ceiling function
logarithms
induction
number theory
prime factorization
number theory unsolved

Show that 2^(A-2k) divides (2^(A-3),k), where k is an integer from [2,A/2].
Here (x,y) means x choose y.

Here are my thoughts:
$ (2^{A\minus{}3},k)\equal{}\frac{(2^{A\minus{}3})!}{k!(2^{A\minus{}3}\minus{}k)!}$
Just use that $ \nu_{2}(m!)\equal{}m\minus{}s(m)$ for any integer $ m$ where $ \nu_{2}(\cdot)$ denotes the exponent of $ 2$ in the prime factorization of $ m!$ and $ s(\cdot)$ denotes the sum of the binary digits of $ m$.  Then,

$ E\equal{}\nu_{2}\bigg(\frac{(2^{A\minus{}3})!}{k!(2^{A\minus{}3}\minus{}k)!}\bigg) \equal{} s(k)\plus{}s(2^{A\minus{}3}\minus{}k)\minus{}s(2^{A\minus{}3})$

and we want to show that $ E\ge A\minus{}2k$.  Let $ (k)_{2}\equal{} d_{1}\ldots d_{t}$ where $ d_{i}$'s are binary digits, $ t\equal{}\lceil\log_{2}k\rceil$, i.e. $ (k)_{2}$ denotes the binary representation for $ k$.  Now, $ 2\le k\le A/2$ and $ \nu_{2}( k)\le t$.  Note $ (2^{A\minus{}3})_{2}\equal{}10\cdots 0$ ($ A\minus{}3$ zeros) and $ (2^{A\minus{}3}\minus{}k)_{2}\equal{}01\cdots 1 d_{2}'\ldots d_{t}'$ (with $ A\minus{}4\minus{}\lceil\log_{2}k\rceil$ consecutive ones and arbitrary digits $ d_{i}'$).  So,

$ E\equal{} s(k)\plus{}s(2^{A\minus{}3}\minus{}k)\minus{}1\ge s(k)\plus{}\big( A\minus{}4\minus{}\lceil\log_{2}k\rceil\big)\minus{}1\ge A\minus{}6\minus{}\log_{2}k.$

Now, the RHS is $ \ge A\minus{}2k$ whenever $ k\ge 4$.

My way was to note that when k=2^n is a power of 2, the number of times 2 divides k! is k-1 [1/2 the numbers not greater than k (2^n-1 numbers) can be divided by 2 0 times, 1/2 of what remains (2^n-2 numbers) can be divided 1 time, 2^n-3 can be divided 2 times, and so on until 2^(n-n) can be divided n-1 times. Then add n times you can divide by 2 for 2^n itself and you have all the times you can divide by 2. Prove by induction that 1x2^(n-2) + 2x2^(n-3) + ... + (n-1)x2^(n-n) + n = (2^n)-1.]

This immediately gives that the number of times 2 divides k, where k is any natural number, is not greater than 2k-3 (actually as you say, k-s(k), or k-1, will do). [If k is not a power of 2, take h the least power of 2 such that h>k. The number of times that 2 divides h is h-1. Suppose the number of times 2 divides k is at least 2k-2, then the number of times 2 divides k is greater than 2(h/2)-2, i.e. greater than h-2, i.e. greater or equal to h-1, but it must be less, which is a contradiction.]

Now we can cancel the 2s from the k! which is the denominator of the expression (2^(A-3),k), and the "front term" (2^(A-3)). Let's say there are L of them. Then L is not greater than 2k-3, and A-2k is a nonnegative integer. The power of 2 which the front term is reduced to will be 2^(A-3-L). This is a multiple of 2^(A-2k) iff A-3-L is not less than A-2k, which is true iff L is not greater than 2k-3, which is true. Then (2^(A-3),k) is equal to 2^(A-3-L) multiplied by the fraction {(2^(A-3)-1)...(2^(A-3)-(k-1))}/Q, where Q=k!/(2^L). The prime divisors of Q are contained in 2^(A-3-L){(2^(A-3)-1)...(2^(A-3)-(k-1))}, and none of them are contained in 2^(A-3-L), so they are all contained in (2^(A-3)-1)...(2^(A-3)-(k-1)), i.e. {(2^(A-3)-1)...(2^(A-3)-(k-1))}/Q is an integer. As 2^(A-3-L) is a multiple of 2^(A-2k), this gives the result.


Does anyone know where to find an good online latex compiler?

I am already familiar with this
[url]http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/latex.php[/url]

Please post an url then.

I have listed a number of them at [url=http://sixthform.info/steve/wordpress/?p=59]Online LaTeX[/url] - you'll probably want the second list there for complete documents. One AoPS user has said he has used the LaTeX to PDF link and is happy about it.

Feedback on usage of these sites would be helpful to others.

I've used this one recently: [url]http://dev.baywifi.com/latex/[/url] and it's pretty good. You can upload pictures as well, and if you haven't got texnic centre (or winedt) to hand then it's useful. I've been using it for most of the day today, and the good thing is that you can complie your document and a preview is given in pdf below (which you can also save). Was a life saver today!

I've also [url]http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/latex.php?sprachauswahl=2&aufruf=22103[/url], and I didn't like it too much. Namely because my document had 'too many errors' so it wouldn't build it! Which is obviously not handy when you need an output for your work.  However, if you're an error-less person then you might like it since you can also build your file as a PS file.  (can also upload images as well).

I haven't used them for 'difficult' things, just to type a report, so I don't know how you might like them. But just now I sort of deleted half of my document and the second link still complains of too many errors! Pft. I'd choose the first link anyday!  :D

EDIT: I forgot to add, that you can't be too clever when doing things using an online compiller. :( With the first link, you can experiment with doing certain things but it's not exactly made for that. But good alternatives though.


Prove that you can express every positive integer using only 1's, multiplication, and the function $ p_n$, where $ p_n$ is the $ n$th prime.  Concatenation of 1's is not allowed.

[quote="i_like_pie"][hide]All positive integers have a unique prime factorization. The exception is 1, which is possible because 1's can be used.[/hide][/quote]

[hide="Ah, but..."] The indices of $ p_n$ might not be prime.  How are they expressed?  ;) [/hide]

I didn't think about that, but can't the process be repeated until they are all prime? For example:

$ 86\equal{}p_1p_{14}\equal{}p_1p_{p_1p_4}\equal{}p_1p_{p_1p_{p_1p_1}}$

[quote="i_like_pie"]I didn't think about that, but can't the process be repeated until they are all prime? For example:

$ 86 \equal{} p_1p_{14} \equal{} p_1p_{p_1p_4} \equal{} p_1p_{p_1p_{p_1p_1}}$[/quote]
I think that's the point.  You have to prove that it is always possible, i.e. that is doesn't ever go on infinitely.

The index of a prime is always at least 1 and always less than the prime itself.
There are only a finite number of integers in between, so there can only be a finite number of steps.

[hide]
Strong induction

Inductive hypothesis: Every natural number at most $ n$ can be written as desired.
Base case: $ n\equal{}1$. $ 1\equal{}1$. Done
Inductive step ($ n\minus{}1 \implies n$)
Case 1: $ n$ is prime:
If $ n$ is the $ k^{\text{th}}$ prime, then $ n>k$, so by the inductive hypothesis, $ k$ can be written, then $ n\equal{}p_k$.
Case 2: $ n \equal{} ab$, $ a,b>1$
$ a,b$ can be written as desired by the inductive hypothesis, thus $ n$ can as well.

This gives the following constructions for the first few numbers:
$ 1 \equal{} 1$
$ 2 \equal{} p_1$
$ 3 \equal{} p_2 \equal{} p_{p_1}$
$ 4 \equal{} 2\cdot2 \equal{} p_1p_1$
$ 5 \equal{} p_3 \equal{} p_{p_2} \equal{} p_{p_{p_1}}$
$ 6 \equal{} 2\cdot3 \equal{} p_1p_2 \equal{} p_1p_{p_1}$
$ 7 \equal{} p_4 \equal{} p_{2\cdot2} \equal{} p_{p_1p_1}$
$ 8 \equal{} 2\cdot4 \equal{} p_1p_1p_1$
$ 9 \equal{} 3\cdot3 \equal{} p_2p_2 \equal{} p_{p_1}p_{p_1}$
$ 10 \equal{} 2\cdot5 \equal{} p_1p_3 \equal{} p_1p_{p_2} \equal{} p_1p_{p_{p_1}}$
$ 11 \equal{} p_5 \equal{} p_{p_{p_{p_1}}}$
[/hide]

Yep, Hamster1800 pretty much has the gist of it.


For $ f(x)\equal{}x^3\plus{}px\plus{}q$ have 3 solution $ a,b,c (q \neq 0)$
prove that:   a, $ a^3 \plus{}b^3\plus{}c^3 \equal{}3abc$
                   b, $ p<0$

$ x^{3} \plus{} px \plus{} q \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)$.

Multiplying out the right-hand-side and comparing coefficients,

$ a \plus{} b \plus{} c \equal{} 0$, $ ab \plus{} bc \plus{} ca \equal{} p$ and $ abc \equal{} q$.

Now $ (a \plus{} b \plus{} c)^{3} \equal{} (a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 3 (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 3abc$.

Substituting, we get $ 0 \equal{} (a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 0 \minus{} 3q$;

that is

$ a^{3} \plus{} b^{3} \plus{} c^{3} \equal{} 3abc$.

Let $ S_n \equal{} a^n \plus{} b^n \plus{}c^n$ and $ s_1 \equal{} a \plus{} b \plus{} c, s_2 \equal{} ab\plus{}bc\plus{}ac, s_3 \equal{} abc$
By Newton Sums
$ S_1 \equal{} s_1 \equal{} 0$
$ S_3 \equal{} S_2s_1 \minus{} S_1s_2 \plus{} 3s_3 \equal{} 0 \minus{} 0 \plus{} 3abc \equal{} 3abc$


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="turs12"]Prove that the center of S[size=75]n[/size] (n>2) is the identity subgroup.[/quote]

$ Z(S_n)$ is a normal subgroup.
The only proper normal subgroup of $ S_n$ with $ n \neq 4$ are $ A_n$ and $ \{e\}$ and $ A_n$ isn't commutative...
With $ n \equal{} 4,\ S_n$ have 3 proper normal subgroup: $ A_n,\ V,\ \{e\}$ but $ V$ and $ A_n$ aren't commutative...

It's easy to determine the conjugate classes in $ S_n$ (i.e. the partition of $ S_n$ by the relation $ \sigma_1 R \sigma_2$ iff there exist $ \tau$ s.t. $ \tau\sigma_1\tau^{\minus{}1} \equal{} \sigma_2$. From this you see what $ Z(S_n)$ is

In particular any two permutations are conjugate iff they have the same cycle structure. Since an element is in the center iff its only conjugate is itself, the only element in the center is the permutation with all cycles of length $ 1$, i.e. the identity permutation.


limit
real analysis
real analysis unsolved

Dear everyone.

Please think of the following.

$ a_{1} > b_{1} > 0$, $ a_{n} \equal{} \frac {a_{n \minus{} 1} \plus{} b_{n \minus{} 1}}{2}$, $ b_{n} \equal{} \sqrt {a_{n \minus{} 1}\cdot b_{n \minus{} 1}}$.

It is well known that $ a_{n}$, $ b_{n}$ converge to the same value. Find the value.

In other words, find the limit $ \lim_{n\to\infty}{a_{n}} \equal{} \lim_{n\to\infty}{b_{n}} \equal{} \alpha$.

Thank you for reading my shoddy post. :)

This is classical. See [url=http://en.wikipedia.org/wiki/Arithmetic-geometric_mean]here[/url], for instance.


If you meant $ \frac{a\plus{}b\plus{}c}{b\plus{}c} \plus{} \frac{b}{c\plus{}a} \plus{} \frac {c}{a\plus{}b}$
then your inequality is wrong, because this is equivalent to
$ 1 \plus{} \frac{a}{b\plus{}c} \plus{} \frac{b}{c\plus{}a} \plus{} \frac {c}{a\plus{}b}$ which is $ \geq \frac{5}{2}$

So, i assume that this was not what you meant, in which case i would request you to learn Latex

If it's $ \geq \frac{5}{2}$ it's also $ \geq 2$  :D

oh yeah :oops: then it is correct. however the one i posted is a stronger one

Nesbitt's inequality:

$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge \frac{3}{2}$

Simply expand both left and right side by +1 and we get the inequality we desire.

[quote="akashram"]a+b+c/b+c +b/c+a +c/a+b>=2
[/quote]

[This is not (a + b + c)/(b + c) + b/(c + a) + c/(a + b) >= 2,
but it was asked about it as if it were.]

[quote="akashram"]gokussj3 I think if it is >=5/2 for the l.h.s of my inequality it will not work.Try a=3/4 b=1/2 c=1 yields619/280 =2.21[/quote]

[quote="gokussj3"]If you meant $ \frac {a \plus{} b \plus{} c}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$
then your inequality is wrong, because this is equivalent to
$ 1 \plus{} \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$ which is $ \geq \frac {5}{2}$
[/quote]

askashram,
your version and gokussj3's version are not true, because
if $ a \equal{} 3, b \equal{} 1, c \equal{} \minus{} 2$, then it becomes:

$ \frac {2}{ \minus{} 1} \plus{} \frac11 \plus{} \frac {\minus{}2}{4} \equal{} \frac {\minus{}3}{2}$.

There is no separate restriction given on the category of the variables.
Other than that, from the inequality itself two or more of the variables
cannot equal zero, or none of the pairs of variables in the denominators
can add to zero.


geometry
3D geometry
sphere
abstract algebra
topology
advanced fields
advanced fields unsolved

[quote="Moubinool"]2/ Prove that $\pi_{2}(\mathbb S^2) = 0$[/quote]

$\pi_2$ or $\pi_1$?

Sorry  I edited
Thank you amfulger and grobber

Oh, you don't need to thank me, I just wanted to know what $\pi_2$ was :).

Let X be a topological space. Let $n\in \mathbb{N}$. Let $S^{n}$ be the n-th dimensional sphere (it lies in $\mathbb{R}^{n+1}$).
Fix $x_0$ a point in X and 1 a point in $S^n$.

Then, as a set $\pi_n(X,x_0)$ is the factor set of the set of continuous maps $f:S^n\to X$ such that $f(1)=x_0$ by the equivalence given by homotopies.

However, I can't find the group structure on it for n>1. I read somewhere that for n>1 $\pi_n$ is "naturally endowed" with a group structure which turns it into an [b]abelian[/b] group. I can't see that group structure.

For n=1 it is easy to give the group structure, by concatenating paths.

Thanks! What's $\pi_n$ usually called? I'd like to look for some info on the Internet.

$\pi_1$ is called the fundamental group.
The $\pi_n$'s  are called homotopy groups.

$\pi_1(S^1) \simeq \mathbb{Z}$ is shown in every book on algebraic topology, and very often in general books that focus on point set topology. $\pi_1(S^n) \simeq 0, \quad n \geq 2$ follows from the fact that $S^n$ is simply connected for those $n$, which in turn you can show by using a simplical approximation or just the simple lemma that since $S^n$ can be written as the union of two simply connected open sets ($S^n - \{x\}, \, S^n - \{y\}$ with $x \neq y$ -- simple because they are homeomorphic to $\mathbb{R}^n$) whose intersection is path connected.

[quote="Moubinool"]1/ Prove that $\pi_{1}(\mathbb S^1) = \mathbb Z$
2/ Prove that $\pi_{1}(\mathbb S^2) = 0$[/quote]

$exp : x\in \mathbb{R} \mapsto exp(ix) \in \mathbb{S}^{1}$ is a universal
etale covering of $\mathbb{S}^{1}$. It is galois, of group $\mathbb{Z}$
thus $\pi_{1}(\mathbb{S}^{1} )= \mathbb{Z}$.
Moreover, $\mathbb{S}^{2} - \{P\}$ is isomorphic to $\mathbb{C}$
with the stereographic projection of pole $P$ . Thus, if we consider a loop $c$ in
$\mathbb{S}^{2}$, we can find a point $P$ which does not belong to $c$.
the image of $c$ in $\mathbb{C}$ is equivalent to $0$ as $\mathbb{C}$ is
simply-connected. then, pulling-back the homotopy in $\mathbb{S}^2$,
we get the fact that $c$ is homotopic to $0$.

Fabrice

[quote="koo"]What's an universal étale covering?[/quote]

A universal covering  $ p \colon X \to S$ is a covering, with
$\pi_{1}(X) = {0}$.

Fabrice


$ xyz\equal{}1,x,y,z\ge0$,then
 \[ \frac{1}{\sqrt{4\plus{}5x}}\plus{}\frac{1}{\sqrt{4\plus{}5y}}\plus{}\frac{1}{\sqrt{4\plus{}5y}}\le1\]
$ a,b,c>0$,then
 \[ \sqrt{\frac{7a}{a\plus{}3b\plus{}3c}}\plus{}\sqrt{\frac{7b}{b\plus{}3c\plus{}3a}}\plus{}\sqrt{\frac{7c}{c\plus{}3a\plus{}3b}}\le3\]

the second can be solved by tangent line.

[quote="arqady"][quote="lovegyz1"]$ xyz \equal{} 1,x,y,z\ge0$,then
\[ \frac {1}{\sqrt {4 \plus{} 5x}} \plus{} \frac {1}{\sqrt {4 \plus{} 5y}} \plus{} \frac {1}{\sqrt {4 \plus{} 5y}}\le1\]
[/quote]
Maybe $ \frac {1}{\sqrt {4 \plus{} 5x}} \plus{} \frac {1}{\sqrt {4 \plus{} 5y}} \plus{} \frac {1}{\sqrt {4 \plus{} 5z}}\le1$ $ ?$ :wink:
If so see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150022[/quote]

Can we use Cauchy? :maybe:


How many different functions can be defined from a domain D={a, b, c, d} onto a range R={1, 2, 3}?

wouldnt it be zero b/c you'd have more than one value in its domain paired with a value in its range, which would contradict the defn. of a function.

unless i misinterpreted the problem.

More generally, we can use the Stirling numbers of the second kind.

[url=http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html]MathWorld Link[/url]

Let $f$ be a function mapping $X$ to some subset of $Y$ (where $X$ and $Y$ are sets).  In symbols, we write $f: X\to Y$.  We say that $f$ maps $X$ onto $Y$ if $f(X)=Y$.  For example, $e^x$ maps $\mathbb{R}$ onto $(0,\infty)$, but does not map $\mathbb{R}$ onto $\mathbb{R}$.

[quote="Boy Soprano II"]To arkmastermind :  No; you're mixing up domain and range.  For example, $f(x)=1$ is a function mapping an infinite set onto a set of cardinality one.

[hide]Since this is an onto function ("onto", "range", and the problem would just be trivial otherwise), each element of the range must be used once, which means that exactly one value of the range must be used exactly twice.  Without loss of generality, let us call it $1$.  If this is the case, then there are $\displaystyle \frac{4!}{2!}=12$ possible functions.  But the element of the range used twice could be anything, so our answer is $12\cdot 3=\mbox{\fbox{36}}$.

Q.E.D.

(Of course, if the function does not have to be onto, then we have $3^{4}=81$.)[/hide][/quote]

oh right. should have read more carefully.  :blush:


[color=brown][size=200]Learning to Members....[/size][/color]
[size=150]Let a,b,c  be positive real numbers,and a+b+c=3. Prove that the following inequality holds[/size]
$ \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2}\geq (\frac {2}{3} \plus{} \frac {\sqrt {2}}{3})(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}) \minus{} \frac {1}{4} \minus{} \frac {\sqrt {2}}{2}.$
BQ

That doesn't make sense. You seemed to make a typo.

Oh, you have just edited the problem.

First you should say ''thank you '', shouldn't you? :wink:

I like you...， Just to see the time，I thought you were Ironically to me.
BQ

Whether it is meaningful，wait testing...

let $ f(t) \equal{} a^t \plus{} b^t \plus{} c^t$,and$ f(1) \equal{} 1$,prove that:
$ f(1) \plus{} f(3) \plus{} f(5)\geq f(2) \plus{} f(4) \plus{} f(6)$
BQ

$ f(u) \plus{} f(v) \plus{} f(w)\geq f(p) \plus{} f(q) \plus{} f(r)$,
if$ p \plus{} q \plus{} r\geq u \plus{} v \plus{} w$?
BQ

Let a,b,c,d be positive real numbers,and let $ f(t) \equal{} a^t \plus{} b^t \plus{} c^t \plus{} d^t$,and$ f(1) \equal{} 1$,prove that:
$ f(1) \plus{} f(3) \plus{} f(5)\geq f(2) \plus{} f(4) \plus{} f(6)$
Can be extended to n variables?

BQ

[quote="kunny"]Don't do Spam! :mad:

Japanese Communities Moderator

kunny[/quote]

Wronged me

BQ

Let a,b,c be positive real numbers,and let $ f(t)\equal{}a^t\plus{}b^t\plus{}c^t$,and$ f(1)\equal{}1$,prove that:
$ f(\minus{}1)\plus{}f(\minus{}3)\plus{}f(\minus{}5)\le f(\minus{}2)\plus{}f(\minus{}4)\plus{}f(\minus{}6)$

Can be extended to n variables?

Our thinking should be open to out,
please see:

$ f(t) \equal{} (1 \minus{} a^t)(1 \minus{} b^t)(1 \minus{} c^t),a \plus{} b \plus{} c \equal{} 1,$
$ f( \minus{} 1) \plus{} f( \minus{} 3) \plus{} f( \minus{} 5) \minus{} f( \minus{} 2) \minus{} f( \minus{} 4) \minus{} f( \minus{} 6)\geq 0$
$ f(1) \plus{} f(3) \plus{} f(5) \minus{} f(2) \minus{} f(4) \minus{} f(6) \le 0,$
$ a, b, c$ of the meaning  Same as above.

BQ

may be Extended to the four variables.
BQ

Stop it!

What is on earth, what do you want to do in this forum?

Sorry, my sequential posting.

This should be locked.

Japanese Communities Moderator

kunny

Dear kunny:
      you see,this inequality is nice:


search
inequalities unsolved
inequalities

Try the general subbing. Official soltion may have been posted....

Bomb

Problems from recent IMO Shortlists are often already posted on ML, so you can just search for them:

http://www.mathlinks.ro/Forum/viewtopic.php?t=30865

  darij


could someone explain me what balls and urns is and where to use it. In words, thx :lol:
i didnt understand in the wiki. And i wanna know if it's useful in MC?

well does MC have questions that can be solved with balls and urns?

Certainly. I put a couple of last year's state targets on that link. They were #2 and #5.


Would it be possible to use the subject class credit from an independent study course towards a problem series course in the winter?

Certainly.  You can also save it and use it on a subject class to be named later.

In the Independent Intermediate Class A description it says that

Assigned problems and readings for this course will come from Volumes 1 + 2 of the Art of Problem Solving and from the Art and Craft of Problem Solving by Paul Zeitz

Is there a paperback edition of the Art and Craft of Problem Solving by Paul Zeitz, or is the $60 hardback the only version available?

Unfortunately there is no paperback option for Art and Craft of Problem Solving.  The book is a little bit pricey, but it's a great book and worth every penny.


rotation
geometry
rectangle
geometric transformation

Source: Sprint 92-93, number 9:

Seven points are equally distributed on a circle.  How many non-congruent quadrilaterals can be drawn with vertices chosen from among the seven points?

And how would I do a quesiton similar to this?

And: Source: Countdown 98-99, number 2:

Twenty switches in an office computer network are to be connected so that each switch has a direct connection to exactly three other switches.  HOw many connections will be necessary?

I can do this using guess and check in 10 minutes... need a way to do it in 45 sec. thanks

Mr.Ocax

I did this by drawing a couple circles but also listing the pattern of consecutive points versus points skipped.

You can use 4 in a row(any four will give you the same quad)

You can use three in a row(if you skip one, you are skipping 2 on the other side, so only one way to do this)

You can do just 2 in a row, and then skip one, and then choose to do two in a row or one and then skip one again.(if you skip 2 after the 1st two, you are forced to do 2 in a row or you are back to the beginning and that looks just like skipping one and choosing 2 in a row anyway)

Those are the only 4 ways.

The second problem I think is very different as they are not in a circle.  Circles change everything :)

I don't know the answer to the second one, but I would say this :

Each of the 20 need three connections, so 20x3 =60, but since connections go both ways you need to divide by 2.  So you need 30 connections.

I think that makes sense :)

yeah, the last one makes great sense.

question on number 1 again, how do you know if it wants the number of shapes you can get, or the total number of figures you can get?

I tried this by doing 7C4, and dividing by 7 (# of ways to rotate it), but this doesn't work in this case, because some figure gets counted twice this way (although this is the traditional method).

[quote]Seven points are equally distributed on a circle.  How many non-congruent quadrilaterals can be drawn with vertices chosen from among the seven points?[/quote]
Let the distance between the points be 1. Each partition of the 7 units makes non-congruent quadrilaterals. So find how many partitions of 7 have exactly 4 members.

[quote]
Twenty switches in an office computer network are to be connected so that each switch has a direct connection to exactly three other switches.  HOw many connections will be necessary?

I can do this using guess and check in 10 minutes... need a way to do it in 45 sec. [/quote]Each switch connects to 3 others. There are 20 switches. Each connection gets counted 2 times. So 3*20/2=30.

i did what monkey did on the first one, but it doesn't seem to work.

And solafidefarms, what do you mean partitions?  are you talking about a rectangle? because that wouldn't work with this one.. they are on circles.  I guess I just don't get what you're trying to point out...

Mr.ocax

Let me explain some more. You have 7 units on the edge of the circle, formed by the 7 points. Any geometric construction with its vertices on the points must necessarily have a integer number of units between its vertices. The numbers of units between the vertices total is 7. Think about it, and try to draw a figure that doesn't have 7 units total between the vertices so long as the vertices are on the evenly spaced points.

If a figure has the same distances between its vertices as another, those 2 are equal, either by flipping or rotation or both. Try it.

We want quadrilaterals, which have 4 vertices, dividing the 7 units into 4 sections. Now if we carefully find the partitions of 7, those are the possible noncongruent figures. We only want partitions that have 4 members, so that there are 4 distances between points, so that we form a quadrilateral.

Billy

wow, ok... woo, that was pretty brilliant.  Thanks!!

Mr.Ocax


integration
induction
calculus
derivative
function
Pascal\u0027s Triangle
algebra unsolved

[u]    [i] Problem [/i][/u] :
 [i]The expansion of
  $ f(x)$ = $ ( 1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{1000}) ^{1000}$

  is the polonomial $ f(x) \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^2 \plus{} ...... \plus{} a_{10^6}x^{10^6}$

   Calculate $ A \equal{} a_{0} \plus{} a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{1000}$[/i]

    [i]Scorpius_119 , Nayel , ....Let  perform [/i]:wink:

  [i] Love An Forever[/i]

  [i]I can't buy a  gift  for my love but i think she likes this proposed problem[/i] :blush:

Nice problem!
[hide="Not very formal beginning"]Since $ A$ is the sum of only the first $ 1000$ elements, the result is not changed if you substitute   $ f(x)$ by
$ g(x) \equal{} (1 \plus{} x \plus{} x^2 \plus{} \dots)^{1000}$.[/hide][hide="Not very formal ending"]The first coefficients of $ (1 \plus{} x \plus{} x^2 \plus{} \dots)^n$ are:
$ n \equal{} 0$: $ 1,0,0,0,0,\dots$
$ n \equal{} 1$: $ 1,1,1,1,1,\dots$
$ n \equal{} 2$: $ 1,2,3,4,5,\dots$
$ n \equal{} 3$: $ 1,3,6,10,15,\dots$
Notice this is Pascal's triangle, and voilà, the solution is
$ \binom{2000}{1000}$[/hide]

[i]    I think the solution of  Lordwings is not nice
        Try a grreater solution :wink: [/i]

More formal $ \neq$ more beautiful.

I'll use that, for $ n\in\mathbb{N}$
$ S_n(x) \equal{} \frac {1}{(1 \minus{} x)^n} \equal{} \sum_{k \equal{} 0}^{\infty}{\binom{n \minus{} 1 \plus{} k}{k}x^k}$.
[hide="Proof by induction"]It's true for $ n \equal{} 1$:
$ S_1(x) \equal{} \frac {1}{1 \minus{} x} \equal{} \sum_{k \equal{} 0}^{\infty}{x^k}$.

Now, for the induction step, we'll prove that
$ S_{n \plus{} 1}(x) \equal{} \frac {1}{(1 \minus{} x)^{n \plus{} 1}} \equal{} \frac {S_n(x)}{1 \minus{} x}$.

$ (1 \minus{} x)S_{n \plus{} 1}(x) \equal{} (1 \minus{} x)\sum_{k \equal{} 0}^{\infty}{\binom{n \plus{} k}{k}x^k} \equal{}$
$ \equal{} \sum_{k \equal{} 0}^{\infty}{\binom{n \plus{} k}{k}x^k} \minus{} \sum_{k \equal{} 0}^{\infty}{\binom{n \plus{} k}{k}x^{k \plus{} 1}} \equal{}$
$ \equal{} \sum_{k \equal{} 0}^{\infty}{\binom{n \plus{} k}{k}x^k} \minus{} \sum_{k' \equal{} 1}^{\infty}{\binom{n \plus{} k' \minus{} 1}{k' \minus{} 1}x^{k'}} \equal{}$
$ \equal{} 1 \plus{} \sum_{k \equal{} 1}^{\infty}{\left(\binom{n \plus{} k}{k} \minus{} \binom{n \plus{} k \minus{} 1}{k \minus{} 1}\right)x^{k}} \equal{}$
$ \equal{} \sum_{k \equal{} 0}^{\infty}{\binom{n \plus{} k \minus{} 1}{k}x^k} \equal{} S_n(x)$,

where I've used the main property in Pascal's triangle:
$ \binom{n \plus{} k \minus{} 1}{k \minus{} 1} \plus{} \binom{n \plus{} k \minus{} 1}{k} \equal{} \binom{n \plus{} k}{k}$.[/hide]

Now,
$ f(x) \equal{} \left(\frac {1 \minus{} x^{1001}}{1 \minus{} x}\right)^{1000} \equal{} (1 \minus{} x^{1001})^{1000}S_{1000}(x) \equal{}$
$ \equal{} (1 \minus{} 1000x^{1001} \plus{} \dots)S_{1000}(x)$.

$ S_{1000}(x) \minus{} f(x) \equal{} x^{1001}(1000 \plus{} \dots)$.
Therefore, they share the first $ 1001$ coefficients:
$ a_i \equal{} \binom{999 \plus{} i}{i}$, $ 0\leq i \leq 1000$.

Finally,
$ A \equal{} \sum_{i \equal{} 0}^{1000}a_i \equal{} \sum_{i \equal{} 0}^{1000}\binom{999 \plus{} i}{i} \equal{} \binom{2000}{1000}$,
using the [url=http://pages.csam.montclair.edu/~kazimir/patterns.html]hockey stick pattern[/url] in Pascal's triangle.

[i] Is there a solution not use generating function?[/i] :maybe:

[quote="aviateurpilot"]$ a_k \equal{} |\{(y_1,y_2,...,y_{1000})\in [|0,1000|]: \ \sum_{i \equal{} 1}^{1000}y_i \equal{} k\}| \equal{} C_{...}^{...}$
gives $ A \equal{} \sum_{k \equal{} 0}^{1000}a_k \equal{} |\{(y_1,y_2,...,y_{1000})\in [|0,1000|]: \ \sum_{i \equal{} 1}^{1000}y_i\le 1000\}|$[/quote]

For a nice solution not using generating functions just continue aviateurpilot's solution:
$ A \equal{} |\{(y_1,y_2,...,y_{1001})\in [|0,1000|]: \ \sum_{i \equal{} 1}^{1001}y_i \equal{} 1000\}|$
(simply set $ y_{1001} : \equal{} 1000 \minus{} y_0 \minus{} \ldots \minus{} y_{1000}$), but the number of integer solutions
of an equation of the form $ y_1 \plus{} \ldots \plus{} y_n \equal{} r$ with $ y_i \ge 0$ is well known to be
$ \binom{n\plus{}r\minus{}1}{n\minus{}1}$. (Represent a solution of the equation by a string of $ y_1$ 0s, a 1,
$ y_2$ 0s, a 1, \ldots, then what you get is a string of $ n\plus{}r\minus{}1$ digits, of which $ n\minus{}1$ are 1s.
so the number of these strings is the above binomial coefficient.)
So the answer is $ A \equal{} \binom{2000}{1000}$.

It is the number of non-negative integer solutions to

$ x_1\plus{}x_2\plus{}...\plus{}x_{1000}\le 1000$...

add an extra parameter...it is $ a\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{1000}\equal{}1000$

which has $ \binom{1001\plus{}1000\minus{}1}{1000}$ solutions.

Think of $ A_i$ = {$ a_{i,1},a_{i,2} \ldots a_{i,1000}$}, We are goint to choose $ i(i\in [0,1000])$ elements. We each time can choose $ x_i$ elements from $ A_{i}$. For the number $ i$, we've $ j(j\in [1,i])$ choices. It's the number of the positive solutions of $ \sum_{k \equal{} 1}^{j}{x_j} \equal{} i$, clearly it's $ \binom{i \minus{} 1}{j \minus{} 1}$. And then we put the elements into 1000 boxes. Obviously it's $ \binom{1000}{j}$. And we should add 1 more as $ a_0$.
So it's

$ \sum_{i \equal{} 1}^{1000}{\sum_{j \equal{} 1}^{i}{\binom{i \minus{} 1}{ i \minus{} j} \binom{1000}{ j}}} \plus{} 1$

$ \equal{} \sum_{i \equal{} 0}^{1000}{999 \plus{} i \choose i}$

$ \equal{} \binom{2000}{1000}$  :lol:

Maybe my solution's ugly :blush:

Lordwings ; your solution is very nice and natural , Congrats

    After learning generating function , I understood your solution completely

  But my initial solution and the official are rather similiar to Altheman's one

  Anyway , thanks you :D

  And Rust : What an outstanding solution :blush


calculus algebra polynomial inequalities search

nice! :D

my solution is almost the same. i use the fact that $n!>2^{n}$, for $n\ge 4$.

[quote="mets501"][hide="Better solution :-)"]
The well-known Maclaurin series for $e^{x}$ is
$e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots$
Substituting 1 for $x$, you get
$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots$

It follows (as $e\approx 2.718 < 2.8$) that
$1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots <2.8$

Q.E.D.
[/hide][/quote]

um that's like saying well i cheated and knew the exact value for e and it's less than 2.8 so i win

[quote="mets501"][quote="13375P34K43V312"]um that's like saying well i cheated and knew the exact value for e and it's less than 2.8 so i win[/quote]
Yes, but that is all the problem is asking.  If the problem was to prove that $\pi$ is greater than $3$, is it not significantly rigorous to write $\pi \approx 3.1415926$, so therefore $\pi > 3$?[/quote]

I think you thought it was showing that the expression on the LHS equals $e$, but the title of this thread is "e<2.8" so it assumes you know that, I guess. But I am understanding your logic as well. What if we change the problem then, to:
[quote] Prove that
\[e < 2.8. \]
[/quote]

[quote="me@home"]In conclusion, this problem is too hard for HSB and the intended solution is of the style mentioned by Nakagawa and (of course) nayel, which are pretty nice. In fact calculating $e$ works on this identity (usually) so it is pointless to replace the lhs with $e$.[/quote]

I disagree
[hide]
Suppose we already know that $e<3$
Let's recall from our calculus that $|R_{n-1}(x)| \leq |\frac{f^{(n)}(x)(x-c)^{n}}{n!}|$ (This is called Lagrange Error if you haven't had calculus yet).
Let's consider the 5th degree polynomial
$R_{5}(1) \leq \frac{e}{5!}< \frac{3}{120}$

Thus, $e < 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{3}{120}< 2.8$

So in other words, we see that if we know a bound on $e$, we can use it to prove a sharper bound on $e$.
[/hide]

But obviously the other solutions are better.

i luv the solutions....beautiful.

but dont u think that nayel's is not for hsb.

idk.

sry. i meant mets501's solution.

great idea, though.

Another method is to use to use the inequality
\[e < \left( 1+\frac{1}{n}\right)^{n+1}, \]
where $n$ is a positive integer.  Then plug in say $n = 20$.

okey the solutions are nice. but i just wanted a simple and elementary proof of the inequality \[1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots <2.8\] without assuming that the $LHS=e$, assuming even that you don't know what $e$ is, because i think everyone knows that. and if you know it, the problem is obvious.

[quote="kunny"][quote="Kouichi Nakagawa"]I was separated from a problem about the Napier Number, but remembered that there were the following problems by a problem about circle ratio.

Prove that pi is greater than 3.05.
(Tokyo University entrance examination/science 2003, Problem 6)

That is to say Prove that $\pi>3.05$

Maybe, This problem may be introduced somewhere.[/quote]

You can see here.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1081288167&t=17871[/url][/quote]

Thank you :)


Where in NJ is everyone from?

Town? County? Don't have to give town lol if you feel that is too specific.

Me?

SB representing.

I'm not from New Jersey but I voted for Mercer anyway because I go to Princeton. :P

I used to live in Fort Lee, Bergen Country for 6 years. Just 5 minutes from NYC. :D

Now I live 30 minutes from SF.

Used to live in Germany, moved to Middlesex when I was 2. Anyone know Dr. Doehner?

[quote="ssslee"]I live in middlesex county and im in a small town called plainsboro.[/quote]
haha, I live in Plainsboro too!  :D However, I used to live Edison,Nj, and before that, in Utah and before that, China.
You'll form your own opinion or Dr.Doehner when you meet here...

[quote="QuantumTiger"]I know Dr. Doehner. :D I live in Princeton.[/quote]
How do you know her?

I went to middle school where she was. She was my math teacher. :P Then I moved to Princeton.

lol, I had her as a teacher last year and I'm having her again next year. xD
Did you go to community or grover?

[quote="neelnal"]How do people vote for "Did I miss any?"

I put all 21 counties!!![/quote]
Maybe they made a new county?


trigonometry
function
LaTeX
ceiling function
logarithms
modular arithmetic
\/closed

Okay you know you have to set the MODE to Degrees when computing trig functions? When I started to graph them in the Degree mode the graphs are linear...when I turn it back to Radians, the graph is perfect...

Anyone help me?

Because Trig functions have to be in radians when grpahing methinks

Try using the ZTrig zoom option. It will change your display window settings so that trig graphs look nice (and it will work for both degrees and radians). You can get to it by pressing the "ZOOM" button at the top of your calculator and then choosing option 7. After you're done with graphing trig functions and you want to go back to the original settings, just set it to ZStandard (option 6).  :)

EDIT: By the way, this isn't quite the right forum for this topic. It's not a big deal, but next time, try the Round Table, or maybe the Games and Fun Factory.

[quote="solafidefarms"]Did you change your Xmin,Xmax so that they range from 0 to 360 instead of 0 to 2pi?[/quote]

Yea I followed the EXACT directions given by my math book (settings and stuff) but it doesn't come out right.

Then during science second period I finally saw the ZTrig option  :blush:  :ewpu:

[quote="chess64"][quote="nutz_for2.718281828"]Try zooming out. Besides, sin(180)=1 :wink:[/quote]

rotfl, no it isn't[/quote]

I think he meant

$\sin(90^\circ)=1$

@chess64, do you think it's possible to get that uber awesome problem I showed u the other day on LaTeX?

this is all i remember...
\[9x\equiv-\left\lceil\left(\sqrt[6]{ \frac{ \left(\sum_{i=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{i}\right) \binom{3\sqrt{9}}{9}\left(\log_{\lceil .476 \rceil}e-1.7\right) }{ (\sin^{2}x+\cos^{2}x)(\sec^{2}x-\tan^{2}x) }}\right)^{\pi_{1337}\right\rceil \pmod{10}}\]

[code]
9x\equiv -\left\lceil\left(\sqrt[6]{
 \frac{
  \left(\sum_{i=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^i\right)
  \binom{3\sqrt{9}}{9}
  \left(\log_{\lceil .476 \rceil} e-1.7\right)
 }{
  (\sin^2 x + \cos^2 x)(\sec^2 x - \tan^2 x)
 }
}\right)^{\pi_{1337}\right\rceil \pmod {10}
[/code]

[quote="chess64"]
[hide]

this is all i remember...
\[9x\equiv-\left(\sqrt[6]{ \frac{ \left({\sum_{i=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{i}}\right) \binom{3\sqrt{9}}{9}}{ (\sin^{2}x+\cos^{2}x)(\sec^{2}x-\tan^{2}x) }}\right)^{\pi_{1337}}\pmod{10}\]
[code]
9x\equiv -\left(\sqrt[6]{
 \frac{
  \left(
   {\displaystyle\sum_{i=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^i}
  \right)
  \binom{3\sqrt{9}}{9}
 }{
  (\sin^2 x + \cos^2 x)(\sec^2 x - \tan^2 x)
 }
}\right)^{\pi_{1337}} \pmod {10}
[/code][/hide][/quote]

so the summation doesn't come out rite?

edit:

$9x\equiv-\left\lceil\left(\sqrt[6]{ \frac{ \left({\sum_{i=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{i}}\right) \binom{3\sqrt{9}}{9}}{ (\sin^{2}x+\cos^{2}x)(\sec^{2}x-\tan^{2}x) }}\right)^{\pi_{1337}}\right\rceil\pmod{10}$

[quote="chess64"]oh yeah didn't you have some log thing in there?[/quote]
yeah

log \lciel .476 \rciel e-1.7.....don't know it in latex lol


[url=http://www.4aunlimited.co.nr]Just go to my website, which has great games. Try it out.[/url]

cool, but isnt it kinda redundantly AoPS-like?

click on the links in the top navigation bar(frame). Then click on old site on the links that are listed. it should take you to a games page.

nice games, but breakout doesn't work, and i lost all 12 races i bet on in crab racer :mad:


Let x,y,z be positive real numbers. what is the maximum area of rectangle $ABCD$given that it has interior point P such that the lenghs of $AP,BP,CP$ is x,y,z,respectively?

Without loss of generality we assume x <= y <= z.
Firstly my concern would be to show that one such construction is possible and then maximize the area of the rectangle. Now trying to maximize the distance between the points [b]A[/b] and [b]B[/b] and then increasing the distance of [b]C[/b] along the  :perp:  at [b]A[/b] would give us a rectangle as shown in the fig 1. In this case the point [b]P[/b] lies on the side [b]AB[/b]. Moving the point [b]P[/b] to a configuration as shown in the fig 2 is the case we are intrested in. Refering to the fig2 we have the area of the rectangle:
 [b]R[/b]= AC.AB= [a+:sqrt:( z2-x2+a2)][ :sqrt:( x2-a2) +  :sqrt:(y2-a2)].
Using the condition for maxima [b][i]dR/da =0[/b][/i] we have.
a= xy/ :sqrt: (y2 +z2). Hence the maximum area.

I hope my arguement is correct. Enjoy problem solving.
Kalyan


Er der nogen der kan hjælpe mig med følgende

1) Vis at $ x^{5}\minus{}5x\plus{}1$ er irreducibelt i $ \mathbb Q[X]$. (Man kan f.eks betrage $ f(x\minus{}1))$. Angiv antallet af reelle rødder(nulpunkter) i $ f(x)$.

2) Lad $ M$ være spaltningslegemet for $ f(x)$ over $ \mathbb Q$. Vis at $ M/\mathbb Q$ er en normal udvidelse og bestem $ Gr(M/\mathbb Q)$. Er $ f(x)$ opløselig ved rodtegn? Det kan bruges at diskriminanten for $ f(x)$ er -3\cdot5^6\cdot17.

3) Vis at $ \mathbb Q(\sqrt{\minus{}51})$ er indeholdt i $ M$ og undersøg, om $ f(x)$ er irreducibelt i $ \mathbb Q(\sqrt{\minus{}51})[X]$.

4) Bestem Galoisgruppen $ Gr(M/\mathbb(\sqrt{\minus{}51}))$.

var ikke dette på en algebra eksamen på UiO?

kanskje du finner et løsningsforslag.


What is the correct name for the compound with the formula $ CrPO_{4}$?

(1) chromium(II) phosphate
(2) chromium(III) phosphate
(3) chromium (II) phosphide
(4) chromium(III) phosphide

$ \bullet$ How would you figure out whether $ PO_{4}$ was phosphate or phosphide?
$ \bullet$ How would you determine the charge of $ PO_{4}$?  I need this to determine whether the name is chromium(II) or chromium(III).  (The charges of the entire compound is supposed to add up to $ 0$.)

Thanks in advance!

In general, I have not seen people make a distinction between "Chromium (III) Phosphate" and "Chromium(III) Phosphate", although I would probably use the first one.

It is important to learn to recognize common [url=http://www.standnes.no/chemix/examples/polyatomic-ions-list.htm]polyatomic ions[/url].

[quote="inteluser"]In general, I have not seen people make a distinction between "Chromium (III) Phosphate" and "Chromium(III) Phosphate", although I would probably use the first one. [/quote]

Oops, Sorry - Edited.

[quote="Carcul"]- the phosphide ion is P3-

- the charge of PO4 is 3-

- therefore, the correct name is chromium(III) phosphate[/quote]

How do we know this?

We know that the charge of phosphate PO4 is -3 is because it's a polyatomic ion. You just need to remember that.
The charge of phosphide is -3 is because of its position on the periodic table.For example, chloride is -1, oxide is -2 and thus, phosphide is -3. However, -3 is just the most common charge of phosphide. it can also have other charges beside -3.

Thanks a lot for everyone's help!  :lol:


@alan: Sorry, my first equation doesn't make sense...if I didn't misled you, you would have got it right in about..30 seconds.. My apologies..

It was really easy..

First you convert quarter-mile and third-mile into just miles and find out costs per mile.
That yields this:
Letting $ B_t\equal{}$Blue taxi and $ Y_t\equal{}$Yellow taxi,

$ B_t\equal{}140$cents per mile

$ Y_t\equal{}90$cents per mile

Then you set up an equation like this: $ B_t\equal{}Y_t$
Changing dollar into cents, and letting $ m$=miles yields:

$ 125\plus{}140m\equal{}175\plus{}90m$

Solve to get $ m\equal{}1$

Then you add $ \frac12$ miles which was given, so the answer is $ \boxed{1 \frac12}$

lol, I don't think so either... We have an Alan on our team.


superior algebra
superior algebra unsolved

Prove that

There is an exact sequence

$ 0 \rightarrow \mathbb{Z}/d \rightarrow \mathbb{Z}/n  \stackrel{m}{\rightarrow} \mathbb{Z}/n \rightarrow \mathbb{Z}/d \rightarrow 0$,

where d = gcd(m, n).

Any hint or help will be appreciated.

You are given one of the maps, and you need to construct two more.  Which one are you having trouble with?


Here is a beautiful theorem:
Every simple group has a cyclic Sylow subgroup.

I suspect that there is no "relatively easy" proof, but I wanted to check in the forum in case I am wrong.

This is a post-classification result.

This review (not the paper) says this explicitly, http://www.ams.org/mathscinet-getitem?mr=1272370

It was listed as an unsolved problem in the 8th Kourovka notebook (1982-1983).

Even a little more is known: whenever a finite simple group has a cyclic Sylow p-subgroup, its Sylow p-subgroups intersect trivially in pairs.  When the Sylow p-subgroup has order p this is trivial, but it is important for larger cyclic groups that occur in groups of Lie type.

Blau, Harvey I. "On trivial intersection of cyclic Sylow subgroups."
Proc. Amer. Math. Soc. 94 (1985), no. 4, 572-576.
http://www.ams.org/mathscinet-getitem?mr=792262
http://dx.doi.org/10.2307/2044865

Thanks JackSchmidt,
So we need the classification results indeed to prove this theorem. Needless to say it is not easy then.


[quote="José"]Find all real numbers $a,b$ such that $a^2+b^2\geq 2ab$[/quote]
Nice joke :rotfl:

Well, it is a "joke" because all the real numbers work! ;)

Indeed : $a^2 + b^2 \ge 2ab \iff (a-b)^2 \ge 0$, and the latter is always true! :)

:rotfl: .
$\frac{a}{b}+\frac{b}{a}\geq2< \Rightarrow  a^2+b^2\geq 2ab$.

I think is it's a joke.Anywhay, here's a new proof.

[quote="José"]why? A friend of my told me the problem....[/quote]
You have really good friends. :rotfl:

Guys how you have liked this joke :lol: ,you have so much discussed it that I desided  to post another joke
[b]Find all real $x$ such that $x^2\geq 0$ [/b]:D  :rotfl: .
I hope you will like this one too. :D

[quote="stancioiu sorin"]Find all positive real numbers x such that -x<0. :rotfl:  :rotfl:[/quote]
Jose,please change the name of this topic and write
[b]Collections of math jokes[/b] :D

Here is a good one:
Find all solution for a and b real:
$a^3+3a^2b+3b^2a+b^3>(a+b)^3-(a-b)^2-0.00000000000000001$
 :rotfl:  :rotfl:

this is great  :rotfl:  :rotfl:
what that man did on IMO?

Well, a new joke:
\[ i^2=-1  \]
have fun! :D

We consider a quadrilateral ABCD with AB=a,BC=b,CD=c,AC=d and BD=e.Find all real positive numbers such that de=<ac+bd. :rotfl:

All the while I was working my way down this page, I was wondering where this one hid all this time. Now I get it. Good job, Arne! But we should not laugh at Jose anyway. He must have missed the point somehow, or is playing a bigger joke on us. Either way, it's prudent to keep straight faces. :|
[hide="oops"] :D sory I could not resist[/hide]

HOW FUNNY YOU ALL ARE!!!!!!!!! PLEASE DON'T MAKE ME LAUGH!!!! I CAN'T STOP LAUGHING SINCE LAST MONTH!!!!

What a coincidence! So can't I!! :lol:
Leaving out the school exams of course though. :(

"Let two points in the plane be non-collinear ..."
"Let three points in the space be non-coplanar ..."

 :rotfl:

EDIT: Don't take it as an offense José, on the 1st page, you can see why your friend's statement is always true.
Now, your initial topic has completely derived, and it is a topic where you put all your math jokes, nothing against you! ;)

Let two circles cut each other at five points. Prove that at least three of these coincide. :P

False. The two circles coincide.

i!/(u!(i-u)!) , pikachu!

[quote="Werwulff"]So u guys like jokes?

This one is not an inequality, but nevertheless killingly funny:

At last years IMO preparation, we were told to calculate \[ \sum_0^\infty\frac{n^2}{2^n}.  \]
NoName (Christian Sattler) seriously proposed solving it this way:
\[ \sum_0^\infty\frac{n^2}{2^n}=\sum_0^\infty n^2-\sum_1^\infty\frac{n^2}{2}-\sum_2^\infty\frac{n^2}{4}-\dots  \]
 :rotfl:[/quote]

Can you explain that, please? I don't really understand the summations, if that's what they're called. Sorry.  :blush:

Find all possible positive integers a,b such that $\frac{|a|}{a} = \frac{|b|}{b}$.

 ;)

[quote="mathnerd314"]Find $\sin^2 38.5$, $\cos^2 38.5$, and then add them together. :rotfl:[/quote]
  The answer is 1. :rotfl:

[quote="qqwref"]I think the first part is sort of nontrivial... but the second part certainly is!

Given equilateral triangle ABC, calculate $AB^2 - 2 AB \, BC + BC^2$.[/quote]

The answer is nought.

Find all integers $(x,y)$ such that $x^2-y^2$ is either odd or a multiple of 4.

Find all real positive numbers such that (1/a+1/b+1/c)(a+b+c)>=9. :rotfl:  :rotfl:

:rotfl: I can't stop laughing!I think this topic will be locked finally.

[quote="mathmanman"]Don't take it as an offense José, nothing against you! [/quote]

Took it as an offense...

[quote="Arne"]Seems like it is time to lock this (I'm not a mod in this subforum so can some other mod do this)?[/quote]

Really, not only they didn't help me, they moreover insulted me

Hey man, don't take it that serious. I'm sure you know that nobody wanted to insult you :)


[quote="TALENTO"]If we delete some digits of the number $ 2001009$ we get the number $ 201 \;\;(2\not{0}01\not{0}\not{0}\not{9} \rightarrow{} 201)$. We can also get the number $ 19\;\; (\not{2}\not{0}\not{0}1\not{0}\not{0}9 \rightarrow{} 19)$ and we can get other sizes too, but there are some numbers that can not be obtained, for example, the number $ 92$.
Alex found the least natural number such that, taking away some of their digits, as explained above, it is possible to obtain all the natural numbers less than $ 2009$. If Alex has $ A$ number of digits, and the sum of the three digits that are on the far left is $ B$, computes $ A \plus{} 3 B$.[/quote]

Since we need to buid numbers $ 1000,1111,1222,1333,1444,1555,1666,1777,1888,1999$, we need at least four $ 1$ and three of each other numbers, so at least $ 31$ numbers.

And such a number with $ 31$ digits exists : $ 1120345678901234567890123456789$, so $ A\equal{}31$ and leftmost digit is $ 1$.

And it's impossible for the fewest such number to begin with something else than $ 112$ :

$ 111....$ will forbid to build $ 211$
$ 110....$ will forbid to build $ 2000$
$ 10.....$ will forbid to build $ 2000$

So $ B\equal{}4$ and $ A\plus{}3B\equal{}43$


geometry
incenter
geometric transformation
reflection
analytic geometry
angle bisector
geometry unsolved

(Inspired by [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=260225[/url])

Let $ I$ be the incentre of triangle $ ABC$, and let $ M_1$, $ M_2$, $ M_3$ be the midpoints of the sides.  Let $ Q_1$ be the point on $ M_2 M_3$ such that $ Q_1 I \perp AI$, and define points $ Q_2$, $ Q_3$ similarly.  Show that $ Q_1$, $ Q_2$, and $ Q_3$ are collinear.

[hide]$ Q_1$, $ Q_2$, and $ Q_3$ lie on the polar (with respect to the incircle) of a well-known point.[/hide]

Incircle $ (I)$ touches $ BC, CA, AB$ at $ D, E, F.$ $ (I_a)$ is the A-excircle, tangent to $ BC$ at $ D_a.$ Midline $ M_2M_3,$ angle bisector $ BI,$ and the intouch triangle sideline $ DE$ concur at a point $ P_a$ [1]. Their poles WRT $ (I)$ are therefore collinear. $ C$ is the pole of $ DE,$ direction perpendicular to $ BI$ is the pole of $ BI,$ therefore the pole of $ M_2M_3$ WRT $ (I)$ is intersection $ K_a$ of perpendicular to $ BI$ through $ C$ with the line $ DI \perp M_2M_3.$ $ M_1$ is also midpoint of $ DD_a$ $ \Longrightarrow$ $ BD_a \equal{} CD$ $ \Longrightarrow$ the right $ \triangle BD_aI_a \cong \triangle CDK_a,$ having parallel sides, are centrally congruent with similarity center $ M_1$ and coefficient -1 $ \Longrightarrow$ $ M_1$ is midpoint of $ I_aK_a.$

Let $ G, M$ be centroid and Nagel point of $ \triangle ABC$ and $ J$ incenter of the medial $ \triangle M_1M_2M_3.$ Since $ I$ is the Nagel point of the medial triangle, the points $ G, I, J, M$ are collinear and $ M$ is reflection of $ I$ in $ J.$ ($ \overline{GM} \equal{} \minus{} 2 \overline{GI}$ and $ \overline{GI} \equal{} \minus{} 2 \overline{GJ}$ $ \Longrightarrow$ $ \overline{JM} \equal{} \minus{} \overline{JI}.$) $ K_a$ is the pole of $ M_2M_3,$ direction perpendicular to $ IQ_1,$ i.e., parallel to $ IA$ is the pole of $ IQ_1,$ therefore parallel $ q_1$ to $ IA$ through $ K_a$ is polar of $ Q_1.$ As $ q_1$ is reflection of $ IA$ in $ M_1,$ midparallel of $ IA \parallel q_1$ is the internal bisector of the angle $ \angle M_3M_1M_2,$ passing through the medial triangle incenter $ J.$ Consequently, $ q_1$ goes through reflection $ M$ of $ I$ in $ J.$ Similarly, polars $ q_2, q_3$ of $ Q_2, Q_3$ WRT $ (I)$ go through $ M,$ therefore their poles $ Q_1, Q_2, Q_3$ are collinear and the line $ Q_1Q_2Q_3$ is the polar of $ M$ WRT $ (I).$

Reference:
[1] [url]http://www.mathlinks.ro/viewtopic.php?t=255001[/url]

We can also use barycentric coordinates WRT $\triangle ABC.$ Orthocenter $O_3$ of $\triangle IAB$ is the pole of the C-midline $M_1M_2$ WRT the incircle $(I).$ Barycentric coordinates of $ O_3$ are given by $O_3 (b \minus{} c: a \minus{} c: c),$ thus polar $\tau_c$ of $ Q_3$ WRT $(I)$ passes through $ O_3$ orthogonally to $ IQ_3,$ i.e. parallel to the C-angle bisector $bx-ay=0.$ Hence, equation of $\tau_c$ is

$\tau_c \equiv (a^2 \plus{} ab \minus{} ac)x \plus{} (\minus{}b^2 \plus{} bc \minus{} ab)y \plus{} (a^2 \minus{} b^2 \plus{} bc \minus{} ac)z \equal{} 0$

We verify that the Nagel point $X_{8}(b \plus{} c \minus{} a: a \plus{} c \minus{} b: a \plus{} b \minus{} c)$ satisfies the equation of $\tau_c.$ By analogous reasoning, we conclude that $ Q_1,$ $Q_2,$ $ Q_3$ lie on the polar of $X_{8}$ WRT $(I)$ and the proof is completed.

Why the barycentric coordinate of orthocenter of IAB is (b-c:a-c:c)

There are different ways to find the the coordinates of $ O_3,$ e.g intersecting the line joining $ I (a: b: c)$ with $ J (0: p \minus{} c: p \minus{} b)$ and the line $\ell$ passing  through $ A (1: 0: 0)$ and the infinite point $P_{\infty}$ of the gradient $\perp IB,$ namely

$P_{\infty} (aS_B \plus{} a^2c: aS_A \minus{} cS_C: \minus{} ac^2 \minus{} cS_B) \Longrightarrow$

$ IJ \equiv (bc^2 \plus{} cS_A)x \minus{} (ac^2 \plus{} cS_B)y \plus{} (bS_B \minus{} aS_A)z \equal{} 0$

$ \ell \equiv (ac^2 \plus{} cS_B)y \plus{} (aS_A \minus{} cS_C)z \equal{} 0$

By solving these two latter equations we get $ O_3 \ (b \minus{} c: a \minus{} c: c).$

Sorry I don't quite understand the "infinite point of the perpendicular gradient"
What is it ?What is the gradient of a line in barycentric?
I feel quite confused. :blush:

Infinite point of any perpendicular line to $ px\plus{}qy\plus{}rz\equal{}0$ is given by

$ (S_B(r\minus{}p)\minus{}S_C(p\minus{}q): S_C(p\minus{}q)\minus{}S_A(q\minus{}r): S_A(q\minus{}r)\minus{}S_C(r\minus{}p))$

$S_A=\frac{b^2+c^2-a^2}{2} \ \ , \ \ S_B=\frac{c^2+a^2-b^2}{2} \ \ , \ \ S_C=\frac{a^2+b^2-c^2}{2}$

Thank you, I think I got to know it.
There is really a lot for me to learn in the barycentric world.
I wonder if there is some useful materials to help me learn the barycentric method?


[quote="archimedes1"]Calculate without a calculator:
$\left(\sqrt{\frac{1}{11-2\sqrt{30}}}-\sqrt{\frac{9}{7-2\sqrt{10}}}-\sqrt{\frac{16}{8+4\sqrt{3}}}\right)^{2}$[/quote]

[hide]$\frac{1}{11-2\sqrt{30}}=11+2\sqrt{30}$.
$\frac{9}{7-2\sqrt{10}}=\frac{9(7+2\sqrt{10})}{9}=7+2\sqrt{10}$.
$\frac{16}{8+4\sqrt{3}}=\frac{4}{2+\sqrt{3}}=4(2-\sqrt{3})=8-4\sqrt{3}$.

Are you sure this has a nice answer? Considering the fact that $(a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2ab-2ac+2bc$ makes it even uglier from here, and none of the above three values seem to have nice radical square roots.[/hide]

:ninja:  I'm positive this has an incredibly nice answer.  There's a semi-nice way to do it, and lots of wrong ways to do it.  Keep trying new methods.

hello, i think the result is zero.
Sonnhard.

[size=75][color=green]LadyKn1ght: Please include a solution with your answer.[/color][/size]

alkjash, you're on the right track.  Check for silly mistakes though.

[quote="alkjash"]Oh, I think I have it!

[hide]
Taking the terms seperately, the first is
$\frac{1}{11-2\sqrt{30}}=11+2\sqrt{30}$

You can factor $11+2\sqrt{30}$ into $(\sqrt{5}+\sqrt{6})^{2}$

Then look at
$\frac{9}{7-2\sqrt{10}}=\frac{9(7+2\sqrt{10})}{9}=7+2\sqrt{10}$

Which can be factored into $(\sqrt{5}+\sqrt{7})^{2}$

The last term,
$\frac{16}{8+4\sqrt{3}}=\frac{4}{2+\sqrt{3}}=4(2-\sqrt{3})=8-4\sqrt{3}$

can be factored into $2(1-\sqrt{3})^{2}$

So plugging into the original equation, it becomes

$(\sqrt{5}+\sqrt{6}-\sqrt{5}-\sqrt{7}-\sqrt{2}-\sqrt{6})^{2}$

Which equals $(-\sqrt{7}-\sqrt{2})^{2}= \boxed{9+2\sqrt{14}}$
[/hide][/quote]

how does $7+2\sqrt{10}=(\sqrt{5}+\sqrt{7})^{2}$?
how about $(\sqrt{5}+\sqrt{2})^{2})$
then try your solution again.
just throwing in 2 for 7, i get...
[hide]$(-2\sqrt{2})^{2}=\boxed{8}$[/hide]
EDIT: man, archimedes beat me to it.

Oh... yeah, i wrote $\sqrt{2}$ down on the paper...  :blush:

Oh, there's a sign problem on the third term, it should be

$(\sqrt{5}+\sqrt{6}-\sqrt{5}-\sqrt{2}-\sqrt{2}+\sqrt{6})^{2}$

But that would give $32-8\sqrt{3}$.

Ok, let me think.. if the answer is zero, the terms must all cancel out.
How is that possible?

[hide="HINT"]think [i]positive square root[/i][/hide]

[quote="Walk Around The River"][quote="archimedes1"]There's still a minor flaw in your solution, Walk Around The River and alkjash.
Dr. Graubner has the correct answer.[/quote]

hey hey hey i just took his stuff and plugged in one differet number.  :o

so its not my fault.  :roll:[/quote]

Wow, thanks a lot.

Oh whoops, sorry guys, I was completely ignorant and tried factoring in the form a+brt10, etc., instead of splitting up the rt10 into rt2 and rt5, since 10 isn't prime.

Nobody has posted a correct answer with solution yet.  See if you can find alkjash and Walk Around the River 's mistake.

Oh, i get it!

$ 2(1-\sqrt{3})^{2}$ is equal to $ 2(\sqrt{3}-1)^{2}$, and if you used that instead, you would get 0, since the terms would now cancel out.

But that means that my other answer was correct, right, just 0 is another one?

[quote="archimedes1"]You're right, those two quantities are equal.  But note that in the problem, the quantity was under a radical.  A radical implies only the [i]positive[/i] square root.
For example, $ \sqrt{4}$ means only $ 2$, not $ -2$.
$ \sqrt{2}(1-\sqrt{3})$ is negative, so we should use $ \sqrt{2}(\sqrt{3}-1)$. There is only one answer.[/quote]

Oh, ugh, never noticed that. Makes sense now.  :wallbash_red:  :wallbash_red:


Ok let's make this forum really popular. We have to surpass Alabama.

*goes and views Alabama Forum*

And what exactly is your plan?

Well Timecat8 just spammed in all our topics and he's making lots of marathons ;)


integration
trigonometry
linear algebra
matrix
limit
function
calculus

Help me to check the solution, thanks.
\[ \text{*  }I\equal{}\int\limits_{0}^{\plus{}\infty }{dy}\int\limits_{0}^{\plus{}\infty }{{{\left( x\plus{}y \right)}^{n}}.{{e}^{\minus{}x\minus{}y}}.dx}\equal{}8.\left( 2n\plus{}3 \right)!\]
Solution:

\[ \text{*  }I\equal{}\int\limits_{0}^{a}{dy}\int\limits_{0}^{a\minus{}x}{{{\left( x\plus{}y \right)}^{n}}.{{e}^{\minus{}x\minus{}y}}.dx}\]

\[ D\equal{}\left\{ \left( x\text{, }y \right)|0\le y\le a\text{, }0\le x\le a\minus{}y \right\}\]

\[ \text{Put x}\equal{}{{\left( r.\cos \theta  \right)}^{2}}\text{, }y\equal{}{{\left( r.\sin \theta  \right)}^{2}}\text{, }D\equal{}\left\{ \left( r\text{, }\theta  \right)|0\le \theta \le 2\pi \text{, }0\le r\le \sqrt{a} \right\}\]

\[ \text{x}\equal{}{{\left( r.\cos \theta  \right)}^{2}}\Rightarrow \frac{\partial x}{\partial r}\equal{}2r.{{\left( \cos \theta  \right)}^{2}}\text{, }\frac{\partial x}{\partial \theta }\equal{}2\cos \theta .\left( \minus{}\sin \theta  \right).{{r}^{2}}\equal{}\minus{}{{r}^{2}}.\sin 2\theta\]

\[ y\equal{}{{\left( r.\sin \theta  \right)}^{2}}\Rightarrow \frac{\partial y}{\partial r}\equal{}2r.{{\left( \sin \theta  \right)}^{2}}\text{, }\frac{\partial y}{\partial \theta }\equal{}2\sin \theta .\cos \theta .{{r}^{2}}\equal{}{{r}^{2}}.\sin 2\theta\]

\[ \frac{\partial \left( x\text{, }y \right)}{\partial \left( r\text{, }\theta  \right)}\equal{}\left| \begin{matrix}
   \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta }  \\
   \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta }  \\
\end{matrix} \right|\equal{}\left| \begin{matrix}
   2r.{{\left( \cos \theta  \right)}^{2}} & \minus{}{{r}^{2}}.\sin 2\theta   \\
   2r.{{\left( \sin \theta  \right)}^{2}} & {{r}^{2}}.\sin 2\theta   \\
\end{matrix} \right|\]

\[ \equal{}2{{r}^{3}}.\sin 2\theta .{{\left( \cos \theta  \right)}^{2}}\plus{}2{{r}^{3}}.\sin 2\theta .{{\left( \sin \theta  \right)}^{2}}\equal{}2{{r}^{3}}.\sin 2\theta\]

\[ \Rightarrow I\equal{}2\int\limits_{0}^{2\pi }{\left| \sin 2\theta .d\theta  \right|}\int\limits_{0}^{\sqrt{a}}{{{r}^{2n\plus{}3}}.{{e}^{\minus{}r}}.dr}\equal{}8\int\limits_{0}^{\sqrt{a}}{{{r}^{2n\plus{}3}}.{{e}^{\minus{}r}}.dr}\]

\[ 2\int\limits_{0}^{2\pi }{\left| \sin 2\theta  \right|.d\theta }\equal{}2\int\limits_{0}^{\pi }{\left| \sin 2\theta  \right|.d\left( 2\theta  \right)}\equal{}2\int\limits_{0}^{\pi /2}{\sin 2\theta .d\left( 2\theta  \right)}\minus{}2\int\limits_{\pi /2}^{\pi }{\sin 2\theta .d\left( 2\theta  \right)}\]

\[ \equal{}2\left( \minus{}\left[ \cos 2\theta  \right]_{0}^{\pi /2}\plus{}\left[ \cos 2\theta  \right]_{\pi /2}^{\pi } \right)\equal{}2\left( 1\plus{}1\plus{}\left( 1\plus{}1 \right) \right)\equal{}8\]

\[ \Rightarrow \underset{a\to \plus{}\infty }{\mathop{\lim }}\,I\equal{}\underset{a\to \plus{}\infty }{\mathop{\lim }}\,\int\limits_{0}^{a}{dy}\int\limits_{0}^{a\minus{}x}{{{\left( x\plus{}y \right)}^{n}}.{{e}^{\minus{}x\minus{}y}}.dx}\equal{}8\int\limits_{0}^{\plus{}\infty }{{{r}^{2n\plus{}3}}.{{e}^{\minus{}r}}.dr}\]

\[ \text{Put }u\equal{}{{r}^{n}}\Rightarrow du\equal{}n.{{r}^{n\minus{}1}}.dr\text{, }dv\equal{}{{e}^{\minus{}r}}.dr\Rightarrow v\equal{}\int{{{e}^{\minus{}r}}.dr}\equal{}\minus{}{{e}^{\minus{}r}}\]

\[ \Rightarrow {{I}_{n}}\equal{}\int\limits_{0}^{\plus{}\infty }{{{r}^{n}}.{{e}^{\minus{}r}}.dr}\equal{}\minus{}\left[ {{e}^{\minus{}r}}.{{r}^{n}} \right]_{0}^{\plus{}\infty }\plus{}n\int\limits_{0}^{\plus{}\infty }{{{r}^{n\minus{}1}}.{{e}^{\minus{}r}}.dr}\equal{}n.{{I}_{n\minus{}1}}\]

\[ {{I}_{0}}\equal{}\int\limits_{0}^{\plus{}\infty }{{{e}^{\minus{}r}}.dr}\equal{}\minus{}\left[ {{e}^{\minus{}r}} \right]_{0}^{\plus{}\infty }\equal{}1\text{, }{{I}_{1}}\equal{}1!\text{, }{{I}_{2}}\equal{}2!\text{, }{{I}_{2n\plus{}3}}\equal{}\left( 2n\plus{}3 \right)!\]

\[ \Rightarrow I\equal{}8\int\limits_{0}^{\plus{}\infty }{{{r}^{2n\plus{}3}}.{{e}^{\minus{}r}}.dr}\equal{}8.\left( 2n\plus{}3 \right)!\]

\[ \Rightarrow \int\limits_{0}^{\plus{}\infty }{dy}\int\limits_{0}^{\plus{}\infty }{{{\left( x\plus{}y \right)}^{n}}.{{e}^{\minus{}x\minus{}y}}.dx}\equal{}8.\left( 2n\plus{}3 \right)!\]

First one is incorrect - a posivite function cannot have zero integral.
Second one is incorrect as well - there should be $ \pi$ somewhere.


*rolls eyes*

Is (a :^2: - b :^2: ) completely factored :P ?

1. Forget my previous answer. I meant [hide="this."]$(a^2+b^2+ab\sqrt{2})(a^2+b^2-ab\sqrt{2})(a^2+b^2)(a+b)(a-b)$[/hide]

And seriously jb05, my answer wasn't even hidden, so you had to see it, and you still put $a^2-b^2$??

[quote="Treething"][quote="Silverfalcon"]
If a line that has x-intercept (5,0) and is perpendicular to line $10-8y = 2x$, find the y-intercept of the line.[/quote]

[hide="Answer"]
I've seen way too many of these perpendicular line problems.

8y = 2x - 10
y = x/4 - doesn't matter
perpendicular line has a slope of -4, so the answer's [b]20[/b]
[/hide][/quote]

that's wrong...
[hide="My Answer"]
the slope of the line is 4, so the answer is -20.[/hide]

[quote="Silverfalcon"]Factor as much as you can.
$a^8 - b^8$.
[/quote]
[hide]
$a^8 - b^8=(a^4)^2 - (b^4)^2$. Let $A=a^4$ and $B=b^4$. Then we have $A^2-B^2=(A+B)(A-B)=(a^4+b^4)(a^4-b^4)=(a^4+b^4)[(a^2+b^2)(a^2-b^2)]$
$(a^4+b^4)[(a^2+b^2)(a^2-b^2)]=(a^4+b^4)(a^2+b^2)(a+b)(a-b)$
At least that's what I get. How did you guys get the square roots and stuff?
[/hide]

[quote="Silverfalcon"]
If a line that has x-intercept (5,0) and is perpendicular to line $10-8y = 2x$, find the y-intercept of the line.[/quote]

[hide]
$10 - 2x = 8y \Rightarrow \frac{5}{4} - \frac{x}{4} = y$. The slope of this is $-\frac{1}{4}$. The slope of the line perpendicular to this is $-(-\frac{1}{4})^{-1} = 4$. Now we have $y=4x+b$. Because the x-intercept is (5,0), we have $0=4(5)+b$, so $b=-20$, which is the y-intercept.
[/hide]

[quote="Silverfalcon"]Factor as much as you can.

$a^8 - b^8$.

If a line that has x-intercept (5,0) and is perpendicular to line $10-8y = 2x$, find the y-intercept of the line.[/quote]

1.

$a^8-b^8=(a^4-b^4)(a^4+b^4)=(a^2-b^2)(a^2+b^2)(a^4+b^4)=(a-b)(a+b)(a^2+b^2)(a^4+b^4)$

2.

the slope of $10-8y = 2x$ is $-\frac{1}{4}$

so the slope of the per. lin is 4.

$y=4(x-5)$

y-int: (0,-20)

[quote="NoSoupForYou"]$a^4+b^4=(a^2+b^2+ab\sqrt{2})(a^2+b^2-ab\sqrt{2})$[/quote]

I still don't see how you got the square roots. Did you memorize that factorization or something?


A uniform disk (I = 1/2MR^2) of mass 8.0kg can rotate without friction on a fixed axis. A string is wrapped around its circumference and is attached to a 6.0 kg mass straight off the edge of a table. The string does not slip. What is the tension in the cord while the mass is falling?

Anyone know how to do this?

Do you want hints or the solution?

Do you know how to balance torques and forces? Keep in mind that the net force is NOT 0, so therefore tension is not equal to the force of gravity.

[hide="Solution"]Let the mass of the disk be $ M$ and that of the falling block be $ m$.
$ mg \minus{} T \equal{} ma$
$ TR \equal{} I\alpha \equal{} Ia/R \Rightarrow T \equal{} Ia/R^2 \equal{} (1/2)(MR^2)(a/R^2) \equal{} Ma/2$

$ mg \minus{} Ma/2 \equal{} ma\Rightarrow mg \equal{} a(m \plus{} M/2)\Rightarrow a \equal{} \frac {mg}{m \plus{} M/2}$

$ T \equal{} Ma/2 \equal{} \frac {mMg}{2m \plus{} M}.$

Now plug in the given values to get $ T \equal{} (6*8*9.8)/(12\plus{}8) \equal{} 23.52 N$.

[/hide]


[quote="Pirkuliyev Rovsen"]evaluate the sums; $ \sum_{k \equal{} 0}^{11} [\frac {x \plus{} 2^k}{2^{k \plus{} 1}} ]$   if $ x \equal{} 2009$   :?:[/quote]
By the third brackets, do you mean the floor function  or something else? :oops:
Sorry for my question......

It's probably the floor function.
[hide="One way"][url=http://en.wikipedia.org/wiki/Hermite%27s_identity]Hermite's identity[/url][/hide]


Find maximum of this funtion: $ y \equal{} \sqrt {a \plus{} cosx} \plus{} \sqrt {a \plus{} sinx}$, with $ a \ge 1$.

[quote="igiul"][hide]$ \left(\frac {\pi}{4},2\sqrt {\frac {1 \plus{} a\sqrt {2}}{\sqrt {2}}}\right)$[/hide][/quote]OK... so? Where's the solution?  :huh:

[hide="Solution"]$ (y/2)^2 \leq a + \frac {\cos x + \sin x}{2} \leq a + \sqrt {\frac {\cos^2 x + \sin^2 x}{2}} = a + \frac {1}{\sqrt {2}}$, where the first inequality is Power Mean with 1/2 and 1, and the second inequality is Power Mean with 1 and 2 (AM-QM).

The equality case for both is at $ \sin x = \cos x \implies x = \frac {\pi}{4} + n \pi$

Since $ y > 0$, $ (y/2)^2$ is a monotone increasing function of $ y$, and thus $ y$ is maximized at $ x = \frac {\pi}{4} + n \pi$ as well.

$ (y/2)^2 \leq a + \frac {1}{\sqrt {2}} \implies y \leq 2 \sqrt {a + \frac {1}{\sqrt {2}}}$, thus, the maxima are at $ \boxed{(x,y) = \left( \frac {\pi}{4} + n \pi, \; 2 \sqrt {a + \frac {1}{\sqrt {2}}} \right)}$[/hide]

[hide]$ |y|^2 = |\sqrt{a + \cos x} + \sqrt{a + \sin x}|^2 \leq 2(|\sqrt{a + \cos x}|^2 + |\sqrt{a + \sin x}|^2)$

$ = 2(a + \cos x + a + \sin x)$

$ = 2\left[2a + \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)\right]$

$ \leq 2(2a + \sqrt{2}) = 4\left(a + \frac{1}{\sqrt{2}}\right)$

and so

$ |y| \leq 2 \sqrt{a + \frac{1}{\sqrt{2}}\right}$.

In this argument I have used the inequality $ |\alpha + \beta|^2 \leq 2(|\alpha|^2 + |\beta|^2)$.[/hide]

[hide]$ |y| \leq 2 \sqrt{a + \frac{1}{\sqrt{2}}\right}$[/hide]


probability
integration
logarithms
conditional probability
probability and stats

A point is chosen randomly inside the interval [0,1] , a second  point is  randomly chosen between the second point and one. ¿What  is the probability that the sum  of the distances from cero  to each point chosen  be  greater  than  1?

English notes: cero=zero, and the appropriate form of "be" in the last sentence is "is".
I assume you mean "randomly" to be uniform on the possible interval.

[hide="Solution"]The distance is long enough given a first point of $ x$ when the second is greater than $ 1\minus{}x$. The conditional probability of this is $ \frac{x}{1\minus{}x}$ when $ 0\le x\le\frac12$ and $ 1$ when $ x>\frac12$. The total probability is then $ \int_0^{\frac12}\frac{x}{1\minus{}x}\,dx\plus{}\int_{\frac12}^1 1\,dx\equal{} \ln 2\minus{}\frac12\plus{}\frac12\equal{}\ln 2$.[/hide]


March 30, 2003 can be expressed numerically with eight total digits as 03/30/2003. However, only three distinct digits are used to express the date in that manner. What was the next date after 03/30/2003 that used exactly three distinct digits when expressed numerically in this way? Write the date in this same manner.

[hide]All other dates in 2003 are invalid as their month will be a different digit. The first such date in 2004 is February 2, 2004. $\boxed{02/02/2004}$[/hide]

Omg we did the exact worksheet I got the problem from yesterday at mc practice!! You weren't supposed to answer that...


Refer to the first post of the second page in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259275]this thread.[/url]

i guess i can replace in the one by AIME15 i think, well anyways the one in the stickies.........

I don't know if it's too late, but I can be a replacement in anything.

3 replacements for FTW tourney 4 can sign up here.

dang, forgot to ask for replacements....

if you want to replace in tourney 8, plz sign up here

sorry i forgot to ask about replacements

would everyone willing to replace unavailable members of the tourney please sign up here.

 :)

replacements sign up here for the Hi all tourney.


$ n\not \equal{} m$ natural number,  . If $ a^{n}\plus{}b^{n}\plus{}c^{n}\equal{}0$ and $ a^{m}\plus{}b^{m}\plus{}c^{m}\equal{}0$ --->  $ prove: abc\equal{}0$

There is no restriction like $ a,b,c \geq0$ ?

I don't believe there needs to be such a restriction.

[hide="Hint"]Consider $ f(x) \equal{} \sqrt[x]{1\plus{}a^x}$ for positive reals $ a$.[/hide]

Anyone post a solution please?It seems that I'missing something.


[quote="vutang"]Given three numbers $\; x, y,z$ in [$0,1$]. Find maximum of $\qquad x^2y+y^2x+z^2x-y^2x-z^2y-x^2z$[/quote]
um, is there a typo in the problem? because it seems odd that the second and fourth terms cancel out right off the bat

It factors as (y-x)(z-y)(x-z).  To make this positive, we have either 3 factors positive (implying y>x>z>y contradiction) or 2 factors negative.  Wlog y>z>x.   Then set y = x+a+b, z = x+a.  The product is ab(a+b), with a maximum of 1/4.

(x-y)(y-z)(x-z)=$\qquad x^2y+y^2z+x^2y-y^2x-x^2z-yz^2$ which is not equal to $\qquad x^2y+y^2x+z^2x-y^2x-z^2y-x^2z$ in the problem

He probably made a typo; I've seen the problem before, with the exact same restrictions, and I'm pretty sure it's supposed to factor as $(x-y)(y-z)(x-z)$.

vutang, please correct the statement of this problem. :)

It seems that the balance of this equality is not keeping from point of view the cycle of $x,y,z.$

kunny

[quote="vutang"]Given three numbers $\; x, y,z$ in [$0,1$]. Find maximum of $\qquad x^2y+y^2z+z^2x-y^2x-z^2y-x^2z$[/quote]
[quote]i'm sorry because i hadn't check after wrote. :) [/quote]


If a,b,c are positive real numbers and are  NOT  sides of a triangle prove that

 1 + x/(y+z-x)  +  y/(z+x-y)  +  z/(x+y-z)  \leq 0

Wlog we may assume that 0 < x \leq  y  \leq  z. (1)
Since x,y,z are not the sides of a triangle, we then have z > x+y. (2)

Let 2a = x+y-z, 2b = x-y+z and 2c = -x+y+z. Thus x = a+b, y = c+a and z = c+b.
Let a' = -a.
From (1) and (2), we deduce that :
1) a < 0 and then a' > 0.
2) c > a' > 0 and b > a' > 0.

In another hand, the desired inequality may written as :
a'/c + c/a' + a'/b+ b/a' - 2 - b/c - c/b  \geq  0
which is equivalent to (b+c)(a' - c)(a'-b)/(a'bc)  \geq  0.

It seems that, if i made no mistake, then the strict inequality holds.

Pierre.


induction
limit
algebra unsolved
algebra

Let $f: R^{+}\rightarrow R^{+}$ be such that

$f(f(x))+x=f(2x)$

Show that $f(x)\geq x$ for all $x$.

Assume $\{u_{n}\}$ is sequence defined by $u_{1}=\frac{1}{2}$ and $u_{n+1}=\frac{u_{n}^{2}+1}{2}(n=1,2,...)$. Then, by induction on $n$ we have $f(x)\geq u_{n}x\forall n\;\forall x$. Therefore $f(x)\geq x\forall x$, because $\lim_{n\to\infty}u_{n}=1$.


Ring Theory
vector
abstract algebra
group theory
superior algebra
superior algebra unsolved

I have this theoretical question: I have two rings, A (commutative) and B. B is an extension of finite degree (let's say n) over A. Then is it true that for every ideal of B (let's say C) there exists an ideal of A (let's say D) such as C is an extension of degree n over D? I really need the answer and a proof, or maybe a counter-example. Sorry I didn't use Latex. :maybe:

well A is a subring in B and |B/A|=n. For example, C is an extension of degree 2 over R.

So the ring $ \mathbb R\times\mathbb C$ over $ \mathbb R\times\mathbb R$ has degree ... 1.5? owk

nope, it's 2. you can check out the wikipedia article http://en.wikipedia.org/wiki/Field_extension.

so $ \mathbb{R} \times \mathbb{R}$ is a field? :-)
@owk: you really should read the wikipedia article, you'll learn a lot ..

i gave the link to the field extension article because the one on ring extensions is a stub, and it's only two lines long. RxR isn't a field of course, but it's a ring, and my initial question was about ring extensions. -oo- were you being sarcastic towards me, or owk?

Anyway...can someone please answer the actual question?

the actual question is
[quote="owk"]What do you mean by "extension of degree n"? owk[/quote]

ok, so did I give the wrong answer to that? what's the right one then?

I already said :"A is a subring in B and |B/A|=n. For example, C is an extension of degree 2 over R."

[quote="v0id"]|B/A|=n[/quote]

What is B/A? What is |...|? For fields $ K\subseteq L$, the degree is the dimension of $ L$ as $ K$-vector space. If $ B$ is a free $ A$-module, you may take the rank. But for general rings (or even ideals), there is no useful notion of degree. owk

B/A is the factor ring. |...| is its number of elements. So n=|B/A|=[B:A]. Maybe I don't know the exact notion in english. If that's the case, pardon me. But anyway, I hope I've made myself clear as to what I meant by "degree n"

If you think that $ |B/A| \equal{} [B: A]$ for fields $ A,B$, you understood it completely wrong. The first doesn't even always make sense (see the next line), the second one is a dimension of a vector space.
The quotient ring $ B/A$ isn't even defined for rings $ A,B$. Instead, $ A$ has to be an ideal.

$ [B: A]$ can be understood as the index of the (additive) subgroup $ A$ in $ B$. but then we have $ [\mathbb{C} : \mathbb{R}] \equal{} \infty$ and your initial statement is false.


You and a friend go to an auction and are required to participate in bidding for at least one item.  Unfortunately, you see nothing you like and there is only one item remaining.  The auctioneer announces he will auction off a [b]one dollar bill[/b].

There is a catch to this final auction.  The second highest bidder must also pay their bid, though they do not get the dollar.

Immediately the bidding opens at a penny.  Your friend quickly bids 2 cents.  The original bidder doesn't yet look out of the game and making a quick buck.  How should you proceed and why?

It seems that you should not bet at all.  Your friend will pay up to 1.02 for that dollar, so you can't get it for less than that.  But if you pay 1.02, you're losing money.

Not bidding violates the rule of the auction.

You should offer to split it with him.  Do you dislike your friend to the point where you would like to make him profit as little as he can from this transaction?  Anyways, if you must bid, I think you pay your 3c.

I don't think your friend can make a positive move after this.  He is in a worse situation if he bets more than 1.02.  You, on the other hand, can cover a bet of 1.03 without being in a worse position than your initial position.

[quote="blahblahblah"]You should offer to split it with him.  Do you dislike your friend to the point where you would like to make him profit as little as he can from this transaction?  Anyways, if you must bid, I think you pay your 3c.
[/quote]

It's good to bid small, but remember that at an auction you can't communicate with other bidders to set the price.  Assume that you cannot talk with your friend.

[quote="probability1.01"]Wait, is it down to you and your friend bidding now?  And wouldn't you have to eventually pass so that someone wins?[/quote]

There must be only one final bidder, yes.

Wouldn't you just bid 99 cents then and hope no one will be malicious just to cost you money?  Or do you want to make a net profit between you and your friend?

[quote="sheepwarrior"]well since it is an auction.. the one dollar bill is probably something special soooo bid a lot of moneY!! cuz its prob worth more than 1 dollar :D[/quote]

It's only worth a dollar.

i don't think this is rite but this is wut i think

If u don't bet right now ur friend will throw in 2c for nothing so instead of that happening y don't u bet 1.01 so it is like u paying 99c and ur friend paying 2c and u can get the dollar while letting u friend pay u back somehow and still let the stupid bidder throw in 1.00 for nothing...
 :D


calculus
integration
function
complex analysis
calculus computations

Knowing the limits of integration, is there a way to integrate a normally non-integrable function into a closed form expression of the upper and lower limits as well as the relative and non-relative variables?

Example: normally this can't be simplified into a general solution
$ \int f(x,z) dx$

If we place upper and lower limits on the integral, a b, can we know for certain that a closed form expression exists? and how do you work around the fact that it's non-integrable?
$ \int^b_a f(x,z)dx$

There are many functions that do not have an elementary antiderivative but definite integrals do indeed have closed forms.  Most notably, the Gaussian and its variants:

$ \int e^{\minus{}x^{2}}\,dx$ does not have an elementary antiderivative.  However, with a little multivariable calculus, it's easy to show that
$ \int_{\minus{}\infty}^{\infty}e^{\minus{}x^{2}}\,dx\equal{}\sqrt{\pi}$

[quote="JRav"]There are many functions that do not have an elementary antiderivative but definite integrals do indeed have closed forms.  Most notably, the Gaussian and its variants:

$ \int e^{ \minus{} x^{2}}\,dx$ does not have an elementary antiderivative.  However, with a little multivariable calculus, it's easy to show that
$ \int_{ \minus{} \infty}^{\infty}e^{ \minus{} x^{2}}\,dx \equal{} \sqrt {\pi}$[/quote]

Good, that's the answer I'm looking for. What method can I use to integrate something like this:

$ \int^b_a \frac{x^{z\minus{}1}}{e^x\plus{}1}\mathrm{d}x$

If you changed that +1 to a -1, you would have an integral resembling that of $ \Gamma(z)\zeta(z)$, if you set $ a\equal{}0$ and $ b\equal{}\infty$.

In these sorts of integrals, you can't have generic upper and lower limits; that's just the same as an antiderivative. You can only have a few special values; for $ \frac{x^{z\minus{}1}}{e^x\plus{}1}$, $ \int_0^{\infty}$ is the only one likely to be possible, since $ \infty$ is always special and $ 0$ is special for powers of $ x$.

Continuing on what jmerry said, if I was going to attempt to evaluate that integral (with limits $ 0$, $ \infty$), I would skip straight to complex analysis and try to apply the residue theorem.


What is the remainder when $2005^3$ is divided by 9?

[hide]
i worked it out(which is a bad method)and got a remainder of 1.[/hide]

that would be correct, but as you know, there's a better way of doing that.  :P

[hide="a big hint"]List out the cubes of the first few numbers and see what their remainders are when they are divided by 9. [/hide]

[hide]
$1^3=1$
when 1 is divided by nine the remaider is 0
$2^3=8$
when 8 is divided by nine the remainder is 1
$3^3=27$
when 27 is divided by nine the remainder is 0
So the pattern is 0, 1, 0, 1, 0, 1, 0, 1, 0, 1...
So 2005 is an odd number so the remainder would be $\boxed{1}$[/hide]

[quote="math92"]$2^3=8$
when 8 is divided by nine the remainder is 1
[/quote]

What?!  :rotfl:

So congruences come in Basic level.... If yes, this problem is quite simple!!!


function
complex analysis
complex analysis unsolved

Suppose f: D->D satisfying Re(f(z)) >= 0 for all z in D and suppose that  f is analytic.

 i) Show that Re(f(z)) >0 for all z in D

 ii) by using an appropriate Mobius transformation prove that if, f(0)=1 then,

                         (1-|z|)/(1+|z|) =< |f(z)| <=  (1+|z|)/(1-|z|)

Well, for the first one just consider the function $ g(z) \equal{} e^{ \minus{} f(z)}$ and note that $ |g(z)|\le 1$ then apply maximum principle :wink:
And for the second one you probably meant that $ f: \mathbb{D} \to \mathbb {C}$, right? Because as written you just get that from the condition of the problem your function is just constant.

No, it may not be a constant

for a) use the fact that $ f$ is analytic so $ f$ is open map to open map, $ f(\mathbb{D})$ is righ-half plane $ \mathbb{H}$, so if there is a point $ p$
 st : $ Re(f(p)) \equal{} 0$ then for any $ \epsilon > 0$  the ball $ B(f(p),\epsilon)\not\subset \mathbb{H}$. Contradiction

for b) let $ g \equal{} \varphi_{1}\circ f\circ \varphi_{ \minus{} 0}$ and apply Swchartz's lemma for $ g$

As written in the first post one has the function from unit disk to itself ($ \mathbb{D}$) and if it's equal to $ 1$ in the interior point then it should be a constant from maximum principle  :)


inequalities
function
inequalities unsolved

a,b,c,d are 4 positive numbers element in the interval (0,1) .Prove the following inequality:
1+ab+bc+da+bd+dc+ca>a+b+c+d

couldn't anybody prove this easy inequality?!?!

Rewrite it as $ (1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)\plus{}abcd\left( \sum \frac{1}{a}\right)\minus{}abcd>0$, which is true.

Can you give a more detailed solution?Please!!! And thanks!!

Is there anything not clear?

$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) > 0$, since $ a,b,c,d$ in $ (0,1)$.
Obviously,
$ abcd\left(\frac1a \plus{} \frac1b \plus{} \frac1c \plus{} \frac1d \minus{} 1\right) > 0$
Just take the sum and QED.

Of course. Because $ a,b,c,d \in (0,1)$, it's sufficient to prove that $ abc\plus{}abd\plus{}acd\plus{}bcd \ge abcd$, but even the inequality $ abc \ge abcd$ holds. Thus, the inequality is proved.

Yeah this solution is very clear, but FelixD before putted a solution using function, and I couldn't get that solution.
Can you post it again?

I edited my post because the solution wasn't true.^^


Anyone knows why friction doesnt depends of area?

If you have any material about this question please post here!!

thanks
Felipe

[quote="Rushil"]I think it's just an experimentally observed fact -that is true only at low velocities!!! :lol:[/quote]

Dont exist a theoretical explication?

This is the same question i used to trouble my tuition sir and friends with but look at this question....

$Q)$If a rough surafce is polished to a very graet extent,even after it becomes smooth enough it is still polished,then what happens to the friction?

Thats' a different thing. On polishing very much, the surfaces become say absolutely smooth and then since all the atoms come in very close proximity, the forces become so strong the friction(wrong word - attraction between surfaces right word) increases so much that we can't separate them!

That's what I have said above. Actually, that 'attraction' has 2 componenet - one perpendicular to surface of contact that is callede Normal Reaction and one parallel to the surface called Friction!

You mean there is NO proof of proving that friction is not dependant on area?

I think so!!! Just like there's no 'proof' for Newton;s Laws!

Sliding friction is proportional to the normal force acting on an object.  If the area of an object is increased (that is, area in contact with the ground), the [i]pressure[/i] on the surface will decrease, since the mass of the object will be spread over a larger area, but the total normal force will not change.  Another way of looking at it is to say that each very small part of the object in contact with the ground experiences a very small normal force and a very small frictional force.  Summing all of these up, we have the total frictional force, which must be proportional to the normal force.

Static friction, on the other hand, [i]does[/i] depend on surface area to an extent, as attractive forces are important.

Reference: http://en.wikipedia.org/wiki/Friction

Yeah,static friction does depend on surface area....Is the fact that rolling friction is less than dynamic friction also due to the attraction thing,if yes then even in this case surface area under contact is important.

Other interesting question would be why friction force dependes of the normal force?

[quote="shadysaysurspammed"]I read friction is attraction between atoms....even riddler told me so.....[/quote]Actually, attraction between atoms doesn't really have anything to do with friction. Friction is caused by irregularities of any given surface at much greater than atom scale. (Usually.) Static friction depends on the area when you take into account adhesion but this is again usually mechanical adhesion which is again caused by irregularities of greater than atom dimensions.

Rolling friction is in fact static friction that is rolling around so it shares many properties with static friction :).

Actually, friction does depend on contact area.  One of the reasons that a train is so cheap to move (in energy terms) is that there is about a dime's worth of contact area between the wheels and the rail for each car, where a truck has about a square foot for each wheel.

You guys can run an experiement to prove this.  Get two different weights (from you sister's set).  Tie one end of a string through the hole in the middle of the smaller weight and the other end through the hold in the middle of the larger weight.  Place one on a carpet and the second weight on top of the first.  Pull the weights by the string holding close to the the bottom weight.  This is case 1.  Reverse the order of the weights and repeat.  This is case 2.

If you can't get the weights from your sister, you can do the same thing with a block of wood and a table.  Cover two different sizes of surfaces of the block with paper (tape and tacks only on the other sides).  Push the block across the table with one side down then the other.  The block weighs the same, so the normal force is the same.  The paper is the same, so the friction coefficient is the same.

Friction is the interaction of the irregularities in both surfaces as they move past each other.  As these irregularities are worn off, the force of friction is less, sometimes much less.  (As one gets an optically flat surface it sticks significantly when placed next to another optically flat surface.)

In the train/truck example, you're talking about rolling friction which we agreed that really does depend on area.

In the carpet example, the friction will depend on area because the friction coefficient changes in the case of soft surfaces with pressure - the surface of carpet will deform.

bus -- You're right the deformation of the surfaces is one of the reasons that the force of friction is dependent on the normal force.  The deformation (pushing the huge, relative to atoms,  irregularties closer together has them interact to a greater degree.

One can think of a pair of velcro pieces gently pushed together and strongly pushed together as an analogy (not great, but same result).  The optically flat surfaces sticking strongly together are where there is significant interaction on the atomic level as the two surfaces are flat enought so a significant number of atoms are close enought to enteract with each other.

I have this confusion about friction,sometimes someone says(I do) that it is because of attraction thing sometimes it is the interlocking of the rough surfaces....does friction keep changing its characteristic properties all teh time :?

Can the attraction theory be used in rough surfaces as well?And would there be something of pressure in that case?

yep , i read in many places that friction is due to inter locking whereas others say that the interlocking thing is the " old " concept whereas attractrion between the atoms is the "modern" theory, dont know which one is actually right though

i would like  to add a question here . Can some one xplain me exactly why we are not able to walk if there's no friction? Many people ( namely my friends) give stupid answers like its due to slipping , no one actually explains it exactly . A full answer would be welcome .

also if there's a box on a surface, when theres' static friction , what actually applies the frictional force ? surface or box ? if either of them , why not the other?

[quote="eminem"]also if there's a box on a surface, when theres' static friction , what actually applies the frictional force ? surface or box ? if either of them , why not the other?[/quote]Both. Remember third Newton's law?

[quote="eminem"]i would like  to add a question here . Can some one xplain me exactly why we are not able to walk if there's no friction? Many people ( namely my friends) give stupid answers like its due to slipping , no one actually explains it exactly . A full answer would be welcome .[/quote]To move (or accelerate), you need to apply some kind of force on something. There's no acceleration without force that would cause it. In our ordinary world, when you walk you apply frictional force on the ground and the ground applies frictional force back on you so you may accelerate. In a world without friction, there would be no way to do this.

[quote="bus"][quote="eminem"]i would like  to add a question here . Can some one xplain me exactly why we are not able to walk if there's no friction? Many people ( namely my friends) give stupid answers like its due to slipping , no one actually explains it exactly . A full answer would be welcome .[/quote]
To move (or accelerate), you need to apply some kind of force on something. There's no acceleration without force that would cause it. In our ordinary world, when you walk you apply frictional force on the ground and the ground applies frictional force back on you so you may accelerate. In a world without friction, there would be no way to do this.[/quote]

That is what all ur "friends" say eminem,so is bus stupid as well?(no offence bus,just between eminem and i :D )

oh yeah!? well shady first of all all my "FRIENDS"  never give a proper answer to any physics question , most of the times "they" and me after a long 1/2 hr discussion reach back to the starting point ( not talking about you shady..... :huh:  :huh:  :|  :rotfl: uhhhh maybe i am )  so there 's no point in asking them any doubts, is there? no that doesnt include bus!

anyway bus , cant i simply say, when i walk i push the ground backward, the ground pushes me forward by the 3rd law , hence i move forward. So where's the concept of friction here?  :huh:  :huh:

pardon if this is dumb , but with "friends" like mine , i never get clear answers( i dont want XTRA- smart comments from you shady! :mad:  :mad: )so i have to ask these doubts here

I'm sure shadysaysurspammed is just teasing you :).

The concept of friction is in that part where you say "I push the ground backwards". If there were no friction, you would [b]not[/b] push the ground backwards. If the ground had some handles that you could hold on to and push them backwards that would be fine, but sice there are usually no handles on the ground, you just use friction to do the same thing for you - friction is the way you hold on to ground with your feet, it's the only way you can interact with it. So no friction = no pushing the ground.


[color=blue]Buenos dias ...
me he comunicado con Orlando (orl)
que es uno de los administradores de:
[b]ReSOURCE SECTION...[/b]

me comunica que como Comunidad en Español que somos
nos pide que publiquemos los eventos que se realizan
en esta region iberoamericana, como son:
[b]
- Olimpiada Iberoamericana
- Olimpiada del Centro
- Olimpiada del Cono Sur
- Olimpiada de Mayo[/b]

En este orden de prioridad ....
esto, para difundir los eventos que se realizan en
cada región del mundo, y como manera de aprovechar
la capacidad de las diversas comunidades en cuanto
a fuentes de informacion y por supuesto el idioma..

Esta seccion sera para coordinar el esfuerzos de los
 miembros que deseen colaborar con esta tarea ..

Prioricemos nuestros esfuerzos en este listado primero..
la idea es:

- Traducir los enunciados
- Colgarlos en la seccion que los moderadores de la
comunidad definiran para este fin....
- Y cuando sean publicados, orl vendra y relinkeara los
enunciados al formato que se maneja para este fin en la
[b]RESOURCE SECTION.[/b]

Los interesdos en enviar un PM a mi cuenta (carlosbr)
y/o a la de Pascual..

Saludos,

Los moderadores.[/color]

Aqui encuentran todos los enunciados

[url=http://www.oma.org.ar/enunciados/index.htm]www.oma.org.ar/enunciados/index.htm[/url]

Colaboremos

[color=blue]OTRAS FUENTES:

[url=http://www.obm.org.br/frameset-eureka.htm]OBM -Revista Eureka[/url]

IBERO 2005 : (YA ESTA PUBLICADO)
IBERO 2004 :Eureka 20 /Pag 11
IBERO 2003 :Eureka 16/Pag 34
estan en portugues..

OK...
Definamos los encargados por examen:.

Carlosbr ---------- IBERO 2004
Carlosbr ---------- IBERO 2003
hucht -------------- IBERO 2002
TIPE ----------------IBERO 2001
JOTAJOTARIM ----- IBERO 2000

----------
PD: Con respecto a las imagenes , he visto q en muchos casos, solo son referenciales, aunque seria bueno que en el momento de postear la traduccion, lo adjunten..

Y cualquier comentario hacerlo por un PM  a mi cuenta (carlosbr)...gracias ..

Los Moderadores ..[/color]

Y donde los subimos.... acá mismo?  :?

$Tipe$

[color=blue]Haber,

para llevar un orde3n, debemos posteear las traducciones al TOPIC:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79001]RESOURCE SECTION[/url]

PD: Cualquier sugerencia enviar un PM.

Los Moderadores.[/color]

[quote="tipe"]Supongo que después [b] orl [/b] se encargará de colocar cada problema en un topic diferente ... no?

Osea, nosotros subimos los seis problemas juntos ?

$Tipe$[/quote]

si ... asi ....


function
number theory
modular arithmetic

A reference book lists a set of annual calenders. For every given year, there is a calender in the set that corresponds to it. How many annual calenders must be included in the set in order to have a corresponding calender for every possible year?

I guess there is a Getting Started level solution to this problem, that is a "solution" that I knew by the time I was in junior high, namely what I will put in hidden text below.

[hide]You can look this up in the World Almanac or any other reference book that has a perpetual calendar. The number of different calendars necessary to show all possible years of combinations of dates and months of the year (according to the Gregorian calendar) is more than seven, but less than seven times twelve. Now that I've told you how to look it up, how would you reason through it?[/hide]

After edit: when I posted the above, I hadn't actually used the method I suggest in the hidden text to check the answer. I agree with the reasoning used in other hidden solutions for determining that that answer is correct.

Here's my solution

[hide]You need one calendar for every day that January 1st is on, which is 7.  THen, you need 7 more calendars to account for leap years, making it a tatoal of 14 calendars.[/hide]

The trick to this question, in my opinion, is leap years.

Solution:

The key to a rigorous proof uses modular arithmetic, which in my opinion, isn't beyond what beginners are capable of.

[hide]Each week has 7 days, so we can consider the entire thing in mod 7. Let's just start with an example. Say your calendar begins with January 1st being Monday in 2000. Since each year, 365  :equiv: 1(mod 7), so the next, year, January 1st would begin on a Tuesday and so on. Just listing these, the next year has Jan. 1st being Wednesday... and all days of the week where January first begins on are covered.

Here's to demonstrate:

Assuming 2000, January 1st was a Monday:

2000(Monday) 2001(Tueday) 2002(Wednesday) 2003(Thursday)
2004(Saturday) 2005(Sunday) 2006(Monday) 2007(Tuesday)
2008(Thursday) 2009(Friday) 2010(Saturday) 2011(Sunday)

So, on non-leap years, January 1st can occur on all days of the week, so that is 7 different calendars.

However, the calendar is altered to include February 29th in leap years.

The leap years are in increments of 5(mod 7), so all of them are covered. You have 5(mod 7) + 5(mod 7) = 10(mod 7) = 3(mod 7), 3(mod 7) + 5(mod 7) = 1(mod 7), 1(mod 7) + 5(mod 7) = 6(mod 7), and so on. That covers all 0 - 6 (mod 7), accounting for all days of the week. So, we have 14 total.[/hide]

i agree the answer is 14  i soved it in my head.  Where can i find a good site for modular arithmetic and the functions in it such as powers and multiplication and division

Excellent explanation white_horse_king88! This is what I am looking for.

Thanks all.


Suppose $A$ and $B$ are non-zero digits and $\underline{A5} \cdot \underline{B1} = 1995$.  Find $A + B$.

Ohh, I remember this problem from practice
[hide]
So we have a foil problem.
We can write $A5$ and $B1$ as
$10A+5$ and $10B+1$
and by using foil we have
$100AB+10A+50B+5=1995$
$100AB+10A+50B=1990$
$10AB+A+5B=199$
so obviously AB must equal 18, 16, or 15

Case 1--
If AB equals 16 then
$160+A+5B=199$
$A+5B=39$
The only solutions are then
$(9,6)$ and $(4,7)$ and neither work so...

Case 2 --
If AB equals 15 then
$150+A+5B=199$
$A+5B=49$
So A and B can equal
$(9,8)$ and $(4,9)$
none work then...

Case 3 --
If AB equals 18 then
$180+A+5B=199$
$A+5B=19$
Then the solutions are
$(9,2)$ and $(4,3)$
So $A$ and $B$ must be $9$ and $2$ respectively
Then $A+B = 11$

 [/hide]

[hide]First, we factor 1995, giving us 5x7x3x19. Then we look for combinations such that the units digit of one # is 1, and the other 5. The only two digit #s which work are 95 and 21, which means a+b=2+9=11.

Answer: 11[/hide]

[hide]Factor 1995 and you get 3*5*7*19, you can make 95 and 21, therefore A and B are 9 and 2 and A+B=11.[/hide]

One heck of an easy problem if you know how to do it.

[hide]The prime factorization of 1995 is 3*5*7*19. 7*3=21 and 19*5=95. 2+9=11.[/hide]

Press alt f4 - that's a good solution, but it won't let you solve the problem nearly quick enough in a contest situation.

Well I work Really fast :D
But you are correct and that may be why i didnt qualify for Nationals :(

Algebra is a wonderful thing but sometimes gets in the way, I've found. Sorry you didn't make it alt f4.

Say, what happens if I press alt F4 anyway? *bleep, screen vanishes*

[hide]B*5 is 5 or 0 therefore A has to be 4 or 9.  45 won't work because for it to work b must be odd, which can't be true because 4 is closest for b.  Therefore A must be 9 and B 2.  So A+B=11.[/hide]
Oh, Press Alt f4, I make the same mistake.  You should have my practice national problems from the coach.  Half my mistakes were obvious errors.


$p$ is a prime number($p=4k+3$) and a is integer. Solve this equation

$3a^2+1=p^2$

[i]moderator remark: see http://www.mathlinks.ro/Forum/viewtopic.php?t=74584[/i]

I got (a,p) = (-4,7) and (4,7) as the only solutions, simply by factoring the p and the 1 and narrowing down into cases where one divisor divides into 3.