modular arithmetic
algebra
polynomial
IMC
college contests

Let $ x$,  $ y$ and $ z$ be integers such that $ S = x^{4}+y^{4}+z^{4}$ is divisible by 29. Show that $ S$ is divisible by $ 29^{4}$.

[quote="Valentin Vornicu"]Oh man I hope it's not that the only $ x,y,z$ with $ x^{4}+y^{4}+z^{4}\equiv 0 \pmod{29}$ are $ x \equiv y \equiv z \equiv 0 \pmod{29}$ :)[/quote]
Judging by day 1 #1 and this problem, the problem selectors don't think the contestants are any good at NT... or maybe there just weren't any appropriate difficulty NT problems suggested.  :P

So now the solution: it's easy to notice that $ x^{28}\equiv 1 \pmod{29}$, so either $ x^{4}\equiv 1 \pmod{29}$ or $ x^{4}$ is a root of the polynomial $ f(x) = x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$ in $ \mathbb Z / 29 \mathbb Z$. That polynomial only has 6 different roots (because $ \mathbb Z / 29 \mathbb Z$ is a field), so it seems natural to start looking at them.

We find $ 2^{4}\equiv 16$, $ 3^{4}\equiv 23$, $ 4^{4}\equiv 16^{2}\equiv 24$, $ 5^{4}\equiv 20$, $ 8^{4}\equiv 24^{2}\equiv 7$ and $ 11^{4}\equiv 25$, so we have the 6 different roots. Now it's just a matter of observing that no matter how we combine the 8 total numbers 0, 1, 7, 16, 20, 23, 24, 25 we cannot obtain 0 modulo 29, unless all 3 are 0, in which case the problem is solved.

In competition  I make some conclusion and than I wrote that if we sum three number from the set {0,1,7,16,-9,-6,-5,-4} we can't obtain 0 or 29
(imposibility of other case like  -29, 58, .. and so on I show in my work). But i didn't write this sums I said that it is easy to make this.
Jury thought that it isn't easy and I get only 10 of 20.I think that it is stupid, because actually i have full solution.

What concerned Valentin's solution he would got only 7!!!

What do you think about this(I think that for such solution in IMO we get 20, I mean perfect score)

As I've heard the Jury said something like "the problem is easy anyway so all there is to do is to sum in all possible ways to notice that it can't be 0 mod 29" so by this argument whoever didn't wrote all possible sums, he/she didn't receive full score.

How I can understand what I should wrote? Actually it is much more hard to find residues of $ n^{4}$. If somebody only catch this residues why he or she didn't write how he multiple this numbers may be he made some mistakes, but residues are the same?   Actually 7 points means that you didn't solve problem!!!
If you look at numbers {0,1,7,16,-9,-6,-5,-4} you understand that  summing any three of them is very easy without using paper and much more easy to show that sum isn't equal 0 or 29.

Also, without loss of generality, one of the residues is 1, if not all of them are zero (just multiply by the inverse^4).

[quote="Xevarion"]The idea is that if you are faced with a really easy problem, you shouldn't blow it off and say "this is trivial", you should give a detailed and careful solution. My solution was a full page. I made a table of residues of $ n$, $ n^{2}$ and $ n^{4}$ mod 29, and then I also explicitly showed that you couldn't make the sums (rather than just stated it wouldn't) by writing out the set of all pairwise sums. That was enough for 20.[/quote]
How I can understand that some problem is really really easy? Did you write multiplication? Why not??? May be you make some mistakes when you do multiplication.
Is it so hard to sum two numbers 10 times, I think that olimpiad isn't  arithmetic exam (where you should only sum some numbers) or calligraphy.
Besides  my question was:
 Do you think that Valetin's solution worth only 7 points?(maximum 20)

Zubr, did you really check all the triple sums in your head?

[quote="kunterbunt"]Zubr, you posted a solution to checking the triples which missed for example 1+1+x and now you seem to have  deleted it again.[/quote]
I delete it because I don't write very fast and I was in a hurry(and may be miss something )
Can you do this in your head without calculator:

First of all we show that we cant't obtain 29:
  16+16+x=29 there isn't such x in out set this is obvios.
  16+7+x  the same
  others sum less than 29.
So now it is enought to show that we cannot obtain 0:
  We can't form 16,7,1 as sum of two numbers from set{0,4,5,6,9} this is evident because
  16: sum less than 16(except 18)
  1 and 7: most of all sums greater than 7(except 0+0,0+4 etc).

  7+1=8, 7+7, 1+1 doesn't belong to {0,4,5,6,9}

[quote="Valentin Vornicu"]Zubr, in this particular case I would grade my post (!) if I would have written it like that during the competition with a 15/20 (or 6/7). [b]However[/b] my post was merely a guideline of the solution, [b]and not[/b] an actual write up of something to get a perfect score in an international competition.[/quote]

Thank you for understanding and for grading your post. I agree with you in this mark. I thoght that somebody can grade it like 15-18/20 (5-7/7), but not 7/20 like it was on competition. The decision made me angry. Besides they said full results only after closing ceremony and forgot to add some points which I take after appeal. So my medal didn't change but place is worse that it should be.


MATHCOUNTS FTW AMC AIME USA(J)MO USAMO Alcumus

[size=75]Okay, so maybe it's not exactly the 42nd, but you get the point. [/size]

...How am I supposed to start this? 5849206328x, Temperal, and I decided to make a full Mock Mathcounts, with sprint, target, team, and masters. We're still debating on whether or not to include countdown. As in the real Mathcounts tests, you all will get in teams of four and first take the sprint and target individually. After everyone on your team has taken the sprint and target, your team captain will private message either me, Temperal, abacadaea, or 5849206328x and your team will be allowed to meet and take the team round. We're still deciding on how qualification for Master's Round will work, as that depends on whether we'll have countdown. Once we've reached a decision, we'll state it in this thread.

As we're still working out the details of how we'll administer the contest, and making sure all of our problems are clear and have no mistakes, we will probably have signups open for about a week.

We realize that there are a lot of other mock tests out in the forums right now. However, we've spent a lot of time reviewing these problems and have made sure that they are well written, clear, at a sufficient difficulty level, and actually involve you to be creative.

5849206328x wrote the sprint and target, I wrote the team, and Temperal wrote the Master's Round. I guess abacadaea is just a grader :P.

To finish this off, there might be one or two things that we'll change sometime in the next week. If we do, one of us will post it in this thread.

...so do we sign up here?  This sounds so much better organized than the rest of these, so I'll join.

as a sidenote, the master's round is ridiculously difficult because
a) I don't like operating solely on the honor system; you can look up the methods,
b) I'm sure the top 4 people can handle it, and
c) I can't think of easy yet elegant problems to save my life.

not that it contains particularly advanced topics; it's just difficult.

EDIT: sure, sign up here if you want... but please make your teams outside the thread.

Yes; [b]please state that you're signing up here in the thread[/b]. Also, we'd prefer that teams be made outside of this thread. For example, if you wish to join person A's team, I would recommend private messaging the team members (or perhaps, just a team member) and asking. This is just to avoid clutter in the thread.

Why not...

I join.

BTW, all mathcounts master rounds can be killed using a huge amount of algebraic number theory, abstract algebra, and analysis.

[quote="pythag011"]
BTW, all mathcounts master rounds can be killed using a huge amount of algebraic number theory, abstract algebra, and analysis.[/quote]

...I'm giving you a perfect score automatically just for that.

Temperal, if you did that, then you'd be fired :P

btw, when you get a team of 4 and everyone is fine with it, post the team here with the captain and a team name

I join....PM me if you want to be in my team...

I'll join.  Can we take this contest as an individual, or do we have to be on a team?

[quote="mathswimmer"]How do we know who is starting a team?...and what time are you planning on doing it?[/quote]

Well, one of us will be posting a list of who's joined, who's already on which team, and who's starting a team. You can see the teams that have already been formed and aren't full and may private message one of the members to ask.

As said, we plan on letting signups continue for the next week or so. The window for taking the contest will most likely be until January 5th or so, since that's when winter break ends for most people. Please note that the aforementioned window is only for the written rounds.

[b]@Discrete_Math[/b]: You can take the written rounds as an individual, but to take the team round, you have to find other individuals to take it with. That's why we recommend you find a team that you think you work well with.

if I have received a meeting time and someone doesn't show up at that time, then the person that did will win

btw, someone (maybe me) announced that the countdown round wasn't going to use FTW questions, so basically we're going to have to determine how to count that one

of course, if ghjk resigns/forfeits/whatever, then you win

Wait what?

Then...wait....how else would we do a countdown?

i made up a lot of countdown type problems

I'm on FTW way too much for my own good, so there's not too many problems with the idea

Oh, I see. Oops.

So was our countdown counted or not? I didn't mem any of the FTW problems that occurred in that countdown so...

[quote="isabella2296"]I didn't mem any of the FTW problems that occurred in that countdown so...[/quote]

:rotfl: :rotfl: :rotfl:

No seriously, they were four of the new problems.

wait, so if we dont use ftw, then how do we do countdown?
how else would we do it? im lost....

Ok so; you pm 5849206328x the time you and the other person will be on. Then, on ftw, he makes a password game with you two. Then, he copy and pastes problems that were made for this countdown round. Then you answer. Yay.

if I heard correctly, someperson01 won 4-0

[u]Matches to be played[/u]
8 isabella
vs
5 someperson01

1 dafjbomb
vs
2 pythag011

[u][b]Still[/b] idle[/u]
AIME_is_hard

yo pythag, you never replied to my pms, when are you available for countdown?
are you open on friday? at arround 9:00 pm est? (im bussy on week days due to swimming)

deadline is [b]tomorrow[/b] for pythag011 and dajfbomb, we still need the match between someperson01 and isabella2296 as done somewhat soon

Ok, this is taking way too long. Using random.org*, pythag wins the match that he was supposed to play. Still need isabella2296 and someperson01 to play, or that'll be random.orged too.

*It's too hard to really come up with a way to decide who wins the countdowns. We don't want to use written scores as they've already been counted and shouldn't have that much influence. I really liked pythag's idea of using AMC/AIME/USAMO scores, but not everybody has taken that and a lot can change in a year. So, we're stuck using random.org.

I sent someperson01 a pm but he's not replying.


trigonometry
induction
algebra
telescoping sums

Prove
\[ \frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.  \]

I was hoping to check my answer but since no one has posted one here; I will propose what I came up with.

[hide]
First change the LHS to summation notation so we have:
$\sum\limits_{n=0}^{88} \frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\cos 1}{\sin^2 1}.$
Seeing $\cos(n)$ and $\cos(n+1)$ suggests an approach using a telescoping series so lets find a partial fraction decomposition for the summation on the LHS.
$\frac{A}{\cos(n)}+\frac{B}{\cos(n+1)}=\frac{1}{{\cos(n)}{\cos(n+1)}}$
Finding the common denominator and setting the numerators equal provides ${A}{\cos(n+1)}+{B}{\cos(n)}=1$ which is the same as $A\cos(n)\cos(1)-A\sin(1)\sin(n)+B\cos(n)=1$ by the cosine sum identity.  Now, remembering that $\sin^2(x)+\cos^2(x)=1$ group the terms with cosine and the terms with sine and factor each out respectively producing $(A\cos(1)+B)(\cos(n))+(-A\sin(1))(\sin(n))=1$
Now we need to find values for A and B such that we obtain our familiar trig identity so
$-A\sin(1)=\sin(n)$ or $A=\frac{-\sin(n)}{\sin(1)}$  Substituting this value for A into $A\cos(1)+B=\cos(n)$ we can solve to find $B=\frac{\cos(n)\sin(1)+\cos(1)\sin(n)}{\sin(1)}$ which we recognize as $B=\frac{\sin(n+1)}{\sin(1)}$ by the sum relation for sine.
Now replacing these values for A and B into the partial fraction decomposition we find that the LHS is equivalent to:
$\sum\limits_{n=0}^{88} \frac{-\sin(n)}{\cos(n)\sin(1)}+\frac{\sin(n+1)}{\cos(n+1)\sin(1)}$
From this we can factor a $\frac{1}{\sin(1)}$ from the summation and use $\tan(x)=\frac{\sin(x)}{\cos(x)}$ to get:
$\frac{1}{\sin(1)}\sum\limits_{n=0}^{88}\tan(n+1)-\tan(n)$
The summation part of this telescopes to $\tan(89)-\tan(0)=\tan(89)$ so the total is $\frac{\sin(89)}{\sin(1)\cos(89)}$ Which is the same as $\frac{cos(1)}{\sin^2(1)}$ since $\sin(90-x)=\cos(x)$ and likewise for sine.
[/hide]

[hide="Just use induction..."]

We prove using induction that $ \sum_{n = 0}^{k}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\tan (k+1)}{\sin 1}.$

For $ k = 0$ we have $ \sum_{n = 0}^{0}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{1}{\cos0\cos1}=\frac{\tan 1}{\sin 1}$

Assume it holds for $ k-1$. Then
\begin{eqnarray*}\sum_{n=0}^{k}\frac{1}{{\cos(n)}{\cos(n+1)}}&=&\frac{\tan k}{\sin 1}+\frac{1}{{\cos(k)}{\cos(k+1)}}\\ &=&\frac{\sin{k}\cos(k+1)+\sin 1}{\cos(k)\cos (k+1)\sin 1}\\ &=&\frac{\cos{k}\sin(k+1)}{\cos(k)\cos(k+1)\sin 1}\\ &=&\frac{\tan(k+1)}{\sin 1}\end{eqnarray*}

Note the use of $ \sin 1 =\sin (k+1-k) =\cos{k}\sin(k+1)-\sin{k}\cos(k+1)$ from line 2 to line 3.

Thus $ \sum_{n = 0}^{88}\frac{1}{{\cos(n)}{\cos(n+1)}}=\frac{\tan{89}}{\sin1}=\frac{\cos{1}}{\sin^{2}{1}}$
[/hide]

[quote="MithsApprentice"]Prove
\[ \frac{1}{\cos 0\dg\cos 1\dg}+\frac{1}{\cos 1\dg\cos 2\dg}+\cdots+\frac{1}{\cos 88\dg\cos 89\dg}=\frac{\cos 1\dg}{\sin^{2}1\dg}.\]
[/quote]
We use : $ \frac{sin1}{cosn.cos(n+1)}=\frac{sin[(n+1)-n]}{cosn.cos(n+1)}=cot(n)-cot(n+1)$
there for:
$ \frac{1}{\cos 0\dg\cos 1\dg}+\frac{1}{\cos 1\dg\cos 2\dg}+\cdots+\frac{1}{\cos 88\dg\cos 89\dg}=\frac{1}{sin1}(cot1)=\frac{cos1}{sin^{2}1}$

Let $ M_k$ denote the point $ (1, \tan k^\circ)$.
[asy]size(200);
defaultpen(1);
pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30);
pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=y/y.x, Z=z/z.x, E=(D+Y)/2;
pair X=(O+E)/2;
draw(O--A--Z);
draw(B--O--C--D--O--Y--Z--O);
label("\(O\)",O,SW);
label("\(M_0\)",A,ESE);
label("\(M_1\)",B,ESE);
label("\(M_2\)",C,ESE);
label("\(M_3\)",D,ESE);
label("\(\vdots\)",E,ESE);
label("\(M_{88}\)",Y,ESE);
label("\(M_{89}\)",Z,ESE);
label("\(\ddots\)",X);[/asy]
Then :
\begin{align*}
OM_k &= 1/\cos k^\circ \\
\text{m} \angle M_a O M_b &= (b-a)^\circ \\
[M_a O M_b ] &= \tfrac{1}{2} \sin (b-a)^\circ \cdot OM_a \cdot OM_b =
\frac{\tfrac{1}{2} \sin (b-a)^\circ}{\cos a^\circ \cos b^\circ} ,
\end{align*}
so
\begin{align*} \sum_{k=0}^{88} \frac{\tfrac{1}{2} \sin 1^{\circ}}{\cos k^\circ \cos (k+1)^\circ} &= \sum_{k=0}^{88} [M_k O M_{k+1} ] \\
&= [M_0 O M_{89} ] = \frac{\tfrac{1}{2} \sin 89^\circ}{ \cos 0^\circ \cos 89^\circ} = \frac{\tfrac{1}{2} \cos 1^\circ}{\sin 1^\circ},
\end{align*}
which proves the desired identity.

This might be a newbie type question, but how did calc rulz know to induct on such an outlandish expression?

What was the motivation?

I think most people in this thread over complicated this problem completely...[hide="Solution"]

Note that multiplying both sides by $\sin(1^\circ)$ gives us that we desire:

$\frac{\sin(1^\circ)}{\cos(0^\circ)\cos(1^\circ)}+\frac{\sin(1^\circ)}{\cos(1^\circ)\cos(2^\circ)}+\cdots \frac{\sin(1^\circ)}{\cos(88^\circ)\cos(89^\circ)}=\frac{\cos(1^\circ)}{\sin(1^\circ)}$.

Writing all the $\sin(1^\circ)$ as $\sin(n-(n-1))^\circ)$, we get $\sin(n)\cos(n-1)-\sin(n-1)\cos(n)$ which when divided by $\cos(n)*\cos(n-1)$ gives us $\tan(n)-\tan(n-1)$.

Therefore, the sum telescopes as follows:  $\sum_{i=1}^{89}(\frac{\sin(1^\circ)}{\cos(i^\circ)\cos(i+1^\circ)}=\sum_{i=1}^{89}(\tan(n)-\tan(n-1))=\tan(89^\circ)-\tan(0^\circ)$ which is $\tan(89^\circ)$.  Then use $\tan(89^\circ)=\cot(1^\circ)=\frac{\cos(1^\circ)}{\sin(1^\circ)}$ and we are done.

[/hide]

seriously? 0 mohs

Multiply both sides by $\sin(1^{\circ})$, then we have $\frac{\sin(1-0)}{\cos(0)\cos(1)}+\frac{\sin(2-1)}{\cos(1)\cos(2)}+....+\frac{\sin(89-88}{\cos(88)\cos(89)},$ since $\tan(a)-\tan(b)=\frac{\sin(a-b)}{\cos(a)\cos(b)},$ we can telescope this to get $\tan(89)-\tan(0)=\cot(1),$ as desired $\blacksquare$

(everyhting in degrees)


search
combinatorics unsolved
combinatorics

In a $ 4\times 4$ board we want to color some unit cells in a way that for every $ 2$ neighboring cells (those who share a side) there is at least one colored cell.

In how many ways we can do this?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=373117819&t=169651

(I searched for the word "adjacent" in a post by me -- I believe this has appeared many times before.  If the source had been included, it might be easier to find.)


Cum se traduce matematic termenul de linie mijlocie?

Multumesc anticipat.


A prismatic bar [i]AB[/i] of weight [i]Q[/i] and length[i] l [/i], is supported at one end [i]B[/i] by a string [i]CB[/i] of length [i]a[/i] and rests at [i]A[/i],vertically below [i]C[/i], against a perfectly smooth vertical wall (check figure).Find the position of the bar ,as defined by the length [i]a[/i] for which equilibrium is possible.

Ans : [img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/b/5/f/b5f6d01c14ee04a4aa0b9267375218749138ba7a.gif[/img]

If F is the force exerted at the end of the rod by the string, F cos C = Q. Balancing torques around the end of the rod abutting the wall, Q sin A = 2 F sin B => a (x^2 + a^2 - l^2)/2ax = 2 x => a^2 - l^2 = 3x^2 => QED.


I got my USAMO form at school today, but I didn't get my email. What should I do?

Did you write down your email when they asked for it?

I think so. Then again it might have been my bad hand writing...

lol I did that in Mathcounts before (bad handwriting).  :lol:

It probably doesn't matter as long as you know your number.  And there's a postal mail.


what's your favorite movie genre?
I like comedy.

and i should probably add favorite movies.

action / comedy

good examples of comedy r the classics: Monthy Python, Three Stooges, etc

comedy, mystery, scifi, old timers, action, those types.


Show that the number of 5-tuples ($a$, $b$, $c$, $d$, $e$) such that $abcde = 5(bcde + acde + abde + abce + abcd)$ is odd

The problem must intend positive integer solutions.

The equation is equivalent to $1/5 = \Sigma (1/a)$.  There is one solution with all variables equal, and the rest can be organized into sets of two by exchanging pairs of unequal variables.  One needs to also show that the number of solutions is finite.

[quote="tµtµ"]I don't understand the problem, since, as it, there is an infinite number of solutions (taking $a=b=c=0$ for instance).[/quote]
 :alien:[b] It seems to be really you dont understand the problem[/b]
    question is about number of 5-tuplesi.e{(a,b,c,d,e)/ a,b,c,d,e satisifies the given condition}
  a= 0 , b = 0 , c = 0 are not seperate solutions .
 Solve problems carefully by taking
 a= 1 , 2, 3,.........9 we get contradictions( suppose a = 1 L.H.S <20,000 where as R.H.S>20,00} hence we conclude  a= 0
after that  0bcde = 5(bcde + ....)  which is possible only when b=c= d= e =0.
 Hence we conclude that only solution is {(0,0,0,0,0,0} is one
 Hence numbe of solutions are odd.
                                                                                Brahmam

I think tutu is right.if the problem is not restricted to positive integers.
and here is proof for the problem is restricted to positive integers.
there is only one solution such  that $a=b=c=d=e$
for $a,b,c,d,e$ are not all equal.change order of $a,b,c,d,e$ you will get even 5-tuples$(a,b,c,d,e)$
so the the number of 5-tuples$(a,b,c,d,e)$ is even+1=odd ;)
$Remark$:by hint of tutu,I think we should prove that the number is finite first,which is easy

[quote="t0rajir0u"]brahmam:

Taking $a = b = c = 0$ we get $0 \times 0 \times 0 \times de = 5(0 \times 0 \times de + 0 \times 0 \times de + 0 \times 0 \times de + 0 \times 0 \times 0 \times d + 0 \times 0 \times 0 \times e)$ which holds true regardless of the values of $d, e$, so we have an infinite number of solutions unless the problem is restricted to positive integers.

Also:

$a = b = c = d = e = 25$ is a solution, so you can't possibly claim $a = b = c = d = e = 0$ is the only solution.[/quote]
 :rotfl:
 Hello what do you understand by abcde ?
 Is   abcde = ax b x c xd xe ?
   no
   abcde = a x 10 power 4  + b x 10 power 3 + c x 1o power2 + d x 10 + e
   brahmam

[quote="tµtµ"][quote="brahmam"]It seems to be really you dont understand the problem[/quote]

It seems to be really you didn't care reading what I wrote :P[/quote]
maybe there you have different understandinf for $abcde$,we know the right form?

Well, as I have not been able to find any similar such problems in the Indian NO list, I think this problem is a little unclear as to which interpretation of $abcde$ it wants  :)

[quote="zhaobin"][quote="ZetaX"]Since it's from India, I think we can trust brahmam's interpretation.[/quote]
 :rotfl:
it is from the earth.
so should trust someone on the earth :rotfl:
(just kidding)[/quote]

Dont be kind towars brahmam that  iam from india
 If my solution is  logically weak discuss thats all.
  [b][u] brahmam[/u][/b]

Check the solution here:

http://olympiads.hbcse.tifr.res.in/uploads/inmo-2004


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

For $ p$ a prime number, let $ \mathbb{Z}_{p^\infty}\equal{}\left\{\frac{\bar{a}}{p^i}: a\in\mathbb{Z}, i\geq 0\right\}$ be an infinite abelian subgroup of $ \mathbb{Q}/\mathbb{Z}$.

How do I show that $ \mathbb{Q}/\mathbb{Z}\cong\oplus_p\mathbb{Z}_{p^\infty}$?

in general, every abelian torsion group $ A$ is the direct sum of the $ Syl_p(A)$, where $ p$ runs through the primes and $ Syl_p(A)$ is the union of the $ p^n$-torsion subgroups of $ A$. that the sum is direct, is immediate, and the existence of the decomposition of $ a \in A$ follows from the chinese remainder theorem applied to $ \langle a \rangle$ or an easy induction on the order of $ a$.

observe $ Syl_p(\mathbb{Q}/\mathbb{Z}) \equal{} \mathbb{Z}/p^\infty$.


prove that $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$ is never an integer for any $n$.

if $S_n= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$
you must proof at first that $S_n=\frac{p_n}{q_n}$, $p_n$ is an odd number and $q_n$ is an even one

We'll consider the number:$\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$

[b][color=darkblue]The lemma:[/color] [/b]In the sequence of numbers: $1,2,3,\cdots,n$ there exists a number $k$, which is divisible by such power of $2$, that does not divide any other
element of the sequence.

[b][color=darkblue]Lemma 2[/color][/b]$lcp(a_1,a_2,\cdots,a_n)=\sum_i p_i^{max(e_1,e_2,\cdots,e_t)}$, where $a_k=p_1^{e_1}p_2^{e_2}\cdots$

Now let

$\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=k$, $k\in \mathbb{Z}$
Write our number using the least common denominator=the least common multiple.
This multiple will look like this :$2^{max(e_1,e_2,e_3,...,e_n)} \cdot something odd$
If we multiply both sides by this number, we will obtain even number on the right side and odd on the left side, because of the first lemma
Contradiction

This one has been posted quite a lot of times...

This , most likely, have been posted before too, but a classic probelm here (and a little harder)
if $p$ is prime (>3)   and $\frac m n = 1 + \frac 1 2 + \frac 1 3 \dots + \frac 1 {p-1}$  (where $m,n$ are coprime) ; prove m is  divisible by:

Q1)  $p$   (Easier)
Q2)  $p^2$ (A little harder to prove)


Let $ a,b,c >0 : a+b+c =1 $ .Prove that :
 $ \sqrt{a^2+abc} + \sqrt{b^2+abc} + \sqrt{c^2+abc} \geq \frac{2}{3}(\sqrt a + \sqrt b + \sqrt c) $

It was discussed in http://www.mathlinks.ro/Forum/viewtopic.php?t=28061


Hey I was wonderin if anybody can help me find the indefinite solution to this problem

tan(x)-sin(x)=1-tan(x)sin(x)

We can factor it as

$(\tan x-1)(\sin x+1) = 0$

Which has roots

$x = \frac{\pi}{4}+\pi k, k \in \mathbb{Z}$

And

$x = \frac{3 \pi}{2}+2 \pi k, k \in \mathbb{Z}$

can u help on how to find the roots please?

The first case occurs when

$\tan x = 1$

Observing the unit circle, this occurs when $x = 45^{\circ}$ or $\frac{\pi}{4}$ radians.  Since $\tan x$ has a period of $180^{\circ}$ or $\pi$ radians, we have

$\tan (x+\pi k) = \tan x$

And therefore

$\tan (\frac{\pi}{4}+\pi k) = \tan \frac{\pi}{4}= 1$

The second case occurs when

$\sin x =-1$

Observing the unit circle, this occurs at $x = 270^{\circ}$ or $\frac{3\pi}{2}$ radians.  Since $\sin x$ has a period of $360^{\circ}$ or $2 \pi$ radians, we have

$\sin (x+2 \pi k) = \sin x$

And therefore

$\sin (\frac{3 \pi}{2}+2 \pi k) = \sin \frac{3 \pi}{2}=-1$

I'm stuck on this problem also, I did the product to sum laws but don't know what to do after.

(sin(4x))/sin(x-pie/4)=2sin(x+pie/4)

$\sin 4x = 2 \sin (x+\frac{\pi}{4}) \sin (x-\frac{\pi}{4})$

Let $y = x-\frac{\pi}{4}$.  Then

$\sin (4y-\pi) =-\sin 4y = 2 \sin (y+\frac{\pi}{2}) \sin y = 2 \sin y \cos y = \sin 2y$

But of course this implies

$-2 \sin 2y \cos 2y = \sin 2y \implies \cos 2y =-\frac{1}{2}, \sin 2y = 0$

Which has the solutions

$2y = 0, \frac{ 2 \pi}{3}, \pi, \frac{4 \pi}{3}+2 \pi k, k \in \mathbb{Z}$

Solving for $x$,

$x = \frac{\pi}{4}, \frac{7 \pi}{12}, \frac{3 \pi}{4}, \frac{11 \pi}{12}+\pi k, k \in \mathbb{Z}$


I'm trying to make a picture that involves using both the olympiad and graph package, but I can't find any way to use both of them for the same picture.

Also, if you see anything I could do more efficienctly in my coding, I'd greatly appreciate it. Some pictures that are seemingly simple to draw on paper can take hours for me using to recreate using asymptote.

[code]import olympiad;
import graph;
size(150);
pair A,B,C,D,O;
A = (2.5,-1.8*cos(2.5)+0.5);
B = (-2.5,-1.8*cos(-2.5)+0.5);
C = (2.5,0);
D = (-2.5,0);
O= (0,0);
xaxis(-3.3,3.3,Arrows);
yaxis(-3.3,3.3,Arrows);
real f(real x)
{
return -1.8cos(x) +0.5 ;
}
draw(graph(f,-3.14,3.14),red);
draw(A--C,blue);
draw(B--D,blue);
draw(A--B,blue);
label("\scriptsize{$0$}",O,SW);
label("\scriptsize{$x$}",(3.3,0),S);
label("\scriptsize{$y$}",(0,3.3),W);
label("\scriptsize{$x$}",C,S);
label("\scriptsize{$-x$}",D,S);
label("\scriptsize{$f(x)$}",midpoint(A--C),E);
label("\scriptsize{$f(-x)$}",midpoint(B--D),W);
dot(A,red);
dot(B,red);
add(pathticks(C--A,1,0.5,1,5));
add(pathticks(A--B,1,0.5,1,5));[/code]

Check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213729]this[/url]. If it doesn't help, let me know (and tell the version of Asymptote you are using) :).

The correction didn't appear to work and I'm using asymptote v 1.43. I'll update later if the newest version fixes it for me. Similarly, my TexnicCentre, GhostScript, etc is about a year old so is there any point in updating those as well?

Edit: Downloaded Asymptote version 1.57, hit alt+A for one of my .asy documents and got the following error message:

fatal error -system shared memory version mismatch detected... this problem is probably due to incompatible versions of the cygwin DLL. Search for cygwin.dll using the windows start/find search facility...

It gave other instructions but the search came up with nothing. So I guess that's confirmation I need to update all my AOPS related programs?

Hmmm... Your code worked perfectly well on my PC and the only change I made to olympiad.asy was exactly the one I indicated in that sticky. I also use 1.43. My attempt to install 1.57 resulted in a path error: Asymptote correctly found the directory containing latex.exe but added /latex instead of \latex to it, so windows failed to recognize it as a valid command and I could do nothing about it (though I didn't try real hard), which means that I cannot tell you much about 1.57 version now :(. If you return to 1.43, what exactly is the error message you are getting?

I can't return to 1.43, I uninstalled it before downloading the newer version and googling for anything but the newest version doesn't yield any findings.

Here is the full error message, in any case.

[code]
7 main ? [40448] C:\Program Files\Asymptote\asy.exe: ***fatal error- system shared memory version mismatch detected - 0x2D1E009C/0x8A88009C.
This problem is probably due to using incompatible versions of the cygwin DLL. Search for cygwin1.dll using the windows Start -> Find/Search facility and delete all but the most recent version. The most recent version *should* reside in x:\cygwin\bin, where 'x' is the drive in which you have installed the cygwin distribution. Rebooting is also suggested if you are unable to find another cygwin DLL.[/code]

As I said before, the start-> search feature doesn't bring up any results, I don't have any drives containing a \cygwin\bin folder and I already tried rebooting my computer (I think). I even tried replacing the cygwin1.dll file with one I downloaded off the net but to no avail. If it's of any importance, I'm using Vista.

You mean you deleted asymptote-1.43-setup.exe right after installing the 1.43 version? That may be not very clever. I have it but the question is how to send it to you (the maximal attachment size in my e-mail is 2MB and the file size is 3MB).

I still have that path problem with 1.57 but otherwise it works. You may try to choose the option "run program in compatibility mode with", say, windows 2000 or windows XP and see if it helps. If the problem persist, post it on the [url=http://sourceforge.net/forum/forum.php?forum_id=409349]asymptote forum on sourceforge.net[/url]

I found one completely idiotic thing about 1.57 but it may be the root of all evil: you need to call asy.bat now, [b]not[/b] asy.exe! I bet your tools in TeXnic center, etc. still call asy.exe directly.

[quote]You mean you deleted asymptote-1.43-setup.exe right after installing the 1.43 version? That may be not very clever. I have it but the question is how to send it to you (the maximal attachment size in my e-mail is 2MB and the file size is 3MB). [/quote]

Filefront? Rapidshare? Megaupload? Torrent? That is, unless it's illegal to send files that way, but asymptote is freeware to my knowledge.

[quote]I found one completely idiotic thing about 1.57 but it may be the root of all evil: you need to call asy.bat now, not asy.exe! I bet your tools in TeXnic center, etc. still call asy.exe directly.[/quote]

I renamed the file and the shortcut and it did change the error message, but now it thinks the file is .exe.

[code]
C:\Program Files\Asymptoe\asy.exe is not recognized as an internal or external command.
Inoperable program or batch file.
Press any key to continue...
[/code]

I'll try posting this on their forum and see what happens. Though, at least this might lead to a wiki update for other members encountering similar problems with the new version or something.

[b]Asymptoe[/b]? :? Check the directory names and look whether you have this misprint in asy.bat (located in the same directory as asy.exe). Also, type [i]C:\Program Files\Asymptote\asy.bat yourfilename[/i] from the command line and see what happens. Make sure that there are no other asy.bat files except the one coming with the distribution of 1.57 (you might create a file with this name yourself for capturing error messages, etc. and I surmise that was the one that was called).

Also, this may be a crazy question, but [b]what file did you rename?[/b] I hope, not asy.exe; asy.exe has to stay what it is. What I meant was that now they have a [b]separate[/b] asy.bat file that first sets an environment variable (that, presumably, asy.exe uses in some way), then goes to the user root directory (or something like that, I do not know Vista well enough) and only then calls asy.exe!

From what you wrote, it seems like you tried to change the name of asy.exe to asy.bat not even telling the file explorer to show the extension first! This is definitely [b]not[/b] the right thing to do...

I finally deduced what the problem is. Asymptote only works if I "run as administrator", otherwise the cygwin error appears. It also appears that asymptote works in accordance with Texnic Centre if I run that as an administrator.

Thanks for the help :D


[b]Understanding Mathematics[/b]
http://www.math.utah.edu/~alfeld/math.html

[b]Martin Gardner[/b]
http://www.g4g4.com/

[b]Books online[/b]
http://www.ams.org/online_bks/online-books-web.html
http://spot.colorado.edu/~dubin/bookmarks/b/1240.html
http://web.math.fsu.edu/Science/Books.html
http://www.geocities.com/alex_stef/mylist.html              http://davidmlane.com/hyperstat/
http://www.lightandmatter.com
http://www.physicsforfree.com
http://www.gotmath.com/notes.html
http://www.theassayer.org/
http://www.thiel.edu/mathproject/atps/
http://www.math.miami.edu/~ec/book/
http://joshua.smcvt.edu/linearalgebra/
http://www.math.cornell.edu/~hatcher/

[b]Competitons[/b]
http://dir.yahoo.com/Science/Mathematics/Education/K_12/Academic_Competitions/
http://www.mathpropress.com/competitions.html
http://mathcounts.org/
http://www.usamts.org/
http://www.mualphatheta.org/
http://www.dbhsmao.com/
http://www.mathleague.com/
http://www.kalva.demon.co.uk/
http://www.enc.org/weblinks/classroom/competitions/
http://my.netian.com/~ideahitme/mo2.html

[b]Miscellaneous (math and science)[/b]
http://eng.mathdb.org/
http://ocw.mit.edu/OcwWeb/Mathematics/index.htm
http://mathforum.org/
http://mathworld.wolfram.com/
http://www.g4g4.com/paul/BOOK.pdf
http://www.enc.org/weblinks/science/
http://www.enc.org/weblinks/math/
http://www.enc.org/weblinks/classroom/problems/
http://www.ams.org/mathweb
http://sciencegems.com/math.html
http://aleph0.clarku.edu/~djoyce/java/elements/toc.html
http://www.ams.org/employment/highschool.html
http://math.about.com/
http://archives.math.utk.edu/
http://www.americatakingaction.com/studyhall/math.htm
http://www.qbyte.org/
http://www-gap.dcs.st-and.ac.uk/~history/
http://www.simonsingh.net/Mathematics_Links.html
http://encarta.msn.com/encnet/departments/homework/
http://www.wtv-zone.com/Angelaruth49/MathLinks.html

[b]Tutorials[/b]
http://www.purplemath.com
http://home.alltel.net/okrebs/
http://www.bymath.com/index.html
http://www-gap.dcs.stand.ac.uk/~history/Miscellaneous/Briggs2/index.html
http://www.ping.be/~ping1339/
http://aleph0.clarku.edu/~djoyce/java/trig/
http://www.number2.com
http://www.aaamath.com/
http://www.dabanasa.com/dave/math/
http://aleph0.clarku.edu/~djoyce/
http://www.calculus.net/ci2/?tag=
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/index.html
http://regentsprep.org/Regents/math/math-a.cfm
http://archives.math.utk.edu/visual.calculus/
http://www.karlscalculus.org/
http://www.mathsrevision.net/alevel/
http://www.mathsrevision.net/gcse/index.php
http://www.math.hmc.edu/calculus/tutorials/
http://www.mathlinks.info/
http://www.thinkquest.org/library/site_sum.html?tname=C0110248&url=C0110248/index.htm
http://www.acts.tinet.ie/trigonometry_645.html
http://www.studyworksonline.com/cda/home/0,,NAV1,00.html
http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book-5-17-03.pdf

[b]Chemistry/physics RELATED[/b]
http://www.chemistrycoach.com/tutorial.htm
http://webbook.nist.gov/chemistry/
http://www.hhmi.org/biointeractive/vlabs/
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
http://www.chemicalforums.com/index.php?op=forum

[b]math formulas[/b]
http://www.math2.org/

[b]Tests/Exams[/b]
http://www.mymathtest.com/
http://www.exambot.com/
http://www.edhelper.com

[b]Large DATABASES[/b]
http://school.discovery.com/homeworkhelp/bjpinchbeck/
http://www.gotmath.com/
http://www.physicsnetwork.com/
http://www.educationplanet.com/

[b]MATH CAREERS[/b]
http://www.coolmath.com/frm_careers.htm

NEED HELP WITH CHEMISTRY?????  WALK YOURSELF OVER TO

http://www.chemicalforums.com
---------------------------------------------------------------------------
Thanks to all contributors: Mathfanatic, splintercelll990

Please feel free to private message me your links to great math webpages so I can organize them accordingly!

Maybe free software could go here too:

Algebra:
http://research.mupad.de/

Geometry:
http://www.mit.edu/~ibaran/kseg.html

Graphing:
http://www.graphcalc.com/

Fractals:
http://www.chaospro.de/

well, i just cannot access some of the links......

Here's another link I find very practical, for times when you need a numerical result:

http://www.akiti.ca/Mathfxns.html

[quote="paladin8"][quote="qxy"]the site  http://www.kalva.demon.co.uk/    doesn't  work :( ?   Has it changed the adress?  If so, who could tell me the new adress?[/quote]

http://www.kalva.demon.co.uk/index.html[/quote]
Still does not work... :(  :?

Readers of the forum might like to explore the Open Directory Project mathematics pages at http://dmoz.org/ The ODP is the largest, most comprehensive human-edited directory of the Web, constructed and maintained by a vast, global community of volunteer editors.

It is hierarchically organised: school-level mathematics resources are listed at http://dmoz.org/Kids_and_Teens/School_Time/Math/ and college- and professional-level resources at http://dmoz.org/Science/Math/

The ODP relies on submissions from the public, which are reviewed by volunteer editors (and yes, I am one of them).

thank you all for these links.  They help alot. :)

Thanks for useful links so far.

Also, for Info on Maths Careers (UK):
[url]http://www.mathscareers.org.uk/default.aspx[/url]

And, GCSE Maths Revision (UK):
[url=http://www.gcse-maths-revision-cards.co.uk/index.html]GCSE Maths Revision Cards[/url]


What is the resonant frequency of a 35-millihenry coil and a 12-picofarad capacitor?


congratulations my dears.

because ure footbal team won the game. :D

Let see if they play against Brazil in the round of 16...........

[quote="José"]I have to congratulate you Italy... fair champ![/quote]

me too. :wink:

I would like to congratulate this Italians because this year they got many honnorable prizes from International competitions :

-Football World Cup 2006
-two gold medals ay IMO  :lol:

so good year for italians

davron latipov


function
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Let $f$ be a real smooth function on $(0,1)$. Is it true that $f$ is analytic on some sub-interval?

no, one can find examples of functions $f$ which have large derivatives at the rationals.

Another approach is to consider the series $\sum_{k\ge 1} 2^{-k^2}\cos 2^k x$. It is $2\pi$-periodic and not analytic on the entire real line (because the coefficients decay too slowly for an analytic function but fast enough for a $C^\infty$-function). On the other hand, it is a sum of an analytic function and a function with arbitrarily small period, so, if the sum is analytic on an interval, it must be analytic  on the entire real line.

Make it $f(x)=\sum_{k=1}^{\infty}2^{-2^{k/2}}\cos(2^kx).$

I actually got a publication out of this example - it's in the [i]Missouri Journal of Mathematics[/i], if you ever want to look for it.

The gist of the argument: let's actually look at

$g(z)=\sum_{k=1}^{\infty}2^{-2^{k/2}}z^{2^k}$ restricted to the unit circle $z=e^{i\theta}.$

If a function is analytic at a point, then it's actually analytic on a neighborhood of that point.

If $g(z)$ were to be analytic at any point on the unit circle, it would be analytic on some small disk cetered at that point. Now (and this is where the fact that it's a lacunary series comes in),  it has the same behavior on an overlapping sequence of balls all the way around the circle - and that would make $g$ analytic on a disk centered at $0$ of radius strictly greater than $1.$ But the radius of convergence of this power series (by the root test) is exactly 1.

That $f$ is $C^{\infty}$ can be shown just by looking at the series for each derivative. So the example is $C^{\infty}$ everywhere, analytic nowhere.

Fedja's argument is essentially the same. So it was a cheap paper!


geometry
circumcircle
3D geometry
sphere
induction
tetrahedron
perpendicular bisector

Hi,

Consider the following statement : Every set $S \in \mathbb{R}^n$ such that $0 < diam(S) < \infty$ can be partitionned in at most $n+1$ sets of diameter smaller than $diam(S)$.

Prove it for $n = 1, 2, 3$.

Enjoy

I'll prove (more like sketch) the cases $n=1,2$, and I'll see what I can do about $n=3$ some other time.

First notice that taking convex hulls and closures does not alter the diameter, so we may as well assume our set $\mathcal K$ is compact and convex.

If $n=1$, the set is just a compact interval. Every division into two non-degenerate subintervals works.

When $n=2$, let $AB$ be a diameter of $\mathcal K$. The aim will be to prove that we can divide $\mathcal K$ through two cuts: one cut along the line $d$, parallel to $AB$ and close enough to it, and another cut along the perpendicular bisector of $AB$ (after the first cut, we only cut the part containing $AB$).

Notice that two different diameters cannot be disjoint. This means that after the first cut, only the piece containing $AB$ can have a diameter. Suppose that no matter how close to $A,B,\ A_1,B_1\in\partial\mathcal K$ are respectively, on a certain side of $AB$, such that $AB\|A_1B_1$, if we cut $\mathcal K$ along $A_1B_1$ we'll always find a diameter in one of the sets we obtain by cutting the piece containing $AB$ again along the perpendicular bisector of $AB$. By a compactness argument it then follows that our convex body contains the equilateral triangle of side $AB$. Repeating this for the other side of $AB$, we find that our body must contain the two equilateral triangles $ABC,ABD$ with side $AB$. This, however, is impossible, because $AB$ is a diameter and $CD>AB$.

The question is posed for compact sets. I'm not certain if convexity can be assumed, since there are different results achieved for smooth curves. I am thinking it should be allowed, but am not certain.

This is a sketch of my proof.

We restrict our attention to 1,2,3- dimensional space.

Let $r_{n}$ be the circumradius of a 'tetrahedron' whose edges have unit length in n-space. Consider an set (referred to as the original set) of diameter 1.

Lemma 1 - Any $n+1$ points with diameter 1 can be covered by a sphere of radius $r_{n}$

This is easy to show for $n=1,2$, harder for $n=3$

Lemma 2 - We have a collection of $k$ $n-$spheres, of which any $n+1$ spheres have a non-empty intersection (possibly at a point). Then all the $k$ spheres must have a non-empty intersection.

After making the assumption that no $n$ of the spheres intersect within another sphere, this can be shown by induction applied to the surface of a sphere.. It holds true for all integers $n$

Lemma 3 - Any region of diameter 1 can be covered by a sphere of diameter 1.

If there are finitely many points, it follows from Lemma 1 and 2, since any $n+1$ spheres of radius $r_{n}$ about these points have a non-empty intersection. Otherwise, take an ordered dense subset that 'approximates' our set.

Lemma 4 - Consider the density of our original set on the surface of our sphere in lemma 3. This density is $< \frac{2}{n+1}$.

Since antipodal points on the surface of our sphere cannot both be in the set, then density is $\leq \frac{1}{2}$. For n=3, a slightly finer argument is needed.

It follows then that there is a tetrahedron of unit length embedded in our sphere in lemma 3, which has only 1 vertex lying in our original set. We $n+1-$sect our tetrahedron by joining the vertices to the center. It can be shown that the maximal distance between any 2 points in each subsection is achieved only if the 2 points are both vertices. [I have only shown this for $n=1,2,3$.] Take an open ball around the other $n$ vertices which does not intersect our original set. Then the $n+1$ partitions complement by these $n$ open balls give us our required $n+1$ partition.

This was written up in an attempt to try and extend the argument to n=4. However, the main problem i face is getting the bound in lemma 4.


Let
\[f(a,b,c)=\parallel \frac{b-a}{ab}|+\frac{a+b}{ab}-\frac{2}{c}|+|\frac{a-b}{ab}|+\frac{a+b}{ab}+\frac{2}{c}. \]
Prove that $f(a,b,c)=4\cdot \max{(\frac{1}{a},\frac{1}{b},\frac{1}{c})}$.

[hide="Two key hints"]1. $\max(u,v)={1\over 2}(|u-v|+u+v)$ - prove it.
2. $\max(u,v,w)=\max(\max(u,v),w)$[/hide]


let p(z) be a polynomial in Z[x]. Let q(z) be either or z^n -1 or z^n+1.
we know that for every root of q(z) such as u we have |p(u)| <=1.
prove that either q(z) | p(z) or there exists a unic k such that 0<= k <n and
q(z) divides p(z) +z ^k or p(z) - z^k :?

quite simple problem. it suffices to consider p(z) in $Z[x]/(q(x))$. we have to find that all polynomials with $|p(u)|\leq 1$ for all zeroes u of q (we main obtain values of p at these zeroes even when working in $Z[x]/(q(x))$). now let $p\neq 0$ be such a polynomial. if $z^k$ is the term of smallest degree in p, we may look at $p(z)/z^k$, which also has the desired property. by multiplying with -1 if necessary, we may assume that the constant term of p is a positive integer. but summing $p(u)$ over all zeroes u of q, we get n times the constant term of p. thus, since $|p(u)|\leq 1$ for all those zeroes, we must have $p(u)=1$ everywhere. in $Z[x]/(q(x))$, polynomials have degree at most n-1, but our polynomial p has n given values, so there is exact one polynomial for which $p(u)=1$ for all zeroes, that is $p=1$. but that was to show.

Peter


Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, their sum is $ 0$.)

$ \textbf{(A)}\ \frac{3}{8} \qquad
\textbf{(B)}\ \frac{1}{2} \qquad
\textbf{(C)}\ \frac{43}{72} \qquad
\textbf{(D)}\ \frac{5}{8} \qquad
\textbf{(E)}\ \frac{2}{3}$

[hide]
If no dice are rolled, the probability is $ 0$.  Otherwise the probability is $ \frac{1}{2}$.  So the total probability is $ \frac{1}{2}\left(1\minus{}\frac{1}{4}\right)\equal{}\frac{3}{8}$.  Answer is $ A$.
[/hide]

Sorry if I am missing something extremely obvious, but how can [b]E^(pi*i)=-1[/b] conclude that the probability is $\tfrac{1}{2}\cdot\left(1-\tfrac{1}{4}\right)$ so quickly? How can we get that from $0$ and $\tfrac{1}{2}$?

Thanks in advance! :)
~ Ahaan.

Well if both die are rolled, the probability the sum is odd is $\frac{1}{2}$ (Out of the $36$ possible sums of two die, half are odd and half are even.) If $1$ die is rolled, the probability of an odd sum is still $\frac{1}{2}$. So as long as a head results, whether it be one or two, the probability of getting an odd sum is $\frac{1}{2}$. The probability of not getting a heads for both coins is $\frac{1}{2}  \cdot \frac{1}{2} = \frac{1}{4}.$. So the desired probability is $\frac{1}{2}(1- \frac{1}{4}) = \frac{3}{8}$.


If P(x) is a polynomial of degree 2007 whose co-efficients are odd integers.Then find the maximum number of rational roots P(x) can have.

max. no. of rational roots ...so we must have 2007 nearest odd prime no.s...we will surely get all the roots rational...i.e. 2007...

[hide="Removed; this is wrong.  :ninja: "]Right; a polynomial such as
\[P(x) = 3^{2007}\cdot \left( x-\frac{5}{3}\right)\left( x-\frac{7}{3}\right)\left( x-\frac{11}{3}\right)\left( x-\frac{13}{3}\right)\left( x-\frac{17}{3}\right) \cdots \]
with the roots having prime numerators satisfies the conditions and has $2007$ rational roots.[/hide]
[b]Follow-up Question:[/b]
How many odd, integer roots can such a polynomial (with odd, integer coefficients) have?

[quote="cincodemayo5590"]Right; a polynomial such as
\[P(x) = 3^{2007}\cdot \left( x-\frac{5}{3}\right)\left( x-\frac{7}{3}\right)\left( x-\frac{11}{3}\right)\left( x-\frac{13}{3}\right)\left( x-\frac{17}{3}\right) \cdots \]
with the roots having prime numerators satisfies the conditions[/quote]

Not true.  $\bmod 2$ this polynomial is congruent to

$Q(x) = (x-1)^{2007}$

And this will have an even coefficient of $x^{k}$ whenever ${2007 \choose k}$ is even (I believe this first occurs when $k = 8$).

Edit:  It also doesn't matter whether the numerators are prime or not, since the coefficients end up having the same value $\bmod 2$.


Simplify: $ 7 \times 10^3 \plus{} 5 \times 10^2 \minus{} 4 \times  10^1 \plus{} 7 \times  10^0$.


Tooken from a math textbook..im not this smart to make them up! Glencoe RULZ!

(-x)^2(-x^2)^3(-x)^3

Try and solve that

(-2)^0(3x^-2y^-2)^-1

Beat That!

8x^2y^-3/x^-2y*(4xy^2)^-1/x^2y

MUAHAHA!!

Solve Equations

(2x-3)(3x+2)=0
4x^2=256
15x^3-240x=0

Try and solve these! I guarantee your brain will go....WOOHOO!

Ummm, assuming the exponents are only the number?

Also, this is clearly HW (Glencoe).

I dont get how u solve the first equations because there is a variable x and u dont give wat the equation is equals too...

[hide]but i can solve the last three questions[/hide]

$ (2x\minus{}3)(3x\plus{}2)\equal{}0$

This is ez because it is already factored out for u. $ x$ must be $ \frac{3}{2}$ or $ \frac{\minus{}2}{3}$

second equation...

$ 4x^2\equal{}256$

if u divide both sides by $ 4$ you get $ x^2\equal{}64$ now square root both sides $ x\equal{},\minus{}8$ and this is your answer

third equation...

$ 15x^3\minus{}240x\equal{}0$

divide both sides by $ 15$ and you get $ x^3\minus{}16x\equal{}0$

from here divide both sides by $ x$ you can already c if you plug in 0 u get a solution.

if u divide it becomes $ x^2\minus{}16\equal{}0$
if you solve this quadratic it will become $ x\equal{}4,\minus{}4$ and the $ 0$ from before

these are quite ez if you know how to solve quadratic equations...

No, its not homework. Im done with this chapter. But i have a cumulative. So i go back to questions i dont know and study them

Your explanations were useless!!! JK JK JK!! UR MY BEST BUDDY MCNAT!!

WHat i mean is to simplify the first 3 questions! J=Kudos to you McNat!!

Now solve

3x-7x^2=0
3(2x+1)=7(5-x)-6

[hide="solution  :lol: "]

$ \minus{}7x^2\plus{}3x\equal{}0$

this would be $ 0, 3/7$

i really DO NOT WANT TO SOLVE ALL OF THESE FOR U lol LaTeX takes 4EVA
lol just look in the back of the book :wink: [/hide]

Sorry, but my teacher took these questions from my the text book so i dont know where.

But can somone please simplify.
8x^2y^-3/x^-2y*(4xy^2)^-1/x^2y ?

Thanks!

It would help if you used $ \text{\LaTeX}$, that problem looks in no way clear to me...

[quote="uanaka"]Sorry, but my teacher took these questions from my the text book so i dont know where.

But can somone please simplify.
$ \frac{\frac{8x^2y^{\minus{}3}}{x^{\minus{}2}y \cdot (4xy^2)^{\minus{}1}}}{{x^{2y}}}$ ?

Thanks![/quote]

[quote="uanaka"]I do not know what you mean by Latex[/quote]

Look on the left sidebar.

I can get everything in $ LateX$ this afternoon when i get home

ok w/e ill do it now lol

BOGRTO, to me it seems like it should be $ \frac {\frac {8 x^2 y^{ - 3}}{x^ - 2 y (4xy^2)^{ - 1}}}{x^2 y}$ instead. (Note the denominator of the big fraction.) This simplifies to $ \frac {32x^3}{y^2}$, and your fraction simplifies to $ \frac {32x^{5 - 2y}}{y}$. (Is this simplified?)

uanaka: See, this is why you should always post in $ \text{\LaTeX}$. It makes things much, much more readable.


I like Luigi with the Dragster. My favorite track is Rainbow Road. If only it did not come right after Bowser's Castle.

[quote="cf249"]I don't have a Wii, but if I did, this would probably be the first game I brought for it.  I am a big Mario Kart fan.

EDIT:  Wow, another case of the malfunctioning poll.[/quote]

Yeah, there are double of everything. Click on the one above when voting.

[quote="Wickedestjr"]Where's Yoshi?[/quote]
I was not allowed to put that many choices. If the choice does not appear, then post your response.

i put Mario instead, because he is my second favorite.

What do you think of Rainbow Road? My best time is 3:03.

I don't remember what my best time is. It is hard. I have only had the game for about a week, and I haven't even finished rainbow road yet.

In grumble volcano time trial, turn left from the start line, then boost onto the rock. Drive around the rock, going through the ridge every time. If you do not go through the ridge, it will not work. Every time you drive around, it is a lap. You can finish it in 1 minute. I still have to acutally do this. I wonder if it works...I just know that if you are aligned with the finish line, it counts, so I thought of this. In time trial, the mushrooms are for a reason. If you use it in the correct place, you will get a better boost than normal. One per lap. Figure out the best place to use a mushroom.

Hmm, interesting, i'll check it out.  :)

I just realized that Nintendo deletes all ghosts that take that shortcut, because it is beyond stategy, it is cheating. So do not do it.

You mean, you can't do it if you are a ghost?

I meant that if you try it in a time trial, Nintendo will see it on the Mario Kart Channel, and they have moderator powers, so they will delete that time. The reason is that it is not strategic, it is cheating. So do not do it. I have not tried it yet, so I am lucky. Just do it the way you normally do it. BTW, the grand prix is kinda luck-based, because you could get lousy items while your opponents could get exellent items. I remember when I was in first place, and I got hit by a blue shell, followed by a red shell, followed by a kart with a star, followed by a bullet bill, followed by a kart using a mega mushroom, followed by a kart using a mushroom boost, and in all, I ended up getting knocked onto the grass, which is an offroad area with maximum friction. To make this worse, this was on the last lap!

I was racing in Waluigi Stadium, and it was the very end of the last lap. At the very end of the course, there is a cliff you have to jump to with the risk of falling inbetween the cliff you are jumping off. I was in 1st place, and I was in the air about to make it onto the cliff, but then that bird thing swooped in and hit me while I was in mid air. This caused me to fall into the hole. Then when I finally came to, I ended up getting last.

[quote="mathelete"]I meant that if you try it in a time trial, Nintendo will see it on the [b]Mario Kart Channel[/b], and they have moderator powers, so they will delete that time. The reason is that it is not strategic, it is cheating. So do not do it. I have not tried it yet, so I am lucky. Just do it the way you normally do it. BTW, the grand prix is kinda luck-based, because you could get lousy items while your opponents could get exellent items. I remember when I was in first place, and I got hit by a blue shell, followed by a red shell, followed by a kart with a star, followed by a bullet bill, followed by a kart using a mega mushroom, followed by a kart using a mushroom boost, and in all, I ended up getting knocked onto the grass, which is an offroad area with maximum friction. To make this worse, this was on the last lap![/quote]

Hm... is this like a satellite?

No, but you can connect the wii to the internet.

YES!  I DONT HAVE INTERNET CONNECTION ON MY WII! :harhar: :harhar: :harhar:

Is that a new emoticon, because you have used it four times today. Counting the three that you just used.

Um, I don't really understand your last post ernie.  :D

They can still see your ghost and delete it.

I prefer Yoshi. I don't know why, but I always win with Yoshi and lose with Luigi and Mario.

My favorite has changed from Yoshi to Mario. I realized that I have only raced with Yoshi one time, and I race with Mario all the time, so consider the votes for Mario one higher than they are now, and the votes for yoshi one less.

[quote="Wickedestjr"]really? Even when it isn't connected to the internet? How is that possible?

I don't really understand the short cut anyway.[/quote]

Forget the shortcut. It only works sometimes, when the game is not glitchy. And Nintendo can see it because in every Mario Kart Wii game there is spyware. It makes sure you do not cheat to become the world champion. PS. I switched from the Dragster to the Wild Wing. I just need to improve my Koopa Cape time by 2 seconds to become world champion.

On Tuesday, November 25, 2008, at 4PM Eastern time, I set a record of 2:38.629 for Koopa Cape, with automatic drift. This means that I could not mini turbo! I used the wild wing. Can anybody beat that?

I've never played time trial before. I actually haven't played Mario Kart in about a month.


geometry
perpendicular bisector
power of a point

In a circle, a certain chord is the perpendicular bisector of a radius.  If the length of the chord is $ 6\sqrt{3}$ what is the area of the circle?

There was a diagram but idk how to put it on the forum.

[hide="Solution"]By the Power of a Point Theorem, we have:

$ (\frac{6\sqrt {3}}{2})^{2} \equal{} \frac {r}{2}\cdot \frac {3r}{2} \\
\Rightarrow 27 \equal{} \frac {3r^2}{4} \\
\Rightarrow 108 \equal{} 3r^2 \\
\Rightarrow 36 \equal{} r^2$

Since the area of a circle is $ \pi r^2$, the area is $ \boxed{36\pi}$[/hide]

Okay, here's how I think you can do it:

Call $ A$ and $ B$ the two endpoints of the chord, $ O$ the center, $ OD$ the radius, $ C$ the point on radius $ OD$ where chord $ AB$ bisects it, and $ E$ the point on the opposite side of the circle to $ D$ (such that $ ED$ is the diameter).  Now since chord $ AB \perp OD$, $ AC\equal{}BC\equal{}3\sqrt{3}$ (from the information given).  There is a formula that states $ AC\cdot BC\equal{}CE\cdot CD$.  Now we know that $ CD\equal{}\frac{r}{2}$ because the problem says that the chord bisects the radius.  Since $ CE\plus{}CD\equal{}2r$, we can conclude that $ CE\equal{}\frac{3r}{2}$.  So now plug everything in: $ 3\sqrt{3}\cdot 3\sqrt{3}\equal{}\frac{3r}{2}\cdot \frac{r}{2}\Rightarrow 27\equal{}\frac{3r^{2}}{4}\Rightarrow r\equal{}6$.  Then from $ A\equal{}\pi r^{2}$, we get $ \boxed{A\equal{}36\pi}$.

Thanks i think i get it but could some one explain the point to a power theorum. thx


is it $\frac{1}{8}$
I will post my solution soon (or later). :)

[quote="deej21"][hide]
$\frac12 (\cos 40-\cos 60 )(\sin 70)$
$\frac12 (\sin 70 \cos 40-\frac12 \sin 70)$
$\frac14 (\sin 110+\sin 30-\sin 70)$
$\frac14 (\sin 110-\sin 70+\frac12)$
$\frac14 (2\sin 20 \cos 90+\frac12)$

$\boxed{ \frac18 }$
[/hide][/quote]
deej21, i don't really understand your solution... :blush:
could you explain in detail??
Thanks,

Yep... two very important bookmarks:
[url=http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html]Sum to product[/url]
[url=http://mathworld.wolfram.com/WernerFormulas.html]Product to sum[/url]

[quote="ShoeFactory"]I never bothered to learn the sum/product trig identities. They're too hard to remember and are rarely used :\[/quote]

They're actually used quite often. If you really don't want to bother, remember the basic ones and derive the rest when you need them.

[quote="drunner2007"]good job  :lol:

kimby, he's just using the sum to product and product to sum formulas. for example, starting at the original expression he notices that $\sin10 \sin50=\frac{1}{2}(\cos(40)-\cos(60))$ from the identity $2 sinx siny=\cos(x-y)-\cos(x+y)$. same type of thing for the a lot of the manipulations later on. just refer to a trig book to see all the identities[/quote]
thanks,
i knew the sum formulas but never thought about working backward.....nice problem.

[hide="A solution without using sum-to-product"]
Using $\sin (90-\theta) = \cos \theta$, we can rewrite $\sin 10\sin 50\sin70$ as $\cos 20\cos 40\cos 80$.

Then using double angle formula for sines,

$\cos 20\cos 40\cos 80$
$=\frac{1}{\sin 20}\sin 20\cos 20\cos 40\cos 80$
$=\frac{1}{2\sin 20}\sin 40\cos 40\cos 80$
$=\frac{1}{4\sin 20}\sin 80\cos 80$
$=\frac{1}{8\sin 20}\sin 160$

Since $\sin 160= \sin (180-20) = \sin 20$, the expression is equivalent to $\frac{1}{8}$.
[/hide]


limit
real analysis
real analysis unsolved

Let $ a\in \mathbb{R}$ , and a sequence $ (x_n)_{n\in N}$ such that $ x_n<a$ $ \forall n \in \mathbb{N}$ and $ \lim_{n\to\infty}x_n=a$ .Is it necessary that $ x_n\leq x_{n+1}$ $ \forall n \geq n_0$ , where $ n_0 \in \mathbb{N}$ ?

No. For example,
$ x_1 \equal{} 0$, $ x_2 \equal{} \frac{1}{2}$
$ x_3 \equal{} \frac{1}{4}$, $ x_4 \equal{} \frac{1}{2}$, $ x_5 \equal{} \frac{3}{4}$
$ x_6 \equal{} \frac{5}{8}$, $ x_7 \equal{} \frac{3}{4}$, $ x_8 \equal{} \frac{7}{8}$
and so on...
it converges to 1, satisfies your condition, but doesn't satisfy your conclusion.


[quote="Interval"]If the length of a rectangular prism is doubled, its width is tripled, and its height remains the same, what is the volume of the new rectangular prism?

(1) double the original volume
(2) triple the original volume
(3) six times the original volume
(4) nine times the original volume[/quote]
[hide]If the original prism's volume is $l*w*h=lwh$, then the new prism's volume will be $2l*3w*h=6lwh$ which is 6 times the original volume (3)[/hide]

[quote="Interval"]If the length of a rectangular prism is doubled, its width is tripled, and its height remains the same, what is the volume of the new rectangular prism?

(1) double the original volume
(2) triple the original volume
(3) six times the original volume
(4) nine times the original volume[/quote]

[hide]
For the first one, you do this:

$\text{Volume}=lwh$.

The new one is $2l3wh=6lwh$.

$6lwh/lwh=6\implies\boxed{\textbf{(C)}}$.
[/hide]


linear algebra
matrix
linear algebra unsolved

Hi,
I dont know if this is the right place or the right name of what Im trying to do here but I saw matrix in the topic description *Smile*

So what I want to do is weigh the values in the middle of a matrix (variable size of course ...wouldnt be that hard otherwise) 2xs and those on the edges 1x.

I came up with:
weight =[( -2/(height+width) )/2] * ( |width/2 -x|+|height/2 -y|/2 ) + 2
where: height =# rows ; width =# cols ; x =column ; y =row

when rows == cols it gives me:
[.5	1	.5]
[?	?	?]
[1	2	1]
[?	?	?]
[.5	1	.5]

when # rows and # cols are different I get ugly values ...that are higher in the middle ~1.6 but extremely low on the corners ~.01 even negatives sometimes

I want (rather rows == cols or not):
[1	1	1]
[1	?	1]
[1	2	1]
[1	?	1]
[1	1	1]

Could anyone help me out with this equation or at least tell me what type of math books would talk about this sort of thing?

BTW
If I could get that first weighted matrix on matrices with varying # rows and # cols I would be happy.

Thanks for your help,
-Jerome BG

In case any1 else is following this thread I got a couple great equations here:
http://www.mymathforum.com/viewtopic.php?p=6114#6114

The resultant matrices are of the form:
[1     1.5     1]
[                  ]
[1.5  2.5  1.5]
[                  ]
[1     1.5     1]

which is better than mine :)


The angle trisectors of a regular pentagon intersect other vertices of the pentagon; i.e.
they are diagonals of the pentagon. What is the smallest n > 5 such that the angle trisectors of a regular n-gon intersect other
vertices of the n-gon?

Hint
[hide]
Inscribed Angle Theorem
[/hide]

[hide]
By Inscribed Angle Theorem, an internal angle of regular n-gon is divided by the $ (n\minus{}3)$ diagonals directed from its vertex into $ (n\minus{}2)$ equal angles. Therefore, to have trisectors in these diagonals, we need $ n\minus{}2\equiv 0 \pmod 3$. The smallest such $ n$ greater than $ 5$ is $ 8$ then.
[/hide]


I think the link of the number theory solution is down [url=http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2008_Tests]here[/url] and i can't find it. Can someone post it, plz?

thanks


A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $ 1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is:

$\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\
\textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\
\textbf{(C)}\  \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\
\textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\
\textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$

[hide]Extend BA to a point P 4 in. from B and 2 inches from A.
Let M be the midpoint of BC.
How are triangles BMA and BCP related?[/hide]

[hide="Solution"]
Using KMST's hint, we quickly see that $ \triangle BMA\sim \triangle BCP$ with a ratio of $ \frac12$, so $ PC\equal{}2\cdot1.5\equal{}3$. Since $ P$ is fixed, the answer is $ \boxed{\textbf{(D)}\ \text{a circle with radius 3 in. and center 4 in. from B along BA}}$.

Alternatively, let $ A\equal{}(0,0)$, $ B\equal{}(2,0)$, and $ M\equal{}(x,y)$, so that $ C\equal{}(2x\minus{}2,2y)$ for all $ x^2\plus{}y^2\equal{}\frac94$. Parametrizing yields $ (x,y)\equal{}\left(\frac32\cos\theta,\frac32\cos\theta\right)\implies(x',y')\equal{}(2x\minus{}2,2y)$
$ \equal{}\left(3\cos\theta\minus{}2,3\sin\theta\right)\implies(x'\plus{}2)^2\plus{}y'^2\equal{}9$, so the desired locus is clearly a circle $ 4$ inches from $ B$ along $ \overline{BA}$ with radius $ 3$ inches, which corresponds to $ \boxed{\textbf{(D)}}$.
[/hide]


let a,b and c be three positive real numbers verifying the condition
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}c}\equal{}2$
prove that:
$ \sqrt{a\plus{}b\plus{}c\plus{}3}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

[quote="sofyane"]let a,b and c be three positive real numbers verifying the condition
$ \frac{1}{a\plus{}b}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}c}\equal{} 2$
prove that:
$ \sqrt{a\plus{}b\plus{}c\plus{}3}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$[/quote]

Are you sure ? . Try $ a\equal{}b\equal{}c\equal{}\frac{3}{4}$  :(

I think that the exercise is wrong sorry

Yeah. One thing to note is that there is no upper bound on $ a\plus{}b\plus{}c$ or $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$ because we can make $ a \equal{} p^{2}, b \equal{}\frac{1}{p}$ and $ c$ also very small which would result in $ AB$ being as large as we want for large $ p$ while the condition on the variables would still hold.

I think our problem must be :

$ a,b,c>0,\sum\frac{1}{a\plus{}b}\equal{}3$.

Prove that :

$ \sqrt{a\plus{}b\plus{}c\plus{}3}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

 :)

[quote="quangpbc"]I think our problem must be :

$ a,b,c > 0,\sum\frac{1}{a\plus{}b}\equal{} 3$.

Prove that :

$ \sqrt{a\plus{}b\plus{}c\plus{}3}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

 :)[/quote]
It's still wrong. :(
Try $ a\equal{}8.82$ and $ b\equal{}c\equal{}0.18.$ :wink:

[quote="quangpbc"]I think our problem must be :

$ a,b,c > 0,\sum\frac{1}{a\plus{}b}\equal{} 3$.

Prove that :

$ \sqrt{a\plus{}b\plus{}c\plus{}3}\ge\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

 :)[/quote]
Please see my post above your post. This will not work even if you put 1000 instead of 3.


Δίνονται σύνολα $ A_1,A_2,\ldots$ με $ |A_i| \equal{} r \geq 2$ για κάθε $ i$ και $ A_i \cap A_j \neq \emptyset$ για κάθε $ i,j$. Να δείξετε ότι υπάρχει σύνολο $ B$, έτσι ώστε $ |B| \equal{} r \minus{} 1$ και $ B \cap A_i \neq \emptyset$ για κάθε $ i$.

[u][b]ΣΧΕΤΙΚΗ ΘΕΩΡΙΑ[/b][/u]
Το πιο κάτω λήμμα των Erdos και Rado λύνει (σχεδόν) αμέσως την άσκηση. Σας ζητάω να αποδείξετε το λήμμα, αλλά και να λύσετε την άσκηση χωρίς το λήμμα.

[u]Sunflower Lemma (Erdos-Rado 1960)[/u]: Δίνεται οικογένεια συνόλων $ \mathcal{F}$ με $ |A| \equal{} r$ για κάθε $ A \in \mathcal{F}$. Άν $ |\mathcal{F}| > r! (k \minus{} 1)^r$, τότε υπάρχει σύνολο $ B$ με $ 0 \leq |B| \leq r \minus{} 1$ και $ A_1,\ldots,A_k \in \mathcal{F}$ έτσι ώστε $ A_i \cap A_j \equal{} B$ για κάθε $ i \neq j$.

[u]Εικασιά (Erdos-Rado 1960)[/u]: Το λήμμα εξακολουθεί να ισχύει αν αντικαταστήσουμε το $ |\mathcal{F}| > r! (k \minus{} 1)^r$ με $ |\mathcal{F}| > C^r$ όπου το $ C$ εξαρτάται μόνο από το $ k$ (και όχι από το $ r$).

[quote="Demetres"]Δίνονται σύνολα $ A_1,A_2,\ldots$ με $ |A_i| \equal{} r \geq 2$ για κάθε $ i$. Να δείξετε ότι υπάρχει σύνολο $ B$, έτσι ώστε $ |B| \equal{} r \minus{} 1$ και $ B \cap A_i \neq \emptyset$ για κάθε $ i$.[/quote]

Ίσως να μην καταλαβαίνω κάτι, αλλά μου φαίνεται ότι κάτι λείπει από την εκφώνηση για τα $ A_i$. Δηλ. αν $ r \equal{} 2$ και $ A_1 \equal{} \{1,2\}, A_2 \equal{} \{11,22\}, A_3 \equal{} \{111,222\}\ldots$, δεν μπορώ να φανταστώ ένα σύνολο $ B$ με ένα στοιχείο που να έχει $ B\cap A_i \neq \varnothing$...

Τι δεν καταλαβαίνω?  :ewpu:  :help:

Cheerio,

Durandal 1707

Έχεις δίκιο. Συγνώμη για το λάθος. Διόρθωσα την εκφώνηση.

Να δώσω μια μικρή βοήθεια.

[hide]
Υπάρχει x το οποίο ανήκει σε άπειρα $ A_i$.
[/hide]

[quote="Demetres"]Δίνονται σύνολα $ A_1,A_2,\ldots$ με $ |A_i| \equal{} r \geq 2$ για κάθε $ i$ και $ A_i \cap A_j \neq \emptyset$ για κάθε $ i,j$. Να δείξετε ότι υπάρχει σύνολο $ B$, έτσι ώστε $ |B| \equal{} r \minus{} 1$ και $ B \cap A_i \neq \emptyset$ για κάθε $ i$.
[/quote]
Ίσως να μην κατάλαβα κάτι καλά στην εκφώνηση αλλά...
Αν πάρουμε $ r \equal{} 3$ και $ A_1 \equal{} \{ 1,2,3 \} ,A_2 \equal{} \{1,22,33 \} ,A_3 \equal{} \{1,222,333 \}....$ τότε $ (\forall i)B \cap A_i \neq \emptyset$ αλλά $ |B| \equal{} 1 \neq 2 \equal{} r \minus{} 1$ :maybe:

[quote="giannis18"]
Ίσως να μην κατάλαβα κάτι καλά στην εκφώνηση αλλά...
Αν πάρουμε $ r \equal{} 3$ και $ A_1 \equal{} \{ 1,2,3 \} ,A_2 \equal{} \{1,22,33 \} ,A_3 \equal{} \{1,222,333 \}....$ τότε $ (\forall i)B \cap A_i \neq \emptyset$ αλλά $ |B| \equal{} 1 \neq 2 \equal{} r \minus{} 1$ :maybe:[/quote]

Σ'αυτήν την περίπτωση πέρνει π.χ. $ B \equal{} \{1,2\}$.

Sorry είπα βλακεία δεν το σκέφτηκα καλά :oops:

[quote="Demetres"]Επειδή βλέπω ότι η άσκηση πάει να ξεχαστεί, θα βάλω την λύση μου σε 1-2 μέρες. Αν κάποιος θέλει περισσότερο χρόνο να την σκεφτεί ας μου το πει και θα περιμένω.[/quote]

Δυστυχώς δεν την έχω κοιτάξει καθόλου λόγω χρόνου... αλλά αυτό δεν το βλέπω να αλλάζει τις επόμενες μέρες, οπότε βάλε τη λύση σου. Κρίμα γιατί φαίνεται πολύ όμορφη η άσκηση και πάει να θαφτεί :(

Αν βρω χρόνο (χλωμό) θα την κοιτάξω. Ωστόσο, και πάλι υπάρχει κάτι που δεν καταλαβαίνω: αν πάρουμε $ r \equal{} 2$ και $ A_1 \equal{} \{1,2\}, A_2 \equal{} \{2,3\}, A_3 \equal{} \{1,3\}$ και μετά κυκλικά (mod 3 τα ίδια σύνολα), τότε έχουμε $ A_i\cap A_j \neq \varnothing$, αλλά σίγουρα δεν υπάρχει σύνολο με ένα μόνο στοιχείο ώστε $ B\cap A_i \neq \varnothing$ για όλα τα $ i$.

Χάνω κάτι?  :|

Cheerio,

Durandal 1707

Κάτι που βρήκα ...
Έστω $ A_i$ ένα σύνολο με $ A_i \equal{}\left\{a_1,...,a_r\right\}$ από την οικογένεια $ F$ . Και έστω ένα τυχαίο σύνολο $ A$ με λιγότερα από $ r$ στοιχεία το οποίο περιέχεται σε άπειρα σύνολα της οικογένειας $ F$ .(τέτοιο σύνολο υπάρχει . Για παράδειγμα το κενό σύνολο) . Έστω τώρα ότι για αυτό το σύνολο $ A_i$ το $ A$ δεν έχει κοινά στοιχεία μαζί του (αν έχει τότε έχουμε τελειώσει καθώς το .$ A_i$ είναι τυχαίο ) . Αφού τώρα το $ A_i$ έχει ένα κοινό στοιχειο με τα υπόλοπα άπειρα σύνολα της συλλογής , θα υπάρχει στοιχείο $ a_j$ που θα περιέχεται σε άπειρα σύνολα της συλλογής . Τώρα όμως το σύνολο $ B_j \equal{} A\cup\left\{a_j\right\}$
Με αυτό τον τρόπο αυξάνουμε τον αριθμό στοιχείων του $ A$ και κάποια στιγμή καταλήγουμε στο $ B$ .

Δουλεύει ? (Την ιδέα την πήρα από το μία απόδειξη στο site tou gowers me just do it ! :wink: )

[quote="Durandal"]
Αν βρω χρόνο (χλωμό) θα την κοιτάξω. Ωστόσο, και πάλι υπάρχει κάτι που δεν καταλαβαίνω: αν πάρουμε $ r \equal{} 2$ και $ A_1 \equal{} \{1,2\}, A_2 \equal{} \{2,3\}, A_3 \equal{} \{1,3\}$ και μετά κυκλικά (mod 3 τα ίδια σύνολα), τότε έχουμε $ A_i\cap A_j \neq \varnothing$, αλλά σίγουρα δεν υπάρχει σύνολο με ένα μόνο στοιχείο ώστε $ B\cap A_i \neq \varnothing$ για όλα τα $ i$.

Χάνω κάτι?  :|

Cheerio,

Durandal 1707[/quote]

Ίσως φταίω λίγο στην εκφώνηση. Σου λέω ότι δίνεται οικογένεια συνόλων $ A_1,A_2,\ldots$. Αυτός είναι ένας άλλος τρόπος να πω σύνολο συνόλων. Εννοείται δηλαδή ότι όλα τα σύνολα είναι διαφορετικά. Με πεπερασμένο αριθμό συνόλων έχεις δίκιο ότι δεν ισχύει.

EDIT:
Τώρα που ξανακοίταξα την εκφώνηση κατάλαβα πως πάλι έχεις δίκιο. Χίλια συγνωμη. Θα προσπαθήσω  :lol:  να είμαι πιο προσεκτικός στο μέλλον.

[quote="silouan"]Κάτι που βρήκα ...
Έστω $ A_i$ ένα σύνολο με $ A_i \equal{} \left\{a_1,...,a_r\right\}$ από την οικογένεια $ F$ . Και έστω ένα τυχαίο σύνολο $ A$ με λιγότερα από $ r$ στοιχεία το οποίο περιέχεται σε άπειρα σύνολα της οικογένειας $ F$ .(τέτοιο σύνολο υπάρχει . Για παράδειγμα το κενό σύνολο) . Έστω τώρα ότι για αυτό το σύνολο $ A_i$ το $ A$ δεν έχει κοινά στοιχεία μαζί του (αν έχει τότε έχουμε τελειώσει καθώς το .$ A_i$ είναι τυχαίο ) . Αφού τώρα το $ A_i$ έχει ένα κοινό στοιχειο με τα υπόλοπα άπειρα σύνολα της συλλογής , θα υπάρχει στοιχείο $ a_j$ που θα περιέχεται σε άπειρα σύνολα της συλλογής . Τώρα όμως το σύνολο $ B_j \equal{} A\cup\left\{a_j\right\}$
Με αυτό τον τρόπο αυξάνουμε τον αριθμό στοιχείων του $ A$ και κάποια στιγμή καταλήγουμε στο $ B$ .

Δουλεύει ? (Την ιδέα την πήρα από το μία απόδειξη στο site tou gowers me just do it ! :wink: )[/quote]

Από ότι κατάλαβα, παίρνεις $ B_0 \equal{} \emptyset$. Υπάρχει $ F_1 \in \mathcal{F}$ ώστε $ B \cap F_1
 \equal{} \emptyset$. Υπάρχει $ a_1 \in F_1$ και άπειρη οικογένεια συνόλων $ \mathcal{F}_1 \subseteq \mathcal{F}$, ώστε $ a_1 \in A$ για κάθε $ A \in \mathcal{F}_1$. Ορίζεις $ B_1 \equal{} B_0 \cup \{a_1\} \equal{} \{a_1\}$ και συνεχίζεις...

Η ιδέα είναι σωστή αλλά αν δεν μου έχεις πει πως ακριβώς συνεχίζεις και γιατί όντως καταλήγεις στο $ B$.

[quote="silouan"]Αυτό που λέω ειναι το εξής : Άμα πάρεις ένα σύνολο $ S$ με πληθικό αριθμό μιρκότερο του $ r$ το οποίο να περιέχται σε άπειρα υποσύνολα της $ F$ . Τότε είτε το $ S$ έχει κοινό σημείο με κάθε σύνολο της οικογένειας (οπότε τελειώνουμε) είτε υπάρχει ένα στοιχείο $ s$ που δεν ανήκει στο $ S$ ώστε το σύνολο $ S\cup\{s\}$  να περιέχεται σε άπειρα υποσύνολα της $ F$ . Μπορούμε να εφαρμόσουμε αυτό το αποτέλεσμα $ r$ φορές και τότε σίγουρα τελειώνουμε αφού ένα σύνολο με $ r$ στοιχεία δεν μπορεί να περιέχεται σε άπειρα υποσύνολα της $ F$ .

Είναι σωστό ?[/quote]

Ναι :clap2: Είχα μπερδευτεί λίγο στο πρώτο σου post και δεν ήμουν σίγουρος αν αυτό εννοούσες.

Τώρα μένει να αποδείξεις και το sunflower lemma  :wink:

Μιας και μου την θύμισε ο Παναγιώτης, να βάλω μια απόδειξη του "λήμματος του ηλιοτροπίου". Λέγεται έτσι διότι μπορούμε να σκεφτούμε το σύνολο $ B$ σαν τον πυρήνα του ηλιοτροπίου και τα σύνολα $ A_i \setminus B$ σαν τα πέταλα του ηλιοτροπίου.  :)

Η απόδειξη είναι με επαγωγή στο $ r$. Για $ r\equal{}1$ είναι προφανές. Ας υποθέσουμε πως ισχύει για $ r$ και θα δείξουμε πως ισχύει για $ r\plus{}1$. Ας πάρουμε  $ A_1,\ldots,A_s \in \mathcal{F}$ ανά δύο ξένα μεταξύ τους με $ s$ μέγιστο. Αν $ s \geqslant k$, τότε τελειώσαμε (με $ B \equal{} \emptyset$).  Μπορούμε λοιπόν να υποθέσουμε πως $ s \leqslant k\minus{}1$. Επειδή $ s$ μέγιστο, για κάθε $ A \in \mathcal{F}$, υπάρχει $ i$ ,ωστε $ A \cap A_i \neq \emptyset$. Επειδή $ |A_1 \cup \cdots \cup A_s| \equal{} s(r\plus{}1) \leqslant (k\minus{}1)(r\plus{}1)$, και $ |\mathcal{F}| > (r\plus{}1)!(k\minus{}1)^{r\plus{}1}$, πρέπει να υπάρχει $ x \in A_1 \cup \cdots \cup A_s$ το οποίο να ανήκει σε περισσότερα από $ \frac{(r\plus{}1)!(k\minus{}1)^{r\plus{}1}}{(k\minus{}1)(r\plus{}1)} \equal{} r!(k\minus{}1)^r$ σύνολα που ανήκουν στο $ \mathcal{F}$.

Έστω λοιπόν $ \mathcal{F}' \{ A \setminus \{x\} : A \in \mathcal{F},x \in A\}$. Τότε $ |\mathcal{F}'| > r!(k\minus{}1)^{r}$. Από την επαγωγική υπόθεση η $ \mathcal{F}'$ περιέχει ένα ηλιοτρόπιο με $ k$ πέταλα. Ομοίως και η $ \mathcal{F}$, προσθέτωντας το $ x$ στον πυρήνα του ηλιοτροπίου της $ \mathcal{F}'$.


Okay i have five questions but im only asking for you to write a solution to problems 4&5, If you feel like it, check my answers for problems 1-3.
(I'm not sure about this but i dont think problem 4 is possible.)
Here are my answers:
1.12
2. No becuase the sum of the first 10 numbers is 55 which is not divisible by 2.
3 1/3
4.s=3 d=5 m=31 f=34 (resolved question)
5.? (still have no clue on)

4 is possible. Answer:

[hide]Son: 3
Daughter: 5
Mother: 31
Father: 34[/hide]

Ok thanks I understand how to do 4 after some time. Basically you just set up two equations, the only tricky part about it is that algebraically the sons age 4 years ago would be -1, if I avoided that mistake and considered it to be 0 then I could have probably gotten it right.
But can anyone post their solution to problem 5?

Problem 5.
[hide]Okay, let uphill = u and downhill = d.
Now, u/12 + d/30 = 180 minutes, and u/30 + d/12 = 170. Solve for u and d, and you get u = 80/3 and d = 70/3. Add these up, and you have 150/3 = 50 miles.[/hide]

so it took 180 minutes from A-B and 170 from B-A so 5 minutes a mile uphill and 2 minutes a mile downhill. if x is time spent uphill and y is time spent downhill,

5x+2y=180
they are switched on way back
2x+5y=170
multiply second by 2.5
 5x+12.5y=425
-5x+2y=180
10.5y=245
y=23 1/3
x=26 2/3
so if 5 minutes a mile uphill or x, they did 5 1/3 miles
if 2 minutes a mile downhill or y, they did 11 2/3 miles
17 miles in all.

@ulvidad, are you sure those are the solutions to those equations?

@Ihatepie, something's wrong with your answer. Try plugging in your x and y.


trigonometry
geometry
geometric transformation
reflection
trapezoid

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$?
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=402[/img]

Let A be any fixed point on the circle (O) and AD a chord of the circle (O) with radius R = 5. The locus of midpoints N of the chords AD is a circle (P) with the diameter AD, i.e., with radius $r = PO = PA = \frac R 2 = 2.5$. Generally, the circle (P) can intersect the chord BC at 2 points, touch it at one point, or it may not have a common point with the chord BC. By the problem condition, the circle (P) touches the chord BC at one point N. Let M be the midpoint of the chord BC, $BM = \frac{BC}{2} = 3$. From the right angle triangle $\triangle OMB$,

$OM = \sqrt{OB^2 - BM^2} = \sqrt{5^2 - 3^2} = 4,\ \ \ \tan \widehat{BOM} = \frac{BM}{OM} = \frac 3 4$

Since the circle (P) passing through the point O is tangentb to the line BC, at the distance OM from the point O,

$OM = PN + PO \cos \widehat{AOM} = r(1 + \cos \widehat{AOM})$

$\cos \widehat{AOM} = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$

$\tan \widehat {AOM} = \frac{\sqrt{1 - \cos^2 \widehat{AOM}}}{\cos \widehat{AOM}} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Finally,

$\angle AOB = \angle AOM - \angle BOM$

$\tan \widehat{AOM} = \frac{\tan \widehat{AOM} - \tan \widehat{BOM}}{1 + \tan \widehat{AOM} \cdot \tan \widehat{AOM}} = \frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$

$\sin \widehat{AOM} = \frac{\tan \widehat{AOM}}{\sqrt{1 + \tan^2 \widehat{AOM}}} = \frac{7}{25}$

I was repeatedly asked in PMs about solution of this problem and I would do it differently now:

Suppose $ A$ is on the minor arc $ BC$ and $ D$ on the major arc $ BC$ of the circle $ (O).$ Since the midpoint $ AD$ is on a circle $ (P)$ with diameter $ AO$ and this circle would then cut the chord $ BC$ at 2 points $ N, N',$ there would be 2 chords $ AD, AD'$ bisected by $ BC$ at $ N, N',$ contrary to the problem condition. So $ A$ is on the major arc $ BC.$

Again, the midpoint of $ AD$ is on a circle $ (P)$ with diameter $ AO$ and this circle must be tangent to $ BC$ at $ N,$ otherwise there would be either 2 chords $ AD, AD'$ bisected by $ BC$ (if $ (P)$ cut $ BC$ at 2 points $ N, N'$) or no chord $ AD$ bisected by $ BC$ (if $ (P)$ had no common point with $ BC$). Since $ (P)$ is tangent to $ (O)$ at $ A,$ which is their external similarity center, tangent of $ (O)$ at $ D$ is parallel to tangent $ BC$ of $ (P)$ at $ N.$ $ D$ is therefore midpoint of the minor arc $ BC$ of $ (O).$ Then $ O$ cuts $ BC$ at its midpoint $ M.$ Power of $ M$ to $ (O)$ is

$ MB^2 \equal{} \overline{MD} \cdot (2R \minus{} \overline{MD})$ $ \Longrightarrow$ $ MD^2 \minus{} 2R \cdot \overline{MD} \plus{} MB^2 \equal{} 0.$

Putting $ R \equal{} 5$ and $ \overline{MB} \equal{} 3$ yields $ \overline{MD} \equal{} 1$ (the smaller root). Let $ E$ be reflection of $ A$ in $ OM$ and let $ K \in OM$ be the midpoint of $ AE.$ Since $ AD \equal{} 2 \cdot ND,$ $ KD \equal{} 2 \cdot MD \equal{} 2.$ Power of $ K$ to $ (O)$ is $ KA^2 \equal{} \overline{KD} \cdot (2R \minus{} \overline{KD}) \equal{} 16$ and $ \overline{KA} \equal{} 4.$

$ ABCE$ is cyclic isosceles trapezoid with parallel bases $ AE \equal{} 8, BC \equal{} 6$ and height $ KM \equal{} MD \equal{} 1.$ If parallel to $ CE$ through $ B$ cuts $ AE$ at $ F,$ then the $ \triangle ABF$ is isosceles with the base $ AF \equal{} AE \minus{} BC \equal{} 2$ and the B-altitude equal to $ KM \equal{} 1,$ therefore the angle $ \angle ABF$ is right, the angle $ \angle CEA \equal{} \angle BFA \equal{} 45^\circ,$ and the angle $ \angle COA \equal{} 2 \angle CEA \equal{} 90^\circ$ is also right. By cosine theorem for the $ \triangle OBC,$

$ \sin \widehat{BOA} \equal{} \cos \widehat{COB} \equal{} \frac {OC^2 \plus{} OB^2 \minus{} BC^2}{2 \cdot OB \cdot OC} \equal{} \frac {2R^2 \minus{} BC^2}{2R^2} \equal{} \frac {7}{25}.$


given $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$
prove that $ abc\plus{}bcd\plus{}cde\plus{}dea\plus{}eab\le5$
where a,b,c,d,e are nonnegative reals

[quote="AbsolutZer0"]given $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 5$
prove that $ abc \plus{} bcd \plus{} cde \plus{} dea \plus{} eab\le5$
where a,b,c,d,e are nonnegative reals[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112980


Let $a,b,c,d,e$ be five real numbers so that the quadratic equation $ax^{2}+(b+c)x+(d+e)=0$ has two roots that are greater than $1$. Prove that the equation $ax^{4}+bx^{3}+cx^{2}+dx+e=0$ has at last one real root.

Interestingly, this appeared already in a Vietnam TST (probably not for IMO though).

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=815937#815937

Can you post other problems from the TST as well?

[quote="Gabriel(fr)"]$T(x)=ax^{4}+bx^{3}+cx^{2}+dx+e=ax^{4}+(b+c)x^{2}+(e+d)+bx^{3}-bx^{2}+dx-d$ then

$T(x)= P(x)^{2}+(x-1)(bx^{2}+d)$. Now let assume that $T(x)$ has no real root, it means that $T(x)$ is always positive or always negative for all real x, WLOG assume that $T(x)>0$ for all real x and $a>0$

Now let $k$ be the root of $T(x)$ consequently if we set $u=+\sqrt{y}$ we get

$T(u) = P(u^{2})+(u-1)(bu^{2}+d)=(u-1)(bu^{2}+d)$ and $T(-u)=-(u-1)(bu^{2}+d)$ but $T(u)$ and $T(-u)$ cannot be positive at the same time yelding a contradition[/quote]

Thank you for your sulution. Bay the way if you read the linked topic in the post #2 you can find it there :wink: .


It's bigger, and has more problems and opportunities. It's also a lot better because they have more rounds and FOOD too. I also think that MATHCOUNTS is better because there are more rounds of difficulty-like chapter, state, nats,...

[quote="styrofoam1994"]See subject description..
How does Mathcounts compare to different middle school mathematics competitions?[/quote]
Mathcounts is a national competition, so in standards, more people like it.  :D
Also, there aren't much other middle school math competitions, other then math olympiad, so mathcounts tends to stand out more then the others. Unless your in MAO though. But Florida tends to focus on MAO a lot more then anything else. For reasons I don't know why.

The only other math competition I've ever run across was Math Derby, but that was elementary school.  Still, I noticed that it wasn't as organized as MathCounts, and there also weren't very many levels of competition.

There's always the AMCs.
Sure, the only "middle school" level is AMC 8, many people still take the 10s and 12s.

Mathcounts is the most nationally accepted competition, with the most organization and acceptance.  So, it is probably the most acclaimed middle school math competition.

Ehh I'm a bit confused by the question but my idea is just make it as important as you want it to be...

mamathbug

[quote="styrofoam1994"]See subject description..
How does Mathcounts compare to different middle school mathematics competitions?[/quote]

This is based on your own opinions, really

Mathcounts is only as important as you want it to be but many of us actually get addicted to it and understand it to be serious but fun and is a learning opportunity. (IF we had letter grades in Mathcounts, i'd probably be glad to make a B )
 :read:

Yes, I totally agree. It's a fun way to be better at math.

Well mathcounts isn't that important compared to high school competitions and the stuff you do doesn't help high school/college math comps that much so you shouldn't stress out too much if you don't do that well. A lot of pro people at my school didn't do mathcounts, and some of the people who did mathcounts aren't so pro.

in middle school, it is among the more major math competitions.

And why compare mathcounts to high school?  There's a pretty big difference in terms of the material.

MathCounts can impact your career very heavily. One of my mom's friend's son placed second in only our State competition but he got a full ride scholarship (along those lines) to CalTech  :wink: .

[quote="pianoforte"]methcounts is the meth (gateway) of math competitions

imo

at least it was for me[/quote]

I would more consider it the marijuana, but whatever.

Anyway, in terms of your actual problem solving skills in pure math, MathCounts can only bring you so far, but it does draw you to not just math, but things like engineering, physics, chemistry, etc. However, people get drawn to high school math without ever doing MathCounts and do just as well as people who have done MathCounts (I mean, the USA is the only country that has such a thing).

MathCounts is easily the most acclaimed middle school math competition. This can easily be seen from its wide array of corporate sponsors (Lockheed Martin, General Motors, Shell Oil, Raytheon, Texas Instruments, the list goes on). In fact, this reason can be used to argue that it is the most acclaimed math competition of any level, because most other math competions are sponsored by organizations such as MathCamnp, MathWorld, and other math competitions. I, however, don't believe this.

Compared to other competitions that are strictly on the middle-school level, MathCounts is probably both the most important and the most acclaimed, as noted above. However, note that even as middle-schoolers, other contests such as the AMC sequence of tests are open to you; indeed, it is not rare to have a middle-school MOPper. Hence, MathCounts is really not important at all and you should move on as soon as you can.


Let $0<a<b$. What is the minimum area of a triangle whose vertices lies on: $(a,b)$, x-axis, and the line $y=x$.

I get this.
[hide]
$\sqrt{2}\cdot \sqrt{a^{2}+b^{2}}$
[/hide]

Is this right?

[hide="Umm,"]
we can just draw any line from the point $(a,b)$, through the line $x=y$, and then through the x-axis. That has the three points, as specified, and area $\boxed{0}$.

Perhaps you meant something else? The [i]perimeter[/i], instead of the area?[/hide]

[quote="cincodemayo5590"]
Perhaps you meant something else? The [i]perimeter[/i], instead of the area?[/hide][/quote]
 :oops:
I am so sorry I mean the perimeter.

---
You can have figured that out because $\sqrt{a^{2}+b^{2}}$ is a linear quantity so it cant possible measure area.


MIT
college
real analysis
real analysis theorems

Where can I find good analysis exams (first year, metric spaces, continuity, ...) ?

Have you tried MIT open courseware?  Try this:

http://ocw.mit.edu/OcwWeb/Mathematics/18-100BFall-2006/Exams/index.htm


Write, if possible, $2005^{2004}$ as the sum of the squares of two natural numbers.

[quote="Maxbl"]What about $(2005^{1002})^2+0^2$?[/quote]

$0$ is not allowed...

$(\frac{4}{5}2005^{1002})^2+(\frac{3}{5}2005^{1002})^2$

Whatever those come out to be

EDIT: Got rid of fraction
EDIT2: Put it back for originality

[hide]$(3*401*2005^{1001})^2 + (4*401*2005^{1001})^2$[/hide]

EDIT: Darn :( Beat by the Zuton Force :)

How'd you do it Zuton? Pythagorean triple?


What is the greatest possible quotient of any two distinct members of the set $ \{\frac{2}{5}, \frac{1}{2},5,10\}$?

If we take

$ \frac{25}{\frac{2}{5}}$

we get a quotient of $ 25$.

Note: The reason why we did that is because to get a larger quotient, we take the largest number and have it divided by the smallest number.

[quote="JRav"]If we take

$ \frac {25}{\frac {2}{5}}$

we get a quotient of $ 25$.[/quote]

That should read $ \frac {10}{\frac {2}{5}}$, which reduces to 25.


[b]Can a sum of five squared primes be a squared prime? [/b]

From all calculations I've made it seems that it is not possible. How can I prove it?

Please do NOT answer for about 9 hours, it's from local competition.  :mad:


calculus
integration
derivative
linear algebra
matrix
vector
topology

HOW MUCH calculus must a student know to answer differential equations questions?  Is it enough to know calculus 1 and 2, for example?  How about calculus 3?

For the usual ordinary differential equations course, Calculus I and Calculus II should be sufficient, with Calculus II being the key prerequisite. One needs to be comfortable with some of the techniques of integration, and power series can play a prominent role. Maybe a partial derivative gets mentioned in discussions of the Existence-Uniqueness Theorem, but there's very little, if any, of the content of Calculus III needed for such a course.

An ordinary differential equations course does have an interface with linear algebra and matrix theory. Sometimes the necessary linear algebra is contained within the differential equations course, and sometimes it is brought in from the outside.

As for partial differential equations - one shouldn't touch those without a good grasp of the tools of vector calculus.

Okay....Thank you.  What is calculus 3 all about in simple terms?  What about linear algebra?  Is linear algebra the same as abstract algebra and classical algebra?  Is topology required in terms of earning a B.A. degree in math?

For an "Introduction to Linear Algebra" course, see my post on [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=320474#p320474]this thread.[/url] Such a course might be a co-requisite with a differential equations course. There's a lot more to linear algebra than that, so there is room for second and third courses.

Linear algebra is concerned with vector spaces, linear operators, and matrices. Abstract algebra mostly concers other algebraic structures, including groups, rings, and fields.

Calculus 3 (in typical American semester usage) concerns functions defined on $\mathbb{R}^2$ and $\mathbb{R}^3$ and functions which take values in $\mathbb{R}^2$ and $\mathbb{R}^3.$ Double and triple integrals, line and surface integrals. The high point of the course is the higher-dimensional generalizations of the Fundamental Theorem of Calculus: the theorems of Green, Gauss, and Stokes.

I certainly can't speak for every university, but our bachelor's degree in mathematics does not require a topology course. However, our "Real Analysis" sequence does contain material on the usual topology of $\mathbb{R}$ and on  general metric spaces.

Thank for your great answers.  I now know what awaits me in the near future.


function
LaTeX
induction
number theory
relatively prime
number theory solved

Find $f: (1,2,3,...) \to (0,1,2,...)$ such that :
 i) $f(mn)=f(m)+f(n)$
 ii) $f(30)=0$
 iii) $f(n) = 0$ if $n \equiv 7 [10]$

  Sorry , this problem is editted

if by $n \equiv 0 [10]$ you mean modulo 10 then it is easy to get $f(2)=...=f(2^n)$ and $f(5)=...=f(5^n)$ fo r all $n$. From $f(2)+f(5)=0$ and  $f(2 \cdot 2)=2f(2)$ we get that $f(2)=f(5)=0$ and from this we get that $0=f(10 \cdot n)=f(n)$ for all $n$

Sorry , $n \equiv 7 [10]$ . When I posted this probelm , there was some problems with Latex  ;)

For any number $m$ relatively prime to $10$, there exists $n$ such that $mn\equiv7\mod{10}$, and hence $0=f(mn)=f(m)+f(n)$, and hence $f(m)=0$.

But similarly, $0=f(30)=f(2)+f(15)$, which gives $f(2)=0$, and $0=f(30)=f(5)+f(6)$, which gives $f(5)=0$.  An easy induction on $a$ and $b$ establishes that $f(2^a5^bc)=0$, where $(c,10)=1$ - i.e. $f(n)=0$ for all $n$.


Define $p(x) = 4x^{3}-2x^{2}-15x+9, q(x) = 12x^{3}+6x^{2}-7x+1$. Show that each polynomial has just three distinct real roots. Let $A$ be the largest root of $p(x)$ and $B$ the largest root of $q(x)$. Show that $A^{2}+3 B^{2}= 4$.

We need to prove that if $A$ is the largest root of $p$ that $B=\frac{\sqrt{3(4-A^{2})}}{3}$ is the root $q$ and that it's the largest one.  As $q(B)=\ldots=(3-\frac{4A^{2}}{3})\sqrt{3(4-A^{2})}+9-2A^{2}$ and as $9-2A^{2}>0$ and $3-\frac{4A^{2}}{3}$ (this from the fact that $A$ is the greatest root of $p$), we only need to show that $(3-\frac{4A^{2}}{3})^{2}3(4-A^{2})=(9-2A^{2})^{2}$. This is equivalent to $16A^{4}-124A^{3}+261A^{2}-81=0$ and that follows from $(4A^{3}-25A)^{2}=(2A^{2}-9)^{2}$ (since $A$ is the root of $p$). Now since the largest root of $q$ ia in the interval $(1/3,1)$ we only need to prove that $\frac{\sqrt{3(4-A^{2})}}{3}$ is in that interval, which is easy.


Simplify:1-csc^2theta/cot^2theta

(A)-1
(B)1
(C)tan^2theta
(D)1/sin^4theta

The answer is choice A.

A friend said that 1 - csc^2 (theta) = [(sin^2x-1)/sin^2x].

My question is how does 1 - scs ^2 (theta) become

[(sin^2x-1)/sin^2x]?

Re-write the numerator as $ \frac{\sin^{2}{\theta}\minus{}1}{\sin^{2}{\theta}}$.  Dividing by $ \cot^{2}{\theta}$ is the same as multiplying by $ \frac{\sin^{2}{\theta}}{\cos^{2}{\theta}}$ so combining the two we get $ \frac{\sin^{2}{\theta}\minus{}1}{\cos^{2}{\theta}}\equal{}\tan^{2}{\theta}\minus{}\sec^{2}{\theta}\equal{}\tan^{2}{\theta}\minus{}(1\plus{}\tan^{2}{\theta})\equal{}\minus{}1$.


The Fibonacci Sequence $ 1,1,2,3,5,8,13,21,\ldots$ starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci Sequence?

$ \textbf{(A)}\  0 \qquad \textbf{(B)}\  4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\  7 \qquad \textbf{(E)}\ 9$

[quote="4everwise"]The Fibonacci Sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?

$\text {(A)} 0 \qquad \text {(B)} 4 \qquad \text {(C)}6 \qquad \text {(D)} 7 \qquad \text {(E)}9$[/quote]

[hide](C) 6

is there any other way than by randomly listing?  :?  :roll:  [/hide]

[hide]You could just list and add last digits, like this (but that's probably what you did):

1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,1,7,8,5...

So far, 1,2,3,4,5,7,8,9,0 have been listed, so the answer is 6.[/hide]

Sorry for reviving, but I don't really understand the problem.

...so we use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to represent our numbers. There are 10 of them. The question asks for which of these is the LAST to appear in the units digit position if we count through the Fibonacci Sequence. We start from 1, 1, 2, 3, 5...and just add units digits to count up till 6, as robinhe explained.

I guess confusion occurs at "ten digits" ("ten" in this case a numerical value) instead of "tens digits" (the place value).

Oh, the problem was badly worded. The ten digits confused me.

[quote=jhu08][hide=Solution]The easiest way for me to approach this problem was by listing out the terms, but only keeping the units digit. This is because the problem asks us to only find the units position, basically taking $mod$ $10$ of each term. Doing so, you can get a sequence of $1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0$. There is no need to continue the sequence, since we have all numbers other than $6$, so our answer is choice [b][u]C[/b][/u][/hide][/quote]

1) Nice necro  :D

2) That's probably what I would do as well

3) How would you solve the problem if the sequence required you to go much more (you would need to go to the $x$th term to find the answer using your method, and $x$ is very large)?

[hide=Solution]Listing gives $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610$. Out of these we can see only $6 \implies \boxed{\text{(C)}}$ does not appear.[/hide]


geometry
inradius
number theory proposed
number theory

The lengths of the altitudes of a triangle are positive integers, and the inradius is a prime number.

Find the lengths of the sides of the triangle.

What is RSA? I think I saw this problem in romanian team selection test and the solution was really horrible. It isn't difficult, but it's really tedious.

RSA = Republic of South Africa (my second home country :))


[quote="Syntax Error"]what is the greatest integer n, where 2^n divides 16!/125[/quote]

oops, i meant where 16!/125 divides 2^n

now when i think about it, im not sure. either way, do that. i got this from a mathcounts thingie we had, but i cant find the paper, so im tryin to do this from memory. sigh.

are you trying to say: 'What is the greatest integer N such that 2^N is a divisor of 16!/125?'?

If you say The converse of what you wrote:What is the greatest integer N such that 16!/125 is a divisor of 2^N?

No solution for this one, since 2^N can't divide 3 and get an integral quotient, so I think you wrote it correctly. MathFiend missed some though...[hide]All the numbers which you add should be the multiple of 2[/hide], can you figure out why? (This might be two big of a hint...)

ugh... the original question stated...

What is the greatest value of n such that 2^n wll evenly divide 16!/125...

its not right

please explain like triple m said

MathFiend wrote:[hide]14[/hide]

Very close...I think you missed [hide]1[/hide], can someone second that?

Yeah. Though it may not be 'missing' [hide]one[/hide], it probably is but you really have to see the working to find out..

[hide]my bad...I did miss one.

16!/125 simplifies to

(16)(3)(14)(13)(12)(11)(2)(9)(8)(7)(6)(1)(4)(3)(2)(1)

  4          1           2         1       3       1       2       1

add those up and get 15. I forgot a two in my first explabnation which caused me to forget it. OOPS! Well, I'm pretty sure the answer's fifteen now. [/hide]

[quote=HamstPan38825]It doesn't really matter about the divide by 125 part as I see it, since it doesn't cancel any 2's...[/quote]

these users are at least adults by now......

MathCounts challenges on State back then are the same level as high School by now.

8+4+2+1=15. wow maathcounts was easy back then.