[quote="Vasc"]If $a,b,c$ are positive numbers such that $abc=1$, then
$\sqrt[3]{\frac a{b+26}}+\sqrt[3]{\frac b{c+26}}+\sqrt[3]{\frac c{a+26}}\geq 1.$[/quote]
Let $a=\frac{x^{2}}{yz},$ $b=\frac{y^{2}}{xz}$ and $c=\frac{z^{2}}{xy},$ where $x,$ $y$ and $z$ are positive and Jensen. :)

[quote="Vasc"]Arqady, does Jensen way works also for the inequality
$\sqrt[3]{\frac{(1+p)a}{b+p}}+\sqrt[3]{\frac{(1+p)b}{c+p}}+\sqrt[3]{\frac{1+p)c}{a+p}}\geq 3$
for any $p \geq 0$ ?[/quote]
No. For $0\leq p<\frac{13}{16}$ it is not work:
For $p=0$ if $a=\frac{x^{2}}{yz}$... then $\sqrt[3]{\frac{(1+p)a}{b+p}}+\sqrt[3]{\frac{(1+p)b}{c+p}}+\sqrt[3]{\frac{1+p)c}{a+p}}\geq 3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\sqrt[3]{\frac{x^{3}}{y^{3}}}\geq3\Leftrightarrow\sum_{cyc}\frac{x}{\sqrt[3]{y^{3}}}\geq3.$
Let $x+y+z=1.$
Hence, $\sum_{cyc}\frac{x}{\sqrt[3]{y^{3}}}\geq\frac{1}{\sqrt[3]{\sum xy^{3}}}.$ And $\frac{1}{\sqrt[3]{\sum xy^{3}}}\geq3\Leftrightarrow$
$\Leftrightarrow(x+y+z)^{4}\geq27(x^{3}z+y^{3}x+z^{3}y),$ which wrong. :D

As Arqady shown, Jensen's way is valid for $p \geq 13/16$.
Do Tien, it seems that your method is valid only for $p=26$.
Thus, I still wait a solution for any $p \geq 0$. :(

Can you show your solution using Jensen arqady. I find that we must prove something like:
$(p+1)(x+y+z)^{4}\geq 27(x^{3}y+y^{3}z+z^{3}x+pxyz(x+y+z))$ and this one not valid for all $p \geq \frac{13}{16}$  :maybe:
I have a nice solution for the problem with $p \geq 0$, Vasc do you have a solution for it ??  :)

[quote="Anh Cuong"]Can you show your solution using Jensen arqady. I find that we must prove something like:
$(p+1)(x+y+z)^{4}\geq 27(x^{3}y+y^{3}z+z^{3}x+pxyz(x+y+z))$ and this one not valid for all $p \geq \frac{13}{16}$  :maybe:
[/quote]
You are right, Anh Cuong. It's true for all $p\geq\frac{473}{256}$
and for best $p=\frac{473}{256}$ equality holds  ( in particular case ) when $x=0,$ $y=3$ and $z=1.$

[quote="pvthuan"]Arqady, very nice result. Can you show me the solution for it?[/quote]
It's simply $y=x+u,$ $z=x+v$ work. ( $x=\min\{x,y,z\}$ )
$(p+1)(x+y+z)^{4}\geq 27(x^{3}y+y^{3}z+z^{3}x+pxyz(x+y+z))\Leftrightarrow$
$\Leftrightarrow A(u,v,p)x^{2}+B(u,v,p)x+(1+p)(u+v)^{4}-27u^{3}v\geq0.$
Enough to find the best $p$ such that
$(1+p)(u+v)^{4}\geq27u^{3}v$ holds for all $u\geq0$ and $v\geq0$ ($p_{best}=\frac{473}{256}$  :wink:  )
since, $A(u,v,p)\geq0$ and $B(u,v,p)\geq0$ for $p\geq\frac{473}{256}.$

[quote="arqady"] It's simply $y=x+u,$ $z=x+v$ work, suppose that  $x=\min\{x,y,z\}$.[/quote]

This technique prove useful for symmetric inequalities, however, it requires a lot of calculation. Do you use Maple, Arqady?

Can anyone show me more inequalities in this forum which can be solved using the same technique?

[quote="pvthuan"] Do you use Maple, Arqady?
[/quote]
No, I do it by myself.
[quote="pvthuan"]Can anyone show me more inequalities in this forum which can be solved using the same technique?[/quote]
See here my inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56040

[quote="Vasc"]Do Tien, it seems that your method is valid only for $p=26$.[/quote]
actually it's valid for all $p\geq 25$ :)

I believe that even this inequality holds:
$\sqrt[4]{\frac{a}{b+p}}+\sqrt[4]{\frac{b}{c+p}}+\sqrt[4]{\frac{c}{a+p}}\geq \frac{3}{\sqrt[4]{p+1}}$  :wink: , what do you think ?

[quote="Anh Cuong"]I believe that even this inequality holds:
$ \sqrt [4]{\frac {a}{b \plus{} p}} \plus{} \sqrt [4]{\frac {b}{c \plus{} p}} \plus{} \sqrt [4]{\frac {c}{a \plus{} p}}\geq \frac {3}{\sqrt [4]{p \plus{} 1}}$  :wink: , what do you think ?[/quote]
See my proof here: http://www.mathlinks.ro/Forum/viewtopic.php?t=139744


trigonometry
geometry
circumcircle
angle bisector

Suppose we have a triangle ABC, and the angle bisector of <CAB intersects BC at D, with E and F similarly defined. Express the lengths BD, CD, CE, AE, AF and BF in terms of the angles and circumradius of ABC in the nicest way you can find.

If $AB=c, BC=a, CA=b$, then by the property of an angle bisector

$BD=\frac{ac}{b+c}, CD=\frac{ab}{b+c}$

$CE=\frac{bc}{a+c}, AE=\frac{ab}{a+c}$

$AF=\frac{ac}{a+b}, BF=\frac{bc}{a+b}$

Then just plug $a=2R\sin\alpha, b=2R\sin\beta, c=2R\sin\gamma$.

Ooops, I think I interpreted the badly worded example sheet question wrongly. Let D be where the incircle meets BC, and E and F defined for CA and AB similarly. The rest of the problem is the same. Sorry.

Well in this case just use the fact that $AE=AF=s-a, BE=BD=s-b, CD=CF=s-c$ where $s=\frac{a+b+c}{2}$ then use $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$ and you are done.

[hide="For the edited problem"]If $a=BC, b=CA, c=AB$, then

$AE=AF=\frac{-a+b+c}{2}$

$BD=BF=\frac{a-b+c}{2}$

$CD=CE=\frac{a+b-c}{2}$

Then just plug $a=2R\sin\alpha, b=2R\sin\beta, c=2R\sin\gamma$[/hide]

Ah, thanks very much. My geom is rather appalling, but then again I am British, so it's tradition :roll:.


After a class has ended, for how long will the forum for that class still be open? Inefinitely?


At the beginning of a storm, a raindrop counter starts with a value of zero. Each time a raindrop hits the roof of a specific building, the counter increases by one. The raindrop tells how many drops of rain have hit the roof after a given amount of time. Does a raindrop counter describe a function? how?

It could, but we don't know what the function will be, unless we have more information about the storm.

Assuming time is the x-value, then it will be a function, because there is no x-value that has more than one corresponding y-value.

Some would argue that it is not a function, however, because the points do not form a continuous curve; i.e. the raindrops are static.

Yeah, it's a function.

And it doesn't matter that it's not continuous.

And you meant that the raindrops are [i]discrete[/i], not static  :P

My bad...I haven't even looked at this stuff in a while.


X = set of positive integer factors of 6.
Y = set of positive integer factors of 9.
Z = set of positive integer factors of 15.

X, Y, and Z represent three sets of numbers, as defined above. What is the set of numbers that belong to set Z but to neither set X nor set Y ?

a){1, 3}
b){1, 5}
c){5, 15}
d){2, 3, 5}
e){1, 3, 5, 15}

$ X\equal{}\left \{ 1,2,3,6 \right \}, Y\equal{}\left \{ 1,3,9 \right \}, Z\equal{}\left \{ 1,3,5,15 \right \}$
$ Z \setminus  (X\cup Y)\equal{}\boxed{\left \{ 5,15 \right \}}$


Let $ P(x) \equal{}\left(\cos{\pi\over 16}\plus{}x\sin{\pi\over 16}\right)^{4}$.

From

$ P(x) \equal{} (x^{2}\plus{}1)Q(x)\plus{}Ax\plus{}B$

by De Moivre we get

$ P(\minus{}i) \equal{}{\sqrt{2}\over 2}\minus{}i{\sqrt{2}\over 2}\equal{}\minus{}Ai\plus{}B$
$ P(i) \equal{}{\sqrt{2}\over 2}\plus{}i{\sqrt{2}\over 2}\equal{} Ai\plus{}B$

which obviously gives $ A \equal{} B \equal{}{\sqrt{2}\over 2}$

Hence the required remainder is $ R(x) \equal{} Ax\plus{}B \equal{}{x\plus{}1\over\sqrt{2}}$


Hello everybody ..
I search algorithm for one of below problem (only one):

1. R|pmtn|Cmax
2. 1|pj=1| Sum wiUi
3. O|zuet|Cmax
4. R||Sum Ci
5. P|rj,pmtn|Lmax
6. F||Cmax, n=2

Of course in deterministic model !!

It can be implementation in C, java or other language...

I'm very grateful for answeare.

Could you translate these problems into some human readable format, please? :D

I'm also not sure what you're asking... :?:


are there any irrationals a,b such that ab is rational?

Solution:
[color=white]consider x= sqrt 2. If xx is rational we are done. Else take a=xx and b=x. Obviously ab=2 which is rational. (from AMM :D) [/color]

Nice, easy and classical: take a=2 and g=log3/log2.

And I am the most stupid person alive. Sorry, take a=sqrt(2) and b=log3/logsqrt(2). We have a^b=3 and it's trivial to prove that a and b are irrational.

take a look at the solution I've hidden. :D (select the text under the word Solution: in my first post).

oh and please prove that log3/ log \sqrt 2 is not rational (it certainly looks like it, but it not that trivial :D).

Yes it is trivial
suppose a/b=log3/log \sqrt 2 where a and b are integers and b<>0. Then
 \sqrt 2 a/b=3 and so 2 a=3 2b, so a=b=0, but b was <>0.

err ... can't you see the beauty of the solution above? :D I give up ...

Sorry, I have just read it and it is absolutely fantastic! And now a question. Isn't x^x irrational? If it is, how do you prove that? I think this is not a trivial problem at all.

well that's the beauty of that solution. you just don't care if xx is irrational or not :D

proving that this number is Transcendent was on the list of 40 (I think they were 40) problems on Hilbert's list. It was solved.

of course that is much more then simpler irrationality.


[quote="Rust"][quote="e.lopes"]it seems very nice, but..

what about x = 5, y = 4, c = 8?  :maybe:[/quote]
$x=5,y=4,c=8$ give $55|71$  :lol:
Let $p|c^{2}-c-1$ prime divisor, then p is odd and $p|(2c-1)^{2}-5$. Therefore $(\frac{5}{p})\ge 0.$
Therefore Jakobi symbol $(\frac{5}{x})=(\frac{x}{5})\ge 0$ and $(\frac{5}{y^{2}-x})=(\frac{y^{2}-x}{5})\ge 0.$
If $c\not \equiv 3(mod 5)$, then $5\not |c^{2}-c-1$, therefore $x=\pm 1(mod 5), y^{2}-x=\pm 1(mod 5)$. It can be if only if $5|y$.[/quote]

rust, my friend, 8² - 8 - 1 = 55. i'm wrong?

Oh, yes.
And I prove $5|y$ if $c\not =3(mod \ 5)$.
If $c=3(mod \ 5)$ can be $5\not |xy$, for example $x=11,y=4,c=8.$

[quote="probability1.01"]Oops, my bad. I guess it should be $5 | xy^{3}-x^{2}y$ instead of $5 | y$.[/quote]
Yes, it is true.


Penni Precisely buys $ \$100$ worth of stock in each of three
companies: Alabama Almonds, Boston Beans, and California
Cauliflower. After one year, AA was up $ 20\%$, BB was down $ 25\%$, and
CC was unchanged. For the second year, AA was down $ 20\%$ from the
previous year, BB was up $ 25\%$ from the previous year, and CC was
unchanged. If $ A$, $ B$, and $ C$ are the final value of the stock,
then \\
\\
A. $ A\equal{}B\equal{}C$ \\
B. $ A\equal{}B<C$ \\
C. $ C<B\equal{}A$ \\
D. $ A<B<C$ \\
E. $ B<A<C$ \\ \\
Select the correct letter.

Let's look at AA first. It went up $ 20\%$, so it's now $ \$120$.

Now take a look at BB. It went down $ 25\%$, so it's now $ \$75$.

Finally, look at CC. It was unchanged, so it's still just $ \$100$.

Now, Year 2. AA went down $ 20\%$. It's value is now $ \$96$.

BB went up $ 25\%$. Now it's $ \$93.75$.

CC remains unchanged.

C has the highest value, A has the next highest, and B has the lowest.

$ C>A>B$ or $ B<A<C$

$ \boxed{\text{E}}$ is our choice.


[quote]Prove that \pi is equal to 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679 .... Mr. Green [/quote]

First of all, it's not (equal to whatever is behind the ..., unless you define the ... to be the decimal representation of pi, in which case it's obvious).  There are a bunch of methods to calculate pi, which btw is not a pre-olympiad problem.

Second of all, in a few minutes you'll probably be blaming this topic on the "hackers" again, eh?

it should not end it 79 i should be rounded its 798...

and its like 1+ 1/1+1/2+1/3+1/4...

[quote="mustafa"]Did those naughty hackers get onto your account again anirudh? :roll:[/quote]
From now on, I shall call him Mr. "Oh don't mind the spam it's the hacker's fault" Anirudh.

[quote="ragnarok23"]Have you not heard? It's not the hackers this time, it's the flying green monkeys that live under his bed![/quote]
Dang, those flying green monkey hackers.

I hear you can scare them with vinegar.

You guys are making very creative replies and all, but I think it'll be best if I lock this topic.  :D

You're welcome to talk about this pi = 3.14... stuff in F&G. However, when spam (like this) comes up in a math forum, please do not make spam replies to it.


Mark your calendar!

On [b]Monday, September 25[/b] at [b]7:30 PM Eastern / 4:30 PM Pacific[/b], we'll be hosting a Math Jam to discuss MIT, the MIT admissions process, and college admissions in general.

Our special guests from MIT will be:

[b]Matt McGann[/b], Associate Director of Admissions, and

[b]Kiran Kedlaya[/b], Assistant Professor of Mathematics.

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for the complete Math Jams schedule[/url]

Just to be precise... it's at 2330 GMT on Monday 25 September.

Reminder -- this Math Jam is coming up on Monday!

If you missed this Math Jam, the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=167]transcript[/url] is available.


I wrote this problem but I'm not exactly sure if there is an answer to this, at least an elegant method to it (I worked on this for a while and couldn't get nowhere).

Find number of integers $ n$ such that $ \frac {n!}{2^n}$ is an integer, for $ n \geq 1$.

Thank you.

 :)

[hide="Solution"]
For any integer $ n$ and any prime $ p$, the number of times $ p$ divides $ n!$ is

$ \mbox{ord}_p(n!) \equal{} \left\lfloor \frac{n}{p} \right\rfloor \plus{} \left\lfloor \frac{n}{p^2} \right\rfloor \plus{} \left\lfloor \frac{n}{p^3} \right\rfloor \plus{} \ldots$

This is simply because there are $ \lfloor n/p \rfloor$ multiples of $ p$ between 1 and $ n$, and then we need to add 1 for every multiple of $ p^2$ of which there are $ \lfloor n/p^2 \rfloor$, and then add 1 for every multiple of $ p^3$, etc. Note that the sum is really a finite sum since at some point $ p^k$ surpasses $ n$.

So $ n!$ is divisible by $ 2^m$ but not $ 2^{m\plus{}1}$ where
$ m \equal{} \left\lfloor \frac{n}{2} \right\rfloor \plus{} \left\lfloor \frac{n}{4} \right\rfloor \plus{} \left\lfloor \frac{n}{8} \right\rfloor \plus{} \ldots$

but this is always strictly less than
$ n \equal{} \frac{n}{2} \plus{} \frac{n}{4} \plus{} \frac{n}{8} \plus{} \ldots$

since this time this is genuinely an infinite series... Therefore $ n!/2^n$ is never an integer. The closest it gets is being half an integer, precisely when $ n$ is a power of 2.

[/hide]

The first problem of [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=181643]this blog post[/url] is a generalization. (for $ p \equal{} 2$ you can determine $ \alpha \equal{} 1$ is the only solution)

Thanks, I thought the answer was 0 (Usually, I can somewhat guess after trying out $ n \equal{} 32$).

But now, what do you mean by Ord? Is that some kind of function? I don't know what it is though.  :maybe:

[quote="randomgraph"]For any integer $ n$ and any prime $ p$, the number of times $ p$ divides $ n!$ is
$ \mbox{ord}_p(n!)$[/quote] This sentence defined the notation ord.

Here's another easy problem: show that $ \frac {(2n)!}{2^n}$ is an integer by giving a combinatorial argument, i.e., show there is are sets $ S_n$ such that the size of $ S_n$ is $ \frac {(2n)!}{2^n}$.  (The result that this is an integer follows easily from what randomgraph wrote (or from a much-simplified version of that argument), of course -- the challenge is to find a different kind of solution.)


[quote="uclabb"]what does this mean?

2005
SIGMA (E)  ((-1)^k(2005 choose k)k)
k=0

i know what 2005 choose k means, just not the sigma part.

thanks![/quote]

ie $\sum_{k=0}^{2005}$ , i think  :wink:


as you can see, there is a bug in the tex program generator here...  but thats not the main point of this post.  the question is still valid


How many three-digit numbers have the units digit larger than the tens digit?

Consider only the last two digits of the number. If the tens digit is 0 then there are 9 possible values for the ones. If the tens digit is 1 then 8 possible values for the ones digit and so on. So simply get the summation of 1 to 9 and multiply it to 9 (since the hundreds can't be 0).

Continuing the solution, we have:

$ 9 (9 \plus{} 8 \plus{} 7 \plus{} 6 \plus{} 5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 1) \equal{} 9 \cdot \frac {9 \cdot 10}{2} \equal{} 9 \cdot 45 \equal{} \boxed{405}$ numbers.

Another solution: When are the tens and ones digits the same? They can be any of the 10 digits, and the hundreds digit can be any of 9 digits, yielding 90 numbers. There are 900 total 3 digit numbers. Half of the numbers with the ones and tens digit different satisfy the problem.

$ \frac{900 \minus{} 90}{2} \equal{} \boxed{405}$


LaTeX
integration
logarithms
Online Math Open

[quote="DIMITRIS T"]ΠΡΟΤΕΙΝΩ ΝΑ ΔΗΜΗΟΥΡΓΗΣΟΥΜΕ ΕΝΑ ΝΕΟ ΤΟΡΙC  ΜΕ ΥΛΗ ΓΙΑ ΜΑΘΗΜΑΤΙΚΟΥΣ ΔΙΑΓΩΝΙΣΜΟΥΣ ΠΑΝΕΠΙΣΤΗΜΙΩΝ(ΠΧ SEEMUS)[/quote]

Πολύ ωραία!
 γράψτε λοιπόν σε αυτό το χώρο όλη τη σχετική αλληλογραφία και τις ασκήσεις που αφορούν αυτούς τους διαγωνισμούς.

 Καλή επιτυχία!
 Μπάμπης

Μια και αναφέρθηκε ο διαγωνισμός SEEMOUS (που παρεπιπτώντως του χρόνου θα τον διοργανωσει η Ελλάδα) θα θέσω το πρώτο πρόβλημα του φετινού διαγωνισμού που έγινε στην Κύπρο. Σε λίγες ημέρες θα στείλω την λύση αν δεν υπάρχει κάποια πρόταση.

[size=150][b][color=red]Πρόβλημα 1[/color][/b][/size]

Δίνεται $ a\in (0,1)\cap \mathbb{Q}$ και έστω $ a\equal{}0,a_1a_2a_3\ldots$ η δεκαδική αναπαράσταση του $ a$. Ορίζουμε $ f_a(x)\equal{}\displaystyle\sum_{n\equal{}1}^{\infty}a_nx^n, \ x\in (0,1)$.

Να δείξετε ότι η $ f_a$ είναι ρητή συνάρτηση.

Αντίστροφα, εαν $ a_k \in \{0,1,2,\ldots , 9\}$ για κάθε $ k\in \{1,2,\ldots \}$ και η $ f_a(x)\equal{}\displaystyle\sum_{n\equal{}1}^{\infty}a_nx^n, \ x\in (0,1)$ είναι ρητή συνάρτηση της μορφής $ f_a(x)\equal{}\displaystyle\frac{P(x)}{Q(x)}$, όπου $ P,Q$ είναι πολυώνυμα με ακέραιους συντελεστές, να δείξετε ότι ο αριθμός $ a\equal{}0,a_1a_2a_3...$ είναι ρητός.

Αλέξανδρος

Cretanman nomizw oti exw mia apodei3h gia to eu8u. to antistrofo den to exw skeftei kala akomh.

An o a einai aplos dekadikos(me peperasmeno plh8os pshfiwn meta thn upodiastolh) tote h askhsh einai profanhs gt an a=0.a1a2...am000000000... tote fa(x)=Σ(anx^n,n=1,m) dld pairnoume P(x)=fa(x) kai Q(x)=1

An o a den einai aplos dekadikos tote einai epanalamvanomenos me periodo m dld a=0.a1a2...ama1a2...am...
opote Σ(anx^n,n=1,oo)=a1x(1+x^m+x^2m+...)+a2x^2(1+x^m+x^2m+...)+...+amx^m(1+x^m+x^2m+...)=(1+x^m+x^2m+...)Σ(anx^n,n=1,m)=1/(-x^m+1)*Σ(anx^n,n=1,m)
opote pairnoume P(x)=Σ(anx^n,n=1,m) kai Q(x)=-x^m+1

diafwneis kapou mhpws?8eleis na e3hghsw kati perissotero?

Perimenw thn apanthsh s!

Filika,
Hlias

Vasika molis m hr8e mia idea k gia to antistrofo.

a=fa(1/10)=P(1/10)/Q(1/10) kai 0<1/10<1

afou P(x),Q(x) anhkoun sto Z[x] P(1/10),Q(1/10) anhkoun sto Q opote o logos tous anhkei sto Q.
to gegonos oti a=0.a1a2.. (kai oxi 3 komma kati gia paradeigma) prokuptei apo to gegonos oti 0<x<1 kai 0<=ai<=9

mia periptwsh p den kaluptetai apo to prwto post k m thn tonise o alexandros_plus einai oti h periodos mporei na arxizei meta apo k dekadika pshfia dld a=0.b1b2...bka1a2...ama1a2...am...
sthn periptwsh auth fa(x)=Σ(bnx^n,n=1,k) +Σ(anx^(k+n),n=1,00)=Σ(bnx^n,n=1,k) +x^kΣ(anx^n,n=1,00)=Σ(bnx^n,n=1,k)+Σ(anx^n,n=1,m)*x^k/(1-x^m)=((1-x^m)Σ(bnx^n,n=1,k)+Σ(anx^n,n=1,m)*x^k)/(1-x^m)

etsi P(x)=(1-x^m)Σ(bnx^n,n=1,k)+Σ(anx^n,n=1,m)*x^k kai Q(x)=(1-x^m)

Πολύ σωστή η λύση σου Ηλία!

Πράγματι το κλειδί είναι να παρατηρήσεις ότι εαν ο $ a$ δεν είναι ρητός με πεπερασμένα δεκαδικά, τότε θα είναι περιοδικός οπότε υπάρχει κάποιος $ n_0$ τέτοιος ώστε $ a_n \equal{} a_{n \plus{} k}$ για $ n > n_0$.

Πριν παραθέσω τη δεύτερη άσκηση να πω για τους ενδιαφερόμενους της Αθήνας ότι ξεκινά στο Πανεπιστήμιο Αθηνών ένα μάθημα με θέμα τις Ολυμπιάδες (κυρίως για φοιτητές) όπου διδάσκουν πολύ γνωστά ονόματα, με αφορμή των διαγωνισμό για τους φοιτητές (SEEMOUS) που θα γίνει του χρόνου στην Ελλάδα. Περισσότερες πληροφορίες καθώς και το πρόγραμμα των μαθημάτων μπορείτε να βρείτε στη σελίδα http://eclass.uoa.gr/courses/MATH206/  Έχει ξεκινήσει ήδη το μάθημα!

Κάτι αντίστοιχο θα ξεκινήσει και στο Πανεπιστήμιο Κρήτης πολύ σύντομα! Θα σας ενημερώσω σχετικά!

Προχωράμε στη δεύτερη άσκηση του περσινού διαγωνισμού

[color=red][b][size=150]ΠΡΟΒΛΗΜΑ 2[/size][/b][/color]

Έστω $ f(x) \equal{} \max_{i} |x_i|$ για το διάνυσμα-στήλη $ x \equal{} (x_1,\ldots,x_n)^{\top} \in\mathbb{R}^n$. Έστω ο  $ n\times n$ πίνακας $ A$ τέτοιος, ώστε $ f(Ax) \equal{} f(x)$ για κάθε $ x\in\mathbb{R}^n$. Να αποδείξετε ότι υπάρχει θετικός ακέραιος $ m$ τέτοιος ώστε $ A^m \equal{} I_n$, όπου $ I_n$ είναι ο μοναδιαίος $ n\times n$ πίνακας.

Αλέξανδρος

[quote="cretanman"]
[color=red][b][size=150]ΠΡΟΒΛΗΜΑ 2[/size][/b][/color]

Έστω $ f(x) \equal{} \max_{i} |x_i|$ για το διάνυσμα-στήλη $ x \equal{} (x_1,\ldots,x_n)^{\top} \in\mathbb{R}^n$. Έστω ο  $ n\times n$ πίνακας $ A$ τέτοιος, ώστε $ f(Ax) \equal{} f(x)$ για κάθε $ x\in\mathbb{R}^n$. Να αποδείξετε ότι υπάρχει θετικός ακέραιος $ m$ τέτοιος ώστε $ A^m \equal{} I_n$, όπου $ I_n$ είναι ο μοναδιαίος $ n\times n$ πίνακας.

Αλέξανδρος[/quote]

Όμορφο :) Για να δούμε...

Προφανώς, η $ f$ αναπαριστά τη μετρική $ \|x\|_\infty \equal{} \max_i\{\vert x_i\vert\}$ του $ \mathbb{R}^n$ και της οποίας η μοναδιαία σφαίρα $ \| x \|_\infty \equal{} 1$ είναι το σύνορο του n-διάστατου κύβου $ C\equal{}[\minus{}1,1]^n$. Υπό αυτή τη μετρική, ο $ A$ είναι μία γραμμική ισομετρία του $ \mathbb{R}^n$ και άρα ανήκει στην ομάδα συμμετρίας του κύβου (o οποίος, προφανώς παραμένει αναλλοίωτος υπό τη δράση ισομετριών).

Αφού η ομάδα συμμετρίας του κύβου είναι πεπερασμένη, έπεται ότι κάθε στοιχείο της θα έχει πεπερασμένη τάξη, δηλ. υπάρχει κάποιος $ m\in\mathbb{N}$ ώστε $ A^m\equal{}I$ (εφόσον ο $ I$ είναι το ουδέτερο στοιχείο).

Πείτε μου αν αυτή η λύση δουλεύει, γιατί δεν έχω βάλει κάτω μολύβι και χαρτί για να τη σκεφτώ καλά :)

Cheerio,

Durandal 1707

Τον τελευταίο καιρό ασχολούμαι με θέματα που έχουν πέσει στον IMC και στον Putnam.Η άσκηση που ακολουθεί είναι η 3η, της 2ης μέρας του IMC,1994.Θα ανεβάσω και τη λύση μου που είναι διαφορετική από την επίσημη για να μου πείτε αν κάπου χάνει.Ελπίζω ότι θα βρίσκω χρόνο να ανεβάζω πού-και-πού κάποια άσκηση καθώς επίσης κι ότι θα καθίσω να δω το LaTex :blush: .

Και η λύση μου...(η επίσημη λύση στη σελίδα του διαγωνισμού http://www.imc-math.org/  )

Ωραίο πρόβλημα :)

Δεν καταλαβαίνω όμως τη λύση σου. Σε ένα σημείο (προτελευταία γραμμή) λες ότι αφού $ \int_a^b [\log(\ldots)]' \textrm{d}x \equal{} \int_a^b 1 \textrm{d}x$ έπεται ότι $ \log(\ldots)' \equal{} 1$ παντού στο $ (a,b)$. Πώς προκύπτει αυτό το συμπέρασμα? Βασικά, εκτός και αν υπάρχει κάποιο επιχείρημα που δεν έγραψες (ή κάτι άλλο προφανές που απλώς αυτή τη στιγμή είμαι εντελώς ηλίθιος και δεν το βλέπω), μου φαίνεται ότι εδώ κάνεις λάθος.

Εγώ σκέφτηκα την εξής λύση:

[hide]
Έστω $ w(x) \equal{} e^{\minus{}x}\sum_{k\equal{}0}^n f^{(k)}(x)$. Τότε, από την εκφώνηση, $ w(b) \equal{} w(a)$ και άρα, από θ. Rolle, υπάρχει $ \xi\in (a,b)$ ώστε $ w'(\xi) \equal{} 0$. Όμως, $ w'(x) \equal{} (f^{(n\plus{}1)}(x)\minus{}f(x))e^{\minus{}x}$ και, συνεπώς, $ f(\xi) \equal{} f^{(n\plus{}1)}(\xi)$.
[/hide]

Ummm, το IMC πώς το λύνει? Δυστυχώς, δεν κατάφερα να μπω στο site (ίσως του κάνουν maintenance?)...

Cheerio,

Durandal 1707

1ον,αυτό ήταν που ήθελα να ελέγξω  :blush: και 2ον αυτή που έκανες είναι και η επίσημη λύση

[hide]Να πάρει κι ήταν ωραία ιδέα :( [/hide]

Αυτό που έγραψα όπως την έλυνα ήταν όπως φαίνεται στο νέο attachment.Είναι ακόμα λάθος,έτσι;(Την έχω πατήσει κι άλλες φορές έτσι :blush: )

[hide]Επιπόλαια,το έγραψα.Μόλις κάθισα και το σκέφτηκα... :blush:  :blush:  :blush:[/hide]

Sorry, φοβάμαι πως ναι, είναι λάθος... το συμπέρασμα ότι έχεις ισότητα σε όλο το $ (a,b)$ παραείναι δυνατό και μπορείς εύκολα να βρεις κάποιο αντιπαράδειγμα  :(

Cheerio,

Durandal 1707

Too good to be true,όντως.Όπως έγραψα και παραπάνω,μόλις συνειδητοποίησα τι έγραψα,εεεε :blush:  :blush:  :blush:

[quote="Fischerman"]Οι φοιτητές που επιλέγονται για την Εθνική Ολυμπιακή Ομάδα με βάση τα αποτελέσματα του διαγωνισμού της 23ης Φεβρουαρίου 2008 είναι οι:

    * Γεωργακόπουλος Αλέξανδρος, Σχολή Εφαρμοσμένων Μαθηματικών και Φυσικών Επιστημών, ΕΜΠ, 2ο έτος
    * Μαστάκας Κωνσταντίνος, Σχολή Εφαρμοσμένων Μαθηματικών και Φυσικών Επιστημών, ΕΜΠ, 2ο έτος
    * Σάββας Μιχαήλ, Σχολή Ηλεκτρολόγων Μηχανικών και Μηχανικών Υπολογιστών ΕΜΠ, 1ο έτος
    * Τζάμος Χρήστος, Σχολή Ηλεκτρολόγων Μηχανικών και Μηχανικών Υπολογιστών ΕΜΠ
    * Χατζάκος Δημήτριος, Τμήμα Μαθηματικών ΕΚΠΑ, 2ο έτος
    * Ψύχας Κωνσταντίνος, Σχολή Ηλεκτρολόγων Μηχανικών και Μηχανικών Υπολογιστών ΕΜΠ, 2ο έτος

http://eclass.uoa.gr/modules/document/document.php?openDir=%2F%C4%E9%E1%E3%F9%ED%E9%F3%EC%FC%F2_23022008[/quote]

syxaritiria stous filous mou!!!syssomo to emp eee?ara dikaiologhmena lene oloi oti einai to koryfaio idryma ths xoras!

[quote="silouan"]To ερώτημα είναι αν το ίδρυμα κάνει τους φοιτητές ή οι φοιτητές το ίδρυμα  :wink:[/quote]
Μάλλον το δεύτερο. :maybe:

[quote="Durandal"]Ωραίο πρόβλημα :)

Δεν καταλαβαίνω όμως τη λύση σου. Σε ένα σημείο (προτελευταία γραμμή) λες ότι αφού $ \int_a^b [\log(\ldots)]' \textrm{d}x \equal{} \int_a^b 1 \textrm{d}x$ έπεται ότι $ \log(\ldots)' \equal{} 1$ παντού στο $ (a,b)$. Πώς προκύπτει αυτό το συμπέρασμα? Βασικά, εκτός και αν υπάρχει κάποιο επιχείρημα που δεν έγραψες (ή κάτι άλλο προφανές που απλώς αυτή τη στιγμή είμαι εντελώς ηλίθιος και δεν το βλέπω), μου φαίνεται ότι εδώ κάνεις λάθος.

Εγώ σκέφτηκα την εξής λύση:

[hide]
Έστω $ w(x) \equal{} e^{ \minus{} x}\sum_{k \equal{} 0}^n f^{(k)}(x)$. Τότε, από την εκφώνηση, $ w(b) \equal{} w(a)$ και άρα, από θ. Rolle, υπάρχει $ \xi\in (a,b)$ ώστε $ w'(\xi) \equal{} 0$. Όμως, $ w'(x) \equal{} (f^{(n \plus{} 1)}(x) \minus{} f(x))e^{ \minus{} x}$ και, συνεπώς, $ f(\xi) \equal{} f^{(n \plus{} 1)}(\xi)$.
[/hide]

Ummm, το IMC πώς το λύνει? Δυστυχώς, δεν κατάφερα να μπω στο site (ίσως του κάνουν maintenance?)...

Cheerio,

Durandal 1707[/quote]
Το IMC το λύνει όπως εσύ Παναγιώτη, πράγμα γνωστό από τα φροντιστηριακά σου χρόνια από τον καθηγητή που σου έκανε ανάλυση χμφ :P

[quote="r_boris"]
Το IMC το λύνει όπως εσύ Παναγιώτη, πράγμα γνωστό από τα φροντιστηριακά σου χρόνια από τον καθηγητή που σου έκανε ανάλυση χμφ :P[/quote]

Χμμμμμ... μάλλον που θα έπρεπε να πρόσεχα περισσότερο τον καθηγητή μου που μου έκανε ανάλυση λοιπόν αντί να ασχολούμαι με τις αιθέριες υπάρξεις των διπλανών θρανίων... ίσως έτσι θα είχα μάθει και τίποτα, eh?!  :P :P

Tsk, tsk, τι να πώ... μ'όποιον δάσκαλο καθήσεις...  :D

Παναγιώτης

[quote="Nick Rapanos"][quote="silouan"]To ερώτημα είναι αν το ίδρυμα κάνει τους φοιτητές ή οι φοιτητές το ίδρυμα  :wink:[/quote]

to thema omos einai oti oloi autoi oi foithtes einai sygentromenoi sto sygekrimeno idryma.... :P[/quote]

Από την άλλη, είναι πραγματικά κρίμα (από μαθηματικής απόψεως) το πόσο λίγοι από τους ταλαντούχους φοιτητές του ΕΜΠ αποφασίζουν τελικά να ασχοληθούν με καθαρά μαθηματική έρευνα...

Μακάρι η (σχετικά καινούρια) ΣΕΜΦΕ να αλλάξει την κατάσταση ;)

Cheerio,

Durandal 1707

Μήπως μπορεί κάποιος να ανεβάσει τα θέματα του SEEMOUS;

Το βρήκα στο site της Ελληνικής Μαθηματικής Εταιρείας

Παραθέτω το πρόβλημα 4 και τη λύση που βρήκα  :)

Έστω $ n$ ένας θετικός ακέραιος και μια συνάρτηση συνεχής $ f: \left[0,1\right] - > R$  τέτοια ώστε

$ \int^1_0 x^kf(x)\,dx = 1$ για κάθε $ k = 0,1,...,n - 1$  . Ndo $ \int^1_0 (f(x))^2\,dx \geq n^2$

Η λύση μου
Αναζητούμε πολυώνυμο (n-1) βαθμού έστω $ p(x) = a_{n - 1}x^{n - 1} + ... + a_0$ τέτοιο ώστε

$ \int^1_0 x^kp(x)\,dx = 1$  για κάθε $ k = 0,1,...,n - 1$ .

Οπότε καταλήγουμε στο σύστημα
$ \frac {a_{n - 1}}{n} + \frac {a_{n - 2}}{n - 1} + ... + \frac {a_0}{1} = 1$

$ \frac {a_{n - 1}}{n + 1} + \frac {a_{n - 2}}{n} + ... + \frac {a_0}{2} = 1$
...
...
...
$ \frac {a_{n - 1}}{2n - 1} + \frac {a_{n - 2}}{2n - 2} + ... + \frac {a_0}{n} = 1$

Το σύστημα αυτό πρφανώς έχει λύση . Θα αναζητήσουμε τώρα το άθροισμα $ a_1 + ... + a_{n - 1}$

Θεωρούμε το πολυώνυμο $ Q(x) = (\frac {a_1}{x + 1} + ... + \frac {a_{n - 1}}{x + n} - 1)(x + 1)(x + 2)...(x + n)$ (1)

Από το σύστημα έχουμε ότι $ Q(0) = Q(1) = ...Q(n - 1) = 0$ και το $ Q(x)$ είναι $ n$ βαθμού άρα πρεπει
$ Q(x) = c(x - 1)(x - 2)(x - 3)...(x - n + 1)$ (2)

Εξισώνοντας τις (1),(2) βρίσκουμε ότι $ c = - 1$ και επιπλέον (συγκρίνοντας τους συντελεστές του $ x^{n - 2}$ )

έχουμε ότι $ S = a_1 + a_2 + ... + a_{n - 1} - \frac {n(n + 1)}{2} = \frac {n(n - 1)}{2}$ άρα $ S = n^2$

Επίσης είναι $ \int^1_0 p(x)(f(x) - p(x))\,dx = 0$

Τώρα έχουμε $ \int^1_0 (f(x) - p(x))^2 \,dx \geq 0$ ή ισοδύναμα

$ \int^1_0 (f(x))^2\,dx \geq - \int^1_0 p(x)(f(x))\,dx + \int^1_0 p(x)f(x)\,dx = + \int^1_0 p(x)f(x)\,dx = S = n^2$

and QED  :D

Σας επισυνάπτω τα Θέματα του διαγωνισμού!
Μάθατε πώς τα πήγαν οι συμμετέχοντες της Ελληνικής ομάδος και γενικά οι έλληνες που συμμετείχαν?

Αλέξανδρος

Από ό,τι πήρε το αυτί μου,

όλη ομάδα πήρε χάλκινο εκτός από τον Τζάμο που πήρε αργυρό!

Συγχαρητήρια σε όλους και ιδιαίτερα στον φίλο μου Μιχαλάκη!  :first:  :clap:  :)

Υπήρχε ένα ασημένιο και 5 χάλκινα για την Ελληνική αποστολή.
Οι βάσεις για τα μετάλλια ήταν
>= 26 --> χρυσό
>= 21 --> αργυρό
>=  8  --> χάλκινο

και οι βαθμολογίες των Ελλήνων

Τζάμος, 22/40
Σάββας, 18
Μάστακας, 16
Γεωργακόπουλος, 13
Ψύχας, 11
Χατζάκος, 10


Why did the MathLinks Olympiad Forum get collapsed? I liked having it separate so that it would be easier to get to the College Playground.

Because we are about to make some modifications to the site\'s layout and this change was needed. Have a little patience ;-)

[i]Posted from a Sony Ericsson K700i[/i]

The network was lagging :) and it didn't show me that I already posted the message, so I though it was not posted ;)

I haven't yet implemented the admin tools on wap mode (ie edit and delete any post  :twisted: ).


[quote="Allnames"]Let $ x,y,x > 0,xyz \equal{} 1$.Prove that
$ \frac {1}{\sqrt [3] {14x^2 \plus{} 12x \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {14y^2 \plus{} 12y \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {14z^2 \plus{} 12z \plus{} 1}}\ge 1$[/quote]

 It 's nice and quite old , my friend :)

Let $ x \equal{} (\frac {ab}{c^2})^3;y \equal{} (\frac {bc}{a^2})^3;z \equal{} (\frac {ca}{b^2})^3$

$ LHS \equal{} \frac {a^4}{\sqrt [3] {14b^6c^6 \plus{} 12b^3c^3a^6 \plus{} a^{12}}} \plus{} \frac {b^4}{\sqrt [3] {14c^6a^6 \plus{} 12c^3a^3b^6 \plus{} b^{12}}} \plus{} \frac {c^4}{\sqrt [3] {14a^6b^6 \plus{} 12a^3b^3c^6 \plus{} c^{12}}}$

Follow Holder:

$ LHS ^3(14\sum a^6b^6 \plus{} 12a^3b^3c^3\sum a^3 \plus{} \sum a^{12}) \ge\ (a^3 \plus{} b^3 \plus{} c^3)^4$

Need prove: $ (a^3 \plus{} b^3 \plus{} c^3)^4 \ge\ 14\sum a^6b^6 \plus{} 12a^3b^3c^3\sum a^3 \plus{} \sum a^{12}$

$ \leftrightarrow \sum x^2y^2(x^4 \plus{} y^4) \ge\ 2\sum x^4y^4$ which is obvious :wink:

We have done! :)

[quote="Allnames"]Let $ x,y,x > 0,xyz \equal{} 1$.Prove that
$ \frac {1}{\sqrt [3] {14x^2 \plus{} 12x \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {14y^2 \plus{} 12y \plus{} 1}} \plus{} \frac {1}{\sqrt [3] {14z^2 \plus{} 12z \plus{} 1}}\ge 1$[/quote]

Another proof:
we have,
$ \sqrt [3] {14x^2 \plus{} 12x \plus{} 1} \le 2x^{\frac {20}{27}} \plus{} 1 \forall x > 0$
And we will move to prove: xyz=1
$ \sum \frac {1}{2x \plus{} 1} \ge 1$

It's Cauchy-Schwarz  :lol:


calculus
function
real analysis
real analysis theorems

Hi
Basically, I'm stuck on the question on the attachment. Could anyone help me out? Thanks!

Introduce a new function $v=u+r^{2}/6$. Since $\nabla^{2}v=0$ and $v$ is radial, it follows that $v=C_{1}+C_{2}/r$. The boundary conditions give two linear equations for $C_{1}$ and $C_{2}$.

[b]tractorboy[/b] waited $1\frac12$ years to post his first question, and it went unanswered... how sad.  :(


function
inequalities
topology
triangle inequality
real analysis
real analysis unsolved

Let Y be a subspace of a metric space X and x in X. We  define :
 $ dist(x,Y) \equal{} inf_Yd_X(x,y)$ . Is it true that given any  Y the function:

f : X to R = f(x)=dist(x,Y) is continouos?

Not only is it continuous, it's Lipschitz (and hence also uniformly continuous). Can you prove that $ |f(x)\minus{}f(y)|\le d_X(x,y)?$

Assume that your main tool is the triangle inequality - after all, that's the most important thing you know about an arbitrary metric.


I've already posted this problem in another forum, but then I realised that it was for this forum. Well, here's the problem:

Find all non-negative integrers $A$ and $B$ such that $(3*[2^A])+1=B^2$.

[hide="solution"]
Equation $3\cdot 2^a+1=b^2$  (where $a\ge 0$,$b\ge 0$) has solutions $(0,2),(3,5),(4,7)$.

[hide="explanation"]
[b]1.[/b] if $a=0$ then equation becomes $3+1=b^2$, hence $b=2$.

[b]2.[/b] if $a\ge 1$ then we have:
$3\cdot 2^a+1=b^2\iff 3\cdot 2^a=(b-1)(b+1)$.
LHS is even so must be RHS, numbers $b-1$ and $b+1$ have same parity, so we may set $b-1=2k$ for some $k\ge 1$.
Hence, $3\cdot 2^a=2k(2k+2)=2^2k(k+1)$ and we see that $a> 2$ so we rewrite equation as $3\cdot 2^{a-2}=k(k+1)$.
Since numbers $k$ and $k+1$ are coprime we have either $k=3$ or $k+1=3$.
$k=3\implies b=7$ and $a=4$.
$k+1=3\implies k=2,b=5$ and $a=3$.
[/hide]

[/hide]


[quote="noneoftheabove"]Do you know anyone who lives in New Orleans? I feel very sorry for them especially because they lost everything although some are at our school[/quote]Yeah some of them are at my school too (in Austin, Texas).  Today in my P.E. class there were two people from New Orleans.

I lived 20 minutes away from New Orleans for 9 years. :( Then I moved to California.

in my school i am going to organize a hurricane katrina relief fund.  We send food, supplies, school items, and lots of other things to the victims in New Orleans.

school items???  theyre problably not having school for a couple of moths...

thats very generous math92

Maybe some will get sent to an operating school out of their area, that would work.

Our high school is having a donation drive where they're asking everyone to donate a dollar, we're hoping to raise $2500.

The worst part of the news reports was that I went to New Orleans with my dad before, and there was this "town central" place that had all this lively stuff. (people playing jazz, people dancing, flags waving). And they showed on the news a picture of it all gray and flooded and it just made me SO sad...  :(

[quote="jli"]my dad has this curse sort of thing because we went on vacation to the WTC and 1 month later, they fell down

then my dad visited New Orleans a month agoa and now its all flooded[/quote]Don't tell your dad where Bucharest is then! :D

PS I am sure New Orleans will be rebuild and all the jazz and lively stuff will be back in some time ;)

YAY!  They have fixed some of the levies and are finally starting to pump out the water :) We helped our neighbors with a bake sale for hurricane katrina, and it has been covered on the media, sometimes it is on 20 different chanels from 0-50, including ESPN sometimes.  Very covered.

Yeah we helped answer phone calls for help at the astrodome (close to where we live)


calculus AMC AIME USA(J)MO USAMO induction email \/closed

Has anyone done an Independent Study that has counted for school credit so far?

I want do an Independent Study for one of my classes next year, but I have a few questions:

1. How do I convince the school that this is a legitimite program? I am sure that they will ask something like that. This might not be a problem at a school like AAST where so many people do math competitions, but my school is bound to give me a hard time about this because I am the only one.

2. How advanced will the course be considered. This is going to sound really bad, but I am going to say it anyway: I might be valedictorian or something at my school and my school weighs classes according to whether they are college preparatory, honors, or advanced placement. I am sure that the content here is more difficult than say AP Calculus, but my school might disagree. The only classes they'll consider in the tier with advanced placement is other classes at local colleges.

3. Will we receive a grade for this class? I am going to need a grade for this class (my school goes from A+ = 96.5+ to F), but how will you assess us? I would fail miserably if I had to take a test that consisted of 15 to 20 AIME or USAMO questions.

I think we would all fail miserably if we had a bunch of AIME/USAMO problems as a test :-)  I got 11 on the AIME and won something, and that's a C!  The median USAMO score was less than 10%!

I dont think that independent study has been offered before.  I'm not sure how you can convince your school.  That is always very hard but if you have a math team coach let them know about this and he/she should help you convince the people needed to let you take this program.  From what I understand it will be pretty advanced.  It really depends on how much you know so I can't tell you for sure.  Have you looked at the description?  It should tell you a lot.  That s all I can say. Mr. Crawford or Mr. Rusczyk could probably help you a lot more though.

Mathnovice,

I will try to answer your questions as best I can.

This coming Fall semester will be the time students will be enrolled in our Independent Study program.  We do expect some students to discuss credit options with their schools, though it will be up to students to work that out with their school systems.  One possibility is to set aside a study period during school and recieve no credit if the school is unwilling to give GPA considerations similar to an AP class.

Question 1: Convincing your school that this program is a reasonable investment of your time might take bringing the website to their attention.  Seeing the level of math here might help convince them that (1) this program is harder than AP calculus and (2) this is a better place for a high school student than a college class that isn't typically designed for advanced high school students.  Hopefully they will recognize that you are such a highly motivated student and capable of recognizing the options that are best for you.  If you put your best foot forward and present yourself as a student that has researched the options and decided on the one that interests you most then it may go over well with an administration.  Administrations don't want to hold you back.  They just want to make sure that they understand the direction you are working in.

Question 2: If you let me know which level of independent study you are interested in taking I can put together enough materials that the head of your math department can recognize the program as more challenging than AP Calculus.  If you get your math teachers on your side then you are in a good position to convince your school to give you AP level credit toward your GPA.

Question 3: If your school is interested in a letter grade coming from us we can make that happen.  We understand the problem with relating this type of class to a typical 100 point scale.  On that scale over 95% of the students failed the USAMO when every student is clearly an A+ student.  We understand this difficulty.  Students who put forth the effort in class and on the problem sets can expect a grade that will absolutely not harm their GPA.  However, it should be noted that we can only provide the school with a grade for students who enroll in the evaluated class.  This is the version in which we grade problem sets and help guide students toward mathematical investigations they might not have explored for a problem idea and also guide development of rigorous communication skills (proof writing).  Again, if you can't get AP credit considerations, you can take the class with for no credit from your school.  This isn't something to be worried about in relation to transcripts because colleges will understand when you write on your resume that you took a class that covered ideas like induction, the pigeon hole priciple, inversions in a plane, and the Cauchy-Schwartz inequality.

Additionally, if administrators are interested in reviewing the AoPS program we are more than happy to take phone calls and emails concerning all our classes and independent study program.  We will do our best to convey what we believe: these courses are designed with highly motivated math students in mind.

If you have any other questions I am happy to answer them.

Another question:

My school's semester 1 doesn't end until the end of January. If I take the Independent Study for the first semester, what would I do for the rest of the two months?

Does the next Independent Study begin immediately after the first one?

What makes the independent study different from just taking two classes?

I also understand that there is assigned reading. Is the reading tailored to what the student has already read? For example, I have already read everything from Volume I and most of Volume II. Would I just reread this and redo the problems?

To answer both your questions:

Yes, the next independent study starts in January.  If you are not taking it, you will still be able to work on the Final Project during January.  We could also make some suggestions at the end of the independent study on what to do next, based on your work during the course.

There are a few additional features of the Independent Study above just the two classes.

First is the guided reading.  If you've finished all of Vol. 1 and Vol. 2 and completed all the problems, you should be looking at the Olympiad course, not the Intermediate.  We will also be augmenting the problems in the books by adding some from other sources.

Second, students in the independent study will have access to a total of around 30 hours of office hours scattered across the independent study duration.  You can use these to ask follow-up questions from the work in the courses, in the texts, or other work.

Third, Independent Study students will complete a final project, which will be an in-depth study of some problem or problems from the course.

[quote="Rep123max"]Will there be problem sets for the independent study?[/quote]

There will be challenging sets of exploratory problems along with problems from the assigned texts.  Don't worry, we'll keep you working.

is it a good sequence for one to take intermediate independent self-study in september, then connect into olympiad level in january( I hope there is one.)?
is amc 12 problem series in november a repeat of the current one, or a whole new problem selection?

I am wondering about a reasonable estimate for hours per week necessary to keep up with enrollment in evaluated independent study at the intermediate level, taking two of the easier classes as part of that enrollment. What's the bare minimum time commitment? What would be an adequate time commitment to ensure steady digestion of the material week by week?

The AMC-12 Problem Series in November will be the same as the current summer course.

As for the best course of study for independent study, it largely depends on where you are now and what your goals are.  We currently expect to have 6 levels of independent study - two each at the introductory, intermediate, and olympiad levels.  The ones we will offer this fall are the first installments of the intermediate and olympiad levels.  We expect to offer both installments of the intermediate and olympiad independent studies next spring, then add the introductory courses next fall.

Please let me know if you have any more questions - if you can give us more specific details about where you are now and your goals (by emailing us at classes@artofproblemsolving.com), we can provide more specific guidance.

[quote="Rep123max"]I just wanna make sure, if someone is positive that they will make AIME, but has a goal of making Olympiad, they should do Intermediate right?[/quote]

Exactly - that's the student the two intermediate independent studies are being designed for.

[quote="Rep123max"]I dont know if Ive asked this, but will the independent study cover a lot of geometry?[/quote]

There will be time spent on many areas, geometry included.  You can always focus on geometry by taking a geometry class.

Well, this seems the right place to post these questions, rather than start a new topic...

I am in a situation similar to the original poster of this thread. For the 2005-2006 school year, I would like to take 1 year's worth of Independent Study classes. But in order for my school to buy it, I have some questions. Alll these questions apply to next year's classes.

1. From what date to what date are the classes (Could not find this on the page where it talked about them)?
2. How will we be evaluated? Someone mentioned effort, and also a project, but what else/more specifically?
3. Will a Fall intermediate course prepare me for a Winter/Spring Olympiad course? I especially want to take Olympiad Geometry. I looked at the diagnostic questions and got 4 of them so far; I'm very close to the 5th...
4. When is a good time to sign up for these courses? Are there enrollment caps?

My qualifications: I think it almost garunteed (SP?)that I will make AIME, but I don't think I make USAMO. I think I will have a good shot at being on the Chicago 2nd-level ARML team this year (To avoid the usual B and C confusion), and a decent shot at Chicago A next year.

One last thing: I would like to thank the makers of this site for all their hard work and help. Since joining the USAMTS and coming to this site, I have learned so much by trying the sample problems already! I would like to let them know that their efforts are very much appreciated.

I will try to answer as thoroughly as possible:

1. We have not yet scheduled the actual dates of the Independent Study courses, but they will run for 14 weeks and will coincide with a normal school semester.  If your school needs exactly 18 weeks of work involved, you can rest assured that the material is challenging enough that there will be material to continue to work on after it is over.  Then again, exams cover the end of such a time period.

2. AoPS instructors review each set of homework and also the final project.  If you email me at crawford@artofproblemsolving.com I can talk with you about the grading system which is favorable to any student who gives a full effor to all the material.

3. Intermediate courses do lead right up to Olympiad courses, though we do not yet offer an Intermediate Geometry course.  Hopefully we will be able to offer one soon.

4. We are currently not capping the Independent Study program because it is not large enough for us to need to do so.  You can sign up for the Independent Study at any point after it is listed.  We typically accept enrollments up to 10 days after the beginning of a course or program.

It sounds like you are currently at a level for our intermediate classes, though you might consider taking an Intro class in any area where you feel you have a deficit relative to other areas.

We are gratified to know you are enjoying our site and that it is helping you improve your problem solving skills.

Feel free to ask any further questions.


[quote="kunny"]Let $ s,\ t,\ u$ be positive constants and $ a,\ b,\ c$ be positive numbers such that $ sa \plus{} tb \plus{} uc \equal{} 1$.
Prove that $ \frac {1}{(1 \minus{} sa)^2} \plus{} \frac {1}{(1 \minus{} tb)^2} \plus{} \frac {1}{(1 \minus{} uc)^2}\geq 5$.[/quote]

But its easy $ 1\minus{}sa\equal{}x,1\minus{}tb\equal{}y,1\minus{}uc\equal{}z$, so $ x\plus{}y\plus{}z\equal{}2$
but $ \sum\frac{1}{x^2}\ge \frac{27}{(x\plus{}y\plus{}z)^2}\equal{}\frac{27}{4}>5$


For every point on the plane, one of $ n$ colors are colored to it such that:

$ (1)$ Every color is used infinitely many times.

$ (2)$ There exists one line such that all points on this lines are colored exactly by one of two colors.

Find the least value of $ n$ such that there exist four concyclic points with pairwise distinct colors.

the second condition is not very clear... ¿each point have a monochromatic line, a bichromatic or there is only one of such lines?

Obviously, $n \ge 4$.
If $n = 4$, let the colors be labeled $1, 2, 3, 4$. Take a certain circle $\Gamma$ and take three points $A, B, C$ on $\Gamma$. Let all points on arc $AB$ (including $A$ but not $B$) be colored $1$, all points on arc $BC$ (including $B$ does not contain $C$) be colored $2$, all points on arc $CA$ (including $C$ but does not contain $A$) be colored $3$, and all other points on the plane be colored $4$. Then both conditions are satisfied, but the only circle containing points with colors $1, 2, 3$ doesn't contain a point with color $4$. Thus, $n \neq 4$, so $n \ge 5$.

If $n = 5$, let the colors be labeled $1, 2, 3, 4, 5$. From condition (2), there exists a line $\ell$ such that all points on it are of only two colors. Let us suppose that there are only points with colors $1, 2$ on line $\ell$, and from condition (1), there exist points $A, B, C \notin \ell$ with colors $3, 4, 5$ respectively. If $A,B,C$ are not collinear, let $\Gamma$ be the circle passing through points $A, B, C$.

If $\Gamma$ has a common point with $\ell$, then there are $4$ concyclic points of different colors, and we are done. If $\Gamma$ and $\ell$ do not intersect, but there is a point colored $1$ or $2$ on $\Gamma$, then there are $4$ concyclic points of different colors, and we are done.

If $\Gamma$ and $\ell$ do not intersect, and there are no points with colors $1$ or $2$ on $\Gamma$, i.e. all points on $\Gamma$ are colored $3, 4$ or $5$, then let $O$ be the center of $\Gamma$. Let $D$ be the point on $\ell$ such that $OD \perp \ell$. WLOG assume $D$ has color $1$. Let $E$ be a point on $\ell$ with color $2$. Let line $OD$ intersect $\Gamma$ at $P$ and $Q$, where $P$ is closer to $\ell$ than $Q$. Then let $QE$ intersect $\Gamma$ at $R$. WLOG assume $P$ has color $3$. If $R$ is not colored $3$ then since $\angle ERP = \angle EDP = 90^{\circ}$, $E,R,P,D$ are concyclic and we are done. So assume $R$ has color $3$. Now let $QB$ intersect $\ell$ at $F$. Note that $B$ has color $4$. If $F$ has color $2$ then since $\angle PBF = \angle PDF = 90^{\circ}$, $P,B,F,D$ are concyclic and we are done. If $F$ has color $1$, then since $QR \cdot QE = QP \cdot QD = QB \cdot QF$, $E,R,B,F$ are concyclic and we are done.

If all points colored $3,4,5$ lie on only one line, let that line be $\ell'$. Call a point on $\ell'$ [i]good[/i] if its color is $3,4$ or $5$. Let $X$ and $Y$ be points on $\ell$ (and not on $\ell'$) colored $1$ and $2$ respectively. Let $P_1$, $P_2$, and $P_3$ (in order from left to right) be good points on $\ell'$ with different colors. WLOG assume the colors of $P_1$, $P_2$, and $P_3$ to be $3$, $4$, and $5$ respectively.

Case 1: There exists a good point $P_4$ not colored $4$ either to the left of $P_1$ or to the right of $P_3$.
Then circles $(P_1P_3X)$ and $(P_2P_4Y)$ intersect and each have at least $3$ distinct colors. Assuming that $Y \notin (P_1P_3X)$ (if not, just pick a different option for $Y$), then an intersection point of those two circles is either colored $1$ or $2$. Either way, one of those two circles will have $4$ distinct colors, and we are done.

Case 2: All good points to the left of $P_1$ or to the right of $P_3$ (if there are any) are colored $4$.
Then from condition (1), there exist points $P_5$ and $P_6$, both between $P_1$ and $P_3$, whose colors are $3$ and $5$ respectively. If $P_5$ is to the left of $P_6$, then $(P_1P_6X)$ and $(P_3P_5Y)$ intersect and each have at least $3$ distinct colors. As shown above, this shows there is a circle with $4$ distinct colors. If $P_5$ is to the right of $P_6$, then we have the following $3$ subcases:

Case 2a: $P_2$ is to the left of $P_6$.
Then $(P_1P_6X)$ and $(P_2P_3Y)$ intersect and each have at least $3$ distinct colors. This shows there is a circle with $4$ distinct colors.

Case 2b: $P_2$ is between $P_6$ and $P_5$.
Then $(P_1P_2X)$ and $(P_5P_6Y)$ intersect and each have at least $3$ distinct colors. This shows there is a circle with $4$ distinct colors.

Case 2c: $P_2$ is to the right of $P_5$.
Then $(P_1P_2X)$ and $(P_3P_5Y)$ intersect and each have at least $3$ distinct colors. This shows there is a circle with $4$ distinct colors.

In all cases, we have shown that there is a circle with $4$ distinct colors for $n=5$, and we are done.


can $ S\equal{}\sum_{k\equal{}1,n}\frac{1}{\sqrt{k}}$ be a rational number?forn>1

[quote="Rust"]Let $ p_0$ is prime suth that $ \frac {n}{2} < p_0\le n$. Exist prime P>n, suth that $ \binom{p}{P} \equal{} 1 \ \forall p\le n, p\not \equal{} p_0$ and $ \binom{p_0}{P} \equal{} \minus{} 1$.
If $ \sum_{k \equal{} 1}^n \frac {1}{\sqrt k } \equal{} \frac {a}{b}$ is rational, then must be $ s \equal{} (\frac {a}{b} \minus{} x)^2 \equal{} \frac {1}{p_0}$ give contradition $ 1 \equal{} \binom{s}{P} \equal{} \binom{p_0}{P} \equal{} \minus{} 1$.[/quote]
It is not true .What is $ s$.
Note :It can be prove that this sum can not is not a integer .
If $ \sum_{i\equal{}1}^n\frac{1}{\sqrt{i}}$ is an algebraic number.

[quote="TTsphn"]It is not true .What is $ s$.
[/quote]
What is not true?
For all $ k\le n , k\not \equal{}p_0$ we define $ x_k\equal{}\frac{1}{\sqrt k}\mod P$, because $ \binom{k}{P}\equal{}1$.
Let \[ x\equal{}\sum_{k\equal{}1,k\not \equal{}p_0}^n x_k\].
If \[ r\equal{}\sum_{k\equal{}1}^n \frac{1}{\sqrt k }\] rational,
then $ s\equal{}(r\minus{}x)^2\mod P$.


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that $ x^{2^n}\plus{}1$ is irreducible in $ \mathbb{Q}[x]$ for $ n \in \mathbb{N}$

Make the substitution $ x\mapsto x\plus{}1$ and apply Eisenstein.


Guess what! With the idea of this problem you can find the easiest solution of one of Mathlinks Contest problems.


algorithm
induction
vector
combinatorics
IMO Shortlist

There are $ n \plus{} 1$ cells in a row labeled from $ 0$ to $ n$ and $ n \plus{} 1$ cards labeled from $ 0$ to $ n$. The cards are arbitrarily placed in the cells, one per cell. The objective is to get card $ i$ into cell $ i$ for each $ i$. The allowed move is to find the smallest $ h$ such that cell $ h$ has a card with a label $ k > h$, pick up that card, slide the cards in cells $ h \plus{} 1$, $ h \plus{} 2$, ... , $ k$ one cell to the left and to place card $ k$ in cell $ k$. Show that at most $ 2^n \minus{} 1$ moves are required to get every card into the correct cell and that there is a unique starting position which requires $ 2^n \minus{} 1$ moves. [For example, if $ n \equal{} 2$ and the initial position is 210, then we get 102, then 012, a total of 2 moves.]

I don't think I'll explain it very well, and I'm not sure it's correct, but I'll try :).

Let $t$ be the card on position $n$ at the beginning. Until we reach the situation when we have to move the card $n$, the game goes on just as if it were a game with cards $1,2,\ldots,n-1$ and positions $0,2,\ldots,n-1$ (i.e. until we have to move card $n$, we have no idea that it's card $n$ and not card $t$). Since such a game can have at most $2^{n-1}-1$ steps (from th induction hypothesis), and when we meet $n$ we need one more move to put it in its place, on position $n$, it means that in at most $2^{n-1}$ moves card $n$ is in its place. We repeat the reasoning: after that, we need at most $2^{n-2}$ moves to put card $n-1$ in its place, and so on, until $2,3,\ldots,n$ are in their places, and we need at most $1$ move to put card $1$ in its place (and then $0$ will be in its place automatically). In total, we need at most $1+2^1+2^2+\ldots+2^{n-1}=2^n-1$ moves.

As for the arrangement, using the argument above and induction, we get that it must be $1,2,\ldots,n,0$.

yes! the secound part is an induction....

Try the following for the first part: To each configuration asign an n tuple $a_2,a_3,...a_{n+1}$ such that $a_i=1$ iff $i$ is on the right place, and $0$ otherwise. Prove no 2 ntuples repeat during the proccess, and thats it.

[quote="Pascual2005"]
Try the following for the first part: To each configuration asign an n tuple $a_2,a_3,...a_{n+1}$ such that $a_i=1$ iff $i$ is on the right place, and $0$ otherwise. Prove no 2 ntuples repeat during the proccess, and thats it.[/quote]

First of all, one thing that confuses me is that we have no $n+1$, so what is $a_{n+1}$? Why did you number the sequence from $2$ to $n+1$? :?

well actually use the nuemration from $1$ to $n$, exclude the card with the $0$ because it is very predictible its behavior...then prove no 2 n tuples repeat.

You know, I did try some stuff with such binary representations (that $2^n$ suggests such an approach), but somehow, I didn't see this :). it looks like if you reverse the binary vector you have described, you get the binary representations of a  the elements in a strictly increasing sequence. Thanks! :)

P.S. I still think my solution is also correct though :)

It is :D  I was just proposing another aproach.. :lol:

Solution by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=3930]seshadri[/url]:

if $ 0$ appears in position $ 0$, then effectively the procedure works on the permutation from $ 1$ to $ n$ and so we can induct, so WLOG $ \pi(0)\neq 0$.
it suffices to show that if $ n$ is in position $ k$, then the number of steps it takes for $ n$ to move to position $ n$ is atmost $ 2^{k}$. (thus when we induct we do so on $ n \plus{} k$).
Suppose $ n$ is in position $ k$ and we are done in all smaller cases. if in fewer than $ 2^{k \minus{} 1}$ steps $ n$ moves forward, then we are done by induction. If this is not the case then in $ 2^{k \minus{} 1}$ steps all changes have occured only in positions $ 0$ through $ k \minus{} 1$. hence it follows that $ \pi(0),\pi(1),\ldots,\pi(k \minus{} 1)\in\{0,1,\ldots,k \minus{} 1\}$ and hence $ \pi|_{\{0,1,\ldots,k \minus{} 1\}}\in \mathcal{S}_k$. By induction we know that these can be sorted in atmost $ 2^{k} \minus{} 1$ steps and so $ n$ can be transferred to position $ n$ in atmost $ 2^k$ steps and that completes the induction.
clearly if the number of steps is to be maximum then again we can argue inductively that the corresponding permutation has to be $ 12\cdots(n \minus{} 1)n1$.


MATHCOUNTS
trigonometry
geometry
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Mingming Tran who beat me at Word Power was Chinese. I think it is genes, but I think it also has to do with the Chinese parents (at least a few of them - NOT ALL!) wanting their child to succeed and finding the opportunities.

Billy

I beleive a more politically correct term would be asian. Many taiwanese and japanese people get mad when people call them chinese.

I thought we [b]were[/b] talking about only the Chinese people?!

Ming-ming is Chinese.

Billy

That doesn't seem too surprising. The Chinese, and the Asian in general, are usually more academically-focused than Caucasians. Also, the Chinese who come to the US are usually the best of the best, so that's why a greater percentage of the Chinese are "smart." (Yeah, I realize that that fact doesn't pertain to the Mathcounts winners, but it's just a side note.)

Intelligence is 99% perspiration and 1% inspiration.  It is hard work that drives anyone to success.

America provides the best opportunities for all and better still equal opportunity for all.   The first generation Asian parents here are very highly academicaly oriented - they bring it from their country of origin.  Take India for example - it is an extremely competitive world out there.  Survival of the fittest is the norm.  Coaching centers open at 4:00 AM and run till 10:00 PM or so.  Admission to top education institutions is solely based on the test administered by the institution.  Parents and kids have their goals set before the kid is through 2nd or 3rd grade.

A Nat. MC girl from IA was saying that in Taiwan(or Korea?), a law was passed to close the coaching centers by midnight.  Otherwise they were running all through the night.

That focus is reducing with every generation though.  The second generation parents are still lucky in that their parents take care of their kids.

Conclusion - the hard working guy/gal with determination and focus is bound to rise to the top. :) The gene factor is found in all races and its contribution to success is a smaller fraction.  Let's not underestimate any race.

[quote="NoSoupForYou"]Have you noticed that the top two in the written rounds were both Eastern European, while the top two in Countdown were both Chinese?
Tells you something, doesn't it?[/quote]

Uhh... that the top two written came from East Europe and the top two countdown came from China?  :?

Penguin!s do well, but there aren't enough competing to dominate.  :(
We only had 1 Penguin! in the entire country. At least that Penguin! made nats and place 16th.

In general, my experience is that the top .1% of people in intelligence are Asians, but the top .001% about that are white.

Yeah, I kind of agree. Not the genes, just the fact that some Asian families seem to push more academically than anything else. (I'm asian)
Eastarn European too it seems.

Hmm...
So that leaves out the US...
I'd say.......Sport inclined???

(of course, there are exceptions to this)

Sergei David and Nathan are all related. 1, 2, 5 in written

Taiwanese is not Cantonese.  It's just another dialect (or some people might say it is a whole different language...)

As for doing well in Mathcounts, the main factors are:

Practice--you won't do well without having done many timed tests
Luck--yep, you need to bring your lucky... whatever it is
Test-taking skill--figure out how to most efficiently/accurately do problems (especially sprint round)
Test day--if you're dead on test day, all of the above was for naught

[quote="kingof21penguins"]NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I AM NOT CHINESE I AM TAIWANESE AND YEA THERE IS A DIFFERENCE I SPEAK TAIWANESE I DONT SPEAK CHINESE !![/quote]
Maybe you should learn to speak Chinese!!! ;)
Congratulations anyway that you did so well at the national Mathcounts.

ok...take out ur yrbook...count the asians...count the white ppl...count the black ppl...does that help answer ur questioN???

[quote="chess64"]I want to do homeschooling, but my dad doesn't want me to  :( . He says there's only a few months of school left anyway. (He said that a few months ago.) It's not fair  :(  :(  :( . He says that you don't get to socialize and stuff[/quote]No you don't want to be homeschooled. No friends?

[quote="Sly Si"]If 3 out of the top 4 is surprising, in 2000 all of the top 4 were Asian, the top non-asian student being Ricky Biggs in 5th.[/quote]

My year, 2003, I think 2 of the top 4 were non-asian... Adam Hesterberg and Brian Lawrence. That must be the highest percentage ever! ;) Adam won Written+Countdown and Brian won Masters, so us asians were left out that day :P

[quote]bob123
PostPosted: Fri May 13, 2005 3:24 pm    Post subject:
kingof21penguins wrote:
Quote
By looking at the national mathcounts final list (both team and individuals), there are 4 Chinese students at TX team, and 3 of top 4 winners (including both #1 and #2) are Chinese as well. (I asked my classmate who is Chinese to verify this for me). Any thoughts about that?

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I AM NOT CHINESE I AM TAIWANESE AND YEA THERE IS A DIFFERENCE I SPEAK TAIWANESE I DONT SPEAK CHINESE !!

Taiwanese as in Cantonese? I never really knew what people meant by Taiwanese...[/quote]
[quote]probability1.01
PostPosted: Fri May 13, 2005 3:32 pm    Post subject:
Taiwanese is not Cantonese. It's just another dialect (or some people might say it is a whole different language...) [/quote][quote]qvc122
PostPosted: Fri May 13, 2005 3:34 pm    Post subject:
kingof21penguins wrote:
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I AM NOT CHINESE I AM TAIWANESE AND YEA THERE IS A DIFFERENCE I SPEAK TAIWANESE I DONT SPEAK CHINESE !!

Maybe you should learn to speak Chinese!!! Wink [/quote]Taiwanese is not a dialect of Chinese it is a whole different country. I know how to speak chinese

Hmmm, this is a sensitive topic. I'm not sure it's related to the whole MathCounts forum anyways...

[quote="nat mc"]Not necessarily.  You could have math friends.  And I'm serious about that.  People that come over to do math and play (video games or computer or movie or something).   And it's better to have friends that like the same thing as you, so math friends are ideal.[/quote]
Heh. Phil, I somehow doubt most people have the same setup that we do. :P And "math friends" is quite a weird name for it....

Anyway, about the original topic.
I really doubt that there is a "math gene" among Asians. I know plenty of non-Asians that are quite gifted and plenty of Asians that aren't. (And vice versa)

But on average, Asian parents do generally push on their children to work more. (My mom is from Taiwan.... my brother and I have to work quite a lot... and my other two siblings.) Regardless of whether the top students are or are not mostly Asian, the percentage of the top students that are Asian is definitely a lot more than the percentage of people in the US that are Asians, which does show something.

Same with parents from several other cultures. (My dad is Russian... also headacheful... too much work... :P)

[url=http://www.srcf.ucam.org/~jsm28/imo-scores/2004/country-scores-order.html]2004 IMO Results, by country[/url].

1 	China [CHN]
2 	United States of America [USA]
3 	Russia [RUS]
4 	Vietnam [VIE]
5 	Bulgaria [BUL]
6 	Taiwan [TWN]
7 	Hungary [HUN]
8 	Japan [JPN]
9 	Iran [IRN]
10 	Romania [ROM]

Yeah but some people are just talented. It's like that in music or athletics too; some people just know what they're doing. They have an intreprid sense of math and numbers.

[quote="09husbri"]This whole topic is ridiculous. As people have said, there is no hereditary gene running through certain "races" making them better or worse at math. It is true that there are some cultural differences contributing to what you call the "Asian Math Gene," but we should not use these vague differences to generalize in an unfair matter. Because work ethic - and work ethic is this "cultural difference" in its simplest form - is really a personal thing, not a cultural thing.
It's true. When we look at these differences, we generalize. Some of us think, "Asians are smarter than white people." That's not true. No amount of statistics can make that true. Some of us think "then white people must be more athletic." That's also BS. Do you know why they make you fill out that race/ethnicity bubble on standardized tests? Not to find out who is smarter. To find out how people of certain races are being treated academically in certain schools, for example, blacks in inner city schools.
Whoopi said it best, "Race is an idea." That's right. Race doesn't even really exist. It is an invention of mankind. What does exist is culture.
If we stop talking about so-called "racial differences," maybe we can see each other as people, for crying out loud, and not as Asians, whites, blacks, and Hispanics. Because that gets in the way.[/quote]

Well, that's not quite true. When you have people separated into different cultures, they evolve in the way that will best survive in that culture. Traditionally, in Africa, people have to do a lot of running; they're good runners. 90% of Australia's population lives within 10 miles of the sea; they're good swimmers. I'm not saying this makes Australians better than anyone else. Everyone is different, and it's how you look at those differences that determines whether or not it gets in the way.

[quote="Phelpedo"]Well, that's not quite true. When you have people separated into different cultures, they evolve in the way that will best survive in that culture. Traditionally, in Africa, people have to do a lot of running; they're good runners. 90% of Australia's population lives within 10 miles of the sea; they're good swimmers. I'm not saying this makes Australians better than anyone else. Everyone is different, and it's how you look at those differences that determines whether or not it gets in the way.[/quote]

What you are talking about is evolution, and athletic qualities like good legs for running and good bodies for swimming are different from good math abilities. One reason for this is that academics are so global now that any math strengths in a culture would not come from "genes" but from cultural differences.

Yes but the point was that cultural differences shape genes.

Okay, you have a point, but I think we should just stop discussing this because the thinking that certain "races" are smarter than others is what the Nazis thought, and look at to where it led them.


i'm anyway satysfied. this is my first solution in this very difficult forum (all are very good solvers!). thanks sinajackson  :lol:

good job girino!!
you are a very good mathlover!


trigonometry
calculus
algebra unsolved
algebra

Can anyone solve the equation cos x - x^2 = 0? or any other equation of this form, like cos x = x? or sin x = x?
(Without using any methods in calculus, like Picard's or Newton's method.)

Such equations can't be solved exactly.  If you want a numerical solution without using techniques such as Newton's method, there's nothing to do except to approximate.


Can anyone prove that if a grid is laid out with 1X1 squares, and a needle, 1 unit long is dropped on the grid...the probability of it landing on a line is 2/pi ?  I found this in a book and found it interesting and tough...

This is the well-known (but still fascinating) Buffon's needle problem.

It can be solved by calculus or by changing the needle into a circle (see "Proofs from THE BOOK" chapter 20) !

I think I see...If we draw a circle of diameter 1 this is where the needle can land without touching lines...then if we fill in the remaining corners with circles the needl can land here also...we can keep doing this infinitely many times...now can someone help me find the area of these circles?

If the answer is 2/pi, then the probability of not landing on the line would be pi-2/pi.
To find the area of the circles, you'll have to use ratios, and find the ratio of the area of a circle with diameter x unit to the area of a x by x unit square.  The ratio will always be the same(pi/4x), so you could make an equation:pi-2/pi=pi/4x, making x be equal to pi^2/4(pi-2).  But then again, the ratio is always the same.  Oh I need help!


let ABCDE is pentagon such that : AB=BC=CD=DE=EA  and \[ <A  \geq <B  \geq  <C  \geq   <D  \geq   <E\]
prove that ABCDE is regular.

What do you mean with
\[ < A \geq < B \geq < C \geq < D \geq < E
\]


Find all positive integer solutions to
\[ a^2\plus{}b^2\equal{}3^n\plus{}2\]

If $ a\le b < 30000$, we have the following solutions for $ (a, b, n)$ (found by computer):

1 2 1
2 5 3
7 14 5
41 1262 13
275 3778 15
314 1223 13
790 985 13
16213 29990 19
17365 29338 19

I don't see any pattern here and I believe there are infinitely many solutions.

fair enough
Probably not, I was just wondering...
Thanks for checking


integration
logarithms
calculus
function
real analysis
real analysis unsolved

Suppose $ f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n)$ with $ f \geq 0$. Define
\[ Sf\left( x \right) \equal{} \int_{\mathbb{R}^n } {\ln \frac{{\left| y \right|}}
{{\left| {x \minus{} y} \right|}}f\left( y \right)dy}.
\]Show that $ Sf(x)$ is finite for all $ x \in \mathbb R^n$ and $ Sf \in L_{loc}^1(\mathbb R^n)$.

 :?:

The general plan in something like this is to start by showing that $ Sf$ is (locally) integrable, from which it will then follow that $ Sf$ is finite [i]a.e.[/i] Have you tried integrating $ \int_BSf(x)\,dx$ for some ball $ B$ just to see what will happen? Expect to interchange the order of integration somewhere (Tonelli's Theorem.)

You are right but how bad the function $ \ln \frac{|y|}{|x\minus{}y|}$ is.  :?:


geometry
area of a triangle
Heron\u0027s formula

[quote="Rep123max"]Am I the only person who hates AXIOMMSs? We are accepting something that could definetly be wrong.[/quote]

Bertrand Russell said that it is only in mathematics where we don't know what we're talking about or whether what we are saying is true.

For your second question, are you assuming a regular polygon?  If not . . . is there only one value the area of a polygon inscribed in a given circle can have?

It would have to be regular, just thinking about triangles theres lots of different areas for different triangles, equilateral being the maximum.

For a triangle, the area is sqrt(s(s-a)(s-b)(s-c)). For a quadrilateral, the area is sqrt((s-a)(s-b)(s-c)(s-d)). I don't know about larger polygons.

[color=cyan]The cases of triangle and quadrilateral are the only cases in which that theorem can work.  (Incidentally, they are known as Hero's or Heron's formula for the triangle, and Brahmagupta's formula for the quadrilateral.)
To see why this is true, take the case of a, say, 16-sided figure.  The thing under the square root will have 16 terms.  Now, double every side.  The area of the figure will go up by a factor of 4 -- area is of dimension 2.  However, each term under the square root will go up by a factor of two, so the whole expression will go up by a factor of 2^16, and when you take the square root, you will find your area to be 2^8 = 256 times the original.  However, we know that this doesn't work.  Thus, the formula cannot be valid.

On the other hand, given the sides of a cyclic polygon, it does have unique area.  So there must exist [i]some[/i] formula, probably a good deal less nice than the ones discussed.[/color]

This problem reasonably simple for triangles and quadrilaterals, but it becomes very complicated for larger sided polygons. There was a discussion sort of similar to this under a problem I posted in the Challenge Others forum - http://www.artofproblemsolving.com/wforum/viewtopic.php?t=193

Also here is an interesting page about cyclic polygons from the Mathworld site - http://mathworld.wolfram.com/CyclicPolygon.html

For the second question, about finding the area of a regular polygon inscribed in a circle of area n, with y sides, I think that the formula is:

[hide]Let x=sqrt(n/Pi), the radius of the circle.

x^2*y*sin(180/y)*cos(180/y)=area of polygon

[/hide]

What does everyone think?

The formula mentioned by rep123max and JBL is Brahmagupta's Formula:
Where s = (a+b+c+d)/2, and A = sqrt((s-a)(s-b)(s-c)(s-d))
Here's more information and a proof of this:
[url]http://jwilson.coe.uga.edu/emt725/brahmagupta/brahmagupta.html[/url]

This is only true of cyclic quadrilaterals (quads. inscribed in a circle).

If one of the sides (a,b,c,or d) is 0, then the formula becomes sqrt((s)(s-a)(s-b)(s-c)) which is Heron's formula for triangles.


LaTeX articles geometry integration \/closed

http://www.mathlinks.ro/Forum/viewtopic.php?t=40150

Check the last two formulas in Flip2004's post. I do believe that the post has been readable before the LaTeX parser got renewed...

  darij

A stripped down version of the code which triggers the weirdness:

\begin{array}{|c|} \hline \\
 [a] \end{array}

Both lines of the array are needed to cause trouble. It displays fine if you switch the order. It goes unparseable if you don't put something between the square brackets, and what the broken code looks like depends on what exactly you put there.

[quote="jmerry"]\begin{array}{|c|} \hline \\
 [a] \end{array}[/quote]That's not a forum problem but a problem with LaTeX syntax which is discussed at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=newlineargs]Start of line goes awry[/url]. If you use this in a document you get the same weird result. The problem is that LaTeX thinks that the code is \\ [x].

To get the correct version, remove the \\
$ \begin{array}{|c|} \hline [a] \end{array}$
or, if that's vital use, \left[x\right]
$ \begin{array}{|c|} \hline \\
\left[a\right] \end{array}$
or the link's suggestion of \relax
$ \begin{array}{|c|} \hline \\
\relax[a] \end{array}$

The problem code in the thread you linked to then could become
\[ \left\{\begin{array}{lll} \displaystyle [x]_q & = & \displaystyle \frac {q^x - 1}{q - 1} \\
& \\
\displaystyle \relax[k]_q \; !\; & = & \prod\limits_{\nu = 1}^{k}[j]_q \; ,\; \; [0]_q!: = 1 \; \\
& \\
\displaystyle \left[\begin{array}{c} n \\
k \end{array}\right]_q & = & \displaystyle \frac {[n]_q !}{[k]_q! [n - k]_q!} \; ,\; \; k\in \{0,1,\ldots ,n\} \\
& & \\
\displaystyle \relax[x;q]_k & = & \displaystyle \prod\limits_{\nu = 0}^{k - 1}[x + \nu]_q \; ,\; \; [x;q]_0 = 1 \end{array}\right. \; .
\]

\[ (III)\; \; \begin{array}{|c|} \hline \displaystyle [x + y;q]_n = \displaystyle \sum\limits_{k = 0}^{n} \left[\begin{array}{c} n \\
k \end{array}\right]_q \; q^{ky}[x;q]_k[y;q]_{n - k} \\
 \\
\hline \end{array}\; .
\]

Thanks a lot to you both! In fact it seems that the LaTeX parser relates the [ to the \\ even though there is a \displaystyle in between. Or are \displaystyle's just edited out before the code is parsed?

  darij

I believe displaystyle is removed as it's not necessary and has no effect as display mode is set by the \ [ ... \ ]. Also there is no need for any arrays either as the code can be made much simpler using the cases environment and the boxed command.

\[ \begin{cases}
[x]_q \equal{}\frac {q^x \minus{} 1}{q \minus{} 1}\\[2ex]
[k]_q \equal{} \prod\limits_{\nu \equal{} 1}^{k}[j]_q & [0]_q!: \equal{} 1 \\[2ex]
\genfrac{[}{]}{0pt}{}{n}{k}_q  \equal{}  \frac {[n]_q !}{[k]_q! [n \minus{} k]_q!} & k\in \{0,1,\ldots ,n\} \\[2ex]
[x;q]_k \equal{} \prod\limits_{\nu \equal{} 0}^{k \minus{} 1}[x \plus{} \nu]_q &[x;q]_0 \equal{} 1
\end{cases}
\](For some reason, the \limits command seems to be ignored in the above code).
\[ (III)\ \boxed{[x \plus{} y;q]_n \equal{} \sum\limits_{k \equal{} 0}^{n} \genfrac{[}{]}{0pt}{}{n}{k}_q q^{ky}[x;q]_k[y;q]_{n \minus{} k}}
\]

[quote="Valentin Vornicu"]So in the end it's not a parser error, but a pilot error right?[/quote]

Theoretically yes. In practice it is a case where the parser behaves formally correct but counterintuitively.

Valentin, are the ML data safe regarding the fires?

  darij

I think that the TeX FAQ should have said that in the amsmath align environment there is no problem with \\. Even using amsmath, eqnarray has a problem:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{eqnarray}
  [a] &=& b \\
  [a] &=& b
\end{eqnarray}
\end{document} [/code]doesn't work and \relax is still needed but
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
  [a] &= b \\
  [a] &= b
\end{align}
\end{document}[/code]is fine and is OK on the forum as well
\begin{align} [a] & = b \\
[a] & = b \end{align}

I don't know what changes there have been in the forum parser but the current version agrees with what happens when you compile a document, which is as it should be. The only thing is that it loses the \limits command which I'm not sure should happen, but that is a very minor point.

In one of the examples above which used the cases environment I used \\[2ex] to spread out the vertical spacing. So LaTeX thinks that
\\
[x]
means \\[x] and thinks that x is some sort of measurement. In the log LaTeX then complains about an [i]illegal unit of measure (pt inserted)[/i] so rewrites the code as \\[0 pt] losing the [ ] round the x and making a mess. So LaTeX tries to guess what you meant and gets it wrong, hence the silly output rather than just an error.

An interesting problem :)

[quote="darij grinberg"]Valentin, are the ML data safe regarding the fires?
  darij[/quote]Yes, the server is not physically located in San Diego area, so there is no danger. We also make daily hard backups (and hourly backups on the server itself).

The problem is with our offices (which are in the city of Alpine, which is a sort of suburb of San Diego, west of which was ordered to execute mandatory evacuation), and my house (which is like 10 minutes away from the fire, but hopefully it seems the winds are slower).

[quote="jmerry"][quote="TeX FAQ"]Note that the amsmath package modifies the behaviour of \\ in maths. With amsmath, the eqnarray example doesn’t need any special measures.[/quote]
I think that might be the change- the forum parser previously treated this as if the amsmath package was in, and now doesn't.[/quote]The amsmath package is still being included.

The modifications were only regarding the string itself (so bringing to a unified syntax). Limits were excluded because \displaystyle is automatically added. Am I wrong with that?

[quote="Valentin Vornicu"]Limits were excluded because \displaystyle is automatically added. Am I wrong with that?[/quote]In some environments (even in display mode) inline style is the default so \limits has to be added to change the way the sum is shown. Compare
\[ \begin{cases}
\sum_{r=1}^{n}r &= \frac{n(n+1)}{2} \\
\sum_{r=1}^{\infty}\frac{1}{r^2} &= \frac{\pi^2}{6}
\end{cases}
\]and
\[ \begin{array}{c}
\sum_{r=1}^{n}r = \frac{n(n+1)}{2} \\
\sum_{r=1}^{\infty}\frac{1}{r^2} = \frac{\pi^2}{6}
\end{array}
\]which both require \limits to be added, with
\begin{align*}
\sum_{r=1}^{n}r &= \frac{n(n+1)}{2} \\
\sum_{r=1}^{\infty}\frac{1}{r^2} &= \frac{\pi^2}{6}
\end{align*}which doesn't.

I see. Which are the exact environments that limits are required in?

That's a question and a half! There are loads of environments, I counted over 50 relevant ones in one book so it would take a lot of work to discover which do what. Worse still I suppose it's possible that they could behave differently when nested which makes the number of possibilities enormous. But even more to the point, it isn't only environments that use inline mode as the default:
\[ 2^{\sum_{r\equal{}1}^n r} \\[2ex]
\frac{\int_{x\equal{}0}^1 x\;dx}{\sum_{r\equal{}1}^n r} \\[2ex]
\cfrac{\prod_{r\equal{}1}^n \frac{r\plus{}1}{r}}{\sum_{r\equal{}1}^n r}
\]
I think that it may mean not removing \limits from the code.

Ok $ 2^{\sum\limits_{k\equal{}1}^n \frac kn}$.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 10
ARML

Here are some of my scores mocking AIME conditions (3 hrs straight, quiet room, etc):
Date- AIME : (score, score if not for silly mistakes; Ex 9*4 +1 is not equal to 33!)
02/21/05- 2000 AIME I: (2,2)
02/24/05- 2000 AIME II: (1,2)
02/25/05- 2001 AIME I: (3,4)
02/27/05- 2001 AIME II: (2,4)
02/28/05- 2002 AIME I: (1,5)
02/28/05- 2002 AIME II: (5,7)
-And I took this in 2 seperate chunks, plus its extremely easy, so this does't really count.
03/01/05- 1986 AIME (Rumors that it was easy are true): (8, 10)

Lately, I've been able to comprehend more and more AIME material; I dunno why. My question is: has anyone ever improved their score to 7/8 (USAMO floor value) while starting out from scratch just a few weeks ago?
-Does anyone have any general tips for me, a person with a mediocre AIME score? Should I even attempt #11-15 on the AIME, or would it just drain my time? I've also noticed I've made a lot of silly mistakes, so what is everyone's checking strategy? Did it work last year?

-[b]Finally, has anyone ever felt they were a victim of burnout[/b]?

i dont think you can raise a score of average 3's to 7/8. It took me the whole mid-winter break and 2 more months to be where i am right now(about 6-8 range). Just keep studying though, do not give up.

[quote="1234567890"]
-[b]Finally, has anyone ever felt they were a victim of burnout[/b]?[/quote]

I only get burned out by meaningless schoolwork. Not math.

So it is FEASIBLE to raise your score by that much, but it isn't too likely. You would probably need to actually learn a lot of new concepts to solve that many more problems on the AIME. But it is definitely a possibility. So don't give up; you can beat the odds!

I've only burned out when I tried to memorize and redo every single mathcounts test 10 times. That was painful.  As long as you move through and stick with problems you enjoy doing, you should be fine.

thats a lot of tests.  it might help you more to just do problems untimed, cause if right now you think you can only get like 4 problems right on the actual test, time probablly won't be an issue.  i think the only reason to do a set for 3 hours straight is to get a feel for your pacing, for how long 3 hrs is, etc.  But by doing that many tests already, you probablly already have a good understanding of those things already.

1234567890, I'm in the same boat as you are--trying to prepare for the AIMEs a week before (since I've never even heard of it before the beginning of this year). I feel that doing problems is important, but I'm also reading the AoPS books like crazy because I feel like I'm short on knowledge also. You should probably decide, if you have a low score, whether it's because of lack of knowledge or lack of experience and plan accordingly.

[quote="paladin8"]I only get burned out by meaningless schoolwork. Not math.[/quote]
I sure feel the same way. How many times have you been asked to prove something on a test, only to lose points because you forgot to put one crucial step *drum roll*

4. AB = AB    |   4. Reflexive Property

[quote="1234567890"] My question is: has anyone ever improved their score to 7/8 (USAMO floor value) while starting out from scratch just a few weeks ago?
[/quote]

Yes, in fact - last year I was getting consistent 3's to 5's before AIME, only understanding at most 8 of them, and then on the actual test I got some sort of insane adrenaline boost and understood how to do 13 of the questions, making 5 stupid mistakes and getting an 8.  Of course, I seem to have a pattern rise to the occasion on the actual test, scoring at least 120% of my practice average.  If this is true for you too, you may have a shot at USAMO.  So don't give up!

Well, if scratch means 0 - 1, then no. In all likelihood you will not get to the USAMO cutoff. However, if scratch is 4 or 5, then there's a chance. Study hard. Do problems. Look at solutions. And if you get lucky on AIME day, you'll get there. :-)

Here are some of my practice scores in last 3 or 4 weeks
(score, score if not for careless mistakes)

1996-- (7, 11)
1992 -- (10, 12)
1994 -- (8, 10)
2002I---(8, 11)
2002II---(10, 11)
2004II---(9, 11)
1995--(10, 11)

BTW only the last 3 or 4 were taking under real AIME conditions, for the first few I usually split it into 1.5 hr first day, 1.5 hr second day.

how much can I improve in a week? I still haven't master the art of being careful :( , I can't do it, not making any careless mistakes!!! :(

Well if you score like that you won't need improvement to make USAMO. :-P

well...i cant really call it improving in one week...but i just got back to doing AIME after a few months...and b4 i  would 4 to 6 but now i get around 7 to 9

[quote="rrusczyk"][quote="beta"]
how much can I improve in a week? I still haven't master the art of being careful :( , I can't do it, not making any careless mistakes!!! :([/quote]

Time to start tapering down, beta.  Don't crush yourself in this last week - it will hurt more than help.  Chill out a bit.  (Easy advice to give, harder to take, I realize.)[/quote]

Ya, if I do as well as I do on practices, I qualify for USAMO(building up self-confidence :) ). Time to relax... Eh one thing I'm a little worried about: I hope I won't freeze or faint or something like that on the real AIME :)

Eh actually, being relaxed on the actual AIME is still difficult, help!

[quote="beta"]

Eh actually, being relaxed on the actual AIME is still difficult, help![/quote]

yeh do NOT be tense
i still feel i did not get perfect score on amc10 because of too much nervousness
and now i will never be able to take the 10 again in my life  :(

its really really hard to not be though, i know

i like to keep myself nervous and tense, because everytime i did that, i did better than i thought.... because when im nervous, i tend to write freely and more widely in terms of problem solving...

Well, I took the 2003 AIME I, and I'm kind of happy, but disappointed as well. I answered 11 questions, 2 with totally wrong approaches, 1 semi-wrong approach, and I made 4 dumb mistakes (as in forgetting a triangle to count), bringing my score down to 4. But in 8 of the problems, I had the right approach, and had them right for a few calculations gone awry. But then I realized that 2003 was a very easy year, with the USAMO cutoff being 8. If I had made no dumb miustakes, I couldv've gotten an 8 or a 9 (This is what I'm pleased about), but I did, unfortunately(I'm not pleased about this.
Anyways, was 2003 really easy? I hope not.
Another question: I checked over my work, but missed the 4 mistakes, even when I worked through them.  How do you avoid that?

Hehe, I'm afraid I can't use any of those techniques, I just got back from the state Matcounts (that means no more Mathcounts studying), but the AIME is on Tuesday! I guess I should just try to do a past AIME, right? I've never even seen one before.

for that, go to Kalva.

if u have never seen em, dont worry. i saw them for the first time last yr, got anxious, got 5 right, (2 away from cufoff) :(

AIME looks hard at first, but it looks easy if you take a careful look. it involves math of no higher than precalculus. so, just take a few AIME, and also study useful books such as AoPS volume1/2(since ur a mathcount competitor, im assuming u are a middle schooler) and some other ones i cannot remember off the top of my head.

good luck.

How much can you improve in one day?  :D

heh....just review material that you already know and dont' try to learn anything new. That way you will have a greater chance of getting problems that you know the theory behind right and not making careless errors.

[quote="filletwho"]In the last three weeks, I brought my score up from 7 to the double digits.[/quote]

WOW!

Eh now I really regret that I made the decision to do AIME 1 :(

[quote="tetrahedr0n"]Hey Phil, I seriously hope you qualify this year. Troy needs something before DCD becomes the premier math school in Michigan.[/quote]

Thanks Fred, but we all know you're going to be the qualifier!

And beta, from the praises i've heard others give you, I don't think you need to worry.

[quote="sponge008"][quote="MithsApprentice"][quote="sponge008"]I actually do have the AoPS books, but so far, I've only done the first few chapters of Book 1. Is the second book actually useful for the AIME? What about the USAMO?[/quote]

While the first book may have techniques handy for MC and the AMCs, the second book is generally toned towards the average-level AIME or ARML competitior. To make 2-digit scores on the AIME consistently, you'd need something a bit beyond the AoPS series. And if you're seriously looking to compete in the USAMO, then you definitely want something beyond the AoPS series. However, if your goal right now is just a "good middle school score" or a low USAMO qualifying score, the AoPS 2 should do the trick. :-) (Supplementing with the AIME Problem Series class wouldn't do any harm.)[/quote]
Just for the future, what would be a good source to study from for USAMOs/IMOs other than the actual competitions? Also, would it be worthwhile for me to go through the rest of the first book, even though I don't have any trouble on AMCs and my Mathcounts days are over?

PS: Wow, I surprised myself by getting 4 of 5 questions from the 2004 AIME A right! (in an hour) Could I possibly have a chance on advancing?  (kicks self for hoping too much, it didn't get me anywhere in Mathcounts) BTW, do the problems generally get harder as you go through the booklet?[/quote]

I'd say you might have a chance to advance, although it'd be pretty tough. The problems get harder as you go on, but strangely, it always seems like there's always at least one problem 12-14 that I can do, and I usually get 5-10. It's sort of weird, always seeing one or two doable problems when you don't expect to be able to do them.

I'm so happy my school has spring break when the A is taken. That gave me three totally free weeks to study. Hopefully it'll get my score a little higher.


trigonometry
geometry
3D geometry
pyramid
ratio
Pythagorean Theorem

1. A conveyor belt rises $2.40$ m. What is the horizontal distance between the ends of the belt for each angle of inclination?
a) $32$
b) $55$
2. Find the measures of the acute angles to 1 decimal place in a right triangle if its sides have lengths $a$, $a+d$, $a+2d$.
3. The volume $V$ of a right square-based pyramid is given by the formula $V=\frac{1}{3}Bh$, where $B$ is the area of the base and $h$ is the height. Express $V$ in terms of the length $l$ along an inclined edge and the angle of inclination $\theta$ of the edge.

I did the 3rd one Ill do the other ones later.
See the figure below.
SOLUTION
[hide]
$b=lcos\theta$
$h=lsin\theta$
So $area of base=2l^2cos^2\theta$
Volume=$\frac{1}{3}*2l^2cos^2\theta*lsin\theta$
[/hide]
I think this is right but sorry if it isnt, Im in a hurry

[quote="danne89"]2.
[hide]Does it really make any seens if (a, a+d, a+2d) isn't a pythagorean triplet?
And that isn't true for all a. You didn't mean sides, a, (a^2 -1)/2 and (a^2-1)/2 + 1 which is always a pythagorean triplet.[/hide]

Edit: BTW, what's a convey belt. Can't find it in my dictionary.[/quote]
I dont know what you're complaining about because i got the answer as
[hide]
By solving this:
$a^2-2ad-3d^2=0$ you can get the $d$ in terms of a which will come as $\frac{a}{3}=d$ and $a=-d$ which is not possible.
So taking the 1st root we get the sides as $a,\frac{4a}{3}$ which gives us $tan\theta=4/3$ which gives the value of one angle as
$53.1$ and the other as $36.9$
[/hide]

[hide="Problem 2"]
In a triangle with side lengths $a$, $a+d$, and $a+2d$
$\sin\theta =\frac{a}{a+2d}=\frac{1}{1+2(\frac{d}{a})}$
Using the Pythagorean Theorem
$\sqrt{a^2+(a+d)^2}=a+2d$
$a^2+a^2+2ad+d^2=(a+2d)^2$
$2a^2+2ad+d^2=a^2+4ad+4d^2$
$a^2=2ad+3d^2$
$1=2(\frac{d}{a})+3(\frac{d}{a})^2$
Let $x$ equal $\frac{d}{a}$
$1=2x+3x^2$
$3x^2+2x-1=0$
$(3x-1)(x+1)=0$
$x=-1,\frac{1}{3}$
Since $x$ cannot be a negative value, $x=\frac{d}{a}=\frac{1}{3}$. Therefore
$\sin\theta=\frac{1}{1+2(\frac{1}{3})}=\frac{1}{\frac{5}{3}}=\frac{3}{5}$
Thus
$\sin ^{-1}(\frac{3}{5})=36.870\approx 36.9$
$90-36.9=53.1$
The measures of the acute angles to 1 decimal place in a right triangle if its sides have lengths $a$, $a+d$, $a+2d$ are $36.9$ and $53.1$

[/hide]
[hide="Problem 3"]
$h$, The Height of the Pyramid, can be expressed as $l\sin\theta$, as $\sin\theta=\frac{h}{l}$.
The base of the pyramid is a square with diagonal $x$. Therefore, its area can be expressed as $\frac{x^2}{2}$. Using the Primary Trigonometric Ratios,
$\cos\theta=\frac{0.5x}{l}$
$0.5x=l\cos\theta$
$x=2l\cos\theta$
Substituting $2l\cos\theta$ in place of $x$,
$B=\frac{(2l\cos\theta)^2}{2}=\frac{4l^2(\cos\theta)^2}{2}=2l^2(cos\theta)^2$
Therefore, the volume $V$ of a right square-based pyramid in terms $l$ and $\theta$ is
$\frac{1}{3}Bh=\frac{1}{3}(2l^2(\cos\theta)^2)(l\sin\theta)=\frac{2}{3}l^3(\cos\theta)^2\sin\theta$
I'll post a diagram later. :P
[/hide]

Ohh, I get #1 now.

[hide="1"]Given a triangle with angle of elevation $\theta$ above the horizontal, and height of the far end is $2.40$ meters, we are trying to find the adjacent side to $\theta$. For a), we say $2.4\cdot \cot 32\approx 3.8408$ meters. For b), we say $2.4\cdot \cot 55\approx 1.6805$ meters.[/hide]


calculus
derivative
integration
calculus computations

Through that I get the first answer right (according to my calculator)

However on the second one, I get
$ f(x) \equal{} \int_{0}^{ln x} \frac {1}{2 \plus{} {sin t}}\, dt \equal{} \frac {1}{{x}(2 \plus{} {sin (ln x)})} \minus{} {\frac {1}{2}}{0} \equal{} \frac {1}{{x}(2 \plus{} {sin (ln x)})}$

thanks for your help,
i wasn't sure if i was right because of my calculators formatting

[quote="kunny"]$ \boxed{\frac {d}{dx} \int_{h(x)}^{g(x)} f(t)\ dt \equal{} f(g(x))g'(x) \minus{} f(h(x))h'(x)}$[/quote]
[hide="proof for those that are interested"]
Suppose $ F(x)$ is the anti-derivative of $ f(x)$.

Then \[ F(g(x))\minus{}F(h(x))\equal{}\int_{h(x)}^{g(x)}f(x)dx\]
Thus, \[ \frac{d}{dx}\int_{h(x)}^{g(x)}f(x)dx\equal{}\frac{d}{dx}\left[F(g(x))\minus{}F(h(x))\right]
\equal{}f(g(x))g'(x)\minus{}f(h(x))h'(x)\]
[/hide]


Sorry for the double post, I posted the wrong poll and couldn't edit.

[quote="Temperal"]Wow. Just... wow, miyomiyo.[/quote]What's wow?


[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/9/7/097049293885fc3e635a8e1e021699dedcd7c221.gif[/img]
http://img212.imageshack.us/my.php?image=d21c05fab0d3eabd6618d17jv2.jpg
Just check up the qns from 2 to 5

[hide]@da , ya ur answer to fourth was correct[/hide]

2) I don't think thats a valid question. You cannot define exponential function for a negative base.

3) $ s_n\equal{}\frac {\cos \frac x2 \minus{} \cos (xn\plus{} \frac x2)}{2\sin \frac x2}$

$ \sum s_i\equal{}\frac {n \cos \frac x2 \minus{} \frac {\cos (x\plus{} \frac {xn}{2}) \sin \frac {nx}{2}}{\sin \frac x2}}{2\sin \frac x2}$

4) $ T(x)\equal{}\frac 12 \cdot 1^2 \sin x$, $ S(x)\equal{}\frac 12 x \cdot 1^2$

5) Say O is centre. and $ \angle AOP \equal{} x$, then $ AP\equal{}AT\equal{}2r\sin \frac x2$, where r is radius od circle. $ \angle TAP \equal{} \frac x2$ and $ \angle ATP \equal{} \frac {\pi}{2}\minus{}\frac x4$
$ AQ\equal{}AT \tan (\frac {\pi}{2}\minus{}\frac x4)\equal{}2r\frac {\sin \frac x2}{\tan \frac x4}$
$ \lim_{x\rightarrow 0} AQ \equal{} 2(2r)$

Was trying limits so thought I would try these:

[hide="1"]Put $ x = ln(e^x)$[/hide]

[hide="2"]a should be R+

Applying L'hospital's rule we get

$ a^{P_{n - 1}} \frac {d (P_n - 1)}{dx} ln (a)$

Putting $ P_{n - 1} = a^{P_{(n - 1)}} - 1$ repeatedly in the differential , and using the fact that $ {P_{n - 1}}$ tends to 0 as x tends to 0 ,we get finally

$ a ln^n(a)$[/hide]