vector
summer program
Mathcamp
function
Functional Analysis
linear algebra
absolute value

This question came up incidentally as part of another problem on my problem set last week, and I'm not entirely convinced that it is very difficult but I wasn't able to solve it:

Given a real Banach space [i]E[/i] and [i]n[/i] linearly independent vectors [tex]\displaystyle{e_1, e_2, \ldots, e_n \in  E}[/tex], let [tex]\displaystyle S(M)=\{(\alpha_1, \alpha_2, \ldots, \alpha_n)[/tex] such that[tex]\displaystyle ||\alpha_1e_1 + \ldots + \alpha_ne_n|| \leq M\}[/tex].

Prove that there exists [tex]\displaystyle (a_1, a_2, \ldots, a_n)[/tex] such that [tex]\displaystyle \forall (\alpha_1, \alpha_2, \ldots, \alpha_n) \in S(M), |\alpha_1| \leq |a_1|, |\alpha_2| \leq |a_2|, \ldots, |\alpha_n| \leq |a_n|[/tex]

Is the dimension of your Banach space known? I think I can do it for finite-dimensional Banach spaces.

I don't know too much about Banach spaces.  But I figured we could assume the space was finite-dimensional by just focusing on the span of the ei.

That might be true, but the solution I think I have does depend on it being finite dimensional.

I don't know that the entire space is finite-dimensional.  As Ravi said, however, we can just consider E' to be our new Banach space with basis e_i.  Actually, this question arose in a circumstance where I chose the e_i to be the basis of a finite-dimensional subspace of the Banach space.  So, if you can just do it for a Banach space of dimension n, that will take care of it for me.

Can anyone explain what a Banach space is? I got bigheaded reading Mathworld's definition. I took a Linear Algebra course at Mathcamp, so I might be able to give this question a shot.

Thanks!

For these purposes, it's a complete normed real vector space.  A vector space has two properties: additivity (if w and v are in the vector space, so is (w + v)) and scalar multiplication -- if r is real and w is in the vector space, rw is in the vector space.  Oh, and it's a group.  I think that's all the important stuff, anyhow.

I think I understand it. If I do, then here's my proof. If not, try not to make me feel too dumb when you explain why I'm stupid :)

I'll just prove why [tex]\alpha_1[/tex] is bounded. The rest follows.

Let [tex]x_2, x_3, \ldots x_n[/tex] be the least-squares approximation to

[tex]x_2 e_2+x_3 e_3 + \cdots + x_n e_n = -e_1.[/tex]

Since the e's are linearly independent, this least-squares approximation is not exactly equal to [tex]-e_1[/tex], so we must have

[tex]\| e_1+x_2 e_2 + \cdots + x_n e_n \| = n > 0.[/tex]

We then get

[tex]\| \alpha_1 e_1 + \cdots + \alpha_n e_n \| \ge \| \alpha_1 x_1 + (\alpha_1 x_2)e_2 + \cdots +  (\alpha_1 x_n)e_2 \| = |\alpha_1| n[/tex].

But if [tex](\alpha_1,\ldots,\alpha_n) \in S(M)[/tex], then this norm must be less than or equal to [tex]M[/tex]. Thus, we must have [tex]|\alpha_1| n \le M[/tex], i.e. [tex]|\alpha_1| < M/n.[/tex] So, [tex]|\alpha_1|[/tex] is bounded as desired.

Here's a start.  I know how to finish, but I'll leave it for you...

[I see that zabelman just posted too.  Oh well.]

Spoiler: [hide]Let a be a vector in Rn.  Define f(a) to be the norm of the sum of ai ei.  It is easy to see that f is continuous.  Let S be the set of vectors in Rn with max norm 1.  It is easy to see that S is compact.  Thus f achieves its minimum on S.  Let c be the minimum value.  It is nonzero because the ei are linearly independent.[/hide]

Okay, can you finish off the proof?

I was just going to pick an orthonormal basis and project all the [tex]e_i[/tex] vectors onto the orthonormal basis since the property clearly holds if the [tex]e_i[/tex]s are basis vectors of an orthonormal basis.

To define orthonormal, don't you need the notion of inner product?  I thought Hilbert spaces have an inner product, but Banach spaces don't necessarily have one.  But I may not know what I am talking about.

You're right, Ravi -- I thought of doing that, too, Simon, but we don't necessarily know that such a thing would exist.

Don't mind my inquisitiveness: ^_^"

1. What's the definition of a compact set that's applicable to this problem?

2. Why did you choose "with max norm 1"?

The map he described is linear -- f(l*x) = l*f(x), for l real and x a vector.  Thus, how the function operates on the set of all vectors with norm 1 tells us basically everything we could want to know about how the function operates everywhere.

The relevant definition of compact is a space in which every sequence of points contains an accumulation point.  It's not too hard to show that this causes every continuous function to achieve both its maximum and minimum on the set.

Right.  (I think you need an absolute value in one of the l's.)

Also, for subsets of Rn, it's nice to know that compact is equivalent to closed and bounded.  Without the condition about norm 1, the set S wouldn't be bounded.

[quote]zabelman, in your proof, how do you know that the least-squares approximation is actually achieved?[/quote]

I don't understand your question.

In your proof, you say that x2, ..., xn is chosen to be the least-squares approximation.  What does that mean?  How do you know that such xi exist?  In particular, how do you know that the errors can't get arbitrarily close to 0?

To put my question in a simpler context, suppose that we wanted to choose a positive number x such that 1/x is as close to 0 as possible.  There is no such number.  You need a way to rule out that kind of scenario.

Ah, I see. I guess I was just assuming there was a 'best approximation' for e1 in Span(e2,...,en). I thought that was a known property... I guess not. [i]My limited linear algebra knowledge from an audited night class at a local college has failed me! NOOOOO![/i] :? Thanks for the tip... I'll see if I can prove it.

I would venture the thought that proving it is a more significant trick than proving the original problem.

zabelman, the least squares approximation that you used only works when you have an inner-product between vectors. The inner product gives you a notion of angles between vectors. But Banach Spaces are more general. In them you have a notion of distance of vectors (i.e. a norm), but not necessarily a notion of angles between vectors (no inner-product).

Your proof does work fine for vector spaces with inner products though.

Here is a little simpler proof.

Given a vector a in Rn, define ||a|| to be the norm of the sum of ai ei.  It's easy to see that this new norm on Rn is indeed a norm.  (Because of linear independence, the only vector with norm 0 is 0 itself.)  But it is well known that all norms on Rn are equivalent.  In particular, this norm and the max norm are within a constant factor of each other.  So we're done.


The official answer key on arml.com says that the answer to #3 is 3:16 AM. However, doesn't 9:16AM work also?

Problems: http://www.arml.com/2005%20Contest/2005%20ARML%20Individuals.pdf

Solutions: http://www.arml.com/2005%20Contest/2005%20ARML%20Individual%20Solutions.doc

[quote="worthawholebean"][quote="Treething"]No, because then it would be PM and not AM anymore...

"...shows a time [i]X[/i] [b]A.M.[/b]"[/quote]
How is 9:16 AM actually PM?[/quote]

If T is 9:16 AM, then X is 2:44 PM, so even though T is AM, X is not AM.


Ok... sorry if this is a little late, but I just decided to start working on these problems.
[hide]Proof: Let $s_{k}=a_{1}+a_{2}+a_{3}+...+a_{k}$.  Plug in $n$ for $k$.  Then we can see that $s_{n}=a_{1}+...+a_{n}.$ Because $a_{2}\leq 1,a_{3}\leq 2,...a_{n}\leq n-1$, and $a_{1}=n$, the sum $s_{n}\leq 1+2+...+n= n(n-1)/2$.  We also know that $s_{n}=nb$, with $b$ being a positive integer.  Because the maximum value of $s_{n}$ is $n(n-1)/2$, $b$'s maximum value is $(n-1)/2<n$.  Since this is true, $a_{n}_{+}_{1}=b$, with $b$ being the only solution because $nb$ is congruent to $-b$ mod $(n+1)$.  Following this, we have that $a_{n}_{+}_{2}=b$ as well, and so on.  Thus, the sequence $a_{1},a_{2},...$ eventually becomes constant.[/hide]
Feedback please ^_^

$1+2+\ldots+n = n(n+1)/2$, not $n(n-1)/2$. :P Doesn't affect the legitimacy of the solution though. That probably would get a 7 unless they take off for silly mistakes like that.

oh right sorry...  :oops:  Thanks though.  Just didn't see any other post with the same solution

yeah that's kind of the same idea I had except worded way differently and I think i got a 6 on mine

I would watch your wording if I were you, it's correct, and I don't think with that wording they would take anything off, but I would suggest that you look at people's solutions, particularly in the olympiad forums (you don't actually have to understand the mathematics involved every time), to pick up on good proof writing. Language is important; yours is decent, but experience will make it better.


trigonometry
logarithms
real analysis
calculus
calculus computations

Determine whether the series converges or diverges.

$ \sum_{n\equal{}1}^{\infty}\left(\frac{1\plus{}\cos n}{2\plus{}\cos n}\right)^{2n\minus{}\ln n}$ :huh:

Hint: If $ \minus{}1<x<1,$ then $ 0<\frac{1\plus{}x}{2\plus{}x}<\frac23.$

The limsup version of the root test should work for this one as well.

so $ \limsup_{n\rightarrow\infty}\sqrt [n]{\left(\frac {2}{3}\right)^{2n \minus{} \ln n}} < 1$

because $ 2n \minus{} \ln n > 0$  and series converges .Right?:maybe:


Hallo

dankzij Peter heb ik het fenomeen IMC leren kennen (een beetje beschaamd dat dat niet eerder gebeurd is)

anyway : is er al iets gekend over de volgende ? data of locatie?

misschien neem ik het wel in overweging? zijn er anderen die er al een mening over hebben

en nog es proficiat aan peter!

normaal is het in Kiev. Zeker een reispasport nodig, misschien zelfs een visum. (hoewel, sinds de laatste presidentsverkiezingen is het land toch een iets westersere toer opgegaan)

Data zal weer ergens in die buurt zijn, maar frederic, waarom zo'n haast? :) 't is nog een vol jaar! als er nieuwe info is, zal je die altijd eerst vinden op http://www.imc-math.org

Anyway zijn er nog vlamingen/nederlanders die willen meegaan naar IMC volgend jaar? Altijd leuker om daar niet alleen te zijn. :)

Ja, 't is goed mogelijk dat ik volgend jaar ook eens meedoe :)

Ik zal het in overweging nemen, maar er is zeker nog niets beslist.

Ik wil de ex-IMO'ers er bij deze toch voor waarschuwen dat ze een iets amateuristischer geval tegemoet gaan dan IMO... :)

Bv. belgische universiteiten willen geen team leader sturen, maar 90% van de mensen die daar zijn komen met een team leader, en...
- de teamleaders brengen de vragen mee
- er wordt de middag voor de competitie vergaderd alle teamleaders kiezen samen de vragen. (er is dus geen apart problem selection committee)
- daarna gaan alle teamleaders gewoon terug naar hun kamer (ze worden niet afgezonderd dus, alles werkt op "erecode")
- tijdens het eerste halfuur van het examen mogen er "taalvragen" gesteld worden aan je eigen teamleader. (...)
- als er fouten optreden (en laat ons zeggen dat toch >1/20 papieren verkeerd gegradeerd worden), mag iemand daarover gaan klagen... en het is geweten dat sommige behendige teamleaders daar misbruik van maken en er toch in slagen hun studenten gemiddeld 20 punten bij te geven :)
- de examenvragen worden verbeterd door teamleaders, en dan hoor je discussie onder de teamleaders van dit moet je zoveel geven, nee dit moet je zoveel geven. :P er staan wel geen namen op, maar ze weten wel allemaal goed wat hun studenten geantwoord hebben.

Enfin, wie niet voor de grand first prizes gaat zal daar wel niet zo veel last van hebben... :) Het blijft in ieder geval even amusant als een IMO, en natuurlijk een mooie prestatie is even eervol :)

Tenzij  je zoals het mongoolse team...[hide]Met de trein komt en een volle week onderweg bent! :rotfl:  :rotfl:  :rotfl: [/hide]


Let $A,B,C \in {\mathbb R}^{n \times n}$ such that $detA,detB,detC >0$,$ABC=kI_n$, where $k \in {\mathbb N}^*$ and let $a,b,c>0$ with $a+b+c=1$. Show that $\sum{det({\frac{1}{a^{\frac{1}{n}}}A^2})} \geq 9k^{\frac{2}{3}}$.

[quote="jianuovidiu"]$... \geq 9k^{\frac{2}{3}}$.[/quote]

Is there n something like $ 9k^{\frac{2n}{3}}$   ?


[quote="ThAzN1"]Let $a,b,c$ be positive reals such that $a^2+b^2+c^2 = a+b+c \geq 3$.  Prove that
\[ \frac{1}{\sqrt{a^4+4}} + \frac{1}{\sqrt{b^4+4}} + \frac{1}{\sqrt{c^4+4}} \geq \frac{3\sqrt{5}}{5}.  \][/quote]

$(a-1)^2+(b-1)^2+(c-1)^2\ge 0\Rightarrow 3\ge a+b+c$ so according to the conditions of the problem we must have $a=b=c=1$ so $\frac{1}{\sqrt{a^4+4}} + \frac{1}{\sqrt{b^4+4}} + \frac{1}{\sqrt{c^4+4}}= \frac{3\sqrt{5}}{5}$

I think I've just found a proof of a generalization:

Let $x_1, x_2, \dots , x_n$ and $p$ be positive reals with $x_1+x_2+\cdots + x_n \leq n$.  Prove that
\[ \sum_{k=1}^n \frac{1}{\sqrt{x_k^4+4p^4}} \geq \frac{n}{\sqrt{1+4p^4}}.  \]
If for instance, we take $p = \sqrt[4]{\frac{3}{4}}$, we get a pretty nice inequality:
\[ \sum_{k=1}^n \frac{1}{\sqrt{x_k^4+3}} \geq \frac{n}{2}.  \]

[quote="ThAzN1"]I think I've just found a proof of a generalization:

Let $x_1, x_2, \dots , x_n$ and $p$ be positive reals with $x_1+x_2+\cdots + x_n \leq n$.  Prove that
\[ \sum_{k=1}^n \frac{1}{\sqrt{x_k^4+4p^4}} \geq \frac{n}{\sqrt{1+4p^4}}.  \][/quote]
Why not rewrite your generalization as:
Let $x_1, x_2, \cdots, x_n$ and $p$ be positive real numbers with $x_1+x_2+\cdots + x_n \leq n$. Prove that
$\sum_{k=1}^n \frac{1}{\sqrt{x_k^4+p}} \geq \frac{n}{\sqrt{1+p}}$.

let $f(x)=\frac 1 {\sqrt{e^{4x}+4}}$
can we prove $f\"(x) \geq 0$
i'm not sure

[quote="ThAzN1"]Let $a,b,c$ be positive reals such that $a+b+c \leq 3$.  Prove that
\[ \frac{1}{\sqrt{a^4+4}} + \frac{1}{\sqrt{b^4+4}} + \frac{1}{\sqrt{c^4+4}} \geq \frac{3\sqrt{5}}{5}.  \][/quote]
Sorri,but I think your problem is wrong ,because if $a\to 0,b\to 0,c\to 3$=>$\frac{1}{\sqrt{a^4+4}} + \frac{1}{\sqrt{b^4+4}} + \frac{1}{\sqrt{c^4+4}}\to  1+1/({85}^{1/2})<\frac{3\sqrt{5}}{5}$ :(


function
inequalities
inequalities unsolved

Let $a, b, c$ be positive reals and $k\ge 4$, prove that

$\displaystyle
\frac{a}{\sqrt[3]{b^{3} + c^{3} + kabc}} +
\frac{b}{\sqrt[3]{c^{3} + a^{3} + kabc}} +
\frac{c}{\sqrt[3]{a^{3} + b^{3} + kabc}}
\ge
\frac{3}{\sqrt[3]{2+k}}.
$

I moved this as a separate topic for those who are interested in this one.

(Edited): the condition on $k$ can be improved to $k\geq\frac{17}{5}$ based on [b]Mildorf[/b]'s analysis.

(I will write [tex]\sum_{C}[/tex] and [tex]\sum_{S}[/tex] for a cyclic and symmetric sum in three variables respectively.) Put [tex]a+b+c = 1[/tex]. Consider the convex function [tex]f(x) = x^{\frac{-1}{3}}[/tex]. We have

[tex]\displaymode{
\sum_{C} \frac{a}{\sqrt[3]{b^3+c^3+kabc}} = \sum_{C}a f(b^3 + c^3 + kabc) \ge f\left(\sum_{S} a^3b + \frac{1}{2}ka^2bc\right) \ge \frac{3}{\sqrt[3]{2+k}}
}[/tex]

Some simple algebra, and this is equivalent to

[tex]\displaymode{
(2+k)(a+b+c)^4 \ge (2+k)\left(\frac{1}{2}\sum_{S} a^4 + 4\sum_{S} a^3b +  9\sum_{S} a^2bc\right) \ge 27\left(\sum_{S} a^3b + \frac{1}{2}ka^2bc}\right)
}[/tex]

I used Muirhead to replace [tex]\sum_{S} a^2b^2[/tex] with [tex]\sum_{S}a^2bc[/tex] when expanding [tex](a+b+c)^4[/tex], but by Schur's with [tex]r=2[/tex], we have

[tex]\displaymode{
\sum_{S} \frac{1}{2}a^4 + \frac{1}{2}a^2bc \ge \sum_{S}a^3b
}[/tex]

So that we need only show

[tex]\displaymode{
(2+k)\left( 5\sum_{S} a^3b + \frac{17}{2} \sum_{S} a^2bc\right) \ge 27\left(\sum_{S} a^3b + \frac{1}{2}ka^2bc\right)
}[/tex]

This is true by Muirhead when [tex]5(2+k) \ge 27[/tex] so we can actually improve the condition on [tex]k[/tex] to [tex]k \ge \frac{17}{5}[/tex].

[quote="Mildorf"]
but by Schur's with [tex]r=2[/tex], we have

[tex]\displaymode{
\sum_{S} \frac{1}{2}a^4 + \frac{1}{2}a^2bc \ge \sum_{S}a^3b
}[/tex]

... so we can actually improve the condition on [tex]k[/tex] to [tex]k \ge \frac{17}{5}[/tex].[/quote]

You are right, I think the improvement of the lower bound on $k$ came from the use of Schur's inequality.

Without using Schur's, I get $(2+k)\cdot \frac{9}{2}\geq 27$, this gives us $k\geq 4$. With Schur's, you get $(2+k)\cdot 5\geq 27$, this gives a better result.

Well done!


[quote="binomial_4eva"]Let $ d(n)$ represent the number of digits of $ n$ in its decimal representation.
evaluate the sum $ \sum_{n \equal{} 1}^{\infty}\frac{1}{d(n)!}$[/quote]
This really belongs in Calculus forum.

[hide="Solution"]
Note that, $ d(1) \equal{} 1,...,d(9) \equal{} 1$ so first we have $ \frac{1}{1!}\plus{}...\plus{}\frac{1}{1!}$ $ 9$ times.

Then, $ d(10) \equal{} 2,...,d(99) \equal{} 2$ so then we have $ \frac{1}{2!}\plus{}...\plus{}\frac{1}{2!}$ $ 90$ times.

Then, $ d(100) \equal{} 3,...,d(999) \equal{} 3$ so then we have $ \frac{1}{3!}\plus{}...\plus{}\frac{1}{3!}$ $ 900$ times.

The pattern is clear.

If we add 'em up we got:
$ \frac{9}{1!}\plus{}\frac{90}{2!}\plus{}\frac{900}{3!}\plus{}... \equal{}\sum_{n \equal{} 1}^{\infty}\frac{9\cdot 10^{n\minus{}1}}{n!}\equal{}\frac{9}{10}\sum_{n \equal{} 1}^{\infty}\frac{10^{n}}{n!}\equal{}\frac{9}{10}(e^{10}\minus{}1)$

Because it is well-known that,
$ e^{x}\equal{}\sum_{n \equal{} 0}^{\infty}\frac{x^{n}}{n!}$ so just let $ x \equal{} 10$ and you get ...
[/hide]

EDIT: Fixed mistake.

[b]Edit:[/b] Sylow, you sneak  :P He's right though, it does assume knowledge of the series for $ e^{x}$. But Sylow, I think you missed an exponent of 10 in your answer.

[hide="Solution"]There are $ 9 (10^{k\minus{}1}) \equal{}\frac{9}{10}10^{n}$ solutions to $ d(n) \equal{} k$ (9 one digit numbers, 90 2 digit numbers, etc.

Our sum is, "for every natural number $ n$, add $ \frac{1}{d(n)!}$. We can rephrase as, "add $ \frac{1}{d(n)!}$ for every $ d(n)$ digit number. This then becomes:

$ \sum_{n \equal{} 1}^{\infty}\frac{1}{d(n)!}\equal{}\sum_{j \equal{} 1}^{\infty}\left(\frac{9}{10}\cdot\frac{10^{j}}{j!}\right) \equal{}\frac{9}{10}\left(\sum_{j \equal{} 0}^{\infty}\left(\frac{10^{j}}{j!}\right)\minus{}\frac{1}{0!}\right) \equal{}\boxed{\frac{9}{10}(e^{10}\minus{}1)}$[/hide]


What is the average of integers $ a$, $ b$ and $ c$ if the average of the two numbers $ a$ and $ 2b$ is 7 and the average of the two numbers $ a$ and $ 2c$ is 8?

$ \frac{a\plus{}2b}{2} \equal{} 7$
$ a\plus{}2b \equal{} 14$

$ \frac{a\plus{}2c}{2} \equal{} 8$
$ a\plus{}2c \equal{} 16$

Now add them both.

$ a\plus{}2b\plus{}a\plus{}2c \equal{} 14 \plus{} 16$
$ 2(a\plus{}b\plus{}c) \equal{} 30$
$ \frac{a\plus{}b\plus{}c}{3} \equal{} \boxed{5}$


geometry
area of a triangle
Heron\u0027s formula

[quote="Pineappleperson"]does anyone know how to find the area of a heptagon? :?:  ;)  :P  :spider:  :huuh:[/quote]

Draw triangles, for every regular $n$ gon, you can simply draw equal triangles and then find out the area of each triangle, and multiply by the number of triangles to find the area of the $n$ gon.

assuming it is a regular polygon, i can think of two formulas that may help:

1)  $A= \frac{1}{2}(a)(s)(n)$, where $a$ is the length of the apothem (or radius of the inscribed circle), $s$ is the side length, and $n$ is the number of sides... in this case 7.

2) $A = \frac{1}{4}s^2(n)\cot{\frac{\pi}{n}}$, where the variables stand for the same thing.

Does anybody know the formula for it when it isn't regular?? I assume it's probably something like Brahmagupta's or Heron's formula....

I always thought it was $\text{hero's formula}$ even in our textbook it was that.

that's for a triangle though... brahmagupta's is for a cyclic quadrilateral

[quote="riddler"]No, you didnt understand what I meant, I asked if it was $\text{Hero's Formula}$ or $\text{Heron's Formula}$????[/quote]

It is indeed Heron's formula, I hope your textbook isn't talking about a different formula. :P

Heron's formula: $\sqrt{s(s-a)(s-b)(s-c)}$

No, It is the same formula, I cant believe they wrote $\text{Hero's Formula}$.?????

ohhhh sorry, i didnt realize what you were saying. both names are used interchangeably for the same formula


geometry
trigonometry
probability
vector
calculus
number theory
group theory

you can make votes by picking the butten and making a message. Also this a never ending poll.

number teory for sure...the one that comes more difficult at least to me....but i einjoy it :lol:

Uh.... I like analysis, particularly measure theory ... I don't think that's what you meant when you put "geometry" so I'm not voting. :P

I like multivariable calculus. No vote from me either.

[quote="Tiks"][quote="7h3.D3m0n.117"][quote="jackdillon"]geometry is the best subject ever![/quote]

i think geometry is too easy.[/quote]

I think you just don't know geometry well enough to understend its beauty and difficulty :P .
GEOMETRY FOREVERE!!!!!!!!!!!!!!!! :D[/quote]

hmm my friends spent half a year learning what they have already learned years ago...I consider that time wasted. however, i love the proofs and postulates....makes math more interesting

Geometry is WAY more complicated than what is taught in school.  Take a look at the Geometry Olympiad forum to see  :D

My vote goes for Number Theory as my favorite; my worst subject is Geometry.

[quote="7h3.D3m0n.117"][quote="jackdillon"]geometry is the best subject ever![/quote]

i think geometry is too easy. i get a better satisfaction finding what $x=$ in algebra then finding the area of a circle...sometimes...for a challenge i try some trig stuff  :wallbash:[/quote]
If you think finding the area of circle is among the hardest things in geometry then you are sadly mistaken and most likely live under a rock. You seriously need to look at harder geometry problems if you think it's too easy.

My forte is combinatorics but I think Number Theory is the most interesting by far.

Why is arithmetic up there?

I guess it could count as Counting & Probability because that option was left off...

Go GEOMETRY!!! and ALGEBRA !!!

I slightly prefer Algebra more, but geometry is cool also.

[b]Number Theory is the Best!!![/b]
Especially after getting the AoPS book.

I can do calc but it is annoying and boring and stuff...unlike Geometry and Algebra. Well that's just my opinion.

I like working with Algebra and Number Theory, but I'm much better at Algebra.  Geometry is too much memorizing for me  :(.

Algebra is the originator of all Mathematics. Algebra is reasoning and logic. mathematics stands on logic and reasoing.if reasoning and logic was not in geometry. it would have been like children drawing something.
so i go for Algebra, number theory,group theory
I am allergic to Statistics and Trigonometry

Algebra lovers vote for Algebra!!!if only I could vote more than once.

Algebra!!! Yes we algebra lovers are winning!

What if i don't like math?

I like algebra.  it's easy and uhh logical

Number Theory, Algebra, Linear Algebra, Abstract Algebra. :)

Personly I love maths absolutly. I am talking about what you like not am good at. There was a disscussion on geometry going on. If you no what you are doing and all the identities then it is easy as making a pie. If otherwise then it is very hard. This is true with olimpiads to.

geometry! cause i don't make stupid mistakes like in algebra(well maybe less) and u get to draw instead of writing stuff.

GO GEOMETRY!!!!
[size=42]and algebra
and number theory because it's hard[/size]


function
modular arithmetic
combinatorics proposed
combinatorics

$ n{}_{C}{}_{p}\equiv \left[\frac{n}{p} \right]mod p$

where  [ ]  denoted greatest integer function  and  p is a prime.

Let $n\equiv r \pmod{p}$ and write $n=pk+r$ with $r<p$ and $k\in \mathbb{N}$.

Now, ${n\choose p}={kp+r \choose p}=\sum_{t=0}^{r} {r\choose t}{kp\choose p-t}$

$\Longrightarrow {n\choose p}={r\choose 0}{kp\choose p}+\sum_{t=1}^{r}{r\choose t}{kp\choose p-t}$

$\Longrightarrow {n\choose p}={kp+r\choose p}\equiv {kp\choose p} \pmod{p} \text{   } \left( \because p\mid {r\choose t}{kp\choose t} \text{ }\forall \text{ } 1\leq t\leq r \right)$

Hence it is enough to show that ${n\choose p}\equiv {kp\choose p}\equiv k\equiv \left[ \dfrac{n}{p}\right] \pmod{p}$

But ${kp\choose p}=\dfrac{\Pi_{t=0}^{p-1}(kp-t)}{p!}\equiv \dfrac{k\Pi_{t=1}^{p-1}(-t)}{(p-1!)}\equiv \dfrac{k(p-1)!}{(p-1!)}\equiv k \pmod{p}$

$\therefore$ $QED$


[quote="N.T.TUAN"][quote="chien than"]Let $x;y;z$ be positive real numbers and satisfying $x+y+z=1$.Prove that:
$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq \sqrt{3}$[/quote]
$x=y\to 0,z\to 1$. Are you sure? :P[/quote]
$x;y;z>0$

[quote="chien than"]Let $x;y;z$ be positive real numbers and satisfying $x+y+z=1$.Prove that:
$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq \sqrt{3}$[/quote]I think the inequality should be the other way around?

If this statement is true then $2\cdot\sqrt{\frac{1}{n}}+\sqrt{\frac{n-2}{n}}\geq \sqrt{3}\;\forall n\geq 3$. But if $n\to\infty$ then $\text{LHS}\to 1$.  :huh:

as towersfreak2006 said the signe schould be reversid iff not here is a conterexample
try with $\frac{1}{6},\frac{1}{2},\frac{1}{3}$. :huh:

i think that with AM-GM we have $\sqrt{x}+\sqrt{y}+\sqrt{z}\leq 3\sqrt{\frac{x+y+z}{3}}=\sqrt{3}$ !!!!

The inequality is reverse

There are 3 methods.

1. R.M.S >= A.M on root(x),root(y),root(z)
2. Chebyshev with both triplets as root(x),root(y),root(z)
3. Cauchy Schwartz on root(x),root(y),root(z) and 1,1,1


[quote="hollandman"]Let $ a,b,c>0$ and $ a\geq b\geq c$. Prove that

$ 2(a\plus{}b\sqrt{\frac{b}{a}}\plus{}c\sqrt[3]{\frac{c^2}{ab}})\geq \sqrt{a^2\plus{}bc}\plus{}\sqrt{b^2\plus{}ca}\plus{}\sqrt{c^2\plus{}ab}$.[/quote]

I think that's what he meant..

[quote="hollandman"]Let $ a,b,c>0$ and $ a\geq b\geq c$. Prove that

$ 2(a+b\sqrt{\frac{b}{a}}+c%Error. "cubert" is a bad command.
{\frac{c^2}{ab}})\geq \sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab}$.[/quote]
It's true for all positives $a$, $b$ and $c$. :wink:


Support
geometry
3D geometry
calculus
Gauss
sphere
analytic geometry

I think the Newton's gravity law is an observational law, that means, it is confirmed by the experiments within a very good precision, but it cannot be proved from other basic laws. (At least at the high school level, but i think this is true also generally in some way.)

Can you imagine how could that law be proven? Can you imagine how could ALL laws be proven somehow theoretically?
Those are some of the deceptive questions that haunt modern physicists / metaphysicists today, I think ... ;-)

[quote="mathematica"]Does any one know the proof for newton's law of gravity, because i looked in a lot of books for this proof but i didn't find it anywhere!!!  :(
Slowly i'm getting a little cautious about what we're easily accepting in school!  :ninja:
Thanks for your reply. :lol:[/quote]

Henry Cavendish proved it using a delicate torsion balance in 1796.

Be careful of seeking purely [i]mathematical[/i] proofs in physics.

"If it doesn't agree with experiment, it's wrong." -- Richard Feynman

The purpose of the [url=http://kossi.physics.hmc.edu/Courses/p23a/Experiments/Cavendish.html]Michell-Cavendash[/url] experiment was to determine the value of G in Newton's Law of Gravity.

The Law itself, though, (as with all Physical Laws) comes originally from either direct observational evidence or reverse engineering observational evidence. Laws in physics have sort of the same role as axioms in mathematics. One of the goals of modern physics is to distill everything down to the most basic of laws.

If you learn more advanced physics you can derive Newton's Law from even more fundamental laws, but those laws are still based (on some level) on hueristic or experimental evidence. To derive Newton's Law from, say, the laws of general relativity would be beyond the scope of high school students.

Cavendish's way of find $\text{G}$ is so marvelous...
How cound he come up with that idea!?

[quote="gauss202"]The purpose of the [url=http://kossi.physics.hmc.edu/Courses/p23a/Experiments/Cavendish.html]Michell-Cavendash[/url] experiment was to determine the value of G in Newton's Law of Gravity.

The Law itself, though, (as are all Physical Laws) comes from observational evidence. Laws in physics have sort of the same role as Axioms in mathematics.[/quote]

That's true, the objective of the experiment was to determine G, however, had Cavendish's data not obeyed an inverse square law, G would not have been constant. Ergo, the physics proves the math and not [i]vice versa[/i].

But there was already better experimental evidence for the validity of Newton's Law (i.e. the motion of the planets) than what came from the Michell-Cavendash experiment.

The experiment may have given some additional support for the law, but its main contribution was an improved value for G. Experimental evidence can never give a "derivation" of a physical law. The best it can do is testify on the Law's behalf.

[quote="gauss202"]I was thinking about this question a little bit more today and I guess it is possible to derive Newton's Law of Gravity from Kepler's Third Law (which was truly based on observation rather than theory).

Kepler's Third Law states that "the squares of the periods of any two planets are to each other as the cubes of their average distances from the Sun." In other words, $\displaystyle \Big(\frac{T_1}{T_2}\Big)^2 = \Big(\frac{R_1}{R_2}\Big)^3$, where $T_1$ and $T_2$ are the periods of two planets orbiting the Sun and $R_1$ and $R_2$ are the respective average distances from the Sun.

Another way of stating Kepler's Third Law is that $\displaystyle \frac{T^2}{R^3}$ = (a constant) for any planet orbiting the Sun. If we call this constant K we can say $T^2 = KR^3$

Now if we assume that the planets move in circular orbits around the Sun (we could do this for general elliptical orbits but the math would be a little bit hairier), then the centripetal force keeping the planets in orbit is $F = \frac{mv^2}{R}$, where m is the mass of the planet and v is its orbital velocity.

The relationship between the orbital velocity, v, and period, T, is $v = \frac{2 \pi R}{T}$. Putting this into the above equation for centripetal force gives $F = \frac{4\pi^2mR}{T^2}$.

Now using the Kepler relationship, $T^2 = KR^3$, we get $F = \frac{4\pi^2m}{KR^2}$
Or, letting $\frac{4\pi^2m}{K} = C$, we can write this as $F = \frac{C}{R^2}$

Newton realized however that the planets don't really orbit around the Sun, but rather both the Sun and the planets orbit around their center of mass. So he reasoned that the constant C should rely symmetrically on the masses of both objects.

Therefore $F = \frac{GMm}{R^2}$ for some constant G.[/quote]

Impressive, [i]Herr[/i] Gauss.

But wait a second how can you prove kepler's third law (obviously, without using newton's law of gravitation)?
And also i am sure that physics' laws are not like mathematical axioms. Math axioms are obvious while newton's gravitational law is clearly not obvious. I would agree on newton's second law to be an axiom because that is our start point on newtonian mechanics you know like euclid's fifth postulate is our start on euclidean geometry. And also for example albert einstein gave mathematical proofs for all the formulas he discovered and everyone knows that he didn't have a labratory. He only used two or three principles (equivalence principle, relativity principle &...) and proved all of his very complex formulas using two or three principles and then how can you say that newton's law of gravitation that is much simpler than general or special relativity doesn't have a proof? Also because relativity contradicts newtonian mechanics i can assure you  that einstein didn't use newtonian mechanics in relativity! :lol:

***Also I don't know you've read the elegant universe or not but in that book it is written that one of the main goals of string theory is to derive all the 20 physical constants in nature mathematically with a few principles and symmetries of nature. So how can we have not a mathematical proof for newton's gravitaional law!!??

**** In this writing whenever i used the word proof, i mean a MATHEMATICAL proof. ;)

When you say kepler got his law from observation, and the prove newton's law based on that it means that newton's law does not have any proof and then the whole coversation starts again. What i mean is didn't anyone try to give a mathematical proof for newton's gravitional law using newton's second law!! like for example, newton himself was first a mathematician then a physicist how could have gotten so deep to prove theorems of calculus but never trying to give a mathematical proof  for the universal gravitaional law that is not very obvious in respect to many mathematical theorems that are very obvious and need a proof.

[quote="mathematica"]But wait a second how can you prove kepler's third law (obviously, without using newton's law of gravitation)?[/quote]

Gauss 202 just showed how. He derived Newton's gravity from Kepler's third, using equations of motion that do not necessarily involve gravity.

[quote="mathematica"]And also i am sure that physics' laws are not like mathematical axioms.[/quote]

That's true, physics laws are not like mathematical axioms; physics laws are more restrictive. Mathematics as it relates to physics only models physics.

 [quote="mathematica"]Math axioms are obvious while newton's gravitational law is clearly not obvious.[/quote]

It depends upon what you mean by obvious and to whom a thing is obvious. I am sure that a great many things were obvious to Newton that are not obvious to me.

[quote="mathematica"]I would agree on newton's second law to be an axiom because that is our start point on newtonian mechanics you know like euclid's fifth postulate is our start on euclidean geometry.[/quote]

I would advise against taking any physical law as an axiom. Axioms are true by definition; physical laws are true by observation.

[quote="mathematica"] And also for example albert einstein gave mathematical proofs for all the formulas he discovered and everyone knows that he didn't have a labratory. He only used two or three principles (equivalence principle, relativity principle &...) and proved all of his very complex formulas using two or three principles and then how can you say that newton's law of gravitation that is much simpler than general or special relativity doesn't have a proof?[/quote]

Einstein made mathematical proof of his ideas. His proof, as all proofs must, started with his assumptions: 1. the speed of light is constant, and 2. the laws of physics are the same in all reference frames.

If either of those assumptions is wrong then Professor Einstein was also wrong.

Newton, on the other hand, started with his observation, an apple falling from a tree, and derived a mathematical expression that would explain that action.

Do you see the difference between the two methods?

Oh so your saying the methods of classical physics are completely different with modern theoretical physics & I agree with you. But I'm not saying how newton derived his gravitational law i'm saying can we find any principles or axioms like in relativity so we can proove everything else with these axioms?? :?:

[quote="mathematica"]Oh so your saying the methods of classical physics are completely different with modern theoretical physics[/quote]

No, not completely different. It may seem that modern physics is all mathematical and far removed from experimental data, but that is not true. Modern physicists still rely heavily on experimental data to form their conjuectures of physical laws.

There are two elements to theoretical physics research. On one hand physicists use experimental data and existing physical laws plus mathematics to postulate new physical laws. On the other hand they use mathematics to derive new consequences of those physical laws so they can be confirmed by experiments.

[quote]But I'm not saying how newton derived his gravitational law i'm saying can we find any principles or axioms like in relativity so we can prove everything else with these axioms?? :?:[/quote]

Whether physics can be (finintely) axiomatized was one of Hilbert's Famous 23 Questions posed in 1900. When you think about it, it is really one of the fundamental questions of physics - to boil everything down to a finite set of fundamental laws. The answer to this question is not known, and may never be known, but I would suspect that most people believe the answer is yes.

Newton's Law of Gravity [i]can[/i] be derived from a more basic set of physical laws (such as those of General Relativity). But I assumed you were asking, can it be derived from other fundamental laws of [i]Newtonian Mechanics[/i]. The answer is no. Not without some additional postulate (like Kepler's Third Law). Therefore Newton's Law of Gravity is often taken as a fundamental law (i.e. not derived from the other Newtonian Laws).

Historically (I believe) Newton used Kepler's Laws as evidence to formulate his General Law of Gravity. He then took his law to be more fundamental and proved that it could be used to imply Kepler's Laws. So although Kepler's Laws came first historically, we often think of Kepler's Laws coming as a [i]consequence[/i] of Newton's General Law of Gravitation.

It turns out that Newton's Law of Gravity is not really perfect. Its only an approximation for the General Realtivistic Laws in "weak" and slow moving gravitational fields. But it is extremely accurate in these conditions, so we use it anyway.

[quote="gauss202"]I was thinking about this question a little bit more today and I guess it is possible to derive Newton's Law of Gravity from Kepler's Third Law (which was truly based on observation rather than theory).

Kepler's Third Law states that "the squares of the periods of any two planets are to each other as the cubes of their average distances from the Sun." In other words, $\displaystyle \Big(\frac{T_1}{T_2}\Big)^2 = \Big(\frac{R_1}{R_2}\Big)^3$, where $T_1$ and $T_2$ are the periods of two planets orbiting the Sun and $R_1$ and $R_2$ are the respective average distances from the Sun.

Another way of stating Kepler's Third Law is that $\displaystyle \frac{T^2}{R^3}$ = (a constant) for any planet orbiting the Sun. If we call this constant K we can say $T^2 = KR^3$

Now if we assume that the planets move in circular orbits around the Sun (we could do this for general elliptical orbits but the math would be a little bit hairier), then the centripetal force keeping the planets in orbit is $F = \frac{mv^2}{R}$, where m is the mass of the planet and v is its orbital velocity.

The relationship between the orbital velocity, v, and period, T, is $v = \frac{2 \pi R}{T}$. Putting this into the above equation for centripetal force gives $F = \frac{4\pi^2mR}{T^2}$.

Now using the Kepler relationship, $T^2 = KR^3$, we get $F = \frac{4\pi^2m}{KR^2}$
Or, letting $\frac{4\pi^2m}{K} = C$, we can write this as $F = \frac{C}{R^2}$

Newton realized however that the planets don't really orbit around the Sun, but rather both the Sun and the planets orbit around their center of mass. So he reasoned that the constant C should rely symmetrically on the masses of both objects.

Therefore $F = \frac{GMm}{R^2}$ for some constant G.[/quote]
no, you can not prove newtons gravitation based upon the keplers third law, because kepler's third law is proved by newtons law of gravitation.
and if both are experimental, then you can not use one, to provel the other one. ;)

Have a look at this and tell me is physics forums better or mathlinks??

http://www.physicsforums.com/showthread.php?t=78111

[quote="amirhtlusa"]no, you can not prove newtons gravitation based upon the keplers third law, because kepler's third law is proved by newtons law of gravitation.
and if both are experimental, then you can not use one, to provel the other one. ;)[/quote]

Actually, it is perfectly acceptable to use one to prove the other. You merely invoke Newton as a theorem to prove Kepler or vice versa.  However when both relations are experimentally and independently derived there is no need to prove either one.

[quote="mathematica"]Does any one of you know any good physics book that has an axiomatic point of view toward newtonian mechanics? I don't mean nessecarilly it should have the law of gravity's proof but it prooves whatever that can be proven with newtonian mechanics? (Also the level of its math shouldn't be higher than Calculus, Vol I, Apostol)
Thanks! :lol:[/quote]

Well you could always go straight to the source and read [url=http://www.amazon.com/exec/obidos/tg/detail/-/0520088174/ref=ase_ericstreasuretroA/102-9659123-9494532?v=glance&s=books]The Principia[/url] by Isaac Newton (trans. by Cohen and Whitman) :)

Aha! I found two webpages that you may like and that may answer some of your questions: http://www.thespacesite.com/space/history/dawn3a.php

http://www.phy6.org/stargaze/Sgravity.htm

When talking about proof for Newton's Law of gravity, there are several issues to tease out:

1) Kepler's equal areas observations for orbital sections, and their intricate relation to velocity and distance of the orbiting body with the body being orbited. (i.e., earth & Sun).

This was one important observation that Newton wanted to explain.

(2) Newton's [b]F = mA[/b] law, which he wanted to universalize for all observable 'forces'.

(3) The observed 'size' of various planets, and the assumption of their probable range of density (and therefore mass),  based upon measurements of 'mass' on earth for smaller objects, and laws of inertia and 'gravity'.

(4) Newton's basic assumption, 1/d^2  (the Inverse Square Law) for gravity was not arbitrary but directly connected to the law of projection, or the law of increasing area for an expanding sphere, which both explained how light dims with distance, (it spreads out), and how a 'force' might be diluted if made to cover, or be spread over a larger area.

These considerations made Newton hope for a simple Inverse Square Law which had a rational basis, rather than some other unknown or 'magical' law.  The other factors, such as mass for each object between which the forces would operate, also were incorporated in a way that suggested itself via the behaviour of objects on earth under gravity, and vis inertia.

(5)  the [b]Gravitational Constant  G[/b].  This constant is itself an assumption.  In Newton's original idea, this 'constant' has no real independant existance, (not like the 'speed of light' or Planck's Constant does.)  It is merely a way to coordinate UNITS of mass with UNITS of distance (apples and oranges!).  Thus it has no profound meaning or value as a 'pure number' but is completely arbitrary and dependant upon choice of units.

(6) So The Cavendish experiment doesn't 'prove' Newton's law at all, but rather (as noted) establishes the  Gravitational Constant  [b]  G[/b], by assuming Newton's equation is true.  The only profound content in 'G' is the more subtle idea that the Inverse Square Law as formulated by Newton (and involving mass) holds over great (all) distances, from the earth's surface to the solar system, to the scale of galaxies.

(7) In fact, the Inverse Square Law actually fails horribly at galaxy-scale distances: The rotational speeds of stars beyond a certain distance shows that the Inverse Square Law is badly out of kilter, and there is currently no plausible or adequate explanation.  (Dark Matter, the only popular hypothesis to explain this currently has no credible substance to it.)

(8) [b]Newton's Gravity Law is well-substantiated at the Solar System scale[/b], and that's all that can be done for it.  Measurements at the earth's surface only establish rough 'estimates' for G since no one knows how to properly account for proximity effects with the earth in order to calculate the 'real'  G.

(9)[b] At molecular distances, Newton's gravitational law becomes hopelessly muddled[/b] (including its most useful approximating methods, the Centre of Mass and the Sphere Theorem).  (a) The Sphere Theorem fails due to discrete spacial quantization of mass, and (b) the Centre of Mass method fails due to proximity effects.   Thankfully it is not a major force compared to Electro-weak forces.

(10) Currently, the Law of Gravity is only accurate to 3 decimal places generally, even where it is known to be valid.  (Some measurements using satellites claim an accuracy of several more decimals, but that is only at the Solar System range.)

I was hoping to do some research and find some axioms which are intuitive and can be easily understood why they are true and basically they don't need a proof. (And obviouslly these axioms will be contradictory with relativity's postulates) and with these axioms i wanted to find a complete proof for newton's law of gravitation.
Do you think these axioms will be found?

[quote]I was hoping to do some research and find some axioms which are intuitive and can be easily understood why they are true and basically they don't need a proof. [/quote]To me, the one given above is an example of just what you are looking for:

[quote] Newton's basic assumption, 1/d^2 (the Inverse Square Law) for gravity was not arbitrary but directly connected to the law of projection, or the law of increasing area for an expanding sphere, which both explained how light dims with distance, (it spreads out), and how a 'force' might be diluted if made to cover, or be spread over a larger area. [/quote]But this 'axiom' is founded on the primary one, that the apparent 'force' of gravity is real, and that the source of the force is a simple unblockable attraction between mass-particles (i.e., acting in straight lines through a Euclidean space.)
What else can one add, except that the forces combine simply according to the parallelogram law like other forces? (accepting Newton's general description of forces.)

[quote="mathematica"]Does any one of you know any good physics book that has an axiomatic point of view toward newtonian mechanics? I don't mean nessecarilly it should have the law of gravity's proof but it prooves whatever that can be proven with newtonian mechanics? (Also the level of its math shouldn't be higher than Calculus, Vol I, Apostol)
Thanks! :lol:[/quote]

Try "Mechanics" by Landau and Lipschitz, or any other book on classical mechanics.

When i was a secondary student,I tried to find the proof of Newton gravitation and Coulomb law,but I failed.

So in Feynman's lecture:He told that:the proof and the way how gravitation occured is really a secret.

and this is:

WE can image that our universe is full of particles,and it flies and collides with our planet (EX our Earth and the sun) ,So:

take n1 as a number of particle collidewith the earth:

n2 is a number of paticle collide with the sphere which we can image
$n1=\frac{n2}{R^{2}}$
take dF as a Force which a particle  brings

so
$F1=\frac{F2}{R^{2}}$

but in F2 ,it is not only the paticle from M but also from m.

so
F2 is propotional to the number of particle .
F2=Kn

but ,the number of particle is propotional to the mass of the planet,the total number is propotional to Mm..... :roll:
so F2=KhMm

so

$F1=\frac{khMm}{R^{2}}$

note that kh=G

so
$F1=\frac{GMm}{R^{2}}$
and we have the proof :D

but :maybe:

this proof is wrong.
especially in which we observed

woww  rogue physicist..after a looong time.Sorry,ignore this post  :D

nitpick police to the rescue

[quote="Dr. No"]2. the laws of physics are the same in all reference frames.[/quote]
I think you mean all [i]inertial[/i] reference frames.


A) The majority of the members of the AoPS/Mathlinks forum, if I am not mistaken, live in the US. However, the forum welcomes everyone. Thus is it inappropriate to post things under the assumption that your readers are from the US? Especially in the Round Table, topics are being posted that speak of "we" and refer to US Americans, say "the government" and refer to the US government.

B) Is this in the appropriate forum?

I must say that I do agree with this point. I often post on issues of the US, but I really try to be very conscious of keeping my wording neutral so that all can participate and I'm not making an assumption about the reader. I do think that assumption is inappropriate in this forum. (On the other hand, it probably isn't intentional, but probably just hasn't been brought to the writer's attention).


Given x,y,z≥0 satisfying x+y+z=1.Prove that
$ \frac{x^2\plus{}1}{y^2\plus{}1}\plus{}\frac{y^2\plus{}1}{z^2\plus{}1}\plus{}\frac{z^2\plus{}1}{x^2\plus{}1}$≤$ \frac{7}{2}$

[quote="NickNafplio"]Inequality is equivalent to:

$ \sum{2x^4x^2 \plus{} 2x^4} \leq \sum{x^2} \plus{} 3\sum{x^2y^2} \plus{} x^2y^2z^2 \plus{} 1 \equal{} \sum{x^2} \plus{} 3\sum{x^2y^2} \plus{} \sum{x} \plus{} x^2y^2z^2$
which is obvious for $ x,y,z \in [0, 1]$ with equality for $ (x,y,z) \equal{} (1,0,0)$ and cyclic[/quote]
So we cannot solve this problem without expanding?

[quote="SUPERMAN2"]So we cannot solve this problem without expanding?[/quote]
Are you sure? You can see my proof here:  :wink:
[quote="nguoivn"]See it here:  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=254080[/quote]


I was teaching math to my sister yesterday and I made up some of these problems...
It's all about decimals and fractions
try!

1. convert following numbers into fractions.

78.414141414....
6.67777777....
0.9999999....
1.3452525252525.....

2.convert following fractions into numbers
7/3
56/97
87/99

3. How do we know if a fraction would turn into infinite decimals when it's converted to numbers..

4.find x

231/999 + 7/999 - 0.015015015.... + 0.776776776... = x

I just thought these were some good practices for Classroom Math...

You can also convert repeating decimals by putting then over 9s. Like if you have a decimal .2727272727272727... you can take one section that repeats, in this case the 27 and put it over 99. So it would be 27/99. That simplfies down to 3/11. :lol:


otro problema  :D :

determine el numero de maneras de ubicar dos alfiles distintos en un tablero de $(m)(n)$ de manera de que estos no se ataquen

Una pregunta, que no se ataquen en una sola jugada, o que nunca se ataken, despues de haber realizados un numero finito de jugadas??

en el problema los alfiles no deben atacarse en el primer paso (¡solo ubicarlos de manera que no se ataquen!)

[color=blue]Tengo una fuerte conjetura, la cual podría asumir por cierta, pero es muy empírica mi demostración: [i]"El alcance del ataque de un alfil llega, al menos, a los bordes más cercanos del tablero."[/i]

¿Alguien podría dar una demostración formal?[/color]

[color=blue]Consideremos que es [i]m[/i] la dimensión menor. Supongamos que son las filas del tablero las que miden [i]m[/i]. Entonces, las columnas se numerarán desde 1 hasta [i]m[/i].

La conjetura dice que un alfil atacará por lo menos una fila de cada columna, considerando también que ataca la casilla sobre la cual éste se encuentre.

Espero haberme explicado bien.[/color]

bueno, una ida creo que seria hacerlo por complemento, me refiero a que si buscamos el numero de formas de colocar 2 alfiles en un tablero de nxm, sin que se ataquen, mejor encontremos el numero total de formas en al cual se ataquen. y como el numero de formas de que se ataquen con las que no se ataquen sera el total de formas de colar las fichas en el tablero, ni bien termino la solucion la posteo.

[color=blue]Es cierto que la mejor forma de solucionar es buscar todos los casos en los cuales los alfiles se ataquen, y descontar luego del total de casos.

Lastimosamente, con los alfiles no es tan sencillo. Por eso quise, al menos, definir la cota inferior de casillas atacadas por un alfil en cualquier posición.

[b]Observación Interesante:[/b] Si hubiesen sido dos torres, en lugar de los dos alfiles, las piezas del enunciado, la solución iba a ser sencilla e inmediata. Exhorto a que comprueben el porqué.[/color]

[color=blue]¿NADIE ha tenido una sóla idea de cómo se puede resolver este problema? Es cierto que ya no he podido dedicar mucho tiempo, pero, al parecer, nadie está muy interesado en resolverlo…[/color]

[color=blue]He notado que un alfil siempre ataca la misma cantidad de casillas indistintamente del lugar que ocupe en el borde menor del tablero. ¿Alguien encuentra algún dato más ineresante?[/color]

Bueno lo que pasa es que es sencillo dar cotas pero las construcciones no lo son.

Por otra parte no todos los alfiles atacan el mismo número de casillas! :wink:

Ok, que hace fácil el problema de las torres? que las torres atacan por filas o columnas y justamente tenemos como representar el tablero en el plano cartesiano por coordenadas de filas y columnas.

La idea es entonces escoger dos cosas: una diagonal hacia abajo y la derecha y una diagonal hacia arriba y la derecha para cada alfil y que estas se intersecten. La condición se traduce entonces en que sus respectivas diagonales no sean las mismas! les dejo la tareita de ver de cuantas formas pueden escoger esto, pero eso si les doy una idea, dividan el tablero en regiones!

[color=blue]En un principio, tenía en mente dividir en cuatro regiones el tablero, y dividir en los distintos casos que podían ofrecerme las paridades de los lados. Pero desistí porque no veía cómo llegar a las conclusiones buscadas (aunque admito que tampoco tuve el tiempo suficiente para dedicarle…)

Quisiera saber, Pascual, si a esta división se refería usted (si es así, no sería la primera vez que veo el camino de la solución y no lo tomo…).[/color]

Contemos el numero de formas de poner 2 torres que se ataquen.
Dos torres se atacan si y solo si estan en la misma diagonal. Entonces toca ver cuantas diagonales hay y que longitud tienen.
Supongamos que $m\ge n$. Supongamos que son $n$ filas $f_{i}$ (enumeradas de abajo hacia arriba) y $m$ columnas $c_{i}$ (enumeradas de derecha a izquierda).
Contemos cuantas diagonales hay que de izquierda a derecha vayan hacia arriba. Hay $m+n-1$ pues escogemos cada una de las casillas de la izquierda y cada una de las casillas de abajo y estas nos determinan una diagonal diferente. La diagonal en $(f_{m},c_{1})$ tiene longitud $1$ en $(f_{m-1},c_{1})$ tiene longitud $2$ y asi hasta la diagonal de $(f_{1},c_{1})$ que tiene longitud $n$ (pues $n\le m$). Ahora en $(f_{1}, c_{2})$ tiene longitud $n$ y asi hasta $(f_{1}, c_{m-n+1})$ que tienen todas $n$. De ahi en adelante empieza a decrecer cada vez 1 hasta llegar a 1 nuevamente.
Las diagonales que de izquierde a derecha que van hacia abajo es la misma cantidad. Entonces solo lo hacemos para las $m+n-1$ diagonales de un sentido y multiplicamos por 2 el resultado.
Si $d$ denota la cantidad de casillas en una diagonal hay $\binom{d}{2}$ formas de poner las dos torres. Por lo que tenemos que el numero total de formas de poner las torres es:
$2(2\sum_{i=2}^{n}\binom{i}{2}+(m-n-1)\binom{n}{2})=2(\sum_{i=1}^{n-1}i(i+1)+(m-n-1)\frac{(n-1)n}{2})= \frac{2(n-1)n(n+1)}{3}+(m-n-1)(n-1)n = \frac{(n-1)n(3m-n-1)}{3}$
Ahora la cantidad de formas de poner dos alfiles en un tablero de $mn$ es:
$\binom{mn}{2}= \frac{mn(mn-1)}{2}$ por lo que tenemos que la numero de forma de poner 2 alfiles sin que se ataquen en un tablero de $mn$ es:
$\frac{mn(mn-1)}{2}-\frac{(n-1)n(3m-n-1)}{3}$ si $m\ge n$.
QED

Me acabo de dar cuenta que en varios pedazos del post hablo de torres cuando deberían ser alfiles. Entonces pues cambienlos.

Por otra parte noten que mi solución es larga y fea. ¿No habrá otra forma de hacer ese conteo? ¿Es posible arreglar esos terminos que me dieron a
mi para obtener una formula mas bonita?
Ahi los dejo con la duda.

JAS

[color=blue]Cuando quise decir que, con dos torres, era fácil solucionar el porblema, fue porque siempre ataca el mismo número de casillas. Entonces una torre ocupa y ataca, a la vez, $m+n-1$ casillas. Si la primera torre tiene $mn$ casillas para elegir, la segunda tendrá $mn-m-n+1$.
Es el número total de combinaciones $\frac{mn(mn-m-n+1)}{2}$ para que dos torres no se ataquen en un tablero.

El problema es que, con dos alfiles, es más complicado, puesto que no siempre ataca uno la misma cantidad de casillas. Seguiré intentando resolverlo.[/color]

Ahi esta con los dos alfiles solo que al escribir me equivoque y puse torres. Pues dice que dos torres se atacan si estan sobre la misma diagonal, que claramente no es el criterio de ataque de las torres.  Son alfiles. Lo que pasa es que el post es muy viejo y no puedo editarlo mas.


These are both from AoPS Volume I:

At the end of a professional bowling tournament, the top 5 bowlers have a playoff.  First #5 bowls #4.  The loser receives fifth prize and the winner bowls #3 in another gamer the loser of this game receives fourth prize and the winner bowls #2.  The loser of this game receives third prize and bowls #1.  The winner of this game gets the final prize and the loser gets second place prize.  In how many orders can bowlers #1 through #5 receive their prizes?

I tried to count the possibilities for each prize...  I'm not sure where I messed up (please point out my error as well as a proper solution)

5th prize - there are two choices here (# 5 or #4)
4th prize - there are three choices here (#5, #4, and #3)
3rd prize - there are four choices here (#5, #4, #3, and #2)
2nd prize - there are five choices here (#5-1)
1st prize - there are five choices here (#5-1)

So in all there would be $ (2 \cdot 3 \cdot 4 \cdot 5 \cdot \ 5) \equal{} 600$  ways

Well, just to show an example that your procedure doesn't work, your method allows for a placement $ 5, 5, 5, 5, 5$.

[hide="Solution?"]There are $ 2$ possibilities for the fifth winner, and given that, there are $ 2$ possibilities for the fourth winner... there are $ 2$ possibilities for any given winner, so in total, there are $ \boxed{2^5 \equal{} 32}$ possibilities.[/hide]

Oh  :blush:  I see that I badly over counted....

Then wouldn't the answer be $ 2^4 \equal{} 16$ because there are $ 4$ "games" each with two possible outcomes?

Yes, each game has two outcomes, and there are 4 games, so 16. If you have trouble visualizing this, make a tree of all of the possibilities.


Fond $ a,b$ such that $ max_{   0\le x\le 16} |\sqrt{x}\plus{}ax\plus{}b|$ is smallest possible.Find thix minnimum value

Let $ g(x)=\sqrt x+ax$. If $ a\ge \frac{-1}{8}$, then $ g(x)$ increase and $ b=-\frac 12 g(16)=-(2+8a)$ and minimal value is $ -b=2+8a\ge 1$. If $ a<-1\8$, then $ x_0=\frac{1}{4a^2}$ and $ b=-\frac{max(0,-g(16))-g(x_0)}{2}.$
If $ -1/8\le a\le -1/4$, then maximal value is $ \frac{-1}{4a}=-2b$. It give $ a=-\frac 14, b=-\frac 12$.
If $ a<\frac{-1}{4}$, then $ g(16)=4+16a<0$ and maximal value for best b is $ -2-\frac{1}{8a}-8a>\frac 12$.
Therefore $ a=\frac{-1}{4},b=\frac{-1}{2}, max |g(x)|=\frac 12$ is best.


( N - ( N (mod 1000) ) ) / 1000? Basically, N minus the remainder when you divide by 1000, then divide that by 1000

Hi could someone help me with this equation, I need to solve for N= 209454202, I hope to remove the last 3 digits of N from this operation?

Hebby :)
It is just removing last three digit from the number.
209454202 - 209454202 mod1000 = 209454202 - 202 = 209454000
And 209454000/1000 = 209454.
Tell me if you want to get something else.

HEY THANKS:)...SO IT REMOVES 3 DIGITS FROM ANY NUMBER?

Well, it removes anything lower than the thounsands digit. In the case of integers, that would just be the hundreds, tens, and units digits, so yes, 3 numbers. But if it had a decimal part, it would also remove those.


Let a,b \in R and 0 < a < b < 2a. Among all triangles with side lengths a^k ,a^k, b^k (k \in R arbitrary) the one with the greatest area. Determine this area and the respective k dependently on a and b.

A bit technical problem. Definitely, it is not an olympiad one.


calculus
function
integration
calculus computations

Hi everybody,

I've been trying to solve this problem. I think I have done part i) (Answer: D) and part ii) (Answer: C), can anyone confirm whether I have got these correct? I am also unsure about how to approach part c), I imagine it's dependent on having gotten the first two parts right?

If you would like to see how I got D and C for my first two answers, I can script up my workings.

Thanks to anyone who can help, I'm quite desperate, it's been bothering me for a while now!

Rory

[img]http://i21.photobucket.com/albums/b256/lando-45/particlequestion.jpg[/img]

You might want to consider attacking part C) with an Energy argument.

No, not energy. We have air resistance; this is not a conservative force.

To rory: Yes, you have the right equation on the way down. There are two approaches from here:
(1) You have a first order differential equation for $ v(t)$ on the way up. Solve it, then integrate to get $ z$ as a function of $ t$, stopping at the zero of $ v$. This gives you the height at the top. Next, solve the differential equation on the way down, and integrate to get $ z$ as a function of $ t$. Stop at the zero of $ z$ and note the velocity there.
(2) Follow the hint, and convert it to a differential equation in for $ v(z)$. Solve it on the way up, stopping at the zero of $ v$. You now know the height at the top. Solve the downward equation for $ v(z)$ with that initial condition, and stop at $ z\equal{}0$ to get the final velocity.

If you've done part (i), you already know how to solve the differential equations. If you have more questions about the details there, please post what you have done in that part.

I've gotten it, thanks. I obtained B as my answer for part c). Basically just involves integration, being careful about the limits, then a bit of algebraic manipulation. Many thanks for your help!


Find all functions $f: R \to R$ that obey the equation: $\large f(x)f(y)+f(x+y)=f(xy)+f(x)+f(y), \forall x,y \in R$

[quote="wenxin"]$f(x)=x$[/quote]
but $f(x)=2$ is also solution

$f(x)=2$
Give $x=y=0$ we have $f(0)=0$ or $f(0)=2$
Give $\neq 0 ; y=0$ we have
 $f(0)f(x)+f(x)=f(0)+f(0)+f(x)$
If $f(0)=0$ then not have functions
If $f(0)=2$ then $f(x)=2$ . That is solution
( :maybe: )
[i]Louis Latin and Vicky[/i]


linear algebra
matrix
induction
IMC
college contests

Let $ n\ge 2$ be an integer. What is the minimal and maximal possible rank of an $ n\times n$ matrix whose $ n^{2}$ entries are precisely the numbers $ 1, 2, \ldots, n^{2}$?

Nice problem. It seems that the answers are: minimal rank 2, and maximal rank $ n^{2}$ for all $ n$.

For minimal rank, let us first notice that there is a prime $ p$, between $ \frac{n^{2}}2 \leq p < n^{2}$ (by Bertrand). Obviously there are no other multiples of $ p$ in the set of elements in the matrix, so any $ 2\times 2$ minor of the determinant containing $ p$ will have a non zero determinant. So the minimal rank is greater than or equal to 2.

An example of matrix with rank 2 is the following \[ A= \left(\begin{matrix}1 & 2 & \ldots & n \\ n+1 & n+2 & \ldots & 2n \\ \vdots & \vdots & \ldots & \vdots \\ n^{2}-n+1 & n^{2}-n+2 & \ldots & n^{2}\end{matrix}\right) .\] To see that, it is easy to check the if we subtract the $ (n-1)-$th column from the $ n-$th column, then the $ (n-2)-$th column from the $ (n-1)-$th column, and so on we obtain a matrix with the same rank \[ A' = \left(\begin{matrix}1 & 1 & \ldots & 1 \\ n+1 & 1 & \ldots & 1 \\ \vdots & \vdots & \ldots & \vdots \\ n^{2}-n+1 & 1 & \ldots & 1 \end{matrix}\right) ,\] which clearly has rank 2.

To prove that the maximal rank is $ n$, we have to provide an example with an arrangement that has the determinant non-zero. I haven't found one yet for the general case :blush:  :maybe:  (examples seem to be plenty for small values of $ n$).

To lorincz: nice solution!
We can prove this statement also by induction (simply after fixation of the first $ n^{2}$ numbers we have       a lot of variants for choising of another $ 2n+1$ numbers, enough to get $ det\ne0$).

Another (elementary) proof for the lower boundary:
Suppose there is the matrix $ (a_{xy})$ with rank 1. Without lost of generality $ a_{kl}=1$. Then there are numbers $ m,n$ such that $ a_{kn}\geq n$ and $ a_{ml}\geq n$. Then $ a_{mn}=\frac{a_{kn}a_{ml}}{a_{kl}}>n^{2}$, else $ a_{kn}=a_{ml}=n$. Contradiction.

[quote="lorincz"]Easier than the first part :). Here's an example of a matrix with a non-zero determinant : put odd numbers on the main diagonal, put even numbers above it, and the rest below. It's easy to see now that if we write the determinant as sums there will be only one odd (the product of the numbers on the diagonal), while the others will be all even, therefore we will have an odd determinant.[/quote]Yeah I just came up with that too, but it seems people already solved it :) At first I tried induction, but it doesn't seem that obvious how to create a general step (without coming back to the even-odd thing which I eventually did). Pavel do you have an induction solution that doesn't use the odd-even argument?

in the same way as pavel kozlov did BUT, observe that $ D= det(a_{i,j})(b_{n+1}-\sum_{k=1}^{n}\alpha_{k}b_{k})$ where the coefficients $ \alpha_{k}$ are chosen so that when we substract the columns 1,2,...,n from the column n+1 we get the first n elements on the column n+1 to be 0.
Now if $ D=0$ then just permute $ b_{n+1}$ with some $ b_{k}$ for which $ \alpha_{k}\neq-1$ and if this determinant is also zero you get that $ b_{n+1}=b_{k}$ which is a contradiction.


What is the background of each inequality?


[quote="jon_voight"]$n$ is a natural number.and $a,b,c$ are non negative numbers.prove that:
    $(a+b+c)^{n}\times (a^{2n}+b^{2n}+c^{2n}) \geq (a^{n}+b^{n}+c^{n}) \times (a^{2}+b^{2}+c^{2})^{n}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=m38+inequality&t=1989
the manlio's proof.


ARML
AMC
AIME
USA(J)MO
USAMO
USAMTS
AMC 10

What are your goals for this year? Here are mine, I am in 8th grade:

Perfect at MC chapter and state
Top 2 Written, top 4 countdown, top team, top masters at MC nationals-notice these are the 4 ways you get money  :P
25 total on the HS Math Leagues
Perfect on the 8th grade math league
Perfect on ASMA
Make MA ARML A team, 7+ individual
Perfect on AMC 8
Perfect on AMC 10A
130+ on AMC 12
8+ on AIME
1 on USAMO  :P

Perfect on AMC 10 (if I take this, which I doubt I can)
130+ on AMC 12
8+ on AIME
nonzero on USAMO

1st in all divisions in KCATM for 9th gr (you're going down jaybird45!)
1st team Mathletics KCATM
1st senior high division KU MAM
3rd+ in Manhattan math olympiad (KSU) (probably 5 ppl will be competing, heh)

You should include something like the likelihood of reaching your goals

Perfect on the AMC 10 (Possible, but would require a lack of stupid mistakes :P

Anything good enough to make USAMO on the AIME (Too early to tell if possible, but I'm at around the 3-5 level right now, so I could make it)

I haven't tried any USAMO's, preferring to make sure that I can get there before doing too much work, so I don't really know what to expect. Making mop would be cool though :rotfl: .

Math League=28

Mandelbrot=30

USAMTS=60

Oh yeah I forgot some

USAMTS-Bronze
MAML Level 1-My goal was 120 but I got 109  :(
MAML Level 2-I don't even know how this is scored...
IMLEM-perfect score

Our school team getting to states (and I think the chances are high, even though our chapter is [b]huge [/b]and full of [b]top [/b]schools)
and if i work really really really really hard (along with prepping for the SAT and HS entrance exam) me getting to nationals

its going to be very hard to choose our team this year.  we have four people who i know are really good (like me,  :lol: ), but lots of new people who show potential and are willing to practice.

oh, and a 25/25 MOEM (i got 24 last year!)
and a >20 AMC 8 (eh, dunno how that'll go)

140+ AMC 10 (Possible if I read more carefully and make no careless errors)
120+ AMC 12 (Definitely possible)
8+ AIME (Possible if I improve efficiency)
7+ USAMO or even MOP (never taken a practice USAMO, so I should sometime during the winter break, not sure if possible)
80+ USAMTS (definitely possible)
5+ ARML
Introduce Mandelbrot and ARML Power Contest to my school by the end of the year

Hmmmm, let me think:
AMC 10 score of 135 or greater
Mathleague, two perfect scores
ARML: attend, use at least half my brain cells :P
AIME: 6 or greater (Unlikely)

Beyond contests:
Volunteer for Mathcounts
Become Math Club VP as a soph

Well, competitions finish very early for us so I have my next year's goals.

Perfect Score for Australian Mathematics Assessments
Medal or Prize for the Westpac Award
Solve all the problems on the September Problem Sets
Be part of the NZIMO
Perfect Score for the Auckland Olympiad
First place for National Bank Mathematics Contest
Be part of my school's Mathex Team
Pass IGCSE Mathematics with A*

I will try to  get 30+/40 in Bucharest olympiad and maybe qualify for NMO. (hard for me :oops: )

AMC 12 - 150 score
AIME - score in the double digits
USAMO - 7+

NYML: 36
USAMTS: gold medal range
AMC12: 150
AIME: 11+
USAMO: Blue MOP (though that might be a pretty ambitious goal...)
ARML: 7-8 ($\ge$ what i got in 2006, that is)

Mandelbrot: though it's my first year, so I don't know what to expect, maybe 45+?  If all goes okay...

NYSML: 5+
AMC 12 A&B:100+
Mathcounts Chapters: 40+
Mathcounts States: 30+
AIME: 3+
HMMT(2006):one of the lower scores, I think... at least...

lol.

- Successfully run the Princeton Math Competition. :D

- Don't noob the Putnam. (Hopefully top 200.)

hmm
AMC12-145+
AIME-9+
USAMO-15+
and MOP!

Realistically, as in not over my head. I have already been practicing almost a month now.

top 10 at vestavia hills math tournament geometry section

duno about AMC 8, never taken it

and top 10 at grissom math tournament and hoover tournament :)

I really hope this will happen. After the last 3 weeks practicing
ma0 stuff etc.

Goals:

AMC 12: 150
AIME: 15
USAMO: 30+ (getting around 20 on practices right now, but I have yet to even read AoPS or similar, so room for improvement exists)
USAMTS: 100
Math league press: Perfect on all contests
State math league: Straight perfect scores (including invitational)
ARML: Help convince DE to send a team, be part of said team, and place in top 5 teams overall

GOALS:

23+AMC8
120+AMC10
100+AMC 12
1+AIME
MC Nationals top 50
MC County countdown winner, first written
MC chapter 1st countdown, top 4 for written
MC state, top 4 written.

Those are the ones I can think of right now...


geometry parallelogram rotation circumcircle ratio Euler vector

Let $ABCD$ be an arbitrary quadrilateral. Let the diagonals $AC$ and $BD$ of this quadrilateral $ABCD$ intersect each other at a point $O$. Let $X$ and $Y$ be the centroids of triangles $AOB$ and $COD$, and let $I$ and $J$ be the orthocenters of triangles $BOC$ and $DOA$. Prove that the lines $XY$ and $IJ$ are perpendicular.

I can say how I would procede to solve the problem.

The parallelograms formed by four hortocenters and four centroids are similar and rotated of 90� each other. This derive from fact that the sides of the former are orthogonal to the diagonals of the quadrilateral  and the sides of the latter are parallel to the diagonals of the quadrilateral.

from this it follows the claim.

PS
One can prove that, also the parallelogram formed by four circumcenters is similar to the above mentioned parallelograms.

I thought about the exact same thing, but that doesn't end it as neatly as one would want: we don't know that the parallelograms are similar. Think about it:if we are given the two directions of the sides of a parallelogram, we can take the distance between two of its parallel sides as large as we want, obtaining infinitely many parallelograms which are pairwisely not similar.

I did complete the proof, but it's not as short and neat as Darij suggested it should be..

[quote="grobber"]I thought about the exact same thing, but that doesn't end it as neatly as one would want: we don't know that the parallelograms are similar...[/quote]

Touche!!

I'm sorry  :blush: .

I propose  a way to close the gap. Considere the circumcenters-parallelogram, this has sides which are orthogonal to the diagonals of ABCD, and [not too difficult to prove]are in the same ratio of AC/BD=r. Also the sides of the centroids-parallegram are in the same ratio r [easy to prove].
Now, given that O's, H's and G's are collinear and GH/OG=2  we have that also the sides of the orthocenters-parallegram are in the same ratio r[non difficult to prove. I already done for another problem in this forum].

[quote="grobber"]Yes, this is exactly what I did :).[/quote]

Actually, this was not true. I had shown (without any theorem of Euler or anything like that) that the ratio of the distances between the pairs of parallel sides was equal to the ratio between the lengths of the diagonals. That is basicly the same thing, but via another route :).

[b]EDIT.[/b] This message was written before the last two posts by Sprmnt21 and Grobber.

[quote="sprmnt21"]I propose  a way to close the gap. Considere the circumcenters-parallelogram, this has sides which are orthogonal to the diagonals of ABCD, and [not too difficult to prove]are in the same ratio of AC/BD=r. Also the sides of the centroids-parallegram are in the same ratio r [easy to prove].
Now, given that O's, H's and G's are collinear and GH/OG=2  we have that also the sides of the orthocenters-parallegram are in the same ratio r[non difficult to prove. I already done for another problem in this forum].[/quote]

Hmm, I am not too sure about this proof. The circumcenters parallelogram and the centroids parallelogram are similar, but the circumcenter and the centroid of the same triangle are not corresponding vertices in the similarity. Probably there should be a special conclusion for this kind of cases, but I don't think you can use the fundamental theorem on similar figures here. Anyway, you can directly prove that the centroids parallelogram and the orthocenters parallelogram are similar by showing their sides to be perpendicular and to stand in equal ratios. This is not difficult.

[b]EDIT.[/b] As the two last posts show, both of you (Sprmnt21 and Grobber) have found this idea, too.

Now some background to the problem:

It is a problem from Arthur Engel. It is usually solved using vectors or complex numbers. But actually there is a quite famous result in geometry which turns out to be equivalent to this problem (or, more precisely, the problem almost trivially follows from this result).

What is this result?
[hide="Hint"]Hint: Draw a figure where the quadrilateral ABCD intersects itself; more precisely, the quadrilateral [b]ADBC[/b] should be convex. Do you see something?  ;)[/hide]

  Darij

Actually this was a problem proposed on ALL Russian olympiad in 1972.
Here is it solution.
(in all solution I use direct segments - vectors)
Let H1 and H2 will be the ortocenters of AOB and COD, G3 and G4
are the gravity centers of BOC and AOD, <W = <AOB=<COD)

OH1 =ctgW(iBA). OH2 = ctgW(iDC) =>  H1H2 =ctgW i(DC-BA).

OG3 =2/3*1/2(OB+OC) OG4 =2/3*1/2(OD+OA)

G3G4 = 1/3 (OD+OA -OB-OC) = 1/3 (CD - AB)

Hence H1H2 = -3ctgW(i G3G4) or H1H2 perpendicular to G3G4.

I don't know about yours proposition.
I check in my mind all famous theorems which I know and nothing.
Maybe you will reveal your conundrum.

[quote="darij grinberg"]But actually there is a quite famous result in geometry which turns out to be equivalent to this problem (or, more precisely, the problem almost trivially follows from this result).

What is this result?[/quote]

Oh, I think I know what Darij is talking about: the Aubert line of a quadrilateral is perpendicular to its Gauss line.

You are right Grobber our problem directly gives that
Aubert line is perpendicular to a Gauss line.
But I never run into this fact (I am not so experience as you guys).
So this is a very nice theorem for me, and a very good conundrum gived by Darij.

[quote="sprmnt21"]I found a direct (without pass through the circumcenters and Euler's theorem) way to prove that the ratio of the sides of the orthocenters-parallelogram is equal  of the ratio of the diagonals of the quadrilateral.
Is not to difficull at all.   Only similarity between rectangular triangles and the fact that if a/b = c/d then (a+b)/b = (c+d)/d.

If it is of some interest I can post it.[/quote]

here is another, more simple and direct, way to prove this fact.

If R, S and T are the orthocenters of ADO, BCO and ABO. then let U and V be points s.t. DRTU and TSCV are parallelograms. Then is easy to see, as UDB ~ VCA, that RT/TS = BD/AC.

Whats that? what are all those special lines you are talking about?

Thanks!

Given a complete quadrilateral, the midpoints of its diagonals lie on its Neuton-Gauss line, and the orthocenters of the four triangles formed by the sides of the quadrilateral lie on its Aubert line.

[quote="sprmnt21"]here is another, more simple and direct, way to prove this fact.

If R, S and T are the orthocenters of ADO, BCO and ABO. then let U and V be points s.t. DRTU and TSCV are parallelograms. Then is easy to see, as UDB ~ VCA, that RT/TS = BD/AC.[/quote]

And triangles UDB and VCA are similar since the points B, S, T, U are collinear and the points A, R, T, V are collinear. Very beautiful simplification of the proof!

Grobber is completely right - the result I meant was that the Aubert line of a complete quadrilateral is perpendicular to its Gauss line. Who is new to this topic of geometry can find an introduction on http://www.mathlinks.ro/Forum/viewtopic.php?t=842 . By the way, instead of "Aubert line", I usually say "Steiner line". In the problem I posted:

[quote="darij grinberg"]Let the diagonals AC and BD of a quadrilateral ABCD intersect at O; let X and Y be the centroids of triangles AOB and COD, and let I and J be the orthocenters of triangles BOC and DOA. Prove that the lines XY and IJ are perpendicular.[/quote]

you can consider the complete quadrilateral formed by the lines AC, BC, BD and DA. The line IJ is the Aubert line of this complete quadrilateral, since it passes through the orthocenters I and J of triangles BOC and DOA. The Gauss line of this complete quadrilateral is the line X'Y', where X' and Y' are the midpoints of the segments BC and DA (this is because the segments BC and DA are two diagonals of our complete quadrilateral). Since the Aubert line is perpendicular to the Gauss line, we thus get IJ $\perp$ X'Y'. Now, in order to prove that IJ $\perp$ XY, it remains only to show that X'Y' || XY. But this is almost trivial: Being the centroids of triangles AOB and COD, the points X and Y divide the medians OX' and OY' of these triangles in the ratio 2 : 1, i. e. we have OX : XX' = 2 and OY : YY' = 2, so that OX : XX' = OY : YY' and X'Y' || XY. Proof complete.

  Darij

[quote="pestich"]Darij,

   Don't they use orthocenters of different triangles for the Aubert-Steiner line?

   Pestich.[/quote]

The Aubert-Steiner line of a complete quadrilateral is the line passing through the orthocenters of the four triangles obtained by taking three of the four lines of the quadrilateral.

In our case, the complete quadrilateral consists of the lines AC, BC, BD and DA (not AB, BC, CD, DA). Two of the triangles formed by taking three of these four lines are the triangles BOC and DOA. The orthocenters of these triangles are I and J, respectively, so that the line IJ is the Aubert-Steiner line of our complete quadrilateral.

Pestich, I can imagine that you were considering another complete quadrilateral. Anyway, in order to make the picture look like a "standard" complete quadrilateral, the quadrilateral ABCD must be self-intersecting, and precisely, the sides AB and CD must intersect each other in their interior.

  Darij

I remember I have seen a solution in an old issue of Gazeta Matematica , before 1980. It was solved with radical axis. I ' ll try to find the problem again.

Babis

[Moderator edit: Some more people posted this problem with the following sources:
geometry in iran and international mathematics olympiad;
New Zealand Mathematical Olympiad 2005]

I change the notations:

[color=blue][b]Problem.[/b] Let $O$ be the point of intersection of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$. Let $G_{1}$ and $G_{2}$ be the centroids, respectively, of the triangles $OAB$ and $OCD$. Let $H_{1}$ and $H_{2}$ be the orthocenters, respectively, of the triangles $OBC$ and $ODA$. Prove that $G_{1}G_{2}$ is perpendicular to $H_{1}H_{2}$.[/color]

We have $3\overrightarrow{G_{1}G_{2}}=\overrightarrow{AC}+\overrightarrow{BD}$.

Then $3\overrightarrow{H_{1}H_{2}}\cdot\overrightarrow{G_{1}G_{2}}=\overrightarrow{H_{1}H_{2}}\cdot\overrightarrow{AC}+\overrightarrow{H_{1}H_{2}}\cdot\overrightarrow{BD}\;\;\;(1)$

On the other hand
$\overrightarrow{H_{1}H_{2}}\cdot\overrightarrow{AC}=\overrightarrow{H_{1}D}\overrightarrow{AC}+\overrightarrow{DB}\overrightarrow{AC}+\overrightarrow{BH_{2}}\overrightarrow{AC}=\overrightarrow{DB}\overrightarrow{AC}\;\;\;(2)$

and $\overrightarrow{H_{1}H_{2}}\overrightarrow{BD}=\overrightarrow{H_{1}A}\overrightarrow{BD}+\overrightarrow{AC}\overrightarrow{BD}+\overrightarrow{CH_{2}}\overrightarrow{BD}=\overrightarrow{AC}\overrightarrow{BD}\;\;\;(3)$

From (1),(2),(3) we obtain $H_{1}H_{2}\perp G_{1}G_{2}$

A simple analytical approach:

Place $ABCD$ to $xy-$plane such that the line $BD$ is the $x-$axis.
Let $D(x_1, 0)$, $A(x_2, y_2)$, $B(x_3, 0)$, $C(x_4, y_4)$.
The slope of $G_1G_2$ is equal to the slope of $M_1M_2$ where $M_1$ is the midpoint of $AD$ and $M_2$ is the midpoint of $BC$.
So the slope of $G_1G_2$ is equal to $m_G = \frac{y_4 - y_2}{x_3 + x_4 - x_1 - x_2}$.

$H_1(x_2, a)$ and $H_2(x_4, b)$.
Since $H_1B \perp AC$ and $H_2D \perp AC$, the slope of two perpendicular lines is
$\frac{a}{x_2 - x_3}\cdot \frac {y_2 - y_4}{x_2 - x_4} = \frac{b}{x_4 - x_1} \frac {y_2 - y_4}{x_2 - x_4} = -1$.

Slope of $H_1H_2$ is equal to $\frac {b-a}{x_4 - x_2}$.
So $b-a = \frac {(x_4-x_1)(x_4-x_2)}{y_2 - y_4} - \frac {(x_2-x_3)(x_4-x_2)}{y_2 - y_4}$

$m_H = \frac {b-a}{x_4 - x_2}= \frac {x_4+x_3 - x_1-x_2}{y_2-y_4}$.

So $m_G \cdot m_H = -1 \Rightarrow H_1H_2 \perp G_1G_2$.$\blacksquare$

Here's my Proof for this problem.
See attachment.

[quote="sunken rock"]Let $M, N$ be the midpoints of $AB, CD$ respectively; obviously, $XY\parallel MN\ (\ 1\ )$.
Let $P, Q$ be the projections of $O$ onto $BC, AD$ respectively; obviously, $IJ\parallel PQ$, but it is a well known fact that $PQ\bot MN$, hence we are done.

Best regards,
sunken rock[/quote]
Your statement is wrong.
$IJ \parallel PQ \Leftrightarrow ABCD $ is cyclic.

[b]@xeroxia[/b]: Yeap, you're right, sorry!

Best regards,
sunken rock

[b]@XmL[/b]: your proof can be shorter:
From $\Delta BOC\implies BI\cdot IK=QI\cdot IC\implies I$ belongs to radical axis of $\odot (M, MA), \odot (N, NC)$.
From $\Delta AOD\implies AJ\cdot JL=JP\cdot JD\implies J$ belongs to the same radical axis, hence $IJ\bot MN$.

Best regards,
sunken rock


When the conditional  expectation of a random variable was defined it was defined conditioning with respect to a sigma algebra i.e. $ E[X|\mathcal{G}]$. But when we define something like $ E[X|X>0]$, should we interpret it as the conditional expectation of $ X$ with respect to the sigma algebra generated by the indicator random variable $ 1_{\{X>0\}}$?

Not quite. In the usual interpretation $ E[X|X > 0]$ is a number, whereas $ E[X|F]$ is a random variable.
Here $ F \equal{} \sigma(1_{\{X > 0\}})$.  The relation between these two objects is
\[ E[X|F] \equal{} E[X|X > 0]1_{\{X > 0\}} \plus{} E[X|X \le 0]1_{\{X \le 0\}}.\]

[quote="solyaris"]\[ E[X|F] \equal{} E[X|X > 0]1_{\{X > 0\}} \plus{} E[X|X \le 0]1_{\{X \le 0\}}.\]
[/quote]
What's $ F$ in the above relation?  Is  $ F \equal{}\sigma(1_{\{X > 0\}})$?
I don't quite understand. How do you define $ E[X|X > 0]$ in general?

If you look at my previous post carefully you will find that I defined $ F$ there.

In general if $ A$ is an event with $ P(A) >0$,
\[ E[X|A] : \equal{} \frac{E[X1_A]}{P(A)}.\]
If $ A$ is an event with $ P(A)\equal{}0$, then $ E[X|A]$ doesn't make much sense.
Again
\[ E[X|\sigma(1_A)]\equal{} E[X|A] 1_A \plus{} E[X|A^c] 1_{A^c},\]
which you can easily check by plugging in the definition of expectations conditioned on events on the right hand side, and use the definition of an expectation conditioned on a $ \sigma$-algebra.

solyaris is right but I prefer (and this the way I have be taught) to always look at conditionnal expectation as a r.v.

The interpretation is acceptable though regarding $ E[x|x>0]$ as it is quite common use.

To complete the remark of solyaris about the fact that you cannot condition over an event of null measure there exists though a correct interpretation of such expression, thank's to Doob' theorem.

Let's say we want to give a meaning to the expression $ E[Y| X\equal{}a ]$ where Y is a positive real valued measurable r.v. and X is also a positive random variable such that $ P(X\equal{}a)\equal{}0$.

As it is possible to define $ Z\equal{}E[Y| X]\equal{} E[Y| \sigma(X)]$ which is $ \sigma(X)$-measurable r.v., then Doob's theorem says that there exists (modulo null sets) a unique measurable function $ g$ such that $ Z\equal{}g(X)$ so you can define $ E[Y| X\equal{}a ]$ as $ g(a)$ (again this definition is uniquely determined up to P null sets).

Regards


[hide="Big hint"] Let those three quantities be equal to $x$.  Then

$a^{2}x+b^{2}x+c^{2}x = 0$ [/hide]

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$, then $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}$.

[quote="drunner2007"]Let $a,b,c$ be nonzero real numbers. Find all ordered pairs $(a,b,c)$ such that

$\frac{2(a-b-c)}{a^{2}}=\frac{4b-a-2c}{b^{2}}=\frac{4c-a-2b}{c^{2}}$[/quote]

[hide="solution based off of the hints"]$\frac{2(a-b-c)}{a^{2}}=\frac{4b-a-2c}{b^{2}}=\frac{4c-a-2b}{c^{2}}=\frac{0}{a^{2}+b^{2}+c^{2}}=0$

$a-b-c=0$
$a-4b+2c=0$
$a+2b-4c=0$

Subtract the first and second equations to get $3b-3c=0$ so $b=c$.

This means $a=2b=2c$ so $\boxed{(2n, n, n)}$ and $n\ne 0$[/hide]


Let $ x<0,y<0,z<0$,and they satisfy $ x\plus{}y\plus{}z<\minus{}3$ and $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$.

1)Show that $ (x\plus{}1)(y\plus{}1)(z\plus{}1)\le 0$

2)Give one example $ (x,y,z)$ of irrationals.

Are you sure the second condition is correct?  [hide="Because as far as I know..."] $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1 \Leftrightarrow \exists \alpha, \beta, \gamma : \alpha \plus{} \beta \plus{} \gamma \equal{} \pi, x \equal{} \cos \alpha, y \equal{} \cos \beta, z \equal{} \cos \gamma$. [/hide]

There are other solutions to that, like $ (x,y,z) \equal{} (\minus{}2,\minus{}2,\minus{}7)$. (complex angles maybe?)

The conditions are correct :wink:

Challenge!! :)

$ \left\{\begin{array}{l} x \plus{} y \plus{} c < \minus{} 3 \\
x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1 \end{array}\right. \iff \exists \alpha, \beta, \gamma : \alpha \plus{} \beta \plus{} \gamma \equal{} 0, \left\{\begin{array}{l} x \equal{} \minus{} \cosh \alpha \\
y \equal{} \minus{} \cosh \beta \\
z \equal{} \minus{} \cosh \gamma \end{array}\right.$

(1) $ x, y, z \le \minus{} 1$ so, trivial.
(2) $ \forall n \in \mathbb{Z}\backslash\{0\} \Rightarrow e^n \in \mathbb{R}\backslash\mathbb{Q}$. Thus, for example $ \left\{\begin{array}{l} x \equal{} \minus{} \cosh 1 \\
y \equal{} \minus{} \cosh 2 \\
z \equal{} \minus{} \cosh ( \minus{} 3) \equal{} \minus{} \cosh 3 \end{array}\right.$


[quote="GameBot"]Given that $ 105\%$ of a number is 63, what is the number?[/quote]

$ \frac{21}{20}x\equal{}63 \implies x\equal{}3 \cdot 20\equal{}60$.


Hai folks, In a short span of time ( in about 3 weeks) , I am going to attempt the General-GRE; I had finished the famous word lists of Barron's.

I am currently practicing model tests; I felt Kaplan followed by Princeton were better when compared with other 'online-tests'.

I ask you ( people of this community who had an experience in securing a very high score on the GRE) to provide some tips/suggestions in securing a 1500+ score out of a possible 1600 on the GRE. ( I feel I can secure above 1500, but I need some advice, suggestions).

I also ask you to provide some advice on attempting the Analytical Writing Assessment section; If (if any) there are people in this forum who have managed to secure a perfect score on the general-GRE, I would love to hear their suggestions/advices.

I THANK YOU ALL IN ADVANCE. :)

[quote="tokenadult"]What do you read for fun? Reading a lot is the key to scoring high on the GRE.[/quote]

Well, he [i]is[/i] writing the GRE in three weeks.  While your advice is good, there are better ways to improve your expected score given the time frame.

@ token.. : I read a lot of newspaper material, books, blogs, excerpts from books, etc; I know that---- reading a lot is the key--- for securing a good verbal score, but I want  some suggestions/ tips that streamline my preparation. Do you have any?

As far as I know the Quant section of the GRE is pretty easy, but the verbal section is relatively tough, so I want tips in securing a good/great verbal score.

@Blahh.... : Okay dude, do you have any suggestions/tips for securing a high/very high ( 1500+) GRE score? If so, do share them with me, I would appreciate your help.

You have taken the GRE test now, right? How did it feel?

[b] Yes, tokenadult [/b]

I apologize for replying so late, I haven't been able to access AOPS forum for the past 3-4 weeks, even when I did manage to access this forum, I might have overlooked your post.

I attempted the GRE and I felt mentally exhausted (to a certain extent) after the test. I would say I felt better after viewing my overall score. I managed to secure 1400 on the GRE, but I felt really disappointed after viewing my Quants score.

That was my GRE experience.  :)


geometry
rectangle
Pythagorean Theorem
geometry proposed

[quote="soruz"]In the  rectangle $ ABCD,E$ is middle of $ AB, FE \perp DE, F \in (BC), FE \cap CD \equal{} \{G \},CD \equal{} 12,AD \equal{} 8$.Find the lenght of $ CG$.[/quote]
This problem is too easy so you should post it in forum middle shool  :)
tan $ \hat{ADE}\equal{}1/2\equal{}1/tan \hat{EDG}$. Then we can calculate $ cos \hat{EDG}$ and by Pythagore 's theorem we can calculate DE ->CG

Yes this problem should be posted in forum middle shool, but your solution is wrong!
By Pythagorean theorem, we can calculate $ DE \equal{} 10$.
$ \bigtriangleup EDA \sim \bigtriangleup GDE \Rightarrow \frac {ED}{DG} \equal{} \frac {DA}{EG} \equal{} \frac {EA}{DE}$
$ \frac {ED}{DG} \equal{} \frac {EA}{DE} \Rightarrow DG \equal{} 16.\overline{6} \Rightarrow CG \equal{} 4.\overline{6}.$


calculus
integration
function
limit
induction
inequalities
real analysis

Let $ f$ be a bounded function on $ [a,b]$, and $ \{\mathcal P_n\}$ a sequence of partitions of $ [a,b]$.

Show that if $ \|\mathcal P_n\|\to 0$ as $ n\to\infty$,

then $ U(f)\equal{}\lim_{n\to\infty} U(f,\mathcal P_n)$  and  $ L(f)\equal{}\lim_{n\to\infty} L(f,\mathcal P_n)$.

It is obvious that $ L(f,\mathcal P_n) \leq U(f,\mathcal P_n)$.

We show that $ U(f,\mathcal P_2)\geq U(f,\mathcal P_3)\geq U(f,\mathcal P_4) \geq \ldots \geq L(f,\mathcal P_2)$.

We show that $ L(f,\mathcal P_2)\leq L(f,\mathcal P_3)\leq L(f,\mathcal P_4) \leq \ldots \leq U(f,\mathcal P_2)$.

Hence we have two monotonic and bounded sequences $ \{U(f,\mathcal P_n)\}$ and $ \{L(f,\mathcal P_n)\}$. Therefore they are convergent.

We aren't given that the partitions are refinements of each other, nor does that argument show that the desired limits are exactly the upper and lower Darboux integrals.  I would suggest looking at a partition $ P$ with $ U(f, P) \ge U(f) \minus{} \epsilon$ and showing that eventually some $ P_n$ is a refinement of $ P$, and similarly for the lower sums.

We have to prove a lemma first:

[b]Lemma. [/b]Let $ f$ be a bounded function on $ [a,b]$, and $ \mathcal P$, $ \mathcal Q$ two partitions of $ [a,b]$ such that $ \mathcal P\subset\mathcal Q$ and $ \mathcal Q\backslash\mathcal P = \{u_1, u_2, \cdots, u_m\}$. Then

$ U(f,\mathcal P) - U(f,\mathcal Q)\leq mh\|\mathcal P\|\;$ and $ \;L(f,\mathcal Q) - L(f,\mathcal P)\leq mh\|\mathcal P\|,$

where $ h = \sup_{x\in[a,b]}f(x) - \inf_{x\in[a,b]}f(x)$.

[b][i]Proof of the lemma.[/i][/b]
Assume that $ \mathcal Q$ has only one more point, say $ u$, than $ \mathcal P$. The general cases can easily be proven with an induction argument. Suppose

$ \mathcal P = \{a = t_0 < t_1 < \cdots < t_n = b\},$

$ \mathcal Q = \{a = t_0 < t_1 < \cdots < t_{k - 1} < u < t_k < \cdots < t_n = b\},$

and let

$ S = \sum_{\substack{i = 1 \\
i\neq k}}^n \sup_{x\in[x_{i - 1},x_i]}f(x)(x_i - x_{i - 1}).$

Then

$ \begin{aligned} U(f,\mathcal P) & = \sup_{x\in[x_{k - 1},x_k]}f(x)(x_k - x_{k - 1}) + S \\
& \leq \sup_{x\in[a,b]}f(x)\|\mathcal P\| + S, \\
 \\
U(f,\mathcal Q) & = \sup_{x\in[x_{k - 1},u]}f(x)(u - x_{k - 1}) + \sup_{x\in[u,x_k]}f(x)(x_k - u) + S \\
& \geq \inf_{x\in[a,b]}f(x)(u - x_{k - 1}) + \inf_{[a,b]}f(x)(x_k - u) + S \\
& \geq \inf_{x\in[a,b]}f(x)\|\mathcal P\| + S. \end{aligned}$

Consequently,

$ U(f,\mathcal P) - U(f,\mathcal Q)\leq\sup_{x\in[a,b]}f(x)\|\mathcal P\| - \inf_{x\in[a,b]}f(x)\|\mathcal P\| = h\|\mathcal P\|.$

The proof for the second inequality is similar. QED

[b][i]Proof of the original proposition.[/i][/b]
For every $ \epsilon > 0$, as $ U(f) = \inf_{\mathcal P} U(f,\mathcal P)$, there exists a partition $ \mathcal P^*$ of $ [a,b]$ such that

${ U(f) > U(f,\mathcal P^*}) - \frac {\epsilon}2.$

Take $ m^* = |\mathcal P^*|$. As $ \lim_{n\to\infty}\|\mathcal P_n\| = 0$, there exists $ N\in\mathbb N$ such that

$ \|\mathcal P_n\| < \frac {\epsilon}{2m^* h} \;\;$ whenever $ \;\; n\geq N,$

where $ h = \sup_{x\in[a,b]}f(x) - \inf_{x\in[a,b]}f(x)$. Let $ m = |(\mathcal P^*\cup \mathcal P_n)\backslash\mathcal P_n|$. It is obvious that $ m\leq m^*$. Then, for all $ n\geq N$, by the lemma we have

$ \begin{aligned} U(f,\mathcal P_n) - U(f) & < U(f,\mathcal P_n) - U(f,{\mathcal P^*}) + \frac {\epsilon}2 \\
& \leq U(f,\mathcal P_n) - U(f,{\mathcal P^*}\cup\mathcal P_n) + \frac {\epsilon}2 \\
& \leq mh\|\mathcal P_n\| + \frac {\epsilon}2 \\
& \leq m^* h\|\mathcal P_n\| + \frac {\epsilon}2 \\
& \leq m^* h\cdot\frac {\epsilon}{2m^* h} + \frac {\epsilon}2 \\
& = \epsilon. \end{aligned}$

This shows $ \lim_{n\to\infty}U(f,\mathcal P_n) = U(f)$. The proof for the second identity is similar. $ \Box$


Let $ ABCD$ be a square of side $ 2a$. An equilateral triangle $ AMB$ is constructed in the plane through $ AB$ perpendicular to the plane of the square. A point $ S$ moves on $ AB$ such that $ SB\equal{}x$. Let $ P$ be the projection of $ M$ on $ SC$ and $ E$, $ O$ be the midpoints of $ AB$ and $ CM$ respectively.
(a) Find the locus of $ P$ as $ S$ moves on $ AB$.
(b) Find the maximum and minimum lengths of $ SO$.

Solution posted at

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.VMO1986Problem4

by Vo Duc Dien


Prove that $ (2^a \minus{} 1) \bmod{(2^b \minus{} 1)} \equal{} 2^{a \bmod{b}} \minus{} 1$ if a and b are positive integers.

[hide="Solution"]First if $ a\le b$ the problem is trivial.

So let $ a>b$
Let $ a\equal{}qb\plus{}r$

We have the following: $ 2^a\minus{}1 \equiv 2^a\minus{}2^b\equiv 2^b\cdot(2^{a\minus{}b}\minus{}1) \equiv 2^{a\minus{}b}\minus{}1\equiv 2^{a\minus{}b}\minus{}2^{b}\mod (2^b\minus{}1)$

We can thus continue to factor $ 2^b \equiv 1\mod (2^b \minus{}1)$  until we reach $ 2^{a\minus{}qb}\minus{}1$.

However, we know that $ a\minus{}qb\equal{}r$, or in other words, $ a \equiv r \mod b$, and so we are done.[/hide]

Let $ a\equal{}qb\plus{}r$, then $ 2^{qb}\minus{}1\equal{}(2^b)^q\minus{}1$ is divisible by $ (2^b)\minus{}1$ by remainder theorem.

Then $ 2^a\minus{}1\equal{}(2^{qb\plus{}r}\minus{}1)\minus{}(2^{qb}\minus{}1)\equal{}[(2^{qb}\minus{}1)\plus{}1](2^r\minus{}1)\equal{}2^r\minus{}1(mod 2^b\minus{}1)$, done.


probability
function
Asymptote
rotation
probability and stats

A coin is tossed repeatedly, heads appearing with probability p on each toss.
(a) What is the probability that the sequence THT occurs before HHT ?
(b) Let X be the number of tosses until the first occasion by which two heads have appeared successively. Find the distribution function of X.
Thanks in advance.

I would post this earlier, but Asymptote was temporarily down and I wished to include the picture.
(a) We are basically talking about the Markov chain with 4 transit states (determined by the outcomes of the last 2 tosses, circled in green) and 2 terminal states (circled in red):
[asy]size(200);

void f(string s,pair A, pen p=green)
{
D(CR(MP(s,A,6,(0,0)),1),p);
}
void g(string s="",pair A, pair B, pair C=N)
{
path P=(0,0)--(1,0), Q=(-0.5,0.15)--(-0.5,-0.15)--(0,0)--cycle;
pair a=A+unit(B-A), b=B-unit(B-A);
P=shift(a)*scale(d(a,b))*rotate(angle(b-a)*180/3.141592)*P;
Q=shift(b)*rotate(b-a)*Q;
fill(Q,blue);
MC(s,8,D(P),C);
}
void h(string s="",pair A, pair C=N)
{
path P=shift(A)*(dir(60)..(1,2)..(0,4)..(-1,2)..dir(120)),
     Q=shift(A+dir(120))*rotate(-60)*((-0.5,0.15)--(-0.5,-0.15)--(0,0)--cycle);
fill(Q,blue);
MC(s,8,D(P),C);
}

f("HT",(0,0));
f("TT",(5,5));
h((5,5));
f("TH",(10,0));
f("THT",(10,-5),red);
f("HH",(15,0));
h((15,0));
f("HHT",(20,0),red);
g((0,0),(5,5));
g((5,5),(10,0));
g((0,0),(10,0));
g("1-p",(10,0),(10,-5),E);
g("p",(10,0),(15,0));
g((15,0),(20,0));[/asy]
The initial distribution (outcomes of the first two tosses) is $ p^2$ on HH and $ 1 - p^2$ on the rest (i.e., the other three transit states together). Note that from HH you can never reach THT before HHT and from the rest you go through the bottleneck at TH that leads to THT with probability $ 1 - p$ and to HH, which inavoidably leads to HHT, with probability $ p$. So, the answer is $ (1 - p)(1 - p^2)$.

I leave (b) to someone else.

BTW, I vaguely remember that I read an article somewhere that gave a nice formula for the probability that one given string appears before another one. Does anybody know the reference?

(a)
Let $ E$ be the event.
$ P(E) \equal{} pP(E|H) \plus{} (1 \minus{} p)P(E|T)$.
But
$ P(E|H) \equal{} pP(E|HH) \plus{} (1 \minus{} p)P(E|TT) \equal{} (1 \minus{} p)P(E|T)$,
$ P(E|T) \equal{} pP(E|TH) \plus{} (1 \minus{} p)P(E|TT) \equal{} pP(E|TH) \plus{} (1 \minus{} p)P(E|T)$,
$ P(E|TH) \equal{} pP(E|THH) \plus{} 1 \minus{} p \equal{} 1 \minus{} p$.
From this three relations we get $ P(E|TH) \equal{} 1 \minus{} p$,$ P(E|T) \equal{} 1 \minus{} p$,$ P(E|H) \equal{} (1 \minus{} p)^2$,$ P(E) \equal{} (1 \plus{} p)(1 \minus{} p)^2$.

(b)
$ P(X\equal{}k)\equal{}f(k)$, it is easy to see that $ f(k)\equal{}(1\minus{}p)f(k\minus{}1)\plus{}p(1\minus{}p)f(k\minus{}2)$, for $ k>2$. Solving this difference equation we get the desired distribution.

[quote="fedja"]BTW, I vaguely remember that I read an article somewhere that gave a nice formula for the probability that one given string appears before another one. Does anybody know the reference?[/quote]The book [i]Concrete Mathematics[/i] by Graham, Knuth, and Patashnik derives a beautiful formula which they credit to Conway.  They call the game "Penney ante".


If $ x \minus{} 2x \plus{} 3x \equal{} 100$, what is the value of $ x$?

we combine like terms on the left side

$ x\minus{}2x\plus{}3x \equal{} 100$
$ 2x \equal{} 100$
$ x\equal{}50$