just wondering. i graduate (hopefully heh heh) in 2008

pinkbubblez, the one in eighth grade,  u would be either a=

eighty
or
eighter

sounds funky to me!

heehee...I am in 8th grade as well, though the school year is drawing to a close. Only 42 days of school left. Then, I will be a big bad freshman...lol. I can't wait to get out of the middle school. I will graduate in 2008 as well.

hmm, 7th for me.  I will graduate as it is right now in 2009, but I hope to skip (heheheh).

eighth grade pride!!! =D and we get to vote for u.s. president when we're a freshman in college........ not that it's necessarily a good thing. X__x

I'm a freshman but since I skipped I get teased a lot about still being an eighth grader and stuff...mrph

I'm not planning to graduate anytime soon so no year. (there is no such system in Japan)

[quote="Scrambled"]how come sixth graders and seventh graders are referred to as weird names and eigth graders arent?[/quote]

That is probably because the majority of us talking about this issue are 8th graders or the age of an 8th grader. It also could be because we are like the seniors of junior high and we rule!!!!!!!!!!

go 8th graders! i'm one too
but i'll never be in high school, i'm just going to transition school next year. then college

I'm a junior...and all alone among the younger ones...

[quote="Scrambled"]how come sixth graders and seventh graders are referred to as weird names and eigth graders arent?[/quote]

how would i know? its just natural. at miller sixers are also called "pixies"...weirdness.

yes, all wiedness....

they also can be called sixies... or shorties......

8th graders cant be called anything cuz nothing goes w/ their name.

[quote="Caroline"][quote="mathfanatic"]9th grade.  Go class of 2007![/quote]

Go us =)[/quote]
ay I can't believe I am now going to graduate with mathfanatic...oh wait Japanese educational system ends in March :P Go [b]us[/b] not in the us! :lol:

[quote="Caroline"][quote="mathfanatic"]9th grade.  Go class of 2007![/quote]

Go us =)[/quote]

On that note, my history teacher calls the freshman class "007", as in James Bond.  It's kinda odd.

But yes, go me, you, Mystic, Miths, and whoever else is a freshman.

I'll be graduating in 2007 too most likely.

[quote="MithsApprentice"][quote="mathfanatic"]

But yes, go me, you, Mystic, Miths, and whoever else is a freshman.[/quote]

lots of AASTers...chahah, lunmu, divran...etc...

i'm sure there's quite a few others that dont post quite as much...[/quote]

Yeah, Mr. Holbrook had a whole bunch of us sign up... Most people whose username is the first three letters of their first name and the first three letters of their last name are from AAST.  Plus some others, like me ;)  (We're not all freshmen though..)

[quote="ComplexZeta"]Well, I'm not really a freshman. I'm actually a junior.[/quote]

how does it work that you're a junior that  graduates in 2007? just asking

i'm a sophomore but i skipped a grade. i don't get teased like tare though, ha

I'm a junior because my college bases things on units rather than number of years. I'm a junior in terms of units, but this is my first year here, and I don't plan to graduate early.

University of California at Santa Barbara, College of Creative Studies.


let $ x$ ,$ y$ and $ z$ are reals numbers  such that :
$ \frac{x}{y\minus{}z}\plus{}\frac{y}{z\minus{}x}\plus{}\frac{z}{x\minus{}y}\equal{}0$
prove that:
$ \frac{x}{(y\minus{}z)^2}\plus{}\frac{y}{(z\minus{}x)^2}\plus{}\frac{z}{(x\minus{}y)^2}\equal{}0$

[quote="greatestmaths"]let $ x$ ,$ y$ and $ z$ are reals numbers  such that :
$ \frac {x}{y \minus{} z} \plus{} \frac {y}{z \minus{} x} \plus{} \frac {z}{x \minus{} y} \equal{} 0$
prove that:
$ \frac {x}{(y \minus{} z)^2} \plus{} \frac {y}{(z \minus{} x)^2} \plus{} \frac {z}{(x \minus{} y)^2} \equal{} 0$[/quote]
[hide="Solution"]
Multiply the first equation by $ \frac {1}{z \minus{} x} \plus{} \frac {1}{y \minus{} z} \plus{} \frac {1}{x \minus{} y}$ to get that
\[ [\frac {x}{(y \minus{} z)^2} \plus{} \frac {y}{(z \minus{} x)^2} \plus{} \frac {z}{(x \minus{} y)^2}] \plus{} [\frac {x}{(y \minus{} z)(x \minus{} y)} \plus{} \frac {x}{(y \minus{} z)(z \minus{} x)} \plus{} \frac {y}{(z \minus{} x)(x \minus{} y)} \plus{} \frac {y}{(z \minus{} x)(y \minus{} z)} \plus{} \frac {z}{(x \minus{} y)(z \minus{} x)} \plus{} \frac {z}{(x \minus{} y)(y \minus{} z)}] \equal{} 0
\]
We divided the left hand side into two brackets and we will prove that the second bracket is equal to $ 0$.  We create a common denominator and expand to get that
\[ [\frac {x}{(y \minus{} z)(x \minus{} y)} \plus{} \frac {x}{(y \minus{} z)(z \minus{} x)} \plus{} \frac {y}{(z \minus{} x)(x \minus{} y)} \plus{} \frac {y}{(z \minus{} x)(y \minus{} z)} \plus{} \frac {z}{(x \minus{} y)(z \minus{} x)} \plus{} \frac {z}{(x \minus{} y)(y \minus{} z)}] \equal{} \frac {zx \minus{} x^2 \plus{} x^2 \minus{} xy \plus{} yx \minus{} y^2 \plus{} y^2 \minus{} zy \plus{} z^2 \minus{} zx \minus{} z^2 \plus{} zy}{(x \minus{} y)(y \minus{} z)(z \minus{} x)} \equal{} 0
\]
Thus, $ [\frac {x}{(y \minus{} z)^2} \plus{} \frac {y}{(z \minus{} x)^2} \plus{} \frac {z}{(x \minus{} y)^2}] \equal{} 0$ and we are done. [/hide]

notice that $ \frac{x}{y\minus{}z}\plus{}\frac{y}{z\minus{}x}\equal{}\frac{z}{y\minus{}x}$
so $ \frac{xz\minus{}x^2\plus{}y^2\minus{}yz}{(y\minus{}z)(z\minus{}x)(y\minus{}x)} \equal{} \frac{z}{(y\minus{}x)^2}$
similarly , we have
$ \frac{xy\minus{}y^2\plus{}z^2\minus{}xz}{(z\minus{}y)(z\minus{}x)(x\minus{}y)} \equal{} \frac{x}{(y\minus{}z)^2}$
$ \frac{\minus{}xy\plus{}x^2\minus{}z^2\plus{}yz}{(y\minus{}z)(x\minus{}z)(x\minus{}y)} \equal{} \frac{y}{(x\minus{}z)^2}$
and now let's sum them up and we are done


$ a,b,c$ are three real strictly positive numbers  such that  $ abc\equal{}1$
prove that
$ \frac{1}{\sqrt{b\plus{}\frac{1}{a}\plus{}\frac{1}{2}}}\plus{}\frac{1}{\sqrt{c\plus{}\frac{1}{b}\plus{}\frac{1}{2}}}\plus{}\frac{1}{\sqrt{a\plus{}\frac{1}{c}\plus{}\frac{1}{2}}}>\equal{}\sqrt{2}$

what do you want to say ?  the problem is  correct :P because  1/2 x 1/2 x 1/4 = 1/16 not 1

yes sorry : i wanted to write :
$ a\equal{}\frac{1}{2}$   and  $ b\equal{}\frac{1}{2}$  and  $ c\equal{}4$   :wink:
now i think the inequality is not correct   :P

no  you can calculate it  you have a mistake ths probleme is true
(i just calculate it  :wink: )

Let $ a\equal{}\frac yx,b\equal{}\frac zy,c\equal{}\frac xz$. Then our inequality is equivalent to
\[ \sum\frac{1}{\sqrt{\frac zy\plus{}\frac xy\plus{}\frac 12}}\ge \sqrt 2\iff \sum\sqrt{\frac{y}{2x\plus{}y\plus{}2z}}\ge 1\]
Verify that
\[ \sqrt{\frac{y}{2x\plus{}y\plus{}2z}}\ge \frac{y}{x\plus{}y\plus{}z}\iff (x\plus{}y\plus{}z)^2\ge y(2x\plus{}y\plus{}2z)\]
Which is clearly true. Thus
\[ \sum\sqrt{\frac{y}{2x\plus{}y\plus{}2z}}\ge \sum\frac{y}{x\plus{}y\plus{}z}\equal{}1\]
Since $ x,y,z>0$ equality cannot hold.

yes good  :lol:
sorry i have a mystake  :P


trigonometry
inequalities proposed
inequalities

$A+B+C=\pi$. Prove that :
$\cos(A-B)\cos(B-C)\cos(C-A)\geq8\cos(A)\cos(B)\cos(C)$

[quote="jarod"]Use the formula : $\frac{\cos(B-C)}{\cos A}= \frac{b+c}{a}$ and AM-GM ineq[/quote]
jarod, $\{A,B,C\}\subset\mathbb R$ and what is this a,b,c?:)

$\frac{\cos(B-C)}{\cos A}=\frac{\sin(B+C) \cos(B-C)}{\sin A \cos A}=\frac{\sin 2B+\sin 2C}{\sin 2A }$
I think A.B.C must be acute-angel

[quote="zhaobin"]
I think A.B.C must be acute-angel[/quote]
No,not necessarily:
$cos(A-B)cos(B-C)cos(C-A)\geq8cosAcosBcosC\Leftrightarrow$
$\Leftrightarrow p^2+(4q^3+5q)p+8q^2\geq0,$ where $p=cos(A-B),q=cos(A+B).$
Let $\{p,q\}\subset[0,1].$ Then $p^2-(4q^3+5q)p+8q^2\geq0.$
Proof:
$(4q^3+5q)^2-4\cdot8q^2\leq0\Rightarrow p^2-(4q^3+5q)p+8q^2\geq0.$
If $(4q^3+5q)^2-4\cdot8q^2>0$ then
$(4q^3+5q)^2-4\cdot8q^2>0\Leftrightarrow q>\frac{\sqrt{\sqrt32-5}}{2}.$ Hence, $\frac{4q^3+5q}{2}>1.$
And $1^2-(4q^3+5q)\cdot1+8q^2\geq0\Leftrightarrow(1-q)(2q-1)^2\geq0.$
Hence, $\forall\{p,q\}\subset[0,1]$  $p^2-(4q^3+5q)p+8q^2\geq0.$
Hence, $\forall\{p,q\}\subset[-1,1]$  $p^2+(4q^3+5q)p+8q^2\geq0.$
Hence, $cos(A-B)\cos(B-C)\cos(C-A)\geq8\cos(A)\cos(B)\cos(C).$  :)

[quote="jarod"]Use the formula : $\frac{\cos(B-C)}{\cos A}= \frac{b+c}{a}$ and AM-GM ineq[/quote]
very nice ;)


Solve
$ x^3 \plus{} y^3 \plus{} 1 \equal{} 3 xy$
Where $ x,y$ are reals

[hide]
$ x^3\plus{}y^3\plus{}1^3\minus{}3xy1 \equal{} (x\plus{}y\plus{}1)(x^2\plus{}y^2\plus{}1^2\minus{}xy\minus{}y\minus{}x)$
Solutions are either $ x\plus{}y\plus{}1\equal{}0$ or $ x\equal{}y\equal{}1$.
[/hide]

You can only apply AM-GM if both $ x, y$ are nonnegative.

[quote="FOURRIER"][quote="JBL"]Up to a change of sign in the variables, this is essentially identical to Putnam 2006 B1.  (Actually, the Putnam problem was a bit trickier.)[/quote]

I  made up this problem from my mind thinking that equality in AM-GM holds only iif the variables are  equal.[/quote]
You can actually find this problem in the first section of Mathematical Olympiad Treasures.

P.S. I am not claiming that you did not create this problem.


After writing the AMC and getting 104.5 (and realizing I was almost entirely lost on how to do most questions of 15-25) I really had to wonder, "How do people do this!?"

The thing I always hear is, "You have the tools to solve this, it's just a matter of applying them." Yet, I'm very often lost on how to apply everything I know to higher-level problems.

My question is, what are common methods of approaching higher-level problems? When I look at a difficult question, how should I start myself, what should I ask myself, etc.?

Any tricks or common methods you use in solving problems, or approaching problems, would be helpful. Thanks.

At where I am now, I can usually solve questions on the level of, say, 1-20 on the AMC in less than a minute, so it's hard to say what goes through my mind when I'm doing them.  On harder questions though, this is how it might go (I just solved Q25 on the Fermat earlier today, so I'll try and walk you through my thought process).  My apologies if this is too difficult of a question, but hopefully you can take something out of it.

See this thread for the question:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27914

1st thing:  We have a line that passes through $(p,q)$.  Well, that's not too difficult, we know that the equation of this line is just $y=m(x-p)+q$.  This doesn't really get us anywhere, but it gives us an equation, which is good.

2nd thing: Well, lets look at this triline thingy.  It's easy to find x,y intercepts:

$y=mx+(q-mp)$ and $y=m(x-p+q/m)$  Add em.  We get: $3m=q-mp-p+q/m$

3rd thing: Lets look at this last equation.  It has linear expressions AND reciprocal expressions, which are awkward to deal with; lets multiply both sides by $m$.

4th thing: Rearranging, we get a quadratic equation.  What do quadratic equations have ? ........

5th thing: TWO SOLUTIONS! (unless the discriminant is zero).  So the discriminant has to be zero, because the problem says that we need EXACTLY ONE triline.

6th thing: Now we have a nifty algebraic expression, which we rearrange to $(p-q)^2=12q$.  This looks really simple, so we can see that we're almost done.

7th thing:  We have restrictions on $q$.  The biggest it can be is something like $120000$.   The other side is a perfect square.....

8th thing: which, upon appropriate choice of $p$, can be ANY perfect square we want.  So we just need the number of $x^2\leq 120000$ which gives us that $x\leq 346$

9th thing:  $12=2^2\cdot 3$.  So we need $x$ to have a factor of at LEAST $2$ and $3$; or equivalently, to be divisible by $6$.  $346/6$ is just larger than $57$, so $57$ is our answer.

You have a similar score to what I had last year...and I had wondered about the same thing too (and still do). But the thing I found that really helped was to PRACTISE! Yea, yea...you're probably thinking...everyone tells me that...but when I say practise...I mean really practise. Set aside some time each day to work through past contest problems, etc. Also highly recommended is the Art of Problem Solving series (both Vol.1 and 2). They explain all the techniques really well and have hundreds of exercises. That's what I did and I've improved a lot...though definitely not at the same level as being able to solve the question above. Still, one step at a time, huh?

Cheers!


probability
real analysis
real analysis solved

I don't know if this is the right subforum, but I think it involves some limits, so I thought I should post it here.

Given any $p\in [0,1]$, devise an experiment using only a fair coin which has success probability $p$.

This looks really awkward. I think it's not hard to find an experiment for $p=\frac k{2^t}$, but what puzzles me is the fact that we regard infinite processes as "experiments". Some infinite processes must be involved because it can't be done in a finite number of steps for $p\in R-Q$, for example.

I think the fact that the set ${\{\frac k{2^t}|k\geq 0,t\geq 0,k\leq 2^t}$ is dense in $[0,1]$ helps.

Yes it does involve some limits, or better said some infinite experiments.

Consider the following experiment:
At step n, denote by $a_n$ the number which is 1 if we got tails and 0 if we got heads.

The probability that the number, written in base 2, $0,a_1a_2\ldots a_n \ldots$ is smaller that p is p.


Source: 1989 NYSML
I-6. Compute the smallest positive angle $x$, in degrees, such that $\tan (4x)=\frac{\cos x-\sin x}{\cos x+\sin x}$.

Based on the above problem, I created my own pbm!

a) Compute the smallest positive angle $x$, such that
$\tan(2x)=\frac{\cos x - \sin x}{\cos x + \sin x}$.

b) Compute the smallest positive angle $x$, such that
$\cot(2x)=\frac{\cos x - \sin x}{\cos x + \sin x}$.

It's interesting that my own pbms requires (i think) a different solving tech. from I-6.

[quote="Jiang"][hide="I-6"]$\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-\sin^2 x}{\sin^2 x+2\sin x\cos x+\cos^2 x}$
                                                      $=\frac{\cos 2x}{1+\sin 2x}$
                                                      $=\frac{2\sin 2x\cos 2x}{2\sin 2x+2\sin^2 2x}$
                                                      $=\frac{\sin 4x}{2\sin 2x+2\sin^2 2x}$
$\tan 4x=\frac{\sin 4x}{\cos 4x}$
              $=\frac{\sin 4x}{1-2\sin^2 2x}$
then,
$2\sin 2x+2\sin^2 2x=1-2\sin^2 2x$
let $z=\sin 2x$ then, $4z^2+2z-1=0$
we get $z=\frac{-1\pm\sqrt{5}}{4}$
since x is smaller if z is positive, take the positive solution and using a calculator, we get $x=9$[/hide][/quote]

If it came out so nice, you'd think you wouldn't need a calculator.

Think Fibonacci.

Convert to $\frac{ \cos x - \cos (90 - x) }{ \cos x + \cos (90 - x) }$ and use sum to product formulas to get
$- \tan 45 \tan (x - 45) = \tan(45 - x) = \tan 4x \implies x = 9$

[quote="Jiang"][hide="I-6"]$\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\cos^2 x-\sin^2 x}{\sin^2 x+2\sin x\cos x+\cos^2 x}$
                                                      $=\frac{\cos 2x}{1+\sin 2x}$
                                                      $=\frac{2\sin 2x\cos 2x}{2\sin 2x+2\sin^2 2x}$
                                                      $=\frac{\sin 4x}{2\sin 2x+2\sin^2 2x}$
$\tan 4x=\frac{\sin 4x}{\cos 4x}$
              $=\frac{\sin 4x}{1-2\sin^2 2x}$
then,
$2\sin 2x+2\sin^2 2x=1-2\sin^2 2x$
let $z=\sin 2x$ then, $4z^2+2z-1=0$
we get $z=\frac{-1\pm\sqrt{5}}{4}$
since x is smaller if z is positive, take the positive solution and using a calculator, we get $x=9$[/hide][/quote]
Interesting.. I tried that method, but gave up, b/c it was too much time consuming....

Another quick way of doing this problem is;
convert $\tan (4x)$ to $\frac{\sin(4x)}{\cos(4x)}$.
$\frac{\sin(4x)}{\cos(4x)}=\frac{\cos x-\sin x}{\cos x+\sin x}$.
Now cross-multiflying and rearranging would yield
$\sin(4x)\cos(x)+\cos(4x)\sin(x)=\cos(4x)\cos(x)-\sin(4x)\sin(x)$.
Therefore, $\sin(5x)=\cos(5x)$ and $\tan(5x)=1$
$5x=45^\circ$ and $x=9^\circ$.

[quote="mna851"]I'm not sure what you mean by think Fibonacci but I divided by $\ \cos x$ on top and bottom to get $\ \frac{1 - \tan x}{1 + \tan x} = \frac{\tan 45 - \tan x}{1 + \tan 45 \tan x}= \ tan(45 - x)$  Then we can say $\ 45 - x + 180k = 4x$  Now since we are trying to find the least positive angle $\ x$ we can let $\ k=0$ and we find $5x=45$ $\ x=9$[/quote]

Oh that is a rather nice way to do it.

Though Golden Ration/Fibonacci angles are the ones which would appear in golden triangles or something like it.

[quote="mna851"]It may seem like adding the $\ 180k$ in my solution was superfluous, and in this example it turned out to be but take this example:

Find the smallest positive integer soltuion to $\ \tan 19x = \frac{\cos96 + \sin96}{\cos96 - \sin96}$[/quote]

[hide]Dividing the numerator and denominator by $\cos 96$, we have

$\tan 19x=\frac{1+\tan 96}{1-\tan 96}=\frac{\tan 45+\tan 96}{1-\tan 45\tan 96}=\tan (45+96)=\tan 141$.

Therefore, $19x=141+180k$.  Since $x$ is an integer, we have $141+180k\equiv 0\pmod{19}$.  Therefore, $8+9k\equiv 0\pmod{19}$.  The smallest possible value of $x$ will result from the smallest possible value of $k$, which is $16$.  Then $19x=141+180\cdot 16=3021$, and $x=\boxed{159}$.

Can anyone find a fast way to solve this from $19x=141+180k$?  I had to check $k=1,2,3\cdots 15$ before I found that $k=16$ was the smallest possible value of $k$.[/hide]

well, both $x$ and $k$ has to be integer, so it's really a diophatine equation:

[hide]
so $19x = 141 + 180k$, moding both sides by 19 gives $0 = 8 + 9k$, so $9k = 19u - 8$.  Mod by 9 again we have $0 = u - 8$, so $u = 9v+8$.  Substitute back we have $9k = 9*19v + 18*8$, or $k = 19v + 16$.  Smallest $k$ implies that $v = 0$, so we have $k = 16$.  This is my slow way of solving two variable diophatine equations, but hey, it works.
[/hide]


pigeonhole principle
analytic geometry
USAMTS
AMC

Given 5 lattice points, no three of which are collinear in the plane. Prove that if I draw all possible line segments, then at least one lattice point (excluding hte endpoints) will lie on those line segments.

[hide] consider parity:

there are only 4 possible combinations of even and odd for the x and y coordinates of the 5 points: (o,o)(e,e)(e,o)(o,e), so two of the points must have x and y coordinates with the same parity. The difference between numbers of the same parity is even, so one of the midpoints of the segments will always be a lattice point. [/hide]

Um, for example, extend this problem to 3-dimensions. What is the minimum number of distinct lattice points in space, such that there will always be lattice point on the planes connecting them. (Or am I not making any sense? Too tired...)

Or....

What if we had n lattice points on line segments drawn between distinct lattice points? Minimally, How many of those lattice points are needed?

I think i'm blabbering, too tired....

[quote]Um, for example, extend this problem to 3-dimensions. What is the minimum number of distinct lattice points in space, such that there will always be lattice point on the planes connecting them. [/quote]

I believe this was a USAMTS problem last yr, and the answer is 9, since there are 8 possible combinations of even and odd vertices. For n dimensions, there will always be lattice point if you have 2^n + 1 points.

I haven't really thought about the other one, but it would make sense that if you need a lattice points on the line segments connecting b points in c dimensions then b must be greater than or equal to 2^c + n. I very well may be wrong about this, since I haven't really worked this, just thought about it briefly...

My answer: [hide]If it's is n-space, and we want c lattice points in between, it would be (c+1)^n+1[/hide]

Hint: [hide] ok, so it's not much of a hint, but just consider mods instead of plain ol' parity[/hide]

           i just felt like putting smileys here.

Definitely getting there.

Like, 99 degrees warm


Compute the sum of $ a$, $ b$, and $ c$ given that $ \frac a2\equal{}\frac b3\equal{}\frac c5$ and $ abc\equal{}1920$.

Let $ r\equal{}\frac a2\equal{}\frac b3\equal{}\frac c5\equal{}\frac{a\plus{}b\plus{}c}{10}$.

Then we want to find $ 10r$, and since $ r^3\equal{}\frac{abc}{30}\equal{}64\implies r\equal{}4$, our answer is $ 10\cdot4\equal{}\boxed{40}$.

Proof that if $ r\equal{}\frac{a_1}{b_1}\equal{}\frac{a_2}{b_2}$, then $ r\equal{}\frac{a_1\plus{}a_2}{b_1\plus{}b_2}$.

This is pretty simple: we have $ a_1\equal{}rb_1$ and $ a_2\equal{}rb_2$, so adding these, $ a_1\plus{}a_2\equal{}r(b_1\plus{}b_2)$, and dividing, $ r\equal{}\frac{a_1\plus{}a_2}{b_1\plus{}b_2}$.

For anyone interested, the three numbers are $ 8$, $ 12$, and $ 20$.

May you clarify your solution? I don't seem to see how what we're doing with $ r$ leads us to the solution.

hello, we get from
$ \frac{a}{2}\equal{}\frac{b}{3}$
$ b\equal{}\frac{3}{2}a$
and from
$ \frac{a}{2}\equal{}\frac{c}{5}$
$ c\equal{}\frac{5}{2}a$
and so we get
$ abc\equal{}a^3\cdot\frac{15}{4}$
and from here you can compute $ a,b,c$.
Sonnhard.

@PhireKaLk6781

As math154 proved, $ r\equal{}\frac{a}{2}\equal{}\frac{b}{3}\equal{}\frac{c}{5}\equal{}\frac{a\plus{}b\plus{}c}{2\plus{}3\plus{}5}\equal{}\frac{a\plus{}b\plus{}c}{10}$.  Since we are trying to find $ a\plus{}b\plus{}c\equal{}(10)\left(\frac{a\plus{}b\plus{}c}{10}\right)\equal{}10r$, we simply need to find $ r$ and multiply it by $ 10$.

We know that $ r^3\equal{}\frac{a}{2}\cdot\frac{b}{3}\cdot\frac{c}{5}\equal{}\frac{abc}{30}$, and are given that $ abc\equal{}1920$.  Thus
\[ r^3\equal{}\frac{abc}{30}\equal{}\frac{1920}{30}\equal{}64\]
So we find that $ r\equal{}4$, and $ a\plus{}b\plus{}c\equal{}10r\equal{}\boxed{40}$.


In the tradition of the past years (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56717]here[/url] for example) let us vote and comment of the possible winners of the IMO 2007 Vietnam.

Again 10 options are available in the poll, ordered by their results in the last IMO, IMO 2006 Slovenia. Notice that near each country there is a sign, meaning what is their performance in the top 10 compared with IMO 2005 Mexico.

- = Stationary
^ = Going up in the top 10 list
^^ = New in top 10
v = Going down in the top 10

[quote="hungkhtn"]Why Vietnam is not in this list? It is so unjust!  :D[/quote]It's always the previous year's top 10 countries (in the order they ranked).

[quote="jmerry"]Maybe "Other" should be an option. It's always possible...[/quote]

you're right!  :)

I see that China has favour in the peoples' eyes right now. :)
Of course, maybe "other" is preparing extremely well...

China probably will pull off another one - but I'll go with Romania (and of course India :D)
  Ashwathj

Most probably china is going to win again.
But if it were like that what i am going to tell will come true...Then i would tell Bangladesh.
But probably not this year :(

[quote="e.lopes"]interesting, the position of moldova's team is a very increasing sequence! 48 - 40 - 30 - 21 - 17 - 9 - (?)[/quote]

But the result is hard to be kept as Iura and Alex graduate from high-school this year and join the Boston fan group.

[quote="jmerry"]Maybe "Other" should be an option. It's always possible...[/quote]I don't think it's ever been done (country not in top 10 previous year, winning the IMO). It's a matter of generation, so in order to beat China you have to have a really good generation that grows on 1-2 IMOs before that (like Romania in 96, or Bulgaria in 03).

[i][b][color=red][size=200]Vietnam [/size][/color][/b][/i]
 :first:  :P

Korean team is the best. Because there are experience participants one of the third times and two of the second times. At least one person will get perfect score.
the other else will get gold medal.

[quote="Valentin Vornicu"]You forgot Romania at the returning scores. We have a returning student (his 4th IMO!). His ML username is dzeta and he's running for the perfect score this year (he's got already 2 gold medals).[/quote]Right you are!

There should be 1 returnee with 34 points.

Since the editing time window has already expired on me, you can edit my post if you feel like it :).

[quote="MathChauffeur"]Okay, those following the IMO are not going to be as passionate, obsessed, crazy, all-consumed, etc., etc., as those following the World Cup. However, that does not mean that we can’t come up with potentially interesting tidbits leading up to the IMO --  just like thousands of reporters would, starting months (years?) before the World Cup.
*snip*[/quote]
Singapore has 2 returnees from '06, not 1, the total score is 19 + 16.  Thanks.

[quote="jiahan"]Singapore has 2 returnees from '06, not 1, the total score is 19 + 16.  Thanks.[/quote]
Thanks for the correction :blush:.

Was the second score 16 or 15? The official Slovenia site has 15 as the score.

[quote="Albanian Eagle"]there's also an albanian student who took 14 last year![/quote]
I found a few more omissions :blush: :blush:. Can somebody check on Azerbaijan and Germany? I am not sure about them.

Country, # Returnees, Total Points
USA	4	125
Korea	3	86
Moldova	3	81
Armenia	4	66
Hong Kong	3	65
Bosnia & Hertegozinia	4	63
Bulgaria	3	59
Macau	5	55
Columbia	4	54
Taiwan	2	53
Israel	3	50
Japan	2	48
Lituania	3	46
Mexico	3	46
Serbia	3	46
Austria	3	45
Germany	2	45
Argentina	3	44
Sweden	3	44
Estonia	3	43
Mongolia	3	41
Ukraine	2	40
Latvia	3	39
Iceland	4	38
Ireland	4	37
Belarus	2	37
China	1	37
Poland	1	36
Singapore	2	35
Norway	3	34
Peru	2	34
Romania	1	34
Georgia	2	32
Azerbaijan	2	31
Finland	2	31
Croatia	3	30
Malaysia	3	30
Brazil	2	30
Italy	1	30
Slovakia	1	30
New Zealand	3	28
South Africa	3	28
India	2	28
Switzerland	1	28
El Salvador	3	27
Paraguay	3	27
Costa Rica	2	27
Panama	3	24
Macedonia	3	23
Iran	1	23
Thailand	1	23
United Kingdom	1	22
Belgium	2	21
Ecuador	3	19
Hungary	1	19
Turkmenistan	1	19
Uzbekistan	1	17
Czech Republic	1	16
France	1	15
Kazakhstan	1	15
Portugal	1	15
Turkey	1	15
Denmark	2	14
Albania	1	14
Greece	1	14
Slovenia	1	14
Trinidad & Tobago	3	11
Kyrgyzstan	2	11
Cypress	2	10
Luxembourg	1	10
Netherlands	1	10
Pakistan	1	10
Venezuela	1	10
Tajikistan	2	8
Bangladesh	1	8
Nigeria	3	7
Puetro Rico	1	2
Australia	0	0
Bolivia	0	0
Canada	0	0
Liechtenstein	0	0
Morocco	0	0
Russia	0	0
Saudi Arabia	0	0
Spain	0	0
Sri Lanka	0	0
Vietnam	0	0

[quote="MathChauffeur"][quote="orl"]Germany has two previous participants:  Friedrich Feuerstein (IMO 2003) and Peter Scholze (since IMO 2004).[/quote]
Thanks! I wasn't sure about Georg Schröter in 2007 and Georg Schönherr in 2006.[/quote]
year it's a bit confusing because we live in the same city and are born in the same year (1989). but we aren't the same person ;)

Naphthalin

Another correction:

Country, # Returnees, Total Points
Austria 4 56

[quote="gighiuhui"]guys guys it's [b]USA[/b]

also, valentin, the U.S. will beat Romania with a probability of 1.01[/quote]

Rush, two comments:

(1) never over estimate yourself and your teams (which you are very good at and do a very often);

(2) one more Zuming story for you to add: he is telling you to work on math instead of chat and waste time. Did you review all the MOSP materails? If not, go to work! See you in Sep.! (And I hope your math imporved a lot by then  :|

Problems 3 and 6 were very Russian-style, so it might be Russia.

Now i know that, Vietnam is in top 4 ! :P

Vietnam is the 3rd team!!Congratulations!!! :roll:  :roll: Bravo!!! :)


vector
geometry
3D geometry
tetrahedron
parameterization
linear algebra

[b]Let A;B;C;D be four points in space, with position vectors a; b; c; d respectively. (We shall assume
the four points do not lie in a plane. They are called the vertices of a tetrahedron.) Show that the
point (1/4)(a + b + c + d) lies on the segment joining A to the centroid of triangle BCD. [/b]

Is the above question the same basic question as this one (as I've solved this one):

[b]Let A;B;C;D be four points on plane, with position vectors a; b; c; d respectively. Show that the
point (1/4)(a + b + c + d) lies on the segment joining A to the centroid A0 of triangle BCD. Let B0,
C0 and D0 be the centroids of the triangles ACD, ABD and ABC respectively. Deduce that all
four lines AA0, BB0, CC0 and DD0 pass through one point.[/b]

And please you will help with this next part please?

[i]Deduce that all four lines joining a vertex of the tetrahedron to the centroid of the opposite triangle pass
through one point.[/i]

Thank You!

A standard parametrization of a segment: The segment between $ P$ and $ Q$ is given by $ \{tP\plus{}(1\minus{}t)Q: 0\le t\le 1\}$.

Now, can you find a position vector for the centroid of a face, and the appropriate $ t$-value for the centroid $ \frac14(A\plus{}B\plus{}C\plus{}D)$ of the tetrahedron?


Prove if a,b,c>0 such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$  then
$ \sum\frac{a^3}{\sqrt{a^4\minus{}b^4\plus{}b^2}}>\equal{}3$

The condition of this ineq didn't show $ a^4\minus{}b^4\plus{}b^2 \ge 0$.But if $ a^4\minus{}b^4\plus{}b^2 \ge0$(and others expression) , we can prove ^^
$ (a^2 \plus{} b^2 \plus{} c^2)LHS^2 \ge (a^2 \plus{} b^2 \plus{} c^2)^3 \equal{} 27$
=>$ LHS \ge 3$
Done


function
real analysis
real analysis unsolved

Construct $ f: \mathbb{R}\mapsto \mathbb{R}$ so that: (i) $ f$ is one-to-one (ii) $ f$ is continous (iii) $ f$ doesn't map closed sets to closed sets. What if we added the constraint (iv) $ f$ is bounded?

$ f(x)\equal{}\arctan x$. The image of the entire real line (which, of course, is closed) is an open interval. I strongly suspect that sylow_theory meant something else but this is an example for what was literally asked :).

[quote="fedja"]$ f(x) \equal{} \arctan x$. The image of the entire real line (which, of course, is closed) is an open interval. I strongly suspect that sylow_theory meant something else but this is an example for what was literally asked :).[/quote]
I think I meant something else too. Its a problem from many months ago that I have seen, and couldn't solve it. I am sure I considered arctan in that problem and it didn't work. I just dont remember the exact details. Maybe you know the exact problem. The only thing I remember it was an excercise trying to show that closed sets dont go to closed sets (and bounded sets) (unlike compact sets) but it has the one-to-one condition on it.

But if $ f$ is continuous,then it is very obviously that $ f$ map closed sets to closed sets.Have I missed something? :?:

[quote="phils"]While we're at it, can you find a function from $ \mathbb{R}$ to $ \mathbb{R}$ that maps open sets to open sets, but is not continuous?[/quote]
Picard's theorem gives examples like this for $ \mathbb{C}$, but it seems tricky for $ \mathbb{R}$

[quote="zhaobin"][quote="RabbitTrails"][quote="zhaobin"]
But if f is continuous,then [...] f map closed sets to closed sets.
[/quote]

Careful, here, zhaobin--[i]closed[/i] maps take closed sets to closed sets; "f is continuous" means that [i]preimages[/i] of closed sets are closed.  The arctan function is a (the?) classic example of a continuous function (check it) that is not closed (as fedja explained above).[/quote]
So does it mean such $ f$ is equivalent to an open and continuous map?[/quote]
I don't know what you are asking exactly, but closed, open and continuous are "independent" conditions and do not imply each other. Only with injectivity can you start making such claims.

Thank you,I misunderstood all thing,I am sorry :blush:

heh... I've only been here a couple of days, but I'd have to say that you have nothing to apologize for--if everyone here knew everything, then the forum wouldn't be very useful!

[quote="phils"]While we're at it, can you find a function from $ \mathbb{R}$ to $ \mathbb{R}$ that maps open sets to open sets, but is not continuous?
[/quote]
For an extreme version, how about a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31493]function[/url] that takes every open interval to the whole real line?


Vieta
quadratics
ratio
algebra
geometric sequence
algebra unsolved

for three given real numbers  a,b,c consider
\[s_{n}= a^{n}+b^{n}+c^{n}\]
suppose that

\[s_{1}= 2\]     \[s_{2}= 6\]      \[s_{3}= 14\]
  prove that

\[\left|{s_{n}^{2}-s_{n-1}s_{n+1}}\right| = 8\]
for all integers n>1

Rust, I don't understand how you deduced one of a,b,c was zero.

\[2(ab+bc+ca)=(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})\]  \[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\]
Substituting the the given values implies 3abc=0

Using the idendity above is one way, but could you explain yours more, (I am relatively new to this olympiad maths so if you could that would be much appreciated)

Thanks

$a+b+c=2,\ ab+bc+ca=\frac{(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})}{2}=\frac{2^{2}-6}{2}=-1$. Let $abc=k\ a,\ b,\ c$ are the roots of the cubic equation $x^{3}-2x^{2}-x-k=0$ that is to say, $a,\ b,\ c$ satisfy the following equations.
\[a^{3}=2a^{2}+a+k\ \cdots (1) \]

\[b^{3}=2b^{2}+b+k\ \cdots (2) \]

\[c^{3}=2c^{2}+c+k\ \cdots (3) \]
$(1)+(2)+(3)$ gives $a^{3}+b^{3}+c^{3}+2(a^{2}+b^{2}+c^{2})+a+b+c+3k$, yielding $14=2\cdot 6+2+3k\Longleftrightarrow k=abc=0$.

Thus $(1)*a^{n}+(1)*b^{n}+(3)*c^{n}$ gives $a^{n+3}+b^{n+3}+c^{n+3}=2(a^{n+2}+b^{n+2}+c^{n+2})+a^{n+1}+b^{n+1}+c^{n+1}$,
yielding the following reccurence for $s_{n}$; $s_{n+3}=2s_{n+2}+s_{n+1}\ \cdots (3)$. Now when $k=0$, by Vieta's formula, $a,\ b,\ c$ are the roots of the cubic equation $x^{3}-2x^{2}-x=0\Longleftrightarrow x(x^{2}-2x-1)=0$. Without loss of generality, we can set $a=0$, so $b,\ c$ will become the roots of the quadratic equation $x^{2}-2x-1=0$, yielding $b+c=2,\ bc=-1$. Rust seems to have solved by using the charactrristic equation for the reccurence (3).
Back to the reccurence, we have $s_{n+1}=2s_{n}+s_{n-1}$, so we can obtain the following the relation of matrix.
\[\left(\begin{array}{cc}s_{n}& s_{n-1}\\ s_{n+1}&s_{n}\end{array}\right)=\left(\begin{array}{cc}0 & 1 \\ 1 & 2 \end{array}\right)\left(\begin{array}{cc}s_{n-1}& s_{n-2}\\ s_{n}&s_{n-1}\end{array}\right) \]
Use the property of determinant, we have $s_{n}^{2}-s_{n-1}s_{n+1}=(0\cdot 2-1\cdot 1)(s_{n-1}^{2}-s_{n-2}s_{n})$, yielding $s_{n}^{2}-s_{n-1}s_{n+1}=-(s_{n-1}^{2}-s_{n-2}s_{n})$, that is to say the sequence $\{s_{n}^{2}-s_{n-1}s_{n+1}\}$ is the geometric sequence with common ratio $-1$. Thus we have $s_{n}^{2}-s_{n-1}s_{n+1}=(-1)^{n-1}(s_{1}^{2}-s_{0}s_{2})=-8(-1)^{n-1}$, yielding $|s_{n}^{2}-s_{n-1}s_{n+1}|=8$.


algebra
polynomial
inequalities
complex numbers
algebra unsolved

suppose that f(x) is an irreducible polynomial with integer cofficients .
we know that f(x) has a root like $ \alpha$ such that $ |\alpha| \succ \frac {3}{2}.$
prove that $ \alpha^3 \plus{} 1$ is not a root of f(x).

sorry for mistake.
prove that if $ f$ has a root with norm greater than $ \frac{3}{2}$
then for every root like $ t$ ,$ t^3\plus{}1$ cant be a root of $ f(x)$

The root with norm greater than 3/2 may be complex, right?

Another fun problem. Sumita, I like your style  :)

[hide="Solution."]
So $ f(x) \in \mathbb{Z}[x]$ is irreducible, and suppose that $ z \in \mathbb{C}$ is a root of $ f(x)$ with the property that $ z^3 \plus{} 1$ is also a root, that is $ f(z^3 \plus{} 1) \equal{} 0$. Now consider the polynomial $ g(x) \equal{} f(x^3 \plus{} 1)$, which also has integer coefficients. We have $ g(z) \equal{} f(z) \equal{} 0$, but $ f(x)$ being irreducible over $ \mathbb{Z}$ is also irreducible over $ \mathbb{Q}$ which implies that it must divide $ g(x)$ over $ \mathbb{Q}$. In other words, $ f(x)$ is the minimal polynomial of $ z$, so any other polynomial over $ \mathbb{Q}$ which has $ z$ as a root must be a multiple of it.

So now we know that there's a polynomial $ q(x) \in \mathbb{Q}[x]$ such that $ f(x^3 \plus{} 1) \equal{} q(x)f(x)$. It follows that, for [b]any[/b] root $ w$ of $ f(x)$, $ f(w^3 \plus{} 1) \equal{} 0$ and so $ w^3 \plus{} 1$ is also a root. Starting with any given root $ w_0$ we now get a sequence of complex numbers $ w_1 \equal{} w_0^3 \plus{} 1$, $ w_2 \equal{} w_1^3 \plus{} 1$ etc., and $ w_n$ is a root of $ f(x)$ for all $ n$.

Let's see what happens if you start with a root $ w_0$ satisfying $ |w_0| \geq \frac {3}{2}$. Then $ |w_1| \equal{} |w_0^3 \plus{} 1| \geq |w_0^3| \minus{} 1 \geq \frac {19}{8} > |w_0|$, and in general $ |w_{n \plus{} 1}| \equal{} |w_n^3 \plus{} 1| \geq |w_n^3| \minus{} 1 \equal{} |w_n|^3 \minus{} 1 > |w_n|$, where the last inequality is easily shown to follow from $ |w_n| \geq \frac {3}{2}$. Therefore the elements of the sequence $ w_n$ are all distinct, which contradicts the fact that they are all roots of $ f(x)$. QED.
[/hide]

thanks.your soloution was the same as mine.


algebra
function
domain
Ring Theory
superior algebra
superior algebra solved

Prove that :The ring Z[i] is the ring such that every ideal in it is main ideal?
(I know it very famous but I dont know the solution.Can anyone help me?)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75045[/url]
this is  related

you probably meant : prove that it is a principal ideal domain


integration
logarithms
real analysis
real analysis solved

Calculate the limes of the infinite series

\sumn=1 \infty n-1(4n)!(2n)!-216-n.

By binomial
[tex] \sum\limits_{n = 0}^{ + \infty } {\frac{{(2n)!}}{{\left( {n!} \right)^2 }}} x^n  = \frac{1}{{\sqrt {1 - 4x} }} [/tex]
Then
[tex] f\left( x \right) = \sum\limits_{n = 1}^{ + \infty } {\frac{{(2n)!}}{{\left( {n!} \right)^2 }}} \frac{{x^{n - 1} }}{{4^n }} = \frac{1}{{x\sqrt {1 - x} }} - \frac{1}{x} = \frac{1}{{\sqrt {1 - x} \left( {1 + \sqrt {1 - x} } \right)}} [/tex]
Integrate  f,
[tex] F\left( x \right) = \int_0^x {f\left( t \right)dt = \sum\limits_{n = 1}^{ + \infty } {\frac{{(2n)!}}{{\left( {n!} \right)^2 }}} \frac{{x^n }}{{n4^n }} = 2\ln \left( {\frac{{1 + \sqrt {1 - x} }}{2}} \right)} [/tex]
and
[tex] F\left( 1 \right) + F\left( { - 1} \right) = \sum\limits_{n = 1}^{ + \infty } {\frac{{\left( {4n} \right)!}}{{n\left( {\left(2n \right)!} \right)^2 16^n }} = 4\ln 2 - 2\ln \left( {1 + \sqrt 2 } \right)} [/tex]

Good ?


Problem $ 1$:
"Let $ S(n)$ represent the sum of the digits of n. For example, $ S(2007) \equal{} 2 \plus{} 0 \plus{} 0 \plus{} 7 \equal{} 9$. Determine the value of the sum $ S(1) \plus{} S(2) \plus{} S(3) \plus{} ... \plus{} S(2007)$."

Problem $ 2$:
"Determine the value of the sum $ \frac {2}{1\cdot{3}} \plus{} \frac {2}{2\cdot{4}} \plus{} \frac {2}{3\cdot{5}} \plus{} ... \plus{} \frac {2}{2007\cdot{2009}}$ in lowest terms."

I know the answers, so I'll know if you're correct.  However, I don't know how to solve the problems myself.

[hide="Problem 1"]
We are looking for the sum of the digits from $ 1$ to $ 2007$

First add up those which appear in the units position.
We have $ 0 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \equal{} 45$ for every $ 10$ and then $ 0 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \equal{} 28$ for the last one. This adds up to $ 45 \times 200 \plus{} 28 \equal{} 9028$

Now consider the 10s digit.

For each hundred there will be $ 10 \times 0 \plus{} 10 \times 1 \plus{} 10 \times 2 \plus{} \cdots \plus{} 10 \times 9 \equal{} 10 \times ( 0 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 ) \equal{} 450$ There are no $ 10$s digits for out final seven. This adds up to $ 450 \times 20 \equal{} 9000$.

Now consider the 100s digit

For each thousand there will be $ 100 \times 0 \plus{} 100 \times 1 \plus{} \cdots \plus{} 100 \times 9 \equal{} 4500$ There are two thosands giving us a total of $ 9000$ (Which isn't over 9000 :P)

The 1000 digits. There will be $ 1000$ 1s and $ 8$ 2s. Givin us $ 1016$

Therefore the total is $ 9028 \plus{} 9500 \plus{} 9000 \plus{} 1016 \equal{} \boxed{28044}$ I apologise for any arithmetic errors.[/hide]

Cheers for the error catch :)

[hide="Problem 2 Hint"]$ \frac{2}{a(a\plus{}2)} \equal{} \frac{1}{a} \minus{} \frac{1}{a\plus{}2}$[/hide]

Good job, SimonM.  You are right, except that $ 450 * 20 \equal{} 9000$, not $ 9500$.  So the final answer is $ 28044$.

And thanks a lot, MellowMelon!  That hint really helps!  I forgot all about partial fractions.

Consider all the numbers 000-999.

Clearly we have the same number of 0s,1s,2s...,9s.

So there are a total of 1000*3/10=300 0s, 1s, 2s..., 9s respectively.

So that gives us a sum of the digits between 000-999 of 300(0+1+2...+9) = 300(45).
Now, consider only 1_ _ _, the last three digits of all numbers between 1000-1999.  Clearly, it also must have sum of 300(45).

Now we must also add the 1000 1s before it so thats 1000(1).

For 2000-2007, there are 8 2s and 1+2+3+...+7 for units digits.

So we have final sum of $ 300(45) \plus{} 300(45) \plus{} 1000(1) \plus{} 2(8) \plus{} 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 7 \equal{} 28044$

[hide="2"]
Each term is of the form $ \frac{2}{(x\plus{}1)(x\minus{}1)}$. Do a PFD. Then each term is of the form $ \frac{1}{x\minus{}1} \minus{} \frac{1}{x\plus{}1}$

After all the terms cancel out, we're left with $ 1\plus{}\frac{1}{2}\minus{}\frac{1}{2008}\minus{}\frac{1}{2009}\equal{}\boxed{\frac{6047091}{4034072}}$
[/hide]


A manufacturer of HDTVs must ship at least 100 TVs to its two West Coast warehouses. Each warehouse holds a maximum of 100 TVs. Northern warehouse already has 25 TVs on hand. Southern warehouse already has 20 TVs on hand. It cost $12 to ship a TV to the nothern warehouse and$10 to ship a TV to the southern warehouse. Union rules require that at least 300 workers be hired. Shipping a TV to northern warehouse requires 4 workers, while shipping a TV to southern warehouse requires 2 workers.

How many TVs should be shipped to each warehouse to minimize the cost? What is the minimum cost?

[hide="Answer"]$50$ TV's to each. The cost is $1100$.[/hide]

[quote="sapphyre571"]@ interval

if you want to denote currency on AoPS, don't just use a single dollar sign - it tells the computer that everything after the dollar sign should be in LaTeX until it sees another dollar sign, so, to type a dollarsign, type this

dollarsign\dollarsigndollarsign
and you get
$\$$[/quote]

Which, when actually typed in, looks like:[code]$ \$ $[/code] and appears as $\$$


Υπάρχει διαφορίσιμη καμπύλη $ \vec x : I \to \mathbb R^2$ όπου $ I \subset \mathbb R$ διάστημα που η εικόνα της να περιέχει έναν τόπο του $ \mathbb R^2$?
Φαντάζομαι πως δε γίνεται :lol: αλλά επειδή είδα στον Rudin που κατασκεύαζε μία συνεχή καμπύλη που είχε ως εικόνα το μοναδιαίο τετράγωνο μου δημιουργήθηκε αυτή η απορία.
Δεν ξέρω αν είναι προφανές και δεν βλέπω κάτι :blush: .Μπορεί κανείς να βοηθήσει;

Η απάντηση στο ερώτημα που θέτεις είναι όχι. Δε γίνεται να υπάρξει διαφορίσιμη συνάρτηση της οποίας η εικόνα να περιέχει ανοικτό. Η ιδέα της απόδειξης είναι η εξής. Αν αυτό ήταν δυνατό, τότε θα μπορούσαμε να έχουμε συνάρτηση $ f: [0,1]\to \mathbb{R}^2$ τέτοια ώστε $ f([0,1])\supset (a,b)\times(c,d)$ για κάποιους αριθμούς $ a,b,c,d$ με $ a<b$ και $ c<d$. Επιπλέον, η $ f$ θα ήταν της μορφής $ f(t)\equal{}(f_1(t),f_2(t))$. Για κάθε $ u\in(a,b)$, το σύνολο: $ \{t\in[0,1]: f_1(t)\equal{}u\}$ θα ήταν υπεραριθμήσιμο και συνεπώς θα είχε σημείο συμπύκνωσης. Σε ένα όμως τέτοιο σημείο συμπύκνωσης η παράγωγος της $ f_1$ θα ήταν μηδέν. Αυτό θα σήμαινε οτι η συνάρτηση $ f_1$ θα είχε την εξής ιδιότητα: η εικόνα του συνόλου των κρίσιμων σημείων της θα περιείχε ανοικτό διάστημα. Αυτό όμως είναι αδύνατο όπως έπεται από ένα θεώρημα του Lusin που είναι παραλλαγή του γνωστού θεωρήματος του Sard.

Νομίζω το θεώρημα στο οποίο αναφέρεσαι είναι του Dubovickii, αν και η απόδειξή του όντως φέρνει σε εφαρμογή Lusin για να εφαρμόσουμε Sard.

Σε κάθε περίπτωση, όπως αναφέρθηκε, είναι αδύνατο, ακόμα και να παραγωγίσιμες συναρτήσεις. Βασικά είναι πολύ πιο εύκολο να αποδείξεις ότι είναι αδύνατο για Lipschitz συναρτήσεις (και άρα $ \mathcal{C}^1$). Το ότι δυσκολεύει στις παραγωγίσιμες μπορείς να το λάβεις ως ένα σημάδι ότι φτάνουμε σε "οριακές" συνθήκες για την ύπαρξη τέτοιας καμπύλης. Πράγματι, δοκίμασε Γιάννη να κατασκευάσεις μια καμπύλη στο $ \mathbb{R}^2$ που είναι [i]σχεδόν[/i] παντού παραγωγίσιμη, και όμως έχει εικόνα με θετικό μέτρο :!: (ενώ όταν είναι παραγωγίσιμη έχουμε μέτρο μηδέν)


If $\frac{(a-b) \cdot (c-d)}{(b-c) \cdot (d-a)} = -\frac{23}{27}$, compute ${\frac{(a-c) \cdot (b-d)}{(a-b) \cdot (c-d)}}$

Find an equation, [in the form of $x^3+ax^2+bx+c = 0$], whose roots are two less than the roots of $x^3+x+1 = 0$...

Man, that was simply solved

I was lookin too much into that....

[quote="Bictor717"][quote="PeNtaGoNaL"]Find an equation, [in the form of $x^3+ax^2+bx+c = 0$], whose roots are two less than the roots of $x^3+x+1 = 0$...[/quote]
[hide]The roots of $\left(x+2\right)^3+\left(x+2\right)+1=0$ are 2 less than the roots of the given equation. Multiply out to get $x^3+6x^2+13x+11=0$.[/hide][/quote]

Nice solution, but could you explain how you got that answer? All you said that was the equation $\left(x+2\right)^3+\left(x+2\right)+1=0$ had roots 2 less than $x^3+x+1 = 0$  :?

The of that equation are written $x = [roots]$.  Then he just added two to those, which can be written as $x + 2$ since the roots are values of $x$ such that $x^3 + x + 1 = 0$.  Then he just substituted $x + 2$ in for $x$, simplified the following equation, and voila!
Good solution, by the way.

Answer in the first problem is $\frac{50}{23}$
because ${\displaystyle \frac{(a-c) \cdot (b-d)}{(a-b) \cdot (c-d)}}=1-\frac{1}{\frac{(a-b) \cdot (c-d)}{(b-c) \cdot (d-a)}}$

My full proof.
\begin{eqnarray*}\displaystyle\because\qquad \frac{(a-c) \cdot (b-d)}{(a-b) \cdot (c-d)}&=& \frac{ab-ad-bc+cd}{ac-ad-bc+bd} \\
 \\
&=& \frac{ab-ad-bc+cd+ac-ac+bd-bd}{ac-ad-bc+bd} \\
 \\
&=& \frac{ac-ad-bc+bd+ab+cd-ac-bd}{ac-ad-bc+bd} \\
 \\
&=& \frac{ac-ad-bc+bd}{ac-ad-bc+bd}+\frac{ab+cd-ac-bd}{ac-ad-bc+bd} \\
 \\
&=& 1+\frac{(b-c) \cdot (a-d)}{(a-b) \cdot (c-d)}
= 1-\frac{(b-c) \cdot (d-a)}{(a-b) \cdot (c-d)}
= 1+\frac{1}{\frac{(a-b) \cdot (c-d)}{(b-c) \cdot (d-a)}}
= \frac{50}{23}
\end{eqnarray*}

       :D


limit
integration
function
calculus
real analysis
calculus computations

Having $ \lim_{n\to\infty}\int_0^1\frac{dx}{1\plus{}x^n},$ clearly the integrand is bounded by $ g(x)\equal{}1,$ which is an integrable function on $ [0,1]$ and $ \frac1{1\plus{}x^n}\to1$ as $ n\to\infty,$ thus DCT applies and the limit is 1.

Is this correct?

Yes, it is (though using the DCT here is a huge overkill).

Thanks fedja!!! I finally got how to apply the DCT!!! Thanks!

One very minor point: $ \frac1{1+x^n}\to 1$ is true for $ x\in[0,1)$ but not for $ x=1.$ Now, all the DCT asks for is convergence almost everywhere, so you're fine and the proof goes though unchanged.

Now, suppose you're in a typical junior-level "intro to analysis" or "advanced calculus" course and you have not studied the Lebesgue integral and do not feel that you have the DCT (or MCT) available as tools:

Choose an $ \epsilon>0.$ We look at $ \int_0^{1-\epsilon}\frac1{1+x^n}\,dx+\int_{1-\epsilon}^1\frac1{1+x^n}\,dx=A_n+B_n.$

As for term $ A_n:$ we do have that $ \frac1{1+x^n}\to1$ uniformly on $ [0,1-\epsilon].$ Uniform convergence on a bounded interval justifies interchanging limit and integral, so we can say that $ \lim A_n=\int_0^{1-\epsilon}1\,dx=1-\epsilon.$

As for term $ B_n:$ we can say that $ 0<\frac1{1+x^n}<1,$ so $ 0<B_n<\epsilon.$

So we can pick $ n$ large enough that $ |A_n-1-\epsilon|<\epsilon,$ which implies that $ |A_n-1|<2\epsilon.$ Then

$ \left|\int_0^1\frac1{1+x^n}\,dx-1\right|=|A_n+B_n-1|\le|A_n-1|+|B_n|<3\epsilon$

for sufficiently large $ n.$ Since $ \epsilon$ was arbitrary, we have that the integral tends to $ 1.$ (If that $ 3\epsilon$ offends your sensibilities, go back to the beginning and replace $ \epsilon$ by $ \frac{\epsilon}3.$ It doesn't really matter.)

I made it this complicated because I couldn't just say that we had uniform convergence on $ [0,1].$

The other thing to note about that argument is that it's the same idea that goes into proving the DCT. We have a broad region on which convergence is good and we use a uniform estimate, plus a small region on which convergence is bad and we just estimate things with the dominating function.

[quote="Kent Merryfield"]One very minor point: $ \frac1{1 \plus{} x^n}\to 1$ is true for $ x\in[0,1)$ but not for $ x \equal{} 1.$ [/quote]

That's why I don't really like the notation $ \int_0^1$ for the Lebesgue integral. It is never clear whether it means $ \int_{(0,1)}$ or $ \int_{[0,1]}$ or one of the two other options and if the measure has point masses it can make quite a difference.

As to the "junior analyst", he would just write $ 1\minus{}x^n\le\frac 1{1\plus{}x^n}\le 1$ and integrate this inequality. ;) On the other hand, you can learn much more from Kent's approach. :)


[quote="fatehi"]salam be hameye bachehah

It is Known that 2cos^3(pi/7)-cos^2(pi/7)-cos(pi/7) is a rational number (addad gooya). Write this rational number in the form p/q, where p and q are integers.[/quote]
in latex
it is known that $2cos^3(\frac{\pi}{7})-cos^2(\frac{\pi}{7})-cos(\frac{\pi}{7})$ is a rational number (addad gooya).write this rational number in the form $\frac{p}{q}$ where $p$ and $q$ are integers.


If $p$ is an odd prime. Prove there exists one and only one such $n$ such that $1\leq n \leq p!$ and $p!|n^{p}+1$


function
Euler
number theory
relatively prime

A sequence is defined as follows: a1 = 3 and

an+1 =

an/2 if an is even
(an+1983)/2 if an is odd.

Prove the sequence is periodic, and find the (minimum) period.

We have 1 <= a(n) < 1983 for all n :

a(n) < 1983 => a(n+1) < 1983 because a(n)/2 and (a(n)+1983)/2 < 1983.

Now there are only finite integer values between 1 and 1982 so after at most 1987 values a(n) becomes periodic. Period is at most 1983. It's also easy to see that a(n) is always divisible by 3. There are only 660 multiple of 3 between 3 and 1982 so period is at most 661.

I'm not sure that it's the minimum period, though.

[quote="belenos"]Now there are only finite integer values between 1 and 1982 so after at most 1987 values a(n) becomes periodic. Period is at most 1983. [/quote]

I have a question about the definition of "periodic" -- does it mean a sequence of numbers should cycle, or is it enough that one number is hit twice somewhere?

For ex:

(1) 1, 2, 3, 4, 1, 2, 3, 4, ...
(2) 1, 2, 3, 4, 1, 4, 2, 1, 4, 3, 2, 3, 3, 1, ...

I know for sure (1) is periodic, but is (2)? Even if there are only a finite number of possible values for each term, does that mean that a sequence must cycle, eventually?

"Periodic" means that the sequence cycles.  For sequences where a_{n+1} = f(a_n), it is sufficient to show that the sequence contains the same value twice to show that it eventually becomes periodic.  (Why?  If a_i = a_j, then a_{i+1} = a_{j+1}, etc..).  However, depending on whom you are talking to, "periodic" may or may not mean the same as "eventually becomes periodic".  The sequence 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ...  (a_{n+1} = {(a_n)/2 if a_n is even,  3a_n + 1 if a_n is odd}) eventually becomes periodic, but we never have a_k = a_1 for some k>1.  I don't think that this sequence would generally be considered "periodic" - the periodicity begins a distance into the sequence - but it depends on whom you ask.

It is (I think) called eventually periodic.

Wyra raised two good points !

It's because that a(n+1) = f(a(n)) for some function f that a(n) eventually become peridic. As soon as we get a previously got value, a(n) become periodic because a(n) depends only on the previous value.

Actually I've only proved that a(n) becomes periodic if we go far enough, not that a(n) is periodic 'stricto sensu' i.e. there exists no such as a(n0)=a(1).

A good thing to try with recursive sequence problems is going backwards. From the formulas, we get a(n-1) = 2 a(n) or a(n-1) = 2 a(n)-1983. Is it uniquely determined? Yes. Since we know that 1 :le: a(n-1) < 1983, and since only one of 2 a(n) and 2 a(n)-1983 is within those bounds, going backwards is uniquely determined also, proving the sequence is [b]strictly periodic[/b].

Since the numbers are between 1 and 1982, they are equal to their remainders modulo 1983 (none of the terms are divisible by 1983, so 0 is excluded). We see that a(n-1)  :equiv:  2 a(n) in modulo 1983. To find the actual period, we need to find n such that a(n+1) = 3, i.e. a(n+1)  :equiv: 3 mod 1983. Since a(n+1)  :equiv:  2 a(n)  :equiv:  2^2 a(n-1)  :equiv: ... :equiv: 2^n a(1)  :equiv: 3*2^n, we need to find n such that 3*2^n  :equiv:  3 mod 1983, i.e.

2^n  :equiv:  1 mod 1983.

At this point i got lazy, and used a calculator to determine that the order of 2 in modulo 1983 is 660, so a(661) = 3 and the period is 660.

Very good, although I can say two things:

Firstly, you can use the Euler phi function to calculate that number 661 easier.

Secondly, you haven't proved that 660 is the minimum period.

I found a small problem with my last post. I went from 3*2^n :equiv: 3 mod 1983 to 2^n :equiv: 1 mod 1983. Since 3 is not relatively prime to the mod, we cannot cancel the three that easily. When we cancel the three, we get three possibilities:

2^n  :equiv:  1        mod 1983
2^n  :equiv:  662    mod 1983
2^n  :equiv:  1323  mod 1983

These can be summed up more simply as 2^n  :equiv:  1 mod 661.

Since  :phi: (661) = 660 (because 661 is prime), the smallest n that works is a divisor of 660. Thus, "all" we have to do is calculate the remainders in mod 660 of 2^(divisors of 660) until we find a 1.

I'll spare you the details: the smallest n turns out to be 660 anyway.

Well, all you have to do is prove that 2^330 is congruent to -1, that will prove it.

Are you sure? What if it's 1(mod 220) and -1(mod 330)? Then 220 is a smaller period than 660.

True. So there are a few values we have to check (660/each prime factor of 660).


geometry
perpendicular bisector
projective geometry
geometry proposed

Let AB and CD be distinc parallel chords of a circle and let P be a point on theire com mon perpendicular bisector. The lines AP and CP meet the crcle again at Q and R respectively.If S is the intersection of the lines BR and QD, Prove that  PS is rarallel to AB.

Let P be any point not on the given circle (otherwise the points P, Q, R, S would coincide and the line PS would be undefined), not necessarily on the common perpendicular bisector of the parallel chords $AB \parallel CD$. By Pascal's theorem, the intersections $M \equiv AB \cap CD,$ $P \equiv AQ \cap CR,$ $S \equiv BR \cap DQ$ of the chord pairs (AB, CD), (AQ, CR), (BR, DQ) are collinear. Since the lines $AB \parallel CD$ are parallel, their intersection M is at infinity, equivalent to the direction of these 2 parallel lines. Since the line PS passes through M, it has this direction as well, hence, $PS \parallel AB$.


geometry
perimeter
floor function
trigonometry
circumcircle
trig identities
Law of Cosines

In convex quadrilateral $ABCD$, $\angle A \cong \angle C$, $AB = CD = 180$, and $AD \neq BC$.  The perimeter of $ABCD$ is 640.  Find $\lfloor 1000 \cos A \rfloor$.  (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x$.)

[hide]

We use Law of Cosines in triangles ABD and CBD to find

$BD^2=180^2+AD^2-2*180*AD\cos A$

$BD^2=180^2+BC^2-2*180*BC\cos C$

Define $\cos A=\cos C=x$.  We can set these two equations equal and simplify to get

$AD^2-360*AD*x=BC^2-360*BC*x\Rightarrow 360x(BC-AD)=BC^2-AD^2$

Factoring the left side and dividing out $(BC-AD)$, we get $360x=BC+AD$.  Since the perimeter of $ABCD$ is 640 and the other two sides are 180, the sum of $BC$ and $AD$ is $640-2(180)=280$.  So we get $360x=280$, and thus

$x=\frac{280}{360}=\frac{7}{9}=.777777...$

$\lfloor 1000 \cos A \rfloor$ is thus 777.    [/hide]

Sorry for the revive, but can anyone give me an example of a diagram when $ AD \neq BC $.

From what I understand of it, the circumradius of triangle $ ABD $ has radius $ \frac{BD}{\sin A} $ and the circumradius of triangle $ CBD $ has radius $ \frac{BC}{\sin C} $. Therefore the circumradii are equal in length.

From here, we can consider several possibilities:

1. $ ABD $ and $ BCD $ have different circumcircles. In that case, angle $ ADB $ = angle $ CBD $. By $ AAS $ congruence, the triangles are congruent and so $ AD = BC $.

2. $ ABD $ and $ ACD $ have same circumcircles.

Subcase 1: $ A $ and $ C $ lie on opposite sides of line $ BD $. The only way angle $ A $ = angle $ C $ is if they are both right angles. Again we have congruent triangles and so $ AD = BC $.

Subcase 2: $ A $ and $ C $ lie on the same side of line $ BD $. This quadrilateral is complex. Contradicts definition of convex. But it can be seen that even in this case $ AD = BC $.

Did I do anything wrong here, and am I missing a case?

This was wayyy too easy for a #12. Agreement?

[hide=sketch]By LoC on BD in both triangles, we rearrange this to (AD^2-BC^2)=360(AD-BC)cosA, which is the same as 280=640-360=AD+BC=360cosA, which yields that floor[1000cosA]=$\boxed{777.}$[/hide]
[b]Remark.[/b] Hmm... The answer is the exact same as aime 1996/15 (yes i am doing problems now) and has the same form of floor[1000something].


Is there a link to a full PDF guide to Asymptote?

Everything you could every possibly want about Asymptote is found on their [url=http://asymptote.sourceforge.net/]home page[/url]. What you're looking for is probably [url=http://asymptote.sourceforge.net/asymptote.pdf]this[/url].

You actually already have a manual on your computer in your Program Files/Asymptote Folder that's the same as the PDF link ilp gave.

For package documentations, visit the respective package's website.

I'm afraid there is no complete documentation of the basics and every package all in one document.


[quote="236factorial"]How much greater than $\sqrt{16+9}$ is $\sqrt{16}+\sqrt9$?[/quote]

[hide]$\sqrt{16+9}=\sqrt{25}=5$

$\sqrt{16}+\sqrt{9}=4+3=7$

It's 2 greater.[/hide]

[quote="236factorial"]How much greater than $\sqrt{16+9}$ is $\sqrt{16}+\sqrt9$?[/quote]
[hide]16+9=25, $\sqrt{25}=5$ and $\sqrt{16}+\sqrt{9}=4+3=7$ so the answer is $7-5=\framebox{2}$[/hide]

[quote="236factorial"]How much greater than $\sqrt{16+9}$ is $\sqrt{16}+\sqrt9$?[/quote]

[hide]$\sqrt{16+9} = \sqrt{25} = 5$

$\sqrt{16} + \sqrt{9} = 4 + 3 = 7$.

$\text{ EDIT: Difference = 2 }$[/hide]

Fixed. :blush:

[hide][quote="hwenterprise"][quote="236factorial"]How much greater than $\sqrt{16+9}$ is $\sqrt{16}+\sqrt9$?[/quote]

[hide]$\sqrt{16+9} = \sqrt{25} = 5$

$\sqrt{16} + \sqrt{9} = 4 + 3 = 7$.

$\text{ Difference = -2 (Wait, that sounds wrong...)}$[/hide][/quote]

You did it back wards you  need to find How much greater $\sqrt{16}+\sqrt9$ is not how much greater  $\sqrt{16+9}$ is...
btw the answer is 2...[/hide]


Still no one? I have a solution with a little calculaitng.If no one can solve it in a week,I 'll post mine. :)

I think I can solve this problem with using threeliner corrdinates but I need to find time to think about it a litter longer than I did :blush: .So,wait a little more and don't post your solution.I will try to find free time and think on this problem,I am rather confidence that I will manage to solve it :P .

I will wait for you.
By the way ,Could  you tell me what is "threeliner corrdinates "?I don't know more in Geometry. :blush: Thanks.

[quote="Hawk Tiger"]Point $E,F,G,H$lie on sides $AB,BC,CD,DA$ of a convex quadrilateral $ABCD$ such that $\frac{AE*BF*CG*DH}{EB*FC*GD*HA}=1$
Points $A,B,C,D$ lie on sides $H_{1}E_{1},E_{1}F_{1},F_{1}G_{1},G_{1}H_{1}$ a convex quadrilateral  $H_{1}E_{1}F_{1}G_{1}$ such that $E_{1}F_{1}//EF,H_{1}E_{1}//HE,F_{1}G_{1}//FG,G_{1}H_{1}//GH$
Prove that $\frac{E_{1}A}{AH_{1}}=\frac{F_{1}C}{CG_{1}}$[/quote]

So :) ,yesterday was my last examination and now I am ready to start practicing math :lol: .At first here is my solution for this problem as I have promesed you dear Hawk :wink: .

[color=blue]Solution:[/color]

From the condition of the problem and Menelaus' theorem it is obvious that $FG$ and $EH$ intersect on line $BD$;let's denote the intersection point $X$.
Also let's note

\[P\in{{F'G'}\cap{E'H'}}\]
\[M\in{{BD}\cap{F'G'}}\]
\[N\in{{BD}\cap{E'H'}}\]
\[X'\in{{GX}\cap{G'H'}}\]
\[X''\in{{HX}\cap{G'H'}}\]
\[O\in{{BD}\cap{AC}}\]

Now again from Menelaus' theorem

\[\frac{CO}{AO}=\frac{CM}{MP}\frac{PN}{NA}\]

and

\[\frac{G'D}{DH'}=\frac{G'M}{MP}\frac{PN}{NH'}\]

On the other hand,we have that $GX\parallel CM$ and $HX\parallel AN$ =>
\[\frac{MG'}{MC}=\frac{XX'}{XG}\]
\[\frac{NH'}{NA}=\frac{XX''}{XH}\]
But we also have that $X'X''\parallel GH$=>
\[\frac{XX'}{XG}=\frac{XX''}{XH}\]
=>
\[\frac{MG'}{MC}=\frac{NH'}{NA}\]
=>
\[\frac{CO}{AO}=\frac{CM}{MP}\frac{PN}{NA}=\frac{G'M}{MP}\frac{PN}{NH'}=\frac{G'D}{DH'}\]
=>
\[\frac{CO}{AO}=\frac{G'D}{DH'}\]

In the end, using a very famous fact that if $\frac{CO}{AO}=\frac{G'D}{DH'}$ then $\frac{F'B}{BE'}=\frac{G'D}{DH'}$ we get the required property.You can similary prove that  $\frac{E'A}{AH'}=\frac{F'C}{CG'}$.
P.S. Just one more thing,in the solution above I replaced $E_{1}F_{1}G_{1}H_{1}$ by$E'F'G'H'$ :blush: .


Let $ a$ and $ b$ are positive numbers such that $ a^{25}\plus{}b^{25}\equal{} 2.$ Prove that:
\[ \frac{a^3}{b^2}\plus{}\frac{b^3}{a^2}\geq2\]

AM-GM
 $ \frac {a^3}{b^2} \plus{} \frac {b^3}{a^2} \ge 2\sqrt {ab} (*)$
  But we know that  $ \sqrt {ab}\le \frac {a \plus{} b}{2}\le \sqrt [25]{\frac {a^{25} \plus{} b^{25}}{2}} \equal{} 1$
  Thus $ (*)$ is equivalent to $ \frac {a^3}{b^2} \plus{} \frac {b^3}{a^2} \ge 2$ Equality holds if and only if a=b=1. Q.E.D

Well,
$ \frac{a^3}{b^2} \plus{} \frac{b^3}{a^2}\geq2\sqrt {ab}$
and we also know that :
$ 1\geq\sqrt {ab}$.
But that doesn't mean:
$ \frac{a^3}{b^2} \plus{} \frac{b^3}{a^2}\geq2$
Look at this example:
$ 4\geq2$
$ 6\geq2$
so,
$ 4\geq6$?

wangsacl, I agree with you!
With your permission I'll ask him in another way.
[quote="broniran"]AM-GM
 $ \frac {a^3}{b^2} \plus{} \frac {b^3}{a^2} \ge 2\sqrt {ab} (*)$
  But we know that  $ \sqrt {ab}\le \frac {a \plus{} b}{2}\le \sqrt [25]{\frac {a^{25} \plus{} b^{25}}{2}} \equal{} 1$
  Thus $ (*)$ is equivalent to $ \frac {a^3}{b^2} \plus{} \frac {b^3}{a^2} \ge 2$ [/quote]
Why $ (*)$ is equivalent to $ \frac {a^3}{b^2} \plus{} \frac {b^3}{a^2} \ge 2$ $ ?$
Explain please. Thank you!

Because $ \sqrt {ab}\le 1$ not $ \sqrt {ab} \ge 1$ , wangsacl. And because $ \sqrt {ab}\le 1$ , so $ 2 \ge 2\sqrt{ab}$. Perhaps, I haven't used the most appropriate word "equvalent".I should have used " obtain".

But that's exactly why you're wrong; you showed that it was at least something that is SMALLER than 2.

Oh, man :blush:  I've made a great mistake :blush:

Yeah, I tried applying Mildorf's lemma, and this inequality was sharper... it reduced to $ a^6\plus{}a^5b\plus{}ab^5\plus{}b^6 \ge 13a^4b^2 \minus{} 22a^3b^3 \plus{} 13a^2b^4$ which unfortunately is false...

[quote="Altheman"]we must equivalently prove

$ (y + 1)^{25} - 2^{24}\cdot y^{10}(y^5 + 1)\ge 0$

[/quote]
I think, this inequality easier to prove like this:
$ (y + 1)^{25} - 2^{24}\cdot y^{10}(y^5 + 1)\ge 0\Leftrightarrow$
$ \Leftrightarrow(y^2+2y+1)^{12}-2^{24}y^{10}(y^4-y^3+y^2-y+1)\geq0\Leftrightarrow$
$ \Leftrightarrow(u+2)^{12}-2^{24}(u^2-u-1)\geq0,$ where $ u=y+\frac{1}{y}.$
Let $ f(u)=12\ln(u+2)-\ln(u^2-u-1)-24\ln2.$
Hence, $ f'(u)=\frac{12}{u+2}-\frac{2u-1}{u^2-u-1}=\frac{5(u-2)(2u+1)}{(u+2)(u^2-u-1)}.$
Id est, $ f(u)\geq f(2)=0.$
What about the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219238


Three faces of a rectangular prism have the areas shown.  What is the number of cubic centimeters in the volume of the rectangular prism?

[asy]import three; size(101);
real x = 10, y = 7, z = 3;
 currentprojection = orthographic(1/2,-1/2,1/2); draw((0,0,0)--(x,0,0)--(x,0,z)--(0,0,z)--cycle3);
draw((x,0,0)--(x,y,0)--(x,y,z)--(x,0,z)); draw((0,0,z)--(0,y,z)--(x,y,z));
label(rotate(-30)*"$30\mbox{ cm}^2$",(x/2,0,z/2),fontsize(8pt));
label("$70\mbox{ cm}^2$",(x/2,y/2,z),fontsize(8pt));
label(rotate(30)*"$21\mbox{ cm}^2$",(x,y/2,z/2),fontsize(8pt));[/asy]

Instead of finding the side lengths individually, express each face area as a product of two dimensions. So $ ab\equal{}70, bc\equal{}21, ac\equal{}30$. Now we can multiply all of them together to get $ a^2b^2c^2\equal{}44100$. But we have to remember what we want! The value of $ abc$. So take the square root of both sides to get $ abc\equal{}\boxed{210}$.


calculus
integration
function
logarithms
complex analysis
complex analysis unsolved

Let $H$ be a path composed by the $3$ lines $[i+\infty,i-1],[i-1,-i-1],[-i-1,-i-\infty]$. For $\mathbb{C}\setminus [0,\infty]$ let $\arg{z}$ be given by $0 < \arg{z}< 2\pi$ and let $\log{z}= \log{|z|}+i\arg{z}$ be the correpsonding branch of the logarithm and $u^{z}= e^{z\log{u}}$.Now show that
$\zeta(s) = \frac{1}{\Gamma(s)(e^{2\pi i s}-1)}\int_{H}\frac{w^{s-1}}{e^{w}-1}dw$

[quote="sylow_theory"]Isnt there an identity?

$\zeta(z)\Gamma(z)=\int_{0}^{\infty}\frac{t^{z-1}}{e^{z}-1}dt$[/quote]

Yes, that's true.


What is the value of the sum $ 2^{\minus{}1}\plus{}2^{\minus{}2}\plus{}2^{\minus{}3}\plus{} \ldots \plus{}2^{\minus{}9} \plus{} 2^{\minus{}10}$? Express your answer as a common fraction.

hello, after a short computation we get
$ \sum_{i\equal{}1}^{10}2^{\minus{}i}\equal{}\frac{1023}{1024}$
Sonnhard.

[hide="Geometric sequence solution"]

The sum of geometric sequences is as follows:

$S_n=a\left({\frac{1-r^n}{1-r}}\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Plugging things in:

$S_{10}={\frac{1}{2}}\left({\frac{1-\left(\frac{1}{2}\right)^{10}}{1-{\frac{1}{2}}}}\right)$

Simplifying a lot:

$S_{10}=1-{\frac{1}{1024}} \implies \boxed{{\frac{1023}{1024}}}$[/hide]

[quote="HYP135peppers"]In these types of problems, the numerator will be 1 less than the denominator. Thus, the denominator for this solution will be 1024, and the numerator is 1023. Therefore, $ {%Error. "division" is a bad command.
{1023}/{1024}} $[/quote]

[hide="Proof"]
We wish to find the value of $2^{-1}+2^{-2}+\cdots+2^{-n}$.

[quote="newb"]
The sum of geometric sequences is as follows:
$S_n=a\left({\frac{1-r^n}{1-r}}\right)$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.
[/quote]

Pluggin in values:

\[ S_n={\frac{1}{2}}\left({\frac{1-\left(\frac{1}{2}\right)^{n}}{1-{\frac{1}{2}}}}\right) .\]

This becomes \[S_n=\dfrac{\frac{2^n-1}{2^n}}{2\cdot\frac{1}{2}}.\] The denominator of the fraction becomes $1$, so we are left with
\[\boxed{S_n=\dfrac{2^n-1}{2^n}}.\]
[/hide]


Nats 1989 question - I think answer they give is wrong, but can't figure out how to do. Please offer a detailed solution. Thanks.

Question: If a typist produces 4 typed pages for every 5 handwritten pages and can produce 14 pages per hour, how many hrs would it take to type 105 hand pages?

I have some questions.

1. What answer does it give?
2. 14 what kind of pages per hour?

Let t = typed and h = handwritten

We have a ratio of $ 4t: 5h$

So if we have 105h,

$ \frac{4t}{5h}\equal{}\frac{xt}{105h}$

$ x \equal{} 84t$

Hence the typist has typed 105+84=189 pages in total, which would take $ \frac{189}{14}\equal{} 13.5$ hours.

to identity and others, yes, I got 13.5 hrs also, but the official ans. key said 6 - but I just went back and realized question in fact said [b]typed[/b]  14 pages, not produced - which changes how to do the problem and does, in fact, yield the ans of 6 -- (sorry about that - but btw, just saying 14 pages makes for a more challenging question, don't you think?) thanks to those whoresponded


analytic geometry
search
combinatorics proposed
combinatorics

Let $ d > 0$ be a [i]rational[/i] number, and let $ G$ be the infinite graph with the vertices as points with rational coordinates. Two vertices of $ G$ are adjacent if and only if the distance between them is $ d$. Prove that $ G$ is bipartite. If $ d$ is irational does $ G$ still remain bipartite?

It is immediate to see that if $ G$ has its vertices in points with [i]real[/i] coordinates, then the $ G$ is not bipartite (This is a simple consequence of the fact that no matter how we colour the plane with two colours, there exists a monochromatic equilateral triangle).

For this problem and related questions about this graph
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=501047893&t=2635

Pierre.

Nice! I was in fact waiting unconsciously a reference from you. Thanks.


function
logarithms
integration
calculus
calculus computations

Compute the value of each of the following series (preferably without using power series):

$ S_{1}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{n(2n\plus{}3)}$

$ S_{2}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{2n^{2}\plus{}5n\plus{}1}{(n\plus{}2)!}$

$ S_{3}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{9n\minus{}4}{3n(3n\minus{}1)(3n\minus{}2)}$

$ S_{4}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{1}{16n^{2}\minus{}1}$

$ S_{5}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}\frac{(\minus{}1)^{n\minus{}1}}{n(3n\minus{}1)}$

$ S_{6}\equal{}\sum_{n\equal{}1}^{\plus{}\infty}(\minus{}1)^{n\minus{}1}\frac{n}{9n^{2}\minus{}1}$

i am too tired to work out everyone of them in detail now (just finished writing tomorrow's lecture on weierstrass' and jacobi's elliptic functions a few minutes ago), so i will show the methods and the answer should then follow, OK ?  :|

$ S_{1}\; \equal{}\;\frac{2}{3}\sum_{n \equal{} 1}^{\infty}\;\left(\frac{1}{2n}\minus{}\frac{1}{2n\plus{}3}\right)\; \equal{}\;\frac{2}{3}\left(1\plus{}\frac{1}{3}\minus{}\left(1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}\frac{1}{5}\minus{}\;\ldots\right)\right)$
$ \equal{}\;\frac{2}{3}\left(\frac{4}{3}\minus{}\ln 2\right)$

$ S_{2}\; \equal{}\;\sum_{n \equal{} 1}^{\infty}\;\frac{\left[2(n\plus{}1)(n\plus{}2)\minus{}(n\plus{}2)\minus{}1\right]}{(n\plus{}2)!}$
$ \equal{}\;2(e\minus{}1)\minus{}(e\minus{}1\minus{}1)\minus{}\left(e\minus{}1\minus{}1\minus{}\frac{1}{2}\right)$
$ \equal{}\;\frac{5}{2}$

$ S_{3}\; \equal{}\;\sum_{n \equal{} 1}^{\infty}\;\left(\frac{1}{3n\minus{}2}\plus{}\frac{1}{3n\minus{}1}\minus{}\frac{2}{3n}\right)$
$ \equal{}\;\sum_{n \equal{} 1}^{\infty}\;\left(\frac{1}{3n\minus{}2}\minus{}\frac{1}{3n}\right)\;\plus{}\;\sum_{n \equal{} 1}^{\infty}\;\left(\frac{1}{3n\minus{}1}\minus{}\frac{1}{3n}\right)$
$ \equal{}\;\left(1\minus{}\frac{1}{3}\plus{}\frac{1}{4}\minus{}\frac{1}{6}\plus{}\frac{1}{7}\minus{}\frac{1}{9}\plus{}\;\ldots\right)\;\plus{}\;\left(\frac{1}{2}\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\frac{1}{6}\plus{}\frac{1}{8}\minus{}\frac{1}{9}\plus{}\;\ldots\right)$
$ \equal{}\;\left(\int_{0}^{1}\;\frac{1\minus{}x^{2}}{1\minus{}x^{3}}\;\textbf dx\right)\;\plus{}\;\left(\int_{0}^{1}\;\frac{x\minus{}x^{2}}{1\minus{}x^{3}}\;\textbf dx\right)$
$ \equal{}\;\left(\int_{0}^{1}\;\frac{1\plus{}x}{1\plus{}x\plus{}x^{2}}\;\textbf dx\right)\;\plus{}\;\left(\int_{0}^{1}\;\frac{x}{1\plus{}x\plus{}x^{2}}\;\textbf dx\right)$
$ \equal{}\;\left(\frac{\pi}{6\sqrt 3}\plus{}\frac{1}{2}\ln 3\right)\;\plus{}\;\left(\frac{1}{2}\ln 3\minus{}\frac{\pi}{6\sqrt 3}\right)$
$ \equal{}\;\ln 3$

$ S_{4}\; \equal{}\;\frac{1}{2}\sum_{n \equal{} 1}^{\infty}\;\left(\frac{1}{4n\minus{}1}\minus{}\frac{1}{4n\plus{}1}\right)\; \equal{}\;\frac{1}{2}\left(\frac{1}{3}\minus{}\frac{1}{5}\plus{}\frac{1}{7}\minus{}\frac{1}{9}\plus{}\;\ldots\right)$
$ \equal{}\;\frac{1}{2}\left(1\minus{}\frac{\pi}{4}\right)$

$ S_{5}\; \equal{}\;3\sum_{n \equal{} 1}^{\infty}\;(\minus{}1)^{n\minus{}1}\left(\frac{1}{3n\minus{}1}\minus{}\frac{1}{3n}\right)$
$ \equal{}\;3\left\{\left(\frac{1}{2}\minus{}\frac{1}{5}\plus{}\frac{1}{8}\minus{}\;\ldots\right)\minus{}\left(\frac{1}{3}\minus{}\frac{1}{6}\plus{}\frac{1}{9}\minus{}\;\ldots\right)\right\}$
$ \equal{}\;3\left\{\left(\int_{0}^{1}\;\frac{x}{1\plus{}x^{3}}\;\textbf dx\right)\minus{}\left(\frac{1}{3}\ln 2\right)\right\}$
$ \equal{}\;3\left\{\left(\frac{\pi}{3\sqrt 3}\minus{}\frac{1}{3}\ln 2\right)\minus{}\left(\frac{1}{3}\ln 2\right)\right\}$
$ \equal{}\;3\left\{\frac{\pi}{3\sqrt 3}\minus{}\frac{2}{3}\ln 2\right\}$

$ S_{6}\; \equal{}\;\frac{1}{6}\sum_{n \equal{} 1}^{\infty}\;(\minus{}1)^{n\minus{}1}\left(\frac{1}{3n\minus{}1}\plus{}\frac{1}{3n\plus{}1}\right)$
$ \equal{}\;\frac{1}{6}\left\{\left(\frac{1}{2}\minus{}\frac{1}{5}\plus{}\frac{1}{8}\minus{}\;\ldots\right)\plus{}\left(\frac{1}{4}\minus{}\frac{1}{7}\plus{}\frac{1}{10}\minus{}\;\ldots\right)\right\}$
$ \equal{}\;\frac{1}{6}\left\{\left(\int_{0}^{1}\;\frac{x}{1\plus{}x^{3}}\;\textbf dx\right)\plus{}\left(1\minus{}\int_{0}^{1}\;\frac{\textbf dx}{1\plus{}x^{3}}\right)\right\}$
$ \equal{}\;\frac{1}{6}\left\{\left(\frac{\pi}{3\sqrt 3}\minus{}\frac{1}{3}\ln 2\right)\plus{}\left(1\minus{}\frac{1}{3}\ln 2\minus{}\frac{\pi}{3\sqrt 3}\right)\right\}$
$ \equal{}\;\frac{1}{6}\left\{1\minus{}\frac{2}{3}\ln 2\right\}$


trigonometry
calculus
calculus computations

If f(x) =x+tan(x). If f is inverse of g. What is g’(x).

Let $ f(x)\equal{}y\equal{}x\plus{}\tan{x}$.

$ \frac{dy}{dx}\equal{}1\plus{}\sec^2{x}$, so $ g'(y)\equal{}\frac{1}{\frac{dy}{dx}}\equal{}\frac{1}{1\plus{}\sec^2{x}}$. Where $ y\equal{}x\plus{}\tan{x}$.

We cannot write more simpler, because $ y\equal{}x\plus{}\tan{x}$ can't be written by the form of $ x\equal{}h(y)$.


Any guesses or predictions who will be the top-performers at IMO'2008?

* There are about 10 participants who already got gold at IMO'2007.

* South-Korea has a 14-years old child prodigy, Soo Hong Lee.
http://www.imo-official.org/year_reg_team.aspx?code=KOR
http://www.imo-official.org/participant_r.aspx?id=15628
Last year, Soo Hong missed gold by a single point.

* Last year, the most challenging problem was the third one.
Of those who performed reasonably well on 2007/3,
Dániel Korándi	(Hungary)
Livia Alexandra Ilie (Romania)
are still around.

Przemek Mazur from Poland was gold medalist twice. He scored 37 in Slovenia (5th place) and 30 in Vietnam (around 25-30th place). I hope that he will make the perfect score this year.

I'm pretty sure we'll have 2 gold medalists from Romania this year (both were in the team last year too, Livia and Mihai). The rest will get silver or even higher if the style of problems suits them.

[quote="Erken"][quote="Amazigh"]I am a small fan of that S-Korean kid (14), maybe he will be the next mister Tao. He is also very cute.
How can a 13 year old kid win a silver on IMO (a rhetorical question :p)[/quote]
If I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.
And I'm pretty sure that there was even younger gold medalists before...[/quote]

Ofcourse there are younger gold medalists in the history of IMO, but it is special every time :)

Inverse of Tao  :wink: ($ Tao^{\minus{}1}$)

[url]http://www.imo-official.org/participant_r.aspx?id=1581[/url]

[quote="Valentin Vornicu"]So funny how with 40/42 you only got a silver medal back in the day :)[/quote]

There even was one IMO, where 41/42 points only got you silver.
Miss one point, and you lose gold...

http://www.imo-official.org/year_individual_r.aspx?year=1987

Generally,all participants from China will get gold medals...

[quote="Erken"][quote="test20"][quote="Erken"]If I am not mistaken there was a 13 or 14 years old gold medalist from Kazakhstan at IMO 2004,his name is Andrei Kim.[/quote]

Yes, but that's actually a sad story.
Andrei Kim's performance became weaker and weaker, year by year.
2004 he made gold, 2005 he made silver, 2006 he made bronze, and in 2007 he did not even manage to qualify for IMO.

http://www.imo-official.org/participant_r.aspx?id=7857[/quote]
Actually he graduated from school at 2006,that's why he didn't participate to IMO 2007 :lol: ...And the reason of his following results was just that he was unlucky,and I know he was very strong matematician even in 11th grade(we have only 11 grade in Kazakhstan)...As he was absolutely first on Balkan Math Olympiad and Chinese Western Olympiad he "beat" all participants from China with unbelievable high score.
Also as you can see on the link above he solved all the problems at IMO 2004,but made a little bugs on almost every problem,but if he was more attentive i'm sure he could easily score 42 points at age 13 or 14.[/quote]

I am also interested in this phenomenon....An inverse of Tao...

China Western Olympiad requires participants coming from western China which is the poorest area.

Lisa Sauermann from germany was 14 years old when she won a silver medal at IMO 2007, so maybe we can expect something great from her - we will see...

Naphthalin

[quote="test20"]
* South-Korea has a 14-years old child prodigy, Soo Hong Lee.
[/quote]

I support him

So bad that in Georgia we finished high school 17 years old :( I could be in the list too, and I would bet on myself :D

and this year it changed, our guys will have extra year at high school, and that surely means better results

And I am sure we can expect 42 this year, and reasonably higher cuttoffs

good luck everyone ;)

i think morocco will surprise  all the contries. i could probably be with the moroccan team but my scholl  it not intersted in this cind of competition !!!!!! go on morocco and break the habit like in IMO 1995 "kana lah ma3akom"


1.Prove that for every party (a group of people), you can find at less two persons such that, those persons have the same number of friends (INTO THE PARTY).

Ok... Since I am assuming you mean friends at the party...

[hide]Contradiction:

We assume that the case where everyone has a different amount of friends is true.

Suppose there are n friends at the party, then each person may have up to n-1 friends.  To fufill the requirement that none have the same number of friends, we must assign 0 friends ,1 friend, 2 friends ,...,n-1 friends to each person.  However, if we have one person with n-1 friends, then we cannot have a person with 0 friends. Contradiction![/hide]

Right?

Yes, you must count only the friends at the party....


To be a member on AoPS, you are required to have a username.

[hide] :rotfl: [/hide]

NO i mean why did you choose "leoxnlin" for ur username

I am quoting a famous theorem derived by Botolemy.

because otherwise i would hav chosen a depressing one like angelofdeath22 or suffer22undead. (i was tempted by my friend's suggestion of xxdeathxx, tho. 'cause it sounds cool.) so be happy i didnt. (i get depressed easily, especially in summer.)

@ 1=2: on this other forum about why did u pick ur numbers u said it was because u needed a short username...

His username is obviously a short username from a famous theorem :D!

I chose mine, because I admire Sesame Street's ernie.  :rotfl:

pacman because pacman is cool

guess 2812!

I am Danny Ham from Texas. What else do you expect?

@Pacman:

Hmm...

Because the stars aligned in the cradle of Draco, and Mars is in the house of Demeter, I think I will be able to guess the meaning of 2812 using the path of Orion's belt during the week after Midsummer's Night.

/me uses complex astrology stuffs.

I've got it! If you add up the digits in 2812, you get 13, which is an unlucky number, but it must be a lucky number because since you were born on :ninja: December 28th :ninja: under the house of Hades, 13 happens to be your lucky number!

So, was I right  :P ?

hmm gee i wonder y i picked my username..... maybe it has to do with the fact that GUINEAPIGGIES R THE AMAZINGEST ANIMALS IN THE WORLD!!!!   :wow:   :yup:  :gleam:

No, hamsters are.   :wink:

For me?  Oh, um... 'Cause I'm Alex?

My initials..duh

I should have picked "SupremeOvelordOfAllYouPunyEarthlingsWhoSuck" But that might have been too long... :D  :lol:

hmm... Math website... Math website with MATHCOUNTS on it.

What are mathcounters called again?

oh yea....

Except I spelled it wrong. >.<  :rotfl:

[quote="chenhsi"]If you search it(my username) on Google, you will probably find me...[/quote]
Yeah... You're on Wikipedia...
http://en.wikipedia.org/wiki/User:Chenhsi
(1+1=10 	This user likes binary.
1+2=10 	This user likes ternary.
2+3=10 	This user likes quinary.
3+3=10 	This user likes senary.
4+4=10 	This user likes octal.
3x3=10 	This user likes nonary.
5+5=10 	This user likes decimal.
6+6=10 	This user likes dozenal.
6×9=42 	This user jokes in base 13.
8+8=10 	This user likes hexadecimal.
4×5=10 	This user likes vigesimal.
3×4×5=10 	This user likes sexagesimal.)
And Kongregate...
http://www.kongregate.com/accounts/chenhsi
And AoPS.

I just randomly thought up the phrase "Here, Fishy, Fishy."

(Yeah, I understand the 6*9=42 joke. Maybe The Earth was programmed in base 13?)

I simply wanted an original (A username that is 1 common word)
I basically had a list of words that I wanted to be, and after trying them one by one, ProtestanT seemed to be the first one on the list that worked.

echen

first initial, last name. Easy to remember and simple. If my name was something like Mymathskillsequalsathousandsouls53132456 I would forget it by tomorow.

[quote="splatyango"]i used my (very goofy) cousins favorite "word", yangosplat, and rearranged it.[/quote]

that's probably the coolest word i have heard all summer

i generally use different names for different things, though. the only site where i also used Lazarus as my nick is, iirc, Kongregate.

I always use my email for a username.... yoyomattnerdt@aim.. i have the "yo yo" part cause in 6th grade me and my friend would always say "yo yo yo" to the teachers and it would make them mad =) the other part,
matt=first name, nerd= duh...., t= last initial

just because it's my real name and I put the dotts  because Ana. and Ana were already someone else's user names

Russian Rocket was the nickname for the AWESOME swimmer :winner_first: [size=200]Alexander Popov[/size]

DRAGONS ARE COOLER THAN GUINEA PIGS AND HAMSTERS AND YELLOW DUCKIES. 96 is for year of birth.

@Ducky, what about your username?

i dunno i guess i just loved the ring and smooth flow of the name ra5249

my neighbor made it up. definition=crazy maniac  yup, that's me :)

[size=150][color=violet]CRAZY ISLAND MANIAC!!!!!! HI KOOKHEAD!!![/color][/size] ur name is awesome!! ur my hero!!

This may be of interest:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79722[/url]

Or this, possibly, [url]http://www.mathlinks.ro/viewtopic.php?search_id=718014543&t=158658[/url].


For positive real numbers $a$, $b$, and $c$, solve the system

\begin{eqnarray*}cy+bz &=& a\\ az+cx &=& b\\ bx+ay &=& c \end{eqnarray*}

Easy problem in matrix inversion. However, because of the answer you get, i feel there must be some geometric way of doing this.


modular arithmetic
arithmetic sequence
number theory
prime numbers
number theory proposed

Giveing the infinite arithmetic progression $11,21,31,...$.Prove that in this arithmetic progression ,we have infinte prime numbers

Note that a number $\equiv 1 \pmod{5}$ is either $\equiv 1 \pmod{10}$ or $\equiv 6 \pmod{10}$. We'll show that there are infinitely many prime numbers of the form $5k+1$. The result follows, since they have to be of one of the above forms, and the latter form gives even numbers.

Replace $5$ by any prime $q$. We'll show that $q+1, 2q+1, ...$ contains infinitely many primes. An observation: if a prime $p$ is such that $p \mid x^q-1$ but $p \nmid x-1$, then $q \mid p-1$. Indeed, if $d_0$ is the smallest natural $d$ for which $p \mid x^d-1$, all other numbers $d$ for which this happens are multiples of $d$. Then $q$ is a multiple of $d$. But since $q$ is a prime, $q=d$. Since $p \mid x^{p-1}-1$, we have $d \mid p-1 \Rightarrow q \mid p-1$. Done.

Now $q \mid p-1 \Rightarrow p-1=qk, p=qk+1$. So we want to show that the set $x^q-1$ for $x=2, 3, ...$ has infinitely many prime divisors. If this is false, there should be some $k$ such that any prime dividing $x^q-1$ for any $x$ divides $x^q-1$ for some $x \leq k$. Now let $p_1, p_2, ..., p_n$ be the union of the prime divisors of $x^q-1$ for $x=2, 3, ..., k$, and let $z=p_1p_2...p_n$. Setting $x=z$, we get an obvious contradiction.

Now there's a detail left: we haven't checked the condition of these primes not coinciding with those of $x-1$. So suppose that for any $x$, the primes dividing $x^q-1$ are either among $p_1, p_2, ..., p_n$ or coincide with those dividing $x-1$. Let $z-1=y$, and consider the binomial expansion of $z^q-1=(y+1)^q-1=y^q+...+qy$. Dividing by $yq$, it can be dealt with easily.

I believe you can find a proof (here on the forum) that given a natural $n$, there are infinitely many primes of the form $nk+1$.

We want infinitely many primes of the form $10k+1$. A number is of this form iff it's both of the form $2u+1$ and $5k+1$. Since all primes (except for $2$, of course :)) are odd anyway, we only need to show that there are infinitely many primes of the form $5k+1$. Assume $p_1,\ldots,p_n$ are such primes of the form $5k+1$. Any prime divisor of $M=\frac{N^5-1}{N-1}$, where $N=5p_1p_2\ldots p_n$ is then a prime of the form $5k+1$ different from $p_i,\ i\in\overline{1,n}$. This is because $\left(M,N-1\right)=(N-1,5)=1$, so for any prime $p|M$ we have $M\not\equiv 1\pmod p,\ M^5\equiv 1\pmod p$, so $M$ has order $5$ modulo $p$, which implies $5|p-1$.

I see Severius was here first :).

Special case of the Dirichlet's theorem.

I think the point was that it's easier than Dirichlet's theorem... I've never seen Severius's proof before.


I'm supposed to [u]prove[/u] that a closed trajectory to an autonomous system is an periodic solution to it. I don't see how this is going to be proved. Isn't this a fact per definition? :huh:

The closed trajectory is in the phase space, right? The statement would be false in the configuration space, or for nonautonomous systems.

"Statement." Suppose that $X(t)$ is a solution of $\.{X}(t)=f(t)$ ($t>0$) with the initial condition $X(0)=X_{0}$. If $X(T)=X(0)$, then $X(t+T)=X(t)$ for all $t\ge 0$.

"Proof." The function $\tilde X(t)=X(t+T)$ also satisfies the equation, as well as the initial condition. Therefore, $\tilde X(t)=X(t)$ by an appropriate uniqueness theorem*.

(*) Assuming $f$ is such that uniqueness holds.


vector
geometry
geometric transformation
geometry proposed

I was asked to post this problem, so here is it:

[b]Original problem.[/b]

Find all natural numbers n satisfying the following assertion: If, on a plane, we have some regular n-gons $P_1$, $P_2$, ..., $P_k$ not intersecting each other, then we can choose one of these n-gons - call it $P_i$ - and a vector $\overrightarrow{v}$ (not the zero vector) such that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \neq i$).

[Of course, this condition - that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \neq i$) - can be rewritten as follows: If the n-gon $P_i$ moves in the direction of the vector $t\overrightarrow{v}$, then it doesn't collide with any of the other n-gons $P_j$.]

The proposed solution shows that every natural n fulfills the above assertion. Even a much more general result is true:

[b]Generalized problem.[/b]

If $P_1$, $P_2$, ..., $P_k$ are k arbitrary convex polygons on the plane, and $\overrightarrow{v}$ is an arbitrary vector, then there exists a number i such that for any positive number t, the image of the polygon $P_i$ under the translation with translation vector $t\overrightarrow{v}$ doesn't intersect the other polygons $P_j$ (with $j \neq i$).

[The proposed solution is at http://kvant.mccme.ru/1989/12/resheniya_zadachnika_kvanta_ma.htm ; but even if you read Russian, you will not necessarily understand it, since it is written up in a very bad style. I needed a day to understand it.]

[b]Variant of the original problem.[/b]

On the other hand, the original problem can be modified:

Find all natural numbers n satisfying the following assertion: If, on a plane, we have some regular n-gons $P_1$, $P_2$, ..., $P_k$ not intersecting each other, then we can choose one of these n-gons - call it $P_i$ - and a vector $\overrightarrow{v}$ (not the zero vector) [i]which is perpendicular to a side of the polygon $P_i$[/i] such that for any positive number t, the image of the n-gon $P_i$ under the translation with translation vector $t\overrightarrow{v}$ doesn't intersect the other n-gons $P_j$ (with $j \neq i$).

In this case, already in the case n = 5, we can find a configuration of regular 5-gons for which such a polygon $P_i$ and such a vector $\overrightarrow{v}$ don't exist.

  Darij

[quote="darij grinberg"]

[b]Generalized problem.[/b]

If $P_1$, $P_2$, ..., $P_k$ are k arbitrary convex polygons on the plane, and $\overrightarrow{v}$ is an arbitrary vector, then ...

Darij[/quote]

If I well understand in this more general situation we have furthermore the arbitrary convex poligon (instead the regular ones) the imposed direction $\overrightarrow{v}$ towards which $P_i$ can "escape".

While in the original problem "... we can choose one of these n-gons - call it Pi - and a vector v (not the zero vector) such that..."

Is it right?

In any case, I will try to give an "intuitive" solution.

Take a generic Pj and tray to "move" it to the given direction $\overrightarrow{v}$. If Pj1 is the first poligon you "meet", then try to "move" this always in the same direction and so on. As the poligons are a finite number they are a finite number in the "trusted" direction so at te end will be a poligon Pjh that can "escape".

And after this, one after the other, all poligons can "escape"  :) .

PS
this problem seems to describe the situation of the people when they tray to get out of the crowded underground.

[quote="sprmnt21"]If I well understand in this more general situation we have furthermore the arbitrary convex poligon (instead the regular ones) the imposed direction $\overrightarrow{v}$ towards which $P_i$ can "escape".[/quote]

Yes, in the generalized problem,

- the shapes of the polygons,
- their mutual position (as long as they don't intersect each other) and
- the vector $\overrightarrow{v}$

can be arbitrary.

[quote="sprmnt21"]Take a generic Pj and tray to "move" it to the given direction $\overrightarrow{v}$. If Pj1 is the first poligon you "meet", then try to "move" this always in the same direction and so on. As the poligons are a finite number they are a finite number in the "trusted" direction so at te end will be a poligon Pjh that can "escape".[/quote]

Exactly. All what you need to complete this solution is to show that cycles are impossible. Hereby, a "cycle" is a subset of the set of polygons {$P_1$, $P_2$, ..., $P_k$} such that: the first polygon of the subset cannot "escape" because it would "meet" the second one, the second one cannot "escape" because it would "meet" yet the third one, etc., and the last one cannot "escape" because it would "meet" the first one. Proving this is a bit of tricky. Hint: Project on an axis perpendicular to the vector $\overrightarrow{v}$.

[quote="sprmnt21"]this problem seems to describe the situation of the people when they tray to get out of the crowded underground.[/quote]

Indeed  :D

  Darij

[quote="sprmnt21"]I think the reason becouse one cannot have cicles is the convexity of poligons. I later will provide an argument on this.[/quote]

Here is the intuitive argument I was referring to.

Let's suppose that A is the point of firts "impact" of poligon Pj0 with poligon Pj1 (see fig. attached). Then Pj1 "reacts" to the "impact" moving towards direction V untill it impact Pj2 in B1. Then Pj2 "reacts" to the "impact" moving towards direction V untill it impact Pj3 in B2.  And so on. Suppose that finilly Pj3 "impact" Pj0 in B3.

Because of the convexity of poligons all segments AiBi stay inside the correspondent poligon.

For this we should have that P0 impacted Pj3 (in our example) before impact Pj1. Contradiction!

[quote="darij grinberg"]Hmm, I must say I really don't understand this...

  Darij[/quote]

I will try to rephrase some sentences.

In the figure attached for each poligons Pji are highlighted only the segments involved in the "meeteing".

Let's to start supposing that A is a point on the boundary of Pj0 and Pj1 where they have the firts "meeting"[actually could be that Pj0 and Pj1 have the first "meeting" in a set of points and segments; in this case let A be the "lowest", in the direction V, point of this set ].

After this "impact" Pj1 try to "escape" alongside direction V and cover the distance AA1 untill meet Pj2 in B1.  A1B1 is a segment of Pj1 when Pj1 "meet" Pj2.

And so on.

If, as we suppose,  some poligon of this chain would "meet" the first one, we could have the situation reppresented  in the figure. Where AB3 is a segment of Pj0 and B2B'3 is a segment of Pj3 [B'3B2 is the translated segment of B3A3]. this configuration, as AB3 intersect B'3B2 and [i]fortiori[/i] Pj0 intersect Pj3, implies that Pj0 would have meet Pj3 before meet Pj1.

This is contradictory with our starting hypothesis.


[hide]

Since $a-\sqrt{a^2-1}=\frac{1}{a+\sqrt{a^2-1}}$ , By letting $a+\sqrt{a^2-1}=y$ and $x^2-2x=p$ we obtain

$y^p+y^{-p}=2a$ $\implies y^{2p}-2ay^p+1=0$ $\implies y^p=a\pm\sqrt{a^2-1}$

Hence $p=\pm 1$ which gives us $x=1,1\pm\sqrt{2}$
[/hide]

[quote="paladin8"][quote="shyong"][hide]

Since $a-\sqrt{a^2-1}=\frac{1}{a+\sqrt{a^2-1}}$ , By letting $a+\sqrt{a^2-1}=y$ and $x^2-2x=p$ we obtain

$y^p+y^{-p}=2a$ $\implies y^{2p}-2ay^p+1=0$ $\implies y^p=a+-\sqrt{a^2-1}$

Hence $p=+-1$ which gives us $x=1,1+-\sqrt{2}$
[/hide][/quote]

BTW, you can use \pm to give the $\pm$ symbol in LaTeX.[/quote]

awesome !Now that really makes it looks nicer .


algebra
polynomial
number theory unsolved
number theory

[quote="ssibseya"]"$ 1^k, 2^k, 3^k, 4^k, \dots$" 's k-th difference is k!
ex) "$ 1^2, 2^2, 3^2, \dots$"'s 2-th diff = $ 3^2 \minus{} 2\cdot 2^2 \plus{} 1^2$ = $ 4^2 \minus{} 2\cdot 3^2 \plus{} 2^2$ = $ 2!$

How to prove?[/quote]
Could you explain a bit more. What is the k-th difference??
But it is easy to prove that $ (n \plus{} 1)^2 \minus{} 2n^2 \plus{} (n \minus{} 1)^2 \equal{} n^2 \plus{} 2n \plus{} 1 \minus{} 2n^2 \plus{} n^2 \minus{} 2n \plus{} 1 \equal{} 2 \equal{} 2!$
But fx $ 3^3 \minus{} 3*2^3 \plus{} 1^3 \equal{} 27 \minus{} 18 \plus{} 1 \equal{} 10 \neq 3!$....

I guess he means http://www.mathlinks.ro/Forum/viewtopic.php?t=70637 .

  darij

$ \binom{n}{0} {(n \plus{} 1)}^n \minus{} \binom{n}{1} {(n)}^n \plus{} \binom{n}{2} {(n \minus{} 1)}^n \minus{} \ldots \plus{} ( \minus{} 1)^{n} \binom{n}{n} {1}^n \equal{} x \ .$
=> wanted
$ \binom{n \plus{} 1}{0} {(n \plus{} 1)}^n \minus{} \binom{n \plus{} 1}{1} {(n)}^n \plus{} \binom{n \plus{} 1}{2} {(n \minus{} 1)}^n \minus{} \ldots \plus{} ( \minus{} 1)^{n} \binom{n \plus{} 1}{n} {1}^n \equal{} 0 \ .$
=> because surjection(n to n+1) = 0
$ \binom{n}{0} {(n)}^n \minus{} \binom{n}{1} {(n \minus{} 1)}^n \minus{} \ldots \minus{} ( \minus{} 1)^{n} \binom{n}{n \minus{} 1} {1}^n \equal{} \ x \ \equal{} n! \ .$
=> first-second = this (because surjection(n to n) = n! and you gave me this)

but
$ \binom{n}{0} {(n \plus{} 1)}^n \minus{} \binom{n}{1} {(n)}^n \plus{} \binom{n}{2} {(n \minus{} 1)}^n \minus{} \ldots \plus{} ( \minus{} 1)^{n} \binom{n}{n} {1}^n \equal{} n! \ .$
$ \binom{n}{0} {(n \plus{} 2)}^n \minus{} \binom{n}{1} {(n \plus{} 1)}^n \plus{} \binom{n}{2} {(n)}^n \minus{} \ldots \plus{} ( \minus{} 1)^{n} \binom{n}{n} {2}^n \equal{} n! \ .$
$ \binom{n}{0} {(n \plus{} 3)}^n \minus{} \binom{n}{1} {(n \plus{} 2)}^n \plus{} \binom{n}{2} {(n \plus{} 1)}^n \minus{} \ldots \plus{} ( \minus{} 1)^{n} \binom{n}{n} {3}^n \equal{} n! \ .$
and so on
why???

You are dealing with $ \Delta^n((x\plus{}c)^n)$. It's a constant polynomial (because $ \Delta$ decreases degree by $ 1$).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=90529 (and http://www.mathlinks.ro/Forum/viewtopic.php?p=657956 for a general explanation).


geometry
ratio
cyclic quadrilateral
power of a point
AMC

Two circles are concentric. A chord AC of the outer circle touches the inner circle at Q. P is the midpoint of AQ. A line through A intersects the inner circle at R and S. The perpendicular bisectors of PR and CS meet at T on the line AC. What is the ratio AT/TC?

help please

I've been working on this one, but frankly, I don't have any good ideas.  I'll keep trying it, but I thought about bringing it back up to the top just so other people see it.  Where did it come from?

Actually, I shouldn't say I don't have any ideas:  I know what the answer is.  And I know a few other things about the setup, just from playing around with it on Geometer's Sketchpad.  But I can't prove anything.

Okay, I've got it!

Here we go:
We have by power of a point that AR * AS = AQ:^2:
AQ:^2: = AQ * AQ = (2AP) * (AC/2) = AP*AC
So AR * AS = AP * AC
AR/AP = AC/AS
angle RAP = angle SAC

So triangle RAP is similar to triangle CAS.  So angle ACS = angle ARP.  angle ARP + angle PRS = 180 degrees (two angles on a line)
angle ACS + angle PRS = 180 degrees (substitution)
angle PCS + angle PRS = 180 degrees (renaming)

Thus PCSR is a cyclic quadrilateral, so the perpendicular bisectors of PR and CS meet at the center of the circumcircle.  If this is on one of the edges, it must bisect that edge (since it is equidistant from the endpoints).  Thus TC = TP = 3AC/8.
AT = 5AC/8.  AT/TC = 5/3.

im dont remember where this came from, but i liked the problem. thanks for the solution.


function
vector
combinatorics proposed
combinatorics

Suppose that ${\mathcal H}_{n}$ is the set of all $n \times n$ matrices $A=\parallel a_{i,j}\parallel $  with $a_{i,j}\in \{0,1\}.$ Let  [b]m[/b]  be the number of all nonidentical functions $F:{\mathcal H}_{n}\to{\mathcal H}_{n}$ such that  $\left(F\left(F(A)\right)\right)= A$   for all  $A \in{\mathcal H}_{n}. \; \;$ Find    [b]m[/b]  .
(In case $n=2$ it's true that  [b]m[/b]$=12392480$ ?)

Ok, as I see this problem, first of all, we have a vector with $n\times n$ entries $\in {0,1}$. Now, there are $2^{n\ldots n}=N$ such vectors. If we consider the number of functions $f:A\to A$ with $f(f(x))=x$ and $|A|=N$ to be $x_n$ we have the well known recurence $x_{n+1}=x_n + n\cdot x_{n-1}$. The reasoning is well known and has been posted on the forum.
Btw, I wonder if someone managed to find a solution for the recurence. I tried to do something with generating functions, but it reduces to a not very beautiful differential equation :(

Hello,
I've foud some references of this problem in a book of Ioan Tomescu "Probleme de combinatorica si teoria grafurilor" , but it's written in Romanian and I don't know if there is a English version of this book.

Here is demonstrated the recurence posted by Dust $\ x_n=x_{n-1}+(n-1) x_{n-2}$and a formula for
 $\ x_n$=$\ n!$$\displaystyle\sum_{i+2j=n}\frac{1}{i!j!2^j}$ and a closed formula for it's generating exponential function i.e $\ F(x)=\displaystyle\sum_{n \ge 0} x_n \frac{t^n}{n!}=exp(t+\frac{t^2} {2}) $. The demonstration is not very hard.


What is $n$ if $(\sin1)(\sin3)(\sin5)...(\sin89)=\frac{1}{2^{n}}$ and its in degrees.

Er... that completely misses the point.  $n$ is nice.  Find it.

[hide]The product equals $\frac{\sin 1 \sin 2 \sin 3 \ldots \sin 89}{\sin 2 \sin 4 \sin 6 \ldots \sin 88}$.
Repeated use of $\cos x=\sin 90-x$ and $\sin x\cos x=\frac{1}{2}\sin 2x$ reduces the numerator to $2^{-44}\frac{\sqrt{2}}{2}\sin 2 \sin 4 \sin 6 \ldots \sin 88$.  The $\sin$ terms cancel out, leaving $2^{-44}\frac{\sqrt{2}}{2}$.
$\frac{\sqrt{2}}{2}=\sqrt{0.5}=\sqrt{2^{-1}}=2^{-\frac{1}{2}}$
So the product is $2^{-44}2^{-\frac{1}{2}}$ or $\frac{1}{2^{\frac{89}{2}}}$.

The answer is $\boxed{\frac{89}{2}}$.[/hide]