induction AMC AIME calculus integration ratio function

I stumbelled upon a question the other day.

Find the last two digits of 3^123456789.

I know this can be done in matlab or mapel, but im not intested it cheating.

Anybody have any idea how we would go abouts this with induction?

[hide="Solution"] It is known that the last two digits of an integer geometric sequence (that is, $3^{0}, 3^{1}, 3^{2}, ...$) repeats with a period dividing $20$.  Ergo,

$3^{123456789}$

Ends in the same last two digits as

$3^{9}= 19683$

Which ends in the digits $\boxed{ 83 }$. [/hide]

[quote="t0rajir0u"][hide="quote"]It is known that the last two digits of an integer geometric sequence (that is, $3^{0}, 3^{1}, 3^{2}, ...$) repeats with a period dividing $20$.[/hide][/quote]
Interesting, is it a trivial result? This would really help on the AIME and such, can you give a proof please?

[hide="Anyway here is my old fashioned method"]
I just use $3^{4}=1 (80)$ so given $=3 (80) =3 (20)$ which is close but we really need more info, at least $mod 25$ and as I write this, I realize a proof for t0r's trick...
phi(25)=20, so $3^{20}=1 (25)$ so big number=$3^{9}(25)=27^{3}=2^{3}=8 mod 25$
combine 8 mod 25 and 3 mod 80 for 83 mod 100[/hide]

To me@home:
I think t0rajir0u's meant to say that "It is well known that the last two digits of an integer geometric sequence, [b]provided that the integer is relatively prime to[/b] $100$, repeats with a period of $20$."

But here's a nice trick relating to such problems.
Say you want to find $a^{n}\mod m$, with $(a,m)=1$.
Then the first positive integer $k$ such that $a^{k}\equiv 1 \mod m$ divides $\phi{(m)}$. So $a^{n}\mod m$ will repeat with a period of $k$.

So in the problem posted, $\phi{(100)}=40$, so you know $3^{n}\mod 100$ will repeat with a period of $k$, where $k$ divides $40$. So $k$ can equal $1,2,4,5,8,10,20,40$.  You can quickly check that $k=1,2,4,5,8,10$ do not work by multiplying out, so you can conclude $k=20$ or $k=40$, so $3^{123456789}\equiv 3^{9}\equiv \boxed{83}\mod 100$.
(Note that I didn't have to find whether $k$ was $20$ or $40$, since $123456789\equiv 9 \mod 40$ and $123456789\equiv 9 \mod 20$. But it's pretty straightforward to check that $3^{20}\equiv 1 \mod 100$.)

Another example: Since $\phi{(21)}=12$, and $(21,2)=1$, $2^{n}\mod 21$ will be periodic either every $1,2,3,4,6$, or $12$ terms. Indeed, we find that $2^{6}=64=3(21)+1$, thus establishing the periodicity.

So this is some pretty cool number theory.  :D

Now back to t0rajir0u's statement. You can prove that every single integral geometric sequence (with the integer relatively prime to $100$) repeats with a period of $40$, since $\phi{(100)}=40$, and I think t0rajir0u just strengthened this somehow.

Another approach is to note that the powers of 7 mod 100 repeat in a pattern of 4.  That's relevant to our problem because
\[3^{123456789}= (-7+10)^{123456789}\, . \]
Now apply the binomial theorem to the last expression.  We only need to examine the first two terms of the binomial expansion, because all other terms will be divisible by 100.

I must say, that is an amazing solution! (Wow.. pulling 7^4=2401 out of thin air is realyl nice... I'll remember the Ravi method for these types of problems!)

In the middle of my solution I noticed the same thing, I guess I should have editted out the question of why it is true...

[quote="1234567890"]I think t0rajir0u's meant to say that "It is well known that the last two digits of an integer geometric sequence, [b]provided that the integer is relatively prime to[/b] $100$, repeats with a period of $20$." [/quote]

The condition you tagged onto my statement is not necessary.  The period divides $20$ regardless of whether the common ratio is relatively prime to $100$ or not.

(Edit:  I should say that the period after a certain number of terms divides $20$, where the first few terms (which do not repeat) are determined by the gcd.)

[quote="white_horse_king88"]If you take it modulo 4 and 25... and then use CRT, it's actually going to cycle modulo 20, which is stronger than 40.[/quote]

Yes; refer to the Carmichael Function,

http://mathworld.wolfram.com/CarmichaelFunction.html

[quote="t0rajir0u"](Edit:  I should say that the period after a certain number of terms divides $20$, where the first few terms [b](which do not repeat)[/b] are determined by the gcd.)[/quote]

Fair enough.

I should mention that the length of the non-repeating segment is at most one less than the highest power to which a prime is raised in the prime factorization of the modulus, so here it is $1$.

I haven't really read through this thread, but another idea that I don't think anyone has noted is:

$3^{2}= 10-1$

Wow, you guys have very creative ideas.

I just noticed a similar problem on the blog...


Show that $ n^n \minus{} n\cdot(n \minus{} 1)^n \plus{} (n\cdot(n \minus{} 1)\cdot(n \minus{} 2)^n)/2 \minus{} ........................... \equal{} n!$
NO INDUCTION  :P
 :)

Consider all functions $ f : [n] \rightarrow [n]$ where $ [n] \equal{} \{1,2,3,\ldots,n\}$. There are $ n^n$ such functions. But we don't want any functions where at least  1 element is not in the image of the function. So we subtract $ n\cdot(n\minus{}1)^n$. But now we realise that we've subtracted those functions which have two elements not in the image twice. So we add that number which is $ n\cdot\frac{(n\minus{}1)}{2}(n\minus{}2)^n$. And so on. And by the principle of inclusion-exclusion we'll end up counting all those functions on $ [n]$where all the elements are in the image of the function. But this means that the functions counted are permutations of $ [n]$. And that number is our dear old $ n!$. Therefore, the result follows.

Yes thats a good one  :lol:
But any other method would be appreciated ( I have another long one in mind ).

Does $ e^x$ remind you of something?  :maybe:


linear algebra
matrix
linear algebra unsolved

Let $ A$ be a square matrix such that $ \text{rank}(A)\equal{}1$. Prove that $ \det(A\plus{}I)\equal{}\text{tr}(A)\plus{}1$.

[b]With[/b] $ Rank(A) = 1$ you can prove that : $ det(A + \lambda I) = \lambda^{n} + \lambda^{n - 1}.Tr(A)$ ($ \lambda \in M_{n} \mathbb{(R)}$


algebra
polynomial
calculus
integration
function
domain
Ring Theory

Show that the polynomial
$ a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_1x\plus{}a_0 \in k[x,a_0,...,a_n]$ is irreducible.
(*k:infinite field)

First, consider this as a polynomial in variables $ a_0, a_1, \ldots, a_n$ over the ring $ R \equal{} k[x]$. As such, it's just a linear form in those variables, so it's clearly irreducible: if

$ p_0(x)a_0 \plus{} p_1(x)a_1 \plus{} \ldots \plus{}p_n(x)a_n \equal{} f(a_i)g(a_i)$

where $ f$ and $ g$ are polynomials in the $ a_i$ over $ R$, then one of them (say, $ f$) must be a constant $ f \equal{} q(x)$ and the other one must be $ g \equal{} q_0(x)a_0 \plus{} q_1(x)a_1 \plus{} \ldots \plus{} q_n(x)a_n$ with $ q(x)q_i(x) \equal{} p_i(x)$.

[If you don't think this is obvious, prove it as a separate exercise: Let $ R$ be an integral domain and let $ z_1, \ldots, z_n$ be indeterminates over $ R$. if $ c_i \in R$ then the polynomial $ c_1z_1 \plus{} \ldots \plus{} c_nz_n$ is irreducible].

Back to our case, since $ p_0(x)\equal{}1$ we must have $ f \equal{} q(x) \equal{} \alpha \in k$, a constant of the underlying field, so the original polynomial is indeed irreducible over $ k$.

I don't think you need to assume that $ k$ is infinite.


consider the function

$ \frac{1}{\epsilon^2 \plus{} z^2}$

So we know that there are two poles, one at $ z \equal{} i \epsilon$, one at $ z \equal{} \minus{} i \epsilon$. So when this function never hits 0 on the real line, how do the singularities affect its behavior on the line?

Can you please state your question more clearly¿ I dont' understand what you want...

Okay, so poles are a subclass of singularities. I think that $ z \equal{} i \epsilon$ and $ z \equal{} \minus{} i \epsilon$ are poles - I may be wrong here. The question is - how do complex singularities (complex poles in this case) affect a function's behavior when the function is plotted on the real line?

Those are poles. Simple poles, to be specific. They're not on the real line, but they are close to that line. We know that $ |f(z)|,|f'(z)|,$ and so on tend to $ \infty$ as $ z$ approaches a pole. That means that at least some of $ f(x),f'(x),$ and so on become fairly large as we pass close to those poles.

I phrased that fairly loosely, because different kinds of behavior are possible. The graph of $ \frac{1}{\epsilon^2\plus{}z^2}$ looks different in detail than the graph of $ \frac{z}{\epsilon^2\plus{}z^2}.$


I have a new one homepage! Look on that link.

Oh my God, I am so famous, they even named an internet magazine like me.. So great of me..  :lol:

[quote="7h3.D3m0n.117"]what...? i don't get ur post...[/quote]

In my profile is a new link as a my homepage. It is even in thi topic in contct line in AoPS as well as in ML design. That is the point.


modular arithmetic
number theory
least common multiple
induction

1)Prove that among any $39$ consecutive natural numbers
it is always possible to find a number whose sum of digits
is divisible by $11$.

2)Find the largest $x$ for which
$4^{27} + 4^{1000} + 4^x$
equals the square of a whole number?

3)Prove that for any prime number $p$ which is $>2$, the numerator $a$ of the fraction

$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2p-1}$

is divisible by $p$

[hide]
my solution for 2 was $x=1972$
[/hide]

[hide="solutionforproblem3"]

It should be noted that p < 2p-1. Therefore, the summation of fractions contains (1/p). This is essential to understanding the next equation.

1/2 + 1/3 + ... + (1/p) + ... + 1/(2p-1) = (px + k)/y, where x,y, and k are integers and k does not have p as a factor. Therefore, the key lies in representing 1 as p/p. This gives:

a/b = (p/p) + ((px + k)/y) = (p(y + px + k))/py.

Therefore, a is divisible by p, Q.E.D.

[/hide]

sen, could you post your solution to #2?

[hide="3"]Call the consecutive integers $a$, $a+1$,.... $a+38$.  Let $f(a+n)$ be the sum of the digits of $a+n$.  Assume that there exists no $a+n$ such that $f(a+n)\equiv0\pmod{11}$.
Note that if $a+n\equiv0\pmod{10}$, for integers $k$ where $0\leq{k}<10$, ${f(a+n)+k=f(a+n+k})$.  (1)
Note that if $a+n\equiv0,10,20,30,40,50,60,70,80\pmod{10}$, $f(a+n+10)=f(a+n)+1$.  (2)

Let $a+m$ be the lowest integer in this set such that $a+m\equiv0\pmod{10}$, where $m$ is a nonnegative integer.  It is also true that $m<10$, because if $m>10$ then $a+m-10\equiv0\pmod{10}$, a contradiction because $m-10<m$.

If $f(a+m)\equiv{j}\pmod{11}$, then $11-j$ is less than 10, and $f(a+m+11-j)\equiv0\pmod{11}$ by (1).  So $f(a+m)\equiv1\pmod{10}$. (3)

If $a+m$ is not equivalent to $90 \pmod{100}$, then $f(a+m+10)\equiv f(a+m)+1\equiv2\pmod{11}$

by (2), and $f(a+m+10+9)\equiv0\pmod{11}$ by (1).  So $a+m$ must be equivalent to $90\pmod{100}$, and $f(a+m+10)\pmod{11}$ can equal any number.

By (3), $f(a+m+10)\equiv1\pmod{11}$.  Because
$a+m\equiv90\pmod{100}\Rightarrow{a+m+10}\equiv0\pmod{100}$,
$f(a+m+20)\equiv f(a+m+10)+1\equiv2$ by (2).
Finally, $f(a+m+20+9)\equiv{f(a+m+20)+9}\equiv0\pmod{11}$ by (1), so the original statement cannot be true.  Because $m<10$, all of these elements are in the set of consecutive integers, and the original assumption is proved false for all such sets of integers.

I know it's messy, but the above explanation covers all the possibilities and proves the statement.[/hide]

[quote="scorpius119"]wait, how could the numerator in #3 possibly be divisible by p? For when this is put over a common denominator, all terms [i]except[/i] the $\frac 1p$ term will be divisible by p, but the $\frac{1}{p}$ won't. :?[/quote]

[hide="spoiler"]I'm assuming the beef you have with the 1/p term not being divisible when put over a common denominator is the integer k that i was using. Remember, i'm excluding the 1 and i'm just adding the fractions starting from 1/2 on to get the (px + k)/y term. I then re-introduce the 1 as p/p, and add these 2 terms together. When you do this, you should see that the numerator now has a factor of p, and is therefore divisible by p.[/hide]

I think that the solution for 3 should be simplified. If not, it would be saying a/b=ap/bp

[quote="JBL"][quote="scorpius119"]wait, how could the numerator in #3 possibly be divisible by p? For when this is put over a common denominator, all terms [i]except[/i] the $\frac 1p$ term will be divisible by p, but the $\frac{1}{p}$ won't. :?[/quote]

This is right.  The simplest way to verify that this is correct is to calculate: $1 + \frac12 + \ldots + \frac15 = \frac{137}{60}$.  3 most certainly does not divide 137.[/quote]

If you represent $1$ as $\frac{3}{3}$ and do the addition you get $\frac{822}{360}$ , which is the unsimplified version of $\frac{137}{60}$, and $822$ is indeed divisible by $3$. The problem never mentioned $\frac{a}{b}$ had to be in simplest form  :D

[quote="quantabe"]The problem never mentioned $\frac{a}{b}$ had to be in simplest form  :D[/quote]

But then we could multiply numerator and denominator arbitrarily!

I think the thing to do is to better define the problem.  For example, that the denominator on the left is the LCM of the denominators on the right.

My first post.. Let's try with number 1
[hide]
We have that a number n divided by 11 could give as a rest 11 different values. We have to find a number $n\equiv 0 \pmod {11}$ among 39 consecutive numbers. We know that after summing up the digits of 10 numbers we will have a sequence as follow(since their are consecutive; we can prove it by induction): $n, n +1 , n+2 , \cdots ,n+8, n+9, n+1$so after 20 numbers we will get a new number called n+9. Now we have a sequence of eleven numbers, but since they are consecutive one of them must be divisible by 11. [/hide]

Can someone show a solution to question 2  :?

According to Mathematica, sen's answer certainly works, but perhaps a proof that it is the largest $x$ could be posted as well?

umm yeh that was my problem, trying to prove it was largest.

well how i did it:

$(a+b)^2 = a^2 + 2ab + b^2$ as we all know.

we have $2^{54} + 2^{2000} + 2^{2x} = n^2$

now we let $(2^{54}, 2^{2000}) = (a^2, 2ab, b^2)$

after going through each case to find 2^{2x}. i found out that the largest possible one was $2^{3944}$

hence $x = 1972$

A good method for a problem like this:  if you want to show that for larger $x$, it isn't a perfect square, show that it lies between two consecutive perfect squares.  (In fact, that hint doesn't work straight-up in this problem -- there's an additional manipulation you need to do first.)

JBL, will you post your solution please?

Sure thing.  First,[hide="The manipulation"]Because $4^{27}$ is a perfect square, $4^{27} + 4^{1000} + 4^x$ is a perfect square for $x \geq 27$ if and only if $1 + 4^{973} + 4^{x-27}$ is a perfect square.  So, it is equivalent to find the largest $x$ such that this last expression will be a square.[/hide]  Second, [hide="the real work"]Note that $4^{x-27}$ is also a perfect square, say $n^2$ for some $n$.  If $n > 2^{1945}$, then $1 + 4^{973} + 4^{x - 27} = 1 + 2\cdot2^{1945} + n^2 < 1 + 2n + n^2 = (n + 1)^2$, so in this case our expression falls in the gap between two consecutive perfect squares, so it cannot itself be a perfect square.  Thus, the largest possibility is $n = 2^{1945}$, when $4^{x - 27} = 4^{1945}$ and $x = 1945 + 27 = 1972$.[/hide]

[quote="JBL"]Note that $4^{x-27}$ is also a perfect square[/quote]

why?

We can show $x > 27$, therefore $4^{x - 27} = 2^{ 2(x - 27) }$ is a square.  :)


geometry
geometric transformation
rotation
geometry unsolved

We have angle with vertex O and point M lies inside this angle. Find points K and L on another sides angle, that |OL|=|OK| and sum |MK|+|ML| is the smallest. I know that I must rotate A, but I don't know how?

[Edited] cause of [b]neworder[/b] sentence.

THis is a problem form an ongoing mathematical contest  :!:  :!:  :!:  (this year's Polish Mathematical Olympiad)

kisiu111 should be banned, this is the second time he posts a problem from an ongoing contest.

I didn't  know that it's a problem of an ongoing contest and I answered it to some User that asked for solution in Private message. :lol:

[Moderator edit: discussion to be continued at http://www.mathlinks.ro/Forum/viewtopic.php?t=169314 - Amir, feel free to post your solution there]


geometry
angle bisector
geometry proposed

Points D and E are chosen on the sides BC and AB respectively of an equilateral triangle ABC so that CD/DB = BE/EA = (\sqrt 5 + 1)/2.  The straight lines AD and CE intersect at O.  M and N are points on the rays OD and OC respectively so that MN is parallel to BC and AN = 2OM.  The line though O parallel to AC intersects the ray MC at the point P.  Show that the ray AP bisects the angle MAN.

wow, that took like half an hour just to make the correct drawing

I suggest you move this to the advanced section :D

Well, I'm not a moderator, so I can't move threads...

Your diagram doesn't need to be [i]that[/i] correct to do it, as long as you can guess a few things out of it and prove them.

The key is to calculate that $O$ is the midpoint of $AD$ (from the given relation) and, therefore, $O$ lies on the median from $D$ of triangle $\triangle ODC$; Menelaos in $\Delta CDM$ with the transversal $OP$, which passes through the midpoint of $CD$ gives $\frac{CP}{MP} = \frac{OD}{OM}$, but $2\cdot OD = AD$ and $2\cdot OM = AN$, so $\frac{PC}{PM}=\frac{AD}{AN}$, but $P$ is on $DN$ as well, so $\frac{DP}{PN}=\frac{PC}{PM}=\frac{AD}{AN}$, done.

[b]Obs:[/b] [b][i]1) if ${F} \equiv BO \cap AC$, the angle bisector of $\hat{A}$ cuts $CE$ at $Q$ and the angle bisector of $\hat{B}$ intersects $AD$ at $R$,  then $F, Q$ and $R$ are collinear[/i].[/b]

[b][i]2) $\angle MAN=45^\circ$ ( how? - I do not know, as yet)[/i] [/b]

[b][i]3) Actually it is enough $AB = AC$, the problem still being true.[/i][/b]

Best regards,
sunken rock


Γειά σας. Μια άσκηση θεωρίας αριθμών παραθέτω την οποία ελπίζω ότι θα θεωρήσετε εύκολη. Καλά να περάσετε το τριήμερο.

Να δείξετε ότι ο αριθμός  $ \sqrt{123^{2^{123}} \plus{} 2^{123^{2}}}$ δεν είναι φυσικός.

[quote="KostasGT"]Γειά σας. Μια άσκηση θεωρίας αριθμών παραθέτω την οποία ελπίζω ότι θα θεωρήσετε εύκολη. Καλά να περάσετε το τριήμερο.

Να δείξετε ότι ο αριθμός  $ \sqrt {123^{2^{123}} \plus{} 2^{123^{2}}}$ δεν είναι φυσικός.[/quote]

an den kano lathos to uporrizo einai isoupoloipo 3 mod 10 opote telos.

Θέλω $ \sqrt{123^{2^{123}} \plus{} 2^{123^{2}}}\equal{}x \leftrightarrow {123^{2^{123}} \plus{} 2^{123^{2}}}\equal{}x^2$ άτοπο διότι παίρνοντας mod3 καταλήγω ότι  $ 123^{2^{123}}\equiv{0(mod3)}$ and $ 2^{123^{2}}\equiv{2(mod3) }$ επομένως θα ισχύει $ x^2\equiv2(mod3)$  άτοπο διότι το 2 δεν είναι τετραγωνικό υπόλοιπο mod3. Επομένως δεν είναι φυσικός ο χ.


Harvard college Stanford MIT Yale probability Princeton

Out of the top 5 colleges in the United States, which school do you want to go the most, given that you will be accepted to all? i.e., what is your #1 choice?

I am currently in very much the state of hesitation: between Harvard, Stanford, and MIT. That's the big portion of the reasoning behind the poll! I guess, if you are hesitating like I am, choose the most high ranked school out of these or else describe your school of choice without voting.

Also, if applicable, please describe your intended major.

Sorry; the poll got messed up during the 'editing.' Please click the top choice of the three polls named the same.

I think you can probably guess how I voted :D

As of now I'm thinking of majoing in 18 (mathematics) and getting either a minor or double major in a course from the following set {14,15,24} (aka {economics, Sloan School of Business, Philosophy}).

I wouldn't use what school [i]we[/i] want to go to for deciding where [i]you[/i] want to go to.  What do you like to do?  What fields interest you?  If you give us some more information we might be able to help you find out which one's a good fit for you.

[quote="manuel"]joml88 dont do 14 or 15 ewwww[/quote]

why is that? I don't see why not  :)

I have an ok argument for this. Think of which school most people think is "the best". That would be harvard. Think about how much you hear of it on the news and how much people don't even know about places like MIT or Caltech. They've probably heard of Yale, but still definately think that Harvard is way better. Therefore, if you're asking for the MOST wanted school, that's Harvard and there's no argument. End of discussion.

-jorian

Well, the title of the thread is "most wanted school" but in his post, the OP asked which school would [i]you[/i] most want to go to.

Now to comment on the side discussion... First, so that others can follow this, 14 = economics and 15=Sloan/business.  Course 15 majors at MIT get very little respect.  This is because course 15 is generally not that hard to complete and yet the average starting salary coming out of it is in the 6 digits.  I'm considering 15 because I'm a little bit interested in starting a business.

Course 14 actually is probably the least of those 3 possibilities I listed for a minor/double.  If I did 24 (philosophy) I would probably just do it as a minor.

nah im kidding. if you are interested in business do 15.

but just in case, im sure we mathematicians can survive (i.e eat) just doing research and teaching..

[quote="joml88"]Now to comment on the side discussion... First, so that others can follow this, 14 = economics and 15=Sloan/business.  Course 15 majors at MIT get very little respect.  This is because course 15 is generally not that hard to complete and yet the average starting salary coming out of it is in the 6 digits.  I'm considering 15 because I'm a little bit interested in starting a business.[/quote]

[quote="jhredsox"]what are all these numbers like ur saying, "do 15" and such?

-jorian

i know they're some sort of classes[/quote]

Last I heard, reading was a good skill.

courses...
let me see if i know them all, let me try because im bored:
1- Civil Eng.
2- Mech. Eng.
3. Materials Science
4. Architecture
5.Chem
6.Electrical Eng. And Comp Sci
7. Biology
8.Physics
9.  dont remember
10. Chemical Eng.
11. Urban Planning
12. forgot
13. dunno
14. Economics
15. Business
16. Aeroastro Eng.
17. dont remember
18. Math yeaaaaa
19. dont know
20.
21. HUmmanities
22.
23.

ah i dont know them

Heh, we're derailing this quickly...

I'll try and fill in some of those (without looking of course):

9. Brain and Cognitive
12. EAPS (Earth, Atmospheric, and Planetary Science)
13. Ocean Engineering (? I think...)
17. Political Science
19. Doesn't exist anymore; used to be Meteorology according to my advisor.  I'll take her word for it since she majored in it ;)
20. Bio Engineering
22. Nuclear
23. Don't think there is one; not sure what it used to be
24. Philosophy

In an attempt to get this topic back on track now, I'll talk about Stanford, MIT, and Harvard.

Well, MIT and Harvard are only about 1.5 miles apart.  Thus in terms of local stuff around the college, they're pretty similar.  There isn't as much directly next to MIT as there is to Harvard.  Harvard Square is a bustling place.  Stanford is in Palo Alto, CA.  I don't really know much more than that.

my best friend is at stanford (philosophy major) and he tells me its really chilled ..  like it its not so intense, they spend a lot of time inside campus because SF is kinda far.
if you like that kind of life stanford is good.

in the other hand. mit is kind of intense, its a urban campus, many people, a really cool city at the other side of the river, you hang out with people from other colleges of the same area..

i've liked mit so far.. except for chemistry and non-math classes..

[quote="paladin8"][quote="joml88"]Now to comment on the side discussion... First, so that others can follow this, 14 = economics and 15=Sloan/business.  Course 15 majors at MIT get very little respect.  This is because course 15 is generally not that hard to complete and yet the average starting salary coming out of it is in the 6 digits.  I'm considering 15 because I'm a little bit interested in starting a business.[/quote]

[quote="jhredsox"]what are all these numbers like ur saying, "do 15" and such?

-jorian

i know they're some sort of classes[/quote]

Oh I knew they were courses, but I just didn't know if they were for certain schools, or if they changed from school to school. Thank you guys. Dibbs doing Business 15 at Brown!

-jorian

Last I heard, reading was a good skill.[/quote]

Huh, those course numbers are just for MIT.  We refer to our Courses, courses and buildings by their respective number.  We still call our friends by their names though ;)

[quote="joml88"]They are people who might go to the school at some point in the future.  So generally younger people.[/quote]

oh gotcha. prospects... wow I am an idiot  :lol:

I wonder what kinds of students these top-notch schools tend to accept.

like strengths, etc.

[quote="jhredsox"]
xevarion, i never said any certain school is better, i was just saying the most WANTED, not the BEST school.
[/quote]

What might be interesting whether this poll is representative for AoPS-/MLer in real life taking pressure from relatives/friends, Harvard's name recognition etc. into account, e.g. Caltech is an excellent graduate school but then MIT/Harvard (yield rate of  ~78 % for Harvard, ~67 % for MIT) look more promising on your CV.

well if you're just looking to go to the best undergrad college, I would have to say Harvard. However, I would pick schools with combined programs (such as BA/MD or BA/JD programs) over Harvard any day since they relieve you from the stress of LSAT, MCAT, etc and guarantee you admission to grad school. If you've made up your mind of exactly what profession you want to go to, I highly recommend looking at those programs (since there are so many for so many different career opportunities) instead of Harvard, Princeton, Yale, etc just because it relieves you from the stress of those standardized tests, resume building, and interviews.

I want to go to MIT, partly because it would be free anyway since my mom works there (assuming she doesn't get layed off in the next 5 years...) I might want to double math/chem, but if it's too much work I'll just stick with math.

I want to get some graduate work done at MIT, so that I actually have a chance of going to Harvard graduate school. But I've heard that real analysis isn't represented well enough for some people at Harvard, so if I really want to do real analysis I'll go to Yale or Stanford graduate school or something.

For me, it would probably be a toss-up between MIT and Caltech. However, I would go to Caltech because of the winter temperatures!

You don't know how awesome it is to go somewhere (South Carolina, for me) over winter break and wear shorts and a t-shirt in 60 degree weather.  Shoot, I've even been thinking about hitting up the beach :D

I would prefer stanford.  :lol:

sunny old california

I voted for Stanford, mainly because of the warm California weather  :P All the schools on that list are really good, so you can't really go wrong with any of them (at least I wouldn't mind attending any of them  :lol: ).

It comes at no suprise that Yale is not among the top chosen universities as it did not get the best recs here on ML/AoPS. But I thought that it might have been a good idea to explicitely represent Duke or UC Berkeley, e.g. Berkeley has a superb statistics department.

As expected MIT is leading with about 1/3 of the votes as they are lots of positive threads about it on the forum, [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=167]MIT Undergraduate Admissions Math Jam,[/url], [url=http://ocw.mit.edu/index.html]MIT OCW[/url] and [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=2212]joml88[/url]'s blog. Regarding MIT and Google it is of course mostly viewed from Cambridge and Boston, but wrt to regions the ranking is India, Pakistan and Singapore. US only follows on position 8. Looking at the language list Korean is first, followed by Thai and English search request are only listed as number 3. For more details look at [url=http://www.google.de/trends?q=Massachusetts+Institute+of+Technology&date=all&geo=all&ctab=0&sa=N]Google Trends.[/url] And Cambridge is on the second place worldwide searching for [url=http://www.google.de/trends?q=happiness&date=all&geo=all&ctab=0&sa=N]happiness.[/url] What does that tell me?

Harvard can forget about its true yield rate but still it is top-notch in humanities and pure math though MIT offers a broader range of choices.

[quote="drunner2007"][quote="manuel"]MIT is pretty goood in analysis...
maybe u should stay there...[/quote]

Does MIT have much research in algebraic topology? What about algebraic geometry?[/quote]

about algebraic geometry I don't know.
in topology mit is pretty strong too.
Munkres is a prof in MIT, and he is really famous around topologists..
i cant say much more..  :maybe:

[quote="now a ranger"][quote="drunner2007"]Go MIT. Come on Caltech

(my top two are bitter rivals)[/quote]

so drunner, what school do you think you will choose for college?[/quote]

I'm still not sure. Finances will have quite an influence.


superior algebra
superior algebra unsolved

Hello,

I am trying to understand coding theory. A q-ary (q does not have to be a prime power!) $(n,M,d)$ code $C$ is a set of rows of length n, each element of which from a certain set $F_{q}$ with cardinality $q$. That set has to contain $M$ rows, and we call those rows in it the "words".
The distance between two elements of $F_{q}^{n}$ is the number of positions with different elements.
$d$ has to be the minimum distance between two different elements of $C$.

$A_{q}(n,d)$ is by definition the largest $M$ for which a q-ary $(n,M,d)$-code exists.

Well, thanks to lineary codes, the Gilbert-Varshamov lower bound gives us that for each PRIME POWER $q$ :

$A_{q}(n,d) \geq q^{k}$ where $q^{k}$ is the largest power of $q$ such that :
$q^{k}<\frac{q^{n}}{\sum_{i=0}^{ d-2}(q-1)^{i}C^{n-1}_{i}}$

Now my syllabus says that for $q>1$ , not necessarily a prime power, this holds as well.  But I can't use linear codes anymore?!
Does anyone know how this can be proven.  My syllabus says it's an "exercise" but .... :|  I don't think it's that easy!

Thank you,
fredbel6

[quote="Peter VDD"]I don't know where you get that from, it's only true for prime powers. The theorem after it is a weaker version which is valid for all q... (by far not as strong).

[and I doubt you'll find much coding theorists here  :P][/quote]

Hi Peter,

I'm afraid I don't get it then.

That other bound is slightly different which I failed to see. :blush:
But why exactly is that true?  And why is it weaker if q is a prime power ?

[quote="Peter VDD"]the difference is the $d-2$ for the linear case, which is a $d-1$ for the nonlinear case (and the $n-1$ vs $n$ in the binomial). A proof for the nonlinear case can be found on our friend [url=http://en.wikipedia.org/wiki/Gilbert-Varshamov_bound]Wikipedia[/url]. ;)[/quote]Jeez, that proof is not that hard!
Okay, I should have checked Wikipedia.  Thanks. :D

I had found it before wikipedia, but wanted to give you back your favorite answer. :)


function
logarithms
calculus
calculus computations

I have read some books and browse the internet regarding function mapping. I quite understand it, though some questions are bothering me. How could you explain the following:

1.The exponential function $R \mapsto R : x \mapsto e^{x}$ is injective.
2. The natural logarithm $ln[0,+\infty] \mapsto R = ln \mapsto ln(x)$ is injective

If you have a problem regarding injections, surjections and bijections, please post it and I'll try to answer it.

Injections, bijections and surjections aren't the main problem for me...the main problem is :
what are your definitions of $e^{x}$ and $ln(x )$?

I have seen at least three different ways to define them, and the proof of what you ask depends on the definitions of course.

(All definitions of course can be proven to be the same, and the careful mathematician want to see them proven to be the same )


Are you sure?
Anyway, it is only an unsignificant part ;)


ARML
calculus
email
limit
floor function
modular arithmetic
LaTeX

The state tournament is a week from today! I propose that no one exchanges scores so that the awards ceremony will be a surprise. It's much more exciting that way (and I know, I'm probably one of the worst offenders when it comes to exchanging scores). Good luck everyone! PM me with your email address if you'd like copies of tests from previous years.

I'm going to start a problem marathon for the tournament. If you solve the problem, post and hide your solution, and post more problems (preferably from GCTM or of a similar difficulty level) if you would like. If you do not know how to solve a problem, you may still post it and hope that someone else can figure it out. Please do everything in LaTeX if possible and continue the numbering system (so the next person to post should start with "#4"). Here are the first problems. They are all from GCTM 2005.

#1. Evaluate $ \lfloor (1 \plus{} 2005^{ \minus{} 2005})^{2005^{2005}}\rfloor$.

#2. Find the smallest possible area of an ellipse passing through the points $ (4,1)$, $ (4, \minus{} 1)$, $ ( \minus{} 4,1)$, $ ( \minus{} 4, \minus{} 1)$.

#3. Evaluate $ \displaystyle\sum_{n\equal{}1}^{\infty}\frac {F_n}{10^{n \plus{} 1}}$ where $ F_n$ is the $ n$th Fibonacci number, i.e., $ F_1 \equal{} 1$, $ F_2 \equal{} 1$, $ F_3 \equal{} 2$, $ F_4 \equal{} 3$, etc.

#2.

Are you allowed to use calculus?
[hide]
Let the equation of the ellipse be
$ \frac {x^{2}}{a^{2}} \plus{} \frac {y^{2}}{b^{2}} \equal{} 1$
$ ab \equal{} f(a) \equal{} \frac {a^{2}}{\sqrt {a^{2} \plus{} 16}}$
setting $ f'(a) \equal{} 0$ gives

$ 16 \minus{} \frac {a^{2}}{2} \equal{} 0$
so
$ a \equal{} 4\sqrt {2}$
$ b \equal{} \sqrt {2}$

Area = $ ab\pi$ = $ 8\pi$[/hide]

Edit: Sorry! Answer is now hidden
Edit2: y changed to f(x)

Calculus is okay, but please hide your answers. Also, it's not advisable to use the same variable twice in a problem (you used "$ y$" as both the y-coordinate and the value of $ ab$ and it confused me briefly). But your answer is correct.
Here is how I did it:
[hide="#2"]Every possible circumscribing of the rectangle corresponds to a circumscribing of a $ (\pm{1},\pm{1})$ square through a scale change of the x-coordinates by a factor of $ \frac {1}{4}$. Clearly, the ellipse with the smallest area that circumscribes the square is a circle with radius $ \sqrt {2}$. This circle corresponds to an ellipse with semi-axises of lengths $ \sqrt {2}$ and $ 4\sqrt {2}$ and thus an area of $ 8\pi$.[/hide]

#4. Find $ x_{20}$ given the recursive relation $ x_{n} \equal{} x_{n \minus{} 1} \plus{} 6x_{n \minus{} 2}$, given $ x_1 \equal{} \minus{}10$ and $ x_2 \equal{} 5$

#5. Find the last three digits of the decimal representation of $ 639^{2005}.$

Did you mean to give values of $ x_1$, $ x_2$, and $ x_3$ for number four?

Here is number five:
[hide="#5"]Since $ \phi{(1000)} \equal{} 400$, $ 639^{2005}\equiv 639^{5} \pmod{1000}$.
At GCTM state, a calculator would be handy here, but without one, the problem is still doable with minimal calculations:
$ 639^5 \equiv ( \minus{} 361)^5 \equal{} \minus{} 361^5 \equiv \minus{} 19^{10} \pmod{1000}$.
Rewrite $ 19^{10}$ as $ (20 \minus{} 1)^{10}$. Only the last three terms are relevant to the last three digits because the fourth to last term will be a multiple of $ 20^3 \equal{} 8000$. It turns out that the third to last term, $ \binom{10}{2}\cdot 20^2 \equal{} 18000$ is not relevant either. The sum of the last two terms is
$ \minus{} \binom{10}{1}\cdot 20 \plus{} 1 \equal{} \minus{} 199$
Since we are looking for $ \minus{} 19^{10}$, our answer is $ \minus{} ( \minus{} 199) \equal{} 199$[/hide]

This is how I would have done that problem.

[hide]We want to evaluate $ 639^{2005}$ in the integers mod 1000.  For this, it suffices to consider the integers mod $ 5^3$ and mod $ 2^3$.  We have
\[ 639^{2005} \equiv (\minus{}1)^{2005} \equiv \minus{}1 \pmod{2^3}, \]
and
\[ 639^{2005} \equiv 14^{2005} \equiv 14^5 \pmod{5^3} . \]
Now, $ 14^2 \equiv 196 \equiv 71 \pmod{5^3}$, so $ 14^4 \equiv 71^2 \equiv 5041 \equiv 41 \pmod{5^3}$, so $ 14^5 \equiv 14 \cdot 41 \equiv 574 \equiv 74 \pmod{5^3}$.  Hence we want a number equivalent to $ \minus{}1 \pmod{8}$ and $ 74 \pmod{5^3}$.  Since $ 125 \equiv 5 \pmod{8}$, and $ 74 \equiv 2 \pmod{8}$, such a number is $ 74 \plus{} 125 \equal{} 199$.[/hide]

Hmm... interesting..
the first thing I thought of for that problem was Binomial...

[hide]
$ (640 \minus{} 1)^{2005}$
Since $ (\stackrel{2005}{2})640^{2}( \minus{} 1)^{2003} \equiv 0$ (mod 1000),

We want to look at

$ (\stackrel{2005}{1})640^{1}( \minus{} 1)^{2004} \plus{} ( \minus{} 1)^{2005}$ (mod 1000)
which is the same as

$ 200 \minus{} 1 \equal{} 199$ (mod 1000)

[/hide]

PS. Sorry for the bad Combination symbols

[quote="iniquitus"]#4. Find $ x_{20}$ given the recursive relation $ x_{n} \equal{} x_{n \minus{} 1} \plus{} 6x_{n \minus{} 2}$, given $ x_1 \equal{} \minus{} 10$ and $ x_2 \equal{} 5$
[/quote]

[hide]Since the two distinct roots of characteristic equation $ r^2 \equal{} r \plus{} 6$ are $ 3$ and $ \minus{} 2$, we know the closed formula is of the form $ x^n \equal{} a\cdot 3^n \plus{} b\cdot ( \minus{} 2)^n$. Plugging $ x_1 \equal{} \minus{} 10$ and $ x_2 \equal{} 5$, $ a \equal{} \minus{} 1, b \equal{} 7/2$, so $ x^{20} \equal{} \minus{} 3^{20} \plus{} 7/2\cdot ( \minus{} 2)^{20}$. My TI-83 tell me this enormous (negative) number is $ \minus{} 3483114385$. Any good way to calculate this? (Though I bet not.)
[/hide]
@iniquitus: try \binom{n}{k} to display $ \binom{n}{k}$ instead of your original code. BTW, all three of your solutions are very good. I was lazy so I just typed it into the calculator...

Blehh my bad
This was from the state tournament 2005
and one of their choices was $ x^{20} \equal{} \minus{} 3^{20} \plus{} 7/2\cdot ( \minus{} 2)^{20}$ i think.

Yea the characteristic equations are much easier than generating functions

Thanks for the \binom tip!

All from GCTM 2001:

#6. ABCDE is a regular pentagon, and BPC is an equilateral triangle. Given that APD is obtuse, find its measure in degrees.

#7. If $ <x>$ is defined as $ x\minus{}[x]$, what is the value of the only positive real number $ x$ for which $ <x>$, $ [x]$, and $ x$ form a geometric progression?

#8. Find the exact values of all ordered pairs of numbers $ (x,y)$ which satisfy the following system of equations:
$ x^2\minus{}xy\plus{}y^2\equal{}7$
$ x\minus{}xy\plus{}y\equal{}\minus{}1$

Hm so I forget what $ [x]$ is - is it the integer part of x, floor(x)?

If so, um.... we have [x]^2=x(x-[x]) or x^2-[x]x-[x]^2=0 or
\[ x\equal{}\frac{[x](1\pm\sqrt{5})}{2}\]
 and obviously we want positive numbers :) so we have x=[x]$ \phi$

and now pulling out our calculator we determine $ [2\phi]$>2, thus [x]=1 and $ x\equal{}\phi$.

[quote="solafidefarms"]Hm so I forget what $ [x]$ is - is it the integer part of x, floor(x)?[/quote]
Yes. I know that the $ \lfloor$ and $ \rfloor$ symbols are more recognizable, but this is how it was written on the actual tournament, so I copied it.

Yes, Howard, I think that we can safely give you a packet early.

The plans as far as I know are for all members of the team to arrive at UGA sometime Thursday afternoon. (You provide transportation.) We will try to hold our "normal" schedule as much as possible. If you need to be back in Atlanta Friday night, you should probably talk to Don Slater from Lassiter.

Hmm, is it just me, or is there a mistake on #32 of the written round for the 2006 test?  Here's the question:

Given the two points A (-2, 0) and B (4, 0), write the equation defining the set of points that are equidistant from A and the perpendicular bisector of AB.

The answer says that the parabola is $ y^2\equal{}6x\plus{}3$, but how can that be when the focus is on its left and its directrix is on its right?

Maybe I'm just being stupid, but I think the real answer should be $ y^2\equal{}\minus{}6x\minus{}3$.  Plz correct me if I'm wrong, which I might be.

Someone else recently inquired into this, and your answer is correct. The question was thrown out in the scoring.

so what were the rankings, for those who couldn't stay for the awards (like myself)

Individual Rankings:
1. Santhosh Karnik
2. Harrison Brown
3. Howard Tong
4. Eric Morphis*
5. Henry Mei*
6. Miles Edwards
7. Scott Brothers
8. Tianqi Wu
9. Steven Rouk
10. Tomas Leon
* After scoring correction

Team Rankings:
1. Walton
2. Rockdale
3. Northview
4. Lassiter*
5. Chamblee*
* After scoring correction

This is all from memory - I could be wrong somewhere. So congratulations to Santhosh, state champion. Also, congratulations to Mr. Fulton who has now coached winning state MathCounts, middle school state, JV state, and varsity state teams! I also jotted down the initials of everyone who made GA ARML. I will try to change them back into names, but someone may correct me if I'm wrong anywhere:

Scott Brothers
Billy Dorminy
Sam Brotherton
Harrison Brown
Miles Edwards
Benjamin Hu
Robert Grosse
Eddy Ferreira
Santhosh Karnik
Ruby Lai
Michael Lai
Nayoon Kim
Eric Morphis
Karthik Narayan
Claus Zheng
Maja Wichrowska
Howard Tong
Steven Rouk
Oliver Huang
Boru Wang
Heng Tong
Henry Mei
Lillian Wang
Arvind Narayan
Sitan Chen
Tianqi Wu
Andrei Markov
Casey Woodrum
Jerry Fong
Michael Qin
Edwin White
Allen Park
Vladimir Smirnov
Harris Enniss

I think Lassiter and Chamblee switched places after another score correction.  Also, you misspelled Maja...haha.

Edited!
Heh, I had to look up the spelling of her last name, and in the process of focusing on that, I messed up her first name. :huh:

21st place and no ARML... a decidedly mediocre showing. Oh well. Congratulations to everyone, and good luck at ARML next month.

Well, good job, everybody, especially Walton and Santhosh.  So, now we have two sets of siblings on Georgia ARML.  It looks to me like we're set up for a few more good years, so let's make the most of this opportunity to do well.  Try to get to know your teammates during practice, so we can work together better as a team.  And work some problems, so you are better as an individual.

Anyhow, remember that even if you didn't make the first two teams, you can still come this year!  I recommend it; it is a good experience.

wait, what's this thing about scoring corrections?

I noticed that my score was off by ten points, so I protested, and after some searching, it was discovered that one of my ciphering answers had been mis-scored. I think the same thing happened to Rohan, thus raising Lassiter's score.

Also, one of Wheeler's pair ciphering teams scored 100 points, but wasn't recorded. This bumped us up to 6th.

so the rankings displayed above are actually correct?

Thank you Eric. I thought that we had a chance going in, but never dreamed that we would actually win. I was just hoping for top 3. Like all coaches, we look like geniuses when we have talent. Chris, Scott, Henry, and Claus all had a great tournament. (Sometimes the best coaching move is just driving the car to get everyone there!)

There is one more title that I would like to add to that list. But we have another month before that one can happen!  :)


geometry
incenter
circumcircle
geometry unsolved

In triangle $ ABC$, let $ I,I_a$ be the incenter and the excenter with respect to side $ BC$. Let $ A',M$ be the intersections of $ II_a$ with $ BC$ and the circumcircle of triangle $ ABC$ respectively. Let $ N$ be the midpoint of arc $ MBA$, and $ S,T$ the intersection points of rays $ NI,NI_a$ with the circumcircle of triangle $ ABC$. Prove that $ S,T,A'$ are collinear.

[quote="Carlez Tevos"]In triangle $ ABC$, let $ I$ be the incenter . Let $ D,M$ be the intersections of $ AI$ with $ BC$ and the circumcircle of triangle $ ABC$ respectively. Let $ N$ be the midpoint of arc $ MBA$, and $ P$ the intersection points of $ NI$ with the circumcircle of triangle $ ABC$. PD meets the circumcircle (ABC) at Q. NQ meets AI at I'. Prove that I' is the excenter with respect to side $ BC$[/quote]
$ \widehat{IAN}\equal{}\widehat{APN}\Rightarrow \triangle ANI\sim\triangle PNA\Rightarrow NA^2\equal{}NI.NP$
$ \widehat{AQN}\equal{}\widehat{NAM}\equal{}\widehat{I'AN}\Rightarrow\triangle QNA\sim\triangle ANI'\Rightarrow NA^2\equal{}NQ.NI'$
$ \Rightarrow NI.NP\equal{}NQ.NI'\Rightarrow P,I,Q,I'$ lie on a circle. Hence, $ DI.DI'\equal{}DP.DQ\equal{}DB.DC$, which shows that $ I'$ is the intersection of $ AI$ with the circumcircle $ (BIC)$. So, $ I'$ is the excenter with respect to side $ BC$.

[b]Solution 1:[/b]
-Let $ \omega$ be a circumcircle of triangle $ \triangle ABC$.Since we will consider only circle $ \omega$,I won't write "with respect to $ \omega$" everytime I use some pole and polar transformations.So let's start:

-Let $ X$ be a pole of $ ST$.So it is enough to show that $ X$ lies on a polar of $ A'$.

-Suppose that $ NA'$ intersects $ \omega$ at $ R$.It is well-known fact that division $ (AA',II_A)$ is harmonic.Therefore,quadrilateral $ ASRT$ is harmonic and it also follows that $ AR$ passes through $ X$.

-Since $ N$ is the midpoint of arc $ MBA$,it follows that tangent line to $ \omega$ and $ N$ is parallel to $ AA'$.Combining it with fact that
$ (II_A,M\infty) \equal{} \minus{} 1$,we conclude that quadrilateral $ NSMT$ is harmonic,as well.Therefore,points $ N,M,X$ are collinear.

-Since quadrilateral $ NARM$ is cyclic and point $ A'$ its diagonals intersection point and $ AR$ intersects $ NM$ at $ X$,we finally conclude that $ X$ lies on the polar of $ A'$.
$ \blacksquare$.

P.S:
"Solution 1" is actually my proof,which I found out during the [url=http://www.mathlinks.ro/viewtopic.php?t=216901]exam.[/url]There exist a more elegant solution with using just some angle chasing and some other simple facts.
Best regards.

[b]Solution 2:[/b]
This is the most beautiful and elegant proof to this problem and I would be very surprised if one would find a shorter one.

So let's start:

It is obviously that $ \widehat{TI_AI}$ is the half of arc $ NT$.And $ \widehat{IST}$ is equal to the half of arc $ NT$ as well,therefore,quadrilateral $ SITI_A$ is cyclic.

Now it is enough to use the well-known fact,which tell us that radical axes of three circles are concurrent,to three circumcircles of
$ \triangle ABC,\triangle SII_A$ and $ \triangle IBC$.

$ \blacksquare$.

Awesome proof,isn't it? :roll:

Dear Erken and Matlinkers,
to go in the shorter way of Erken without using angles, denote Tn the tangent to the circumcircle at N, applied a version of the Reim's theorem : S, I T Ia are coyclic. (http://perso.orange.fr/jl.ayme  ,see A propos)
Then you have an enterely synthetic proof.
Sincerely
Jean-Louis

To [b]jayme[/b]:
Brilliant comment,thank you.
But in my opinion,using any theorem in a such easy stuff makes a solution to lose its beauty,furthermore,angle-chasing here is just a half of an one line.
Best regards.

[quote="Carlez Tevos"] [color=darkred]Let $ ABC$ be a triangle with the incenter $ I$ and the $ A$ - excenter $ I_a$ . Denote the intersections

$ A'$ , $ M\ne A$ of $ II_a$ with $ BC$ and the circumcircle $ w$ of $ \triangle ABC$ respectively. Let $ N$ be the

midpoint of arc $ MBA$ of $ w$ . Denote $ S\in w\cap (NI$ , $ T\in w\cap (NI_a$ . Prove that $ A'\in ST$ .[/color] [/quote]

[color=darkblue][b]I"ll present below an easy generalization of this [u]very nice[/u] problem :[/b][/color]

[quote="Carlez Tevos & Virgil Nicula"][color=darkred][b][u]EXT.[/u][/b] Let $ ABC$ be a triangle. For two points $ D\in (BC)$ , $ P\in (AD)$ ([b]two degrees of freedom[/b]) define :

the second intersections $ M$ , $ R$ of $ AD$ with the circumcircles of  the triangles $ ABC$ , $ BPC$ respectively ;

the midpoint $ N$ of the arc $ ABM$ in the circumcirle $ w$ of $ \triangle ABC$ ; the second intersections $ S$ , $ T$

of the circle $ w$ with the lines $ NP$ , $ NR$ respectively. Prove that $ D\in ST$ .[/color][/quote]
[color=darkblue][b][u]The nice Erken's proof.[/u][/b] Prove easily that $ \widehat {PRT}\equiv\widehat {NAT}\equiv\widehat {PST}$ , i.e. the quadrilateral $ PTRS$ is cyclically.

Now is enough to use the well-known fact which tell us that radical axes $ BC$ , $ PR$ , $ ST$ of the circumcircles

for the cyclical quadrilaterals $ BTCS$ , $ BPCR$ , $ PTRS$ are concurrently. Since $ D\in BC\cap PR$ obtain $ D\in ST$ .[/color]

[color=darkblue][b][u]Remark.[/u][/b] For $ P: \equal{} I$ obtain the our problem. Study and another particular cases.[/color] [color=darkred][b]Thank you, [u]Carlez Tevos[/u] ![/b][/color]

[quote="Virgil Nicula"][color=darkred][b][u]An equivalent enunciation of EXT.[/u][/b] Let $ ABC$ be an $ A$ - isosceles triangle with the circumcircle $ w$ .

For a mobile point $ M\in (BC)$ denote the second intersection $ P\ne A$ of the line $ AM$ with the circle $ w$ .

For a mobile point $ N\in w$ for which the line $ AC$ separates $ N$ , $ B$ define the point $ R\in AN\cap BC$ .

Denote the intersection $ S\ne A$ of the circle $ w$ with the circumcircle of the triangle $ MAR$ .

Ascertain the geometrical locus of the intersection $ L\in AS\cap NP$ ([b]two degrees of freedom[/b]).[/color][/quote]


trigonometry
inequalities
inequalities unsolved

$ \sum\limits_{i=1}^{n}\frac{\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i=1}^{n}\cot(b_{i})$

$ \sum\limits_{i=1}^{n}\frac{\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i=1}^{n}\frac{\cos{(b_{i})}}{\sin{(b_{i})}}$

$ \sum\limits_{i=1}^{n} \cos(a_{i})\le\sum\limits_{i=1}^{n}{\cos{(b_{i})}}$

I don't understand why there isn't equality here though..

[quote="modularmarc101"]$ \sum\limits_{i = 1}^{n}\frac {\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i = 1}^{n}\cot(b_{i})$

$ \sum\limits_{i = 1}^{n}\frac {\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i = 1}^{n}\frac {\cos{(b_{i})}}{\sin{(b_{i})}}$

$ \sum\limits_{i = 1}^{n} \cos(a_{i})\le\sum\limits_{i = 1}^{n}{\cos{(b_{i})}}$

I don't understand why there isn't equality here though..[/quote]
your solution is not true :D
remember that:
$ \sum\limits_{i = 1}^{n}\frac {\cos{(b_{i})}}{\sin{(b_{i})}} >=<\frac{\sum\limits_{i = 1}^{n} \cos(a_{i})}{\sum\limits_{i = 1}^{n} \cos(b_{i})}$
equality holds when $ a_{i}=b_{i}$

[quote="tuandokim"][quote="modularmarc101"]$ \sum\limits_{i = 1}^{n}\frac {\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i = 1}^{n}\cot(b_{i})$

$ \sum\limits_{i = 1}^{n}\frac {\cos(a_{i})}{\sin(b_{i})}\le\sum\limits_{i = 1}^{n}\frac {\cos{(b_{i})}}{\sin{(b_{i})}}$

$ \sum\limits_{i = 1}^{n} \cos(a_{i})\le\sum\limits_{i = 1}^{n}{\cos{(b_{i})}}$

I don't understand why there isn't equality here though..[/quote]
your solution is not true :D
remember that:
$ \sum\limits_{i = 1}^{n}\frac {\cos{(b_{i})}}{\sin{(b_{i})}} > = < \frac {\sum\limits_{i = 1}^{n} \cos(a_{i})}{\sum\limits_{i = 1}^{n} \cos(b_{i})}$
equality holds when $ a_{i} = b_{i}$[/quote]
Just prove that for n =2,3,the inequality is right ,then n=k is right ... and note that :
$ cot (\sum b_i) = -cot (\pi -\sum b_i)$
Let do it yourself .Good luck :)
P/s:
[hide] post cái bài song ánh lên ý , bài này post phí giấy =.='[/hide]


geometry
circumcircle
trapezoid
parallelogram
vector
trigonometry
angle bisector

In an acute triangle $\triangle{ABC}$, let $AE$ and $BF$ be highs of it, and $H$ its orthocenter. The symmetric line of $AE$ with respect to the angle bisector of $\sphericalangle{A}$ and the symmetric line of $BF$ with respect to the angle bisector of $\sphericalangle{B}$ intersect each other on the point $O$. The lines $AE$ and $AO$ intersect again the circuncircle to $\triangle{ABC}$ on the points $M$ and $N$ respectively.

Let $P$ be the intersection of $BC$ with $HN$; $R$ the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$. Show that $AHSO$ is a paralelogram.

Since AO, BO are isogonals of AH, BH in the triangle $\triangle ABC$, O is its circumcenter. AN is a circumcircle diameter, hence $NM \perp AM$ and $NM \parallel BC$. M is a refelection of the orthocenter H in the side BC, hence E is the midpoint of HM. BC is the midline of the right angle triangle $\triangle HNM$ parallel to its side NM, hence P is the midpoint of HN. OP connects the the midpoints O, P of the hypotenuses AN, HN of the right angle triangles $\triangle ANM, \triangle HNM$, hence $OP \perp NM$ and $OP \perp BC$. Thus P is also the midpoint of BC and $AH \equiv AM \parallel OP \equiv OS$. BC is the perpendicular bisector of the base HM of the trapezoid HMSO passing through its diagonal intersection R, which means that this trapezoid is isosceles and P is also the midpoint of its other base OS. As a result, the diagonals HN, OS of the quadrilateral HNSO cut each other in half at P and it is a parallelogram, i.e., $AN \equiv ON \parallel HS$.

i found one of these days that the problem was still true if $ H$ and $ O$ were substituted by an arbitrary internal point $ P$ and its isogonal conjugate (if that's the name) of $ P$. i have a solution with vectors, but it comes to be a bit long and tired, does anybody have a better solution?

[quote="campos"]i found one of these days that the problem was still true if H and O were substituted by an arbitrary internal point P and its isogonal conjugate (if that's the name) of P.[/quote]
Indeed, the result is still true for two isogonal conjugates WRT ABC.

[b]Proposition:[/b] Let $ P$ and $ Q$ be two isogonal conjugates WRT $ \triangle ABC.$ Rays $ AP,AQ$ cut its circumcircle $(O)$ at $P',Q'.$ Define the points $ M \equiv BC \cap QP',$ $N \equiv PQ' \cap BC$ and $ L \equiv PM \cap QN.$ Then $ APLQ$ is a parallelogram.

$ \triangle A'B'C'$ is pedal triangle of $ P$ WRT $ \triangle ABC.$ Thus, $ QA \perp B'C',$ $QB \perp C'A'$ $\Longrightarrow$

$ \angle AQB \equal{} 180^{\circ} \minus{} \angle A'C'B' \equal{} 180^{\circ} \minus{} (\angle PBA' \plus{} \angle PAB') \equal{} 180^{\circ} \minus{} \angle PBP'$

$ \frac {\overline{P'M}}{\overline{MQ}} \equal{} \frac {\overline{BP'}}{\overline{BQ}} \cdot \frac {\sin \widehat{P'BC}}{\sin \widehat{CBQ}} \ \ , \ \ \frac {\overline{P'P}}{\overline{PA}} \equal{} \frac {\overline{BP'}}{\overline{BA}} \cdot \frac {\sin \widehat{PBP'}}{\sin \widehat{ABP}}$

By sine law for $ \triangle ABQ$ we get: $  \frac {\overline{BA}}{\overline{BQ}} \equal{} \frac {\sin \widehat{AQB}}{\sin \widehat{BAQ}}$

Combining with the two latter expressions, keeping in mind that $ \angle AQB \equal{} 180^{\circ} \minus{} \angle PBP',$ $\angle BAQ \equal{} \angle P'BC$ and  $ \angle CBQ \equal{} \angle ABP,$ we get $ \frac {_{\overline{P'M}}}{^{\overline{MQ}}} \equal{} \frac {_{\overline{P'P}}}{^{\overline{PA}}}$ $ \Longrightarrow$ $ PM \parallel AQ.$ Mutatis mutandis, $ QN \parallel AP.$

Let $P'$ be the feet of perpendicular dropped from $O$ to $BC$. It's known that $2OP=AH$
Hence, $\frac{OP}{AH}=\frac{OP}{AN}$
$\implies H,P',N$ collinear.Hence, $P=P'$
So, $OP \parallel AH$ . It suffices to prove $OP=PS$
Since $HE=EM$ the result follows quite easily.


AMC
USA(J)MO
USAMO
geometry
geometric transformation
homothety
induction

Ok, well, first of all, I'm not sure if I'm going to take either class (because the cost is somewhat steep).  But if I am, I'd like your opinions on whether AMY or WOOT is better.  They both cost the same for this coming school year.  I hear that both are really good from people who have taken them.
Of course, AMY is not really as renowned as WOOT (10 out of 12 USAMO winners in 2007 took WOOT), but that's not going to be a determinant factor in my decision.  It'll mostly be your opinions and mine.

Please respond!

Well first of all, all MOSP participants get into WOOT for free so that 10 of 12 thing is pretty meaningless.  It is true that it is the smartest possible peer group. Yet, I found some of those students to be quite obnoxious since they would post solutions that were not understandable at all and be like OMG I PWND THIS PROBLEM OMG.

I took WOOT last year and I haven't taken AMY but I've gotten three of the packets from AMY to help with writing some lectures for this summer.

I guess that I will give you my thoughts about learning and how WOOT/AMY work with those.

One thing that I think that WOOT does is that the teachers do a very good job of collecting problems. Since I pretty much self studied this year, I realized that it is somewhat of a pain to collect good problems (there are lots of tedious or contrived problems that can waste your time). WOOT is good for giving you an overview of problem solving. Also it is nice to have a sort of incentive to do problems because WOOT has a lecture each month so that is better than self studying.

I think that WOOT has some really good teachers and you should definitely attend the lectures. If you think that you will miss the lectures, you can make them up with transcripts but I felt like when I did that, I lost a lot. It is very different to be engaging yourself in class; I still remember many things from the Oly. Geo classes that I took like three years ago. The AoPS teachers are great.

The MOCK tests are nice because that forces you to actually put yourself in test conditions. The problems on the MOCK tests are pretty good (not great) in my opinion. But then again, you can always just tell yourself to take an old test and that isn't really any different.

Overall, the strengths of WOOT are that you don't have to collect your own problems to study, you get some extra motivation to work, and the teachers are great.

Now I'll talk as much as I can about AMY. There are three packets from AMY that I have looked at. The first one is Circles (by Zuming Feng). I really liked this packet a lot. The main thing that I liked about it was that it was organized in the same way that I would have liked to teach it. First, the packet talked about some general tips about how you should approach the learning process for this packet. I think that these comments are the most important thing that you can get from teachers. It is easy to learn facts (like just read a book), but it is rare for teachers to explain to you how they organize the information. I feel that this is the most important thing you can get from a teacher.

Also the packet started off with leading the student through a discussion of the fundamentals (for example, how do you derive the inscribed angle theorems, power of a point, etc). This is how I learned geometry and I feel that it is the best way to learn.  You need a strong base in the topic before you can understand the harder topics. I was happy to see this because that is how I would have liked to teach the topic.

Then the packet got much more difficult. Another thing that I liked about all the proofs is that they were very rigirous yet understandable. One of the things that I feel like most solutions don't talk about in geometry is: how do we know that a certain point exists if we are constructing it for our solution? The solutions addressed this clearly so that is another thing that I liked.

Then of course, there was a large collection of problems in the AMY segment. Since I didn't take it officially, I cannot comment on what the feedback is like. But this still has the advantage of making you work because you are supposed to submit solutions for a number of problems in each segment.

Another lecture that I got to look at was one on transformations (pole polar, homothety, etc). Again, I liked this packet a lot. I had one of the authors of the packet as a teacher at AMSP and I liked how he introduced the topic. Instead of sort of breezing over some things, he went through the proofs of the simpler facts also. Another particular thing that I liked is that he had a new way (new to me anyway) of showing that homothety preserves tangency...(when I learned it in the AoPS class, we didn't get as nice a proof).

Then I also got a packet on induction when I went to AMSP last year. Again, it had a large collection of problems and it introduced a lot of techniques that were useful.

So I would say the strengths of AMY are that they give a very large collection of problems, the packets are such that they give you the way that the teacher organizes his thoughts on the topic, and that you again get some motivation to work harder on math than if you were working alone.

I'm not sure what the peer group is like for AMY. Although they have no official collaboration, I'm sure that you could contact other students who are doing it and talk with them on AIM or something. This would be just as good as the collaboration that takes place on the WOOT forum in my opinion.

So yea those are my thoughts. If I were to choose one, I guess I'd do AMY because I am a pretty independent learner and I can collect the materials that woot gives, it just takes me time. AMY is more indepth but it covers less material so if you are younger, WOOT might be good just to get a larger survey of problem solving topics. But for me, AMY would be better because of the focus and depth of the packets and the way that they convey problem solving strategies is better.

Let me offer my two cents:

AMY offers a large number of problems, many of which are extremely challenging. You'll have about a month to work on these problems and then send in your solutions to be graded. You'll get scores and comments on your proofs, although not in great detail, and you'll get a solution packet to all the problems.

Contrary to what Altheman said, I didn't find the AMY solutions to offer much intuitive insight. Mostly, the proofs are very formal and very detailed from the beginning, and I found them often tedious to read. It was the kind of formal proof you would write on the USAMO, but not the kind that modeled your thought process on the problem. I think the proofs would have been more helpful if the writer had outlined the argument in the first paragraph before launching into the details.

The WOOT lectures, by contrast, guide you along the thought process, to some degree. The mock tests are usually pretty nice. On the other hand, the problems generally aren't as deep or difficult as the AMY problems.

Now, a word of caution: both of these resources will require tremendous input of time. If you aren't ready to commit 5-6 hours a week to either, you'll quickly find yourself reading transcripts instead of attending class and reading solutions before spending time to tackle the problems.

Well I suppose that we might be talking about different packets. I was talking about "Circles" from AMY 06-07 by Zuming Feng and Yufei Zhao. I don't know about the solutions for the problems, but there were like 40 pages of examples that were well done in my opinion. I suppose that other teachers might have a different way of writing the packets...I guess you would know better since I have not taken it for a whole year.

[quote="Altheman"]Also the packet started off with leading the student through a discussion of the fundamentals (for example, how do you derive the inscribed angle theorems, power of a point, etc). This is how I learned geometry and I feel that it is the best way to learn.  You need a strong base in the topic before you can understand the harder topics. I was happy to see this because that is how I would have liked to teach the topic. [/quote]

Ahem, seems I remember you learning some geometry ;)  Pretty good at it, you were  :yoda:

[quote="archimedes1"]Just a word about WOOT:  It appears (according to the class page) that WOOT is being beefed up this year, with 50% more practice olympiads and six problem sets.[/quote]

We have 6 olympiads, but will not have formal problem sets like we did last time (we're replacing the problem sets with the olympiads).  The trouble with the problem sets was simply that students didn't have the time to do them, but they'd find the time to do the olympiads.  We'll still have the message board problems after each class, for extra practice.

What you describe in terms of WOOT v. the old Olympiad classes is true of any of these sorts of year-round things.  There's only so much time, and if you want to hit a lot of topics (and this true of both WOOT and AMY, and of any similar survey course), you definitely won't get the depth you had in the old Olympiad courses (if you do the work) or in a more immersive experience like a summer program.

(As for inversion, it's not really clear to me that *any* time needs to be spent on it if you're really just training for olympiads rather than just learning math for the fun of it.  Same goes for all the really fancy stuff.  They're deadly traps for most people.)

The formal style proofs vs. the guided solutions are a good comparison between the AoPS books and the Titu/Zuming books in general. I'm not sure yet which one style is better for me though...

How big are the AMY packets? The WOOT handouts were about 10 pages on average, IIRC.

AMY seems to go in depth in a few broad topics, and I like this as well as WOOT. However, could someone tell me how in depth they go? :)

About WOOT classes: they are very good. However, I am pretty lazy and don't always do the problems as the teachers present/discuss them (you should!) Of course, maybe it's the "dude I am on the computer" thing... but oh well.

However, sometimes the problems presented in class are a bit easy (AIME level stuff...) but I guess there are the people who take WOOT for the AIME level stuff. But really, it should be more Olympiad-ish :( (due to its name) Also, due to the chatroom-style classes, sometimes the problems are a bit hard to follow if you aren't writing the stuff down.

The Olympiads are great. I solved USAMO #2 only because (well maybe not, but it helped A LOT) Practice Olympiad 4 #8 (or something, the one with the symmedian.) That WOOT Olympiad problem was half my solution...

Also, the handouts and packets were well thought-out and organized. Some of them were very cool (mean geometry!) The problem sets were good too, but too long (so I'm glad they cut them out... kind of... because there's a part of me that says "problem sets are useful." haha)

The main problem with WOOT is the cost. As a MOPper, I don't have to pay, so... yeah...

I don't have much to say about AMY because I never did it after AMP 06... I didn't even finish the AMP packets T.T

But Zuming, Titu, etc. are VERY VERY good teachers. Like... good.

[quote="shtsxc12"]Sorry, I am new to this area, what is AMY ? AwesomeMath Camp or books from AwesomeMath ?[/quote]

AMY- AwesomeMath Year-Round. It is a set of packets on different mathematical topics that you get throughout the year. It consists of pages of theory, theorems, proofs, and other material on a particular topic followed by problems.

I took AMY this year and thought it was really good. I thought that the comprehensive guides on particular topics that these packets offered were great for reference later. In addition, some of the problems were interesting. The comments were helpful in finding faults with arguments.

Hmm lena how are you getting internet haha

wait we arent allowed to get internet???

funny how the topics get completely off-topic :) Why would AoPS be banned at a math camp?

[quote="Altheman"]
The MOCK tests are nice because that forces you to actually put yourself in test conditions.[/quote]

but the schedule says that you have 5 days to complete each assignment as opposed to 3 hours for an AIME or 9 hours for a USAMO, is that because students probably don't have time during the school year to sit down for 3 or 9 straight hours and work on the tests?

You are supposed to sit down for 3 hours on practice olympiads and AIMEs, as in test conditions. It's not that people don't have enough time to spend doing these things during the school year, more often than not it's people spend their time doing pointless things.

i am still confused as to why one is given 5 days for each assignment when they are supposed to be done in 3 hours. i'm not saying its bad, in fact i think it is good that one is given that much time. it is more convenient.

You're given 5 days during which to find 3 hours to sit down and do the test.  This is to help students who might have time conflicts if we specified a specific 3 hours during which to take the test.

How do solution submissions for WOOT work? What do we write up for grading and how are they sent in?

For like a practice olympiads, there are answer sheets, where you write your solutions. Then you can send them through email or fax.

There is a lot of material given for you to practice on. It is not mandatory, but you are expecting to get better then you need to do it. So, yes there will be a lot of homework [i]available[/i]; however, the amount of work done is your own decision.

Does anyone know if you recieve solutions to the problems in AMY?

Hmm...I don't know. BTW, are we allowed to discuss problems via forum?

[quote="stupidityismygam"]In AMY, there were two different styles of packets.

If it was written by Titu, Ivan, or Iurie then there would be 20-30 pages of information and then a set of Easy, Medium, and Hard questions. The information would be full of examples, at times difficult to understand, but would cover the topic thoroughly. In each subdivision of questions there was around 25 questions. So for each topic, you would have ~75 questions. All of the problems were proof based.

If it was written by Zuming then there would be a few sections of information, and then a group of problems over that material. Then, a few more sections of information and another group of problems etc. At the beginning most of the problems were computational (AIME level), but the later problems were all proofs (Olympiad level). There were ~50 questions per packet.

As an example of the depth, in Zuming's Law of Sines/Cosines packet he introduced/proved Law of Sines, Law of Cosines, Ptolemy's theorem, Ceva's Theorem, Menelaus' theorem, Stewart's theorem, Brahmagupta's formula, Heron's formula, Brocard Points, and Erdos-Mordell inequality.

In Ivan Borsenco's Graph Theory he introduced/proved paths and cycles, Konig's Theorem, Mader's Theorem, Hall's Theorem (Marriage Lemma), Turan's and Reiman's Theorems, Ramsey's Theorem, Euler's formula, Five Color Theorem, and Fary's Theorem.[/quote]

Will someone explain some of these last few graph theorems- i searched but the first few werent on aopswiki

Some of the last ones:

Turan's is about the graph with the most edges among all graphs with n vertices that have no k vertices such at least one pair of those 2 vertices is NOT connected by an edge.

Marriage Lemma(hall's) is about when you can split the graph into two sets such that they can "marry" each other perfectly. Like you could have vertices {a, B, C} and {D, E, F} and A and D are connected, B and E are connected, and C and F are connected, then A could marry D, B could marry E, and C could marry F. And the graph would have a perfect matching.

Five Color Thereom is that you can color every map with 5 colors. (In fact, you can do it with 4!

What do you mean when you talk about vertices of a graph?

[quote="Altheman"]haha yes I certainly learned a lot from Olympiad Geo ! Since I feel like typing, I'll explain my geometry studying.

So basically, the time that I was just starting to get into problem solving was in freshman year when I was taking geometry. Anyway, before I found the AoPS books, I just did lots and lots of problems in my geometry book. I remember that I like did every single challenge problem in the book and I learned all the proofs of the basic things very well.

This prepared me very well for taking the Olympiad Geometry class at AoPS. This class was [i]pivotal[/i] in transitioning to olympiad level math for me. I still remember the lectures on cyclic quadrilaterals and similar triangles and power of a point. These things I sorta knew but I really learned them well.

When I took this class, again and again, I saw myself coming back to these things and I learned an appreciation for the fundamentals of Olympiad geometry. Furthermore, Mr. Rusczyk is an awesome teacher because he sorta tells you all the tricks that he uses to work through problems. Although I have developed my own sort of geometry algorithm, I still come back to many of the things I learned from him. Ex. focus on the hard parts of the diagram, what information have you not used, identify when a technique is useful, work backwards, draw a good diagram and make conjectures, etc.

Then I have done a lot of learning on my own by just reading posts on AoPS and studying specific topics like projective geometry and such.

I felt like the Oly. Geo. class was better in some senses than WOOT because there was more time to get in depth into topics. The thing is that there is just so much to learn in olympiad math and the WOOT classes don't necessarily have time to like spend a whole lecture on inversion or something like that really needs to have that much attention given to it.

I guess it isn't really my place to say this, but I feel like the WOOT classes could be improved my making them more specialized. So what I mean by this is: I feel like one of the most important things to teach a student is how to learn. It is not as important to teach a whole lot of factual information, but you want to get a sense of how you can approach a topic yourself. In this spirit, I think that it would be best to go into a particular topic more deeply instead of giving a broad overview of something like number theory. I feel that this way, the teacher can show the student how to learn how to learn more instead of rushing to get a lot of problems in. I certainly feel like the teachers say their little techniques for learning in passing, but I feel like more focus could be given to them.[/quote]
I think we actually have far more time than we think if we try to find it and I think there is actually enough time for the WOOT lectures to cover much more than they do right now material.


Let $a, b, c \in \mathbb{C}, a \neq b \neq c$ such that

$a^{3}+a = 4$
$b^{3}+b = 4$
$c^{3}+c = 4$

Find $a+b+c$.

[hide="Hint 1"] Don't try to look for a manipulative answer. [/hide] [hide="Hint 2"] Consider polynomials instead. [/hide]

[hide="..."]it is a third degree polynomial x^3+x-4...each variable can take a different root, then by vieta's sums, the answer is 0[/hide]

Provided, of course, that $a\neq b\neq c\neq a$ was what [b]t0rajir0u[/b] had in mind.


geometry
combinatorics unsolved
combinatorics

suppose that a square is partitioned into $n$ triangles of same area. Prove that $n$ is even !

We should stop posting this one over and over again:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80762[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80752[/url]

:blush:  Sorry, I don't go very often in the geometry section  :D  anyway, now I understand why I couldn't solve it !! (and why they couldn't solve it on the other forum :lol:  )


[quote="paganinio"]This time I have to point out the 'about' link on top right corner is dead link.[/quote]I know about that one and will be fixed at some point.


Find the bounds of
(a) $\cos{A}\cos{B}\cos{C}$
(b) $\cos{A}+\cos{B}+\cos{C}$
where $A,B,C$ are angles in a triangle

[quote="shobber"][hide]
$1 \leq \cos{A}+\cos{B}+\cos{C} \leq \frac{3}{2}$[/hide][/quote]
Can you show?

(a)Since, $|\cos a| \leq 1$ for all $a$, the $|\prod \cos a| = \prod |\cos a| \leq 1\cdot 1 \cdot 1 \cdots 1 = 1$ which tells us that the product of cosines always lies in the range $[-1,1]$ no matter how many cosines are being taken together. So, our product can be no smaller than $-1$.
For the lower bound, say $a = \delta, b = \delta, c = 180 - 2\delta$ for a very small $\delta$. As $\delta$ approaches $0$, $a$ and $b$ approach $0$ and $c$ approaches $180$. So, $\displaystyle\lim_{\delta\to 0}\cos \delta = 1, \displaystyle\lim_{\delta\to 0}\cos (180-2\delta) = -1$ and with that we get $\displaystyle\lim_{\delta\to 0}\cos a \cos b \cos c = \left[\displaystyle\lim_{\delta\to 0}\cos \delta\right]\cdot\left[\displaystyle\lim_{\delta\to 0}\cos \delta\right]\cdot\left[\displaystyle\lim_{\delta\to 0}\cos (180-2\delta)\right]$

$ = (1)(1)(-1) = (-1) \Rightarrow  -1 < \cos a \cos b \cos c$

We can have the product $\cos a \cos b \cos c$ be infinitesimally close to $-1$ but it can never reach it if the angles are to be those of a triangle.

A decently well-known and easily provable identity says that:
$\cos^2 a + \cos^2 b + \cos^2 c = 1 - 2\cos a \cos b \cos c$

$\Rightarrow \cos a \cos b \cos c = \frac{1-\displaystyle\sum_{cyc}\cos^2 a}{2}$

By Cauchy-Shwarz:
$(\cos^2 a + \cos^2 b + \cos^2 c)(1^2 + 1^2 + 1^2) \geq (\cos a \cdot 1 + \cos b \cdot 1 + \cos c \cdot 1)^2$

I have proved below that:
$\cos a + \cos b + \cos c > 1 \Rightarrow \frac{(\cos a + \cos b + \cos c)^2}{3} > \frac{1^2}{3}$

$\Rightarrow \cos^2 a + \cos^2 b + \cos^2 c \geq \frac{(\cos a + \cos b + \cos c)^2}{3} > \frac{1}{3}$

$\Rightarrow -\displaystyle\sum_{cyc}\cos^2 a < -\frac{1}{3}$

$\Rightarrow \frac{1-\displaystyle\sum_{cyc}\cos^2 a}{2} < \frac{1-\frac{1}{3}}{2} = \frac{1}{3}$

So, we get $\boxed{-1 < \cos a \cos b \cos c < \frac{1}{3}}$**

(b)Well, there is a nice little identity that says:
$\displaystyle\sum_{cyc}\cos a=4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}+1$

Proofs can be found on a thread I started [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=187198&highlight=#187198]here.[/url]

With that identity:
Since $a,b,c$ are angles of a triangle we know $0^o<a,b,c<180^o \Rightarrow 0^o<\frac{a}{2},\frac{b}{2},\frac{c}{2}<90^o$ since sine is increasing on the interval $(0,90)$ we know that $\sin j$ can come arbitrarily close to $0$ for any $j$ in this interval, but can't reach it, so $0\cdot 0 \cdot 0 < \sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}$

$\Rightarrow  4\cdot 0 + 1 < 4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2} + 1$

$\Rightarrow 1 < \displaystyle\sum_{cyc}\cos a$

Now for the upper bound:
By AM-GM

$\displaystyle\frac{\sin\frac{a}{2}+\sin\frac{b}{2}+\sin\frac{c}{2}}{3} \geq \sqrt[3]{\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}}$

Since sine is increasing in the interval the angles lie in, by Jensen's

$\displaystyle\frac{\sin\frac{a}{2}+\sin\frac{b}{2}+\sin\frac{c}{2}}{3} \leq  \sin\frac{\frac{a}{2}+\frac{b}{2}+\frac{c}{2}}{3} = \sin\frac{180^o}{6} = \sin 30^o = \frac{1}{2}$

$\Rightarrow \sqrt[3]{\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}} \leq \displaystyle\frac{\sin\frac{a}{2}+\sin\frac{b}{2}+ \sin\frac{c}{2}}{3} \leq \frac{1}{2}$

$\Rightarrow \sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2} \leq \frac{1}{8}$

$\Rightarrow 4\sin\frac{a}{2}\sin\frac{b}{2}\sin\frac{c}{2}  + 1 \leq 4 \cdot \frac{1}{8} + 1$

$\Rightarrow \displaystyle\sum_{cyc}\cos a \leq \frac{3}{2}$

So, we have the upper and lower bounds:
$\boxed{1 < \cos a + \cos b + \cos c \leq \frac{3}{2} }$.

**Could someone please help verify my upper bound on (a). I am not sure it is correct and I am not very good with Mathematica and I was trying to maximize it.

Hmm... I'm not sure about the upper bound of (a) either.

And what about $sinAsinBsinC$ and $sinA+sinB+sinC$  :D  :D

So, from your other post, the sharp upper bound of the product is \frac{1}{8}  :)


[from the [b]Manhattan Mathematical Olympiad, grades 7-8[/b] [url]http://www.math.ksu.edu/~soibel/olympiad.html[/url]]

1. Start with a six digit whole number X, and form a new whole number Y, by moving the first three digits of X after the last three digits. ( for example, is X = 154,377, then Y = 377,154.) Show that, when divided by 27,  both X and Y give the same remainder.

2. Integer numbers x, y, and z satisfy the equation
$x^3+y^3=z^3$
Prove that at least one of them is divisible by 3

I'll post my solution for number 1 later.

EDIT:NEW

3. Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent a triangle. Give an example which shows that  if only six segments are used, then such a choice may be impossible.

this might be too difficult for this forum, no proofs,

[quote="G-UNIT"]Isnt number 2 impossible?  Fermats last theorm...... mabye they meant to the second power?....[/quote]
For number 2, you ASSUME there are integers that satisfy the equation and then PROVE that at least one of them is divisible by 3. The problem did not ask the existence of such integers satisfying the equation, in which we know the answer by Fermat's Last Theorem.

actually, fermat's last theorem is only for positive integers, it says nothing about 0 or nonegative integers, a solution is (-n,n,0) for any integer n

[quote="pianoforte"][from the [b]Manhattan Mathematical Olympiad, grades 7-8[/b] [url]http://www.math.ksu.edu/~soibel/olympiad.html[/url]]

2. Integer numbers x, y, and z satisfy the equation
$x^3+y^3=z^3$
Prove that at least one of them is divisible by 3

I'll post my solution for number 1 later.[/quote]

$(x,y,z)=(2,0,2)$?

Maybe they consider 0 to be divisible by 3. After all, $\frac{0}{3} \in \mathbb{Z}$...

[quote="tetrahedr0n"]Given the grade level, I think we should be ok with this post here.[/quote]

It got moved from GS. ;)

[quote="chess64"]It got moved from GS.  [/quote]

Aha. That explains a lot.

[quote="pianoforte"]

2. Integer numbers x, y, and z satisfy the equation
$x^3+y^3=z^3$
Prove that at least one of them is divisible by 3

[/quote]

We can prove that at least one of them is divisable by d, for any d! :lol:

[quote="seamusoboyle"]A mod 9 argument works very well[/quote]

Why do you need, or want to use, mod 9?  There are only two possible residues mod 3, and the answer comes out very cleanly.

You can get every possible remainder (mod3). You could have $x^3 \equiv y^3 \equiv 1 \mod{3}$  and  $z^3 \equiv 2 \mod{3}$

Err, right on.  In my defense, it was 2:56 AM my time when I posted that.


There exist a positive integer $ n$ such that $ 2n \plus{} 1$ is not a prime and $ \displaystyle \phi(2n \plus{} 1) \mid 2n$ ?

(Especially in case the case $ \phi(2n\plus{}1)\equal{}n$ I have found no solution up to $ n<10^8$)

[size=75]I hope it is a well-known question..[/size]


now really :) among my list of things that I should be ashamed of, being lazy is on top. I hate it, but I just can't help it sometimes :D:D. What about you? What's the lazy-iest thing you've ever heard? :)

"C'mon dude, let's play Diablo 2 Hardcore Ironman for four straight days, I bet ya you character will die very soon...blabla" Seriously, I wonder if the 'dude' who said this is alive or not...He is too much into games and almost got me into it (lazy from studying math, not that everyday laziness  :D )

This one is even better: A bathroom shared by four roommates and the bathing stall (no tub  :( ) was not washed for 2 months because no one knew who's turn was it to wash it. It was full of .... you know...

a bedroom with WC inside is best for u  :lol:

wow thats lazy

im so lazy that i got 16 detentions for not doing my homework this year

You get detention for not doing your homework???  In my math class, our teacher rarely even bothers to see if you have done your homework.  Of course, you can easily get detention for talking...

well theres this stupid thing in our school that if your teachers a ***** and you don't do your homework, you can get detention.
so yeah, i probably didn't do my homework at least 300 times this year, but not detention for all of them

[quote="eryaman"]One person at my school always tries to cheat.  Problem is, he's too lazy to even copy someone else's work, he tries to pay people to copy for him.[/quote]

    Amazing .... I recal a similar story in which a man (of some 50 or 60 years) "posed" to be suffering from a fatal illness in order to avoid having to purchase gifts for his relative's wedding.  Of course one can agrue this is more an act of selfishness than sheer laziness, but it makes a great story regardless, especially considering the fact he made a "miraculous recovery" from the "fatal ilness" soon after the wedding was over.

[quote="eryaman"]One person at my school always tries to cheat.  Problem is, he's too lazy to even copy someone else's work, he tries to pay people to copy for him.[/quote]

 :rotfl:  :rotfl:  :rotfl:

well i have been opn AOPS for around 3.5 hours once,not at all doing math problems,just seeing them and going throuh the answers :D I don't know what the heck was i doing then.

I don't do my math hw, physics hw, or history reading.  And somehow I do well.  Not a surprise in physics or math, but somehow i pick up most things in class for history.

[quote="biffanddoc"]one time I spent 10 hours on AoPS straight looking at old posts  :o[/quote]

You win

[quote="karthik_k"]The laziest thing I've heard ?? It's a short joke -- When some guy was asked to write an essay on laziness, he wasted all his time and then just before giving the paper wrote -- "This is laziness"[/quote]

Here's another one: there's a sign in front of a building saying "Procrastinators' meeting - POSTPONED" (more of a visual humor thing) :D

Something I don't understand: in today's world, laziness is more favored than hard-workingness. I know people who will actively work to be lazy or at least seem lazy.

I started a Facebook group called "The Procrastinator's Paradise" lol
The most popular one at my school's Facebook right now. :D

How about hit the books(I DON'T MEAN LITERALY.) for 14 days straight. Hey! I'm going to try it now.


http://www.artofproblemsolving.com/Forum/viewtopic.php?p=523663#p523663

Here, read the last paragrah.

[quote="Karth"]I heard recently that a couple of new AoPS books are going to be published. What topics are they going to be about, and when are they coming out? Also, since they are going to cover a specific topic, how long will they be (much shorter than AoPS?) and how specific will they be (more specific than AoPS?)?

*cough cough an 'intermediate counting and probability book would be really nice'* ;)

Just curious. Any replies would be appreciated  :)[/code][/quote]

In early 2007, we hope to release Intro Algebra, Intermediate Algebra, and Intermediate Counting/Probability.  The last will be around the same length as the original AoPS, the first two will be considerably longer.

Will Intermediate Number Theory be there, by any chance?

[quote="rrusczyk"][quote="surge"]Maybe some Intermidiate Trigonometry/Complex Numbers?  ;)[/quote]

The Trig/complex numbers will be spread across our Intermediate Algebra and Intermediate Geometry/Trigonometry books.

Intermediate Number Theory and Intermediate Geometry/Trig are likely going to have to wait until 2008.[/quote]

 :(  I'm going to have graduated that year :(


inequalities
function
inequalities proposed

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}1.$ Prove that
\[ \frac{1\minus{}4a^2}{1\plus{}3a\minus{}3a^2}\plus{}\frac{1\minus{}4b^2}{1\plus{}3b\minus{}3b^2}\plus{}\frac{1\minus{}4c^2}{1\plus{}3c\minus{}3c^2}\leq1\]

From this inequality follows the Vasc's known inequality: $ \sum_{cyc}\sqrt{1\plus{}\frac{48a}{b\plus{}c}}\geq15.$ :wink:

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
\[ \frac {1 \minus{} 4a^2}{1 \plus{} 3a \minus{} 3a^2} \plus{} \frac {1 \minus{} 4b^2}{1 \plus{} 3b \minus{} 3b^2} \plus{} \frac {1 \minus{} 4c^2}{1 \plus{} 3c \minus{} 3c^2}\leq1
\]
From this inequality follows the Vasc's known inequality: $ \sum_{cyc}\sqrt {1 \plus{} \frac {48a}{b \plus{} c}}\geq15.$ :wink:[/quote]

I think we can using Cauchy-Schwarz inequality:

ineq <=>$ \frac {3a \plus{} a^2}{1 \plus{} 3a \minus{} 3a^2} \plus{} \frac {3b \plus{} b^2}{1 \plus{} 3b \minus{} 3b^2} \plus{} \frac {3c \plus{} c^2}{1 \plus{} 3c \minus{} 3c^2}\geq2$

LHS $ \ge\frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} 3a \plus{} 3b \plus{} 3c)^2}{\sum(a \plus{} 3a)(1 \plus{} 3a \minus{} 3a^2)} \ge2$

I think that mixing variables are good enough ...Here is a solution by this method

Let $ a$ be the min of {a,b,c} . Then $ f(a,b,c)\leq f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2})$ is equivalent to

$ (b \minus{} c)^2(36bc(b \plus{} c) \plus{} 3c^2 \plus{} 3b^2 \plus{} 12ab \plus{} 27c \plus{} 27b \minus{} 16)\geq 0$ which holds since $ a\leq\frac {1}{3}$

So it remains to prove that $ f(a,\frac {1 \minus{} a}{2},\frac {1 \minus{} a}{2})\leq 1$ which is equivalent to

$ \frac {a(3a \minus{} 1)^2(5 \minus{} 3a)}{(1 \plus{} 3a \minus{} 3a^2)(7 \minus{} 3a^2)}\geq 0$ which is clearly true and we are done  :)

[quote="silouan"]So it remains to prove that $ f(a,\frac {1 \minus{} a}{2},\frac {1 \minus{} a}{2})\leq 1$ which is equivalent to

$ \frac {(3a \minus{} 1)^2(5 \minus{} 3a)}{(1 \plus{} 3a \minus{} 3a^2)(7 \minus{} 3a^2)}\geq 0$ which is clearly true and we are done  :)[/quote]
I think there is something wrong in your solution, silouan. Note that the original inequality have equality for $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ and $ a \equal{} b \equal{} \frac {1}{2},c \equal{} 0$. But I cannot see the second point in your solution.

Anyway, we can prove this inequality using hyberbolic function technique. :)

It was a typo sorry but now is fixed and now i think my solution is correct . Isn't it?

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that
\[ \frac {1 \minus{} 4a^2}{1 \plus{} 3a \minus{} 3a^2} \plus{} \frac {1 \minus{} 4b^2}{1 \plus{} 3b \minus{} 3b^2} \plus{} \frac {1 \minus{} 4c^2}{1 \plus{} 3c \minus{} 3c^2}\leq1
\]
From this inequality follows the Vasc's known inequality: $ \sum_{cyc}\sqrt {1 \plus{} \frac {48a}{b \plus{} c}}\geq15.$ :wink:[/quote]

Can you tell me why？


logarithms
trigonometry
integration
modular arithmetic
complex numbers
calculus
calculus computations

Caculate  sum of sequence :
$ \sum_{n \equal{} 0}^{ \plus{} \inf} \frac {cosnx}{n}$
 solve equation :
$ yy'\plus{}x\plus{}y'ln \frac{1}{y'}\equal{} 0$

[hide="Number 2"]Here's a possible approach. We have $ yy' \minus{} y' \ln(y') \plus{} x \equal{} 0 \equal{}> y \equal{} \ln(y') \minus{} \frac{x}{y'}$. Let's now make $ p \equal{} \frac{dy}{dx}$ to get $ y \equal{} \ln p \minus{} \frac{x}{p}$. Now differentiate both sides to get $ dy \equal{} \frac{dp}{p} \minus{} \frac{pdx \minus{} xdp}{p^2}$. But $ dy \equal{} p dx$ and so $ pdx \equal{} \frac{dp}{p} \minus{} \frac{pdx \minus{} xdp}{p^2} \equal{}> p^3dx \equal{} pdp \minus{} pdx \plus{} xdp$
$ \equal{}> (p^3 \plus{} p)dx \plus{} (\minus{}x\minus{}p)dp \equal{} 0$ which is a differential eq. of the form $ M(x,p)dx \plus{} N(x,p)dp \equal{} 0$. It is not, however, an exact differential eq., so let's see if we can find an integrating factor. We have

$ \frac{\partial M}{\partial p} \equal{} 3p^2 \plus{} 1$, $ \frac{\partial N}{\partial x} \equal{} \minus{}1$, $ \frac{\partial M}{\partial p} \minus{} \frac{\partial N}{\partial x} \equal{} 3p^2 \plus{} 2$, and $ \frac{1}{M}\left(\frac{\partial M}{\partial p} \minus{} \frac{\partial N}{\partial x}\right) \equal{} \minus{} \left(\minus{} \frac{3p^2 \plus{} 2}{p^3\plus{}p}\right) \equal{} \minus{} f(p)$.

Therefore, our integrating factor will be $ e^{\int f(p)\,dp}$. But:

$ \int f(p)\,dp \equal{} \minus{} \int \frac{3p^2 \plus{} 2}{p^3\plus{}p}\,dp \equal{} \minus{} \int \frac{3p^2 \plus{} 1 \plus{} 1}{p^3\plus{}p}\,dp$

$ \equal{} \minus{} \int \frac{3p^2 \plus{} 1}{p^3\plus{}p}\,dp \minus{} \int \frac{1}{p(1\plus{}p^2)}\,dp \equal{} \ln \left|\frac{1}{p^2 \sqrt{1\plus{}p^2}}\right|$

after you do all the computations. Hence, the integrating factor will be $ \frac{1}{p^2 \sqrt{1\plus{}p^2}}$ and so we are now to solve the exact differential eq.

$ \frac{\sqrt{1\plus{}p^2}}{p} dx \plus{} \frac{\minus{}x\minus{}p}{p^2 \sqrt{1\plus{}p^2}} dp \equal{} 0$.[/hide]

[quote="Carcul"][hide="Number 1"]The sum should be from n=1 and we have $ S \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty} \frac {\cos(nx)}{n} \equal{} \minus{} \ln\left[2 \sin\left(\frac {x}{2}\right)\right]$, as long as I have made no mistake.[/hide][/quote]
  i can find it  by maple but i need a full solution. thanks

As a complex power series, $ \sum_{n\equal{}1}^{\infty}\frac{z^n}{n}\equal{}\minus{}\text{Log}(1\minus{}z)$ wherever it converges. (That is the complex logarithm: $ \text{Log}(w)\equal{}\ln|w|\plus{}i\,\text{arg}\,w.$) It can be shown that it converges for all $ |z|\le 1$ except $ z\equal{}1.$

In particular, that means that for $ \theta\not\equiv 0\pmod{2\pi},$ \[ \sum_{n\equal{}1}^{\infty}\frac{e^{in\theta}}{n}\equal{}
\minus{}\text{Log}(1\minus{}e^{i\theta}).\]
Take the real part of that:
\[ \sum_{n\equal{}1}^{\infty}\frac{\cos n\theta}{n}\equal{}\minus{}\ln|1\minus{}e^{i\theta}|.\]
Getting from there to the form that Carcul wrote it in is a matter of standard trigonometric identities.


Euler`s formula for distance between centres or circumcircle ( radius R) and inscribed circle (radius r) for triangle, only in terms of r and R, is well-known.  Also, I know the same formula for quadrilateral (if it can be inscribed in circle and if another circle can be inscribed in quadrilateral). Now, I want to find these  formulas for distances between centres of circles for 5-gon, 6-gon, 7-gon, 8-gon, 9-gon (of course for poligon that can be inscribed in circle and  another circle can be inscribed in poligon). Does somebody know these solutions or  where I can find  solutions?

Do you mean [url=http://mathworld.wolfram.com/BicentricPolygon.html]this[/url]?

Oh, it's really interesting problem. I found the formula for 5-gon, but it is so complicated, that I can't write it here.
For polygons with more sides it's going to be more and more complicated... :cursing:


Hi, I'm newcomer here, I need some help with this problem:
Let n be a positive integer. We are given an n*n chessboard. We call the intersection of k adjacent rows and l adjacent colomns a k*l subboard, and
say that k+l is the semiperimeter of subboard. Assume that some subboard with semiperimeters of n cover the main diagonal of the board. Prove that the subboard cover at least half of the unit squares of the board.

I'm sorry but I don't think I properly understand your problem:

What do you mean by "cover the main diagonal", you mean intersecting it or containing it? (Certainly can't be containing, otherwise it's the whole board...so intersecting it?)

Now, worse, suppose the subboard has dimensions a x b. Then a+b=n by your definition. So ab<=n/4, how can this cover "half of the unit squares of the board"?

I am confused. :?

It simply means that the union of the squares of the subboards contains the squares of the diagonal. Think of this example: assume $n=2k$. Take a subboard to be the intersection of the first $k$ rows and first $k$ columns, and take another subboard to be the intersection of the last $k$ rows and last $k$ columns. These two cover the diagonal, and they contain exactly half the squares on the large chessboard.

The subboards don't have to be connected, so they [b]can[/b] cover the diagonal without covering the entire board. Take, for example, the intersection of lines $1,3$ with rows $2,5$. This is also a subboard.

Just an idea. A subboard of dimensions a anb b with a<b "kills" maximun a diagonal squares. Now it is easy to see that if we want to cover the less area we must have disjoint subboards. Let a1,a2,a3.....,an be their smaller sides. Then a1+a2+a3...+an=n and Area= :Sigma: ai(n-ai)=n^2-:Sigma:ai^2>=n^2/2 if and only if (:Sigma:ai)^2>=2:Sigma:ai^2. Now Using that ai*n=ai^2+ai(n-ai)>=2ai^2 because ai<n-ai and suming for all i we obtain the desired inequality.

Excuse me, but it looks like solution than the idea
But how you prove that to cover the less area the subboards should be disjoint?

It is very intuitive, i ill try to think about the right way to express it and write it


If one uses only the information $ 10^3\equal{}1000$, $ 10^4\equal{}10000$, $ 2^{10}\equal{}1024$, $ 2^{11}\equal{}2048$, $ 2^{12}\equal{}4096$, $ 2^{13}\equal{}8192$, what are the largest $ a$ and the smallest $ b$ that one can prove $ a<\log_{10}2<b$?

[hide="brief solution"]We see that $ 1000<1024$, so $ \log10^3<\log2^{10}\Leftrightarrow\frac{3}{10}<\log2$, which is the best we can do.

We also see that $ 10000>8192$, so $ \log10^4>\log2^{13}\Leftrightarrow\frac{4}{13}>\log2$, which is the best we can do. Thus, $ a\equal{}\frac{3}{10}$ and $ b\equal{}\frac{4}{13}$.[/hide]


probability
linear algebra
matrix
limit
linear algebra unsolved

Find the probability $p_n$ of the event that an arbitrary matrix consisting of 1 and 0 has the non-zero determinant. Find $\lim\limits_{n\to\infty}p_n$.

[i](13 points)[/i]

I assume that the matrix is supposed to be an n-by-n matrix, whose entries are chosen randomly from 0 and 1.

Hmm, this problem seems pretty tough for a contest.  The first part seems to ask for a closed form for $p_n$, which I thought was not known.  The limit is 1, as proved by Komlos around 1967.  The proof I have seen relies on Sperner's theorem for posets, and is not simple.

He-he. Jury likes to give even unsolved open questions on DMM Olympiad. ;)
What do you think about problem with [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33772]density[/url]?

The density problem didn't seem easy.  I'll try to think about it more, but first you have to assure me that it is not an unsolved open question...

:lol:
You know that I can't, unfortunately.
In the same time why they ask for $\liminf$, not $\lim$? Is there some background for that?  :?

I also wondered whether "lim inf" could be replaced by "lim".


function
calculus
integration
complex numbers
complex analysis
complex analysis unsolved

Is there an integer series such that $ f(2^n)=(-1)^n$ for all $n$.?

I'm not sure what you mean by an "integer series".  Do you mean a power series with integer coefficients?

[quote="harazi"]No, just a power series with complex coefficients. How do you call that in english? Entire function?[/quote]

 Yes, Entire function or Integral function :D

Yes. Actually there is a truly magnificent result: that if $ \{a_n\}$ is a sequence of complex numbers that goes to infinity in complex plane, and $ \{b_n\}$ is an arbitrary sequence of complex numbers, then there is an entire function $ f$ such that $ f(a_n) \equal{} b_n$ for all $ n$.  :o


algebra
polynomial
inequalities
algebra proposed

Well, it happens that I remembered this problem while reading George Cain's proof of the fundamental theorem of algebra (this is essentially the method sketched by fleeting_guest in my last post on this topic). The solution follows immediately from it.

[quote="julien_santini"]Well, it happens that I remembered this problem while reading George Cain's proof of the fundamental theorem of algebra .
The solution follows immediately from it.[/quote]
Thank you, but...[b] where is the solution of my proposed question ? [/b]

Well, had you bothered reading fleeting_guest's answer to my previous post, as I indicated, it would have saved me some time (except if I made a big mistake, in which case I apologize  :D )... anyway, I do it: calling $p(z)$ your polynomial, then $|p(z)| \geq |z|^{5}.(1-\frac{2}{3|z|}-\frac{7}{3|z|^{2}}-\frac{6}{|z|^{3}}-\frac{35}{3|z|^{4}}-\frac{58}{3|z|^{5}})$, provided that $\frac{2}{3|z|}+\frac{7}{3|z|^{2}}+\frac{6}{|z|^{3}}+\frac{35}{3|z|^{4}}+\frac{58}{3|z|^{5}}\leq 1$. But this is the case for $|z| \geq 3$, and the inequality is even strict. This solves the question.

[quote="julien_santini"].....This solves the question......[/quote]
Yes, now I understand ! Nice solution,Alex