inequalities
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Since I was asked about it I'm posting it.

We begin with a definition:

[b]1.[/b] Let $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ be arbitrary real numbers satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$. Now, if we have all the following conditions fulfilled:

  $ x_1 \geq y_1$;
  $ x_1 \plus{} x_2 \geq y_1 \plus{} y_2$;
  $ x_1 \plus{} x_2 \plus{} x_3 \geq y_1 \plus{} y_2 \plus{} y_3$;
  ...
  generally $ x_1 \plus{} x_2 \plus{} ... \plus{} x_k \geq y_1 \plus{} y_2 \plus{} ... \plus{} y_k$ for any natural k with $ 1 \leq k \leq n \minus{} 1$;
  and $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} y_1 \plus{} y_2 \plus{} ... \plus{} y_n$

(mind the equality sign in the last condition; it is not a $ \geq$ sign!), then we say that [i]the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ [b]majorizes[/b] the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$[/i]. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Now, if instead of

  $ x_1 \geq y_1$;
  $ x_1 \plus{} x_2 \geq y_1 \plus{} y_2$;
  $ x_1 \plus{} x_2 \plus{} x_3 \geq y_1 \plus{} y_2 \plus{} y_3$;
  ...
  generally $ x_1 \plus{} x_2 \plus{} ... \plus{} x_k \geq y_1 \plus{} y_2 \plus{} ... \plus{} y_k$ for any natural k with $ 1 \leq k \leq n \minus{} 1$;
  and $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} y_1 \plus{} y_2 \plus{} ... \plus{} y_n$,

we have the conditions

  $ x_1 \leq y_1$;
  $ x_1 \plus{} x_2 \leq y_1 \plus{} y_2$;
  $ x_1 \plus{} x_2 \plus{} x_3 \leq y_1 \plus{} y_2 \plus{} y_3$;
  ...
  generally $ x_1 \plus{} x_2 \plus{} ... \plus{} x_k \leq y_1 \plus{} y_2 \plus{} ... \plus{} y_k$ for any natural k with $ 1 \leq k \leq n \minus{} 1$;
  and $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} y_1 \plus{} y_2 \plus{} ... \plus{} y_n$

fulfilled (i. e., the same conditions with all $ \geq$'s replaced by $ \leq$'s), then we say that [i]the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ [b]minorizes[/b] the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$[/i]. We write this in the form $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Of course, this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

[b]2.[/b] Thus we have defined the notions "majorizes" and "minorizes" only for non-increasing arrays (i. e. for arrays satisfying $ x_1 \geq x_2 \geq ... \geq x_n$ and $ y_1 \geq y_2 \geq ... \geq y_n$). Now, assume that $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are just arbitrary real numbers, without any conditions. Then, let $ \left(X_1;\;X_2;\;...;\;X_n\right)$ be the [b]non-increasing permutation[/b] of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$, i. e. the permutation of the array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ that satisfies $ X_1 \geq X_2 \geq ... \geq X_n$. Similarly, let $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ be the non-increasing permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, i. e. the permutation of the array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ that satisfies $ Y_1 \geq Y_2 \geq ... \geq Y_n$. The number arrays $ \left(X_1;\;X_2;\;...;\;X_n\right)$ and $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$ are both non-increasing, and hence we have defined majorization for such arrays.

Then, [i]we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ [b]majorizes[/b] the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ majorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$[/i]. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \succ \left( y_1;\;y_2;\;...;\;y_n\right)$.

Similarly, [i]we say that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ [b]minorizes[/b] the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if and only if the number array $ \left(X_1;\;X_2;\;...;\;X_n\right)$ minorizes the number array $ \left(Y_1;\;Y_2;\;...;\;Y_n\right)$[/i]. In this case, we write $ \left( x_1;\;x_2;\;...;\;x_n\right) \prec \left( y_1;\;y_2;\;...;\;y_n\right)$. Again this is equivalent to $ \left( y_1;\;y_2;\;...;\;y_n\right) \succ \left( x_1;\;x_2;\;...;\;x_n\right)$.

[b]3.[/b] So we have defined the terms "majorize" and "minorize" for any two number arrays. It should be noted that [i]majorization is a partial order on the set of number arrays[/i], not a total order - i. e., not for every pair of two number arrays $ \left(x_1;\;x_2;\;...;\;x_n\right)$ and $ \left(y_1;\;y_2;\;...;\;y_n\right)$ one can say that either the first one majorizes the second one, or the second one majorizes the first one. It often happens that none of the arrays majorizes or minorizes the other one. But sometimes when you have some special arrays, you can prove that one of them majorizes the other one.

[b]4.[/b] Now, the [b]Karamata inequality[/b], also called the [b]Majorization Inequality[/b] or the [b]Hardy-Littlewood inequality[/b], states that if $ x_1$, $ x_2$, ..., $ x_n$, $ y_1$, $ y_2$, ..., $ y_n$ are $ 2n$ reals from an interval $ I\subseteq\mathbb{R}$ such that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$, and $ f: I\to\mathbb{R}$ is any convex function, then

  $ f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right) \geq f\left( y_1\right) \plus{} f\left( y_2\right) \plus{} ... \plus{} f\left( y_n\right)$.

If $ f: I\to\mathbb{R}$ is a concave function instead, then we instead have

  $ f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right) \leq f\left( y_1\right) \plus{} f\left( y_2\right) \plus{} ... \plus{} f\left( y_n\right)$.

If the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ minorizes the number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ instead of majorizing it, then both inequalities are reversed.

[b]5.[/b] The Jensen inequality for real numbers is a special case of the Karamata inequality. In fact, if $ m \equal{} \frac {x_1 \plus{} x_2 \plus{} ... \plus{} x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$, then it is easy to show that the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

  $ f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

  $ \frac {f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

  $ f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

  $ \frac {f\left( x_1\right) \plus{} f\left( x_2\right) \plus{} ... \plus{} f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality for $ n$ reals.

[b]6.[/b] The definition of majorizing and minorizing number arrays given above is somewhat unsatisfying from an intuitive point of view, since it does not help one to imagine how an array majorizing another array looks like. Unfortunately, this is partly inherent to the notion of majorization, which indeed [i]is[/i] quite unintuitive. For a - rather facile - visualization of the notion, you can imagine that a number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes a number array $ \left(y_1;\;y_2;\;...;\;y_n\right)$ if the two arrays have the same sum of numbers, but the numbers of the first array are set wider apart than those of the second array, while those of the second array lie closer together. From this intuitive viewpoint, it is clear why the number array $ \left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $ \left(m;\;m;\;...;\;m\right)$, where $ m \equal{} \frac {x_1 \plus{} x_2 \plus{} ... \plus{} x_n}{n}$ is the arithmetic mean of the numbers $ x_1$, $ x_2$, ..., $ x_n$: In fact, the two number arrays have the same sum of elements, but the elements of the second number array lie nearer to each other (they are all equal). Alas, this viewpoint does not help one to really understand what majorization is about.

Well, I know there is more to say. For instance, the Karamata inequality has a kind of converse, but I am not sure how it is formulated, so I leave this to the other MathLinkers more used to inequalities.

  Darij

[quote="ADMann94"]Where can I find a proof of this?

Alex[/quote]

Probably in this [url=http://www.amazon.com/exec/obidos/tg/detail/-/0521358809/qid=1093735967/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-7979132-1252055?v=glance&s=books&n=507846]book[/url].

I heard that Karamata inequality is indeed Littlewood-Hardy inequality so which name is correct or maybe both [the question is: which name shall I pull out during the contest ]?

Best refer to it as "Karamata Majorization Inequality". As for Littlewood and Hardy, it's true that they have discovered it independently from Karamata, but I personally have never seen anybody naming it for Littlewood and Hardy.

  Darij

Let $ {\mathbf x}=(x_1,x_2,...,x_n)  , {\mathbf y}=(y_1,y_2,...,y_n) $
and $ {\mathbf A}=||a_{i,j}|| $ a $n\times n $  matrix having properties:
    1) all $a_{i,j} \ge 0 $ ,
    2)  $ \sum\limits_{i=1}^na_{i,j}= \sum\limits_{j=1}^{n}a_{i,j}=1 $ for $ i,j=1,2,...,n.$

If   $ {\mathbf x} $ is given in $ {\mathbf R}^n $ and $ {\mathbf y}^T=A{\mathbf x}^T $ ( where ${\mathbf x}^T $ denotes the transpose of ${\mathbf x}$)  , what  we can say about $ {\mathbf y} $ ?

Hi all,
here is a [b]very interesting paper[/b] related to discussed subject:

[*] A. OSTROWSKI  , [i] Sur quelques applications des fonctions convexes et concaves au sens de I. Schur,[/i] J.Math.Pures.Appl., (9) 31 (1952) 253-292.

[b]Other possible references :[/b]

[1] G.H.Hardy , J.E.Littlewood, G.P\'olya , [i]Some simple inequalities satisfied by convex functions,[/i] Messenger Math., 58 (1928/29) 145-152.
[2] J.Karamata , [i]Sur une in\'egalit\'e relative aux fonctions convexes,[/i] Publ.Math.Univ. Belgrade 1 (1932) 145-158.
[3] T.Popoviciu , [i] Notes sur les fonctions convexes d'ordre sup\'erieur ,III,[/i]Mathematica (Cluj) 16 (1940) 74-86.
[4] T.Popoviciu , [i] Notes sur les fonctions convexes d'ordre sup\'erieur ,IV,[/i]Disquisitiones Mathematicae 1(1940) 163-171.
[5] T.Popoviciu ,[i] Les fonctions convexes,[/i] Actualit\'esci.Ind. No.992,Paris,1945.

I have a very simple question, which does not let me understand many things. So, can anyone explain me, what functions are called convex and what functions are called concave?

Let $f:[a,b]\to \mathbb{R}$. Then $f$ is called convex function iff
\[k\cdot f(x)+(1-k)\cdot f(y)\geq f(kx+(1-k)y)\quad \forall x,y\in [a,b],\ \forall k\in[0,1].\]

If function $f$ is differentiable then $f$ is convex iff $f''(x)\geq 0$ for all $x\in[a,b]$.

Another question: which functions are differentable? Can you give an example of non-differentable function?

Take $f(x)=|x|$. It is convex non-differentiable function.
To know what is differentiable functions you need any analysis book.

[quote="darij"]Well, I know there is more to say. For instance, the Karamata inequality has a kind of converse, but I am not sure how it is formulated, so I leave this to the other MathLinkers more used to inequalities.[/quote]

anyone know what is this converse?

I think it is the following (although I see absolutely no use of it):

If $\displaystyle\sum f(x_i)\geqslant\sum f(y_i)$ for every convex function $f$ then $(x)\succ(y)$.

Could somebody post a counter example or (even better) prove this?

The converse is proved by considering the inequality for:
constant functions (conclusion: vectors $x$ and $y$ have equal number of components),
linear functions (conclusion: $\Sigma x = \Sigma y$), and
functions of the form $\max(0,x-a)$ (conclusion: the majorization conditions).

Any convex $f$ is (on the given finite set of points) equal to a positive linear combination of these 3 types of functions.

[quote="darij grinberg"]Best refer to it as "Karamata Majorization Inequality". As for Littlewood and Hardy, it's true that they have discovered it independently from Karamata, but I personally have never seen anybody naming it for Littlewood and Hardy. [/quote]

Karamata published later than Hardy,Littlewood and Polya, and in any case majorization was independently discovered many times before and after all these publications.   There is a history in Marshall and Olkin's book on majorization.  In the Anglo-Saxon countries it is called "majorization inequality" or "Hardy-Littlewood-Polya" majorization inequality".

[quote="Bojan Basic"]I think it is the following (although I see absolutely no use of it):

If $\displaystyle\sum f(x_i)\geqslant\sum f(y_i)$ for every convex function $f$ then $(x)\succ(y)$.[/quote]

There is a stonger version of this: If $\sum f\left(x_i\right)\geq\sum f\left(y_i\right)$ holds for every function f of the form f(x) = |x - u| (with u constant), then $\left(x\right)\succ\left(y\right)$. This is actually Lemma 1 in http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 post #11.

  darij

[quote="Peter VDD"][quote="darij grinberg"]
[b]5.[/b] The Jensen inequality is a special case of the Karamata inequality. In fact, if $\displaystyle m=\frac{x_1+x_2+...+x_n}{n}$ is the arithmetic mean of the numbers $x_1$, $x_2$, ..., $x_n$, then you can easily show (little exercise!) that the number array $\left(x_1;\;x_2;\;...;\;x_n\right)$ majorizes the number array $\left(m;\;m;\;...;\;m\right)$. Hence, the Karamata inequality yields:

If f(x) is any convex function, then

  $f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \geq n f\left( m\right)$,

i. e.

  $\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \geq f\left( m\right)$.

If f(x) is a concave function instead, then we instead have

  $f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right) \leq n f\left( m\right)$,

i. e.

  $\displaystyle \frac{f\left( x_1\right) + f\left( x_2\right) + ... + f\left( x_n\right)}{n} \leq f\left( m\right)$.

Of course, this is exactly the Jensen inequality.[/quote]
[quote="Myth"]Actually, true Jensen inequality is
\[a_1f(x_1)+a_2f(x_2)+\dots+a_nf(x_n)\geq f(a_1x_1+\dots+a_nx_n)\],
where $f$ is convex function, $a_i\in[0,1]$ and $a_1+\dots+a_n=1$.
$a_i=\frac{1}{n}$ is a particular case only.[/quote]

Ok, then is there a way to deduce "real" jensen ineq from karamata? :)

It looks like we could extend to rationals and then to reals, but I wonder if someone had a nicer proof, or someone knew sort of a "weighed karamata"?[/quote]

I think Darij posted something known as "Fuchs" in theorems & formula section

Hmm.Guys do we have the Karamata for the productive case to :

Like this:
let $X=(x_1,...,x_n),Y=(y_1,...,y_n)$That  $x \succ Y$ and $f$ be a convex function.
then
$f(x_1)*f(x_2)*...*f(x_n) \ge f(y_1)*f(y_2)*...*f(y_n)$

[hide="An idea"]I think for reaching this kind of productive inequality ,we should change the difination of majorizing to something like this:
$1.x_1\ge x_2 \ge...\ge x_n,y_1\ge y_2\ge...\ge y_n$
$2.\prod_{i=1}^k x_i \ge \prod_{i=1}^k y_i$, for every $1 \le k \le n$
$3.\prod_{i=1}^n x_i = \prod_{i=1}^n y_i$[/hide]

this kind of majorization exists, and the inequality you mentioned exists, even for the weighted version I think it exists. But then, you can just take log and it all becomes the normal one.

[quote="darij grinberg"]Well, we can say that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $. And conversely, if we have two vectors $\left\{ \mathbf{x}\right\} $ and $\left\{ \mathbf{y}\right\} $ such that $\left\{ \mathbf{x}\right\} \succ \left\{ \mathbf{y}\right\} $, then we can find a matrix $\mathbf{A}$ with the properties 1) and 2) such that $ {\mathbf y}^T=\mathbf{A}{\mathbf x}^T $. This is a result found by Karamata.

  Darij[/quote]

can you show a proof?

i have two questions:

[b](1)[/b] Can Karamata be proven by Jensen?

[b](2)[/b] How do you prove the Fuchs inequality?

[quote="perfect_radio"][b](2)[/b] How do you prove the Fuchs inequality?[/quote]I wonder about that one too... however I'm afraid that it's not quite elementary.

PS: Can one prove weighted jensen/muirhead from fuchs

[quote="lomos_lupin"]

Hmm.Guys do we have the Karamata for the productive case to :

Like this:
let $X=(x_1,...,x_n),Y=(y_1,...,y_n)$That  $x \succ Y$ and $f$ be a convex function.
then
$f(x_1)*f(x_2)*...*f(x_n) \ge f(y_1)*f(y_2)*...*f(y_n)$

[/quote]

My guess (although it's only a guess) would be that you'd need a stronger condition on $f$ for this to hold - probably something like $g(x)=\ln f(x)$ would need to be convex, to convert it back to the additive form.

So is Muirhead simply a special case of Karamata?

[quote="k2c901_1"]Is there a continuous or integral analog to this inequality?[/quote]
I also wonder if the following is true:
Let $\varphi$ and $\psi$ be two functions defined on interval $\langle a,b\rangle$ satisfying:
[b]1.[/b] $\int_{a}^{c}\varphi(t)dt\geq \int_{a}^{c}\psi(t)dt$ for any $c\in \langle a,b\rangle$;
[b]2.[/b] $\int_{a}^{b}\varphi(t)dt= \int_{a}^{b}\psi(t)dt$.
Then for any convex function $f$ we have \[\int_{a}^{b}f\left(\varphi(x)\right)\geq \int_{a}^{b}f\left(\psi(x)\right).\] I hope someone can prove it or give a counterexample.
Maybe some majorization theory theorems work also for majorization defined in [b]1.[/b] and [b]2.[/b]? Maybe continuous Muirhead?

Peter VDD proved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108308]here[/url] something that reminds me of Karamata for two variables.

[quote="Peter VDD"][b]Theorem.[/b]
Let $f(t)$ be a smooth nonnegative function which is convex on $[a,b]$ and let $[x,y]\subset[a,b]$ such that $x+y=a+b$. Then we have \[\frac{\dsp\int^{b}_{a}f(t) dt}{b-a}\ge \frac{\dsp\int^{y}_{x}f(t) dt}{y-x}.\] [/quote]

But it's not really what you want. It would be very nice if your conjecture was true :)

The inequality Rzeszut proposed exists, and I think even Muirhead for integrals exists. I will post them as soon as I find them.

[quote="bodan"]Look [url=http://projecteuclid.org/Dienst/Repository/1.0/Disseminate/euclid.pjm/1102992404/body/pdf?userid=&password=]here[/url].[/quote]
Thanks a lot.  :coolspeak:

On Olympiad competition, if you use Karamata, do you need justification? or can you just cite it.

For example, I heard that often markers discourage the use of Muirhead.

[quote="Megus"]Ok - I asked beacuse of guys who laughed at me when had heard Karamata  :D[/quote]
And we will always laugh!  :heli:

Can someone give me links with problems solved by Karamata ineq , please  :oops: ?

Eg

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=166411

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=163149

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1726203183&t=160484

Please use Search tool, you can find a lots of ex about Karamata inequality :wink:

http://www.mathlinks.ro/Forum/search.php

Thank you very much , quangpbc :thumbup:

i'm new in the forum and because of that i didn't used the search function  :blush:
next time i'll know abaut it and i'll use it  . thanks for that , too :D

I post a more general version of Karamata's inequality in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=605332]here.[/url]


inequalities
induction
number theory proposed
number theory

Let $ p_1, p_2,...,p_n$  be the first m odd primes. Prove that:
$ p_1. p_2...p_n \ge {p^2}_{n\plus{}1}$  for all $ m \geq  4$
--------------------
Is this the Bonse-Posa inequality? :D

The inequality holds for $ n \equal{} 4$, and we show by induction it holds for $ n > 4$.  If for some $ k\geq 4$ we have $ \prod_{j \equal{} 1}^k{p_j} > p_{k \plus{} 1}^2$ then
\[ \prod_{j \equal{} 1}^{k \plus{} 1}p_j > p_{k \plus{} 1}^3 > \frac {1}{8}p_{k \plus{} 2}^3
\]
where the last inequality follows from [url=http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate]Bertrand's postulate[/url].  Since $ \frac {1}{8}p_{k \plus{} 2} > 1$, we are done.


analytic geometry
conics
parabola
parameterization

a) [hide="HIDE YOUR SOLUTIONS!"]$\frac{4ac-b^{2}}{4a}= 50$ feet.[/hide]

The maximum x value is $\frac{-b}{2a}$
[hide="(a)"] Then x=30 and plugging in x=30, you get y=50. So 50 feet is the answer for a.[/hide] :P

It seems to me that your second question is the same as the first... The analytic expression for the ordinate of the vertex of the parabola is the one rnwang2 wrote you in his post. The maximum height reached from the ball IS the ordinate of the vertex, so, changing the values of the parameters a,b,c in the equation of the parabola, obviously also the height changes...
Hope I've been exhaustive, although I'm not sure about having understood properly your question...  :roll:

[quote="kryms3n"]The path of a ball is given by y=(-1/20)x^2 + 3x + 5 , where y is the height in feet and x is the horizontal distance in feet.

(a) Find the maximum height of the ball

(b) Which term determines the height at which the ball was thrown? Does changing this term change the coordinates of the maximym height of the ball? Explain.[/quote]

[hide](a)
$-\frac{b}{2a}$, is the x-coordinate of the point so the answer is

$y=(-1/20)*30^{2}+3*30+5=50\text{feet}$

(b)
The term that determines the height is the constant, the one with no x-term.  Changing this term while keeping other values the same will change it since the upward/downward shift will cause the shift in the highest point on the graph, and hence the shift in the maximum height.[/hide]


There are 30 equally-weighted questions on Mr. Daven's math final.  If a student must score 68% or greater to pass, what is the minimum number of questions that must be answered correctly to pass?

[hide]  30* .68= 20.4

Thus the answer is 21 [/hide]


Galois Theory
superior algebra
superior algebra unsolved

The [url=http://en.wikipedia.org/wiki/Artin-Zorn_theorem]Artin-Zorn theorem[/url] (or, if you prefer, Wedderburn's little theorem) makes it somewhat difficult to define finite analogues of the quaternions and the octonions, even though there's some sense to the analogy that if $ \mathbb{R}$ is $ \mathbb{F}_p$ where $ p$ is a prime congruent to $ 3 \bmod 4$ then $ \mathbb{C}$ is $ \mathbb{F}_p[i] \simeq \mathbb{F}_{p^2}$.  It seems in order to get any structure that isn't just a finite field you not only have to drop associativity, you also have to drop alternativity.  How weak do the hypotheses have to be before you get some kind of object of size $ p^4$ containing $ \mathbb{F}_{p^2}$ that isn't a field and behaves like the quaternions in some way?  The octonions?  What do natural geometric constructions (projective planes, etc.) over these objects look like?

I don't understand noncommutative algebra well enough in order to really know what you mean by "behaving like quaternions", but these are probably the two main obstructions to finite quaternion-like objects:
- The Wedderburn theorem which states that any finite division ring is a field.
- The [url=http://en.wikipedia.org/wiki/Chevalley-Warning_theorem]Chevalley-Warning theorem[/url], which heavily limits the possibilities to construct a norm. For instance, quaternions are so interesting because they are a division algebra, and they are a division algebra because their norm is never zero unless the quaternion itself is zero. In $ \mathbb{F}_q$, the sum of four squares can be zero though the squares themselves are nonzero, and this approach is doomed.
This makes me wonder whether there is a way to prove Wedderburn through Chevalley-Warning by considering the division ring as an $ \mathbb{F}_p$-algebra for some prime $ p$, constructing a norm (as the determinant of the left multiplication, as in Galois theory) and somehow concluding from Chevalley-Warning that it must be zero somewhere. Any ideas?

  darij

Wikipedia claims that there is exactly one octonion algebra over any finite or algebraically closed field $ F$: http://en.wikipedia.org/wiki/Octonion_algebra.  Also, [url=http://www.maths.qmw.ac.uk/~raw/talks_files/WWW-O2.pdf]these seminar notes[/url] claim that you can do this for finite fields in a very natural way for odd characteristic, but claims that the characteristic 2 case is "a lot more complicated."


Write a program to find the smallest positive integer m such that :
-All digits in the hexadecimal form of m are mutually distinct
-The sum of all digits in the decimal form of m is equal to n  (n is the positive integer was entered from the keyboard)
 Thanks :)

i have a few observations and an incomplete solution:
m in hex can only be 16 digits long, so it is less than $ {16}^{16} \equal{} 18446744073709551616$
therefore n is less than $ 9*20digits \equal{} 180$
the number of decimal numbers that sum up to 180 is ~~ 180C20 ~~ 175142105857592248012292655
ok.. too large
the number of hexadecimal numbers with unique digits is ~~ 20922789888000
darn.
so obviously a completely exhaustive search is useless.
however, you can search through the hexadecimal numbers with unique digits and hope that the decimal form sums up to n (it should work for small n)
i was able to solve the problem for n<=80 without going over a second


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

Let $ H$ be a subgroup of finite index of an infinite group $ G$. Prove that $ G$ has a normal subgroup of finite index which is contained in $ H$.

Second Isomorphism Theorem: Suppose $ H$, $ J$ are subgroups of $ G$ with $ J$ normal in $ G$. Then $ HJ$ is a subgroup of $ G$, $ H\cap J$ is a normal subgroup of $ H$, and the quotient groups $ HJ/J$, $ H/H\cap J$ are isomorphic.

I was thinking, if $ J\equal{}[G, G]$, the commutator subgroup of $ G$, then $ H\cap [G, G]\lhd H$. So $ G$ has a normal subgroup which is contained in $ H$, but then what do I do to show that this normal subgroup is of finite index? Do I even use the fact that $ G$ is an infinite group?

I don't think your approach works. The simplest proof involves looking at all the conjugates of $ H$, that is all the groups of the form $ g^{\minus{}1}Hg$ (how many different such groups exist?), and looking at their intersection. If you can't figure it out, look [url=http://www.mathlinks.ro/viewtopic.php?search_id=28824910&t=217995]here[/url](but I suggest you try to figure it out first).


number theory
greatest common divisor
probability

Prove or Disprove whether $\mathrm{P}(\mathrm{gcd}(a_1,...,a_n)=1)=\frac{1}{\zeta(n)}$.

$a_1,...,a_n\in\mathbb{Z}$ and are randomly chosen.
$\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$
$P(X)$ is the probability of event $X$ happening, or in this case, $a_1,...,a_n\in\mathbb{Z}$ to be relatively prime.
:)

Masoud Zargar

When you're posing a problem, define the terms you're dealing with.  I at least have an idea of what you're saying; others might not be so fortunate.  Also, you NEED to specify the exact conditions on the $a_i$'s.


function
real analysis
real analysis unsolved

$ f(x)$ is a differentiable function and $ g(x)$ is a double differentiable function such that $ |f(x)|\leqslant 1$ and $ f'(x)\equal{}g(x)$. If $ (f(0))^2\plus{}(g(0))^2\equal{}9$. Prove that there exists some      $ c\in (\minus{}3,3)$ such that $ g(c).g''(c)<0$

Let $ I$ be one of the intervals $ [0,3]$ or $ [\minus{}3,0]$.

Here is a sketch, which I think will work (I have left out the details):

We consider 3 cases.

i) $ g(x)$ is of the the same sign on $ I$. This implies that $ f(x)$ is monotonic and the fact that $ |f'(0)| \ge 2\sqrt(2)$ would imply that one of $ |f(3)|$ or $ |f(\minus{}3)|$ is $ > 1$.

ii) $ g'(x)$ is never $ 0$. Then we have that $ g'(x) > 0 \forall x$ or $ g'(x) < 0 \forall x$. Thus $ g(x)$ is monotonic and would fall into case i).

iii) $ g'(a) \equal{} 0$ for some $ a$. Also assume that $ g(x)$ is not a function of the kind describe in case i). If $ g(a)<0$ and $ g''(a)<0$ then $ a$ is a local maxima. If $ g$ changes sign, there must a local minima c in I such that $ g(c) < g(a)$. Similary consider the case when $ g(a) > 0$. $ g(a) \equal{} 0$ is probably the trickiest case.

By drawing a graph, we should get an idea of how to prove it rigorously I think.


It is observed that allylic chloride on reaction with $ \text{HI}$ gives Markonikoff product while on reaction with $ \text{HOCl}$ gives anti-Markonikoff product .Give Reasons

Give the structure of the product obtained with hypochlorite.

i find no reason why the product would be that.
in fact i would reason the product [b] may[/b] be anti-markonikoff wth $ \text{HI}$ due to steric repulsion between $ \text}I},\text{Cl}$ if at all one of the reactions had to be anti-markonikofff.
i see no reason in why the intermediate carbocation is on the most destabilized carbon


[u]Arithmetic progressions[/u]

Consider the following infinite table :

04, 07, 10, 13, 16, 19, ...
07, 12, 17, 22, 27, 32, ...
10, 17, 24, 31, 38, 45, ...
13, 22, 31, 40, 49, 58, ...
16, 27, 38, 49, 60, 71, ...
19, 32, 45, 58, 71, 84, ...
...

The pattern is quite obvious I think.
Each row/column contains an infinite arithmetic progression.

Prove or disprove :

The number k doesn't occur in this table if and only if 2k + 1 is prime.

Just curious: do you have a soln of this one, Arne?

I think it's true:

We can identify each number in the table by the number of steps we go downwards and the number of steps we have to go to the right, so all the numbers t in the table must have form 4+3*k+(3+2k)*n where k and n are any 2 natural numbers (includding 0). This means that for all t in the table 2t+1=9+6k+6n+4kn=(2k+3)(2n+3), so I guess this shows that t isn't in the table iff 2t+1 can't be written as (2k+3)(2n+3) with k,n>=0, which is equivalent to saying that 2t+1 can't be written as the product of 2 factors >1 iff 2t+1 is a prime.

Yeah, thats correct. When I first saw this I thought, hey, that can't be right, theres no method of finding primes like that. But then of course writing out all these APs would take far too much time to get to primes anywhere relatively large. Quite nice problem though.

Yeah, of course I found the solution, it is not difficult at all.
Nice and simple :)

I asked if you had a solution because, at the beginning, I thought it was hard, and I was wondering if you didn't post some sort of tough, unsolved problem. When I found the soln I realized I was wrong :D

amm "The number k doesn't occur in this table if and only if 2k + 1 is prime" I understand that if 2k+1 is prime it can't appear in the table, but if it isn't prime it may not appear in the table right?


Let $ n\geq 4$ be integer. There are $ n$ people who has each a blank card. Each card will be wrote down one number among $ 1,\ 2,\ 3$ at random. Collect $ n$ cards, if there is same numbered card, then you will abondon the cards. For example, for $ n \equal{} 6$, if the number of 6 cards are $ 1,\ 1,\ 2,\ 2,\ 2,\ 3$, then the card of number 3 is only remaining. Denote $ p_k(n)$ the probability such that $ k$ cards of $ n$ cards only will be remaining. Find $ p_1(n),\ p_2(n)$, then find the value of $ n$ for which $ p_1(n)$ is maximal.

$ p_1(n) \equal{} n(2^{n\minus{}1}\minus{}2n)3^{\minus{}n}$ for $ n\ge 4$. It differs slightly for $ n\equal{}1,2,3$.

Maybe, I forgot to multiply by three.

Anyway, the EGF of card sequences which lead to only card 1 remaining is
\[ F(z) \equal{} (e^z \minus{} z)^2 z \equal{} z (e^{2z} \plus{} z^2 \minus{} 2z e^z)\]
(We have labelled product of two labelled sets of size not 1 and one set of size 1).

So for $ n\neq 3$
\[ n![z^n] F(z) \equal{} n![z^{n \minus{} 1}](e^{2z} \minus{} 2z e^z) \\
\equal{} n!(2^{n \minus{} 1}/(n \minus{} 1)! \minus{} 2/(n \minus{} 2)!) \equal{} n(2^{n \minus{} 1} \minus{} 2(n \minus{} 1)).\]
So the probability of only card 1 remaining is $ n(2^{n \minus{} 1} \minus{} 2n \plus{} 2)3^{ \minus{} n}$ for $ n\neq 3$, consequently, $ p_1(n) \equal{} n(2^{n \minus{} 1} \minus{} 2n \plus{} 2)3^{ \minus{} n \plus{} 1}$ for $ n\neq 3$. So my mistake was that I did not consider empty sets which also lead to zero cards. And yes, I forgot to multiply by 3.

There is, of course, a simple combinatorial argument, but I like GFs more.


All coefficients of $f(x)$ =$x^{n}+a_{n-1}x^{n-1}+.................+a_{1}x+1$ are positive real numbers

If the n roots of $f(x)=0$ are all real prove that $f(2)$ $\geq$ $3^{n}$


Given a quadrilateral ABCD and points E, F on AB such that AE=EF=FB and points G,H on CD such that DG=GH=HC, show that $[CHAE]=[EFGH]=\frac{1}{3}[ABCD]$.

[quote="Farenhajt"][hide="Proof"]$[AEC]={1\over 3}[ABC],[AHC]={1\over 3}[ACD]$, hence $[AECH]={1\over 3}[ABCD]$. Similar for $[FBGD]$, and then the middle one is also one third of $[ABCD]$.[/hide][/quote]

I should think that you swapped $D$ and $C$ or i don't know... but in what you'd written it doesn't come directly that the "middle one" is also one third of $[ABCD]$, because the other two cross that..

Oops!  :blush: You're right, I was too quick.

[hide="Correct proof"]$2\overrightarrow{[ABCD]}=\overrightarrow{DA}\times\overrightarrow{AB}+\overrightarrow{BC}\times\overrightarrow{CD}$

\begin{eqnarray*}2\overrightarrow{[EFHG]}&=& \overrightarrow{GE}\times\overrightarrow{EF}+\overrightarrow{FH}\times\overrightarrow{HG}\\ &=& (\overrightarrow{GD}+\overrightarrow{DA}+\overrightarrow{AE})\times{1\over 3}\overrightarrow{AB}+(\overrightarrow{FB}+\overrightarrow{BC}+\overrightarrow{CH})\times{1\over 3}\overrightarrow{CD}\\ &=&{1\over 9}(\overrightarrow{CD}\times\overrightarrow{AB})+{1\over 3}(\overrightarrow{DA}\times\overrightarrow{AB})+\overrightarrow{0}+{1\over 9}(\overrightarrow{AB}\times\overrightarrow{CD})+{1\over 3}(\overrightarrow{BC}\times\overrightarrow{CD})+\overrightarrow{0}\\ &=&{1\over 9}(\overrightarrow{CD}\times\overrightarrow{AB}+\overrightarrow{AB}\times\overrightarrow{CD})+{1\over 3}(\overrightarrow{DA}\times\overrightarrow{AB}+\overrightarrow{BC}\times\overrightarrow{CD})\\ &=&{1\over 3}(\overrightarrow{DA}\times\overrightarrow{AB}+\overrightarrow{BC}\times\overrightarrow{CD})\\ &=&{2\over 3}\overrightarrow{[ABCD]}\end{eqnarray*}

Hence $[EFHG]={1\over 3}[ABCD]$[/hide]


invariant
geometry
rhombus
rectangle
trigonometry
inradius
geometry unsolved

[quote="yetti"][quote="armpist"]ABCD is a circumscribed quadrilateral.  Take side AB and drop perpendiculars

  AA' and BB' from vertices A and B onto opposite side CD.

  Prove that XX' * sin(X) + YY' * sin(Y)  is invariant for the sides XY  = AB, BC, CD. DA

  of ABCD.

  Thank you.

  M.T.[/quote]

Wow!! Do you believe (i. e., take as truth something you cannot verify) that a general rhombus might be circumscribed, while a rectangle definitely is not  ? :?[/quote]

i think there is some misunderstanding here, as the result is definitely correct. it is equivalent to $AB\cdot\sin A\sin B+CD\cdot\sin C\sin D=BC\cdot\sin B\sin C+DA\cdot\sin D\sin A$. it is not particularly hard to prove, i planed to post a proof when i will have some spare time again (means in end-april). just to show the main idea: it is enough to prove that $BC\cdot\sin B\sin C=\frac{2\cdot\left|XYZ\right|}{r}$, where $\left|P_1P_2P_3\right|$ means the area of any triangle $P_1P_2P_3$, and r is the inradius of the quadrilateral ABCD, and X, Y, Z are the points where the incircle of quadrilateral ABCD touches its sides AB, BC, CD. but this is basicly a triangle identity and thus straightforward.

greetings from bad homburg and sorry for no capitals (they would take twice as long to type here),
darij

[quote="armpist"]hello Darij,

  If you want to discuss something different from the stated problem, I think

  you should start a new thread.  Problem statement stays as initially specified and

  one typical bright response (solution maybe) is logged ( for posterity? )

   T.Y.

   M.T.[/quote]

Armpist, seeing your problem, I really hoped you started to post interesting and maybe even useful things on this forum. Now it seems that everything is "kak prezhde". I expected that somebody like you, who is able to recognize much more arcane and unexpected (and sometimes nonexistent) connections, would find it not worth a further explanation that the statement of your problem is equivalent to $AB\cdot\sin A\sin B+CD\cdot\sin C\sin D=BC\cdot\sin B\sin C+DA\cdot\sin D\sin A$ (hint: what the hell are XX' and YY' ?). On the other hand, it looks like you are trying to provoke me to write a full solution to your problem in my typical fashion (in order to subsequently criticize it?), something I have to decline since I have more important stuff to do.

  Darij

Hello darij,

  if you substitute in the expression XX' * sin(X) + YY' * sin(Y)   A for X  and B for Y,

  you will get      AA' * sin(A) + BB' * sin(B)   where A' and B' are feet of

  perpendiculars from A and B to opposite side CD.

   if you take side BC   the expression will be

        BB' * sin(B) + CC' * sin(C)   where B' and C' are feet of

  perpendiculars from B and C to the opposite side DA.

  Similar for sides CD  and DA.

   It is and was clear from the statement.

  What were you talking about?

  As for Yetti's response, I did not understand it, so I assumed it to be a sophisticated

  solution, which is good enough with me.

   T.Y.

   M.T.

Dear MLs,

.....  bump  .....

Thank you.
M.T.


algebra
polynomial
parameterization
ratio
Vieta
geometric sequence
geometric series

Let $x_{1},x_{2},x_{3}$ be the roots of the equation $x^3-ax^2+bx-8=0$ with real parameters $a,b$ and $A=\{(a,b)\in\mathbb{R}^2  | x_{1},x_{2},x_{3}$ are numbers in a geometric sequence$\}$. Find $S=\sum_{(a,b)\in A}{(a+b)}$.

Right now, I am just getting $ S \geq 18 \ or\ \ \leq -6 $

Oh, sorry!!!

The above estmate is not for $S$ but for a particluar $a+b$

err im kinda confused as to what some of the things in this question mean

first off does real parameters $a,b$ just mean real coefficients?

then what does it mean when it says $a,b$ are elements of the $\mathbb{R}^2$?

then im going to assume that the $|$ notation is just to set everything off, and it's just saying like as a sidenote that $x_1, x_2, x_3$ are in geometric series?

so finally its just asking for the sum of all the possible coefficients that are elements in set A right?

yup, it's basically asking for all possible $a$ and $b$

[hide]
Well since the product of the roots is 8 and they are in geometric progression, we can call the smallest root $x$ and ratio $r$, we get $xr=2$.  So $a = x + 2r + 2$ and $b = 2(x + 2r + 2)$.  So there are infinite real values for $a$ and $b$.  And unfortunately it does not simplify.  Since the polynomial is also valid when $x$ and $r$ are both negative, we can compare the sum $a+b$ of the two polynomials with $x$ and $r$ merely being the opposite of the other.  So the sum is $3(2 + p) + 3(2 - p)$ where $p = x + 2r$, and we see that the sum is always $12$, so the answer is infinity.
[/hide]

This has to be wrong.  Where did I go wrong?

My solution involves just a trick(nothing big)
take the nos. to be $ \frac{A}{r} , A , Ar $
thus , from the product of roots , we get that $ A = 2$
Thus $a = 2(1 + r + \frac{1}{r} ) $
and $ b = 4(1 + r + \frac{1}{r} $
Thus $ S = \sum (a+b) = \sum {3a} $
Now using that $ r+ \frac{1}{r} \geq 2 or \leq -2 $
I get that $ 3a \geq 18 \ or\ \leq -6 $

And now , I believe i'm stuck!!!

I did that too. I obtained $b=2a$ and from Vieta's we have $x_{1}x_{2}x_{3}=8$ and because $x_{1}=\frac{x}{q}, x_{2}=x, \ x_{3}=xq$ we get $x^3=8\Rightarrow x=2$ or $x_{2}=2$. From this point i'm stucked.

Yes that's what I did too.  But there are an infinite amount of pairs $(x_1, x_3)$ that satisfy all our criteria.  There needs to be another exception, something like $a$ and $b$ must be integers or something.

I believe the answer will be infinte... but I 'm still working!!!

[quote="Slizzel"]$a,b\in\mathbb{R}$ and what I forgot to write $x_{1},x_{2},x_{3}\in\mathbb{N}$ different by $0$  :oops:[/quote]

I have a hard time remembering what the capital letters represent.  I know $\mathbb R$ is all reals, what is $\mathbb N$?  And what does it mean the roots are different by 0?

$\mathbb{N} $ refers to Natural nos.

Also , I believe it should be differs [i]from[/i] zero. Which implies that none of $ x_i $ are zero!!!! :rotfl:

That doesn't make sense.  Of course none of the roots can be zero, else there would be no geometric progression, and the product could never be 8.  That is trivial.  Although I can't think of anything else it could mean.

[quote="Rushil"]
Now using that $ r+ \frac{1}{r} \geq 2 or \leq -2 $
I get that $ 3a \geq 18 \ or\ \leq -6 $

And now , I believe i'm stuck!!![/quote]

Could you explain that part a lil more?

But yea assuming what you did is correct, then $a \ge 6$ or $a \le -6$

But since your being asked for $\displaystyle\sum_{(a,b)\rightarrow\mathbb{R}} 3a = 3\sum_{(a,b)\rightarrow\mathbb{R}} a$

You are just adding up all the possible values of $a$ and multiplying it by $3$, so since $a$ can be anything $\ge 6$ and anything $\le -6$, when adding all the possibilities don't all the $a's$ just cross out?

giving you $3(0) = 0$ ?

I believe he means [tex]a\in(-\infty,-2]\cup [6,+\infty)[/tex] (that's what I also came up with), but since [tex]x_1,x_2,x_3\in\mathbb{Z}[/tex] and [tex]a=x_1+x_2+x_3[/tex] thus [tex]a,b\in\mathbb{Z}[/tex]. Since all [tex]|a|\ge 6[/tex] cancel we get three times sum of [tex]a\in\{-2,-3,-4,-5\}[/tex] and that gives [tex]S=-42[/tex].

edit: Slizzel did you say [tex]x_1,x_2,x_3\in\mathbb{N}[/tex] :?: But then [tex]a,b\in\mathbb{N},\, a\ge 6,\, b\ge 12[/tex] and the sum is infinite... :?:

Yes, it should be integers...

But , how do we know that the conditions we have arrived at are sufficient to ensure the roots are in GP. ONly if we prove that this(our result)  is the necessary and sufficient condition can we use your method which basically says that $\forall a $ acc. to our restriction, the roots will be in GP...


Let $ a,b,c$ be positive real numbers .

Prove that :

$ \frac{1}{(a\plus{}6b)^2}\plus{}\frac{1}{(b\plus{}6c)^2}\plus{}\frac{1}{(c\plus{}6a)^2}\geq \frac{9}{49(ab\plus{}bc\plus{}ca)}$

Use can_hang's old result, we can prove the general:
$ \frac {1}{(a \plus{} kb)^2} \plus{} \frac {1}{(b \plus{} kc)^2} \plus{} \frac {1}{(c \plus{} ka)^2} \geq\ \frac {9}{(k \plus{} 1)^2(ab \plus{} bc \plus{} ca)}$
With $ k \geq\ \frac {49 \plus{} 9\sqrt {17}}{32} \equal{} 2.69$ :)


MATHCOUNTS geometry rectangle probability ratio trapezoid perimeter

i was doing Mathcounts problems, and i dont get 2 problems.

1)

A rectangle XYZW has XY=6, and XW=8. If a point M is selected randomly in the interior of this rectangle, what is the probability that triangle XMW has an area equal to or greater than 16 square units? Express your answer as a common fraction.

I got 3/7, but the answer says that its 1/3.

2)

Massaru has 3 pencils in his bookbag. Two pencils are blue and one is red. If he randomly selects two pencils, what is the probability that they are the same color? Express your answer as a common fraction.

no idea... the answer says 1/3.

Help me... :(

1st question:

[hide]OK. The triangle has a determined base of length $ 8$. Thus, in order for the triangle to have an area greater than $ 16$, we must have a height of $ > 4$. Since we are looking for probability, and we know that anything less than or equal to $ 4$ will result in a triangle of area less than $ 16$, we compare the heights, resulting in the final answer of: $ \frac {2}{6} \equal{} \frac {1}{3}$[/hide]

2nd question:

[hide]The only possibility is to have two blue pencils. The chance of that is $ \frac{2}{3}\cdot\frac{1}{2}\equal{}\boxed{\frac{1}{3}}$.[/hide]

The only way they can be the same color is if they are both blue.  There are three choices of what he picks.  Let B1 and B2 be the two blue pencils, and R1 is the red pencil.

B1, B2
B2, R1
B1, R1.
There are 3 choices possible, and one satisfies the conditions, thus it is 1/3.

[quote]
1)

A rectangle XYZW has XY=6, and XW=8. If a point M is selected randomly in the interior of this rectangle, what is the probability that triangle XMW has an area equal to or greater than 16 square units? Express your answer as a common fraction.
[/quote]

It says 'triangle XMW has an area EQUAL to or greater than 16 square units'.

Also, a random point M can land on the side XW or YZ.

OK, but think about it this way. I meant that it was greater or equal to 16, but in any case, it wouldn't matter. Lines have no area. They aren't compatible. Let's think about it this way. Draw the line that m being on will result in an area of 16. Then we know that selecting any other point above or on that line would result in an area of 16+. Then you can imagine two rectangles. It's now just the chance of randomly selecting a point inside the smaller rectangle.

[quote="math154"]OK, but think about it this way. I meant that it was greater or equal to 16, but in any case, it wouldn't matter. Lines have no area. They aren't compatible. Let's think about it this way. Draw the line that m being on will result in an area of 16. Then we know that selecting any other point above or on that line would result in an area of 16+. [b]Then you can imagine two rectangles.[/b] It's now just the chance of randomly selecting a point inside the smaller rectangle.[/quote]

I think it's one triangle and one trapezoid.

also, the ratio of area of triangle : area of trapezoid is 1/2.

[quote="math154"]Maybe you aren't getting what I'm saying...if only I knew Asymptote....[/quote]

yea i dont think im getting it lol

but think about this:

if i didn't give you the answer, could you still figure out that the answer was 1/3?

Yes. It's a concept...that happens to be easier to explain with Asymptote.lol :P

edit: btw, isn't my avatar so 1337???

edit2: Planetary Nebula...sobs...thought it would attract people with my signature....(sigh) lol jk

edit3: How are we the only people posting here. It's been about 1-2 hours.

[quote="math154"]Yes. It's a concept...that happens to be easier to explain with Asymptote.lol :P

edit: btw, isn't my avatar so 1337???[/quote]

lol whats your avatar??

[quote="Ihatepie"]I'll try explaining it.

[hide]

so draw rectangle XYZW with XW as the base. You have to find the probability of XMW having an area greater than or equal to 16 square units. In order for the triangle to have an area of 16 or greater, The triangle must have a height of 4 or more since the area is 1/2*bh. If the point is anywhere in the bottom 2/3 (an 8 by 4 rectangle) it would have a height of less than 4, making the area of triangle XMW less than 16. If the point is anywhere in the top 1/3 (an 8 by 2 rectangle), The height would be 4 or greater, making the area of triangle XMW greater than or equal to 16. So there is a rectangle of area 16 that would be favorable, and a total of area 48. so it is 16/48, or 1/3.[/hide][/quote]

ahh thank you Ihatepie :D  i understood what math154 was saying.

but the problem says 'interior of the rectangle', not just any point inside the rectangle.

points/lines have no area. =P. just let it fudge right now. the problem is that we show points and lines, giving the misconception to some that they have some area.

Think about it this way: is the area of a rectangle affected because you include the perimeter?

here's how I solved:

the whole perimeter of the rectangle is 28, and the length that M can land is 2+8+2=12.

so i got $ \frac{3}{7}$.

why is this solution wrong?

the point can be anywhere inside of the rectangle, not just on the edge.

[quote="FantasyLover"]
1)

A rectangle XYZW has XY=6, and XW=8. If a point M is selected randomly in the INTERIOR of this rectangle, what is the probability that triangle XMW has an area equal to or greater than 16 square units? Express your answer as a common fraction.

[/quote]

GAH IT SAYS INTERIOR

if the point could be anywhere inside of the rectangle, it wouldn't be Mathcounts Nationals problem.

ok i think there was a problem in my brain... :blush:

i thought interior = out...

anyways thank you for ppl who solved my prob... lol

It seems that I have been very off, along with many others recently...I wonder what it is???

It seems that he misunderstood the problem since he'd thought that in Nat, it must have been extremely difficult or something like that. :lol:

Yeah I know what you're saying. :P  :P  :P But the best tests aren't the ones that are perfect(ly) hard/easy. lol

...I thought the point M is landing on the perimeter of the rectangle, not inside.

[quote="mewto55555"]I think there is some confuision. If you choose in random point in OR on the rectangle, the chance of it being on the perimeter is 0.[/quote]
[b]
I THINK[/b] that you should read the whole thread before posting (of course, this doesn't apply to the Mathcounts Marathon).


One uncreated page is "inverse," as in the inverse of a function.  We already have a section on function inverses as part of the Function article (section 5, in fact).  Is it possible to turn the "inverse" page into a redirect to section 5 of the Function page?  (I have tried and failed to make this work.)

Updated question:  I've made the page "Inverse" a redirect to "Function#The_Inverse_of_a_Function."  If I visit the page of Inverse and click on its redirect link, I get sent to the appropriate section.  However, if I instead let it automatically re-route me, by searching for "inverse" or by clicking on one of the pre-existing links to "inverse," I get sent to the top of the "Function" page rather than to the appropriate section.  Thoughts?

I don't like the idea of inverse redirecting to any one page.  Since inverse is used in several different ways, I would prefer a disambiguation.  Certainly ours won't need to be as large, but take a look at Wikipedia's entry on inverse:

http://en.wikipedia.org/wiki/Inverse

That should at least serve as a guide.

[quote="JBL"]That's fine -- do we have a disambiguation template?[/quote]

Not yet.  I'll task joml88 with that job.

Okay, cool.  There are already a couple of pages (Euler's Formula, Inverse) which have been turned into informal disambiguation pages.

The disambiguation template is now up.  Just put {{disambig}} at the bottom of disambiguation pages.


geometry
rotation
analytic geometry
trigonometry
calculus
integration
parameterization

hey every one , i need help with this :

A circle the radius r is rolling without sliding along a fixed circle of radius
R and outside it. Assuming that R=r is an integer, and the area bounded by the curve
described by some point of the moving circle.

Please try to type your problems without confusing misprints if you want your posts to be answered. I strongly believe that R=r should be R/r and "and" after the comma should be "find".

One way to do it is to parametrize the curve $ C$ in question and then honestly integrate $ x\,dy$ along the curve. If we assume that the starting point is $ (R\plus{}2r,0)$, then, when the line joining the origin with the center of the small circle makes angle $ \theta$ with the horizontal line, the small circle rotates by the angle $ n\theta$ where $ n\equal{}1\plus{}\frac Rr$, so the point gets the coordinates $ ((R\plus{}r)\cos\theta\plus{}r\cos n\theta,(R\plus{}r)\sin\theta\plus{}r\sin n\theta)$. Now, setting up the integral $ \int_C x\,dy$ using this parametrization, we get
\[ \begin{aligned}
&\int_0^{2\pi} [(R\plus{}r)\cos\theta\plus{}r\cos n\theta][(R\plus{}r)\cos\theta\plus{}nr\cos n\theta]\,d\theta\\&\equal{}\pi[(R\plus{}r)^2\plus{}nr^2]\equal{}\pi[R^2\plus{}3Rr\plus{}2r^2]
\end{aligned}\]


[hide]im quite sure this is wrong but: 1.0  and  2. 2[/hide]

[hide]Well, for each of them, there's at least one answer. For 32, you can have:

-31, -30, -29, ... 29, 30, 31, 32

For 72:

-71, -70, -69... 69, 70, 71, 72.

I haven't really tried any others.... takes too long.  :)  :lol: [/hide]

woops, i thought it only meant positive integers.............

[hide]That's what I got--My hint--Triangular numbers.[/hide]

#1 is 1, because 32 has only 1 odd factor---1

I didn't get 5 for #2. Hint: the number of ways a number can be written as a sum of consecutive [b]positive[/b] integers is the number of odd factors it has. Sorry I didn't say positive before.  See http://www.artofproblemsolving.com/Resources/Papers/LaurendiPartitions.pdf for a proof.

What is the solution to number one? What consecutive positive numbers is it?

Is it 32 = 32 ? That's a strange solution.

For #1 I get 1 way: 32, for #2 I get 3 ways: 23+24+25, 12+...+4 and 72 but doesn't the word "integer[b]s[/b]" indicate only solutions with more than one integer?

For those who are interested:  The 2001 National Masters Round was based on a similar theme as this question.

[quote="Treething"][quote="chess64"]1. How many ways can 32 be expressed as a sum of consecutive integers?

2. How many ways can 72 be expressed as a sum of consecutive integers?[/quote]

[hide="Number 1"]
You can't. It's a power of two.
[/hide][/quote]

Don't forget the odd factor [b]1[/b]. That's why every positive integer can be expressed as a sum of consecutive integers.

So what's the solution chess64? One way but how?

It's 32 = 32.
\begin{eqnarray*}
32 &=& r + (r + 1) + \cdots + (r + k)\\
32 &=& \frac{(2r + k)(k + 1)}{2}\\
64 &=& (2r + k)(k + 1)
\end{eqnarray*}
Because $(2r + k)$ and $(k + 1)$ have opposite parity, we need to find the number of ways 64 can be expressed as a product of an even number and an odd number. The only odd factor of 64 is 1, so:
\begin{eqnarray*}
64 &=& (2r + k)\\
1 &=& (k + 1)\\
\end{eqnarray*}

Thus $k=0$ and $r=32$, so $32=32$.

[hide="Problem 2"]
We need to find the number of odd factors of $72=2^3 \times 3^2$. A factor is odd only if it doesn't have a factor of 2 in it, so the factors are the powers of 3: 1, 3, and 9. There's 3, so that's the answer.
[/hide]

[quote="Treething"]Notice the word "integer[b]s[/b]"? That plural might be why it's zero ;) .[/quote]

Huh? "[size=59]integer[/size][size=200][b]s[/b][/size]"? I don't see any plural...  :D

Actually, according to J. Laurendi, who wrote an article about partitions and posted in the Articles section, the answer is 1.

It should be integar(s) making it less obvious.


Ring Theory
abstract algebra
algebra
polynomial
superior algebra
superior algebra solved

Knowing that $k[x,y]$ is UFD, $k$ is a field. Can we characterize all prime ideals of $k[x,y]$ ?

the ring is UFD and noetherian, hence a PID. so the question is equivalent to determine the irreducible polynomials in two variables, which is impossible

[quote="-oo-"]the ring is UFD and noetherian, hence a PID.[/quote]

I didn't get it. Why is it true? Say, $\mathbb{R}[x,y]$ is UFD, noetherian, but still isn't PID, for example $(x,y)$ can't be generated by one element.

[quote="-oo-"]sorry, my idea was wrong. never the less, there's no way to characterize the principal prime ideals in general.[/quote]

Well, i didn't say that i want to characterize the principal prime ideals, i want to characterize prime ideals. For example for $a,b \in k$, the ideal $(x-a,y-b)$ is maximal, thus prime, and not principal.

It seems that there only 3 possibilities for the ideal in $k[z,y]$ to be prime, namely:

1).The zero ideal $(0)$.
2). The ideal $(f)$, where $f$ is irreducible in $k[x,y]$.
3). The ideal $(x-a,y-b)$ for some $a,b \in k$.

But i'm not sure whether this classification is true.

I was thinking about the same thing, but haven't proved it.

I think you will encounter difficulties unless you assume that the field k is algebraically closed. But with this assumption there is a one-to-one correspondence between prime ideals in k[x,y] and irreducible varieties in $k^{2}$. These are presumably just the irreducible curves, determined by a single irreducible polynomial, and the points.

Let $R$ be a PID.

Let $p$ be a non-zero prime ideal of $R[X]$. Let $q=p\cap R$.
Then $q=(a)$ for some $a\in R$. $a$ must be a prime element in $R$ or $0$.

Assume first $a\neq 0$.
Consider the natural projection $R[X]\to (R/aR)[X]$. It is easy to see that its kernel is $aR[X]$. Therefore $R[X]/aR[X]\simeq (R/aR)[X]$.
Let $\bar p$ be the image of $p$ in $(R/aR)[X]$. This is a prime ideal of the last ring. Let $k_{a}=R/aR$. Then $k_{a}[X]$ is a PID because $k_{a}$ is a field with $R$ being a PID and $a$ a nonzero prime. There exists $\hat f\in k_{a}[X]$, irreducible or $\hat 0$, such that $\bar p=(\hat f)$. Then it is easy to prove that $p=(a,f)$ with $f$ being any representative for $\hat f$. It may happen that $\hat f=\hat 0$.

Assume that $a=0$.
Let $f$ be a non-zero polynomial in $p$ of minimal degree and such that its leading coefficient determines a maximal ideal among the ideals generated by leading coefficients of polynomials in $p$ of the same degree as $f$. Actually we can take $f$ to have content $1$ which is possible because $p$ is prime and does not contain scalars. Let $K$ be the fraction field of $R$.
Let $u\in p$, $u\neq 0$. Then $u=cf+r$ for some $c,r\in K[X]$ and $deg r< def f$. We can multiply this by some non-zero $A\in R$ such that $Ac,Ar\in R[X]$. Then $(Au)=(Ac)f+(Ar)$ and we get $Ar\in a$. But $deg r<deg f$, $A\neq 0$ and $f$ was chosen of minimal degree in $p$, so $r=0$. We have $u=cf$. We multiply by some $a\in R$ such that $ac\in R[X]$ and it has content $1$. Then $au=(ac)f$, so the content of $au$ is $1$. But this is $a\times (\mbox{the content of }u)$, so $a=1$. This means that $u$ is a multiple of $f$. We have proved that $p=(f)$. It also follows that $f$ is irreducible because $p$ is prime.

To summarise:
The prime ideals of $R[X]$ are:
$(0)$
$(a)$ with $a$ a prime element of $R$.
$(f)$ with $f$ an irreducible polynomial of $R[X]$.
$(a,f)$ with $a$ a prime in $R$ and $f$ is irreducible mod $a$.

Now take $R=K[Y]$ for some field $K$ and see what you get!

P.S. This follows almost word by word the description with proof of the prime ideals of $\mathbb Z[X]$ and marks one of the simmilitudes between $\mathbb Z$ and $K[X]$ for a field $K$.


...and with it comes the promise of another year of ARML-ing. Would anybody happen to know how many of the current MN ARML posse is actually on this site thingy? Seems like a good way for info to get disseminated and problems / other sundry topics to be discussed.

- Niedermaier

Well, I am on AoPS, but I don't really check it so often. If we got a bunch of people on here and could actually work on problems in this forum it could be very helpful. I am quite excited for ARML as well...

-Zachary Sethna

Well, I'm hear. Just joined the team

-- Matt Bunday


quadratics
modular arithmetic
number theory unsolved
number theory

Prove that a positive integer u is a perfect square if and only if at least one of (u+1)2-u, (u+2)2-u, ..., (u+v)2-u is divisible by v for each positive integer v. Anyone have any ideas on this?

In other words,
\[ (u \plus{} x)^2 \equiv u\pmod{v}\]
is solvable with respect to $ x$ for every positive integer $ v$.

If $ u$ is a square, say, $ u \equal{} w^2$ then the above congruence has a solution $ x \equal{} w \minus{} u$. On the other hand, if $ u$ is not a square then there exists a prime $ p$ such that $ u$ is not a square modulo $ p$, and hence the above congruence is not solvable for $ v \equal{} p$.


An expensive perfume is packaged in a box that has an octagonal base with area of ten square inches. The box is a prism. Given that the height of the box is four inches, what is the number of cubic inches in its volume.

Any prism's volume is the area of its base times its height. Therefore, this prism's volume is $ 4 \times 10 \equal{} \boxed{40}$.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ R$ be unital and commutative ring and $ f(x)\in R[x]$.Prove that $ f(x)$ is zero divisor if and only if $ \exists a\not\equal{} 0 \in R$ such that :
$ af(x)\equal{}0$.

Does the thread actually prove the claim?  The only thing offered that I saw was:

 [quote="-oo-"]let $ g \equal{} b_m x^m \plus{} ... \plus{} b_0$ some non-zero polynomial of smallest degree $ m$ that annihilates $ h \equal{} r_n x^n \plus{} ... \plus{} r_0$. if $ m \equal{} 0$, we're done. otherwise $ m \geq 1$. this yields $ b_m h \neq 0$, thus $ b_m r_i \neq 0$ for some $ i$ and therefore $ r_i g \neq 0$. choose some maximal $ j$ with $ r_j g \neq 0$. now we have $ 0 \equal{} hg \equal{} (r_j x^j \plus{} ... \plus{} r_0)g$, so $ r_j b_m \equal{} 0$. but then $ r_j g$ has degree smaller than $ m$ and annihilates $ h$, a contradiction.[/quote]

I don't think that this contradicts $ m \geq 1$ but rather $ b_m h \neq 0$.  Assume $ m$ as above and that $ gh\equal{}0$.  Then $ b_mr_n\equal{}0$ and
 $ b_m (hg)\equal{}(b_m h)g\equal{}0$.  But $ \deg(b_mh) < m$ and minimality of $ m$ implies $ b_m h \equal{}0$.

[quote="Satyr"]Does the thread actually prove the claim?  The only thing offered that I saw was:

 [quote="-oo-"]let $ g \equal{} b_m x^m \plus{} ... \plus{} b_0$ some non-zero polynomial of smallest degree $ m$ that annihilates $ h \equal{} r_n x^n \plus{} ... \plus{} r_0$. if $ m \equal{} 0$, we're done. otherwise $ m \geq 1$. this yields $ b_m h \neq 0$, thus $ b_m r_i \neq 0$ for some $ i$ and therefore $ r_i g \neq 0$. choose some maximal $ j$ with $ r_j g \neq 0$. now we have $ 0 \equal{} hg \equal{} (r_j x^j \plus{} ... \plus{} r_0)g$, so $ r_j b_m \equal{} 0$. but then $ r_j g$ has degree smaller than $ m$ and annihilates $ h$, a contradiction.[/quote]
[/quote]@ -oo-: Why? I cannot see any non-zero polynomial with degree $ <m$ annihilating $ h$.[quote="Satyr"]
I don't think that this contradicts $ m \geq 1$ but rather $ b_m h \neq 0$.  Assume $ m$ as above and that $ gh \equal{} 0$.  Then $ b_mr_n \equal{} 0$ and
 $ b_m (hg) \equal{} (b_m h)g \equal{} 0$.  But $ \deg(b_mh) < m$ and minimality of $ m$ implies $ b_m h \equal{} 0$.[/quote]@ Satyr: Hmm, but $ h$ is fixed and $ m$ was chosen with respect to a minimality condition on $ g$!?! What can be deduced is $ b_mg\equal{}0$, by the minimality condition. But where is that leading us to? I could think of a proof using associated primes.


What is the ratio of the length of the diagonal of a square to the length of the hypotenuse of the isosceles right triangle having the same area?

[hide="solution"] let the side of the square be x and the side of the triangle be y. Then, $ 2x^2\equal{}y^2, y\equal{}x\sqrt{2}$. Thus the ratio is just $ \frac{x\sqrt{2}}{2x}\equal{}\frac{\sqrt{2}}{2}$.[/hide]


Can someone please post an asymptote tutorial lists all the commands there are. I don't think the  one in Aopswiki is that good...


modular arithmetic
floor function
arithmetic sequence
number theory
prime numbers
number theory unsolved

Let n be natural number and n>4. P1,P2,.....Pn is arithmetic progression with length n such that P1,P2,…Pn all  are prime number. suppose that  P2-P1=d .prove that d is divisible by any prime number less than or equal to n.

It is easy to simplify the problem in the case where $ p_1\equal{}n$ but for the moment I cannot think of any aproach. Interesting question

A really nice problem. I hope I didn't make any mistakes:

First of all it is trivial to check that $ d \equiv 0 \pmod 6$, and $ p_1 \geq 5$. Indeed, if $ n > 4$, then if $ d$ is odd, two of the primes are even, contradiction. Also if $ d\neq 0 \pmod 3$, then one of the $ n$ primes must be divisible by 3. It obviously must be $ p_1$. But then $ p_4$ is also divisible by 3, contradiction. So $ d\equiv 0 \pmod 6$, and $ p_1\geq 5$.

Now let us consider a prime $ q$, $ 2\leq q \leq \dfrac n2$. And let's assume that $ d\neq 0 \pmod q$. Then in the sequence $ p_1, p_2,\ldots, p_{q}$, there are $ q$ distinct residues modulo $ q$, so one of them must be 0, say $ p_j \equiv 0 \pmod q$. This means that $ q \equal{} p_j$. But then $ p_{j \plus{} q} \equal{} p_j \plus{} qd \equiv 0 \pmod p$, and as $ j\leq q$, $ j \plus{} q \leq n$ so we obtain a contradiction.

This means that $ d$ is divisible by any prime $ q\leq \dfrac n2$. In particular from [[Bertrand's postulate]] we know that there must exist a prime between $ \left\lfloor \dfrac n4 \right\rfloor \plus{} 1$ and $ 2 \cdot \left\lfloor \dfrac n4 \right\rfloor$, let's call it $ r$. This happens if $ \left\lfloor \dfrac n4 \right\rfloor \geq 3$, or $ n\geq 12$ and in this case $ r > 3$ (we will treat the cases $ n\leq 11$ at the end.) This means that $ d \equal{} 6rd' \geq 6\left\lfloor \dfrac n4 \right\rfloor$.

Now let us consider a prime $ p$, $ \dfrac n2 < p \leq n$. Suppose that $ d\neq 0 \pmod p$. As above there exists an index $ 1\leq k\leq p$ such that $ p_k \equiv 0 \pmod p$. This means that $ p_k \equal{} p$, or $ p_1 \plus{} (k \minus{} 1)d \equal{} p \leq n$, so $ (k \minus{} 1)d \leq n \minus{} 5$. But $ (k \minus{} 1)d\geq 1 \cdot 6 \left\lfloor \dfrac n4 \right\rfloor$, so we would have
\[ n \minus{} 5 \geq 6 \left\lfloor \dfrac n4 \right\rfloor \ \Leftrightarrow \ \dfrac {n \plus{} 1}{6} \geq \left\lfloor \dfrac n4 \right\rfloor \plus{} 1
\]
which would imply that
\[ \dfrac {n \plus{} 1} 6 > \dfrac n4 \ \Leftrightarrow 2 > n ,
\]
contradiction. Therefore $ d \equiv 0 \pmod p$, for all $ p\leq n$.

We are left with the cases $ n \equal{} 5,6,7,8,9,10,11$. We have already proved that $ d \equiv 0 \pmod 6$, so we need $ 5\mid d$. Assume that $ d\neq 0\pmod 5$. Then one of $ p_1,p_2,\ldots, p_5$ must be 5. But that can only be $ p_1$. But then if $ n\geq 6$, then $ p_6 \equal{} p_1 \plus{} 5d \equiv 0 \pmod 5$ contradiction. If $ n \equal{} 5$ then this is where we find our counterexample: $ p_1 \equal{} 5$, $ d \equal{} 6$, and the arithmetic sequence of 5 primes is $ 5,11,17,23,29$. So let's assume that author meant $ n > 5$. In particular this implies that $ p_1 > 5$ (as $ d\equiv 0 \pmod {30}$.)

If $ n \equal{} 8,9,10$, then we take the first 7 primes. One of them has to be 7 or $ d\equiv 0 \pmod 7$. If one of them is 7, it has to be $ p_1$. But then $ p_8$ would have to be 7 too contradiction.

This leaves $ n \equal{} 7$ and $ n \equal{} 11$. If $ n \equal{} 7$, then we showed that $ p_1\geq 7$. If $ d\neq 0 \pmod 7$, then $ p_1 \equal{} 7$ (same reasoning as above.) As it turns out this doesn't work either, because of the sequence $ 7,157,307,457,607,757,907$. For 11 again we get that $ p_1 \equal{} 11$. As with $ n \equal{} 7$, $ n \equal{} 11$ is a negative (so maybe the author should have formulated the problem $ n\geq 12$ to begin with.)
The sequence is $ p_1 \equal{} 11$, $ d \equal{} 2^4 \cdot 3\cdot 5 \cdot 7 \cdot 13 \cdot 37 \cdot 1901 \equal{} 1536160080$. You can verify that this sequence of 11 prime numbers is valid at http://www.onlineconversion.com/prime.htm

Well.. Maybe the problem should have been. "any prime number less than n."? Then it's easy, but also true:)

Valentin: How do you obtain $ (k \minus{} 1)d\geq 1 \cdot 6 \left\lfloor \dfrac n4 \right\rfloor$? What if $ k \equal{} 1$?

Hmm.. I'm in doubt wether the statement is correct or not.. I'll see if i can find a sequence with $ n \equal{} 13$ :)

Edit: There is no such sequence in the interval $ [0;2,309,997,703]$...

Yeah I was just about to correct my statement (I "solved" the problem at 2,30 AM :D) The problem appears indeed when $ p_1 \equal{} p \equal{} n$ ... and there is no clear way to solve that, and it's very very likely that in that case the problem is not true.

Check the table "Smallest AP-k with minimal start" at http://hjem.get2net.dk/jka/math/aprecords.htm
Also this topic (primes in arithmetic progression) is what got Terrence Tao his fields medal 4 years ago, so I doubt we can wrap up something on the spot  :roll:


On the sunny side of Venus air pressure is 9x$10^{6}$ Pa.when temperature 740K.
On earth air  pressure is $10^{5}$ Pa  and the temperature on the sunny side can reach 320K.
This data suggests that the atmosphere on Venus is denser than that on Earth.
Is this true?

(Komal September 2004)

Yes, the atmoshphere of Venus has the highest density of all the terrestrial planets, and is mosltly carbon dioxide.


geometry
3D geometry
search
algebra unsolved
algebra

I dont know if it's here where i can post this , sorry if it's not :
can we construct a cube 2005^3 with only the cubes 2^3,3^3 and 7^3  ?

What he means is that if we can build a $2005^{3}$ cube using cubes of sizes $2^{3},3^{3},7^{3}$.

[quote="yassinus"]yes , thank you Spix ,I mean can we build a cube ...., or can we fill the cube of side 2005 by small cubes with sides 2,3 and 7 .I still searching without succes , thank's for intersting in this problem  :)[/quote]

I search too. It seems impossible (but without proof).

But it is possible to fill a big cube (for example $2005*2005*2005$) with $4$ classes of little cubes whose sides are little primes (for example 2,3,7 and 11).

--
Patrick

[quote="yassinus"]I dont see how !!
 I'm not sure that is possible but i found how to make a surface 2005² with 2² , 3² and 7² but i didn't succes to build the big cube[/quote]

In order to fill a big cube $A^{3}$ with 4 little cubes $p_{1}^{3}$, $p_{2}^{3}$, $p_{3}^{3}$ and $p_{4}^{3}$, I suggest the following steps :

1) Take $a\geq 0$ and $b\geq 0$ such that $ap_{1}+bp_{2}=A$. This is possible since $p_{1}$ and $p_{2}$ are coprime and, for a solution with nonnegative $a,b$, we need $A$ big enough ($A>p_{1}p{2}$).
If you put on x-axis such $a$ $p_{1}^{3}$ cubes along with $b$ $p_{2}^{3}$ cubes and if you add, on y-axis and z-axis $p_{2}-1$ $p_{1}^{3}$ cubes above each $p_{1}^{3}$ cube and $p_{1}-1$ $p_{2}^{3}$ cubes above each $p_{2}^{3}$ cube, you have built a bloc $A*p_{1}p_{2}*p_{1}p_{2}$ and then a bloc $B1$ $A*p_{1}p_{2}*p_{1}p_{2}p_{3}$

2) With the same mechanism, you can build a bloc $B2$ $A*p_{1}p_{3}*p_{1}p_{2}p_{3}$ and a bloc $B3$ $A*p_{2}p_{3}*p_{1}p_{2}p_{3}$

3)Then take $(u,v,w)$ such that $A=u*p_{1}p_{2}+v*p_{1}p_{3}+w*p_{2}p_{3}$. Using $u$ blocs $B1$, along with $v$ blocs $B2$ and $w$ blocs $B3$, you can build a large bloc $C1$ $A*A*p_{1}p_{2}p_{3}$

4) with the same mechanism, you can build a large bloc $C2$ $A*A*p_{1}p_{2}p_{4}$, a large bloc $C3$ $A*A*p_{1}p_{3}p_{4}$ and a large bloc $C4$ $A*A*p_{2}p_{3}p_{4}$

5) Take then $x,y,z,t$ such that  $A=x*p_{1}p_{2}p_{3}+y*p_{1}p_{2}p_{4}+z*p_{1}p_{3}p_{4}+t*p_{2}p_{3}p_{4}$. Using $x$ blocs $C1$, along with $y$ blocs $C2$, $z$ blocs $C3$ and $t$ blocs $C4$, you can build the large cube $A*A*A$

--
Patrick

very smart idea Patrick,so i think if we searsh for a number q ( witch will replace p4) so that q is prime with the others pi, and so that we can build a cube of q whit small cubes with sides in {p1,p2 and p3} it will be done  :wink:

Still dont found a solution ! can any one help me to continu :(


[quote="math92"]Find the slope of a line that contains the points B9-5, 2)and C(7, -4)[/quote]

[hide]$\frac{2-(-4)}{-5-7}$

$\frac{6}{-12}$

$-\left(\frac{1}{2}\right)$[/hide]

Sorry, I forgot to add the negative sign!  :blush:

[quote="Drunken_Math"]$\frac{6}{-12}$

$\left(\frac{1}{2}\right)$[/quote]

No....


Some unit cubes are assembled to form a larger cube and then [b]some[/b] of the faces of the larger cube are painted. After the paint dries, the larger cube is disassembled into the unit cubes and it is found that [b]96[/b] of these have no paint on any of their faces. How many faces of the larger cube were painted?

If any faces are painted, that removes an entire layer of unpainted cubes.  Thus, the number of unpainted faces can be expressed as $(x-a)(x-b)(x-c)$, where x is the length of an edge of the cube, and a, b, and c, are equal to 0, 1, or 2 (the number of layers removed; if neither a face nor its opposite are painted, the number is 0; if a face is painted but not its opposite is painted, the number is 1; if a face and its opposite are painted, the number is 2.)  96 is between 4 and 5 cubed, so I just guessed that the edge length is 5, 6 or 7, to start. We can rule out 5 and 7, because 96 is not a multiple of 5 or 7:if it was 5, then all of the faces would be no more than 4, which is impossible; if it was 7, then all three numbers would have to be 6, which is also impossible.  We can then factor 96 into three numbers approximately equal to 5, so we get 6, 4, and 4, or $(6-0)(6-2)(6-2).$  Thus, two pairs of faces are painted, and we have our answer, 4.

yes, the correct answer is 4
but , my solution is quite long so i only post some hint :D
case 1 : if there is only 1 face of the larger cubes were painted
then the following equation must have natuaral root $a^{3}-a=96$
case 2:   if there are 2 face of the larger cubes were painted
then we have the same thing , but now we get 2 equation (because two faces can be opposite to each other)
$a^{3}-2a+a=96 or a^{3}-2a=96$
case 3: the equations are $a^{3}-3a+2a =96$ or $(a-1)^{3}=96$
case 4: the equations are $a^{3}-4a^{2}+4a=96$ or $a^{3}-4a^{2}+3a-2=96$
...and so on ...

Here is my solution :)
We have to prove the equation for natural roots:
$(n-2)^{3}+a(n-2)^{2}+b(n-2)+c=96$
with $n$ is the side length of the larger cube, $a$ is the faces that were unpainted, $b$ is the same sides of the unpainted faces (if any) and $c$ is the unit cubes at the vertexs of the larger cube if there are unpainted faces with the same vertex.
From the equation, we have $n \in{5;6}$
Replace it to the equation, we have the desired results :D
(Remember that $b,c$ can be equal to zero)


function
group theory
algebra
Fixed point
IMO
IMO 1973

$G$ is a set of non-constant functions $f$. Each $f$ is defined on the real line and has the form $f(x)=ax+b$ for some real $a,b$. If $f$ and $g$ are in $G$, then so is $fg$, where $fg$ is defined by $fg(x)=f(g(x))$. If $f$ is in $G$, then so is the inverse $f^{-1}$. If $f(x)=ax+b$, then $f^{-1}(x)= \frac{x-b}{a}$. Every $f$ in $G$ has a fixed point (in other words we can find $x_f$ such that $f(x_f)=x_f$. Prove that all the functions in $G$ have a common fixed point.

The hypothesis simply says that for every $a$, there is at most one function $f$ of the form $ax+b$ in our group ($G$ is a group). On the other hand, it's clear that the conclusion we want to reach is that if $f_a\equiv ax+b,\ f_c\equiv cx+d,\ a,c\ne 1$, then $\frac{b}{a-1}=\frac{c}{d-1}$.

Suppose the contrary. Then $f_af_c\equiv acx+ad+b,\ f_cf_a=cax+bc+d$. By assumption, we have $ad+b\ne bc+d$, which means that $f_af_c,f_cf_a$ are two distinct functions in our group with leading coefficient $ac$, and this contradicts the hypothesis.

Let $f_1=ax+b,f_2=ax+c$
Then $f^{-1}_1=\frac{x-b}{a} \in G$
$g=f_2(f^{-1}_1)=a \frac{x-b}{a}+c=x-b+c \in G$
But $g(x)=x$ has solution only for $b=c$

Solution from [i]Twitch Solves ISL[/i]:

Note that the only element of $G$ of the form $x \mapsto x+\lambda$ is the identity.
Now consider any two non-identity functions of $G$, say \begin{align*} f(x) &= ax+b \\ g(x) &= cx+d \end{align*} so that $a \neq 1$ and $c \neq 1$. The (unique) fixed point of $ax+b$ is $\frac{b}{a-1}$ and the (unique) fixed point of $cx+d$ is $\frac{d}{c-1}$, so we just need to show \[ \frac{b}{a-1} = \frac{d}{c-1}. \]
Note the inverses $f^{-1}(x) = \frac{x-b}{a}$ and $g^{-1}(x) = \frac{x-d}{c}$ should also be in $G$. Then consider the composition \begin{align*} g(f(g^{-1}(f^{-1}(x)))) &= c \cdot \left[ a \cdot \frac{\frac{x-b}{a}-d}{c} + b \right] + d \\ &= x - b - ad + bc + d. \end{align*} Since this is an element of $G$, as we commented before we must have \[ b - ad + bc + d = 0 \] which was what we wanted.


On a recent math test, the mean score for Mr. Bill's first and second period classes combined was $ 74\%$. The mean score in first period was $ 70\%$ and a mean score in second period was $ 80\%$. There was a total of 50 students in the two classes. How many students were in first period?

Let $ x$ be the number of students in first period, $ y$ in second period. We get the following system of equations:

$ x\plus{}y\equal{}50$
$ 70x\plus{}80y\equal{}50(74)\Rightarrow7x\plus{}8y\equal{}370$

Multiplying the first equation by 7 gives

$ 7x\plus{}7y\equal{}350$

Subtracting that from the second equation gives $ y\equal{}20$, so $ x\equal{}50\minus{}20\equal{}30$. So there are $ \boxed{30}$ students in first period.

You could also do it this [hide="way"]

Using the same variables as gauss had used, we can set up the equations
\[ x \plus{} y \equal{} 50\]
and
\[ (74 \minus{} 4)x \plus{} (74 \plus{} 6)y \equal{} 50(74)\implies74x \minus{} 4x \plus{} 74y \plus{} 6y \equal{} 50(74)\implies\]

\[ 74(x \plus{} y) \minus{} 4x \plus{} 6y \equal{} 50(74)\]
Now substituting $ 50$ in for $ x \plus{} y$:
\[ 74(50) \minus{} 4x \plus{} 6y \equal{} 50(74)\implies6y \equal{} 4x\implies y \equal{} \frac {2}{3}x\]
Substituting back into the first equation:
\[ x \plus{} \frac {2}{3}x \equal{} 50\implies x \equal{} \boxed{30}\]
Which now yields the same answer that gauss found.  What [u][i][b]may [/b][/i][/u]make this faster (for me at least), is that we can skip most of this work, and get right to the fact that $ 6y \equal{} 4x$ simply because we know that $ 80$ is $ 6$ more than $ 74$ and $ 70$ is $ 4$ less than $ 74$.[/hide]


Prove that for $ a, b, c, d, n \ge 0$, $ ac \equal{} bd$, each of the inequalities

1. $ \sum_{cyc}a^7b^4 \ge \sum_{cyc}abc^9$ and
2. $ \sum_{cyc}a^nb \ge \sum_{cyc}ab^n$

is true exactly when the biggest term on the left hand side is bigger than the biggest term on the right hand side.

zeb your second inequality can be rewritten as
$ \frac{a^n\minus{}c^n}{a\minus{}c}\geq \frac{b^n\minus{}d^n}{b\minus{}d}\geq 0$
assume $ a\geq d$ and $ b\geq c$
if $ \frac{c}{a}\equal{}\frac{d}{b}\equal{}k$ then $ \left(a^{n\minus{}1}\minus{}b^{n\minus{}1}\right)\left(\frac{k^n\minus{}1}{k\minus{}1}\right)\geq 0$
so the inequality is true if and only if $ a\geq b$ (so the biggest term on the LHS $ a^nb$ is greater than $ ab^n$)


If we have the pair of numbers (3,4), can we obtain the pair(35,1998) by making a finite number of transformations as below:
(x,y)->(x+1,2x+y+1);
(x,y)->(x/y,1/y) if y is not 0;
(x,y)->(ax,a2y ) where a is a real number, not 0?

The problem was given at the National Olympiad of Moldavian republic in 1998.(Does anyone have the solutions form that olympiad?)

First of all, we can see that the second transformation is a particular case of the third, for a=1/y.
The answer to the question is negative.Well,in  the  pair (3,4) we have
y<x^2, and its not difficult to see that, from a pair with y<x^2 we can only obtain pairs with the same property.But (35,1998) doesnt have this property so the problem is done.


Let function $f: \mathbb{Q}\to\mathbb{R}$ satisfy:
$(i). \forall x,y \in \mathbb{Q}: f(x+y)=f(x)+f(y)+xy$
$(ii). f(2000)=200100$
Calculate $f(\frac{1998}{1999})$.

I proceeded slightly differently from Rust.
First prove $f ( n ) = n \cdot f ( 1 ) + \frac{n \cdot ( n - 1 )}{2}$ for all positive integer $n \geq 1$.
Use this to obtain, $f ( 1 ) = \frac{200100 - 1000 \cdot 1999}{2000}$
Then prove that $f \left( \frac{1}{n} \right) = \frac{f ( 1 ) - \frac{n - 1}{2 n}}{n} = \frac{2 nf ( 1 ) - n + 1}{2 n^2}$
Hence $f \left( - \frac{1}{n} \right) = \frac{1}{n^2} - \frac{2 nf ( 1 ) - n + 1}{2 n^2}$.
Therefore $f \left( 1 - \frac{1}{n} \right) = f ( 1 ) + f \left( - \frac{1}{n} \right) - \frac{1}{n}$
Whence $f \left( 1 - \frac{1}{n} \right) = f ( 1 ) + \frac{1}{n^2} - \frac{2 n f ( 1 ) - n + 1}{2 n^2} - \frac{1}{n}$.
Plugging in $n = 1999$, gives $f \left( \frac{1998}{1999} \right) = - \frac{35924060979}{39960010}$.
As you guess the complete solution is quite tedious and boring, homever I can post it if you want.


[quote="iura"]Bill, do you know the results of the contest?[/quote]
Did the Canadians participate in the USAMO? :)

[quote="Valentin Vornicu"]Did the Canadians participate in the USAMO? :)[/quote]

Yes we did. :)

This is the first year since long ago that the Americans decided to officially invite Canadians to write the USAMO. But even if we weren't invited, we would still be supposed to do it for the purpose of Canadian team selection.

As far as results are concerned, I think they'll be released on May 5.

I'm glad to see we can write officially again; we were banned starting in 1992, I think.  The reason we were banned was because Canadians were doing so well, a little too well for the Americans' taste.  There was Ravi Vakil who came in around third one year, and I think J.P. Grossman won in 1991.  So, Go Canada Go!


geometry
geometric transformation
rotation

Five people are to be seated around a round table, one seat with Tea, the second with coffee, the third with coke, the fourth with milk and the fifth water. How many ways are there of arraning the five people around the table???

nope sorry its not 4! ....i thought i would trick some of u....

[quote="chess64"]Pre-olympiad? :huh:

[hide]
$5!/5=4!=\boxed {24}$?
[/hide][/quote]

  Gahhh  :lol:  Placing distinct objects around the table means all the rotations are too.  It would be funnier if I didn't fall for it as well...

lol..yeh its 5!...i guess its not relli pre-olympiad but oh well....u neva know

Well, you wouldn't include rotations if the problem was placing them around the table in unique ways with respect to each other. (The problem is a little ambiguous unless it states this one way or the other.)


algebra
polynomial
trigonometry
function
complex analysis
complex analysis unsolved

Let $ p(\cos \theta,\sin \theta)$ be a polynomial in $ \cos \theta$ and $ \sin \theta$.  Show that if $ p(\cos \theta,\sin \theta)\equal{}0$ holds for all $ \theta \equal{} \frac{\pi}{n}$ for all $ n \in N$, then $ p(\cos \theta,\sin \theta)\equal{}0$ for all $ 0\leq \theta \leq 2\pi$.

$ g(z) \equal{} p(\cos(z),\sin(z))$ is zero on a set with an accumulation point in $ \mathbb{C}$, hence is identically zero.

I'm only in the first term of complex variables...  how is that the fact that $ g(z)$ is identically zero follows so easily from it being zero on a set with an accumulation point?

It's a little simpler than that.  $ p \left( \frac {2t}{1 \plus{} t^2}, \frac {1 \minus{} t^2}{1 \plus{} t^2} \right) \equal{} 0$ for $ t \equal{} \tan \frac {\pi}{2n}$, but the above (with denominators cleared) is a polynomial of finite degree in $ t$ with a finite number of roots, hence is identically zero.  We can say the same of $ p \left( \frac{u \plus{} \frac{1}{u}}{2}, \frac{u \minus{} \frac{1}{u}}{2} \right)$ for $ u \equal{} e^{ i \frac{\pi}{n} }$.

[quote="t0rajir0u"]$ p \left( \frac {2t}{1 \plus{} t^2}, \frac {1 \minus{} t^2}{1 \plus{} t^2} \right) \equal{} 0$ for $ t \equal{} \tan \frac {\pi}{2n}$, but the above (with denominators cleared) is a polynomial of finite degree in $ t$ with a finite number of roots, hence is identically zero.[/quote]

I'm a little bit confused about this part. I see that the polynomial in $ t$ is the zero polynomial, but I don't see how that implies that the polynomial in two variables is zero.

For example, $ p(u,v) \equal{} u \minus{} v$ is the zero polynomial in $ t$ if we set $ u \equal{} v \equal{} t$.

Edit: What I say is irrelevant to the problem. Sorry about that.

We are not asked to prove that $ p(u, v)$ is the zero polynomial.  Reread the problem statement.

[quote="ambilatrix"]how is that the fact that  is identically zero follows so easily from it being zero on a set with an accumulation point?[/quote]
It's a standard theorem. An analytic function is defined by its power series, so we can find the Taylor series at an accumulation point $ p$. Suppose it's not zero: we get $ f(z)\equal{}a_k(z\minus{}p)^k\plus{}a_{k\plus{}1}(z\minus{}p)^{k\plus{}1}\plus{}\cdots$, where $ a_k$ is the first nonzero coefficient. This power series has some radius of convergence $ r>0$, so $ |a_n|<A\left(\frac{2}{r}\right)^n$ for some $ A>0$ and all $ n$. If $ 0<|z\minus{}p|<s$, we have $ \left|a_{k\plus{}1}(z\minus{}p)^{k\plus{}1}\plus{}a_{k\plus{}2}(z\minus{}p)^{s\plus{}1}\plus{}\cdots\right|< A\frac{(2s/r)^{k\plus{}1}}{1\minus{}(2s/r)}$. By choosing $ s$ small enough that $ A\frac{2s/r}{1\minus{}2s/r}<|a_k|$, we get that $ |f(z)\minus{}a_k(z\minus{}p)^k|<|a_k(z\minus{}p)^k|$ and $ f(z)$ is nonzero. This contradicts the assumption that $ p$ was an accumulation point, so the Taylor series must be identically zero, and $ f$ is identically zero in a neighborhood of $ p$. This easily extends to the entire connected region containing $ p$ on which $ f$ is defined.

For specific functions, we can often find a simpler argument, such as t0rajir0u's.


(a) Let $f: \mathbb{R}^{n\times n}\rightarrow\mathbb{R}$ be a linear mapping. Prove that $\exists ! C\in\mathbb{R}^{n\times n}$ such that $f(A)=Tr(AC), \forall A \in \mathbb{R}^{n\times n}$.

(b) Suppose in addtion that $\forall A,B \in \mathbb{R}^{n\times n}: f(AB)=f(BA)$. Prove that $\exists \lambda \in \mathbb{R}: f(A)=\lambda Tr(A)$

This is the first time I see this characterization of the trace, but I think that it's fairly well-known.

(a) Let $E_{i,j}$ be the matrices with $1$ in the $(i,j)$ position and $0$ everywhere else. These matrices are a basis of $\mathbb R^{n^2}$, so by the linearity of $f$ and $\textrm{tr}$, it will suffice to consider only this case.

We see that $c_{i,j} = f(E_{i,j})$ works and it's clear that $C$ is unique.

(b) I think that these are true: $E_{i,k} = E_{i,j} \cdot E_{j,k}$ and $0 = E_{j,k} \cdot E_{i,j}$ for all $i \neq j \neq k \neq i$. We get that $f(E_{i,k}) = 0$ for $i \neq k$, so $C$ is diagonal. Now we just need to write the diagonal elements of $AB, \, BA$, from where we deduce the conclusion.


let $ (w_{n}) n\in N$ a sequence such that:
$ w_{n}\equal{}\Large \prod_{k\equal{}1}^{n}{(1\plus{}\frac{1}{k^3})}$
prove that $ (\forall n \in N$*$ )w_{n}<3$

[quote="greatestmaths"]let $ (w_{n}) n\in N$ a sequence such that:
$ w_{n} \equal{} \Large \prod_{k \equal{} 1}^{n}{(1 \plus{} \frac {1}{k^3})}$
prove that $ (\forall n \in N$*$ )w_{n} < 3$[/quote]

$ w_n$ is a strictly increasing sequence.
For $ n>1$, we have $ \ln(w_n)\equal{}\sum_{k\equal{}1}^n\ln(1\plus{}\frac{1}{k^3})$. And so :

$ \ln(w_n)\equal{}\ln(2)\plus{}\sum_{k\equal{}2}^n\ln(1\plus{}\frac{1}{k^3})$

$ \ln(w_n)\leq\ln(2)\plus{}\sum_{k\equal{}2}^n\frac{1}{k^3}$

$ \ln(w_n)\leq\ln(2)\minus{}1\plus{}\sum_{k\equal{}1}^n\frac{1}{k^3}$

$ \ln(w_n)<\ln(2)\minus{}1\plus{}\sum_{k\equal{}1}^{\plus{}\infty}\frac{1}{k^3}$

$ \ln(w_n)<\ln(2)\minus{}1\plus{}\zeta(3)$

$ w_n<e^{ln(2)\minus{}1\plus{}1.202}<2.45<3$

Q.E.D.

[quote="greatestmaths"]i think we can do it such a  inequality. :huh:[/quote]

 :?:

I don't understand what you mean.

[quote="greatestmaths"]let $ (w_{n}) n\in N$ a sequence such that:
$ w_{n} \equal{} \Large \prod_{k \equal{} 1}^{n}{(1 \plus{} \frac {1}{k^3})}$
prove that $ (\forall n \in N$*$ )w_{n} < 3$[/quote]
$ w_{n} \equal{} \Large \prod_{k \equal{} 1}^{n}{(1 \plus{} \frac {1}{k^3})} \equal{} 2\Large \prod_{k \equal{} 2}^{n}{(1 \plus{} \frac {1}{k^3})} \equal{} 2\Large \prod_{k \equal{} 2}^{n}{(\frac {k^3 \plus{} 1}{k^3})} < 2\Large \prod_{k \equal{} 2}^{n}{(\frac {k^3 \plus{} 1}{k^3 \minus{} 1})} \equal{} 2\Large \prod_{k \equal{} 2}^{n}{(\frac {k \plus{} 1}{k}\times \frac {k}{k \minus{} 1}\times\frac {k^2 \minus{} k \plus{} 1}{(k \plus{} 1)^2 \minus{} (k \plus{} 1) \plus{} 1})} \equal{} 2\times\frac {n \plus{} 1}{2}\times\frac{n}{1}\times\frac {2^2 \minus{} 2 \plus{} 1}{(n \plus{} 1)^2 \minus{} (n \plus{} 1) \plus{} 1} \equal{} 3\times\frac {n^2 \plus{} n}{n^2 \plus{} n \plus{} 1} < 3$


During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane.  On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc.  If the candidate went $n^2/2$ miles on the $n^{\text{th}}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\text{th}}$ day?

[hide]Let's solve this problem on a Cartesian plane. Let the man start at the origin. We see that after $40$ days, he will be somewhere in quadrant III. To solve the problem, we need only find the point that he'll be at in $40$ days and apply the distance formula. Now, let's consider the cases whre $n$ is even or is odd seperately. The odd n's contribute to the $x$ coordinate, and the even n's contribute to the $y$ coordinate.

Odd: The final $x$ coordinate is $\frac12 (1^2-3^2+5^2-7^2+\cdots-39^2)=-4(1+3+5+\cdots+19)=-400$.

Even: The final $y$ coordinate is $\frac12(2^2-4^2+6^2-8^2+\cdots-20^2)=-2(3+7+11+\cdots+39)=-420$.

Therefore, our answer is $20\sqrt{20^2+21^2}=\boxed{580}$.[/hide]


Diophantine equation
number theory proposed
number theory

Find positive integers $x, y, z$ such that $x > z > 1999 \cdot 2000 \cdot 2001 > y$ and $2000x^{2}+y^{2}= 2001z^{2}.$

i'm wrong?

let y = 1. the pell equation 2001z² - 2000x² = 1 have the solution (1,1), so, have infinite solutions suficiently bigger!

e.lopes seems to be right though... all conditions seem to be satisfied  :maybe:

[quote="N.T.TUAN"]You not see condition of $x,y,z$. Try again  :wink:[/quote]

hm.. i think that i'm right, my friend!  :blush:

Maybe  :oops: But problem said that find ALL :D

Therefore answer of problem original is...?


calculus
integration
trigonometry
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Given the integral: $ \int_1^2\int_{\frac{3}{\sqrt{x}}}^{\sqrt{3}x}{{(x^2\plus{}y^2)}^{\frac{3}{2}}}dy \; dx$.  Convert to polar and evaluate.

[b]2. Relevant equations[/b]
$ r\equal{}\sqrt{(x^2\plus{}y^2)}$

[b]3. The attempt at a solution[/b]
Ok, I've gotten bounds on $ \theta$, $ \frac{\pi}{6} \le \theta \le \frac{\pi}{3}$.  I'm not sure what the bounds for r should be, otherwise I have the integral:
$ \int_\frac{\pi}{6}^\frac{\pi}{3}\int_{?}^{?}{r^3}dr \; d\theta$

you should make use of

$ x\equal{}r \cos \theta, y\equal{}r\sin \theta$

$ y\equal{}\sqrt 3 x \to r \sin \theta \equal{} \sqrt 3 r \cos \theta \to \theta \equal{} \frac{\pi}{3}\plus{}k\pi$

$ y\sqrt 3 \equal{}x \to \theta \equal{} \frac{\pi}{6}\plus{}k\pi$

for a fixed $ \theta \in [\frac{\pi}{6},\frac{\pi}{3}]$, we integrate over $ r \in [ \sec \theta ,2\sec \theta]$

this is because $ r_{min} \cos \theta \equal{} x_{min}, r_{max} \cos \theta \equal{} x_{max}$

so altogether,
$ \int_1^2\int_{\frac{3}{\sqrt{x}}}^{\sqrt{3}x}{{(x^2\plus{}y^2)}^{\frac{3}{2}}}dy \; dx$

$ \equal{}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\int_{\sec \theta}^{2\sec \theta}r^3(r\sin \theta) dr d\theta$

hopefully


A ball falls into snow and makes a hole of radius $ 12cm$ and of $ 8cm$ height. What's the radius of the ball?

[hide]
Call the radius $ r$. Then the height from the real center of the ball the the bottom of the impression is also $ r$, and the distance between the center of the ball and the "snow level" is $ r - 8$. Therefore, we have a right triangle with leg lengths $ r - 8$ and $ 12$, and hypotenuse $ r$. Pythagorean Theorem, anyone?

\begin{align*} 144 + (r - 8)^2 & = r^2 \\
208 & = 16r \\
r & = 13 \end{align*}

Therefore, the radius of the ball is $ \boxed{13}$.
[/hide]

[quote="The Zuton Force"]:P [b]Follow-up Question:[/b] A ball falls into the snow and makes a hole of radius 8 cm and depth 12 cm. What is the radius of the ball?[/quote]

Obviously, if the depth is longer than the radius, then the radius of the ball is the radius of the impression... $ \boxed{8}$.


Consider a convex quadrilateral $ABCD$. Denote $M, \ N$ the points of tangency of the circle inscribed in $\triangle ABD$ with $AB, \ AD$, respectively and $P, \ Q$ the points of tangency of the circle inscribed in $\triangle CBD$ with the sides $CD, \ CB$, respectively. Assume that the circles inscribed in $\triangle ABD, \ \triangle CBD$ are tangent. Prove that:

a) $ABCD$ is circumscriptible.

b) $MNPQ$ is cyclic.

c) The circles inscribed in $\triangle ABC, \ \triangle ADC$ are tangent.

[img]8502[/img]

The two tangents from a poin P to a circle are equal.

a) $AD+BC = AN+DN+BP+PC = AM+DT+BT+CQ = AM+DQ+BM+CQ = (AM+MB)+(QC+QD) = AB+CD$
Therefore the sum of opposite sides is constant. This is known as a sufficient condition for a convex quadrilateral to be circumscribed.

b) Consider the circles centered at B,D with radii $BM=BT=BP,DN=DT=DQ$. Their distance is the sum of their radii, then they are tangent. Also the two incircles are tangent. And the circles centered at B,D are ortogonal to the incircles. Therefore an inversion with centre T sens this four circles to a rectangle. But M,N,P,Q are the points of the intersection of our circles distinct from T. Then M,N,P,Q are the vertices of a rectangle (after inversion), which is of course cyclic.

c) In order to use the same figure, we suppose that ABCD is circumscribed and prove that the incircles of $%Error. "teignale" is a bad command.
BDA,%Error. "teigngle" is a bad command.
BDC$ are tangent. $\triangle BDA, \triangle BDC$ touch BD recpectively at T,T'
We have $DT-BT = DN-BM = DN+NA-BM-MA = AD-AB = AD-AB+(DC+AB)-(AD+BC) = DC-BC = DQ-BP = DT'-BT'$.
We have $DT-BT=DT'-BT'$. We add to both sides the same quantity $DT+BT = DT'+BT'$ and get $2DT = 2DT'$, or $DT = DT'$: the circles are tangent.

Your solution to b) is very interesting but you kill a fly with a super-laser bazoukas  :)
Here is a simpler one .
We have that $DN=DQ$ , $AN=AM$ , $BM=BP$ , $CP=CQ$ .
The bisectors of the angles $A,\ B,\ C,\ D$ pass through the same point since the quadrilateral is incriptible .But the bisectors of these angles are also perpendicular bisectors of $NQ,\ QP,\ PM,\ MN$ so we are done .

I whish I had a super-laser bazouka now to kill the fly that is buzzing in my room  :D

By the way, with that trick you can kill also IMO 1996 A2.

I found a angle-chasing solution to b
Let R be the point where the circles are tangent
We have  $\angle{RDP}=180-2\angle{RQP}$
Also $DN=DR=DN$ therefore $\angle{PNR}=90-\angle{RQP}$
Similar we have $\angle{MQN}=90-\angle{MNR}$
Hence we get $\angle{MQP}+\angle{MNP}=180$


let T be a triangle with side lengths of 13,14, and 15.  Let S be a set of points inside T that do not lie within a distance of 2 from any side of T.  Find the area of S.

You mean like S is the set of all points inside...?

[hide]Let $I$ be the incenter of $\triangle ABC$.

Let $A',B',C' \in IA,IB,IC$ respectively such that $d(AB,A'B') = d(BC,B'C') = d(CA,C'A') = k$
$(A',B',C'$ inside $\triangle ABC)$

Then
$IAB \sim IA'B'$
$IBC \sim IB'C'$
$ICA \sim IC'A'$

and the ratio of the similarity is $\frac{k}{r}$, where $r$ is the inradius of $\triangle ABC$

Then $\frac{(A'B'C')}{(ABC)}= \left(\frac{k}{r}\right)^{2}$

Now, applying the Heron formula we find that $(ABC)=\sqrt{s(s-a)(s-b)(s-c)}=84$

Hence $r = \frac{84}{\text s}=4$

And $\left(\frac{k}{r}\right)^{2}= \left(\frac{2}{4}\right)^{2}= \frac{1}{4}\Rightarrow $

$(A'B'C') = \frac{1}{4}\cdot 84 = 21$

Note
$(X)$ denotes the area of $X$[/hide]


Patratul ABCD are AB =a, intersectia dintre AC si BD este O.
Cercul cu centru D si de raza DO intersecteaza latura [CD] in U, cercul de centru C si de raza CO intersecteaza latura [BC] in H si latura [CD] in T; cercul de centru B este tangent cercului de centru C(D-T-U-C). Tangenta comuna cercurilor tangente intersecteaza prelungirea laturii [BC] in M.
a) Sa se arate ca lungimea segmentului [BM] este jumatate din lungimea diagonalei [AC].
b) Stiind ca UR este paralela cu BD, unde R apartine laturii [BC], sa se arate ca patrulaterul TORC este inscriptibil.
c) Aratati ca segmentul [TU] reprezinta latura unui octogon regulat cu varfurile pe laturile patratului. Exprimati lungimea segmentului [TU], functie de latura patratului.

Succes!
Va doresc un 2007 plin de succese! Sa va bucurati, asa cum ma bucur eu de fiecare clipa ca si cand maine ar fi ultima zi din viata

asha

Curaj :lol: .

O sa scriu cateva repere:

Cerinta 1). Construiti tangenta pe partea din afara patratului, este mai usor de vizualizat. Razele duse in punctele de tangenta sunt perpendiculare pe tangenta si deci, veti avea doua triunghiuri asemenea cu varful M comun....

Cerinta 2) Una din conditiile ca un patrulater sa fie inscriptibil este legata de suma unghiurilor opuse. Daca veti arata ca unghiul TOR are 90^o, conditia rezulta imediat, deoarece masura unghiului UCR este 90^o.

Cerinta 3)Aici, o metoda ar fi sa completati figura cu cercurile de centre B si A, ambele, cu razele egale cu jumatate din diagonala patratului. Veti avea patru cercuri, toate au punctul O comun. Punctele de intersectie ale acestor patru cercuri cu laturile patratului reprezinta varfurile octogonului regulat ... .(se mai poate si altfel)
Pentru a determina lungimea TU:
Din triunghiul UCR ,de exemplu, folosind teorema lui Pitagora, obtineti
UR^2 = ?   a = 2x + y, unde y = TU si x = UC ...

Domnul Nicula a postat o rezolvare la High School, Pre-Olimpiad,
An interesting geometry problem.

Asha