soln for this one. Can you post another soln??

Indeed, a hard problem. This is the same with the following problem of Klamkin:
  In any triangle we have $ (a^2+b^2+c^2)(am_a+bm_b+cm_c)\geq 4m_am_bm_c(a+b+c) $. A tough inequality.

A reference for Klamkin's inequality above :

SIAM review 22 (1980), p.509-511 (problem 79-19)

Pierre.

[quote="harazi"]This is the same with the following problem of Klamkin:
  In any triangle we have $ (a^2 \plus{} b^2 \plus{} c^2)(am_a \plus{} bm_b \plus{} cm_c)\geq 4m_am_bm_c(a \plus{} b \plus{} c)$. A tough inequality.[/quote]

This was discussed on

  http://www.mathlinks.ro/Forum/viewtopic.php?t=6109

This is the thread [i]Advanced Section / Geometry / Solved Problems / perimeter ...[/i].

It is actually equivalent to the fact that the triangle formed by the points (different from A, B, C) where the medians of triangle ABC meet the circumcircle has a perimeter greater or equal to the perimeter of triangle ABC.

Both Fagot and Hxtung gave nice proofs for this fact.

  Darij


Candy sales of Booster Club for January through April are shown.  What were the average sales per month in dollars?

$ \textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$

[hide]

The average is going to be all the candy added up, or $ 320$, divided by the number of months, or $ 4$, for an answer of $ \boxed{\textbf{(D)}\ 80}$[/hide]


vector
algebra
linear equation
linear algebra

could somebody solve this problem?

Given the vectors

        1                      1                        3
A1=  1               A2= 0                 A3= 1
        1                      1                        3

          1
Is  b= 2   in the span of A1, A2, and A3 ?
          0

onse againe A1, A2, A3, and b are vectors
thanks

No, because $A_{1}+2*A_{2}= A_{3}$

We need to see whether b is in the span of $A_{1}and A_{2}$

But in both of them, the 1st and the 3rd elements are the same, and for $b$, it's not. So b cannot be in the span of the three.

(I am sure you can actually solve the linear equation, but.. laziness taking over here  :D )


calculus
integration
calculus computations


calculus
integration
function
geometry
calculus computations

If $ f(a)\equal{}b,f(b)\equal{}a,a<b,f(x)$ is continuous and exists
inverse function $ f^{\minus{}1}$ on $ \left [a,b\right ]$,show that $ \int _a^bf(x)dx$
equals to $ \int _a^b f^{\minus{}1}(x) dx$

 :roll:

if $ f(a)\equal{}b \Rightarrow f^{\minus{}1}(b)\equal{}a$ and we know that$ f(b)\equal{}a \Rightarrow f(b)\equal{}f^{\minus{}1}(b),f(a)\equal{}f^{\minus{}1}(a)$ and assume that :$ \int^b_a f(x)\,dx\equal{}F(b)\minus{}F(a),\int^b_a f^{\minus{}1}(x)\,dx\equal{}G(b)\minus{}G(a)$ and we have $ F'(x)\equal{}f(x) ,G'(x)\equal{}f^{\minus{}1}(x) \Rightarrow F(x)\equal{}G(x)\plus{}C \Rightarrow \int^b_af(x)\,dx\equal{}\int^b_a f^{\minus{}1}(x)\,dx$

[quote="parinaz.sa"]if $ f(a) \equal{} b \Rightarrow f^{ \minus{} 1}(b) \equal{} a$ and we know that$ f(b) \equal{} a \Rightarrow f(b) \equal{} f^{ \minus{} 1}(b),f(a) \equal{} f^{ \minus{} 1}(a)$ and assume that :$ \int^b_a f(x)\,dx \equal{} F(b) \minus{} F(a),\int^b_a f^{ \minus{} 1}(x)\,dx \equal{} G(b) \minus{} G(a)$ and we have $ F'(x) \equal{} f(x) ,G'(x) \equal{} f^{ \minus{} 1}(x) \Rightarrow F(x) \equal{} G(x) \plus{} C \Rightarrow \int^b_af(x)\,dx \equal{} \int^b_a f^{ \minus{} 1}(x)\,dx$[/quote]

When
$ f(b) \equal{} f^{ \minus{} 1}(b),f(a) \equal{} f^{ \minus{} 1}(a)$

How can  we know $ f(x) \equal{} f^{ \minus{} 1}(x),\forall x\in [a,b]$ ?

 :rotfl:

Kent Merryfield and Virgil Nicula separately solved this problem a while ago.

[quote="JoeBlow"]It's essentially just the geometric fact that computing areas by looking at thin vertical slices is the same as computing areas by taking horizontal slices.  Drawing a picture will be very helpful.[/quote]

Would you show your solutions?


There are two ants on opposite corners of a cube. On each move, they can travel along an edge to an adjacent vertex. What is the probability that they both return to their starting point after 4 moves.

Can you fix your typo?  It says after ____ moves.  Whats ___?  you never put a number.

If I understand the question correctly, we can calculate the probability for one ant and then just square it.
[hide]
Let's break the cube into two points and two layers. The starting point is point 1 and the opposite point is point 2. The 3 points adjacent to point 1 form layer 1 and the remaining three points form layer 2.

Starting on point 1, we must first move to layer 1 (3 ways). We can then either return to point 1 (1 way) or go to layer 2 (2 ways).

If we returned to point one, we must go back to layer 1 on turn 3 (3 ways) and then return back (1 way), for a total of 3x1x3x1=9 paths.

If we went on to layer 2, we must go back to layer 1 (2 ways) and then return to point 1, for a total of 3x2x2x1=12 ways.

Thus, there are a total of 21 paths for the ant. Since there are 3 options each move, there are 3^4 = 81 paths, so the probability for one ant is 21/81 = 7/27.

Squaring, the probability for two ants is 49/729.[/hide]


calculus
real analysis
real analysis unsolved

Newbie here and I don't know if this is in the correct forum.

Feel free to move it.

I know perm's and comb's are pre-calculus but here we go...

I was at a trade show yesterday and there was this contest where you can win his and her Ferrari's.  [b]You simply have to guess a 7-digit code.  Numbers can be repeated.[/b]  Of course I lost but I was wondering [b]what are the odds of guessing the code and how can I figure it out.[/b]

I'm at work right now and it's really bugging me.

The first person to reply to this will probably have the answer.

Thanks in advance.

TJ

Mods will move this to an appropriate subforum. And the answer is $ \frac {1}{10^7} \equal{} 1/10000000$. First number can be guessed with chances one to ten you choose one of ten digits), and all others with the same odds. Multiplying it ($ \frac {1}{10}\cdot\frac {1}{10}\cdot\frac {1}{10}\cdot\frac {1}{10}\cdot\frac {1}{10}\cdot\frac {1}{10}\cdot\frac {1}{10}$) you get the answer above.


Let $a,b,c$ be three positive real numbers which satisfy the condition
\[\frac{1}{a^{2}+b+c}+\frac{1}{b^{2}+c+a}+\frac{1}{c^{2}+a+b}\geq 1. \]
Prove that $a+b+c\leq 3$.

and think the constent in the first inequality should be 1.
and if $a+b+c=3$ ,
we know that $f(a)=\frac{1}{a^{2}-a+3}$ is concave if $a<2$
so if $a,b,c<2$ then $f(a)+f(b)+f(c)\leq 3f(\frac13)=1$
if $a>2$ then $f(a)\leq \frac{1}{5}$ and $f(b),f(c)\leq \frac{4}{11}$(because $b^{2}-b\geq-\frac{1}{4}$
so $f(a)+f(b)+f(c)\leq 1$

Sorry, I have edited. In my solution I use a one-line Cauchy. I will post it if nobody writes it.

Nice inequality andiciup . Here is my solution . By Cauchy-Swartz

$(a^{2}+b+c)\geq\frac{(a+b+c)^{2}}{(1+b+c)}$

So
$\frac{2(a+b+c)+3}{(a+b+c)^{2}}\geq 1$ so
$(a+b+c+1)(3-a-b-c)\geq 0$

so $a+b+c\leq 3$

 :lol:


Hello everyone, I'm a budding theoretical physicist and mathematician, all throughout my education I've been taught about mathematical objects, relation between objects, Proofs, etc. Never have I been taught HOW to actually learn math.

I've put together a website with all the lessons on how to learn math your's truly has learned the hard way hopefully it can help someone learn the easy way.

[color=red][url=http://www.geocities.com/deloprator2000/learnmath.html]How to Learn Math[/url][/color]

Please visit my website and if you have any suggestions feel free to sign my guestbook. If you find the website useful please pass it on to someone else

Thank You

John G.

I'm Almost Like Vanilla Ice "If you gotta math problem yo I'll solve it"

Thanks for the site, the resources you mentioned are great. i just want to add one thing that i probally overlooked in your site but just incase.

The Moore Method

 I naturally stumbled on this method back when i was in 9th grade, and it is the sole reason i was able to go from a F in Algebra I to skipping 2 years of math in 50 days. Its not the easiest thing to do but it is rewarding.

http://www.discovery.utexas.edu/rlm/reference/mahavier1.html

p.s.- Mr.Ice i have a few problems for you : http://www.claymath.org/millennium/

Just a suggestion, but I'd recommend adding The Art of Problem Solving to your list of links.  There are clearly some great resources for learning math here.

Thank you for your suggestions, I will place links to them on my website.

John G.


integration
calculus
real analysis
real analysis unsolved

[color=darkblue]Ascertain (over $ \mathcal R$) $ \{\begin{array}{cc}
\blacktriangleright & A_n=\int\frac {x^{3n-1}}{x^{4n}+x^{2n}+1}\ \mathrm {dx}\\\\
\blacktriangleright & B_n=\int\frac {x^{n-1}}{x^{4n}+x^{2n}+1}\ \mathrm {dx}\\\\
\blacktriangleright & C_n=\int\sqrt {\frac {x^n}{1+x^{n+2}}}\ \mathrm {dx}\\\\
\blacktriangleright & \int\frac {2x+3}{x(x+1)(x+2)(x+3)+1}\ \mathrm {dx}\end{array}$, where $ n\in \mathcal N^*\ .$[/color]


inequalities
search
function
inequalities solved

Do not post multiple times.  You only have to post your problem once.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=57683 and http://www.mathlinks.ro/Forum/viewtopic.php?t=57552 (can anyone merge the two threads together?).

Generally, I believe that all the inequalities you have posted were already discussed on MathLinks. Maybe you have them from some person who has them from MathLinks? I advise you to use the [url=http://www.mathlinks.ro/Forum/search.php]search function[/url], it helps you to find proofs much faster than we can post them ;) .

  Darij

Because I am new member, so I don't know the inequalities which I posted  were discussed on MATHLINKS.


function
Asymptote
algebra
rational function

In my algebra textbook, in a section involving the graphing of rational functions, one example gave the function: f(x)= x/ (x^2 - x - 2)
I found out that the y intercept and the x intercept of the graph is (0,0). The two vertical asymptotes are x = -1 and x = 2. The horizontal asymptote is y=0. The graph of that function as shown in my textbook shows three curves, separated by the two vertical asymptotes. But only the far left and far right curves approach the horizontal asymptotes, as the middle curve simply passes through it. How is that possible? Shouldn't the middle curve also approach the horizontal asymptote?

[quote="rkrk"]Shouldn't the middle curve also approach the horizontal asymptote?[/quote]

Asymptotes are generally limit behavior - what the function does as it approaches large negative values or large positive values (that is, non-vertical asymptotes).  What happens in between is not relevant.  For example, you can also find continuous segments of rational functions which pass through an asymptote before approaching it.

But how would I know whether or not if the curve will approach the asymptote or not? For example, the two curves on the ends approach the horizontal asymptote. All the curves belong to the same function, so what is telling me that the middle curve will appear the way that it does, approaching the vertical asymptotes and not the horizontal one. Sorry if my question doesn't make sense or has an obvious answer, but I am pretty new to this concept and am learning this by myself for a summer assignment, so it does seem pretty confusing at first.

[quote="rkrk"]so what is telling me that the middle curve will appear the way that it does, approaching the vertical asymptotes and not the horizontal one.[/quote]

Well, if one of the vertical asymptotes goes to positive infinity and the other to negative infinity, then the middle portion of the curve takes on all real values and hence must pass through any horizontal asymptotes.  Otherwise, only a calculation will tell you for sure, and for purposes of roughly graphing a rational function this is not necessary.

you can set the function equal to the asymptote and see if you get a real solution
(might not work for quintics or higher though)


[quote="chien than"]Let $a;b;c$ be positive real numbers .Find max:
$M=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$[/quote]
Fix $b,c$ , let $a\to 0$. Not exist max.


ok...the problem is $ u_t\plus{}\sqrt{3}u_x\equal{}0$ with $ u(x,o)\equal{}sin ^2(x)$

so $ u\equal{}sin^2(x\minus{}t\sqrt{3})$

the next part asks to skeptch the characteristics in the x--t plane. so would that be the line $ x_0\equal{}x\minus{}t\sqrt{3}$?

Thanks
Jli

For a first-order PDE, the method of characteristics discovers lines (called characteristic lines or characteristics) along which the PDE becomes an ordinary differential equation (ODE). Once the ODE is found, it can be solved along the characteristic lines and transformed into a solution for the original PDE.

~~> http://en.wikipedia.org/wiki/Method_of_characteristics

Depends on how you choose your characteristics.

$ u(x,t)$

Using

$ u(x,t(x))$

Will of course give you characteristics that are lines.

Using

$ u(x(s),t(s))$

Will give you characteristics that are not necessarily lines.

edit: After looking at your link I would like to point out the following. The second choice with getting characteristics that are not necessarily lines is for cases where you do not only have constant coefficients in the PDE. This wiki is only for constant coefficient PDEs.


#1 Show that $f(x)=O(|x-x_{0}|^{2})$ as $x\to x_{0}$ implies $f(x)=o(|x-x_{0}|)$ as $x\to x_{0}$, but give an example to show that the converse is not true.

#2 Show that $f(x) = O(|x-x_{0}|^{k})$ and $g(x)=O(|x-x_{0}|^{k})$ imply that $(f+g)(x) = O(|x-x_{0}|^{k}).$

#3 Show that $f(x)=O(|x-x_{0}|^{k})$ implies $f(x)/(x-x_{0}) =O(|x-x_{0}|^{k-1})$ if $k \geq 1$

As a counterexample for the converse of #1: let $f(x)=|x-x_{0}|^{3/2}.$ The others are similar.

These proofs are largely matters of following the definitions.


geometry
3D geometry
algebra
factorization
sum of cubes

What is the smallest natural number that can be expressed as the sum of two distinct squares, and also two distinct cubes?

do you mean that cannot be expressed as a sum or squares, or cubes?

[quote="Hexaditidom"]No...

Take for example 13. It can be written as $2^2 + 3^2$.
Also, 407 can be written as $4^3 + 7^3$.

What I want is the smallest natural number (not necessarily perfect square or cube) that has both those properties.[/quote]

1
$0^3+1^3=1$ and $0^2+1^2=1$

Okay, [i]second[/i] smallest number...  :roll:

Here is my solution.

[hide="Click"]

The problem asks for the integer $k$ such that $k = m^2+n^2 = o^3+p^3$.

I'm not sure whether you meant by distinct as $m \neq n,o,p$ or just $m \neq n$. If the second one is the case, I have the answer. Fix $m = o = 1$ because $1^2 = 1^3$. Now, we want $n^2 = p^3$. The smallest number that satisifes this is 64 so $k = 1+64 = 65$. [/hide]

If you meant $m \neq n,o,p$, then it's going to take sometime (maybe or maybe not)

I have got a mean question:
How would you [b]prove[/b] that that is the smallest such number?

i think u forgot that 0^2+8^2 and 0^3+4^3 =64.
However i think the second smallest is 0^2+3^2 and 1^3+2^3 = 9.

[quote="Hexaditidom"]Could someone define NATURAL NUMBER for me? I thought it was positive integers, which does not include 0. Maybe I should have said that.[/quote]

According to the problem's wording, only $k$ (look for my post above to see what $k$ is) has to be natural number, which is an integer that is greater than 0. The problem didn't say that the $m,n,o,p$ has to be natural number. In fact, it didn't have any restriction on it (not even integer!).

So, I just assumed $m,n,o,p > 0$ and solved the problem.

[quote="Arne"]How would you prove that that is the smallest such number?[/quote]

Hmm.. Tricky question.. How would I prove that 64 is the minimum of value that there exists $m^2+n^2 = o^3+p^3$.. Hmm... I'm not sure.

Arne, do you have any specific proof in your mind?  :huh:

No, haven't thought about it yet, but I will.

[quote="Silverfalcon"]

[quote="Arne"]How would you prove that that is the smallest such number?[/quote]

Hmm.. Tricky question.. How would I prove that 64 is the minimum of value that there exists $m^2+n^2 = o^3+p^3$.. Hmm... I'm not sure.

Arne, do you have any specific proof in your mind?  :huh:[/quote]

Well, it isn't such a big number, is it? You can try to find which smaller numbers can be written as sum of cubes, then try to write them as sum of squares  :)


What is the sum of the two whole numbers that have a product of 36 and a difference of 5?

We can use guess and check because they're whole numbers.  The numbers are $ 9$ and $ 4$, so the sum is $ \boxed{13}$.

Or, algebraic manipulation yields $ \sqrt {5^2 \plus{} 4\cdot36} \equal{} \sqrt {169} \equal{} \boxed{13}$. (This is slower, but whatever.)


I'm not too sure about this, I'm bad at probability.

[hide="Parts 1 and 2"]1-1
1-2
1-3
2-1
3-1

5 ways to sum to 4 or less, out of $ 6\cdot 6\equal{}36$ total ways, so $ \frac{5}{36}$ chance that you pay 20 and $ \frac{31}{36}$ chance of paying the other fixed amount (let's call it $ p$).

Then the expected value is $ \frac{5}{36}\cdot 20\plus{}\frac{31}{36}\cdot p\equal{}\frac{25}{9}\plus{}\frac{31p}{36}$. Setting this equal to $ 5$ and solving, we get $ p\equal{}\frac{36}{31}\cdot\frac{20}{9}\equal{}\frac{80}{31}\approx 2.58$.[/hide]

You forgot 2-2. The answer turns out to be 2.

Cakewalk.

$ (1)$ You have $ 6*6\equal{}36$ total possibilities, of which $ 6$ sum to $ 4$ or less.  Therefore, the probability that you get a sum of less than or equal to $ 4$ is $ \frac{1}{6}$.  Therefore, the probability of getting a number greater than $ 4$ is $ \frac{5}{6}$.  Let the amount of money you get for this value be $ m$.  Therefore, the expected earnings is $ E\equal{}\frac{20}{6}\plus{}\frac{5m}{6}$.  Since $ E$ must equal $ 5$, we can solve the equation for $ m$, yielding $ \boxed{m\equal{}2}$.
$ (2)...$
$ (3)$  Similarly, there is a $ \frac{1}{6}$ chance of getting a sum of less than or equal to $ 4$, and a $ \frac{5}{6}$ chance of getting a sum greater than $ 4$.  We thus have that: $ E\equal{}\frac{20}{6}\plus{}\frac{5m}{6}$, where $ E$ is the expected earning.  If this must equal $ Y$, then substituting $ Y$ for $ E$ yields: $ \boxed{m\equal{}\frac{6Y\minus{}20}{5}}$.

[quote="jeez123"]You are the customer. Instead of paying 5 bucks weekly, you decide to offer your newspaper carrier a different payment method.

The carrier will roll a pair of dice. If the sum is 4 or less, the carrier will get 20 bucks. If the sum is more than 4, the carrier will get some other fixed amount.

1. What should you choose for the other fixed amount (that is, the amount that the carrier gets for anything over 4) in order for this payment to be equivalent, in the long run, to 5 bucks per week payment?

2. Describe in words how you fund the answer in question 1.

3. Now suppose that the normal weekly cost of the paper was Y bucks, but the carrier still gets 20 bucks for a sum of 4 or less. What should you choose for the other fixed amount that the carrier gets (for any sum over 4) in order to make the payoff fair in the long run? (Hint: work with some specific values for Y.

Please help me with this problem! Thanks![/quote]

you just take probabilty $ (s\leq 4)20$+ prob$ (s > 4)x$= $ 5d$llars

Yeah, it's all from the definition of expected value.  Namely,
$ E\equal{}p_{1}e_1\plus{}p_2e_2\plus{}p_3e_3\plus{}...\plus{}p_{n\minus{}1}e_{n\minus{}1}\plus{}p_ne_n$,
where $ E$ is the expected value, $ p$ is the probability of an event, and $ e$ is the "value" of the event.


algebra
polynomial
geometry
3D geometry
function
inequalities
inequalities open

Does there exist a homogenous symmetrical polynomial in $n$ variables $f : \mathbb{R}^n \to \mathbb{R}$ such that $f$ is nonnegative at all the vertices of the box $[0,1]^n$ but is negative somewhere inside the box?

[Edit][b]Extension:[/b]

For each integer $n \geq 2$, what is the greatest integer $d_n$, such that:

If $f : \mathbb{R}^n \to \mathbb{R}$ is a homogenous symmetrical polynomial of at most $d_n$ degrees, and $f$ is nonnegative at all vertices of the box $[0,1]^n$, then $f$ is non-negative everywhere within the box.

For odd $n$ you can just take $\prod_1^n x_i(x_i-1)$. It clearly equals $0$ at the vertexes of the cube $[0,1]^n$ and is negative inside the cube(for odd $n$!!!).

Misha

Well, it should be homogeneous. Of course, with non-homogeneous polynomials it is too easy: Take the square of the distance of the point to the center of the cube...

  darij

Correct me if I'm wrong, but that can't be homogeeous, can it?
Edit: Darij wrote the above simultaneously.

Here is an example of one in two variables: $f(a,b)=a^4+b^4+16a^2b^2-9ab(a^2+b^2)$. Now $f(0,0)=f(1,1)=0, f(0,1)=f(1,0)=1$, but $f(0.4,0.8) < 0$. (There are of course many similar examples). I'm fairly sure similar polynomials could be devised for greater numbers of variables.

For $n$ variables we can just take your polynomial and add up $x_3^4+...+x_n^4$. It is still non-negative at the vertices and we can choose the same $a$ and $b$ as you did and the othe variables low enough and the result is still negative.

Am I missing something again?

Misha

Yeah, Fiachra's example work, but Misha123's addon is still not symmetrical.

However, using the same idea, we see that the following polynomial

\[
(x_1^n + x_2^n + \cdots + x_n^n - \lambda x_1x_2 \cdots x_n) \prod_{1 \leq i<j \leq n} (x_i - x_j)^2
\]
works for $\lambda > n$.

$\hline$

Here's an extension:

For each integer $n \geq 2$, what is the greatest integer $d_n$, such that:

If $f : \mathbb{R}^n \to \mathbb{R}$ is a homogenous symmetrical polynomial of at most $d_n$ degrees, and $f$ is nonnegative at all vertices of the box $[0,1]^n$, then $f$ is non-negative everywhere within the box.

No one tried this? :(

Quite to my suprise, I found that $d_n = 3$ for all $n \geq 2$.

For $n=2$, the polynomial $(x^2 - 3xy + y^2)(x-y)^2$ serves as a counter example.

For $n \geq 2$, the polynomial
\[
f(x_1, x_2, \cdots, x_n)=
\left( \sum_{i=1}^{n} x_i^2 - \frac{2}{n-1} \prod_{i<j}x_ix_j \right) \left( \sum_{i=1}^{n} x_i^2 - \frac{2}{n-2} \prod_{i<j}x_ix_j \right)
\]
serves as a counter example. It is easy to check that $f$ is nonnegative at all vertices. Now $\frac{1}{n} \sum x^2 \geq (\frac{1}{n} \sum x)^2$ implies that the left bracket is always nonnegative. And the function
\[
\frac{\sum_{i=1}^{n} x_i^2}{\prod_{i<j}x_ix_j}
\]
takes the value $\frac{2}{n-1}$ at $(1,1,\cdots 1)$ and value $\frac{2}{n-2}$ at $(1, 1, \cdots 1, 0)$. So it must take some value in between. So $f$ can be negative.

On the other hand, suppose that the degree of $f$ is at most 3. Since $f$ is a symmetrical homogeneous polynomial, it can be written as a polynomial in terms of $\sigma_1, \sigma_2, \sigma_3$ where
$\sigma_1 = \sum_{i=1}^n x_i$
$\sigma_2 = \sum_{i<j} x_ix_j$
$\sigma_3 = \sum_{i<j<k} x_ix_jx_k$ (N/A when $n=2$)
Furthermore, if we set the constraint $\sigma_1 = 1$, then $f$ becomes a linear combination of $\sigma_2$ and $\sigma_3$. As shown by Naoki Sato [1], such function attains its extremes when a subset of the variables equal and the rest are zero. This is what we want.

This result implies if we can write an inequality as a symmetrical homogeneous polynomial of degree no more than 3, then we're done after checking the vertices. However, if the degree is higher than 3 or if it's not a homogeneous symmetrical polynomial, then further analysis is needed.

[b]Reference:[/b]
[1] Sato, Naoki, [i]Symmetric Polynomial Inequalities[/i], Crux with Mayhem, 27(2001), 529-533.


search
number theory proposed
number theory

Let a; b; c; d be integers. Show that the product
(a -b)(a - c)(a - d)(b -c)(b - d)(c - d)
is divisible by 12.

plz search before posting( esp.if u know the source  :) )
see PEN A11

And additionally to searching, you should use better titles. "nt" is even worse than "Nice Problem"...; what about something like "(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) is divisible by 12" next time¿


A,B are sets of real numbers such that  A c B + x.Z , B c A + x.Z for every positive real number x. Here, K + x.Z consists in every number of the form k + xn, where k E K and n is a integer.

If B is a limited set (exists m E R such that |y| <= m for every y E B), is true that A = B?

Hi Lopes!

It is a time to learn $\LaTeX$!

See how your problem could be written:
Let $A,B$ be sets of real numbers such that $A \subset B + x\cdot \mathbb Z$, $B \subset A + x\cdot \mathbb Z$ for every positive real number $x$. Here, $K + x\cdot \mathbb Z$ consists of every number of the form $k + xn$, where $k \in K$ and $n$ is a integer.

If $B$ is a bounded set (exists $m \in R$ such that $|y|\leq m$ for every $y \in B$), is true that $A = B$?

:D  :D
thanks myth..
i really need learn latex.. and english

Let $B$ is bounded and $|b|< m$ for all $b\in B$. Since $A\subset B+x\mathbb Z$ for all $x>0$ we conclude $A\subset \cap_{k\in \mathbb{N}} (B+2mk\mathbb Z)=B$. So $A$ is a bounded set too. In the similar way we conclude $B\subset A$. Therefore $A=B$.

I´m not understand why A c  :cap: k E N (B + 2mkZ)..  :(

If $X\subset Y$ and $X\subset Z$ then $X\subset Y\cap Z$. And I chose $x$ of the form $2mk$.

Myth.. and..
why     :cap: k E N (B + 2mkZ) = B?

 :(


I don't get this: Why does rounding the atomic weight and subtracting the atomic number equal to the number of neutrons?

I get the idea, but since one proton is ~1.007 amu and a neutron is ~1.008 amu, then wouldn't elements of atomic number greater than or equal to 100 have skewed rounded atomic mass numbers from the rounded atomic mass?

Does this not matter because the atomic mass on the periodic table is the average relative atomic mass anyway? So calculated neutron number would be an "averaged" number as well?

people believe that Atomic mass are defined only by protons, that is because protons are about $1837$times heavier than electorns. but $AMU$ (atomic mass unit)is defined as following:
one unified atomic mass unit is $1.66053873 * 10^{-27}$ kilogram, with a one-standard deviation uncertainty of ±$0.000 00013*10^{-27}$ kilogram.
now notice mass of proton is:
$m_p=1.6726 *10^{-27}$
do you see the differnce between the $AMU$ and $m_p$? they are not equal :)

[quote="breez"]I get the idea, but since one proton is ~1.007 amu and a neutron is ~1.008 amu, then wouldn't elements of atomic number greater than or equal to 100 have skewed rounded atomic mass numbers from the rounded atomic mass?[/quote]

Add to that mass deficit due to binding energy and the results is even less precise.

That's only approximate thing, it may be that it gives correct results for every isotope - by accident :)


calculus
inequalities
function
algebra solved
algebra

Let x, y, z are positive real numbers satisfying  2xyz = 2x + 4y + 7z. Find the minimum value of  P = x + y + z. Try to solve without calculus.

I have a nice solution for this problem. I want to get different ideas.

Namdung

I found this problem very nice and quite difficult. My solution is based on the arithmetic- geometric mean inequality.
  First, it is clear that x+y+z=2/5*(5x/2)+1/3*(3y)+4/15*(15z/4)>= (5x/2)^2/5*(3y)^1/3*(15z/4)^4/15 (because 2/5+1/3+4/15=1).
  Thus, (x+y+z)^2>=(5/2)^4/5*3^2/3*(15/4)^8/15*x^4/5*y^2/3*z^8/15.
   After that, we have in the same manner that 2x+4y+7z>=10^1/5*12^1/3*15^7/15*x^1/5*y^1/3*z^7/15. Hence
 (x+y+z)^2(2x+4y+7z)>=225/2 xyz. Now, 2x+4y+7z=2xyz, from where we find that (x+y+z)^2>=225/4, which means x+y+z>=15/2. So the minimum is 15/2, attained when x=3, y=5/2 and z=2. A very beautiful problem with a very ugly solution (this one).

Hi all

Harazi solution is very nice. This problem was used in 2001 Vietnamese IMO Team selection Test (but the wording is a little bit another). Some of contestants used the same method of weighted AM-GM inequality to solve it. Harazi, can you explain how did you choose the weights? Or firstly you used Lagrange's method to find point of minimum? I have an another approach. I will send it later.

Namdung

Now, I didnt't find first the minimum with the Lagrange method because I don't know it. All I did was to find u,v,w and a,b,c  with a+b+c=1=u+v+w such that:
  x+y+z=a*x/a+b*y/b+c*z/c>= (x/a)^a*(y/b)^b*(z/c)^c and
2x+4y+7z=u*(2x/u)+v*(4y/v)+w*(7z/w)>=(2x/u)^u*(4y/v)^v*(7z/w)^w. Having also xyz involved in this problem, I thought that there must be an inequality of the form (x+y+z)^2*(2x+4y+7z)>=rxyz or (x+y+z)(2x+4y+7z)^2>=sxyz. I treated the first one and I made the product of the two above inequalities. Because we had to have xyz, I had to put 2a+u=2b+v=2c+w=1, from where I deduced a,b,c,u,v,w.

Please your solution, Namdung.

Thank you very much.

My method based on techniques of elimination

We have (2xy - 7)z >= 2x+4y. So 2xy - 7 > 0 and z >= (2x+4y)/(2xy - 7). Now, P= x+ y + z >= x + y + (2x+4y)/(2xy - 7) = x + 11/2x + y - 7/2x + (2x + 14/x)/(2xy-7) >= x + 11/2x + 2sqrt(1+7/x^2). Now, using Cauchy-Buniakovsky inequality we have
    2sqrt(1+7/x^2) = (1/2)sqrt((9+7)(1+7/x^2) >= (3+7/x)/2.  (*)
Putting to the last inequality, we have
    P >= x + 9/x + 3/2 >= 6 + 3/2 = 15/2. The equality holds when x = 3, y = 5/2, z = 2.

All the things are clear except the line (*) when we use C-B ineq with weigths. The question is: How can I get those weight. In another hand, the expression x + 11/2x + 2sqrt(1+7/x^2) is an one-variable function and we can compute its minimum by calculus (but it's not fair!). The main idea here is take the weights in such a way that the equalities in two application (C-B in (*) and AM-GM below) hold at the same point x0.

Namdung


Considering the caracteristics of a Pythagorean triple, find all solutions of the equation that satisfies the equation x² + y² = 2z² and x > y > 0 and (x, y) = 1


geometry
3D geometry
analytic geometry
vector

What is the size of the largest square that can fit inside a unit cube ?

Peek only if you ...

[hide]We can use the general case of a square in a $n$ - dimensional  cube.
The answer obviously  for the side of such a square is:

$\frac{ \sqrt{8n-3(1-(-1)^{n})}}{4}$

(We can check for $n=2$ case , the answer comes out (correctly) as $1$)
Putting $n=3$ we get $\frac{\sqrt{18}}{4}= \frac{3 \sqrt 2}{4}$

(A  8 dimensional  unit cube would be able to get a square of side 2!! )

As I warned before, Peek only if you have mathematical sense of humor..:) in solving  problems while showing off :)

[/hide]

[quote="Gyan"]The answer obviously  for the side of such a square is:

$\frac{ \sqrt{8n-3(1-(-1)^{n})}}{4}$[/quote]This is so "obvious"  :lol:  that it needs explanation.  Anybody?

I got the same thing as Gyan. I haven't proved that it's the biggest possible area, but it's the biggest number I could get.

[hide="Meh."]
[image may take a few moments to load]
[img]http://img.photobucket.com/albums/v194/mukmasterxps/cpic.jpg[/img]

Alright, so I figured if you put in a square like that, it'll be pretty big, if not the biggest possible.
So now you just need $j$ so that it's actually a square and not just some random rectangle.

So then you have $j^{2}+j^{2}= 1^{2}+(1-j)^{2}+(1-j)^{2}\implies j = \frac{3}{4}\implies A = s^{2}= j^{2}+j^{2}= \frac{9}{8}$

Which is the same as having side-length $\frac{3}{4}\sqrt{2}$
[/hide]

Well $2j^{2}$ has to equal $1+2(1-j)^{2}$ or else its not actually a square.

oh ok, I actually mean orientin a vertex of the square on the vertex of the cube and then another vertex on the mid point of the opposin side and repeatin that aain.

so if we use coordinates on (0,0,0) (0,0,1) (1,1,1) etc the square I am referin to is on (1,0,0) (0,0,1/2) (0,1,1) and (1,1,1/2)

[b]@Gyan:[/b] So I just realized you were calculating the side length and I was calculating the area. So we have the same answer for 3D. Cool.

Not sure how you got that generalization, I got $(D-1)j^{2}=1+(j-1)^{2}(D-1)$ $\implies A = (D-1) j^{2}= \frac{D^{2}}{4(D-1)}$, where $D$ is the number of dimensions, through non-intensive speculation.

Both of ours work for 0, 2, and 3 dimensions, and fail for 1 dimension. Yours is probably correct (or more than mine), but please share how you got it.

[b]@Agolsme:[/b] Sides being equal does not make it a square.

Try connecting $(0,0,1/2)$ and $(1,1,1/2)$. The resulting segment has length $\sqrt{2}$. Now connect each of those points to $(1,0,0)$; the resulting segments each have length $\sqrt{5}/2$.

$(\sqrt{5}/2)^{2}+(\sqrt{5}/2)^{2}$ does not equal $(\sqrt{2})^{2}$; thus, your quadrilateral fails the Pythagorean Theorem.

Few Comments.

One easy way to get the upper limit is to see that the diagonal of the square   must be less than the long diagonal.

[hide="hint"]That is for the  problem posted in the original post, , the answer  must be less than $\sqrt{(3/2)}$ (Actual answer is slightly less than that – For even number of dimensions this happens to be  the true answer – see my comments below.  [/hide]

BTW, the problem  is classic.  [hide="details"]  It is said that  Prince Rupert (1600’s), a nephew of England's King Charles I,  won a wager that  such a hole could be made  I read somewhere that first publication of the solution was Wallis or  Nieuwland . The problem is known as “Prince Rupert’s problem and to my knowledge a general case of n dimensional   (m x m x m x … “super solid” passing through n dimensional super solid)   remains unsolved - at least up to mid 90's)
[/hide]
The solution I gave, for 2 dimensional cube in n dim –cube is known and as my math prof used to say *any* known result in math is  “trivial” (or  “obvious”) (Prompting Feynman to say once “ Mathematician can only solve trivial problems”

The Zuton Force : [hide]I don’t understand  how you are arriving at this.  But one very easy way to get the  answer in any  dimensional space. Is given below:

Since the diagonal of the square should fit in, the diagonal of the square can not be more that $\sqrt n$, for even number of dimension this  happens to be correct.  But for odd number of dimensions , it doesn’t quite fit (  One can not have three mutually  perpendicular lines all joining “opposite faces”) hence the answer is little less.

(That is if you have a four dimensional  unit  super cube, its longest  diagonal is $\sqrt{4}= 2$  hence the length of the square would be $\sqrt(2)$ )

If we assume the center of the square is at the center of the super-solid,
-	For even number of dimension, one end of the square could be at (1,1,1 ….)
-	For  odd number of dimension, best one can do is (1/2,1,1, …)
-	Hence the result I gave above.
 [/hide]

One 'easy' way if you have done vectors is to:

[hide]Find three mutually perpendicular vectors originating from origin (center of the cube) and the other point is on the cube , one can prove (fairly easily) that all three can not be more than length (3/2) - and  one indeed has a length of (3/2) for (1/4,1/2,1/2)  Hence th answer.[/hide]
For more details see, for example;

[url=http://mathworld.wolfram.com/CubeSquareInscribing.html]see link[/url]

Yeah, I was just being silly. That's pretty cool, though.


linear algebra
matrix
linear algebra unsolved

Suppose $A\in\mathbb{R}^{n\times n}$ satisfies $A^T=A$. And let \[ C(A)=\{X\in\mathbb{R}^{n\times n} \: |\; AX=XA\}, \] which is a subspace of $\mathbb{R}^{n\times n}$. Prove that, \[ \dim C(A) \ge n, \] and it becomes an equation if and only if $A$ has $n$ different eigenvalues.

I know that the more advanced members have recently been discussing the commutant, and apparently its dimension can be written as the sum of squares of numbers that sum to $n$; however, I will be more naïve in my approach.

Since $A$ is symmetric, we know that it can be diagonalized and that all of its eigenvalues are real. So let $A = UDU^{-1}, D \in \mathbb{R}^{n \times n}$. Then we note that $AX = XA \Leftrightarrow D (U^{-1}XU) = {U^{-1}XU) D}$, so we can equivalently find the dimension of the subspace of matrices that commute with $D$ (as $D$ is a real matrix, these matrices will also be real as they should be). Certainly, just by using the rules of matrix multiplication, it is evident that any diagonal matrix commutes with $D$ and that this subspace has $\dim n$, so the inequation is proven. To prove the equation, suppose that $A$ has two equal eigenvalues, i.e. $D$ has two equal elements, say $d_{ii}$ and $d_{jj}$. Then, I think that the subspace of matrices with nonzero elements at the $ij$th and $ji$th positions is also included in the commutant, and thus the commutant has dimension at least $n + 2$, in fact; this serves to prove that equality holds only if $A$ has $n$ distinct eigenvalues. Finally, if all eigenvalues are distinct, any matrix that commutes with $D$ cannot have a nonzero nondiagonal element (suppose $e$ were some nondiagonal element; then in the corresponding place, we would have $\lambda_i e = \lambda_j e$), so the proof is complete.

you're right MysticTerminator. Because if a matrix is diagonalizable, the dimension of $C(A)$ is exactly $\sum dim(E_i)^2$ where the $E_i$ are the proper subspace. It is because $B$ commutes with $A$ iff the proper subspaces of $A$ are $B$-stable.

ah, that makes sense - when you say proper subspace, that's the same thing as the eigenspaces, right?

[quote="MysticTerminator"]the commutant, apparently its dimension can be written as the sum of squares of numbers that sum to $n$.[/quote]
That sounds to be a beautiful conclusion which I haven't heard before. So, $\dim C(A) \ge n$ holds for arbitrary $A\in M_n(\mathbb F)$, right?

[quote="MysticTerminator"]Since $A$ is symmetric, we know that it can be diagonalized and that all of its eigenvalues are real. So let $A = UDU^{-1}, D \in \mathbb{R}^{n \times n}$. Then we note that $AX = XA \Leftrightarrow D \left(U^{-1}XU\right) = \left(U^{-1}XU\right) D$, so we can equivalently find the dimension of the subspace of matrices that commute with $D$ (as $D$ is a real matrix, these matrices will also be real as they should be). Certainly, just by using the rules of matrix multiplication, it is evident that any diagonal matrix commutes with $D$ and that this subspace has $\dim n$, so the inequation is proven. [/quote]
Yes, I see. $\sigma_U: M_n(\mathbb R)\to M_n(\mathbb R),\; X\mapsto U^{-1}XU$ is an isomorphic transformation. So $\mathcal S\cong U^{-1}\mathcal SU$ where $\mathcal S$ denotes any subspace of $M_n(\mathbb R)$, and hence they have the same dimension.


Let a_1,a_2,...,a_n be positive real numbers, $a_{1}\ge a_{2}\ge...\ge a_{n}$. Compare $P$ and $0$ with
\[P=\frac{a_{1}-a_{2}}{a_{1}^{2}+a_{2}^{2}}+\frac{a_{2}-a_{3}}{a_{2}^{2}+a_{3}^{2}}+...+\frac{a_{n}-a_{1}}{a_{n}^{2}+a_{2}^{2}}.\]

Hi all mathlinkers again! The final exam of University make me unable to reach Mathlinks for a long time. Now I finish the first most important constest, so I've gone here. Good luck!


how would you solve this problem?

The functions $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$. If the equations $f(x)=0$ has exactly four distinct real roots, then what is the sum of these roots.

Think of how the general graph of this function would look.  Do you see any symmetry?

[hide]The roots of the equation can be expressed on the form $2 + m, 2 - m, 2 + n, 2 - n$ such that $m,n \in \mathbb{R}$.
Their sum is $2 + m + 2 - m + 2 + n + 2 - n = 2 + 2 + 2 + 2 = \boxed{8}$[/hide]

[quote="maokid7"]im not sure what you mean.[/quote]

Basically, this graph is symmetric about the line x=2 (do you see why?)  Therefore, $f(n) = f(4-n)$  So given two roots $r,s$ you should know that $4-r$ and $4-s$ will also be roots.  The sum of these is 8.  Does this make somewhat more sense?

it makes a little more sense now. ill think about it some more, im pretty sure ill undertand it in a while.

[quote="10000th User"]This exact problem was asked before. Please use the SEARCH feature of this site. If I can find it, you can do it too.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=57568]http://www.mathlinks.ro/Forum/viewtopic.php?t=57568[/url][/quote]

Thank you. I was about to say the same thing.

Mod, can you kindly please lock this topic. Thanks.  :)


geometry
3D geometry
sphere
ratio
calculus
derivative
trigonometry

What is the ratio of the volume of a sphere to the maximum volume of a cone inscribed in a sphere?

Volume of sphere = 4/3*pi*r^3
Volume of cone = 1/3*pi*r^2*h
let sqrt(r^2 - x^2) = radius of cone
where r is the radius of sphere
then the height of cone = r+x
volume = 1/3*pi*(r^3 + x*r^2 - r*x^2  - x^3)
taking the derivative and setting to 0
x = -r/2+r*sqrt(3)/2
x+r = r/2+r*sqrt(3)/2
x^2 = r^2 - r^2*sqrt(3)/2
r^2 - x^2 =  r^2*sqrt(3)/2
volume of cone = 1/3*pi*r^2*sqrt(3)/2*(r/2+r*sqrt(3)/2)=
1/12*pi*(r^3*sqrt(3) + 3* r^3)
ratio = 4/3*pi*r^3/(1/12*pi*(r^3*sqrt(3) + 3* r^3)) = 16/(sqrt(3)+3) =
(24 - 8*sqrt(3))/3

I got a different answer
[hide]
We look at a cross section through the middle of the sphere with radius $ r$, so the cross section of cone would be a triangle, with its apex located on $ (\minus{}r,0)$. We get the equation for the area of the C-S of the cone to be $ A\equal{}(r\plus{}x)\sqrt{r^2\minus{}x^2} \Rightarrow A'\equal{}\sqrt{(r\plus{}x)^3(r\minus{}x)}\equal{}0 \Rightarrow x\equal{}r/2$. So the height of the triangle is $ r/2\plus{}r\equal{}3r/2$. Therefore, the area of triangle is $ \frac{3\sqrt{3}}{8}*r^2$, so the ratio of sphere to cone is $ \frac{8}{3\sqrt{3}}$[/hide]

I got a different answer  :lol:
[hide]
Assuming the radius to be 1, I connected the sphere's centre to its apex, then drew 2 lines to the circular base (like a tripod), then from the centre perpendicular down to the base. I called the nested angle between this perpendicular and the legs to the base $ \theta$ (I dunno how to draw, sorry).

Then:

$ V_{cone} \equal{} \frac{1}{3}\pi\sin^{2}\theta(\cos{\theta}\plus{}1)$
Assuming $ \sin{x} \neq 0$, which is evident,
$ \frac{dV}{d\theta} \equal{} 3\cos^{2}\theta\plus{}2\cos\theta\minus{}1$
Set this equal to 0 to get
$ \cos\theta \equal{} \frac{1}{3}$,
$ \sin^{2}\theta \equal{} 1\minus{}\cos^{2}\theta \equal{} \frac{8}{9}$.

Thus, the ratio of the volumes is

$ \frac{\frac{4}{3}\pi}{\frac{1}{3}\times\pi\times\frac{8}{9}\times\frac{4}{3}} \equal{} \frac{27}{8}$[/hide]

[hide]WLOG let the radius of the sphere be 1.
[asy]pair O,V,A,B,C,H;
O=(0,0); V=(0,1); A=(-.6,-.8); B=(.6,-.8); C=(0,-1); H=(0,-.8);
draw(Circle(O,1));
draw(V--A--B--cycle);
draw(V--C);
label("$V$",V,N);
label("$C$",C,S);
label("$A$",A,SW);
label("$B$",B,SE);
label("$H$",H,NE);
label("$h$",(V+H)/2,E);
label("$r$",(A+H)/2,N);[/asy]
Let $ h$ be the height of the cone and $ r$ its radius.  Thus $ VH=h\Rightarrow HC=2-h\Rightarrow r=AH=\sqrt{h(2-h)}$.  The volume of the cone is $ V=\frac{1}{3}\pi h^2(2-h)$.  We want to know when this achieves a maximum, so we need to know when $ \frac{dV}{dh}=0$.

\begin{eqnarray*}
\frac{dV}{dh} = \frac{1}{3}\pi(-3h^2+4h) &=& 0 \\
-3h^2+4h &=& 0 \\
h=0 & \text{or} & h=\frac{4}{3}
\end{eqnarray*}

Clearly we want $ h=\frac{4}{3}$.  Thus the volume is $ \frac{1}{3}\pi\left(\frac{4}{3}\right)^2\left(\frac{2}{3}\right)=\frac{32}{81}\pi$.  Thus the desired ratio is $ \frac{\frac{4}{3}\pi}{\frac{32}{81}\pi}=\boxed{\frac{27}{8}}$.
[/hide]


1) ( [b]National Technical University of Athens, entry exams, 1949[/b] )

Δίνεται τετράπλευρο $ABCD$ , του οποίου η πλευρά $AB$ είναι σταθερή σε θέση και μέγεθος, η δε $CD$ σταθερή σε μέγεθος και οι οποίες τέμνονται στο σταθερό σημείο $O$. Να βρεθεί ο γεωμετρικός τόπος της τομής $M$ των $AD$ και $BC$, καθώς και ο γεωμετρικός τόπος της τομής $N$ των διαγωνίων $AC$ και $BD$, όταν η πλευρά $CD$ στρέφεται γύρω από το $O$, έτσι ώστε το $OD$ να είναι σταθερό.

2) ( [b]Mathematics department of Athens, entry exams, 1963[/b] )

Δίνονται δύο περιφέρειες $(K)$ και $(L)$, που εφάπτονται εσωτερικά και έστω $A$ το σημείο της επαφής τους. Από το $A$ φέρουμε χορδή $ADB$ που τέμνει την εσωτερική στο $D$ και την εξωτερική στο $B$, ώστε η από το $B$ κάθετη στην διάκεντρο να τέμνει την εσωτερική στο $E$. Από το $D$ φέρουμε κάθετη στη διάκεντρο που τέμνει την εξωτερική στο $C$. Να αποδειχθεί ότι $AE=AC$.

Υπόδειξη λύσης.:Μήπως με Μενέλαο στα ΟΑΓ και ΟΒΓ και διατέμνουσες αντίστοιχα ΑΜΔ και ΝΑΔ οι λόγοι ΑΜ/ΜΓ και ΝΑ/ΑΔ προκύπτουν σταθεροί και μετά ομοιοθεσία των τόπων των Γ και Δ (κυκλοι)?

Σωστά κύριε Μπόρη, οι γεωμετρικοί τόποι είναι κύκλοι.

Η λύση μου: (Σχήμα δεν έχω βάλει επειδή δεν έχω χρόνο)

[hide]α) Εφόσον το $O$ είναι σταθερό σημείο και τα μήκη $OD, OC$ είναι σταθερά σαν δεδομένα, προκύπτει αμέσως ότι οι γεωμετρικοί τόποι των $C$ και $D$ είναι αντίστοιχα δύο σταθερές περιφέρειες $(C_{1}),(C_{2})$, με κέντρο $O$. Εφαρμόζοντας το θεώρημα του Μενελάου στο $\triangle ODA$, με $MCB$ διατέμνουσα παίρνουμε: $\frac{DM}{MA}\cdot\frac{AB}{BO}\cdot\frac{OC}{OD}=1$. (1)

Οι λόγοι όμως $\frac{AB}{BO}, \frac{OC}{CD}$ είναι σταθεροί και γνωστοί από τα δεδομένα. Επομένως ο λόγος $\frac{DM}{MA}$ είναι σταθερός και γνωστός. Έστω $\frac{DM}{MA}=\lambda$ (2), όπου $\lambda=\frac{BO}{AB}\cdot\frac{CD}{OC}$. Η σχέση (2) γράφεται ισοδύναμα: $\frac{DM}{AM}=\lambda\implies\frac{AM-AD}{AM}=\lambda\implies\frac{AD}{AM}=1-\lambda\implies \frac{AM}{AD}=\frac{1}{1-\lambda}$ (3).

Η σχέση (3) σημαίνει ότι, όταν $D$ γράφει την περιφέρεια $(C_{2})$, το $M$ θα γράφει το ομοιόθετο της $(C_{2})$ ως προς κέντρο ομοιοθεσίας $A$ και λόγο $\frac{1}{1-\lambda}$.
β) Για το γεωμετρικό τόπο του σημείου $N$, εφαρμόζουμε το θεώρημα του Μενελάου στο $\triangle OAC$ με $DNB$ διατέμνουσα και έχουμε $\frac{CD}{DO}\cdot\frac{OB}{BA}\cdot\frac{AN}{NC}=1$. Οι λόγοι $\frac{CD}{DO}, \frac{OB}{BA}$ είναι γνωστοί και σταθεροί από τα δεδομένα του προβλήματος και επομένως ο λόγος $\frac{AN}{NC}$ λόγω της (1) θα είναι σταθερός και γνωστός. Επομένως $\frac{AN}{NC}=\frac{DO}{DC}\cdot\frac{AB}{OB}$.
Από αυτή τη σχέση έχουμε ότι όταν το $C$ γράφει την περιφέρεια $(C_{1})$, το $N$ θα γράφει μία περιφέρεια ομοιόθετη της $(C_{1})$ ως προς κέντρο ομοιοθεσίας $A$ και λόγο αυτόν που αναφέρθηκε παραπάνω.[/hide]

Επειδή έχει μείνει πολύ καιρό άλυτη στέλνω την λύση (ΕΥΚΟΛΗ-ΚΑΛΥΤΕΡΗ ΗΤΑΝ Η ΠΡΩΤΗ ΜΕ ΤΟΥΣ ΤΟΠΟΥΣ ΝΟΜΙΖΩ)


Without using the standard definition of scalar multiplication and addition in $R^{2}$ , define the operations

1)$(x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})$
$a(x_{1},y_{1})=(ax_{1},y_{1})$

2)$(x_{1},y_{1})+(x_{2},y_{2})=(x_{1},0)$
$a(x_{1},y_{1})=(ax_{1},ay_{1})$

Justify whether they  are a vector space using above definition .

I dont have idea how to start it since i just started to learn vector space thing . Any help ?

The first violates the distributive law $(a+b)x=ax+bx$.

The second also violates the (other) distributive law. It's also not a group under addition; there is no zero element. (For most $x$, there is no solution $y$ to $x+y=x$)

Look at the algebraic axioms for a vector space. Which axioms should you be suspicious of? With oddball definitions of the two operations, perhaps you should start by being suspicious of the two distributive laws:

$\alpha(v+w)=\alpha v+\alpha w$ and $(\alpha+\beta)v=\alpha v+\beta v$

For #1, I especially suspect that it doesn't satisfy the first of those two distributive laws; for #2, the other one.

Try fooling around with examples to see if you can construct counterexamples.


calculus
derivative
limit
function
calculus computations

Why is the derivative of a constant defined as 0? It doesn't fit the definition of a derivative...

It fits the definition. Let $f(x)=a$ where $a$ is a constant. Then $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{a-a}{h}$ $=\lim_{h\to 0}0=0$.

I don't think we need a special definition for the derivative of a constant. There is just a general definition of derivative, $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. (HilbertThm90 typed it slightly incorrectly). Using this definition, you [b]prove[/b] that the derivative of a constant function is zero.


Prove that for $a$, an integer $> 1$: \[ \mbox{Si} \  a \notin \mathbb{P}, \  \exists p \in Div(a), \  p \in \mathbb{P}, \  p^2 \leq a.\] I think it isn't hard but... :blush:  :(

[quote="grobber"]As far as I can understand, you want to show that every non-prime $n$ has a prime divisor which is at most $\sqrt n$. If $n$ is not a prime, then we can write $n=km,\ k,m>1$. Since one of $k,m$ is $\le\sqrt n$, and none of them is $1$, assume $k\le\sqrt n$, and take any prime $p|k\Rightarrow p|n,\ p\le\sqrt n$. That's all..

Did I misunderstand? Were you asking something else? :?[/quote] Exactly, my problem returns to that. Thank you very much. :)


hello
$ 3^{\frac{4}{n}}\equal{}3$
$ n\equal{}4,n^2\equal{}16$
thank u


if
$ \sum_{n\equal{}1}^{n} (a\minus{}2n\plus{}2)^2\equal{}19600$

find a

[quote="tony123"]if
$ \sum_{n \equal{} 1}^{n} (a \minus{} 2n \plus{} 2)^2 \equal{} 19600$

find a[/quote]

Something is wrong with the notation. You can't have the same variable used as the index and the upper limit. :!:


Who here accesses internet behind a firewall?

Why exactly are you interested in whether or not people have a firewall?  Just seems like an odd question to me...

Well, I'm currently behind a firewall which checks and tries to block nearly everything, so I wondered who else had one,what they thought about it, and whether it was really worthwhile to have one if I already have virus and spyware protection.

Oh yes, it's definitly worth it. Peace of mind. Besides, antivirus and antispyware programs do nothing for hackers, plus if a virus/spyware isn't caught, then when it tries to send out stuff, you see it.

Ok, thanks for your reply.  I guess I'll keep my firewall then.  BTW, which firewall do you use?

Personal firewalls are kind of annoying, it keeps on popping up.  So I just use the firewall my router provides.  And since most people use windows XP you probably have the Windows Service pack 2 firewall.

[quote="stupidkid"]Personal firewalls are kind of annoying, it keeps on popping up.[/quote]
Exactly ;)

[quote="stupidkid"]you probably have the Windows Service pack 2 firewall.[/quote]
Actually, I have Kerio.

Actually, Windows XP's firewall, is useless, becasue it is not for blocking inbound traffic. Your router's firewall is true SPI firewall and blocks stuff, but when you connect remotely to another's computer, you are unprotected. You should always install a personal firewall, even if it's annoying. You can download FREE ZoneAlarm Firewall [url=http://www.zonelabs.com]here[/url].

Thanks for the information.  I'll give that a try.

i disabled firewall cuz it interferes with online gameplay :lol:
however, i have spybot s&d just in case...

[quote="M"]I think it's improtant to use a firewall and if you configure it correctly everything is working fine.

I've a personal firewall, and it seldomly pops up, because the rules for most of the programs are clearly set (Opera: allow; Explorer: deny :D )[/quote]

explorer is dangerous. only opera and firefox are safe.  :)


[quote="sri340"]In a math class, 12 out of 15 girls are freshmen and 11 out of 15 boys are freshmen. What is the probability that in a randomly selected group of five students from the class, there will be two freshmen girls and three freshmen boys? Express your answer as a decimal to the nearest thousandth.[/quote]

We know there are 30 total students, and $ 30 \times 29 \times 28 \times 27 \times 26 \equal{} 17100720$ ways in total to select five students.

There are $ 12 \times 11 \times 11 \times 10 \times 9$ to choose 2 girls and 3 boys, which is a total of $ 130680$.

Therefore, the probability is $ 0.008$.

Isabella , for this problem you would use combinations, because order does not matter.

Meh real solution:
[hide]
Total ways is $30$ choose $5$. Which is $142506$.
The number of ways we can get $3$ freshman boys is $11$ choose $3$. Which is $165$.
The number of ways we can get $2$ freshman girls is, clearly, $12$ choose $2$. Which is $66$.
We multiply $66*165=10890$.
Our answer is $\boxed{\frac{605}{7917}}$.
[/hide]


Prove that: If $ a,b,c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then:
\[ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \geq a^2 \plus{} b^2 \plus{} c^2
\]
I tried to do this problem by Chebyshev Inequality and S.O.S method but I can't. Plz help me !


Michael Jordan's probability of hitting any basketball shot is three times greater than mine (which ranges from 0 to $\frac{1}{3}$).  Jordan and I will each attempt a shot of my choice once, and I will win if I hit the shot and he misses.  What is the maximum probability of winning which I can attain?

[hide="Very inelegant solution"]I dunno if you are allowed to graph using calculator, but I used graphing calculator to find maximum of X - 3X 2 where X is probability of you hitting shot and 1 - 3X is MJ prob. Answer = 1/12

your probability = (1/6)
Mj's prob = (1/2)

Someone else should find more elegant solution[/hide]

[hide]
Probability that you make your basket = x
Probability that MJ doesn't make his basket=1-3x
$P(A and B)=P(A)*P(B)$
$P(win)=x(1-3x)=x-3x^2$
This is a parabola, whose maximum is its vertex, given by -b/2a
-1/(2*-3)=1/6.
Putting this into the equation, we obtain Max P(win)=1/12[/hide]

Okay in defence of truth let us   drop the name of Michael Jordan ( Value of $p$ for him, even if he is sleeping on the court will be pretty close to $1$...:)

So .. let us call him just Mr. M.  and assume that probablity for him making the basket is $p$

Interesting thing is:   it does not matter  if he is 3 times better than you  or  a hundred times, best chance for you to win  with above condition is when [hide="figure out yourself before clicking"]  when  $p= \frac 1 2 $ [/hide]

[hide=" Why intutively "]If p is close to one, that is Mr. M is too  good, he is not gonna miss, and thus you would loose.
If p is close to zero, you are very bad too, and you are not gonna make the basket - again you loose.

So for you to win best value is when p is somewhat in between.
[hide="If you really want math"]let us assume   $p = \frac 1 2 + x $
So probablity of your  making basket is  $\frac{(1/2)+x }{3} $
And him missing is  $(1/2)-x)$
So for a win  the prob is $\frac {((1/2)+x)((1/2)-x)}3$ = $\frac {(1/2)^2 - x^2} 3$
Obviously $x^2$ is always positive, so max value will reach when $x$ is $0$ hence $p =(1/2)$  and thus the probablity that you would win is $(1/6)(1/2)$[/hide] [/hide]

Hope it is helpful...


calculus
integration
trigonometry
calculus computations

yes. You can also substitute 4x=u and solve it.

Or writing: $ \int \! \frac {1}{9 \plus{} 16x^2} \, dx \equal{} \int \! \frac {1}{3^{2} \plus{} (4x){^2}} \, dx.$, which is a well known form.

hello, we have $ \int\frac{1}{a^2\plus{}b^2x^2}\,dx\equal{}\frac{1}{ab}\arctan\left(\frac{b}{a}x\right)\plus{}C$
Sonnhard.

Hi :) !!

wa can also take this forme :

$ \int \frac{dx}{ax^2 \plus{} bx \plus{}c} \equal{} \frac{2}{\sqrt{\minus{}\Delta}} arctan \left( \frac{2a}{\sqrt{\minus{} \Delta}} x \minus{} \frac{b}{\sqrt{\minus{} \Delta}} \right) \plus{} Cte$ with obvously $ \Delta < 0$


number theory
least common multiple
number theory unsolved

let $n \geq a_{1}> \cdots > a_{k}$ be positive integers such that $lcm(a_{i}, a_{j}) \leq n$ for all $i, j$.
 Prove that $ia_{i}\leq n$ for $i=1, 2, \cdots, k$.

Have a look here for a proof and more :
http://www.mathlinks.ro/Forum/viewtopic.php?t=103571

Pierre.


geometry
conics
parabola
real analysis
real analysis unsolved

Find the maximum area of a frustrum bounded by a paroloid and line y=0. (constraining parabola =-x^2 + 16.)


quadratics
algebra
polynomial
number theory
Diophantine equation
number theory unsolved

This just came to my mind. Given a fixed $n$, the diophantine equation $\sum_{k=1}^{n}\frac{1}{a_k}=1$ always have a solution on the positive integers. It is also true that for any of such solutions $\sum_{k=1}^{n}a_k\geq n^2$, but, is there any quadratic polynomial $p(x)$ such that $\sum_{k=1}^{n}a_k\leq p(n)$? If it does, give such a $p$. If it does not, what is the smaller degree of sucha $p$?

I know that this is not exactly a regular olympiad number theory question, but I wasn't sure where else it could belong.

There is no such polynomial.

The maximum occurs for $a_1=2,a_2=3,a_3=7,\dots,a_{k+1}= a_k^2-a_k+1,\dots,a_n=a_{n-1}^2-a_n$. The sum is dominated by the final term, and $a_n$ grows like $c^{2^n}$ for some $c$.


geometry
trigonometry
inequalities
advanced fields
advanced fields unsolved

Assume that $ C$ is a given circle and $ n\geq 3$ is a fixed integer.

1. Prove that among the cyclic $ n$-gons inscribed in $ C$ there exists one, say $ P$, with maximal area.
2. Prove that $ P$ is regular.

Partition the $ n$-gon into triangles by drawing lines from the center of the circle to the vertices of the $ n$-gon.  Let the interior angles formed be $ \theta_1, ... \theta_n$.  Then the area of the $ n$-gon is (given a radius of $ 1$)

$ \frac{1}{2} \sum_{i\equal{}1}^{n} \sin \theta_i$.

If the $ \theta_i$ are all less than $ 180^{\circ}$ then $ \sin \theta$ is concave and Jensen's inequality applies.  Otherwise, the $ n$-gon takes up at most half the area of the circle and it is clear that this isn't maximal.


Να αποδειχθεί ότι οι ακολουθίες

$ u_{n \plus{} 1} \equal{} \frac {u_{n}v_{n}(u_{n} \plus{} v_{n})}{u_{n}^{2} \plus{} v_{n}^{2}}$
$ v_{n \plus{} 1} \equal{} \frac {u_{n} \plus{} v_{n}}{2}$

με $ 0 < u_{0} < v_{0}$ ισοσυγκλίνουν και να βρεθεί το κοινό τους όριο...
Πιστεύω ότι μεγαλύτερη σημασία έχει η εύρεση του ορίου μιας κι έχω βρει έναν τρόπο που δεν τον είχα ξαναδεί...

[hide]
$ U_{n \plus{} 1} \equal{} \frac {U_{n}V_{n}(U_{n} \plus{} V_{n})}{U_{n}^{2} \plus{} V_{n}^{2}}$

$ V_{n \plus{} 1} \equal{} \frac {U_{n} \plus{} V_{n}}{2}$

με $ 0 < U_{0} < V_{0}$

Υποθέτουμε ότι $ 0 < U_n < V_n$. Θέτουμε $ a_n \equal{} \frac {1}{U_n}$ και $ b_n \equal{} \frac {1}{V_n}$. Τότε $ 0 < b_n < a_n$

[hide="Εύρεση αναδρομικών τύπων για τα a_n, b_n"]
$ \frac {1}{U_{n \plus{} 1}} \equal{} \frac {U_{n}^{2} \plus{} V_{n}^{2}}{U_{n}V_{n}(U_{n} \plus{} V_{n})} \Rightarrow$

$ a_{n \plus{} 1} \equal{} \frac {\left(\frac {1}{a_n}\right)^2 \plus{} \left(\frac {1}{b_n}\right)^2} {\frac {1}{a_n}\frac {1}{b_n}(\frac {1}{a_n} \plus{} \frac {1}{b_n})} \equal{} \frac {\left(\frac {1}{a_n}\right)^2 \plus{} \left(\frac {1}{b_n}\right)^2} {2\frac {1}{a_n}\frac {1}{b_n}}\cdot \frac {2}{\frac {1}{a_n} \plus{} \frac {1}{b_n}} \equal{}$

$ \equal{} \left(\frac {\left(\frac {1}{a_n} \plus{} \frac {1}{b_n}\right)^2} {2\frac {1}{a_n}\frac {1}{b_n}} \minus{} 1\right)\cdot \frac {2}{\frac {1}{a_n} \plus{} \frac {1}{b_n}} \equal{} \frac {\frac {1}{a_n} \plus{} \frac {1}{b_n}} {\frac {1}{a_n}\frac {1}{b_n}} \minus{} \frac {2}{\frac {1}{a_n} \plus{} \frac {1}{b_n}} \equal{} (a_n \plus{} b_n) \minus{} \frac {2}{\frac {1}{a_n} \plus{} \frac {1}{b_n}}$

$ \frac {1}{V_{n \plus{} 1}} \equal{} \frac {2}{U_{n} \plus{} V_{n}}\Rightarrow$

$ b_{n \plus{} 1} \equal{} \frac {2}{\frac {1}{a_n} \plus{} \frac {1}{b_n}}$

Παρατηρούμε ότι $ a_{n \plus{} 1} \plus{} b_{n \plus{} 1} \equal{} a_n \plus{} b_n$, δηλαδή το άθροισμα διατηρείται. Άρα  $ a_n \plus{} b_n \equal{} a_0 \plus{} b_0 \equal{} S,\ \ \forall n$

Επειδή ο $ b_{n \plus{} 1}$ είναι ο αρμονικός μέσος των θετικών $ a_n,b_n$ θα έχουμε $ 0 < b_n < b_{n \plus{} 1} < \frac {a_n \plus{} b_n}{2} \Rightarrow 0 < b_{n \plus{} 1} < (a_n \plus{} b_n) \minus{} b_{n \plus{} 1} \equal{} a_{n \plus{} 1}$

Έτσι, με την υπόθεση $ 0 < U_n < V_n$ καταλήξαμε στο $ 0 < b_{n \plus{} 1} < a_{n \plus{} 1} \Rightarrow 0 < U_{n \plus{} 1} < V_{n \plus{} 1}$, επαγωγικά ότι αυτή η διάταξη θα ισχύει για κάθε $ n$[/hide]
________________________________________________________________________________________

Έχουμε λοιπόν τις ακολουθίες $ 0 < b_n < a_n$ με $ a_0 \plus{} b_0 \equal{} S$ και

$ b_{n \plus{} 1} \equal{} \frac {2a_nb_n}{S}, \ \ \ a_{n \plus{} 1} \equal{} S \minus{} b_{n \plus{} 1}$

Ορίζω τώρα τις ακολουθίες:

$ x_n \equal{} a_n \minus{} \frac {S}{2} \ \ \ y_n \equal{} b_n \minus{} \frac {S}{2}$

[hide="Εύρεση αναδρομικών τύπων για τα y_n"]
$ b_{n \plus{} 1} \equal{} \frac {2a_nb_n}{S} \Rightarrow$

$ y_{n \plus{} 1} \plus{} \frac {S}{2} \equal{} 2\frac {\left(x_n \plus{} \frac {S}{2}\right)\left(y_n \plus{} \frac {S}{2}\right)}{S}$, και αφού $ x_n \plus{} y_n \equal{} 0\Rightarrow$

$ y_{n \plus{} 1} \plus{} \frac {S}{2} \equal{} 2\frac {\left( \minus{} y_n \plus{} \frac {S}{2}\right)\left(y_n \plus{} \frac {S}{2}\right)}{S}$

$ y_{n \plus{} 1} \plus{} \frac {S}{2} \equal{} 2\frac {\frac {S^2}{4} \minus{} y_n^2}{S}$

$ y_{n \plus{} 1} \equal{} \minus{} \frac {2}{S}\cdot y_n^2$[/hide]
____________________________________________________________________________________________

$ y_{n \plus{} 1} \equal{} \frac {1}{\Phi} \cdot y_n^2 \ \ (1)$, όπου $ \Phi \equal{} \minus{} \frac {S}{2}$

ΘΔΟ $ y_n < y_{n \plus{} 1} < 0, \ \ \forall n$

Αντί να αποδείξω ότι η ακολουθία $ \{y_n\}$ είναι γν.αύξουσα και φραγμένη και να ψάχνω το όριο, προτιμώ να βρω το γενικό τύπο της $ y_n$

$ \blacksquare$ ΘΔΟ $ y_n \equal{} \Phi \left(\frac {y_0}{\Phi}\right)^{2^n}$

[hide="Απόδειξη"]
$ \bullet$ Για $ n \equal{} 0$ ισχύει, αφού $ \Phi \left(\frac {y_0}{\Phi}\right)^{1} \equal{} y_0$

$ \bullet$ Υποθέτω ότι ισχύει για $ n$ και θδο $ y_{n \plus{} 1} \equal{} \Phi \left(\frac {y_0}{\Phi}\right)^{2^{n \plus{} 1}}$

$ y_{n \plus{} 1} \equal{} \frac {1}{\Phi}y_n^2 \equal{} \frac {1}{\Phi} \left[\Phi \left(\frac {y_0}{\Phi}\right)^{2^n}\right] ^2 \equal{} \Phi \left(\frac {y_0}{\Phi}\right)^{2^{n \plus{} 1}}$
[/hide]
___________________________________________________________________________________________________

Είναι

$ 0 < b_0 < \frac {S}{2} \iff$

$ \minus{} \frac {S}{2} < b_0 \minus{} \frac {S}{2} < 0 \iff$

$ \Phi < y_0 < 0 \iff$

$ 1 > \frac {y_0}{\Phi} > 0$

Άρα η ακολουθία $ y_n \equal{} \Phi \left(\frac {y_0}{\Phi}\right)^{2^n}$ τείνει στο $ 0$

$ b_n \equal{} y_n \plus{} \frac {S}{2} \Rightarrow \{b_n\} \to \frac {S}{2}$

$ a_n \equal{} S \minus{} b_n \Rightarrow \{a_n\} \to \frac {S}{2}$

$ U_{n} \equal{} \frac {1}{a_n} \Rightarrow \{U_n\} \to \frac {2}{S}$

$ V_{n} \equal{} \frac {1}{b_n} \Rightarrow \{V_n\} \to \frac {2}{S}$

όπου $ S \equal{} a_0 \plus{} b_0 \equal{} \frac {1}{U_0} \plus{} \frac {1}{V_0} \Rightarrow \frac {2}{S} \equal{} \frac {2U_0V_0}{U_0 \plus{} V_0}$[/hide]

Υ.Γ. Ελπίζω να υπάρχει απλούστερη λύση  :D

Λουκάς

Για την  άσκηση που έδωσα θα δώσω μόνο επιγραμματικά τι κάνω διότι μερικά σημεία είναι εύκολα να αποδειχθούν...

Αποδεικνύω τα εξής:

$ 0 < u_{n} < v_{n}$ για κάθε $ n\in R$
$ u_{n < u_{n + 1}}$ και $ v_{n + 1} < v_{n}$ για κάθε $ n\in R$

Οπότε οι δύο ακολουθίες συγκλίνουν.
Ύστερα αποδεικνύω ότι

$ v_{n + 1} - u_{n + 1} = \frac {1}{2}(v_{n} - u_{n})$ και με απαγωγή βγαίνει ότι $ v_{n + 1} - u_{n + 1} = (\frac {1}{2})^{n + 1}(v_{0} - u_{0})$,άρα οι δύο ακολουθίες ισοσυγκλίνουν οπότε έχουν κοινό όριο το $ l$.

Τέλος θέτω όπως εσύ τις ακολουθίες $ a_{n} = \frac {1}{u_{n}}$ και $ b_{n} = \frac {1}{v_{n}}$ και μετά από λίγες πράξεις βγαίνει ότι

$ a_{n + 1} + b_{n + 1} = a_{n} + b{n} = .... = a_{0} + b_{0}$
Παίρνοντας το όριο για την τελεταία σχέση έχουμε ότι
$ \frac {1}{l} + \frac {1}{l} = \frac {1}{u_{0}} + \frac {1}{v_{0}} < = > l = \frac {2v_{0}u_{0}}{v_{0} + u_{0}}$ που ουσιαστικά αποδεικνύει το ζητούμενο....

Δεν ξέρω αν είναι απλούστερη αλλά τουλάχιστον δεν χρειάστηκε να γράψω πολλά σε Latex....


The digits of a number in base $ q$ are represented by $ a_{n}a_{n\minus{}1}a_{n\minus{}2}...a_{3}a_{2}a_{1}a_{0}$.

(i) Using sigma notation, find the number and the sum of its digits.

(ii) Assuming that the number has two digits, find a eform for bases where the standard divisibility test for $ 3$ works and then prove it for all numbers.

I think its something like this:

(i)
N = $ a_{n} . q^n \plus{} a_{n\minus{}1} . q^{n\minus{}1}...a_{0} . q^0$
I don't know how to use sigma notation, but the sum of digits is clearly an + an-1 ... + a0

Take $ 1337$, for instance in $ base 10$:
number of digits = 4, our powers are 0,1,3

$ 1337 \equal{} 1 . 10^3 \plus{} 3 . 10^2 \plus{} 3 . 10 \plus{} 7 . 10^0$

(ii)
A number is divisible for 3 when the sum of their digits equals a multiple of 3, that is, for 2 digits:
$ a_{1} \plus{} a\plus{}{0} \equal{} 3|c$

Now what means 3|c ? I have no idea, but let me guess again:

c is multiple of 3, that is c = 3*q + r, for some q and r in Z, or something like that...

PS: Sorry for all the mistakes above and for the horrible presentation (I am going to learn latex later today)


arithmetic sequence
Ramsey Theory
combinatorics unsolved
combinatorics

The points on a circle are colored in three different colors. Prove there exist infinitely many isoceles triangles with vertices on the circle of the same color.

Pascual

Recall Van der Waerden's theorem which states that for each every positive integers $k,r$, there exists an integer $N=W(k,r)$ such that for any $r$-coloring of $\{1,2,...,N\}$ there exists a monochromatic arithmetic progression of length $k$.

Now, suppose for a contradiction that your coloring pf the circle produces only a finite number of isocele triangles, say $k$ (note I don't care about the fact that the coloring uses 3 or more colors, just a finite $r$ number). Now let's choose $N=W(k+3,r)$ and choose a regular $N$-gon inscribed in the circle, say $A_1A_2...A_N$.
The coloring of the circle gives a coloring of the $N$-gon, which induces a coloring of the integers $1,2,...,N$.
By VdW theorem there exist a monochromatic arithmetic progression  of length $k+1$, say $a_1,...,a_{k+3}$. Now it turns that for $i=1,...k+1$, each of the tiangles $A_iA_{i+1}A{i+2}$ is isocele at $A_{i+1}$ and has its three vertices with the same color, a contradiction.

Pierre.

Nice, when i tried to solve the problem i wanted to prove also i could find a number N such that in a regular n-agon there was an arithmetic progression of lenght at least 3, but i didnt know it was a theorem and it had a generalization, I think  that W(2,3)=9, at leas it is a Bulagarian problem from 1998, and i know how to prove it, but do you know a proof of the general statement???

Thanx...

PAscual

The proof of the general fact is long and tedious. I read a proof once, but I was pretty confused and I definitely can't remember anything right now. Try looking it up on the Internet, pascual.

Well, I know several proofs of VdW theorem, some are very clear, but they are very long so I've to apologize but I will not write one here.
You should better look on some classical books on Ramsey theory as
R.L.Graham, B.L.Rothschild, J.H.Spence, ''Ramsey Theory'', Wiley
which is the best on the subject (very clear),
Or look after
A.Y.Khinchin, ''Three pearls of number theory'', Dover.

In another hand, few exact values of VdW numbers W(k,r) are known.
Clearly W(1,r) = 1, W(k,1) = k, and W(2,r) = r+1. As far as I know, the only nontrivial known values are W(3,2) = 9, W(3,3) = 27, W(3,4) = 76
W(4,2) = 35 and W(5,2) = 178.

Pierre.

i know a different solution of this problem. it takes a regular 13-gon and shows that it consists a monochromatic isosceles triangle. it isn't short but it works.


what do you think?
i expect the third option to have the unanimous vote

I can't understand why she should die in that terrible way. Why don't give she a mercy killing? I will prefer to be beheaded if I got only two choices.
Let Terri die honourably, what an altisonant excuse!  Make a mummy before she die?

[quote="bubala"]Lol you spelled schiavo wrong!

Anyhow, I agree that the whole thing is BS.  It's just another thing the Republicans are doing to create a wedge issue and please the evangelical Christian wrong, as I call it.

And yes, the women definitely has the right-to-die, but I'm not sure she felt any pain for the person who said he/she would've preferred a beheading.  Remember: she was in a persistent vegetative state.[/quote]

Please the Evangelical Christin wrong? Care to explain?

Something is strange here. I thought she already died several days ago, don't understand how you are discussing the issue as if it didn't happend yet.

i think noone has a a right to kill another one

[quote="A+MATH"]Ugh, lets please not start a discussion about her here.[/quote]
why would the topic be posted if ppl weren't supposed to discuss it.
i feel using technology to sustain a dying person is not right. and by letting her die, it was not murder. although lethal injection might have been nicer than starvation.

[quote="bubala"]Yes, she did die several days ago.

The evangelical Christian wrong was a joke.
It's called the Christian right because many evangelicals are very conservative, and conservative = right, liberal = left.
However, I replaced right by wrong, because I feel that those people are not right, but wrong!!!!

(And yes, I'm liberal if you were wondering)[/quote]

Librelism is a mental disorder. :b

She should be allowed to live.

Many docters and nurses said that she was not braindead.

And if you give one life is it not wrong to take the saame life?

[quote="tetrahedr0n"][quote]Many docters and nurses said that she was not braindead. [/quote]

Yes, but somehow this happened only after the case reached national acclaim. Of course numerous doctors and politicians will bastardize their true beliefs to grab their fifteen minutes of fame.[/quote]
Well actually nurses came forward with information that she was crying after Michale Schiavo had come out of her room and the nurse claims to have found needle marks in Terri's body.

IMO, we should just respect her death. All anyone wants is to die with dignity. What she endured (these past few months) I think everyone could agree was not what she would have wanted. I certainly know that I would not have wanted a hoopla surrounding my suffering. At this point, I feel that all we can, or should, do is to mourn for her family, and to learn from this. If this is not what you want for your life, TELL YOUR FAMILY THAT. It's never too early. I told my mother when I was 14 that if something happened to me, I wanted all of my organs donated, and then I wanted to be cremated. If you are old enough to have an opinion, make it known! Bring it up NOW!

[quote="shobber"]who's Terri Shiavo?? :?[/quote]

a woman in florida, USA who was appearantly braindead and her husband wanted to take her off life support (including a feeding tube). she got taken off of life support and later died (i'm guessing of starvation). there was a lot of political controversy over the issue.

[quote="bleumoose"]I told my mother when I was 14 that if something happened to me, I wanted all of my organs donated, and then I wanted to be cremated. If you are old enough to have an opinion, make it known! Bring it up NOW![/quote]

Unfortunately, that is insufficient. You need it in writing, and to be close to safe, you have to have the infamous "living will" done. Even that is arguably insufficient.

[quote="shobber"]
who's Terri Shiavo??[/quote]

A litte more information: She fell into her state (no brain activity, not a coma) about 15 years ago. It is believed to be caused to heart failure, as she suffered from an eating disorder. Her husband, and thus her official guardian, spent five years and huge sum of money on her recovery (he went as far as getting a nursing degree to better take care of her.) However, after these five years and repeated verdicts of complete, unrecoverable brain death, he gave up on a potential recovery. He claims that after the painful death witnessed by both of them of a relative, she said she wouldn't want to be in this state. Thus, he requested the feeding tube be removed. The parents however disagree, and a messy lawsuit resulted, followed by media and political interest in the case.

[quote="furious"]she got taken off of life support and later died (i'm guessing of starvation).[/quote]

Dehydration, actually.

[quote="tetrahedr0n"]What you said above does not in any way contrast with what I wrote.[/quote]
she came out with this information two years ago.

[quote="icantpeeloranges"]
And if you give one life is it not wrong to take the saame life?
[/quote]

Of course, everything is flipped upside down if you are a criminal, or a soldier.

Also, I find it hard to believe that Fundamentalist doctors and nurses where not influenced by their religious beliefs when they made their conclusions.

well, i don't think they should take off her life support. Since there will be still chances for her to wake up again. but they killed her, which means they give up the last hope. it was really not a very good way.

[quote]Since there will be still chances for her to wake up again.[/quote]

What makes you think this is true?

She was in a completely vegetative state. All truly professional [i]independent[/i] opinions said thus, so she wasn't going to wake again.

I truly believe that both the husband and the parents had the best interests of poor Terri at heart. The only selfish nimrods in this issue were the Republicans in Congress who made it an issue of national morals. This case had no place being on the news 24/7.

All truly independent?

No one is independent.

And I'm a thinking that the ones who said she was still capable of living weren't all conservatives. But if they were it's just because conservatives like innocent people living.

Conservatives like innocent people living?

NO NO NO NO NO!!!!!! You DON'T GET IT!!!!! Let me help you.

1) Conservatives use social issues with right-to-life mumbo jumbo to attract the religious right.
2) Conserrvaties couldn't care less about innocent people living!

Conservatives save fetuses (anti-abortion), people in persistent vegetative states (i.e. Terri Schiavo), and frozen embryos that are about to be thrown out (anti-stem cell research).

These "pro-life" positions attract the religious right, and is a main factor in conservative control  of the Bible Belt.

On the other hand, conservatives care nothing for actual people!  They are warmongers (Pres. Bush, Dick Cheney, Donald Rumsfeld, etc.) who bascially sent hundreds of thousands of soldiers off to Iraq, where 1500 will not return and counting.  DONALD RUMSFELT STAMP SIGNED LETTERS TO FAMILIES OF SOLDEIRS WHO DIED!! Oh yeah, conservatives are really compassionate and caring people, and they support our troops a lot.

By being anti stem-cell research, conservatives are saving frozen embryos that are about to be thrown out anyway.  Why not use them to potentially develop cures to dieases that kill millions and millions of people throughout our world each year, you ask?  Well, because they want to take the "pro-life" stance to attract the religious right.  Throw out the embryos, don't let science use them!!! Who cares about curing disease

Well, yesterday I got a paper cut, and a few of my blood cells died.  Want to arrest me?  I guess should've been more pro-life and avoided the paer cut.

Lastly, conserveratives are pro-gun to attract farmers and other rural Americans who like to hunt.  GUNS KILL TENS OF THOUSANDS OF AMERICANS EACH YEAR.  Many other countries have strict gun-control laws.  But no, even though Democrats are fighting to get dangerous guns off our streets to keep us safer, conservatives shout "Second Amendment rights" without even bothering to read the FIRST HALF OF THE FREAKING AMENDMENT!

Through these "pro-life" but really anti-life positions, conservatives convince millions of people to vote against their own economic self-interests.  Basically, they convince people to agree to keep their high gas prices, cut taxes for the rich, keep having no health insurance coverage, keep living poorly, etc.  all so that they can be "pro-life."

It's ridiculous!!! But you know what, I'll take the tax cut.  It's given us rich people a LOT of money.  And I do like those nice vacations...

On the other hand, I didn't need the money.  The middle-class and the poor needed the money.  There are a lot of problems in America and in the world.  Saving fetuses and people in a vegetative states should NOT be our first priority.  Please.

[quote="bubala"]Conservatives couldn't care less about innocent people living![/quote]

I'm going to take this as a joke and just ignore it.

[quote="bubala"]These "pro-life positions attract the religious right, and is a main factor in the conservative control of the Bible Belt[/quote]

So they do. Does that mean they're wrong?

[quote="bubala"]On the other hand, conservatives care nothing for actual people! They are warmongers (Pres. Bush, Dick Cheney, Donald Rumsfeld, etc.) who bascially sent hundreds of thousands of soldiers off to Iraq, where 1500 will not return and counting. DONALD RUMSFELT STAMP SIGNED LETTERS TO FAMILIES OF SOLDEIRS WHO DIED!! Oh yeah, conservatives are really compassionate and caring people, and they support our troops a lot.[/quote]

So Terry Schiavo's not an actual person. Cool.

Also, somebody could turn that around very easily and say that if troops weren't in Iraq, more people would be suffering greatly each day and it would be uncompassionate and not caring to not send troops. But, that's another issue. Search for it on the forums. I'm sure it's up there somewhere.

Side note- I love your spelling of Rumsfeld the second time, lol.

[quote="bubala"]Well, yesterday I got a paper cut, and a few of my blood cells died. Want to arrest me? I guess should've been more pro-life and avoided the paer cut.[/quote]

I get paper cuts quite often. They don't really hurt, but they're very annoying. Haven't gotten one since school ended last year, though. Maybe it's just a middle school thing.

[quote="bubala"]Through these "pro-life" but really anti-life positions, conservatives convince millions of people to vote against their own economic self-interests. Basically, they convince people to agree to keep their high gas prices, cut taxes for the rich, keep having no health insurance coverage, keep living poorly, etc. all so that they can be "pro-life." [/quote]

So abortions raise gas prices. Hm, never knew that. Maybe we could use some of those stem-cells to limit the price of gasoline...

If it interests you at all, I'm not a Conservative. I actually have more liberal views. But partly because about 2/3 of my school is liberal, I don't like being liberal very much. Frankly, I find liberals in my school and many others that I know annoying. They talk about how politics should be and bash the right wing and sometimes, christians in general. A ton of it is just plain BS. They seriously think all Republicans are stupid and all evangelical christians are dumbasses. I hear that stuff so much, it's like they're always looking for something to complain about. So, I've heard the stuff you said about a thousand times this year. It's that kind of stuff that makes me just not care about politics. I'll look into the issues and decide for myself whatever the heck I want to do, but I'm not going to try to persuade others or to join in on any "discussions".

Your accusations are baseless.

[quote]bubala wrote:
These "pro-life positions attract the religious right, and is a main factor in the conservative control of the Bible Belt

So they do. Does that mean they're wrong? [/quote]

I was about to discuss why it was wrong, which I did do later in the post.

[quote]bubala wrote:
On the other hand, conservatives care nothing for actual people! They are warmongers (Pres. Bush, Dick Cheney, Donald Rumsfeld, etc.) who bascially sent hundreds of thousands of soldiers off to Iraq, where 1500 will not return and counting. DONALD RUMSFELT STAMP SIGNED LETTERS TO FAMILIES OF SOLDEIRS WHO DIED!! Oh yeah, conservatives are really compassionate and caring people, and they support our troops a lot.

So Terry Schiavo's not an actual person. Cool. [/quote]

Pardon me for placing the value of thousands and thousands of human beings over the value of one person who was in a persistent vegetative state and could never become conscious again.  Look, I know that all life is precious, but some life is more precious than others.  In other words, I wouldn't murder my next-door neighbor if he bit me.  I might murder a bug if it bit me.

I'm very sorry that you have fallen victim to Bush/Cheney propaganda.
[quote]somebody could turn that around very easily and say that if troops weren't in Iraq, more people would be suffering greatly each day and it would be uncompassionate and not caring to not send troops[/quote]

Look, right, we went into Iraq to look for weapons of mass destruction.

BUT THERE WERE NO WMD!!

Then, it was because of the link between Saddam and Al Qaeda

BUT THE 9/11 COMMISSION, CHAIRED BY A REPUBLICAN WHO BUSH APPOINTED HIMSELF, SAID THERE WAS NO CONNECTION BETWEEN SADDAM and Al Qaeda.

Then, we were liberating the people and bringing freedom to the middle east.

Saddam tortured his people, yes, but we did, too.  Remember ABU GHARIB!!!

AND WE WERE GOING TO BE RECEIVED AS LIBERATORS

Oh, shoot, we underestimated the size of the insurgency.

And the Iraqi oil is going to pay for the reconstruction, except for all the hundreds and hundreds of billions of dollars that the Congress has given the President to pay for the war in Iraq that could easily have gone to improving our schools or other bettter causes.

Btw, you totally missed the point of the paper cut thing!  I meant that it's stem cell research is just like killing an embryo that will be thrown out in 2 wks anyway (just like a red blood cell) to help save millions of lives.

And no, abortions don't raise gas prices.  Bush's policies raise gas prices!  Why doesn't he work toward an energy-independent America?  Oh, well, his Saudi friends might get mad. But, let's save Terri Schiavo and kill all the poor animals in Alaska because we need to drill more U.S. oil even if it only lasts us 6 months, right?

Bubala, I think I speak for several of us when I say that you're really coming across as (forgive me for saying so) a jerk.  Take a chill pill and try actually reading what other people are saying before you assume the fetal position.  I don't think anyone accused you of disrespecting the troops or disrespecting life.

The "paper cut" point doesn't work.  Science has shown that not all stem cells are created equal in terms of their ability to be used for genetics and medical research.  Embryonic stem cells are far more useful than anything you can get with a simple paper cut.  I don't support the Republican stance on stem cell research, but they're simply trying to avoid a world where we harvest mass quantities of embryos in a hubris-driven effort to create "the perfect human."  Supporters of stem cell research have enough faith in our doctors' sense of medical ethics to believe this isn't going to happen.  I think that's the difference between the positions, not that Republicans have a bizarre emotional attachment to a single discarded stem cell.

The idea that "some life is more precious than others," while obviously a value statement, is a rather grim window into the way some people think.  I hope you're not actually implying that a fully-able person's life is more precious than a mentally disabled person's life.  The former may be more "useful" to society in terms of material productivity, but I think that we desecrate our own essential humanity when we say that some lives are more worth protecting than others.

Finally:  I'm of the sincere belief that any intelligent, discerning American should have serious concerns about the Bush administration's sense of ethical conduct.  That's why I voted against him and will continue to vote against legislators who support him.  I was angry when he won the election, but what made me even more pissed off was that Bush voters then went out of their way to shove their victory in the faces of people who, in good conscience, refused to vote for Bush.  Similarly, I think it's really crass and classless for liberals to talk down to opponents and naysayers as though they're third graders who don't have the mental faculties to comprehend the subtleties of politics.  The absurd thing about all this is that you went to such trouble over someone who quite clearly said that he/she isn't conservative.  I don't mean to embarrass you, but I just think the last post was a little mean-spirited, considering the respondent had some good questions to ask.

I feel sorry for her, as well as to her husband. It was right for her to be dead(no offense), because nobody wants to see her suffer any longer, especially her husband, who according to the whole controversy seemed to have calm and unmoving poise. I think that Terri Shiavo's parents mishandled the business. They misunderstood what was best for her. And i dont like the whole political and court case thing that is going on. Person's life shouldnt be controversy enough to put a lot of people and a lot of money at the court. ONce again, best wishes to the Shiavo family.

Mathjoker, I agreee with you.  I was just in a very bad mood last night.  I edited the post.  Sorry.

First off, you misread me! I do not think that a fully abled person is more important than a mentallly disabled person!! I was saying that a fully-abled person is more important than a person in a vegetative state.  If you feel that I'm wrong, I'm sorry.
I just think that a person in a vegetative state is never going to become conscious and wake up, so instead of trying to prevent him/her from dying, we should try and prevent other conscious people from dying.  If you came up to me tomorrow and shot me, and my family sued yours and you were arrested, etc.  (not that you would ever do that) it might be on the front page of the newspaper for 1 day.

On the other hand, the Terri Schiavo case was on the front page of our newspapers for a week?  Why?  Because Congress and the President were taking ridiculous actions to save her.  Why wouldn't they do that for people like you and me, who are fully conscious?  (There's this joke that the Republicans don't care about you no matter how sick you are, unless you're in that bad condition, and then and only then they'll save your life.  )

Anyhow, I'm well aware that yif man 12 said he wasn't conservative.  It's just that I found some of his post to be
obnoxious.  And the combination of that and me being in a bad mood made my post obnoxious.  Sorry again. But for example,

[quote]So Terri Schiavo's not a real person.  Cool.[/quote]

[quote]
I'm going to take this as a joke and just ignore it.
[/quote]

[quote]I get paper cuts quite often. They don't really hurt, but they're very annoying. Haven't gotten one since school ended last year, though. Maybe it's just a middle school thing. [/quote]

And look at this last one.  [quote]So abortions raise gas prices. Hm, never knew that. Maybe we could use some of those stem-cells to limit the price of gasoline... [/quote]

I mean, come on!  Being in a bad mood, when I saw this (and you've got to agree that it's very condescending too) I just flew off the handle.   I apologize for that. Don't worry about embarassing me, I totally agree.

Btw, you also missed the point of my paper cut thing.  Both a paper cut and an embryo are similar in that
1) both will die in a few weeks, ( the embryo because the IV clinic will throw it out)
2) both a just a few cells that will never grow into a person.

However, the embryo can potentially be used to develop a cure for major diseases and save millions of lives.  So why not use it?

Thank you for voting against President Bush.  I commend you for it.  I would have, but I'm not old enough yet.

I totally agree with the proposition that the enormous amount of media attention devoted to the Schiavo case was outrageous and uncalled-for.  I think that it was caused largely by the extremely childish conflict between the Schiavos and the Schindlers, which ultimately forced the courts and even the government to intervene.  Each side had a legitimate interest in the case and a reason to be upset at the other side, but they handled the whole mess in an incredibly stupid way.  This stupidity impoverished both families (unless Terri's medical trust fund covered all the legal expenses), and lined the pockets of greedy lawyers who love cases like this.  I think each lawyer's fees (Gibbs and Felos) came to something like half a million dollars.  A mathematician would have to prove the Riemann Hypothesis, the Birch-Swinnerton-Dyer Conjecture, or some other extremely major problem to make that much.  Wow.

I don't think any human has a right to decide that one human life deserves more protection than another.  All we can decide is what lives WE personally choose to protect, and what protections we wish to place on our own lives.  Personally, I would not wish to be kept alive artificially if I were in a persistent vegetative state.  But I would put that in writing for the benefit of people who do not believe they can in good conscience make that decision for me.

And as I say, I agree with you on stem cell research.  The embryos have to be used after a certain amount of time or get tossed out - absolutely.  I think the Republican disagreement with this comes from a fundamental misunderstanding of how the research works, and a general distrust of the ethics of science research (at least where humans are concerned).  But I also think that Republicans fear the possibility that if we let stem cell research remain unregulated now, we'll end up going to greater lengths to acquire embryos for research purposes.  At that point, we may be harvesting embryos from places other than IV clinics that won't use them.  I think it's a concern that at least deserves to be addressed.

So you're saying there is no humor/sarcasm allowed in the Round Table?  :?

No, not at all!!! I just think that you went over the edge.  I did too, originally.  However, mathjoker reprimanded me for doing so, and I changed my post.  I was just pointing out that your post was also very condescending , and that was one of the factors that led me to post in a condescending manner.  So yes, I think that human and sarcasm is fine, but if you really want to know, I believe you crossed the line.  But no worries though!!!! :)

As to the which human life deserves more protection than others, I still stand by what I said.  I believe that a fully abled person deserves more important protection than a person in a persistent vegetative state.  I think that many people have this belief that all human life should be protected equally because it's a religious teaching.  At the same time, they totally disrespect all other kinds of life.   In other words, conservatives care nothing for the environment.  They want to drill for oil in Alaska, they don't care about pollution, they don't care about deforestation and desertification, etc. yet the want to save a person who cannot think, who is not conscious, who can never wake up, and if given food and water, will basically just exist for years on end just like a vegetable?  w/e, I mean, if you value Terri Schiavo's life as much as your next -door neighbor's life, that's fine with me.  I just think you're wrong.  In fact, many people believe in practice that some life deserves more protection than others.  For example, if you're grandparents are still alive, I'm sure they would die to save your life because they know they will die very soon and you have your whole life ahead of you.  Terri Schiavo had ZERO life ahead of her, so what was the use of protecting it?  DON'T take this to mean that I think we should kill alll people who have nothing to contribute.  NOT AT ALL.  What I am saying is that if we have a choice between protecting a fully abled person and protecting a person in a persistent vegetative, I would hope that we would choose to protect the fully abled person.  Apparently this is not the case.


Prove that if $a,b,c>0, a+b+c=1$ then
\[
8(1-a)(1-b)(1-c)\&\leq (1-a^2)^2+(1-b^2)^2+(1-c^2)^2\\
\&\qquad \qquad \leq (1+a)(1+b)(1+c)
\]
(These were proposed and solved by Pham Van Thuan, Hanoi University of Science, Vietnam)

$\displaystyle 2+8D+2C \geq 8B-8D = LHS$

This is true by Shur :)


This is from the 2000 AMC 10.  problem# 12.
Figures 0,1,2, and 3 consist of 1,5,13, and 25 nonoverlapping unit squares, respectively.  If the pattern were continued, how many non-overlapping unit squares would there be in figure 100.
I dont have a scanner, so I couldn't copy the drawing.  I understood the solution, but how do I come up with a whole formula?  How do you figure the pattern?

Thanks

from the first few, the differences from adjacent terms are 4,8,12, so that being an arithmetic progression, it should be easy to solve, i.e. $a_n-a_{n-1}=4n$

how do you find figure 100 with the formula?

Thanks  :)

Using the formula, start writing the equations:
\begin{eqnarray} a_{100} - a_{99} &=& 400 \nonumber \\ a_{99} - a_{98} &=& 396 \nonumber \\ &\dots& \nonumber \\ a_{2} - a_{1} &=& 4 \nonumber \end{eqnarray}

Now add all of these equations together and see that all of the middle terms cancel and you are left with:
$a_{100} - a_1 = 400+396+392+\dots+8+4$

Factor out a four on the right and you are left with
$a_{100} - a_1 = 4(100+99+98+\dots+2+1)$

Since the sum of the first $n$ consecutive integers is $\frac{n(n+1)}{2}$,
$a_{100} - a_1 = 4*\frac{100(101)}{2}$
$a_{100} - 1 = 20200$
$a_{100} = 20201$


geometry
circumcircle
Euler
ratio
vector
trigonometry
similar triangles

Given triangle $ ABC$ with orthocenter $ D$ and circumcenter $ O$ and point $ O'$ such that $ OB \equal{} O'C,OC \equal{} O'B$ ($ O\neq O'$), prove that $ AD \equal{} OO'$.

edit: typo fixed

Hello, there is a typo here. We have two points C

There are up to two points that have the property by which you define $ O'$, so the question is ambiguous. I used Geometer's Sketchpad and found that neither distance $ OO'$ is equal to the circumradius. :(

sorry, i got the circumcenter and orthocenter mixed up (and have corrected it)... probably i was too tired yesterday's night from the 3-day outdoor camping...

Can you prove it without using the properties of euler line? Actually this fact of $ AD\equal{}OO'$ was used in proving $ GD\equal{}2GO$ so it would be circular reasoning if you use the fact to be proven in the proof :D

Well, I think it is required to have $ GD\equal{}2GO$ in order to know the ratio of the two similar triangles, and thus deducing $ AD\equal{}2OM$. However, is it possible to prove the problem statement without using the fact $ GD\equal{}2GO$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=668022#668022]Here[/url] is the proof (using complex numbers) that I was talking about...

In order to prove $ h\minus{}a\equal{}b\plus{}c$, it is easy to see these two vectors are parallel, but their norms being equal would not be that obvious... (in that proof, using the ratio of circumcenter, centroid, and orthocenter to prove the equality of the norms would indeed be circular reasoning)

$ OB = OC = r$ and $ O'C = O'B = r$
Thus $ O'$ is symmetric of $ O$ wrt. $ a$
So if $ X \in BC$ with $ OX \perp BC$
$ OX^2 = r^2 - \frac {a^2} {4}$
$ OX = \sqrt {\frac {4r^2 - a^2} {2}}$

$ OO' = 2.OX$
        $ = \sqrt {4r^2 - a^2}$

In $ \triangle ADC$
By sine relation
$ \frac {AD} {\cos A} = \frac {b} {\sin B}$
$ AD = \frac {b\cos A} {\sin B}$

If $ AD = OO'$ then,
$ \sqrt {4r^2 - a^2} = \frac {b\cos A} {\sin B}$
$ 4r^2 - a^2 = \frac {b^2\cos^2 A} {\sin^2 B}$
Since $ r = \frac {b} {2\sin B}$
$ \frac {b^2} {\sin^2 B} - a^2 = \frac {b^2(1 - \sin^2 A)} {\sin ^2 B}$
$ a^2 = \frac {b^2\sin^2 A} {\sin^2 B}$
$ \frac {a} {\sin A} = \frac {b} {\sin B}$ which is true.