Compute the smallest possible $a+b$ for which $a^{2}+b^{2}\equiv 1255 (\text{mod }2007)$.

I have literally no practices with mods, but here is my solution:

[hide]
$a^{2}+b^{2}\equiv 1255 (\text{mod }2007)$

To minimize the value of a and b, we assume

$a^{2}+b^{2}\equiv 1255$

Note that this equation mirrors a pythagorean triple. Also note that 1255 factors into 5 and 251. I dont recall any pythagorean triple with 251, but I do remember a pythagorean triple with 5; namely, (3, 4, 5). Since 1255 is 5 multiplied by a factor of 251, a and b are both 3 and 4 multiplied by a factor of 251. Therefore, (a + b) = (3 + 4) x 251 = 1757.

Which is odd, because this is supposedly a fake AIME problem, so there should be at most 3 digits. I have a feeling my answer is wrong..
[/hide]

[quote="p4fn2w"]
Which is odd, because this is supposedly a fake AIME problem, so there should be at most 3 digits. I have a feeling my answer is wrong..
[/quote]
1255 is not a perfect square, so pythagorean triples cannot be applied here.

[quote="bigboy82892"]
1255 is not a perfect square, so pythagorean triples cannot be applied here.[/quote]

It is true that 1255 is not a perfect square, but pythagorean triples dont always have to be perfect squares. For example, (6, 8, 10) is a triple but neither 6, 8, nor 10 are perfect squares.

edit: nevermind, I got what you mean. Sorry!

Can a mod move this problem to Intermediate or Pre-Olympiad?


Prove that if $a,b \in \mathbb{R}^+$, \[ \frac{b}{a+b} < \sum_{i=1}^b \frac{1}{a+i} < \frac{b}{a} \]

Rewrite $\frac{a}{a+b}$ as $\frac{1}{a+b}+\frac{1}{a+b}+...+\frac{1}{a+b}$ and $\frac{b}{a}$ as $\frac{1}{a}+\frac{1}{a}+...+\frac{1}{a}$ and note that the denominators of $\sum_{i=1}^b \frac{1}{a+i}$ lie between a+b and b, so each one of the terms of $\sum_{i=1}^b \frac{1}{a+i}$ is less than or equal to $\frac{1}{a}$ and greater than or equal to $\frac{1}{a+b}$, so $\frac{b}{a+b} < \sum_{i=1}^b \frac{1}{a+i} < \frac{b}{a}$


arithmetic sequence
algebra proposed
algebra

[quote="digitalboy1"]Show that any representation of 1 as the sum of distinct reciprocals of numbers drawn from the arithmetic progression $\{2,5,8,11,...\}$ such as given in the following example must have at least eight terms: \[1=\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{20}+\frac{1}{41}+\frac{1}{110}+\frac{1}{1640}\] [/quote]
suppose$1=\sum \frac{1}{a_{l}}$
$a_{l}\equiv 2(\mod 3)$
then $a_{1}a_{2}\dots a_{l}=\prod_{k=1}{l}{\frac{a_{1}a_{2}\dots a_{l}}{a_{k}}}$
then$2^{l}\equiv 2^{l-1}*l(\mod 3)$
then $l\equiv 2(mod 3)$
on the other hand,$1\geq \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}$
so $l\geq 8$


function
number theory unsolved
number theory

Find all functions $ f : Z_\plus{} \mapsto Z_\plus{}$ such that :
 $ f(1)\equal{}1$ and $ \frac{2mn}{m^2\plus{}n^2} \geq |f(n)\minus{}f(m)|$ for all $ m,n \in Z_\plus{}$

Obviosly: if $ n\not \equal{}m$, then $ |\frac{2mn}{m^2\plus{}n^2}|<1$, therefore $ f(n)\equal{}f(m)$. It mean $ f(x)\equal{}const$.


inequalities
geometry
power of a point
radical axis
geometry unsolved

Let  $AB , AC$ the tangent segments to a circle. From a point E on the line ,which comes through the midpoints of segments $AB , AC$, we draw the tangent $ED$ to the circle. Prove that  $EA = ED$

  babis

Well, the source is
                      http://www.mathlinks.ro/Forum/viewtopic.php?t=54479

   and not an indian problem.This problem is a part of the above topic.Sorry!

[u] Babis[/u]

its a very easy problem. :D then why do u insist ion solving it with out using radical axis? :(

I think it was easier that should be in own problems.
it also could be solved with the concept of power of point!!!
if you want i can also post the solution. :roll:  :huh:

i think this problem is easy for you (ashegh and pooyan_seva)becasue you are so cool in geometry.
 :D

[i]To ashegh[/i]

     I would prefer  an elementary solution to this beautiful problem - not only to this part, but to the problem with the initial form from you . It is not the best to consinder a point as a circle with radius zero and to use radical axis!. To me it is good , but some pupils of no olympiad level find this extreme unusual.If you have any idea , please write it down. Otherwise , I ' ll do it alone later, if I' ll find some time.

   [u]babis[/u]

here is my solution for this:
let $o$ be the center of the circle and $M$ be the midpoint of $AC$.$AO$ intersects $EM$ at point $F$ and $CB$ intersects $AO$ in point $H$.
\[ OH %Error. "TIMES" is a bad command.
OA=R^2 ,FH=AF \rightarrow (OF-FX) \times (OF+FA)=R^2 \rightarrow OF^2-FA^2=R^2 \rightarrow FA^2=OF^2-R^2 \rightarrow FO^2+EF^2-R^2=EF^2+AF^2 \rightarrow EO^2-R^2=AF^2+EF^2 \rightarrow ED=AE \] AND We R done. :D

Thank you Jensen !

   You saved me valuable time. I'll study in two minutes your solution.

[u]   Babis[/u]


function
inequalities
real analysis
real analysis unsolved

Hi,

this problem was in my book:
Let $f$ be convex on $I$.  Show that if $g$ is convex and increasing on an interval containing $f(I)$, then $g \circ f$ is convex on $I$.

This seems way too easy, so maybe there's a problem with what I'm doing... but to prove this don't you only need to apply $g$ to the inequality on $f$ given by convexity, which is preserved because $g$ is increasing?

Applying $g$ to $f(ta+(1-t)b)\le tf(a)+(1-t)f(b)$, you get $(g\circ f)(ta+(1-t)b)\le g(tf(a)+(1-t)f(b))$, which is not quite the same thing as convexity of $g\circ f$.

Of course, convexity of $g$ finishes the problem from there.

right, with what jmerry said is how i did the problem.  it just seemed too easy so i wanted to make sure i wasn't missing something.


If the square of a number of two digits is decreased by the square of the number formed by reversing the digits, then the result is not always divisible by:

$ \textbf{(A)}\ 9 \qquad
\textbf{(B)}\ \text{the product of the digits}\qquad
\textbf{(C)}\ \text{the sum of the digits}\qquad
\textbf{(D)}\ \text{the difference of the digits}\qquad
\textbf{(E)}\ 11$

So the two numbers are (10x+y)^2-(10y+x)^2. This simplifies to 99x^2-99y^2. This can be factored to 11*9*x+y*x-y so the answer is B

I think you meant $ 11\times9\times(a\plus{}b)\times(a\minus{}b)$ in that last expression. I agree with the rest of the solution.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt{\frac{2(8a^2\plus{}bc)}{b^2\plus{}c^2}} \ge \frac{47a^2\minus{}b^2\minus{}c^2\plus{}2ab\plus{}2ac\plus{}14bc}{5(a^2\plus{}b^2\plus{}c^2)\plus{}2(ab\plus{}bc\plus{}ca)}\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1),(0,1,1),(1,1,0).$

PS: From this inequality, we obtain the following nice statement
\[ \sum \sqrt{\frac{8a^2\plus{}bc}{b^2\plus{}c^2}} \ge \frac{9}{\sqrt{2}}\]
With equality if $ (a,b,c) \sim (1,1,1), (1,1,0)$.
:)


see that  $245 + 754 = 999$  and  $999 \mid 245754$
can we find all integers satisfay this proprièté .

[quote="Arne"]The question isn't clear. Which property are you talking about?

Please state the question formally.[/quote]

[color=red]245[/color] +[color=green]754[/color] =999
and [color=red]245[/color][color=green]754[/color] is divisible by 999
the property is find all integers  $x ,y , z$   $x+y=z$ and [u]xy[/u] is divisible by $z$
i think that it's clear now  ;)

This problem is very nicely. Understood I properly ?

Ascertain $x,y\in N^*$ such that $\left(\exists \right) n\in N^*$ for which $0<x<10^n,\ (x+y)\ |\ (x+y\cdot 10^n)\ \ (a|b \equiv$ "a divides b").

I thing that it mean:
Ascertain $x,y\in N^*$ such that $\left(\exists \right) n\in N^*$ for which $10^{n-1}<x<10^n,\ (x+y)\ |\ (x+y\cdot 10^n)\ \ (a|b$mean "a divides b").

and I guess the root is
        $x=a.k$
        $y=a.l$
 with  $10^n-1\vdots (k+l)$ and k,l are positive integer

And shall we solve the problem as opposed to defining it? ;)


This is a problem that's a real stumper, and I need to ask this. This actually isn't from mc, but I think its close enough...

A and B are playing a game. Both guess integers between 1 and 5 (inclusive). Then they take turns guessing each other's numbers. A starts; however B has twice the chance of getting A's number on any given turn. (I'm not quite sure what the heck that's supposed to imply) What is the probability that B wins the game?

This exact problem has been posted before.  Here is the adress: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128436


geometry
Ring Theory
complex numbers
advanced fields
advanced fields theorems

Could anyone tell me why the notation Spec(k[ x,y ])\{(0,0)} designes D(x) union D(y)?

The only point in R^2 at which both x and y vanish is (0,0); so it is the only point in R^2 = spec k[x,y] not covered by D(x) union D(y).

There is just no reason to use an R  :wink:

Tks!
I'm a very beginner and I don't fully understand.
Spec(R[x,y]) is a set of prime ideals.
(0,0) is a point of R².
Why can you identify R² with Spec(R(x,y])

Sorry, I didn't realize that you are talking about $ \mathrm{Spec}(k[x,y])$ and not $ k^2$ in your starting post.
You can't identify $ k^2$ with $ \mathrm{Spec}(k[x,y])$ (for no $ k$, there is a natual way). But you can see $ k^2$ as a subset, where the point $ (a,b) \in k^2$ belongs to the ideal $ < x \minus{} a,y \minus{} b > \in \mathrm{Spec}(k[x,y])$.
Now for your problem, is can also stated as: no prime ideal except $ < x,y >$ contains both $ x$ and $ y$. To see this, note that $ < x,y >$ is a maximal ideal, so no other ideal can contain it.

It also makes it a lot better if $ k$ is algebraicly closed so the residue fields are isomorphic to it but whatever. Also, in some thing like $ spec(\mathbb{C}[x])$ the closed points are of the form $ (x\minus{}a)$ with the generic point $ (0)$, but in the case of $ spec(\mathbb{R}[x])$, we have additional stuff like $ (x^2\plus{}1)$.

This question may be a little bit off topic, but is there any way to visualize spectrums of rings like $ \mathbb{R}[x]$?

The only non-closed point is the generic point $ (0)$. The closed points correspond to reals and pairs of conjugated complex numbers; so it's like folding the real plane at a line. Even more, the canoical morphism $ \mathrm{Spec}\mathbb C[x] \to \mathrm{Spec}\mathbb R[x]$ corresponds to that folding as it maps exactly the conjugated complex numbers to a single point (and the generic point to the generic point).

And can we describe spec C[x,y] or even spec C[x,y,z]...?
Is there some famous spectrum?

They are $ \mathbb{C}^2$ and $ \mathbb{C}^3$ with generic points.
More generally Hilbert's Nullstellensatz gives you the visualization of Spec of finitely generated $ k$-algebra where $ k$ is an algebraically closed field.


I want to study for AP Physics B and Physics C by myself.
Do you know what textbooks are useful? I dont have a tutor or whatever.
Im gonna study them both by myself.
Which textbooks are for AP physics b? and physics c?

Hmm...there's a lot of textbooks that over material in physics B/C. I think it might be best to talk to the teacher who teaches these subjects at your school and then check out the book from school. This way, you don't have to spend hundreds of dollars.

You should check out the collegeboard website for more information.

Specifically, the following pages have a list of text books
[url=http://www.collegeboard.com/student/testing/ap/sub_physc.html?physicsc]Physics C[/url]
[url=http://www.collegeboard.com/student/testing/ap/sub_physb.html?physicsb]Physics B[/url]

Also, you should check out the course description for AP Physics B and C
[url=http://www.collegeboard.com/prod_downloads/ap/students/physics/AP_Physics_Course_Description.pdf]Course Description[/url]

Also, there are sample free response questions on the following pages
[url=http://www.collegeboard.com/student/testing/ap/physics_c/samp.html?physicsc]Physics C[/url]
[url=http://www.collegeboard.com/student/testing/ap/physics_b/samp.html?physicsb]Physics B[/url]

Hope this helps!  :lol:


linear algebra
matrix
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let Mn(R) denote the ring of nxn matrices with entries in the ring R.
Assume the ring R contains 1.
Prove that for each ideal I of R, the set Mn(I) is an ideal of Mn(R).
Conversely, prove that every ideal J of this ring Mn(R) is of the form Mn(I) for some ideal I of R.

It is clear that $Mat_n(I)$ is an ideal of $Mat_n(R)$ for each ideal $I$ of $R$. This is a simple caluclation by the definitions. Now, let $J$ be an ideal of $Mat_n(R)$. Consider
\[ I = \{r \in R : r\mbox{~is the entry of some matrix in~} J\}  \]
Now, the point is that $r$ occurs at every position of an appropriate matrix in $J$, once it occurs anywhere in a matrix in $J$. This can be seen by multiplication of permutation matrices and the ideal-condition of $J$. So, $I$ is a subgroup of $R$, and multiplication by $r*I_n$ shows that is even an ideal of $R$. We want so show $Mat_n(I) = J$. The inclusion $J \subseteq Mat_n(I)$ is trivial. Now let $a$ be an arbitrary matrix in $Mat_n(I)$. Then
\[ a = \sum\limits_{1 \leq i,j \leq n} \begin{pmatrix} 0 & \cdots & 0 & \cdots & 0 \\ & \ddots & & \\ \vdots & & a_{ij} & & \vdots \\ & & & \ddots & \\ 0 & \cdots & 0 & \cdots & 0 \end{pmatrix}  \]
and $a_{ij}$ is the entry of some matrix $B_{ij} \in J$, i.e. $a_{ij} = (B_{ij})_{kl}$. If $E_{kl}$ denotes the matrix which is only $1$ at the entry $(k,l)$ and $0$ elsewhere, then $E_{lk} B_{ij} E_{lk}$ has the entry $a_{ij}$ at $(k,l)$ and is elsewhere $0$. So, for appropriate permutation matrices $P,Q$ we have
\[ \begin{pmatrix} 0 & \cdots & 0 & \cdots & 0 \\ & \ddots & & \\ \vdots & & a_{ij} & & \vdots \\ & & & \ddots & \\ 0 & \cdots & 0 & \cdots & 0 \end{pmatrix} = P E_{lk} B_{ij} E_{lk} Q \in J  \]
which implies $a \in J$.


the problem is:
let $x,y,z>0$,and $x^2+y^2+z^2=1$.Then
$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}} \leq 3\sqrt[4]{3}$

Sorry my friend, but it is not true.
For the contradiction check this $x=\frac{1}{\sqrt{2}}$, $y=\frac{1}{\sqrt{3}}$, $z=\frac{1}{\sqrt{6}}$

davron

I think that you mean

$\displaystyle\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\ge 3\sqrt[4]{3}$,

but yeah, this is true.  Let $a=\sqrt{x}$, $b=\sqrt{y}$, and $c=\sqrt{z}$.  Then we have $a^4+b^4+c^4=1$.

By the General Means Inequality, we have

$\displaystyle\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \le \displaystyle\left(\frac{a^4+b^4+c^4}{3}\right)^{\frac{1}{4}}=\left(\frac{1}{3}\right)^{\frac{1}{4}}$

which implies

$\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt[4]3$.


find all relatif solution

$ \displaystyle \frac{x}{y} \equal{} \frac{(x^2 \minus{} y^2)\frac{y}{x} \plus{} 1}{(x^2 \minus{} y^2)\frac{y}{x} \minus{} 1}$.

hello, simplifying your expression we get
$ \frac{(x\plus{}y)(x^2y\minus{}x\minus{}2xy^2\plus{}y^3)}{y(x^2y\minus{}y^3\plus{}x)}$
Sonnhard.

Please Dr. Sonnhard let's see some solutions not just answers.


[quote="vitalislife"]A person carries a block of wood in one hand and a bucket of water in the other. The block of wood of is now put on the water( it floats) without spilling the water. Does the person now carry less load,more load or same load as before? Explain[/quote]
[hide]

The person is carrying the same load. No mass has been lost, and the total mass times the acceleration of gravity remains the same.  The weight of the water that the wood block has displaced, pushed out of its way, is greater than the wood's own weight. The water pushes back on the wood with this same force and therefore the wood floats.[/hide]


Sorry if this is a stupid question, but is there a formula that tells you the $n^{th}$ term?

With the convention $F_{0}= 0, F_{1}= 1$, we have the identity

$F_{n}= \frac{1}{ \sqrt{5}}\left( \left( \frac{1+\sqrt{5}}{2}\right)^{n}-\left( \frac{1-\sqrt{5}}{2}\right)^{n}\right)$

A slightly nicer form:  Let $\phi, \varphi$ be the two roots of the quadratic equation $x^{2}= x+1$.  Then

$F_{n}= \frac{\phi^{n}-\varphi^{n}}{\phi-\varphi}$

Thanks.

[size=42][i]The message is too small. Please make the message longer before submitting.[/i][/size]

This was in the AOPS Volume 2 book :lol:

In case your teacher cares, it's called Binet's Formula.

I don't remember exactly how you prove it, but I know that you can verify that $F_{0}= 0, F_{1}= 1$, and $F_{n}+F_{n+1}= F_{n+2}$ implying that Binet's Formula generates the entire sequence.

Hard problem: Find $\sum_{i=1}^{\infty}\left(F_{n}\phi-F_{n+1}\right)^{2}$ where $\phi = \frac{1+\sqrt{5}}{2}$ defined in the normal manner.


Hey, I don't think that's fair! :) Just because we study for MathCounts doesn't mean we do so an hour a day or anything! I'm sure everybody here does a bunch of other stuff. For example - I'm a pianist, tennis player, and I read voraciously.

I play piano for 30-60 mins. I play violin but rarely practice at home. I usually read 1hour-2hours a day and homework takes another 1-2hours.

[quote="236factorial"]As we usually know, everything takes up some quantity of time. Studying for math takes up quite a bit of time because you need to practice with a lot of problems. Therefore, math freaks usually have little time to do anything else. Do any of you here have an exception to this statement? Do you play any sports or musical instruments?[/quote]

I honestly disagree a lot with you. May be is that I am too old to give my opinions here, because it has been seven years since I graduated from High School. Nevertheless, I did pretty well in Math, and in Math Olympiad in my contry (unfortunatelly a little late to do pretty well in international olympiads  :( ), and I used to do a lot of stuff out of math. I use to read a lot, to write a little (I was never really good at it, but I did it pretty often) and I use to spend quite a lot of time dancing (Salsa, Merengue, etc). Unfortunatelly, even when I tried to learn to play the Guitar several times, I must confess that I was not given Music or Plastic Arts skills at all. And all of that, I am talking about HS times!!

That perseption that if you like math and are pretty good at it, then you are a [b]freak[/b] is a total misconseption. There are people who really like soccer and are great at it, and spend several hours a day playing. Would you say that they don't usually have time for anything else?

Thanks for getting that message across. I agree 100%.

Most people that do math [b]do[/b] have a life.

I do have a life. Somewhat. Maybe. Ok, not really.

I participate in my school's volleyball, basketball, track (although I'm not going to nationals so that doesn't interfere), and Science Bowl programs (which we didn't win state so that's over with), along with MATHCOUNTS. I spend a lot of time doing homework, but that's mainly because I get bored with it and/or procrastinate.  :lol:

(Plus I have this science teacher who gave us this impossible project. We're supposed to view the moon twice a night/morning for 10 days at the same time each day. That is humanly impossible since she wants us to view it at least an hour apart, and now the moon is rising at like 11:00 p.m. and getting later. Oh well. Phew, that feels better. I really don't like that teacher. Grrr...) :(

Whether or not I have a life, the one thing I do know is that MATHCOUNTS should always come first.  :first:

I have a life!
(sort of)
AoPSing!
(kind of)
Being weird!
(hope not)
Piano!
(when i'm not practising)
TKD!
(soem times...)
Computer Programming!
(just trying to figure it out...)
Posting weird posts!
(I didn't say that)

I play piano (.50000000000001 hrs a day)
do Math
hang out on AoPS
Study for MathCounts Nats
Study for State Geography Bee
Study Math
Do homework
Study Math
Freetime:
Math!!!!!!
Yea!!!!!! :lol:

Wow...you're not helping me prove my point  :)

[quote="09husbri"]Nah, I'm just poking fun at you :)[/quote]

On the other hand, the guy is being totally honest. Look his nickname.

I play Basketball, and Track at my School, and I play Football, and Basketball in a league. I also play the trumpet for 50 min. a day.

Trumpet-10 secs a day
Piano-30.0123456789-min a day
Cross Country-fall
Track-spring
Baseball-spring
Tennis-spring
Running-alot (3+ miles a day)
Reading-alot (2+ hrs a day)
TV-less (.501 hrs a day)
Math-1 hr a day

I like that schedule a lot, nice and balanced. I play baseball and tennis too - although baseball in the summer.

Many afternoons I go to the nearby fitness club to workout or play basketball and/or swim. I also play piano for at least one hour each day. Plus, I play chess a lot and have friends over sometimes.

I find it easier to work about 2-3 three hours each day on math. That saves plenty of time for me and still is good study.

[quote="chess64"]hey where are the chessplayers?  :)

chess - 1-2 hours/day
math - 2-3 hours/day
homework - 15-30 min/day
soccer - fall and winter[/quote]
I play chess just not that much.

[quote="Phelpedo"][quote="236factorial"]As we usually know, everything takes up some quantity of time. Studying for math takes up quite a bit of time because you need to practice with a lot of problems. Therefore, math freaks usually have little time to do anything else. Do any of you here have an exception to this statement? Do you play any sports or musical instruments?

I play the piano (3/2 hours a day), violin, and swimming (2 hours every other day). I still have time to do the entire mathcouts handbook in one week or less.[/quote]

I think that the best mathletes, like the National Champions, spend most of their time on math (cuz other wise they wouldn't be the national champion haha!), but the rest of us do indeed have a life. That anyone who makes it to Nationals is a geek with big glasses and pi memorized to the thousandth digit is just a stereo type.

BTW, what are some of ur swimming times?[/quote]
Yeah I agree with you. I made nationals and I'm not a geek and I have no glasses and have  :pi: memorized to only the 8th digit. 3.14159265....

nothing really.

After all they'll be owning the big businesses in the future.

[quote="biffanddoc"][quote="GoBraves"][quote="biffanddoc"]nothing[/quote]

I disagree[/quote]Why???[/quote]

1. They make me look dumb...(its ok at first, but then...)
2. They always correct me when I say something with scientific explanations
3. I find them annoying

Sorry if this offends anyone. The above are characteristics of people I know, not anyone here.

Oh. I thought I was a geek I guess I'm not.

[quote="biffanddoc"][quote="GoBraves"][quote="biffanddoc"]nothing[/quote]

I disagree[/quote]Why???[/quote]

OK, let's start. According to the [i]The American Heritage® Dictionary of the English Language, Fourth Edition[/i],

[quote="The American Heritage® Dictionary of the English Language"][b]1.[/b]A person regarded as foolish, inept, or clumsy.
[b]2.[/b]A person who is single-minded or accomplished in scientific or technical pursuits but is felt to be socially inept.
[/quote]

I hope that the first definition is bad enough to say that there are bad things about being a geek (or neerd), but let's talk about the second one. To start, if someone is single-minded is like having an ice cream store who only sells a single variety of vanilla ice cream. Most of the people likes Vanilla ice cream, nevertheless, it probably would go to the bank rupcy pretty fast, because people eventually get bored. Same thing with a geek person. It is only that, if they don't have variety, they could be the best on their field, but finally people get bored of them because of their lack of variety.

On the other hand, every person that I have seen who is really consider a geek, is also a person that I think has a very low selfsteem and can't feel good unless he/she shows someone else that there is one aspect where he is better.

The type of people I hate are the ones who think they know a lot, and act like they know a lot, but then actually don't know a lot... That really annoys me sometimes...

[quote="elementsofmathkid"][hide="track is painful"]Awe, gosh. About the only life I have anymore is track. It started yesterday and we practice 5 days a week for about an hour and a half, or something like that. I am so frickin sore. We have a different athletic director, but the person who does all of the leading of the warmup and everything is the same. I think they're out to kill us or something. We start with a half mile warm up, do all of our stretches, and then do the calisthenics (which includes lots of lunges, butt kicks, skipping, bounding, accelerating, kareoke, leaping, and high knees. WoW). After all of that, we go to our events (I sprint, as I would never do distance and survive. My best official mile time at school was 8:28 on hilly ground, but I felt sick afterwards.)

And plus, the person who does sprints is the same person who has us do all of the warmups. Yesterday wasn't too bad, we just ran a bunch of short sprints, but today, relays were the death of me. I guess they were probably about 200 meters each, and you ran to the next person, and waited for someone to come back to you. (there were 4 people in a group.[/hide]

Does this sound painful, or am I just a big baby? I think it'll get better though, as this week we don't have a meet, so it's probably more conditioning.

What do the other people who run track do at their practices?
 :)  :)  :)[/quote]

Wow, track must stink... unless those warm-ups are warm-ups from the MATHCOUNTS handbook and the sprints are practice sprint rounds from MATHCOUNTS as well. :D

my schedule

weekdays --
1-2 hours eating
9-10 hours sleeping
8 hours school
1 hour homework
1 hour math
1-2 hours being bored

weekends
1-2 hours eating
10-11 hours sleeping
4 hours video games
2 hours math
2 hours watching adult swim on Cartoon Network
2 hours exercise
1 hour goofing off

anyone else play video games ?

The video discussion has been moved to F&G. Here is the link: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32755]Video games debate[/url]

Billy

Probably you guys are not like some people I know, who study to 1:30 am about I don't know what. these people quit every hobby they have to free up time for studying.

I said I play 3/2 hours of piano everyday, but I do play 2-3 hours on weekends. A total nightmare.

swimming takes 3 hours every other day instead or 2 hours. Maybe with all this reduced time, I won't do all the mathcounts problems in the world in as long as I thought.


I was working on a simple set of linear congruences.

The system was as follows:

$x\equiv 1\pmod{5}$ and $x\equiv 8\pmod{9}$.

So I just followed the usual way:

$5j+1=9k+8$

In mod 5,

$1\equiv 4k+8\pmod{5}$

so $4k\equiv-7\equiv-2\equiv 3\pmod{5}$

then $4k=5c+3$, so I just plugged it in like the Volume II Number theory thing tells you and got

$9k+8=2(4k)+k+8=2(5c+3)+k+8=10c+14+k$.  The question is, well I almost did it but how would you get rid of the k?

You don't, just as yet.

From $4k=5c+3$ you go on - take $\mod 4$ to get $c\equiv-3\equiv 1\pmod{4}$ or $c=4d+1$. Plug that back to get $4k=5(4d+1)+3=20d+8\implies k=5d+2$, then from that $x=9k+8=9(5d+2)+8=45d+26$, which is obviously $\equiv 1\pmod{5}$ and you're done.

Generally, you keep taking mods until you get the left-side coefficient of $1$.

[quote="Kurt Gödel"]Hmm, you don't really have to take mods until the left-side coefficients are 1. For example, you got $4k\equiv 3\mod{5}$. Multiply both sides with 4. You'll get $k\equiv 16k\equiv 12\equiv 2\mod{5}$. Tadaaaa. (We call 4 the [url=http://mathworld.wolfram.com/ModularInverse.html]modular inverse[/url] of 4 modulo 5.)[/quote]

Sure thing, but taking mods till $1$ is a no-brainer and surefire, albeit sometimes a bit longer.

[hide="Solution"]$x\equiv 1\bmod 5\implies x=1+5n$

$x\equiv 8\bmod 9\implies n\equiv \frac{7}{5}\equiv 5\bmod 9$

$n=5+9k\implies x=1+25+45k=\boxed{26+45k}$[/hide]

[quote="Farenhajt"]Sure thing, but taking mods till $1$ is a no-brainer and surefire, albeit sometimes a bit longer.[/quote]I wish you luck mod 1001.

What do you mean?

For example, $1001x=6y+3$ gives $5x\equiv 3\pmod{6}\iff 5x=6t+3$, and then $t+3\equiv 0\pmod{5}\iff t=5u-3\iff 5x=6(5u-3)+3=30u-15$ which gives $x=6u-3$, and then $6y=1001x-3=1001(6u-3)-3=6006u-3006\iff y=1001u-501$ and we're done.


I think there is a statement's mistake ..
I guess $a_{n}=\frac{n^{2}+1}{\sqrt{n^{4}+4}}$
 :cool:

[quote="ashwath.rabindranath"]Let $a_{n}=\frac{n^{2}+1}{\sqrt{n^{4}+4}}$ for $n = 1,2,3...$
  Define $b_{n}$ as $a_{1}a_{2}...a_{n}$ for some natural $n$.
  Prove that $\frac{b_{n}}{\sqrt{2}}= \frac{\sqrt{n^{2}+1}}{\sqrt{n^{2}+2n+2}}$[/quote]

factoring the denominator of $a_{n}$, we get:
$a_{n}=\frac{n^{2}+1}{\sqrt{((n+1)^{2}+1)\cdot((n-1)^{2}+1)}}$.
and i think $a_{1}a_{2}\cdot \cdot\cdot a_{n}$ telescopes....


geometry
circumcircle
cyclic quadrilateral

ABC is an equilateral triangle and D is a point on minor arc AB of the circumcircle
of ABC such that BD = 2005 and CD = 2006. Compute AD.

I don't really know the terminology. Does "minor arc" mean semicircle?

I think Minor Arc means the smaller arc of the circle..

[hide="Hint"]Pt[/hide]
Yep, that's the hint (any more is a giveaway)
[hide="Solution"]
Ptolemy!
We have $ AD \cdot BC \plus{} BD \cdot AC \equal{} CD \cdot AB$. Since $ AB \equal{} BC \equal{} CA$, it follows $ AD \equal{} 1$.
See [url]http://en.wikipedia.org/wiki/Ptolemy's_theorem[/url] corollary 6 (why won't the link work?)[/hide]

[hide="solution"]$ Ptolmeys  Theorem$ states that in a cyclic quadrilateral $ ABCD$, $ AB \cdot CD + BD \cdot AC = CB \cdot AD$ We see by drawing the figure that $ ABCD$ is cyclic, so by using $ Ptolmeys Theorem$, we can write the equation:
$ BD\cdot AC + BC \cdot AD = AB \cdot CD$.
plugging in the values given we get:
 $ 2005 \cdot AC + BC \cdot AD = AB \cdot 2006$
We know that $ AC=BC=AB$, so we can rewrite the equation as
$ 2005 \cdot AC + AC\cdot AD= AC \cdot 2006$.
Then divide by the common length $ (AC)$ to get:
 $ 2005+AD=2006$ and ${ AD= \boxed{\boxed{\boxed 1}}}$  [/hide]

This is basically what dgreenb801 said but more in depth in case you didnt get it.


Can someone please explain Pick's Theorem for me? Also please provide some example problems....AOPSWiki didn't really help.

picks theorem says that given any polygon on a grid of points, the area of that polygon can be found by the formula $ i \plus{} \frac {b}{2} \minus{} 1$ where $ i$ is the number of interior grid points of the polygon and $ b$ is the number of grid points on the boundary of the polygon.

This seems like a pretty unimportant formula to memorize though (correct me if Im wrong). You should probably memorize the shoelace theorem instead.

Actually it can help if your given 4 coordinates and are told to find the area.................

[quote="Thunder365"]This seems like a pretty unimportant formula to memorize though (correct me if Im wrong). You should probably memorize the shoelace theorem instead.[/quote]

I believe quite on the contrary. It certainly is possible to find the area of [hide="this"][img]http://img22.imageshack.us/img22/46/picks.png[/img][/hide] by counting squares and triangles, or subtracting the area from outside, but it's much easier to use Pick's Theorem to find it.

If you wanted the area of this figure:
[asy]import graph;
for (int i=0; i<9; ++i)
{
 for (int j=0; j<9; ++j)
 {
   dot((i,j),green);
  }
}
draw((0,0)--(8,0),Arrow);
draw((0,0)--(0,8),Arrow);
pair a=(7,3);
pair b=(3,2);
pair c=(1,7);
dot(a,red);
dot(b,red);
dot(c,red);
dot((4,5),red);
dot((3,3),blue);
dot((3,4),blue);
dot((3,5),blue);
dot((2,5),blue);
dot((2,6),blue);
dot((4,3),blue);
dot((4,4),blue);
dot((5,3),blue);
dot((5,4),blue);
dot((6,3),blue);
draw(a--b--c--cycle,blue);[/asy]

We can use the formula:
$ i \plus{} \frac {b}{2} \minus{} 1 \equal{} 10 \plus{} \frac {4}{2} \minus{} 1 \equal{} \boxed{11}$
I think that is faster than using the distance formula or shoelace

[quote="batteredbutnotdefeated"]Pick's theorem is very useful in geoboard questions because it lets you compute the area extremely quickly. If you were given a question with 5 answer choices and each of the answer choices had a diagram similar to RoFl's diagram, and if you wanted to evaluate the area of each of the diagrams, then Pick's would be the smartest way.[/quote]

I think there was a question on the 2000 AMC8(forgot which year) similar to that.  It's also useful on the last page of the MC School round, which has a problem like that usually.

[quote="SuperNerd123"]If you wanted the area of this figure:
[asy]import graph;
for (int i=0; i<9; ++i)
{
 for (int j=0; j<9; ++j)
 {
   dot((i,j),green);
  }
}
draw((0,0)--(8,0),Arrow);
draw((0,0)--(0,8),Arrow);
pair a=(7,3);
pair b=(3,2);
pair c=(1,7);
dot(a,red);
dot(b,red);
dot(c,red);
dot((4,5),red);
dot((3,3),blue);
dot((3,4),blue);
dot((3,5),blue);
dot((2,5),blue);
dot((2,6),blue);
dot((4,3),blue);
dot((4,4),blue);
dot((5,3),blue);
dot((5,4),blue);
dot((6,3),blue);
draw(a--b--c--cycle,blue);[/asy]

We can use the formula:
$ i \plus{} \frac {b}{2} \minus{} 1 \equal{} 10 \plus{} \frac {4}{2} \minus{} 1 \equal{} \boxed{13}$
I think that is faster than using the distance formula or shoelace[/quote]

$ i \plus{} \frac {b}{2} \minus{} 1 \equal{} 10 \plus{} \frac {4}{2} \minus{} 1 \equal{} 10 \plus{} 2 \minus{} 1 \equal{} \boxed{11}$


Hi everyone. So I was trying to study for the upcoming test and I  stumbled on these two questions on my book that puzzled me.

1. The function f(x)=x^4+ax^3-4x^2+bx+6 has x-intercepts (2,0) & (-3,0). Find the values of a & b.

2. The functions y=(a-2b)x^4-3x-2 & y=x^4-x^3+(a+5b)x^2-5x+7 both have an x-intercept of 1. Find the value of a & b.

I've tried solving it but I messed up in the process. I also have this parabola question from my previous test that I can't seem to answer perfectly:

The parabola has a turning point at (z,-8); it intersects the y-axis at y=10 and one of the x intercepts is x=5. Find the value of z and equation of parabola.

So, I really hope people are eager to help me solving my seemingly simple questions. I am a responsible student and I learn fast. Thanks!

Hint for 2nd part
[hide]Let the parabola be $ y \equal{} ax^2 \plus{} bx \plus{} c$. You know $ c \equal{} 10$. Since $ y$ value of the turning point is $ \minus{} 8$, this must be equal to $ c \minus{} \frac {b^2}{4a}$ (Why is this?). Plug in the other point $ (5, 0)$ and now you have 2 equations on $ a$ and $ b$. This gives you the equation for the parabola. You should now be able to find $ z$.[/hide]

[quote="123456789"][hide="Number One Hint"]You have the function $ f(x) \equal{} x^4 \plus{} ax^3 \minus{} 4x^2 \plus{} bx \plus{} 6$
Plugging in your two points ($ (2,0)$ and $ ( \minus{} 3,0)$), we have
\[ 0 \equal{} {(2)}^4 \plus{} a{(2)}^3 \minus{} 4{(x)}^2 \plus{} b(x) \plus{} 6 \\
0 \equal{} {( \minus{} 3)}^4 \plus{} a{( \minus{} 3)}^3 \minus{} 4{( \minus{} 3)}^2 \plus{} b( \minus{} 3) \plus{} 6
\]
Now you have two equations and two variables. Can you solve for $ a$ and $ b$?[/hide][/quote]
Yes. I can.
[hide="Solution"]
Simplifying the top we have:
$ 0\equal{}16\plus{}8a\minus{}16\plus{}2b\plus{}6$, or
$ \minus{}6\equal{}8a\plus{}2b$
The bottom one simplifies to:
$ 81\minus{}27a\minus{}36\minus{}3b\plus{}6$ or
$ 17\equal{}9a\plus{}b$
Now we have a [i]simplified[/i] system of equations.
$ \minus{}6\equal{}8a\plus{}2b$
$ 17\equal{}9a\plus{}b$
Solving these yields $ \boxed {a\equal{}4,y\equal{}\minus{}19}$[/hide]


[b][u]Directions[/u][/b]

[u][b]Solve For X[/b][/u]

Heres what I did so far

X- 1/x
_____    =1

x+1- 2/x

I find the LCD and it is X amd I want to get rid of the fraction.

(X) (X-1/x)
__________ =1    I multiply X with all the terms and then it becomes (x^2-1x)
                                                                                _________ =1
(X) (x-2/x)                                                                                (x^2+1x-2x/x)

I also get rid of the denominator's X with the numerator's x

x^2-1
_____

X^2+1x-2

I simplified

(x-1) (x+1)
_________
(x-1) (x+2)

and I cross out the likes which were x-1 in the numerator and the denominator.

       X+1
X= _____ =1
        X+2

This is where I am confused,.[/u]

[quote="jnsoccer66"]$\frac{x-\frac{1}{x}}{x+1-\frac{2}{x}}=1$
$x-\frac{1}{x}=x+1-\frac{2}{x}$
$x^{2}-1=x^{2}+x-2$
$x-2=-1$
$x=1$.
 :roll:[/quote]

Thanks pal.

How were you able to get the problem in such nice form, as you can see, mine gets a little messed up whim I type it.

[quote="jnsoccer66"]$\frac{x-\frac{1}{x}}{x+1-\frac{2}{x}}=1$
$x-\frac{1}{x}=x+1-\frac{2}{x}$
$x^{2}-1=x^{2}+x-2$
$x-2=-1$
$x=1$.
 :roll:[/quote]

Isn't that faulty? Substitute 1 back into the original equation and you're gonna divide by 0.

You can use the typesetting system LaTeX to make your math look nice. This site offers downloads and a good guide.

[hide="About the problem"]
$\frac{x-\frac{1}{x}}{x+1-\frac{2}{x}} = 1$
$x-\frac{1}{x}=x+1-\frac{2}{x}$
$x^2 - 1 = x^2 + x - 2$
$(x-1)(x+1) = (x-1)(x+2)$
Clearly $LS \neq RS$
Does this mean that the equation has no real solutions?
:huh:
[/hide]

Yeah, at x=1 there's a removable discontinuity. To clean up the original equation, multiply top and bottom by x. Then you can see the bottom factors and the whole thing reduces, removing the discontinuity (and yielding the solution that doesn't work). However, you could say that $\lim_{x \to 1} \frac{x-\frac{1}{x}}{x+1-\frac{2}{x}} = 1$

[color=green][size=75]mod edit: AHHH NO CALCULUS!! OH WELL[/size][/color]

I having trouble with a word problem, could someone give me tips on how I would solve a problem like this? please

 The soviets, in drilling the worlds deepest hole in 1984, found that the temperature X kilometer below the surface of the earth was giving by

 T= 30+25(x-3)3[u]<[/u]x[u]<[/u]15

where T is temperature in degrees celsius. At what depth will the temperature be between 200 degrees celsius and 300 degrees celsius inclusive.

[hide]Because we have\[ T=30+25(x-3) \]
and \[ 200<T<300 \]
we've got the inequation \[ 200<30+25(x-3)<300 \]
after a little algebra we've got \[ 9,8<x<13,8 \]
(substracting $30$ , dividing by $25$  and adding $3$)
Conclusion:after soviets calculs by about $10km$  deep it's [b]very hooooooooot![/b] :rotfl: [/hide]


$\alpha+\beta+\gamma+2=\alpha\beta\gamma$  $\forall$ $\alpha,\beta,\gamma\in\Re+$

Prove: $\alpha\beta\gamma\geq 8$

[hide]
$\alpha\beta\gamma = \alpha+\beta+\gamma+2 \ge 4\sqrt[4]{2\alpha\beta\gamma}$ by AM-GM

$(\alpha\beta\gamma)^4 \ge 2 \cdot 4^4 \alpha\beta\gamma$

$(\alpha\beta\gamma)^3 \ge 8^3$

$\alpha\beta\gamma \ge 8$.[/hide]

[hide="1"]
Let $P=\alpha \beta \gamma$. By AM-GM, $\frac{P-2}{3}\geq \sqrt[3]{P}\Rightarrow (P-2)^3\geq 27P$.

This is equivalent to $P^3-6P^2-15P-8\geq 0$. Since $P^3-6P^2-15P-8=(P-8)(P+1)^2$, $P\geq 8$.
[/hide]
[hide="2"]
Let $a=\alpha +1$, $b=\beta +1$, $c=\gamma +1$ because when $a,b,c$ are substituted into the equation the constant terms and the 1st-degree terms will cancel out. So this becomes
\[ a+b+c-1=(a-1)(b-1)(c-1)\Rightarrow ab+bc+ca=abc\Rightarrow \frac 1a +\frac 1b +\frac 1c =1 \]
By Cauchy, $(a+b+c)\left( \frac 1a +\frac 1b +\frac 1c \right) \geq 9\Rightarrow a+b+c\geq 9\Rightarrow \alpha +\beta +\gamma \geq 6\Rightarrow \alpha \beta \gamma \geq 8$.

This method even proves the inequality true for all $\alpha ,\beta ,\gamma >-1$.
[/hide]

$\alpha +\beta +\gamma \geq 6\Rightarrow \alpha \beta \gamma \geq 8$ is this true?

what if $\alpha = 4, \beta = 1, \gamma=1 \implies \alpha \beta \gamma = 4 < 8$?

It is from the given condition: $\alpha\beta\gamma = \alpha + \beta + \gamma + 2 \geq 6 + 2 = 8$.


[img]http://img.photobucket.com/albums/v603/Slippy6/math012.jpg[/img]

How do I do the bonus? I can't seem to solve it[/img]

[hide="Solution"]
We notice that
\[ \cos 2A \equal{} \cos^2 A \minus{} \sin^2 A \equal{} (\cos^2 A \plus{} \cos^2 A) \minus{} (\cos^2 A \plus{} \sin^2 A)
\]

\[ \cos 2A \equal{} 2\cos^2 A \minus{} 1\rightarrow \sec 2A \equal{} \frac {1}{2\cos^2 A \minus{} 1}
\]
and
\[ \tan 2A \equal{} \frac {\tan A \plus{} \tan A}{\tan A(\tan A)} \equal{} \frac {2\tan A}{1\minus{}\tan^2 A}
\]
So,
\[ \frac {2\tan A}{1\minus{}\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {1}{\cos 2A} \plus{} \frac {\sin 2A}{\cos 2A} \equal{} \frac {1 \plus{} \sin 2A}{\cos 2A}
\]

\[ \frac {2\tan A}{1\minus{}\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {1 \plus{} 2\cos A\sin A}{\cos^2 A \minus{} \sin^2 A} \equal{} \frac {\sin^2 A \plus{} \cos^2 A \plus{} 2\sin A\cos A}{\cos^2 A \minus{} \sin^2 A}
\]

\[ \frac {2\tan A}{1\minus{}\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {(\sin A \plus{} \cos A)^2}{(\cos A \minus{} \sin A)(\sin A \plus{} \cos A)} \equal{} \frac {\sin A \plus{} \cos A}{\cos A \minus{} \sin A}
\]

\[ \frac {2\tan A}{1\minus{}\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {\cos A \plus{} \sin A}{\cos A \minus{} \sin A}
\]
[/hide]
EDIT: Oh, sorry, AstroPhys  :blush:  I edited it.  It doesn't affect the solution, however, because we wanted to prove that $ \frac {2\tan A}{1\minus{}\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {\cos A \plus{} \sin A}{\cos A \minus{} \sin A}$, not $ \frac {2\tan A}{\tan^2 A} \plus{} \frac {1}{2\cos^2 A \minus{} 1} \equal{} \frac {\cos A \plus{} \sin A}{\cos A \minus{} \sin A}$

The QuattoMaster 6000 , is not it $ 1\minus{}\tan^{2} A$?


For which natural numbers n does there exist a convex n-gon which can be cut into some (not necessarily equal) regular polygons?

  Darij

I suppose it sjould be cut in more than one piece, right?

Yeah, the proposer seems to have considered this as evident  :D ... and I have not really thought about the problem; the solution in Kvant is 4 pages long and very messy, I just wanted to share the problem with some other MathLinkers...

  Darij


Hello,  I created this problem. I have a solution already, but I'm looking for an elegant solution, one without much algebra and expansion.

If $x,y,z$ are positive reals and $xyz=1$ then show that:
$\frac{1}{\sqrt{x^8+8x^5}} +\frac{1}{\sqrt{y^8+8y^5}} +\frac{1}{\sqrt{z^8+8z^5}} \geq 1$

[quote="zhaobin"][quote="Soarer"]I think you've reversed the sign of chebyshev.[/quote]
no,I think you've... :P[/quote] :blush:


Prove that there is a function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x))=-x$, for any $x\in\mathbb{R}$ ;)

Very well, too bad that it's already posted.

I apologise, I didn't know that. :blush:

There are many such fuctions. However, there's an interesting geometric observation: rotating the graph of such $f$ with $90^{\circ}$  around the origin we obtain the same graph (I think this is an old russian problem).
Here's an example:


New Berland consists of N (1<= N<= 100) islands, some of them are connected by bridges. There can be no more than one bridge between any pair of islands. Mr. President issued a law to paint all bridges. A bridge can be painted white or black. Any island must have at least one white bridge and at least one black (of course if an island has more than one bridge). You need to find if there exist such painting.

idea is augmenting path of max flow.

first color all the edges black. then find an augmenting path.

Can you explain it. I didin't understand.

Start with all bridges black.
Choose some island which has all its bridges black. Look at its neighbours. If not all their bridges are white (except the one going back), you can flip one of them. If all their bridges are white (except the one going back), look at their neighbours. If all their bridges are black (except the one going back), you can flip the two bridges you have followed so far. Etc.


It is quite hard . See here http://www.mathlinks.ro/Forum/viewtopic.php?t=52376&start=20&sid=71b62b1511b9df6e03d445ae3a0234e2


Well, maybe I'm a little late since IMO 2002 was held in Scotland ([url]http://www.imo2002.com[/url]) - but maybe that means that some of you reading this will already know what a wonderful country it is :)

Er.. there's the whole haggis thing (they have three legs and only go one way around hillsides, and they are ritually stabbed on January 25th).  There's also the Loch Ness Monster, bagpipes, kilts (can anyone fail to love a man in a kilt??), and many other very strange traditions.

There are lots of famous Scottish mathematicians and scientists: Taylor, Maclaurin, Kelvin, Clerk Maxwell, Hodge, Napier, Stirling, Mary Somerville, D'Arcy Thompson, ET Bell, Simson, Stewart...

So come to Scotland.  :)

didn't knew that Taylor, Stewart and Simson are scotish :D

Hey, Erica, this might seem stupid, but I always wanted to know: WHAT IS haggis?  :D

here is a link to what haggis are and also.. haggis hunt.. poor fellows! :(

http://www.haggishunt.com/

cheers! :D

[quote="grobber"]I always wanted to know: WHAT IS haggis?  :D[/quote]

aah, well!

the haggis is a small animal which lives on scottish hillsides.  they only have three legs, and one leg is shorter than the others so that it doesnt fall over (because the hills are very steep round here!)

hence they may only proceed in one direction around the hill.  rumour has it that males and females go in different directions, this being the only way to tell them apart from a distance - however, i suspect there would be some physical difficulties in reproduction (it is suggested that this is why haggis are so rare... and definitely it doesnt make sense evolutionarily)

the mysteries of the haggis are such that it remains a favourite topic of conversation and much research.  the annual haggis-hunt is celebrated on Burns' Night (January 25th) with a ritual stabbing of the beast, accompanied by poetry recitations.

come to scotland and we'll show you :)

Thanks a lot! I'm not joking you know, this really is amazing! I had dever heard of the haggis being a real animal before!

Bye! :D

i've been to scotland last year at the IMO and haven't seen any haggish-es :D

Is that REALLY the plural for haggis? :D  :D  :D

here is a noce site for those of you who would like to go to scotland.. interesting things can be found here! :D :D

http://www.mv.com/ipusers/becky/scottish.htm

cheers! :D

erica wrote:

[quote]  kilts (can anyone fail to love a man in a kilt??) [/quote]

then.. gonna buy me some kilts! :D :D :D .... and probably all the Romanians would  say that I have gone mad! :D :D

hehe cheers! :D

Now that you mentioned it, where could one buy a kilt ?  :D

there were only a few shops in Edinburgh at which one could buy kilts (at least that's what my friend Duni told me when we met there in 2002 - he was there for about 3 months and told me that at the festival there were many discounts and off-prices at kilts, and the offer is bigger).

I 've never heard of haggis. I can't imagine what it looks like. What a strange animal. But they live only in Scotland,don't they ? :D

My sister and brother-in-law live in Edinburgh... unfortunately, I've never been there (though we do plan to, if I don't get stuck doing something over the summer...)

Anyway, their description of haggis... well, I wouldn't exactly call them *amusing*, though it does say something about scottish people.

It may be late a bit, but I want to turn back to first message about Scotland. I can say that scotland was beautiful. glasgow is an old and interesting city. But I can't say the same about London(ENG). we had a travelling of 6 tiring hours for going and turning back, but it was not what I was waiting in London.


quadratics
modular arithmetic
number theory unsolved
number theory

let p be a prime number and let $a$ be the smallest nature number such that:$a^{\frac{p-1}{2}} \equiv -1 (\mod p)$
prove that:$a<\sqrt{p}+1$

[quote="grodij"]($\cdots$)such that:$a^{\frac{p-1}{2}}\equiv -1 (mod p)$
prove that:$a<\sqrt{p}+1$[/quote]
Is it now the right question?

[i]Edited[/i] because I made a mistake  :(

Hmm.. I'm sure I'm doing something wrong here, but it doesn't look that hard (or easy :)).

We want to show that the smallest quadratic non-residue is smaller than $\sqrt p-1$. $a(a-2)\ge p$. For each natural $x$, let $\bar x$ be the smallest positive integer congruent to $x\pmod p$. The sequence $\overline{a(a-2)},\overline{a(a-3)},\ldots,\overline a$ must contain a positive integer less than $a$, and we're done (sicne each of those numbers is a quadratic non-residue, and a positive integer smaller than $a$ in that sequence would mean a quadratic non-residue smaller than $a$).

Is there anything wrong with it? :?

[quote="grobber"]$a(a-2)>(a-1)^2>p$
[/quote]
Are you sure ?  ;)

That was wrong, I have edited it, but that was not a problem in itself, just a slight typo which did not invalidate the proof, if everything else is Ok.

let $b$  the smallest number such that $ab>p \Rightarrow a(b-1)\leq p$ , and we know that $a\geq 1$  then $0<ab-p<b$

now we have ${{\left(\frac{ab-p}{p}\right)}=\left(\frac{ab}{p}\right)}=1$, but ${\left(\frac{a}{p}\right)}=-1$

 therefore ${\left(\frac{b}{p}\right)}=-1$ from this we can get $b\leq a$ and now:

$(a-1)^2<(a-1)b\leq (b-1)a<p$ $\Longrightarrow$  $a<\sqrt{p}+1$

-Amir


[hide]um i think this is wrong but is it 0?[/hide]

[quote="236factorial"]What is the remainder when 1234567891011.......112211231124 is divided by 4?[/quote]

[hide]Using the divisibility by 4 rule...

24 is divisible by 4... so the entire number is divisible by 4...

The remainder is 0. Nonie is correct.

b4k4ka... how'd you get your answer? [/hide]

[quote="236factorial"]Challenge: What is its remainder when it is divided by 8?  :D[/quote]

[hide]The law for divisibility by 8 is that if the last three digits are divisible by 8, the entire thing is.

124 is NOT divisible by eight, so the entire number ISN'T divisible by 8.[/hide]

[quote="Drunken_Math"][hide]Since it's not divisible by 8, and it IS divisible by 4, the remainder would be 4.[/hide][/quote]

Good notice there. :10:


geometry
symmetry
analytic geometry
calculus
calculus computations

Is there an uncountable family $ S \subset \mathcal{P}\mathbb{N}$ such that $ A \cap B$ is finite for all distinct $ A, B \in S$?

Q: What is the meaning of "family"?

A [url=http://en.wikipedia.org/wiki/Family_of_sets]family[/url] of sets.

So it refers to a collection of subsets of $ \mathbb{N}$ then. Does $ \mathcal{P}\mathbb{N}$ have the same cardinality as the reals?

I don't think we can have such an uncountable family though

Yes, we can. Replace $ \mathbb{N}$ with $ \mathbb{Z} \times \mathbb{Z}$ or $ \mathbb{Q}$ (why can we do this?). Intuitive examples of such families can be found in both cases.

[quote="CommandoGuard"]Yes, we can. Replace $ \mathbb{N}$ with $ \mathbb{Z} \times \mathbb{Z}$ or $ \mathbb{Q}$ (why can we do this?)[/quote]

Is this because all these sets are isomorphic in terms of being countable? We can find bijections between all of these.

I don't find that intuitive though - any hints?  :maybe: I'm still hazy about all this countable/uncountable/aleph/infinite ladder business!

With $ \mathbb{Q}$, take for each $ \alpha \in \mathbb{R}$ an arbitrary sequence of rational numbers converging to $ \alpha$ and denote the set of its elements by $ A_{\alpha}$. Then $ \{ A_{\alpha} \}_{\alpha \in \mathbb{R}}$ will do the job.

With $ \mathbb{Z} \times \mathbb{Z}$ you can try to turn the assumption on finite cardinalities of intersections to an assumption on finite areas of intersections (of figures in the Euclidean plane).

[quote="CommandoGuard"]With $ \mathbb{Z} \times \mathbb{Z}$ you can try to turn the assumption on finite cardinalities of intersections to an assumption on finite areas of intersections (of figures in the Euclidean plane).[/quote]

What does this mean o.0

You can identify $ \mathbb{Z}\times\mathbb{Z}$ with the cartesian plane, with lattice points representing elements of $ \mathbb{Z}\times\mathbb{Z}$.  You can write out the elements of $ \mathbb{Z}$ as a number line, with the points being the integers.  Now take the cartesian product with another copy of $ \mathbb{Z}$ and you get the cartesian plane, with just lattice points.

At least, that's how I interpret CommandoGuard's comment.

Yup. My suggestion was to consider infinite strips of fixed "thickness" (say, 2) whose axis of symmetry is $ \{ (x,ax) : x \in \mathbb{R} \}$ for $ 0 < a < \pi$ and in each such strip take its intersection with the integer lattice. The intersection of two such strips is a finite-area polygon and thus contains only a finite number of lattice points.

Here is a similar problem that I myself have not solved: prove that $ \mathbb{N}$ contains uncountably many infinite subsets $ (N_{\alpha})_{\alpha\in\mathbb{R}}$ such that $ N_{\alpha}\cap N_{\beta}$ is finite if $ \alpha\neq\beta$.  I would love to see a solution.

It's the same question  :maybe:  (Since there are only countably many finite subsets of $ \bf N$, they're irrelevant.)

Thanks, may I ask where you guys read about this stuff? E.g. countable/uncountable subsets, power sets, infinite cardinalities, etc. Any good websites/books to recommend?

[quote="JRav"][quote="aidan"]Does $ \mathcal{P}\mathbb{N}$ have the same cardinality as the reals? [/quote]
Yes, assuming the continuum hypothesis.  That is, assuming there is no cardinality "in between" $ \aleph_{0}$ and $ \frak{c}$.[/quote]

No, assuming the continuum hypothesis is completely unnecessary.

Also, for your similar problem, identify $ \mathbb{N}$ with $ \mathbb{Q}$, and for each $ \alpha\in\mathbb{R}$, let $ N_\alpha$ be a set consisting of the members of a sequence of rational numbers converging to $ \alpha$.

Another very nice solution here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315245


Only some state problems and nationals problems are too hard to do. The only reason for low scores is because of the speed you need.


Prove that
$ 1 \plus{} \frac{3}{4} \plus{} \frac{3*5}{4*8} \plus{} \frac{3*5*7}{4*8*12} \plus{} \frac{3*5*7*9}{4*8*12*16} \plus{} ... \equal{} \sqrt{8}$

Rewrite this in summation notation.

What are you referring to with summation notation?


Problem 140
 ||,
latexify this plz :(

a=$ 4.5^{99}\minus{}\sum_{r\equal{}0}^{99}5^{r}$
b=$ 5^{99}\plus{}\sum_{r\equal{}0}^{99}5^{r}$
matrix:
a a-1
b b+1

thr mayb a bug in d final ans bt the method is
just note the sequence
1st row- 4, 19,94,562..
2nd row- 5, 25,257
find the 100th term das all :P

No i guess u ve gotten the answer terribly wrong :!:

Well genl term is
$ a_{11,n} \equal{} 5 \cdot a_{11,n \minus{} 1} \minus{} 1$
$ a_{12,n} \equal{} 5 \cdot a_{12,n \minus{} 1} \minus{} 2$
$ a_{21,n} \equal{} 5 \cdot a_{21,n \minus{} 1} \plus{} 1$
$ a_{22,n} \equal{} 5 \cdot a_{22,n \minus{} 1} \plus{} 2$

Where $ a_{ij}$ represents the element in the ith row and the jth column :)

ya thats the same thing i got. :maybe:
i said my final ans cud be wrong- carelesness i guess :oops:


integration
trigonometry
calculus
function
logarithms
calculus computations

Computing the [b]first[/b] and [b]third[/b] integral:

$SS=\int_{0}^{\frac{\pi}{2}}\sin x\sqrt{\sin 2x}\, dx = \int_{0}^{\frac{\pi}{2}}\cos x\sqrt{\sin 2x}\, dx$ ...($\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx$ )

$\Rightarrow 2SS = \int_{0}^{\frac{\pi}{2}}(\sin x+\cos x)\sqrt{1-(1-\sin 2x)}\, dx$
$\Rightarrow 2SS = \int_{0}^{\frac{\pi}{2}}(\sin x+\cos x)\sqrt{1-(\sin x-\cos x)^{2}}\, dx$

Use the substitution $y = \sin x-\cos x$ to get
$2SS = \int_{-1}^{1}\sqrt{1-y^{2}}\, dy$
$\Rightarrow SS = \int_{0}^{1}\sqrt{1-y^{2}}\, dy$ ... (The function to be integrated is an even function)

Now, make one last substitution $y= \sin \theta$ to get
$SS = \int_{0}^{\frac{\pi}{2}}\cos^{2}\theta \, d\theta = \frac12 \int_{0}^{\frac{\pi}{2}}(1+\cos 2\theta)\, d\theta$
$\Rightarrow SS = \boxed{\frac{\pi}{4}}$

Also, note that from our first step, the [b]third[/b] integral $CS = \boxed{\frac{\pi}{4}}$

In addition, $SC = CC.$ So only one of the two needs to be computed.

-----------------------------------------------

Okay,here is the method for evaluating $SC.$

$SC = \int_{0}^{\frac{\pi}{2}}\sin x\sqrt{|\cos 2x|}\,dx = \int_{0}^{\frac{\pi}{4}}\sin x\sqrt{\cos 2x}\,dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\sqrt{-\cos 2x}\,dx$
$SC = \int_{0}^{\frac{\pi}{4}}\sin x\sqrt{2\cos^{2}x-1}\,dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\sqrt{1-2\cos^{2}x}\,dx$

Now, we make the obvious substitution $y = \cos x$ and from there on make one last substitution, which also must be obvious, for each integral.

It is posible to find primitives also. For the 1st one, i.e., make de substitution

$x=\pi/4-u$,

thus

$\sin 2x = \sin(\pi/2-2u) = \cos 2u$

$\sin x = \sin (\pi/4-u) = \frac{\sqrt{2}}{2}(\cos u-\sin u)$,

$\cos u = \cos (\pi/4-x) = \frac{\sqrt{2}}{2}(\cos x+\sin x)$,

$\sin u = \sin (\pi/4-x) = \frac{\sqrt{2}}{2}(\cos x-\sin x)$

and

$\int \sin x \sqrt{\sin 2 x}\cdot dx = \frac{\sqrt{2}}{2}\int \left( \sin u \cdot \sqrt{\cos 2u}-\cos u \cdot \sqrt{\cos 2 u}\right) \cdot du =$

$= \frac{\sqrt{2}}{2}\int \left( \sin u \cdot \sqrt{2 \cos^{2}u-1}-\cos u \cdot \sqrt{1-2 \sin^{2}u}\right) \cdot du =$

$=-\frac{1}{2}\sqrt{\left( \sqrt{2}\cos u\right)^{2}-1}\, \cdot d \left( \sqrt{2}\cos u \right)-\frac{1}{2}\int \sqrt{1-\left( \sqrt{2}\sin u \right)^{2}}\, \cdot d \left( \sqrt{2}\sin u \right) =$

$=-\frac{\sqrt{2}}{4}\cos u \cdot \sqrt{2 \cos^{2}u-1}+\frac{1}{4}\ln \left| \sqrt{2}\cos u+\sqrt{2 \cos^{2}u-1}\right|-$

$-\frac{\sqrt{2}}{4}\sqrt{1-2 \sin^{2}u}-\frac{1}{4}\arcsin \left( \sqrt{2}\sin u \right)+C =$

$=-\frac{1}{4}(\cos x+\sin x) \sqrt{\sin 2x}+\frac{1}{4}\ln \left| \cos x+\sin x+\sqrt{\sin 2x}\right|-$

$-\frac{1}{4}(\cos x-\sin x) \sqrt{\sin 2x}-\frac{1}{4}\arcsin (\cos x-\sin x)+C =$

$=C-\frac{1}{2}\cos x \cdot \sqrt{\sin 2x}\,+\, \frac{1}{4}\ln \left| \cos x+\sin x+\sqrt{\sin 2x}\right|-\frac{1}{4}\arcsin (\cos x-\sin x)$


An alphabet consists of three letters. Some sequences of letters of length two or more are prohibited, and all of the prohibited sequences have different lengths. A words is a sequence of letters of any length. A correct word does not contain any prohibited sequences. Prove that there are correct words of any length.

[hide] This seems to be an easy induction on the fact that there are at least two correct words of any length, but I may quite likely be missing something. [/hide]

are only finitely many sequences prohibited?

I think not necessarily. But remember that prohibited sequences have distinct lengths.


For all  $x,y,z$ . Show that inequality holds.

$(x^2+y^2+z^2+xy+yz+zx)^2 + 12 \geq 6 (2xyz +x^2y+y^2x+x^2z+z^2x+y^2z+z^2y)$

Davron

Nice proof Ramanujan !  :)

I have one question
@all: if there is given that $a+b+c+d=0$
then WLOG can we say that

$a=x+y+z$
$b=-x$
$c=-y$
$d=-z$

Davron

You just rename the variables.

If you say that $b=-x ,c =-y , d= -z$ from the condition $a+b+c+d=0$ you have

$a= x+y+z$ . It seems ok   :)

Davron I think you take the problem from here

http://www.mathlinks.ro/Forum/viewtopic.php?t=71330

 :D


number theory
relatively prime
number theory unsolved

Prove that there are infinitely many pairs of natural numbers $ a$ and $ b$ such that $ a^8\plus{}b^4\plus{}1$ is divisible by $ ab$.

[hide="Solution"]
Notice that $ \frac {a^8 \plus{} b^4 \plus{} 1}{ab}$ can only be an integer if $ a$ and $ b$ are relatively prime.  Thus, $ b^4 \plus{} 1 \equiv 0\bmod a$ and $ a^8 \plus{} 1\equiv 0\bmod b$.  Letting $ b \equal{} \frac {a^8 \plus{} 1}{k}$ for some integer $ k$, we see that $ k$ must be relatively prime to $ a$.  Thus, $ (\frac {a^8 \plus{} 1}{k})^4 \plus{} 1 \equiv (\frac {1}{k})^4 \plus{} 1 \bmod a$.  Since this is congruent to $ 0\bmod a$, we have that $ k^4 \plus{} 1 \equiv 0\bmod a$.  Similarly, we can let $ a \equal{} \frac {b^4 \plus{} 1}{l}$ for some integer $ l$.  We get that $ a\equiv \frac {1}{l}\bmod b$, so $ a^8 \plus{} 1 \equiv \frac {1}{l^8} \plus{} 1\bmod b$.  Since this is equivalent to $ 0\bmod b$, we have that $ l^8 \plus{} 1\equiv 0\bmod b$.

Notice that $ (a, b) \equal{} (1, 2)$ works.  Now, we will show that for any pair $ (a,b)$ that works, we can find a pair $ (x,y)$ so that $ x > a$ or $ y > b$.  If $ a > \sqrt {b^2 \plus{} 1}$, then $ b < a$, so switch $ b$ with $ \frac {a^8 \plus{} 1}{b} > b$, so $ b$ becomes $ k$, and $ (a, k)$ works since $ a^8 \plus{} 1 \equiv 0\bmod k$ and $ k^4 \plus{} 1 \equiv 0\bmod a$.  Now, if $ a < \sqrt {b^2 \plus{} 1}$, then $ a\le b$.  Thus, we may replace $ a$ with $ \frac {b^4 \plus{} 1}{a} > a$.  Then, $ a$ becomes $ l$.  We have that $ b^4 \plus{} 1 \equiv 0\bmod l$ and $ l^8 \plus{} 1 \equiv 0\bmod b$, so $ (l, b)$ works and $ l > a$.  Either way, we have found a "larger" pair, so we are done.  [/hide]


If $ a,b,c$ are positive integers such that $ a\plus{}b\plus{}c\equal{}1$, prove that
\[ a\sqrt[3]{1\plus{}b\minus{}c}\plus{}b\sqrt[3]{1\plus{}c\minus{}a}\plus{}c\sqrt[3]{1\plus{}a\minus{}b} \le 1
\]

[quote="bunhiacovski"][quote="Johan Gunardi"]If $ a,b,c$ are positive integers such that $ a \plus{} b \plus{} c \equal{} 1$, prove that
\[ a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b} \le 1
\]
[/quote]
by holder $ (a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b})^3\le 9(a^3 \plus{} b^3 \plus{} c^3)$
we need prove that $ 9(a^3 \plus{} b^3 \plus{} c^3)\le 1$ but it's right  :wink:[/quote]
It is right i dont think so.Because With Holder ineq we have $ 9(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b \plus{} c)^3 \equal{} 1$
Here is my proof
With Holder ineq $ (\sum \sqrt [3]{a}\sqrt [3]{a}\sqrt [3]{a \plus{} ab \minus{}ac})^3\leq (a \plus{} b \plus{} c)^2(\sum (a \plus{} ab \minus{} ac)) \equal{} 1$

[quote="Allnames"][quote="bunhiacovski"][quote="Johan Gunardi"]If $ a,b,c$ are positive integers such that $ a \plus{} b \plus{} c \equal{} 1$, prove that
\[ a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b} \le 1
\]
[/quote]
by holder $ (a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b})^3\le 9(a^3 \plus{} b^3 \plus{} c^3)$
we need prove that $ 9(a^3 \plus{} b^3 \plus{} c^3)\le 1$ but it's right  :wink:[/quote]
It is right i dont think so.Because With Holder ineq we have $ 9(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b \plus{} c)^3 \equal{} 1$
Here is my proof
With Holder ineq $ (\sum \sqrt [3]{a}\sqrt [3]{a}\sqrt [3]{a \plus{} ab \minus{} ac})^3\leq (a \plus{} b \plus{} c)^2(\sum (a \plus{} ab \minus{} ac)) \equal{} 1$[/quote]
yes, thank you very much :maybe:I wrong (^.-) a proof stupid

[quote="Johan Gunardi"]If $ a,b,c$ are positive integers such that $ a \plus{} b \plus{} c \equal{} 1$, prove that
\[ a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b} \le 1
\]
[/quote]
By am-gm ; \[ a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b} \le \sum_{cyc}a(\frac{1\plus{}b\minus{}c\plus{}1\plus{}1}{3}) \equal{} \frac{3(a\plus{}b\plus{}c)}{3} \equal{} 1\]

$ LHS\equal{}\sum a\sqrt[3]{a\plus{}2b}\leq \frac {\sum a(1 \plus{} 1 \plus{} a \plus{} 2b)}{3} \equal{} \frac {2(a \plus{} b \plus{} c) \plus{} (a \plus{} b \plus{} c)^2}{3} \equal{} 1$


a,b,c are positive real numbers such that: a+b+c+abc=4 , Prove that:

$ \frac{{(a \minus{} b)^2  \plus{} (b \minus{} c)^2  \plus{} (a \minus{} c)^2 }}{{48}} \plus{} 1 \geq \frac{{a \plus{} b \plus{} c}}{{ab \plus{} ac \plus{} bc}}$

That is not correct. Putting $ a\rightarrow 3 \ b\equal{}1 \ c\rightarrow 0$ we get:
$ \frac{5}{4}\geq \frac{4}{3}$ which is clear nonsense :wink:

thank you very much my dear friend, what about this:
(a,b,c)>0 such that: a+b+c=2 ,Prove that:

$ \sqrt {a \plus{} b \minus{} 2ab}  \plus{} \sqrt {a \plus{} c \minus{} 2ac}  \plus{} \sqrt {b \plus{} c \minus{} 2bc}  \geqslant 2$


function
limit
calculus
calculus computations

Consider the following sequence of continuous functions defined on $ [0,\ 1]$.

$ a_{n}\equal{}\left\{\frac{x}{1\plus{}nx}\right\}_{n\in\mathbb{N}}\ \ \ \ \ b_{n}\equal{}\ \left\{\frac{nx}{1\plus{}n^{2}x^{2}}\right\}_{n\in\mathbb{N}}$.

(1) Prove that $ \lim_{n\to\infty}a_{n}$ is uniformly convergent to 0 on $ [0, 1]$.

(2) Prove that $ \lim_{n\to\infty}b_{n}$ isn't uniformly convergent to 0 on $ [0, 1]$.

[quote="kunny"]Consider the following sequence of continuous functions defined on $ [0,\ 1]$.

$ a_{n}=\{\frac{x}{1+nx}\}_{n\in\mathbb{N}}\ \ \ \ \ b_{n}=\ \{\frac{nx}{1+n^{2}x^{2}}\}_{n\in\mathbb{N}}$.
[/quote]

[quote]
(1) Prove that $ \lim_{n\to\infty}a_{n}$ is uniformly convergent to 0 on $ [0, 1]$.[/quote]
First we show that $ \lim a_{n}(x) = 0$ on $ [0,1]$. This is true because if $ x=0$ then certainly, and if $ x\not = 0$ then the numerator is bounded by $ |x|$ and denominator is going to $ \infty$.

Now,
$ |\frac{x}{1+nx}|\leq\frac{x}{nx}=\frac{1}{n}$ if $ x\in (0,1]$.
But trivially, at $ x=0$ we have $ a_{n}(x) <\frac{1}{n}$.
Thus, for any $ x\in [0,1]$ we have $ a_{n}(x)\leq\frac{1}{n}$.
This means we can chose $ N =\frac{1}{\epsilon}$ to prove uniform convergence.

[quote]
(2) Prove that $ \lim_{n\to\infty}b_{n}$ isn't uniformly convergent to 0 on $ [0, 1]$.[/quote]
It certainly converges pointwise to $ 0$ on $ [0,1]$.

Chose $ \epsilon = 1/2$ then,
$ |\frac{nx}{1+n^{2}x^{2}}| =\frac{nx}{1+n^{2}x^{2}}<\frac{1}{2}$ for $ n\geq N\in\mathbb{N}$ and all $ x\in [0,1]$.
In particular if $ n=N$ we must have,
$ \frac{Nx}{1+N^{2}x^{2}}<\frac{1}{2}$ for all $ x\in [0,1]$.
Now, $ \frac{1}{N}\in [0,1]$ but,
$ \frac{N(\frac{1}{N})}{1+N^{2}(\frac{1}{N})^{2}}=\frac{1}{2}<\frac{1}{2}$.
A contradiction!
So the convergence cant be uniform.
Q.E.D.


If $ n_{1},n_{2},n_{3}$ and $ n_{4}$ are the $ n$th,$ (n\plus{}1)$th,$ (n\plus{}2)$th and $ (n\plus{}3)$th terms respectively in the expansion of $ (x\plus{}y)^{n}$ where $ n$ is a natural number, then the following always holds:

$ \frac{(n_{2})^{2}\minus{}n_{1}n_{3}}{(n_{3})^{2}\minus{}n_{2}n_{4}}\equal{}\frac{(n\plus{}2)n_{1}}{nn_{3}}$.

Prove or disprove this!

[quote="math_07"][quote="manjil"]If $ n_{1},n_{2},n_{3}$ and $ n_{4}$ are the $ n$th,$ (n \plus{} 1)$th,$ (n \plus{} 2)$th and $ (n \plus{} 3)$th terms respectively in the expansion of $ (x \plus{} y)^{n}$ where $ n$ is a natural number, then the following always holds:

$ \frac {(n_{2})^{2} \minus{} n_{1}n_{3}}{(n_{3})^{2} \minus{} n_{2}n_{4}} \equal{} \frac {(n \plus{} 2)n_{1}}{nn_{3}}$.

Prove or disprove this![/quote]
I think that the problem is wrong (I mean not correctly stated) as we can get just get $ n \plus{} 1$ terms expanding the polynomial $ P(x) \equal{} (x \plus{} y)^n$[/quote]

Yes the problem is worng, it should have been!
If $ n_{1},n_{2},n_{3}$ and $ n_{4}$ are the $ n$th,$ (n \plus{} 1)$th,$ (n \plus{} 2)$th and $ (n \plus{} 3)$th terms respectively in the expansion of $ (x \plus{} y)^{r}$ where $ r$ is a natural number,and $ r$ is greater than $ n$ then the following always holds:

$ \frac {(n_{2})^{2} \minus{} n_{1}n_{3}}{(n_{3})^{2} \minus{} n_{2}n_{4}} \equal{} \frac {(n \plus{} 2)n_{1}}{nn_{3}}$.

Prove or disprove this!

[quote="ZetaX"]Problem description FAIL.[/quote]

Can you please explain?


i don't know if i'd have a right answer or not but i need an so i can check over, thanks!
1) show that an nxn mx A is invertible exactly when 0 is not an eigenvalue of A

2) Let A be the nXn mx aI+B (a = scale, I = identity), where B is strictly upper triangular. Show that A is diagonalizable exactly when B = 0

3) Let A = ( 0 1 0 )
( 1 1 0 )
(0 1 -1)

Is there an invertible 3x3 mx C with C^-1AC a diagonal mx? explain.

4) Let A b rxr, B sxs and C be sxr, show that
| 0 A |
| B C | = -+ |A||B|

5) show that A and C^-1AC (A nxn, C nxn invertible) have the same eigenvalues. must they have the same eigenvectors?

thanks guys

The High School forums are not an appropriate place for Linear Algebra.  You want this forum:

http://www.artofproblemsolving.com/Forum/index.php?f=353

wow, im so screwed, is there any lifesaver out there please  :(

[b]1)[/b]  $\lambda$ is an eigenvalue of a matrix $A$ if and only if $\det (A - \lambda ) =0$ .
Now $0$ is an eigenvalue of $A$ if and only if $\det (A ) =0$ hence $A$ is invertible if and only if $0$ is not an eigenvalue of $A$ .

 [b]2)[/b] Say that $A$ is diagonalizable. Then there exist invertible matrix $C$ such that $C^{-1}AC =D$ where $D$ a diagonal matrix .
Now

$\\ C^{-1}AC = C^{-1} ( aI +B ) C \Rightarrow D = aI + C^{-1}BC \Rightarrow \\ C^{-1}BC = D-aI \Rightarrow B = C (D-aI) C^{-1}$

But $B$ is nilpotent i.e. there exist $k \in \mathbb{N}$ such that $B^k = O$
Then $O =B^k = (C (D-aI) C^{-1} )^k = C(D-aI)^{k} C^{-1} \Rightarrow (D-aI)^{k} =O$
But since $D-aI$ is diagonal $(D-aI)^{k} =O %Error. "Rightleftarrow" is a bad command.
D-aI =O$ or $B=O$.

 [b]3)[/b] $A$ is diagonalizable since it has three distinct eigenvalues $\{ \frac{1 + \sqrt{5} }{2} , \frac{1 - \sqrt{5} }{2} , -1 \}$
The invertible matrix $C$ you are looking for is the matrix with columns the eigenvectors corresponding to the above eigenvalues .

$C = \left(\begin{array}{ccc} \frac{1 + \sqrt{5} }{2}&\frac{1 - \sqrt{5} }{2}&0\\\ \frac{3 + \sqrt{5} }{2}&\frac{3 - \sqrt{5} }{2}&0\\\ 1&1&1 \end{array}\right)$ .

 [b]4)[/b] Let $D = \left(\begin{array}{cc} O&A\\\ B&C \end{array}\right)$ . By induction .For $s=r=1 , \, n=r+s=2$ we have a $2 \times 2$ matrix $\left(\begin{array}{cc} 0&a\\\ b&c \end{array}\right)$

and we know that $\left|\begin{array}{cc} 0&a\\\ b&c \end{array}\right| =-ab$. Suppose that the claim is true for $n-1$ . I'll prove that it is true for $s+r=n$ .

If $r=1$ we expand $\det(D)$ by the first row : $\det(D) = (-1)^{s+r+1} a \det(B) = +- \det(A) \det(B)$

If $r> 1$ we expand $\det(D)$ by the first row .The nonzero entries are in columns $s+1$ to $s+r$  and the matrix $\widetilde {D}_{1j}$ is a $(n-1) \times (n-1)$ matrix .

It is  $\det(D) = \sum_{j=s+1}^{s+r} (-1)^{j+1} d_{1j} \det (\widetilde {D}_{1j} ) = \sum_{j=s+1}^{s+r} (-1)^{j+1} d_{1j} \det (\widetilde {A}_{1j} ) \det(B)$ (by induction )

and $\sum_{j=s+1}^{s+r} (-1)^{j+1} a_{1j} \det (\widetilde {A}_{1j} ) \det(B) = +- \det(A) \det(B)$

 [b]5)[/b] [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=86921[/url]

I am not sure for 2 and the indices in 4 . Check them again. Also check computations in 3 .  :D

I hope it is mostly correct and will help you  :)


limit
trigonometry
integration
calculus
calculus computations

\[ \begin{gathered}
  \mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{x \minus{} \cos ^2 x}}
{{x \plus{} \sin x}}}  \hfill \\
  \mathop {\lim }\limits_{h \to 0} \int\limits_{ \minus{} 1}^1 {\frac{h}
{{h^2  \plus{} x^2 }}} dx \hfill \\
  \mathop {\lim }\limits_{x \to \infty } \left[ {x\sqrt {x^2  \plus{} a^2 }  \minus{} \sqrt {x^4  \plus{} a^4 } } \right] \hfill \\
\end{gathered}
\]

$ \mathop {\lim }\limits_{h \to 0} \int\limits_{ \minus{} 1}^1 {\frac{h}
{{h^2 \plus{} x^2 }}} dx \equal{} 2\mathop {\lim }\limits_{h \to 0} \int\limits_{ 0}^1 {\frac{h}
{{h^2 \plus{} x^2 }}} dx \equal{} 2 \lim_{h \to 0} h. \frac{1}{h} [\tan^{\minus{}1} (\frac{1}{h}) \minus{} \tan^{\minus{}1} (0)] \equal{} \pi$.

$ \mathop {\lim }\limits_{x \to \infty } \left[ {x\sqrt {x^2 \plus{} a^2 } \minus{} \sqrt {x^4 \plus{} a^4 } } \right]$
$ \equal{} \mathop {\lim }\limits_{x \to \infty } \left[ {x\sqrt {x^2 \plus{} a^2 } \minus{} \sqrt {x^4 \plus{} a^4 } } \right] \frac{\left[ {x\sqrt {x^2 \plus{} a^2 } \plus{} \sqrt {x^4 \plus{} a^4 } } \right]}{\left[ {x\sqrt {x^2 \plus{} a^2 } \plus{} \sqrt {x^4 \plus{} a^4 } } \right]}$
$ \equal{} \mathop {\lim }\limits_{x \to \infty }\frac{a^{2}(x^{2} \minus{} a^{2})}{\left[ {\sqrt {x^4 \plus{} x^2a^2 } \plus{}\sqrt {x^4 \plus{} a^4 } } \right]} \equal{} \frac{a^2}{2}$.

$ \lim_{x\rightarrow\infty}\sqrt{\frac{x\minus{}\cos^{2} x}{x\plus{}\sin x}}\equal{}\lim_{x\rightarrow\infty}\sqrt{\frac{1\minus{}\frac{\cos^{2} x}{x}}{1\plus{}\frac{\sin x}{x}}}\equal{}1$

Why $ \lim_{x\to\infty } \frac {\cos ^ 2 x}{x} \equal{} 0?$ :wink:

[quote="kunny"]Why $ \lim_{x\to\infty } \frac {\cos ^ 2 x}{x} = 0?$ :wink:[/quote]

\begin{align*}
0 \leq &\cos^2 (x) \leq 1\\
\frac{0}{x} \leq &\frac{\cos^2 (x)}{x} \leq \frac{1}{x}\\
\end{align*}

Now note that:

\[ \lim_{x \to \infty} \frac{0}{x} = 0 = \lim_{x \to \infty} \frac{1}{x}\]

Hence, by the squeeze theorem:

\[ \boxed{\lim_{x \to \infty} \frac{\cos^2 x}{x} = 0}\]

[quote="manipalmps"]i have a problem with the post of Mr timestopper  :?: !
How can we use

lim x->infinity   sin x /x =1
it is only valid when x->0

so i think the work od Mr timestopper is wrong

BETTER USE THE EXPANSIONS TO GET THE DESIRED RESULT[/quote]

No, Mr timestopper is right. $ \lim_{x\to\infty} \frac{\sin x}{x}\equal{}0$. :wink:

yup kunny
u r right in saying that the value tends to zero
i thought 1  :(
sorry 4 the inconvenience

but expansions r also a way to do this question :)


Given a triangle ABC which intersects a circle O in six points, D, D', E, E', F, and F' such that F and F' are on AB, E and E' are on CA and D and D' are on BC.  If AD, BE, and CF are concurrent, prove AD', BE', and CF' are concurrent.

This is unexpectedly easy. *blink*

[hide]By Ceva's theorem,

(AF / FB) * (BD / DC) * (CE / EA) = 1

Rearrange:

(AF / AE) * (BD / BF) * (CE / CD) = 1 (*)

Now, by power of a point,

AE' * AE = AF' * AF

Rearrange to get AF / AE = AE' / AF'.

Similarly, BD / BF = BF' / BD', CE / CD = CD' / CE'. Sub into (*):

(AE' / AF') * (BF' / BD') * (CD' / CE') = 1

(AE' / E'C) * (CD' / D'B) * (BF' / F'A) = 1

Result follows by Ceva.[/hide]

Well, I didn't put it in Advanced, did I?  Good job.


1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

Anyone care to try some of these ARML problems?  Problem 8, if I recall correctly, is from the ARML that I went to in 1979!

*tries*

1.
[hide]$N(N+1)-N^2=N$
$1+...+N=210$
$N(N+1)=420$
$N=20$[/hide]

2.
[hide]max distance of innermost square = $10\sqrt{2}$
min distance of outermost square = $\frac{30}{\sqrt{2}}=15\sqrt{2}$
difference = $5\sqrt{2}$[/hide]

3.
[hide]$7x+7z=kx+ky$
$11x+11z=kz-ky$
$18(x+z)=k(x+z)$
$18=k$[/hide]

[hide="3."]It's well-known that $\frac{a}{b}=\frac{c}{d}=\frac{a\pm c}{b\pm d}$.
Apply to first and third to get $\frac{k}{x+z}=\frac{18}{x+z}$.

k=18[/hide]

alan, yes those are fine solutions for the first 3.

Bictor717, yes, you have a slick way to solve #3.  I think what you meant is that if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a\pm c}{b\pm d}$ is equal to both of them too.

*tries some more*
4.
[hide]list out terms
$12.08\rightarrow 1.30, 15.10\rightarrow 4.32, 18.12\rightarrow 7.34, ...$
it is of the form $(1+3n).(30+2n), $ so we want $1+3n=30+2n-1$
$n=28, 1+3n=85, 30+2n=86$
$85.86+10.78=96.64$[/hide]

5.
[hide]i did it the long way, and ended up with a really weird $\frac{1971}{3757},$ which i think is wrong :(
there has to be a better way than just brute force...[/hide]

6.
[hide]that last senior iml is still fresh in my mind :lol: with that said, i assume that the problem can be solved, so the term under the radical must be a square.
$\sqrt{21+12\sqrt{3}}=x+y\rightarrow 21=x^2+y^2, 2xy=12\sqrt{3}$
$x\sqrt{(21-x^2)}=6\sqrt{3}\rightarrow x^2(21-x^2)=108$
$x^4-21x^2+108=0\rightarrow (x^2-9)(x^2-12)=0$
$x$ and $y$ are interchangeable, so $\sqrt{21+12\sqrt{3}}=3+2\sqrt{2}$
$\sqrt{4+2\sqrt{3}}=m+n$
$4=m^2+n^2, 2mn=2\sqrt{3}$ (note: at this point, we can get the answer by inspection)
$m^2(4-m^2)=3\rightarrow m^4-4m^2+3=0$
$(m^2-3)(m^2-1)=0\rightarrow \sqrt{4+2\sqrt{3}}=1+\sqrt{3}$
$a,b= (1,3)$[/hide]

For number 5 Alan, you're telling me there's a better-than-1/2 chance of getting a king and a heart on two picks?  The chance of getting a heart on two picks isn't even that good!  (Also, how did you get a denominator larger than 52*51?)

[hide="Hint"]Calculate 1 - P.[/hide]


[quote="anirudh"]Los Angeles Airport bears 140 degrees from San Francisco Airport and is 540 km away. A pilot is planning a direct flight from San Francisco Airport to Los Angeles Airport to leave at 2 P.M. The plane's air speed will be 640 km/h, and there will be a 60 km/h wind blowing from 290 degrees. What should the compass heading be, and what is the plane's estimated time of arrival (ETA) to the nearest minute?[/quote]

I got that the bearing the plain should go is approximately 142.6867242. And the approximate time of arrival is.....2:55:51


Ok all you video game lovers out there, I want to know what you're looking forward to the most in the Next-Gen console showdown.  Represent your system!

PS3 (especially if it weren't so expensive, i find it the best by far)

NOTHING!!!!!!!!!!!! I am against video games. They teach bad habits!


geometry
incenter
vector
geometry proposed

[quote="mr.danh"]Let a triangle ABC and I is its incenter. The segemnts AI,BI,CI meet the incircle (I) at A',B',C' respectively. M,N,P are the midpoints of BC,CA,AB respectively. Prove that A'M,B'N,C'P are concurrent.[/quote]

If you've got a synthetic proof of this, please post it - it appears for the fifth or so time on this forum without a solution...

  darij

OK. I have two solutions, one is short with verorial method, one is longer but it is geometrical.
[hide="First proof"][b]Lemma:[/b] The vector u passes through the vectex A of triangle ABC. If $ \vec{u} \equal{} m\vec{AB} \plus{} n\vec{AC}$, then $ \vec{u}$ cuts BC at a point D satisfying $ \frac {DC}{DB} \equal{} \frac {m}{n}$
The tangent of (O) at A' meets AB,AC at D,E. A' is the midpoint of  DE.
$ \vec{MA'} \equal{} \frac {\vec{BD} \plus{} \vec{CE}}{2} \equal{} \frac {1}{2}(\frac {BD}{BA}\vec{BA} \plus{} \frac {CE}{CA}\vec{CA})$
$ \equal{} \frac {BD}{BA}\vec{MN} \plus{} \frac {CE}{CA}\vec{MP}$
If MA' meets NP at K, then $ \frac {KP}{KN} \equal{} \frac {BD}{CE}.\frac {AC}{AB} \equal{} \frac {IB}{IC}.\frac {AC}{AB}$ (as lemma)
Since Ceve's theorem in triangle MNP, hence result.[/hide]

I will post second proof in a week if nobody gives more one.

This is beautiful! Only remark: (O) should be (I).

  darij

[quote="jayme"]Dear Mr. Danh,
I am waiting your synthetical proof.
Sincerely
Jean-Louis[/quote]
I don't know whether it is a synthetic proof. My second proof is longer, but it doesn't contain vectors  :)

Dear Mathlinkers,
two synthetic proofs of this known result originated from Clark Kimberling has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis

[quote="mr.danh"]$ \vec{MA'} \equal{} \frac {\vec{BD} \plus{} \vec{CE}}{2} \equal{} \frac {1}{2}(\frac {BD}{BA}\vec{BA} \plus{} \frac {CE}{CA}\vec{CA})$
$ \equal{} \frac {BD}{BA}\vec{MN} \plus{} \frac {CE}{CA}\vec{MP}$
If MA' meets NP at K, then $ \frac {KP}{KN} \equal{} \frac {BD}{CE}.\frac {AC}{AB} \equal{} \frac {IB}{IC}.\frac {AC}{AB}$ (as lemma)
[/quote]

i have a stupid question: why $ \frac {BD}{CE} \equal{} \frac {IB}{IC}$ ? :blush:


This was a problem chess64 posted in Intermediate, but no one solved it besides me, and I wanted to know if my answer was correct. Thanks!
[quote="chess64"][size=150]Problem 5[/size] :D

How many of the first 1000 positive integers can be expresssed in the form

\[ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor ?  \]

(ACoPS)[/quote]

[quote="Alfred"][hide]
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$

My solution seems a bit to easy. Maybe I have missed something. I suppose $x$ is an integer. If $x$ is positive, we have that every number inside is positive as well. Therefore we can write it as:
$2x+4x+6x+8=20x$.
If $x$ is negative, we have:
$-2x-4x-6x-8x=-20x$
Every number divisible with twenty can be written in this way, wich is every 20th number. $1000/20=50$. So I guess fifty numbers can be written in this way.
[/hide]

What was your solution? Where was the previous discussion of the problem?[/quote]
Alfred, I'm pretty sure there are more than that. Ex: $x=1.5, y=30$My answer was 550. See the Intermediate forum dloor functions thread.

[hide]It looks like an other problem in the algebra section (from IMO longlist ?)

Each time x increases by $1/8$ or $1/6$, $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$ increases at least by $1$ but of course it may increases also of $2$ or $4$.

I find $600$ positive numbers.[/hide]

[quote="chess64"][quote="Alfred"][hide]
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$

My solution seems a bit to easy. Maybe I have missed something. I suppose $x$ is an integer. If $x$ is positive, we have that every number inside is positive as well. Therefore we can write it as:
$2x+4x+6x+8=20x$.
If $x$ is negative, we have:
$-2x-4x-6x-8x=-20x$
Every number divisible with twenty can be written in this way, wich is every 20th number. $1000/20=50$. So I guess fifty numbers can be written in this way.
[/hide]

What was your solution? Where was the previous discussion of the problem?[/quote]

....Those are floor functions, not absolute value. $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Ex: $\lfloor 5.5 \rfloor = 5$.[/quote]

Thanks, I had not heard of floor functions before.  :blush:

[hide]
Well we check the numbers up to 20 since when $x = 1$ we get 20.

So we find each time the value steps up, these are

$x = \frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, 1$.

These give values of 1, 2, 4, 5, 6, 10, 11, 12, 14, 15, 16, 20. So $\frac{12}{20} = \frac{3}{5}$ of the numbers are reached... meaning $\frac{3}{5} \cdot 1000 = 600$.[/hide]

[quote="chess64"][quote="paladin8"][hide]$x = \frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, 1$.[/hide][/quote]

How did you find these numbers? :? Did you just check multiples of $\frac{1}{8}$ and $\frac{1}{6}$?[/quote]

Yeah, you know it'll only increment when the denominator is 2, 4, 6, 8. So we just use those.

Oh, i miscounted and got 11 instead of 12, so I got 550


I'm having trouble seeing how:

$ \dfrac {1}{3}(x^2 \plus{} 4)^{\minus{}2/3} \left [\dfrac {3(x^2 \plus{} 4) \minus{} 2x^2} {(x^2 \plus{} 4)^{2/3}} \right ]$

simplifies to:

$ \dfrac {x^2 \plus{} 12} {3(x^2 \plus{} 4)^{4/3}}$

Can anyone help with the initial step/s please? Thanks very much!

Hi,

[hide="Hint"]
First, note the negative 2/3 exponent can go on the bottom as a +2/3 exponent, combining with the +2/3 already there.

In addition, the 3 goes on the bottom.

Now, for the numerator, just expand 3(x^2+4)=3x^2+12 and subtract.
[/hide][/hide]

Thanks for that!  I thought the numerator had to multiply the numerator
though - what rule is this where you are multiplying the denominator instead please?
(And I thought when you multiplied you added exponents, which would seem to
add up to a zero exponent in this case.)  Obviously my algebra is too weak for this problem :(

You could split up the fraction like so:
$ \dfrac {1}{3}\cdot (x^2 + 4)^{ - 2/3} \cdot \frac {1}{(x^2 + 4)^{2/3}} \cdot \left [{3(x^2 + 4) - 2x^2\right ]}$

And then combing the middle two fractions, you obtain $ (x^2 + 4)^{ - 2/3 - 2/3} = (x^2 + 4)^{ - 4/3} = \frac {1}{(x^2 + 4)^{4/3}}$. Then you can combine this new fraction with the other two fractions and then I would suggest that you distribute and try to simplify from there.

A few important exponential properties:

$ m^xm^y\equal{}m^{x\plus{}y}$
$ m^{\minus{}x}\equal{}\frac1{m^x}$
$ \frac{m^x}{m^y}\equal{}m^{x\minus{}y}$
$ (m^x)^y\equal{}m^{xy}$
$ (mn)^x\equal{}m^xn^x$

Don't memorize these--many are obvious and redundant if you think about them.

Oops! I see what I missed - the neg. exponent means I'll be
multiplying denominators :) Thanks for the help guys!

No don't get yourself into that mistake.

You will actually be subtracting the exponents

$ \frac {a^x}{a^y} \equal{} \frac {1}{a^{y \minus{} x}}$

In this case $ y \equal{} \frac {2}{3}$ and $ x \minus{} \frac {2}{3}$

edit:nvm, I read the last post wrong.


vector
geometry
parallelogram
search
Pythagorean Theorem

I know that this shouldn't be here but I want mathematically rigorous proofs that physicists do not care about.

I have thought physics to be impure since I knew more maths (and picked up my first AoPS book). Maths is absolute. Pure maths is better.

Thus I often require a proof to accept it. I am quite skeptical.

[list]Most physics textbooks (like mine) simply say that a vector is a arrow or something like that. To the mathematician, this is totally incorrect. What is the mathematical definition of a vector?[/list]
[list]Prove the Triangle Rule/Law of Vector Addition (if there exists a proof)[/list]
[list]Prove the Parallelogram Rule/Law of Vector Addition (if there exists a proof)[/list]

$ maths >> physics$

Most simply, vectors are objects that possess magnitude (size) and direction. Therefore, you physics book is not totally incorrect about an arrow being a vector. There is definately more to talk about as far as defining vectors, but for our purposes this definition shoul suffice. As for the proofs, I'm not sure what you mean by the triangle rule (perhaps the Pythagorean Theorem?). However, the parrallelogram rule is what we use to add vectors. As for the proof, just search this exact phrase in google: "parallelogram law proof" and the first link will be planetmath.org. There will be two proofs.

[/hide]

I think the confusion with the arrow was partly because the symbol for a vector is an arrow.

An arrow represents a vector well because it can show both magnitude and direction. But a vector is not an arrow. It is only represented by one.

Obviously the "vector" you are thinking about is the one we in Three dimensional euclidan space.
The definition for this one is quite simple:
The vector is an object which transforms according to the auto representation of the SO(3), three dimensional special orthogonal group.

The "triangle" rule, and "paralelogram" rule cannot be proved, since they are definitions. This is how we DEFINE the addition of vectors.


geometry
circumcircle
parallelogram
geometry proposed

Let P and Q be points on the base AB of isosceles triangle ABC [where C > pi/3] such that PCQ = BAC. A circle is tangent externally to the circumcircle of PQC and is also tangent to the rays CA and CB at M and N. Prove that MN = 2AB.

Let R be the point of tangency of the two circles. As 180-<NRM = <CMN = <CAB = 180-<PRQ, it holds that <NRM = <PRQ. Then,   as parallel and proportional chords of two homotetic circles, PQ and NM are homotetic throug R. This means that M, R and Q are collinears, as are P, R and N.
As <PRM = <PCQ = <PAC, we have that AMRP is ciclyc. Similarly one can prove that BQRN is ciclyc too.  From here is easy to verify that AMN ~ BPN, or that AQ*BP = AM*BN = AM^2 = BN^2.
On the other hand, by angle chasing, it is easy to see that AQC ~ BPC, or that AQ*BP = AC*BC = AC^2 = BC^2.
From these facts one can easily derive the claim.

PS
The configurazion has a lot of nice properties. For instance:

a) if S is the second point common to c(AMRP) and c(BQRN ), S is collinear with C and R;

b) the lines AS and BR are parallel wrt CQ and BP respectively;

c) if U = AS^MN and V = BS^MN, then U lies on c(BQRN ) and V lies on c(AMRP);

d) AUQC and BCPV are parallelograms; SUV and CQP are hometics.

post my graph
I think we could make a new problem,a more challenging problem

let isosceles triangle ABC AC=BC
P and Q are points on line AB,directed angle BAC=dtrected angle PCQ
find the locus of the point O and circle O
circle O is the circumcircle of triangle ABC

now the graph

What do you mean? :|  :?

And SAMO = South African Mathematical Olympiad

if we do not use the GSP,we may not have the idea to the problem I changed.


http://video.google.com/videoplay?docid=-2134266654801392897&q=alkali+metals&hl=en

just watch through beginning, its dumb.

[quote="Franz Joseph"]I hope this is not true. But in our school there is a rumour (well it has become somewhat of a legend) that some years ago a teacher of Chemistry (at our school) wants to demonstrate the reactivity of alkali metals. So he took the class down to the field by the sea and fed some sodium to a sea gull. The sea gull blew up............[/quote]

it is true but I bet the seagull part isn't real. That's illegal.


inequalities
Ring Theory
inequalities proposed

1. $(a+b+c)(ab+bc+ca) \ge 9abc$, therefore, it suffices to show that $a^{2}+b^{2}+c^{2}+abc+2 \ge a+b+c+ab+bc+ca$
This is the consequence of the following inequalities
$\frac{1}{2}(a^{2}+b^{2}+c^{2}+3) \ge a+b+c$
$\frac{1}{2}(a^{2}+b^{2}+c^{2}+2abc+1) \ge ab+bc+ca$

[quote="hungkhtn"]Prove the following non-homogeneous inequality for all $a,b,c$ non-negative

(i). $(a+b+c-1)(ab+bc+ca-1)+a^{2}+b^{2}+c^{2}\ge 8abc-1.$

(ii). $a(a-1)^{2}+b(b-1)^{2}+c(c-1)^{2}+1 \ge \frac{1}{2}\left(abc-(a+b-c)(b+c-a)(c+a-b)\right).$

(iii). $(a+b)(c-1)^{2}+(c+a)(b-1)^{2}+(a+b)(c-1)^{2}\le a^{3}+b^{3}+c^{3}+3abc+2.$

(iv). $a^{3}+b^{3}+c^{3}+2(ab+bc+ca)+1 \ge 2abc+(1-k)M+kN,$ in which
\[k\le\frac{4\sqrt{6}}{9}, M=(a+b-c)(b+c-a)(c+a-b), N=abc. \]
[/quote]
All of this problem can be verified by PID theorem with assumtion $a+b+c=t$ where $t \in \mathbb{R^{+}}$

I don't think so because they are non-homogeneous inequality...  :?:
For (ii), by expending expression we have:
$(a^{3}+b^{3}+c^{3}+\sum ab(a+b)+2(a+b+c)+2\ge 3abc+4(a^{2}+b^{2}+c^{2})$
But I can't find when the equal hold ...  :blush:

For (iii), similar to (ii)  :!: Maybe I am stupid or I have made mistake in expanding express... (I am hungrying, maybe everything will better if I have lunch :P). Okie, I will try more ...  :wink:

What about the following? The soul?  :P
\[a(a-1)^{2}+b(b-1)^{2}+c(c-1)^{2}+d(d-1)^{2}+2\left(1+\sqrt{\frac{32}{27}}\right) \ge \sum a^{2}-\sum abc. \]
When does the equality holds?

For $(iv)$, there are three cases of inequality. It is very interesting. Moreover, it is stronger than the problem of Sung Yoon Kim post here (for $k=1$)

http://www.mathlinks.ro/Forum/viewtopic.php?t=105462

For the problem of 4 variables, I solved by CID theorem only. It is very hard to find another one, but I still wait for it. Notice that the equality holds for $a=b=1+\sqrt{\frac{2}{3}},c=d=0$ or permutations.


function
calculus
integration
real analysis
real analysis unsolved

Suppose that $ f: [a,b] \rightarrow \mathbb{R}$ is continuous and $ f(x) \geq 0$ for all $ x \in [a,b]$. Prove that if $ \int^b_a f(x)dx\equal{}0$, then $ f(x)\equal{}0$ for all $ x \in [a,b]$.
Attempt
I had attempted to do this problem by contradiction, except I did not understand how to finish the problem. I would appreciate a few helpful hints on this one.

Suppose there exists $ p$ such that $ f(p) \equal{} \epsilon > 0$.  What can you say about the behavior of $ f$ in a neighborhood of $ p$?  More specifically, what is the preimage of an open ball of radius $ \frac{\epsilon}{2}$ about $ \epsilon$?

Suppose there exists some $ p \in [a,b]$ for which f(p) = k > 0 [just like t0rajir0u suggests; he has already provided all that is needed for this problem].
By continuity of f on a [i]closed[/i] and [i]bounded[/i] interval: [a,b], if we are given $ \epsilon > 0$, there exists some $ \delta > 0$ such that $ |f(x) \minus{} f(p)| < \epsilon$ whenever $ |x \minus{} p| < \delta$ and $ x \in [a,b]$. So for $ \epsilon \equal{} \frac {k}{2}$, we have:
 $ |f(x) \minus{} f(p)| < \frac {k}{2} \implies f(p) \minus{} \frac {k}{2} < f(x) < f(p) \plus{} \frac {k}{2} \implies \frac {k}{2} < f(x) < \frac {3k}{2}\implies \int_{a}^{b}\frac {k}{2}dx < \int_{a}^{b}f(x) dx < \int_{a}^{b} \frac {3k}{2}dx \implies \frac {k}{2} < 0 < \frac {3k}{2}$.


limit
logarithms
real analysis
real analysis unsolved

Maybe this idea would work:

We have, since $\frac{b}{ab+1}\geq \frac{1}{a+1}$  and $\frac{ab}{ab+1}<1$, that

$n\sum_{a=1}^{n}\frac{1}{a+1}\leq \sum_{b=1}^{n}\sum_{a=1}^{n}\frac{b}{ab+1}=\sum_{a=1}^{n}\frac{1}{a}\sum_{b=1}^{n}\frac{ba}{ab+1}\leq n\sum_{a=1}^{n}\frac{1}{a}$
Thus the limit is equivalent to the calculation of the simple limit:

$\lim\frac{1+\frac{1}{2}+\cdots+\frac{1}{n+1}}{\ln n}=1$ which equals 1 based on Cesaro Stoltz lemma.

Squeeze Theorem is a powerfull theorem :)

correct solution didilica :D


Prove that

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

[quote="mathwizarddude"]Prove that

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \ge \frac {3}2$[/quote]
It's wrong. Try $ a\equal{}1$ and $ b\equal{}\minus{}1.$ :wink:

Of course you have to assume that $ a,b,c\geq 0$.

Note that $ (a\plus{}b\plus{}c)^2\minus{}\frac{3}{2}(a(b\plus{}c)\plus{}b(c\plus{}a)\plus{}c(a\plus{}b)) \equal{} \sum \frac{1}{2}(a\minus{}b)^2 \geq 0 \ (*)$, whence it follows that

$ \sum \frac{a}{b\plus{}c} \equal{} \sum \frac{a^2}{a(b\plus{}c)} \geq \frac{(a\plus{}b\plus{}c)^2}{a(b\plus{}c)\plus{}b(c\plus{}a)\plus{}c(a\plus{}b)} \geq \frac{3}{2}$

where the first inequality is cauchy-schwartz and the second is $ (*)$.

This is not the correct forum for inequalities this simple, you should post this in the intermediate or pre-olympiad forums instead.

[quote="mathwizarddude"]Prove that

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \ge \frac {3}2$[/quote]

With $ a,b,c\ge0$,that's Nesbitt inequality. See here
http://www.mathlinks.ro/viewtopic.php?t=207221

are you joking?
that is nesbit inequality


If I hold a barbell over my head and my ELBOWS ARE LOCKED (hence bb is stationery), how would I draw the free body diagram for the BARBELL?

Do i have to include both "lift force" and "weight of bb" or ONLY the weight of bb

Assuming the only forces are in the y-direction, you need to draw an arrow going down, which is $ \vec{F}\equal{}mg$, (weight), and an arrow going up, which represents the normal force, or the force supplied by you to keep it up.

[quote="JRav"]Assuming the only forces are in the y-direction, you need to draw an arrow going down, which is $ \vec{F} \equal{} mg$, (weight), and an arrow going up, which represents the normal force, or the force supplied by you to keep it up.[/quote]

if there is no friction between bb and hand