vector
function
linear algebra
linear algebra unsolved

Let V be the subspace of the vector space of all real-valued functions that is spanned by the set S={(cos(x)^2), (sin(x)^2), cos(2x)}. Show that V is isomorphic to R^2.

I have to prove this. This is my logic... since cos(2x) is a linear combination of (cos(x)^2) and (sin(x)^2), then I can form a new set S'={(cos(x)^2), (sin(x)^2)} which also spans the vector space of all real-valued functions. Obvisously, S' is linearly independent. (I used the Wronskian to show this.) Ok, so then S' forms a basis for the vector space of all real-valued functions.

Now, since V is a subspace of the vector space of all real-valued functions, then S' also forms a basis for V. Now, the dim of V is 2 since the basis S' has 2 vectors.

Obviously dim(R^2) is 2. So, there is a theorem that two vector spaces are isomorphic to each other if and only if they have the same dimension. So, since dim(V)=dim(R^2)=2, then V is isomorphic to R^2.

Is my logic and thus proof correct? I would appreciate it if someone could verify this. Thanks!

I deem it correct yet do not understand what you mean by "S' forms a basis for the vector space of all real-valued functions. ". You mean by this V?


Let $ n \geq 2$ be an integer. Find the largest real number $ \lambda$ such that the inequality \[ a^2_n \geq \lambda \sum^{n\minus{}1}_{i\equal{}1} a_i \plus{} 2 \cdot a_n.\] holds for any positive integers $ a_1, a_2, \ldots a_n$ satisfying $ a_1 < a_2 < \ldots < a_n.$

I will show that $MAX(\lambda)=\frac{2n-4}{n-1}$
[hide="Proof"]
Now $1\leq a_1<a_2<\cdots <a_n$
So $a_m\leq a_n+(m-n)$ where $1\leq m\leq n$ and $n\leq a_n$
So $\sum^{n-1}_{i=1}a_i\leq (n-1)a_n-\sum^{n-1}_{i=1} i =(n-1)a_n-\frac{n(n-1)}{2}$
So $\frac{2n-4}{n-1}\sum^{n-1}_{i=1}a_i\leq 2(n-2)a_n-n(n-2)=(n-2)(2a_n-n)$
$n\leq a_n$ so $a_n-n+2>a_n-n\geq 0$ so $(a_n-n+2)(a_n-n)\geq 0$
So $a_n^2-2a_n\geq (n-2)(2a_n-n)$
So $a_n^2\geq \frac{2n-4}{n-1}\sum^{n-1}_{i=1}a_i +2a_n$

So $MAX(\lambda)\geq\frac{2n-4}{n-1}$

Consider $a_m=m$ where $1\leq m\leq n$
Then $n(n-2)=n^2-2n\geq \lambda \cdot \frac{n(n-1)}{2}$
Then $\lambda \leq\frac{2n-4}{n-1}$

So $MAX(\lambda)\leq\frac{2n-4}{n-1}$

Hence $\frac{2n-4}{n-1}\leq MAX(\lambda)\leq\frac{2n-4}{n-1}$
Hence $MAX(\lambda)=\frac{2n-4}{n-1}$
[/hide]


Prove that for every positive $ n$, the set of integers {$ 1, 2, 3, ... , 2n\minus{}1, 2n$} can be partitioned into $ n$ pairs:

{$ a_1, b_1$}, {$ a_2, b_2$}, ... , {$ a_n, b_n$}

so that $ a_i \plus{} b_i$ is prime for every $ i$.

Please wait for a while :
http://mathcentral.uregina.ca/mp/current/

Pierre.


Given $ a, b, c \geq\ 0$ satisfy $ a + b + c = 5$. Prove that: ${ 10 + ab^2 + bc^2 + ca^2 \geq\ \frac {7}{8}(a^2b + b^2c + c^2a)}$
When does equality occurs ?
We can prove it easily by Am-Gm  :)

Case 1: $ a \geq c \geq b$ then $ LHS\minus{}RHS\equal{}80\plus{}7(a\minus{}b)(b\minus{}c)(c\minus{}a)\plus{}ab^2\plus{}bc^2\plus{}ca^2 \geq 0$ Obvious
Case 2: $ a \geq b \geq c$  then
$ LHS\minus{}RHS \geq 80 \minus{}7ab(a\minus{}b)\plus{}ab^2 \equal{}80\minus{}\frac{ 5a.20b(7a\minus{}8b)}{100} \geq 80 \minus{}\frac{(12a\plus{}12b)^3}{27.100} \geq 0$.
Equality occurs if and only if: $ a\equal{}4,b\equal{}1,c\equal{}0$

[quote="Hong Quy"]Case 1: $ a \geq c \geq b$ then $ LHS \minus{} RHS \equal{} 80 \plus{} 7(a \minus{} b)(b \minus{} c)(c \minus{} a) \plus{} ab^2 \plus{} bc^2 \plus{} ca^2 \geq 0$ Obvious
Case 2: $ a \geq b \geq c$  then
$ LHS \minus{} RHS \geq 80 \minus{} 7ab(a \minus{} b) \plus{} ab^2 \equal{} 80 \minus{} \frac { 5a.20b(7a \minus{} 8b)}{100} \geq 80 \minus{} \frac {(12a \plus{} 12b)^3}{27.100} \geq 0$.
Equality occurs if and only if: $ a \equal{} 4,b \equal{} 1,c \equal{} 0$[/quote]
Very nice,Nam.Here is my proof:
I prove that
\[ f(a,b,c)\equal{} 80 \plus{} 7(a \minus{} b)(b \minus{} c)(c \minus{} a) \plus{} ab^2 \plus{} bc^2 \plus{} ca^2 \geq 0
\]
Now,Assume $ c\equal{}min(a,b,c)$;$ t$ is a nonnegative number such that $ t\le c$
.It is obivious to see
\[ f(a,b,c)\ge f(a\minus{}t;b\minus{}t;c\minus{}t)
\]
Hence we only need to  prove that the inequality when $ c\equal{}0$
Now is the time for AMGM inequality,which is very easy as nguoivn said  :lol:


I would like to know the answer to the question below.

What is the best way to help students who post questions here?

I notice that some of you answer my questions but do not provide an easy explanation or detail for your answers.

Why?

Take this sample:

x + 4 = 20

Most of you say:  x = 16 and NOTHING MORE.

Very few of you take time to make sure that students fully grasp what is going on.

What makes a good math tutor or teacher?

Also, what about [color=red]HINTS?  [/color]

Some of the hints are worst than the language used in my textbook.

Isn't the goal to make things clear for people who are obviously lost in mathematics.

I expect many replies here.

Some of you complain that students that do not show an effort when posting questions.  What most of you math tutors do not seem to understand is the fact that there are questions that are simply too complex to even begin an approach.

I cannot speak for other students.  In MY case, the questions posted here are questions that I personally tried answering for over 30 minutes.  There is a small quiz at the end of the study book that I own with 25 questions.

I simply post questions after getting most of the quiz questions right.  The questions here are the ones I considered hard and thus, should be approached by people who are gifted in math or better prepared for upper level high school mathematics.

What is your perspective?

Thanks!
interval

Some people come to this site for problems that they can't solve or for practice for a test or just for a challange! But I think that people should show all their steps. If someone needs help with a problem, they should post some things that they tried. One problem I posted got to $x^{3}-6x-9=0$ and I didn't know how to factor it or solve it and then the next step after that was $x=3$ (the only real number that solves it) and then I had to say, how did you get that? and even then, I didn't know how to solve it...

EDIT: what if the problems require multiple steps? or are questions that the posters haven't learned yet?

[b]we are not math tutors[/b]...where did you get that from??? we are just kids who are interested in math

i am against your posting of problems...these types of problems don't belong on the forum because they follow a very simple pattern, there is no thinking involved, there is no unique twist to the problems, if you look at the amc problems, they are not just the run of the mill problems that you are posting,

if you spend 30 mins on some of these problems, you are not approaching your study correctly, part of mathematics is a way of thinking, you have to learn to find patterns and algorithms to solve the problems, you learn these processes from an example in a book, then you use the practice tests to test if you know the process or not,

Yeah-why should we waste time posting trivial steps for trivial problems?

I'm a bit skeptical of some of the things you're saying. First, you refer to "tutors." The Art of Problem Solving is not a tutoring site, it is a site for discussion of interesting problems (that have to do with areas of problem solving, not textbook problems). Yes, if you spend a lot of time on a problem and want a hint, posting a problem is the thing to do. But you ask for hints, many people post hints very often. Also, you always complain about problems that do not have a sufficient explanation. However, the solutions written are very well-explained, you don't seem to be trying hard enough to understand them. Yes, if they are using theorems or concepts that you don't know much about, it would be the right thing to do to ask about them. But many of the solutions that are posted to your problems use very elementary methods that are easy to understand. They are explained fine. Yes, it is annoying to just see an answer to a problem you don't know how to approach, but almost always when you request a better explained solution, the solution is self-explanatory. I've been through this experience before, mind you, when I first read AoPS Volume 2. I thought the solutions to their examples were way too vague, and I never picked them up at first. But I realized what I was doing wrong, I was just trying to read. They leave "in-between," more trivial steps out, because the reader should fill them in themselves. This is a good tactic, I believe Zeitz calls it "active reading."

As a side note, I have noticed a pattern in your posts. After people post solutions that you like for a problem, you almost always make a post like "I like the replies," "Thanks for the tips," "Very well explained," or "Nice work." As I have notified you before via pm, this is generally spam. If you have a compliment for someone, by all means give it, but not in a thread, but via pm.

(edit) man, Altheman beat me with what I wanted to say.

[quote="Altheman"][b]we are not math tutors[/b]...where did you get that from??? we are just kids who are interested in math
[/quote]
yes, this isn't for tutoring people-if i wanted to tutor i'd go over to the library and charge people lots of money per hour to be tutored

the lowest level here is middle school math and..at least from where i am...middle schoolers would understand a x-4=16 or whatever problem
perhaps not quadratics(though most do!) but we shouldn't have to cover each trivial step if we post a solution(sorry if others feel differently)
read every step and see how it works...research easy things yourself

personally, most teachers(and tutors) can easily, very easily be replaced now
what makes a good teacher, tutor, etc. is that they help but don't give you the answer-you have to discover some things for your self

I also strongly suggest that you move your posts to the Classroom Math forum, many people are getting annoyed with your posts clogging the forums when they want to solve challenging problems for fun.

These questions are ridiculously easy, even for Classroom Math.

No offense.

And yeah, if you think this is a tutoring, you are sadly mistaken.

Forums are meant to talk about things, not necessarily tutors.

Yes, I think that is something that you need to realize.
We are NOT OBLIGATED to answer your questions. We come here for fun because we enjoy solving problems that are challenging. I think most of your questions could be solved by consulting an algebra textbook, no offense meant.
Most of us don't show work because most of the problems you post are considered trivial by most of AoPS.

[quote="bpms"][quote="mathnerd314"]Take it easy, guys.

Interval, if you are looking for a math tutoring site, here are a couple that I found:

http://mathforum.org/dr.math/
http://www.fliegler.com/mathman.htm

I hope I have been of assistance.[/quote]
I think most people are tired of him flooding the forum even though several people have said that he should mostly post in the classroom math forum.

This could discourage a new member who is interested in solving challenging and unique contest problems, but then sees these posts and thinks AoPS is a website for school math problems.[/quote]

he should either:

post in Classroom Math

or: don't post in this HSB

it's getting spammy sort of full of easy problems


One-half of the people in a room have two brown eyes, one-fourth
have two blue eyes and one-sixth have two green eyes.
The other 9 people in the room have two hazel eyes. How
many people are in the room?

LOL i like how they specify they have $ 2$ hazel eyes. What if someone had one of each???

[b]Another way to solve this problem:[/b]

$ \frac{1}{2} \plus{} \frac{1}{4} \plus{} \frac{1}{6} \equal{} \frac{11}{12}$

The rest of the people have hazel eyes, and there's $ 9$ of them, so that means that $ 9$ is $ \frac{1}{12}$ of the total number of people. So we divide:

$ 9 \div \frac{1}{12} \equal{} 9 \times 12 \equal{} 108$

There are $ \boxed {108 \text{ people}}$

A 3rd way of solving the problem. Some people could have more than 2 eyes, and as such, since 9 have 2 hazel eyes and we know about at least half of all people, then $ 2\times{9} = 18$ for a minimum of 18, therefore:

$ 18\LEQ{\text{number of people in the room}}\leq{108}$ As it wants a set number of people, we average these two and find the average number of people in the room is $ \frac {18 + 108}{2} = 63$ Unfortunately, we are overlooking a few problems, such as "What defines people?" and "What constitutes the difference between brown and hazel?" Also, FTW does not accept 63 as the answer, so I must assume that this problem is glitched with a "corrcect" answer of $ 108$

[quote="mewto55555"]A 3rd way of solving the problem. Some people could have more than 2 eyes, and as such, since 9 have 2 hazel eyes and we know about at least half of all people, then $ 2\times{9} \equal{} 18$ for a minimum of 18, therefore:

$ \geq{18}\text{number of people in the room}\leq{108}$ As it wants a set number of people, we average these two and find the average number of people in the room is $ \frac {18 \plus{} 108}{2} \equal{} 63$ Unfortunately, we are overlooking a few problems, such as "What defines people?" and "What constitutes the difference between brown and hazel?" Also, FTW does not accept 63 as the answer, so I must assume that this problem is glitched with a "corrcect" answer of $ 108$[/quote]

 :rotfl:  :rotfl:  :rotfl:


Let $ x,y,z \in \mathbb{R}$ such that $ x \plus{} y \plus{} z \equal{} 0$. Knowing there exists positive reals $ a,b,c \in (0;1)$ such that $ a \plus{} b \plus{} c \equal{} 1.5$ and $ ax \plus{} by \plus{} cz \equal{} 1.5$ then prove that:
\[ p^x \plus{} p^y \plus{} p^z \ge \frac {3}{2}(p \plus{} p^{ \minus{} 1}), \quad \forall p \in \mathbb{R}_ \plus{}
\]

If it's too easy try the generalized version:
Let $ x_1,...,x_n \in \mathbb{R}$ such that $ x_1 \plus{} ... \plus{} x_n \equal{} 0$. Knowing there exists positive reals $ a_1,a_2,...,a_n \in (0;1)$ such that $ a_1 \plus{}... \plus{} a_n \equal{} \frac{n}{2}$ and $ a_1x_1 \plus{} ... \plus{} a_nx_n \equal{} \frac{n}{2}$ then prove that:
\[ p^{x_1} \plus{} ... \plus{} p^{x_n} \ge \frac {n}{2}(p \plus{} p^{\minus{}1}), \quad \forall p \in \mathbb{R}_ \plus{}
\]

[hide="Strong hint"]Use weighted AM-GM!![/hint][/hide]

Since no one could solve it, I'd better provide a solution :)

First by weighted AM-GM:
$ \frac{a_1p^{x_1} \plus{} ... \plus{} a_np^{x_n}}{a_1 \plus{} ... \plus{} a_n} \ge \sqrt[a_1 \plus{} ... \plus{} a_n]{p^{a_1x_1 \plus{} ... \plus{} a_nx_n}} \equal{} p$
But also:
$ \frac{(1\minus{}a_1)p^{x_1} \plus{} ... \plus{} (1\minus{}a_n)p^{x_n}}{n\minus{}a_1 \plus{} ... \plus{} a_n} \ge \sqrt[n\minus{}a_1 \plus{} ... \plus{} a_n]{p^{(1\minus{}a_1)x_1 \plus{} ... \plus{} (1\minus{}a_n)x_n}} \equal{} p^{\minus{}1}$

Addition and multiplying by $ \frac{n}{2}$ gives the desired. QED


Let $ a,b,c$ be nonnegative real numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$, then
\[ a\plus{}b\plus{}c \ge abc\plus{}2\]
:)

Let me think! :lol:
Ok! This is the answer for your problem:
By squaring both sides, we get
$ \left(a\plus{}b\plus{}c\right) \geq \left(abc\plus{}2\right)$
$ \leftrightarrow a^2\plus{}b^2\plus{}c^2 \plus{} 2(ab\plus{}bc\plus{}ca)\geq (abc)^2\plus{}4abc\plus{}4$
$ \leftrightarrow 2(ab\plus{}bc\plus{}ca) \geq (abc)^2\plus{}4abc$
Using Cauchy- Schwartz, we get
$ 2(ab\plus{}bc\plus{}ca) \geq 6\sqrt[3]{a^2b^2c^2}$
From the condition given, we easily deduct the following inequality
$ abc \leq 1$
That is what we need!


geometry
parallelogram
angle bisector
geometry unsolved

I am a little stumped on this problem. any help is appreciated.

Two circles X and Y meet at point P.
Let tangent line to X at P...meet Y at point B.
Let tangent line to Y at P...meet X at point A.
Let circle Z pass through P,B,A.
Let tangent line to Z at P...meet X at C and Y at D.

Prove that PC=PD.
(hint: draw lines joining p and centers of three circles. look for a parallelogram).

It's rather simple. Denote O,O',E be the centers of circles X,Y,Z.
$ OE\perp PA\Rightarrow OE\parallel{}O'P$, $ O'E\perp PB\Rightarrow O'E\parallel{}OP$, thus OPO'E is a parallelogram.
Draw the diameters PM,PN of (O),(O') respectively. Then E is the midpoint of MN.
$ MC\perp CD, ND\perp CD, EP\perp CD, EM\equal{}EN$, hence PC=PD.
P.s. Your diagram maybe rather bad  :| .

△AO1P∽△PO2B  => △PO1O3∽△APB
∠O3PO2+∠O2PD=90=∠PBD+∠O2PD   =>   ∠PBD=∠O3PO2=∠ABP
Analogously, ∠PAC=∠O1PO3=∠PAB
Using sin law
PC=sin∠PAB*AP/sin∠ACP
PD=sin∠PBA*BP/sin∠BDP
=>  PC=PD

Using the above notations ( mr.dan's): after having seen that POEO’ was a parallelogram,  hence $ \triangle POE \sim  \triangle APB$, we shall get $ \triangle ACP\sim  \triangle PDB\sim  \triangle APB$, that is, $ \angle ACP \equal{} \angle BDP$ ( 1 ), $ \angle CAP \equal{} \angle BAP$ ( 2 ) and $ \angle ABP \equal{} \angle DBP$ ( 3 ).
If {F} = AC∩BD, from (1) we get $ \triangle CDF$ isosceles, while from (2) and (3) P is the F excircle of  $ \triangle ABF$, i.e. PF is the angle bisector of $ \angle CFD$, hence the result.

Best regards,
sunken rock


rotation
geometry
geometric transformation
function
modular arithmetic
combinatorics proposed
combinatorics

A circular disk is partitioned into $ 2n$ equal sectors by $ n$ straight lines through its center. Then, these $ 2n$ sectors are colored in such a way that exactly $ n$ of the sectors are colored in blue, and the other $ n$ sectors are colored in red. We number the red sectors with numbers from $ 1$ to $ n$ in counter-clockwise direction (starting at some of these red sectors), and then we number the blue sectors with numbers from $ 1$ to $ n$ in clockwise direction (starting at some of these blue sectors).

Prove that one can find a half-disk which contains sectors numbered with all the numbers from $ 1$ to $ n$ (in some order). (In other words, prove that one can find $ n$ consecutive sectors which are numbered by all numbers $ 1$, $ 2$, ..., $ n$ in some order.)

[hide="Problem 8 from CWMO 2007"]$ n$ white and $ n$ black balls are placed at random on the circumference of a circle.Starting from a certain white ball,number all white balls in a clockwise direction by $ 1,2,\dots,n$. Likewise number all black balls by $ 1,2,\dots,n$ in anti-clockwise direction starting from a certain black ball.Prove that there exists a chain of $ n$ balls whose collection of numbering forms the set $ \{1,2,3\dots,n\}$.[/hide]

Here's an idea that might (I'm not sure) work:

We regard the sectors as $2n$ colored and numbered points. The problem asks us to show there is a line which contains points numbered $1,2,..,n$ on each of its sides.

We draw $n$ segments, each having $2$ points numbered the same as its extremities. It's obvious that no $2$ such segments intersect. I think it's easy to derive from here that there must be a line cutting all $n$ segments, and that solves the problem.

Think it's Ok?

I do not see why two segments won't intersect  :(

Pierre.

You are right: it is possible for them to intersect :). As I said in that message, it just seemed Ok. Sorry.

Now that you broughtit up, does anyone have a really nice and short solution?

I think I managed to solve it, but I doubt ever having the patience to write it here, since it's a bit hard to explain (even though it's not long). I have to define all sorts of orientations and stuff :(. A diagram would help a lot, but I don't know what to use to draw one which is fairly accurate.

I have one i found when the tst was pusblished but i think its the same as you, because it is imposible to write, lot of considerations and things, i will try to do it later

True source of this problem:

20th Tournament of Towns, Spring 1999, Juniors, hard variant, problem 4.

Author of the problem: V. Proizvolov.

See here:

In Russian: http://www.turgor.ru/20/turnir20.php .

In English: http://servus.matematik.su.se/matcir/turgor/tg20vo.htm .

Is the official solution not so ugly to be given?

Pierre.

Ok Pierre, I will post the official solution soon.

I don't know the official solution, but here my solution is.

Let us consider two auxiliary disks consisting of n sectors, one over another. One for blue sectors (we will call it "blue disk") and another for red sectors (we will call it "red disk").

Consider some half-disk on the original disk. There are some blue sectors in it and some red sectors. Consider corresponding sequences of sectors on the blue and red disks. If we rotate the half-disk there are 4 cases of behaviour these sequences:

1. Red sequence rotates clockwise by one sector.
2. Blue sequence rotates counter-clockwise by one sector.
3. One sector from the end of red sequence moves to the head of blue sequence.
4. One sector from the end of blue sequence moves to the head of red sequence.

Observe that in all cases the distance between the heads of sequnces always decreases by 1 or allways increases by 1. (Distance is a number of sectors between these heads.) WLOG we may assume that in decreases (otherwise we may consider the distance between the ends of the sequences). Distance is a number from 0 to n-1. Therefore after at most n rotations we will have that such a distance will be 0. It is easy to see that this case corresponds the case that the half-disk contains all numbers 1, ..., n.

It looks pretty much like what I did :). I guess it wasn't as nasty as I thought :?.

Ah... :? I didn't understand what you meant with the blue/red circles and what is a head and the distance, and how it uses the values... :(
Hope Valentin's proposal would be clearer... :blush:

Pierre.

Hi!!
Maybe the easiest consideration (also the official one :D) is to consider the 2 numbers that are equal, of different colors, and closest to each other among all other such pairs. Some checking will help you see that somewhere there lies the sequence we are looking for! :)

Ok, I think I got it. :)
Thanks Dust.

Pierre.

Here is my version:
If all the blue sectors lie in a demi-disk, then all the integers are also in it.
Suppose that there is no demi-disk with only one color, and suppose that all the integers of the blue sectors are placed. The position of the integers on red sectors is only determined by the position of 1. So, there are n ways to place the integers on red sectors in the counter clockwise direction.
Now, I define a bijection between the n diameters, and the n possible positions for the 1 on red sector.
A diameter separates the disk in 2 demi-disks, and there is only one position for the red 1 so that all the integers from 1 to n are on these 2 demi-disks (here we use the hypothesis that there is no demi-disk with only one color). It is quite easy to see that this function is an injection (this is again due to the fact that there is no unicolor demi-disk). As the two sets have the same number of elements, the function is also a surjection, which means that for every position of 1, there is a diameter which splits the disk in two demi-disks which contain all the numbers from 1 to n.

[quote]
There is a circle, which is divided into 2n sectors. n of them are black and n are white. Now you have to number the white ones consecutively clockwise and the black ones counterclockwise.

You have to proof that there is always a way to divide the circle into to halves so you have the numbers from 1 to n in each of them, no matter in which way the black and white sectors are arranged and where you start to number them.[/quote]

Does this work?

First of all suppose that there is no line that divides the circle in a black and a white half; in that case that line would verify the assertion. Then order the sectors clockwise and call f(i) the number of sector i, where 0<=i<=2n-1 and "i" is taken modulo 2n and f(i) modulo n. Define also g(i) belonging to {0,1} as the colour of the corresponding sector, where g(i)=0 means "white" and g(i)=1 means "black". Define the k-th "half of the circle" ( k modulo 2n) as the one made up by the n sectors k+1, k+2,... k+n. Consider the sequence f(k+1), f(k+2).. f(k+n); it can be partitioned in 2 nonempty (by the initial assumption) subsequences, the one made up by the f(i) for which g(i)=0, and the one made up by the f(i) for which g(i)=1. The first subsequence is an increasing sequence of consecutive residues modulo n ( in the sense a(i+1)=a(i)+1 (mod n) ) which has as its initial and bottom element f(j) for the "leftmost" (in the sense "conterclockwise-most") j in the k-th half of the circle such that g(j)=0. We will call that element W(k). The second subsequence, on the contrary, is a decreasing sequence of consecutive residues modulo n which has as its initial term, which we will call B(k), f(j) for the "leftmost j" in the k-th half of the circle such that g(j)=1.
The fact that the k-th half contains all the n numbers is then equivalent to B(k)=W(k)-1 or D(k)=W(k)-B(k)=1 (mod n).When we pass from k-th half to the k+1-th one. There are 2 possibilities: for the leftmost j in the k-th half g(j)=0 and therefore W(k+1)=W(k)+1 and B(k+1)=B(k), or g(j)=1 and therefore W(k+1)=W(k) and B(k+1)=B(k)-1. In both cases D(k+1)=D(k)+1 (all mod n). Now it becomes obvious thar proceeding clockwise we will reach a k such that D(k)=1, and we are done.

Sorry for any mistakes.

Just an addendum: it's problem 1 of 2. round of Bundeswettbewerb Mathematik 2006, too.

I hope that this works.

[b]Solution to CWMO Problem:[/b] Number the balls in clockwise order on the circle as $\omega_1, \omega_2, \dots, \omega_{2n}$ where $\omega_1$ is chosen arbitrarily. All indices of balls will be taken in $\pmod{2n}$. Let $a_{i}$ denote the number on the white ball either equal to $\omega_i$ or closest to $\omega_i$ in a clockwise direction around the circle. Let $b_{i}$ denote the number on the black ball either equal to $\omega_i$ or closest to $\omega_i$ in a clockwise direction around the circle. Let $c_i = a_i - b_i$. Now, considering the cases corresponding to the colours of consecutive balls, it follows that $c_{i+1} \equiv c_{i} + 1 \pmod{n}$. This implies that varying the ball $\omega_i$ causes $c_i$ to assume all values in $\pmod{n}$. Hence there exists a ball $\omega_{j}$ such that $c_{j} \equiv 1 \pmod{n}$ which implies that $a_j = b_j +1=m$ for some $m$. Now consider the balls $\omega_j, \omega_{j+1}, \dots, \omega_{j+n-1}$. If this set of balls contains $k$ white balls, then these have numbers $m, m+1, \dots, m+k-1 \pmod{n}$ and the blacks balls have numbers $m-1, m-2, \dots, m+k-n \pmod{n}$. Since this forms a complete residue set in $\pmod{n}$, the numbers on the balls $\omega_j, \omega_{j+1}, \dots, \omega_{j+n-1}$ are $\{ 1, 2, \dots, n \}$ in some permutation.

Label the sectors $A_0, A_2, \dots A_{2n-1}$ in circular order. Take the integer $i$ such that the two sectors with number $i$ are closest together over all choices of $i.$ WLOG, the two sectors in question are $A_0$ and $A_k$ where $k \le n.$ Sectors $A_1, A_2, \dots A_{k-1}$ must be all the same color. Suppose WLOG sectors $A_0$ through $A_{k-1}$ are one color and $A_k$ is another, then choosing $A_1, A_2, \dots A_{n}$ works.


How many integers from 1 to 2003 inclusive have an odd number of distinct positive divisors?

[quote]Only perfect squares have odd number of divisors.[/quote]

Is it possible to prove it?

I just want to know how you know it. :)

[quote="Silverfalcon"][quote]Only perfect squares have odd number of divisors.[/quote]
Is it possible to prove it?[/quote]
No, it is impossible. It is an open question, and if you prove it you will win $10^6$ dollars :D

[quote]I just want to know how you know it.[/quote]
I think it is not my business to solve problems from Getting Started Section ;)

[quote="Myth"]
No, it is impossible. It is an open question, and if you prove it you will win $10^6$ dollars :D
[/quote]

Math humor is funny.

Just wondering: why is it impossible to prove?
Wouldn't it be enough to say that if there are an odd number of divisors, then there exists a factor $x$ such that $x^2$ is equal to the number, making it a perfect square? If there are an even number of divisors, then each factor $a$ corresponds to another factor $b$ such that $a$ and $b$ are distinct and $ab=c$. Or is that not a very rigorous proof? This problem seems quite interesting.

[quote="nolachrymose"]Just wondering: why is it impossible to prove?[/quote]
It was a joke!

done how do you knwo the answer is 44 did you use a math program or some sort or equation?

Well, we know that only perfect squares have an odd number of divisors. Therefore, all he had to do was find the largest perfect square less than or equal to 2003. Since 45^2=2025, he knows the largest square is 44^2, so the answer is 44.

i guess i learned something new.  Thanks for that idea

[quote="1+1=3 yepthatzme"][quote="blahblahblah"]Well, strictly speaking, every natural number can be written in the form:

$\displaystyle\prod_{p_i\in\mathbb{P}} {p_i}^{r_i}$, where the $p_i$'s are prime.

Now, the number of divisors of such a number is given by:

$\displaystyle\prod (r_i+1)$

A number is a perfect square iff all of the $r_i$'s are even.  If one of the $r_i$'s is odd, then $\displaystyle\prod (r_i+1)$ will contain a factor of two and hence be even.  Hence a number contains an odd number of divisors iff it is a perfect square.[/quote]

WHAZZZAH???? please explain in further detail!! im not THAT smart "1+1=3 yepthatzme" get the picture?? lol just kiddin![/quote]

I ll try to explain those in layman's terms.

First $\prod$ means the product of a sequence.  It is used just like $\sum$ but you multiply everything.  For example $\prod_1^4{x}=1\cdot 2\cdot 3\cdot 4=24.$

$\prod p_i^{r_i}$ where the $p_i$ are primes.

This means that every number can be prime factorized.  It can also be written as $p_1^{r_1}\cdot p_2^{r_2}\cdots p_n^{r_n}.$

Next the number of divisors of that number is given by the formula:
$\prod (r_i+1)$.

So if we wanted the number of divisors of $150$ we say that $150=2\cdot3\cdot5^2.$  So the number of divisors is $(1+1)(1+1)(2+1)=12$.

(You can use a counting argument to prove this...try it)

Next, he basically says that a number is a perfect square if and only if all the exponents when the number is factorized are even.

Hope this helps!

[quote="Done"]That's a good hint Myth, thanks. I also think that the trick in the question is the word 'distinct'.[/quote]I am just curious why do you think that the trick in the question is the word 'distinct'?
It seems to me that the trick was knowing that the only numbers with an odd number of divisors are perfect squares (we are assuming here that we are talking about positive divisors), so the word distinct is superfluos in the statement.

[quote="Done"]Take for example number 4.

4 = 2 * 2.

We say that number 4 has a total of 4 factors: 1, 4, and two prime factors. Or, it has 3 factors: 1, 4, and a single distinct prime factor.

So the number of factors differs, depending whether we are talking about total factors or total distinct factors.

Same applies for, say, 25.

This is why the question specifically includes 'distinct'.[/quote]You are wrong. First of all if we are talking about decompositions into prime factors the correct form is $ 4 =2^2$, or $36= 2^2\cdot 3^2$, and so on. There is only one prime factor in the decomposition of 4, namely 2. There are 2 prime factors in the decomposition of 36, namely 2 and 3. And so on.

Further more, I hope that you weren't thinking that when we say the sum of divisors of 4 we mean $1+2+2+4 = 9$ (notice that I haven't used the distinct attribute), because it's wrong. In fact saying distinct divisors is superfluous. Just if you are not convinced yet, think about what sense does the "non-distinct" divisors expression have? ;)

I wonder how we should approach this question then,

For any positive integer n, n>1, the "length" of n is the number of its positive primes whose product is n.

What is the greatest possible length of a positive integer less than 1000?

a) 10
b) 9
c) 8
d) 7
e) 6

note that I haven't used the word 'distinct' here.

[quote="Done"]I wonder how we should approach this question then,

For any positive integer n, n>1, the "length" of n is the number of its positive primes whose product is n.
[/quote]The definition of the "length" is ambiguous, due to a gramatical flaw. Please either correct it or rephrase the question :)


inequalities
inequalities proposed
highschoolmath

$ S\equal{}\frac {1 \plus{} a}{1 \minus{} a} \plus{} \frac {1 \plus{} b}{1 \minus{} b} \plus{} \frac {1 \plus{} c}{1 \minus{} c}\equal{}\frac {2a\plus{}b\plus{}c}{b\plus{}c} \plus{} \frac {2b\plus{}a\plus{}c}{a\plus{}c} \plus{} \frac {2c\plus{}a\plus{}b}{a\plus{}b}\equal{}3\plus{}2\sum_{cyc}\frac{a}{b\plus{}c}$
we have $ (a\plus{}b)^2\geq4ab$
$ \leftrightarrow \frac{2}{a\plus{}b}\leq\frac{1}{2}\left(\frac{1}{a}\plus{}\frac{1}{b}\right)$
$ \leftrightarrow \frac{2c}{a\plus{}b}\leq\frac{1}{2}\left(\frac{c}{a}\plus{}\frac{c}{b}\right)$
therefore:
$ 2\sum_{cyc}\frac{a}{b\plus{}c}\leq\left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \right)$
then: $ S\leq 2\left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \right)$

Thank you but i don't understand your solution
[quote="peine"]$ S \equal{} \frac {1 \plus{} a}{1 \minus{} a} \plus{} \frac {1 \plus{} b}{1 \minus{} b} \plus{} \frac {1 \plus{} c}{1 \minus{} c} \equal{} \frac {2a \plus{} b \plus{} c}{b \plus{} c} \plus{} \frac {2b \plus{} a \plus{} c}{a \plus{} c} \plus{} \frac {2c \plus{} a \plus{} b}{a \plus{} b} \equal{} 3 \plus{} 2\sum_{cyc}\frac {a}{b \plus{} c}$
we have $ (a \plus{} b)^2\geq4ab$
$ \leftrightarrow \frac {2}{a \plus{} b}\leq\frac {1}{2}\left(\frac {1}{a} \plus{} \frac {1}{b}\right)$
$ \leftrightarrow \frac {2c}{a \plus{} b}\leq\frac {1}{2}\left(\frac {c}{a} \plus{} \frac {c}{b}\right)$
therefore:
[color=red](Here!!!!!!!!!)[/color] [color=red]$ 2\sum_{cyc}\frac {a}{b \plus{} c}\leq\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right)$
then: $ S\leq 2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \right)$[/color][/quote]
Can you show me how you do it?

Sorry for my bad English!! ^^
Best Regarts!

Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac {1 \plus{} a}{1 \minus{} a} \plus{} \frac {1 \plus{} b}{1 \minus{} b} \plus{} \frac {1 \plus{} c}{1 \minus{} c} \leq 2(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c})$

I think apply Bunyacovski  inequality is the best way to solute  this inequality!

$ RHS \equal{} 3\plus{}2\sum_{a,b,c}\frac{a}{1\minus{}a} \leq 2(\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c})
\Leftrightarrow 2\sum_{a,b,c}a(\frac{1}{c}\minus{}\frac{1}{1\minus{}a}) \geq 3
\Leftrightarrow \sum_{a,b,c}a\frac{1\minus{}a\minus{}c}{c(1\minus{}a)} \geq \frac{3}{2}
\Leftrightarrow \sum_{a,b,c}\frac{ab}{c(b\plus{}c)} \geq \frac{3}{2}$
Apply Bunyacovski inequality we have:
$ (\sqrt{\frac{ab}{c}}\plus{}\sqrt{\frac{bc}{a}}\plus{}\sqrt{\frac{ca}{b}})^2 \equal{}
(\sqrt{\frac{ab}{c(b\plus{}c)}(b\plus{}c)}\plus{}\sqrt{\frac{bc}{a(c\plus{}a)}(c\plus{}a)}\plus{}\sqrt{\frac{ca}{b(a\plus{}b)}(a\plus{}b)})^2 \leq
2.\sum_{a,b,c}\frac{ab}{c(b\plus{}c)}$
Because $ (a\plus{}b\plus{}c)^2 \geq 3(ab\plus{}bc\plus{}ca)$
$ \Rightarrow (\sqrt{\frac{ab}{c}}\plus{}\sqrt{\frac{bc}{a}}\plus{}\sqrt{\frac{ca}{b}})^2 \geq 3(a\plus{}b\plus{}c) \equal{} 3$
Completed!
^^

I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you!
Let $ a, b, c>0$. Prove that: $ \frac{a}{bc(c\plus{}a)}\plus{}\frac{b}{ca(a\plus{}b)}\plus{}\frac{c}{ab(b\plus{}c)} \geq \frac{27}{2(a\plus{}b\plus{}c)^2}$

[quote="Ronald Widjojo"]I think this is an JBMO Problem, isn't it ?
My solution just use AM-GM[/quote]
^^ No! It's from MO Romanian 2004. Can you show your solution?

[quote="D.I.Culianop"]I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you!
Let $ a, b, c > 0$. Prove that: $ \frac {a}{bc(c \plus{} a)} \plus{} \frac {b}{ca(a \plus{} b)} \plus{} \frac {c}{ab(b \plus{} c)} \geq \frac {27}{2(a \plus{} b \plus{} c)^2}$[/quote]
this is my solution
we have with AM-GM:
$ S\equal{}\frac {a}{bc(c \plus{} a)} \plus{} \frac {b}{ca(a \plus{} b)} \plus{} \frac {c}{ab(b \plus{} c)} \geq 3\sqrt[3]{\frac{1}{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$
$ \leftrightarrow S\geq3\sqrt[3]{\frac{1}{\left(\frac{a\plus{}b\plus{}c}{3}\right)^3.\left(\frac{2a\plus{}2b\plus{}2c}{3}\right)^3)}}$
then: $ S\geq\frac{27}{2(a\plus{}b\plus{}c)^2}$

[quote="gagaga"]my solution
let $ abc \equal{} 1$

we must prove

$ \frac{a}{bc(c \plus{} a)} \plus{} \frac{b}{ca(a \plus{} b)} \plus{} \frac{c}{ab(b \plus{} c)} \geq \frac{27}{2(a \plus{} b \plus{} c)^2}$

$ \Leftrightarrow$ $ \frac{a^2}{c \plus{} a} \plus{}$ $ \frac{b^2}{a \plus{} b} \plus{}$ $ \frac{c^2}{b \plus{} c}$ $ \geq$ $ \frac{27}{2(a \plus{} b \plus{} c)^2}$

 CBS $ \frac{a^2}{(c \plus{} a)} \plus{} \frac{b^2}{(a \plus{} b)} \plus{} \frac{c^2}{(b \plus{} c)} \geq \frac{a \plus{} b \plus{} c}{2} \geq \frac{27}{(a \plus{} b \plus{} c)^2}$[/quote]
^^ Thank you gagaga. I think we don't need let $ abc \equal{} 1$. You use Schwarz first and than we can use Cauchy to sovle. It's a beautiful solution!

[quote="peine"][quote="D.I.Culianop"]I have two solution apply CBS for this unit but i don't know a different solution! Can you show me! Thank you!
Let $ a, b, c > 0$. Prove that: $ \frac {a}{bc(c \plus{} a)} \plus{} \frac {b}{ca(a \plus{} b)} \plus{} \frac {c}{ab(b \plus{} c)} \geq \frac {27}{2(a \plus{} b \plus{} c)^2}$[/quote]
this is my solution
we have with AM-GM:
$ S \equal{} \frac {a}{bc(c \plus{} a)} \plus{} \frac {b}{ca(a \plus{} b)} \plus{} \frac {c}{ab(b \plus{} c)} \geq 3\sqrt [3]{\frac {1}{abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$
$ \leftrightarrow S\geq3\sqrt [3]{\frac {1}{\left(\frac {a \plus{} b \plus{} c}{3}\right)^3.\left(\frac {2a \plus{} 2b \plus{} 2c}{3}\right)^3)}}$
then: $ S\geq\frac {27}{2(a \plus{} b \plus{} c)^2}$[/quote]

Thank you very much, peine! ^^  :oops: You're intelitence! I think your solution with AG is very nice. This is my solution:

$ (\sqrt {\frac {a}{bc}} \plus{} \sqrt {\frac {b}{ca}} \plus{} \sqrt {\frac {c}{ab}})^2 \equal{} (\sqrt {\frac {a}{bc(c \plus{} a)}(c \plus{} a)} \plus{} \sqrt {\frac {b}{ca(a \plus{} b)}(a \plus{} b)} \plus{} \sqrt {\frac {c}{ab(b \plus{} c)}(b \plus{} c)})^2 \leq 2(a \plus{} b \plus{} c)(\sum_{a,b,c}\frac {a}{bc(c \plus{} a)}) \equal{} 2(a \plus{} b \plus{} c)M$
Because $ (a \plus{} b \plus{} c)^2 \geq 3(ab \plus{} bc \plus{} ca)$
We have: $ (\sqrt {\frac {a}{bc}} \plus{} \sqrt {\frac {b}{ca}} \plus{} \sqrt {\frac {c}{ab}})^2 \geq 3(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$
$ \rightarrow M \geq \frac {3(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}{2(a \plus{} b \plus{} c)} \geq \frac {27}{2(a \plus{} b \plus{} c)^2}$
$ \leftrightarrow \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq \frac {9}{a \plus{} b \plus{} c}$ (true by Schwarz)  :D

[quote="D.I.Culianop"]^^ Let $ x, y, z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 4$. Prove that:
$ \frac {1}{2x \plus{} y \plus{} z} \plus{} \frac {1}{x \plus{} 2y \plus{} z} \plus{} \frac {1}{x \plus{} y \plus{} 2z} \leq 1$[/quote]
From CS; \[ LHS \le \sum{\frac{1}{4}(\frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z})} \equal{} \frac{1}{2}(\sum{\frac{1}{x\plus{}y}}) \leq \frac{1}{2}\sum\frac{1}{4}(\frac{1}{x}\plus{}\frac{1}{y}) \equal{} 1\]

[quote="apollo"][quote="D.I.Culianop"]^^ Let $ x, y, z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 4$. Prove that:
$ \frac {1}{2x \plus{} y \plus{} z} \plus{} \frac {1}{x \plus{} 2y \plus{} z} \plus{} \frac {1}{x \plus{} y \plus{} 2z} \leq 1$[/quote]
From CS;
\[ LHS \le \sum{\frac {1}{4}(\frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z})} \equal{} \frac {1}{2}(\sum{\frac {1}{x \plus{} y}}) \leq \frac {1}{2}\sum\frac {1}{4}(\frac {1}{x} \plus{} \frac {1}{y}) \equal{} 1
\]
[/quote]
Your solution apply Schwarz inequality ! ^^ Perfect! Thank you!

Back to the original problem, the following inequalities hold for postive reals $ a,\ b,\ c$.
\[ \frac {b \plus{} c}{a \plus{} b} \plus{} \frac {c \plus{} a}{b \plus{} c} \plus{} \frac {a \plus{} b}{c \plus{} a}\leq \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}
\]

\[ \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {c \plus{} a}{a \plus{} b}\leq \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}
\]


calculus
integration
topology
invariant
analytic geometry

problem in Algebraic Topology
Hey guys, i am studying cohomology by hatcher's. Could anyone provide me some ideas on these problems? Thank you all!

Let f : S2n-1 -> Sn denote a continuous map. Let Xf = D2n union f Sn be the space obtained by attaching a 2n-dim cell to Sn using the map f.
i). Calculate the integral homology of Xf .
ii). Find an n and an f such that H*(Xf ;Z=2) has a non-trivial product. Justify it.

I'm going to assume for now that $ n > 1$.

i) You have a single 0-cell, a single n-cell and a single (2n)-cell. As $ n > 1$ you can't have any boundary homomorphisms, and so you get homology $ \mathbb{Z}$ in dimensions 0,n,2n; no other homology exists.
ii) Since the homology is torsion-free you get the same groups appearing as cohomology; $ \mathbb{Z}$ in 0,n,2n; nothing else. If $ x \in H^n(Xf)$ is the nontrivial generator, you may get a cup product $ x \smile x \in H^{2n}(Xf);$ it will be a multiple $ h$ of the generator in that dimension - $ h$ is called the Hopf invariant of $ f$.

A simple example: let $ Xf \equal{} \mathbb{C}P^2$, which is formed with attaching map $ f \equal{} \eta$ (the Hopf map -- same Hopf, no surprise.) Then since $ H^*(\mathbb{C}P^2) \equal{} \mathbb{Z}[x]/(x^3)$ we get that nontrivial product $ x^2$ in dimension four, and the map has hopf invariant one. The multiplication will be nontrivial no matter what coefficient group you are using, as squaring in this cases gives an isomorphism $ H^2(\mathbb{C}P^2) \to H^4(\mathbb{C}P^2)$.

Taking care of the case $ n \equal{} 1$ shouldn't be too hard, since then the map is $ f : S^1 \to S^1$, which is entirely classified up to homotopy by degree, and the homotopy type of $ Xf$ only on the homotopy class of $ f$.

Thank you very much! hopf map is so hard. Do you mean that CP2=D6 Uf S3 with hopf f? Do you have an easy example? Thanks anyway.

No, I mean $ \mathbb{C}P^2 \equal{} S^2 \cup_\eta D^4;$ the space has cells in even dimensions, and the hopf map is $ \eta : S^3 \to S^2$. It's very easy to write down, since $ S^2 \equal{} \mathbb{C}P^1$ canonically, and $ S^3 \subset \mathbb{R}^4 \equal{} \mathbb{C}^2$. The hopf map is given by $ \eta(z,w) \equal{} [z,w] \in \mathbb{C}P^1$.

sorry,  i have no idea about this operation

Oh sorry $ [x,y]$ is just the homogeneous coordinate in $ \mathbb{C}P^1$. The space $ \mathbb{C}P^1$ consists of all pairs $ [a,b]$ where $ a,b \ne 0$, under the equivalence relation $ [a,b] \sim [c,d]$ whenever $ ad \minus{} bc \equal{} 0$. You can identify such a point $ [a,b]$ with $ b/a \in \mathbb{C} \cup \{\infty\}$ (in fact $ \mathbb{C}P^1 \equal{} S^2$ is the Riemann sphere.)


which expression is the same as (6+4) X 8?
a)   8 X 6 + 8 X 4
b)   8 X 6 + 4
c)   8 X 4 + 6
d)   8 X 4 X 8 X 6

This is known as the distributive property which says:

$ a(b \plus{} c) \equal{} ab \plus{} ac$

I think you can figure out the rest, and once again wrong forum.

I really never went to school and now I'm  trying hard to help my son
we moved recently to the US

could you also tell me the right forum I was supposed to post to

have a great evening
Emanuele

Classroom Math, it is the first forum on the forums page, its in the middle school section.


Find all the ploynominals $p(x)$ satisfying :
$(\forall x \in \mathbb R) : (x-8)p(2x)=8(x-1)p(x)$

Well, puting $x=1$ we obtain $p(2)=0$, now put $x=2$, and we have $p(4)=0$. It is obvious that $p(8)=0$. Hence we can write $p$ as $p(x)=(x-2)(x-4)(x-8)q(x)$ for some polinomial $q$. Replacing with this in the given identity yields $8(x-1)(x-2)(x-4)(x-8)q(2x) = 8(x-1)(x-2)(x-4)(x-8)q(x)$ therefor for all $x \notin {1,2,4,8}$ we have that $q(2x)=q(x)$, and if $\deg{ q} >0$ then for some $k$ we will have $|q(2x)|>|q(x)|$ $ \forall x>k$. So $q$ is constant, and it is easy to see that the polinomial $p(x)=(x-2)(x-4)(x-8)k$ works for any $k \in \mathbb R$


geometry
factorial
number theory
prime factorization

Do you know any useful formulas for MOEMS that you want to share?

Here's one: To find the number of positive factors of a number, take the prime factorization of a number, add 1 to each of the exponents of each prime, and multiply these new exponents together. For example, take 150. The prime factorization is $2^13^15^2$. Add 1 to each exponent to get $2^23^25^3$. Multiply all the exponents together and get 2*2*3=12 factors of 150.

Also, make sure you know how to manipulate and work with area and perimeter.

Also, to determine the number of zeroes at the end of the factorial of a number (let's say N) take N and divide it by powers of 5 starting from 1 and take the integer part and finally summing it all up.

:huh: looks complicated. I'd better give an example. let's say 200!

200/5=40, 200/25=8, 200/125=1.6

Summing up all the integer parts: 40+80+1=49 zeroes at the end.

[quote="math92"]just to let you guys know, this topic is really helping me so keep posting formulas!!!!! :lol:  :lol:[/quote]
Learning the technique/method is much better than memorizing formulas. Maybe some come in handy for timed competitions but you should spend time learning the technique rather than using the time to learn the formulas. Sorry to break the fun out :( Maybe this didn't help, I am not familiar with MOEMS...
[Edit]: Ack, I posted in Middle School...

These formulas are given to you with a warning from me. Be careful on how you use them, as it is the application of these formulas which is essential. Often times, you will have to manipulate one of these formulas to solve a problem, and without understanding where a formula is from, it is very difficult to think in that sense. So, take heed, and happy problem solving.


A semicircle with center O has a radius of 9cm. What is the number of centimeters in the length of $ \overline{RQ}$, a diagonal of the rectangle shown?
[asy]pair O,R,Q;
O = (0,0); R = (-3,0); Q = (0,4);
draw((0,5)--Q--R--Q+R--Q--O); draw((-5,0)--(5,0));
draw(arc(O,5,0,180));
label("O",O,S); label("R",R,S); label("Q",Q,E);[/asy]

Well, since the figure is a rectangle, the diagonals are equal.  One can notice that if the other diagonal is drawn, it's a radius.  So the answer is $ \boxed{9}$.


Discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=5190 , also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=60519 .

  darij

ha ha ha ha...

again inversion shows its power...

the problem will be solved very nice by inversion... ;)


Hi, have a fixed pulley and a string in tension placed over it in contact with the pulley over an angle A. Coefficient of friction between the pulley and string is u. The pulley is then rotated and slips with respect to the string, show that the ration of the tensions in the two parts of the string is e^(uA). Assuming theres an integral to find but where do you get it from?

I don't think this is the correct forum for that question.

What forum should I move this to? Is it physics maybe?

perhaps you should try
http://www.physicsforums.com. I am quite hopeful that it would be a helpful place, I know that personally. :)

Thanks, sorry wasn't sure. I suppose was off a physics training sheet.

Now that this is in the correct part of the forums can anyone be of assistance?

[quote="Inspired By Nature"]I know this sounds dumb but what does uA stand for :?:[/quote]

$u$ is coefficient of friction and $A$ is the angle


videos
geometry
rotation
geometric transformation
induction
puzzles

Some medic I use damned God as sad Ogden made suicide memos.

  This sentence highlights a mthematical principle.  What is it?

palindrome!  That's a good one... never seen it.

There are only so many mathematical principles that can be represented by a bunch of strange words...  ;) Nice one, anyway.

its really good ,thanks for it,I am going to give it right away for the bulletein board
Ofcource I could not crack it

no sir, away a papaya war is on.
lol, that's longest palidrome i know.

Look at this palindrome. It's so long! :o

[b]Star? Not I! Movie – it too has a star in or a cameo who wore mask – cast are livewires.

Soda-pop straws are sold, as part-encased a hot tin, I saw it in mad dog I met. Is dog rosy? Tie-dye booths in rocks.

All ewes lessen ill. I see sheep in Syria? He, not I, deep in Syria, has done. No one radio drew old one.

Many moths – I fondle his; no lemons are sold. Loot delis, yob, moths in a deli bundle his tin. Pins to net a ball I won – pins burst input. I loot to get a looter a spot paler. Arm a damsel – doom a dam. Not a base camera was in a frost, first on knees on top spot. Now a camera was a widened dam.

Ask: Cold, do we dye? No, hot – push tap, set on to hosepipe. Nuts in a pod liven.

A chasm regrets a motto of a fine veto of wars. Too bad – I all won. A sadist sent cadets – a war reign a hero derides. A bad loser, a seer, tossed a cradle – he begat to cosset – a minaret for Carole, Beryl, Nora. We’re not as poor to self.

I risk cold as main is tidal. As not one to delay burden, I don’t set it on “hot”. A foot made free pie race losses runnier. As draw won pull, eye won nose. Vile hero saw order it was in – even a moron saw it – no, witnessed it: Llama drops – ark riots. Evil P.M. in a sorer opus enacts all laws but worst arose. Grab a nosey llama – nil lesser good, same nicer omen.

In pins? No, it is open. If a top spins, dip in soot.

Madam, as I desire, dictates: Pull aside, damsels, I set a rag not for a state bastion. A test I won e.g. a contest I won.

Kidnap, in part, an idle hero. Megastars, red, rosy, tied no tie. Blast! A hero! We do risk a yeti’s opposition!

He too has a wee bagel still up to here held.

Demigods pack no mask, cap nor a bonnet, for at last a case is open – I left a tip – it wets. A dog wets too. Radios to help pay my tip, pull a tip.

Ale, zoo beer, frets yon animal. Can it? New sex arose but, we sots, not to panic – it’s ale – did I barrel? Did I lose diadem, rare carrot in a jar of mine? Droop as tops sag – unseen knots.

A cat ate straw as buck risk cud; evil foe, nil a red nag ate? Bah! Plan it – silage. Model foot in arboreta.

I, dark Satanist, set fire – voodoo – to slat. I design a metal as parrot, I deem it now. One vast sum is no ten in set – amen! Indeed, nine drag a yam, nine drag a tie. Dame nabs flower; can we help man? Woman is worse nob.

Mud level rose, so refill a rut. A nag of iron I made to trot I defied – I risk leg and its ulnae. Can a pen I felt to bid dollar or recite open a crate, open a cradle, his garret?

Sample hot Edam in a pan. I’m a rotten digger – often garden I plan, I agreed; All agreed? Aye, bore ensign; I’d a veto – I did lose us site. Wool to hem us? No, cotton. Site pen in acacias or petals a last angel bee frets in.

I met a gorilla (simian); a mate got top snug Noel fire-lit role. Manet, Pagnol, both girdle his reed bogs.

Flan I reviled, a vet nods to order it, Bob, and assign it. Totem users go help mates pull as eye meets eye. Son – mine – pots a free pie, yes? No. Left a tip? Order a dish to get. A ring is worn – it is gold. Log no Latin in a monsignor, wet or wise. Many a menu to note carrot.

Cat in a boot loots; As I live, do not tell! A bare pussy, as flat on fire, I know loots guns, fires a baton, nets a hero my ale drop made too lax.

If it is

to rain, a man is a sign; I wore macs, no melons rot. I use moths if rats relive, sir, or retire.

Vendor pays: I admire vendee, his pots net roe. Nine dames order an opal fan; I’ll ask cold log fire vendor to log igloo frost. Under Flat Six exist no devils.

Marxist nods to Lenin. To Lenin I say: “Mama is a deb, besides a bad dosser.”

Gen it up to get “ova” for “egg”. I recall a tarot code: yell at a dessert side-dish sale. Yes/nos a task cartel put correlate: E.S.P. rocks a man. I am a man, am no cad, I’m aware where it’s at!

Fire! Its an ogre-god to help, man, as I go. Do not swap; draw, pull a troll!

It’s not a cat I milk – calf, for a fee, sews a button - knit or tie damsel over us. Mined gold lode I fill until red nudes I met in a moor-top bar can. I sit, I fill a diary – trap nine men in ten-part net – oh, sir, I ask, cod nose? No, damp eel.

So, to get a name! I say, Al! I am Al! Last, I felt, to breed, deer begat.

To can I tie tissue – damp – or deliver Omani artist – a man of Islam.

In a den mad dogs lived on minis a signor who lived afore targets in at. As eremites pull, I, we, surf, fantasise, mend a bad eye. No hero met satyr; Tony, as I stressed, won’t, so cosset satyr.

A vet on isles made us sign it, a name. Foe man one sub.

Aside no dell I fret a wallaby; metal ferrets yodel, like so. On a wall I ate rye. Bored? No, was I rapt! One more calf? O.K., calf, one more, bossy! No! Lock cabin, rob yam, sip martini. Megastar was in a risk.

Cat? No, I’m a dog; I’m a sad loyal pet. A design I wore – kilts (a clan); if net drawn, I put it up. Royal spots snag – royal prevents rift.

Composer, good diet, are both super, God – label it a love of art, lustre. Video bored, no wise tale e.g. a mini tale – no sagas seen. Knack: cede no foes a canal.

Pay – as I sign I lie; clear sin it is; e.g. “Amadeus” sign I – lira for ecu, decimal – sin as liar.

Trad artistes pull a

doom, a drawer won’t.

Is it sold loot? No, I suffered loss. A man is god; Amen! I came nice Tahiti (sic).

It’s ale for a ban if for a fast – is role to help mash turnip? Use zoo? No - grasp order – use no zoos. Warts on time did sag.

No grade “X” “A” Level? Oh, “A”! I’d a “B” or a “C”. So – pot? No, we lop. Date? Take no date! Bah! Play L.P.

Miss (a lass, all right?) flew to space in NASA era. Rose no (zero) cadets ate raw. As a wise tart I fined rags red Lenin, we help pay bet – a risk – cash to Brian. I put a clam in a pool – a pool wets.

Mahdi puts a stop to harem – miss it in one vote, lost in one, veto of none. Post-op, no tonsil; I ate; no tastier, eh? We sleep at noon time so I dare not at one; no time stops as I time tides. A bed: under it, roll; in a mania, panic!

In a pond I did as Eros as Lee felt tenrec. “Ink” – list it under “I”. Termites put pen in a way. Democrats wonder, I too. To slay moths a dog did.

I saw elf; elf, far now, is a devilish taboo, rag-naked. I hid a bootleg disc. I, saboteur, toss it in. Oops! No legs! Laminated, a cask, conker in it, negates all if it is simple.

Hot pages are in a mag, nor will I peer, familiar tat, so lewd, native rot. Toner, ewe wore no trace; vagabond ewes do. Oh, Ada! Have pity! A pitiable eel – “Oh wet am I!” - to save, note: bite gill as I do.

Call a matador minor, eh? As I live, don’t! Is torero no rigid animal debaser if tipsy? Ale drew esteem in a matador. A bolero, monks I rate play or go dig rocks; a can I step on.

Go! Gas – it evades a bedsit – set a roost on fire. Boss sent a faded eclair to green imp or dog, I’d don a belt to boot it; if Ada hid a boot, panic.

I mock comic in a mask, comedian is a wit if for eventide. Vole no emu loved is not a ferret, so pet or witness a weasel if not. I hired less, am not so bossy, as yet amateur.

To stir evil, Edna can impugn a hotel: bad loos, hot on Elba: I may melt. Tart solicits it rawer, gets it rare. Push crate open; I ram buses, use no trams.

Did I say, not to idiot nor a bare ferret, to trap rat, strap loops rat? Stewpot was on. Hot? I was red! Lessen it! Fine man on pot? No, pen inside by a bad law. So I made rips – nine delays.

Some Roman items in a.m. ordered “Is room for a ban?” “It is,” I voted: I sat pews in aisle. Beryl, no tiro to my burden, made off for a contest, I won kiss. I may raid fine dales. I raid lochs if I to help am.

Forecast for Clare v. Essex: If no rain, a man is ref. Fusspots net foxes.

Senor is a gnome, latinos’ bad eyesore. Help misses run to border, Casanova, now, or drab hotel.

Ma has a heron; I sleep, pet’s on nose, sir! Rev. I rag loved art live – fine poser. Ultra-plan: I feign, I lie: cedar to disperse – last one? No, last six. Enamel bonnet for a dark car to toss a snail at. In it all, Eve lost; Seth’s a hero slain on a trap – Rise, Sir Ogre Tamer.

Upon Siamese box I draw design. I, knight able to help, missed an alp seen in Tangier of fine metal pots. Tin I mined rages – order nine, melt ten. Tone radios; tones are not to concur. Ten-tone radar I bomb – best fire-lit so hostel side meets eerie mini red domicile. A gulf to get is not a rare tale; no time to nod.

Row on, evil yobs, tug, pull. If dogs drowse, fill a rut. An era’s drawers draw. Put in mid-field in a band I dig a tub deep. Staff on a remit did refill a minaret.

Sam’s a name held in a flat, or, sir, bedsit. I wonder, is it illicit ore? No ties? A bit under? Retarded? Is ‘owt amiss? I’m on pot; not so Cecil, a posh guy a hero met. A red date was not to last so Cecil sat.

Tip? An iota to pay, a dot; sad, I drop item. I’d ask, call, Odin, a Norseman’s god: “Pay payee we owe radio dosh o.n.o.” I to me? No, I to media.

Peril in golf – is ball a “fore”? K.O.!

Vexed I am re my raw desires. Alto has eye on nose but tone-muser pianist is level-eyed. I lost a tie. Blast! In uni no grades are musts. Avast! Never port! Sea may be rut.

Part on rose? - It’s a petal. Define metal:

Tin is … (I gulp!) can!

I am a fine posse man, I pull a ton. Ron, a man I put on, I made suffer of evil emu’s sadism. Leo’s never a baron - a bad loss but evil – topple him, Leo’s lad. Assign a pen, can I? A pal is note decoding.

Is damp mule tail-less? No, ill; I breed for its tone. Radio speed, to grower, grew. Open a lot? No, stamp it; if for a free peso – not ecu -deign it. Times ago stone rates, e.g. at Scilly, display a wont.

No wish to get a design I, Sir Des, I’ve let? No bus sees Xmas fir. O.K. – cab – tart it up; tie lots – diamond, log or tinsel; first end errata edit. So “le vin (A.C.)”, Martini, Pils lager, one tonic.

I pegged a ball up to here when I got a top star role, Beryl. Gun is too big – won’t I menace? Yes? No?

Ill? A cold? Abet icecap’s nip. U.S.A. meets E.E.C. inside tacit sale – see! Beg a cotton tie, ma! No trial, so dodo traps exist. Arabs under-admire card label good hood stole.

In rage erupted Etna. Will a rotunda, bare villa, to tyro. Lack car? Non-U! Get a mini! My, my, Ella, more drums per gong; get a frog – nil less. Rod, never ever sneer. Got to?

I disperse last pair of devils (ah!) here today or else order cash to breed emus. Said I: “Are both superlative?” C.I.D. assign it lemon peel still. I wore halo of one bottle from a ref (football) – a tip; so hit last ego slap a mate got.

Late p.m. I saw gnu here (non-a.m.) or an idea got a dog to nod – I made felt to boot.

Fill in a lad? Nay, not all, Edna – lash to buoy. Did you biff one Venus? Not I! “Broth, girl!” ladies ordered – “No, with gin!” – a fine plate, maybe suet; no carton I made rots in it.

Med: a hill, Etna, clears in it. Ali, Emir, to slap in/slam in. All in all I made bad losers sign it – alibi. Set a lap for a level bat.

A bed, sir, eh? To put cat now? Drat! Such an idyll of a dog’s lair! That`s it, open it – a cage! Big nit sent rat! Some day (A.D.) send ewe. No, draw a pot now, do! Of wary rat in a six ton tub.

Edna, ask satyr: “Tel. a.m.?” No, tel. p.m.; Israeli tuner is damp. Use item: “Anna Regina”. No! Dye main room (“salle”) red!

Nice caps for a sea cadet in U.S.A. – Now I, space cadet, am it, sea vessel rep. Pin it on Maria, help Maria fondle her fine hotpot. No! Meet; set up to net, avoid a lesion. Set acid arena: Bruno one, Reg nil. Like it to sign in? Even I am nine-toed! I vote votes.

Oh, can a nose-rut annoy? No, best is Dorset. I know, as liar, to snoop, malign. “I’ll order it to get a bedroom door,” began a miser I fed.

Am I to peer, fan? Is a door by metal? Ere sun-up, drowse, nod, lose magnet. Food? Buns? I’ll ask. Corn? I’ll ask. Corn – I snack. Cats snack (cold rat). Sum for a bag: nil. First, is remit “traps in net”? Yes, on a par. Coots yell over a dam I made. Bared nudist went a foot, I made roots. I tip a canon: “Row, sir, at same tide; man one: row tug.”

Sewer of denim axes a wide tail – a terror recipe to hero made manic. I, to resign? I ? Never!

“OFT I FELT ITS SENSUOUSNESS” – title fit for evening is erotic; I named a more hot epic – error retaliated – I was examined for ewe’s gut, wore no named item.

A star is worn on a cap, it is too red. Am I too fat? Newts I’d under a bed. Am I mad? Are volleys too crap? A nosey tennis part-timer sits rifling a bar of mustard.

Lock cans, stack cans in rocks, all in rocks, all I snub. Do often games, old ones, word-pun use; relate, my brood, as in a free pot I made fires, I manage brood. Moor debate got tired rolling, I lampoon, so trail saw on kites.

Rod sits, ebony on nature, so Nana chose to veto video. Ten in main evening is O.T.T. i.e. killing; Ere noon, urban eradicates noise, lad, I ovate not. Put esteem on top (to hen, if reheld).

No fair ample hair – am not I nipper-less? Eva estimated ace caps I won as united. A Caesar of space, Cinderella’s moor, Niamey Don (a Niger-an name), ties up mad sire, nut! I, Lear, simpleton male, try tasks “A” and “E”

but not “XI”. Sanitary raw food won top award one Wednesday – a demo.

Start nesting, I beg a cat. I? Nepotist? Ah, trials, God! A folly, Dinah, custard won’t act up; other is debatable. Velar: of palate; sibilating is “s”.

Resold: a bed, a mill, an ill animal – snip, also trim. Eilat in Israel can tell I had ‘em. Tin I stored (am I not raconteuse?) by a metal pen. If a night, I wondered, rose, I’d all right orbit on sun, even off.

I buoy, did you? Both Sal and Ella, Tony and Alan (“Ill if too bottle-fed, am I?”) do not. God! A toga! Ed in a Roman one, rehung! Was I, M.P. et al., to get a map? Also get salt? I, hospital lab to offer, am, or felt to be, no fool – a hero.

Will it sleep? No, melting is sad ice. Vital re-push to be raid, I assume. Deer, both sacred roes, Leroy (a doter, eh?) has lived for. I, apt sales rep’s idiot to greens, revere vendors selling or fat egg-nog reps.

Murder O’Malley, my mini mate – gun on rack. Calory total: liver, a bad nut or all I wanted (“et puree garnie”): lots. “Do, oh do, ogle bald racer,” I’m dared – N.U.S. bar at six.

Esparto, dodo’s lair to name it, not to cage bees, elasticated, is nice. Esteem, as up in space, cite bad local lions, eye can emit now. G.I. boots in ugly rebel or rat’s potato gin (eh?) were hot. Pull a bad egg – epic, I note, no regal slip in it. Ram can … (I’ve lost idea!)

Tarred nets, rifles, nitro, gold – no maid stole it. Put it, rat, back or if Sam (“X”) sees sub on televised rising, I sedate Goths. I won’t – no way.

Alps, idyllic stage set, are not so gas-emitting, I educe. To nose, peer, far off, I tip mats onto lane. Power grew or got deep so I dare not stir. Of deer, billions sell. I ate lump – mad sign, I do cede – tonsil a pain, acne pang is sad also. Elm I help pot, live – tub’s sold; a ban or a bar, even so, elms, I’d assume, live for. Effused am I not, up in a manor, not all up in a mess.

Open if a main A.C. plug is in it.

Late men I fed late – pasties or not. “Rapture” by a maestro prevents a vast sum erased.

Argon in units, albeit at solid eye level, sits in a … (I presume not) … tube, son. No eyes: a hot laser – is Ed wary?

Mermaid, ex- evoker of all A.B.s, I flog. Nil I repaid. Emotion! Emotion, oh so do I dare, woe!

Wee yap-yap dog’s name’s Ron. An idol lacks a dime tip, or did, as today a potato in a pitta slice costs a lot – tons. A wet adder ate more hay. Ugh! So, pal, ice cost on top? No, miss, I’m a two-sided rat, erred nut, I base it on erotic ill; It is I, red now; it is debris, rot.

Alf, an idle he-man as “master animal lifer” did time, ran off at speed, but a G.I. did nab an idle if dim nit. Upwards rewards are natural life’s words, God. Fill up guts, boy, live now or do not emit one later. A rat on site got flu.

Gaelic, I’m odd Erin, I’m Eire, esteemed islet. So hostile rifts ebb. Mob, I.R.A., dare not net R.U.C. – no cotton. Erase not, so I dare not nettle men in red rose garden – I’m in it.

Stop late men if foreign at nine. Esplanades, simple hotel, bath, gin – king is Edward IX; obese; Ma is no pure mater. Go! Rise, sir; part anon.

I also rehash tests – ‘O’ Level Latin, Italian. S.A.S., so, to track radar. Often nobleman exists alone – not sales reps – I do. Trade ceiling, i.e. final part, lures open if evil trade.

Volga River rises on no steppe. Elsinore has a hamlet – Oh, Bard, row on Avon!

A sacred robot nurses simple hero’s eye; dabs on it a lemon. Gas, iron, Essex often stops, suffers in a mania. Ron fixes several crofts, acer of maple. Hot, I fish; cold, I arise laden; if diary amiss, I know it set no car off. Foe-damned ruby motor, it only rebels.

Ian I swept aside to visit, in a bar of moorside red, Romanis met in a more mossy ale den. Inspired am I, Oswald. A bay bed is nine p on top. No name, niftiness- elder saw it. Oh no! Saw top wet star’s pool – part star, part otter. Refer a baron to idiot, Tony, as I did.

Smart ones use submarine.

Poet, arch-super-artiste, grew artistic. I lost rattle; my amiable, not oh so old, able to hang up, mina, can deliver it, so true. “Ta, matey!” – says so Boston (Mass.) elder I hit.

On file S.A.E. was sent – I wrote poster re fat on side, volume one – loved it, never off it, I was in. Aide mocks a manic; I mock comic, I nap: too bad I had a fit, I too. Bottle ban odd, I go drop mine, ergo trial ceded a fatness, sober if not so, or a test is debased.

A vet is agog – no pet’s in a cask – corgi dog, royal pet, a risk no more.

Lob a rod at a man I meet. Sewer delays pit fires – a bedlam in a dig – iron ore rots it. No devil is a hero – Nimrod.

At a mall a cod is all I get. I bet on Eva, so Tim ate whole eel bait, I pay tip, Eva had a hood sewed. No B.A. gave car to Nero, we were not to rev it and we lost a trail; I’m a free pill, I wrong a man. I erase gap; to help miss it, I fill a set. A gent in ire knocks a cadet.

Animals’ gel on spoon – it is so true to basics – I’d gel; too bad I hide kangaroo baths – I lived as I won raffle, flew as I did go, dash, to my, also too tired now, star comedy: A wan, inept, upset I’m retired, nut; its ilk, nicer. Nettle feels a sore; sad, I did no panic in a pain, am an ill or tired, nude, based item; it is a spot.

Semitone, not a tone, radios emit; no, on tape; elsewhere it’s a tone.

Tail is not on; pots open on foot, even on it, so let oven (on, it is) simmer – a hotpot’s a stupid ham stew.

Loop a loop, animal – cat up in air.

Both sacks I rate by apple hewn in elder’s garden if it rates, I was aware – tasted a core.

Zones or areas, Annie, cap, so twelfth girl, lass, alas, simply (alpha beta) done, Kate. Tadpole won top Oscar, Obadiah, “O” Level axed.

Argon gas did emit no straw, so ozone sure drops argon, oozes up in Ruth’s ample hotel or sits afar off in a bar – of elastic, is it?

I hate cinema; cinema dogs in a mass. Older effusion to old – lost, is it now? Reward: a mood.

All upsets it.

Radar trails an Islamic educer of a riling issue, damages it in Israel. Ceiling is, I say, a plan, a case of one deck. Can knees sag as one Latin image elates, I wonder?

Oboe diverts ultra foe, volatile bald ogre – push to berate; I’d do, ogre. So, p.m., Oct. first, never play organ’s stops – lay or put it up in ward ten.

Final cast like rowing – I sedate play, old as am I, God! Am I! On tacks I ran; I saw rats. A Gemini tramp is May born.

I back colony’s sober omen of lack of lace. Rome, not Paris, a wonder.

Obey retail law – a noose killed oyster. Reflate my ball, a water-filled one. Disabuse no name of emanating issue.

Damsels, I note, vary tastes so cost now desserts. I say no! Try taste more honeyed. A bad nemesis at naff ruse will upset. I, mere Satanist, e.g. rater of a devil – (Oh wrong is a sin!) – I’m no devil’s god, damned.

Animals, if on a mat, sit. Rain, a more vile drop, made us site it in a cottage. Breed deer – bottle fits a llama.

I lay, as I emanate, go to sleep, mad ones on docks – air is hot. Entrap, net, nine men in party raid - all if it is in a crab-pot room, an itemised, under-lit, nullified old log den – I’m sure voles made it rot in knot.

Tubas we see far off lack limit. A cat on still or tall upward paws to no dog is an ample hot-dog, ergo nastier if tastier, eh? We, raw amid a conman, a mama in a mask, corpse et al., err.

Octuple tracks at a son’s eyelash side distressed a tall eye doctor, a tall ace, rigger of a vote: got put in egress; odd, abased, is ebbed, as I am, Amy, asinine lot! Nine lots! Don’t six rams live? Don’t six exist?

Alfred, nuts or fool gigolo, trod never if gold locks all in a flap on a red rose; made nine or ten stops.

I heed never, I’m Daisy, a prod never, I terrorise viler starfish. To me suitors, no lemons, came rowing. Is a sin a mania? Rot!

Sit! I fix a looted amp or delay more, hasten not. A baser if snug stool, wonkier, if not - Alf says - super, a ballet to no devil, is a stool too. Ban it, actor, race to no tune.

May names I wrote wrong (Is no man in it, a long old log?) sit in row, sign irate Goths; I dare drop it. At felon’s eye I peer, fast open – I’m nosey, esteem eyes. All upset, ample hogs resume totting. Is sad nabob tired? Roots don’t evade liver in Alf’s gob.

Deers I held right; oblong, apt enamel or tile rifle on gun spot to get a man – aim is all. I rogate, minister. Feeble gnats, alas late, prosaic, a canine pet is not to consume hot.

Loo, wet, issues old idiot; evading, I sneer, obey a deer, gall a deer, gain alpine dragnet for egg I’d net to ram in a pan I made to help master. Rags I held, arcane poet, arcane poetic error, all odd; I bottle fine panacean lust. I’d nag elks I ride if editor toted a minor. I fog a natural life.

Roses, or level dumb ones – rows in a mown, ample, hewn acre. Wolfsbane made it a garden in May, a garden indeed.

Nine mates, nine tons I must save now on time – editor raps a late man. G.I.s edit also, too. Do over if tests in a task radiate. Rob ran; I, too, fled.

“Omega” – list in alphabet.

A gander, a line of live ducks, irk cubs. A wart, set at a cast on knee, snug as spots.

A poor denim for a janitor, racer, armed aide, solid idler – rabid; I’d elastic in a pot, tons to sew.

Tubes or axes went in a clam, in an oyster. Free booze – lap it all up. Pity, my apple hot, so I’d a root stew. God, a stew! Tip it at feline! Posies, a cat’s altar often, no baron packs. A monk caps dog – I meddle here – hot? Pull its leg! A bee was a hoot, eh?

No, it is opposite. Yaks I rode wore hats, albeit on deity’s orders. Rats age more held in a trap, nip and I know it – set no cage now.

It’s eta; no, it’s a beta – Tsar of Tonga rates isles. Mad Ed is all upset at cider, is Ed? Is a madam too? Snip? I’d snip, spot a fine position, snip nine more cinemas.

Do ogres sell in a mall? Yes, on a barge so rats row tubs.

Wall last canes up or Eros, an imp, lives to irk, rasp or dam all tides sent. I won’t – I was no Roman – even I saw tired row – a sore. He lives on. “No!” we yell.

Up, now! Wards are in nurses’ sole care. I, peer, fed, am too fat? Oh, not I, test no dined ruby ale; dote not on salad it’s in – I am sad.

Locks I rifle so troops atone re war. Only rebel or a crofter animates so cottage beheld arcades, so trees are sold, abased. I redo, rehang, I err – a wasted act; nests I’d – as an owl – laid. A boot’s raw foot, even if a foot to master, germs (ah!) can evil do.

Pan is tune-pipe – so hot notes, paths up to honeydew.

Odd locks, a maddened (I was aware) macaw on top, spot no seen knots, rifts or fan, I saw. Are maces a baton, madam? Oodles, madam? Rare laptops are too late – got too lit up.

Nits rub – snip now, I’ll abate, not snip, nits I held.

Nubile Danish tomboys I led to old loser as no melons I held; no fish to my name. Nod lower, do I dare? No, one nods a hairy snipe. (Edit: one hairy snipe, eh?) See silliness, else we’ll ask cornish to obey deity’s or god’s item. I, God, damn it! I was in it! To Hades, acne trap, sad loser! As warts pop, a dosser I – we – vile rat, sack! Same row, oh woe! Macaroni, rats, as a hoot, tie. I vomit on rats. [/b]

Wow thats amazing!!! :bruce: :bruce: :bruce: :bruce:

Does anyone want to check whether this is indeed a palindrome? :D

this, indeed, is a palindrome, if we dont count all the quotations, exclamation/question marks, periods, etc.

He got it from: "Source: http://www.palindromelist.com/longest.htm", as he stated above.  And if you really think about, this is just numerous palindromes piled together, they really have no flow, or mutual topic....

[quote="gr8sk8r"]:wacko: :wacko:  :wacko:  :wacko:  :wacko:

where did you get that one? or did you make it up? which i highly doubt (no offense).[/quote]
None taken. The person who made this must been really bored. :rotfl:

[quote="mathnerd314"]And, of course, the old joke that fear of palindromes is

[size=200]AIBOHPHOBIA[/size][/quote]

now I know why it's called that  :D

kayak, radar
Ten animals I slam in a net :!:

A palindrome in [url=http://users.utu.fi/mitaja/pitkpali.txt]Finnish[/url]. ;)

[quote="Sathej"]All forgot about the original question, or what?
I feel its The Principle of Mathematical Induction??
Sanjith[/quote]

What? The answer to the original question was that it's a palindrome.

Does the finnish palindrome actually make sense? If it does, that is so cool! :)

Do you people know about sentences like this: "In this sentence, there are three letters "i", three letters "n"", etc., except the fact that the sentence describes the number of letters for each letter in the alphabet!
I know some in Russian. they are really cool.
It's actually very hard to come up with those.

The palindrome has no proper meaning. For example, here is a part of the palindrome and its translation. It's little hard to translate it to English since the palindrome is not good Finnish. Eino and Riku are Finnish male names as well as Veli-Matti. And Vellu is a nickname of Veli-Matti.

Kaskelotti: ovi ehätti nenään. Ei, Eino! Poni,
poni vei Vellun. Isäs sätti käsiä, Riku mielisteli. -Enon akku katosi,
muksusi sanoi.

A sperm whale: the door had time to go to nose. Eino, no! A pony, pony took Vellu. Your father was scloding hands, Riku flattered. -Uncle's battery disappeared said your kid.

[quote="me@home"]When I was younger my teacher had a book called
"Go hang a salami! I'm a lasagna hog!"
which is a book of illustrated palindromes (which is always funny, as they aren't meant to be comprehended)[/quote]

I have that book!! :D I loved the illustrations...


search
combinatorics proposed
combinatorics

Noah has 8 species of animals to fit into 4 cages of the ark. He plans to put species in each cage. It turns out that, for each species, there are at most 3 other species with which it cannot share the accomodation. Prove that there is a way to assign the animals to their cages so that each species shares with compatible species.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=907693445&t=212214

I searched for "noah cage" to find that thread.  The problem was also apparently reused at the 2005 Pan African Olympiad.


Is it possible to have a comment environment in $ \text{\LaTeX}$? Such as the code tags are on this forum.

For example instead of

[code]% bla bla
% asd asd
% tt tt
[/code]

one would use

[code]\begin{comment}
bla bla
asd asd
tt tt
\end{comment}
[/code]

You can have [i]exactly[/i]
[code]\begin{comment}
bla bla
asd asd
tt tt
\end{comment}[/code] All you need is to add the comment package with \usepackage{comment} in the preamble.

Awesome! Thanks for the quick reply Steve.


[quote="Ezbakhe Yassin"]Let $a,b,c>0$ that satisfy $abc=1$. Prove that:

\[ 2\cdot \frac{(a+b+c)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3  \][/quote]

Hi Yassin!

Cauchy's ineq yields $3(a+b+c)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$

so $\text{LHS}\geq 2 \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^3}{9}$

Calling $\sqrt{a}+\sqrt{b}+\sqrt{c}=A$ we just need to prove that

$2A^3\geq 9A+27$ which is equivalent to $(A-3)(2A^2+6A+9)\geq 0$

obviously true due to AM-GM: $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq 3\sqrt[6]{abc}=3$

Hope I'm right and unterstandable  :)  :D


probability
function
topology
real analysis
real analysis theorems

a) Since there are infinite many reals and rationals, is it true that there are "more" reals than rationals? and how do you prove that?

b) From Archimedean property of real - for every real number x, there exists a natural number n, where n>x.
I know both reals and natural numbers are NOT bounded above, but am I correct to interpret the above as this - [i]no matter how large a real number is selected, I can always find a larger natural number, hence the largest natural number is bigger than the largest real number (I know both do not exist) [/i]?

Thanks for the help.

"More" in what sense?

There are three senses for what you might mean by "more".

1. The crudest version of the word "more" is given by cardinality. Two sets have the same cardinality if and only if they they can be put into a one-to-one correspondence with each other. By this criterion, by cardinality, there are exactly as many even integers are there are integers. In fact, there are exactly as many rational numbers as there are integers. In fact, there are exactly as many algebraic numbers as there are integers. (Examples of algebraic numbers include $ \sqrt{2},\sqrt[5]{8\plus{}\sqrt[3]{17}},$ the real root of $ x^7\plus{}x^3\plus{}6\equal{}0,$ and so on.) All of these sets are countably infinite - all that means is that they can be put into one-to-one correspondence with $ \mathbb{N},$ the set of natural numbers.

On the other hand, the real numbers are uncountable - which means they cannot be put into a one-to-one correspondence with $ \mathbb{N}.$ And that means that the irrational numbers are also uncountable - that there as many of them as the real numbers. This cardinality - the cardinality of the reals, the cardinality of the continuum - is strictly greater than the countable cardinality. So there are, by cardinality, more irrationals than rationals.

For a proof of the uncountability of $ \mathbb{R},$ look up "Cantor's diagonal argument" or something like that. For a theorem that guarantees that you can actually compare cardinalities, see the "Cantor-Schroeder-Bernstein Theorem" (or any two of those names).

2. Or you can compare sets by measure. Probability is a specific version of measure. Choose a real number in $ [0,1]$ randomly according to the uniform distribution on that interval. The probability that the number you choose is rational is zero. The probability that the number you choose is irrational is one. By that criterion, there are more irrationals than rationals.

3. Baire category. The rationals can be written as a countable union of nowhere dense sets. The real numbers, being a complete metric space, cannot be written as a countable union of nowhere dense sets, meaning that the irrationals cannot be written as a countable union of nowhere dense sets. So by Baire category, there are more irrationals than rationals.

As for what you said in part (b): there is no such thing as the largest real number. Or the largest natural numbers. Given any real, I can find a larger integer. But given any integer I can find a larger real. Any set which is not bounded above has no largest element. But even some sets that are bounded above, such as the open interval $ (0,1),$ also do not have a largest element.

As always, I appreciate your detail response, Prof. Merryfield.

a) For "more", I am referring to Cardinality. Your explanation is clear, I will have to do some reading on countability.

b) On my other question, I knew something was wrong, I am aware that they are not bounded above and there is no largest element but I couldn't escape my fallacy from misinterpreting the "n>x" in Archimedean property. Furthermore, what you said
[quote]... But given any integer I can find a larger real...[/quote]
implies that given any n, there exists a real number x, where x>n. How can both "n>x" and "x>n" be true? or perhaps, it is more like "n>x" and "y>n" where x,y are real numbers. I am confused.

Sorry for another push for detail  :)

They are indeed not the same instances of those variable names. We reuse names all the time.

It looks like you're not really getting the quantifiers; in "for any $ n$, there exists $ x$ such that...", that $ x$ doesn't exist independently of $ n$, and actually choosing one would be a function $ x(n)$.

Yes, I see my problem now, hopefully it sticks, thanks for pointing that out  :)


what do bots do exactly? i've looked at--for example--MSNBot's profile and they never posted a thing!! is there a point to them?

Well, I know GameBot posts people's questions when you click "Ask on Forum". As for the other bots, no clue.

Please use the following links:
http://www.mathlinks.ro/viewtopic.php?search_id=1205226179&t=99267
http://en.wikipedia.org/wiki/Internet_bot


inequalities
algebra
polynomial
algebra unsolved

schow that if the polynomiale $ax^{3}+bx^{2}+cx+d$ have three positif roots then the condition $bc\leq3ad$ is nessecére.

$a\not =0$, as $bx^{2}+cx+d$ has no three roots.
$x_{1},x_{2},x_{3}>0$ distinct roots of $ax^{3}+bx^{2}+cx+d$, then
Then [url=http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas]Viète's formulas[/url] gives
$x_{1}+x_{2}+x_{3}=-{b\over a}$,
$x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}={c\over a}$ and
$x_{1}x_{2}x_{3}=-{d\over a}$.
The [url=http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means]AM-GM inequality[/url] then gives
$-{b\over 3a}={x_{1}+x_{2}+x_{3}\over 3}>\sqrt[3]{x_{1}x_{2}x_{3}}=\sqrt[3]{-{d\over a}}$ and
${c\over 3a}={x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}\over 3}>\sqrt[3]{x_{1}x_{2}x_{1}x_{3}x_{2}x_{3}}=\sqrt[3]{(-{d\over a}})^{2}$,
and multiplying those inequalities gives
$-{bc\over 9a^{2}}>-{d\over a}$.
Multiplying with $-9a^{2}<0$ gives $bc<9ad$.
Now $ad=-x_{1}x_{2}x_{3}a^{2}<0$, so $9ad<3ad$ and the result follows.

Now you mentioned roots of cubics here a very simple but quite useful result.

Excluding the cubic $f(x)=x^{3}$.

Let $y_{1}, y_{2}$ be the $y-value$ of the two stationary points of a cubic polynomial

1. If $y_{1}y_{2}< 0$ then the cubic has $3$ distinct real roots

2. If $y_{1}y_{2}= 0$ then the cubic has $2$ real roots where one is a root of multiplicity 2 (i.e a double root)

3. If $y_{1}y_{2}> 0$ then the cubic has $1$ real root

These can be proved graphically.


prove that there is not n that:
[b]7l2^n+1[/b]

there are three expressions for each positive integer like $n$:
$n = 3k \rightarrow 2^{3k}+1 \equiv 8^{k}+1 \equiv 2 \;( mod\;7 )$
$n = 3k+1 \rightarrow 2^{3k+1}+1 \equiv 8^{k}\times 2+1 \equiv 3 \;( mod\;7 )$
$n = 3k+2 \rightarrow 2^{3k+2}+1 \equiv 8^{k}\times 4+1 \equiv 5 \;( mod\;7 )$
As a consequence there's no positive integer $n$ that $7 | 2^{n}+1$ .

Essentially the same thing, but $2^{n}\equiv 1, 2, 4 \mod{7}$, and an $n$ which satisfies the condition required would require $2^{n}\equiv 6\mod{7}$.  Where is this from?  I don't think this would be an olympiad problem.


Hey i found this problem and i was trying to brute force it but it wouldn't come out so yea please help

Find all real triples (a,b,c) of real numbers such that x is a real number that satisfies

$ \sqrt {2x^2 + ax + b}  > x-c $

$ x\le 0$ and $x> 1$

yea i got it down to something but there were so many combinations so im not sure how to get this

I think you switched around the greater than and less than signs.

yea probably otherwise i dun see how u could get a finite amount of (a,b,c)
oh well i found it on this kid's site
http://www.xanga.com/osdes
so maybe he wrote it wrong

But its a 2002 MSOP homework problem  :?

If and only if x<=0 and x>1
edit:  oops, yeah If and only if x<=0 or x>1

And I have a question.  I know it means x can't be between 0  and 1, but does it also mean that x [i]must[/i] be true for [i]all[/i] values {x: x<=0  :cup: x>0}

[quote="djshowdown2"]But its a 2002 MSOP homework problem  :?

If and only if x<=0 and x>1[/quote]

Maybe you mean 'or' instead of 'and' :)


Solve this absolute value inequality:

$|x+3|<|x-8|$.

[hide]
by a simple graphic interpretation, $x<\frac{5}{2}$, finding this value from the equation $-x+8=x+3$
[/hide]

Or if this was harder, i.e it had $2x$ or something in it, just square both sides to make them positive, and solve normally.
$x<\frac{5}{2}$

$|x+3|<|x-8|\Longleftrightarrow |x+3|^2<|x-8|^2$, yielding $x<\frac{55}{22}.$ The answer is $x<\frac{5}{2}.$

[quote="kunny"]yielding $x<\frac{55}{22}.$ The answer is $x<\frac{5}{2}.$[/quote]

How did you make that transition?

Yes, you probably can.
Just like you can change $|x|=k \Longrightarrow -k=x=k$
into $|x|=|k| \Longrightarrow -k=x=k$

[quote="236factorial"][quote="kunny"]You can reduce by 11.[/quote]

Oh  :rotfl:  that was stupid of me.

On another thought, can you use the theorem

$|x|<k$ if and only if $-k<x<k$

In this problem?[/quote]

Yes. You can rewrite as $|x+3|<|x-8|\Longleftrightarrow \left|\frac{x+3}{x-8}\right|<1.$


show that (p-1)! = -1 (mod p) if and only if p is a prime number.

(the symbol "=" means congruent in this problem)

This is the Wilson theorem, you can use the Fermat' theorem or the propriety of [tex]a<b<p[/tex] such that [tex]a \cdot b\equiv 1[/tex] to proof it.

Hi ev'rybody, this is my first post here, I hope to learn a lot from you. Bye


A rectangle of white squares, each one unit in area, is surrounded by a single layer of unit squares, all black. Find the minimum number of white and black squares such that the number of white squares equals the number of black squares.

Hi,

[hide]Is it  number of white squares = number of black squares = 24 ?[/hide]

Sambit

[quote="JBL"]One normally provides a solution in addition to just an answer, so others can see where the answer came from.  After all, the answer might be correct while the method is flawed.[/quote]

Now what? :? I think the crew here should first know what they want before complaining

If we post a solution, it gets removed to give others a chance, if we don't, you complain we didn't post one?

Please, any mod should know that step 1 is to agree on the rules and only step 2 is to execute them :roll:

Anyway, here goes the full solution:

[hide]The surface of the  rectangle is .

The border around it is .

 can be rewritten as

Since we may suppose WLOG that , the only natural solutions are:

In the first case we need 30 squares, in the second 24, so the minimum amount is 24 squares.[/hide]

Peter

Hi,

I am sorry about not posting my solution. Actually, I got the solution by logically arguing about the rectangle. I didn't do it exactly the way PeterVDD did. Even though both our ideas are similar, still his method is definitely better in the sense its mathematically concrete, while I just argued based on the diagram. (In fact, I missed the case of 30 squares.)

I didn't post my solution, for two reasons.
First, I wanted to be sure that my method was correct. So, if my answer was correct, then I would have posted my solution; on the other hand, if its wrong, then I would have tried to find out the error in my technique.

Secondly, (assuming I correctly solved the problem) I thought its always good to allow others to solve the problem, so that one can try on his own (without even a hint) and then possibly tally the answers with what other people got. If there is still some problem, then people can sort that out with extensive explanations. But, one should try out a problem first before looking at the solution.

On the other hand, if the question asks to prove something, then one can't post any solution except either a hint to the proof, or the entire proof itself. That's a different case.

Agreed, one can post one's solution in Spoiler and thus allow others to have a go at the problem themselves. But, I thought just posting the answer will be a better way.

Anyway, I am really sorry if I shouldn't have done that. In fact, I used to post full solutions initially, till I found one day that I got irritated if I saw the solution to some problem (when the page is loading) and thus couldn't try it fully on my own.

I guess this has been too long. I wrote this just to say that giving others a chance was the only reason why I didn't give my solution.

Thanks for reading all this.

Sambit

[quote="sambitnayak"]In fact, I used to post full solutions initially, till I found one day that I got irritated if I saw the solution to some problem (when the page is loading) and thus couldn't try it fully on my own. [/quote]
Don't worry about doing that, it is an art just to focus on the question, not on the posts around it. Not just not to know the answer, but as a style of solving.

Self-discipline is a very valuable characteristic if you're training on yourself, and not looking at the solutions even while they're on your screen for a second, is really a good step towards more self-discipline. (really, this has been proven)

[quote="sambitnayak"]I guess this has been too long. I wrote this just to say that giving others a chance was the only reason why I didn't give my solution.[/quote]
Don't worry sambitnayak, you're free to solve any and all questions you want, in the way you like. No reason to feel bad about what you've posted ;)


AMC USA(J)MO USAMO AIME MATHCOUNTS probability geometry

It really is slow in the MC board, at least for the time around nats. There are no comps. It seems like someone not going to nats should do one. I'll do it if people want. There's only about 2 weeks till then, so it may have to be fast. So, here's how it works. Every round there will be 3 problems, and 2 days to do them in. If you miss one, you're out. Last one perfect wins. It's different than most that have been here before. Anyone interested?

Count me in  :D

This'll definitely be good practice for everyone who's goin to nats.

Sounds cool

maybe instead of three per two days, one problem per day?

Just tell me when and where, and I'll be there!

[quote="13375P34K43V312"][quote="perfect628"]Yeah, but then if you didn't come on during a certain day, you'd be out. That doesn't seem fair.[/quote]

heh but what if you're out two days in a row
like i was during the usamo

o.O[/quote]

Two days is your own fault. One day is understandable. It seems lke this will happen, so I'll elaborate a little more. The rounds will be of increasing difficulty. Once there are about 10-15 people, I'll put up the first round, but people can sign up after that. Again, two days for 3 problems. Only answers are needed. Of course, everything in simplest radical or fractional form, etc.

edit:nvm-don't sign up, i don't need it. Just send in answers.

How hard will you make the problems? I'll probably be out in the first round, but what if 2 or 3 people get all of them right?

It'll start easy, but, depending on how long it goes, they could get pretty tough. If there are only a few left after 1 or 2 rounds, I'll give everyone left some sort of pass to miss a question and everyone who got out on the last round will be back in. I may post the problems as soon as tomorrow, but probably not until saturday.

Suggestion:
Maybe if someone PM's you in advance, they are allowed to miss a few days?

... just a suggestion. :maybe:

sure i guess ill try, but i might not go on the forum everyday :maybe:

how hard are your questions?

im pretty dumb, so u'll probably have to start with some easy ones  :blush:

[quote="vamathletes"]What if its like 15:12, though?[/quote]
Read the question carefully.

I will do the Alternate Round 1.
For #1, do you mean the area above the chord (the smaller part of the circle),
the area of the triangle,
the radius of the circle ,
or the length of the chord

The area of the circle-you have only half an hour for Alt1 and R2.

Ah--thanks for reminding me that the two capital letters after "8:00" are the important ones.
Okay, about the card conjecture problem:
Are two cards face up and the rest face down?  Could the blue card I see possibly also be the queen of sqades, or are the five cards described one by one?  How do you define "checking a card":  Do you pull cards from the deck or flip over the five you already have?  Is the conjecture for all cards in the deck or just the 5 that have been dealt?  Sorry about all the questions, but I'm not sure what you're trying to say. Is it just me?

For the clock problem, I'm assuming that "8:7" and "4:9" are Not legitimate times, because you'd never see them on a clock.

[quote="perfect628"]I'm extending this round until tomorrow, at 4:00 ET. I got very few responses, and would like to see if time was the issue.[/quote]

Im not responding because I dont understand what  number 1 is trying to say?  can i do alt rd 1 as round 2?

I think he meant alternate round question 1.
[hide="question 1 rephrased"]
There is a circle O, with a chord.  The [b]perpendicular distance[/b] from the chord to the center is 5.  If the chord is 24 units long, what is the area of the circle?[/hide]

EDIT: Oooookay, maybe he didn't mean Alt. R1.
I rephrased regular round 1 question 1 too, take a look at that.

[quote="perfect628"]No-1 is asking what is the prob. that when the minutes and hour of a randomly selected time from 1:00-12:59 are switched, you still have a time before 8:20 that is a real time.[/quote]

what?  but that means 1:9 isnt a time right?-like what archemedies said because thats exactly what I did in my solution to you.

Always assume 0 can be placed in front of a number to make it a time-so that one is 1:09. Sorry if that was unclear. Oh, and Circlesquared, the distance from the point to a line is defined as the perp. distance. No need to help people with definitions. :wink:

AM I the only one who doesn't get Alt.1 #3?  Can someone who understands it please explain it to me?

I'll clarify.  BTW sorry perfect628, I won't help with definitions anymore.
[hide]
There are 2 decks.  One has red on the back, and the other has blue.  You see 5 cards total from these 2 decks on a table.  Your friend Jim conjectures that all cards on the table with red backs have spades on the other side.  You look at the 5 cards on the table.

2 are facedown red-backed cards.
1 is a facedown blue-backed card.
One is a faceup queen of spades of unknown back color.
One is a faceup 2 of hearts of unknown back color.

How many minimum do you have to flip over to see for sure if Jim's conjecture is true or false?[/hide]

r u gonna post the answers perfect?  It's like 7:00 or something now.

I got 3 responses, so I'm ending this. A mod can lock this if it doesn't seem anyone is saying anything.

No wonder no one responded, the questions were not like any mathcounts questions I have ever done.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="fermatprime371"]Let $ p$ be a prime, $ G$ a finite group, and $ P$ a $ p$-Sylow subgroup of $ G$.  Let $ M$ be any subgroup of $ G$ which contains $ N_G(P)$.  Prove that $ [G: M]\equiv 1$ (mod $ p$). (Hint: look carefully at Sylow's Theorems.)[/quote]

[hide="Solution"]By Sylow's Theorem Part 3, we have that $ [G: N(P)] \equiv 1 \mod{p}$. Then $ P$ is a $ p$-Sylow group of $ M$, and $ N(P)$ is the normalizer of $ P$ in $ M$, so $ [M: N(P)] \equiv 1 \mod{p}$. Thus $ [G: M] \equal{} \frac {[G: N(P)]}{[M: N(P)]} \equiv 1 \mod{p}$.[/hide]

[quote="fermatprime371"]To be precise should this be:

By Sylow's Theorem Part 3, we have that $ [G: N_G(P)] \equiv 1 \mod{p}$. Then $ P$ is a $ p$-Sylow group of $ M$, and $ N_M(P)$ is the normalizer of $ P$ in $ M$, so $ [M: N_M(P)] \equiv 1 \mod{p}$. Thus $ [G: M] \equal{} \frac {[G: N_G(P)]}{[M: N_M(P)]} \equiv 1 \mod{p}$.[/quote]

It's important to note though that $ N_M(P)\equal{}N_G(P)$.


calculus
integration
logarithms
real analysis
real analysis unsolved

Compute the following double integral:

$\int_{0}^{1}\int_{0}^{1}{(xy)^{p-1}(1-xy)^{n}}dxdy$, where $p\in \mathbb{R}, n\in \mathbb{N}$


Show that of all the quadrilaterals with fixed side lengths, the one of maximum area is cyclic. Find the maximum area in terms of the side lengths.

ok, but try proving it will have maximum area when it's cyclic without using brahmagupta's formula

you can use lagrange multipliers  :wink:


The lengths of two sides of triangle ABC form the ratio AB:BC=1:2. Find the ratio of the altitude from C to the altitude from A.

[hide]
This is basically equating areas..
$BC = 2 AB$
$\frac{1}{2}AB\cdot h_{c}= [ABC] = \frac{1}{2}BC\cdot h_{a}$
$\therefore h_{c}: h_{a}= 2 : 1$
[/hide]

inom, it would actually be 2 because it is the ratio of the h from
C to the h from A. you did the opposite.


Just eliminate the logs and it just becomes a simple system...

[quote="me@home"]yes, all is written in base four so rewrite as:
$\begin{cases}\frac{(x^2+y^2)}{(2x+1)}=(x+y)\\\frac{(xy+1)}{(4x^2+2y-2x+4)}=\frac{x}{y}-1\end{cases}$
I guess you could solve this with a little insight and a lot of algebra...[/quote]

I would say that RHS in the second equation is ${x\over 4y}$

I see your reasoning. So the equations are:
$\begin{cases}\frac{(x^2+y^2)}{(2x+1)}=(x+y)\\\frac{(xy+1)}{(4x^2+2y-2x+4)}=\frac{x}{4y}\end{cases}$

It seems like there is a nice factorization for the bottom and stuff like that, does anyone want to figure out the solution?

Man, please give sol, because I hate system and equation very much, and I've never been interested in them. This problem is, too (I don't want to think about it) :)

[quote="me@home"]yes, all is written in base four so rewrite as:
$\begin{cases}\frac{(x^2+y^2)}{(2x+1)}=(x+y)\\\frac{(xy+1)}{(4x^2+2y-2x+4)}=\frac{x}{4y}\end{cases}$
have no proof.


[quote="GameBot"]By increasing the sales tax by $ \frac12$ percent, the state of Euphoria expects to raise an extra $ \$\frac12$ million. What is the dollar amount of their expected sales, in millions?[/quote]

Let the sales now be $ x$. Thus,

$ \frac{51x}{50}\equal{}x\plus{}\frac{1}{2}$
$ x\equal{}25$

They expect to make 25.5 million dollars.

Half a percent is equal to half a million dollars. So, 100% would be $\boxed{100}$ million dollars.


[quote="zozozo123"]For all $a,b,c,\ x,y,z \in \mathbb{R}$, show that:
\[ ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}+\frac23 (a+b+c)(y+z) \ge 0  \][/quote]
I think it should be:

[quote]For all $a,b,c,\ x,y,z \in \mathbb{R}$, show that:
\[ ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}+\frac23 (a+b+c)x(x+y+z) \ge 0  \][/quote]

I think it should be:
For all $a,b,c,\ x,y,z \in \mathbb{R}$, show that:
\[ ax+by+cz+\sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}\geq\frac23 (a+b+c)(x+y+z)  \]
This is beautiful vasc's inequality.